Hibbeler, R.C. 2015. Solutions Manual, Fluid Mechanics. Pearson

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1–1. Represent each of the following quantities with combinations of units in the correct SI form, using an appropriate prefix: (a) GN # mm, (b) kg >mm, (c) N>ks2, (d) kN>ms.

SOLUTION

a) GN # mm = (109)N(10-6)m = 103 N # m = kN # m 3

-6

Ans.

9

b) kg>mm = (10 )g>(10 )m = 10 g>m = Gg>m

Ans.

c) N>ks2 = N>(103 s)2 = 10-6 N>s2 = mN>s2

Ans.

3

-6

9

Ans.

d) kN>ms = (10 )N>(10 )s = 10 N>s = GN>s

Ans: a) kN # m b) Gg>m c) µN>s2 d) GN>s 1

1–2. Evaluate each of the following to three significant  figures, and express each answer in SI units using an  appropriate prefix: (a) (425 mN)2, (b) (67 300 ms)2, (c) 3 723 ( 106 ) 4 1>2 mm.

SOLUTION a) (425 mN)2 =

3 425 ( 10-3 ) N 4 2

b) (67 300 ms) = 3 67.3 ( 10 2

c)

3 723 ( 106 ) 4 1>2 mm

3

=

= 0.181 N2

Ans.

)( 10 ) s 4 = 4.53 ( 10 ) s -3

2

3 723 ( 106 ) 4 1>2 ( 10-3 ) m

3

2

Ans. Ans.

= 26.9 m

Ans: a) 0.181 N2 b) 4.53 ( 103 ) s2 c) 26.9 m 2

1–3. Evaluate each of the following to three significant  figures, and express each answer in SI units using an  appropriate prefix: (a) 749 mm>63 ms, (b) (34 mm) (0.0763 Ms)>263 mg, (c) (4.78 mm)(263 Mg).

SOLUTION a) 749 mm>63 ms = 749 ( 10-6 ) m>63 ( 10-3 ) s = 11.88 ( 10-3 ) m>s Ans.

= 11.9 mm>s b) (34 mm)(0.0763 Ms)>263 mg =

3 34 ( 10-3 ) m4 3 0.0763 ( 106 ) s4 > 3 263 ( 10-6 )( 103 ) g4

= 9.86 ( 106 ) m # s>kg = 9.86 Mm # s>kg

c) (4.78 mm)(263 Mg) =

3 4.78 ( 10 ) m 4 3 263 ( 10 ) g 4 -3

Ans.

6

= 1.257 ( 106 ) g # m = 1.26 Mg # m

Ans.

Ans: a) 11.9 mm>s b) 9.86 Mm # s>kg c) 1.26 Mg # m 3

*1–4. Convert the following temperatures: (a) 20°C to degrees Fahrenheit, (b) 500 K to degrees Celsius, (c) 125°F to degrees Rankine, (d) 215°F to degrees Celsius.

SOLUTION a) TC =

5 (T - 32) 9 F

20°C =

5 (T - 32) 9 F Ans.

TF = 68.0°F b) TK = TC + 273 500 K = TC + 273

Ans.

TC = 227°C c) TR = TF + 460

Ans.

TR = 125°F + 460 = 585°R d) TC = TC =

5 (T - 32) 9 F 5 (215°F - 32) = 102°C 9

Ans.

4

1–5. Mercury has a specific weight of 133 kN>m3 when the temperature is 20°C. Determine its density and specific gravity at this temperature.

SOLUTION g = rg 133 ( 103 ) N>m3 = rHg ( 9.81 m>s2 ) rHg = 13 558 kg>m3 = 13.6 Mg>m3 SHg =

rHg rw

Ans.

3

=

13 558 kg>m 1000 kg>m3

Ans.

= 13.6

Ans: rHg = 13.6 Mg>m3 SHg = 13.6 5

1–6. The fuel for a jet engine has a density of 1.32 slug>ft 3. If the total volume of fuel tanks A is 50 ft3, determine the weight of the fuel when the tanks are completely full. A

SOLUTION The specific weight of the fuel is g = rg = ( 1.32 slug>ft 3 )( 32.2 ft>s2 ) = 42.504 lb>ft 3 Then, the weight of the fuel is W = g V = ( 42.504 lb>ft 3 )( 50 ft 3 ) = 2.13 ( 103 ) lb = 2.13 kip

Ans.

Ans: g = 42.5 lb>ft 3 W = 2.13 kip 6

1–7. If air within the tank is at an absolute pressure of 680  kPa and a temperature of 70°C, determine the weight of the air inside the tank. The tank has an interior volume of 1.35 m3.

SOLUTION

From the table in Appendix A, the gas constant for air is R = 286.9 J>kg # K. p = rRT

680 ( 10

3

) N>m2 = r(286.9 J>kg # K)(70° + 273) K r = 6.910 kg>m3

The weight of the air in the tank is W = rg V = ( 6.910 kg>m3 )( 9.81 m>s2 )( 1.35 m3 ) Ans.

= 91.5 N

Ans: 91.5 N 7

*1–8. The bottle tank has a volume of 1.12 m3 and contains oxygen at an absolute pressure of 12 MPa and a temperature of 30°C. Determine the mass of oxygen in the tank.

SOLUTION

From the table in Appendix A, the gas constant for oxygen is R = 259.8 J>kg # K. p = rRT

12 ( 10

6

) N>m2 = r(259.8 J>kg # K)(30° + 273) K r = 152.44 kg>m3

The mass of oxygen in the tank is m = rV = ( 152.44 kg>m3 )( 0.12 m3 ) Ans.

= 18.3 kg

8

1–9. The bottle tank has a volume of 0.12 m3 and contains oxygen at an absolute pressure of 8 MPa and temperature of 20°C. Plot the variation of the pressure in the tank (vertical axis) versus the temperature for 20°C … T … 80°C. Report values in increments of ∆T = 10°C.

SOLUTION TC (°C) p(MPa)

20

30

40

50

60

70

80

8.00

8.27

8.55

8.82

9.09

9.37

9.64

p(MPa) 10

From the table in Appendix A, the gas constant for oxygen is R = 259.8 J>(kg # K). For T = (20°C + 273) K = 293 K, p = rRT

9 8 7 6

8 ( 106 ) N>m2 = r3259.8 J>(kg # K) 4(293 K)

5

r = 105.10 kg>m3

4

Since the mass and volume of the oxygen in the tank remain constant, its density will also be constant. p = rRT

3 2 1

p = ( 105.10 kg>m3 ) 3259.8 J>(kg # K) 4(TC + 273)

0

p = (0.02730 TC + 7.4539) ( 106 ) Pa

p = (0.02730TC + 7.4539) MPa where TC is in °C.

10 20 30 40 50 (a)

60 70 80

TC (ºC)

The plot of p vs TC is shown in Fig. a.

Ans: p = (0.0273 Tc + 7.45) MPa, where Tc is in C° 9

1–10. Determine the specific weight of carbon dioxide when the temperature is 100°C and the absolute pressure is 400 kPa.

SOLUTION

From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J>kg # K. p = rRT

400 ( 10

3

) N>m2 = r(188.9 J>kg # K)(100° + 273) K r = 5.677 kg>m3

The specific weight of carbon dioxide is g = rg = ( 5.677 kg>m3 )( 9.81 m>s2 ) = 55.7 N>m3

Ans.

Ans: 55.7 N>m3 10

1–11. Determine the specific weight of air when the temperature is 100°F and the absolute pressure is 80 psi.

SOLUTION

From the table in Appendix A, the gas constant for the air is R = 1716 ft # lb>slug # R. p = rRT 80 lb>in2 a

12 in. 2 b = r(1716 ft # lb>slug # R)(100° + 460) R 1 ft r = 0.01200 slug>ft 3

The specific weight of the air is g = rg = ( 0.01200 slug>ft 3 )( 32.2 ft>s2 ) = 0.386 lb>ft 3

Ans.

Ans: 0.386 lb>ft 3 11

*1–12. Dry air at 25°C has a density of 1.23 kg>m3. But if it has 100% humidity at the same pressure, its density is 0.65% less. At what temperature would dry air produce this same density?

SOLUTION For both cases, the pressures are the same. Applying the ideal gas law with r1 = 1.23 kg>m3, r2 = ( 1.23 kg>m3 ) (1 - 0.0065) = 1.222005 kg>m3 and T1 = (25°C + 273) = 298 K, p = r1 RT1 = ( 1.23 kg>m3 ) R (298 K) = 366.54 R Then p = r2RT2;

366.54 R = ( 1.222005 kg>m3 ) R(TC + 273) Ans.

TC = 26.9°C

12

1–13. The tanker carries 1.5(106) barrels of crude oil in its hold. Determine the weight of the oil if its specific gravity is 0.940. Each barrel contains 42 gallons, and there are 7.48 gal>ft 3.

SOLUTION The specific weight of the oil is go = Sogw = 0.940 ( 62.4 lb>ft 3 ) = 58.656 lb>ft 3 Weight of one barrel of oil: Wb = goV = ( 58.656 lb>ft 3 ) (42 gal>bl) a = 329.4 lb>bl

1 ft 3 b 7.48 gal

Total weight: W = 1.5 ( 106 ) bl(329.4 lb>bl) = 494 ( 106 ) lb

Ans.

Ans: 494 (106 ) lb 13

1–14. Water in the swimming pool has a measured depth of 3.03 m when the temperature is 5°C. Determine its approximate depth when the temperature becomes 35°C. Neglect losses due to evaporation.

9m

4m

SOLUTION From Appendix A, at T1 = 5°C, 1rw 2 1 = 1000.0 kg>m3. The volume of the water is V = Ah. Thus, V1 = (9 m)(4 m)(3.03 m). Then m m (rw)1 = ; 1000.0 kg>m3 = 2 V1 36 m (3.03 m) m = 109.08 ( 103 ) kg At T2 = 35°C, (rw)2 = 994.0 kg>m3. Then (rw)2 =

m ; V2

994.0 kg>m3 =

109.08 ( 103 )

( 36 m2 ) h

h = 3.048 m = 3.05 m

Ans.

Ans: 3.05 m 14

1–15. The tank contains air at a temperature of 15°C and an absolute pressure of 210 kPa. If the volume of the tank is 5 m3 and the temperature rises to 30°C, determine the mass of air that must be removed from the tank to maintain the same pressure.

SOLUTION

For T1 = (15 + 273) K = 288 K and R = 286.9 J>kg # K for air, the ideal gas law gives p1 = r1RT1;

210 ( 103 ) N>m2 = r1(286.9 J>kg # K)(288 K) r1 = 2.5415 kg>m3

Thus, the mass of air at T1 is m1 = r1V = ( 2.5415 kg>m3 )( 5 m3 ) = 12.70768 kg

For T2 = (273 + 30) K = 303 K and R = 286.9 J>kg # K p2 = r1RT2;

210(103) N>m2 = r2(286.9 J>kg # K)(303 K) r2 = 2.4157 kg>m3

Thus, the mass of air at T2 is m2 = r2V = ( 2.4157 kg>m3 )( 5 m3 ) = 12.07886 kg Finally, the mass of air that must be removed is ∆m = m1 - m2 = 12.70768 kg - 12.07886 kg = 0.629 kg

Ans.

Ans: 0.629 kg 15

*1–16. The tank contains 2 kg of air at an absolute pressure of 400 kPa and a temperature of 20°C. If 0.6 kg of air is added to the tank and the temperature rises to 32°C, determine the pressure in the tank.

SOLUTION

For T1 = 20 + 273 = 293 K, p1 = 400 kPa and R = 286.9 J>kg # K for air, the ideal gas law gives p1 = r1RT1;

400(103) N>m2 = r1(286.9 J>kg # K)(293 K) r1 = 4.7584 kg>m3

Since the volume is constant. Then V =

m2 m1 m2 r1 = ;r = r1 r2 2 m1

Here m1 = 2 kg and m2 = (2 + 0.6) kg = 2.6 kg r2 = a

2.6 kg 2 kg

b ( 4.7584 kg>m3 ) = 6.1859 kg>m3

Again applying the ideal gas law with T2 = (32 + 273) K = 305 K

p2 = r2RT2 = ( 6.1859 kg>m3 ) (286.9 J>kg # k)(305 K) = 541.30 ( 103 ) Pa = 541 kPa

16

Ans.

1–17. The tank initially contains carbon dioxide at an absolute pressure of 200 kPa and temperature of 50°C. As more carbon dioxide is added, the pressure is increasing at 25  kPa>min. Plot the variation of the pressure in the tank (vertical axis) versus the temperature for the first 10 minutes. Report the values in increments of two minutes.

SOLUTION p(kPa) TC(°C)

200 50.00

225 90.38

250 130.75

275 171.12

300 211.50

325 251.88

From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J>(kg # K). For T = (50°C + 273) K = 323 K, p = rRT

200 ( 103 ) N>m2 = r3188.9 J>(kg # K) 4(323 K) r = 3.2779 kg>m3

Since the mass and the volume of carbon dioxide in the tank remain constant, its density will also be constant. p = rRT

p = ( 3.2779 kg>m3) 3188.9 J>(kg # K)4(TC + 273) K p = (0.6192 TC + 169.04) ( 103 ) Pa

p = (0.6192 TC + 169.04) kPa where TC is in °C

The plot of p vs TC is shown in Fig. a p(kPa) 350 300 250 200 150 100 50 0

50

100

150

200

250

300

TC (ºC)

(a)

Ans: p = (0.619 Tc + 169) kPa, where Tc is in C° 17

1–18. Kerosene has a specific weight of gk = 50.5 lb>ft 3 and benzene has a specific weight of gb = 56.2 lb>ft 3. Determine the amount of kerosene that should be mixed with 8 lb of benzene so that the combined mixture has a specific weight of g = 52.0 lb>ft 3.

SOLUTION The volumes of benzene and kerosene are given by gb =

Wb ; Vb

56.2 lb>ft 3 =

8 lb Vb

Vb = 0.1423 ft 3

gk =

Wk ; Vk

50.5 lb>ft 3 =

Wk Vk

Vk = 0.019802 Wk

The specific weight of mixture is g =

Wm ; Vm

52.0 lb>ft 3 =

Wk + 8 lb 0.1423 ft 3 + 0.019802 Wk Ans.

Wk = 20.13 lb = 20.1 lb

Ans: 20.1 lb 18

1–19. The 8-m-diameter spherical balloon is filled with helium that is at a temperature of 28°C and a pressure of 106 kPa. Determine the weight of the helium contained in the 4 balloon. The volume of a sphere is V = pr 3. 3

SOLUTION

For Helium, the gas constant is R = 2077 J>kg # K. Applying the ideal gas law at T = (28 + 273) K = 301 K, 106(103) N>m2 = r(2077 J>kg # K)(301 K)

p = rRT;

r = 0.1696 kg>m3 Here V =

4 3 4 256 pr = p(4 m)3 = p m3 3 3 3

Then, the mass of the helium is

Thus,

M = r V = ( 0.1696 kg>m3 ) a

256 p m3 b = 45.45 kg 3

W = mg = ( 45.45 kg )( 9.81 m>s2 ) = 445.90 N = 446 N

Ans.

Ans: 446 N 19

*1–20. Kerosene is mixed with 10 ft3 of ethyl alcohol so that the volume of the mixture in the tank becomes 14 ft3. Determine the specific weight and the specific gravity of the mixture.

SOLUTION From Appendix A, rk = 1.58 slug>ft 3 rea = 1.53 slug>ft 3 The volume of kerosene is Vk = 14 ft 3 - 10 ft 3 = 4 ft 3 Then the total weight of the mixture is therefore W = rk g Vk + rea g Vea = (1.58 slug>ft 3)(32.2 ft>s2)(4 ft 3) + (1.53 slug>ft 3)(32.2 ft>s2)(10 ft 3) = 696.16 lb The specific weight and specific gravity of the mixture are gm =

W 696.16 lb = = 49.73 lb>ft 3 = 49.7 lb>ft 3 V 14 ft 3

Ans.

Sm =

49.73 lb>ft 3 gm = = 0.797 gw 62.4 lb>ft 3

Ans.

20

1–21. The tank is fabricated from steel that is 20 mm thick. If it contains carbon dioxide at an absolute pressure of 1.35 MPa and a temperature of 20°C, determine the total weight of the tank. The density of steel is 7.85 Mg>m3, and the inner diameter of the tank is 3 m. Hint: The volume of 4 a sphere is V = a bpr 3. 3

SOLUTION From the table in Appendix A, the gas constant for carbon dioxide is R = 188.9 J>kg # K. p = rRT

1.35 ( 10

6

) N>m2 = rco(188.9 J>kg # K)(20° + 273) K rco = 24.39 kg>m3

Then, the total weight of the tank is W = rst g Vst + rco g Vco W =

3 3 4 3.04 3.00 b(p) c a mb - a mb d 3 2 2 3 4 3.00 + ( 24.39 kg>m3 )( 9.81 m>s2 ) a b(p)a mb 3 2

3 7.85 ( 103 ) kg>m3 4 ( 9.81 m>s2 ) a

W = 47.5 kN

Ans.

Ans: 47.5 kN 21

1–22. What is the increase in the density of helium when the pressure changes from 230 kPa to 450 kPa while the temperature remains constant at 20°C? This is called an isothermal process.

SOLUTION Applying the ideal gas law with T1 = (20 + 273) K = 293 K, p1 = 230 kPa and R = 2077 J>(kg # k), p1 = r1RT1;

230 ( 103 ) N>m2 = r1(2077 J>(kg # K))(293 K) r1 = 0.3779 kg>m3

For p2 = 450 kPa and T2 = (20 + 273) K = 293 K, p2 = r2RT2;

450 ( 103 ) N>m2 = r2(2077 J>(kg # k))(293 K) r2 = 0.7394 kg>m3

Thus, the change in density is ∆r = r2 - r1 = 0.7394 kg>m3 - 0.3779 kg>m3 = 0.3615 kg>m3 = 0.362 kg>m3

Ans.

Ans: 0.362 kg>m3 22

1–23. The container is filled with water at a temperature of 25°C and a depth of 2.5 m. If the container has a mass of 30 kg, determine the combined weight of the container and the water.

1m

2.5 m

SOLUTION From Appendix A, rw = 997.1 kg>m3 at T = 25°C. Here the volume of water is V = pr 2h = p(0.5 m)2(2.5 m) = 0.625p m3 Thus, the mass of water is Mw = rwV = 997.1 kg>m3 ( 0.625p m3 ) = 1957.80 kg The total mass is MT = Mw + Mc = (1957.80 + 30) kg = 1987.80 kg Then the total weight is WT = MT g = (1987.80 kg) ( 9.81 m>s2 ) = 19500 N = 19.5 kN

Ans.

Ans: 19.5 kN 23

*1–24. The rain cloud has an approximate volume of 6.50 mile3 and an average height, top to bottom, of 350 ft. If a cylindrical container 6 ft in diameter collects 2 in. of water after the rain falls out of the cloud, estimate the total weight of rain that fell from the cloud. 1 mile = 5280 ft.

350 ft

SOLUTION

6 ft

The volume of rain water collected is Vw = p(3 ft)

2

1

2 12 ft

2

3

= 1.5p ft . Then, the

weight of the rain water is Ww = gwVw = ( 62.4 lb>ft 3 )( 1.5p ft 3 ) = 93.6p lb. Here, the volume of the overhead cloud that produced this amount of rain is Vc ′ = p(3 ft)2(350 ft) = 3150p ft 3 Thus, gc =

W 93.6p lb = = 0.02971 lb>ft 3 Vc ′ 3150p ft 3

Then Wc = gcVc = a0.02971

lb 52803 ft 3 b c (6.50) a bd 1 ft 3

= 28.4 ( 109 ) lb

24

Ans.

1–25. If 4 m3 of helium at 100 kPa of absolute pressure and 20°C is subjected to an absolute pressure of 600 kPa while the temperature remains constant, determine the new density and volume of the helium.

SOLUTION

From the table in Appendix A, the gas constant for helium is R = 2077 J>kg # K, p1 = r1RT1

100 ( 10

3

) N>m3 = r(2077 J>kg # K)(20° + 273) K r1 = 0.1643 kg>m3 T1 = T2 r1RT1 p1 = p2 r2RT2 p1 r1 = r2 p2

0.1643 kg>m3 100 kPa = r2 600 kPa r2 = 0.9859 kg>m3 = 0.986 kg>m3

Ans.

The mass of the helium is m = r1V1 = ( 0.1643 kg>m3 )( 4 m3 ) = 0.6573 kg Since the mass of the helium is constant, regardless of the temperature and pressure, m = r2V2 0.6573 kg = ( 0.9859 kg>m3 ) V2 V2 = 0.667 m3

Ans.

Ans: r2 = 0.986 kg>m3, V2 = 0.667 m3 25

1–26. Water at 20°C is subjected to a pressure increase of  44 MPa. Determine the percent increase in its density. Take EV = 2.20 GPa.

SOLUTION m>V2 - m>V1 ∆r V1 = = - 1 r1 m>V1 V2 To find V1 >V2, use EV = - d p > ( dV>V ) .

dp dV = V EV

V2

p

2 dV 1 = dp LV1 V EV Lp1

V1 1 ∆p ln a b = V2 EV V1 = e ∆p>EV V2

So, since the bulk modulus of water at 20°C is EV = 2.20 GPa, ∆r = e ∆p>EV - 1 r1 = e (44 MPa)>2.20 GPa) - 1 Ans.

= 0.0202 = 2.02,

Ans: 2.02, 26

1–27. A solid has a specific weight of 280 lb > ft3. When a pressure of 800 psi is applied, the specific weight increases to 295 lb > ft3. Determine the approximate bulk modulus.

SOLUTION g = 280 lb>ft 3 EV = -

dp dV V

Since V = Thus EV =

Therefore

dg W , dV = - W 2 g g

-dp dp = dg dg c -W 2 n( W>g ) d g g

800 lb>in2

EV = a

295 lb>ft 3 - 280 lb>ft 3 280 lb>ft 3

= 14.9 ( 103 ) lb>in2

Ans.

b

Note: The answer is approximate due to using g = gi. More precisely,

EV =

L

dp

dg L g

=

800 = 15.3 ( 103 ) lb>in2 ln (295>280)

Ans: 14.9 ( 103 ) lb>in2 27

*1–28. If the bulk modulus for water at 70°F is 319 kip > in2, determine the change in pressure required to reduce its volume by 0.3%.

SOLUTION Use EV = - dp> ( dV>V ) . dp = - EV

dV V

pf

∆p =

Lpi

Vf

dp = - EV

dV LVi V

= - ( 319 kip>in2 ) ln a

= 0.958 kip>in2 (ksi)

V - 0.03V b V

Ans.

28

1–29. Sea water has a density of 1030 kg>m3 at its surface, where the absolute pressure is 101 kPa. Determine its density at a depth of 7 km, where the absolute pressure is 70.4 MPa. The bulk modulus is 2.33 GPa.

SOLUTION Since the pressure at the surface is 101 kPa, then ∆ p = 70.4 - 0.101 = 70.3 MPa. Here, the mass of seawater is constant. M = r0V0 = rV r = r0 a

To find V0 >V, use EV = - dp> ( dV>V ) .

V0 b V

dV 1 = dp EV L L V

ln a

V 1 b = ∆p V0 EV V0 = e ∆p>EV V

So, r = r0e ∆p>EV = ( 1030 kg>m3 ) e (70.3 MPa>2.33 GPa) = 1061.55 kg>m3 = 1.06 ( 103 ) kg>m3

Ans.

Ans: 1.06 ( 103 ) kg>m3 29

1–30. The specific weight of sea water at its surface is 63.6 lb>ft 3, where the absolute pressure is 14.7 lb>in2. If at a point deep under the water the specific weight is 66.2 lb>ft 3, determine the absolute pressure in lb>in2 at this point. Take EV = 48.7 ( 106 ) lb>ft 2.

SOLUTION Use EV = - dp> ( dV>V ) and the fact that since the mass and therefore the weight of the seawater is assumed to be constant, mg = g1V1 = g2V2, so that V2 >V1 = g1 >g2. L

dp = - EV

dV LV

V2 b V1 g1 = - EV ln a b g2

∆p = - EV ln a

p = p0 + ∆p = 14.7 lb>in2 = 13.6 ( 103 ) psi

3 48.7 ( 106 ) lb>ft2 4 a

63.6 lb>ft 3 1 ft 2 b ln a b 12 in. 66.2 lb>ft 3

Ans.

Ans: 13.6 ( 103 ) psi 30

1–31. A 2-kg mass of oxygen is held at a constant temperature of 50° and an absolute pressure of 220 kPa. Determine its bulk modulus.

SOLUTION EV = -

dp dpV = dV>V dV

p = rRT dp = drRT EV = r =

m V

dr = EV =

drRT V drpV = dV rdV

mdV V2 mdV pV 2

V (m>V)dV

Ans.

= P = 220 kPa

Note: This illustrates a general point. For an ideal gas, the isothermal (constanttemperature) bulk modulus equals the absolute pressure.

Ans: 220 kPa 31

*1–32. At a particular temperature the viscosity of an oil is m = 0.354 N # s>m2. Determine its kinematic viscosity. The specific gravity is So = 0.868. Express the answer in SI and FPS units.

SOLUTION The density of the oil can be determined from ro = Sorw = 0.868 ( 1000 kg>m3 ) = 868 kg>m3 0.354 N # s>m2 mo yo = r = = 0.4078 ( 10-3 ) m2 >s = 0.408 ( 10-3 ) m2 >s o 868 kg>m3

Ans.

In FPS units,

yo = c 0.4078 ( 10-3 )

2 m2 1 ft dc d s 0.3048 m

= 4.39 ( 10-3 ) ft 2 >s

Ans.

32

1–33. The kinematic viscosity of kerosene is y = 2.39 ( 10-6 ) m2 >s. Determine its viscosity in FPS units. At  the temperature considered, kerosene has a specific gravity of Sk = 0.810.

SOLUTION The density of kerosene is rk = Sk rw = 0.810 ( 1000 kg>m3 ) = 810 kg>m3 Then, mk = yrk = 3 2.39 ( 10-6 ) m2 >s 4 ( 810 kg>m3 ) =

3 1.9359 ( 10-3 ) N # s>m2 4 a

= 40.4 ( 10-6 ) lb # s>ft 2

0.3048 m 2 lb ba b 4.4482 N 1 ft

Ans.

Ans: 40.4 ( 10 - 6 ) lb # s>ft 2 33

1–34. An experimental test using human blood at T = 30°C indicates that it exerts a shear stress of t = 0.15 N>m2 on  surface A, where the measured velocity gradient at the surface is 16.8 s-1. Since blood is a non-Newtonian fluid, determine its apparent viscosity at the surface.

A

SOLUTION Here

du = 16.8 s-1 and t = 0.15 N>m2. Thus dy t = ma

du ; dy

0.15 N>m2 = ma ( 16.8 s-1 ) ma = 8.93 ( 10-3 ) N # s>m2

Ans.

Realize that blood is a non-Newtonian fluid. For this reason, we are calculating the apparent viscosity.

Ans: 8.93 ( 10 - 3 ) N # s>m2 34

1–35. Two measurements of shear stress on a surface and the rate of change in shear strain at the surface for a fluid have been determined by experiment to be t1 = 0.14 N>m2, (du>dy)1 = 13.63 s-1 and t2 = 0.48 N>m2, (du>dy)2 = 153 s-1. Classify the fluid as Newtonian or non-Newtonian.

SOLUTION Applying Newton’s Law of viscosity, t1 = m1a

du b ; dy 1

t2 = m2 a

du b ; dy 2

0.14 N>m2 = m1 ( 13.63 s-1 )

m1 = 0.01027 N # s>m2

0.48 N>m2 = m2 ( 153 s-1 )

m2 = 0.003137 N # s>m2

Since m1 ≠ m2 then m is not constant. It is an apparent viscosity. The fluid is non-Newtonian. Ans.

Ans: non-Newtonian 35

*1–36. When the force of 3 mN is applied to the plate the line AB in the liquid remains straight and has an angular rate of rotation of 0.2 rad > s. If the surface area of the plate in contact with the liquid is 0.6 m2, determine the approximate viscosity of the liquid.

B u

A

SOLUTION The shear stress acting on the fluid contact surface is t =

3 ( 10-3 ) N N P = = 5 ( 10-3 ) 2 A 0.6 m2 m

Since line AB′ is a straight line, the velocity distribution will be linear. Here, the velocity gradient is a constant. The velocity of the plate is # U = au = (0.004 m)(0.2 rad>s) = 0.8 ( 10-3 ) m>s Then, t = m

du dy

5 ( 10-3 ) N>m2 = m a

0.8 ( 10-3 ) m>s 0.004 m

m = 0.025 N # s>m2

b

Ans.

Alternatively, t = m

du dt

5 ( 10-3 ) N>m2 = m(0.2 rad>s)

m = 0.025 N # s>m2

36

3 mN

B¿ 4 mm

1–37. When the force P is applied to the plate, the velocity profile for a Newtonian fluid that is confined under the plate is approximated by u = (12y1>4) mm>s, where y is in  mm. Determine the shear stress within the fluid at y = 8 mm. Take m = 0.5 ( 10-3 ) N # s>m2.

24 mm/s 16 mm y

P

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = 12y1>4 du = 3y-3>4 dy At y = 8 mm, t = m

du = 0.5 ( 10-3 ) N # s>m2 3 3(8 mm)-3>4 s-1 4 dy

t = 0.315 mPa

Ans.

du S ∞ , so that t S ∞. Hence the equation can not be used at Note: When y = 0, dy this point.

Ans: 0.315 mPa 37

1–38. When the force P is applied to the plate, the velocity profile for a Newtonian fluid that is confined under the plate is approximated by u = ( 12y1>4 ) mm>s, where y is in  mm. Determine the minimum shear stress within the fluid. Take m = 0.5 ( 10-3 ) N # s>m2.

24 mm/s 16 mm y

P

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = 12y1>4 du = 3y-3>4 dy The velocity gradient is smallest when y = 16 mm. Thus, t min = m

du = dy

3 0.5 ( 10-3 ) N # s>m2 4 3 3(16 mm)-3>4 s-14

t min = 0.1875 mPa

Ans.

du S ∞, so, that t S ∞. Hence the equation can not be used at Note: When y = 0, dy this point.

Ans: 0.1875 mPa 38

1–39. The velocity profile for a thin film of a Newtonian fluid that is confined between a plate and a fixed surface is defined by u = (10y - 0.25y2) mm>s, where y is in mm. Determine the shear stress that the fluid exerts on the plate and on the fixed surface. Take m = 0.532 N # s>m2.

36 mm/s

4 mm y

P

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = ( 10y - 0.25y2 ) mm>s du = (10 - 0.5y) s-1 dy At the plate tp = m

du = ( 0.532 N # s>m2 ) 3 10 - 0.5(4 mm) s-1 4 = 4.26 Pa dy

Ans.

At the fixed surface

tfs = m

du = ( 0.532 N # s>m2 ) 3 (10 - 0) s-1 4 = 5.32 Pa dy

Ans.

Ans: tp = 4.26 Pa, tfs = 5.32 Pa 39

*1–40. The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u = (10y - 0.25y2) mm>s, where y is in mm. Determine the force P that must be applied to the plate to cause this motion. The plate has a surface area of 5000 mm2 in contact with the fluid. Take m = 0.532 N # s>m2.

36 mm/s

4 mm y

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = ( 10y - 0.25y2 ) mm>s du = (10 - 0.5y) s-1 dy At the plate tp = m

du = ( 0.532 N # s>m2 ) 310 - 0.5(4 mm)4 s-1 = 4.256 Pa dy

P = tpA =

3 (4.256) N>m2 4 3 5000 ( 10-6 ) m2 4

Ans.

= 21.3 mN

40

u

P

1–41. The velocity profile of a Newtonian fluid flowing over p a fixed surface is approximated by u = U sin a yb. 2h Determine the shear stress in the fluid at y = h and at y = h>2. The viscosity of the fluid is m.

U u

h y

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = U sin a

p yb 2h

du p p = U a b cos a yb dy 2h 2h

At y = h,

t = m t = 0;

du p p = mU a b cos (h) dy 2h 2h

Ans.

At y = h>2, t = m t =

du p p h = mU a b cos a b dy 2h 2h 2

0.354pmU h

Ans.

Ans: At y = h, t = 0; At y = h>2, 41

t =

0.354pmU h

1–42. If a force of P = 2 N causes the 30-mm-diameter shaft to slide along the lubricated bearing with a constant speed of 0.5 m > s, determine the viscosity of the lubricant and the constant speed of the shaft when P = 8 N. Assume the lubricant is a Newtonian fluid and the velocity profile between the shaft and the bearing is linear. The gap between the bearing and the shaft is 1 mm.

50 mm

0.5 m/s P

SOLUTION Since the velocity distribution is linear, the velocity gradient will be constant. t = m

du dy

0.5 m>s 2N = ma b 32p(0.015 m)4(0.05 m) 0.001 m

m = 0.8498 N # s>m2

Ans.

Thus, 8N v b = ( 0.8488 N # s>m2 ) a 32p(0.015 m)4(0.05 m) 0.001 m v = 2.00 m>s

Ans.

Also, by proportion, a a

2N b A

8N b A

=

v =

ma

0.5 m>s t v ma b t

b

4 m>s = 2.00 m>s 2

Ans.

Ans: m = 0.849 N # s>m2 v = 2.00 m>s 42

1–43. The 0.15-m-wide plate passes between two layers, A  and B, of oil that has a viscosity of m = 0.04 N # s>m2. Determine the force P required to move the plate at a constant speed of 6 mm > s. Neglect any friction at the end supports, and assume the velocity profile through each layer is linear.

6 mm

A

P B

4 mm 0.20 m

SOLUTION

FA P

The oil is a Newtonian fluid. Considering the force equilibrium along the x axis, Fig. a, + S ΣFx = 0;

FB (a)

P - FA - FB = 0 P = FA + FB

Since the velocity distribution is linear, the velocity gradient will be constant. 6 mm>s du = ( 0.04 N # s>m2 ) a b = 0.04 Pa dy 6 mm 6 mm>s du tB = m = ( 0.04 N # s>m2 ) a b = 0.06 Pa dy 4 mm

tA = m

P = ( 0.04 N>m2 ) (0.2 m)(0.15 m) + ( 0.06 N>m2 ) (0.2 m)(0.15 m) Ans.

= 3.00 mN

Ans: 3.00 mN 43

*1–44. The 0.15-m-wide plate passes between two layers A and B of different oils, having viscosities of mA = 0.03 N # s>m2 and mB = 0.01 N # s>m2. Determine the force P required to move the plate at a constant speed of 6 mm > s. Neglect any friction at the end supports, and assume the velocity profile through each layer is linear.

6 mm

A

P B

4 mm 0.20 m

SOLUTION

FA P

The oil is a Newtonian fluid. Considering the force equilibrium along the x axis, Fig. a, ΣFx = 0;

FB (a)

P - FA - FB = 0 P = FA + FB

Since the velocity distribution is linear, the velocity gradient will be constant. tA = m tB = m

6 mm>s du = ( 0.03 N # s>m2 ) a b = 0.03 Pa dy 6 mm

6 mm>s du = ( 0.01 N # s>m2 ) a b = 0.015 Pa dy 4 mm

P = ( 0.03 N>m2 ) (0.2 m)(0.15 m) + ( 0.015 N>m2 ) (0.2 m)(0.15 m) Ans.

= 1.35 mN

44

1–45. The tank containing gasoline has a long crack on its side that has an average opening of 10 mm. The velocity through the  crack is approximated by the equation u = 10 ( 109 ) 310(10-6y - y2) 4 m>s, where y is in meters, measured upward from the bottom of the crack. Find the shear stress at the bottom, at y = 0 and the location y within the crack where the shear stress in the gasoline is zero. Take mg = 0.317(10-3) N # s>m2.

10 !m

SOLUTION

y(m)

Gasoline is a Newtonian fluid.

10(10–6)

u

10(109)[10(10–6)y – y2] m s

The rate of change of shear strain as a function of y is du = 10 ( 109 ) 3 10 ( 10-6 ) - 2y 4 s-1 dy

At the surface of crack, y = 0 and y = 10 ( 10-6 ) m. Then

du ` = 10 ( 109 ) 3 10 ( 10-6 ) - 2(0) 4 = 100 ( 103 ) s-1 dy y = 0

or

u(m s) (a)

du ` = 10 ( 109 ) 5 10 ( 10-6 ) - 2 3 10 ( 10-6 ) 46 = -100 ( 103 ) s-1 dy y = 10 (10-6) m

Applying Newton’s law of viscosity, ty = 0 = mg

du ` = dy y = 0

t = 0 when

3 0.317 ( 10-3 ) N # s>m2 4 3 100 ( 103 ) s-1 4

= 31.7 N>m2

Ans.

du = 0. Thus dy

du = 10 ( 109 ) 3 10 ( 10-6 ) - 2y 4 = 0 dy

10 ( 10-6 ) - 2y = 0

y = 5 ( 10-6 ) m = 5 mm

Ans.

Ans: ty = 0 = 31.7 N>m2 t = 0 when y = 5 µm 45

1–46. The tank containing gasoline has a long crack on its side that has an average opening of 10 mm. If the velocity profile through the crack is approximated by the equation u = 10(109) 310(10-6y - y2) 4 m>s, where y is in meters, plot both the velocity profile and the shear stress distribution for the gasoline as it flows through the crack. Take mg = 0.317(10-3) N # s>m2.

10 !m

SOLUTION y ( 10-6 m ) u(m>s)

0 0

1.25 0.1094

2.50 0.1875

3.75 0.2344

6.25 0.2344

7.50 0.1875

8.75 0.1094

10.0 0

5.00 0.250

y(10–6 m) 10.0 7.50 5.00 2.50 0

0.10

0.20

u(m s)

0.30

(a)

Gasoline is a Newtonian fluid. The rate of change of shear strain as a function of y is du = 10 ( 109 ) 3 10 ( 10-6 ) - 2y 4 s-1 dy

Applying Newton’s law of viscoscity, t = m

du = dy

3 0.317 ( 10-3 ) N # s>m2 4 5 10 ( 109 ) 3 10 ( 10-6 )

t = 3.17 ( 106 ) 3 10 ( 10-6 ) - 2y 4 N>m2

- 2y 4 s-1 6

The plots of the velocity profile and the shear stress distribution are shown in Fig. a and b respectively. y ( 10-6 m )

0

1.25

2.50

3.75

5.00

t ( N>m2 )

31.70 6.25 - 7.925

23.78 7.50 - 15.85

15.85 8.75 - 23.78

7.925 10.0 - 31.70

0

y(10–6 m) 10.0 7.50 5.00 2.50

–40

–30

–20

–10 0

10

20

30

40

τ(Ν m2)

(b)

46

Ans: y = 1.25 ( 10 - 6 ) m, u = 0.109 m>s, t = 23.8 N>m2

1–47. Water at A has a temperature of 15°C and flows along the top surface of the plate C. The velocity profile is approximated as mA = 10 sin (2.5py) m>s, where y is in meters. Below the plate the water at B has a temperature of 60°C and a velocity profile of u B = 4 ( 103 )( 0.1y - y2 ) , where y is in meters. Determine the resultant force per unit length of plate C the flow exerts on the plate due to viscous friction. The plate is 3 m wide.

y

100 mm

A

100 mm

B

C

SOLUTION Water is a Newtonian fluid.

Water at A, T = 15°C. From Appendix A m = 1.15 ( 10-3 ) N # s>m2. Here du A 5p 5p 5p = 10 a b cos a yb = a25p cos yb s-1 dy 2 2 2

At surface of plate C, y = 0. Then

du A 5p ` = 25p cos c (0) d = 25p s-1 dy y = 0 2

Applying Newton’s law of viscosity tA & y = 0 = m

du A ` = dy y = 0

3 1.15 ( 10-3 ) N # s>m2 4 ( 25p s-1 )

= 0.02875p N>m2

Water at B, T = 60°C. From Appendix A m = 0.470 ( 10-3 ) N # s>m2. Here du B = dy

3 4 ( 103 ) (0.1

At the surface of plate C, y = 0.1 m. Then

- 2y) 4 s-1

du B ` = 4 ( 103 ) 30.1 - 2(0.1) 4 = -400 s-1 dy y = 0.1 m

Applying Newton’s law of viscosity, tB & y = 0.1 m = m

du B ` = dy y = 0.1 m

3 0.470 ( 10-3 ) N # s>m2 4 (400 s-1)

= 0.188 N>m2

Here, the area per unit length of plate is A = 3 m. Thus

F = ( tA + tB ) A = ( 0.02875p N>m2 + 0.188 N>m2 ) (3 m) Ans.

= 0.835 N>m

Ans: 0.835 N>m 47

*1–48. Determine the constants B and C in Andrade’s equation for water if it has been experimentally determined that m = 1.00 ( 10-3 ) N # s>m2 at a temperature of 20°C and that m = 0.554 ( 10-3 ) N # s>m2 at 50°C.

SOLUTION The Andrade’s equation is m = Be C>T

At T = (20 + 273) K = 293 K, m = 1.00 ( 10-3 ) N # s>m2. Thus 1.00 ( 10-3 ) N # s>m2 = Be C>293 K

ln 3 1.00 ( 10-3 ) 4 = ln ( Be C>293 )

- 6.9078 = ln B + ln e C>293 - 6.9078 = ln B + C>293 (1)

ln B = - 6.9078 - C>293 At T = (50 + 273) K = 323 K, m = 0.554 ( 10

-3

) N # s>m . Thus, 2

0.554 ( 10-3 ) N # s>m2 = Be C>323

ln 3 0.554 ( 10-3 ) 4 = ln ( Be C>323 )

-7.4983 = ln B + ln e C>323 -7.4983 = ln B +

C 323

ln B = - 7.4983 -

C 323

(2)

Equating Eqs. (1) and (2) -6.9078 -

C C = - 7.4983 293 323

0.5906 = 0.31699 ( 10-3 ) C Ans.

C = 1863.10 = 1863 K Substitute this result into Eq. (1) B = 1.7316 ( 10-6 ) N # s>m2 = 1.73 ( 10-6 ) N # s>m2

Ans.

48

1–49. The viscosity of water can be determined using the empirical Andrade’s equation with the constants B = 1.732 ( 10-6 ) N # s>m2 and C = 1863 K. With these constants, compare the results of using this equation with those tabulated in Appendix A for temperatures of T = 10°C and T = 80°C.

SOLUTION The Andrade’s equation for water is m = 1.732 ( 10-6 ) e 1863>T At T = (10 + 273) K = 283 K,

m = 1.732 ( 10-6 ) e 1863 K>283 K = 1.25 ( 10-3 ) N # s>m2

Ans.

From the Appendix at T = 10°C,

m = 1.31 ( 10-3 ) N # s>m2

At T = (80 + 273) K = 353 K,

m = 1.732 ( 10-6 ) e 1863 K>353 K = 0.339 ( 10-3 ) N # s>m2

Ans.

From the Appendix at T = 80°C,

m = 0.356 ( 10-3 ) N # s>m2

Ans: At T = 283 K, m = 1.25 (10 - 3) N # s>m2 At T = 353 K, m = 0.339 (10 - 3) N # s>m2 49

1–50. Determine the constants B and C in the Sutherland equation for air if it has been experimentally determined that at standard atmospheric pressure and a temperature of 20°C, m = 18.3 ( 10-6 ) N # s>m2, and at 50°C, m = 19.6 ( 10-6 ) N # s>m2.

SOLUTION The Sutherland equation is m =

BT 3>2 T + C

At T = (20 + 273) K = 293 K, m = 18.3 ( 10-6 ) N # s>m2. Thus, 18.3 ( 10-6 ) N # s>m2 =

B ( 2933>2 ) 293 K + C

B = 3.6489 ( 10-9 ) (293 + C) At T = (50 + 273) K = 323 K, m = 19.6 ( 10 19.6 ( 10-6 ) N # s>m2 =

-6

(1)

) N # s>m . Thus 2

B ( 3233>2 ) 323 K + C

B = 3.3764 ( 10-9 ) (323 + C)

(2)

Solving Eqs. (1) and (2) yields B = 1.36 ( 10-6 ) N # s> ( m2 K2 ) 1

Ans.

C = 78.8 K

50

Ans: 1 B = 1.36 ( 10 - 6 ) N # s> ( m2 # K 2 2 , C = 78.8 K

1–51. The constants B = 1.357 ( 10-6 ) N # s> ( m2 # K1>2 ) and C = 78.84 K have been used in the empirical Sutherland equation to determine the viscosity of air at standard atmospheric pressure. With these constants, compare the results of using this equation with those tabulated in Appendix A for temperatures of T = 10°C and T = 80°C.

SOLUTION The Sutherland Equation for air at standard atmospheric pressure is m =

1.357 ( 10-6 ) T 3>2 T + 78.84

At T = (10 + 273) K = 283 K, m =

1.357 ( 10-6 )( 2833>2 ) 283 + 78.84

= 17.9 ( 10-6 ) N # s>m2

Ans.

From Appendix A at T = 10°C,

m = 17.6 ( 10-6 ) N # s>m2

At T = (80 + 273) K = 353 K, m =

1.357 ( 10-6 )( 3533>2 ) 353 + 78.84

= 20.8 ( 10-6 ) N # s>m2

Ans.

From Appendix A at T = 80°C,

m = 20.9 ( 10-6 ) N # s>m2

Ans: Using the Sutherland equation, at T = 283 K, m = 17.9 (10-6) N # s>m2 at T = 353 K, m = 20.8 ( 10-6 ) N # s>m2 51

*1–52. The read–write head for a hand-held music player has a surface area of 0.04 mm2. The head is held 0.04 µm above the disk, which is rotating at a constant rate of 1800 rpm. Determine the torque T that must be applied to the disk to overcome the frictional shear resistance of the air between the head and the disk. The surrounding air is at standard atmospheric pressure and a temperature of 20°C. Assume the velocity profile is linear.

8 mm T

SOLUTION Here Air is a Newtonian fluid. v = a1800

rev 2p rad 1 min ba ba b = 60p rad>s. min 1 rev 60 s

Thus, the velocity of the air on the disk is U = vr = (60p)(0.008) = 0.48p m>s. Since the velocity profile is assumed to be linear as shown in Fig. a, 0.48p m>s du U = = = 12 ( 106 ) p s-1 dy t 0.04 ( 10-6 ) m For air at T = 20°C and standard atmospheric pressure, m = 18.1 ( 10-6 ) N # s>m2 (Appendix A). Applying Newton’s law of viscosity, t = m

du = dy

3 18.1 ( 10-6 ) N # s>m2 4 3 12 ( 106 ) p s-1 4

= 217.2p N>m2

Then, the drag force produced is

FD = tA = ( 217.2p N>m2 ) a

0.04 2 m b = 8.688 ( 10-6 ) p N 10002

The moment equilibrium about point O requires a+ ΣMO = 0;

T -

3 8.688 ( 10-6 ) p N 4 (0.008 m)

= 0

T = 0.218 ( 10-6 ) N # m = 0.218 mN # m

Ans.

0.008 m

u

t = 0.04(10–6) m

0 U

0.48 m/s T

y

FD

(a)

52

8.688(10–6)

N (b)

1–53. Disks A and B rotate at a constant rate of vA = 50 rad>s and vB = 20 rad>s respectively. Determine the torque T required to sustain the motion of disk B. The gap, t = 0.1 mm, contains SAE 10 oil for which m = 0.02 N # s>m2. Assume the velocity profile is linear.

100 mm

vA A t B

SOLUTION

vB ! 20 rad/s

Oil is a Newtonian fluid.

T

The velocities of the oil on the surfaces of disks A and B are UA = vAr = (50r) m>s and UB = vBr = (20r) m>s . Since the velocity profile is assumed to be linear as shown in Fig. a,

y

UA - UB du 50r - 20r = = = 300 ( 103 ) r s-1 dy t 0.1 ( 10-3 ) m

UA

50r

Applying Newton’s Law of viscosity, t = m

t = 0.1(10–3) m

du = ( 0.02 N # s>m2 ) 3 300 ( 103 ) r 4 = (6000r) N>m2 dy

u

The shaded differential element shown in Fig. b has an area of dA = 2pr dr. Thus, dF = tdA = (6000r)(2pr dr) = 12 ( 103 ) pr 2 dr. Moment equilibrium about point O in Fig. b requires a+ ΣMO = 0;

T T -

L0

L

r dF = 0

UB

20r

(a)

0.1 m

0.1 m

r 3 12 ( 103 ) pr 2 dr 4 = 0 T =

L0

r 0.1 m

12 ( 103 ) pr 3 dr

= 12 ( 103 ) p a

0 T

r 4 0.1 m b` 4 0

= 0.942 N # m

dr dF

!dA

Ans. (b)

Ans: 0.942 N # m 53

1–54. If disk A is stationary, vA = 0 and disk B rotates at vB = 20 rad>s, determine the torque T required to sustain the motion. Plot your results of torque (vertical axis) versus the gap thickness for 0 … t … 0.1 m. The gap contains SAE10 oil for which m = 0.02 N # s>m2. Assume the velocity profile is linear.

100 mm

vA A t B

SOLUTION

vB ! 20 rad/s

t ( 10-3 ) m T(N # m)

0 ∞

0.02 3.14

0.04 1.57

0.06 1.05

0.08 0.785

T

0.10 0.628 y

T(N.m) 3.5 3.0 2.5 t

2.0 1.5

u

1.0 0.5 0

UB

(a) 0.02

0.04

0.06 (C)

0.08

0.10

20r

t(10–3m)

y of disks A and B are Oil is a Newtonian fluid. The velocities of the oil on the surfaces UA = vAr = 0 and UB = vBr = (20r) m>s. Since the velocity profile is assumed to be linear as shown in Fig. a,

0.1 m

r

UA - UB du 0 - 20r 20r -1 = = = abs dy t t t

0

Applying Newton’s law of viscosity,

du 20r 0.4r t = m` ` = ( 0.02 N # s>m2 ) a b = a b N>m2 dy t t

54

dr dF

T

(b)

!dA

1–54. (continued)

The shaded differential element shown in Fig. b has an area of dA = 2pr dr. Thus, 0.4r 0.8p 2 dF = tdA = a b(2pr dr) = a br dr. Moment equilibrium about point O t t in Fig. b requires a+ ΣMO = 0;

T T T =

L

r dF = 0

L0

0.1 m

L0

0.1 m

T = a T = c

r ca a

0.8p 2 br dr d = 0 t

0.8p 3 br dr t

0.8p r 4 0.1 m ba b ` t 4 0

20 ( 10-6 ) p t

The plot of T vs t is shown Fig. c.

d N#m

Ans.

where t is in m

Ans: T = c 55

20 ( 10-6 ) p t

d N # m, where t is in m

1–55. The tape is 10 mm wide and is drawn through an applicator, which applies a liquid coating (Newtonian fluid) that has a viscosity of m = 0.830 N # s>m2 to each side of the tape. If the gap between each side of the tape and the applicator’s surface is 0.8 mm, determine the torque T at the instant r = 150 mm that is needed to rotate the wheel at 0.5 rad>s. Assume the velocity profile within the liquid is linear.

30 mm 0.5 rad/s

T

r ! 150 mm

SOLUTION

P = 2F = 2(0.02334 N)

Considering the moment equilibrium of the wheel, Fig. a, ΣMA = 0;

W

T - P(0.15 m) = 0

0.15 m

Since the velocity distribution is linear, the velocity gradient will be constant. P = t(2A) = m(2A)

T

du dy

P = ( 0.830 N # s>m ) (2)(0.03 m)(0.01 m) a 2

0.5 rad>s(0.15 m) 0.0008 m

P = 0.04669 N

0x

b

Thus T = (0.04669 N)(0.15 m)

= 7.00 mN # m

Ans.

0y (a)

Ans: 7.00 mN # m 56

*1–56. The very thin tube A of mean radius r and length L is placed within the fixed circular cavity as shown. If the cavity has a small gap of thickness t on each side of the tube, and is filled with a Newtonian liquid having a viscosity m, determine the torque T required to overcome the fluid resistance and rotate the tube with a constant angular velocity of v. Assume the velocity profile within the liquid is linear.

T t

t

r

L

SOLUTION Since the velocity distribution is assumed to be linear, the velocity gradient will be constant. t = m = m

du dy (vr) t

F=

T = 2(m) T =

T - 2tAr = 0 (vr) t

2 µ r 2L t

T

Considering the moment equilibrium of the tube, Fig. a, ΣM = 0;

A

O

(2prL)r

r

4pmvr 3L t

Ans.

57

(a)

1–57. The shaft rests on a 2-mm-thin film of oil having a viscosity of m = 0.0657 N # s>m2. If the shaft is rotating at a constant angular velocity of v = 2 rad>s, determine the shear stress in the oil at r = 50 mm and r = 100 mm. Assume the velocity profile within the oil is linear.

v ! 2 rad/s

T

100 mm

SOLUTION Oil is a Newtonian fluid. Since the velocity distribution is linear, the velocity gradient will be constant. At r = 50 mm, t = m

du dy

t = ( 0.0657 N # s>m2 ) a

(2 rad>s)(50 mm) 2 mm

t = 3.28 Pa

b

Ans.

At r = 100 mm, t = ( 0.0657 N # s>m2 ) a t = 6.57 Pa

(2 rad>s)(100 mm) 2 mm

b

Ans.

Ans: At r = 50 mm, t = 3.28 Pa At r = 100 mm, t = 6.57 Pa 58

1–58. The shaft rests on a 2-mm-thin film of oil having a viscosity of m = 0.0657 N # s>m2. If the shaft is rotating at a  constant angular velocity of v = 2 rad>s, determine the torque T that must be applied to the shaft to maintain the motion. Assume the velocity profile within the oil is linear.

v ! 2 rad/s

T

100 mm

SOLUTION Oil is a Newtonian fluid. Since the velocity distribution is linear, the velocity gradient will be constant. The velocity of the oil in contact with the shaft at an arbitrary point is U = vr. Thus, t = m

mvr du = dy t

dr

Thus, the shear force the oil exerts on the differential element of area dA = 2pr dr shown shaded in Fig. a is mvr 2pmv 2 dF = tdA = a b(2pr dr) = r dr t t

Considering the moment equilibrium of the shaft, Fig. a, a+ ΣMO = 0;

Lr

T =

p a0.0657

r

T O

(a)

dF - T = 0 T =

Substituting,

dF

Lr

dF =

2pmv R 3 r dr t L0

pmvR4 2pmv r 4 R a b` = = t 4 0 2t N#s b(2 rad>s)(0.1 m)4 m2 = 10.32 ( 10-3 ) N # m = 10.3 mN # m 2(0.002 m)

Ans.

Ans: 10.3 mN # m 59

1–59. The conical bearing is placed in a lubricating Newtonian fluid having a viscosity m. Determine the torque T required to rotate the bearing with a constant angular velocity of v. Assume the velocity profile along the thickness t of the fluid is linear.

v

T R

SOLUTION

u

t

Since the velocity distribution is linear, the velocity gradient will be constant. The velocity of the oil in contact with the shaft at an arbitrary point is U = vr. Thus, t = m

mvr du = dy t

From the geometry shown in Fig. a, z =

r tan u

dz =

dr tan u

T

(1)

R ds

Also, from the geometry shown in Fig. b,

r

dz

(2)

dz = ds cos u

dF

Equating Eqs. (1) and (2), dr = ds cos u tan u

φ

dr ds = sin u

(a)

The area of the surface of the differential element shown shaded in Fig. a is 2p dA = 2prds = rdr. Thus, the shear force the oil exerts on this area is sin u dF = tdA = a

z

mvr 2pmv 2 2p ba rdr b = r dr t sin u t sin u

dz

θ ds (b)

Considering the moment equilibrium of the shaft, Fig. a, ΣMz = 0;

T -

L

rdF = 0 2pmv R 3 r dr t sin u L0

L 2pmv r 4 R a b` = t sin u 4 0

T =

=

rdF =

pmvR4 2t sin u

Ans.

Ans: T = 60

pmvR4 2t sin u

*1–60. The city of Denver, Colorado, is at an elevation of 1610 m above sea level. Determine how hot one can prepare water to make a cup of coffee.

SOLUTION At the elevation of 1610 meters, the atmospheric pressure can be obtained by interpolating the data given in Appendix A. patm = 89.88 kPa - a

89.88 kPa - 79.50 kPa b(610 m) = 83.55 kPa 1000 m

Since water boils if the vapor pressure is equal to the atmospheric pressure, then the boiling temperature at Denver can be obtained by interpolating the data given in Appendix A. Tboil = 90°C + a

83.55 - 70.1 b(5°C) = 94.6°C 84.6 - 70.1

Note: Compare this with Tboil = 100°C at 1 atm.

61

Ans.

1–61. How hot can you make a cup of tea if you climb to the top of Mt. Everest (29,000 ft) and attempt to boil water?

SOLUTION At the elevation of 29 000 ft, the atmospheric pressure can be obtained by interpolating the data given in Appendix A patm = 704.4 lb>ft 2 - a = a659.52

704.4 lb>ft 2 - 629.6 lb>ft 2 30 000 lb>ft 2 - 27 500 lb>ft 2

lb 1 ft 2 ba b = 4.58 psi ft 2 12 in

b(29 000 ft - 27 500 ft)

Since water boils if the vapor pressure equals the atmospheric pressure, the boiling temperature of the water at Mt. Everest can be obtained by interpolating the data of Appendix A Tboil = 150°F + a

4.58 psi - 3.72 psi 4.75 psi - 3.72 psi

Note: Compare this with 212°F at 1 atm.

b(160 - 150)°F = 158°F

Ans.

Ans: patm = 4.58 psi, Tboil = 158 °F 62

1–62. The blades of a turbine are rotating in water that has a temperature of 30°C. What is the lowest water pressure that can be developed at the blades so that cavitation will not occur?

SOLUTION From Appendix A, the vapor pressure of water at T = 30°C is py = 4.25 kPa Cavitation (boiling of water) will occur if the water pressure is equal or less than py. Thus Ans.

pmin = py = 4.25 kPa

Ans: 4.25 kPa 63

1–63. As water at 40°C flows through the transition, its pressure will begin to decrease. Determine the lowest pressure it can have without causing cavitation.

SOLUTION From Appendix A, the vapor pressure of water at T = 40°C is py = 7.38 kPa Cavitation (or boiling of water) will occur when the water pressure is equal to or less than py. Thus, Ans.

p min = 7.38 kPa

Ans: 7.38 kPa 64

*1–64. Water at 70°F is flowing through a garden hose. If the hose is bent, a hissing noise can be heard. Here cavitation has occurred in the hose because the velocity of the flow has increased at the bend, and the pressure has dropped. What would be the highest absolute pressure in the hose at this location in the hose?

SOLUTION From Appendix A, the vapor pressure of water at T = 70°F is py = 0.363 lb>in2 Cavitation (boiling of water) will occur if the water pressure is equal or less than py. pmax = py = 0.363 lb>in2

Ans.

65

1–65. Water at 25°C is flowing through a garden hose. If the hose is bent, a hissing noise can be heard. Here cavitation has occurred in the hose because the velocity of the flow has increased at the bend, and the pressure has dropped. What would be the highest absolute pressure in the hose at this location in the hose?

SOLUTION From Appendix A, the vapor pressure of water at T = 25°C is py = 3.17 kPa

Cavitation (boiling of water) will occur if the water pressure is equal or less than py. Ans.

pmax = py = 3.17 kPa

Ans: 3.17 kPa 66

1–66. A stream of water has a diameter of 0.4 in. when it begins to fall out of the tube. Determine the difference in pressure between a point located just inside and a point just outside of the stream due to the effect of surface tension. Take s = 0.005 lb>ft.

SOLUTION Consider a length L of the water column. The free-body diagram of half of this column is shown in Fig. a.

0.4 in.

ΣF = 0 2(s)(L) + po(d)(L) - pi(d)(L) = 0 2s = ( pi - po ) d pi - po = ∆p =

2s d

2(0.005 lb>ft) (0.4 in.>12) ft

= 0.300 lb>ft 2 = 2.08 ( 10-3 ) psi

Ans.

z Po

d

y

x

Pi

!

L

! (a)

Ans: 2.08 ( 10 - 3 ) psi 67

1–67. Steel particles are ejected from a grinder and fall  gently into a tank of water. Determine the largest average diameter of a particle that will float on the water with a contact angle of u = 180°, if the temperature is 80°F. Take gst = 490 lb>ft 3 and s = 0.00492 lb>ft. Assume that 4 each particle has the shape of a sphere where V = pr 2. 3

SOLUTION

!

W

!

The weight of a steel particle is 245p 3 4 d 3 d W = gstV = ( 490 lb>ft 3 ) c p a b d = 3 2 3

d r =2 (a)

Force equilibrium along the vertical, Fig. a, requires + c ΣFy = 0;

d 245p 3 (0.00492 lb>ft) c 2p a b d d = 0 2 3 0.00492pd =

245p 3 d 3

d = 7.762 ( 10-3 ) ft = 0.0931 in.

Ans.

Ans: 0.0931 in. 68

*1–68. When a can of soda water is opened, small gas bubbles are produced within it. Determine the difference in pressure between the inside and outside of a bubble having a diameter of 0.02 in. The surrounding temperature is 60°F. Take s = 0.00503 lb>ft.

SOLUTION

= 0.00503 lb ft

The FBD of a half a bubble shown in Fig. a will be considered. Here A is the projected area. Force equilibrium along the horizontal requires + S ΣFx = 0;

0.02 pout A + (0.00503 lb>ft) c p a ft b d - pinA = 0 12

( pin - pout ) c

2 p 0.02 a ft b d = 8.3833 ( 10-6 ) p lb 4 12

pin - pout

1 ft 2 = ( 12.072 lb>ft ) a b 144 in2

69

Fout = poutA

Fin = pinA

(a)

2

= 0.0838 psi

0.01 in

Ans.

1–69. Determine the distance h that a column of mercury in the tube will be depressed when the tube is inserted into the mercury at a room temperature of 68°F. Set D = 0.12 in.

D

h

SOLUTION

50!

Using the result h =

2s cos u rgr

From the table in Appendix A, for mercury r = 26.3 slug>ft 3 and s = 31.9 ( 10-3 ) 2c 31.9 ( 10-3 ) h =

=

a26.3

slug ft

3

lb d cos (180° - 50°) ft

ba32.2

ft 1 ft bd b c (0.06 in.)a 2 12 in. s

3 -9.6852 ( 10-3 ) ft 4 a

= -0.116 in.

lb . ft

12 in. b 1 ft

Ans.

The negative sign indicates that a depression occurs.

Ans: 0.116 in. 70

1–70. Determine the distance h that the column of mercury in the tube will be depressed when the tube is inserted into the mercury at a room temperature of 68°F. Plot this relationship of h (vertical axis) versus D for 0.05 in. … D … 0.150 in. Give values for increments of ∆D = 0.025 in. Discuss this result.

D

h

SOLUTION

50!

0.05 - 0.279

d(in.) h(in.)

0.075 -0.186

0.100 -0.139

0.125 -0.112

0.150 0.0930

h(in.) 0.025 0.05 0.075 0.100 0.125 0.150 0

d(in.)

−0.1

−0.2

−0.3

From the table in Appendix A, for mercury at 68°F, r = 26.3 slug>ft 3, and s = 31.9 ( 10-3 ) lb>ft. Using the result h =

2s cos u rgr

h = £ h = a

2 3 31.9 ( 10-3 ) lb>ft 4 cos (180° - 50°)

( 26.3 slug>ft )( 32.2 ft>s ) 3(d>2)(1 ft>12 in)4 3

-0.01395 b in. d

2

§a

12 in b 1 ft

where d is in in.

The negative sign indicates that a depression occurs.

Ans: d = 0.075 in., h = 0.186 in. 71

1–71. Water in the glass tube is at a temperature of 40°C. Plot the height h of the water as a function of the tube’s inner diameter D for 0.5 mm … D … 3 mm. Use increments of 0.5 mm. Take s = 69.6 mN>m.

D

h

SOLUTION When water contacts the glass wall, u = 0°. The weight of the rising column of water is p 1 W = gwV = rwg a D2hb = prwgD2h 4 4

σ w

The vertical force equilibrium, Fig. a, requires + c ΣFy = 0;

s(pD) -

σ

1 pr gD2h = 0 4 w h =

4s rwgD

h

From Appendix A, rw = 992.3 kg>m3 at T = 40°C . Then h =

4(0.0696 N>m)

( 992.3 kg>m3 )( 9.81 m>s2 ) D

=

28.6 ( 10-6 ) D

D

m

(a) h(mm)

For 0.5 mm … D … 3 mm

60

D(mm) h(mm)

0.5 57.2

1.0 28.6

1.5 19.07

2.0 14.3

2.5 11.44

3.0 9.53

50 40

The plot of h vs D is shown in Fig. b.

30 20 10 0

0.5 1.0

1.5 2.0 2.5 (b)

3.0

D(mm)

Ans: D = 1.0 mm, h = 28.6 mm 72

*1–72. Many camera phones now use liquid lenses as a means of providing a quick auto-focus. These lenses work by electrically controlling the internal pressure within a liquid droplet, thereby affecting the angle of the meniscus of the droplet, and so creating a variable focal length. To analyze this effect, consider, for example, a segment of a spherical droplet that has a base diameter of 3 mm. The pressure in the droplet is 105 Pa and is controlled through a tiny hole at the center. If the tangent at the surface is 30°, determine the surface tension at the surface that holds the droplet in place.

30!

3 mm

SOLUTION Writing the force equation of equilibrium along the vertical by referring to the FBD of the droplet in Fig. a + c ΣFz = 0;

N a105 2 b 3 p(0.0015 m)2 4 - (s sin 30°) 32p(0.0015 m) 4 = 0 m

s = 0.158 N>m

30º

r = 0.0015m

30º

Ans. 105 N m2 (a)

73

1–73. The tube has an inner diameter d and is immersed in water at an angle u from the vertical. Determine the average length L to which water will rise along the tube due to capillary action. The surface tension of the water is s and its density is r.

d L u

SOLUTION The free-body diagram of the water column is shown in Fig. a. The weight of this prgd 2L d 2 column is W = rg V = rgc p a b L d = . 2 4

x "

For water, its surface will be almost parallel to the surface of the tube (contact angle ≈ 0°). Thus, s acts along the tube. Considering equilibrium along the x axis,

L

2

ΣFx = 0;

s(pd) -

prgd L sin u = 0 4 L =

4s rgd sin u

Ans.

# d W=

N

!Pgd2L 4

(a)

Ans: L = 4s>(rgd sin u) 74

1–74. The tube has an inner diameter of d = 2 mm and is immersed in water. Determine the average length L to which the water will rise along the tube due to capillary action as a function of the angle of tilt, u. Plot this relationship of L (vertical axis) versus u for 10° … u … 30°. Give values for increments of ∆u = 5°. The surface tension of the water is s = 75.4 mN>m, and its density is r = 1000 kg>m3.

d L u

SOLUTION 10 88.5

u(deg.) L(mm)

15 59.4

20 44.9

25 36.4

30 30.7

x

= 0.0754 N m

L(mm) L 100 80

˜

N

60 0.002m

40 20 0

W = [9.81(10–3) h] N (a)

5

10

15

20

25

30

The FBD of the water column is shown in Fig. a. The weight of this column is W = rg V = ( 1000 kg>m3 )( 9.81 m>s2 ) c

p (0.002 m)L d = 4

3 9.81 ( 10-3 ) pL 4 N.

For water, its surface will be almost parallel to the surface of the tube (u ≅ 0°) at the point of contact. Thus, s acts along the tube. Considering equilibrium along x axis, ΣFx = 0;

(0.0754 N>m) 3 p(0.002 m) 4 L = a

0.0154 bm sin u

3 9.81 ( 10-3 ) pL 4 sin u

where u is in deg.

= 0 Ans.

The plot of L versus u is shown in Fig. a.

Ans: L = (0.0154>sin u) m 75

1–75. The marine water strider, Halobates, has a mass of 0.36 g. If it has six slender legs, determine the minimum contact length of all of its legs to support itself in water having a temperature of T = 20°C. Take s = 72.7 mN>m, and assume the legs are thin cylinders that are water repellent.

SOLUTION The force supported by the legs is P =

3 0.36 ( 10-3 ) kg 4 3 9.81 m>s2 4

= 3.5316 ( 10-3 ) N

Here, s is most effective in supporting the weight if it acts vertically upward. This requirement is indicated on the FBD of each leg in Fig. a. The force equilibrium along vertical requires + c ΣFy = 0;

P = 3.5316(10 –3) N l

l

3.5316 ( 10-3 ) N - 2(0.0727 N>m)l = 0 l = 24.3 ( 10-3 ) m = 24.3 mm

Ans.

Note: Because of surface microstructure, a water strider’s legs are highly hydrophobic. That is why the water surface curves downward with u ≈ 0°, instead of upward as it does when water meets glass.

(a)

Ans: 24.3 mm 76

*1–76. The ring has a weight of 0.2 N and is suspended on the surface of the water, for which s = 73.6 mN>m. Determine the vertical force P needed to pull the ring free from the surface. Note: This method is often used to measure surface tension.

P

SOLUTION

50 mm

The free-body diagram of the ring is shown in Fig. a. For water, its surface will be almost parallel to the surface of the wire (u ≈ 0°) at the point of contact, Fig. a. + c ΣFy = 0;

P - W - 2T = 0 P - 0.2 N - 2(0.0736 N>m) 32p(0.05 m) 4 = 0

P = 0.246 N

P

Ans.

T = 7.36(10−2)! N

T = 7.36(10−2)! N

W = 0.2 N (a)

77

1–77. The ring has a weight of 0.2 N and is suspended on the surface of the water. If it takes a force of P = 0.245 N to lift the ring free from the surface, determine the surface tension of the water.

P

SOLUTION

50 mm

The free-body diagram of the ring is shown in Fig. a. For water, its surface will be almost parallel to the surface of the wire (u ≈ 0°) at the point of contact, Fig. a. + c ΣFy = 0;

0.245 N - 0.2 N - 23s(2p(0.05 m)) 4 = 0

P = 0.245 N

s = 0.0716 N>m = 0.0716 N>m

Ans.

T = 0.1

T = 0.1 W = 0.2 N (a)

Ans: 0.0716 N>m 78

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–1. Show that Pascal’s law applies within a fluid that is accelerating, provided there is no shearing stresses acting within the fluid.

SOLUTION Consider the free-body diagram of a triangular element of fluid as shown in Fig. 2–2b. If this element has acceleration components of ax, ay, az, then since dm = rdV the equations of motion in the y and z directions give ΣFy = dmay;

py(∆x)(∆s sin u) -

ΣFz = dmaz;

pz(∆x)(∆s cos u) -

3 p(∆x∆s) 4 sin u

3 p(∆x∆s) 4 cos u

1 = r a ∆x(∆s cos u)(∆s sin u) bay 2

1 1 - g c ∆x(∆s cos u)(∆s sin u) R = r a ∆x(∆s cos u)(∆s sin u) b az 2 2

Dividing by ∆x∆s and letting ∆s S 0, so the element reduces in size, we obtain py = p pz = p

By a similar argument, the element can be rotated 90° about the z axis and ΣFx = dmax can be applied to show px = p. Since the angle u of the inclined face is arbitrary, this indeed shows that the pressure at a point is the same in all directions for any fluid that has no shearing stress acting within it.

79

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–2. The water in a lake has an average temperature of 15°C. If the barometric pressure of the atmosphere is 720 mm of Hg (mercury), determine the gage pressure and the absolute pressure at a water depth of 14 m.

SOLUTION From Appendix A, T = 15°C. rw = 999.2 kg>m3 pg = rwgh = ( 999.2 kg>m3 )( 9.81 m>s2 ) (14 m) = 137.23 ( 103 ) Pa = 137 kPa

Ans.

patm = rHg gh = ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.720 m) = 95.71 kPa pabs = patm + pg = 95.71 kPa + 137.23 kPa Ans.

= 233 kPa

Ans: pg = 137 kPa, pabs = 233 kPa 80

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–3. If the absolute pressure in a tank is 140 kPa, determine the pressure head in mm of mercury. The atmospheric pressure is 100 kPa.

SOLUTION pabs = patm + pg 140 kPa = 100 kPa + pg pg = 40 kPa From Appendix A, rHg = 13 550 kg>m3. p = gHg hHg 40 ( 10

3

) N>m2 = ( 13 550 kg>m3 )( 9.81 m>s2 ) hHg Ans.

hHg = 0.3009 m = 301 mm

Ans: hHg = 301 mm 81

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–4. The oil derrick has drilled 5 km into the ground before it strikes a crude oil reservoir. When this happens, the pressure at the well head A becomes 25 MPa. Drilling “mud” is to be placed into the entire length of pipe to displace the oil and balance this pressure. What should be its density so that the pressure at A becomes zero?

A

5 km

SOLUTION Consider the case when the crude oil is pushing out at A where pA = 25 ( 106 ) Pa, Fig. a. Here, ro = 880 kg>m3 (Appendix A) hence po = rogh = ( 880 kg>m3 )( 9.81 m>s2 ) (5000 m) = 43.164 ( 106 ) Pa pb = pA + po = 25 ( 106 ) Pa + 43.164 ( 106 ) Pa = 68.164 ( 106 ) Pa It is required that pA = 0, Fig. b. Thus pb = pm = rmgh 68.164 ( 106 )

N = rm ( 9.81 m>s2 ) (5000 m) m2 rm = 1390 kg>m3

Ans.

pA = 0

pA = 25(106) Pa

5000 m

5000 m

pm

po

pb

pb = 45.664(106) Pa

(a)

(b)

82

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–5. In 1896, S. Riva-Rocci developed the prototype of the current sphygmomanometer, a device used to measure blood pressure. When it was worn as a cuff around the upper arm and inflated, the air pressure within the cuff was connected to a mercury manometer. If the reading for the high (or systolic) pressure is 120 mm and for the low (or diastolic) pressure is 80 mm, determine these pressures in psi and pascals.

SOLUTION Mercury is considered to be incompressible. From Appendix A, the density of mercury is rHg = 13 550 kg>m3. Thus, the systolic pressure is pS = rHgghs = ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.12 m) = 15.95 kPa = 16.0 ( 103 ) Pa pS = c 15.95 ( 103 )

The diastolic pressure is

2

2

N 1 lb 0.3048 m 1ft da ba b a b = 2.31 psi 2 4.4482 N 1 ft 12 in. m

Ans. Ans.

pd = rHgghd = ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.08 m) = 10.63 ( 103 ) Pa = 10.6 kPa pd = c 10.63 ( 103 )

2

2

N 1 lb 0.3048 m 1 ft da ba b a b = 1.54 psi 2 4.4482 N 1 ft 12 in. m

Ans. Ans.

Ans: ps = 16.0 kPa = 2.31 psi pd = 10.6 kPa = 1.54 psi 83

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–6. Show why water would not be a good fluid to use for a barometer by computing the height to which standard atmospheric pressure will elevate it in a glass tube. Compare this result with that of mercury. Take gw = 62.4 lb>ft 3, gHg = 846 lb>ft 3.

SOLUTION For water barometer, Fig. a, pw = gwhw = patm a62.4

lb lb 12 in. 2 bhw = a14.7 2 b a b 3 1 ft in. ft

Ans.

hw = 33.92 ft = 33.9 ft

For mercury barometer, Fig. b, pHg = gHg hHg = patm 847

lb lb 12 in. 2 h = a14.7 b a b Hg 1 ft in.2 ft 3 hHg = (2.4992 ft) a

12 in. b = 30.0 in. 1 ft

Ans.

A water barometer is not suitable since it requires a very long tube.

hw

hHg

pHg

pw

patm = 14.7 psi (a)

patm = 14.7 psi (b)

Ans: hw = 33.9 ft hHg = 30.0 in. 84

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–7. The underground storage tank used in a service station contains gasoline filled to the level A. Determine the gage pressure at each of the five identified points. Note that point B is located in the stem, and point C is just below it in the tank. Take rg = 730 kg>m3.

1m A 1m

B C

2m D

E

SOLUTION Since the tube is open-ended, point A is subjected to atmospheric pressure, which has zero gauge pressure. Ans.

pA = 0

The pressures at points B and C are the same since they are at the same horizontal level with h = 1 m. pB = pC = ( 730 kg>m3 )( 9.81 m>s2 ) (1 m) = 7.16 kPa

Ans.

For the same reason, pressure at points D and E is the same. Here, h = 1 m + 2 m = 3 m. pD = pE = ( 730 kg>m3 )( 9.81 m>s2 ) (3 m) = 21.5 kPa

Ans.

Ans: pA = 0 pB = pC = 7.16 kPa pD = pE = 21.5 kPa 85

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–8. The underground storage tank contains gasoline filled to the level A. If the atmospheric pressure is 101.3 kPa, determine the absolute pressure at each of the five identified points. Note that point B is located in the stem, and point C is just below it in the tank. Take rg = 730 kg>m3.

1m A 1m

B C

2m D

SOLUTION Since the tube is open-ended, point A is subjected to atmospheric pressure, which has an absolute pressure of 101.3 kPa. pA = patm + pg pA = 101.3 ( 103 ) N>m2 + 0 = 101.3 kPa

Ans.

The pressures at points B and C are the same since they are at the same horizontal level with h = 1 m. pB = pC = 101.3 ( 103 ) N>m2 + ( 730 kg>m3 )( 9.81 m>s2 ) (1 m) Ans.

= 108 kPa For the same reason, pressure at points D and E is the same. Here, h = 1 m + 2 m = 3 m. pD = pE = 101.3 ( 103 ) N>m2 + ( 730 kg>m3 )( 9.81 m>s2 ) (3 m)

Ans.

= 123 kPa

86

E

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–9. The field storage tank is filled with oil. This standpipe is connected to the tank at C, and the  system is open to the atmosphere at B and E. Determine the maximum pressure in the tank is psi if the oil reaches a level of F in the pipe. Also, at what level should the oil be in the tank, so that the absolute maximum pressure occurs in the tank? What is this value? Take ro = 1.78 slug>ft 3.

D B

4 ft E

10 ft C

SOLUTION

8 ft

F 4 ft

A

Since the top of the tank is open to the atmosphere, the free surface of the oil in the tank will be the same height as that of point F. Thus, the maximum pressure which occurs at the base of the tank (level A) is (pA)g = gh = ( 1.78 slug>ft 3 )( 32.2 ft>s2 ) (4 ft) = 229.26

lb 1 ft 2 b = 1.59 psi a ft 2 12 in.

Ans.

Absolute maximum pressure occurs at the base of the tank (level A) when the oil reaches level B. (pA)

abs max

= gh = ( 1.78 slug>ft 3 )( 32.2 ft>s2 ) (10 ft) = 573.16 lb>ft 2 a

1 ft 2 b = 3.98 psi Ans. 12 in.

Ans: ( pA)g = 1.59 psi Absolute maximum pressure occurs when the oil reaches level B. ( pA)abs = 3.98 psi max

87

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–10. The field storage tank is filled with oil. The standpipe is connected to the tank at C and open to the atmosphere at E. Determine the maximum pressure that can be developed in the tank if the oil has a density of 1.78 slug>ft3. Where does this maximum pressure occur? Assume that there is no air trapped in the tank and that the top of the tank at B is closed.

D B

4 ft E

10 ft

SOLUTION

C

Level D is the highest the oil is allowed to rise in the tube, and the maximum gauge pressure occurs at the base of the tank (level A).

8 ft

F 4 ft

A

(p max )g = gh = ( 1.78 slug>ft 3 )( 32.2 ft>s2 ) (8 ft + 4 ft) = a687.79

lb 1 ft 2 b = 4.78 psi b a ft 2 12 in.

Ans.

Ans: ( pmax )g = 4.78 psi 88

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–11. The closed tank was completely filled with carbon tetrachloride when the valve at B was opened, slowly letting  the carbon tetrachloride level drop as shown. If the valve is then closed and the space within A is a vacuum, determine the pressure in the liquid near valve B when h = 25 ft. Also, determine at what level h the carbon tetrachloride will stop flowing out when the valve is opened. The atmospheric pressure is 14.7 psi.

A

h

B

SOLUTION From the Appendix, pct = 3.09 slug>ft 3. Since the empty space A is a vacuum, pA = 0. Thus, the absolute pressure at B when h = 25 ft is (pB)abs = pA + gh = 0 + ( 3.09 slug>ft 3 )( 32.2 ft>s2 ) (25 ft)

The gauge pressure is given by

= a2487.45

lb 1 ft 2 b = 17.274 psi ba 2 12 in. ft

(pB)abs = patm + (pB)g 17.274 psi = 14.7 psi + (pB)g Ans.

(pB)g = 2.57 psi

When the absolute at B equals the atmospheric pressure, the water will stop flowing. Thus, (pB)abs = pA + gh a14.7

lb 12 in. 2 b = 0 + ( 3.09 slug>ft 3 )( 32.2 ft>s2 ) h b a 1 ft ft 2 h = 21.3 ft

Ans.

Note: When the vacuum is produced, it actually becomes an example of a Rayleigh– Taylor instability. The lower density fluid (air) will migrate up into the valve B and then rise into the space A, increasing the pressure, and pushing some water out the valve. This back-and-forth effect will in time drain the tank.

Ans: ( pB)g = 2.57 psi h = 21.3 ft 89

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–12. The soaking bin contains ethyl alcohol used for cleaning automobile parts. If h = 7 ft, determine the pressure developed at point A and at the air surface B within the enclosure. Take gea = 49.3 lb>ft 3.

1 ft

3 ft

B

SOLUTION h

The gauge pressures at points A and B are

2 ft

lb 1 ft 2 b = 1.71 psi = a246.5 2 b a 12 in. ft

Ans.

pB = gea hB = ( 49.3 lb>ft 3 ) (7 ft - 6 ft) = a49.3

6 ft A

lb pA = gea hA = a49.3 3 b (7ft - 2ft) ft

lb 1 ft 2 b = 0.342 psi ba 2 12 in. ft

90

Ans.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–13. The soaking bin contains ethyl alcohol used for cleaning automobile parts. If the pressure in the enclosure is pB = 0.5 psi, determine the pressure developed at point A and the height h of the ethyl alcohol level in the bin. Take gea = 49.3 lb>ft 3.

1 ft

3 ft

B

h

6 ft A

SOLUTION

2 ft

The gauge pressure at point A is (pA)g = pB + gea hBA = 0.5 psi + a49.3

lb 1 ft 2 b b(6 ft 2 ft) a 12 in. ft 3

Ans.

= 1.869 psi = 1.87 psi

The gauge pressure for the atmospheric pressure is (patm)g = 0. Thus, (pB)g = (patm)g + gea hB a0.5

lb 12 in. 2 lb b = 0 + a49.3 3 b(h - 6) b a 2 1 ft in. ft

Ans.

h = 7.46 ft

Ans: ( pA)g = 1.87 psi h = 7.46 ft 91

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–14. The pipes connected to the closed tank are completely filled with water. If the absolute pressure at A is 300 kPa, determine the force acting on the inside of the end caps at B and C if the pipe has an inner diameter of 60 mm.

B 1.25 m

A 0.5 m

0.75 m C

SOLUTION Thus, the force due to pressure acting on the cap at B and C are FB = pBA =

3 292.64 ( 103 ) N>m2 4 3 p ( 0.03 m ) 2 4

= 827.43 N = 827 N FC = pCA =

3 312.26 ( 10 ) N>m 4 3 p ( 0.03 m ) 4 3

2

Ans.

2

= 882.90 N = 883 N

Ans.

Ans: FB = 827 N FC = 883 N 92

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–15. The structure shown is used for the temporary storage of crude oil at sea for later loading into ships. When it is not filled with oil, the water level in the stem is at B (sea level). Why? As the oil is loaded into the stem, the water is displaced through exit ports at E. If the stem is filled with oil, that is, to the depth of C, determine the height h of the oil level above sea level. Take ro = 900 kg>m3, rw = 1020 kg>m3.

A h B 40 m

1m C

SOLUTION

D

The water level remains at B when empty because the gage pressure at B must be zero. It is required that the pressure at C caused by the water and oil be the same. Then

E

5m 1m

10 m

(pC)w = (pC)o rwghw = rogho

( 1020 kg>m3 ) (g)(40 m) = ( 900 kg>m3 ) g(40 m + h) Ans.

h = 5.33 m

Ans: 5.33 m 93

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–16. If the water in the structure in Prob. 2–15 is displaced with crude oil to the level D at the bottom of the cone, then how high h will the oil extend above sea level? Take ro = 900 kg>m3, rw = 1020 kg>m3.

A h B 40 m

1m C

SOLUTION It is required that the pressure at D caused by the water and oil be the same.

D

(pD)w = (pD)o

E 10 m

rwghw = rogho

( 1020 kg>m ) (g)(45 m) = ( 900 kg>m3 ) (g)(45 m + h) 3

Ans.

h = 6.00 m

94

5m 1m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–17. The tank is filled with aqueous ammonia (ammonium hydroxide) to a depth of 3 ft. The remaining volume of the tank contains air under absolute pressure of 20 psi. Determine the gage pressure at the bottom of the tank. Would the results be different if the tank had a square bottom rather than a curved one? Take ram = 1.75 slug>ft 3. The atmospheric pressure is ratm = 14.7 lb>in2.

3 ft

SOLUTION The gage pressure of the air in the tank is (pair)abs = patm + (pair)g 20

lb lb = 14.7 2 + (pair)g in.2 in.

(pair)g = a5.3

lb 12 in. 2 lb b = 763.2 2 ba 2 1 ft in. ft

Using this result, the gage pressure at the bottom of tank can be obtained. (pb)g = (pair)g + gh = 763.2

slug lb + a1.75 3 b ( 32.2 ft>s2 ) (3 ft) 2 ft ft

lb 1 ft 2 b = 6.47 psi b a ft 2 12 in. No, it does not matter what shape the bottom of the tank is. = a932.25

95

Ans.

Ans: No, it does not matter what shape the bottom of the tank is. 1pb 2 g = 6.47 psi

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–18. A 0.5-in.-diameter bubble of methane gas is released from the bottom of a lake. Determine the bubble’s diameter when it reaches the surface. The water temperature is 68°F and the atmospheric pressure is 14.7 lb>in2.

20 ft

SOLUTION Applying the ideal gas law, p = rRT of which T is constant in this case. Thus, p = constant r

Since r =

m , where m is also constant, then V p = constant m>V (1)

pV = constant At the bottom of the lake, the absolute pressure is pb = patm + gWhW = a14.7

lb 12 in. 2 ba b + ( 62.4 lb>ft 3 ) (20 ft) = 3364.8 lb>ft 2 2 1 ft in.

At the surface of the lake, the absolute pressure is ps = patm = a14.7

Using Eq. (1), we can write

lb 12 in. 2 ba b = 2116.8 lb>ft 2 2 1 ft in.

pbVb = psVs 4 3

( 3364.8 lb>ft 2 ) c p a

ds 3 0.5 in. 3 4 b d = ( 2116.8 lb>ft 2 ) c p a b d 2 3 2 d s = 0.5835 in. = 0.584 in.

Ans.

Ans: d s = 0.584 in. 96

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–19. The Burj Khalifa in Dubai is currently the world’s tallest building. If air at 40°C is at an atmospheric pressure of 105  kPa at the ground floor (sea level), determine the absolute pressure at the top of the tower, which has an elevation of 828 m. Assume that the temperature is constant and that air is compressible. Work the problem again assuming that air is incompressible.

SOLUTION

For compressible air, with R = 286.9 J>(kg # K) (Appendix A), T0 = 40°C + 273 = 313 K, z0 = 0, and z = 828 m, p = p0e -(g>RT0)(z - z0) p = (105 kPa)e - 39.81>286.9(313)4(828 - 0) = 95.92 kPa

Ans.

For incompressible air, with r = 1.127 kg>m3 at T = 40°C (Appendix A), p = p0 - rgh = 105 ( 103 ) N>m2 - ( 1.127 kg>m3 )( 9.81 m>s2 ) (828 m) Ans.

= 95.85 kPa

Ans: For compressible air, p = 95.92 kPa For incompressible air, p = 95.85 kPa 97

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–20. The Burj Khalifa in Dubai is currently the world’s tallest building. If air at 100°F is at an atmospheric pressure of 14.7 psi at the ground floor (sea level), determine the absolute pressure at the top of the building, which has an elevation of 2717 ft. Assume that the temperature is constant and that air is compressible. Work the problem again assuming that air is incompressible.

SOLUTION

For compressible air, with R = 1716 ft # lb>(slug # R) (Appendix A), T0 = (100 + 460)°R = 560°R p = p0e -(g>RTo)(z - zo) p = (14.7 psi)e - 332.2>1716(560)4(2717 - 0)

Ans.

= 13.42 psi

3

For incompressible air, with r = 0.00220 slug/ft at T = 100°F (Appendix A), p = p0 - gh = 14.7 lb>in2 - ( 0.00220 slug>ft 3 )( 32.2 ft>s2 ) (2717 ft) a = 13.36 psi

1 ft 2 b 12 in.

98

Ans.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–21. The density r of a fluid varies with depth h, although its bulk modulus EV can be assumed constant. Determine how the pressure varies with depth h. The density at the surface of the fluid is r0.

SOLUTION The fluid is considered compressible.

However, V =

EV = -

m . Then, r

p=0

dp dV>V

- ( m>r2 ) dp dr dV = = r V m>r

z=h z

Therefore, EV =

dp dr>r

p At the surface, where p = 0, r = r0, Fig. a, then

(a)

p dr EV dp = Lr0 r L0 r

or

EV ln a

r b = p r0

r = r0ep>EV Also, p = p0 + rgz dp = rgdz dp = gdz r Since the pressure p = 0 at z = 0 and p at z = h, Fig. a. p

dp

L0 r0e p>E EV r0

11

V

=

L0

h

gdz

- e - p>EV 2 = gh

1 - e - p>EV =

r0gh EV

p = - EV ln a1 -

r0gh b EV

Ans.

Ans: p = -EV ln a1 99

rogh b EV

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–22. Due to its slight compressibility, the density of water varies with depth, although its bulk modulus EV = 2.20 GPa (absolute) can be considered constant. Accounting for this compressibility, determine the pressure in the water at a depth of 300 m, if the density at the surface of the water is r = 1000 kg>m3. Compare this result with assuming water to be incompressible.

SOLUTION The water is considered compressible. Using the definition of bulk modulus, dp dV>V

EV = However, V =

m . Then r

- ( m>r2 ) dr dp dV = = r V m>r

Therefore, dp dr>r

EV =

At the surface, p = 0 and r = 1000 kg>m3, Also, EV = 2.20 Gpa. Then p

dr = dp 3 r L0 L1000 kg>m r

3 2.20 ( 109 ) N>m2 4

p = 2.20 ( 109 ) ln a p

r = 1000 e 2.20(10 )

r b 1000

(1)

9

Also, dp = rgdz dp = 9.81dz r

(2)

Substitute Eq. (1) into (2). dp p

1000e 2.20(10 ) 9

= 9.81dz

Since the pressure p = 0 at z = 0 and p at z = 300 m P

dp

L0 1000e 2.20(10 ) p

9

L0

=

p

- 2.2 ( 106 ) e - 2.20(10 ) `

p

9

0

300 m

9.81dz

= 9.81z `

300 m 0

100

2–22. Continued

p

- 2.2 ( 106 ) 3 e - 2.20(10 ) - 1 4 = 2943 9

p

e - 2.20(10 ) = 0.9987 9

p

ln e - 2.20(10 ) = ln 0.9987 9

-

p 2.20 ( 109 )

= - 1.3386 ( 10-3 )

Compressible: p = 2.945 ( 106 ) Pa = 2.945 MPa

Ans.

If the water is considered incompressible, p = rogh = ( 1000 kg>m3 )( 9.81m>s2 ) (300 m) = 2.943 ( 106 ) Pa = 2.943 MPa

Ans.

Ans: Incompressible: p = 2.943 MPa Compressible: p = 2.945 MPa 101

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–23. As the balloon ascends, measurements indicate that the temperature begins to decrease at a constant rate, from T = 20°C at z = 0 to T = 16°C at z = 500 m. If the absolute pressure and density of the air at z = 0 are p = 101 kPa and r = 1.23 kg>m3, determine these values at z = 500 m.

z

SOLUTION We will first determine the absolute temperature as a function of z. T = 293 - a

293 - 289 bz = (293 - 0.008z) K 500

Using this result to apply the ideal gas law with R = 286.9 J>(kg # K) p p r = p = rRT; = RT 286.9(293 - 0.008z) =

p 84061.7 - 2.2952z

dp = -gdz = -rgdz dp = -

p(9.81)dz 84061.7 - 2.2952z

dp 9.81dz = p 84061.7 - 2.2952z When z = 0, p = 101 ( 103 ) Pa. Then p

z dp dz = - 9.81 p 3 84061.7 - 2.2952z L101(10 ) L0

ln p ` ln c

p 101(103)

p

101 ( 103 )

ln c

p 101 ( 10

3

d = 4.2741 ln a

)

p 101 ( 10

3

= ( - 9.81) c -

)

d = ln c a = a

z 1 ln (84061.7 - 2.2952z) d ` 2.2952 0

84061.7 - 2.2952z b 84061.7

84061.7 - 2.2952z 4.2741 b d 84061.7

84061.7 - 2.2952z 4.2741 b 84061.7

p = 90.3467 ( 10-18 ) (84061.7 - 2.2952z)4.2741

102

2–23. Continued

At z = 500 m, p = 90.3467 ( 10 - 18 ) 3 84061.7 - 2.2952(500) 4 4.2741 = 95.24 ( 103 ) Pa = 95.2 kPa

Ans.

From the ideal gas law; p = rRT;

p = R = constant rT

Thus, p1 p2 = r1T1 r2T2 Where p1 = 101 kPa, r1 = 1.23 kg>m3, T1 = 293 k, p2 = 95.24 kPa, T2 = 289 K. Then 101 kPa

( 1.23 kg>m ) (293 K) 3

=

95.24 kPa r2(289 K)

r2 = 1.176 kg>m3 = 1.18 kg>m3

Ans.

Ans: p = 95.2 kPa r = 1.18 kg>m3 103

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–24. As the balloon ascends, measurements indicate that the temperature begins to decrease at a constant rate, from T = 20°C at z = 0 to T = 16°C at z = 500 m. If the absolute pressure of the air at z = 0 is p = 101 kPa, plot the variation of pressure (vertical axis) verses altitude for 0 … z … 3000 m. Give values for increments of u ∆z = 500 m.

z

SOLUTION We will first determine the absolute temperature as a function of z T = 293 - a

293 - 289 bz = (293 - 0.008z)k 500

Using this result to apply this ideal gas law with R = 286.9 J>kg # k p = rRT;

r =

p p = RT 286.9(293 - 0.008z) =

p 84061.7 - 2.2952z

dp = -gdz = -rgdz dp = -

p(9.81)dz 84061.7 - 2.2952z

dp 9.81dz = p 84061.7 - 2.2952z When z = 0, p = 101 ( 103 ) Pa, Then p

z dp dz = -9.81 3 L101(10 ) p L0 84061.7 - 2.2952z

ln p `

p 101(10 ) 3

p

ln c

101 ( 103 )

ln c

101 ( 10

d = ln a

)

p 101 ( 10

z 1 ln (84061.7 - 2.2952z) d ` 2.2952 0

d = 4.2741 ln a

p 3

= (-9.81) c -

3

)

= a

84061.7 - 2.2952z b 84061.7

84061.7 - 2.2952z 4.2741 b 84061.7

84061.7 - 2.2952z 4.2741 b 84061.7

p = c 90.3467 ( 10-18 ) (84061.7 - 2.2952z)4.2741 d Pa Where z is in m

104

2–24. Continued

The plot of p vs. z is shown in Fig. a z(m) p (kPa)

0

500

1000

1500

2000

2500

3000

101

95.2

89.7

84.5

79.4

74.7

70.1

p(kPa) 110

100

90

80

70

0

z(m) 500

1000

1500

2000

2500

3000

(a)

105

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–25. In the troposphere, the absolute temperature of the air varies with elevation such that T = T0 - Cz, where C is a constant. If p = p0 at z = 0, determine the absolute pressure as a function of elevation.

SOLUTION dp = - gdz = - rgdz Since the ideal gas law gives p = rRT or r =

p , RT

g dz dp = p RT Since p = p0 at z = 0, integrating this equation gives p dp

Lp0 p

ln p ` ln Therefore,

= p p0

=

g z dz R L0 T0 - Cz

z g 1 c ln (T0 - Cz) d R C 0

g p T0 - Cz = ln a b p0 RC T0 p = p0 a

T0 - Cz g>RC b T0

Ans.

Ans: p = p0a 106

T0 - Cz g>RC b T0

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–26. In the troposphere the absolute temperature of the air varies with elevation such that T = T0 - Cz, where C is  a constant. Using Fig. 2–11, determine the constants, T0 and C. If p0 = 101 kPa at z0 = 0, determine the absolute pressure in the air at an elevation of 5 km.

SOLUTION From Fig. 2–11, T0 = 15°C at z = 0. Then 15°C = T0 - C(0) Ans.

T0 = 15°C Also, TC = - 56.5°C at z = 11.0 ( 10

3

) m. Then

-56.5°C = 15°C - C 3 11.0 ( 103 ) m 4 C = 6.50 ( 10-3 ) °C>m

Ans.

Thus, TC =

3 15

The absolute temperature is therefore T = (15 - 6.50 z) + 273 =

- 6.50 ( 10-3 ) z 4 °C

3 288

- 6.50 ( 10-3 ) z 4 K

Substitute the ideal gas law p = rRT or r = dp = -

(1)

p into dp = -gdz = - rgdz, RT

p gdz RT

g dp dz = p RT From the table in Appendix A, gas constant for air is R = 286.9 J>(kg # K). Also, p = 101 kPa at z = 0. Then z 9.81 m>s2 dp dz = -a b # 3 p 286.9 J>(kg K) L0 288 - 6.50 ( 10-3 ) z L101(10 ) Pa p

ln p `

ln c At z = 5 ( 10

3

p

101(103) Pa

p

101 ( 103 )

= 5.2605 ln 3 288 - 6.50 ( 10-3 ) z 4 `

d = 5.2605 ln c

p = 101 ( 103 ) c

) m,

p = 101 ( 10 ) W 3

288 -

288 - 6.50 ( 10-3 ) z 288

288 - 6.50 ( 10-3 ) z 288

3 6.50 ( 10-3 ) 4 3 5 ( 103 ) 4 288

= 53.8 ( 103 ) Pa = 53.8 kPa

d

z 0

d

5.2605

5.2605



Ans. Ans: T0 = 15°C C = 6.50 1 10 - 3 2 °C>m p = 53.8 kPa

107

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–27. The density of a nonhomogeneous liquid varies as a function of depth h, such that r = (850 + 0.2h) kg>m3, where h is in meters. Determine the pressure when h = 20 m.

SOLUTION Since p = rgh, then the liquid is considered compressible. dp = rg dh Integrating this equation using the gage pressure p = 0 at h = 0 and p at h. Then, L0

P

h

(850 + 0.2 h) (9.81) dh L0 p = ( 8338.5 h + 0.981 h2 ) Pa

dp =

At h = 20 m, this equation gives p =

3 8338.5(20)

+ 0.981 ( 202 ) 4 Pa

= 167.16 ( 103 ) Pa = 167 kPa

Ans.

Ans: 167 kPa 108

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–28. The density of a non-homogeneous liquid varies as a  function of depth h, such that r = (635 + 60h) kg>m3, where h is in meters. Plot the variation of the pressure (vertical axis) versus depth for 0 … h 6 10 m. Give values for increments of 2 m.

SOLUTION 0 0

h(m) p(kPa)

2 13.6

4 29.6

6 48.0

8 68.7

10 91.7

The liquid is considered compressible. Use dp = rgdh Integrate this equation using the gage pressure p = 0 at h = 0 and p at h. Then L0

p

h

(635 + 60h)(9.81) dh L0 p = 3 9.81 ( 635h + 30h2 ) 4 Pa

dp =

p =

3 0.00981 ( 635h

+ 30h2 ) 4 kPa where h is in m.

The plot of p vs h is shown in Fig. a. p(kPa) 100

80

60

40

20

0

h(m) 2

4

6

8

10

(a)

109

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–29. In the troposphere, which extends from sea level to  11 km, it is found that the temperature decreases with altitude such that dT>dz = - C, where C is the constant lapse rate. If the temperature and pressure at z = 0 are T0 and p0, determine the pressure as a function of altitude.

SOLUTION First, we must establish the relation between T, and z using T = T0 at z = 0, T

LT0 L0 T - T0 = - Cz dT = -c

z

dz

T = T0 - Cz Applying the ideal gas lan p p = RT R ( T0 - Cz ) dp = -gdz = - rgdz

p = rRT ;

r =

dp = -

gpdz R ( T0 - Cz )

-g dp dz = ¢ ≤ p R T0 - Cz Using p = p0 at z = 0, p dp

Lp0 p ln p `

p p0

=

- g z dz R L0 T0 - Cz

= -

z g 1 c a - b ln (T0 - Cz) d ` R C 0

g p T0 - Cz ln = ln a b p0 CR T0 ln

p T0 - Cz g>CR = ln c a b d p0 T0

p T0 - Cz g>CR = a b p0 T0 p = p0 a

T0 - Cz g>CR b T0

Ans.

Ans: p = p0a 110

T0 - Cz g>RC b T0

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–30. At the bottom of the stratosphere the temperature is assumed to remain constant at T = T0. If the pressure is p = p0, where the elevation is z = z0, derive an expression for the pressure as a function of elevation.

SOLUTION p = rRT0 dp = -rgdz =

- pg dz RT0

g dp = dz p RT0 ln p = At z = z0, p = p0, so that

zg + C RT0

p = e -(z - z0)g>RT0 p0 p = p0 e -(z - z0)g>RT0

Ans.

Ans: p = p0e -(z - z0)g>RT0 111

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–31. Determine the pressure at an elevation of z = 20 km into the stratosphere if the temperature remains constant at T0 = -56.5°C. Assume the stratosphere beings at z = 11 km as shown in Fig. 2–11.

SOLUTION Within the Troposphere TC = T0 - Cz, and Fig. 2-7 gives TC = 15°C at z = 0. Then 15°C = T0 - C(0) T0 = 15°C Also, TC = - 56.5°C at z = 11.0 ( 103 ) m. Then - 56.5°C = 15°C - C 3 11.0 ( 103 ) m 4 C = 6.50 ( 10 -3 ) °C>m

Thus TC =

3 15

- 6.50 ( 10-3 ) z 4 °C

The absolute temperature is therefore

T = 15 - 6.50 ( 10-3 ) z + 273 =

Substitute the ideal gas law p = rRT or r = dp = -

3 288

- 6.50 ( 103 ) z 4 k

(1)

p into Eq. 2–4 dp = -gdz = - rgdz, RT

p gdz RT

g dp = dz p RT

(2)

From table in Appendix A, gas constant for air is R = 286.9 J>kg # K. Also, p = 101 kPa at z = 0. Then z 9.81 m>s2 dp dz = -a b # 3 p 286.9 J>kg K L101(10 ) Pa L0 288 - 6.50 ( 10-3 ) z p

p

ln p `

ln c

101(103) Pa

p

101 ( 103 )

= 5.2605 ln 3 288 - 6.50 ( 10-3 ) z 4 `

d = 5.2605 ln c

p = 101 ( 103 ) c

288 - 6.50 ( 10-3 ) z 288

288 - 6.50 ( 10-3 ) z 288

z 0

d

d

5.2605

112

2–31. Continued

At z = 11.0 ( 103 ) m, p = 101 ( 103 ) W

288 -

3 6.50 ( 10-3 ) 4 3 11.0 ( 103 ) 4 288

5.2605



= 22.51 ( 103 ) Pa

Integrate Eq. (2) using this result and T = -56.5°C + 273 = 216.5 K p

z 9.81 m>s2 dp = -c d dz # (286.9 J>kg K)(216.5 K) L11(103) m L22.51(103) Pa p p

ln p ` ln

22.51(103) Pa

p

22.51 ( 103 )

At z = 20 ( 103 ) m,

= - 0.1579 ( 10-3 ) z `

z

11(103) m

= 0.1579 ( 10-3 ) 3 11 ( 103 ) - z 4

-3 3 p = 3 22.51(103)e 0.1579(10 )311(10 ) - z4 4 Pa

-3 3 3 p = 3 22.51 ( 103 ) e 0.1579(10 )311(10 ) - 20(10 )4 4 Pa

= 5.43 ( 103 ) Pa = 5.43 kPa

Ans.

Ans: 5.43 kPa 113

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–32. The can, which weighs 0.2 lb, has an open end. If it is inverted and pushed down into the water, determine the force F needed to hold it under the surface. Assume the air in the can remains at the same temperature as the atmosphere, and that is 70°F. Hint: Account for the change in volume of air in the can due to the pressure change. The atmospheric pressure is patm = 14.7 psi.

1 ft F

SOLUTION 0.5 ft

When submerged, the density of the air in the can changes due to pressure changes. According to the ideal gas law, pV = mRT 0.25 ft

Since the temperature T is constant, mRT is also constant. Thus, (1)

p1V1 = p2V2 When p1 = patm

F

lb 12 in. 2 lb = a14.7 2 b a b = 2116.8 2 , V1 = p(0.125 ft)2(0.5 ft) 1 ft in. ft

p3 = 2179.2

0.5 ft – ∆h

= 7.8125 ( 10-3 ) p ft 3.

lb ft2

0.2 lb

When the can is submerged, the water fills the space shown shaded in Fig. a. Thus, p2 = patm + gw h = a2116.8 = (2210.4 - 62.4∆h)

lb ft 2

lb lb b + a62.4 3 b(1.5 ft - ∆h) ft 2 ft

V2 = p(0.125 ft)2(0.5 ft - ∆h) = Substituting these values into Eq. (1), a2116.8

3 0.015625p(0.5 ft

(a)

- ∆h) 4 ft 3

lb lb b 3 7.8125 ( 10-3 ) p ft 3 4 = c (2210.4 - 62.4∆h) 2 d 2 ft ft 62.4∆h2 - 2241.6∆h + 46.8 = 0

3 0.015625p(0.5

- △h) ft 3 4

Solving for the root 6 0.5 ft, we obtain ∆h = 0.02089 ft Then p2 = 2210.4 - 62.4(0.02089) = 2209.10

lb ft 2

The pressure on top of the can is p3 = patm + gwh = a2116.8

lb lb lb b + a62.4 3 b(1 ft) = 2179.2 2 ft 2 ft ft

Considering the free-body diagram of the can, Fig. b, + c ΣFy = 0;

a2209.10

p2 = 2209.10

∆h

lb lb b 3 p(0.125 ft)2 4 - 0.2 lb - a2179.2 2 b 3 p(0.125 ft)2 4 - F = 0 ft 2 ft Ans.

F = 1.27 lb

114

(b)

lb ft2

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–33. The funnel is filled with oil and water to the levels shown. Determine the depth of oil h′ that must be in the funnel so that the water remains at a depth C, and the mercury is at h = 0.8 m from the top of the funnel. Take ro = 900 kg>m3, rw = 1000 kg>m3, rHg = 13 550 kg>m3.

0.2 m h

A h¿

Oil

D

B Water 0.4 m Mercury

SOLUTION

0.1 m

C

Referring to Fig. a, hCD = 0.2 m + h′ + 0.4 m - 0.8 m = h′ - 0.2 m. Then the manometer rule gives pA + roghAB + rwghBC - rHgghCD = pD 0 + ( 900 kg>m ) gh′ + ( 1000 kg>m3 ) g(0.4) - ( 13 550 kg>m3 ) g(h′ - 0.2 m) = 0 3

Ans.

h′ = 0.2458 m = 246 mm

0.2 m

A

hAB = h′

h = 0.8 m

B

hBC = 0.4 m C

D

hCD

(a)

Ans: 246 mm 115

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–34. The funnel is filled with oil to a depth of h′ = 0.3 m and water to a depth of 0.4 m. Determine the distance h the mercury level is from the top of the funnel. Take ro = 900 kg>m3, rw = 1000 kg>m3, rHg = 13 550 kg>m3.

0.2 m h

A h¿

Oil

D

B Water 0.4 m Mercury

Referring to Fig. a, hCD = 0.2 m + 0.3 m + 0.4 m - h = 0.9 m - h. Then the manometer rule gives pA + roghAB + rwghBC - rHgghCD = pD 0 + ( 900 kg>m ) g(0.3 m) + ( 1000 kg>m ) g(0.4 m) - ( 13 550 kg>m ) (g)(0.9 m - h) = 0 3

C

0.1 m

SOLUTION

3

3

Ans.

h = 0.8506 m = 851 mm

0.2 m hAB = 0.3 m

hBC = 0.4 m

A

h D

B hCD C

(a)

Ans: 851 mm 116

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–35. The 150-mm-diameter container is filled to the top with glycerin, and a 50-mm-diameter thin pipe is inserted within it to a depth of 300 mm. If 0.00075 m3 of kerosene is then poured into the pipe, determine the height h to which the kerosene rises from the top of the glycerin.

50 mm

h

300 mm

SOLUTION The height of the kerosene column in the pipe, Fig. a, is hke =

Vke pr

2

=

A

0.00075 m3 1.2 = a bm p p(0.025 m)2

h

From Appendix A, rke = 814 kg>m3 and rgl = 1260 kg>m3 writing the manometer equation from A S B S C by referring to Fig. a, patm + rkeghke - rglghgl = patm

Thus,

hgl = a

hke =

1.2

m

hgl

814 kg>m3 rke 1.2 bhke = ° ¢a mb = 0.2468 m rgl p 1260 kg>m3

h = hke - hgl =

C

1.2 m - 0.2468 m = 0.1352 m = 135 mm p

B

Ans.

(a)

Ans: 135 mm 117

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–36. The 150-mm-diameter container is filled to the top with glycerin, and a 50-mm-diameter thin pipe is inserted within it to a depth of 300 mm. Determine the maximum volume of kerosene that can be poured into the pipe so it does not come out from the bottom end. How high h does the kerosene rise above the glycerin?

50 mm

h

300 mm

SOLUTION From Appendix A, rke = 814 kg>m3 and rgl = 1260 kg>m3. The kerosene is required to heat the bottom of the tube as shown in Fig. a. Write the manometer equation from A S B S C, patm + rkeghke - rglghgl = patm rgl hke = h rke gl Here, hke = (h + 0.3) m and hgl = 0.3 m. Then (h + 0.3) m = °

1260 kg>m3 814 kg>m3

¢(0.3 m) Ans.

hke = 0.1644 m = 164 mm Thus, the volume of the kerosene in the pipe is Vke = pr 2hke = p(0.025 m)2(0.1644 m + 0.3 m) = 0.9118 ( 10-3 ) m3 = 0.912 ( 10-3 ) m3

A kerosene h C

0.3 m

B

glycerin

(a)

118

Ans.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–37. Determine the pressures at points A and B. The containers are filled with water. 2 ft

Air C

3 ft B

4 ft

SOLUTION pA = gAhA = ( 62.4 lb>ft 3 ) (2ft + 4ft) = a374.4 pB = gBhB = ( 62.4 lb>ft 3 ) (3 ft) = a187.2

1 ft 2 lb ba b = 2.60 psi 2 12 in. ft

1 ft 2 lb ba b = 1.30 psi 2 12 in. ft

A

Ans. Ans.

Ans: pA = 2.60 psi, pB = 1.30 psi 119

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–38. Determine the pressure at point C. The containers are filled with water. 2 ft

Air C

3 ft B

4 ft

A

SOLUTION pC = ghC = ( 62.4 lb>ft 3 ) (2 ft) = a124.8

lb 1 ft ba b = 0.870 psi ft 2 12 in.

Ans.

Ans: 0.870 psi 120

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–39. Butyl carbitol, used in the production of plastics, is stored in a tank having the U-tube manometer. If the U-tube is filled with mercury to level E, determine the pressure in the tank at point A. Take SHg = 13.55, and Sbc = 0.957.

A 300 mm 50 mm C

250 mm

D

E 120 mm 100 mm

B

Mercury

SOLUTION Referring to Fig. a, the manometer rule gives pE + rHg ghDE - rbcg(hCD + hAC) = pA 0 + 13.55 ( 1000 kg>m3 )( 9.81 m>s2 ) (0.120 m) - 0.957 ( 1000 kg>m3 )( 9.81 m>s2 ) (0.05 m + 0.3 m) = pA pA = 12.67 ( 103 ) Pa = 12.7 kPa

Ans.

A

hAC = 0.3 m E

C

hDE = 0.12 m D

hBC = 0.25 m

hCD = 0.05 m

B (a)

Ans: 12.7 kPa 121

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–40. Butyl carbitol, used in the production of plastics, is stored in a tank having the U-tube manometer. If the U-tube is filled with mercury to level E, determine the pressure in the tank at point B. Take SHg = 13.55, and Sbc = 0.957.

A 300 mm 50 mm C

250 mm

SOLUTION pE + rHgghDE + rbcg( - hCD + hBC) = pB

0 + 13.55 ( 1000 kg>m3 )( 9.81 m>s2 ) (0.120 m) + 0.957 ( 1000 kg>m3 )( 9.81 m>s2 ) ( -0.05 m + 0.25 m) = pB pB = 17.83 ( 103 ) Pa = 17.8 kPa

Ans.

A

hAC = 0.3 m E

C

hDE = 0.12 m D

hBC = 0.25 m

hCD = 0.05 m

B (a)

122

120 mm 100 mm

B

Referring to Fig. a, the manometer rule gives

D

E

Mercury

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–41. Water in the reservoir is used to control the water pressure in the pipe at A. If h = 200 mm, determine this pressure when the mercury is at the elevation shown. Take rHg = 13 550 kg>m3. Neglect the diameter of the pipe.

E

D

h

A

400 mm B

SOLUTION

C 150 mm 100 mm 200 mm Mercury

Referring to Fig. a with h = 0.2 m, the manometer rule gives pA + rwghAB - rHg ghBC - rwg(hCD + hDE) = pE

E

pA + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.25 m) - ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.1 m) - ( 1000 kg>m

3

pA = 18.20 ( 10

3

hDE = h

)( 9.81 m>s ) (0.55 m + 0.2m) = 0 2

) Pa = 18.2 kPa

D

Ans.

hCD = 0.55 m

A hAB = 0.25 m

C

B

hBC = 0.1 m (a)

Ans: 18.2 kPa 123

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–42. If the water pressure in the pipe at A is to be 25 kPa, determine the required height h of water in the reservoir. Mercury in the pipe has the elevation shown. Take rHg = 13 550 kg>m3. Neglect the diameter of the pipe.

E

D

h

A

400 mm B

SOLUTION Referring to Fig. a, the manometer rule gives

C 150 mm 100 mm 200 mm Mercury

pA + rwghAB - rHgghBC - rwg(hCD + hDE) = pE 25 ( 103 ) N>m2 + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.25 m) - ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.1 m) - ( 1000 kg>m3 )( 9.81 m>s2 ) (0.550 m + h) = 0 Ans.

h = 0.8934 m = 893 mm E hDE = h D hCD = 0.55 m

A hAB = 0.25 m

C

B

hBC = 0.1 m (a)

Ans: 893 mm 124

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–43. A solvent used for plastics manufacturing consists of cyclohexanol in pipe A and ethyl lactate in pipe B that are being transported to a mixing tank. Determine the pressure in pipe A if the pressure in pipe B is 15 psi. The mercury in the manometer is in the position shown, where h = 1 ft. Neglect the diameter of the pipe. Take Sc = 0.953, SHg = 13.55, and Sel = 1.03.

A

B

C D Mercury

2 ft h

3 ft

0.5 ft

SOLUTION Referring to Fig. a, the manometer rule gives pA + gcghAC + gHghCD - gelghBD = pB pA + 0.953 ( 62.4 lb>ft 3 ) (1.5 ft) + (13.55) ( 62.4 lb>ft 3 ) (1 ft) - (1.03) ( 62.4 lb>ft 3 ) (0.5 ft) = a15 pA = 1257.42

lb 1 ft 2 a b = 8.73 psi ft 2 12 in.

Ans.

lb 12 in. 2 ba b 2 1 ft in.

A

C

B hBD = 0.5 ft

D

hAC = 1.5 ft hCD = 1 ft

(a)

Ans: 8.73 psi 125

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–44. A solvent used for plastics manufacturing consists of cyclohexanol in pipe A and ethyl lactate in pipe B that are being transported to a mixing tank. If the pressure in pipe A is 18 psi, determine the height h of the mercury in the manometer so that a pressure of 25 psi is developed in pipe B. Neglect the diameter of the pipes. Take Sc = 0.953, SHg = 13.55, and Sel = 1.03.

A

B

C D

h

Mercury

SOLUTION Referring to Fig. a, the manometer rule gives

18 lb 12 in. 2 b + 0.953 ( 62.4 lb>ft 3 ) (2.5 ft - h) + 13.55 ( 62.4 lb>ft 3 ) (h) a 1 ft in.2 25 lb 12 in. 2 b a - (1.03) ( 62.4 lb>ft 3 ) (0.5 ft) = 1 ft in.2 h = 1.134 ft = 1.13 ft

0.5 ft

C B

Ans.

hCD = h

hBD = 0.5 ft

D

(a)

126

3 ft

hAC = 2.5 ft – h

A

pA + gchAC + gHghCD - gelhBD = pB

2 ft

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–45. The two pipes contain hexylene glycol, which causes the level of mercury in the manometer to be at h = 0.3 m. Determine the differential pressure in the pipes, pA - pB. Take rhgl = 923 kg>m3, rHg = 13 550 kg>m3. Neglect the diameter of the pipes.

0.1 m

D B h

0.1 m C

SOLUTION

Mercury

Referring to Fig. a, the manometer rule gives A

pA + rhglghAC - rHgghCD - rhglghBD = pB pA + ( 923 kg>m3 )( 9.81 m>s2 ) (0.1 m) - ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.3 m) - ( 923 kg>m3 )( 9.81 m>s2 ) (0.1 m) = pB pA - pB = 39.88 ( 103 ) Pa = 39.9 kPa

Ans.

B hBD = 0.1 m

D h = 0.3 m A CD C hAC = 0.1 m (a)

Ans: 39.9 kPa 127

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–46. The two pipes contain hexylene glycol, which causes the differential pressure reading of the mercury in the manometer to be at h = 0.3 m. If the pressure in pipe A increases by 6 kPa, and the pressure in pipe B decreases by  2 kPa, determine the new differential reading h of the manometer. Take rhgl = 923 kg>m3, rHg = 13 550 kg>m3. Neglect the diameter of the pipes.

0.1 m

D B h

0.1 m C

SOLUTION

Mercury

As shown in Fig. a, the mercury level is at C and D. Applying the manometer rule, A

pA + rhglghAC - rHgghCD - rhglghDB = pB pA + ( 923 kg>m3 )( 9.81 m>s2 ) (0.1 m) - ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.3 m) - ( 923 kg>m3 )( 9.81 m>s2 ) (0.1 m) = pB (1)

pA - pB = 39 877.65 Pa

When the pressure at A and B changes, the mercury level will be at C′ and D′, Fig. a. Then, the manometer rule gives (pA + ∆pA) + rhglghAC′ - rHgghC′D′ - rhglghD′B = (pB - ∆pB)

3 pA

+ 6 ( 103 ) N>m2 4 + ( 923 kg>m3 )( 9.81 m>s2 ) (0.1 m + ∆h) - ( 13.550 kg>m3 )( 9.81 m>s2 ) (0.3 m + 2∆h)

- ( 923 kg>m3 )( 9.81 m>s2 ) (0.1 m - ∆h) =

pA - pB = 31 877.65 + 247741.74 ∆h

3 pB

- 2 ( 103 ) N>m2 4

(2)

Equating Eqs. (1) and (2), we obtain 39 877.65 = 31 877.65 + 247741.74 ∆h ∆h = 0.03229

Thus,

h′ = 0.3 m + 2∆h = 0.3 m + 2(0.03229 m) = 0.36458 m = 365 mm B



D′ hCD = 0.3 m hCD = 0.1 m A

Ans.

hBD = 0.1 m D ′



C C′ ′

(a)

Ans: 365 mm 128

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–47. The inverted U-tube manometer is used to measure the difference in pressure between water flowing in the pipes at A and B. If the top segment is filled with air, and the water levels in each segment are as indicated, determine this pressure difference between A and B. rw = 1000 kg>m3.

75 mm C D

225 mm A

300 mm

150 mm

SOLUTION

B

Notice that the pressure throughout the air in the tube is constant. Referring to Fig. a, And

pA = (pw)1 + pa = rwg(hw)1 + pa pB = (pw)2 + pa = rwg(hw)2 + pa

Therefore, pB - pA = 3rwg(hw)2 + pa 4 - 3rwg(hw)1 + pa 4 = rwg3(hw)2 - (hw)1 4

= ( 1000 kg>m3 )( 9.81 m>s2 ) (0.3 m - 0.225 m) Ans.

= 735.75 Pa = 736 Pa Also, using the manometer equation, pA - rw ghAC + rw ghDB = pB pB - pA = rwg3hDB - hAC 4 pa pa

(hw)2 = 0.3 m

(hw)1 = 0.225 m

(pw)2

(pw)1

pA

pB (a)

Ans: 736 Pa 129

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–48. Solve Prob. 2–47 if the top segment is filled with an oil for which ro = 800 kg>m3.

75 mm C D

225 mm A

300 mm

150 mm B

SOLUTION Referring to Fig. a, write the manometer equation starting at A and ending at B, pA - rwg(hw)1 + roilghoil + rwg(hw)2 = pB pB - pA = rwg3(hw)2 - (hw)1 4 + roilghoil

= ( 1000 kg>m3 )( 9.81 m>s2 ) (0.3 m - 0.225 m) + ( 800 kg>m3 )( 9.81 m>s2 ) (0.075 m) = 1.324 ( 103 ) Pa = 1.32 kPa

Ans. Oil

0.075 m hoil = 0.075 m (hw)1 = 0.225 m (hw)2 = 0.3 m

+ 0.15 m

A

B Water

+

(a)

130

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. Oil

2–49. The pressure in the tank at the closed valve A is 300 kPa. If the differential elevation in the oil level in h = 2.5 m, determine the pressure in the pipe at B. Take ro = 900 kg>m3.

0.5 m C

0.75 m 2m

h

Water A

D

Water

3.5 m B

SOLUTION Referring to Fig. a, the manometer rule gives

hAC = 2.75 m

pA - rwghAC + roghCD + rwghBD = pB 300 ( 103 ) N>m2 - ( 1000 kg>m3 )( 9.81 m>s2 ) (2.75 m) + ( 900 kg>m3 )( 9.81 m>s2 ) (2.5 m) + ( 1000 kg>m

3

C

)( 9.81 m>s ) (3.5 m) = pB 2

hCD = 2.5 m

pB = 329.43 ( 103 ) Pa = 329 kPa

Ans.

A

D hBD = 3.5 m

B

(a)

Ans: 329 kPa 131

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. Oil

2–50. The pressure in the tank at B is 600 kPa. If the differential elevation of the oil is h = 2.25 m, determine the pressure at the closed valve A. Take ro = 900 kg>m3.

0.5 m C

0.75 m 2m

Water A

h D

Water

SOLUTION

3.5 m B

Referring to Fig. a, the manometer rule gives pA - rwghAC + roghCD + rwghBD = pB pA - ( 1000 kg>m3 )( 9.81 m>s2 ) (2.75 m) + ( 900 kg>m3 )( 9.81 m>s2 ) (2.25 m) + ( 1000 kg>m3 )( 9.81 m>s2 ) (3.5 m) = 600 ( 103 ) N>m2 pA = 572.78 ( 103 ) Pa = 573 kPa

Ans.

hAC = 2.75 m

C hCD = 2.25 m A

D hBD = 3.5 m

B

(a)

Ans: 573 kPa 132

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–51. The two tanks A and B are connected using a manometer. If waste oil is poured into tank A to a depth of h = 0.6 m, determine the pressure of the entrapped air in tank B. Air is also trapped in line CD as shown. Take ro = 900 kg>m3, rw = 1000 kg>m3.

h

Air

Oil

D E

2m

Water

C

1m

Air B

1.25 m 0.5 m

1.25 m Water

1.5 m

SOLUTION Referring to Fig. a, the manometer rule gives pA + roghAE + rwghCE + rwghBD = pB 0 + ( 900 kg>m3 )( 9.81 m>s2 ) (0.6 m) + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.25 m) + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.75 m) = pB pB = 15.11 ( 103 ) Pa = 15.1 kPa

Ans.

hAE = 0.6 m A D E

hCE = 0.25 m

B

C

1m

0.25 m hBD = 0.75 m (a)

Ans: 15.1 kPa 133

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

*2–52. The two tanks A and B are connected using a manometer. If waste oil is poured into tank A to a depth of h = 1.25 m, determine the pressure of the trapped air in tank B. Air is also trapped in line CD as shown. Take ro = 900 kg>m3, rw = 1000 kg>m3.

h

Air

Oil

D E

2m

Water

C

Referring to Fig. a, the manometer rule gives pA + roghAE + rwghCE + rwghBD = pB 0 + ( 900 kg>m

3

)( 9.81 m>s2 ) (1.25 m) + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.25 m) + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.75 m) = pB pB = 20.846 ( 103 ) Pa = 20.8 kPa

Ans.

hAE = 1.25 m A

D

E

B hCE = 0.25 m

1m

C 0.25 m hBD = 0.75 m (a)

134

B

1.25 m 0.5 m

SOLUTION

1m

Air

1.25 m Water

1.5 m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–53. Air is pumped into the water tank at A such that the pressure gage reads 20 psi. Determine the pressure at point B at the bottom of the ammonia tank. Take ram = 1.75 slug>ft3.

A Water

1 ft 4 ft

Ammonia

2 ft

3 ft

6 ft C

B

2 ft

SOLUTION

Ammonia

Referring to Fig. a, the manometer rule gives

A

pA + gwhAC + gAmhBC = pB 20 lb 12 in. 2 b + ( 62.4 lb>ft 3 ) (5 ft) + ( 1.75 slug>ft 3 )( 32.2 ft>s2 ) (1 ft) = pB a 1 ft in2 pB = 3248.35

lb 1 ft 2 b = 22.6 psi a ft 2 12 in.

hAC = 5 ft C

Ans. B hBC = 1 ft (a)

Note: This is actually an example of a Rayleigh–Taylor instability. The lower density fluid (ammonia) will actually migrate up into the water tank pushing some of the water below the ammonia. This will take place over time.

Ans: 22.6 psi 135

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–54. Determine the pressure that must be supplied by the pump so that the air in the tank at A develops a pressure of 50 psi at B in the ammonia tank. Take ram = 1.75 slug>ft3.

A Water

1 ft 4 ft

Ammonia

2 ft

3 ft

6 ft C

B

2 ft Ammonia

SOLUTION Referring to Fig. a, the manometer rule gives

A

pA + gwhAC + gAmhBC = pB pA + ( 62.4 lb>ft 3 ) (5 ft) + ( 1.75 slug>ft 3 )( 32.2 ft>s2 ) (1 ft) = a50

lb 12 in. ba b 1 ft in2

hAC = 5 ft

2

C

lb 1 ft 2 pA = a6831.65 2 ba b = 47.4 psi 12 in. ft

Ans.

B hBC = 1 ft (a)

Note: This is actually an example of a Rayleigh–Taylor instability. The lower density fluid (ammonia) will actually migrate up into the water tank pushing some of the water below the ammonia. This will take place over time.

Ans: 47.4 psi 136

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–55. The micro-manometer is used to measure small differences in pressure. The reservoirs R and upper portion of the lower tubes are filed with a liquid having a specific weight of gR, whereas the lower portion is filled with a liquid having a specific weight of gt, Fig. (a). When the liquid flows through the venturi meter, the levels of the liquids with respect to the original levels are shown in Fig. (b). If the cross-sectional area of each reservoir is AR and the crosssectional area of the U-tube is At, determine the pressure difference pA - pB. The liquid in the Venturi meter has a specific weight of gL.

A

A

B

h1

gL R

R

d

d gR

h2

e gt

(a)

SOLUTION

(b)

Write the manometer equation starting at A and ending at B, Fig. a pA + gL ( h1 + d ) + gR ah2 - d + -gR ah2 -

B

A

e b - gte 2

e + db - gL ( h1 - d ) = pB 2

h1

B

d

d

(1)

pA - pB = 2gRd - 2gLd + gte - gRe

Since the same amount of liquid leaving the left reservoir will enter into the left tube, e ARd = At a b 2 Substitute this result into Eq. (1),

d = a

h2

At be 2AR

e

At At be - 2gL a be + gte - gRe 2AR 2AR At At = ec a bgR - a bgL + gt - gR d AR AR

e 2

pA - pB = 2gR a

= ec gt - a1 -

At At bg - a bgL d AR R AR

Ans. (a)

Ans: pA - pB = ec gt - a1 137

At At bg - a bgL d AR R AR

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–56. The Morgan Company manufactures a micromanometer that works on the principles shown. Here there are two reservoirs filled with kerosene, each having a crosssectional area of 300 mm2. The connecting tube has a crosssectional area of 15 mm2 and contains mercury. Determine h if the pressure difference pA - pB = 40 Pa. What would h be if water were substituted for mercury? rHg = 13 550 kg>m3, rke = 814 kg>m3. Hint: Both h1 and h2 can be eliminated from the analysis.

B A

h2

h1 h

SOLUTION Referring to Fig. a, write the manometer equation starting at A and ending at B. pA + r1gh1 - r2gh - r1g(h1 - h + h2) = pB (1)

pA - pB = r2gh - r1gh + r1gh2

Since the same amount of liquid leaving the left reservoir will enter the left tube h2 h AR = a b = At a b 2 2

Substitute this result into Eq. (1)

h2 = a

At bh AR

h1

At pA - pB = r2gh - r1gh + r1g a bh AR pA - pB = hc r2g - a1 -

At br g d AR 1

40 N>m2 = hc ( 13 550 kg>m3 )( 9.81 m>s2 ) - a1 -

Mercury: h = 0.3191 ( 10-3 ) m = 0.319 mm 3

B

h2 2 h

Liquid 2 initial level

Liquid 2

15 b ( 814 kg>m3 )( 9.81 m>s2 ) d 300

Ans.

3

15 b ( 814 kg>m3 )( 9.81 m>s2 ) d 300

Ans.

From the results, we notice that if r2 W r1, h will be too small to be read. Hence, when choosing the liquid to be used, r2 should be slightly larger than r1 so that the sensitivity of the micromanometer is increased.

138

h2 Liquid 1

(2)

When r1 = rke = 814 kg>m , r2 = rw = 1000 kg>m and PA - PB = 40 Pa 40 N>m2 = hc ( 1000 kg>m3 )( 9.81 m>s2 ) - a1 -

A

h 2

When r1 = rke = 814 kg>m3, r2 = rHg = 13550 kg>m3 and pA - pB = 40 Pa,

Water: h = 0.01799 m = 18.0 mm

Liquid 1 initial level

(a)

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. Water

2–57. Determine the difference in pressure pB - pA between the centers A and B of the pipes, which are filled with water. The mercury in the inclined-tube manometer has the level shown SHg = 13.55.

A

B 100 mm C

250 mm

Mercury 40!

D

SOLUTION Referring to Fig. a, the manometer rule gives

hBD = 0.25 m A

pA + rwghAC + rHgghCD - rwghDB = pB pA + ( 1000 kg>m

3

)( 9.81 m>s ) (0.1 m) + 13.55 ( 1000 kg>m )( 9.81 m>s ) (0.15 m) 2

3

2

hAC = 0.1 m

+

B

+

- ( 1000 kg>m3 )( 9.81 m>s2 ) (0.250 m) = pB pB - pA = 18.47 ( 103 ) Pa = 18.5 kPa

Ans.

C D hCD = 0.15 m (a)

Ans: 18.5 kPa 139

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–58. Trichlorethylene, flowing through both pipes, is to be added to jet fuel produced in a refinery. A careful monitoring of pressure is required through the use of the inclined-tube manometer. If the pressure at A is 30 psi and the pressure at B is 25 psi, determine the position s that defines the level of mercury in the inclined-tube manometer. Take SHg = 13.55 and St = 1.466. Neglect the diameter of the pipes.

B

A D C 18 in.

12 in.

30! 14 in.

s Mercury

SOLUTION Referring to Fig. a, the manometer rule gives pA + gthAC - gHghCD - gthBD = pB 30 lb 12 in. 18 14 ba b + 1.466 ( 62.4 lb>ft 3 ) a ft - s b sin 30° - 13.55 ( 62.4 lb>ft 3 ) a ft - s sin 30° b 1 ft 12 12 in2 12 lb 12 in. 2 ft b = 25 2 a b 12 1 ft in 12 in. s = 0.7674 ft a b = 9.208 in = 9.21 in. 1 ft

1.466 ( 62.4 lb>ft 3 ) a

Ans.

B

+

+

12 12

1.5 ft – s C

s sin 30˚

hBD =

D

)

hAC = 18 ft – s sin 30˚ 12 A

)

a

2

s 30˚

hCD =

14 ft – s sin 30˚ 12

(a)

Ans: 9.21 in. 140

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–59. Trichlorethylene, flowing through both pipes, is to be added to jet fuel produced in a refinery. A careful monitoring of pressure is required through the use of the inclined-tube manometer. If the pressure at A is 30 psi and s = 7 in., determine the pressure at B. Take SHg = 13.55 and St = 1.466. Neglect the diameter of the pipes.

B

A D C 18 in.

12 in.

30! 14 in.

s Mercury

SOLUTION Referring to Fig. a, the manometer rule gives pA + gthAC - gHghCD - gthBD = pB

30 lb 12 in. 2 11 7 12 a b + 1.466 ( 62.4 lb>ft 3 ) a ft b - 13.55 ( 62.4 lb>ft 3 ) a ft b - 1.466 ( 62.4 lb>ft 3 ) a ft b = pB 1 ft 24 8 12 in.2 pB = 3530.62 lb>ft 2 a

1 ft 2 b = 24.52 psi = 24.5 psi 12 in.

Ans.

B

+ hAC = 11 sin 30˚ ft = 11 ft 24 12 A

+

s=

7 ft 12

D

hBD =

12 ft 12

11 ft 12 C

30˚ 7 sin 30˚ ft 12

hCD = 14 ft – 7 ft = 7 ft 8 12 24

(a)

Ans: 24.5 psi 141

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–60. The vertical pipe segment has an inner diameter of 100 mm and is capped at its end and suspended from the horizontal pipe as shown. If it is filled with water and the pressure at A is 80 kPa, determine the resultant force that must be resisted by the bolts at B in order to hold the flanges together. Neglect the weight of the pipe but not the water within it.

A

2m

B

2m

SOLUTION

C

The forces acting on segment BC of the pipe are indicated on its free-body diagram, Fig. a. Here, FB is the force that must be resisted by the bolt, Ww is the weight of the water in segment BC of the pipe, and PB is the resultant force of pressure acting on the cross section at B. + c ΣFy = 0;

FB - Ww - pBAB = 0 FB = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m)(p)(0.05 m)2 +

3 80 ( 103 ) N>m2

= 937 N

+ 1000 kg>m3 ( 9.81 m>s2 ) (2 m) 4 p(0.05 m)2 Ans.

FB

pB

Ww

(a)

142

100 mm

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–61. Nitrogen in the chamber is at a pressure of 60 psi. Determine the total force the bolts at joints A and B must resist to maintain the pressure. There is a cover plate at B having a diameter of 3 ft.

B 5 ft

3 ft

SOLUTION The force that must be resisted by the bolts at A and B can be obtained by considering the free-body diagrams in Figs. a and b, respectively. For the bolts at B, Fig. b, + ΣFx = 0; S

pBAB - FB = 0 FB = pBAB = ( 60 lb>in2 ) a = 61073 lb = 61.1 kip

For the bolts at A, Fig. a, + ΣFx = 0; pAAA - FA = 0 S

12 in. 2 b 3 (p)(1.5 ft)2 4 1 ft

pA = 60 psi

2.5 ft

FA

2.5 ft

Ans. (a)

12 in. 2 FA = pAAA = ( 60 lb>in2 ) a b 3 (p)(2.5 ft)2 4 1 ft = 169646 lb = 170 kip

pB = 60 psi

Ans.

1.5 ft

FB

1.5 ft (b)

Ans: FB = 61.1 kip, FA = 170 kip 143

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–62. The storage tank contains oil and water acting at the depths shown. Determine the resultant force that both of  these liquids exert on the side ABC of the tank if the side has a width of b = 1.25 m. Also, determine the location of this resultant, measured from the top of the tank. Take ro = 900 kg>m3.

A 0.75 m B

1.5 m

SOLUTION Loading. Since the side of the tank has a constant width, then the intensities of the distributed loading at B and C, Fig. 2–28b, are

C

wB = roghABb = ( 900 kg>m3 )( 9.81 m>s2 ) (0.75 m)(1.25 m) = 8.277 kN/m

A

wC = wB + rwghBCb = 8.277 kN>m + ( 1000 kg>m3 )( 9.81 m>s2 ) (1.5 m)(1.25 m)

0.75 m

= 26.77 kN/m

B

8.277 kN/m

Resultant Force. The resultant force can be determined by adding the shaded triangular and rectangular areas in Fig. 2–28c. The resultant force is therefore 1.5 m

FR = F1 + F2 + F3 =

1 1 (0.75 m)(8.277 kN>m) + (1.5 m)(8.277 kN>m) + (1.5 m)(18.39 kN>m) 2 2

= 3.104 kN + 12.42 kN + 13.80 kN = 29.32 kN = 29.3 kN

C

Ans.

(a)

26.67 kN/m

As shown, each of these three parallel resultants acts through the centroid of its respective area. y1 =

2 (0.75 m) = 0.5 m 3

y2 = 0.75 m +

1 (1.5 m) = 1.5 m 2

y3 = 0.75 m +

2 (1.5 m) = 1.75 m 3

The location of the resultant force is determined by equating the moment of the resultant above A, Fig. 2–28d, to the moments of the component forces about A, Fig. 2–28c. We have, yPFR = ΣyF;

yP (29.32 kN) = (0.5 m)(3.104 kN) + (1.5 m)(12.42 kN) + (1.75 m)(13.80 kN) Ans.

yP = 1.51 m

A

A y1 = 0.5 m y2 = 1.5 m y3 = 1.75 m

B

F1 = 3.104 kN 8.277 kN/m

yP

F2 = 12.42 kN F3 = 13.80 kN

FR = 29.32 kN

P

C 8.277 kN/m 26.67 kN/m – 8.277 kN/m = 18.39 kN/m (b)

144

(c)

Ans: FR = 29.3 kN, yP = 1.51 m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 3 ft

2–63. Determine the weight of block A if the rectangular gate begins to open when the water level reaches the top of the channel, h = 4 ft. The gate has a width of 2 ft. There is a smooth stop block at C.

A B h ! 4 ft C

SOLUTION Since the gate has a constant width of b = 2 ft, the intensity of the distributed load at C can be computed from wC = gwhCb = ( 62.4 lb>ft 3 ) (4 ft)(2 ft) = 499.2 lb>ft

1 1 F = wC hC = (499.2 lb>ft)(4 ft) = 998.4 lb 2 2

2 (4 ft) 3 4 ft F = 998.4 lb

Referring to the free-body diagram of the gate, Fig. a, 2 998.4 lbc (4 ft) d - WA (3 ft) = 0 3

FC = 0

Ans.

WA = 887.47 lb = 887 lb

WA

3 ft

Bx

The resultant triangular distributed load is shown on the free-body diagram of the gate, Fig. a, and the resultant force of this load is

a+ ΣMB = 0;

By

wC = 499.2 lb ft (a)

Ans: 887 lb 145

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 3 ft

*2–64. Determine the weight of block A so that the 2-ft-radius circular gate BC begins to open when the water level reaches the top of the channel, h = 4 ft. There is a smooth stop block at C.

A B h ! 4 ft C

SOLUTION Since the gate is circular in shape, it is convenient to compute the resultant force as follows. FR = gwhA

By

WA

3 ft

Bx

F = ( 62.4 lb>ft ) (2 ft)(p)(2 ft) = 499.2p lb 3

2

The location of the center of pressure can be determined from yp =

=

Ix yA

a

yp = 2.5 ft 4 ft

+ y p(2 ft)4 4

b

(2 ft)(p)(2 ft)

2

FC = 0

+ 2 ft = 2.50 ft

(a)

Referring to the free-body diagram of the gate, Fig. a, a+ ΣMB = 0;

499.2p lb(2.5 ft) - WA (3 ft) = 0 Ans.

WA = 1306.90 lb = 1.31 kip

146

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–65. The uniform rectangular relief gate AB has a weight of 8000 lb and a width of 4 ft. Determine the minimum depth h of water within the canal needed to open it. The gate is pinned at B and rests on a rubber seal at A.

h

B

6 ft 30!

A

SOLUTION Here hB = h - 6 sin 30° = (h - 3) ft and hA = h. Thus, the intensities of the distributed load at B and A are wB = gwhBb = ( 62.4 lb>ft 3 ) (h - 3 ft)(4 ft) = (249.6h - 748.8) lb>ft wA = gwhAb = ( 62.4 lb>ft 3 ) (h)(4 ft) = (249.6h) lb>ft. Thus,

( Fp ) 1 = 3 (249.6h - 748.8 lb>ft) 4 (6 ft) = (1497.6h - 4492.8) lb 1 2

( Fp ) 2 = 3 (249.6h lb>ft) - (249.6h - 748.8 lb>ft) 4 (6 ft) = 2246.4 lb

If it is required that the gate is about to open, then the normal reaction at A is equal to zero. Write the moment equation of equilibrium about B, referring to Fig. a, a+ ΣMB = 0; 3 (1497.6h - 4492.8 lb) 4 (3 ft) + (2246.4 lb)(4 ft) - (8000 lb) cos 30°(3 ft) = 0

h = 5.626 ft = 5.63 ft

Ans.

3 ft By

Bx

8000 lb

3 ft

30˚ wB

1 (6) = 3 ft 2

2 (6) = 4 ft 3

(Fp)1 (Fp)2

wA

(a)

Ans: 5.63 ft 147

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–66. The uniform swamp gate has a mass of 4 Mg and a width of 1.5 m. Determine the angle u for equilibrium if the water rises to a depth of d = 1.5 m.

2m

A

u

d

SOLUTION Since the gate has a constant width of b = 1.5 m, the intensity of the distributed load at A can be computed from wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.5 m)(1.5 m) = 22.07 ( 103 ) N>m The resulting triangular distributed load is shown on the free-body diagram of the gate, Fig. a. F =

16.554 ( 103 ) 1 1.5 m 1 wAL = c 22.07 ( 103 ) N>m d a b = 2 2 sin u sin u

Referring to the free-body diagram of the gate, Fig. a, a+ ΣMA = 0;

£

16.554 ( 103 ) sin u

§c

1 1.5 m a b d - 3 4000(9.81) N 4 cos u(1 m) = 0 3 sin u

sin 2 u cos u = 0.2109

Solving numerically, u = 29.49° or 77.18° 1.5 m = 1.54 m 6 2 m and sin 77.18° solution is valid.

1.5 m = 3.05 m 7 2 m only one sin 29.49°

Since

Ans.

u = 77.2° Note: This solution represents an unstable equilibrium.

4000(9.81) N

1m

16.554 10 sin

Ax

1.5 m sin

Ay wA = 22.07 103 N/m

3

1 1.5 m 3 sin (a)

Ans: 77.2° 148

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–67. The uniform swamp gate has a mass of 3 Mg and a width of 1.5 m. Determine the depth of the water d if the gate is held in equilibrium at an angle of u = 60°.

2m

A

u

d

SOLUTION Since the gate has a constant width of b = 1.5 m, the intensity of the distributed load at A can be computed from wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (d)(1.5 m) = 14 715d N>m The resulting triangular distributed load is shown on the free-body diagram of the gate, Fig. a. F =

1 1 d 7357.5 2 w L = (14 715d)a d b = 2 A 2 sin 60° sin 60°

Referring to the free-body diagram of the gate, Fig. a, a+ ΣMA = 0;

Since

a

7357.5 2 1 d d bc a b d - 3 3000(9.81) N 4 cos 60°(1 m) = 0 sin 60° 3 sin 60°

Ans.

d = 1.6510 m = 1.65 m

1.6510 m = 1.906 m 6 2 m, this result is valid. sin 60°

Note: This solution represents an unstable equilibrium. The gate is “held” in place by small external stabilizing forces.

3000(9.81) N

60°

1m

d sin 60° Ax F=

7357.5 d2 sin 60°

Ay

d 1 3 sin 60° wA = 14715 d (a)

Ans: 1.65 m 149

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2 ft

*2–68. Determine the critical height h of the water level before the concrete gravity dam starts to tip over due to water pressure acting on its face. The specific weight of concrete is gc = 150 lb>ft3. Hint: Work the problem using a 1-ft width of the dam.

12 ft

h

SOLUTION We will consider the dam as having a width of b = 1 ft. Then the intensity of the distributed load at the base of the dam is

B

A 4 ft

wB = gwhb = ( 62.4 lb>ft 3 ) (h)(1 ft) = 62.4h lb>ft The resulting triangular distributed load is shown on the free-body diagram of the dam, Fig. a. F =

w1 = 3600 lb

1 1 w h = (62.4h)h = 31.2h2 2 B 2

w2 = 1800 lb

It is convenient to subdivide the dam into two parts. The weight of each part is

3 ft

2 (2 ft) 3

W1 = gCV1 = ( 150 lb>ft 3 ) 3 2 ft(12 ft)(1 ft) 4 = 3600 lb

F = 31.2 h2

1 2

W2 = gCV2 = ( 150 lb>ft 3 ) c (2 ft)(12 ft)(1 ft) d = 1800 lb

The dam will overturn about point A. Referring to the free-body diagram of the dam, Fig. a, a+ ΣMA = 0;

h 2 31.2h2 a b - (3600 lb)(3 ft) - (1800 lb) c (2 ft) d = 0 3 3 h = 10.83 ft = 10.8 ft

150

h 3 Ax wB = 62.4 h Ay

Ans.

(a)

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2 ft

2–69. Determine the critical height h of the water level before the concrete gravity dam starts to tip over due to water pressure acting on its face. Assume water also seeps under the base of the dam. The specific weight of concrete is gc = 150 lb>ft3. Hint: Work the problem using a 1-ft width of the dam.

h

SOLUTION We will consider the dam having a width of b = 1 ft . Then the intensity of the distributed load at the base of the dam is

12 ft

B

A 4 ft

wB = gwhbb = ( 62.4 lb>ft 3 ) (h)(1) = 62.4 h lb>ft The resultant forces of the triangular distributed load and uniform distributed load due the pressure of the seepage water shown on the FBD of the dam, Fig. a, are F1 =

1 1 w h = (62.4 h) h = 31.2 h2 2 B 2

F2 = wBLB = 62.4 h(4 ft) = 249.6 h It is convenient to subdivide the dam into two parts. The weight of each part is w1 = gCV1 = ( 150 lb>ft 3 ) 3 (2 ft)(12 ft)(1 ft) 4 = 3600 lb

1 w2 = gCV2 = ( 150 lb>ft 3 ) c (2 ft)(12 ft)(1 ft) d = 1800 lb 2 The dam will overturn about point A. Referring to the FBD of the dam, Fig. a, a+ ΣMA = 0;

h 2 31.2 h2 a b + 249.6 h(2 ft) - (3600 lb)(3 ft) - (1800 lb) c (2 ft)d = 0 3 3 10.4 h3 + 499.2 h - 13200 = 0

Solve numerically, Ans.

h = 9.3598 ft = 9.36 ft

w1 = 3600 lb 3 ft 2 (2 ft) 3 w2 = 1800 lb

F1 = 31.2 h2 h 3

Ax

wB = 62.4 h Ay

2 ft

F2 = 249.6 h (a)

Ans: 9.36 ft 151

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–70. The gate is 2 ft wide and is pinned at A and held in place by a smooth latch bolt at B that exerts a force normal to the gate. Determine this force caused by the water and the resultant force on the pin for equilibrium.

3 ft A

SOLUTION Since the gate has a width of b = 2 ft, the intensities of the distributed loads at A and B can be computed from

3 ft

B

wA = gwhAb = ( 62.4 lb>ft 3 ) (3 ft)(2 ft) = 374.4 lb>ft

3 ft

wB = gwhBb = ( 62.4 lb>ft 3 ) (6 ft)(2 ft) = 748.8 lb>ft The resulting trapezoidal distributed load is shown on the free-body diagram of the gate, Fig. a. This load can be subdivided into two parts. The resultant force of each part is

3 ft

F1 = wALAB = (374.4 lb>ft) 1 322 ft 2 = 1123.222 lb

F2 =

1 1 (w - wA)LAB = (748.8 lb>ft - 374.4 lb>ft) 1 322 ft 2 = 561.622 lb 2 B 2

Considering the free-body diagram of the gate, Fig. a, a + ΣMA = 0;

1 2 1123.222 lb a 322 ft b + 561.622 lb a 322 ft b - NB 1 322 ft 2 = 0 2 3 Ans.

NB = 1323.7 lb = 1.32 kip

ΣFx = 0; a + ΣFy = 0;

Ax = 0 1323.7 lb - 1123.222 lb - 561.622 lb + Ay = 0 Ay = 1058.96 lb = 1.059 kip

Thus, FA = 2(0)2 + (1.059 kip)2 = 1.06 kip wA = 374.4 lb/ft

F1 = 1123.2√2 lb

Ans.

Ax

2 (3√2 ft) 3 1 (3√2 ft) 2

Ay

F2 = 561.6√2 lb 3√2 ft NB

wB = 748.8 lb/ft y

x

Ans: NB = 1.32 kip FA = 1.06 kip

(a)

152

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–71. The tide gate opens automatically when the tide water at B subsides, allowing the marsh at A to drain. For the water level h = 4 m, determine the horizontal reaction at the smooth stop C. The gate has a width of 2 m. At what height h will the gate be on the verge of opening?

D B A 6m

SOLUTION

h

Since the gate has a constant width of b = 2 m, the intensities of the distributed load on the left and right sides of the gate at C are

3.5 m C

(wC)L = rwghBC (b) = ( 1000 kg>m3 )( 9.81 m>s2 ) (4 m)(2 m) = 78.48 ( 103 ) N>m (wC)R = rwghAC (b) = ( 1000 kg>m3 )( 9.81 m>s2 ) (3.5 m)(2 m) = 68.67 ( 103 ) N>m The resultant triangular distributed load on the left and right sides of the gate is shown on its free-body diagram, Fig. a, FL = FR =

1 1 (w ) L = a78.48 ( 103 ) N>mb(4 m) = 156.96 ( 103 ) N 2 C L BC 2

1 1 (w ) L = a68.67 ( 103 ) N>mb(3.5 m) = 120.17 ( 103 ) N 2 C R AC 2

These results can also be obtained as follows

FL = ghLAL = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m) 3 (4 m)(2 m) 4 = 156.96 ( 103 ) N

FR = ghRAR = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.75 m) 3 3.5 m(2 m) 4 = 120.17 ( 103 ) N

Referring to the free-body diagram of the gate in Fig. a, a+ ΣMD = 0;

3 156.96 ( 103 ) N 4 c 2 m

2 (4 m) d 3

+

FC = 25.27 ( 103 ) N = 25.3 kN

3 120.17 ( 103 ) N 4

c 2.5 m +

Ans.

2 (3.5 m) d - FC(6 m) = 0 3

When h = 3.5 m, the water levels are equal. Since FC = 0, the gate will open. Ans.

h = 3.5 m

Dy 2 m) 2 m + —(4 3

Dx

2 2.5 m + —(3.5 m) 3

6m

FL = 156.96(103) N

FR = 120.17(103) N FC

wL = 78.48(10 ) N/m 3

wR = 68.67(103) N/m

153

Ans: FC = 25.3 kN h = 3.5 m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–72. The tide opens automatically when the tide water at B subsides, allowing the marsh at A to drain. Determine the horizontal reaction at the smooth stop C as a function of the depth h of the water level. Starting at h = 6 m, plot values of h for each increment of 0.5 m until the gate begins to open. The gate has a width of 2 m.

D B A 6m h

SOLUTION

3.5 m

Since the gate has a constant width of b = 2 m, the intensities of the distributed loads on the left and right sides of the gate at C are

C

(WC)L = rwghBCb = ( 1000 kg>m3 )( 9.81 m>s2 ) (h)(2 m) = 19.62 ( 103 ) h (WC)R = rwghACb = ( 1000 kg>m3 )( 9.81 m>s2 ) (3.5 m)(2 m) = 68.67 ( 103 ) N>m The resultant forces of the triangular distributed loads on the left and right sides of the gate shown on its FBD, Fig. a, are FL = FR =

1 1 (w ) h = 3 19.62 ( 103 ) h 4 h = 9.81 ( 103 ) h2 2 C L BC 2

1 1 (w ) h = 3 68.67 ( 103 ) N>m 4 (3.5 m) = 120.17 ( 103 ) N 2 C R AC 2

Consider the moment equilibrium about D by referring to the FBD of the gate, Fig. a,

3 9.81 ( 103 ) h2 4 a6 m - h +

a+ ΣMD = 0;

2 2 hb - 120.17 ( 103 ) c2.5 m + (3.5 m)d 3 3 - FC (6 m) = 0

58.86 ( 10 ) h - 3.27 ( 10 ) h - 580.83 ( 103 ) - 6FC = 0 3

2

3

3

FC = ( 9.81 h2 - 0.545 h3 - 96.806 )( 103 ) N FC = ( 9.81 h2 - 0.545 h3 - 96.8 ) kN where h is in meters

Ans.

The gate will be on the verge of opening when the water level on both sides of the gate are equal, that is when h = 3.5 m. The plot of FC vs h is shown in Fig. b. FC (kN) 140 Dy

120 100

Dx

80 60

6m

2 h 6m–h+— 3

FL = 9.81(103) h2

2 2.5 m + —(3.5 m) 3

40

FR = 120.17(103) N

20 h(m)

FC

0

(a)

h(m) FC(kN) 154

1

3.5 0

2

3

4 25.3

4.5 52.2

4

5.0 80.3

5

5.5 109.3

6

6.0 138.6

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. D

2–73. The bin is used to store carbon tetrachloride, a cleaning agent for metal parts. If it is filled to the top, determine the magnitude of the resultant force this liquid exerts on each of the two side plates, AFEB and BEDC, and the location of the center of pressure on each plate, measured from BE. Take rct = 3.09 slug>ft3.

E 2 ft

C F B

3 ft

6 ft 2 ft

SOLUTION Since the side plate has a width of b = 6 ft, the intensities of the distributed load can be computed from

2 ft

A

2 ft

C

2√2 ft

wB = rghBb = ( 3.09 slug>ft 3 )( 32.2 ft>s2 ) (2 ft)(6 ft) = 1193.976 lb>ft wA = rghAb = ( 3.09 slug>ft 3 )( 32.2 ft>s2 ) (5 ft)(6 ft) = 2984.94 lb>ft

P

The resulting distributed load on plates BCDE and ABEF are shown in Figs. a and b, respectively. For plate BCDE, FBCDE =

1 1 (w )L = (1193.976 lb>ft) 1 222 ft 2 = 1688.54 lb = 1.69 kip 2 B BC 2

Ans.

And the center of pressure of this plate from BE is d = For ABEF,

1 1 222 ft 2 = 0.943 ft 3

wB = 1193.976 lb ft

FBCDE d (a)

Ans. wB = 1193.976 lb ft B

F1 = wBLAB = (1193.976 lb>ft)(3 ft) = 3581.93 lb 1 1 F2 = (wA -wB)LAB = (2984.94 lb>ft - 1193.976 lb>ft)(3 ft) = 2686.45 lb 2 2 FABEF = F1 + F2 = 3581.93 lb + 2686.45 lb = 6268.37 lb = 6.27 kip

Ans.

The location of the center of pressure measured from BE can be obtained by equating the sum of the moments of the forces in Figs. b and c. a+ MRB = ΣMB;

B

1 2 (6268.37 lb)d′ = (3581.93 lb)c (3 ft) d + (2686.45 lb) + a (3 ft) b 2 3

1 ft) —(3 2 ft) 2 —(3 3 F1 F2 A wA = 2984.94 lb ft (b)

Ans.

d′ = 1.714 ft = 1.71 ft

B d′ P FABEF

A (c)

Ans: FBCDE = 1.69 kip, d = 0.943 ft FABEF = 6.27 kip, d′ = 1.71 ft 155

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–74. A swimming pool has a width of 12 ft and a side profile as shown. Determine the resultant force the pressure of the water exerts on walls AB and DC, and on the bottom BC.

A

D 3 ft

C

8 ft B 20 ft

SOLUTION I Since the swimming pool has a constant width of b = 12 ft, the intensities of the distributed load at B and C can be computed from

A

wB = ghABb = ( 62.4 lb>ft 3 ) (8ft)(12ft) = 5990.4 lb>ft wC = ghDCb = ( 62.4 lb>ft 3 ) (3 ft)(12 ft) = 2246.4 lb>ft

8 ft

Using these results, the distributed loads acting on walls AB and CD and bottom BC are shown in Figs. a, b, and c. 1 1 wBhAB = (5990.4 lb>ft)(8 ft) = 23 962 lb = 24.0 kip 2 2

Ans.

1 1 = wChCD = (2246.4 lb>ft)(3 ft) = 3369.6 lb = 3.37 kip 2 2

Ans.

FAB = FDC

1 1 (w + wC)LBC = 3 5990.4 lb>ft + 2246.4lb>ft 4 (20 ft) 2 B 2 = 82 368 lb = 82.4 kip

FAB

B wB = 5990.4 lb/ft (a)

FBC =

Ans.

D

SOLUTION II The same result can also be obtained as follows. For wall AB, FAB = ghABAAB = ( 62.4 lb>ft 3 ) (4 ft) 3 8 ft(12 ft) 4 = 23 962 lb = 24.0 kip

Ans.

3 ft

FCD

For wall CD,

FCD = ghCDACD = ( 62.4 lb>ft 3 ) (1.5 ft) 3 3 ft(12 ft) 4 = 3369.6 lb = 3.37 kip Ans.

C wC = 2246.4 lb/ft

For floor BC,

FBC = ghBCABC = ( 62.4 lb>ft 3 ) (5.5 ft) 3 20 ft(12 ft) 4 = 82 368 lb = 82.4 kip Ans.

(b)

FBC wB = 5990.4 lb/ft

wC = 2246.4 lb

C B

20 ft (c)

Ans: FAB = 24.0 kip FDC = 3.37 kip FBC = 82.4 kip 156

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–75. The pressure of the air at A within the closed tank is  200 kPa. Determine the resultant force acting on the plates BC and CD caused by the water. The tank has a width of 1.75 m.

A

2m

B

C 1.5 m

SOLUTION

D

pC = pB = pA + rghAB

1.25 m

= 200 ( 103 ) N>m2 + ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m) = 219.62 ( 103 ) Pa pD = pA + rghAD = 200 ( 103 ) N>m2 + ( 1000 kg>m3 )( 9.81 m>s2 ) (3.5 m) = 234.335 ( 103 ) Pa

1.25 m

Since plates BC and CD have a constant width of b = 1.75 m, the intensities of the distributed load at points B (or C) and D are

C

B

wC = wB = pBb = ( 219.62 ( 103 ) N>m2 ) (1.75m) = 384.335 ( 103 ) N>m wD = pDb = ( 234.335 ( 103 ) N>m2 ) (1.75 m) = 410.086 ( 103 ) N>m

wB = 384.335 (103) N/m

Using these results, the distributed loads acting on plates BC and CD are shown in Figs. a and b, respectively. FBC FCD

FBC

N = wBLBC = c 384.335 ( 10 ) d (1.25 m) = 480.42 ( 103 ) N = 480 kN Ans. m 1 = (FCD)1 + (FCD)2 = wCLCD + (wD - wC)LCD 2 N 1 N N = c 384.335 ( 103 ) d (1.5 m) + c 410.086 ( 103 ) - 384.335 ( 103 ) d (1.5 m) m 2 m m 3

= 595.82 ( 103 ) N = 596 kN

(a)

wC = 384.335 (103) N/m

Ans.

1.5 m

(FCD)1

(FCD)2

wD – wC wD = 410.086 (103) N/m (b)

Ans: FBC = 480 kN, FCD = 596 kN 157

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–76. Determine the smallest base length b of the concrete gravity dam that will prevent the dam from overturning due to water pressure acting on the face of the dam. The density of concrete is rc = 2.4 Mg>m3. Hint: Work the problem using a 1-m width of dam.

9m

SOLUTION

b

If we consider the dam as having a width of b = 1 m, the intensity of the distributed load at the base of the dam is wb = rgh(b) = ( 1000 kg>m3 )( 9.81 m>s2 ) (9 m)(1 m) = 88.29 ( 103 ) N>m The resultant force of the triangular distributed load shown on the free-body diagram of the dam, Fig. a. is F =

1 1 w h = 3 88.29 ( 103 ) N>m 4 (9 m) = 397.305 ( 103 ) N 2 b 2

The weight of the dam is given by W = rC g V =

3 2.4 ( 103 ) kg>m3 4 ( 9.81 m>s2 ) c

= 105 948b

1 (9 m)(1 m)b d 2

The dam will overturn about point O. Referring to the free-body diagram of the dam, Fig. a, 1 2 a+ ΣM0 = 0; 3 397.305 ( 103 ) N 4 c (9 m) d - 105 948b a bb = 0 3 3 Ans.

b = 4.108 m = 4.11 m

w = 105948b

F = 397.305(103) N/m 1 (9 m) 3 Ox wb = 88.29(103) N/m 2b 3 Oy (a)

158

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–77. Determine the smallest base length b of the concrete gravity dam that will prevent the dam from overturning due to water pressure acting on the face of the dam. Assume water also seeps under the base of the dam. The density of concrete is rc = 2.4 Mg>m3. Hint: Work the problem using a 1-m width of dam.

9m

SOLUTION

b

If we consider the dam having a width of b = 1 m, the intensity of the distributed load at the base of the dam is wb = rghb = ( 1000 kg>m3 )( 9.81 m>s2 ) (9 m)(1 m) = 88.29 ( 103 ) N>m The resultant forces of the triangular distributed load and the uniform distributed load due to the pressure of the seepage water shown on the FBD of the dam, Fig. a is 1 1 F1 = wbh = 3 88.29 ( 103 ) N>m 4 (9 m) = 397.305 ( 103 ) N 2 2 F2 = wbb = 3 88.29 ( 103 ) N>m 4 b = 88.29 ( 103 ) b

The weight of the dam is given by W = rC g V =

3 2.4 ( 103 ) kg>m3 4 ( 9.81 m>s2 ) c

= 105948b

1 (b)(9 m)(1 m) d 2

The dam will overturn about point O. Referring to the FBD of the dam, Fig. a, a+ ΣM0 = 0;

3 397.305 ( 103 ) N 4 c

1 (9 m) d + 3

3 88.29 ( 103 ) b 4 a

b = 6.708 m = 6.71 m

b 2 b - 105948 b a bb = 0 2 3 Ans.

w = 105948b

2 — b 3 F1 = 397.305(103) N 1 — (9 m) 3

Ox wb = 88.29(103) N/m

Oy

b — 2

F2 = 88.29(103) b

(a)

Ans: 6.71 m 159

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–78. Determine the placement d of the pin on the 2-ft-wide rectangular gate so that it begins to rotate clockwise (open) when waste water reaches a height h = 10 ft. What is the resultant force acting on the gate?

SOLUTION

A

Since the gate has a constant width of b = 2 ft, the intensity of the distributed load at A and B can be computed from

C

d

3 ft

B

h

wA = gwhAb = ( 62.4 lb>ft 3 ) (3 ft)(2 ft) = 374.4 lb>ft wB = gwhBd = ( 62.4 lb>ft 3 ) (6 ft)(2 ft) = 748.8 lb>ft

4 ft

The resultant trapezoidal distributed load is shown on the free-body diagram of the gate, Fig. a. This load can be subdivided into two parts for which the resultant force of each part is F1 = wALAB = 374.4 lb>ft(3 ft) = 1123.2 lb 1 1 ( w - wA ) LAB = (748.8 lb>ft - 374.4 lb>ft)(3 ft) = 561.6 lb 2 B 2 Thus, the resultant force is

1.5 ft

d – 1.5 ft Cx 2 ft – d

F2 = 561.6 lb

FB = 0

F1 = 1123.2 lb

(561.61b)(2 ft - d) - (1123.2 lb)(d - 1.5 ft) = 0 Ans.

d = 1.67 ft

d

Ans.

When the gate is on the verge of opening, the normal force at A and B is zero as shown on the free-body diagram of the gate, Fig. a. a+ ΣMC = 0;

wA = 374.4 lb/ft FA = 0

F2 =

FR = F1 + F2 = 1123.2 lb + 561.6 lb = 1684.8 lb = 1.68 kip

2 (3 ft) = 2 ft — 3

wB = 748.8 lb/ft (a)

Ans: FR = 1.68 kip d = 1.67 ft 160

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–79. Determine the placement d of the pin on the 3-ftdiameter circular gate so that it begins to rotate clockwise (open) when waste water reaches a height h = 10 ft. What is the resultant force acting on the gate? Use the formula method.

A C

d

3 ft

B

SOLUTION

h

4 ft

Since the gate is circular in shape, it is convenient to compute the resultant force as follows. FR = gwhA = ( 62.4 lb>ft 3 ) (10 ft - 5.5 ft)(p)(1.5 ft)2 = 1984.86 lb = 1.98 kip

Ans.

The location of the center of pressure can be determined from yP =

=

Ix yA

+ y °

p(1.5 ft)4 4

¢

(10 ft - 5.5 ft)(p)(1.5 ft)2

+ (10 ft - 5.5 ft) = 4.625 ft

When the gate is on the verge of opening, the normal force at A and B is zero as shown on the free-body diagram of the gate, Fig. a. Summing the moments about point C requires that FR acts through C. Thus, Ans.

d = yp - 3 ft = 4.625 ft - 3 ft = 1.625 ft = 1.62 ft x 3 ft yp = 4.625 ft FA = 0 d Cx

FR

Cy

FB = 0

(a)

Ans: FR = 1.98 kip d = 1.62 ft 161

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–80. The container in a chemical plant contains carbon tetrachloride, rct = 1593 kg>m3, and benzene, rb = 875 kg>m3. Determine the height h of the carbon tetrachloride on the left side so that the separation plate, which is pinned at A, will remain vertical.

SOLUTION

h

B

1m

CT

1.5 m

CT

Assume 1 m width. The intensities of the distributed load shown in Fig. a are, w2 = rbghbb = ( 875 kg>m3 )( 9.81 m>s2 ) (1 m)(1 m) = 8.584 ( 103 ) N>m w1 = w2 + rCTg(hCT)Rb =

3 8.584 ( 103 ) N>m 4

+ ( 1593 kg>m3 )( 9.81 m>s2 ) (1.5 m)(1 m)

= 32.025 ( 103 ) N>m

w3 = rCTg(hCT)Lb = ( 1593 kg>m3 )( 9.81 m>s2 ) h (1 m) =

3 15.63 ( 103 ) h 4 N>m

Thus, the resultant forces of these distributed loads are F1 = F2 = F3 = F4 =

1 3 32.025 ( 103 ) N>m - 8.584 ( 103 ) N>m 4 (1.5 m) = 17.58 ( 103 ) N 2

3 8.584 ( 103 ) N>m 4 (1.5 m)

= 12.876 ( 103 ) N

1 3 8.584 ( 103 ) N>m 4 (1 m) = 4.292 ( 103 ) N 2 1 3 15.63 ( 103 ) h N>m 4 (h) = 2

And they act at y1 =

3 7.814 ( 103 ) h2 4 N

1.5 m 1.5 m 1m = 0.5 m y2 = = 0.75 m y3 = 1.5 m + = 1.8333 m 3 2 3 h y4 = 3

For the plate to remain vertical, a+ ΣMA = 0; 3 17.58 ( 103 ) N 4 (0.5 m) + +

3 12.876 ( 103 ) N 4 (0.75 m)

3 4.292 ( 103 ) N 4 (1.8333 m)

-

h = 2.167 m = 2.16 m

3 7.814 ( 103 ) h2 N 4 a

F3 w2 h

F2

F4

y3

y1

y4 Ax

w3

F1

y2

w1

Ay

162

h b = 0 3

Ans.

A

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–81. The tapered settling tank is filled with oil. Determine the resultant force the oil exerts on the trapezoidal clean-out plate located at its end. How far from the top of the tank does this force act on the plate? Use the formula method. Take ro = 900 kg>m3.

0.75 m 4m

1m

0.75 m

SOLUTION Referring to the geometry of the plate shown in Fig. a

1m

1 A = (1 m)(1.5 m) + (1.5 m)(1.5 m) = 2.625 m2 2 y =

1 (3.25 m) 3(1 m)(1.5 m) 4 + (3 m) c (1.5 m)(1.5 m) d 2 2.625 m2

= 3.1429 m

1 (1 m)(1.5 m)3 + (1 m)(1.5 m)(3.25 m - 3.1429 m)2 12 1 1 + (1.5 m)(1.5 m)3 + (1.5 m)(1.5 m)(3.1429 m - 3 m)2 36 2

Ix =

= 0.46205 m4 The resultant force is FR = roghA = ( 900 kg>m3 )( 9.81 m>s2 ) (3.1429 m) ( 2.625 m2 ) = 72.84 ( 103 ) N = 72.8 kN

Ans.

And it acts at yP =

Ix 0.46205 m4 + 3.1429 m = 3.199 m = 3.20 m + y = yA (3.1429 m) ( 2.625 m2 )

1 4 – —(1.5) = 3.25 m 3

— — y=h

Ans.

2 4 – —(1.5) = 3 m 3

0.75 m

0.75 m 1m

c2

c2 1.5 m c1

(a)

Ans: FR = 72.8 kN yP = 3.20 m 163

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–82. The tapered settling tank is filled with oil. Determine the resultant force the oil exerts on the trapezoidal clean-out plate located at its end. How far from the top of the tank does this force act on the plate? Use the integration method. Take ro = 900 kg>m3.

0.75 m 4m

1m

0.75 m

SOLUTION With respect to x and y axes established, the equation of side AB of the plate, Fig. a is y - 2.5 4 - 2.5 = ; x - 1.25 0.5 - 1.25

2x = 5 - y x

Thus, the area of the differential element shown shaded in Fig. a is dA = 2xdy = 5 - y dy. The pressure acting on this differential element is p = r0gh = ( 900 kg>m3 )( 9.81 m>s2 ) y = 8829y. Thus, the resultant force acting on the entire plate is FR =

LA

1m

2.5 m 1.25 m

4m

8829y(5 - y)dy L2.5 m

pdA = 2

3

= 22072.5y - 2943y `

x

1.25 m

A

h=y

x

1.5 m

4m

dy

2.5 m

= 72.84 ( 103 ) N = 72.8 kN

Ans.

And it acts at

B 0.5 m 0.5 m y (a)

yP = =

LA

ypdA FR

=

1 72.84 ( 103 )

1 72.84 ( 103 ) N

1 14715y3

= 3.199 m = 3.20 m

`

4m

y(8829y)(5 - y)dy

2.5 m

- 2207.25y4 2 `

4m 2.5 m

Ans.

Ans: FR = 72.8 kN yP = 3.20 m 164

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–83. Ethyl alcohol is pumped into the tank, which has the shape of a four-sided pyramid. When the tank is completely full, determine the resultant force acting on each side, and its location measured from the top A along the side. Use the formula method. rea = 789 kg>m3.

A

6m

SOLUTION

C

The geometry of the side wall of the tank is shown in Fig. a. In this case, it is convenient to calculate the resultant force as follows.

2m

2m

B

4m

2 1 FR = gea hA = ( 789 kg>m3 )( 9.81 m>s2 ) c (6 m) d a b(4 m) 1 240 2 3 2 = 390.1 ( 103 ) N = 390 kN

The location of the center of pressure can be determined from yP =

Ix yA

+ y

1 (4 m) 1 240 m 2 3 36 2 = + a b 1 240 m 2 3 2 1 1 240 m 2 a (4 m) 1 240 m 2 b 3 2 = 4.74 m

2 √40 m y=— 3 A

Ans.

A

yp

√(6 m)2 + (2 m) 11 √40 m

c p 2 h = — (6 m) 3 =4m

2m

2m

(a)

Ans: FR = 390 kN yP = 4.74 m 165

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–84. The tank is filled to its top with an industrial solvent, ethyl ether. Determine the resultant force acting on the plate ABC, and its location on the plate measured from the base AB of the tank. Use the formula method. Take gee = 44.5 lb>ft3.

C 12 ft

60!

SOLUTION

B A

The resultant force is 1 FR = gee hA = ( 44.5 lb>ft 3 ) (8 sin 60° ft) c (10 ft)(12 ft) d 2 = 18.498 ( 103 ) lb = 18.5 kip

Ix =

1 1 bh3 = (10 ft)(12 ft)3 = 480 ft. Then 36 36

yP =

Ix + y = yA

Thus,

Ans.

480 ft + 8 ft = 9 ft 1 (8 ft) c (10 ft)(12 ft) d 2 Ans.

d = 12 ft - yP = 12 ft - 9 ft = 3 ft

yp

12 ft h = 8 sin 60˚ ft y = 8 ft d

1 —(12) = 4 ft 3 y (a)

166

5 ft

5 ft

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–85. Solve Prob. 2–84 using the integration method. C 12 ft

SOLUTION

60!

With respect to x and y axes established, the equation of side AB of the plate, Fig. a, is y - 0 12 - 0 = ; x - 0 5 - 0

B A

5 ft

5 ft

5 y 12

x =

Thus, the area of the differential element shown shaded in Fig. a is dA = 2xdy = 5 5 2 a ybdy = ydy. The pressure acting on this differential element is p = gh = 12 6 ( 44.5 lb>ft 3 )( y sin 60° ) = 38.54 y. Thus, the resultant force acting on the entire plate is FR =

A L

pdA =

= 10.71 y3 `

L0

12 ft

12 ft

5 (38.54y) a ydyb 6

0

= 18.50 ( 10 ) lb = 18.5 kip 3

Ans.

And it acts at

yP =

LA

y pdA =

FR

=

1 18.50 ( 10

3

) L0

1 18.50 ( 10

3

= 9.00 ft

)

12 ft

5 y (38.54y) a ydyb 6

( 8.03y4 ) `

12 ft 0

Thus, d = 12 ft - yp = 3.00 ft

Ans.

x

A y dy

h = y sin 60˚

12 ft

x x B

P 5 ft

60˚ 5 ft

Ans: FR = 18.5 kip d = 3.00 ft

y

(a)

167

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–86. Access plates on the industrial holding tank are bolted shut when the tank is filled with vegetable oil as shown. Determine the resultant force that this liquid exerts on plate A, and its location measured from the bottom of the tank. Use the formula method. rvo = 932 kg>m3.

4m 4m 1.5 m 2.5 m 1m

5m

SOLUTION

2m

B

A

Since the plate has a width of b = 1 m, the intensities of the distributed load at the top and bottom of the plate can be computed from wt = rvo ghtb = ( 932 kg>m3 )( 9.81 m>s2 ) (3 m)(1 m) = 27.429 ( 103 ) N>m wb = rvo ghbb = ( 932 kg>m3 )( 9.81 m>s2 ) (5 m)(1 m) = 45.715 ( 103 ) N>m The resulting trapezoidal distributed load is shown in Fig. a, and this loading can be subdivided into two parts for which the resultant forces are F1 = wt(L) = F2 =

3 27.429 ( 103 ) N>m 4 (2 m)

= 54.858 ( 103 ) N

1 1 (w - wt)(L) = 3 45.715 ( 103 ) N>m - 27.429 ( 103 ) N>m 4 (2 m) = 18.286(103) N 2 b 2

Thus, the resultant force is

FR = F1 + F2 = 54.858 ( 103 ) N + 18.286 ( 103 ) N = 73.143 ( 103 ) N = 73.1 kN Ans. The location of the center of pressure can be determined by equating the sum of the moments of the forces in Figs. a and b about O. a +(MR)O = ΣMO;

3 73.143 ( 103 ) N 4 d

=

3 54.858 ( 103 ) N 4 (1 m)

d = 0.9167 m = 917 mm

+

3 18.286 ( 103 ) N 4 c Ans.

1 (2 m) d 3

wt = 27.429(103) N/m

F2 = 54.858(103) N FR = 73.143(103) N

1m (2 m) d O wb = 45.715(103) N/m

O

F2 = 18.286(10 ) N 3

(a)

(b)

Ans: FR = 73.1 kN d = 917 mm 168

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–87. Access plates on the industrial holding tank are bolted shut when the tank is filled with vegetable oil as shown. Determine the resultant force that this liquid exerts on plate B, and its location measured from the bottom of the tank. Use the formula method. rvo = 932 kg>m3.

4m 4m 1.5 m 2.5 m 1m

5m

SOLUTION

2m

Since the plate is circular in shape, it is convenient to compute the resultant force as follows FR = gvo hA = ( 932 kg>m3 )( 9.81 m>s2 ) (2.5 m) 3 p(0.75 m)2 4 = 40.392 ( 103 ) N = 40.4 kN

A

Ans.

The location of the center of pressure can be determined form

h = 2.5 m

yp

(0.75 m)4

yP =

B

p Ix 4 + 2.5 m + y = yA (2.5 m)(p)(0.75 m)2

FR

= 2.556 m

2.5 m

d

From the bottom of the tank, Fig. a, Ans.

d = 5 m - yP = 5 m - 2.556 m = 2.44 m

(a)

Ans: FR = 40.4 kN d = 2.44 m 169

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–88. Solve Prob. 2–87 using the integration method.

4m 4m 1.5 m 2.5 m 1m

5m

SOLUTION

2m

A

With respect to x and y axes established, the equation of the circumference of the circular plate is x2 + y2 = 0.752;

B

y

x = 20.752 - y2

Thus, the area of the differential element shown shaded in Fig. a is dA = 2xdy = 220.752 - y2 dy. The pressure acting on this differential element is p = rvogh = ( 932 kg>m3 )( 9.81 m>s2 ) (2.5 - y) = 9142.92(2.5 - y). Thus, the resultant force acting on the entire plate is FR =

LA

0.75 m

pdA =

9142.92(2.5 - y) c 220.752 - y2 dy d L-0.75 m

h = 2.5 – y x

x

0.75 m

dy

L-0.75 m

(2.5 - y) a20.752 - y2 b dy

= 18285.84

= 22857.3c y20.752 - y2 + 0.752 sin-1 + 6095.282 ( 0.752 - y2 ) 3 `

= 40.39 ( 10

3

0.75 m

y

0.75 m y d` 0.75 -0.75 m

x

-0.75 m

) N = 40.4 kN

Ans.

And it acts at

yP = =

LA

(a)

(2.5 - y)pdA FR 0.75 m

(2.5 - y) 3 9142.92(2.5 - y) 4 a220.752 - y2 dyb 40.39 ( 103 ) L-0.75 m 1

0.75 m

= 0.4527

(6.25 + y2 - 5y )a20.752 - y2 bdy L-0.75 m

y 0.75 m y = 1.4147c y20.752 - y2 + 0.752 sin-1 d ` + 0.4527c - 2 ( 0.752 - y2 ) 3 a -0.75 m 4 +

0.75 m0.75 m y 0.752 ay20.752 - y2 + a2 sin-1 bd ` + 0.75450 ` 8 0.75 -0.75 m -0.75 m

= 2.5562 m

From the bottom of tank is d = 5 m - yp = 5 m - 2.5562 m = 2.44 m

170

Ans.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–89. The tank truck is filled to its top with water. Determine the magnitude of the resultant force on the elliptical back plate of the tank, and the location of the center of pressure measured from the top of the tank. Solve the problem using the formula method.

y2 –––––– ! x2 " 1 0.75 m2

0.75 m x 1m

SOLUTION Using Table 2-1 for the area and moment of inertia about the centroidal x axis of the elliptical plate, we get F = rwghA = ( 1000 kg>m3 )( 9.81 m>s2 ) (0.75 m)(p)(0.75 m)(1 m) Ans.

= 17.3 kN The center of pressure is at yP =

Ix + y yA

1 c p(1 m)(0.75 m)3 d 4 = + 0.75 m (0.75 m)p(1 m)(0.75 m) Ans.

= 0.9375 m = 0.938 m

Ans: F = 17.3 kN yP = 0.938 m 171

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–90. Solve Prob. 2–89 using the integration method.

y2 –––––– ! x2 " 1 0.75 m2

0.75 m x 1m

SOLUTION By integration of a horizontal strip of area dF = p dA = ( 1000 kg>m3 )( 9.81 m>s2 ) (0.75 m - y)(2x dy) 0.75 m

F = 19 620

L-0.75 m

(0.75 - y)a1 -

y2 (0.75)

0.75 m

= 19 620 c = =

L-0.75 m

2(0.75)2 - y2 dy -

0.75 m

1 y2(0.75)2 - y2 dy d 0.75 L-0.75 m

0.75 y 0.75 19 620 1 19 620 c y 2(0.75)2 - y2 + (0.75)2 sin-1 d c - 2 ( (0.75)2 - y2 ) 3 d 2 0.75 -0.75 0.75 3 -0.75

19 620p(0.75)2 2 L-0.75 m

Ans.

- 0 = 17 336 N = 17.3 kN

0.75 m

19 620 yP =

1 2

b dy 2

y(0.75 - y)a1 17 336 N

y2 (0.75)

1 2

b dy 2

= -0.1875 m Ans.

yP = 0.75 m + 0.1875 m = 0.9375 m = 0.938 m

Ans: F = 17.3 kN yP = 0.938 m 172

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–91. The tank truck is half-filled with water. Determine the magnitude of the resultant force on the elliptical back plate of the tank, and the location of the center of pressure measured from the x axis. Solve the problem using the formula method. Hint: The centroid of a semi-ellipse 4b measured from the x axis is y = 3p .

y2 –––––– ! x2 " 1 0.75 m2

0.75 m x 1m

SOLUTION From Table 2-1, the area and moment of inertia about the x axis of the half-ellipse plate are A = Ix =

p p ab = (1 m)(0.75 m) = 0.375p m2 2 2

1 p 3 1 p a ab b = c (1 m)(0.75 m)3 d = 0.05273p m4 2 4 2 4

Thus, the moment of inertia of the half of ellipse about its centroidal x axis can be determined by using the parallel-axis theorem. Ix = Ix + Ad y2 0.05273p m4 = Ix + (0.375 p) c 4

Ix = 0.046304 m

Since h =

4(0.75 m) 3p

4(0.75 m) 3p

d

2

= 0.3183 m, then

FR = ghA = ( 1000 kg>m3 )( 9.81 m>s2 ) (0.3183 m) ( 0.375p m2 ) = 3.679 ( 103 ) N = 3.68 kN

Ans.

Since y = h = 0.3183 m, yP = =

Ix + y yA 0.046304 m4 (0.3183 m) ( 0.375p m2 )

+ 0.3183 m Ans.

= 0.4418 m = 442 mm

Ans: FR = 3.68 kN yP = 442 mm 173

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

*2–92. Solve Prob. 2–91 using the integration method.

y2 –––––– ! x2 " 1 0.75 m2

0.75 m x 1m

SOLUTION Using a horizontal strip of area dA, dF = pdA dF = ( 1000 kg>m3 )( 9.81 m>s2 ) ( - y)(2x dy) 0

F = - 19 620

L-0.75 m

(y) a1 -

0

= =

1 2

y2

b dy 0.752

19 620 y20.752 - y2 dy 0.75 L-0.75 m

0 26 160 c 2 ( 0.752 - y2 ) 3 d 3 -0.75 m

= 3.679 ( 103 ) N = 3.68 kN 0

- 19 620 yP = -

L-0.75 m 3678.75 N

y(y)a1 -

y2

Ans. 1 2

b dy 0.752

= 0.4418 m = 442 mm

174

Ans.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–93. The trough is filled to its top with carbon disulphide. Determine the magnitude of the resultant force acting on the parabolic end plate, and the location of the center of pressure measured from the top. rcd = 2.46 slug>ft3. Solve the problem using the formula method.

1 ft

y ! 4x2

4 ft x

SOLUTION From Table 2-1, the area and moment of inertia about the centroidal x axis of the parabolic plate are

With h =

A =

2 2 bh = (2 ft)(4 ft) = 5.3333 ft 2 3 3

Ix =

8 8 bh3 = (2 ft)(4 ft)3 = 5.8514 ft 4 175 175

2 2 h = (4 ft) = 1.6 ft, 5 5 FR = ghA = ( 2.46 slug>ft 3 )( 32.2 ft>s2 ) (1.6 ft) ( 5.3333 ft 2 ) Ans.

= 675.94 lb = 676 lb Since y = h = 1.6 ft, yP = =

Ix + y yA 5.8514 ft 4 (1.6 ft) ( 5.3333 ft 2 )

+ 1.6 ft Ans.

= 2.2857 ft = 2.29 ft

Ans: FR = 676 lb yP = 2.29 ft 175

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–94. Solve Prob. 2–93 using the integration method.

1 ft

y ! 4x2

4 ft

SOLUTION

x

Using a horizontal strip of area, FR =

LA

L0

pdA =

= 158.424

L0

= 79.212 a

4 ft

L0

4 ft

( 2.46 slug>ft 3 )( 32.2 ft>s2 ) (4 - y)2x dy

1 1 (4 - y)a b ( y2 ) dy 2

4 ft

3

1

( 4y2 - y2 ) dyb

8 3 2 5 4 ft = 79.212 a y2 - y2 b ` 3 5 0

Ans.

= 675.94 lb = 676 lb

FR(d) =

LA

y(pdA) = 158.424

(675.94 lb)(d) = 158.424 = 79.212

L0

L0

4 ft

4 ft

L0

4 ft

1 1 y(4 - y)a b ( y2 ) dy 2

1 1 y(4 - y)a b y2 dy 2 3

5

( 4y2 - y2 ) dy

2 7 4 ft 8 5 = 79.212 a y2 - y2 b ` 3 5 0 = 1158.76 lb # ft

d =

1158.76 lb # ft = 1.7143 ft 675.94 lb

yP = 4 ft - d = 4 ft - 1.7143 ft Ans.

= 2.2857 ft = 2.29 ft

Ans: FR = 676 lb yP = 2.29 ft 176

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–95. The tank is filled with water. Determine the resultant force acting on the triangular plate A and the location of the center of pressure, measured from the top of the tank. Solve the problem using the formula method.

2m 2m A C

0.3 m 0 .6 m

The resultant force is

0.5 m

0.75 m

0.6 m

SOLUTION

B

0.75 m 0.3 m

1 FR = rwghA = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.75 m) c (0.75 m)(0.75 m) d 2

Ans.

= 4828.36 N = 4.83 kN

1 1 ( bh3 ) = (0.75 m)(0.75 m)3 = 8.7891 ( 10-3 ) m4 Ix = 36 36 yP =

Ix + y = yA

8.7891 ( 10-3 ) m4 1 (1.75 m) c (0.75 m)(0.75 m) d 2

+ 1.75 m Ans.

= 1.768 m = 1.77 m

y = h = 2 – 0.25 = 1.75 m

1 (0.75) = 0.25 m — 3

Ans: FR = 4.83 kN yP = 1.77 m 177

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–96. Solve Prob. 2–95 using the integration method.

2m 2m A C

0.3 m

0.75 m

0 .6 m

From the geometry shown in Fig. a y 0.375 m - x = 0.75 m 0.375 m

0.5 m

0.75 m

0.6 m

SOLUTION

B

0.3 m

x = (0.375 - 0.5y) m

Referring to Fig. b, the pressure as a function of y can be written as p = rwgh = ( 1000 kg>m3 )( 9.81m>s2 ) (2m - y) = 39810(2 - y) 4 N>m2

0.75 m

This pressure acts on the element of area dA = 2xdy = 2(0.375 - 0.5y)dy. Thus,

3 9810(2

dF = pdA =

y

- y) N>m2 4 32(0.375 - 0.5y)dy4

= 19 620 1 0.5y2 - 1.375y + 0.75 2 dy

Then FR =

L

dF = 19 620

= 19 620c

L0

0.75 m

1 0.5y2

x 0.375 m (a)

- 1.375y + 0.75 2 dy

0.75 m 0.5y3 1.375y2 + 0.75y d ` 3 2 0

Ans.

= 4828.36 N = 4.83 kN

2–y

And

y

yp =

=

L

(2 - y)dF FR

L0

0.75 m

=

(2 - y) 3 19 620 1 0.5y - 1.375y + 0.75 2 dy 4 2

y

x

x

4828.36

19 620 =

dy

p

L0

0.75 m

1 - 0.5y3

0.375 m

+ 2.375y2 - 3.5y + 1.5 2 dy

(b)

4828.36

19 620 1 - 0.125y4 + 0.79167y3 - 1.75y2 + 1.5y 2 `

0.75 m 0

4828.36

= 1.768 = 1.77 m

178

Ans.

0.75 m x

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–97. The tank is filled with water. Determine the resultant force acting on the semicircular plate B and the location of the center of pressure, measured from the top of the tank. Solve the problem using the formula method.

2m 2m A C

0.5 m

0.75 m

0.6 m 0.3 m 0 .6 m

SOLUTION

B

0.75 m 0.3 m

The resultant force is FR = rwghA = ( 1000 kg>m3 )( 9.81 m>s2 ) c a2 = 6887.26 N = 6.89 kN

Ix = 0.1098 r = 0.1098(0.5 m) = 6.8625 ( 10 4

Then yp =

Ix + y = yA

4

6.8625 ( 10-3 ) m4 2 1 c a2 b m d c p(0.5 m)2 d 3p 2

-3

2 1 b m d c p(0.5 m)2 d 3p 2

Ans.

)m

+ a2 -

4

2 b m = 1.798 m 3p Ans.

= 1.80 m

( 0.5 m

Ans: FR = 6.89 kN yP = 1.80 m 179

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–98. Solve Prob. 2–97 using the integration method.

2m 2m A C

0.3 m

p = rwgh = ( 1000 kg>m

3

0.75 m

0 .6 m

Referring to Fig. a, the pressure as a function of y can be written as

0.5 m

0.75 m

0.6 m

SOLUTION

B

0.3 m

)( 9.81m>s ) (2m - y) = 3 9810(2 - y) 4 N>m 2

2

1

This pressure acts on the strip element of area dA = 2xdy. Here, x = ( 0.25 - y2 ) 2. Thus, 2 12 dA = 2 ( 0.25 - y ) dy. Then dF = pdA = 9810(2 - y) 3 2 1 0.25 - y2 ) 2 dy 4 1

= 19 620 3 2 ( 0.25 - y

2

Then FR =

L

dF = 19 620

L0

0.5 m

2–y

) - y ( 0.25 - y ) 4 dy 1 2

2

1 2

y x2 + y2 = 0.52

3 2 ( 0.25 - y2 ) - y ( 0.25 - y2 ) 4 dy 1 2

1

= 19 620c y ( 0.25 - y2 ) 2 + 0.25 sin-1

1 2

dy p

0.5 m y 3 1 + ( 0.25 - y2 ) 2 d 0.5 3 0

y

x

x x (a)

= 6887.26 N Ans.

= 6.89 kN And

yP =

=

L

(2 - y) dF FR

L0

0.5 m

(2 - y) 5 19 620 3 2 ( 0.25 - y2 ) 2 - y ( 0.25 - y2 ) 2 4 dy 6 1

1

6887.26

19 620 =

L0

0.5 m

3 4 ( 0.25

- y2 ) 2 - 4y ( 0.25 - y2 ) 2 + y2 ( 0.25 - y2 ) 2 4 dy 1

1

1

6887.26

y 3 4 + 1 0.25 - y2 2 2 0.5 3 0.5m y y 1 1 + 1 0.25 - y2 2 2 + sin - 1 d` 32 128 0.5 0 6887.26

=

19 620c 2y 1 0.25 - y2 2 2 + 0.5 sin - 1

=

12380.29 = 1.798 m = 1.80 m 6887.26

1

+

y 3 ( 0.25 - y2 ) 2 4

Ans. Ans: FR = 6.89 kN yP = 1.80 m 180

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–99. The tank is filled with water. Determine the resultant force acting on the trapezoidal plate C and the location of the center of pressure, measured from the top of the tank. Solve the problem using the formula method.

2m 2m A C

0.3 m

Referring to the geometry shown in Fig. a,

0 .6 m

1 A = (0.6 m)(0.6 m) + (0.6 m)(0.6 m) = 0.54 m2 2

y = Ix =

1 (1.7 m)(0.6 m)(0.6 m) + (1.8m) c (0.6 m)(0.6 m) d 2 0.54 m2

0.5 m

0.75 m

0.6 m

SOLUTION

B

0.75 m 0.3 m

= 1.7333 m

1 (0.6 m)(0.6 m)3 + (0.6 m)(0.6 m)(1.7333 m - 1.7 m)2 12 +

1 1 (0.6 m) ( 0.6 m ) 3 + (0.6 m)(0.6 m)(1.8 m - 1.7333 m)2 36 2

= 0.0156 m4 The resultant force is FR = rwghA = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.7333 m) ( 0.54 m2 ) = 9182.16 N Ans.

= 9.18 kN And it acts at yP =

Ix 0.0156 m4 + 1.7333 m = 1.75 m + y = yA (1.7333 m)(0.54 m)

y=h

Ans.

1.7 m 1.8 m

C1

0.6 m C2 0.3 m

C2 0.3 m

0.3 m

0.3 m

Ans: FR = 9.18 kN yP = 1.75 m

(a)

181

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–100. Solve Prob. 2–99 using the integration method.

2m 2m A

SOLUTION

B

C

0.6 m - y x - 0.3 m = ; 0.6 m 0.3 m

0.3 m

x = (0.6 - 0.5y) m

1 1000 kg>m3 2 1 9.81m>s2 2 (2

- y) m =

3 9810(2

0.75 m

0 .6 m

Referring to Fig. b, the pressure as a function of y can be written as p = rwgh =

0.75 m

0.6 m

Referring to the geometry shown in Fig. a,

0.5 m

0.3 m

(0.6 – y) m

- y) 4 N>m2

2(0.6 - 0.5y)dy =

This pressure acts on the element of area dA = 2xdy = (1.2 - y)dy. Thus

0.6 m

dF = pdA = 9810(2 - y)(1.2 - y)dy

(x – 0.3) m y

= 9810 ( y - 3.2y + 2.4 ) dy 2

x

Then FR =

L

dF = 9810

L0

= 9810 a

0.6 m

0.3 m

( y - 3.2y + 2.4 ) dy

0.6 m y3 - 1.6y2 + 2.4yb ` 3 0

= 9182.16 N = 9.18 kN

0.3 m

(a)

2

Ans.

And it acts at

yP =

=

L

=

y

FR L0

0.6 m

(2 - y) 3 9810 1 y2 - 3.2y + 2.4 2 dy 4 9182.16

9810 =

2–y

(2 - y) dF

L0

0.6 m

9810 a -

1 - y3

dy p y

+ 5.2y2 - 8.8y + 4.8 2 dy

x

x x

9182.16

0.6 m y4 + 1.7333y3 - 4.4y2 + 4.8yb ` 4 0 9182.16

(b)

Ans.

= 1.75 m

182

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–101. The open wash tank is filled to its top with butyl alcohol, an industrial solvent. Determine the magnitude of the resultant force on the end plate ABCD and the location of the center of pressure, measured from AB. Solve the problem using the formula method. Take gba = 50.1 lb>ft3.

2 ft

2 ft

2 ft

B 3 ft

A 8 ft

D

C

SOLUTION First, the location of the centroid of plate ABCD, Fig. a, measured from edge AB must be determined. 1 (1.5 ft) 3 3 ft(2 ft) 4 + (1 ft) c (4 ft)(3 ft) d 2 y = = = 1.25 ft ΣA 1 3 ft(2 ft) + (4 ft)(3 ft) 2 Σy∼ A

Then, the moment of inertia of plate ABCD about its centroid x axis is Ix = c

1 1 1 (2 ft)(3 ft)3 + 2 ft(3 ft)(1.5 ft - 1.25 ft)2 d + c (4 ft)(3 ft)3 + (4 ft)(3 ft)(1.25 ft - 1 ft)2 d = 8.25 ft 4 12 36 2

The area of plate ABCD is

A = 3 ft(2 ft) +

1 (4 ft)(3 ft) = 12 ft 2 2

Thus, FR = ghA = ( 50.1 lb>ft 3 ) (1.25 ft) ( 12 ft 2 ) = 751.5 lb = 752 lb yP =

Ix 8.25 + y = + 1.25 = 1.80 ft yA 1.25(12)

2 ft

2 ft

Ans. Ans.

2 ft B

A Ct

Cr

D 1 (3 ft) = 1 ft 3 (a)

Ct

C

3 ft

1 (3 ft) = 1.5 ft 2

Ans: FR = 752 lb yP = 1.80 ft 183

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–102. The control gate ACB is pinned at A and rest on the smooth surface at B. Determine the amount of weight that should be placed at C in order to maintain a reservoir depth of h = 10 ft. The gate has a width of 3 ft. Neglect its weight.

3 ft C A

h

1.5 ft

SOLUTION

B 4 ft

The intensities of the distributed load at C and B shown in Fig. a are wC = gw hC b = ( 62.4 lb>ft 3 ) (6 ft)(3 ft) = 1123.2 lb>ft wB = gw hBb = ( 62.4 lb>ft 3 ) (7.5 ft)(3 ft) = 1404 lb>ft Thus, F1 = (1123.2 lb>ft)(3 ft) = 3369.6 lb F2 = (1123.2 lb>ft)(1.5 ft) = 1684.8 lb F3 =

1 3 (1404 - 1123.2 lb>ft) 4 (1.5 ft) = 210.6 lb 2

Since the gate is about to be opened, NB = 0. Write the moment equation of equilibrium about point A by referring to Fig. a, a+ ΣMA = 0;

(3369.6 lb)(1.5 ft) + (1684.8 lb)(0.75 ft) + (210.6 lb)(1 ft) - wC(3 ft) = 0 Ans.

WC = 2176.2 lb = 2.18 kip WC 1.5 ft

1.5 ft

wC

Ax

2 (1.5) = 1 ft 3

F2 Ay

F1

(a)

1 (1.5) = 0.75 ft 2

F3

wB

Ans: 2.18 kip 184

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–103. The control gate ACB is pinned at A and rest on the smooth surface at B. If the counterweight C is 2000 lb, determine the maximum depth of water h in the reservoir before the gate begins to open. The gate has a width of 3 ft. Neglect its weight.

3 ft C A

h

1.5 ft B 4 ft

SOLUTION The intensities of the distributed loads at C and B are show in Fig. a wC = gw hCb = ( 62.4 lb>ft 3 ) (h - 4 ft)(3 ft) = wB = gwhBb = ( 62.4 lb>ft ) (h - 2.5 ft)(3 ft) 3

Thus,

3 187.2(h - 4) 4 lb>ft = 3 187.2(h - 2.5) 4 lb>ft

F1 = (187.2(h - 4) lb>ft)(3 ft) = 561.6(h - 4) lb F2 = (187.2(h - 4) lb>ft)(1.5 ft) = 280.8(h - 4) lb F3 =

1 3 187.2(h - 2.5) lb>ft - (187.2(h - 4) lb>ft 4 (1.5 ft) = 210.6 lb 2

Since the gate is required to be opened NB = 0. Write the moment equation of equilibrium about point A by referring to Fig. a a+ ΣMA = 0;

3 561.6(h

- 4) lb 4 (1.5 ft) +

3 280.8(h

- 4) lb 4 (0.75 ft)

+ (210.6 lb)(1 ft) - (2000 lb)(3 ft) = 0 1053(h - 4) = 5789.4

Ans.

h = 9.498 ft = 9.50 ft 2000 lb

1.5 ft

1.5 ft

wc

Ax 2 (1.5) = 1 ft 3

F2 Ay

F1

F3 1 —(1.5) = 0.75 ft 2

(a) wB

Ans: 9.50 ft 185

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–104. The uniform plate, which is hinged at C, is used to control the level of the water at A to maintain its constant depth of 12 ft. If the plate has a width of 8 ft and a weight of 50 ( 103 ) lb, determine the minimum height h of the water at B so that seepage will not occur at D.

A 4 ft C B 8 ft h

SOLUTION

D

Referring to the geometry in Fig. a x h = ; 10 8

x =

6 ft

5 h 4

The intensities of the distributed load shown in Fig. b are w1 = gwh1b = ( 62.4 lb>ft 3 ) (4 ft)(8 ft) = 1996.8 lb>ft

10 ft

w2 = gwh2b = ( 62.4 lb>ft 3 ) (12 ft)(8 ft) = 5990.4 lb>ft w3 = gwh3b = ( 62.4 lb>ft 3 ) (h)(8 ft) = (499.2h) lb>ft

8 ft

x

Thus, the resultant forces of these distributed loads are

h

F1 = (1996.8 lb>ft)(10 ft) = 19968 lb F2 =

1 (5990.4 lb>ft - 1996.8 lb>ft)(10 ft) = 19968 lb 2

F3 =

1 5 (499.2h lb>ft)a hb = ( 312h2 ) lb 2 4

6 ft (a)

and act at d1 =

10 ft = 5 ft 2

d2 =

2 (10 ft) = 6.667 ft 3

50000 lb

1 5 d 3 = 10 ft - a hb = (10 - 0.4167h) ft 3 4

C x d3

For seepage to occur, the reaction at D, must be equal to zero. Referring to the FBD of the gate, Fig. b, a+ ΣMC = 0;

3 (50000 lb)a b(5 ft) + ( 312h2 lb ) (10 - 0.4167h) ft 5

w1

5 ft 4

C y

5

3

- (19968 lb)(5 ft) - (19968 lb)(6.667 ft) = 0

F3

F1

- 130 h3 + 3120 h2 - 82960 = 0 Solving numerically,

d2

w3

Ans.

h = 5.945 ft = 5.95 ft 6 8 ft

F2 w2 (b)

186

d1

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–105. The bent plate is 1.5 m wide and is pinned at A and rests on a smooth support at B. Determine the horizontal and vertical components of reaction at A and the vertical reaction at the smooth support B for equilibrium. The fluid is water.

1m A

4m

SOLUTION Since the gate has a width of b = 1.5 m, the intensities of the distributed loads at A and B can be computed from

B

3m

2m

wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (1 m)(1.5 m) = 14.715 ( 103 ) N>m wB = rwghBb = ( 1000 kg>m3 )( 9.81 m>s2 ) (5 m)(1.5 m) = 73.575 ( 103 ) N>m 2.5 m

Using these results, the distributed load acting on the plate is shown on the freebody diagram of the gate, Fig. a. Ax

F1 wA = 14.715(103) N/m F4

F1 = wALAB = ( 14.715 ( 103 ) N>m ) (5 m) = 73.575 ( 103 ) N F2 =

1 1 (w - wA)LBC = (73.575 ( 103 ) N>m - 14.715 ( 103 ) N>m)(4 m) 2 B 2

= 117.72 ( 10

3

Ay

)N

2m+

2 3

C

(3 m)

2 2 m 3 (4 m) F3 F2

F3 = wALBC = (14.715 ( 103 ) N>m)(4 m) = 58.86 ( 103 ) N F4 on the free-body diagram is equal to the weight of the water contained in the shaded triangular block, Fig. a.

wB = 73.575(103) N/m

5m (a)

1 F4 = rwg V = ( 1000 kg>m3 )( 9.81 m>s2 ) c (3 m)(4 m)(1.5 m) d = 88.29 ( 103 ) N 2

NB

Considering the free-body diagram of the gate, Fig. a. a+ ΣMA = 0;

2 NB(5 m) - 73.575 ( 103 ) N(2.5 m) - 58.86 ( 103 ) N(2 m) - 117.72 ( 103 ) N a (4 m) b 3 2 3 - 88.29 ( 10 ) Na2 m + (3 m) b = 0 3

NB = 193.748 ( 103 ) N = 194 kN + S ΣFx = 0;

Ax - 58.86 ( 10

3

Ans.

) N - 117.72 ( 10 ) N = 0 3

Ax = 176.58 ( 103 ) N = 177 kN + c ΣFy = 0;

- Ay - 73.575 ( 10

3

Ans.

) N - 88.29 ( 10 ) N + 193.748 ( 10 ) N = 0 3

3

Ay = 31.88 ( 103 ) N = 31.9 kN

Ans.

Ans: NB = 194 kN Ax = 177 kN Ay = 31.9 kN 187

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–106. The thin quarter-circular arched gate is 3 ft wide, is pinned at A, and rests on the smooth support at B. Determine the reactions at these supports due to the water pressure. 6 ft

SOLUTION

B

Referring to the geometry shown in Fig. a, p (6 ft)2 = (36 - 9p) ft 2 4

AADB = (6 ft)(6 ft) -

x =

(3 ft)[(6 ft)(6 ft)] - a

8 p ft b c (6 ft)2 d p 4

(36 - 9p) ft 2

= 4.6598 ft

The horizontal component of the resultant force acting on the gate is equal to the pressure force on the vertically projected area of the gate. Referring to Fig. b NB = gwhBb = ( 62.4 lb>ft 3 ) (6 ft)(3 ft) = 1123.2 lb>ft Thus, Fh =

1 (1123.2 lb>ft)(6 ft) = 3369.6 lb 2

The vertical component of the resultant force acting on the gate is equal to the weight of the column of water above the gate (shown shaded in Fig. b). Fv = gwV = gwAADBb = ( 62.4 lb>ft 3 ) 3 (36 - 9p) ft 2 4 (3 ft) = 1446.24 lb

Considering the equilibrium of the FBD of the gate in Fig. b, a + ΣMA = 0;

(1446.24 lb)(4.6598 ft) + (3369.6 lb)(4 ft) - NB(6 ft) = 0 Ans.

NB = 3369.6 lb = 3.37 kip + ΣFx = 0; S

3369.6 lb - Ax = 0

+ ΣFy = 0; S

3369.6 lb - 1446.24 lb - Ay = 0

Ans.

Ax = 3369.6 lb = 3.37 kip

4.6598 ft

Fv

Ay = 1923.36 lb = 1.92 kip Ans. Ax 2 (6) = 4 ft 3

6 ft

x D

(b)

Fh

A

6 ft

C 6 ft

Ay

=

C1

6 ft



wB 6 ft

C2 NB

B

3 ft (a)

Ans: NB = 3.37 kip Ax = 3.37 kip Ay = 1.92 kip 188

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–107. Water is confined in the vertical chamber, which is 2 m wide. Determine the resultant force it exerts on the arched roof AB. 6m

SOLUTION Due to symmetry, the resultant force that the water exerts on arch AB will be vertically downward, and its magnitude is equal to the weight of water of the shaded block in Fig. a. This shaded block can be subdivided into two parts as shown in Figs.  b and c. The block in Fig. c should be considered a negative part since it is a hole. From the geometry in Fig. a, u = sin

2m a b = 30° 4m

A

B 4m

2m

2m

-1

h = 4 cos 30° m

Then, the area of the parts in Figs. b and c are AOBCDAO = 6 m(4 m) + AOBAO = Therefore,

1 (4 m)(4 cos 30° m) = 30.928 m2 2

60° 60° ( pr 2 ) = 3 p(4 m)2 4 = 2.6667p m2 360° 360°

FR = W = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 ( 30.928 m2 - 2.6667p m2 ) (2 m) 4 = 442.44 ( 103 ) N = 442 kN

Ans.

2m 2m D

4m C

D

C

6m

6m B

A h

A

A

B

B

4m

4m O

O

O h = 4 cos 30o m

(a)

(b)

(c)

Ans: 442 kN 189

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–108. Determine the horizontal and vertical components of reaction at the hinge A and the normal reaction at B caused by the water pressure. The gate has a width of 3 m. 3m

SOLUTION

B

The horizontal component of the resultant force acting on the gate is equal to the pressure force on the vertically projected area of the gate. Referring to Fig. a,

3m

wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (6 m)(3 m) = 176.58 ( 103 ) N>m wB = rwghBb = ( 1000 kg>m3 )( 9.81 m>s2 ) (3 m)(3 m) = 88.29 ( 103 ) N>m Thus, (Fh)1 = (Fh)2 =

3 88.29 ( 103 ) N>m 4 (3 m)

= 264.87 ( 103 ) N = 264.87 kN

1 3 176.58 ( 103 ) N>m - 88.29 ( 103 ) N>m 4 (3 m) = 132.435 ( 103 ) N = 132.435 kN 2

They act at

1 ∼ y 1 = (3 m) = 1.5 m 2

1 ∼ y 2 = (3 m) = 1 m 3

The vertical component of the resultant force acting on the gate is equal to the weight of the imaginary column of water above the gate (shown shaded in Fig. a) but acts upward

( Fv ) 1 = rwgV1 = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (3 m)(3 m)(3 m) 4 = 264.87 ( 103 ) N = 264.87 kN p 4

( Fv ) 2 = rwgV2 = ( 1000 kg>m3 )( 9.81 m>s2 ) c ( 3 m ) 2(3 m) d = 66.2175p ( 103 ) N = 66.2175p kN

They act at

1 ∼ x 1 = (3 m) = 1.5 m 2

~ x1

∼ x2 =

4(3 m) 3p

4 = a bm p

(Fv)1

wB

NB

(Fh)1 3m ~ y1 Ax

x~2

wA (Fv)2

(Fh)2

~ y2

Ay (a)

190

A

2–108. Continued

Considering the equilibrium of the FBD of the gate in Fig. a a+ ΣMA = 0;

(264.87 kN)(1.5 m) + (132.435 kN)(1 m) + (264.87 kN)(1.5 m) + (66.2175p kN)a

4 mb - NB(3 m) = 0 p Ans.

NB = 397.305 kN = 397 kN + ΣFx = 0; S

397.305 kN - 264.87 kN - 132.435 kN - Ax = 0 Ans.

Ax = 0 + c ΣFy = 0;

264.87 kN + 66.2175p kN - Ay = 0 Ans.

Ay = 472.90 kN = 473 kN

191

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–109. The 5-m-wide overhang is in the form of a parabola, as shown. Determine the magnitude and direction of the resultant force on the overhang.

y

3m

1 y ! — x2 3

SOLUTION

3m

x

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the wall. Referring to Fig. a, wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (3 m)(5 m) = 147.15 ( 103 ) N>m Thus, Fh =

1 1 w h = 3 147.15 ( 103 ) N>m 4 (3 m) = 220.725 ( 103 ) N = 220.725 kN 2 A A 2

The vertical component of the resultant force is equal to the weight of the imaginary column of water above surface AB of the wall (shown shaded in Fig. a) but acts upward. The volume of this column of water is V =

2 2 ahb = (3 m)(3 m)(5 m) = 30 m3 3 3

Thus, Fv = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 )( 30 m3 ) = 294.3 ( 103 ) N = 294.3 kN The magnitude of the resultant force is FR = 2Fh 2 + Fv 2 = 2(220.725 kN)2 + (294.3 kN)2 = 367.875 kN = 368 kN Ans.

Its direction is

u = tan-1a y

Fv 294.3 kN b = tan-1a b = 53.13° = 53.1° Fh 220.725 kN 1 x2 — y=— 3

— x

b

Ans.

B

C

3m F —h y x

A Fv

3m

wA FR

Ans: FR = 368 kN u = 53.1° b

(a)

192

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–110. Determine the resultant force that water exerts on the overhang sea wall along ABC. The wall is 2 m wide. 1.5 m

B A 2m C

SOLUTION

2.5 m

Horizontal Component. Since AB is along the horizontal, no horizontal component exists. The horizontal component of the force on BC is (FBC)h = gwhA = ( 1000 kg>m3 )( 9.81 m>s2 ) a1.5 m +

1 (2 m) b(2 m(2 m)) = 98.1 ( 103 ) N 2

Vertical Component. The force on AB and the vertical component of the force on BC is equal to the weight of the water contained in blocks ABEFA and BCDEB (shown 2

shaded in Fig. a), but it acts upwards. Here, AABEFA = 1.5 m(2.5 m) = 3.75 m and p ABCDEB = (3.5 m)(2 m) - (2 m)2 = (7 - p) m2. Then, 4 FAB = gwVABEFA = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 ( 3.75 m2 ) (2 m) 4

2.5 m F

(FBC)v FAB E

D

1.5 m A

B (FBC)h

2m C (a)

= 73.575 ( 103 ) N = 73.6 kN

(FBC)v = gwVBCDEB = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (7 - p) m2(2 m) 4 = 75.702 ( 103 ) N

Therefore, FBC = 2 ( FBC ) h2 + ( FBC ) v2 = 2 3 98.1 ( 103 ) N 4 2 + = 123.91 ( 103 ) N = 124 kN

FR = 2 ( FBC ) h2 +

3 FAB

+ ( FBC ) v 4 2

= 2 3 98.1 ( 103 ) N 4 2 +

3 73.6 ( 103 ) N

= 178.6 ( 103 ) N = 179 kN

3 75.702 ( 103 ) N 4 2

+ 75.702 ( 103 ) N 4 2

Ans.

Ans: 179 kN 193

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–111. Determine the magnitude and direction of the resultant hydrostatic force the water exerts on the face AB of the overhang if it is 2 m wide.

C

A 2m

B

SOLUTION Horizontal Component. The intensity of the distributed load at B is wB = rwghBb = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m)(2 m) = 39.24 ( 103 ) N>m

A

Then, (FR)h =

1 1 (w )h = (39.24 ( 103 ) N>m)(2 m) = 39.24 ( 103 ) N 2 B B 2

C 2m

(FR)h

Vertical Component. This component is equal to the weight of the water contained in the block shown shaded in Fig. a, but it acts upwards. Then p (FR)v = rwgVABCA = ( 1000 kg>m3 )( 9.81 m>s2 ) c (2 m)2(2 m) d 4 = 61.638 ( 103 ) N c

B (a)

Thus, the magnitude of the resultant force is FR = 2 ( FR ) h 2 + ( FR ) v2 = 2 3 39.24 ( 103 ) N 4 2 + = 73.07 ( 103 ) N = 73.1 kN

3 61.638 ( 103 ) N 4 2

FR

Ans.

And its direction, Fig. b, is defined by u = tan-1 £

(FR)v (FR)h

≥ = tan-1 C

61.638 ( 103 ) N 39.24 ( 103 ) N

S = 57.5°

a

Ans. (b)

Ans: FR = 73.1 kN u = 57.5° a 194

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

*2–112. The 5-m-wide wall is in the form of a parabola. If the depth of the water is h = 4 m, determine the magnitude and direction of the resultant force on the wall.

y2 ! 4x

h

x

SOLUTION The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the wall. Referring to Fig. a. wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (4 m)(5 m) = 196.2 ( 103 ) N>m Thus, Fh = It acts at

1 1 w h = 3 196.2 ( 103 ) N>m 4 (4 m) = 392.4 ( 103 ) N = 392.4 kN 2 A A 2

1 1 4 ∼ y = hA = (4 m) = m 3 3 3 The vertical component of the resultant force is equal to the weight of the column of water above surface AB of the wall (shown shaded in Fig. a). The volume of this column of water is V =

1 1 ahb = (4 m)(4 m)(5 m) = 26.67 m3 3 3

Thus, Fr = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 ) (26.67 m>s) = 261.6 ( 103 ) N = 261.6 kN It acts at 3 3 6 ∼ x = a = (4 m) = m 10 10 5 The magnitude of the resultant force is FR = 2F h2 + F v2 = 2(392.4 kN)2 + (261.6 kN)2 = 471.61 kN = 472 kN

Fv

Ans.



x

And its direction is

Fv 261.6 kN u = tan a b = tan-1a b = 33.69° Fh 392.4 kN

FR

-1

C

B

Ans.

y2 = 4x

4m Fh —

y

wA

x

A 4m (a)

195

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–113. The 5-m wide wall is in the form of a parabola. Determine the magnitude of the resultant force on the wall as a function of depth h of the water. Plot the results of force (vertical axis) versus depth h for 0 … h … 4 m. Give values for increments of ∆h = 0.5 m.

y2 ! 4x

h

x

SOLUTION The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the wall. Referring to Fig. a,

Fv y

wA = rwghAb = ( 1000 kg>m3 )( 9.81m>s2 ) (h)(5 m) = 49.05 ( 103 ) h Thus, Fh =

B

1 1 w h = 3 49.05 ( 103 ) h 4 h = 24.525 ( 103 ) h2 2 A A 2

FR h

The vertical component of the resultant force is equal to the weight of the column of water above surface AB of the wall (shown shaded in Fig. a) The volume of this column of water is V = Thus,

1 1 h2 5 3 ahb = a b(h)(5 m) = h 3 3 4 12

wA

4

2

Then the magnitude of the resultant force is 2F h2

+

F v2

FR = 2 3 24.525 ( 103 ) h2 4 2 +

x

A h2

5 Fv = rwg V = ( 1000 kg>m )( 9.81 m>s ) a h3 b = 4087.5 h3 12 3

FR =

Fh

h(m) FR(kN)

3 4087.5 h3 4 2

FR = 2601.48 ( 106 ) h4 + 16.71 ( 106 ) h6

(a)

0

0.5

1.0

1.5

2.0

2.5

3.0

0 3.5 347.8

6.15 4.0 471.6

24.9

56.9

103.4

166.1

246.8

FR(kN)

The plot of FR vs h is shown in Fig. b

FR = c 2601 ( 106 ) h4 + 16.7 ( 106 ) h6 d N

500

where h is in m.

400 300 200 100 h(m) 0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

(b)

Ans: FR = c 2601 1 106 2 h4 + 16.7 1 106 2 h6 d N

where h is in m 196

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–114. Determine the resultant force the water exerts on AB, BC, and CD of the enclosure, which is 3 m wide.

y2 ! 2x A

D

2m

B

x

C 2.5 m

2m

SOLUTION Horizontal Component. The horizontal component of the force CD is the same as the force on AB. Its magnitude can be determined from FAB = ( FCD ) h = gwhA = ( 100 kg>m3 )( 9.81 m>s2 ) (1 m)(2 m(3 m)) = 58.86 ( 103 ) N = 58.9 kN

Ans.

Vertical Component. The force on BC and the vertical component of the force on CD is equal to the weight of the water contained in blocks ABCEA and  CDEC (shown shaded in Fig. a). Here, AABCEA = 2 m(2.5 m) = 5 m2 and 1 1 ACDEC = bh = (2 m)(2 m) = 1.3333 m2 (Table 2-1). Then, 3 3 FBC = gwVABCEA = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 5 m2(3 m) 4 = 147.15 ( 103 ) N = 147 kN

(FCD)V = gwVCDEC = ( 1000 kg>m

3

= 39.24 ( 103 ) N

Ans.

)( 9.81 m>s ) 3 1.3333 m (3 m) 4 2

2

Therefore, FCD = 2 ( FCD )h2 + ( FCD )v2 = 2 3 58.86 ( 103 ) N 4 2 + = 70.74 ( 103 ) N = 70.7 kN

FBC 2.5 m

A

3 39.24 ( 103 ) N 4 2

Ans.

(FCD)v 2m E

D

2m B

C FAB (FCD)h (a)

Ans: FAB = 58.9 kN FBC = 147 kN FCD = 70.7 kN 197

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–115. Determine the magnitude of the resultant force the water exerts on the curved vertical wall. The wall is 2 m wide.

A 4m 45!

45!

B

SOLUTION Horizontal Component. This component can be determined by applying

( FAB ) h = gwhA = ( 1000 kg>m3 )( 9.81m>s2 ) (4 sin 45°) 3 2(4 sin 45° m)(2 m) 4 = 313.92 ( 103 ) N

Vertical Component. The downward force on BD and the upward force on AD is equal to the weight of the water contained in blocks ACDBA and ACDA, respectively. Thus, the net downward force on ADB is equal to the weight of water contained in block p 1 ADBA shown shaded in Fig. a. Here, AADBA = (4 m)2 - 2 c (4 sin 45°)(4 cos 45°)d 4 2 = (4p-8) m2. Then, (FAB)v = gwVADBA = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (4p - 8)m2(2 m) 4 = 89.59 ( 103 ) N

Then, FAB = 2 ( FAB )h2 + ( FAB )v2 = 2 3 313.92 ( 103 ) N 4 2 + = 326.45 ( 103 ) N = 326 kN

A

O

3 89.59 ( 103 ) N 4 2 Ans.

C

4m D

B (a)

Ans: 326 kN 198

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

*2–116. Gate AB has a width of 0.5 m and a radius of 1 m. Determine the horizontal and vertical components of reaction at the pin A and the horizontal reaction at the smooth stop B due to the water pressure.

A

x

y ! "x2

1m 1m B

SOLUTION Vertical Component. This component is equal to the weight of the water contained in the block shown shaded in Fig. a, but it acts upward. This block can be subdivided into parts (1) and (2) as shown in Figs. b and c. Part (2) is a hole and should be considered as a negative part. Thus, the area of the block, Fig. a, is p ΣA = (1 m)(1 m) - (1 m)2 = 0.2146 m2 and the horizontal distance measured 4 form its centroid to point A is

x =

ΣxA = ΣA

0.5 m(1 m)(1 m) -

4(1 m) p c (1 m)2 d 3p 4

0.2146 m2

= 0.7766 m

The magnitude of the vertical component is (FR)v = rwgV =

1 1000 kg>m3 2 1 9.81 m>s2 2 3 0.2146 m2(0.5 m) 4

= 1052.62 N

Horizontal Component. The intensity of the distributed load at B is wB = rwghBb = ( 1000 kg>m3 )( 9.81 m>s2 ) (1 m)(0.5 m) = 4.905 ( 103 ) N>m Then, (FR)h =

1 1 w h = 3 4.905 ( 103 ) N>m 4 (1 m) = 2452.5 N 2 B B 2

4(1 m) — x

Considering the free-body diagram of the gate in Fig. d,

2 a + ΣMA = 0; (2452.5 N) c (1 m) d + (1052.62 N)(0.7766 N) - FB(1 m) = 0 3 FB = 2452.5 N = 2.45 kN

+ ΣFx = 0; S

A

1

A 1m

=1m

A –

C1

2 1m C2

Ans. 1m

2452.5 N - 2452.5 N - Ax = 0

(a)

(b)

(c)

Ans.

Ax = 0 + c ΣFy = 0;

0.5 m

x = 0.7766 m

1052.62 N - Ay = 0 Ans.

Ay = 1052.62 N = 1.05 kN

Ax

(FR)v = 1052.62 N

Ay

2 —(1 m) 3

1m FB wA = 4.905(103) N/m

(FR)h = 2452.5 N (d)

199

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. C

2–117. A quarter-circular plate is pinned at A and tied to the tank’s wall using the cable BC. If the tank and plate are 4 ft wide, determine the horizontal and vertical components of reaction at A, and the tension in the cable due to the water pressure.

B

6 ft A

SOLUTION Referring to the geometry shown in Fig. a p AADB = (6 ft)(6 ft) - (6 ft)2 = (36 - 9p) ft 2 4

x =

(3 ft) 3 (6 ft)(6 ft) 4 - c a6 -

8 p bft d c (6 ft)2 d p 4

(36 - 9p) ft 2

= 1.3402 ft

The horizontal component of the resultant force acting on the shell is equal to the pressure force on the vertically projected area of the shell. Referring to Fig. b wA = gwhAb = ( 62.4 lb>ft 3 ) (6 ft)(4 ft) = 1497.6 lb>ft Thus, Fh =

1 (1497.6 lb>ft)(6 ft) = 4492.8 lb 2

The vertical component of the resultant force acting on the shell is equal to the weight of the imaginary column of water above the shell (shown shaded in Fig. b) but acts upwards. Fv = gwV = gwAADBb =

1 62.4 lb>ft3 2 3 (36

- 9p) ft 2 4 (4 ft) = 1928.33 lb

Write the moment equation of equilibrium about A by referring to Fig. b, a + ΣMA = 0; TBC(6 ft) - (1928.33 lb)(1.3402 ft) - (4492.8 lb)(2 ft) = 0

Ans.

TBC = 1928.33 lb = 1.93 kip + ΣF = 0 S x

-Ax + 4492.8 lb - 1928.32 lb = 0 Ans.

Ax = 2564.5 lb = 2.56 kip

c + ΣFy = 0 1928.33 lb - Ay = 0 Ans.

Ay = 1928.33 lb = 1.93 kip

Fv

–x B

TBC

3 ft

D 6 ft

C

6 ft

Fh =

A

1.3402 ft

6 ft C1

C2 6 ft

1 (6) = 2 ft — 3

wA

Ax Ay

6 ft (b)

(a)

200

Ans: TBC = 1.93 kip Ax = 2.56 kip Ay = 1.93 kip

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–118. The bin is 4 ft wide and filled with linseed oil. Determine the horizontal and vertical components of the force the oil exerts on the curved segment AB. Also, find the location of the points of application of these components acting on the segment, measured from point A. glo = 58.7 lb>ft 3.

y 3 ft B 3 ft

SOLUTION The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of curve AB of the bin. Referring to Fig. a, wB = gohBb = ( 58.7 lb>ft 3 ) (3 ft)(4 ft) = 704.4 lb>ft wA = gohAb = ( 58.7 lb>ft 3 ) (6 ft)(4 ft) = 1408.8 lb>ft Then, (Fh)1 = (704.4 lb>ft)(3 ft) = 2113.2 lb 1 (Fh)2 = (1408.8 lb>ft - 704.4 lb>ft)(3 ft) = 1056.6 lb 2 Thus, Fh = (Fh)1 + (Fh)2 = 2113.2 lb + 1056.6 lb = 3169.8 lb = 3.17 kip

Ans.

1 1 Here, ∼ y1 = (3 ft) = 1.5 ft and ∼ y2 = (3 ft) = 1 ft. The location of the point of 2 3 application of Fh can be determined from y =

(1.5 ft)(2113.2 lb) + (1 ft)(1056.6 lb) ΣyF = = 1.3333 ft ΣF 3169.8 lb

The vertical component of the resultant force is equal to the weight of the imaginary column of water above curve AB of the bin (shown shaded in Fig. a) but acts upwards (Fv)1 = gwV1 = ( 58.7 lb>ft 3 ) 3 (3 ft)(3 ft)(4 ft) 4 = 2113.2 lb

Thus,

(Fv)2 = gwV2 = ( 58.7 lb>ft 3 ) c

p (3 ft)2(4 ft) d = 1659.70 lb 4

Fv = (Fv)1 + (Fv)2 = 2113.2 lb + 1659.70 lb = 3772.90 lb = 3.77 kip

Ans.

4(3 ft) 1 4 (3 ft) = 1.5 ft and ∼ = ft. The location of the point of x2 = p 2 3p application of Fv can be determined from Here, x∼1 =

ΣxF = x = ΣF

4 ft b(1659.70 lb) p = 1.4002 ft 3772.90 lb

(1.5 ft)(2113.2 lb) + a

The equation of the line of action of FR is given by Fv y - y = - (x - x ) Fh y - 1.3333 = -

3772.9 (x - 1.4002) 3169.8

y = -1.1903x + 3

201

A

x

2–118. Continued

Use substitution to find the intersection of this line and the circle x2 + (y - 3)2 = 9: x2 +

3 ( - 1.1903x

+ 3) - 3 4 2 = 9

2

2.4167x = 9

Ans.

x = 1.9298 m = 1.93 m Back-substituting,

y = - 1.1903(1.9298) + 3 Ans.

= 0.7035 m = 0.704 m

~ x1

(Fv)1

(Fv)2

B

~ x2

wB

3 ft

(Fh)1

~ y1

3 ft A wA

(Fh)2

~ y2

(a)

Ans: Fh = 3.17 kip Fv = 3.77 kip x = 1.93 m y = 0.704 m 202

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–119. If the water depth is h = 2 m, determine the magnitude and direction of the resultant force, due to water pressure acting on the parabolic surface of the dam, which has a width of 5 m.

y

y ! 0.5x2

h

SOLUTION

x

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the dam. Referring to Fig. a wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m)(5 m) = 98.1 ( 103 ) N>m Thus, Fh =

1 1 wAhA = 3 98.1 ( 103 ) N>m 4 (2 m) = 98.1 ( 103 ) N = 98.1 kN 2 2

The vertical component of the resultant force is equal to the weight of the column of water above the dam surface (shown shaded in Fig. a). The volume of this column of water is V =

2 2 ahb = (2 m)(2 m)(5 m) = 13.33 m3 3 3

Thus, Fv = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 )( 13.33 m3 ) = 130.80 ( 103 ) N = 130.80 kN The magnitude of the resultant force is FR = 2Fh 2 + Fv 2 = 2(98.1kN)2 + (130.80 kN)2 = 163.5 kN

Ans.

and its direction is u = tan-1

Fv 130.80 kN = tan-1a b = 53.1° Fh 98.1 kN y x

Ans. Fv

C

B y = 0.5 x2 FR

2m Fh y x

A 2m

Ans: FR = 163.5 kN u = 53.1° c 203

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–120. Determine the magnitude of the resultant force due to water pressure acting on the parabolic surface of the dam as a function of the depth h of the water. Plot the results of force (vertical axis) versus depth h for 0 … h … 2 m. Give values for increments of ∆h = 0.5 m. The dam has a width of 5 m.

y

y ! 0.5x2

h

SOLUTION

x

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the dam. Referring to Fig. a, Fv

y

wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (h)(5 m) = 49.05 ( 103 ) h Thus,

C

1 1 Fh = wAhA = 3 49.05 ( 103 ) h 4 h = 24.525 ( 103 ) h2 2 2

FR

The vertical component of the resultant force is equal to the weight of the column of water above the dam surface (shown shaded in Fig. a). The volume of this column of water is 2 2 V = ahb = 1 22 h 2 (h)(5) = 4.7140 h3>2 3 3

Thus,

h

Fh

Fv = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 )( 4.7140 h3>2 ) = 46.2447 ( 103 ) h3>2

wA

x

A —

Then the magnitude of the resultant force is FR = 2Fh 2 + Fv 2

√2h

FR = 2 3 24.525 ( 103 ) h2 4 2 +

(a)

3 46.2447 ( 103 ) h3>2 4 2

FR(kN)

FR = 2601.476 ( 10 ) h + 2.13858 ( 10 ) h 6

4

9

3

The plot of FR vs h is shown in Fig. b. FR = where h is in m. h(m) FR(kN)

3 2601 ( 106 ) h4

+ 2.14 ( 109 ) h3 4 N

0

0.5

1.0

1.5

2.0

0

17.5

52.3

101.3

163.5

Ans.

180 160 140 120 100 80 60 40 20 0

0.5

1.0

1.5 (b)

204

2.0

h(m)

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–121. The canal transports water and has the cross section shown. Determine the magnitude and direction of the resultant force per unit length acting on wall AB, and the location of the center of pressure on the wall, measured with respect to the x and y axes.

B

9 ft

SOLUTION

1 y ! — x3 3

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of curve AB of the canal. Referring to Fig. a, wA = gwhAb = ( 62.4 lb>ft 3 ) (9 ft)(1 ft) = 561.6 lb>ft

x

A

Thus,

3 ft

1 Fh = (561.6 lb>ft)(9 ft) = 2527.2 lb 2 And it acts at y =

1 (9 ft) = 3 ft 3

The vertical component of the resultant force is equal to the weight of the column of water above curve AB of the canal Fv = However, y =

L

dF =

L

gw dV = gw

1 1 1 3 x or x = 3 3 y 3 . Then 3

Fv = 62.4

L0

9 ft

L

bdA = gw

L

(1)(xdy)

1 1 1 3 4 9 ft 33 y 3 dy = 62.4a33b c y3 d ` = 1263.6 lb 4 0

The location of its point of application can be determined from

Fv

y

∼ x dF x L x = where ∼ x = and dF = gwxdy = 62.4x dy Fv 2 Thus, x =

L0

9 ft

B

x a b(62.4xdy) 2 1263.6

x

9 ft

x2dy L0 1263.6

31.2 =

9 ft

31.2 =

L0 1263.6 2

9 ft Fh

dy ~ x

2

33 y 3 dy

_ y

2 3 5 9 ft 31.2a33 ba y3b ` 5 0 = 1263.6

wA

A (a)

= 1.20 ft The magnitude of the resultant force is FR = 2Fh 2 + Fv 2 = 2(2527.2 lb)2 + (1263.6 lb)2 = 2825.50 lb = 2.83 kip Ans. 205

x

2–121. Continued

And its direction is defined by u = tan-1 a

Fv 1263.6 lb b = tan-1a b = 26.67° Fh 2527.2 lb

Ans.

The equation of the line of action of FR is y - y = m(x - x); y - 3 = - tan 26.57°(x - 1.20) y = - 0.5x + 3.6 The intersection point of the line of action of FR and surface AB can be obtained by solving simultaneously this equation and that of AB. 1 3 x = - 0.5x + 3.6 3 1 3 x + 0.5x - 3.6 = 0 3 Solving numerically Ans.

x = 1.9851 ft = 1.99 ft when x = 1.9851 ft, y =

1 ( 1.98513 ) = 2.61 ft 3

Ans.

Ans: FR = 2.83 kip u = 26.6° c x = 1.99 ft y = 2.61 ft 206

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. y

2–122. The settling tank is 3 m wide and contains turpentine having a density of 860 kg>m3. If the parabolic shape is defined by y = ( x2 ) m, determine the magnitude and direction of the resultant force the turpentine exerts on the side AB of the tank.

2m A

y ! x2

4m

SOLUTION

B

x

The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of surface AB. Referring to Fig. a, wB = rtghBb = ( 860 kg>m3 )( 9.81 m>s2 ) (4 m)(3 m) = 101.24 ( 103 ) N>m Thus, 1 3 101.24 ( 103 ) N>m 4 (4 m) = 202.48 ( 103 ) N = 202.48 kN 2 The vertical component of the resultant forced is equal to the weight of the column of turpentine above surface AB of the wall (shown shaded in Fig. a). The volume of this column of water is 2 2 V = ahb = (2 m)(4 m)(3 m) = 16 m3 3 3 Thus, Fh =

Fv = rtgV = ( 860 kg>m3 )( 9.81 m>s2 )( 16 m3 ) = 134.99 ( 103 ) N = 134.99 kN The magnitude of the resultant force is FR = 2Fh 2 + Fv 2 = 2(202.48 kN)2 + (134.99 kN)2 = 243.35 kN = 243 kN Ans.

And its direction as defined by u = tan-1a

Fv 134.99 kN b = tan-1a b = 33.79° Fh 202.48 kN

Ans.

y

_ x

Fv

y = x2

FR

4m

Fh _ y

wB

x 2m

Ans: FR = 243 kN u = 33.7° c 207

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. C

2–123. The radial gate is used to control the flow of water over a spillway. If the water is at its highest level as shown, determine the torque T that must be applied at the pin A in order to open the gate. The gate has a mass of 5 Mg and a center of mass at G. It is 3 m wide.

4m T G

A

60! 4.5 m

B

SOLUTION Horizontal Component. This component can be determined by applying (FBC)h = gwhA = ( 1000 kg>m3 )( 9.81 m>s2 ) (4.5 sin 30° m) 3 2(4.5 sin 30° m)(3 m) 4 = 297.98 ( 103 ) N

Vertical Component. The upward force on BE and downward force on CE is equal to the weight of the water contained in blocks BCDEB and CEDC, respectively. Thus, the net upward force on BEC is equal to the weight of the water contained in block BCEB shown shaded in Fig. a. This block can be subdivided into parts (1) and (2), Figs. a and b, respectively. However, part (2) is a hole and should be considered as a negative part. The area of block BCEB is p 1 ΣA = c (4.5 m)2 d - (4.5 m)(4.5 cos 30° m) = 1.8344 m2 and the horizontal 6 2 distance measured from its centroid to point A is ΣxA x = = ΣA

a

2 9 p 1 mb c (4.5 m)2 d - (4.5 cos 30° m) c (4.5 m)(4.5 cos 30° m) d p 6 3 2 1.8344 m2

= 4.1397 m

)

The magnitude of the vertical component is

When the gate is on the verge of opening, NB = 0. Referring to the free-body diagram of the gate in Fig. d, 2 3 5000(9.81) N 4 (4 m) + 3 297.98 ( 103 ) N 4 c (4.5 m) - 2.25 m d 3 -

3 53.985 ( 10 ) N 4 (4.1397 m) 3

T = 196.2 ( 103 ) N # m = 196 kN # m

- T = 0

3 5000(9.81) N 4 (4 m)

E 60°

B

! 30°

A

C2 "

4.5 m 4.5 cos 30° m

(a)

(b)

2.25 m

(c)

5000(9.81) N/m 4m

Ay T

2 —(4.5 m) 3

- T = 0

T = 196.2 ( 103 ) N # m = 196 kN # m

A

C1

A

Ans.

This solution can be simplified if one realizes that the resultant force will act perpendicular to the circular surface. Therefore, FBC will act through point A and so produces no moment about this point. Hence, a + ΣMA = 0;

2 = — m —(4.5 cos 30° m) 3

D C

= 53.985 ( 103 ) N

a + ΣMA = 0;

)

2 — 3

_ x

(FBC)v = gwVBCEB = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 1.8344 m2(3 m) 4

Ax

Ans. (FBC)h = 297.98(103) N/m 4.1397 m (FBC)v = 53.985(103) N/m (d)

Ans: 196 kN # m 208

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. C

*2–124. The radial gate is used to control the flow of water over a spillway. If the water is at its highest level as shown, determine the horizontal and vertical components of reaction at pin A and the vertical reaction at the spillway crest B. The gate has a weight of 5 Mg and a center of gravity at G. It is 3 m wide. Take T = 0.

4m T G

B

A

60! 4.5 m

SOLUTION Horizontal Component. This component can be determined from (FBC)h = gwhA = ( 1000 kg>m3 )( 9.81 m>s2 ) (4.5 sin 30° m) 3 2(4.5 sin 30° m)(3 m) 4 = 297.98 ( 103 ) N

Vertical Component. The upward force on BE and downward force on CE is equal to the weight of the water contained in blocks BCDEB and CEDC, respectively. Thus, the net upward force on BEC is equal to the weight of the water contained in block BCEB shown shaded in Fig. a. This block can be subdivided into parts (1) and (2), Fig. a and b, respectively. However, part (2) is a hole and should be considered as a negative part. The area of block BCEB is p 1 ΣA = c (4.5 m)2 d - (4.5 m)(4.5 cos 30° m) = 1.8344 m2 and the horizontal 6 2 distance measured from its centroid to point A is

x =

ΣxA = ΣA

a

2 9 p 1 mb c (4.5 m)2 d - (4.5 cos 30° m) c (4.5 m)(4.5 cos 30° m) d p 6 3 2 1.8344 m2

= 4.1397 m

)

2 = — m —(4.5 cos 30° m) 3

D C E

A

C1

A 60°

B

Thus, the magnitude of the vertical component is

)

2 — 3

_ x

! 30°

A

C2 "

4.5 m 4.5 cos 30° m

(FBC)v = gwVBCEB = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 1.8344 m2(3 m) 4

(a)

= 53.985 ( 103 ) N

(b)

(c)

Considering the free-body diagram of the gate in Fig. d, a + ΣMA = 0;

3 5000(9.81) N 4 (4 m) -

+

3 297.98 ( 103 ) N 4 c

2 (4.5 m) - 2.25 m d 3

3 53.985 ( 103 ) N 4 (4.1397 m)

81) N/m 5000 (9.81) N

4m Ay

2 — (4.5 m) 2.25 m .5 ) 3

-NB(4.5 cos 30° m) = 0

NB = 50.345 ( 103 ) N = 50.3 kN + c ΣFy = 0;

50.345 ( 103 ) N + 53.985 ( 103 ) N - 5000(9.81) N - Ay = 0 Ay = 55.28 ( 10

3

+ ΣFx = 0; S

Ax

= 29 hAns.

) N = 55.3 kN

Ans.

297.98 ( 103 ) N - Ax

4.1397 m (FBC)h = 297.98(103) N/m

(F

=

)

(FBC) = 53.985(103) N

NB

Ax = 297.98 ( 103 ) N = 298 kN

Ans.

209

/

4.5 cos 30° m

(d)

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–125. The 6-ft-wide plate in the form of a quarter-circular arc is used as a sluice gate. Determine the magnitude and direction of the resultant force of the water on the bearing O of the gate. What is the moment of this force about the bearing?

45!

D

O

45!

SOLUTION

12 ft

B

Referring to the geometry in Fig. a, AADB =

p 1 ( 12 ft2 ) - 3 (2)(12 sin 45° ft)(12 cos 45° ft) 4 = 41.097 ft2 4 2 2 12 sin 45° ft ∼ x1 = a b = 7.2025 ft 3 p>4 2 ∼ x2 = (12 cos 45° ft) = 5.6569 ft 3

(7.2025 ft) c

x =

p 1 (12 ft)2 d - (5.6569 ft) c (2)(12 sin 45° ft)(12 cos 45° ft) d 4 2 41.097 ft

~

x~

2

A D

= 9.9105 ft The horizontal component of the resultant force is equal to the pressure force on the vertically projected area of the gate. Referring to Fig. b

45° O 45° 12 ft

D

A 12 ft

45° O 45° C1 12 ft

B

45° O

Fv

45°

C2 B

B (a)

wB = gwhBb = ( 62.4 lb>ft ) (16.971 ft)(6 ft) = 6353.78 lb>ft Fh =

x2

A

3

Thus,

~

x1

12 ft

9.9105 ft

d

1 (6353.78 lb>ft)(16.971 ft) = 53.9136 ( 103 ) lb = 53.9136 kip 2

O

Fh _ y

It acts at 1 y = (16.971 ft) = 5.657 ft 3 d = 12 sin 45° ft - 5.657 ft = 2.8284 ft

wB

2(12 sin 45°) = 16.971 ft (b)

The vertical component of the resultant force is equal to the weight of the block of water contained in sector ADB shown in Fig. a but acts upward. Fv = gwVADB = gwAADBb = ( 62.4 lb>ft 3 )( 41.097 ft 2 ) (6 ft) = 15.3868 ( 103 ) lb = 15.3868 kip Thus, the magnitude of the resultant force is FR = 2Fh 2 + Fv 2 = 2(53.9136 kip)2 + (15.3868 kip)2 = 56.07 kip = 56.1 kip

Ans.

Its direction is

u = tan-1a

15.3868 kip Fv b = tan-1a b = 15.93° = 15.9° Fh 53.9136 kip

a

Ans.

By referring to Fig. b, the moment of FR about O is a + (MR)O = ΣMO; (MR)O = (53.9136 kip)(2.8284 ft) - (15.3868 kip)(9.9105 ft) Ans.

= 0

This result is expected since the gate is circular in shape. Thus, FR is always directed toward center O of the circular gate.

210

Ans: FR = 56.1 kip u = 15.9° a

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–126. The curved and flat plates are pin connected at A, B, and C. They are submerged in water at the depth shown. Determine the horizontal and vertical components of reaction at pin B. The plates have a width of 4 m.

3m B 3m

SOLUTION A

C

The horizontal component of the resultant force is equal to the pressure force on the vertically projected area of the plate. Referring to Fig. a

4m

wB = rw ghB b = ( 1000 kg>m3 )( 9.81 m>s2 ) (3 m)(4 m) = 117.72 ( 103 ) N>m wA = wC = rwghCb = ( 1000 kg>m3 )( 9.81 m>s2 ) (6 m)(4 m) = 235.44 ( 103 ) N>m Thus, (Fh)AB1 = (Fh)BC1 = (Fh)AB2 = (Fh)BC2 = They act at

3 117.72 ( 103 ) N>m 4 (3 m)

= 353.16 (103) N = 353.16 kN

1 3 235.44 ( 103 ) N>m - 117.72 ( 103 ) N>m 4 (3 m) = 176.58 ( 103 ) N = 176.58 kN 2

1 1 ∼ y2 = ∼ y4 = (3) = 1.5 m ∼ y1 = ∼ y3 = (3 m) = 1 m 2 3

The vertical component of the resultant force is equal to the weight of the column of water above the plates shown shaded in Fig. a (Fv)AB1 = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (3 m)(3 m)(4 m) 4 = 353.16 ( 103 ) N = 353.16 kN 1 (Fv)AB2 = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 ) c p(3 m)2(4 m) d = 88.29p ( 103 ) N = 88.29p kN 4

(Fv)BC1 = rwgV = ( 1000 kg>m3 ) (9.81 m>s2) 3 (3 m)(4 m)(4 m) 4 = 470.88 ( 103 ) N = 470.88 kN

1 (Fy)BC2 = rwgV = ( 1000 kg>m3 )( 9.81 m>s2 ) c (3 m)(4 m)(4 m) d = 235.44 ( 103 ) N = 235.44 kN 2

They act at

1 ∼ x1 = (3 m) = 1.5 m 2

4(3 m) 4 ∼ x2 = = m p 3p

1 ∼ x3 = (4 m) = 2 m 2

Referring to Fig. a and writing the moment equations of equilibrium about A and C a + ΣMA = 0;

Bx(3 m) - By(3 m) - (353.16 kN)(1.5 m) - (88.29p kN)a

4 mb p

- (353.16 kN)(1.5 m) - (176.58 kN)(1 m) = 0 (1)

Bx - By = 529.74

211

1 4 ∼ x4 = (4 m) = m 3 3

2–126. Continued

4 (470.88 kN)(2 m) + (235.44 kN)a mb + (353.16 kN)(1.5 m) 3

a + ΣMC = 0;

+ (176.58 kN)(1 m) - Bx(3 m) - By(4 m) = 0 (2)

3Bx - 4By = 1962 Solving Eq. (1) and (2) Bx = 582.99 kN = 583 kN

Ans.

By = 53.2 kN

Ans.

(Fv)AB

~ x1

~ x

1

1

3

2

(Fv)BC

_ wB x2 (Fh)AB

(Fv)BC

(Fv)AB

Bx

~ x4

Bx

2

wB (Fh)BC

1

3m

_ y2 _ y1 (Fh)AB

wA 2

3m

_ y4

By

By

Ax

wC 4m

3m

Ay

1

Cx

_ y3

(Fh)BC

2

Cy (a)

Ans: Bx = 583 kN By = 53.2 kN 212

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–127. The stopper in the shape of a frustum is used to plug the 100-mm-diameter hole in the tank that contains amyl acetate. If the greatest vertical force the stopper can resist is 100 N, determine the depth d before it becomes unplugged. Take raa = 863 kg>m3. Hint: The volume of a cone is V =

1 3

pr 2h.

d

100 mm

SOLUTION

40 mm

The vertical downward force on the conical stopper is due to the weight of the liquid contained in the block shown shaded in Fig a. This block can be subdivided into parts (1), (2), and (3), shown in Figs. b, c, and d, respectively. Part (2) is a hole and should be considered as a negative part. Thus, the volume of the shaded block in Fig. a is

20!

20!

V = V1 - V2 + V3 1 1 p(0.05 m)2(0.13737 m) + p(0.03544 m)2(0.09737 m) 3 3

= p(0.05 m)2d =

3 2.5 ( 10-3 ) pd

- 0.2316 ( 10-3 ) 4 m3

The vertical force on the stopper is required to be equal to 100 N. Then, F = rwgV 100 N = ( 863 kg>m3 )( 9.81 m>s2 ) 3 2.5 ( 10-3 ) pd - 0.2316 ( 10-3 ) 4 d = 1.5334 m = 1.53 m

Ans.

0.05 m tan 20°

1

= 0.13737 m 20°

d 20°

0.04 m

2

d

0.05 m (a)

0.05 m (b)

0.05 m (c)

3

20°

0.1373 m − 0.04 m = 0.09737 m

0.09737 tan 20° m = 0.03544 m (d)

Ans: 1.53 m 213

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–128. The stopper in the shape of a frustum is used to plug the 100-mm-diameter hole in the tank that contains amyl acetate. Determine the vertical force this liquid exerts on the stopper. Take d = 0.6 m and raa = 863 kg>m3. Hint: The volume of a cone is V =

1 2 3 pr h.

d

100 mm 40 mm

SOLUTION The vertical downward force on the conical stopper is due to the weight of the liquid contained in the block shown shaded in Fig. a. This block can be subdivided into parts (1), (2), and (3), shown in Figs. b, c, and d, respectively. Part (2) is a hole and should be considered as a negative part. Thus, the volume of the shaded block in Fig. a is V = V1 - V2 + V3 = p(0.05 m)2(0.6 m) -

1 1 p(0.05 m)2(0.13737 m) + p(0.03544 m)2 (0.09737 m) 3 3

= 4.4808 ( 10-3 ) m3 Then, F = rwgV = ( 863 kg>m3 )( 9.81 m>s2 ) 3 4.4808 ( 10-3 ) m3 4

Ans.

= 37.93 N = 37.9 N

1

m 0.13737 m 0.6 m

2.6 m

20°

20°

3

0.13737 m − 0.04 m = 0.09737 m

2

0.04 m

0.09737 tan 20° 0.05 m

0.05 m

(a)

(b)

0.05 m (c)

= 0.03544 m (d)

214

20!

20!

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–129. The steel cylinder has a specific weight of 490 lb>ft 3 and acts as a plug for the 1-ft-long slot in the tank. Determine the resultant force the bottom of the tank exerts on the cylinder when the water in the tank is at a depth of h = 2 ft.

h 0.35 ft

SOLUTION

A

The vertical downward force and the vertical upward force are equal to the weight of the water contained in the blocks shown shaded in Figs. a and b, respectively. The volume of the shaded block in Fig. a is V1 = c 2.35 ft(0.7 ft) -

B

0.5 ft

p (0.35 ft)2 d (1 ft) = 1.4526 ft 3 2

The volume of the shaded block in Fig. b is V2 = 2 e 0.1 ft(2.35 ft) + c = 0.5037 ft 3

44.42° 1 (p)(0.35 ft)2 - (0.25 ft)(0.2449 ft) d f(1 ft) 360° 2

Then, F = gw(V1 - V2) = ( 62.4 lb>ft 3 )( 1.4526 ft 3 - 0.5037 ft 3 ) = 59.21 lb T The weight of the cylinder is W = gstVC = ( 490 lb>ft 2 ) 3 p(0.35 ft)2(1 ft) 4 = 188.57 lb. Considering the free-body diagram of the cylinder, Fig. c, we have + c ΣFy = 0;

N - 59.21 lb - 188.57 lb = 0

Ans.

N = 247.78 lb = 248 lb

0.1 ft

w = 188.57 lb

0.1 ft

2.35 ft 2.35 ft F = 59.21 lb

(

0.25 ft 0.35 ft

= 44.42° 0.35 sin 44.42° = 0.2449 ft 0.35 ft

0.35 ft (a)

)

N (c)

0.25 ft (b)

Ans: 248 lb 215

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–130. The steel cylinder has a specific weight of 490 lb>ft 3 and acts as a plug for the 1-ft-long slot in the tank. Determine the resultant force the bottom of the tank exerts on the cylinder when the water in the tank just covers the top of the cylinder, h = 0.

h 0.35 ft

A

SOLUTION

B

0.5 ft

The vertical downward force and the vertical upward force are equal to the weight of the water contained in the blocks shown shaded in Figs. a and b, respectively. The volume of the shaded block in Fig. a is V1 = c 0.35 ft(0.7 ft) -

p (0.35 ft)2 d (1 ft) = 0.05258 ft 3 2

The volume of the shaded block in Fig. b is V2 = 2e 0.35 ft(0.1 ft) + c = 0.10372 ft 3

44.42° 1 (p)(0.35 ft)2 - (0.25 ft)(0.2449 ft) d f(1 ft) 360° 2

Then, F = gw(V1 - V2) = ( 62.4 lb>ft 3 )( 0.10372 ft 3 - 0.05258 ft 3 ) = 3.192 lb c The weight of the cylinder is W = gstVC = ( 490 lb>ft 2 ) 3 p(0.35 ft)2(1 ft) 4 = 188.57 lb. Considering the force equilibrium vertically by free-body diagram of the cylinder, Fig. c, we have + c ΣFy = 0;

N + 3.192 lb - 188.57 lb = 0 Ans.

N = 185.38 lb = 185 lb

) 0.1 ft

= 44.42°

0.35 ft

0.25 ft 0.35 ft

)

0.1 ft

w = 188.57 lb

0.35 ft 0.35 sin 44.42° = 0.2449 ft F = 3.192 lb

0.35 ft 0.25 ft (a)

(b)

(c)

Ans: 185 lb 216

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–131. The sluice gate for a water channel is 1.5 m wide and in the closed position, as shown. Determine the magnitude of the resultant force of the water acting on the gate. Solve the problem by considering the fluid acting on the horizontal and vertical projections of the gate. Determine the smallest torque T that must be applied to open the gate if its weight is 30 kN and its center of gravity is at G.

2m 2m T G

20! 20! 1.5 m

SOLUTION The horizontal component of the resultant force is equal to the pressure force acting on the vertically projected area of the gate. Referring to Fig. a w1 = rwgh1b = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m)(1.5 m) = 29.43 ( 103 ) N>m w2 = rwgh2b = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m + 2m sin 40°)(1.5 m) = 48.347 ( 103 ) N>m Then (Fh)1 =

3 29.43 ( 103 ) N>m 4 (2 sin 40°m)

= 37.834 ( 103 ) N = 37.834 kN

1 3 (48.347 - 29.43) ( 103 ) N>m 4 (2 sin 40°m) = 12.160 ( 103 ) N = 12.160 kN 2 Fh = (Fh)1 + (Fh)2 = 37.834 ( 103 ) N + 12.160 ( 103 ) N = 49.994 ( 103 ) N = 49.994 kN

(Fh)2 =

Also 1 2 ∼ ∼ y1 = (2 m sin 40°) = 0.6428 m and y2 = (2 m sin 40°) = 0.8571 m 2 3 The vertical component of the resultant force is equal to the weight of the imaginary column of water above the gate (shown shaded in Fig. a) but acts upward. The volume of this column of water is V = c (2 m - 2 m cos 40°)(2 m) + = 2.0209 m3

1 40° 1 (2 m)2 a p radb - (2 m cos 40°)(2 m sin 40°) d (1.5 m) 2 180° 2

Fv = rwg V = ( 1000 kg>m3 )( 9.81 m>s2 )( 2.0209 m3 ) = 19.825 ( 103 ) N = 19.825 kN Referring to Fig b and c r =

2 2 m sin 20° = 1.3064 m ≥ 3£ 20 p 180

~ x2 = 1.3064 m cos 20° = 1.2276 m

2 m - 2 m cos 40° ∼ x1 = 2 m cos 40° + a b = 1.7660 m 2 2 ∼ x3 = (2 m cos 40°) = 1.0214 m 3 Thus, Fv acts at 1 40 1 (1.7660 m)(2 m - 2 m cos 40°)(2 m) + (1.2276 m)c (2 m)2 a p radb d - (1.0214 m) c (2 m cos 40°)(2 m sin 40°) d 2 180 2 x = 1 40 1 (2 m - 2 m cos 40°)(2 m) + (2 m)2 a p radb - (2 m cos 40°)(2 m sin 40°) 2 180 2 = 1.7523 m

217

O

2–131. Continued

The magnitude of the resultant force is FR = 2Fh2 + Fv 2 = 2(49.994 kN)2 + (19.825 kN)2 = 53.78 kN = 53.8 kN Ans.

Referring to the FBD of the gate shown in Fig d, a + ΣM0 = 0;

(30 kN)(1.5 cos 20°m) + (37.834 kN)(0.6428 m) + (12.160 kN)(0.8571 m) -(19.825 kN)(1.7524 m) - T = 0

T = 42.29 kN # m = 42.3 kN # m

Ans.

Note that the resultant force of the write acting on the give must act normal to its surface, and therefore it will pass through the pin at O. Therefore it produces moment about the pin. (2 - 2 cos 40˚) m

w1

y1

x1

=

20˚ 20˚

40˚

(Fh)1

2m

(Fh)2 w2

x3

x2

2m

– 2m

2m

r~

y2

(b) (a)

30 kN

(c)

1.5 cos 20˚ m Oy

0.8571 m

0.6428 m T

37.834 kN

Ox

12.160 kN 1.7524 m 19.825 kN (d)

Ans: FR = 53.8 kN T = 42.3 kN # m 218

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–132. Solve the first part of Prob. 2-131 by the integration method using polar coordinates. 2m 2m T G

Referring to Fig a, h = (2 + 2 sin u) m. Thus, the pressure acting on the gate as a function of u is p = rwgh = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 + 2 sin u) m = [19620(1 + sin u)] N>m2 This pressure is acting on the element of area dA = bds = 1.5 ds = 1.5 (2 du) = 3 du. Thus,

2m 2m

dF = pd A = 19620(1 + sin u)(3 du). = 58.86 ( 10 ) (1 + sin u) du 3

P

The horizontal and vertical components of dF are

ds

(dF )h = 58.86 ( 103 ) (1 + sin u) cos u du

(a)

= 58.86 ( 103 ) (cos u + sin u cos u) du (dF )v = 58.86 ( 103 ) (1 + sin u) sin u du = 58.86 ( 103 )( sin u + sin 2 u ) du Since sin 2u = 2 sin u cos u and cos 2u = 1 - 2 sin2 u, then (dF)h = 58.86 ( 103 ) acos u + (dF )v = 58.86 ( 103 ) asin u +

1 sin 2u b du 2

1 1 - cos 2u b du 2 2

The horizontal and vertical components of the resultant force are L

(dF )h = 58.86 ( 103 )

= 58.86 ( 103 ) c sin u -

L0

2p 9

acos u + 2p

1 cos 2u d ` 9 4 0

1 sin 2u b du 2

= 49.994 ( 103 ) N = 49.994 kN Fv =

L

20! 1.5 m

SOLUTION

Fh =

20!

2p

(dF)v = 58.86 ( 10

= 58.86 ( 103 ) a - cos u +

3

)

L

9

0

asin u +

1 1 - cos 2u b du 2 2

2p 1 1 u - sin 2u b d ` 9 2 4 0

= 19.825 ( 103 ) N = 19.825 kN

Thus, the magnitude of the resultant force is FR = 2Fh 2 + Fv 2 = 2(49.994 kN)2 + (19.825 kN)2 = 53.78 kN = 53.8 kN Ans. 219

O

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–133. A flat-bottomed boat has vertical sides and a bottom surface area of 0.75 m2. It floats in water such that its draft (depth below the surface) is 0.3 m. Determine the mass of the boat. What is the draft when a 50-kg man stands in the center of the boat?

SOLUTION Equilibrium requires that the weight of the empty boat is equal to the buoyant force. Wb = Fb = rwgVDisp = ( 1000 kg>m3 )( 9.81m>s2 )( 0.75 m2 ) (0.3 m) = 2207.25 N Thus, the mass of the boat is given by mb =

Wb 2207.25 N = = 225 kg g 9.81 m>s2

Ans.

When the man steps into the boat, the total mass is mb + m = 225 kg + 50 kg = 275 kg. Then Wb + m = mb + mg = 275 kg ( 9.81 m>s2 ) = 2697.75 N. Under this condition, the boat will sink further to create a greater buoyancy force to balance the additional weight. Thus, Wb + m = rwgV′Disp 2697.75 N = ( 1000 kg>m3 )( 9.81 m>s2 )( 0.75 m2 ) (h) Ans.

h = 0.367 m

Ans: mb = 225 kg h = 0.367 m 220

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–134. The raft consists of a uniform platform having a mass of 2 Mg and four floats, each having a mass of 120 kg and a length of 4 m. Determine the height h at which the platform floats from the water surface. Take rw = 1Mg>m3.

h

0.25 m

SOLUTION

h

Each float must support a weight of 0.25 m

1 W = c (2000 kg) + 120 kg d 9.81 m>s2 = 6082.2 N 4 For equilibrium, the buoyant force on each float is required to be + c ΣFy = 0;

Fb - 6082.2 N = 0

0.25 m (a)

Fb = 6082.2 N

Therefore, the volume of water that must be displaced to generate this force is Fb = gV;

6082.2 N = ( 1000 kg>m3 )( 9.81 m>s2 ) V V = 0.620 m3

1 Since the semicircular segment of a float has a volume of (p)(0.25 m)2(4 m) = 2 3 3 0.3927 m 6 0.620 m , then it must be fully submerged to develop Fb. As shown in Fig. a, we require 0.620 m3 =

1 (p)(0.25 m)2(4 m) + (0.25 m - h)(0.5 m)(4 m) 2

Thus, Ans.

h = 0.136 m = 136 mm

Ans: 136 mm 221

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–135. Consider an iceberg to be in the form of a cylinder of arbitrary diameter and floating in the ocean as shown. If the cylinder extends 2 m above the ocean’s surface, determine the depth of the cylinder below the surface. The density of ocean water is rw = 1024 kg>m3, and the density of the ice is ri = 935 kg>m3.

2m

d

SOLUTION The weight of the iceberg is W = rigVi = ( 935 kg>m3 )( 9.81 m>s2 ) 3 pr 2(2 + d) 4

The buoyant force is

Fb = rswgVsub = ( 1024 kg>m3 )( 9.81 m>s2 )( pr 2d )

w

Referring to the FBD of the iceberg, Fig. a, equilibrium requires, + c ΣFy = 0;

r

Fb - w = 0 2m

( 1024 kg>m )( 9.81 m>s2 )( pr 2d ) - ( 935 kg>m3 )( 9.81m>s2 ) 3 pr 2(2 + d) 4 = 0 3

1024 d = 935(2 + d)

Ans.

d = 21.01 m = 21.0 m

d

Fb (a)

Ans: 21.0 m 222

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–136. The cylinder floats in the water and oil to the level shown. Determine the weight of the cylinder. ro = 910 kg>m3. 100 mm

200 mm

SOLUTION

200 mm

The buoyant force fuel to the submerging in oil and water are (Fb)oil = roil g(Vsub)oil = ( 910 kg>m3 )( 9.81 m>s2 ) 3 p(0.1m)2(0.1 m) 4 = 8.9271 p N (Fb)N = rwg(Vsub)w = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 p(0.1 m)2(0.2 m) 4 = 19.62 p N

Referring to the FBD of the cylinder, Fig. a, equilibrium requires, + c ΣFy = 0;

8.9271 p N + 19.62 p N - W = 0

(Fb)oil 0.1 m

Ans.

W = 89.68 N = 89.7 N

0.1 m w 0.2 m

(Fb)W (a)

223

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 25 mm

2–137. A glass having a diameter of 50 mm is filled with water to the level shown. If an ice cube with 25 mm sides is placed into the glass, determine the new height h of the water surface. Take rw = 1000 kg>m3 and rice = 920 kg>m3. What will the water level h be when the ice cube completely melts?

25 mm

h

100 mm

50 mm

SOLUTION

50 mm

Since the ice floats, the buoyant force is equal to the weight of the ice cube which is Fb = Wi = riVi g = ( 920 kg>m3 ) (0.025 m)3 ( 9.81 m>s2 ) = 0.1410 N This buoyant force is also equal to the weight of the water displaced by the submerged ice cube with at a depth hs. Fb = rwgVs;

0.1410 N = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (0.025 m)2hs 4

hs = 0.023 m Referring to Fig. a,

V1 = V2 - V3

3 p(0.025 m)2 4 (0.1 m)

=

3 p(0.025m)2 4 h

- (0.025m)2 (0.023 m) Ans.

h = 0.1073 m = 107 mm

The mass of ice cube is Mi = riVi = ( 920 kg>m3 )( 0.025 m ) 3 = 0.014375 kg. Thus, the nice in water level due to the additional water of the melting ice cubs: can be determined from Mi = rwVw;

0.014375 kg = ( 1000 kg>m3 ) 3 p(0.025 m)2 ∆h 4 ∆h = 0.007321

Thus, Ans.

h′ = 0.1 m + 0.007321 m = 107 mm Note The water level h remains unchanged as the cube melts.



0.1 m

1

0.05 m

=

2

3

h3 = 0.02375 m

h

0.05 m (a)

Ans: 107 mm 224

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–138. The wood block has a specific weight of 45 lb>ft 3. Determine the depth h at which it floats in the oil–water system. The block is 1 ft wide. Take ro = 1.75 slug>ft 3.

1 ft

h 1 ft

0.5 ft

SOLUTION The weight of the block is W = gbVb = ( 45 lb>ft 3 ) 3 (1ft)3 4 = 45 lb

Assume that block floats in both oil and water, Fig, a. Then, the volume of the water and oil being displaced is (Voil)Disp = 0.5 ft(1 ft)(1 ft) = 0.5 ft 3 (Vw)Disp = (0.5 ft - h)(1 ft)(1 ft) = (0.5 - h) ft 3 Thus, the buoyancy forces on the block due to the oil and water are (Fb)oil = goil(Voil)Disp = ( 1.75 slug>ft 3 )( 32.2 ft>s2 )( 0.5 ft 2 ) = 28.175 lb (Fb)w = gw(Vw)Disp = ( 62.4 slug>ft 3 ) (0.5 - h) ft 3 = (31.2 - 62.4h) lb Considering the free-body diagram in Fig. b, + c ΣFy = 0;

(31.2 - 62.4h) lb + 28.175 lb - 45 lb = 0 Ans.

h = 0.2304 ft = 0.230 ft Since h 6 0.5 ft, the assumption was correct and the result is valid.

w = 45 lb

1 ft h 0.5 ft

Oil

0.5 ft – h

Water

(Fb)oil = 28.175 (Fb)w = (31.2 – 62.4 h)

(a)

(b)

Ans: 0.230 ft 225

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–139. Water in the container is originally at a height of h = 3 ft. If a block having a specific weight of 50 lb>ft 3 is placed in the water, determine the new level h of the water. The base of the block is 1 ft square, and the base of the container is 2 ft square.

1 ft

1 ft

h

SOLUTION The weight of the block is Wb = gbVb = ( 50 lb>ft 3 ) 3 (1ft)3 4 = 50 lb

Equilibrium requires that the buoyancy force equal the weight of the block, so that Fb = 50 lb. Thus, the displaced volume is Fb = gwVDisp

50 lb = ( 62.4 lb>ft 3 ) VDisp VDisp = 0.8013 ft 3

The volume of the water is Vw = 2 ft(2 ft)(3 ft) = 12 ft 3 When the level of the water in the container has a height of h, Vw = V′ - VDisp 12 ft 3 = 4 h ft 3 - 0.8013 ft 3 Ans.

h = 3.20 ft

Ans: 3.20 ft 226

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–140. The cross section of the front of a barge is shown. Determine the buoyant force acting per foot length of the hull when the water level is at the indicated depth.

30!

30!

4 ft

25 ft

SOLUTION Referring to the geometry shown in Fig. a, the volume of the water displaced per foot length of the hull is

25 ft

1 VDisp = 25 ft(4 ft) + 2c (4 tan 30° ft)(4 ft) d = 109.24 ft 3 >ft 2

4 ft

Thus, the buoyancy force acting per foot length of the hull is

Fb = gwVb = ( 62.4 lb>ft 3 )( 109.24 ft 3 >ft )

227

30˚

30˚ (a)

Ans.

= 6816 lb>ft = 6.82 kip>ft

4 tan 30˚ ft

4 tan 30˚ ft

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–141. The cone is made of wood having a density of rwood = 650 kg>m3. Determine the tension in rope AB if the cone is submerged in the water at the depth shown. Will this force increase, decrease, or remain the same if the cord is shortened? Why? Hint: The volume of a cone is V = 13pr 2h.

1m

3m

A 0.5 m

SOLUTION The weight of the wooden cone is

0.5 m B

1 W = rwood gVc = ( 650 kg>m3 )( 9.81 m>s2 ) c p(0.5 m)2(3 m) d = 1594.125p N 3

The volume of water that is displaced is the same as the volume of the cone. Thus, the buoyancy force is 1 Fb = rwgVc = ( 1000 kg>m3 )( 9.81 m>s2 ) c p(0.5 m)2(3 m) d = 2452.5p N 3

Considering the free-body diagram of the cone in Fig. a, + c ΣFy = 0;

2452.5p N - 1594.125p N - TAB = 0 Ans.

TAB = 2696.66 N = 2.70 kN

The tension in rope AB remains the same since the buoyancy force does not change. For a fully submerged body, the buoyancy force is independent of the depth to which the body is submerged. Ans.

Remains the same

TAB (a)

Ans: TAB = 2.70 kN Remains the same 228

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–142. The hot-air balloon contains air having a temperature of 180°F, while the surrounding air having a temperature of 60°F. Determine the maximum weight of the load the balloon can lift if the volume of air it contains is 120(103) ft 3. The empty weight of the balloon is 200 lb.

SOLUTION From the Appendix, the densities of the air inside the balloon where T = 180° F and outside the balloon where T = 60° F, are ra ' T = 60° F = 0.00237 slug>ft 3 ra ' T = 180° F = 0.00193 slug>ft 3

wa

Thus, the weight of the air inside the balloon is Wa ' T = 180° F = ra ' T = 180° FgV = ( 0.00193 slug>ft 3 = 7457.52 lb

)( 32.2 ft>s2 ) 3 120 ( 103 ) ft 3 4

The buoyancy force is equal to the weight of the displaced air outside of the balloon. This gives

Fb = 9157.68 lb

Fb = ra ' T = 60° FgV = ( 0.00237 slug>ft 3 )( 32.2ft>s2 ) 3 120 ( 103 ) ft 3 4

w = 200 lb

= 9157.68 lb

Considering the free-body diagram of the balloon in Fig. a, + c ΣFy = 0;

wL

9157.68 lb - 7457.52 lb - 200 lb - WL = 0 Ans.

WL = 1500.16 lb = 1.50 kip

(a)

Ans: 1.50 kip 229

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–143. The container with water in it has a mass of 30 kg. Block B has a density of 8500 kg>m3 and a mass of 15 kg. If springs C and D have an unstretched length of 200 mm and 300 mm, respectively, determine the length of each spring when the block is submerged in the water. C

kC ! 2 kN/m

B

SOLUTION The volume of block B is,

D

15 kg mB VB = = 1.7647 ( 10-3 ) m3 = r 8500 kg>m3

kD ! 3 kN/m

Thus, the bouyant force is Fb = rwgVsub = ( 1000 kg>m3 )( 9.81 m>s2 )( 1.7647 ( 10-3 ) m3 ) = 17.31 N Referring to the FBD of block B, Fig. a, + c ΣFy = 0;

(Fsp)c + 17.31 N -

3 15(9.81) N 4

= 0

(Fsp)c = 129.84 N

= 0

(Fsp)D = 311.61 N

Referring to the FBD of the container, Fig. b, + c ΣFy = 0;

(Fsp)D - 17.31 N -

3 30(9.81) N 4

Thus, the deformations of springs C dand D are (Fsp)C 129.84 N dC = = = 0.06492 m = 64.92 mm kC 2000 N>m dD =

(Fsp)D kD

=

311.61 N = 0.1039 m = 103.87 mm 3000 N>m

Thus lC = (lo)C + dc = 200 mm + 64.92 mm = 264.92 mm = 265 mm

Ans.

lD = (lo)D + dD = 300 mm - 103.87 mm = 196.13 mm = 196 mm

Ans.

Fb (Fsp)C

30(9.81) N Fb

15(9.81) N (a)

Ans: 265 mm 196 mm

(Fsp)D (b)

230

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–144. An open-ended tube having an inner radius r is placed in a wetting liquid A having a density rA. The top of the tube is just below the surface of a surrounding liquid B, which has a density rB, where rA 7 rB. If the surface tension s causes liquid A to make a wetting angle u with the tube wall as shown, determine the rise h of liquid A within the tube. Show that the result is independent of the depth d of liquid B.

B

u

u

d h

A

SOLUTION The volume of the column of liquid A in the rise is V = pr 2h. Thus, its weight is W = gAV = rAg ( pr 2h ) = pgrAr 2h The force on the top and bottom of the column is ptA = rB(d - h) ( pr 2 ) and rB(d) ( pr 2 ) . The difference in these forces is the bouyont force.

pt A

FB = gBV = rBg ( pr 2h ) = pgrBr 2h The force equilibrium along the vertical, Fig. a, requires. + c ΣFy = 0;

s(2pr) cos u + pgrBr 2h - pgrAr 2h = 0 2prs cos u + pgr 2h(rB - rA) = 0

pb A

2prs cos u = (rA - rB)pgr 2h h =

2s cos u gr(rA - rB)

Ans.

W (a)

Note that the result is independent of d.

231

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–145. A boat having a mass of 80 Mg rests on the bottom of the lake and displaces 10.25 m3 of water. Since the lifting capacity of the crane is only 600 kN, two balloons are attached to the sides of the boat and filled with air. Determine the smallest radius r of each spherical balloon that is needed to lift the boat. What is the mass of air in each balloon if the water temperature is 12oC? The balloons are at an average depth of 20 m. Neglect the mass of air and of the balloon for the calculation required for the lift. The volume of a sphere is V = 43pr 3.

600 kN

r

r

SOLUTION The bouyant force acting on the boat and a balloon are (Fb)B = rwg(VB)sub = ( 1000 kg>m3 )( 9.81 m>s2 )( 10.25 m3 ) = 100.55 ( 103 ) N = 100.55 kN 4 (Fb)b = rwg(Vb)sub = ( 1000 kg>m3 )( 9.81 m>s2 ) c pr 3 d = 13.08pr 3 ( 103 ) N 3 = 13.08pr 3 kN

Referring to the FBD of the boat, Fig. a, + c ΣFy = 0;

2T + 100.55 kN + 600 kN T = 42.124 kN

3 80(9.81) kN 4

= 0

Referring to the FBD of the balloon Fig. b + c ΣFy = 0;

13.08pr 3 - 42.125 kN = 0 Ans.

r = 1.008 m = 1.01 m

Here, p = patm + rwgh = 101 ( 10 ) Pa + ( 1000 kg>m )( 9.81 m>s ) (20 m) = 297.2 ( 103 ) Pa and T = 12° C + 273 = 285 K. From Appendix A, R = 286.9 J>kg # K. Applying the ideal gas law, 3

p = rRT;

r =

3

2

297.2 ( 103 ) N>m2 p = = 3.6347 kg>m3 RT (286.9 J>kg # K)(285 K)

Thus, 4 m = rV = ( 3.6347 kg>m3 ) c p(1.008 m)3 d = 15.61 kg = 15.6 kg 3

Ans.

600 kN

(Fb)b = 13.08!r3 80 (9.81) kN

T

T

+

T = 42.125 kN

Ans: r = 1.01 m m = 15.6 kg

(b) (Fb)B = 100.55 kN (a)

232

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–146. The uniform 8-ft board is pushed down into the water so it makes an angle of u = 30° with the water surface. If the cross section of the board measures 3 in. by 9 in., and its specific weight is gb = 30 lb>ft 3, determine the length a that will be submerged and the vertical force F needed to hold its end in this position.

3 in. 8 ft

F

30!

a

SOLUTION The weight of the board is W = gbVb = ( 30 lb>ft 3 ) c a

3 9 ft ba ft b(8 ft) d = 45 lb 12 12

Fb = gwVsub = ( 62.4 lb>ft 3 ) c a

W 30˚

3 9 ft ba ft ba d = 11.7a lb 12 12

F

Referring to the FBD of the board, Fig. a, equilibrium requires, a+ ΣMo = 0;

a 11.7 a (cos 30°)a b - (45 cos 30° lb)(4 ft) = 0 2 a = 5.547 ft = 5.55 ft

+ c ΣFy = 0;

Ans.

O 30˚

11.7(5.547) - 45 lb - F = 0 Ans.

F = 19.90 lb = 19.9 lb

Fb

a 2 4 ft

Ans: a = 5.55 ft F = 19.9 lb 233

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 75 mm

2–147. The cylinder has a diameter of 75 mm and a mass of 600 g. If it is placed in the tank, which contains oil and water, determine the height h above the surface of the oil at which it will float if maintained in the vertical position. Take  r0 = 980 kg/m3.

h 50 mm

150 mm

SOLUTION Since the cylinder floats, the buoyant force is equal to the weight of the cylinder. Fb = (0.6 kg) ( 9.81 m>s2 ) = 5.886 N Assuming that the cylinder is submerged below the oil layer, then, the buoyant force produced by the oil layer is (Fb)oil = roil g(Vs)oil = ( 980 kg>m3 )( 9.81 m>s2 ) 3p(0.0375 m)2(0.05 m) 4

(O.K!)

= 2.124 N 6 Fb

The buoyant force produced by the water layer is (Fb)w = rwg(Vs)w = ( 1000 kg>m3 )( 9.81 m>s2 ) 3p(0.0375 m)2(0.15 m - 0.05 m - h) 4 = 43.339 (0.1 - h)

We require Fb = (Fb)oil + (Fb)w 5.886 N = 2.124 N + 43.339(0.1 - h) Ans.

h = 13.2 mm

Ans: h = 13.2 mm 234

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–148. When loaded with gravel, the barge floats in water at the depth shown. If its center of gravity is located at G, determine whether the barge will restore itself when a wave causes it to roll slightly at 9°.

G O

SOLUTION

6m

When the barge tips 9°, the submerged portion is trapezoidal in shape, as shown in Fig. a. The new center of buoyancy, Cb ′, is located at the centroid of this area. Then 1 (0)(6)(1.0248) + (1) c (6)(0.9503) d 2 = 0.3168 m x = 1 (6)(1.0248) + (6)(0.9503) 2 1 1 1 (1.0248)(6)(1.0248) + c 1.0248 + (0.9503) d c (6)(0.9503) d 2 3 2 = 0.7751 m y = 1 (6)(1.0248) + (6)(0.9503) 2 The intersection point M of the line of action of Fb and the centerline of the barge is the metacenter, Fig. a. From the geometry of triangle MNCb ′ we have MN =

x 0.3168 = = 2m tan 9° tan 9°

Also, GN = 2 - y = 2 - 0.7751 = 1.2249 m Since MN 7 GN, point M is above G. Therefore, the barge will restore itself.

W M

G 0.5 m 9˚ 9˚ C′b

6 tan 9˚ = 0.9

2m

1.5 m

1.5 m

N

y

3m

3m 3m Fb

x (a)

235

1.5 – 3 tan 9˚ = 1.0248 m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–149. When loaded with gravel, the barge floats in water at the depth shown. If its center of gravity is located at G determine whether the barge will restore itself when a wave causes it to tip slightly.

G O 2m

1.5 m

6m

SOLUTION The barge is tilted counterclockwise slightly and the new center of buoyancy Cb ′ is located to the left of the old one. The metacenter M is at the intersection point of the center line of the barge and the line of action of Fb, Fig. a. The location of Cb ′ can be obtained by referring to Fig. b. 1 (1 m) c (6 m)(6 tan f m) d 2 x = = 2 tan f m (1.5 m)(6 m) Then d = x cos f = 2 m tan f cos f = (2 m)a Since f is very small sin f = f, hence

sin f b(cos f) = (2 sin f) m cos f (1)

d = 2f m From the geometry shown in Fig. a

(2)

d = MCb sinf = MCbf Equating Eqs. (1) and (2) 2f = MCbf MCb = 2 m

Here, GCb = 2 m - 0.75 m = 1.25 m. Since MCb 7 GCb, the barge is in stable equilibrium. Thus, it will restore itself if tilted slightly. W

W M G

5m

f

G

1.5 m

C′b

Cb 3m

Cb f

3m Fb

Fb

(a) 6 tan f m Cb

x

f

=

(b)

C2

1 m) = 1 m 2 m) – —(6 —(6 2 3 f Cb C1 6m

Ans: It will restore itself. 236

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–150. The barrel of oil rests on the surface of the scissors lift. Determine the maximum pressure developed in the oil if the lift is moving upward with (a) a constant velocity of 4 m> s, and (b) a constant acceleration of 2 m>s2. Take ro = 900 kg>m3. The top of the barrel is open to the atmosphere.

1.25 m

a

SOLUTION a) Equilibrium p = rogh = 900 kg>m3 ( 9.81 m>s2 ) (1.25 m) Ans.

= 11.0 kPa b) p = rogh a1 +

aC b g

p = 900 kg>m3 ( 9.81 m>s2 ) (1.25 m)a1 +

2 m>s2 9.81 m>s2

p = 13.3 kPa

b

Ans.

Ans: a) 11.0 kPa b) 13.3 kPa 237

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–151. The truck carries an open container of water as shown. If it has a constant acceleration 2 m>s2, determine the angle of inclination of the surface of the water and the pressure at the bottom corners A and B.

1m

a

2m A

B

SOLUTION 5m

The free surface of the water in the accelerated tank is shown in Fig. a. 2 m>s 2 ac tan u = = g 9.81 m>s2 From the geometry in Fig. a.

Ans.

u = 11.52° = 11.5°

∆hA = ∆hB = (2.5 m) tan 11.52° = 0.5097 m Thus, pA = rwghA = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m + 0.5095 m) = 24.6 kPa

Ans.

pB = rwghB = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m - 0.5095 m) Ans.

= 14.6 kPa

2.5 m

2.5 m

hA 2 m hB

A

2m

B (a)

Ans: u = 11.5° pA = 24.6 kPa pB = 14.6 kPa 238

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–152. The truck carries an open container of water as shown. Determine the maximum constant acceleration it can have without causing the water to spill out of the container.

1m

a

2m A

B

5m

SOLUTION When the tank accelerates, the water spill from the left side wall. The surface of the water under this condition is shown in Fig. a. tan u =

ac 1m = 2.5 m 9.81 m>s2

2.5 m

1m

ac = 3.92 m>s2

Ans.

(a)

239

2.5 m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 18 ft

2–153. The open rail car is 6 ft wide and filled with water to the level shown. Determine the pressure that acts at point B both when the car is at rest and when the car is moving with a constant acceleration of 10 ft>s2. How much water spills out of the car?

A

a 9 ft

7.5 ft B

T

SOLUTION When the car is at rest, the water is at the level shown by the dashed line shown in Fig. a At rest:

pB = gwhB = ( 62.4 lb>ft 3 ) (7.5 ft) = 468 lb>ft 2

Ans.

When the car accelerates, the angle u the water level makes with the horizontal can be determined. tan u =

10 ft>s2 ac = ; g 32.2 ft>s2

u = 17.25°

Assuming that the water will spill out. Then the water level when the car accelerates is indicated by the solid line shown in Fig. a. Thus, h = 9 ft - 18 ft tan 17.25° = 3.4099 ft The original volume of water is V = (7.5 ft)(18 ft)(6 ft) = 810 ft 3 The volume of water after the car accelerate is V′ =

1 (9 ft + 3.4099 ft)(18 ft)(6 ft) = 670.14 ft 3 6 810 ft 3 2

(OK!)

Thus, the amount of water spilled is ∆V = V - V′ = 810 ft 3 - 670.14 ft 3 = 139.86 ft 3 = 140 ft 3

Ans.

The pressure at B when the car accelerates is With acceleration:

pB = gwhB = ( 62.4 lb>ft 3 ) (9 ft) = 561.6 lb>ft 2 = 562 lb>ft 2 Ans.

ac = 10 ft/s2

9 ft 7.5 ft

B

h

Ans: At rest: pB = 468 lb>ft 2 With acceleration: ∆V = 140 ft 3 pB = 562 lb>ft 2

18 ft (a)

240

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–154. The fuel tank, supply line, and engine for an airplane are shown. If the gas tank is filled to the level shown, determine the largest constant acceleration a that the plane can have without causing the engine to be starved of fuel. The plane is accelerating to the right for this to happen. Suggest a safer location for attaching the fuel line.

150 mm

300 mm

SOLUTION

900 mm

100 mm

If the fuel width of the tank is w, the volume of the fuel can be determined using the fuel level when the airplane is at rest indicated by the dashed line in Fig. a. Vf = (0.9 m)(0.3 m)w = 0.27w It is required that the fuel level is about to drop lower than the supply line. In this case, the fuel level is indicated by the solid line in Fig. a. 1 (0.45 m - 0.1 m)(0.9 m - b)w = 0.27w 2

Vf = (0.9 m)(0.45 m)w -

b = 0.1286 m Thus, tan u =

0.45 m - 0.1 m = 0.4537 0.9 m - 0.1286 m

And so tan u =

ac ; g

0.4537 =

ac 9.81 m>s2

ac = 4.45 m>s2 The safer location for attaching the fuel line is at the bottom of the tank.

Ans.

b ac

0.45 m

0.3 m 0.1 m

0.9 m

Ans: ac = 4.45 m>s2 The safer location is at the bottom of the tank. 241

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 0.2 m

2–155. A large container of benzene is transported on the truck. Determine the level in each of the vent tubes A and B if the truck accelerates at a = 1.5 m>s2. When the truck is at rest, hA = hB = 0.4 m.

a

hA

3m A 0.7 m

0.2 m B

hB

SOLUTION The imaginary surface of the benzene in the accelerated tank is shown in Fig. a. tan u =

1.5 m>s2 ac = g 9.81 m>s2

u = 8.6935° Then, ∆h = (1.5 m) tan 8.6935° = 0.2294 m Thus, h′A = hA - ∆h = 0.4 m - 0.2294 m = 0.171 m

Ans.

h′B = hB + ∆h = 0.4 m + 0.2294 m = 0.629 m

Ans.

1.5 m

1.5 m

h′B hA = 0.4 m

h′A

Imaginary free surface

hB = 0.4 m

(a)

Ans: h′A = 0.171 m h′B = 0.629 m 242

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 0.2 m

*2–156. A large container of benzene is being transported by the truck. Determine its maximum constant acceleration so that no benzenes will spill from the vent tubes A or B. When the truck is at rest, hA = hB = 0.4 m.

a

SOLUTION The imaginary surface of the benzene in the accelerated tank is shown in Fig. a. Under this condition, the water will spill from vent B. Thus, ∆h = h′B - hB = 0.7 m - 0.4 m = 0.3 m. tan u =

ac 0.3 m = 0.2 = 1.5 m g

ac = 0.2 ( 9.81 m>s2 ) = 1.96 m>s2

1.5 m

1.5 m

h′B = 0.7 m Imaginary free surface

h′B = 0.4 m

(a)

243

Ans.

hA

3m A 0.7 m

0.2 m B

hB

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 5m

2–157. The closed cylindrical tank is filled with milk, for which rm = 1030 kg>m3. If the inner diameter of the tank is 1.5 m, determine the difference in pressure within the tank between corners A and B when the truck accelerates at 0.8 m > s2.

A

0.8 m/s2 B

1.5 m

SOLUTION The imaginary surface of the milk in the accelerated tank is shown in Fig. a. tan u =

Imaginary free surface

0.8 m>s2

ac = = 0.08155 g 9.81 m>s2

B

A

Then, ∆hAB = LAB tan u = (5 m)(0.08155) = 0.4077 m

5m

Finally,

(a)

∆pAB = rmg∆hAB = ( 1030 kg>m3 )( 9.81 m>s2 ) (0.4077 m) = 4.12 ( 103 ) Pa = 4.12 kPa

Ans.

Ans: 4.12 kPa 244

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–158. Determine the water pressure at points B and C in the tank if the truck has a constant acceleration ac = 2 m>s2. When the truck is at rest, the water level in the vent tube A is at hA = 0.3 m.

3m a

B

hA

1m A 2m C

SOLUTION The water level at vent tube A will not change when the tank is accelerated since the water in the tank is confined (no other vent tube). Thus, the imaginary free surface must pass through the free surface at vent tube A. tan u =

aC ; g

tan u =

2 m>s2

u = 11.52°

9.81 m>s2

From the geometry in Fig. a, ∆hB = (3 m) tan 11.52° = 0.6116 m ∆hC = (1 m) tan 11.52° + 0.3 m = 0.5039 m Then, hB = - ( ∆hB - 0.3 m ) = - (0.6116 m - 0.3 m) = - 0.3116 m hC = ∆hC + 2 m = 0.5039 m + 2 m = 2.5039 m Thus, pB = rwghB = ( 1000 kg>m3 )( 9.81 m>s2 ) ( - 0.3116 m) = - 3.057 ( 103 ) Pa = - 3.06 kPa pC = rwghC = ( 1000 kg>m

3

Ans.

)( 9.81 m>s ) (2.5039 m) 2

= 24.563 ( 103 ) Pa = 24.6 kPa

B

Ans.

0.3 m 2m 1m

3m

(a)

C

Ans: pB = -3.06 kPa pC = 24.6 kPa 245

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–159. If the truck has a constant acceleration of 2 m > s2, determine the water pressure at the bottom corners A and B of the water rank.

1m 1m

a A

B

SOLUTION The imaginary free surface of the water in the accelerated tank is shown in Fig. a.

2m

3m

2

tan u =

2 m>s aC = = 0.2039 g 9.81 m>s2

From the geometry in Fig. a, ∆hA = (1 m) tan u = (1 m)(0.2039) = 0.2039 m ∆hB = (1 m + 3 m) tan u = (4 m)(0.2039) = 0.8155 m Then hA = 2 m + ∆hA = 2 m + 0.2039 m = 2.2039 m hB = 2 m - ∆hB = 2 m - 0.8155 m = 1.1845 m Finally, pA = rwghA = ( 1000 kg>m3 )( 9.81 m>s2 ) (2.2039 m) = 21.62 ( 103 ) Pa = 21.6 kPa pB = rwghB = ( 1000 kg>m

3

Ans.

)( 9.81 m>s ) (1.1845 m) 2

= 11.62 ( 103 ) Pa = 11.6 kPa

1m

2m

Ans.

1m 1m B

A

3m

(a)

Ans: pA = 21.6 kPa pB = 11.6 kPa 246

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–160. If the truck has a constant acceleration of 2 m > s2, determine the water pressure at the bottom corners B and C of the water tank. There is a small opening at A.

A a

1m 1m C

B

SOLUTION Since the water in the tank is confined, the imaginary free surface must pass through A as shown in Fig. a. We have tan u =

2 m>s2 aC = = 0.2039 g 9.81 m>s2

From the geometry in Fig. a, ∆hC = (2 m) tan u = (2 m)(0.2039) = 0.4077 m ∆hB = (3 m) tan u = (3 m)(0.2039) = 0.6116 m Then hC = 2 m + ∆hA = 2 m + 0.4077 m = 2.4077 m hB = 2 m - ∆hB = 2 m - 0.6116 m = 1.3884 m Finally, pC = rwghC = ( 1000 kg>m3 )( 9.81 m>s2 ) (2.4077 m) = 23.62 ( 103 ) Pa = 23.6 kPa

Ans.

pB = rwghB = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.3884 m) = 13.62 ( 103 ) Pa = 13.6 kPa

Ans.

C

2m

C

B

2m

3m

247

2m

3m

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. a

2–161. The cart is allowed to roll freely down the inclined plane due to its weight. Show that the slope of the surface of the liquid, u, during the motion is u = f.

u

SOLUTION

f

Referring to the free-body diagram of the container in Fig. a, +ΣFx′ = max′ b w sin f =

w a g

w

a = g sin f

dx

Referring to Fig. b,

a

ax = - (g sin f) cos f

dy

x′

ay = -(g sin f) sin f We will now apply Newton’s equations of notation, Fig. c. + ΣFx = max; S

N (a)

g(dxdydz) 0px - apx + dxbdydz + px dydz = ax 0x g

dpx = -

y

gdx a g x

In y direction, + c Σ Fy = may;

ax

0 py

gdxdydz pydxdz - apy + dybdxdz - gdxdydz = ay 0y g ay dpy = - gdy a1 + b g

x′

(b)

px dydz

dy g sin f cos f sin f cos f sin f ax = = = = = tan f dx g + ay g - g sin f sin f cos f cos2 f

ay

a = g sin f

At the surface, p is constant, so that dpx + dpy = 0, or dpx = -dpy. ay gdx ax = -gdy a1 + b g g

x

f

(

px +

dpx dx

dx

Since at the surface,

(

py +

dy = - tan u dx

dpy dy

then py dx dz

tan u = tan f or

(c)

Q.E.D.

u = f

248

(

dy dx dz

(

dx dy dz

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. a

2–162. The cart is given a constant acceleration a up the plane, as shown. Show that the lines of constant pressure within the liquid have a slope of tan u = (a cos f)>(a sin f + g).

u

f

SOLUTION As in the preceding solution, we determine that

y

dy ax = dx g + ay

(1) ay

a

Here, the slope of the surface of the liquid, Fig. a, is dy = - tan u dx

(2)

(a)

Equating Eqs. (1) and (2), we obtain tan u =

ax g + ay

(3)

By establishing the x and y axes shown in Fig. a, ax = a cos f

ay = a sin f

Substituting these values into Eq. (3), tan u =

ax

a cos f a sin f + g

Q.E.D

249

x

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–163. The open railcar is used to transport water up the 20o incline. When the car is at rest, the water level is as shown. Determine the maximum acceleration the car can have when it is pulled up the incline so that no water will spill out.

18 ft

0.5 ft 9 ft

20!

SOLUTION The volume of the water can be determined by using the water level when the car is at rest, indicated by dashed line in Fig. a. Here a = 0.5 ft + 18 tan 20° ft = 7.0515 ft If the width of the car is w Vw =

1 (0.5 ft + 7.0515 ft)(18 ft)w = 67.9632 w 2

It is required that the water is about to spill out. In this case, the water is indicated by the solid line in Fig. a, Vw =

1 (9 ft)(b) w = 67.9632 w 2 (O.K.)

b = 15.1029 ft < 18 ft Then, u = tan-1 a

9 ft b - 20° = 10.7912 ° 15.1029 ft

Consider the vertical block of water of weight dw = gwhdA shown shaded in Fig. a + c ΣFy = may;

pdA - gwhdA =

gwhdA a sin 20° g

p =

gwh a sin 20° + gwh g

p =

gwh (a sin 20° + g) g

(1)

Consider the horizontal block of water of weight dw = gwxdA shown shaded in Fig. a + ΣFx = max; S

p2 dA - p1dA = p2 - p1 =

gwxdA a cos 20° g

gwx a cos 20° g

(2)

However, from Eq. (1), p2 is at h2 and p1 is at h1, so that gw p2 - p1 = (h - h1)(a sin 20° + g) g 2 Substituting this result into Eq. (2), we have gw gwx (h2 - h1)(a sin 20° + g) = a cos 20° g g h2 - h1 a cos 20° = x a sin 20° + g 250

2–163. Continued

However, tan u =

h2 - h1 . Thus x tan u =

a cos 20° a sin 20° + g

Here, u = 10.7912°. Then tan 10.7912° =

a cos 20° a sin 20° + 32.2 ft>s2

a = 7.02 ft>s2

Ans.

y 18 ft a 20˚ x

h1 h

9 ft

a

h2

0.5 ft

20˚

x

p2

A

p1

A

b pA (a)

Ans: 7.02 ft>s2 251

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–164. The open railcar is used to transport water up the 20o incline. When the car is at rest, the water level is as shown. Determine the maximum deceleration the car can have when it is pulled up the incline so that no water will spill out.

18 ft

0.5 ft 9 ft

SOLUTION The volume of water can be determined using the water level when the car is at rest indicated by the dashed line in Fig. a. Here,

20!

a = 0.5 ft + 18 ft tan 20° = 7.0515 ft If the width of the car is w, Vw =

18 ft

1 (0.5 ft + 7.0515 ft)(18 ft)w = 67.9632 w 2

w

It is required that the water is about to spill out. In this case, the water level is indicated by the solid line in Fig. a

h2

h1

1 Vw = (9 ft)(b)(w) = 67.9632 w 2 (O.K) a

b = 15.1029 ft < 18 ft

p

x 20˚

Consider the vertical block of water of weight dw = gwhdA shown shaded in Fig. a.

b

(a) y

gwh p = (g - a sin 20°) g

20˚

(1)

Consider the horizontal block of water of weight dw = gwxdA shown shaded in Fig. a + ΣFx = max; S

p1dA - p2dA = p2 - p1 =

a

gwxdA ( - a cos 20°) g

x

gwx a cos 20° g

(2)

However, from Eq. (1), since p1 is at h1 and p2 is at h2. gw p2 - p1 = (h - h1)(g - a sin 20°) g 2 Substituting this result into Eq. (2) gw gwx (h2 - h1)(g - a sin 20°) = a cos 20° g g h2 - h1 a cos 20° = x g - a sin 20° However, tan u =

h2 - h1 . Thus x tan u =

a cos 20° g - a sin 20°

Here, u = 50.7912°. Then tan 50.7912° =

a cos 20° 32.2 ft>s2 - a sin 20°

a = 29.04 ft>s2 = 29.0 ft>s2

Ans. 252

20˚

0.5 ft

p

2A

9 ft b + 20° = 50.7912° u = tan-1 a 15.1029 ft gwhdA p dA - gwhdA = ( - a sin 20°) g

9 ft

pA

1A

Then

+ c ΣFy = may;

h

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–165. A woman stands on a horizontal platform that is rotating at 1.5 rad>s. If she is holding a cup of coffee, and the center of the cup is 4 m from the axis of rotation, determine the slope angle of the coffee’s surface. Neglect the size of the cup.

SOLUTION Since the coffee cup is rotating at a constant velocity about the vertical axis of rotation, then its acceleration is always directed horizontally toward the axis of rotation and its magnitude is given by ar = v2r = ( 1.5 rad>s ) (4 m) = 9 m>s2 2

Thus, the slope of coffee surface is m = tan u =

9 m>s2 ar = 0.917 = g 9.81 m>s2 Ans.

u = 42.5°

Ans: 42.5° 253

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 0.25 m

2–166. The drum is filled to the top with oil and placed on the platform. If the platform is given a rotation of v = 12 rad>s, determine the pressure the oil will exert on the cap at A. Take ro = 900 kg>m3.

0.35 m

A

SOLUTION We observe from Fig. a that h = hA at r = 0.25 m. hA =

v2 2 r 2g

hA = £

v

(12 rad>s)2 2 ( 9.81 m>s2 )

= 0.4587 m

§ (0.25 m)2

pA = roghA = ( 900 kg>m3

Imaginary surface

)( 9.81 m>s2 ) (0.4587 m)

= 4.05 ( 103) Pa Ans.

= 4.05 kPa

hA A 0.25 m (a)

Ans: 4.05 kPa 254

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 0.25 m

2–167. The drum is filled to the top with oil and placed on the platform. Determine the maximum rotation of the platform if the maximum pressure the cap at A can sustain before it opens is 40 kPa. Take ro = 900 kg>m3.

0.35 m

A

SOLUTION It is required that pA = 40 kPa. Thus, the pressure head for the oil is pA hA = = gO

40 ( 103 ) N>m2

( 900 kg>m3 )( 9.81 m>s2 )

v

= 4.531 m Imaginary surface

We observe from Fig. a that h = hA at r = 0.25 m. hA =

2

v 2 r 2g

4.531 m = £

v2 2 ( 9.81 m>s2 )

v = 37.7 rad>s

hA

§ (0.25 m)2

A

Ans.

0.25 m (a)

Ans: 37.7 rad>s 255

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–168. The beaker is filled to a height of h = 0.1 m with kerosene and place on the platform. What is the maximum angular velocity v it can have so that no kerosene spills out of the beaker?

0.15 m

0.2 m

SOLUTION

h

When the kerosene is about to spill out of the beaker, Fig. a, h>2 = 0.1 m or h = 0.2 m. h =

v2 2 r 2g

0.2 m =

v2 2 ( 9.81 m>s2 )

v = 26.4 rad>s

a

0.15 m 2 b 2

R = 0.075 m

Ans.

h — = 0.1 m 2

h — = 0.1 m 2 (a)

256

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 0.15 m

2–169. The beaker is filled to a height of h = 0.1 m with kerosene and placed on the platform. To what height h = h′ does the kerosene rise against the wall of the beaker when the platform has an angular velocity of v = 15 rad>s?

0.2 m h

SOLUTION H = H =

v2 2 r 2g (15 rad>s)2 2 ( 9.81 m>s2 )

= 0.0645 m

a

0.15 m 2 b 2

R = 0.075 m

From Fig. a, we observe that h′ = 0.1 m +

0.0645 m 2

H — 2

Ans.

h′ = 0.132 m = 132 mm

H

h′

0.1 m (a)

Ans: 132 mm 257

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–170. The tube is filled with water to the level h = 1 ft. Determine the pressure at point O when the tube has an angular velocity of v = 8 rad>s.

v

h ! 1 ft

O

SOLUTION 2 ft

The level of the water in the tube will not change. Therefore, the imaginary surface will be as shown in Fig. a. H =

2 ft

(8 rad>s)2(2 ft)2 v2R2 = = 3.9752 ft 2g 2 ( 32.2 ft>s2 )

R = 2 ft

We observe from Fig. a that hO = H - 1 ft = 3.9752 ft - 1 ft = 2.9752 ft Finally, the pressure at O must be negative since it is 2.9752 ft above the imaginary surface of the liquid.

1 ft

pO = ghO = ( 62.4 lb>ft 3 ) ( - 2.9752 ft) = a -185.65

= -1.29 psi

O H

lb 1 ft 2 ba b 2 12 in. ft

hO Imaginary surface

Ans.

(a)

Ans: -1.29 psi 258

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 2–171. The sealed assembly is completely filled with water such that the pressures at C and D are zero. If the assembly is given an angular velocity of v = 15 rad>s, determine the difference in pressure between points C and D.

D

C

0.6 m

A

SOLUTION H = =

0.5 m

v2R2 2g

B

v

( 15 rad>s ) 2(0.5 m)2 2 ( 9.81 m>s2 )

= 2.867 m Imaginary surface

From Fig. a, ∆h = H = 2.867 m. Then, ∆p = pD - pC = rwg∆h = ( 1000 kg>m3 )( 9.81 m>s2 ) (2.867 m) = 28.13 ( 103 ) Pa = 28.1 kPa

Ans. C

D R = 0.5 m (a)

Ans: 28.1 kPa 259

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–172. The sealed assembly is completely filled with water such that the pressures at C and D are zero. If the assembly is given an angular velocity of v = 15 rad>s, determine the difference in pressure between points A and B.

D

C

0.6 m

A

SOLUTION H = =

0.5 m

v2R2 2g

v

(15 rad>s)2(0.5 m)2 2 ( 9.81 m>s2 )

= 2.867 m From Fig. a, ∆h = hB - hA = H = 2.867 m. Then, ∆p = pB - pA = rwg∆h = ( 1000 kg>m3 )( 9.81 m>s2 ) (2.867 m) = 28.13 ( 103 ) Pa = 28.1 kPa

Imaginary surface

Ans.

H

hB

hA

0.6 m

A

R = 0.5 m (a)

260

B

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–173. The U-tube is filled with water and A is open while B is closed. If the axis of rotation is at x = 0.2 m, determine the constant rate of rotation so that the pressure at B is zero.

v B 1m

SOLUTION

0.6 m

Since points A and B have zero gauge pressure, the imaginary free surface must pass through them as shown in Fig. a. h′A =

v2(0.4 m)2 v2rA2 = = 0.008155v2 2g 2 ( 9.81 m>s2 )

h′B =

v2(0.2 m)2 v2rB2 = = 0.002039v2 2g 2 ( 9.81 m>s2 )

C 0.6 m

x

From Fig. a, 1 m - h′A = 0.6 m - h′B 1 m - 0.008155v2 = 0.6 m - 0.002039v2 Ans.

v = 8.09 rad>s

Imaginary surface

rA = 0.4 m

rB = 0.2 m

A B

hA 1m

0.6 m

hB

(a)

Ans: 8.09 rad>s 261

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–174. The U-tube is filled with water and A is open while B is closed. If the axis of rotation is at x = 0.2 m and the tube is rotating at a constant rate of v = 10 rad>s, determine the pressure at points B and C.

v B 1m

SOLUTION

0.6 m

Since point A has zero gauge pressure, the imaginary free surface must pass through this point as shown in Fig. a. H =

(10 rad>s)2(0.4 m)2 v2R2 = 0.8155 m = 2g 2 ( 9.81 m>s2 )

h′ =

(10 rad>s)2(0.2 m)2 v2r 2 = 0.2039 m = 2g 2 ( 9.18 m>s2 )

C 0.6 m

x

and

From Fig. a, a = 1 m - H = 1 m - 0.8155 m = 0.1845 m Then, hB = -(0.6 m - h′ - a) = - (0.6 m - 0.2039 m - 0.1845 m) = -0.2116 m hC = h′ + a = 0.2039 m + 0.1845 m = 0.3884 m Finally, pB = rwghB = ( 1000 kg>m3 )( 9.81 m>s2 ) ( - 0.2116 m) = - 2.076 ( 103 ) Pa Ans.

= -2.08 kPa pC = rwghC = ( 1000 kg>m3 )( 9.81 m>s2 ) (0.3884 m) = 3.81 ( 103 ) Pa

Ans.

= 3.81 kPa Imaginary surface

R = 0.4 m

r = 0.2 m

B H

hB

1m

0.6 m

h hC

a

a C (a)

Ans: pB = -2.08 kPa pC = 3.81 kPa 262

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. A

2–175. The U-tube is filled with water and A is open while B is closed. If the axis of rotation is at x = 0.4 m and the tube is rotating at a constant rate of v = 10 rad>s, determine the pressure at points B and C.

v B 1m 0.6 m

SOLUTION

C

Since point A has zero gauge pressure, the imaginary free surface must pass through this point as shown in Fig. a. 2

0.6 m

x

2

H =

(10 rad>s) (0.4 m) v2R2 = = 0.8155 m 2g 2 ( 9.81 m>s2 )

h′A =

(10 rad>s)2(0.2 m)2 v2rA2 = = 0.2039 m 2g 2 ( 9.81 m>s2 )

and

From Fig. a, a = 1 m - h′A = 0.7961 m Then, hC = H + a = 0.8155 m + 0.7961 m = 1.6116 m hB = hC - 0.6 m = 1.6116 m - 0.6 m = 1.0116 m Finally, pB = rwghB = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.0116 m) = 9.924 ( 103 ) Pa Ans.

= 9.92 kPa pC = rwghC = ( 1000 kg>m3 )( 9.81 m>s2 ) (1.6116 m) = 15.81 ( 103 ) Pa = 15.8 kPa

R = 0.4 m

Imaginary Surface

Ans.

rA = 0.2 m H

hA

hB

hC B

1m

a

a

0.6 m

C (a)

263

Ans: pB = 9.92 kPa pC = 15.8 kPa

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. *2–176. The cylindrical container has a height of 3 ft and a diameter of 2 ft. If it is filled with water through the hole in its center, determine the maximum pressure the water exerts on the container when it undergoes the motion shown.

6 ft/s2 2 ft

3 ft

SOLUTION

10 rad/s 2

The container undergoes an upward acceleration of aC = 6 ft>s , the maximum pressure occurs at the bottom of the container. (pac)max = gha1 +

6 ft>s2 aC b = a b = 222.08 lb>ft 2 g 32.2 ft>s2

Since the container is fully filled and the pressure at the center O, of the lid is atmospheric pressure, the imaginary parabolic surface above the lid will be formed as if there were no lid. Fig. a h =

v2 2 r ; 2g

The maximum pressure is

ho = c

(10 rad>s)2 2 ( 32.2 ft>s

2

)

1 ft

d (1 ft)2 = 1.5528 ft

r

hO

pmax = (pac)max + (pw)max = 222.08 lb>ft 2 + ( 62.4 lb>ft 3 )( 1.5528 ft ) = ( 31 8.97 lb>ft 2 ) a

1 ft 2 b = 2.22 psi 144 in2

Ans.

O hA 3 ft

(a)

264

A

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft 3. Also, assume all pressures are gage pressures. 300 mm

2–177. The drum has a hole in the center of its lid and contains kerosene to a level of 400 mm when v = 0. If the drum is placed on the platform and it attains an angular velocity of 12 rad>s, determine the resultant force the kerosene exerts on the lid.

200 mm

400 mm

SOLUTION The volume of the air contained in the paraboloid must be the same as the volume of air in the drum when it is not rotating. Since the volume of the paraboloid is equal to one half the volume of the cylinder of the same radius and height, then VAir = Vpb 1 ( pri2h) 2 ri 2h = 0.036

p(0.3 m)2(0.2) =

(1)

Then h =

w2 2 r ; 2g

h = c

122 d ri2 = 7.3394ri2 2(9.81)

Solving Eqs. (1) and (2), h = 0.5140 m

ri = 0.2646 m

From Section 2–14, beneath the lid, gv2 2 br + C 2g Since g = rg, this equation becomes p = a

At r = ri, p = 0. Then

p = a

rv2 2 br + C 2

rv2 2 ri + C 2 rv2 2 ri C = 2 0 =

Thus, p =

rv2 2 ( r - ri2) 2

265

(2)

v

2–177. Continued

Then the differential force dF acting on the differential annular element of area dA = 2prdr shown shaded in Fig. a is rv2 2 ( r - ri2) (2prdr) dF = pdA = 2

ro = 0.3 m

= prv2 ( r 3 - ri2r) dr F =

L

2

dF = prv

Lri

= prv2 °

ro

ri

air

( r - ri r) dr 3

2

0.2 m 4

2

ro

ri 2 r r ¢` 4 2 ri

h 0.4 m

ro4 ri2 ro2 ri4 = prv2 ° + ¢ 4 2 4

(a)

p 2 4 rv ( ro - 2ri2 ro2 + ri4) 4 p = rv2 ( ro2 - ri2) 2 4 =

Here r = rke = 814 kg>m3, v = 12 rad>s, ro = 0.3 m and ri = 0.2646 m p F = ( 814 kg>m3 ) (12 rad>s)2 3 (0.3 m)2 - (0.2646 m)2 4 2 4 = 36.69 N = 36.7 N

dr r

ri

Ans.

ro (b)

Ans: 36.7 N 266

3–1. A marked particle is released into a flow when t = 0, and the pathline for a particle is shown. Draw the streakline, and the streamline for the particle when t = 2 s and t = 4 s.

4m t!3s 6m

t!4s

t!2s

4m

pathline 60"

t!0

SOLUTION Since the streamlines have a constant direction for the time interval 0 … t 6 3 s, the pathline and streakline coincide with the streamline when t = 2 s as shown in Fig. a. The pathline and streakline will coincide with the streamline until t = 3 s, after which the streamline makes a sudden change in direction. Thus, the streamline of the marked particle and the streakline when t = 4 s will be as shown in Fig. b.

marked particle streamline 4m

streakline 60˚ t=2s (a) marked particle

streamline

streakline

60˚ 4m t=4s (b)

267

6m

3–2. The flow of a liquid is originally along the positive x axis at 2 m>s for 3 s. If it then suddenly changes to 4 m>s along the positive y axis for t 7 3 s, draw the pathline and streamline for the first marked particle when t = 1 s and t = 4 s. Also, draw the streaklines at these two times.

SOLUTION Since the streamlines have a constant direction along the positive x axis for the time interval 0 … t 6 3 s, the pathline and streakline coincide with the streamline when t = 1 s as shown in Fig. a.

t=0

The pathline and streakline will coincide with the streamline until t = 3 s, after which the streamline makes a sudden change in direction. Thus, the streamline and pathline of the first marked particle and the streakline when t = 4 s will be as shown in Fig. b.

pathline t=1s streamline x

2m

first marked particle

streakline

(a) y

streamline streakline

first marked particle t=4s

4m

t=0

x t=3s 6m pathline t=4s (b)

268

3–3. The flow of a liquid is originally along the positive y axis at 3 m>s for 4 s. If it then suddenly changes to 2 m>s along the positive x axis for t 7 4 s, draw the pathline and streamline for the first marked particle when t = 2 s and t = 6 s. Also, draw the streakline at these two times.

SOLUTION Since the streamlines have a constant direction along the positive y axis, 0 … t 6 4 s, the pathline and streakline coincide with the streamline when t = 2 s as shown in Fig. a. The pathline and streakline will coincide with the streamline until t = 4 s, when the streamline makes a sudden change in direction. The pathline, streamline, and streakline are shown in Fig. b.

y pathline t=2s

streamline first marked particle streakline

6m

t=0 (a) y t=4s

pathline t = 6 s streamline first marked particle

12 m

streakline

x t=0

4m (b)

269

*3–4. A two-dimensional flow field for a fluid can be described by V = 5 (2x + 1)i - (y + 3x)j 6 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at (2 m, 3 m), and its direction measured counterclockwise from the x axis.

y

3m

x

SOLUTION

2m

The velocity vector for a particle at x = 2 m and y = 3 m is

Vx = 5 m/s

"

V = 5 (2x + 1)i - (y + 3x)j 6 m>s

!

= [2(2) + 1]i - [3 + 3(2)]j

The magnitude of V is

= 5 5i - 9j 6 m>s

V = 2V x2 + V y2 = 2 ( 5 m>s ) 2 +

( -9 m>s ) 2 = 10.3 m>s

Ans.

As indicated in Fig. a, the direction of V is defined by u = 360° - f, where

Thus,

f = tan-1 a

Vy Vx

b = tan-1 a

9 m> s 5 m>s

Ans.

270

V (a)

b = 60.95°

u = 360° - 60.95° = 299°

Vy = 9 m/s

x

3–5. A two-dimensional flow field for a liquid can be described by V = 5 ( 5y2 - x ) i + (3x + y)j 6 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at (5 m, -2 m), and its direction measured counterclockwise from the x axis.

SOLUTION The velocity vector of a particle at x = 5 m and y = -2 m is V = = The magnitude of V is

5( 5y2 - x ) i + 3 5( - 2)2 - 5 4 i

Vy = 13 m/s

V

(3x + y)j6 m>s

+ 33(5) + ( - 2)4j

= 515i + 13j 6 m>s

! Vx = 15 m/s

V = 2V x2 + V y2 = 2 ( 15 m>s ) 2 + ( 13 m>s ) 2 = 19.8 m>s

Ans.

x

(a)

As indicated in Fig. a, the direction of V is defined by u = tan - 1 a

Vy Vx

b = tan - 1 a

13 m>s 15 m>s

b = 40.9°

Ans.

Ans: V = 19.8 m>s u = 40.9° 271

3–6. The soap bubble is released in the air and rises with a velocity of V = 5 (0.8x)i + ( 0.06t 2 ) j 6 m>s, where x is meters and t is in seconds. Determine the magnitude of the bubble’s velocity, and its direction measured counterclockwise from the x axis, when t = 5 s, at which time x = 2 m and y = 3 m. Draw its streamline at this instant.

v u

SOLUTION The velocity vector of a particle at x = 2 m and the corresponding time t = 5 s is

5(0.8x)i

V =

3 0.8(2)i

=

Vy = 1.5 m s

+ ( 0.06t ) j6 m>s + 0.06(5)2 j4

= 5 1.6i + 1.5j 6 m>s

The magnitude of V is

V

2

!

V = 2V x2 + V y2 = 2 ( 1.6 m>s ) 2 + ( 1.5 m>s ) 2 = 2.19 m>s

Vx = 1.6 m s

Ans.

x

(a)

As indicated in Fig. a, the direction of V is defined by u = tan-1 a

Vy Vx

b = tan-1 a

1.5 m>s 1.6 m>s

b = 43.2°

Using the definition of the slope of the streamline and initial condition at x = 2 m, y = 3 m. dy v dy 0.06t 2 = ; = dx u dx 0.8x Note that since we are finding the streamline, which represents a single instant in time, t = 5 s, t is a constant. y

L3 m

dy t

2

x

=

L2 m

0.075dx x

1 x (y - 3) = 0.075 ln 2 2 t y = a0.075t 2 ln

When t = 5 s,

y(m)

x + 3b m 2

6 5 4

x y = 0.075 ( 52 ) ln a b + 3 2

3

x y = c 1.875 ln a b + 3 d m 2

2 1

x(m)

0.5

1

2

3

4

5

6

y(m)

0.401

1.700

3

3.760

4.300

4.718

5.060

0 0.5 1

2

3

4

5

6

x(m)

(a)

The plot of the streamline is shown in Fig. a

Ans: V = 2.19 m>s u = 43.2° 272

3–7. A flow field for a fluid is described by u = (2 + y) m>s and v = (2y) m>s, where y is in meters. Determine the equation of the streamline that passes through point (3 m, 2 m), and find the velocity of a particle located at this point. Draw this streamline.

SOLUTION As indicated in Fig a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

y v = 2y m/s

dy = tan u dx

"

dy v = dx u

u = (2 ! y) m/s

streamline

y

dy 2y = dx 2 + y

x x

2 + y dy = dx L 2y L ln y +

V

(a)

1 y = x + C 2

At point (3 m, 2 m), we obtain ln(2) +

1 (2) = 3 + C 2

C = - 1.31 Thus, ln y +

1 y = x - 1.31 2

ln y2 + y = 2x - 2.61

Ans.

At point (3 m, 2 m) u = (2 + 2) m>s = 4 m>s S v = 2(2) = 4 m>s c The magnitude of the velocity is V = 2u2 + v2 = 2 ( 4 m>s ) 2 + ( 4 m>s ) 2 = 5.66 m>s

Ans.

and its direction is

4 m>s v u = tan-1 a b = tan-1 a b = 45° u 4 m>s

Ans.

Ans: ln y2 + y = 2x - 2.61 V = 5.66 m>s u = 45° a 273

*3–8. A flow field is described by u = 1 x2 + 5 2 m>s and v = ( -6xy) m>s. Determine the equation of the streamline that passes through point (5 m, 1 m), and find the velocity of a particle located at this point. Draw this streamline.

SOLUTION y

As indicated in Fig a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore, dy = tan u dx

streamline u = (x2 + 5) m/s

dy -6xy v = = 2 dx u x + 5 dy

L y

= -6

x dx 2 x L + 5

v = (6xy) m/s (a)

At x = 5 m, y = 1 m. Then, ln 1 = - 3 ln 3 (5)2 + 5 4 + C C = 3 ln 30

ln y = - 3 ln ( x2 + 5 ) + 3 ln 30 ln y + ln ( x2 + 5 ) 3 = 3 ln 30 ln 3 y ( x2 + 5 ) 3 4 = ln 303 y ( x2 + 5 ) 3 = 303 y =

x x

ln y = - 3 ln ( x2 + 5 ) + C

Thus

!

y

27 ( 103 )

Ans.

( x2 + 5 ) 3

At point (5 m, 1m), u = ( 52 + 5 ) m>s = 30 m>s S v = - 6(5)(1) = - 30 m>s = 30 m>s T The magnitude of the velocity is V = 2u2 + v2 = 2 ( 30 m>s ) 2 + ( 30 m>s ) 2 = 42.4 m>s

Ans.

And its direction is

30 m>s v u = tan-1 a b = tan-1 a b = 45° u 30 m>s

Ans.

274

V

3–9. Particles travel within a flow field defined by V = 5 2y2i + 4j 6 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m), and find the velocity of a particle located at this point. Draw this streamline.

SOLUTION As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

y

v = 4 m/s

dy = tan u dx

y2 dy =

L

x

2dx

x (a)

1 3 y = 2x + C 3

At x = 1 m, y = 2 m. Then

streamline

y

dy v 4 = = 2 dx u 2y L

V

! u = 2y2 m/s

1 3 (2) = 2(1) + C 3 C =

2 3

Thus, 1 3 2 y = 2x + 3 3 y3 = 6x + 2

Ans.

At point (1 m, 2 m) u = 2 ( 22 ) = 8 m>s S v = 4 m>s c The magnitude of the velocity is V = 2u2 + v2 = 2 ( 8 m>s ) 2 + ( 4 m>s ) 2 = 8.94 m>s

Ans.

And its direction is

v 4 u = tan-1 a b = tan-1 a b = 26.6° u 8

Ans.

Ans: y3 = 6x + 2 V = 8.94 m>s u = 26.6° a 275

3–10. A balloon is released into the air from the origin and carried along by the wind, which blows at a constant rate of u = 0.5 m>s. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = (0.8 + 0.6y) m>s. Determine the equation of the streamline for the balloon, and draw this streamline.

y

v u

x

SOLUTION As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

y v = (0.8 + 0.6y) m/s V streamline

dy = tan u dx

u = 0.5 m/s

dy 0.8 + 0.6y v = = = 1.6 + 1.2y dx u 0.5

y x

Since the balloon starts at y = 0, x = 0, using these values,

x

y

lna

!

x dy dx = L0 1.6 + 1.2y L0 y 1 ln(1.6 + 1.2y) ` = x 1.2 0

1.6 + 1.2y 1.6 lna1 +

(a) y(m)

b = 1.2x

y=

4 1.2x ( e – 1) m 3

3 yb = 1.2x 4

1 +

3 y = e 1.2x 4 4 y = ( e 1.2x - 1 ) m 3

streamline

Ans.

Using this result, the streamline is shown in Fig. b.

x(m) (b)

Ans: y = 276

4 1.2x (e - 1) 3

3–11. A balloon is released into the air from point (1 m, 0) and carried along by the wind, which blows at a rate of u = (0.8x) m>s, where x is in meters. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = (1.6 + 0.4y) m>s, where y is in meters. Determine the equation of the streamline for the balloon, and draw this streamline.

y

v u

x

SOLUTION

y

As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

v = (1.6 + 0.4y) m/s V

dy = tan u dx

streamline

! u = (0.8x) m/s

dy 1.6 + 0.4y v = = dx u 0.8x

y x

The balloon starts at point (1 m, 0).

x

y

x dy dx = 1.6 + 0.4y 0.8x L0 L1

(a) y

y x 1 1 ln(1.6 + 0.4y) ` = ln x ` 0.4 0.8 0 1

y = 4(x½ – 1) m

1.6 + 0.4y 1 1 lna b = ln x 0.4 1.6 0.8 lna1 + a1 +

1 2 yb = ln x 4

streamline

1 2 yb = x 4

x

y = 4 ( x1>2 - 1 ) m

Ans.

Using this result, the streamline is shown in Fig. b.

1m (b)

Ans: y = 4 ( x1>2 - 1 ) 277

*3–12. A flow field is defined by u = (8y) m>s and v = (6x) m>s, where x and y are in meters. Determine the equation of the streamline that passes through point (1 m, 2 m). Draw this streamline.

SOLUTION As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

y

dy = tan u dx dy v 6x = = dx u 8y L

8y dy = 2

L

v = (6x) m/s

V

! u = (8y) m/s streamline

y

6x dx

x

2

4y = 3x + C x

At x = 1 m, y = 2 m. Then 4(2)2 = 3(1)2 + C

(a)

C = 13 Thus 4y2 = 3x2 + 13

Ans.

278

3–13. A flow field is defined by u = (3x) ft>s and v = (6y) ft>s, where x and y are in feet. Determine the equation of the streamline passing through point (3 ft, 1 ft). Draw this streamline.

SOLUTION As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

y v = (6y) ft/s

dy = tan u dx dy 6y v = = dx u 3x

V

streamline

! u = (3x) ft/s

y

x

dy

dx = L 2y L x

x (a)

1 ln y = ln x + C 2 At x = 3 ft, y = 1 ft . Then 1 ln y = ln x - ln3 2 1 x ln y = ln 2 3 x 2 ln y = lna b 3 y =

x2 9

Ans.

279

Ans: y = x2 >9

3–14. A flow of water is defined by u = 5 m>s and v = 8 m>s. If metal flakes are released into the flow at the origin (0, 0), draw the streamline and pathline for these particles.

SOLUTION Since the velocity V is constant, Fig. a, the streamline will be a straight line with a slope. dy = tan u dx dy v 8 = = dx u 5

y

v = 8 m/s V !

streamline pathline

u = 5 m/s

y = 1.6x + C At x = 0, y = 0. Then

x

C = 0 (a)

Thus Ans.

y = 1.6x

Since the direction of velocity V remains constant so does the streamline, and the flow is steady. Therefore, the pathline coincides with the streamline and shares the same equation.

Ans: y = 1.6x 280

3–15. A flow field is defined by u = 3 8x> ( x2 + y2 ) 4 m>s and v = 3 8y> ( x2 + y2 ) 4 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 1 m). Draw this streamline.

SOLUTION As indicated in Fig. a, the velocity V of a particle on the streamline is always directed along the tangent of the streamline. Therefore,

y

dy = tan u dx y dy v = = = dx u x 8x> ( x2 + y2 ) L y

=

) x 8y+ y ) m/s V 2

2

!

8y> ( x2 + y2 )

dy

v=

u=

y

) x 8y+ y ) m/s 2

2

streamline x

dx L x

x (a)

y ln x = C y = C′ x At x = 1 m, y = 1 m. Then C′ = 1 Thus, y = 1 x y = x

Ans.

Ans: y = x 281

*3–16. A fluid has velocity components of u = 3 30>(2x + 1) 4 m>s and v = (2ty) m>s, where x and y are in meters and t is in seconds. Determine the pathline that passes through point (2 m, 6 m) at time t = 2 s. Plot this pathline for 0 … x … 4 m.

SOLUTION Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform flow. Because we are finding a pathline, t is not a constant but a variable. We must first find equations relating x to t and y to t, and then eliminate t. Using the definition of velocity x

t

L2 m

dx 30 = u = ; dt 2x + 1

(2x + 1)dx = 30

( x2 + x ) `

x 2m

2

dt L2 s

= 30 t `

t 2s

x + x - 6 = 30(t - 2)

t = dy = v = 2ty; dt

1 2 ( x + x + 54 ) 30

(1)

y

t dy = 2 tdt L6 m y L2 s

ln y ` ln

y 6m

= t2 `

t

y(m)

2s

y = t2 - 4 6 y 2 = et - 4 6 2

y = 6e t

50 40

-4

(2)

Substitute Eq. (1) into Eq. (2), y = 6e 900 (x 1

2

+ x + 54)2 - 4

Ans.

30 20 10

The plot of the pathline is shown in Fig. a. x(m)

0

1

2

3

4

y(m)

2.81

3.58

6.00

13.90

48.24

282

0

1

2

3 (a)

4

x(m)

3–17. A fluid has velocity components of u = 3 30>(2x + 1) 4 m>s and v = (2ty) m>s where x and y are in meters and t is in seconds. Determine the streamlines that passes through point (1 m, 4 m) at times t = 1s, t = 2 s, and t = 3 s. Plot each of these streamlines for 0 … x … 4 m.

SOLUTION Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. The slope of the streamline is dy v = ; dx u

dy 2ty 1 = = ty(2x + 1) dx 30>(2x + 1) 15 y

x dy 1 = t (2x + 1)dx 15 L1 m L4 m y

ln y ` ln

y 4m

=

x 1 t ( x2 + x ) ` 15 1m

y 1 = t ( x2 + x - 2 ) 4 15 y = 4e t(x

+ x - 2) >15

y = 4e (x

+ x - 2) >15

2

For t = 1 s, 2

For t = 2 s,

y = 4e 2(x

2

For t = 3 s,

Ans.

+ x - 2) >15

Ans.

y(m) 45

y = 4e

(x2 + x - 2)>5

Ans.

40

The plot of these streamlines are shown in Fig. a

35

For t = 1 s

30

x(m)

0

1

2

3

4

y(m)

3.50

4

5.22

7.79

13.3

x(m)

0

1

2

3

4

y(m)

3.06

4

6.82

15.2

44.1

t=3s

t=2s

25 20

For t = 2 s

15 10

t=1s

5

For t = 3 s x(m) y(m)

0

1

2

3

4

2.68

4

8.90

29.6

146

0

1

2

3

4

x(m)

(a)

Ans: 2 For t = 1 s, y = 4e (x + x - 2)>15 2 ( For t = 2 s, y = 4e 2 x + x - 2)>15 (x2 + x - 2)>5 For t = 3 s, y = 4e 283

3–18. A fluid has velocity components of u = 3 30>(2x + 1) 4 m>s and v = (2ty) m>s, where x and y are in meters and t is in seconds. Determine the streamlines that pass through point (2 m, 6 m) at times t = 2 s and t = 5 s. Plot these streamlines for 0 … x … 4 m.

SOLUTION y(m)

For t = 2 s

t=5s

x(m) y(m)

0 2.70

1 3.52

2 6.00

3 13.35

4 38.80

50 40

For t = 5 s

30

x(m) y(m)

0 0.812

1 1.58

2 6.00

3 44.33

t=2s

4 638.06

20 10

Since the velocity components are a function of time and position the flow can be classified as unsteady nonuniform. The slope of the streamline is dy v = ; dx u

dy 2ty 1 = = ty(2x + 1) dx 30>(2x + 1) 15

0

1

2

3

4

x(m)

(a)

Note that since we are finding the streamline, which represents a single instant in time, either t = 2 s or t = 5 s, t is a constant. y

x dy 1 = t (2x + 1)dx 15 L2 m L6 m y

ln y `

y 6m

ln

=

x 1 t ( x2 + x ) ` 15 2m

y 1 = t ( x2 + x - 6 ) 6 15 y = 6e 15 t(x 1

2

+ x - 6)

For t = 2 s, y = 6e 15 (x 2

2

+ x - 6)

Ans.

For t = 5 s, y = 6e 3 (x 1

2

+ x - 6)

Ans.

The plots of these two streamlines are show in Fig. a.

Ans: 2 For t = 2 s, y = 6e 21x + x - 6)>15 1x2 + x - 6) >3 For t = 5 s, y = 6e 284

3–19. A particle travels along a streamline defined by y3 = 8x - 12. If its speed is 5 m>s when it is at x = 1 m, determine the two components of its velocity at this point. Sketch the velocity on the streamline.

SOLUTION x(m) y(m)

0 -2.29

1 - 1.59

1.5 0

2 1.59

3 2.29

4 2.71

5 3.04

The plot of the streamline is shown in Fig. a. Taking the derivative of the streamline equation, 3y2

dy = 8 dx

dy 8 = tan u = 2 dx 3y When x = 1 m, y3 = 8(1) - 12;

y = -1.5874

Then dy 8 ` = tan u ` = ; dx x = 1 m 3( - 1.5874)2 x=1 m

u 0 x = 1 m = 46.62°

Therefore, the horizontal and vertical components of the velocity are u = ( 5 m>s ) cos 46.62° = 3.43 m>s

Ans.

v = ( 5 m>s ) sin 46.62° = 3.63 m>s

Ans.

y(m) 3 3 3 2 1

0 –1 –2

1

2

v

3

4

5

x(m)

5 m/s u

46.62°

–3

Ans: u = 3.43 m>s v = 3.63 m>s

(a)

285

*3–20. A flow field is defined by u = (0.8t) m>s and v = 0.4 m>s, where t is in seconds. Plot the pathline for a particle that passes through the origin when t = 0. Also, draw the streamline for the particle when t = 4 s.

SOLUTION Here, u =

dx . Then, dt dx = udt

Using x = 0 when t = 0 as the integration limit, L0

Also, v =

x

t

3 (0.8t) m>s 4 dt L0 x = 0.4t 2

dx =

dy . Then dt

(1)

dy = vdt Using y = 0 when t = 0 as the integration limit, L0

y

dy =

L0

t

( 0.4 m>s ) dt (2)

y = 0.4t Eliminating t from Eqs. (1) and (2) y2 = 0.4x

This equation represents the pathline of the particle. The x and y values of the pathline for the first five seconds are tabulated below. t 1 2 3 4 5

x 0.4 1.6 3.6 6.4 10

y 0.4 0.8 1.2 1.6 2

y(m)

A plot of the pathline is shown in Fig. a. From Eqs. (1) and (2), when t = 4 s, x = 0.4 ( 42 ) = 6.4 m

y = 0.4(4) = 1.6 m 1

Using the definition of the slope of the streamline, dy v = ; dx u y

t

L1.6 m

dy =

t(y - 1.6) = y = c

y2 = 0.4x

2

dy 0.4 = dx 0.8t

0

x

1 dx 2 L6.4 m

2

4

6

–1

1 (x - 6.4) 2

–2

1 (x - 6.4) + 1.6 d m 2t

(a)

286

8

10

x(m)

*3–20. (continued)

When t = 4 s,

y(m)

1 y = ( x - 6.4) + 1.6 2(4) y =

y=

1 x + 0.8 8

1 x + 0.8 8

0.8

The plot of the streamline is shown is Fig. b.

x(m)

(b)

287

3–21. The velocity for an oil flow is defined by V = 5 3y2 i + 8 j 6 m>s, where y is in meters. What is the equation of the streamline that passes through point (2 m, 1 m)? If a particle is at this point when t = 0, at what point is it located when t = 1 s?

SOLUTION Since the velocity components are a function of position only, the flow can be classified as steady nonuniform. Here, u = 1 3y2 2 m>s and v = 8 m>s . The slope of the streamline is defined by dy v = ; dx u

dy 8 = 2 dx 3y y

x

L1 m

3y2dy = 8

y3 `

y

= 8x `

1m

3

dx L2 m

x 2m

y - 1 = 8x - 16 y3 = 8x - 15

(1)

Ans.

From the definition of velocity dy = 8 dt y

L1 m

dy =

y`

y 1m

L0

= 8t `

1s

8 dt 1s 0

y - 1 = 8 Ans.

y = 9m Substituting this result into Eq. (1) 93 = 8x - 15

Ans.

x = 93 m

Ans: y3 = 8x - 15, y = 9 m x = 93 m 288

3–22. The circulation of a fluid is defined by the velocity field u = (6 - 3x) m>s and v = 2 m>s, where x is in meters. Plot the streamline that passes through the origin for 0 … x 6 2 m.

SOLUTION x(m)

0

0.25

0.5

0.75

1

y(m)

0

0.089

0.192

0.313

0.462

x(m)

1.25

1.5

1.75

2

y(m)

0.654

0.924

1.386



Since the velocity component is a function of position only, the flow can be classified as steady nonuniform. Using the definition of the slope of a streamline, dy v = ; dx u

dy 2 = dx 6 - 3x L0

y

x

dy = 2

dx 6 3x L0

x 2 y = - ln (6 - 3x) ` 3 0

6 - 3x 2 y = - ln a b 3 6

y = The plot of this streamline is show in Fig. a

2 2 ln a b 3 2 - x

Ans.

y(m) 1.5

1

0.5

0

0.25

0.5

0.75

1.0

1.25

1.5

1.75

2

x(m)

(a)

Ans: y = 289

2 2 ln a b 3 2 - x

3–23. A stream of water has velocity components of u = -2 m>s, v = 3 m>s for 0 … t 6 10 s; and u = 5 m>s, v = -2 m>s for 10 s 6 t … 15 s. Plot the pathline and streamline for a particle released at point (0, 0) when t = 0 s.

SOLUTION Using the definition of velocity, for 0 … t 6 10 s dx = u; dt

y(m)

dx = -2 dt L0

B

x

dx = -2

L0

30

t

dt (1)

x = ( - 2t) m

10

When t = 10 s, x = - 2(10) = -20 m dy = v; dt

C

20

A –20

dy = 3 dt y

L0

dy = 3

L0

–15

–10

–5

0

5

x(m)

(a)

t

dt (2)

y = (3t) m When t = 10 s, y = 3(10) = 30 m

The equation of the streamline can be determined by eliminating t from Eq. (1) and (2). 3 y = - x Ans. 2 For 10 6 t … 15 s. dx = u; dt

dx = 5 dt x

L-20 m

dx = 5

t

dt L10 s

x - ( - 20) = 5(t - 10) (3)

x = (5t - 70) m At t = 15 s, x = 5(15) - 70 = 5 m dy = v; dt

dy = -2 dt y

L30 m

dy = - 2

t

dt L10 s

y - 30 = - 2(t - 10) (4)

y = ( -2t + 50) m When t = 15 s, y = - 2(15) + 50 = 20 m Eliminate t from Eqs. (3) and (4), 2 y = a - x + 22b 5

Ans.

The two streamlines intersect at ( -20, 30), point B in Fig. (a). The pathline is the path ABC. 290

Ans: 3 For 0 … t 6 10 s, y = - x 2 2 For 10 s 6 t … 15 s, y = - x + 22 5

*3–24. A velocity field is defined by u = (4x) m>s and v = (2t) m>s, where t is in seconds and x is in meters. Determine the equation of the streamline that passes through point (2 m, 6 m) for t = 1 s. Plot this streamline for 0.25 m … x … 4 m.

SOLUTION x(m)

0.25

0.5

0.75

1

2

3

4

y(m)

4.96

5.31

5.51

5.65

6

6.20

6.35

Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. Using the definition of the slope of the streamline, dy v = ; dx u

dy 2t t = = dx 4x 2x y

L6 m

dy =

x

t dx 2 L2 m x

t x ln 2 2 t x y = ln + 6 2 2

y - 6 =

1 x y = a ln + 6bm 2 2 The plot of this streamline is shown in Fig. a.

Ans.

For t = 1 s,

y(m) 7 6 5 4 3 2 1

0 0.25 0.5 0.75 1

2

3

4

(a)

291

x(m)

3–25. The velocity field is defined by u = (4x) m>s and v = (2t) m>s, where t is in seconds and x is in meters. Determine the pathline that passes through point (2 m, 6 m) when t = 1 s. Plot this pathline for 0.25 m … x … 4 m.

SOLUTION x(m)

0.25

0.50

0.75

1

2

3

4

y(m)

5.23

5.43

5.57

5.68

6

6.21

6.38

Since the velocity components are a function of time and position, the flow can be classified as unsteady nonuniform. Using the definition of velocity, x

t

dx dt = L2 m 4x L1 S

dx = u = 4x; dt

x t 1 ln x ` = t` 4 2m 1s

1 x ln = t - 1 4 2 1 x ln + 1 4 2

t = y

dy = v = 2t; dt

(1)

t

L6 m

dy =

2t dt L1 s

y - 6 = t2 `

t 1s

2

(2)

y = t + 5

Substitute Eq. (1) into (2), 2 1 x y = a ln + 1b + 5 4 2

y = a

1 2x 1 x ln + ln + 6b 16 2 2 2

Ans.

The plot of this pathline is shown in Fig. (a) y(m) 7 6 5 4 3 2 1 0 0.25 0.5 0.75 1

2

3

4

x(m)

Ans: y =

(a)

292

1 2x 1 x ln + ln + 6 16 2 2 2

3–26. The velocity field of a fluid is defined by u = ( 12 x ) m>s, v = ( 18 y2 ) m>s for 0 … t 6 5 s and by u = ( - 14 x2 ) m>s, v = ( 14 y ) m>s for 5 s 6 t … 10 s, where x and y are in meters. Plot the streamline and pathline for a particle released at point (1 m, 1 m) when t = 0 s.

SOLUTION Using the definition of velocity, for 0 … t 6 5 s, dx 1 = x dt 2

dx = u; dt x

t

dx 1 = dt L1 m x L0 2 1 t 2

ln x =

1

When t = 5 s,

1

x = e 2 (5) = 12.18 m dy = v; dt

x = ae 2 t b m

(1)

dy 1 = y2 dt 8 y

dy

L1 m y

2

t

=

1 dt L0 8

1 y 1 - a b` = t y 1m 8 1 -

1 1 = t y 8

y - 1 1 = t y 8 y a1 -

When t = 5 s, y =

y = a

8 = 2.667 m 8 - 5

1 tb = 1 8 8 bm 8 - t

t ≠ 8s

(2)

The equation of the streamline and pathline can be determined by eliminating t from Eqs. (1) and (2)

x(m)

1

3

5

y(m)

1

1.38

1.67

y = a

8 bm 8 - 2 ln x

7

9

11

12.18

1.95

2.22

2.50

2.67

For 5 s < t … 10 s, dx 1 = - x2 dt 4

dx = u; dt x

t

dx 1 = dt 2 4 L5 s L12.18 m x

293

3–26. (continued)

- a

x = a

When t = 10 s,

x =

1 1 1 b = - (t - 5) x 12.18 4

1 t = - 1.1679 x 4

4 bm t - 4.6717

t ≠ 4.6717 s

(3)

4 = 0.751 m 10 - 4.6717 dy = v; dt

dy 1 = y dt 4 y

t dy 1 = dt 4 L5 s L2.667 m y

y 1 = (t - 5) 2.667 4 y 1 = e 4 (t - 5) 2.667

ln

1

y = c 2.667e 4 (t - 5) d m

1

(4)

y = 2.667e 4 (10 - 5) = 9.31 m

When t = 10 s,

Eliminate t from Eqs. (3) and (4), y = 2.667e 4 34( x + 1.1679) - 54 1

1

= c 2.667e ( x - 0.08208) d m 1

x(m)

0.751

1

3

y(m)

9.31

6.68

3.43

5

7

9

11

12.18

3.00

2.83

2.75

2.69

2.67

The two streamlines intersect at (12.18, 2.67), point B in Fig. (a). The pathline is the path ABC. y(m) 10 C 9 8 7 6 5 4 3 2 1 A 0

B

1 2 3 4 5 6 7 8 9 10 11 12 13 0.751 12.18

Ans: x(m)

For 0 … t 6 5 s, y =

8 8 - 2 ln x

For 5 s 6 t … 10 s, y = 2.67e 11>x - 0.0821) 294

3–27. A two-dimensional flow field for a liquid can be described by V = 5( 6y2 - 1 )i + (3x + 2) j 6 m>s, where x and y are in meters. Find the streamline that passes through point 16 m, 2 m2 and determine the velocity at this point. Sketch the velocity on the streamline.

SOLUTION We have steady flow since the velocity does not depend upon time.

y

u = 6y2 - 1

20 m s

v = 3x + 2 dy v 3x + 2 = = 2 dx u 6y - 1 L2

y

( 6y2 - 1 ) dy =

L6

2

(3x + 2)dx 6

2y3 - y ` = 1.5 x2 + 2x ` 2

2y - y -

3 2(2)

3

23 m/s

x

y

3

30.5 m s

- 2 4 = 1.5x + 2x 2

2y3 - 1.5x2 - y - 2x + 52 = 0

x

x 6

3 1.5(6)2

+ 2(6) 4

Ans.

At (6 m, 2 m) u = 6(2)2 - 1 = 23 m>s S v = 3(6) + 2 = 20 m>s c V = 2 ( 23 m>s ) 2 + ( 20 m>s ) 2 = 30.5 m>s

Ans.

Ans: 2y3 - 1.5x2 - y - 2x + 52 = 0 V = 30.5 m>s 295

*3–28. A flow field for a liquid can be described by V = 5 (2x + 1) i - y j 6 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at points 13 m, 1 m2. Sketch the velocity on the streamline.

SOLUTION We have steady flow since the velocity does not depend upon time.

y

u = 2x + 1 v = -y

1

dy -y v = = dx u 2x + 1 -

3

dy dx = L y L (2x + 1) - ln y =

1 ln (2x + 1) 2

y

1

-y ` = (2x + 1) 2 ` 1

1 2

x 3

- y + 1 = (2x + 1) -

3 2(3) 1

y = 3.65 - (2x + 1)2

+ 1 42 1

Ans.

u = 2(3) + 1 = 7 m>s v = - 1 m>s V = 2 ( 7 m>s ) 2 +

7 m/s 1m s

( - 1 m>s ) 2 = 7.07 m>s

296

Ans.

7.07 m/s x

3–29. Air flows uniformly through the center of a horizontal duct with a velocity of V = ( 6t 2 + 5 ) m>s, where t is in seconds. Determine the acceleration of the flow when t = 2 s.

SOLUTION Since the flow is along the horizontal (x axis) v = w = 0. Also, the velocity is a function of time t only. Therefore, the convective acceleration is zero, so that u

0V = 0. 0x 0V 0V + u 0t 0x = 12t + 0

a =

= (12t) m>s2 When t = 2 s, a = 12(2) = 24 m>s2

Ans.

Note: The flow is unsteady since its velocity is a function of time.

Ans: a = 24 m>s2 297

3–30. Oil flows through the reducer such that particles along its centerline have a velocity of V = (4xt) in.>s, where x is in inches and t is in seconds. Determine the acceleration of the particles at x = 16 in. when t = 2 s.

24 in.

x

SOLUTION Since the flow is along the x axis, v = w = 0 a =

0u 0u + u 0t 0x

= 4x + (4xt)(4t) = 4x + 16xt 2 = When t = 2 s, x = 16 in. Then a =

3 4(16) 3 1

3 4x ( 1

+ 4t 2 ) 4 in.>s2

+ 4 ( 22 ) 4 4 in.>s2 = 1088 in.>s2

Note: The flow is unsteady since its velocity is a function of time.

Ans: 1088 in.>s2 298

3–31. A fluid has velocity components of u = (6y + t) ft>s and v = (2tx) ft>s where x and y are in feet and t is in seconds. Determine the magnitude of the acceleration of a particle passing through the point (1 ft, 2 ft) when t = 1 s.

SOLUTION For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar component of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 1 + (6y + t)(0) + (2tx)(6) = (1 + 12tx) ft>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 2x + (6y + t)(2t) + (2tx)(0) = ( 2x + 12ty + 2t 2 ) ft>s2 When t = 1 s, x = 1 ft and y = 2 ft, then ax = 31 + 12(1)(1) 4 = 13 ft>s2

ay =

3 2(1)

+ 12(1)(2) + 2 ( 12 ) 4 = 28 ft>s2

Thus, the magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 13 ft>s2 ) 2 + ( 28 ft>s2 ) 2 = 30.9 ft>s2

Ans.

Ans: 30.9 ft>s2 299

*3–32. The velocity for the flow of a gas along the center  streamline of the pipe is defined by u = ( 10x2 + 200t + 6 ) m>s, where x is in meters and t is in seconds. Determine the acceleration of a particle when t = 0.01 s and it is at A, just before leaving the nozzle.

A x 0.6 m

SOLUTION a = 0u = 200 0t a =

3 200

When t = 0.01 s, x = 0.6 m. a =

5 200

+

= 339 m>s2

0u 0u + u 0t 0x 0u = 20 x 0x

+ ( 10x2 + 200t + 6 ) (20x) 4 m>s2

3 10 ( 0.62 )

+ 200(0.01) + 6 4 320(0.6) 4 6 m>s2

300

Ans.

3–33. A fluid has velocity components of u = ( 2x2 - 2y2 + y ) m>s and v = (y + xy) m>s, where x and y are in meters. Determine the magntiude of the velocity and acceleration of a particle at point (2 m, 4 m).

SOLUTION Velocity. At x = 2 m, y = 4 m, u = 2 ( 22 ) - 2 ( 42 ) + 4 = - 20 m>s v = 4 + 2(4) = 12 m>s The magnitude of the particle’s velocity is V = 2u2 + v2 = 2 ( - 20 m>s ) 2 + ( 12 m>s ) 2 = 23.3 m>s

Ans.

Acceleration. The x and y components of the particle’s acceleration, with w = 0 are ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 - 2y2 + y ) (4x) + (y + xy)( - 4y + 1) At x = 2 m, y = 4 m, ax = - 340 m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x2 - 2y2 + y ) (y) + (y + xy)(1 + x) At x = 2 m, y = 4 m, ay = - 44 m>s2 The magnitude of the particle’s acceleration is a = 2ax2 + ay2 = 2 ( - 340 m>s2 ) 2 +

( - 44 m>s2 ) 2 = 343 m>s2

Ans.

Ans: V = 23.3 m>s a = 343 m>s2 301

3–34. A fluid velocity components of u = ( 5y2 - x ) m>s and v = ( 4x2 ) m>s, where x and y are in meters. Determine the velocity and acceleration of particles passing through point (2 m, 1 m).

SOLUTION Since the velocity components are a function of position only the flow can be classified as steady nonuniform. At point x = 2 m and y = 1 m, u = 5 ( 12 ) - 2 = 3 m>s v = 4 ( 22 ) = 16 m>s The magnitude of the velocity is V = 2u2 + v2 = 2 ( 3 m>s ) 2 + ( 16 m>s ) 2 = 16.3 m>s

Ans.

Its direction is 16 m>s v b = 79.4° uv = tan-1a b = tan-1a u 3 m>s

Ans.

For two dimensional flow, the Eulerian description is a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axis ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 5y2 - x ) ( - 1) + 4x2(10y) = ( x - 5y2 ) + 40x2y ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 5y2 - x ) (8x) + 4x2(0) = 8x ( 5y2 - x ) At point x = 2 m and y = 1 m, ax =

32

- 5 ( 12 ) 4 + 40 ( 22 ) (1) = 157 m>s2

ay = 8(2) 3 5 ( 12 ) - 2 4 = 48 m>s2

The magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 157 m>s2 ) 2 + ( 48 m>s2 ) 2 = 164 m>s2

Ans.

Its direction is ua = tan-1a

ay ax

b = tan-1 °

48 m>s2 157 m>s2

Ans.

¢ = 17.0°

302

Ans: V = 16.3 m>s uv = 79.4° a a = 164 m>s2 ua = 17.0° a

3–35. A fluid has velocity components of u = ( 5y2 ) m>s and v = ( 4x - 1 ) m>s, where x and y are in meters. Determine the equation of the streamline passing through point 11 m, 1 m2 . Find the components of the acceleration of a particle located at this point and sketch the acceleration on the streamline.

SOLUTION Since the velocity components are independent of time but are a function of position, the flow can be classified as steady nonuniform. The slope of the streamline is dy v = ; dx u

4

dy 4x - 1 = dx 5y2

y

y(m)

3

x

L1 m

5y2dy =

L1 m

(4x - 1)dx

2 1

1 ( 6x2 - 3x + 2 ) where x is in m 5 For two dimensional flow, the Eulerian description is y3 =

ax = 30 m/s2 0

0V 0V 0V + u + v 0t 0x 0y

a =

ay = 20 m/s2 a = 36.1 m/s2

1

2

3 (a)

4

5

x(m)

Writing the scalar components of this equation along x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 5y2(0) + (4x - 1)(10y) = 40xy - 10y ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 5y2(4) + (4x - 1)(0) = 20y2 At point x = 1 m and y = 1 m, ax = 40(1)(1) - 10(1) = 30 m>s2 ay = 20 ( 12 ) = 20 m>s2 The magnitude of the acceleration is a = 2ax2 + ay2 = 2 ( 30 m>s2 ) 2 + ( 20 m>s2 ) 2 = 36.1 m>s2

Ans.

Its direction is

u = tan-1a

ay ax

b = tan-1 °

20 m>s2 30 m>s2

Ans.

¢ = 33.7°

The plot of the streamline and the acceleration on point (1 m, 1 m) is shown in Fig. a. x(m) y(m)

0 0.737

0.5 0.737

1 1

2 1.59

3 2.11

4 2.58

5 3.01

Ans: a = 36.1 m>s2 u = 33.7° a 303

*3–36. Air flowing through the center of the duct has been found to decrease in speed from VA = 8 m>s to VB = 2 m>s in a linear manner. Determine the velocity and acceleration of a particle moving horizontally through the duct as a function of its position x. Also, find the position of the particle as a function of time if x = 0 when t = 0.

B A

VA ! 8 m/s

VB ! 2 m/s

x 3m

SOLUTION Since the velocity is a function of position only, the flow can be classified as steady nonuniform. Since the velocity varies linearly with x, V = VA + a

VB - VA 2 - 8 bx = 8 + a bx = (8 - 2x) m>s LAB 3

Ans.

For one dimensional flow, the Eulerian description gives a =

0V 0V + V dt dx

= 0 + (8 - 2x)( -2) = 4(x - 4) m/s2 using the definition of velocity, dx = V = 8 - 2x; dt -

x

Ans.

t

dx = dt 8 2x L0 L0 x 1 ln(8 - 2x) ` = t 2 0

1 8 ln a b = t 2 8 - 2x ln a

8 b = 2t 8 - 2x

8 = e2 t 8 - 2x

x = 4 ( 1 - e -2 t ) m

Ans.

304

3–37. A fluid has velocity components of u = ( 8t 2 ) m>s and v = (7y + 3x) m>s, where x and y are in meters and t is in seconds. Determine the velocity and acceleration of a particle passing through point (1 m, 1 m) when t = 2 s.

SOLUTION Since the velocity components are functions of time and position the flow can be classified as unsteady nonuniform. When t = 2 s, x = 1 m and y = 1 m. u = 8 ( 22 ) = 32 m>s v = 7(1) + 3(1) = 10 m>s The magnitude of the velocity is V = 2u2 + v2 = 2 ( 32 m>s ) 2 + ( 10 m>s ) 2 = 33.5 m>s

Ans.

Its direction is

10 m>s v b = 17.4° uv = tan-1a b = tan-1a u 32 m>s

Ans.

uv

For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 16t + 8t 2(0) + (7y + 3x)(0) = (16t) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 8t 2 ) (3) + (7y + 3x)(7) =

3 24t 2

When t = 2 s, x = 1 m and y = 1 m.

ax = 16(2) = 32 m>s2

+ 7(7y + 3x) 4 m>s2

ay = 24 ( 22 ) + 737(1) + 3(1) 4 = 166 m>s2

The magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 32 m>s2 ) 2 + ( 166 m>s2 ) 2 = 169 m>s2

Ans.

Its direction is

ua = tan-1a

ay ax

b = tan-1a

166 m>s2 32 m>s2

b = 79.1°

Ans.

ua

Ans: V = 33.5 m>s uV = 17.4° a = 169 m>s2 ua = 79.1° a 305

3–38. A fluid has velocity components of u = (8x) ft>s and v = (8y) ft>s, where x and y are in feet. Determine the equation of the streamline and the acceleration of particles passing through point (2 ft, 1 ft). Also find the acceleration of a particle located at this point. Is the flow steady or unsteady?

SOLUTION Since the velocity components are the function of position but not the time, the flow is steady (Ans.) but nonuniform. Using the definition of the slope of the streamline,

y

dy 8y y = = dx 8x x

dy v = ; dx u

streamline v = (8y) ft/s

y

x dy dx = L1 ft y L2 ft x

ln y `

y 1 ft

= ln x `

u = (8x) ft/s x

y

x

x

2 ft

x ln y = ln 2 y =

V

!

(a)

1 x 2

Ans.

For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 8x(8) + 8y(0) = (64x) ft>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + (8x)(0) + 8y(8) = (64y) ft>s2 At x = 2 ft, y = 1 ft . Then ax = 64(2) = 128 ft>s2

ay = 64(1) = 64 ft>s2

The magnitude of the acceleration is a = 2ax2 + ay2 = 2 ( 128 ft>s2 ) 2 + ( 64 ft>s2 ) 2 = 143 ft>s2

Ans.

Its direction is

u = tan-1a

ay ax

b = tan-1a

64 ft>s2 128 ft>s2

b = 26.6°

Ans.

u

Ans: y = x>2, a = 143 ft>s2 u = 26.6° a 306

3–39. A fluid velocity components of u = ( 2y2 ) m>s and v = (8xy) m>s, where x and y are in meters. Determine the equation of the streamline passing through point (1 m, 2 m). Also, what is the acceleration of a particle at this point? Is the flow steady or unsteady?

SOLUTION Since the velocity components are the function of position, not of time, the flow can be classified as steady (Ans.) but nonuniform. Using the definition of the slope of the streamline,

y v = (8xy) m/s

dy 8xy 4x = = dx y xy2

dy v = ; dx u

y

L2 m y2 y 2

`

! u = (2y2) m/s

x

y dy =

2m

L1 m

streamline

4x dx

= 2x2 `

V

y

x 1m

x

y2 - 2 = 2x2 - 2 2

x (a)

y2 = 4x2 Ans.

y = 2x (Note that x = 1, y = 2 is not a solution to y = -2x.) For two dimensional flow, the Eulerian description gives. a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 2y2(0) + 8xy(4y) = ( 32xy2 ) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 2y2(8y) + (8xy)(8x) = ( 16y3 + 64x2y ) m>s2 At point x = 1 m and y = 2 m, ax = 32(1) ( 22 ) = 128 m>s2 ay =

3 16 ( 23 )

The magnitude of the acceleration is

+ 64 ( 12 ) (2) 4 = 256 m>s2

a = 2ax2 + ay2 = 2 ( 128 m>s2 ) 2 + ( 256 m>s2 ) 2 = 286 m>s2

Ans.

Its direction is

u = tan-1a

ay ax

b = tan-1a

256 m>s2 128 m>s2

b = 63.4°

Ans.

u

307

Ans: y = 2x a = 286 m>s2 u = 63.4° a

*3–40. The velocity of a flow field is defined by V = 5 4 yi + 2 xj 6 m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point (2 m, 1 m). Find the equation of the streamline passing through this point, and sketch the velocity and acceleration at the point on this streamline.

SOLUTION The flow is steady but nonuniform since the velocity components are a function of position, but not time. At point (2 m, 1 m) u = 4y = 4(1) = 4 m>s v = 2x = 2(2) = 4 m>s Thus, the magnitude of the velocity is V = 2u2 + v2 = 2 ( 4 m>s ) 2 + ( 4 m>s ) 2 = 5.66 m>s

Ans.

For two dimensional flow, the Eulerian description gives 0V 0V 0V + u + v 0t 0x 0y

a =

Writing the scalar components of this equation along the x and y axes 0u 0u 0u + u + v 0t 0x 0y

ax =

= 0 + 4y(0) + (2x)(4) = (8x) m>s2 0v 0v 0v + u + v 0t 0x 0y

ay =

= 0 + 4y(2) + 2x(0) = (8y) m>s2 At point (2 m, 1 m), ax = 8(2) = 16 m>s2 ay = 8(1) = 8 m>s2 The magnitude of the acceleration is a = 2ax2 + ay2

= 2 ( 16 m>s ) 2 + ( 8 m>s ) 2 = 17.9 m>s2

Using the definition of the slope of the streamline, dy v = ; dx u

dy 2x x = = dx 4y 2y

y

x

L1 m

2y dy =

y2 `

y 1m

=

y2 - 1 = y2 =

x dx L2 m x2 x ` 2 2m

x2 - 2 2 1 2 x - 1 2 308

Ans.

*3–40.

(continued)

The plot of this streamline is shown is Fig. a x(m) y(m)

22 0

2

3

4

5

6

±1

±1.87

±2.65

±3.39

±4.12

y(m)

y(m)

4

4 V = 5.66 m/s

3

3

v = 4 m/s

2

2 45º

1

0

ay = 8 m/s2

1

u = 4 m/s 1

2

3

4

5

a = 17.9 m/s2

ax = 16 m/s2 6

x(m)

0

–1

–1

–2

–2

–3

–3

–4

–4

(a)

309

1

2

3

4

5

6

x(m)

3–41. The velocity of a flow field is defined by V = 54xi + 2j6 m>s, where x is in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point (1 m, 2 m). Find the equation of the streamline passing through this point, and sketch these vectors on this streamline.

SOLUTION Since the velocity components are a function of position but not time, the flow can be classified as steady nonuniform. At point (1 m, 2 m), u = 4x = 4(1) = 4 m>s

y(m)

v = 2 m>s The magnitude of velocity is V = 2u2 + v2 = 2 ( 4 m>s ) 2 + ( 2 m>s ) 2 = 4.47 m>s

Ans.

For two dimensional flow, the Eulerian description gives 0V 0V 0V + u + v 0t 0x 0y

a =

2

u = 4 m/s

1

Writing the scalar components of this equation along the x and y axes ax =

V = 4.47 m/s

v = 2 m/s

3

0

1

2

3

4

x(m)

5

0u 0u 0u + u + v 0t 0x 0y

= 0 + 4x(4) + 2(0) = 16x ay =

m 0v 0v 0v + u + v 0t 0x 0y V

y(m)

3

= 0 + 4x(0) + 2(0) = 0 At point (1 m, 2 m),

a = 16 m/s2

2

ax = 16(1) = 16 m>s2

ay = 0

1

Thus, the magnitude of the acceleration is 2 a0 = ax 1= 16 m>s 2

Ans.

0

1

2

3

4

5

x(m)

Using the definition of the slope of the streamline,

dy 2 1 = = dx 4x 2x

dy v = ; dx u

y

x

L2 m

1 dx 2 L1 m x

y - 2 =

1 ln x 2

dy =

1 y = a ln x + 2b 2

Ans.

The plot of this streamline is shown in Fig. a x(m)

e -4

1

2

3

4

5

y(m)

0

2

2.35

2.55

2.69

2.80 Ans: V = 4.47 m>s, a = 16 m>s2 1 y = ln x + 2 2 310

3–42. The velocity of a flow field is defined by u = ( 2x2 - y2 ) m>s and v = ( - 4xy) m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 11 m, 1 m2 . Find the equation of the streamline passing through this point, and sketch the velocity and acceleration at the point on this streamline.

SOLUTION Since the velocity components are a function of position but not time, the flow can be classified as steady but nonuniform. At point (1 m, 1 m), u = 2x2 - y2 = 2 ( 12 ) - 12 = 1 m>s v = -4xy = -4(1)(1) = - 4 m>s The magnitude of the velocity is V = 2u2 + v2 = 2 ( 1 m>s ) 2 +

( - 4 m>s ) 2 = 4.12 m>s

Ans.

For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 - y2 ) (4x) + ( - 4xy)( - 2y) = 4x ( 2x2 - y2 ) + 8xy2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x2 - y2 ) ( - 4y) + ( - 4xy)( - 4x) = - 4y ( 2x2 - y2 ) + 16x2y At point (1 m, 1 m), ax = 4(1) 3 2 ( 12 ) - 12 4 + 8(1) ( 12 ) = 12 m>s2

ay = - 4(1) 3 2 ( 12 ) - 12 4 + 16 ( 12 ) (1) = 12 m>s2

The magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 12 m/s2 ) 2 + ( 12 m/s2 ) 2 = 17.0 m>s2

Ans.

Using the definition of the slope of the streamline, dy v = ; dx u

dy 4xy = - 2 dx 2x - y2

( 2x2 - y2 ) dy = - 4xydx 2x2dy + 4xydx - y2dy = 0 However, d ( 2x2y ) = 2 ( 2xydx + x2dy ) = 2x2dy + 4xydx. Then d ( 2x2y ) - y2dy = 0

311

3–42. (continued)

Integrating this equation, 2x2y -

y3 = C 3

with the condition y = 1 m when x = 1 m, 2 ( 12 ) (1) -

13 = C 3 y(m)

5 3

C =

3.0

Thus, 2x2y -

y3 3

=

2.5

5 3

2.0

6x2y - y3 = 5 x2 =

1.5

y3 + 5 6y

1.36 1.0

Taking the derivative of this equation with respect to y 6y ( 3y dx = 2x dy

2

v = 4 m/s

) - ( y + 5 ) (6) 2y - 5 = 6y2 ( 6y ) 2 3

3

V = 4.12 m/s

0.5

0

2y3 - 5 dx = dy 12xy2 Set

u = 1 m/s

Ans.

0.5

1.0

1.5

2.0

x(m)

0.960

dx = 0; dy y(m 2y3 - 5 = 0

y(m)

y = 1.3570 m

3.0

The corresponding x is

2.5

x2 = 1.3573 + 5

a = 17.0 m/s2

2 2.0 ay = 12 m/s

x = 0.960 0m y(m) 0.25 0.5 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 x(m) 1.83 1.31 1.10 1.00 0.963 0.965 0.993 1.04 1.10 1.17 1.25 1.33 = /s

1.5 1.0 ax = 12 m/s2 0.5

0

0.5

1.0

1.5

2.0

x(m)

Ans: V = 4.12 m>s a = 17.0 m>s2 y3 + 5 x2 = 6y 312

3–43. The velocity of a flow field is defined by u = (-y>4) m>s and v = (x>9) m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point (3 m, 2 m). Find the equation of the streamline passing through this point, and sketch the velocity and acceleration at the point on this streamline.

SOLUTION Since the velocity components are a function of position but not time, the flow can be classified as steady nonuniform. At point (3 m, 2m) u =

-y 2 = - = - 0.5 m>s 4 4

v =

x 3 = = 0.3333 m>s 9 9

The magnitude of the velocity is V = 2u2 + v2 = 2 ( - 0.5 m>s ) 2 + ( 0.333m>s ) 2 = 0.601 m>s

Ans.

For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y are + 2 ax = 0u + u 0u + v 0u 1S 0t 0x 0y = 0 + a

= a-

( + c ) ay =

-y x 1 b(0) + a ba - b 4 9 4

1 xb m>s2 36

0v 0v 0v + u + v 0t 0y 0y

= 0 + a = c-

-y 1 x ba b + a b(0) 4 9 9

1 y d m>s2 36

At point (3 m, 2 m),

ax = -

1 (3) = - 0.08333 m>s2 36

ay = -

1 (2) = - 0.05556 m>s2 36

The magnitude of the acceleration is a = 2ax2 + ay2

= 2 (- 0.08333 m>s2 ) 2 + (- 0.05556 m>s2 ) = 0.100 m>s2

313

Ans.

3–43. (continued)

Using the definition of slope of the streamline, x>9 dy 4x = = dx - y>4 9y

dy v = ; dx u y

9

x

L2 m

ydy = - 4

xdx L3 m

x 9y2 y ` = - ( 2x2 ) ` 2 2m 3m

9y2 - 18 = - 2x2 + 18 2 9y2 + 4x2 = 72 y2 x2 + = 1 72>4 72>9 y2 x2 + = 1 (4.24)2 (2.83)2

Ans.

This is an equation of an ellipse with center at (0, 0). The plot of this streamline is shown in Fig. a

y(m) 3 !" 2 m

y 3 !" 2 m

ax = 0.0833 m/s2

v = 0.333 m/s

V = 0.601 m/s

ay = 0.0556 m/s2 2 2 !" 2 m

2 u = 0.5 m/s 3

x(m)

3 a = 0.100 m/s2

2 !" 2 m

x

(a)

314

Ans: V = 0.601 m>s a = 0.100 m>s2 x2 > 14.242 2 + y2 > 12.832 2 = 1

*3–44. The velocity of gasoline, along the centerline of a tapered pipe, is given by u = (4tx) m>s, where t is in seconds and x is in meters. Determine the acceleration of a particle when t = 0.8 s if u = 0.8 m>s when t = 0.1 s.

SOLUTION The flow is unsteady nonuniform. For one dimensional flow, a = Here, u = (4 tx) m>s. Then

0u 0u + u 0t 0x

0u 0u = 4x and = 4t. Thus, 0t 0x

a = 4x + (4tx)(4t) = ( 4x + 16t 2x ) m>s2 Since u = 0.8 m>s when t = 0.1 s, 0.8 = 4(0.1) x

x = 2m

The position of the particle can be determined from dx = u = 4 tx; dt

x

t

dx t dt = 4 L2 m x L0.15 ln x `

x 2m

= 2t 2 `

t 0.15

x ln = 2t 2 - 0.02 2 x 2 e 2t - 0.02 = 2 2

x = 2e 2t

- 0.02

x = 2e 2(0.8 ) - 0.02 = 7.051 m 2

Thus, t = 0.8 s, a = 4(7.051) + 16 ( 0.82 ) (7.051) = 100.40 m>s2 = 100 m>s2

Ans.

315

3–45. The velocity field for a flow of water is defined by u = (2x) m>s, v = (6tx) m>s, and w = (3y) m>s, where t is in seconds and x, y, z are in meters. Determine the acceleration and the position of a particle when t = 0.5 s if this particle is at (1 m, 0, 0) when t = 0.

SOLUTION The flow is unsteady nonuniform. For three dimensional flow, a =

0V 0V 0V 0V + u + v + w 0t 0t 0t 0t

Thus, 0u 0u 0u 0u + u + v + w 0t 0x 0y 0t

ax =

= 0 + 2x(2) + (6tx)(0) + 3y(0) = (4x) m>s2 0v 0v 0v 0v + u + v + 0t 0x 0y 0t

ay =

= 6x + 2x(6t) + 6tx(0) + 3y(0) = (6x + 12tx) m>s2 0w 0w 0w 0w + u + v + w 0t 0x 0y 0z

az =

= 0 + 2x(0) + 6tx(3) + 3y(0) = (18tx) m>s2 The position of the particle can be determined from x

t

dx = 2 dt L1 m x L0

dx = u = 2x; dt

ln x = 2t x = ( e 2t ) m dy = v = 6tx = 6te 2t ; dt

L0

y

dy = 6

y = y =

L0

t

te 2tdt

t 3 ( 2te 2t - e 2t ) ` 2 0

3 ( 2te 2t - e 2t + 1 ) 2

dz 9 = w = 3y = 3 2te 2t - e 2t + 1 4 ; dt 2 L0

t

z

dz =

z =

9 ( 2te 2t - e 2t + 1 ) dt 2 L0

t 9 2t 1 1 c te - e 2t - e 2t + t d ` 2 2 2 0

z =

9 2t ( te - e 2t + t + 1 ) m 2

316

3–45. (continued)

When, t = 0.5 s, x = e 2(0.5) = 2.7183 m = 2.72 m y = Thus, z =

Ans.

3 3 2(0.5)e 2(0.5) - e 2(0.5) + 14 = 1.5 m 2

Ans.

9 3 0.5e 2(0.5) - e 2(0.5) + 0.5 + 14 = 0.6339 m = 0.634 m 2

Ans.

ax = 4(2.7183) = 10.87 m>s2

ay = 6(2.7183) + 12(0.5)(2.7183) = 32.62 m>s2 az = 18(0.5)(2.7183) = 24.46 m>s2 Then a = 510.9i + 32.6j + 24.5k6 m>s2

Ans.

317

Ans: x = 2.72 m y = 1.5 m z = 0.634 m a = 5 10.9i + 32.6j + 24.5k6 m>s2

3–46. A flow field has velocity components of u = -(4x + 6) m>s and v = (10y + 3) m>s where x and y are in meters. Determine the equation for the streamline that passes through point (1 m, 1 m), and find the acceleration of a particle at this point.

SOLUTION Since the velocity components are the function of position but not of time, the flow can be classified as steady but nonuniform. Using the definition of the slope of the streamline, dy 10y + 3 = dx - (4x + 6)

dy v = ; dx u y

x dy dx = 10y + 3 4x + 6 L1 m L1 m y x 1 1 ln(10y + 3) ` = - ln(4x + 6) ` 10 4 1m 1m

10y + 3 1 1 10 lna b = lna b 10 13 4 4x + 6 1

ln a

1

4 10y + 3 10 10 b = ln a b 13 4x + 6

a

1

1

4 10y + 3 10 10 b = a b 13 4x + 6

5

2 10y + 3 10 b = a 13 4x + 6

y = c

411

Ans.

- 0.3 d m (4x + 6)5>2 For two dimensional flow, the Eulerian description gives dV dV dV + u + v dt dx dy Writing the scalar components of this equation along the x and y axes, a =

ax =

du du du + u + v dt dx dy

= 0 + 3 -(4x + 6)( -4) 4 + (10y + 3)(0) = 34(4x + 6) 4 m>s2

ay =

dv dv dv + u + v dt dx dy

= 0 + 3 -(4x + 6)(0) 4 + (10y + 3)(10) At point (1m, 1m),

= 310(10y + 3) 4 m>s2 ax = 434(1) + 64 = 40 m>s2 S ay = 10310(1) + 34 = 130 m>s2 c

318

3–46. (continued)

The magnitude of acceleration is a = 2ax2 + ay2 = 2 ( 40m>s2 ) 2 + ( 130m>s2 ) 2 = 136 m>s2

Ans.

And its direction is

u = tan-1 a

ay ax

b = tan-1 a

130 m>s2 40 m>s2

b = 72.9°

Ans.

u

Ans: y =

411 14x + 62 5>2

a = 136 m>s2 u = 72.9° a 319

- 0.3

3–47. A velocity field for oil is defined by u = (100y) m>s, v = ( 0.03 t 2 ) m>s, where t is in seconds and y is in meters. Determine the acceleration and the position of a particle when t = 0.5 s. The particle is at the origin when t = 0.

SOLUTION Since the velocity components are a function of both position and time, the flow can be classified as unsteady nonuniform. Using the defination of velocity, dy = v = 0.03t 2; dt

L0

y

t

t 2 dt L0 y = ( 0.01t 3 ) m

dy = 0.03

When t = 0.5 s, y = 0.01 ( 0.53 ) = 0.00125 m = 1.25 mm dx = u = 100y = 100 ( 0.01t 3 ) = t 3; dt

L0

y

dx =

Ans. L0

t

t 3 dt

1 x = a t4b m 4

When t = 0.5 s, x =

1 ( 0.54 ) = 0.015625 m = 15.6 mm 4

Ans.

For a two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Write the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + (100y)(0) + ( 0.03t 2 ) (100) = ( 3t 2 ) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0.06t + (100y)(0) + 0.03t 2(0) = (0.06t) m>s2 When t = 0.5 s, ax = 3 ( 0.52 ) = 0.75 m>s2 S ay = 0.06(0.5) = 0.03 m>s2 c The magnitude of acceleration is a = 2ax2 + ay2 = 2 ( 0.75 m>s2 ) 2 + ( 0.03 m>s2 ) 2 = 0.751 m>s2

Ans.

And its direction is

u = tan-1 a

ay ax

b = tan-1a

0.03 m>s2 0.75 m>s2

b = 2.29°

Ans.

u

320

Ans: y = 1.25 mm x = 15.6 mm a = 0.751 m>s2 u = 2.29° a

*3–48. If u = ( 2x2 ) m>s and v = ( - y) m>s where x and y are in meters, determine the equation of the streamline that passes through point (2 m, 6 m), and find the acceleration of a particle at this point. Sketch the streamline for x 7 0, and find the equations that define the x and y components of acceleration of the particle as a function of time if x = 2 m and y = 6 m when t = 0.

SOLUTION Since the velocity components are a function of position but not time, the flow can be classified as steady but nonuniform. Using the definition of the slope of the streamline, dy -y = dx 2x2

dy v = ; dx u y

x dy 1 dx = 2 L2 m x2 L6 m y

ln y `

y 6m

ln ln

=

1 1 x a b` 2 x 2m

y 1 1 1 = a - b 6 2 x 2 y 2 - x = 6 4x y 2-x = e 1 4x 2 6 y = c 6e 1 4x 2 d m 2-x

Ans.

The plot of this streamline is shown in Fig. a. For two dimensional flow, the Eulerian description gives. a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 ) (4x) + ( - y)(0) = ( 8x3 ) m>s2 ay =

dv dv dv + u + v dt dx dy

= 0 + ( 2x2 ) (0) + ( - y)( - 1) = (y)m>s2 At point (2 m, 6 m), ax = 8 ( 23 ) = 64 m>s2 S ay = 6m>s2 c The magnitude of acceleration is a = 2ax2 + ay2 = 2 ( 64 m>s2 ) 2 + ( 6m>s2 ) 2 = 64.3m>s2

Ans.

And its direction is

u = tan-1a

ay ax

b = tan-1a

6 m>s2 64 m>s2

b = 5.36°

Ans.

u

321

3–48. (continued)

Using the definition of the velocity,

y(m)

dx = u; dt

dx = 2x2 dt x

7

t

dx = dt 2 L2 m 2x L0

6

1 1 x - a b` = t 2 x 2m

5 4

1 1 1 - a - b = t 2 x 2

3

x - 2 = t 4x

x = a

dy = v; dt

2

2 bm 1 - 4t

dy = -y dt

y

-

1

ln

y 6m

= t

6 = t y

6 = et y y = ( 6e -t ) m

Thus,

Then,

u = 2x2 = 2 a ax =

ay =

1

2

3

4 (a)

t dy = dt y L0

L6 m

- ln y `

0

2 2 8 b = c d m>s and v = - y = 1 - 4t ( 1 - 4t ) 2

du 64 = -16(1 - 4t)-3( - 4) = c d m>s2 dt (1 - 4t)3 dv = ( 6e -t ) m>s2 dt

( - 6e -t ) m>s

Ans.

Ans.

x(m)

0.5

1

2

3

4

5

6

y(m)

12.70

7.70

6.00

5.52

5.29

5.16

5.08

322

5

6

x(m)

3–49. Airflow through the duct is defined by the velocity field u = ( 2x2 + 8 ) m>s v = ( - 8x) m>s, where x is in meters. Determine the acceleration of a fluid particle at the origin (0, 0) and at point (1 m, 0). Also, sketch the streamlines that pass through these points.

y

x

x!1m

SOLUTION Since the velocity component are a function of position but not time, the flow can be classified as steady but nonuniform. For two dimensional flow, the Eulerian description gives 0V 0V 0V + u + v 0t 0x 0y

a =

Writing the scalar components of this equation along the x and y axes ax = = = ay =

0u 0u 0u + u + v 0t 0x 0y

30

+ ( 2x2 + 8 ) (4x) + ( -8x)(0) 4

3 4x ( 2x2

+ 8 ) 4 m>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x2 + 8 ) ( - 8) + ( - 8x)(0) = At point (0, 0),

3 - 8 ( 2x2

ax = 4(0) 3 2 ( 02 ) + 8 4 = 0

+ 8 ) 4 m>s2

ay = - 8 3 2 ( 02 ) + 8 4 = -64 m>s2 = 64 m>s2 T

Thus,

a = ay = 64 m>s2 T

Ans.

At point (1 m, 0), ax = 4(1) 3 2 ( 12 ) + 8 4 = 40 m>s2 S

ay = - 8 3 2 ( 12 ) + 8 4 = - 80 m>s2 = 80 m>s2 T

The magnitude of the acceleration is

a = 2ax2 + ay2 = 2 ( 40 m>s2 ) 2 + ( 80 m>s2 ) 2 = 89.4 m>s2

Ans.

And its direction is

u = tan-1 a

ay ax

b = tan-1 a

80 m>s2 40 m>s2

b = 63.4°

cu

Ans.

Using the definition of the slope of the streamline, dy v - 8x = = ; dx u 2x2 + 8

L

dy = - 8

x dx 2 L 2x + 8

y = - 2 ln ( 2x + 8 ) + C 2

323

3–49. (continued)

For the streamline passing through point (0, 0), 0 = - 2 ln3 2 ( 02 ) + 8 4 + C

Then y = c 2 ln a

C = 2 ln 8

8 bd m 2x2 + 8

Ans.

For the streamline passing through point (1 m, 0), 0 = - 2 ln3 2 ( 12 ) + 8 4 + C

C = 2 ln 10

y = c 2 ln a

For point (0, 0)

10 bd m 2x2 + 8

Ans.

x(m)

0

±1

±2

±3

±4

±5

y(m)

0

-0.446

-1.39

-2.36

-3.22

-3.96

For point (1 m, 0) x(m)

0

±1

±2

±3

±4

±5

y(m)

0.446

0

-0.940

-1.91

-2.77

-3.52

y(m)

1

–4

–3

–2

2

3

4

–1

y = 2 ln

5

x(m)

10 2x2 + 8

!

–2 –3 y = – 2 ln

!

–5

1 0

!

–1

8 2x2 + 8

!

–4

Ans: At point (0, 0), a = 64 m>s2w At point (1 m, 0), a = 89.4 m>s2, u = 63.4° c For the streamline passing through point (0, 0), 8 bd m 2x2 + 8 For the streamline passing through point (1 m, 0), y = c 2 ln a y = c 2 ln a 324

10 bd m 2x2 + 8

3–50. The velocity field for a fluid is defined by u = 3 y> ( x2 + y2 ) 4 m>s and v = 3 4x> ( x2 + y2 ) 4 m>s, where x and y are in meters. Determine the acceleration of a particle located at point (2 m, 0) and a particle located at point (4 m, 0). Sketch the equations that define the streamlines that pass through these points.

SOLUTION Since the velocity components are a function of position but not time, the flow can be classified as steady but nonuniform. For two dimensional flow, the Eulerian description gives a =

0V 0V 0V + u + v 0t 0x 0y

Write the scalar components of this equation along x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + a = £

ay =

y 2

2

x + y

4x3 - 6xy2

( x2 + y2 ) 3



( x2 + y2 ) (0) - y(2x) ( x2 + y2 ) (1) - y(2y) 4x § + a 2 b£ § 2 2 2 2 x + y (x + y ) ( x2 + y2 ) 2

§ m>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + a = £

y 2



2

x + y

4y3 - 36x2y

( x2 + y2 ) 3

( x2 + y2 ) (4) - 4x(2x) ( x2 + y2 ) (0) - 4x(2y) 4x § + a 2 b£ § 2 2 2 2 x + y (x + y ) ( x2 + y2 ) 2

§ m>s2

At point (2 m, 0) ax = ay =

4 ( 23 ) - 6(2) ( 02 )

( 22 + 02 ) 3

= 0.5 m>s2 S

4 ( 03 ) - 36 ( 22 ) (0)

( 22 + 02 ) 3

= 0

Thus a = ax = 0.5 m>s2 S

Ans.

325

3–50. (continued)

At point (4 m, 0) ax = ay =

4 ( 43 ) - 6(4)(0)

( 42 + 02 ) 3

= 0.0625 m>s2 S

4 ( 03 ) - 36 ( 42 ) (0)

( 42 + 02 ) 3

= 0

Thus a = ax = 0.0625 m>s2 S

Ans.

Using the definition of the slope of the streamline, 4x> ( x2 + y2 ) dy v 4x = = = ; 2 2 dx u y y> ( x + y )

ydy = 4 xdx L L y2 = 2x2 + C′ 2 y2 = 4x2 + C

For the streamline passing through point (2 m, 0), 02 = 4 ( 22 ) + C Then

C = -16

y2 = 4x2 - 16 y = { 24x2 - 16

Ans.

x Ú 2m

For the streamline passes through point (4 m, 0) 02 = 4 ( 42 ) + C

C = -64

Then y2 = 4x2 - 64 y = { 24x2 - 64

x Ú 4m

326

3–50. (continued)

For the streamline passing through point (2 m, 0) x(m) y(m)

2 0

3 ±4.47

4 ±6.93

5 ±9.17

6 ±11.31

7 ±13.42

8 ±15.49

9 ±17.55

For the streamline passing through point (4 m, 0) x(m) y(m)

4 0

5 ±6.00

6 ±8.94

7 ±11.49

8 ±13.86

9 ±16.12

y(m) 20 15

y = !4x2 – 16

10 5

0

1

2

3

4

5

6

7

8

9

x(m)

–5 y = !4x2 – 64 –10 –15 –20

Ans: For point (2 m, 0), a = 0.5 m>s2 y = { 24x2 - 16 For point (4 m, 0), a = 0.0625 m>s2 y = { 24x2 - 64 327

3–51. As the valve is closed, oil flows through the nozzle such that along the center streamline it has a velocity of V = 3 6 ( 1 + 0.4x2 ) (1 - 0.5t) 4 m>s, where x is in meters and t  is in seconds. Determine the acceleration of an oil particle at x = 0.25 m when t = 1 s.

A B

x 0.3 m

SOLUTION Here V only has an x component, so that V = u. Since V is a function of time at each x, the flow is unsteady. Since v = w = 0, we have ax = = =

0u 0u + u 0t 0x 0 0x

3 6( 1

3 6( 1

+ 0.4x2 ) (1 - 0.5t) 4 +

+ 0.4x2 ) (0 - 0.5) 4 +

3 6( 1

3 6( 1

Evaluating this expression at x = 0.25 m, t = 1 s, we get

+ 0.4x2 ) (1 - 0.5t) 4

0 3 6 ( 1 + 0.4x2 ) (1 - 0.5t) 4 0x

+ 0.4x2 )( 1 - 0.5t ) 4 3 6(0 + 0.4(2x))(1 - 0.5t) 4

as = - 3.075 m>s2 + 1.845 m>s2 = -1.23 m>s2

Ans.

Note that the local acceleration component ( - 3.075 m>s2 ) indicates a deceleration since the valve is being closed to decrease the flow. The convective acceleration ( 1.845 m>s2 ) is positive since the nozzle constricts as x increases. The net result causes the particle to decelerate at 1.23 m>s2.

Ans: - 1.23 m>s2 328

*3–52. As water flows steadily over the spillway, one of its particles follows a streamline that has a radius of curvature of 16 m. If its speed at point A is 5 m>s which is increasing at 3 m>s2, determine the magnitude of acceleration of the particle.

A

SOLUTION

n

The n - s coordinate system is established with origin at point A as shown in Fig. a. Here, the component of the particle’s acceleration along the s axis is

A

an streamline

as = 3 m>s2 Since the streamline does not rotate, the local acceleration along the n axis is zero, 0V so that a b = 0. Therefore, the component of the particle’s acceleration along 0t n the n axes is an = a

0V V2 b + 0t n R

= 0 +

( 5 m>s ) 2

16 m Thus, the magnitude of the particle’s acceleration is a = 2as2 + an2 = 2 ( 3 m>s2

= 1.5625 m>s2

) 2 + ( 1.5625 m>s2 ) 2

= 3.38 m>s2

Ans.

329

as S (a)

16 m

3–53. Water flows into the drainpipe such that it only has a radial velocity component V = ( - 3>r) m>s, where r is in meters. Determine the acceleration of a particle located at point r = 0.5 m, u = 20°. At s = 0, r = 1 m. s

r ! 0.5 m u

SOLUTION Fig. a is based on the initial condition when s = 0, r = rD. Thus, r = 1 - s. Then the radial component of velocity is V = -

3 3 = ab m>s r 1 - s

This is one dimensional steady flow since the velocity is along the straight radial line. The Eulerian description gives a =

0V 0V + V 0t 0s

= 0 + a= c

3 3 bcd 1 - s (1 - s)2

9 d m>s2 (1 - s)3

When 1 - s = r = 0.5 m, this equation gives a = a

9 b m>s2 = 72 m>s2 0.53

Ans.

The positive sign indicates that a is directed towards positive s. Note there is no normal component for motion along a straightline. s r

r0 = 1 m (a)

Ans: 72 m>s2 330

3–54. A particle located at a point within a fluid flow has velocity components of u = 4 m>s and v = - 3 m>s, and acceleration components of ax = 2 m>s2 and ay = 8 m>s2. Determine the magnitude of the streamline and normal components of acceleration of the particle.

SOLUTION

y

V = 2 ( 4 m>s ) 2 +

( - 3 m>s ) 2 = 5 m>s

a = 2 ( 2 m>s2 ) 2 + ( 8 m>s2 ) 2 = 8.246 m>s2

4 m/s

a = 2i + 8j u = tan-1

x

! V

3 = 36.870° 4

n

3 m/s

ûs

u s = cos 36.870°i - sin 36.870°j S

= 0. 8i - 0.6j

as = a # us = (2i + 8j) # (0.8i - 0.6j) as = - 3.20 m>s2 as = 3.20 m>s2

Ans.

a = 2as2 + an2

(8.246)2 = (3.20)2 + an2

an = 7.60 m>s2

Ans.

Ans: as = 3.20 m>s2 an = 7.60 m>s2 331

3–55. A particle moves along the circular streamline, such that it has a velocity of 3 m>s, which is increasing at 3 m>s2. Determine the acceleration of the particle, and show the acceleration on the streamline.

3 m/s

4m

SOLUTION The normal component of the acceleration is an =

V2 = r

( 3 m>s ) 2 4m

! = 36.9º

= 2.25 m>s2 4m

Thus, the magnitude of the acceleration is a = 2as2 + an2 = 2 ( 3 m>s2 ) 2 + ( 2.25 m>s2 ) 2 = 3.75 m>s2 2.25 m>s2 an b = tan-1a b = 36.9° as 3 m>s2

a = 3.75 m/s2

Ans. (a)

And its direction is u = tan-1a

as = 3 m/s2

an = 2.25 m/s2

Ans.

The plot of the acceleration on the streamline is shown in Fig. a.

Ans: a = 3.75 m>s2 u = 36.9° c 332

*3–56. The motion of a tornado can, in part, be described by a free vortex, V = k>r where k is a constant. Consider the steady motion at the radial distance r = 3 m, where V = 18 m>s. Determine the magnitude of the acceleration of a particle traveling on the streamline having a radius of r = 9 m.

r=9m

SOLUTION Using the condition at r = 3 m, V = 18 m>s . V =

k ; r

18 m>s =

k 3m

k = 54 m2 >s

Then V = a

54 b m>s r

54 b m>s = 6 m>s . Since the velocity is constant, the streamline 9 component of acceleration is

At r = 9 m, V = a

as = 0 The normal component of acceleration is an = a

Thus, the acceleration is

0V V2 b + = 0 + 0t n r

( 6 m>s ) 2 9m

= 4 m>s2

a = an = 4 m>s2

Ans.

333

3–57. Air flows around the front circular surface. If the steady-steam velocity is 4 m>s upstream from the surface, and the velocity along the surface is defined by V = (16 sin u) m>s, determine the magnitude of the streamline and normal components of acceleration of a particle located at u = 30°.

V 4 m/s u 0.5 m

SOLUTION The streamline component of acceleration can be determined from as = a

0V 0V b + V 0t s 0s

However, s = ru. Thus, 0 s = r 0 u = 0.5 0 u. Also, the flow is steady, a 0V 0V 0V = = 2 = 2(16 cos u) = 32 cos u. Then 0s 0.5 0 u 0u

0V b = 0 and 0t s

as = 0 + 16 sin u(32 cos u) = 512 sin u cos u = 256 sin 2u When u = 30°, as = 256 sin 2(30°) = 221.70 m>s2 = 222 m>s2

Ans.

V = (16 sin 30°) m>s = 8 m>s The normal component of acceleration can be determined from ( 8 m>s ) 2 0V V2 an = a b + = 0 + = 128 m>s2 0t n R 0.5 m

Ans.

Ans: as = 222 m>s2 an = 128 m>s2 334

3–58. Fluid particles have velocity components of u = (8y) m>s v = (6x) m>s, where x and y are in meters. Determine the magnitude of the streamline and the normal components of acceleration of a particle located at point (1 m, 2 m).

SOLUTION

us

6 m/s

At x = 1 m, y = 2 m u = 8(2) = 16 m>s

!

16 m/s

v = 6(1) = 6 m>s 6 = 20.56° 16 u s = cos 20.56° i + sin 20.56° j u = tan-1

= 0.9363i + 0.3511j Acceleration. With w = 0, we have ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 8y(0) + 6x(8) = 48x ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 8y(6) + 6x(0) = 48y At x = 1 m, and y = 2 m, ax = 48(1) = 48 m>s2 ay = 48(2) = 96 m>s2 Therefore, the acceleration is And its magnitude is

a = 5 48i + 96j 6 m>s2

a = 2ax2 + ay2 = 2 ( 48 m>s2 ) 2 + ( 96 m>s2 ) 2 = 107.33 m>s2

Since the direction of the s axis is defined by us, the component of the particle’s acceleration along the s axis can be determined from as = a # us = 348i + 96j 4 # 30.9363i + 0.3511j 4 = 78.65 m>s2 = 78.7 m>s2

Ans.

The normal component of the particle’s acceleration is an = 2a2 - as2 = 2 ( 107.33 m>s2 ) 2 - ( 78.65 m>s2 ) 2 = 73.0 m>s2

Ans.

Ans: as = 78.7 m>s2 an = 73.0 m>s2 335

3–59. Fluid particles have velocity components of u = (8y) m>s and v = (6x) m>s, where x and y are in meters. Determine the acceleration of a particle located at point (1 m, 1 m). Determine the equation of the streamline, passing through this point.

SOLUTION Since the velocity components are independent of time, but a function of position, the flow can be classified as steady nonuniform. For two dimensional flow, (w = 0), the Eulerian description is 0V 0V 0V a = + u + v 0t 0x 0y Writing the scalar components of this equation along the x and y axes, 0u 0u 0u + u + v 0t 0x 0y

ax =

= 0 + 8y(0) + 6x(8) = 48x 0v 0v 0v + u + v 0t 0x 0y

ay =

= 0 + 8y(6) + 6x(0) = 48y At point x = 1 m and y = 1 m ax = 48(1) = 48 m>s2 ay = 48(1) = 48 m>s2 The magnitude of the acceleration is a = 2ax2 + ay2 = 2 ( 48 m>s2 ) 2 + ( 48 m>s2 ) 2 = 67.9 m>s2

Ans.

Its direction is

u = tan-1a

ay ax

b = tan-1a

48 m>s2 48 m>s2

a

b = 45°

Ans.

The slope of the streamline is dy v = ; dx u

dy 6x = dx 8y

y

x

L1 m

8ydy =

4y2 `

2

(1)

y 1m 2

6xdx L1 m

= 3x2 `

x 1m

Ans.

4y - 3x = 1

Ans: a = 67.9 m>s2 u = 45° a 4y2 - 3x2 = 1 336

*3–60. A fluid has velocity components of u = ( 2y2 ) m>s and v = (8xy) m>s, where x and y are in meters. Determine the magnitude of the streamline and normal components of acceleration of a particle located at point (1 m, 2 m).

SOLUTION dy 8xy 4x = = 2 dx y 2y

dy r = ; dx u y

x

L2 m

y dy =

4x dx L1 m

x y2 y ` = 2x2 ` 2 2m 1m

y2 - 2 = 2x2 - 2 2 y2 = 4x2 y = 2x (Note that x = 1, y = 2 is not a solution y = -2x.) Two equation from streamline is y = 2x (A straight line). Thus, R S ∞ and since the flow is steady, V2 = 0 R 0u 0u 0u + u + v ax = 0t 0x 0y

Ans.

an =

= 0 + 2y2(0) + (8xy)(4y) = ( 32xy2 ) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 2y2(8y) + (8xy)(8x) = ( 16y3 + 64x2y ) m>s2 At (1 m, 2 m), ax = 32(1) ( 22 ) = 128 m>s2 ay =

3 16 ( 23 )

+ 64 ( 12 )( 2 ) 4 = 256 m>s2

a = as = 2ax2 + ay2 = 2 ( 128 m>s2 ) 2 + ( 256 m>s2 )

as = 286 m>s2

337

Ans.

3–61. A fluid has velocity components of u = ( 2y2 ) m>s and v = (8xy) m>s, where x and y are in meters. Determine the magnitude of the streamline and normal components of the acceleration of a particle located at point (1 m, 1 m). Find the equation of the streamline passing through this point, and sketch the streamline and normal components of acceleration at this point.

SOLUTION x(m) y(m)

23>2 0

1.0

2.0

3.0

4.0

5.0

1.0

3.61

5.74

7.81

9.85

Since the velocity component is independent of time and is a function of position, the flow can be classified as steady nonuniform. The slope of the streamline is dy 8xy 4x = = dx y 2y2

dy v = ; dx u y

x

L1 m

y dy =

y2 2

`

y 1m

4x dx L1 m

= 2x2 `

x

y(m)

1m

4x2 - y2 = 3 where x and y are in m

Ans. 10

For two dimensional flow (w = 0), the Eulerian description is a =

0V 0V 0V + u + v 0t 0x 0y

9

Writing the scalar components of this equation along the x and y axes, ax =

0u 0u 0u + u + v 0t 0x 0y

8 7

= 0 + ( 2y2 ) (0) + (8xy) (4y)

6

= ( 32xy2 ) m>s2

5

0v 0v 0v ay = + u + v 0t 0x 0y

as = 85.4 m/s2

4

= 0 + ( 2y2 ) (8y) + (8xy) (8x)

3

= ( 16y3 + 64x2y ) m>s2

a = 18.2 m/s2

At point x = 1 m and y = 1 m

2

!a = 68.2°

ax = 32(1) ( 12 ) = 32 m/s2 ay = 16 ( 13 ) + 64 ( 12 ) (1) = 80 m>s2

1 an = 11.6 m/s2

0

Thus, a = 5 32i + 80j 6 m>s2

338

1

2

3 (a)

4

5

x(m)

3–61. (continued)

The magnitude of the acceleration is a = 2ax2 + ay2 = 2 ( 32 m>s2 ) 2 + ( 80 m>s2 )

= 86.16 m>s2 = 18.2 m>s2

and its direction is

At point (1 m, 1 m),

ay 80 m>s2 ua = tan-1a a b = tan-1a b = 68.2° x 32 m>s2 tan u =

4(1) dy = = 4; u = 75.96° ` 1 dx x = 1 m y=1 m

Thus, the unit vector along the streamline is

u = cos 75.96°i + sin 75.96°j = 0.2425i + 0.9701j Thus, the streamline component of the acceleration is as = a # us = (32i + 80j) # (0.2425i + 0.9701j) = 85.37 m>s2 = 85.4 m>s2

Ans.

Then an = 2a2 - as2 = 2 ( 86.16 m>s2 ) 2 - ( 85.37 m>s2 ) 2 = 11.64 m>s2 = 11.6 m>s2

Ans.

Ans: 4x2 - y2 = 3 as = 85.4 m>s2 an = 11.6 m>s2 339

4–1. Water flows steadily through the pipes with the average velocities shown. Outline the control volume that contains the water in the pipe system. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

C 1.5 m/s

2 m/s

B

4 m/s A

SOLUTION Since the flow is steady, no local change occurs. However, the water flows in and out of the control volume through the open (inlet and outlet) control surfaces. Thus, convective changes take place.

AC

VB = 2 m s

AB

VC = 1.5m s

Outlet control surfaces

Inlet control surface

AA

VA = 4m s

340

4–2. Water is drawn steadily through the pump. The average velocities are indicated. Select a control volume that contains the water in the pump and extends slightly past it. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

8 m/s 12 m/s

SOLUTION Since the flow is steady, there is no local change. However, the water flows in and out of the control volume through the open (inlet and outlet) control surfaces. Thus, convective changes take place.

Inlet control surface

Vin = 8 m s

Ain

Outlet control surface

Aout

Vout = 12 m s

341

4–3. The average velocities of water flowing steadily through the nozzle are indicated. If the nozzle is glued onto the end of the hose, outline the control volume to be the entire nozzle and the water inside it. Also, select another control volume to be just the water inside the nozzle. In each case, indicate the open control surfaces, and show the positive direction of their areas. Specify the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

2.5 m/s A

SOLUTION Since the flow is steady, no local change occurs. However, the water flows in and out of the control volume through the opened (inlet and outlet) control surfaces. Thus, convective changes take place.

Outlet control surface

Outlet control surface

VA = 2.5 m s

VA = 2.5 m s AA

AB

VB = 8 m s

AA

Inlet control surface

AB

VB = 8 m s

Inlet control surface

342

B

8 m/s

*4–4. Air flows through the tapered duct, and during this time heat is being added that changes the density of the air within the duct. The average velocities are indicated. Select a control volume that contains the air in the duct. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume the air is incompressible.

2 m/s

SOLUTION Since the density of the air within the control volume changes with time due to the increased heating, local changes occur. Also, air flows/in and out of the control volume through the opened (inlet and outlet) control surfaces. This causes a convective change to take place.

Inlet control surface

Vin = 2 m s Ain

Outlet control surface

Vout = 7 m s

Aout

Note: If the heating is constant, then no local changes will occur.

343

7 m/s

4–5. The tank is being filled with water at A at a rate faster than it is emptied at B; the average velocities are shown. Select a control volume that includes only the water in the tank. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

A 2 m/s

SOLUTION Since the volume of the control volume changes with time, local changes occur. Also, the water flows in and out of the control volume through the open (inlet and outlet) control surfaces. This causes convective changes to take place.

Inlet control surface

AA

VA = 6 m s

VB = 2 m s AB Outlet control surface

344

B

6 m/s

4–6. The toy balloon is filled with air and is released. At the instant shown, the air is escaping at an average velocity of 4 m>s, measured relative to the balloon, while the balloon is accelerating. For an analysis, why is it best to consider the control volume to be moving? Select this control volume so that it contains the air in the balloon. Indicate the open control surface, and show the positive direction of its area. Also, indicate the direction of the velocity through this surface. Identify the local and convective changes that occur. Assume the air to be incompressible.

2 m/s

SOLUTION The control volume is considered moving so that the Reynolds transport theorem can be applied using relative velocities. Since the volume of the control volume (ballon) changes with time, local changes occur. Also, air flows/out from the control volume through the open (outlet) control surface. This causes convective changes to take place.

Outlet control surface

(Va cv)out = 4 m s Aout

345

4–7. Compressed air is being released from the tank, and at the instant shown it has a velocity of 3 m>s. Select a control volume that contains the air in the tank. Indicate the open control surface, and show the positive direction of its area. Also, indicate the direction of the velocity through this surface. Identify the local and convective changes that occur. Assume the air to be compressible.

3 m/s

SOLUTION Since the density of the air changes with time, local changes occur. Also, the air flows out of the control volume through the open (outlet) control surface. This causes convective changes.

Vout = 3 m s

Outlet control surface

Aout

346

*4–8. The blade on the turbine is moving to the left at 6 m>s. Water is ejected from the nozzle at A at an average velocity of 2 m>s. For the analysis, why is it best to consider the control volume as moving? Outline this moving control volume that contains the water on the blade. Indicate the open control surfaces, and show the positive direction of their areas through which flow occurs. Also, indicate the magnitudes of the relative velocities and their directions through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

2 m/s A

SOLUTION The control volume is considered moving with the blade so that the flow can be considered steady as measured relative to the control volume. No local changes occurs. Also, the water flows in and out through the open (inlet and outlet) control surfaces. This causes convective changes.

(Vw cs)out = 8 m s Outlet control surface

Aout

(Vw cs)in = 8 m s Ain

Inlet control surface

347

6 m/s

4–9. The jet engine is moving forward with a constant speed of 800 km>h. Fuel from a tank enters the engine and is mixed with the intake air, burned, and exhausted with an average relative velocity of 1200 km>h. Outline the control volume as the jet engine and the air and fuel within it. For an analysis, why is it best to consider this control volume to be moving? Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the magnitudes of the relative velocities and their directions through these surfaces. Identify the local and convective changes that occur. Assume the fuel is incompressible and the air is compressible.

800 km/h

SOLUTION The control volume is considered moving, so that the Reynolds transport theorem can be applied using relative velocities since the masses are conserved within the control volume the flow is steady, no local changes occur. Also, mass flows in and out from the control volume through the open (inlet and outlet) control surfaces. This causes convective changes to take place.

(Vf cs)in Inlet control surface Ain

(Va cs)in = 800 km h

Outlet control surface

Aout

Ain

(Ve cs)out = 1200 km h

Inlet control surface

348

4–10. The balloon is rising at a constant velocity of 3 m>s. Hot air enters from a burner and flows into the balloon at A at an average velocity of 1 m>s, measured relative to the balloon. For an analysis, why is it best to consider the control volume as moving? Outline this moving control volume that contains the air in the balloon. Indicate the open control surface, and show the positive direction of its area. Also, indicate the magnitude of the velocity and its direction through this surface. Identify the local and convective changes that occur. Assume the air to be incompressible.

3 m/s

A

SOLUTION The control volume is considered moving so that the Reynolds transport theorem can be applied using relative velocities. Since the volume of the control volume (ballon) changes with time, local changes occur. Also, the air flows in the control volume through the opened (inlet) control surface. This causes convective changes to take place.

Inlet control surface Ain

(Va cs)in = 1 m s

349

4–11. The hemispherical bowl is suspended in the air by the water stream that flows into and then out of the bowl at the average velocities indicated. Outline a control volume that contains the bowl and the water entering and leaving it. Indicate the open control surfaces, and show the positive direction of their areas. Also, indicate the direction of the velocities through these surfaces. Identify the local and convective changes that occur. Assume water to be incompressible.

3 m/s 3 m/s

SOLUTION Since the flow is steady, there is no local change. However, the water flows in and out of the control volume through the open (inlet and outlet) control surface, thus convective changes take place.

Aout

Outlet control surface Vout = 3 m s

Ain

Vout = 3 m s Inlet control surface

Vin = 3 m s

350

*4–12. Water flows along a rectangular channel having a width of 0.75 m. If the average velocity is 2 m>s, determine the volumetric discharge. 0.5 m

SOLUTION The vertical cross section of the flow of area A = (0.75 m)(0.5 m) = 0.375 m2 is shown shaded in Fig. a. The component of velocity perpendicular to this cross section is V# = (2 m>s) cos 20° = 1.8794 m>s. Thus, Q = V#A

= (1.8794 m>s) ( 0.375 m2 ) = 0.705 m3 >s

Ans.

0.5 m 20˚ 20˚ V=2ms

0.75 m (a)

351

20!

4–13. Water flows along the triangular trough with an average velocity of 5 m>s. Determine the volumetric discharge and the mass flow if the vertical depth of the water is 0.3 m.

60!

30!

0.3 m

SOLUTION From the geometry shown in Fig. a, the top width w and height h of the cross section perpendicular to the flow are w = 2[0.3 m(tan 30°)] = 0.3464 m and h = (0.3 m)(cos 30°) = 0.2598 m. Thus, the cross-sectional area of this cross section 1 is A = (0.3464 m)(0.2598 m) = 0.045 m2. Thus, 2 Q = V#A = ( 5 m>s )( 0.045 m2 ) = 0.225 m3 >s

Ans.

# m = rQ = ( 1000 kg>m3 )( 0.225 m3 >s )

Ans.

= 225 kg>s

w h 30˚ 30˚

0.3 m

30˚ 30˚

(a)

Ans: Q = 0.225 m3 >s # m = 225 kg>s 352

4–14. Air enters the turbine of a jet engine at a rate of 40  kg>s. If it is discharged with an absolute pressure of 750  kPa and temperature of 120°C, determine its average velocity at the exit. The exit has a diameter of 0.3 m.

0.3 m

SOLUTION

From Appendix A, R = 286.9 J>(kg # K) for air. p = rRT 750 ( 103 ) N>m2 = r ( 286.9 J>(kg # K) ) (120° + 273) r = 6.6518 kg>m3 # m = rVA 40 kg>s = ( 6.6518 kg>m3 ) V(p)(0.15 m)2 Ans.

V = 85.1 m>s

Ans: 8.51 m>s 353

4–15. Determine the mass flow of CO2 gas in a 4-in.diameter duct if it has an average velocity of 20 ft>s. The gas has a temperature of 70°F, and the pressure is 6 psi.

4 in.

SOLUTION ft # lb for CO 2 . Here, the absolute pressure slug # R lb 12 in. 2 + pg = 14.7 psi + 6 psi = a20.7 2 ba b = 2980.8 lb>ft 2 and 1 ft in

From Appendix A, R = 1130 is

p = patm

T = 70°F + 460 = 530°R

p = rRT (6 psi + 14.7 psi)a

12 in. 2 b = r ( 1130 ft # lb>slug # R ) (70° + 460) 1 ft r = 0.004977 slug>ft 3 # m = rVA

= ( 0.004977 slug>ft 3 )( 20 ft>s ) (p)a = 8.69 ( 10-3 ) slug>s

2 2 ft b 12

Ans.

Ans: 8.69 ( 10 - 3 ) slug>s 354

*4–16. Carbon dioxide gas flows through the 4-in.diameter duct. If it has an average velocity of 10 ft>s and the gage pressure is maintained at 8 psi, plot the variation of mass flow (vertical axis) versus temperature for the temperature range 0°F … T … 100°F. Give values for increments of ∆T = 20°.

4 in.

SOLUTION From Appendix A, R = 1130 ft # lb>(slug # R) for CO 2. Here the absolute pressure lb 12 in. 2 is p = patm + pg = 14.7 psi + 8 psi = a22.7 2 ba b = 3268.8 lb>ft 2 and 1 ft in T = TF + 460. p = rRT

3268.8 lb>ft 2 = r ( 1130 ft # lb>(slug # R) ) (TF + 460) r = a

The mass flow is

2.8927 b slug>ft 3 TF + 460

# m = rVA 2 # 2.8927 2 m = ca b slug>ft 3 d 1 10 ft>s 2 c p a ft b d TF + 460 12

# 2.5244 m = a b slug>s where TF is in °F. TF + 460

# The plot of m vs TF is shown in Fig. a. T(°F) # m 1 10 - 3 2 slug>s

0

 20

40

60

80

100

5.49

5.26

5.05

4.85

4.67

4.51

m(10–3) slug s 6

5

4

3

2

1

0

T(˚F) 20

40

60

80

100

(a)

355

4–17. Water flowing at a constant rate fills the tank to a height of h = 3 m in 5 minutes. If the tank has a width of 1.5 m, determine the average velocity of the flow from the 0.2-m-diameter pipe at A.

A

h

2m

SOLUTION The volume of water in the tank when t = 5(60) s = 300 s is V = (3 m)(2 m)(1.5 m) = 9 m3 Thus, the discharge through the pipe at A is V 9 m3 = = 0.03 m3 >s t 300 s Then, the average velocity of the water flow through A is Q =

Q = VA 0.03 m3 >s = V(p)(0.1 m)2

Ans.

V = 0.955 m>s

Ans: 0.955 m>s 356

4–18. Water flows through the pipe at a constant average velocity of 0.5 m > s. Determine the relation between the time needed to fill the tank to a depth of h = 3 m and the diameter D of the pipe at A. The tank has a width of 1.5 m. Plot the time in minutes (vertical axis) versus the diameter 0.05 m … D … 0.25 m. Give values for increments of ∆D = 0.05 m.

A

h

2m

SOLUTION The volume of water filled to h = 3 m in time t is V = (2 m)(1.5 m)(3 m) = 9 m3 Thus, the discharge through the pipe is Q = Applying

9 V = a b m3 >s t t 9 p = ( 0.5 m>s ) c ( D2 ) d t 4

Q = VA; t = ca t = a

22.9183 1 min b sda b 2 60 s D

0.382 b min where d is in m D2

Ans.

The plot of t vs D is shown in Fig. a D(m) t(min)

0.05

0.10

0.15

0.20

0.25

153

38.2

17.0

9.55

6.11

t(min) 160 140 120 100 80 60 40 20

0

D(m) 0.05

0.15

0.10

0.20

0.25

(a)

357

Ans: t = a

0.382 b min, where d is in m D2

4–19. Determine the mass flow of air in the duct if it has an average velocity of 15 m>s. The air has a temperature of 30°C, and the (gage) pressure is 50 kPa. 0.3 m 0.2 m

SOLUTION

From Appendix A, R = 286.9 J>kg # K for air. p = rRT (50 + 101) ( 103 ) N>m2 = r(286.9 J>(kg # K)(30° + 273) r = 1.737 kg>m3 # m = rVA = ( 1.737 kg>m3 ) (15 m>s)(0.3 m)(0.2 m) Ans.

= 1.56 kg>s

Ans: 1.56 kg>s 358

*4–20. Air flows through the duct at an average velocity of 20 m>s. If the temperature is maintained at 20°C, plot the variation of the mass flow (vertical axis) versus the (gage) pressure for the range of 0 … p … 100 kPa. Give values for increments of ∆p = 20 kPa. The atmospheric pressure is 101.3 kPa.

0.3 m 0.2 m

SOLUTION

From Appendix A, R = 286.9 J>(kg # K) for air. Here, the absolute pressure is p = pg + patm = ( pg + 101.3 ) kPa p = rRT

( pg + 101.3 )( 103 ) N>m2 = r ( 286.9 J>(kg # K) ) (20 + 273) K r =

3 0.01190 ( pg

+ 101.3 )4 kg>m3

The mass flow is # m = rVA # m = 3 0.01190 ( pg + 101.3 ) kg>m3 4 ( 20 m>s ) 3 0.3 m(0.2 m) 4 # m = [0.01428 ( pg + 101.3 ) ]kg>s where pg is in kPa # The plot of m vs pg is shown in Fig. a. Pg(kPa) # m (kg>s)

0

20

40

60

80

100

1.45

1.73

2.02

2.30

2.59

2.87

m(Kg s) 

3

2

1

0

20

40

60

80

100

Pg(kPa)

(a)

359

4–21. A fluid flowing between two plates has a velocity profile that is assumed to be linear as shown. Determine the average velocity and volumetric discharge in terms of Umax. The plates have a width of w.

b –– 2

Umax

b –– 2

SOLUTION

y

The velocity profile in Fig. a can be expressed as U max - 0 v - 0 = ; h h y 0 2 2

V

v = U max a1 -

dy

2 yb h

h 2 V

The differential rectangular element of the thickness dy on the cross section will be considered. Thus, dA = wdy. Q =

LA

= 2

(a)

v = dA

L0

h 2

c U max a1 -

= 2wU max

L0

Umax

h 2

a1 -

2 yb d (wdy) h 2 ybdy h h

y2 2 = 2wU max ay - b ` h 0 =

wU max h 2

Ans.

Also, Q = =

LA

v # dA = volume under velocity diagram

wU max h 1 (h)(w) ( U max ) = 2 2

Ans.

wU max h U max Q = = A 2(w)(h) 2

Ans.

Therefore, V =

Ans:

wU maxh 2 Umax V = 2

Q =

360

4–22. The velocity profile of a liquid flowing through the pipe is approximated by the truncated conical distribution as shown. Determine the average velocity of the flow. Hint: The volume of a cone is V = 13 pr 2h.

0.3 ft 0.6 ft

5 ft/s

0.3 ft

SOLUTION Q =

V1

LA

V # dA =

V2

5 ft s

the volume enclosed by the velocity profile

Here, this volume is equal to the volume of cone (1) minus that of cone (2) in Fig a. From the geometry of cone (1), V2 0.3 ft = ; 5 ft>s + V2 0.6 ft

0.6 ft

0.3 ft 1

V2 = 5 ft>s V2

Then, V1 = 5 ft>s + V2 = 10 ft>s

0.3 ft

1 The volume of the cone can be computed by V = pr 2h. Then, 3 Q =

2

1 1 p(0.6 ft)2 ( 10 ft>s ) - p (0.3 ft)2 ( 5 ft>s ) 3 3

(a)

= 1.05p ft 3 >s

The average velocity is V =

Q 1.05p = = 2.92 ft>s A p(0.6 ft)2

Ans.

Ans: 2.92 ft>s 361

4–23. The velocity profile of a liquid flowing through the pipe is approximated by the truncated conical distribution as shown. Determine the mass flow if the liquid has a specific weight of g = 54.7 lb>ft 3. Hint: The volume of a cone is V = 13 pr 2h.

0.3 ft 0.6 ft

SOLUTION Q =

5 ft/s

0.3 ft

V1

LA

V # dA =

5 ft s

the volume enclosed by the velocity profile

Here, this volume is equal to the volume of cone (1) minus that of code (2) in Fig. a. From the geometry of cone (1), V2 0.3 ft = ; 5 ft>s + V2 0.6 ft

0.6 ft

0.3 ft 1

V2 = 5 ft>s

Then,

V2

V1 = 5 ft>s + V2 = 10 ft>s The volume of the cone can be computed by V = Q =

V2

0.3 ft

1 2 pr h. Then, 3

2

1 1 p(0.6 ft)2 ( 10 ft>s ) - p(0.3 ft)2 ( 5 ft>s ) 3 3

(a)

= 1.05p ft 3 >s

Then the mass flow is

54.7 lb>ft 3 # m = rQ = a b ( 1.05p ft 3 >s ) = 5.60 slug>s 32.2 ft>s2

Ans.

Ans: 5.60 slug>s 362

*4–24. Determine the average velocity V of a very viscous fluid that enters the 8-ft-wide rectangular open channel and eventually forms the velocity profile that is approximated by u = 0.8 1 1.25y + 0.25y2 2 ft>s, where y is in feet.

y

V

Here, dA = bdy. Thus, the discharge is LA

vdA =

L0

6 ft

0.8 ( 1.25y + 0.25y2 ) (8dy)

= 0.8(8) ( 0.625y2 + 0.08333y3 ) `

6 ft 0

= 32.4(8) The average velocity is V =

u y

SOLUTION

Q =

6 ft

32.4(8) Q = = 5.40 ft>s A 6(8)

Ans.

363

4–25. Determine the mass flow of a very viscous fluid that enters the 3-ft-wide rectangular open channel and eventually forms the velocity profile that is approximated by  u = 0.8 1 1.25y + 0.25y2 2 ft>s, where y is in ft. Take g = 40 lb>ft 3.

y

V

6 ft

u y

SOLUTION Here, dA = bdy = 3dy. Thus, the mass flow is # m =

LA

rvdA = a

40 lb>ft 3 2

32.2 ft>s

b

L0

6 ft

0.8 ( 1.25y + 0.25y2 ) (3dy)

= 2.9814 ( 0.625y2 + 0.08333y3 ) `

6 ft 0

Ans.

= 120.75 slug>s = 121 slug>s

Ans: 121 slug>s 364

4–26. The velocity field for a flow is defined by u = (6x) m>s and v = 1 4y2 2 m>s, where x and y are in meters. Determine the discharge through the surface at A and at B.

z

300 mm 100 mm

200 mm

y

B A

v u

V

x

SOLUTION The velocity perpendicular to and passing through surface A, where x = 0.2 m, is VA = u = (6x) m>s = [6(0.2)] m>s = 1.2 m>s The velocity perpendicular to and passing through surface B, where y = 0.3 m, is VB = v = ( 4y2 ) m>s = c 4 ( 0.32 ) d m>s = 0.36 m>s

The areas of surfaces A and B are AA = (0.3 m)(0.1 m) = 0.03 m2 and AB = (0.2 m) (0.1 m) = 0.02 m2. Thus, QA = VAAA = ( 1.2 m>s) ( 0.03 m2 ) = 0.036 m3 >s

QB = VBAB = ( 0.36 m>s )( 0.02 m

2

) = 0.0072 m >s 3

365

Ans. Ans.

Ans: QA = 0.036 m3 >s QB = 0.0072 m3 >s

4–27. The velocity profile in a channel carrying a very viscous liquid is approximated by u = 3 1 e 0.2y - 1 2 m>s, where y is in meters. If the channel is 1 m wide, determine the volumetric discharge from the channel.

y u ! 3 (e 0.2y " 1) m/s

0.4 m

SOLUTION The differential rectangular element of thickness dy on the cross section will be considered, Fig. a. Since the channel has a constant width of 1 m, then the area of this element is dA = (1 m)dy = dy. Q = =

LA L0

= 3a

V # dA

0.4 m

3 3 ( e 0.2y

- 1 ) 4 (dy)

0.4 m e 0.2y - yb ` 0.2 0

= 0.0493 m3 >s

Ans.

366

Ans: 0.0493 m3 >s

*4–28. The velocity profile in a channel carrying a very viscous liquid is approximated by u = 3 1 e 0.2y - 1 2 m>s, where y is in meters. Determine the average velocity of the flow. The channel has a width of 1 m.

y u ! 3 (e 0.2y " 1) m/s

0.4 m

SOLUTION

y

V = 3(e0.2y–1) m s

The differential rectangular element of thickness dy on the cross section will be considered, Fig. a. Since the channel has a constant width of 1 m, then the area of this element is dA = (1 m)dy = dy. Q = =

LA L0

= 3a

The average velocity is V =

V # dA

0.4 m

3 3 ( e 0.2y

0.4 m

- 1 ) 4 (dy)

dy V

0.4 m e 0.2y - yb ` 0.2 0

(a)

= 0.04931 m3 >s

0.04931 m3 >s Q = = 0.123 m>s = 12.3 mm>s A 0.4 m(1 m)

367

Ans.

4–29. The velocity profile for a fluid within the circular pipe  for fully developed turbulent flow is modeled using Prandtl’s one-seventh power law u = U 1 y>R 2 1> 7. Determine the average velocity for this case.

y

U R

y

SOLUTION

dr

L

R

udA = = = = =

u =

L

y U a b 2prdr, R L0 0

R LR

2pU 1 7

R 7 L0

2pU

r = R - y

1

y7(R - y)( -dy) R

1 7

R

r

8 7

aRy - y bdy

1

8 7 15 R 2pU 7 c Ry7 y7 d 1 15 0 R7 8

y

49 2 pR U 60

udA

L

1 7

= dA

a

49 bpR2U 60 2

pR

=

49 U 60

Ans.

Ans: 49 U 60 368

4–30. The velocity profile for a fluid within the circular pipe for fully developed turbulent flow is modeled using Prandtl’s seventh-power law u = U 1 y>R 2 1> 7. Determine the mass flow of the fluid if it has a density r.

y

U R

y

SOLUTION Referring to Fig. a, r = R - y. Thus dr = - dy. Also the area of the differential element shown shaded is dA = 2prdr. Thus, the mass flow rate is # m = = =

R

0

2prU

R 7 LR 1

2prU R

1 7

L0

r

1

y 7(R - y)( - dy)

R

1

y

8

aRy4 - y7 b dy

dr

2prU 7 8 7 15 R a Ry7 y7 b ` = 1 8 15 0 R7 = =

R

1

y 7 rvdA = r U a b (2prdr) R LA L0

2prU 1 7

R

a

(a)

49 15 R7 b 120

49p rUR2 60

Ans.

Ans: 49p # m = rUR 2 60 369

4–31. The velocity profile for a liquid in the channel is determined experimentally and found to fit the equation u = 1 8y1>3 2 m>s, where y is in meters. Determine the volumetric discharge if the width of the channel is 0.5 m.

y

1 ––

u ! (8y 3 ) m/s

0.5 m

x

SOLUTION

y

The differential rectangular element of thickness dy on the cross section will be considered, Fig. a. Since the channel has a constant width of 0.5 m, the area of this element is dA = (0.5 m)dy = 0.5dy. Q = Q =

LA L0

V # dA

0.5

1 8y 2 (0.5 m)dy 1 3

Q = 1.19 m3 >s

1

( (

dy V = 8y 3 m s

0.5 m

4

= 3y 3 `

0.5

V

0

Ans. (a)

370

Ans: 119 m3 >s

*4–32. The velocity profile for a liquid in the channel is determined experimentally and found to fit the equation u = 1 8y 1>3 2 m>s, where y is in meters. Determine the average velocity of the liquid. The channel has a width of 0.5 m.

y

0.5 m

1 ––

u ! (8y 3 ) m/s x

SOLUTION The differential rectangular element of thickness dy on the cross section will be considered, Fig. a. Since the channel has a constant width of 0.5 m, the area of this element is dA = (0.5 m) dy = 0.5 dy. Q = Q =

LA L0

V # dA

0.5 m

The average velocity is V =

Pg(kPa) # m (kg>s)

1 8y 2 (0.5 m)dy 1 3

4

= 3y 3 `

0.5 0

= 1.191 m3 >s

1.191 m3 >s Q = = 4.76 m>s A (0.5 m)(0.5 m)

Ans.

0

20

40

60

80

100

1.45

1.73

2.02

2.30

2.59

2.87

m(Kg s) 

3

2

1

0

20

40

60

80

100

(a)

371

Pg(kPa)

4–33. A very viscous liquid flows down the inclined rectangular channel such that its velocity profile is approximated by u = 4 1 0.5y2 + 1.5y 2 ft>s, where y is in feet. Determine the mass flow if the channel width is 2 ft. Take g = 60 lb>ft 3.

y

1.5 ft u

4(0.5y2

1.5y) ft/s

SOLUTION The differential rectangular element of thickness dy on the cross section will be considered, Fig. a. Since the channel has a constant width of 2 ft, the area of this element is dA = (2 ft)dy. Thus, # m =

LA

= a

rV # dA

60 slug>ft 3 b 32.2 L0

= 14.91

L0

= 14.91a

1.5 ft

1.5 ft

3 4 ( 0.5y2

( 0.5y2 + 1.5y ) dy

+ 1.5y ) ft>s 4 3 (2 ft)dy 4

1.5 ft 0.5y3 + 0.75y2 b ` 3 0

Ans.

= 33.5 slug>s

(

1.5 ft

(

u = 4 0.5y2 + 1.5y ft s

y

dy u

(a)

Ans: 33.5 slug>s 372

4–34. A very viscous liquid flows down the inclined rectangular channel such that its velocity profile is approximated by u = 4 1 0.5y2 + 1.5y 2 ft>s, where y is in feet. Determine the average velocity of the liquid if the channel width is 2 ft.

y

1.5 ft u

4(0.5y2

1.5y) ft/s

SOLUTION The differential rectangular element of thickness dy on the cross-section will be considered, Fig. a. Since the channel has a constant width of 2 ft, the area of this element is dA = 2dy. Thus, Q =

LA

= 8

udA =

L0

= 8a

1.5 ft

L0

1.5 ft

4 ( 0.5y2 + 1.5y ) (2dy)

( 0.5y2 + 1.5y ) dy

1.5 ft 0.5 3 y + 0.75y2 b ` 3 0

= 18 ft 3 >s

Thus, the average velocity is V =

Q 18 = = 6.00 ft>s A 1.5(2)

(

(

u = 4 0.5y2 + 1.5y ft s

y

1.5 ft

Ans.

dy u

(a)

Ans: 6.00 ft>s 373

4–35. Determine the volumetric flow through the 50-mm-diameter nozzle of the fire boat if the water stream reaches point B, which is R = 24 m from the boat. Assume the boat does not move.

30! B

A

3m R

SOLUTION The xy coordinate system with origin at A is established as shown in Fig. a. Horizontal Motion. +) (d

sx = (s0)x + (VA)x t 24 = 0 + (VA cos 30°)t t =

27.7128 VA

(1)

Vertical Motion.

( + c)

sy = (s0)y + (VA)y t +

1 2 at 2 c

- 3 = 0 + (VA sin 30°)t +

1 ( - 9.81 m>s2 ) t 2 2

4.905t 2 - 0.5VAt - 3 = 0

(2)

Solving Eqs. (1) and (2), t = 1.854 s VA = 14.95 m>s Using the result of VA, Q = VA AA = ( 14.95 m>s ) c = 0.0294 m3 >s

p (0.05 m)2 d 4

ac = 9.81 m s2

Ans.

y

VA 30˚

A

Sy = 3 m Sx = 24 (a)

374

Ans: 0.0294 m3 >s

*4–36. Determine the volumetric flow through the 50-mm-diameter nozzle of the fire boat as a function of the distance R of the water stream. Plot this function of flow (vertical axis) versus the distance for 0 … R … 25 m. Give values for increments of ∆R = 5 m. Assume the boat does not move.

30! B

A

3m R

SOLUTION

ac = 9.81 m s2

The x-y coordinate system with origin at A is established as shown in Fig. a. Horizontal Motion +) (d

VA

Sy = 3 m

sx = (s0)x + (VA)x t

30˚

R = 0 + (VA cos 30°)t t = Vertical Motion

( + c)

A

R 2R = VA cos 30° 23VA

(1) Sx = R (a)

1 2 at 2 c

sy = (s0)y + (VA)y t +

- 3 m = 0 + (VA sin 30°)t +

1 ( - 9.81 m>s2 ) t 2 2

Q(m3 s) 0.03

2

(2)

4.905t - 0.5VAt - 3 = 0 Substitute Eq (1) into (2) 4.905 a

Q = £

2

2R 23VA

Thus, the discharge is Q = VA;

b - 0.5VAa

VA = a

23VA

b - 3 = 0

0.00502R 1

(0.5774R + 3)2

0

1

2 6.54R2 b 0.5774R + 3

Q = a

0.02 0.01

2R

1

§ m3 >s where R is in m

R(m)

0

5

10

15

20

25

3

0

0.0103

0.0170

0.0221

0.0263

0.0301

375

5

10

15 (b)

2 6.54R2 b 3 p(0.025 m)2 4 0.5774R + 3

The plot of Q vs R is shown in Fig. b

Q(m >s)

y

Ans.

20

25

R(m)

4–37. For a short time, the flow of carbon tetrachloride through the circular pipe transition can be expressed as Q = (0.8t + 5) 1 10-3 2 m3 >s, where t is in seconds. Determine the average velocity and average acceleration of a particle located at A and B when t = 2 s.

75 mm 50 mm A

B B

SOLUTION When t = 2 s, Q =

3 0.8(2)

+ 5 4 ( 10-3 ) m3 >s = 6.6 ( 10-3 ) m3/s

Thus, the velocities at A and B are VA =

6.6 ( 10-3 ) m3 >s Q = = 3.36 m>s AA p(0.025 m)2

Ans.

VB =

6.6 ( 10-3 ) m3 >s Q = = 1.49 m>s AB p(0.0375 m)2

Ans.

Here, VA = VB = Thus

(0.8 t + 5) ( 10-3 ) m3 >s 1.6(0.8 t + 5) Q = = c d m>s 2 p AA p(0.025 m)

(0.8 t + 5) ( 10-3 ) m3 >s 0.7111(0.8 t + 5) Q = = c d m>s p AB p(0.0375 m)2 aA =

dVA 1.6 = (0.8) = 0.407 m>s2 p dt

Ans.

aB =

dVB 0.7111 = (0.8) = 0.181 m>s2 p dt

Ans.

Ans: VA = 3.36 m>s, VB = 1.49 m>s aA = 0.407 m>s2, aB = 0.181 m>s2 376

4–38. Air flows through the gap between the vanes at 0.75 m3 >s. Determine the average velocity of the air passing through the inlet A and the outlet B. The vanes have a width of 400 mm and the vertical distance between them is 200 mm.

20!

40!

200 mm VB

VA

40!

B

20!

A

SOLUTION The discharge can be calculated using Q =

Va # dA Lcs

Here, the average velocities will be used. Referring to Fig. a, QA = VA # AA; QB = VB # AB;

- 0.75 m3 >s = (VA cos 140°) 3 (0.2 m)(0.4 m) 4 VA = 12.2 m>s

0.75 m3 >s = (VB cos 20°) 3 (0.2 m)(0.4 m) 4 VB = 9.98 m>s

Ans.

Ans.

140˚ AA

20˚

40˚

AB VB

VA

(a)

Ans: VA = 12.2 m>s, VB = 9.98 m>s 377

4–39. The human heart has an average discharge of 0.1 1 10-3 2 m3 >s, determined from the volume of blood pumped per beat and the rate of beating. Careful measurements have shown that blood cells pass through the capillaries at about 0.5 mm>s. If the average diameter of a capillary is 6 µm, estimate the number of capillaries that must be in the human body.

SOLUTION n is the number of the capillaries in the human body. From the discharge of the blood from heart, Q = nAV;

0.1 ( 10-3 ) m3 >s = n 5 p 3 3.0 ( 10-6 ) m 4 2 6 3 0.5 ( 10-3 ) m>s 4

n = 7.07 ( 109 )

Ans.

Ans: 7.07 ( 109 ) 378

*4–40. Rain falls vertically onto the roof of the house with an average speed of 12 ft>s. On one side the roof has a width of 15 ft and is sloped as shown. The water accumulates in the gutter and flows out the downspout at 6 ft 3 >min. Determine the amount of falling rainwater in a cubic foot of air. Also, if the average radius of a drop of rain is 0.16 in., determine the number of raindrops in a cubic foot of air. Hint: The volume of a drop is V = 43pr 3.

8 ft

6 ft

SOLUTION The discharge QDP of the downspout is equal to the discharge of the rain water contained in the air of volume equal to that of the volume shown in Fig. a. Here, QDP = ( 6 ft 3 >min ) a

And the volume of the air is

1 min b = 0.1 ft 3 >s 60 s v = 12 ft s

Qa = (8 ft)(15 ft) ( 12 ft>s ) = 1440 ft 3 >s

In another words, 1440 ft 3 of air contains 0.1 ft 3 of rain water. Therefore, for 1 ft 3 of air it contains Vw = a

0.1 ft 3 b ( 1 ft 3 ) = 69.44 ( 10-6 ) ft 3 = 69.4 ( 10-6 ) ft 3 1440 ft 3

Ans.

Then the number of rain drops in 1 ft 3 of air is n =

69.44 ( 10-6 ) ft 3 3 4 0.16 pa ft b 3 12

Ans.

= 6.994 = 7

379

15 ft 8 ft (a)

4–41. Acetate flows through the nozzle at 2 ft 3 >s. Determine the time it takes for a particle on the x axis to pass through the nozzle, from x = 0 to x = 6 in. if x = 0 at t = 0. Plot the distance-versus-time graph for the particle.

6 in. 0.5 in.

2 in.

x

x

SOLUTION Since the flow is assumed to be one dimensional and incompressible, its velocity can be determined using

1.5 ft 12

Q u = A

r x

0.5 ft 12

0.5 – x (a)

From the geometry shown in Fig. a, the radius r of the nozzle’s cross-section as a function of x is r =

0.5 ft 12

0.5 0.5 - x 1.5 1 ft + a ba ft b = (4 - 6x) ft 12 0.5 12 24

Thus, the cross-sectional area of the nozzle is A = pr 2 = pc Then u =

2 1 p (4 - 6x) d = (4 - 6x)2 ft 2 24 576

2 ft 3 >s Q 1152 = = c d ft>s p A p(4 - 6x)2 (4 - 6x)2 ft 2 576

Using the definition of velocity and the initial condition of x = 0 at t = 0, dx 1152 = u = dt p(4 - 6x)2 t

x(ft) 1 2

x

p dt = (4 - 6x)2dx 1152 L0 L0

5 12 1 3

x

t =

p ( 36x2 - 48x + 16 ) dx 1152 L0

p ( 12x3 - 24x2 + 16x ) 1152 px ( 3x2 - 6x + 4 ) t = 288

1 4

t =

1 12

6 ft = 0.5 ft, When x = 12 t =

p(0.5) 288

3 3 ( 0.52 )

1 6

(1)

t(ms)

- 6(0.5) + 4 4 = 9.5448 ( 10-3 )

0

1

2

3

4

5

6

7

8

9

10

(b)

Ans.

= 9.54 ms

Using Eq (1), the following tabulation can be computed and the plot of x vs t is shown in Fig. b. x(ft)

0

t(ms)

0

1 12 3.20

1 6 5.61

1 4 7.33

1 3 8.48

5 12 9.18

1 2 9.54 Ans: 9.54 ms 380

4–42. Acetate flows through the nozzle at 2 ft 3 >s. Determine the velocity and acceleration of a particle on the x axis at x = 3 in. When t = 0, x = 0.

6 in. 0.5 in.

2 in.

x

x

SOLUTION Since the flow is assumed to be one dimensional and incompressible, its velocity can be determined using Q u = A

0.5 ft 12

From the geometry shown in Fig. a, the radius r of the nozzle’s cross-section as a function of x is r = Thus,

r

0.5 ft 12

x

0.5 – x (a)

0.5 - x 1.5 0.5 1 ft + a ba ft b = (4 - 6x) ft 12 0.5 12 24

A = pr 2 = pc Then, u =

Thus, when x = a

1.5 ft 12

2 1 p (4 - 6x) d = (4 - 6x)2 ft 2 24 576

2 ft 3 >s Q 1152 = = c d ft>s p A p(4 - 6x)2 (4 - 6x)2 ft 2 576

3 b ft = 0.25 ft 12 u =

1152

p 3 4 - 6(0.25) 4 2

= 58.67 ft>s = 58.7 ft>s

Ans.

The acceleration can be determined using a = Since Q is constant

Then

0u = 0, that is, there is no local change at the print. Since, 0t

0u 1152 13824 1 = 3 ( - 2) 3 (4 - 6x)-3 4 ( -6) 4 = c d p p 0x (4 - 6x)3 a = 0 + u =

When x =

0u 0u + u 0t 0x

2 ft = 0.25 ft, 12 a =

0u 0x

13824 u c d ft>s2 p (4 - 6x)3

u = 58.67 ft>s . Then

58.67 13824 W ¶ = 16523 ft>s2 p 3 4 - 6(0.25) 4 3

381

Ans.

Ans: u = 58.7 ft>s, a = 16 523 ft>s2

4–43. The tapered pipe transfers ethyl alcohol to a mixing tank such that a particle at A has a velocity of 2 m>s. Determine the velocity and acceleration of a particle at B, where x = 75 mm.

200 mm

60 mm

20 mm A

B

x

2 m/s

x

SOLUTION From the geometry shown in Fig. a, the radius r of the pipe as a function of x is

0.02 m

0.2 - x b(0.02 m) = (0.03 - 0.1x) m r = 0.01 m + a 0.2

x

Thus, the cross-sectional area of the pipe as a function of x is 2

A = pr =

3 p(0.03

- 0.1x)

2

r 0.2 – x

0.01 m

(a)

4m

2

The flow rate is constant which can be determined from

Q = u A AA = (2 m>s) 3 p(0.03 m)2 4 = 1.8p ( 10-3 ) m3 >s

Thus, the velocity of the flow as a function of x is u = At x = 0.075 m,

1.8p ( 10-3 ) m3 >s 1.8 ( 10-3 ) Q = £ § m>s = 2 2 A p(0.03 - 0.1x) m (0.03 - 0.1x)2

u =

1.8 ( 10-3 )

3 0.03

- 0.1(0.075) 4 2

= 3.556 m>s = 3.56 m>s

Ans.

The acceleration of the flow can be determined using a =

0u 0u + u 0t 0x

0u Since the flow rate is constant there is no local changes, so that at the point, = 0. 0t Here 0u = 1.8 ( 10-3 ) ( - 2)(0.03 - 0.1x)-3( - 0.1) 0x =

0.36 ( 10-3 ) (0.03 - 0.1x)3

Thus a = 0 + u = c

0u 0x

0.36 ( 10-3 ) u (0.03 - 0.1x)3

When x = 0.075 m, u = 3.556 m>s . Then a =

0.36 ( 10-3 ) (3.556)

3 0.03

- 0.1(0.075) 4 3

d m>s2

= 112.37 m>s2 = 112 m>s2

Ans.

Ans: u = 3.56 m>s, a = 112 m>s2 382

*4–44. The tapered pipe transfers ethyl alcohol to a mixing tank such that when a valve is opened, a particle at A has a velocity at A of 2 m>s, which is increasing at 4 m>s2. Determine the velocity of the same particle when it arrives at B, where x = 75 mm.

200 mm

60 mm

20 mm A

B

x

2 m/s

x

SOLUTION From the geometry shown in Fig. a, the radius r of the pipe as a function of x is 0.2 - x b(0.02 m) = (0.03 - 0.1x) m r = 0.01 m + a 0.2

Thus, the cross-sectional area of the pipe as a function of x is 2

A = pr =

3 p(0.03

- 0.1x)

2

uA

L2 m>s

du =

L0

4m

t

4 dt

u A - 2 = 4t u A = (4t + 2) m>s Thus, the flow is Q = u A AA =

3 (4 t

+ 2) m>s 4 3 p(0.03 m)2 4

= p(0.0036t + 0.0018) m3 >s

Then, the velocity of the flow as a function of t and x is u =

p(0.0036 t + 0.0018) m3 >s Q = A p(0.03 - 0.1 x)2 m2 = c

Using the definition of velocity, dx = u; dt L0

x

L0

x

0.0036 t + 0.0018 d m>s (0.03 - 0.1 x)2

dx 0.0036 t + 0.0018 = dt (0.03 - 0.1 x)2

(0.03 - 0.1x)2dx =

L0

t

(0.0036 t + 0.0018)dt

( 0.01x2 - 0.006x + 0.0009 ) dx =

L0

r x

0.2 – x (a)

2

The velocity of a particle passing through the cross-section at A can be determined from du = a; dt

0.02 m

t

(0.0036 t + 0.0018)dt

0.003333x3 - 0.003x2 + 0.0009x = 0.0018t 2 + 0.0018t

383

0.01 m

*4–44. Continued

When x = 0.075 m, 0.003333 ( 0.0753 ) - 0.003 ( 0.0752 ) + 0.0009(0.075) = 0.0018t 2 + 0.0018t t 2 + t - 0.02890625 = 0 t = 0.02812 s When x = 0.075 m, t = 0.02812 s, u =

0.0036(0.02812) + 0.0018

3 0.03

- 0.1(0.075) 4 2

Ans.

= 3.7555 m>s = 3.76 m>s

384

4–45. The radius of the circular duct varies as r = 1 0.05e -3x 2 m, where x is in meters. The flow of a fluid at A is Q = 0.004 m3 >s at t = 0, and it is increasing at dQ>dt = 0.002 m3 >s2. If a fluid particle is originally located at x = 0 when t = 0, determine the time for this particle to arrive at x = 100 mm.

r 200 mm

50 mm x B A

SOLUTION

x

The discharge as a function of time t is Q = 0.004 m3 >s + ( 0.002 m3 >s2 ) t = (0.004 + 0.002 t) m3 >s

The cross-sectional Area of the duct as a function of x is A = pr 2 = p ( 0.05e -3x ) 2 = ( 0.0025p e -6x ) m2 Thus, the velocity of the flow is u =

(0.004 + 0.002 t) m3 >s Q = A ( 0.0025p e -6x ) m2 =

1.6 6x 0.8 t 6x e + e p p

=

4 6x e (t + 2) 5p

Using the definition of velocity, dx = u; dt x

L0 e

dx 6x

dx 4 6x = e (t + 2) dt 5p t

=

4 (t + 2)dt 5p L0

x t 4 t2 1 a + 2tb ` - a e -6x b ` = 6 5p 2 0 0

1 2 ( 1 - e -6x ) = ( t 2 + 4t ) 6 5p When x = 0.1 m,

1 2 3 1 - e -6(0.1) 4 = ( t 2 + 4t ) 6 5p t 2 + 4t - 0.5906 = 0

t =

- 4 { 242 - 4(1) ( -0.5906) 2(1)

7 0 Ans.

t = 0.143 s

Ans: 0.413 s 385

4–46. The radius of the circular duct varies as r = 1 0.05e -3x 2 m, where x is in meters. If the flow of the fluid at A is Q = 0.004 m3>s at t = 0, and it is increasing at dQ>dt = 0.002 m3>s2, determine the time for this particle to arrive at x = 200 mm.

r 200 mm

50 mm x B A

SOLUTION

x

The discharge as a function of time t is Q = 0.004 m3 >s + ( 0.002 m3 >s2 ) t = (0.004 + 0.002t) m3 >s

The cross-sectional area of the duct as a function of x is A = pr 2 = p ( 0.05e -3x ) 2 = ( 0.0025p e -6x ) m2 Thus, the velocity of the flow is (0.004 + 0.002t) m3 >s Q = A ( 0.0025pe -6x ) m2

u =

=

1.6 6x 0.8t 6x e + e p p

=

4 6x e (t + 2) 5p

Using the definition of velocity, dx 4 6x = e (t + 2) dt 5p

dx = u; dt x

L0 e -

dx 6x

t

=

4 (t + 2)dt 5p L0

t 1 -6x x 4 t2 (e ) ` = a + 2tb ` 6 5p 2 0 0

1 2 2 ( 1 - e -6x ) = ( t + 4t ) 6 5p At point E, x = 0.2 m.

1 2 3 1 - e -6(0.2) 4 = ( t 2 + 4t ) 6 5p

t 2 + 4t - 0.9147 = 0 t =

-4 { 242 - 4(1)( -0.9147) 2(1)

7 0 Ans.

t = 0.217 s

Ans: 0.217 s 386

4–47. Water flows through the pipe at A at 300 kg>s, and then out the double wye with an average velocity of 3 m>s through B and an average velocity of 2 m>s through C. Determine the average velocity at which it flows through D.

250 mm

B 350 mm

150 mm C

A D

SOLUTION # m = rVA AA

250 mm

300 kg>s = ( 1000 kg>m3 ) (VA) 3 p(0.175 m)2 4 VA = 3.118 m>s

Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume. Continuity Equation. Since the fluid is water which has a constant density, then 0 rd V + rV # dA = 0 0t L L cv

cs

- VAAA + VBAB + VCAC + VDAD = 0 - ( 3.118 m>s ) 3 p(0.175 m)2 4 + ( 3 m>s ) 3 p(0.125 m)2 4 + ( 2 m>s ) 3 p(0.075 m)2 4 + VD 3 p(0.125 m)2 4 = 0

Ans.

VD = 2.39 m>s

VB = 3 m s AB VA

AC

VC = 2 m s

AA AD VD (a)

Ans: 2.39 m>s 387

*4–48.

If water flows at 150 kg>s through the double wye at  B, at 50 kg>s through C, and at 150 kg>s through D, determine the average velocity of flow through the pipe at A.

250 mm

B 350 mm

150 mm C

A D

250 mm

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume. # Continuity Equation. Since m = rV # A, then 0 rd V + rV # dA = 0 dt Lcv Lcs # # # # 0 - mA + mB + mC + mD = 0 # - mA + 150 kg>s + 50 kg>s + 150 kg>s = 0 # mA = 350 kg>s # mA = rVAAA 350 kg>s = ( 1000 kg>m3 ) (VA) 3 p(0.175 m)2 4 VA = 3.64 m>s

VB AB

AC VA

VC

AA AD VD (a)

388

Ans.

4–49. Air having a specific weight of 0.0795 lb>ft 3 flows into the duct at A with an average velocity of 5 ft>s. If its density at B is 0.00206 slug>ft 3, determine its average velocity at B.

2 ft

1 ft

A

B

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume. Continuity Equation. The density of the air at A and B is different. The density at A 0.0795 lb>ft 3 gA is rA = = = 0.002469 slug>ft 3. Then, g 32.2 ft>s2

VA = 5 ft s AA

AB

VB

(a)

0 rd V + rV # dA = 0 0t L L cv

cs

0 - rAVAAA + rBVBAB = 0 - ( 0.002469 slug>ft 3 )( 5 ft>s ) 3 p(0.5 ft)2 4 + ( 0.00206 slug>ft 3 ) (VB) 3 p(1 ft)2 4 = 0 Ans.

VB = 1.50 ft>s

Ans: 1.50 ft>s 389

4–50. An oscillating water column (OWC), or gully generator, is a device for producing energy created by ocean waves. As noted, a wave will push water up into the air chamber, forcing the air to pass through a turbine, producing energy. As the wave falls back, the air is drawn into the chamber, reversing the rotational direction of the turbine, but still creating more energy. Assuming a wave will reach an average height of h = 0.5 m in the 0.8-m-diameter chamber at B, and it falls back at an average speed of 1.5 m>s determine the speed of the air as it moves through the turbine at A, which has a net area of 0.26 m2. The air temperature at A is TA = 20°C, and at B it is TB = 10°C.

A 0.8 m h

B

SOLUTION Here, the control volume contains the air in the air chamber from A to B. Thus, it is a fixed control Volume, 0 rd V + rV # dA = 0 0t Lcv Lcs Since the flow is steady and the volume of the control volume does not change with time, there are no local changes. Thus, 0 - rAVAAA + rBVBAB = 0 From Appendix A, at TA = 20°C, rA = 1.202 kg>m3 and at TB = 10°C, rB = 1.247 kg>m3. Thus, - ( 1.202 kg>m3 ) (VA) ( 0.26 m2 ) + ( 1.247 kg>m3 )( 1.5 m>s ) 3 p(0.4 m)2 4 = 0

Ans.

VA = 3.01 m>s

Ans: 3.01 m>s 390

4–51. An oscillating water column (OWC), or gully generator, is a device for producing energy created by ocean waves. As noted, a wave will push water up into the air chamber, forcing the air to pass through a turbine, producing energy. As the wave falls back, the air is drawn into the chamber, reversing the rotational direction of the turbine, but still creating more energy. Determine the speed of the air as it moves through the turbine at A, which has a net open area of  0.26 m2, if the speed of the water in the 0.8-m-diameter chamber is 5 m>s. The air temperature at A is TA = 20°C, and at B it is TB = 10°C.

A 0.8 m h

B

SOLUTION Here, the control volume contains the air in the air chamber from A to B. Thus, it is a fixed control volume. 0 rd V + rV # dA = 0 0t Lcv Lcs Since the flow is steady and the volume of the control volume does not change with time, there are no local changes. Thus 0 + rAVAAA + rB ( - VBAB ) = 0 From Appendix A, at TA = 20°C, rA = 1.202 kg>m3 and at TB = 10°C, rB = 1.247 kg>m3. Thus,

( 1.202 kg>m3 )( VA )( 0.26 m2 ) + ( 1.247 kg>m3 ) e -VB 3 p(0.4 m)2 4 f = 0 VA = 2.0057 VB

Here, VB = 5 m>s, so that Ans.

VA = 2.0057(5 m>s) = 10.0 m>s

Ans: 10.0 m>s 391

*4–52. A jet engine draws in air at 25 kg>s and jet fuel at  0.2 kg>s. If the density of the expelled air–fuel mixture is 1.356 kg>m3, determine the average velocity of the exhaust relative to the plane. The exhaust nozzle has a diameter of 0.4 m.

SOLUTION Control Volume. If the control volume moves with the plane, the flow is steady if viewed from the plane. No local changes occur within this control volume. # Continuity Equation. Since m = rVf>cs # A, then 0 rd V + rVf>cs # dA = 0 0t L L cv cs # # 0 - ma - mf + rVe>csAe = 0 - 25 kg>s - 0.2 kg>s + ( 1.356 kg>m3 )( Ve>cs ) 3 p(0.2 m)2 4 = 0 Ve>cs = 148 m>s

392

Ans.

Ae (a)

Ve cs

4–53. Carbon dioxide flows into the tank at A at VA = 4 m>s, and nitrogen flows in at B at VB = 3 m>s. Both enter at a gage pressure of 300 kPa and a temperature of 250°C. Determine the steady mass flow of the mixed gas at C.

VC 0.2 m

VA

C

0.2 m A

VB

B 0.15 m

SOLUTION From Appendix A, the values of the gas constants for CO 2 and nitrogen are RCO2 = 188.9 J>kg # K and RN = 296.8 J>kg # K. p = rRT

(300 + 101.3) ( 10

3

) = rCO2(188.9 J>kg # K)(250° + 273) K

VC

rCO2 = 4.062 kg>m3 and

AC

VA = 4 m s AA

(300 + 101.3) ( 103 ) = rN(296.8 J>kg # K)(250° + 273) K AB

rN = 2.585 kg>m3 Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume.

VB = 3 m s (a)

Continuity Equation. Since the densities of the fluids through the open control surfaces are different but of constant value, then 0 rd V + rV # dA = 0 0t L L cv cs # 0 - rCO2VAAA - rNVBAB + mm = 0 # - ( 4.062 kg>m3 ) (4 m>s) c p(0.1 m)2 d - ( 2.585 kg>m3 ) (3 m>s) c p(0.075 m)2 d + mm = 0 # mm = 0.647 kg>s

Ans.

Ans: 0.647 kg>s 393

4–54. Carbon dioxide flows into the tank at A at VA = 10 m>s, and nitrogen flows in at B with a velocity of VB = 6 m>s. Both enter at a pressure of 300 kPa and a temperature of 250°C. Determine the average velocity of the mixed gas leaving the tank at a steady rate at C. The mixture has a density of r = 1.546 kg>m3.

VC 0.2 m

VA

C

0.2 m A

VB

B 0.15 m

SOLUTION From Appendix A, the values of the gas constants for CO 2 and nitrogen are RCO2 = 188.9 J>kg # K and RN = 296.8 J>kg # K. p = rRT

(300 + 101.3) ( 103 ) = rCO2(188.9 J>kg # K)(250 + 273) K rCO2 = 4.062 kg>m3 and

VC AC

VA = 10 m s AA

AB VB = 6 m s

(300 + 101.3) ( 103 ) = rN(296.8 J>kg # K)(250 + 273) K

(a)

rN = 2.585 kg>m3 Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume. Continuity Equation. 0 rd V + rV # dA = 0 0t L L cv

cs

0 - rCO2VA AA - rNVB AB + rmVC AC = 0 - ( 4.062 kg>m3 ) (10 m>s) c p(0.1 m)2 d - ( 2.585 kg>m3 ) (6 m>s) c p(0.075 m)2 d + ( 1.546 kg>m3 ) (VC) c p(0.1 m)2 d = 0 VC = 31.9 m>s

Ans.

Ans: 31.9 m>s 394

4–55. The flat strip is sprayed with paint using the six nozzles, each having a diameter of 2 mm. They are attached to the 20-mm-diameter pipe. The strip is 50 mm wide, and the paint is to be 1 mm thick. If the average speed of the paint through the pipe is 1.5 m>s, determine the required speed V of the strip as it passes under the nozzles.

1.5 m/s

V

2.5 m

SOLUTION Since the flow is steady, there is no local change. Also, r for the point is constant and the average velocities will be used. Thus, the continuity equation reduces to 0 rd V + rV # dA = 0 0t Lcv Lcs 0 - VpAp + 6Vno Ano = 0 However, Qno = Vno Ano. Then -Vp Ap + 6Qno = 0 Qno =

Vp Ap 6

=

(1.5 m>s) 3 p(0.01 m)2 4 6

= 25(10-6)p m3 >s

1 1 10-3 2 m

= 0.1667 1 10-3 2 m 6 thickness of paint on the strip. Thus, the volume of paint required is

Each nozzle has to cover 0.5 m of length and put

Thus,

Vp = (0.5 m)(0.05 m) 3 0.1667 ( 10-3 ) m 4 = 4.1667 ( 10-6 ) m3 Vp = Qno t;

4.1667 ( 10-6 ) m3 = c 25 ( 10-6 ) p m3 >s d t t = 0.05305 s

Then the required speed of the strip is V =

0.5 m = 9.42 m>s 0.05305 s

Ans.

Ans: 9.42 m>s 395

*4–56. The flat strip is sprayed with paint using the six nozzles, which are attached to the 20-mm-diameter pipe. The strip is 50 mm wide and the paint is to be 1 mm thick. If the average speed of the point through the pipe is 1.5 m>s, determine the required speed V of the strip as it passes under the nozzles as a function of the diameter of the pipe. Plot this function of speed (vertical axis) versus diameter for 10 mm … D … 30 mm. Give values for increments of ∆D = 5 mm.

1.5 m/s

V

2.5 m

SOLUTION Since the flow is steady, there is no local change. Also, r for the paint is constant and the average velocities will be used. Thus, 0 rd V + rV # dA = 0 0t Lcv Lcs 0 - VpAp + 6VnoAno = 0 However, Qno = VnoAno. Then

Qno =

Vp Ap

=

6

p D 2 b d 4 1000 = 6

( 1.5 m>s ) c a

3 62.5 ( 10-9 ) p D2 4 m3 >s 1 ( 10-3 ) m

= 0.1667 ( 10-3 ) m 6 thickness of paint on the strip. Thus, the volume of paint required is Each nozzle has to cover 0.5 m of length and put

Thus,

Vp = (0.5 m)(0.05 m) 3 0.1667 ( 10-3 ) m 4 = 4.1667 ( 10-6 ) m3 Vp = Qno t;

4.1667 ( 10-6 ) m3 = c 62.5 ( 10-9 ) p D2 m3 >s d t t = a

Then the required speed of the strip is

21.22 bs D2 0.5 m

V =

( 21.22>D2 ) s V = ( 0.0236 D2 ) m>s where D is in mm

Ans.

The plot of V vs D is shown in Fig. a. D(mm)

10

15

20

25

30

V(m>s)

2.36

5.30

9.42

14.7

21.2

V(m s)

30

20

10

0

5

10

15

20

25

30

(a)

396

D(mm)

4–57. Pressurized air in a building well flows out through the partially opened door with an average velocity of 4 ft>s. Determine the average velocity of the air as it flows down from the top of the building well. Assume the door is 3 ft wide and u = 30°.

V 3 ft

4 ft 4 ft/s

3 ft

u

7 ft 4 ft/s

SOLUTION The control volume contains the air in the building well and in the circular sector of the door opening. We use a circular sector because the velocity must be normal to the area through which it flows. It can be considered fixed. Also, since in this case the air is assumed to be incompressible, the flow is steady, thus there is no local change. Here, the density of the air remains constant and average velocities will be used. 0 rd V + rV # dA = 0 0t L L cv

cs

0 - VinAin + ( Vout ) 1 ( Aout ) 1 + ( Vout ) 2 ( Aout ) 2 = 0

(1)

Here, the entrance open control surface is the cross-section of the building well above the door. Ain = (4 ft)(3 ft) = 12 ft 2 The exit open control surfaces are the top and side of the door opening. 1 2

1 2

( Aout ) 1 = r 2u = (3 ft)2 a ( Aout ) 2 = r uh = (3 ft)a

30° p radb = 0.75p ft 2 180°

30° p radb(7 ft) = 3.5p ft 2 180°

Substituting these values into Eq. (1), - Vin(12 ft 2) + (4 ft>s) ( 0.75p ft 2 ) + (4 ft>s) ( 3.5p ft 2 ) = 0 Ans.

Vin = 4.45 ft>s

Ans: 4.45 ft>s 397

4–58. Pressurized air in a building well flows out through the partially opened door with an average velocity of 4 ft>s. Determine the average velocity of the air as it flows down from the top of the building well as a function of the door opening u. Plot this function of velocity (vertical axis) versus u for 0° … u … 50°. Give values for increments of ∆u = 10°.

V 3 ft

4 ft 4 ft/s

SOLUTION

3 ft

The control volume contains the air in the building well and in the circular sector of the door opening. We use a circular sector because the velocity must be normal to the area through which it flows. It can be considered fixed. Also, since in this case the air is assumed to be incompressible, the flow is steady, thus there is no local change. Here, the density of the air remains constant and average velocities will be used.

u

7 ft 4 ft/s

0 rd V + rV # dA = 0 0t L L cv

cs

0 - VinAin + ( Vout ) 1 ( Aout ) 1 + ( Vout ) 2 ( Aout ) 2 = 0

(1)

Here, the entrance open control surface is the cross-section of the building well above the door. Ain = (4 ft)(3 ft) = 12 ft 2 The exit open control surfaces are the top and side of the door opening. 1 2

1 2

( Aout ) 1 = r 2 u = (3 ft)2 a

( Aout ) 2 = r uh = (3 ft) a

u p radb = (0.025p u) ft 2 180°

u p radb(7 ft) = (0.1167p u) ft 2 180°

Substituting these values into Eq. (1), - Vin(12 ft 2) + (4 ft>s) ( 0.025p u ft 2 ) + (4 ft>s) ( 0.1167p u ft 2 ) = 0 Ans.

Vin = (0.0472p u) ft>s where u is in degrees. The plot of Vin vs u is shown in Fig. a. u(deg.)

0

10

20

30

40

50

Vin(ft>s)

0

1.48

2.97

4.45

5.93

7.42

10

20

Vin(ft s ) 8 7 6 5 4 3 2 1 0

30

40

50

(deg.)

Ans: Vin = (0.0472pu) ft>s, where u is in degrees 398

4–59. Drilling fluid is pumped down through the center pipe of a well and then rises up within the annulus. Determine the diameter d of the inner pipe so that the average velocity of the fluid remains the same in both regions. Also, what is this average velocity if the discharge is 0.02 m3 >s? Neglect the thickness of the pipes.

Vout

Vin

d

200 mm

SOLUTION The control volume considered is the volume of drilling fluid in pipe which is fixed. Here, the flow is steady, thus there are no local changes. Also, the density of the fluid is constant and the average velocities will be used. 0 rd V + rV # dA = 0 0t L L cv

cs

(1)

0 - VinAin + VoutAout = 0 Here, it is required that Vin = Vout. Also, Ain =

p 2 p p d and Aout = (0.2 m)2 - d 2. 4 4 4

Then p p p - V a d 2 b + V c (0.2 m)2 - d 2 d = 0 4 4 4 2d 2 = 0.04

Ans.

d = 0.1414 m = 141 mm Considering the flow in the center pipe, Q = VA;

0.02 m3 >s = V c

V = 1.27 m>s

p (0.1414 m)2 d 4

Ans.

Ans: d = 141 mm V = 1.27 m>s 399

*4–60. Drilling fluid is pumped down through the center pipe of a well and then rises up within the annulus. Determine the velocity of the fluid forced out of the well as a function of the diameter d of the inner pipe, if the velocity of the fluid forced into the well is maintained at Vin = 2 m>s. Neglect the thickness of the pipes. Plot this velocity (vertical axis) versus the diameter for 50 mm … d … 150 mm. Give values for increments of ∆d = 25 mm.

Vout

200 mm

SOLUTION The control volume is the volume of the drilling fluid in the pipe which is fixed. Here, the flow is steady thus there is no local change. Also the density of the fluid is constant and average velocities will be used. 0 rd V + rV # dA = 0 0t L L cv

cs

0 - VinAin + VoutAout = 0 Here,

Ain = Aout =

p d 2 a b = 0.25 ( 10-6 ) p d 2 and 4 1000

p d 2 p c (0.2 m)2 - a b d = 3 0.04 - ( 10-6 ) d 2 4 4 1000 4

-(2 m>s) 3 0.25(10-6)p d 2 4 + Vout e Vout = c

2 ( 10-6 ) d 2

0.04 - ( 10-6 ) d 2

p 3 0.04 - (10-6)d 2 4 f = 0 4

d m>s where d is in mm

Ans.

The plot of Vout vs d is shown in Fig. a. 50

75

100

125

150

Vout(m>s)

0.133

0.327

0.667

1.28

2.57

25

50

d(mm)

Vout (m s) 3

2

1

d(mm) 0

75

100

125

150

(a)

400

Vin

d

4–61. The unsteady flow of glycerin through the reducer is such that at A its velocity is VA = ( 0.8 t 2 ) m>s, where t is in  seconds. Determine its average velocity at B, and its average acceleration at A, when t = 2 s. The pipes have the diameters shown.

0.1 m 0.3 m B A

SOLUTION When t = 2 s, the velocity of the flow at A is

VA

VA = 0.8(2)2 = 3.20 m>s Control Volume. The fixed control volume is shown in Fig. a. Since the volume of the control volume does not change over time, no local changes occur within this control volume.

AB

AA

VB

(a)

Continuity Equation. Since the water has a constant density, then 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VAAA + VBAB = 0 - ( 3.20 m>s ) c p a

0.3 m 2 0.1 m 2 b d + VB c p a b d = 0 2 2 VB = 28.8 m>s

Ans.

With u = VA and v = w = 0, we have aA =

0VA 0t

= 1.6t 0 t = 2 s = 3.20 m>s2

Ans.

Ans: VB = 28.8 m>s aA = 3.20 m>s2 401

4–62. Oil flows into the pipe at A with an average velocity of 0.2 m>s and through B with an average velocity of 0.15 m>s. Determine the maximum velocity Vmax of the oil as it emerges from C if the velocity distribution is parabolic, defined by vC = Vmax ( 1 - 100r 2 ) , where r is in meters measured from the centerline of the pipe.

Vmax

C 300 mm

A

200 mm

B 0.15 m/s

0.2 m/s

SOLUTION

200 mm

The control volume considered is fixed as it contain the oil in the pipe. Also, the flow is steady and so no local changes occur. Here, the density of the oil is constant. Then 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VAAA + VBAB +

L

Vc dA = 0

A

V = Vmax (1 – 100r2)

- ( 0.2 m>s ) 3 p(0.15 m)2 4 + ( 0.15 m>s ) 3 p(0.1 m)2 4 +

L0

0.1 m

V max ( 1 - 100r 2 ) (2prdr) = 0

r

- 3 ( 10-3 ) pm3 + 5 ( 10-3 ) pV max = 0

(a)

Ans.

V max = 0.6 m>s

Note: The integral in the above equation is equal to the volume under the velocity profile, while in this case is a paraboloid. L A

Vc dA =

1 2 1 pr h = p(0.1 m)2(V max ) = 5 ( 10-3 ) pV max . 2 2

VC AC

AA AB

VA = 0.2 m s

VB = 0.15 m s

(b)

Ans: 0.6 m>s 402

4–63. The unsteady flow of linseed oil is such that at A it has a velocity of VA = (0.7t + 4) m>s, where t is in seconds. Determine the acceleration of a fluid particle located at x = 0.2 m when t = 1 s. Hint: Determine V = V(x, t), then use Eq. 3–4.

0.5 m 0.3 m 0.1 m

B A x

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. Since the volume does not change over time, no local changes occur within this control volume. Continuity Equation. Referring to the geometry shown in Fig. a, the radius of the pipe at an arbitrary distance x is r - 0.05 0.5 - x = ; 0.1 0.5

r = (0.15 - 0.2x) m

0.5 m 0.1 m

r

0.05 m VA

A

AA

V 0.5 – x

x

Then,

(a)

0 rdV + rV # dA = 0 0t L L cv

cs

0 - VAAA + VA = 0 - (0.7t + 4) 3 p(0.15 m)2 4 + V 3 p(0.15 - 0.2x)2 4 = 0 V =

0.0225(0.7t + 4) (0.15 - 0.2x)2

For A differential control volume at x = 0.2 m, with u = V and v = w = 0, a = =

=

0V 0V + V 0t 0x 0.0225(0.7) (0.15 - 0.2x)

2

+ a

0.0225(0.7t + 4) (0.15 - 0.2x)

2



0.0225 3 (0.15 - 0.2x)2 4 (0) - (0.7t + 4)(2)(0.15 - 0.2x)( - 0.2) 4 (0.15 - 0.2x)4

¢

0.0225(0.7t + 4) 0.009(0.7t + 4) 0.01575 + £ §£ § 2 (0.15 - 0.2x) (0.15 - 0.2x)2 (0.15 - 0.2x)3

For t = 1 s, x = 0.2 m, a = 1.3017 + 8.7397(31.781) = 279 m>s2

Ans.

Ans: 279 m>s2 403

*4–64. The unsteady flow of linseed oil is such that at A it has a velocity of VA = ( 0.4t 2 ) m>s, where t is in seconds. Determine the acceleration of a fluid particle located at x = 0.25 m when t = 2 s Hint: Determine V = V(x, t), then use Eq. 3–4.

0.5 m 0.3 m 0.1 m

B A x

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. Since the volumes does not change over time, no local changes occur within this control volume. Continuity Equation. Referring to the geometry shown in Fig. a, the radius of the pipe at an arbitrary distance x is r - 0.05 0.5 - x = ; 0.1 0.5

r = (0.15 - 0.2x) m

Then, 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VAAA + VA = 0 - ( 0.4t 2 ) 3 p(0.15 m)2 4 + V 3 p(0.15 - 0.2x)2 4 = 0 V =

0.009t 2 (0.15 - 0.2x)2

For A differential control volume at x = 0.25 m, with u = V and v = w = 0, a =

=

=

0V 0V + V 0t 0x

( 0.15 - 0.2x)2(0) - ( 0.009t 2 ) (2)(0.15 - 0.2x)( -0.2) 0.018t 0.009t + £ §£ § 2 2 (0.15 - 0.2x) (0.15 - 0.2x) (0.15 - 0.2x)4 0.018t 0.009t 2 0.0036t 2 + £ §£ § 2 2 (0.15 - 0.2x) (0.15 - 0.2x) (0.15 - 0.2x)3

For t = 2 s, x = 0.25 m,

a = 3.6 + 3.6(14.4) = 55.4 m>s2

Ans.

404

4–65. Water flows through the nozzle at a rate of 0.2 m3 >s. Determine the velocity V of a particle as it moves along the centerline as a function of x.

8!

40 mm

x 100 mm

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume. Continuity Equation. From the geometry shown in Fig. a,

0.02 m

VA

r = 0.02 m - x tan 8° = (0.02 - 0.1405x) m

0.1 m 8˚

r A

AA

V

x

Realizing that QA = VAAA = 0.2 m3 >s,

(a)

0 rdV + rV # dA = 0 0t L L cv

cs

0 - QA + VA = 0 - 0.2 m3 >s + V 3 p(0.02 - 0.1405x)2 4 = 0 V =

0.0637 (0.02 - 0.141x)2

Ans.

Ans: V =

405

0.0637 (0.02 - 0.141x)2

4–66. Water flows through the nozzle at a rate of 0.2 m3 >s. Determine the acceleration of a particle as it moves along the centerline as a function of x.

8!

40 mm

x 100 mm

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. Since the flow is steady, there is no change in volume, and therefore no local changes occur within this control volume. VA

Continuity Equation. From the geometry shown in Fig. a,

0.1 m

0.02 m



r A

AA

V

x

r = 0.02 m - x tan 8° = (0.02 - 0.1405x) m

(a)

Realizing that QA = VAAA = 0.2 m3 >s,

0 rdV + rV # dA = 0 0t L L cv

cs

0 - QA + VA = 0 -0.2 m >s + V 3 p(0.02 - 0.1405x)2 4 = 0 3

V =

0.06366 (0.02 - 0.1405x)2

Since the flow is one dimensional, the acceleration can be determined using 0V 0V a = + V 0t 0x Here,

0V = 0 and 0t (0.02 - 0.1405x)2(0) - 0.06366(2)(0.02 - 0.1405x)( - 0.1405) 0V = 0x (0.02 - 0.1405x)4 =

0.01789 (0.02 - 0.1405x)3

Thus, a = 0 + £ = £

0.06366 0.01789 §£ § (0.02 - 0.1405x)2 (0.02 - 0.1405x)3

1.14(10-3) (0.02 - 0.141x)5

§ m>s2

Ans.

Ans: £ 406

1.14 110 - 3 2

(0.02 - 0.141x)5

§ m>s2

4–67. The cylindrical plunger traveling at Vp = 1 0.004t 1>2 ) m>s, where t is in seconds, injects a liquid plastic into the mold to make a solid ball. If d = 50 mm, determine the amount of time needed to do this if the volume of the ball is V = 43pr 3.

Vp

d y 10 mm

SOLUTION The control volume considered is the volume of the liquid plastic contained in the plunger. Its volume changes with time, Fig. a. The volume V0 of the lower portion of the control volume is constant.

75 mm

0 r dV + rpVp>cs # dA = 0 0t L p L cv

cs

Since rp is constant, it can be factored out of the integrals. Also, the average velocity will be used. Thus, the above equation becomes dV rp + rpVAAA = 0 dt Since, QA = VAAA, dV + QA = 0 (1) dt The volume of the control volume is

0.05 m

y

V = p(0.025 m)2y + V0 = 0.625 ( 10-3 ) py + V0

A

dy dV = 0.625(10-3)p dt dt

V0

(a)

dy 1 = - Vp = ( - 0.004t 2 ) m>s. The negative sign indicates that Vp is directed dt in the opposite sense to positive y. 1 1 dV = 0.625 ( 10-3 ) p( -0.004t 2 ) = 3 - 2.5 ( 10-6 ) pt 2 4 m3 >s dt The negative sign indicates that the volume is decreasing. However,

1

- 2.5 ( 10-6 ) pt 2 + QA = 0 1

QA = ( 2.5 ( 10-6 ) pt 2 ) m3 >s

The volume of the ball is 4 4 Vs = pr 3 = p(0.075 m)3 = 0.5625 ( 10-3 ) p m3 3 3 The time required to fill up the mold is given by L L0

T

QAdt = Vs

1

2.5 ( 10-6 ) pt 2 dt = 0.5625 ( 10 - 3 ) p L0

T

1

t 2 dt = 225 2 3 T 2 = 225 3 2

T = (337.5) 3 = 48.5 s

Ans. Ans: 48.5 s

407

*4–68. The 1 cylindrical plunger traveling at Vp = 1 0.004 t 2 2 m>s, where t is in seconds, injects a liquid plastic into the mold to make a solid ball. Determine the time needed to fill the mold as a function of the plunger diameter d. Plot the time needed to fill the mold (vertical axis) versus the diameter of the plunger for 10 mm … d … 50 mm. Give values for increments of ∆d = 10 mm. The volume of the ball is V = 43 pr 3.

Vp

d y 10 mm

SOLUTION The control volume is the volume of the liquid plastic contained in the plunger for which its volume changes with time, Fig. a. The volume V0 of the lower portion of the control volume is constant.

75 mm

0 r dV + rpVp>cs # dA = 0 0t L p L cv

cs

Since rp is constant, it can be factored out of the integral. Also, the average velocity will be used. Thus, the above equation becomes 0V rp + rpVAAA = 0 0t Since QA = VAAA,

0.05 m

The volume of the control volume is p V = a d 2 by + V0 4

y

0V p 2 0y = d 0t 4 0t

0y 1 However, = -Vp = ( - 0.004 t 2 ) m>s. The negative sign indicates that Vp is 0t directed in the opposite sense to that of positive y.

d(mm)

10

t(s)

414

1 1 0V p = d 2( - 0.004t 2 ) = ( -0.001pd 2t 2 ) m3 >s 0t 4

20

30

40

50

164

88.9

60.6

45.0

t (s)

400 350 300 250 200 150 100 50 d (mm) 0

10

20

30

40

50

(b)

408

A (a)

V0

*4–68. Continued

The negative sign indicates that the volume is decreasing. Substituting into Eq (1), 1

= - 0.001pd 2t 2 + QA = 0 1

QA = ( 0.001pd 2t 2 ) m3 >s

The volume of the sphere (mold) is Vs =

4 3 4 pr = p(0.075 m)3 = 0.5625 ( 10-3 ) p m3 3 3

The time to fill up the sphere is dt =

Vs ; QA

dt = Lo

t

0.5625 ( 10-3 ) p m3 1

0.001pd 2t 2

1

t 2 dt = 0.5625d - 2 2 3 t 2 = 0.5625d - 2 3 4

t = 0.8929d - 3

Ans.

409

4–69. The pressure vessel of a nuclear reactor is filled with boiling water having a density of rw = 850 kg>m3. Its volume is 185 m3. Due to failure of a pump, needed for cooling, the pressure release valve A is opened and emits steam having a density of rs = 35 kg>m3 and an average speed of V = 400 m>s. If it passes through the 40-mm-diameter pipe, determine the time needed for all the water to escape. Assume that the temperature of the water and the velocity at A remain constant.

A

V

SOLUTION The steam has a steady flow and the density of the water in the pressure vessel is constant since the temperature is assumed to be constant. Here, the control volume is changing since it contains the water in the vessel. 0 r dV + rsV # dA = 0 0t L w L cv

cs

Since rw and rs are constant, they can be factored out from the integrals. Also, the average velocity of the steam will be used. Then

L

V # dA = Vs A.

cs

rw

0 dV + rsVs A = 0 0t L cv

rw

0V + rsVs A = 0 0t

(35 kg>m )(400 m>s) 3 p(0.02 m) rsVs A 0V = = rw 0t 850 kg>m3 3

2

= - 0.02070 m3 >s

4

The negative sign indicates that the volume of water is decreasing. Thus, the time needed for all the water to escape is t =

V 185 m3 1 hr = (8938 s)a = b = 2.48 hr 3 0 V>0t 3600 s 0.02070 m >s

Ans.

Ans: 2.48 hr 410

4–70. The pressure vessel of a nuclear reactor is filled with boiling water having a density of rw = 850 kg>m3. Its volume is 185 m3. Due to failure of a pump, needed for cooling, the pressure release valve is opened and emits steam having a density of rs = 35 kg>m3. If the steam passes through the 40-mm-diameter pipe, determine the average speed through the pipe as a function of the time needed for all the water to escape. Plot the speed (vertical axis) versus the time for 0 … t … 3 h. Give values for increments of ∆t = 0.5 h. Assume that the temperature of the water remains constant.

A

SOLUTION The steam has a steady flow and the densities of the water in the pressure vessel and the steam are constant since the temperature is assumed to be constant. Here the control volume is changing since it contains the water in the vessel. 0 r dV + rsVs # dA = 0 0t L w L cv

cs

Since rw and rs are constants, they can be factored out from the integrals. Also the average velocity of the steam will be used. Then L

Vs # d𝚨 = Vs A.

cs

rw

0 dV + rsVs A = 0 0t L cv

( 35 kg>m3 ) (Vs) 3 p(0.02 m)2 4 rsVsA 0V = = rw 0t 850 kg>m3

0V = 3 - 51.74 ( 10-6 ) Vs 4 m3 >s 0t The negative sign indicates that the volume of water is decreasing. Thus, the time needed for all the water to escape is t = -

V = 0 V>0t

t = e J

185 m3

3 51.74 ( 10-6 ) Vs 4 m3 >s

3.5753 ( 106 ) Vs

R s fa

1 hr b 3600

411

V

4–70. Continued

t =

993.14 Vs

Vs = a

993 b m>s where t is in hrs t

Ans.

The plot of Vs vs t is shown in Fig. a t(hr)

0

0.5

1.0

1.5

2.0

2.5

3.0

Vs(m>s)



1986

993

662

497

397

331

Vs (m s)

2100 1800 1500 1200 900 600 300 t (hr)

0

0.5

1.0

1.5

2.0

2.5

3.0

(a)

Ans: Vs = a 412

993 b m>s, where t is in hrs t

4–71. The wind tunnel is designed so that the lower pressure outside the testing region draws air out in order to reduce the boundary layer or frictional effects along the wall within the testing tube. Within region B there are 2000 holes, each 3 mm in diameter. If the pressure is adjusted so that the average velocity of the air through each hole is 40 m>s, determine the average velocity of the air exiting the tunnel at C. Assume the air is incompressible.

1.30 m

1.10 m

15 m/s

A

B

C

SOLUTION The control volume is fixed since it contains the air in the tunnel. Since the flow is steady, no local changes take place. Also, the density of air is constant (incompressible) and the average velocities will be used. Thus, 0 rdV + rV # dA 0t Lcv Lcs 0 - VAAA + 2000 VBAB + VC AC = 0 -(15 m>s) 3 p(0.65 m)2 4 + 2000(40 m>s) 3 p(0.0015 m)2 4 + VC 3 p(0.55 m)2 4 = 0

Ans.

VC = 20.4 m>s

Ans: 20.4 m>s 413

*4–72. Water flows through the pipe such that it has a parabolic velocity profile V = 3 ( 1 - 100r 2 ) m>s, where r is in meters. Determine the time needed to fill the tank to a depth of h = 1.5 m if h = 0 when t = 0. The width of the tank is 3 m.

200 mm r

h

SOLUTION

2m

The control volume is the volume of the water in the tank. Thus, its volume changes with time 0 rwdV + rwV # dA = 0 0t Lcv Lcs Since rw is constant (incompressible), it can be factor out of the integrals. rw

0V + rw V # dA = 0 0t Lcs

0V + Qout - Qin = 0 0t

(1)

Here, Qout = 0 and Qin =

LA

vdA =

L0

0.1m

3 ( 1 - 100r 2 ) (2prdr) = (0.015p) m3 >s

vdA can also be determined by computing the volume under the LA velocity profile, which in this case is a paraboloid.

The integral

LA

vdA =

1 2 1 pr h = p(0.1 m)2 (3 m>s) = (0.015p) m3 >s 2 2

Also, the volume of the control volume at a particular instant is V = (2m)(3m)(h) = 6h Thus, dV dh = 6 dt dt Substituting these results into Eq (1) 6

dh - 0.015p = 0 dt dh = 0.0025p dt

L0

1.5 m

dh = 0.0025p

L0

t

dt

1.5 = 0.0025pt t = (190.99 s)a = 3.18 min

1 min b 60 s

Ans.

414

4–73. Ethyl alcohol flows through pipe A with an average velocity of 4 ft>s, and oil flows through pipe B at 2 ft>s. Determine the average density at which the mixture flows through the pipe at C. Assume uniform mixing of the fluids occurs within a 200 in3 volume of the pipe assembly. Take rea = 1.53 slug>ft 3 and ro = 1.70 slug>ft 3.

4 in.

C

A

6 in.

B 3 in.

SOLUTION The fluids are assumed to be incompressible, and so their volumes remain constant. Also the volume within the pipe is constant. Therefore - VAAA - VBAB + VCAC = 0 - (4 ft>s) c pa

2 2 2 2 1.5 3 ft b d - (2 ft>s) c pa ft b d + VC c pa ft b d = 0 12 12 12

VC = 2.278 ft>s

Applying the conservation of mass for steady flow. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - reaVAAA - roVB AB + rCVCAC = 0 Thus rC =

rC =

( 1.53 slug>ft 3 ) (4 ft>s ) c pa

rC = 1.57 slug>ft 3

reaVAAA + roVBAB VcAc 2 2 2 1.5 ft b d + ( 1.70 slug>ft 3 ) (2 ft>s ) c p a ft b d 12 12

(2.278 ft>s)ap a

2 3 ft b b 12

Ans.

Ans: 1.57 slug>ft 3 415

4–74. Ethyl alcohol flows through pipe A at 0.05 ft 3 >s, and oil flows through pipe B at 0.03 ft 3 >s. Determine the average density of the two fluids as the mixture flows through the pipe at C. Assume uniform mixing of the fluids occurs within a 200 in3 volume of the pipe assembly. Take rea = 1.53 slug>ft 3 and ro = 1.70 slug>ft 3.

4 in.

C

A

6 in.

B 3 in.

SOLUTION The fluids are assumed to be incompressible, so their volumes remain constant. Also, the volume within the pipe is constant. Therefore - QA - QB + QC = 0 - 0.05 ft 3 >s - 0.03 ft 3 >s + QC = 0 QC = 0.08 ft 3 >s

Applying the conservation of mass for steady flow 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - rea QA - roQB + rmQC = 0 0 - ( 1.53 slug>ft

3

)( 0.05 ft3 >s ) - ( 1.70 slug>ft 3 )( 0.03 ft3 >s ) + rC ( 0.08 ft3 >s ) = 0 rC = 1.59 slug>ft 3

Ans.

Ans: 1.59 slug>ft 3 416

4–75. Water flows into the tank through two pipes. At A the flow is 400 gal>h, and at B it is 200 gal>h when d = 6 in. Determine the rate at which the level of water is rising in the tank. There are 7.48 gal>ft 3.

3 ft

d 8 in.

B

A

SOLUTION Control volume. The deformable control volume shown in Fig. a will be considered. If the initial control volume is V0, then its volume at any given instant is V = V0 + p(1.5 ft)2y = Continuity Equation. Realizing that Q = QA

3 V0

+ 2.25py 4 ft 3

y Initial water level

V # dA and

Lcs

VA

3

gal

1.5 ft

1 ft 1h = a400 ba ba b = 0.01485 ft 3 >s h 7.48 gal 3600 s

AB

AA

VB

(a)

1 ft 3 1h QB = a200 ba ba b = 0.007427 ft 3 >s h 7.48 gal 3600 s gal

Since the density of water is constant, rw c

0 dV + V # dA d = 0 0t Lcv Lcs

0 ( V + 2.25py ) - 0.01485 ft3 >s - 0.007427 ft3 >s = 0 0t 0 2.25p

0y = 0.02228 0t

0y = 3.15 ( 10-3 ) ft>s 0t

Ans.

Ans: 3.15 ( 10 - 3 ) ft>s 417

*4–76. Water flows into the tank through two pipes. At A the flow is 400 gal>h. Determine the rate at which the level of water is rising in the tank as a function of the discharge of the inlet pipe B. Plot this rate (vertical axis) versus the discharge for 0 … QB … 300 gal>h. Give values for increments of ∆QB = 50 gal>h. There are 7.48 gal>ft 3.

3 ft

d 8 in.

B

A

SOLUTION The deformable control volume shown in Fig. a will be considered. If the initial volume of this control volume is V0, then its volume at any given instant is

1.5 ft y

V = V0 + p(1.5 ft)2y = (V0 + 2.25py) ft 3

Initial water level

The discharges at A and B are QA = a400 aQB

gal h

ba

gal h

ba

1 ft 3 1h ba b = 0.01485 ft 3 >s 7.48 gal 3600 s

1 ft 3 1h ba b = 7.48 gal 3600 s

Since the density of water is constant and Q =

Lcs

V # dA,

0y = 0t

3 2.10 ( 10-3 )

The plot of

0y ( 10 - 3 ft>s ) 0t

(a) y t

3

0V - QA - QB b = 0 0t

2 1

0y = 0.01485 + 37.14 ( 10-6 ) QB 0t

0

50

100

150 (b)

+ 5.25 ( 10-6 ) QB 4 ft>s where QB is in gal>h

Ans.

0y vs QB is shown in Fig. b. 0t

QB(gal>h)

VB

4

0 ( V + 2.25 py ) = QA + QB 0t 0 2.25p

AB

AA

3 37.14 ( 10-6 ) QB 4 ft3 >s

0 rdV + rV # dA = 0 0t Lcv Lcs ra

VA

0

50

100

150

200

250

300

2.10

2.36

2.63

2.89

3.15

3.41

3.68

418

200

250

300

QB(gal h)

4–77. The piston is traveling downwards at Vp = 3 m>s, and as it does, air escapes radially outward through the entire bottom of the cylinder. Determine the average speed of the escaping air. Assume the air is incompressible. Vp 50 mm 2 mm

SOLUTION

Initial air level

Control Volume. The deformable control volume shown in Fig. a will be considered. If the initial control volume is V0, then its volume at any given instant is V = V0 - p(0.025 m)2y =

3 V0

y

- 0.625 ( 10-3 ) py 4 m3

Continuity Equation. Since the air is assumed to be incompressible, its density is constant. rc

A

V

(a)

0 dV + V # dA d = 0 0t Lcv Lcs

0 3 V0 - 0.625 ( 10-3 ) py 4 + V 32p(0.025 m)(0.002 m)4 = 0 0t - 0.625 ( 10-3 ) p

0y + 0.1 ( 10-3 ) pV = 0 0t

V = 6.25 However,

dy dt

dy = 3 m>s. Then dt V = 6.25 ( 3 m>s ) = 18.8 m>s

Ans.

Ans: 18.8 m>s 419

4–78. The piston is travelling downwards with a velocity Vp, and as it does, air escapes radially outward through the entire bottom of the cylinder. Determine the average velocity of the air at the bottom as a function of Vp. Plot this  average velocity of the escaping air (vertical axis) versus the velocity of the piston for 0 … Vp … 5 m>s. Give values for increments of ∆Vp = 1 m>s. Assume the air is incompressible.

Vp 50 mm 2 mm

SOLUTION The deformable control volume shown in Fig. a will be considered. If the initial control volume is V0, then its volume at any given instant is V = V0 - p(0.025 m)2y =

3 V0

- 0.625 ( 10-3 ) py 4 m3

Initial air level y

Since the air is assumed to be incompressible, its density is constant. 0 rc dV + V # dA d = 0 0t Lcv Lcs

0 3 V0 - 0.625 ( 10-3 ) py 4 + V 3 2p(0.025 m)(0.002 m) 4 = 0 0t

A

V

(a)

0y + 0.1 ( 10-3 ) pV = 0 0t 0y V = 6.25 0t

-0.625 ( 10-3 ) p

However,

0y = Vp. Then 0t

Ans.

V = (6.25 Vp) m>s

The plot of V vs Vp is shown in Fig. b Vp(m>s)

0

1

2

3

4

5

V(m>s)

0

6.25

12.5

18.75

25.0

31.25

V(m s ) 40 30 20 10

0

1

2

3

4

5

Vp (m s)

(b)

Ans: V = (6.25Vp) m>s 420

4–79. The cylindrical syringe is actuated by applying a force on the plunger. If this causes the plunger to move forward at 10 mm>s, determine the average velocity of the fluid passing out of the needle.

10 mm/s 1.5 mm

20 mm

SOLUTION Control Volume. The deformable control volume is shown in Fig. a. If the volume of the control volume is initially V0 then at any instant its volume is p V = V0 - (0.02 m)2x = 3 V0 - 0.1 ( 10-3 ) px 4 m3 4 Continuity Equation. With the fluid assumed to be incompressible, r is constant. since VA p and AA are in the same sense, QA = VAAA = VA c (0.0015 m)2 d = 0.5625 ( 10-6 ) pVA. 4

AA VA A

x (a)

0 rdV + pV # dA = 0 0t Lcv Lcs 0 rw c (V) + VA AA d = 0 0t

0 3 V - 0.1 ( 10-3 ) px 4 + 0.5625 ( 10-6 ) pVA = 0 0t 0 -0.1 ( 10-3 ) p

However,

0x + 0.5625 ( 10-6 ) pVA = 0 0t

0x = 10 mm>s = 0.01 m>s 0t Then

3 0.1 ( 10-3 ) p 4 (0.01)

+ 0.5625 ( 10-6 ) pVA = 0 Ans.

VA = 1.78 m>s

Ans: 1.78 m>s 421

*4–80. Water enters the cylindrical tank at A with an average velocity of 2 m>s , and oil exits the tank at B with an average velocity of 1.5 m>s. Determine the rates at which the top level C and interface level D are moving. Take ro = 900 kg>m3.

C

200 mm

B D

150 mm A

SOLUTION We will consider two control volumes separately namely one contains water and the other contains oil in the tank their volume changes with time, Fig. a. Here, the densities of water and oil are constant and the average velocities will be used.

1.2 m

0 rdV + rV # dA = 0 0t Lcv L

oil

For water,

yt

0Vw rw - rwVA AA = 0 0t

ys

0Vw - VAAA = 0 0t

(1)

water

Here,

3 p(0.6 m)2 4 ys

Vw =

1.2 m

= 0.36 pys

0ys 0Vw = 0.36p 0t 0t

(a)

Substitute this result into Eq (1) 0.36 p

0ys - ( 2 m>s ) 3 p(0.075 m)2 4 = 0 0t Vp =

0ys = 0.0312 m>s 0t

Ans.

Positive sign indicates that the separation level is rising. For the oil, 0V0 + roVB AB = 0 0t 0V0 + VB AB = 0 0t

ro

Here

V0 =

3 p(0.6 m)2 4 (yt

(2)

- ys)

= 0.36p(yt - ys)

0yt 0ys 0V0 = 0.36p a b 0t 0t 0t = 0.36 p a

Substituting this result into Eq. (2), 0.36 p a

0yt - 0.03125 m>s b 0t

0yt - 0.03125 b + ( 1.5 m>s ) 3 p(0.1 m)2 4 = 0 0t

VC =

0yt = - 0.0104 m>s 0t

The negative sign indicates that the top level descends. 422

Ans.

yt – ys

4–81. The tank contains air at a temperature of 20°C and absolute pressure of 500 kPa. Using a valve, the air escapes with an average speed of 120 m>s through a 15@mm-diameter nozzle. If the volume of the tank is 1.25 m3, determine the rate of change in the density of the air within the tank at this instant. Is the flow steady or unsteady?

SOLUTION

From Appendix A, the gas constant for air is R = 286.9 J>(kg # K)

V = 120 m s

p = rRT

500 ( 103 ) N>m2 = r ( 286.9 J>(kg # K) ) (20°C + 273)

A

r = 5.948 kg>m3 Control Volume. The control volume is shown in Fig. a. The control volume does not change, but the density of the air changes and therefore results in local changes. Continuity Equation. 0 rdV + r V # dA = 0 0t Lcv Lcv

(a)

0r (V) + rVA = 0 0t 0r ( 1.25 m3 ) + ( 5.948 kg>m3 )( 120m>s ) 3 p(0.0075 m)2 4 = 0 0t 0r = - 0.101 kg> ( m3 # s ) 0t

Ans.

The negative sign indicates that the density of the air is decreasing. Flow is unsteady, since the pressure within the tank is decreasing and this affects the flow.

Ans: - 0.101 kg> ( m3 # s ) , unsteady 423

4–82. The natural gas (methane) and crude oil mixture enters the separator at A at 6 ft 3 >s and passes through the mist extractor at B. Crude oil flows out at 800 gal>min through the pipe at C, and natural gas leaves the 2-in-diameter pipe at D at VD = 300 ft>s. Determine the specific weight of the mixture that enters the separator at A. The process takes place at a constant temperature of 68°F. Take ro = 1.71 slug>ft 3, rme = 1.29 1 10-3 2 slug>ft 3. Note 1 ft 3 = 7.48 gal.

D

VD

B

A

C

SOLUTION The control volume is fixed which is the volume of the crude oil and natural gas contained in the tank. Here, the flow is steady. Thus, no local changes take place. Also, the densities of the gas oil mixture, gas and oil separation are constant, and the average velocities will be used. 0 rdV + r V # dA = 0 0t Lcv Lcs (1)

0 - rmixVAAA + rcoVC AC + rmeVD AD = 0

From Appendix A, rco = 1.71 slug>ft and rm = 1.29 ( 10 ) slug>ft when gal 1 ft 3 1 min T = 68°F. Also, QC = VCAC = a800 ba ba b = 1.783 ft 3 >s and min 7.48 gal 60 s -3

3

3

QA = VAAA = 6 ft 3 >s. Substituting these results into Eq. (1), +

- rmix(6 ft 3 >s) + ( 1.71 slug>ft 3 )( 1.783 ft 3 >s )

3 1.29 ( 10-3 ) slug>ft3 4 ( 300 ft>s ) c p a rmix = 0.5094 slug>ft 3

2 1 ft b d = 0 12

gmix = rmix g = ( 0.5094 slug>ft 3 )( 32.2 ft>s2 ) = 16.4 lb>ft 3

Ans.

Ans: 16.4 lb>ft 3 424

4–83. The natural gas (methane) and crude oil mixture having a density of 0.51 slug>ft 3 enters the separator at A at 6 ft 3 >s, and crude oil flows out through the pipe at C at 800 gal>min. Determine the average velocity of the natural gas that leaves the 2-in.-diameter pipe at D. The process takes place at a constant temperature of 68°F. Take ro = 1.71 slug>ft 3, rme = 1.29 1 10-3 2 slug>ft 3. Note 3 1 ft = 7.48 gal.

D

VD

B

A

C

SOLUTION The control volume is fixed which is the volume of the mixture of crude oil and natural gas contained in the tank. Here, the flow is steady. Thus, no local changes take place. Also the densities of the oil gas mixture, gas and oil separation, are constant and average velocities will be used. 0 rdV + rV # dA = 0 0t Lcv Lcs (1)

0 - rmixVA AA + rCOVC AC + rmeVD AD = 0

From Appendix A, rCO = 1.71 slug>ft 3 and rme = 1.29 ( 10-3 ) slug>ft 3 at T = 68°F 1 ft 3 1 min ba b = 1.783 ft 3 >s and QA = VAAA = 6 ft 3 >s. min 7.48 gal 60 s Substituting these results into Eq 1, Also, QC = a800

gal

ba

- ( 0.51 slug>ft 3 )( 6 ft 3 >s ) + ( 1.71 slug>ft 3 )( 1.783 ft 3 >s ) +

3 1.29 ( 10-3 ) slug>ft3 4 ( VD ) c p a

2 1 ft b d = 0 12

Ans.

VD = 422 ft>s

Ans: 422 ft>s 425

*4–84. The cylindrical storage tank is being filled using a pipe having a diameter of 3 in. Determine the rate at which the level in the tank is rising if the flow into the tank at A is 40 gal>min . Note 1 ft 3 = 7.48 gal.

10 ft

A

15 ft h

SOLUTION The control volume is the volume of oil contained in the tank, which changes with time. Here, the density of the oil is constant and the average velocity will be used 0 r dV + r V # dA = 0 0t Lcv Lcs r Since QA = VAAA = a40

gal min

ba

0V - rVAAA = 0 0t

1 ft 3 1 min ba b = 0.08913 ft 3 >s. Then 7.48 gal 60 s 0V = 0.08913 0t

(1)

Here, the volume of the control volume at a particular instant is V = pr 2h = p(5 ft)2h = 25ph 0V 0h = 25p 0t 0t Substituting this result into Eq (1) 0h 25p = 0.08913 0t 0h = 1.13 ( 10-3 ) ft>s 0t

Ans.

426

4–85. The cylindrical storage tank is being filled using a pipe having a diameter of D. Determine the rate at which the level is rising as a function of D if the velocity of the flow into the tank is 6 ft>s. Plot this rate (vertical axis) versus the diameter for 0 … D … 6 in. Give values for increments of ∆D = 1 in.

10 ft

A

15 ft

SOLUTION h

The control volume is the volume of oil contained in the tank of which its volume changes with time. Here, the density of the oil is constant and the average velocities will be used. 0 rdV + rV # dA = 0 0t Lcv Lcs ra

0V - VAAA b = 0 0t

p D 2 0V = VAAA = ( 6 ft>s ) c a b d 0t 4 12

0V = 0.03272 D2 0t Here, the volume of the control volume at a particular instant is

(1)

V = pr 2h = p(5 ft)2h = 25ph 0V 0h = 25p 0t 0t Substituting this result into Eq (1), 25p 0h = 0t The plot of

0h = 0.03272 D2 0t

3 0.417 ( 10-3 ) D2 4 ft>s where D is in inches.

0h vs D is shown in Fig. a. 0t

D(in.)

0

1

2

3

4

5

6

0h ( 10 - 3 ) ft>s 0t

0

0.417

1.67

3.75

6.67

10.4

15.0

Ans. h –3 (10 ) ft s t 16 14 12 10 8 6 4 2

0

1

2

3

4

5

6

D (in.)

(a)

Ans: 0h = 0t 427

3 0.417 ( 10 - 3 ) D 2 4 ft>s, where D is in in.

4–86. Air is pumped into the tank using a hose having an inside diameter of 6 mm. If the air enters the tank with an average speed of 6 m>s and has a density of 1.25 kg>m3, determine the initial rate of change in the density of the air within the tank. The tank has a volume of 0.04 m3.

SOLUTION Control Volume. The control volume is shown in Fig. a. The control volume does not change but the density of the air changes and therefore results in local changes.

V=6m s

Continuity Equation.

A

0 rdV + rV # dA = 0 0t Lcv Lcs 0r (V) - rVA = 0 0t 0r ( 0.04 m3 ) - ( 1.25 kg>m3 )( 6 m>s ) 3 p(0.003 m)2 4 = 0 0t 0r = 0.00530 kg> ( m3 # s ) 0t

(a)

Ans.

Ans: 0.00530 kg> ( m3 # s ) 428

4–87. As air flows over the plate, frictional effects on its surface tend to form a boundary layer in which the velocity profile changes from that of being uniform to one that is parabolic, defined by u = 3 1000y - 83.33 1 103 2 y2 4 m>s, where y is in meters, 0 … y 6 6 mm. If the plate is 0.2 m wide and this change in velocity occurs within the distance of 0.5 m, determine the mass flow through the sections AB and CD. Since these results will not be the same, how do you account for the mass flow difference? Take r = 1.226 kg>m3.

3 m/s

3 m/s A

u

C y

B

6 mm

D

0.5 m

SOLUTION Mass Flow Rate. For section AB, since r is constant and the velocity has a constant magnitude, # mAB = rVABAAB = ( 1.226 kg>m3 ) (3 m>s) 30.006 m(0.2 m) 4 = 0.00441 kg>s = 4.41 g>s

Ans.

For section CD, since the velocity is a function of y, a differential element of thickness dy, which has an area dA = bdy = (0.2 m)dy, is chosen. Thus, # mCD = r

L

( VAC )avg. V=3 VAB = 3 m s

AAC

AAB

(VCD )avg. ACD

(a)

udA

= ( 1.226 kg>m3 ) a

L0

0.006 m

3 1000y

= 0.2452 3 500y2 - 27.78 ( 103 ) y3 4 `

- 83.33 ( 103 ) y2 4 m>s b(0.2 m)dy

0.006 m 0

Ans. # # To satisfy continuity the difference between mAB and mCD requires that mass flows through the control surface AC as indicated on the control volume in Fig. a. = 0.00294 kg>s = 2.94 g>s

What this means is that the streamlines in fact cannot be horizontal, as the figure implies. The fluid velocity must have a vertical component in addition to the horizontal one.

Ans: 2.94 g>s 429

*4–88. Kerosene flows into the rectangular tank through pipes A and B, at 3 ft>s and 2 ft>s, respectively. It exits at C at a constant rate of 1 ft>s. Determine the rate at which the surface of the kerosene is rising. The base of the tank is 6 ft by 4 ft.

2 ft/s

4 in.

3 ft/s

B

A

8 in.

6 ft C y

SOLUTION

4 ft

The control volume is the volume of the kerosene in the tank. Thus its volume changes with time. 0 r dV + rkeV # dA = 0 0t Lcv ke Lcs Since rke is constant (incompressible), it can be factored out of the integral. rke

0V + rke V # dA = 0 0t Lcs

Here, we will use the average velocities. dV - VAAA - VB AB + VC AC = 0 dt 2 2 dV p 4 p 8 p - (3 ft>s) c a ft b d - ( 2 ft>s ) c a ft b d + (1 ft>s) c (1 ft)2 d = 0 dt 4 12 4 12 4

dV = 0.1745 ft 3 >s dt

(1)

The volume of the control volume at a particular instant is V = (6 ft)(4 ft)y = (24y) ft 3 Thus 0y 0V = 24 0t 0t Substituting this result into Eq (1), 24

0y = 0.1745 0t 0y = 0.00727 ft>s 0t

Ans.

430

12 in.

1 ft/s

4–89. Kerosene flows into the 4-ft-diameter cylindrical tank through pipes A and B, at 3 ft>s and 2 ft>s, respectively. It exists at C at a constant rate of 1 ft>s. Determine the time required to fill the tank if y = 0 when t = 0.

2 ft/s

4 in.

3 ft/s

B

A

8 in.

6 ft C

12 in.

1 ft/s

y

SOLUTION

4 ft

The control volume is the volume of the kerosene in the tank. Thus, its volume changes with time. 0 r dV + rkeV # dA = 0 0t Lcv ke Lcs Since rke is constant (incompressible), it can be factored out of the integral 0V + rke V # dA = 0 0t Lcs Here, we will use the average velocities rke

dV - VAAA - VB AB + VC AC = 0 dt 2 2 p 4 p 8 p dV - ( 3 ft>s ) c a ft b d - ( 2 ft>s ) c a ft b d + ( 1 ft>s ) c (1 ft)2 d = 0 dt 4 12 4 12 4

dV = 0.1745 ft 3 >s dt

(1)

The volume of the control volume at a particular instant is p V = (4 ft)2y = (4py) ft 3 4 Thus, dy dV = 4p dt dt Substituting this result into Eq (1), 4p L0

dy = 0.1745 dt

6 ft

dy = 0.01389

L0

t

dt

6 = 0.01389t t = (432.5) a

1 min b = 7.20 min. 60 s

Ans.

Ans: 7.20 min 431

4–90. The conical shaft is forced into the conical seat at a constant speed of V0. Determine the average velocity of the liquid as it is ejected from the horizontal section AB as a function of y. Hint: The volume of a cone is V = 13pr 2h.

V

R

B

A H

H

y

SOLUTION V

Control Volume. The deformable control volume shown in Fig. a will be considered. y r R = ;r = y R H H Then, the volume of the control volume at any instant is V =

A cos

A y

H

1 1 R 2 pR2 ( H3 - y3 ) pR2H - p a yb y = 3 3 H 3H2

Continuity Equation. Since the density of the liquid is constant, rc

R

0 dV + V # dA d = 0 0t Lcv Lcs

r

0 V + V(A cos u) = 0 0t

H

y

0 pR2 R 2 ( H3 - y3 ) d + V c pc R2 - a yb d cos u d = 0 c 2 0t 3H H 2

(a)

2

dy pR pR ( - 3y2 ) + V c 2 ( H2 - y2 ) d cos u = 0 2 dt 3H H V = a

However,

y2

dy b (sec u) H - y2 dt 2

dy 2H2 + R2 = V0 and sec u = Then, dt H V = a

y2 2

2

H - y

V = V0

bV0

2H2 + R2 H

y2 2H2 + R2

Ans.

H(H2 - y2)

Ans: V = V0

432

y2 2H2 + R2

H 1 H2 - y2 2

4–91. The 0.5-m-wide lid on the barbecue grill is being closed at a constant angular velocity of v = 0.2 rad>s, starting at u = 90°. In the process, the air between A and B will be pushed out in the radial direction since the sides of the grill are covered. Determine the average velocity of the air that emerges from the front of the grill at the instant u = 45° rad. Assume that the air is incompressible.

0.4 m

A v ! 0.2 rad/s u B

SOLUTION The flow is considered one dimensional since its velocity is directed in the radial direction only. The control volume is shown in Fig. a, and its volume changes with time. At a particular instant it is V =

0.5 m

1 2 1 r ub = (0.4 m)2u(0.5 m) = 0.04u m3 2 2

Thus, dV du = 0.04 dt dt

0.4 m

However, (a)

du = v = - 0.2 rad>s . dt Then dV = - 0.008 m3 >s dt

Notice that negative sign indicates that the volume is decreasing with time. The opened control surface is shown shaded in Fig. a. Its area is A = rub = (0.4 m)u(0.5 m) = 0.2u Thus, 0 r dV + raV # dA = 0 0t Lcv a Lcs Since here air is assumed to be incompressible, ra is constant. Also, the average velocity of the air is directed radially outward. Thus, it always acts perpendicular to the opened control surface. Hence, the above equation becomes ra

0V + raVaA = 0 0t 0V = - VaA 0t

Va = -

0V>0t A

= -

-0.008 0.04 = m>s 0.2u u

When u = 45° = VA =

p rad, 4

0.04 = 0.0509 m>s p>4

Ans.

433

Ans: 0.0509 m>s

*4–92. The cylinder is pushed down into the tube at a rate of V = 5 m>s. Determine the average velocity of the liquid as it rises in the tube. V

150 mm

y

200 mm

SOLUTION Control Volume. The deformable control volume shown in Fig. a will be considered. If the initial water level in the tube is y0, Fig. b, then the control volume at any instant is V = p(0.1 m)2 ( y0 - y1 ) + p(0.1 m)2 ( y1 + y2 ) - p(0.075 m)2 ( y1 + y2 )

0.075 m

= p ( 0.01y0 - 0.005625y1 + 0.004375y2 ) Continuity Equation.

y2

0 r dV + rwVf>cs # dA = 0 0t Lcv w Lcs

y1

Since no water enters or leaves the control volume at any instant, Then,

rwVf>cs # dA = 0.

Lcs

y0 – y1

0 r dV = 0 0t Lcv w

(b)

0 rw V = 0 0t 0 3 p ( 0.01y0 - 0.005625y1 + 0.004375y2 ) 4 = 0 0t -0.005625

0y1 0y2 + 0.004375 = 0 0t 0t

0y2 0y1 = 1.2857 0t 0t However,

y0

0y1 = Vr = 5 m>s . Then 0t 0y2 = 1.2857(5 m>s) = 6.43 m>s 0t

Ans.

434

0.1 m (a)

4–93. Determine the speed V at which the cylinder must be pushed down into the tube so that the liquid in the tube rises with an average velocity of 4 m>s. V

150 mm

y

200 mm

SOLUTION Control Volume. The deformable control volume shown in Fig. a will be considered. If the initial water level in the tube is y0, Fig. b, then the control volume at any instant is V = p(0.1 m)2 ( y0 - y1 ) + p(0.1 m)2 ( y1 + y2 ) - p(0.075 m)2 ( y1 + y2 ) = p ( 0.01y0 - 0.005625y1 + 0.004375y2 ) Continuity Equation.

0 r dV + rwVf>cs # dA = 0 0t Lcv w Lcs

Since no water enters or leaves the control volume at any instant, Then,

rwVf>cs # dA = 0. Lcs

0 r dV = 0 0t Lcv w rw

0 V = 0 0t

0 3 p ( 0.01y0 - 0.005625y1 + 0.004375y2 ) 4 = 0 0t

- 0.005625

0y1 0y2 + 0.004375 = 0 0t 0t 0y1 0y2 = 0.7778 0t 0t

However,

0y1 0y2 = Vr and = 4 m>s . Then 0t 0t Ans.

Vr = 0.7778(4 m>s) = 3.11 m>s

Ans: 3.11 m>s 435

4–94. The tank originally contains oil. If kerosene having a mass flow of 0.2 kg>s enters the tank at A and mixes with the oil, determine the rate of change of the density of the mixture in the tank if 0.28 kg>s of the mixture exits the tank at the overflow B. The tank is 3 m wide.

200 mm B 150 mm

2.5 m

A

3m

SOLUTION Control Volume. The control volume is shown in Fig. a. The control volume does not change but the density of the mixture changes and therefore results in local changes. Continuity Equation. Since the liquids are incompressible and the volume is constant, we have 0 rdV + rV # dA = 0 0t Lcv Lcv

AB

0r (3 m)(3 m)(2.5 m) - 0.2 kg>s + 0.28 kg>s = 0 0t 0r = - 0.00356 kg> ( m3 # s ) 0t

VB

Ans.

The negative sign indicates that the density of the mixture is decreasing. VA

AA

(a)

Ans: - 0.00356 kg>(m3 # s) 436

4–95. Benzene flows through the pipe at A with an average velocity of 4 ft>s, and kerosene flows through the pipe at B with an average velocity of 6 ft>s. Determine the required average velocity VC of the mixture from the tank at C so that the level of the mixture within the tank remains constant at y = 3 ft. The tank has a width of 3 ft. What is the density of the mixture leaving the tank at C? Take rb = 1.70 slug>ft 3 and rke = 1.59 slug>ft 3.

4 ft

0.3 ft VA A y

0.4 ft

0.2 ft

VC

VB C

B

SOLUTION The density of the mixture can be determined from rm =

rbQA + rkeQB QA + QB

Here, QA = VA AA = (4 ft>s) 3 p(0.15 ft)2 4 = 0.09p ft 3 >s

QB = VB AB = (6 ft>s) 3 p(0.1 ft)2 4 = 0.06p ft 3 >s

Then rm =

AA VA = 4 ft s

( 1.70 slug>ft 3 )( 0.09p ft3 >s ) + ( 1.59 slug>ft 3 )( 0.06p ft3 >s )

= 1.656 slug>ft 3

0.09p ft 3 >s + 0.06p ft 3 >s

VC AC AB V = 6 ft s B

Ans.

Here, the control volume is fixed since it contains the mixture of which the volume does not change. The flow is steady thus there are no local changes. Here, we will use the average velocities,

4 ft

0 rdV + rV # dA = 0 dt Lcv Lcs 0 - rbVAAA - rkeVBAB + rmVcAc = 0 - ( - 1.70 slug>ft 3)(0.09 pft 3 >s) - (1.59 slug>ft 3 )( 0.06p ft 3 >s ) + (1.656 slug>ft 3)(VC) 3 p(0.2 ft)2 4 = 0

Ans.

VC = 3.75 ft>s

Ans: rm = 1.656 slug>ft 3 VC = 3.75 ft>s 437

*4–96. Benzene flows through the pipe at A with an average velocity of 4 ft>s, and kerosene flows through the pipe at B with an average velocity of 6 ft>s. If the average velocity of the mixture leaving the tank at C is VC = 5 ft>s, determine the rate at which the level in the tank is changing. The tank has a width of 3 ft. Is the level rising or falling? What is the density of the mixture leaving the tank at C? Take rb = 1.70 slug>ft 3 and rke = 1.59 slug>ft 3.

4 ft

0.3 ft VA A y

0.4 ft

0.2 ft

VC

VB C

B

SOLUTION The density of the mixture can be determined from rm =

rbQA + rkeQB QA + QB

Initial mixture level

Here, QA = VA AA = (4 ft>s) 3 p(0.15 ft)2 4 = 0.09p ft 3 >s

4 ft (a)

( 1.70 slug>ft 3 )( 0.09p ft3 >s ) + ( 1.59 slug>ft 3 )( 0.06p ft3 >s )

= 1.656 slug>ft 3

0.09p ft 3 >s + 0.06p ft 3 >s

Ans.

Here, the volume of the control volume changes with time since it contains the mixture in the tank. Its volume is V = (4 ft)(3 ft)y = 12y 0y 0V = 12 0t 0t Here, the densities of the liquids are constant and the average velocity will be used. rm

0V - rbVAAA - raVBAB + rmVCAC = 0 0t 0y 0t

( 1.656 slug>ft 3 ) a12 b - ( 1.70 slug>ft 3 )( 0.09p ft3 >s )

- ( 1.59 slug>ft 3 )( 0.06p ft 3 >s ) + ( 1.656 slug>ft 3 )( 0.2p ft 3 >s ) = 0 0y = -0.0131 ft>s 0t

The negative sign indicates the level of the mixture is falling.

438

AB V = 6 ft s B

VC = 6 ft s

QC = VC AC = (5 ft>s) 3 p(0.2 ft)2 4 = 0.2p ft 3 >s

rm =

AA VA = 4 ft s

AC

QB = VB AB = (6 ft>s) 3 p(0.1 ft)2 4 = 0.06p ft 3 >s

Then

y

Ans.

4–97. The three pipes are connected to the water tank. If the average velocities of water flowing through the pipes are VA = 4 ft>s, VB = 6 ft>s, and VC = 2 ft>s, determine the rate at which the water level in the tank changes. The tank has a width of 3 ft.

4 ft 4 ft/s

6 ft/s

A 0.3 ft

B

0.5 ft

0.4 ft 2 ft/s C

SOLUTION Control Volume. The deformable control volume shown in Fig. a will be considered. V = (4 ft)(3 ft)y = (12y) ft 3

4 ft VA = 4 ft s

Continuity Equation. Since water has a constant density, 0 dV + V # dA d = 0 rw c 0t Lcv Lcs

AA

0 V - VA AA - VB AB + VC AC = 0 0t 0 (12y) - (4 ft>s) 3 p(0.15 ft)2 4 - (6ft>s) 3 p(0.25 ft)2 4 + (2 ft>s) 3 p(0.2 ft)2 4 = 0 0t 0y 12 = 1.2095 0t

0y = 0.101 ft>s 0t

VB = 6 ft s Initial water level

AB

y

AC VC = 2 ft s (a)

Ans.

Ans: 0.101 ft>s 439

4–98. The 2-m-diameter cylindrical emulsion tank is being filled at A with cyclohexanol at an average rate of VA = 4 m>s and at B with thiophene at an average rate of VB = 2 m>s. Determine the rate at which the depth increases as a function of depth h.

40 mm VA

A

6m

60 mm B VB

2m h

SOLUTION The control volume considered is the volume of liquid mixture contained in the sector of the tank (shown shaded in Fig. a) which changes with time. The volume of this control volume at a particular instant is 1 1 u u V = e (1 m)2u - 3 2(1 m) sin (1 m) cos 4 f(6 m) = 3(u - sin u) 2 2 2 2 0u 0u 0u 0V = 3a - cos u b = 3(1 - cos u) 0t 0t 0t 0t

1m

u u 1 - h - 1 and cos = = 1 - h. Thus 2 2 1 cos u = 2(1 - h)2 - 1. Then

However, cos u = 2 cos2

0V = 351 0t

Here,

3 2( 1

- h)

cos

2

du du - 14 6 = 6 ( 2h - h2 ) dt dt

2

1m–h

2 h

(1) (a)

u = 1 - h 2

1 u du dh a - sin b = 2 2 dt dt du 2 = dt sin

However, sin

u 2

dh dt

212 - (1 - h)2 u = = 22h - h2 . Thus, 2 1 du 2 dh = 2 dt dt 22h - h

Substituting this result into Eq. (1),

dV 2 dh dh = 6 ( 2h - h2 ) ° ¢ = 1222h - h2 2 dt dt dt 22h - h

(2)

Since the liquids are assumed incompressible, there volume remain the same. Thus, a

dV b = VAAA + VBAB = (4 m>s)(p)(0.02 m)2 + (2 m>s)(p)(0.03 m)2 dt

From Eq. (2),

= 0.010681 m3 >s

0.010681 = 1222h - h2

dh dt

0.890 ( 10-3 ) dh = m>s dt 22h - h2

Ans.

Ans: 0.890 ( 10 - 3 ) 22h - h2

440

m>s

4–99. The 2-m-diameter cylindrical emulsion tank is being filled at A with cyclohexanol at an average rate of VA = 4 m>s and at B with thiophene at an average rate of VB = 2 m>s. Determine the rate at which the depth of the mixture is increasing when h = 1 m. Also, what is the average density of the mixture? Take rcy = 779 kg>m3, and rt = 1051 kg>m3.

40 mm VA

A

6m

60 mm B VB

2m h

SOLUTION See the solution to part 4–98. When h = 1 m, 0.890 ( 10-3 ) dh = = 0.890 ( 10-3 ) m>s 2 dt 22(1) - 1

Ans.

The average density of the mixture is ravg = =

rcy(QA) + rt(QB) QA + QB 779 kg>m3(p)(0.02 m)2 ( 4 m>s ) + 1051 kg>m3(p)(0.03 m)2 ( 2 m>s ) p(0.02 m)2 ( 4 m>s ) + p(0.03 m)2 ( 2 m>s )

= 923 kg>m3

Ans.

Ans: dh = 0.890 110 - 32 m>s dt ravg = 923 kg>m3 441

*4–100. Hexylene glycol is flowing into the trapezoidal container at a constant rate of 600 kg>min. Determine the rate at which the level is rising when y = 0.5 m. The container has a constant width of 0.5 m. rhg = 924 kg>m3.

0.4 m

30!

30!

A

y

V

SOLUTION The volume of the control volume considered changes with time since it contains the hexylene glycol in the tank, Fig. a. Its volume is

y tan 30˚

1 V = (0.4 m + 2y tan 30° + 0.4 m)(y)(0.5 m) 2

0.4 m

y tan 30˚

= ( 0.2y + 0.5 tan 30°y2 ) m3 dy 0y 0y 0V = 0.2 + tan 30° y = ( 0.2 + tan 30°y ) 0t dt 0t 0t

y

0V r dV + rhgV # dA = 0 0t Lcv hg Lcs

(a)

Since rhg is constant, it can be factored out from the integrals. Also, the average velocity will be used. Thus, this equation reduces to rhg

0V - rhgVAAA = 0 0t

#

Here rhgVAAA = mhg = a600

kg min

ba

( 924 kg>m3 ) (0.2 + tan 30°y)

When

30˚

30˚

Thus,

1 min b = 10 kg>s. Then 60 s

0y - 10 kg>s = 0 0t

dy 5 = c d m>s. dt 462 ( 0.2 + tan 30°y ) y = 0.5 m dy 5 = 0.0221 m>s = dt 462 3 0.2 + tan 30°(0.5) 4

442

Ans.

4–101. Hexylene glycol is flowing into the container at a constant rate of 600 kg>min. Determine the rate at which the level is rising when y = 0.5 m. The  container is in the form of a conical frustum. Hint: the volume of a cone is V = 13 pr 2h. rhg = 924 kg>m3.

0.4 m

30!

30!

A

y

V

SOLUTION Since the control volume contains hexylene glycol in the tank, Fig. a, its volume is V = =

0.2 1 0.2 1 p(0.2 + y tan 30°)2 + ay + mb - p(0.2 m)2a mb 3 tan 30° 3 tan 30°

0.2 m y tan 30˚

1 1 p a y3 + 0.223y2 + 0.12yb 3 3

0y 0y 0y 0V 1 = p ay2 + 0.423y + 0.12 b 0t 3 0t 0t 0t =

30˚

y

0.2 m

0.2 m tan 30˚

0y 1 p ( y2 + 0.423y + 0.12 ) 3 0t

Thus, 0 r dV + rhgV # dA = 0 0t Lcv hg Lcs Since rhg is constant, it can be factored out from the integrals. Also, the average velocity will be used. Thus, the equation reduces to rhg

#

Here, rhgVAAA = mhg = a600 1 3

(a)

0V - rhgVAAA = 0 0t

kg min

ba

1 min b = 10 kg>s .Then 60 s 0y 0t

( 924 kg>m3 ) c p ( y2 + 0.423y + 0.12 ) d - 10 kg>s = 0

When y = 0.5 m 0y = 0t

0y 15 = c d m>s 2 0t 462p ( y + 0.423y + 0.12 ) 15 462p 3 0.52 + 0.4 23 (0.5) + 0.12 4

= 0.0144 m>s

Ans.

Ans: 0.0144 m>s 443

4–102. Water in the triangular trough is at a depth of y = 3  ft. If the drain is opened at the bottom, and water flows out at a rate of V = 1 8.02y1>2 2 ft>s, where y is in feet, determine the time needed to fully drain the trough. The trough has a width of 2 ft. The slit at the bottom has a crosssectional area of 24 in2.

30!

30!

y " 3 ft

SOLUTION Control Volume. The deformable control volume shown in Fig. a. will be considered. Its volume at any instant is V = 2c

1 ( y tan 30° ) y d (2 ft) = ( 1.1547y2 ) ft 3 2

Continuity Equation. Since water has a constant density. rw c

30˚ 30˚

0 dV + V # dA d = 0 0t Lcv Lcs

y

A

0 V + VA = 0 0t

V

1 0 24 2 ( 1.1547y2 ) + ( 8.02y2 ) a ft b = 0 0t 144

2.3094y

y tan 30˚

(a)

0y 1 = -1.3367y 2 0t

0y 1 = -0.5788y- 2 0t Integrating, L0

t

dt =

L3

0 ft

1

-1.7277y 2 dy

2 3 0 t = - 1.7277a y 2 b ` 3 3 ft

Ans.

t = 5.99 s

Ans: 5.99 s 444

4–103. Water in the triangular trough is at a depth of y = 3  ft. If the drain is opened at the bottom, and water flows out at a rate of V = 1 8.02y1>2 2 ft>s, where y is in feet, determine the time needed for the water to reach a depth of y = 2 ft. The trough has a width of 2 ft. The slit at the bottom has a cross-sectional area of 24 in2.

30!

30!

y " 3 ft

SOLUTION Control volume. The deformable control volume shown in Fig. a will be considered. Its volume at any instant is V = 2c

1 ( y tan 30° ) y d (2 ft) = ( 1.1547y2 ) ft 3 2

Continuity Equation. Since water has a constant density. rw c

30˚ 30˚

0 dV + Vf>cs # dA d = 0 0t Lcv Lcs

y

A

0 V + VA = 0 0t

V

1 0 24 2 ( 1.1547y2 ) + ( 8.02y2 ) a ft b = 0 0t 144

2.3094y

y tan 30˚

(a)

0y 1 = -1.3367y2 0t

0y 1 = -0.5788y- 2 0t Integrating, L0

t

dt =

2 ft

L3 ft

1

-1.7277y 2 dy

2 3 2 ft t = - 1.7277a y 2 b ` 3 3 ft

Ans.

t = 2.73 s

Ans: 2.73 s 445

*4–104. As part of a manufacturing process, a 0.1-m-wide plate is dipped into hot tar and then lifted out, causing the tar to run down and then off the sides of the plate as shown. The thickness w of the tar at the bottom of the plate decreases with time t, but it still is assumed to maintain a linear variation along the plate as shown. If the velocity profile at the bottom of the plate is approximately parabolic, such that u = 3 0.5 ( 10-3 ) (x>w)1>2 4 m>s, where x and w are in meters, determine w as a function of time. Initially, when t = 0, w = 0.02 m.

x

0.3 m

0.5(10!3) m/s w

SOLUTION The flow is considered one dimensional since its velocity is directed downward. The control volume is shown in Fig. a and its volume changes with time. At a particular instant it is 1 V = w(0.3 m)(0.1 m) = (0.015w) m3 2 Thus, 0V 0w = 0.015 0t 0t The opened control surface is shown shaded in Fig. a. The differential area element is dA = bdx. 0 r dV + rtV # dA = 0 0t Lcv t Lcs Since the tar is assumed to be incompressible, rt is constant. Also, the velocity of the tar is always directed perpendicular to the opened surface. Hence the above equation reduces to 0V rt + rt udA = 0 0t Lcs

0.1 m

0V = - udA 0t Lcs

w dx

The negative sign indicates that V is decreasing with time. Then w

0.015

1 2

dw x = 0.5 ( 10-3 ) a b (0.1dx) dt w L0

0.015

0.5 ( 10 ) (0.1) 2 3 w dw = a x2b ` 1 dt 3 0 w2 dw = - 3.333 ( 10-5 ) w dt

udA is equal to the volume of the parabolic block under the Lcs 2 velocity profile, ie, udA = 3 0.5 ( 10-3 ) 4 w(0.1) = 0.0333 ( 10-3 ) w 3 Lcs Or, the integral

dw = -2.222 ( 10-3 ) w dt w

-

L0.02 m 2.222 ( 10-3 ) w dw

=

L0

x (a)

-3

0.015

0.3 m

t

dt

w = t 0.02 w t ln = 0.02 450 w = e -t>450 0.02

-450 ln

w = ( 0.02e -t>450 ) m

Ans. 446

4–105. The cylindrical tank in a food-processing plant is filled with a concentrated sugar solution having an initial density of rs = 1400 kg>m3. Water is piped into the tank at A at 0.03 m3 >s and mixes with the sugar solution. If an equal flow of the diluted solution exits at B, determine the amount of water that must be added to the tank so that the density of the sugar solution is reduced by 10% of its original value.

1m A

2m

B

SOLUTION The control volume considered here is the volume of the tank. It is a fixed control volume since its volume does not change throughout the mixing. 0 rdV + rVds # dA = 0 0t Lcv Lcs V

0r + rQ - rwQ = 0 0t V

0r = Q ( rw - r ) 0t

0r Q t 0t = V L0 Lrs rw - r r

- ln ( rw - r ) ` - ln a t =

r

= rs

Q t V

rw - r Q b = t rw - rs V

rw - rs V ln a b rw - r Q

Here, V = p(0.5 m)2(0.2 m) = 0.5p m3, and it is required that r = 0.9rs = 0.9 ( 1400 kg>m3 ) = 1260 kg>m3. Then t = °

1000 kg>m3 - 1400 kg>m3 0.5p m3 ¢ ln° ¢ 0.03 m3 >s 1000 kg>m3 - 1260 kg>m3

= 22.556

The amount of water to be added is Vw = Qt = ( 0.03 m3 >s ) (22.556) = 0.677 m3

Ans.

Ans: 0.677 m3 447

4–106. The cylindrical pressure vessel contains methane at an initial absolute pressure of 2 MPa. If the nozzle is opened, the mass flow depends upon the absolute pressure and is # m = 3.5 1 10-6 2 p kg>s, where p is in pascals. Assuming the temperature remains constant at 20°C, determine the time required for the pressure to drop to 1.5 MPa.

6m

2m

SOLUTION

From Appendix A, the gas constant for Methane is R = 518.3 J>(kg # K). Using the ideal gas law with T = 20°C + 273 = 293 K which is constant throughout, p = r ( 518.3 J>(kg # k) ) (293 k)

p = rRT;

p = 151861.9r r = 6.5849 ( 10-6 ) p

(1)

The control volume considered is the volume of tank which contains Methane. Since the tank is fully filled at all times, the control volume can be classified as fixed. d rdV + rV # dA = 0 dt Lcv Lcs Here

# dV = V (fixed control volume) and m =

Lcv

rV # A. Then Lcv

dr # + m = 0 dt # Since, V = p(1 m)2(6 m) = 6p m3 and m = 3 3.5 ( 10-6 ) p 4 kg>s, then Eq (1), V

(2)

dr dp = 6.5849 ( 10-6 ) dt dt

Substitute these results into Eq. (2), 6pc 6.5849 ( 10-6 )

dp d + 3.5 ( 10-6 ) p = 0 dt

dp = - 0.02820p dt p

t dp = - 0.02820 dt L0 Lp0 p

lna

p b = - 0.02820t p0 t = - 35.46 lna

Here p0 = 2 Mpa. Thus when p = 1.5 Mpa, t = - 35.46 lna = 10.2 s

p b p0

1.5 MPa b 2 MPa

Ans.

Ans: 10.2 s 448

4–107. The cylindrical pressure vessel contains methane at an initial absolute pressure of 2 MPa. If the nozzle is opened, the mass flow depends upon the absolute pressure and # is m = 3.5 1 10-6 2 p kg>s, where P is in pascals. Assuming the temperature remains constant at 20°C, determine the pressure in the tank as a function of time. Plot this pressure (vertical axis) versus the time for 0 … t … 15 s. Give values for increments of ∆t = 3 s.

6m

2m

SOLUTION

P(MPa)

From Appendix A, the gas constant for Methane is R = 518.3 J>kg # K using the 2.0 ideal gas law with T = 20°C + 273 = 293 K, which is constant throughout, p = r(518.3 J>kg # k)(293 k)

p = rRT;

1.5

p = 151861.9r r = 6.5849 ( 10-6 ) p

(1)

The control volume considered is the volume of tank which contains Methane. Since the tank is fully filled at all times, the control volume can be classified as fixed.

0.5

d rdV + rVds # dA = 0 dt Lcv Lcs Here

Lcv

# dV = V (fixed control volume) and m =

1.0

t(s)

rV # dA. Then

Lcv

0

dr # + m = 0 (2) dt # Since, V = p(1 m)2(6 m) = 6p m3 and m = 3 3.5 ( 10-6 ) p 4 kg>s, then from Eq (1),

3

6

9

12

15

(a)

V

dr dp = 6.5849 ( 10-6 ) dt dt

Substitute these results into Eq (2), 6pc 6.5849 ( 10-6 )

0p d + 3.5 ( 10-6 ) p = 0 0t

dp = -0.02820p dt p

t dp = - 0.02820 dt L0 Lp0 p

lna

p b = - 0.02820t p0 p = e -0.02820t p0

p = p0e -0.02820t Here p0 = 2MPa, then p = ( 2e -0.02820t ) MPa, where t is in seconds

Ans.

The plot of p vs t is shown in Fig. a t(s) p(MPa)

0

3

6

9

12

2.0

1.84

1.69

1.55

1.43

15 1.31 Ans: p = ( 2 e - 0.0282t ) MPa, where t is in seconds 449

*4–108. As nitrogen is pumped into the closed cylindrical # tank, the mass flow through the tube is m = 1 0.8r-1>2 2 slug>s. Determine the density of the nitrogen within the tank when t = 5 s from the time the pump is turned on. Assume that initially there is 0.5 slug of nitrogen in the tank.

4 ft

2 ft

SOLUTION Control Volume. The fixed control volume is shown in Fig. a. This control volume has a constant volume of

V

A

V = p(1 ft)2(4 ft) = 4p ft 3 The density of the nitrogen within the control volume changes with time and therefore contributes to local changes. # Continuity Equation. Realizing that m =

rV # dA, Lcs

0 rdV + rV # dA = 0 0t Lcv Lcs 0r # V - m = 0 0t 4p

0r -1 - 0.8r 2 = 0 0t

0r 0.2 -1 = r2 p 0t Integrating, L0

t

r

Lro

dt =

1

5pr2 dr

2 3 r t = 5pa r 2 b ` 3 ro t =

3 10p 3 ( r2 - ro2 ) 3 2

Here, ro =

0.5 slug 4p ft 3

=

r = a

3 3 3t + ro2 b slug>ft 3 10p

0.125 slug>ft 3. Then, when t = 5 s, p 3(5)

3

2

0.125 2 3 r = c + a b d slug>ft 3 p 10p = 0.618 slug>ft 3

Ans.

450

(a)

4–109. As nitrogen is pumped into the closed cylindrical tank, the mass flow through the tube is m = 1 0.8r-1>2 2 slug>s. Determine the density of the nitrogen within the tank when t = 10 s from the time the pump is turned on. Assume that initially there is 0.5 slug of nitrogen in the tank.

4 ft

2 ft

SOLUTION Control Volume. The fixed volume is shown in Fig. a. This control volume has a constant volume of

V

A

V = p(1 ft)2(4 ft) = 4p ft 3 The density of the nitrogen within the control volume changes with time and therefore contributes to local changes. # Continuity Equation. Realizing that m = rV # dA, Lcs

(a)

0 rdV + rV # dA = 0 0t Lcv Lcs 0r # V - m = 0 0t 4p

0r -1 - 0.8r 2 = 0 0t

0r 0.2 -1 = r2 p 0t Integrating, L0

t

r

dt =

Lro

1

5pr2dr

2 3 r t = 5pa r2 b ` 3 ro t =

3 10p 3 ( r2 - ro2 ) 3 2

Here, ro =

0.5 slug 4p ft 3

r = a

=

3 3 3t + ro2 b slug>ft 3 10p

0.125 slug>ft 3. Then, when t = 10 s, p r = c

3(10) 10p

3

+ a

2

0.125 2 3 b d slug>ft 3 p

= 0.975 slug>ft 3

Ans.

Ans: 0.975 slug>ft 3 451

4–110. Water flows out of the stem of the funnel at an average speed of V = 1 3e -0.05t 2 m>s, where t is in seconds. Determine the average speed at which the water level is falling at the instant y = 100 mm. At t = 0, y = 200 mm.

200 mm

200 mm y

10 mm

SOLUTION The control volume is the volume of the water in the funnel. This volume changes with time. d r dV + rwV # dA = 0 dt Lcv w Lcs

r

Since rw is constant (in compressible), it can be factored out from the integrals rw

0.1 m

dV + rw V # dA = 0 dt Lcs

y

Here, the average velocity will be used. Then dV + VA = 0 dt dV p + ( 3e -0.05t ) c (0.01 m)2 d dt 4

(a)

dV + 75 ( 10-6 ) pe -0.05t = 0 dt

(1)

The volume of the control volume at a particular instant is V =

1 2 pt y 3

From the geometry shown in Fig. a, r 0.1 m = ; y 0.2 m

Then V =

r =

1 y 2

1 1 2 1 p a yb y = a py3 b m3 3 2 12

dy dV 1 = p a3y2 b dt 12 dt dy dV 1 = py2 dt 4 dt

Substitute into Eq. (1), 1 2 dy py + 75 ( 10-6 ) pe -0.05t = 0 4 dt y2

dy = - 0.3 ( 10-3 ) e -0.05t dt

(2)

0.3 ( 10-3 ) e -0.05t dy = dt y2

(3)

452

0.2 m

4–110. Continued

Separating the variables of Eq. (2) 0.1 m

L0.2 m

y2dy = -0.3 ( 10-3 )

L0

t

e -0.05tdt

y3 0.1 m e -0.05t t ` = - 0.3 ( 10-3 ) a b` 3 0.2 m -0.05 0

-2.3333 ( 10-3 ) = 6 ( 10-3 ) e -0.05t ` - 2.3333 ( 10

-3

) = 6 ( 10 )( e -3

t

0

-0.05t

- 1)

e -0.05t = 0.6111 Substitute this value and y = 0.1 m into Eq. (3), 0.3 ( 10-3 ) (0.6111) dy = = -0.0183 m>s dt 0.12

Ans.

The negative sign indicates that y is decreasing, ie., the water level is falling.

Ans: - 0.0183 m>s 453

4–111. A part is manufactured by placing molten plastic into the trapezoidal container and then moving the cylindrical die down into it at a constant speed of 20 mm>s. Determine the average speed at which the plastic rises in the form as a function of yc. The container has a width of 150 mm.

20 mm/s

30!

30! 100 mm

yc

150 mm

SOLUTION The control volume segments shown shaded in Fig. a can be consider fixed at a particular instant. At this instant, no local changes occur since the molten plastic is incompressible. Also, its density is constant. If we use average velocities. Also then 0 rdV + rV # dA = 0 0t Lcv Lcs (1)

0 + VBAB - QA = 0 Here, AB = (2yc tan 30° + 0.15 m)(0.15 m) - p(0.05 m)2 = (0.3 tan 30° yc + 0.01465) m2 VB =

0yc 0t

The volume of the die submerged in the plastic is Vd = p(0.05 m)2yd = ( 0.0025pyd ) m3 Realizing that QA =

dyd = Vd = 0.02 m>s, then dt

dy2 dV2 = 0.0025p = (0.0025p)(0.02 m>s) = 50 ( 10-6 ) p m3 >s dt dt

Substitute these results into Eq (1),

dyo (0.3 tan 30°yc + 0.01465) - 50 ( 10-6 ) p = 0 dt 0.157 ( 10-3 ) dyc = a b m>s dt 0.173yc + 0.0146

Ans.

2yc tan 30˚ + 0.15 m 0.1 m B

B

yd yc

A

A

30˚

30˚

0.15 m (a)

Ans: 0.157 110 - 32 dyc = a b m>s dt 0.173yc + 0.0146 454

5–1. Water flows in the horizontal pipe. Determine the average decrease in pressure in 4 m along a horizontal streamline so that the water has an acceleration of 0.5 m>s2. 4m

SOLUTION 1 dp + as + g sin u = 0 r ds a

∆p 1 ba b + 0.5 m>s2 + 0 = 0 3 4m 1000 kg>m ∆p = - 2000 Pa = - 2 kPa

Ans.

The negative sign indicates that the pressure drops as the water flows from A to B.

Ans: -2 kPa 455

5–2. The horizontal 100-mm-diameter pipe is bent so that its inner radius is 300 mm. If the pressure difference between  points A and B is pB - pA = 300 kPa, determine the volumetric flow of water through the pipe.

400 mm A

B

300 mm

SOLUTION

0.4 m

Referring to the coordinate system shown in Fig. a, we notice that dn = -dR since the n and R axes are opposite in sense. Also, dz = 0. Since the pipe is lying on the horizontal plane, then rV dp dz - rg = dn dn R

2

LpA

dp = rV

2

0.3 m nA

B

R

rV 2 dp = dR R pB

S

(a)

0.4 m

L0.3 m

dR R

pB - pA = rV 2 ln

4 3

300 ( 103 ) N>m2 = ( 1000 kg>m3 ) V 2 ln

4 3

V = 32.29 m>s The volumetric flow is Q = VA = ( 32.29 m>s ) 3 p(0.05 m)2 4 = 0.254 m3 >s

456

Ans.

Ans: 0.254 m3 >s

5–3. Air at 60°F flows through the horizontal tapered duct. Determine the acceleration of the air if on a streamline the pressure is 14.7 psi and 40 ft away the pressure is 14.6 psi. 40 ft

SOLUTION The pressure is approximately 1 atmosphere along the length in question. From Appendix A, the density of air at T = 60° F is r = 0.00237 slug>ft 3. This density will be used, since r will change only slightly with the small change in pressure. Since the duct is level, sin u = 0. 1 dp + as + g sin u = 0 r ds

a

1 b≥ 0.00237 slug>ft 3

lb 12 in. 2 da b 1 ft in2 ¥ + as + 0 = 0 40 ft

c ( - 0.1 lb>in2 )

as = 151.90 ft>s2 = 152 ft>s2

Ans.

Also, realizing that zB = zA = 0, since the duct is level, pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 ∆p = pB - pA = °

VA2 - VB2 ¢r 2

For constant acceleration VB2 = VA2 + 2ac(sB - sA) or Then

VA2 - VB2 = - as(sB - sA). 2

∆p = - r(sB - sA)as

as =

- ∆p = r(sB - sA)

( - 0.1 lb>in2 ) a

12 in 2 b 1 ft

( 0.00237 slug>ft 3 ) (40 ft)

as = 152 ft>s2

Ans.

Ans: 152 ft>s2 457

*5–4. Air at 60°F flows through the horizontal tapered duct. Determine the average decrease in pressure in 40 ft, so that the air has an acceleration of 150 ft>s2. 40 ft

SOLUTION From Appendix A, the density of air at T = 60° F is r = 0.00237 slug>ft 3. This density will be used on the assumption that the change in p will be small, leading to only a small change in r. Since the duct is level, sin u = 0. 1 dp + as + g sin u = 0 r ds a

∆p 1 ba b + 150 ft>s2 + 0 = 0 0.00237 slug>ft 3 40 ft ∆p = -14.22

lb = -14.2 lb>ft 2 ft 2

Ans.

Also, realizing that zB - zA = 0, since the duct is level, pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 ∆p = pB - pA = °

VA2 - VB2 ¢r 2

For constant acceleration VB = VA2 + 2as(sB - sA) or Then

VA2 - VB2 = -as(sB - sA). 2

∆p = - r(sB - sA)as = - ( 0.00237 slug>ft 3 ) (40 ft) ( 150 ft>s2 ) ∆p = -14.2

lb ft 2

Ans.

The negative sign indicates that the pressure drops as the air flows from A to B. Note: 14.2 lb>ft is less then 0.1 psi, so indeed the change in pressure is small compared to p ≈ 1 atm = 14.7 psi.

458

5–5. An ideal fluid having a density r flows with a velocity V through the horizontal pipe bend. Plot the pressure variation within the fluid as a function of the radius r, where ri … r … ro and ro = 2ri.

V

ri

r ro

SOLUTION Since the fluid is inviscid (ideal fluid) and the flow is steady (constant V) and along the circular bend, Euler’s differential equation in the n-direction can be applied -

rV 2 dp dz - rg = dn dn R

Since the pipe lies in the horizontal plane, the elevation term (second term on the left) can be excluded. Also the n axis and r are opposite in sense thus, dn = -dr. with R = r, rV 2 dp = dr r L

r

dp = rV 2

dr Lri r

∆p = rV 2 ln

r ri

The tabulation for ri … r … 2ri is calculated below. r

ri

1.25 ri

1.50 ri

1.75 ri

2 ri

∆p

0

0.223 rV 2

0.405 rV 2

0.560 rV 2

0.693 rV 2

The plot of this relation is shown in Fig. a P 0.8 V 2 0.6 V 2 0.4 V 2 0.2 V 2 r 0

ri

1.25ri

1.5ri

1.75ri

459

2ri

Ans: ∆p = rV 2 ln1r>ri 2

5–6. Water flows through the horizontal circular section with a uniform velocity of 4 ft>s. If the pressure at point D is 60 psi, determine the pressure at point C.

4 ft/s

4 ft/s B

A

1 ft D

1.5 ft

C

SOLUTION

1 ft

Referring to the coordinate systems shown in Fig. a, we notice that dn = - dR since the n and R axes are opposite in sense. Also, dz = 0 since the pipe is lying on the horizontal plane. Thus, dp rV 2 dz - rg = dn dn R

LpD

dP = rV 2

1.5 ft

L1 ft

1.5 ft

n D

S

C (a)

rV 2 dp = dR R pC

R

dR R

pC - pD = rV 2 ln 1.5 pC = rV 2 ln 1.5 + pD = °

62.4 lb>ft 3 32.2 ft>s2

= a8652.57

¢ ( 4 ft>s ) 2(ln 1.5) + ( 60 lb>in2 ) a

12 in. 2 b 1 ft

lb 1 ft 2 b = 60.09 psi = 60.1 psi ba 2 12 in. ft

Ans.

Ans: 60.1 psi 460

5–7.

Solve Prob. 5–6 assuming the pipe is vertical.

4 ft/s

4 ft/s B

A

1 ft D

1.5 ft

C

SOLUTION

Z

Referring to the coordinate systems shown in Fig. a, we notice that dn = dz and dn = - dR. Thus, rV 2 dp dz - rg = dn dn R

LpD

dp = rV

2

1.5 ft

L1 ft

R

1.5 ft

n D

rV 2 dp - g = dR R pC

1 ft

S

C (a)

1.5 ft

dR + g dR R L1 ft

2

pC - pD = rV ln 1.5 + g(0.5 ft) pC = rV 2 ln 1.5 + g(0.5 ft) + pD = °

62.4 lb>ft 3 2

32.2 ft>s

= 8683.77

¢ ( 4 ft>s ) 2(ln 1.5) + a62.4

lb 1 ft 2 a b = 60.3 psi ft 2 12 in.

lb lb 12 in. 2 b(0.5 ft) + a60 2 ba b 3 1 ft in ft Ans.

Ans: 60.3 psi 461

*5–8. By applying a force F, a saline solution is ejected from the 15-mm-diameter syringe through a 0.6-mm-diameter needle. If the pressure developed within the syringe is 60 kPa, determine the average velocity of the solution through the needle. Take r = 1050 kg>m3.

15 mm F

SOLUTION The saline solution can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. Applying this equation between a point in the syringe and the other point at the tip of the needle of which both points are on the central streamline, ps pn Vs 2 Vn 2 + gzs = + gzn + + r r 2 2 Vs 2 is negligible. Since the tip of the needle is exposed 2 to the atmosphere, pn = 0. Here, the datum will coincide with the central stream line. Then 60 ( 103 ) N>m2 Vn 2 + 0 + 0 = 0 + + 0 2 1050 kg>m3

Since Vs 6 6 6 6Vn, the term

Ans.

Vn = 10.69 m>s = 10.7 m>s Using the continuity equation, -VsAs + Vn An = 0 - Vs 3 p(0.0075 m)2 4 + ( 10.69 m>s ) 5 p 3 0.3 ( 10-3 ) m 4 2 6 = 0 Vs = 0.0171 m>s

This result proves that Vs is indeed very small as compared to Vn. Therefore, the solution is acceptable.

462

5–9. By applying a force F, a saline solution is ejected from the 15-mm-diameter syringe through a 0.6-mm-diameter needle. Determine the average velocity of the solution through the needle as a function of the force F applied to the plunger. Plot this velocity (vertical axis) as a function of the force for 0 … F … 20 N. Give values for increments of ∆F = 5 N. Take r = 1050 kg>m3.

15 mm F

SOLUTION The saline solution can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. Applying this equation between a point in the syringe and the other at the tip of the needle of which both points are on the central streamline, ps pn Vs 2 Vn 2 + gzs = + gzn + + r r 2 2 Vs 2 is negligible. Since the tip of the needle is exposed 2 F F to the atmosphere, pn = 0. Here, Ps = = = 5.659 ( 103 ) F and the As p(0.0075 m)2 datum will coincide with the control streamline. Then Since Vs 6 6 6 6Vn the term

5.659 ( 103 ) F 3

1050 kg>m

+ 0 + 0 = 0 +

Vn 2 + 0 2

Vn = ( 3.2831 1F ) m>s

Vn = ( 3.283 1F ) m>s where F is in N

Ans.

The plot of V vs F is shown in Fig. a. Using the continuity equation,

- Vs As + Vn An = 0 -Vs 3 p(0.0075 m)2 4 + ( 3.2831 1F ) 5 p 3 0.3 ( 10-3 ) m 4 2 6 = 0 Vs = 0.005251F

This result shows that Vs is indeed very small as compared to Vn. Therefore, the solution is acceptable.

463

5–9. Continued

This result shows that Vs is indeed very small as compared to Vn. Therefore, the solution is acceptable. F(N)

0

5

10

15

20

Vn ( m>s )

0

7.34

10.4

12.7

14.7

Vn (m s) 15

10

5

F(N) 0

5

10

15

20

(a)

Ans: Vn = 464

1 3.283 2F 2 m>s, where F is in N

5–10. An infusion pump produces pressure within the syringe that gives the plunger A a velocity of 20 mm>s. If the saline fluid has a density of rs = 1050 kg>m3, determine the pressure developed in the syringe at B.

20 mm/s

40 mm 1 mm B

A

SOLUTION The saline solution can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. Applying this equation between a point in the syringe and the other at the tip of the needle of which both points are on the central streamline, ps pn Vs 2 Vn 2 + gzs = + gzn + + r r 2 2 Since the tip of the needle is exposed to the atmosphere, pn = 0. Here, the datum will coincide with the central streamline. ps Vn 2 Vs 2 + 0 = 0 + + 0 + r 2 2 r ( Vn 2 - Vs 2 ) 2

ps =

(1)

Applying the continuity equation, - Vs As + Vn An = 0 - 3 20 ( 10-3 ) m>s 4 3 p(0.02 m)2 4 + Vn 5 p 3 0.5 ( 10-3 ) m 4 2 6 = 0 Vn = 32.0 m>s

Substituting this value into Eq. (1), ps = °

1050 kg>m3 2

¢ 5 ( 32.0 m>s ) 2 -

= 537.60 ( 103 ) Pa

3 20 ( 10-3 ) m>s 4 2 6 Ans.

= 538 kPa

Ans: 538 kPa 465

5–11. If the fountain nozzle sprays water 2 ft into the air, determine the velocity of the water it leaves the nozzle at A.

B

2 ft

A

SOLUTION Bernoulli Equation. Since the water jet is in the open atmosphere at A and B, pA = pB = 0. Also, vB = 0 since the jet achieves its maximum height at B. If the datum is set at A, zA = 0 and zB = 2 ft. pB pA VA2 VB2 + gzA + + gzB + + r r 2 2 0 +

VA2 + 0 = 0 + 0 + ( 32.2 ft>s2 ) (2 ft) 2 Ans.

VA = 11.3 ft>s

Ans: 11.3 ft>s 466

*5–12. The jet airplane is flying at 80 m>s in still air, A, at an altitude of 3 km. Determine the absolute stagnation pressure at the leading edge B of the wing.

80 m/s C

A B

SOLUTION Bernoulli Equation. If the flow of the air is viewed from the plane, it will be a steady flow. If we observe the air from the plane, the still air at A will have VA = 80 m>s and the air at B has the same velocity as the plane, VB = 0. From Appendix A, (pA)abs = 70.12 kPa and r = 0.9092 kg>m3 at an altitude of 3 km. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 70.12 ( 103 ) N>m2 3

0.9092 kg>m

+

(80 m>s)2 2

+ 0 =

(pB)abs 0.9092 kg>m3

+ 0 + 0

( pB ) abs = 73029.44 Pa = 73.0 kPa

Ans.

467

5–13. The jet airplane is flying at 80 m>s in still air, A, at an altitude of 4 km. If the air flows past point C near the wing at 90 m>s, determine the difference in pressure between the air near the leading edge B of the wing and point C.

80 m/s C

A B

SOLUTION Bernoulli Equation. If the flow of the air is viewed from the plane, it will be a steady flow. Thus, from the plane, the air at B is VB = 80 m>s and at C, VC = 90 m>s. From Appendix A, r = 0.8194 kg>m3 at an altitude of 4 km. pB pC VC2 VB2 + gzB = + gzC + + r r 2 2 pB 3

0.8194 kg>m

+

( 80 m>s ) 2 2

+ 0 =

pC 3

0.8194 kg>m

+

( 90 m>s ) 2 2

+ 0 Ans.

pB - pC = 3.32 kPa

Ans: 3.32 kPa 468

5–14. A river flows at 12 ft>s and then turns and drops as a waterfall, from a height of 80 ft. Determine the velocity of the water just before it strikes the rocks below the falls.

SOLUTION Bernoulli Equation. If the datum is set at the rocks, zA = 80 ft (before the flow drops), zB (just before it strikes the rocks) = 0. Since the flow from A to B in the open atmosphere, pA = pB = 0. From A to B, pA pB VA2 VB 2 + gzA = + gzB + + r r 2 2 0 +

( 12 ft>s ) 2 2

+ ( 32.2 ft>s2 ) (80 ft) = 0 +

VB 2 2

+ 0 Ans.

VB = 72.8 ft>s

Ans: 72.8 ft>s 469

5–15. Water is discharged through the drain pipe at B from the large basin at 0.03 m3 >s. If the diameter of the drainpipe is d = 60 mm, determine the pressure at B just inside the drain when the depth of the water is h = 2 m.

A

d

h

B

SOLUTION QB = VB AB 0.03 m3 >s = VB c p(0.03 m)2 d VB = 10.61 m>s

Bernoulli Equation. Since the water is discharged from a large source, VA ≅ 0. Here, pA = 0 since surface A is exposed to the atmosphere. If we set the datum along the base of the basin, zB = 0 and zA = 2 m. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 0 + 0 + ( 9.81 m>s2 ) (2 m) =

pB 3

1000 kg>m

+

( 10.61 m>s ) 2 2

pB = - 36.67 ( 103 ) Pa = - 36.7 kPa

+ 0 Ans.

Ans: -36.7 kPa 470

*5–16. Water is discharged through the drain pipe at B from the large basin at 0.03 m3 >s. Determine the pressure at B just inside the drain as a function of the diameter d of the drainpipe. The height of the water is maintained at h = 2 m. Plot the pressure (vertical axis) versus the diameter for 60 mm 6 d 6 120 mm. Give values for increments of ∆d = 20 mm.

A

d B

SOLUTION The discharge requirement is 0.03 m3 >s = VB £

Q = VBAB;

VB = £

p d 2 a b § 4 1000

38.197 ( 103 ) d2

§ m>s

The water can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore Bernoulli’s equation is applicable. Applying this equation between A and B realizing that VA _ 0 (the water is discharged from a large reservoir), pA = 0 (surface A is exposed to the atmosphere). Also if we set the datum along the base of the basin zB = 0 and zA = 2 m. pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 0 + 0 + ( 9.81 m>s2 ) (2 m) =

pB + 1000 kg>m

pB = °19.62 -

0.7295 ( 109 ) d4

The plot of pB vs. d is shown in Fig. a pA = £ 19.6 d(mm)

60

pB(kPa) -36.7

0.730(109) d4

3 38.197 ( 103 ) >d 2 4 2 2

¢ ( 103 )

Ans.

§ kPa where d is in mm

80

100

120

1.79

12.3

16.1

pB(kPa) 30 0 –30

20

40

60

80

100

+ 0

120

d(mm)

–60 –90 –120 –150 –180 –210 –240 –270

471

h

5–17. A fountain is produced by water that flows up the tube at Q = 0.08 m3 >s and then radially through two cylindrical plates before exiting to the atmosphere. Determine the velocity and pressure of the water at point A.

200 mm 200 mm

5 mm

A

V

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s Equation is applicable. Writing this equation between points A and B on the radial streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Since point B is exposed to the atmosphere, pB = patm = 0. Here, points A and B have the same elevation since the cylindrical plates are in the horizontal plane. Thus, zA = zB = z. pA 1000 kg>m3

+

V0 2 VA2 + gz = 0 + + gz 2 2

pA = 500 ( VB2 - VA2 )

(1)

Continuity requires that Q = VA AA;

0.08 m3 >s = VA [2p(0.2 m)(0.005 m)]

Ans.

VA = 12.73 m>s = 12.7 m>s

Q = VB AB;

0.08 m3 >s = VB[2p (0.4 m)(0.005 m)] VB = 6.366 m>s

Substituting these results into Eq. (1), pA = 500 ( 6.3662 - 12.732 ) = - 60.79 ( 103 ) Pa = - 60.8 kPa

Ans.

The negative sign indicates that the pressure at A is a partial vacuum.

Ans: V = 12.7 m>s, p = -60.8 kPa 472

5–18. A fountain is produced by water that flows up the tube at Q = 0.08 m3 >s and then radially through two cylindrical plates before exiting to the atmosphere. Determine the pressure of the water as a function of the radial distance r. Plot the pressure (vertical axis) versus r for 200 mm … r … 400 mm. Give values for increments of ∆r = 50 mm.

200 mm 200 mm

V

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s Equation is applicable. Writing this equation between points A and B on the radial streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Since point B is exposed to the atmosphere, pB = patm = 0. Here points A and B have the same elevation since the cylindrical plates are in the horizontal plane. Thus zA = zB = z. p 3

1000 kg>m

p = Continuity requires that Q = VA AA;

Q = VB AB;

5 mm

A

+

VB2 V2 + gz = 0 + + gz 2 2

3 500 ( VB2

0.08 m3 >s = V c 2p a V = a

- V 2 ) 4 Pa

(1)

r b(0.005 m) d 1000

2546.48 b m>s r

0.08 m3 >s = VB [2p (0.4 m)(0.005 m)] VB = 6.366 m>s

473

5–18. Continued

Substituting these results into Eq. (1), p = 500 £ 6.3662 - a p = 500 £ 40.5 p = 0.5£ 40.5 -

2546.48 2 b § Pa r

6.48 ( 106 ) r2 6.48 ( 106 ) r2

§ Pa Ans.

§ kPa where r is in mm

r(mm)

200

250

300

350

400

p(kPa)

-60.8

-31.6

-15.8

-6.20

0

P(kPa) 0

50

100

150

200

250

300

350

400 r(mm)

–50 –100 –150 –200 –250 –300 –350

Ans: p = 0.5c 40.5 -

6.48 1 106 2

where r is in mm. 474

r2

d kPa,

5–19. The average human lung takes in about 0.6 liter of air with each inhalation, through the mouth and nose, A. This lasts for about 1.5 seconds. Determine the power required to do this if it occurs through the trachea B having a cross-sectional area of 125 mm2. Take ra = 1.23 kg>m3. Hint: Recall that power is force F times velocity V, where F = pA.

A B

SOLUTION Assume that air is incompressible and inviscid and the flow is steady. Then, Bernoulli’s equation can be applied between points A and B on the central streamline along the trachea shown in Fig. a.

A

pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2

Trachea

Since the density of air is small, the elevation terms can be neglected. Since the air is taken in from the atmosphere, which is a large reservoir, pA = 0 and VA = 0. Then

B

pB VB2 + 0 + 0 = ra 2 VB = Using this result, the volumetric flow is Q = VB AB;

Q = ° pB = -

A

-

2 pB ra Lung

2 pB ¢ AB A ra -

ra Q ° ¢ 2 AB

(a)

2

The negative sign indicates that the pressure in the trachea is in partial vacuum. Then

2

The power of F is

r a QB ° ¢ AB F = pBAB = 2 AB 2

ra Q ° ¢ (ABVB) P = FVB = 2 AB =

Here ra = 1.23 kg>m3, AB =

Q = a p =

1 ra Q 3 ° ¢ 2 AB2

1 125 mm2 2 a

2 1m b = 0.125 1 10-3 2 m2 and 1000 mm

0.6 L 1 m3 b° ¢ = 0.4 1 10-3 2 m3 >s. Then 1.5 s 1000 L

3 -3 3 3 1 ( 1.23 kg>m ) 3 0.4 ( 10 ) m >s 4 £ § = 0.00252 W = 2.52 mW 2 3 0.125 ( 10-3 ) m2 4 2

475

Ans.

Ans: 2.52 mW

*5–20. Water flows from the hose at B at the rate of 4 m>s when the water level in the large tank is 0.5 m. Determine the pressure of air that has been pumped into the top of the tank at A.

A C

0.5 m B

SOLUTION Bernoulli Equation. Since the water is discharged from a large tank, VC = 0. Also, the water is discharged into the atmosphere at B, thus pB = 0. If the datum is at B, zC = 0.5 m and zB = 0. pC pB VC 2 VB 2 + gzC = + gzB + + r r 2 2 pC 3

1000 kg>m

+ 0 +

1 9.81 m>s2 2 (0.5 m)

= 0 +

(4 m>s)2 2

+ 0 Ans.

pC = 3095 Pa = 3.10 kPa

476

5–21. If the hose at A is used to pump air into the tank with a pressure of 150 kPa, determine the discharge of water at the end of the 15-mm-diameter hose at B when the water level is 0.5 m.

A C

0.5 m B

SOLUTION Bernoulli Equation. Since the water is discharged from a large tank, VC = 0. Also, the water is discharged into the atmosphere at B, thus pB = 0. If the datum is set at B, zC = 0.5 m and B, zB = 0. pC pB VB2 VC2 + gzC = + gzB + + r r 2 2 N VB 2 m2 2 ( ) + 0 + 9.81 m>s (0.5 m) = 0 + + 0 2 1000 kg>m3

150 ( 103 )

VB = 17.601 m>s Q = VB AB = ( 17.601 m>s ) 3 p(0.0075 m)2 4 = 3.11 ( 10-3 ) m3 >s

477

Ans.

Ans: 3.11 1 10 - 3 2 m3 >s

5–22. Piston C moves to the right at a constant speed of 5 m>s, and as it does, outside air at atmospheric pressure flows into the circular cylinder through the opening at B. Determine the pressure within the cylinder and the power required to move the piston. Take ra = 1.23 kg>m3. Hint: Recall that power is force F times velocity V, where F = pA.

5 m/s

B

50 mm C

SOLUTION Assume that air is incompressible and inviscid and the flow is steady. Then Bernoulli’s equation can be applied between points A and B on the central streamline. pA VA2 pC VC2 + + gzA = + + gzC ra ra 2 2 Since the density of air is small, the elevation terms can be neglected. Since the air is taken in from the atmosphere, which is a large reservoir, pA = patm = 0 and VA = 0. Then 0 + 0 = pC = - ra

pC VC2 + ra 2

(5 m>s)2 VC 2 = - ( 1.23 kg>m3 ) ° ¢ = - 15.375 Pa 2 2 = - 15.4 Pa

Ans.

Then F = pBAC = a15.375

The power of F is

#

N b 3 p(0.025 m)2 4 = 0.030189 N m2

W = (0.030189) ( 5 m>s ) = 0.151 W = 151 mW

Ans.

C

A

B

(a)

Ans: pC = -15.4 Pa

#

W = 151 mW 478

5–23. A fountain ejects water through the four nozzles, which have inner diameters of 10 mm. Determine the pressure in the pipe and the required volumetric flow through the supply pipe so that the water stream always reaches a height of h = 4 m.

F

60 mm

h E

A

B

C

D

SOLUTION Bernoulli Equation. Since the water stream DF flows in the open atmosphere, pD = pF = 0. Also, when the stream achieves its maximum height at F , VF = 0. If we set the datum along the streamline at the centerline of the pipe, zD = zE = 0 and zF = 4 m. From D to F, pD VD2 pF VF2 + gzD = + gzF + + r r 2 2 0 +

VD2 + 0 = 0 + 0 + ( 9.81 m>s2 ) (4 m) 2 VD = 8.859 m>s

From E to F, pF pE VE2 VF 2 + gzE = + gzF + + r r 2 2 pE 1000 kg>m3

+

VE2 + 0 = 0 + 0 + ( 9.81 m>s2 ) (4 m) 2 pE = 39240 - 500VE2

(1)

Continuity Equation. Take the final control volume to be the water within the pipe. Since there are four nozzles, 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VEAE + 4VDAD = 0 - VE 3 p(0.03 m)2 4 + 4 ( 8.859 m>s ) 3 p(0.005 m)2 4 = 0 VE = 0.9843 m>s

The flow at E is Q = VE AE = ( 0.9843 m>s ) 3 p(0.03 m)2 4 = 2.78 ( 10-3 ) m3 >s

Ans.

Substituting the result of VE into Eq. (1),

pE = 39240 - 500 ( 0.98432 ) = 38.76 ( 103 ) Pa = 38.8 kPa

Ans.

Ans: Q = 2.78 110 - 3 2 m3 >s pE = 38.8 kPa 479

*5–24. A fountain ejects water through the four nozzles, which have inner diameters of 10 mm. Determine the maximum height h of the water stream passing through the nozzles as a function of the volumetric flow rate into the 60-mm-diameter pipe at E. Also, what is the corresponding pressure at E as a function of h?

F

60 mm

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Since the water stream DF flow in the open atmosphere, pD = pF = 0. Also, when the stream achieves its maximum height at F, VF = 0. If we set the datum along the streamline coinciding with the centerline of the pipe, zD = zE = 0 and zF = h. Writing between points D and F, pD pF VD2 VF 2 + gzD = + gzF + + r r 2 2 0 +

VD2 + 0 = 0 + 0 + ( 9.81 m>s2 ) h 2 VD =

Between E and F,

( 119.62 h ) m>s

pF pE VE 2 VF 2 + gzE = + gzF + + r r 2 2 pE 3

1000 kg>m

+

VE 2 + 0 = 0 + 0 + ( 981 m>s2 ) h 2

pE = °9.81 h -

VE 2 ¢ ( 103 ) Pa 2

(1)

Take the fixed control volume to be the water contained in the pipe, since there are four nozzles, 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VEAE + 4VDAD = 0 -VE 3 p(0.03 m)2 4 + 4 ( 119.62 h ) 3 p(0.005 m)2 4 = 0 VE = 0.4922 2h

The discharge is

Q = VE AE;

Q = ( 0.49221h ) 3 p(0.03 m)2 4 h =

3 516 ( 103 ) Q2 4 m where Q is in m3 >s

Ans.

Substituting the result of VE into Eq. (1), pE = C 9.81h -

1 0.4922 2h 2 2 2

S ( 103 )

pE = (9.69h) ( 103 ) Pa Ans.

pE = (9.69h) kPa where h is in meters

480

h E

A

B

C

D

5–25. Determine the velocity of water through the pipe if the manometer contains mercury held in the position shown. Take rHg = 13 550 kg>m3.

V

B

A

100 mm 50 mm 50 mm

200 mm

SOLUTION

h

AC

Bernoulli Equation. Since point B is a stagnation point, VB = 0. If the datum is along the horizontal streamline connecting A and B, zA = zB = 0. pB pA VA2 VB 2 + gzA = + gzB + + r r 2 2 pA 1000 kg>m3

= 0.15 m A

B h

BD

C

= 0.2 m

D

2

+

pB VA + 0 + 0 + 0 = 2 1000 kg>m3

h

CD

pA - pB = - 500VA2

= 0.05 m (a)

Manometer Equation. Referring to Fig. a, pA + rwghAC + rHg ghCD - rwghBD = pB pA + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.15 m) +

( 13 550 kg>m3 ) ( 9.81 m>s2 ) (0.05 m) - ( 1000 kg>m3 )( 9.81 m>s2 ) (0.2 m) = pB pA - pB = - 6155.78 Solving Eqs. (1) and (2) Ans.

VA = 3.51 m>s

Ans: 3.51 m>s 481

5–26. A water-cooled nuclear reactor is made with plate fuel elements that are spaced 3 mm apart and 800 mm long. During an initial test, water enters at the bottom of the reactor (plates) and flows upwards at 0.8 m>s. Determine the pressure difference in the water between A and B. Take the average water temperature to be 80°C.

B

800 mm

SOLUTION The water can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. Applying this equation between A and B, where both points are on the central streamline, A

pA pB VA2 VB 2 + + gzA = + + gzB rW rW 2 2 Since the inlet (A) and outlet (B) control surfaces have the same cross-sectional area, continuity requires that VA = VB = V. Here, the datum will be set through point A. Then,

3 mm 3 mm

60 mm

pA pB V2 V2 + + 0 = + + gh rW rW 2 2 pA - pB = rw gh From Appendix A, rw = 971.6 kg>m3 at T = 80° C. Here, h = 0.8 m. Then pA - pB = ( 971.6 kg>m3 )( 9.81 m>s2 ) (0.8 m) = 7.625 ( 103 ) Pa = 7.63 kPa

Ans.

Ans: 7.63 kPa 482

5–27. Blood flows from the left ventricle (LV) of the heart, which has an exit diameter of d 1 = 16 mm, through the stenotic aortic valve of diameter d 2 = 8 mm, and then into the aorta A having a diameter of d 3 = 20 mm. If the cardiac output is 4 liters per minute, the heart rate is 90 beats per minute, and each ejection of blood lasts 0.31 s, determine the pressure change over the valve. Take rb = 1060 kg>m3.

LV d 1 d2

d3

A

SOLUTION The volume of blood pumped per heartbeat is V =

4 L>min 90 beat>min

= ( 0.04444 L/beat ) °

Thus, the discharge of blood by LV is Q =

1 m3 ¢ = 44.44 ( 10-6 ) m3 >beat 1000 L

44.44 ( 10-6 ) m3 >beat V = = 0.1434 ( 10-3 ) m3 >s t 0.31 s>beat

Then, the average velocities of the blood flow from the LV and into the Aorta; V1 and V3, respectively are 0.1434 ( 10-3 ) m3 >s = V1 3 p ( 0.008 m ) 2 4

Q = V1A1;

V1 = 0.7131 m>s

0.1434 ( 10

Q = V3 A3:

-3

) m >s = V3 3 p ( 0.01 m ) 2 4 3

V3 = 0.4564 m>s

Writing Bernoulli’s equation between the two points, p1 p3 V3 2 V1 2 + + gz1 = + + gz3 rb rb 2 2 p1 1060 kg>m3

+

(0.7131 m>s)2 2

+ 0 =

p3 1060 kg>m3

+

(0.4564 m>s)2 2

∆ p = p3 - p1 = 159 Pa

+ 0 Ans.

Ans: 159 Pa 483

*5–28. Air enters the tepee door at A with an average speed of 2 m>s and exits at the top B. Determine the pressure difference between these two points and find the average speed of the air at B. The areas of the openings are AA = 0.3 m2 and AB = 0.05 m2. The density of the air is ra = 1.20 kg>m3.

B

SOLUTION Since the air can be considered as an ideal fluid (incompressible and inviscids) and the flow is steady, the Bernoulli’s equation is applicable. Consider the control volume to be the air within the nozzle. Continuity requires 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VA AA + VB AB = 0 - ( 2 m>s )( 0.3 m2 ) + VB ( 0.05 m2 ) = 0 Ans.

VB = 12 m>s Applying the Bernoulli equation between points A and B, rA pB VA2 VB 2 + + gzA = + + gzB ra ra 2 2 Since the density of the air is small, the elevation terms can be neglected. pA pB VA2 VB 2 + = + ra ra 2 2 ∆p = pB - pA =

∆p =

( 1.20 kg>m3 ) 2

= -84.0 Pa

ra ( V 2 - VA2 ) 2 B

3 (2 m>s)2

- ( 12 m>s ) 2 4

484

Ans.

A

5–29. One method of producing energy is to use a tapered channel (TAPCHAN), which diverts sea water into a reservoir as shown in the figure. As a wave approaches the shore through the closed tapered channel at A, its height will begin to increase until it begins to spill over the sides and into the reservoir. The water in the reservoir then passes through a turbine in the building at C to generate power and is returned to the sea at D. If the speed of the water at A is VA = 2.5 m>s, and the water depth is hA = 3 m, determine the minimum height of the channel to prevent water from entering the reservoir.

B A

C

VA ! 2.5 m/s D

SOLUTION The sea water can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. Apply this equation between points A and B along the streamline on the water surface, pA pB VA2 VB 2 + + gzA = + + gzB rsw rsw 2 2 The datum is set along the base of the channel, then zA = hA = 3 m, zB = hB. Since points A and B are on the water surface, pA = pB = patm = 0. We require VB = 0. 0 +

( 2.5 m>s ) 2 2

+ ( 9.81 m>s2 ) (3 m) = 0 + 0 + ( 9.81 m>s2 ) hB Ans.

hB = 3.32 m

Ans: 3.32 m 485

5–30. The large tank is filled with gasoline and oil to the depth shown. If the valve at A is opened, determine the initial discharge from the tank. Take rg = 1.41 slug>ft 3 and ro = 1.78 slug>ft 3.

C Gasoline

2 ft

B

Oil 0.5 ft

4 ft

A

SOLUTION Bernoulli Equation. Since the oil is discharged from a larger tank, VB = 0. The lb pressure at B is pB = rg ghBC = ( 1.41 slug>ft 3 )( 32.2 ft>s2 ) (2 ft) = 90.804 2 . Since ft the oil is discharged to the atmosphere, pA = 0. If the datum is at A, zA = 0 and zB = 4 ft. pB pA VB 2 VA2 + + + gzB = + gzA ro ro 2 2 lb VA2 ft 3 2 ( ) + 0 + 32.2 ft>s (4 ft) = 0 + + 0 2 1.78 slug>ft 3 90.804

VA = 18.96 ft>s Discharge. Q = VAAA = ( 18.96 ft>s ) 3 p(0.25 ft)2 4 = 3.72 ft 3 >s

Ans.

486

Ans: 3.72 ft 3 >s

5–31. Determine the air pressure that must be exerted at the top of the kerosene in the large tank at B so that the initial discharge through the drain pipe at A is 0.1 m3 >s once the valve at A is opened.

B

4m A 0.1 m

SOLUTION Since the kerosene can be considered as an ideal fluid (incompressible and inviscids) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points B and A, pB pA VB2 VA2 + gzB = + gzA + + rkc rkc 2 2 Since the kerosene tank is a large reservoir, VB ≃ 0. Also, the drain pipe is exposed to the atmosphere, pA = patm = 0. From Appendix A, rkc = 814 kg>m3. Here the datum is set through point A and so zB = 4 m and zA = 0. pB 814 kg>m3

+ 0 + ( 9.81 m>s2 ) (4 m) = 0 +

VA2 + 0 2

pB = 407 ( VA2 - 78.48 ) Pa

(1)

From the given discharge, Q = VA AA;

0.1 m3 >s = VA 3 p(0.05 m)2 4 VA = 12.73 m>s

Substituting this result into Eq. (1), pB = 407 ( 12.732 - 78.48 ) = 34.04 ( 103 ) Pa Ans.

= 34.0 kPa

Ans: 34.0 kPa 487

*5–32. If air pressure at the top of the kerosene in the large tank is 80 kPa, determine the initial discharge through the drainpipe at A once the valve is opened. B

4m A 0.1 m

SOLUTION Since the kerosene can be considered as an ideal fluid (incompressible and inviscids) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points B and A, pB pA VB 2 VA2 + + gzB = + + gzA rkc rkc 2 2 Since the kerosene tank is a large reservoir, VB = 0. Also the drain pipe is exposed to the atmosphere pA = patm = 0. From the Appendix A, rkc = 814 kg>m3. Here, the datum is set to pass through point A and so zB = 4 m and zA = 0. 80 ( 103 ) N>m2 3

814 kg>m

+ 0 + ( 9.81 m>s2 ) (4 m) = 0 +

VA 2 + 0 2

VA = 16.58 m>s Thus, the discharge through the drain pipe is Q = VAAA = ( 16.58 m>s ) 3 p(0.05 m)2 4 = 0.130 m3 >s

488

Ans.

5–33. Water flows up through the vertical pipe such that when it is at A, it is subjected to a pressure of 150 kPa and has a velocity of 3 m>s. Determine the pressure and its velocity at B. Set d = 75 mm.

B

d

2m

A 100 mm

SOLUTION Continuity Equation. d e dV + eV # dA = 0 dt Lcv Lcs 0 - VAAA + VBAB = 0 - ( 3 m>s ) 3 p(0.05 m)2 4 + VB 3 p(0.0375 m)2 4 = 0

Ans.

VB = 5.333 m>s = 5.33 m>s

Bernoulli Equation. If we set the datum at point A, ZA = 0 and zB = 2 m. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 £

150 ( 103 ) N>m2 1000 kg>m3

§ +

(3 m>s)2 2

+ 0 =

pB

( 1000 kg>m3 )

+

(5.333 m>s)2 2

pB = 120.66 ( 103 ) Pa = 121 kPa

+ ( 9.81 m>s2 ) (2 m) Ans.

Ans: VB = 5.33 m>s pB = 121 kPa 489

5–34. Water flows through the vertical pipe such that when it is at A, it is subjected to a pressure of 150 kPa and has a velocity of 3 m>s. Determine the pressure and velocity at B as a function of the diameter d of the pipe at B. Plot the pressure and velocity (vertical axis) versus the diameter for 25 mm … d … 100 mm. Give values for increments of ∆d = 25 mm. If dB = 25 mm, what is the pressure at B? Is this reasonable? Explain.

B

d

2m

A 100 mm

SOLUTION The fixed control volume contains the water in the pipe, d rdV = r V # dA = 0 dt Lcv Lcs 0 - VA AA + VB AB = 0 - ( 3 m>s ) 3 p(0.05 m)2 4 + VB £ VB = £

30 ( 103 ) dB 2

p dB 2 a b § = 0 4 1000

§ m>s where d B is in mm

Ans.

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation between points A and B. If we set the datum through point A, zA = 0 and zB = 2 m, then pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 d B(mm)

25

50

75

100

VB ( m>s )

48.0

12.0

5.33

3.00

VB (m s) 50

40

30

20

10

dB (mm) 0

25

50

75

100

(a)

490

5–34. Continued

£°

150 ( 103 ) N>m2 1000 kg>m3

¢§ +

( 3 m>s ) 2 2

pB = £ 134.88 pB = £ 135 -

pB

+ 0 =

450 ( 106 ) dB4

450 ( 106 ) dB4

1000 kg>m3

+

3 30 ( 103 ) >dB24 2 2

+ ( 9.81 m>s2 ) (2 m)

§ ( 103 ) Pa

§ kPa where d B is in mm

Ans.

The plot of VB VS d B and PB VS d B are shown in Fig. a and b respectively. Realistically, gage pressures less than - 101 kPa are physically impossible, and water will cavitate a few kPa above that. d B(mm)

25

50

75

100

pB(kPa)

- 1017

62.9

121

130

pB(kPa) 200 100

0

25

50

75

100

dB(mm)

–100 –200 –300 –400 –500 –600 –700 –800

Ans:

–900

VB = £

–1000 (b)

30 1 103 2 dB2

pB = £ 135 491

§ m>s, where dB is in mm.

450 1 106 ) dB4

§ kPa, where dB is in mm.

5–35. If the velocity of water changes uniformly along the transition from VA = 10 m>s to VB = 4 m>s, determine the pressure difference between A and x.

2m x

10 m/s

4 m/s A B

SOLUTION

6ms V(x)

Bernoulli Equation. Referring to Fig. a, V(x) = 4 m>s + a

6 m>s 2m

VA = 10 m s

b(2 m - x) = (10 - 3x) m>s

If the datum is set along the horizontal streamline, zA = z(x) = 0. 2

p(x) V (x) pA VA2 + gzA = + gzx + + r r 2 2 pA 3

1000 kg>m

+

( 10 m>s ) 2

2

+ 0 =

p(x)

( 1000 kg>m ) 3

+

3 (10

p(x) - pA = ( 30x - 4.5x2 )( 103 ) Pa

VB = 4 m s A

x

2m–x 2m

B

(a)

- 3x) m>s 4 2

= ( 30x - 4.5x2 ) kPa

2

+ 0

Ans.

Ans: p(x) - pA = 492

1 30x

- 4.5x2 2 kPa

*5–36. If the velocity of water changes uniformly along the transition from VA = 10 m>s to VB = 4 m>s, find the pressure difference between A and x = 1.5 m.

2m x

10 m/s

4 m/s A B

SOLUTION Bernoulli Equation. Referring to Fig. a, VC = 4 m>s +

0.5 m ( 6 m>s ) = 5.5 m>s 2m

If the datum is set along the horizontal streamline, zA = zC = 0. pC pA VC2 VA2 + gzA = + gzC + + r r 2 2 pA 3

1000 kg>m

+

( 10 m>s ) 2 2

+ 0 =

pC 3

1000 kg>m

+

( 5.5 m>s ) 2 2

pC - pA = 34.875 ( 103 ) Pa = 34.9 kPa

Ans.

6ms VC V A = 10 m s

VB = 4 m s A

x = 1.5 m 2m C

B

+ 0

0.5 m

(a)

493

5–37. Water flows up through the vertical pipe. Determine the pressure at A if the average velocity at B is 4 m>s.

B

40 mm

4 m/s 500 mm

A

120 mm VA

SOLUTION Since the water can be considered as ideal fluid (incompressible and inviscids) and the flow is steady, the Bernoulli Equation is applicable. Writing this equation between points A and B on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Since point B is exposed to the atmosphere, pB = patm = 0. Here the datum is set through A, then zA = 0 and zB = 0.5 m. pA 1000 kg>m3

+

VA2 + 0 = 0 + 2 pA =

( 4 m>s ) 2

3 500 ( 25.81

2

+ ( 9.81 m>s2 ) (0.5 m)

- VA2 ) 4 Pa

(1)

Consider the control volume to be the water in the pipe. The continuity requires that d rdV + eV # dA = 0 dt Lcv Lcs 0 - VAAA + VBAB = 0 - VA 3 p(0.06 m)2 4 + ( 4 m>s ) 3 p ( 0.02 m ) VA = 0.4444 m>s

2

4

= 0

Substitute this result into Eq. (1), PA =

3 500 ( 25.81

- 0.44442 ) 4 Pa

= 12.81 ( 103 ) Pa = 12.8 kPa

Ans.

Ans: 12.8 kPa 494

5–38. Water flows along the rectangular channel such that after it falls to the lower elevation, the depth becomes h = 0.3 m. Determine the volumetric discharge through the channel. The channel has a width of 1.5 m.

0.5 m A

1m

B

h

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscids) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation between points A and B on the streamline along the water surface, pA pB VA VB2 + + gzA = + + gzB pw 2 pw 2 Since the surface of the water is exposed to the atmosphere, pA = pB = patm = 0. Here the datum is set through points B, then zA = 1 m + 0.5 m - 0.3 m = 1.2 m and zB = 0 0 +

VA2 VB2 + ( 9.81 m>s2 ) (1.2 m) = 0 + + 0 2 2 VB2 - VA2 = 23.544

(1)

Consider the control volume to be the water from A to B. Continuity requires that 0 edV + eV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - VA[(1.5 m)(0.5 m)] + VB[(1.5 m)(0.3 m)] = 0 (2)

VA = 0.6VB solving Eqs. 1 and 2, VB = 6.065 m>s

VA = 3.639 m>s

Thus, the discharge is Q = VBAB = ( 6.065 m>s ) [(1.5 m)(0.3 m)] = 2.73 m3 >s

Ans.

495

Ans: 2.73 m3 >s

5–39. Water flows at 3 m>s at A along the rectangular channel that has a width of 1.5 m. If the depth at A is 0.5 m, determine the depth at B.

0.5 m A

1m

B

h

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation between points A and B on the streamline along the water surface, pA VA2 pB VB2 + + gzA = + + gzB pw 2 pW 2 Since the water surface is exposed to the atmosphere, pA = pB = patm = 0. Here the datum is through points B, then zA = 1 m + 0.5 m - h = (1.5-h) m and zB = 0. 0 +

( 3 m>s ) 2 2

+ ( 9.81 m>s2 ) (1.5- h) = 0 +

VB2 + 0 2

VB2 = 38.43 - 19.62h

(1)

The control volume considered contains the water from section A and B which is fixed. Continuity requires that 0 rdV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - (3 m>s)[1.5 m(0.5 m)] + VB [(1.5 m)h] = 0 VB =

1.5 h

(2)

substituting Eq. 2 into 1, a

1.5 2 b = 38.43 - 19.62h h

38.43h2 - 19.62h3 - 2.25 = 0 solving numerically, Ans.

h = 0.2598 m = 0.260 m

Ans: 0.260 m 496

*5–40. Air at a temperature of 40°C flows into the nozzle at 6 m>s and then exits to the atmosphere at B, where the temperature is 0°C. Determine the pressure at A.

300 mm 100 mm 6 m/s B A

SOLUTION Assume that air is an ideal fluid (incompressible and inviscids) and the flow is steady. Then Bernoulli’s equation is applicable. Writing this equation between points A and B on the central streamline, pa pB VA2 VB2 + + gzA = + + gzB (pa)A 2 (pa)B 2 From Appendix A, (pa)A = 1.127 kg>m3 ( T = 40°c) and (pa)B = 1.292 kg>m3 (T = 0°c). Since point B is exposed to the atmosphere pB = patm = 0. Here the datum coincides with central streamline. Then ZA = ZB = 0. pA

+

3

1.127 kg>m

pA =

(6 m>s)2 2

+ 0 = 0 +

3 0.5635 (VB2

VB2 + 0 2

- 36) 4 pa

(1)

Consider the control volume to be the air within the nozzle. For steady flow, the continuity condition requires 0 edV + eV # dA = 0 0t Lcv Lcs 0 - (pa)AVAAA + (pa)B VBAB = 0 - ( 1.127 kg>m3 )( 6 m>s ) 3 p(0.15 m)2 4 + ( 1.292 kg>m3 ) (VB) 3 p(0.05 m)2 4 = 0 VB = 47.10 m>s

substituting this result into Eq. 1 pA = 0.5635 ( 47.102 - 36 ) = 1229.98 Pa Ans.

= 1.23 kPa

497

5–41. Water flows through the pipe at A with a velocity of 6 m>s and at a pressure of 280 kPa. Determine the velocity of the water at B and the difference in elevation h of the mercury in the manometer.

150 mm 150 mm 150 mm

A

B 200 mm

6 m/s

h

SOLUTION

6 m/s

B

Continuity Equation. Consider the water within the pipe and transition.

hBC = 0.2 m

0 rd V + rV # dA = 0 0t Lcv Lcs

hCD = h

0 - VAAA + VBAB = 0 - ( 6 m>s ) 3 p(0.15 m)

2

4

D

C

+ VB 3 p(0.075 m)

2

VB = 24.0 m>s

4

= 0

(a)

Bernoulli Equation. If we set the datum to coincide with the horizontal streamline connecting A and B, zA = zB = 0. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 280 ( 103 ) N>m2 3

1000 kg>m

+

( 6 m>s ) 2 2

+ 0 =

pB 3

1000 kg>m

+

( 24.0 m>s ) 2 2

+ 0

pB = 10 ( 103 ) Pa Manometer Equation. Referring to Fig. a, pB + rwghBC - rHg ghCD = 0 10 ( 103 ) Pa + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.2 m) - ( 13 550 kg>m3 )( 9.81 m>s2 ) h = 0 Ans.

h = 0.090 m = 90 mm

Ans: 90 mm 498

5–42. In order to determine the flow in a rectangular channel, a 0.2-ft-high bump is added on its bottom surface. If the measured depth of flow at the bump is 3.30 ft, determine the volumetric discharge. The flow is uniform, and the channel has a width of 2 ft.

A 4 ft

B 3.30 ft 0.2 ft

SOLUTION Bernoulli Equation. Since surfaces A and B are exposed to the atmosphere, pA = pB = 0. If we set the datum at the base of the channel, zA = 4 ft and zB = 3.30 ft + 0.2 ft = 3.5 ft. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 0 +

VA2 VB2 + ( 32.2 ft>s2 ) (4 ft) = 0 + + ( 32.2 ft>s2 ) (3.50 ft) 2 2 VB2 - VA2 = 32.2

(1)

Continuity Equation. Consider the water from A to B as the control volume. 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - VA[(4 ft)(2 ft)] + VB(3.30 ft)(2 ft) = 0 (2)

VA = 0.825VB Solving Eqs. (1) and (2) yields VA = 8.284 ft>s VB = 10.04 ft>s Discharge. Q = VAAA = ( 8.284 ft>s ) [(4 ft)(2 ft)] = 66.3 ft 3 >s

Ans.

499

Ans: 66.3 ft 3 >s

5–43. As water flows through the pipes, it rises within the piezometers at A and B to the heights hA = 1.5 ft and hB = 2 ft. Determine the volumetric flow. hB

hA A

B

0.75 ft 1.25 ft

SOLUTION Continuity Equation. Consider the water within the transition from A to B to be the control volume. 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VBAB + VAAA = 0 - VB £ p a

1.25 ft 2 0.75 ft 2 b § + VA £ p a b § = 0 2 2 VA = 2.778VB

(1)

Bernoulli Equation. The static pressure at A and B is given by pA = rwghA = rw ( 32.2 ft>s2 ) (1.5 ft) = 48.3rw pB = rwghB = rw ( 32.2 ft>s2 ) (2 ft) = 64.4rw If we set the datum to coincide with the horizontal streamline connecting A and B, zA = zB = 0. pB pA VB2 VA2 + gzB = + gzA + + r r 2 2 64.4rw 48.3rw VB2 VA2 + + 0 = + + 0 rw rw 2 2 VA2 - VB2 = 32.2

(2)

Solving Eqs. (1) and (2) yields VA = 6.082 ft>s VB = 2.190 ft>s Discharge. Q = VAAA = ( 6.082 ft>s ) £ p a = 2.69 ft 3 >s

0.75 ft 2 b § 2

Ans.

500

Ans: 2.69 ft 3 >s

*5–44. The volumetric flow of water through the transition is 3 ft 3 >s. Determine the height it rises in the piezometer at A if hB = 2 ft. hB

hA A

B

0.75 ft 1.25 ft

SOLUTION The control volume considered contains the water within the transition from A to B which is fixed. Using the discharge, Q = VAAA;

3 ft 3 >s = VA £ pa

0.75 ft 2 b § 2

Q = VBAB;

3 ft 3 >s = VB £ pa

1.25 ft 2 b § 2

VA = 6.7906 ft>s

VB = 2.4446 ft>s

Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation between points B and A, pB pA VB2 VA2 + + gzB = + + gzA rw rw 2 2 The static pressures at A and B are pA = rwghA = rw ( 32.2 ft>s2 ) hA = (32.2 rw hA) lb>ft 2 pB = rwghB = rw ( 32.2 ft>s2 ) (2 ft) = (64.4 rw) lb>ft 2 If we set the datum to coincide with the horizontal streamline connecting A and B, then zA = zB = 0. Therefore, 64.4 rw + rw

( 2.4446 ft>s ) 2 2

+ 0 =

32.2 rwhA + rw

( 6.7906 ft>s ) 2 2

+ 0 Ans.

hA = 1.377 ft = 1.38 ft

501

5–45. Determine the flow of oil through the pipe if the difference in height of the water column in the manometer is h = 100 mm. Take ro = 875 kg>m3.

300 mm 150 mm A

B

h

SOLUTION Bernoulli Equation. Since point B is a stagnation point, VB = 0. If we set the datum to coincide with the horizontal line connecting A and B, zA = zB = 0. pA pB VA2 VB2 + + gzA = + + gzB ro ro 2 2 pA 3

875 kg>m

+

pB VA2 + 0 = + 0 + 0 2 875 kg>m3

pB - pA = 437.5VA2

(1)

Manometer Equation. Referring to Fig. a, pA + ro ghAC + rwghCD - poghBD = pB pA + ( 875 kg>m3 )( 9.81 m>s2 ) (a) + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.1 m) - ( 875 kg>m3 )( 9.81 m>s2 ) (a + 0.1 m) = pB (2)

pB - pA = 122.625 Equating Eqs. (1) and (2), 437.5 VA2 = 122.625 VA = 0.5294 m>s Discharge. Q = VAAA = ( 0.5294 m>s ) 3 p(0.150 m)2 4 = 0.0374 m3 >s A

Ans.

B

hAC = a

hBD = a + 0.1 D

C

hCD = 0.1 m

(a)

502

Ans: 0.0374 m3 >s

5–46. Determine the difference in height h of the water column in the manometer if the flow of oil through the pipe is 0.04 m3 >s. Take ro = 875 kg>m3.

300 mm 150 mm A

B

h

SOLUTION Since oil can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation between points A and B, pA pB VA2 VB2 + gzA = + gzB. + + ro ro 2 2 Since point B is a stagnation point, VB = 0. If we set the datum to coincide with the horizontal streamline connecting A and B, then zA = zB = 0. Also, from the discharge 0.04 m3 >s = VA 3 p(0.15 m)2 4

Q = VAAA; Therefore pA

( 875 kg>m3 )

+

( 0.5659 m>s ) 2 2

+ 0 =

VA = 0.5659 m>s

pB

( 875 kg>m3 )

+ 0 + 0 (1)

pB - pA = 140.10 Writing the manometer equation with reference to Fig. a, pA + poghAC + rwghCD - roghBD = pB pA + ( 875 kg>m3 )( 9.81 m>s2 ) (a) + ( 1000 kg>m3 )( 9.81 m>s2 ) (h) - ( 875 kg>m3 )( 9.81 m>s2 ) (a + h) = pB

(2)

pB - pA = 1226.25 h Equating Eqs. (1) and (2) 1226.25 h = 140.10

Ans.

h = 0.1142 m = 114 mm

A

B

hAC = a hBD = a + h

C

hCD = h

D

(a)

Ans: 114 mm 503

5–47. Air at 60°F flows through the duct such that the pressure at A is 2 psi and at B it is 2.6 psi. Determine the volumetric discharge through the duct.

8 in. B 8 in.

6 in. A 6 in.

SOLUTION Assume that air is an ideal fluid (incompressible and inviscid) and the flow is steady. Then, the Bernoulli equation is applicable. Writing this equation between points A and B on central streamline, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Here, pA = a2

lb 12 in 2 lb 12 in 2 2 ba b = 288 lb>ft and p = a2.6 ba b = 374.4 lb>ft 2 B 1 ft 1 ft in2 in2

From Appendix A, at 60°F, ra = 0.00237 slug>ft 3. Here, the datum coincides with the central streamline. Then, zA = zB = 0 288 lb>ft 2 0.00237 slug>ft 3

+

374.4 lb>ft 2 VA2 VB2 + 0 = + 2 2 0.00237 slug>ft 3

VA2 - VB2 = 72.911 ( 103 )

(1)

Consider the air within the dust as the control volume. Continuity condition requires that 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VA AA + VB AB = 0 -VA £ a

6 6 8 8 ft ba ft b § + VB £ a ft ba ft b § = 0 12 12 12 12 VB = 0.5625VA

(2)

Solving Eqs 1 and 2 VA = 326.59 ft>s

VB = 183.71 ft>s

The discharge is Q = VAAA = ( 326.59 ft>s ) £ a = 81.6 ft 3 >s

6 6 ft ba ft b § 12 12

504

Ans.

Ans: 81.6 ft 3 >s

*5–48. Air at 100°F flows through the duct at A at 200 ft>s under a pressure of 1.50 psi. Determine the pressure at B.

8 in. B 8 in.

6 in. A 6 in.

SOLUTION Assume that air is an ideal fluid (incompressible and inviscid) and the flow is steady. Then, the Bernoulli equation is applicable. Writing this equation between points A and B on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 lb 12 in 2 ba b = 216 lb>ft 2. From Appendix A, at 100 °F, ra = 1 ft in2 0.00220 slug>ft 3. Here, the datum coincides with the central streamline. Then, zA = zB = 0.

Here, pA = a1.50

216 lb>ft 2 0.00220 slug>ft 3

1200 ft>s 2 2

+

2

+ 0 =

pB 0.00220 slug>ft 3

+

VB2 + 0 2

pB = 0.00110 3 236.36 ( 103 ) - VB2 4 lb>ft 2

(1)

Consider the air within the dust as the control volume. Continuity requires that 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VA AA + VB AB = 0 - 1200 ft>s 2 £ a

6 6 8 8 ft ba ft b § + VB £ a ft ba ft b § = 0 12 12 12 12 VB = 112.5 ft>s

Substituting this result into Eq 1 pB = 0.001100 3 236.36 ( 103 ) - 112.52 4 = a246.08

lb 1 ft 2 ba b = 1.71 Psi 2 12 in ft

Ans.

505

5–49. Carbon dioxide at 20°C flows past the Pitot tube B such that mercury within the manometer is displaced 50 mm as shown. Determine the mass flow if the duct has a crosssectional area of 0.18 m2.

A

B

50 mm

SOLUTION Assume that carbon dioxide is an ideal fluid (incompressible and inviscid) and the flow is steady. Thus, the Bernoulli equation is applicable. Writing this equation between points A and B on the central streamline

A B

pA pB VA2 VB2 rCO2 + 2 + gzA = rCO2 + 2 + gzB. Here VB = 0 since it is a stagnation point. From Appendix A, rCO2 = 1.84 kg>m3 at T = 20°C. Here, the datum coincides with the central streamline. Then zA = zB = 0. pA 1.84 kg>m3

+

VA2 2

+ 0 =

pB 1.84 kg>m3

h = 0.05 m

+ 0 + 0 (a)

pB - pA = 0.92 VA2

(1)

Write the manometer equation, where the pressure contribution of the CO 2 is negligible compared to that of the mercury. pA + rHg ghHg = pB pB - pA = rHg ghHg pB - pA = ( 13550 kg>m3 )( 9.81 m>s2 ) 10.05 m2

pB - pA = 6646.275

(2)

Substitute Eq 2 into 1 0.92 VA2 = 6646.275 VA = 84.995 m>s The mass flow rate is

#

m = rCO2 VAAA = ( 1.84 kg>m3 ) 184.995 m>s 2 ( 0.18 m2 ) = 28.2 kg>s

Ans.

Ans: 28.2 kg>s 506

5–50. Oil flows through the horizontal pipe under a pressure of 400 kPa and at a velocity of 2.5 m>s at A. Determine the pressure in the pipe at B if the pressure at C is 150 kPa. Neglect any elevation difference. Take ro = 880 kg>m3.

100 mm

75 mm

B

A C

25 mm

SOLUTION Bernoulli Equations. Since the flow occurs in the horizontal plane, the elevation terms can be ignored. From A to C, pA pC VC2 VA2 + gzA = + gzC + + r r 2 2 400 ( 103 ) N>m3 880 kg>m3

+

12.5 m>s 22 2

+ 0 =

150 ( 103 ) N>m3

+

880 kg>m3

VC2 + 0 2

VC = 23.97 m>s From A to B, pA pB VA2 VB2 + zA = + zB + + r r 2 2 400 ( 103 ) N>m3 880 kg>m3

+

12.5 m>s 2 2 2

+ 0 =

pB 880 kg>m3

+

VB2 + 0 2

pB = 880 ( 457.67 - 0.5VB2 )

(1)

Continuity Equation. Consider the oil in the pipe as the control volume. 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCAC = 0 - 12.5 m>s 2 3 p10.05 m2 2 4 + 123.97 m>s 2 3 p10.0125 m2 2 4 + VB 3p10.0375 m2 24 = 0 VB = 1.781 m>s

Substituting the result of VB into Eq. (1), pB = 401.35 ( 103 ) Pa = 401 kPa

Ans.

Ans: 401 kPa 507

5–51. Oil flows through the horizontal pipe under a pressure of 100 kPa and a velocity of 2.5 m>s at A. Determine the pressure in the pipe at C if the pressure at B is 95 kPa. Take ro = 880 kg>m3.

100 mm

75 mm

B

A C

25 mm

SOLUTION Bernoulli Equations. Since the flow occurs in the horizontal plane, the elevation terms can be ignored. From A to B, pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 100 ( 103 ) N>m2 3

880 kg>m

+

( 2.5 m>s ) 2 2

+ 0 =

95 ( 103 ) N>m2 3

880 kg>m

+

VB2 + 0 2

VB = 4.197 m>s From A to C, pA pC VC2 VA2 + gzA = + gzC + + r r 2 2 100 ( 103 ) N>m3 880 kg>m3

+

( 2.5 m>s ) 2 2

+ 0 =

pC 880 kg>m3

+

VC2 + 0 2

pC = 880 ( 116.76 - 0.5VC2 )

(1)

Continuity Equation. Consider the oil in the pipe are the control volume. 0 rdV + rV # dA = 0 0t L L cs cv 0 - VAAA + VBAB + VC AC = 0 - ( 2.5 m>s ) 3 p(0.05 m)

2

4

+ ( 4.197 m>s ) 3 p(0.0375 m)2 4 + VC 3 p(0.0125 m)2 4 = 0 VC = 2.228 m>s

Substituting the result of VC into Eq. (1), pC = 100.57 ( 103 ) Pa = 101 kPa

Ans.

Ans: 101 kPa 508

*5–52. Water flows through the pipe transition at a rate of 6 m>s at A. Determine the difference in the level of mercury within the manometer. Take rHg = 13 550 kg>m3.

150 mm 75 mm B

A

h

SOLUTION Continuity Equation. Consider the water in the pipe as the control volume. 0 rdV + rV # dA = 0 0t L L cs cv 0 + VAAA - VBAB = 0

( 6 m>s ) 3 p(0.0375 m)2 4 - VB 3 p(0.075 m)2 4 = 0 VB = 1.5 m>s

Bernoulli Equation. If we set the datum to coincide with the horizontal line connecting points A and B, zA = zB = 0. pB pA VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA 1000 kg>m3

+

( 6 m>s ) 2 2

+ 0 =

pB 1000 kg>m3

+

( 1.5 m>s ) 2 2

+ 0 (1)

pB - pA = 16 875 Pa Manometer Equation. Referring to Fig. a, pA + rwghAC + r Hg ghCD - rwghBD = pB

pA + ( 1000 kg>m3 )( 9.81 m>s2 ) (a) + ( 13 550 kg>m3 )( 9.81 m>s2 ) (h) - ( 1000 kg>m3 )( 9.81 m>s2 ) (a + h) = pB (2)

pB - pA = 123 115.5h Equating Eqs. (1) and (2), 16 875 = 123 115.5h

Ans.

h = 0.1371 m = 137 mm

A

B

hAC = a hBD = a + h C

hCD = h D

(a)

509

5–53. Due to the effect of surface tension, water from a faucet tapers from a diameter of 0.5 in. to 0.3 in. after falling 10 in. Determine the average velocity of the water at A and at B.

A 0.5 in.

10 in.

B

0.3 in.

SOLUTION Continuity Equation. Consider the water stream as the control volume. 0 rdV + rV # dA = 0 0t L L cs cv 0 - VAAA + VBAB = 0 - VA c

p p (0.5 in.)2 d + VB c (0.3 in.)2 d = 0 4 4

(1)

VA = 0.36VB

Bernoulli Equation. Since the water flows in the open atmosphere, pA = pB = 0. 10 If we set the datum at B, zA = ft and zB = 0. 12 pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 0 +

VA2 VB2 10 + ( 32.2 ft>s2 ) a ft b = 0 + + 0 2 12 2 VB2 - VA2 = 53.667

(2)

Solving Eqs. (1) and (2) yields VA = 2.83 ft>s

Ans.

VB = 7.85 ft>s

Ans.

Ans: VA = 2.83 ft>s VB = 7.85 ft>s 510

5–54. Due to the effect of surface tension, water from a faucet tapers from a diameter of 0.5 in. to 0.3 in. after falling 10 in. Determine the mass flow in slug>s.

A 0.5 in.

10 in.

B

0.3 in.

SOLUTION Continuity Equation. Consider the water stream as the control volume. 0 rdV + rV # dA = 0 0t L L cs cv 0 - VAAA + VBAB = 0 p p - VA c (0.5 in.)2 d + VB c (0.3 in.)2 d = 0 4 4

(1)

VA = 0.36VB

Bernoulli Equation. Since the water flow in the open atmosphere, pA = pB = 0. 10 If we set the datum at B, zA = ft and zB = 0. 12 pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 0 +

VA2 VB2 10 + ( 32.2 ft>s2 ) a ft b = 0 + + 0 2 12 2 VB2 - VA2 = 53.667

(2)

Solving Eqs. (1) and (2) yields VA = 2.827 ft>s VB = 7.852 ft>s Mass Flow Rate. 2 # 62.4 p 0.5 slug>ft 3 b ( 2.827 ft>s ) c a ft b d m = rVAAA = a 32.2 4 12

= 0.00747 slug>s

Ans.

Ans: 0.00747 slug>s 511

5–55. Air at 15°C and an absolute pressure of 275 kPa flows through the 200-mm-diameter duct at VA = 4 m>s. Determine the absolute pressure of the air after it passes through the transition and into the 400-mm-diameter duct B. The temperature of the air remains constant.

200 mm

A

400 mm

B

SOLUTION Here, the flow is steady. The control volume can be classified as fixed since its volume does not change with time (contains the air in the 200-mm diameter duct, the transition and 400-mm diameter duct). 0 rdV + rV # dA = 0 0t L L cv cs Since the flow is steady and the control volume is fixed, there are no local changes. Thus, the above equation becomes L cs

rV # dA = 0

Using the average velocities, and treating the air as incompressible so that rA = rB, (1)

-VAAA + VBAB = 0

Using the ideal gas law with R = 286.9 J>kg # K for air (Appendix A) 275 ( 103 )

pA = rARTA;

N = rA ( 286.9 J>kg # k ) (15 + 273) K m2

rA = 3.3282 kg>m3 Substitue into Eq 1 - ( 4 m>s ) c

p p (0.2 m)2 d + (VB) c (0.4 m)2 d = 0 4 4 VB = 1 m>s

Applying the Bernoulli equation since the air is assumed incompressible and no temperature change occurs, we have pA pB VA2 VB2 + + gzA = + + gzB rA rB 2 2 275 ( 103 ) N>m2 3.3282 kg>m3

+

( 4 m>s ) 2 2

+ 0 =

pB 3.3282 kg>m3

+

( 1 m>s ) 2 2

+ 0 Ans.

pB = 275.025 kPa

The very small change in pressure is consistent with our assumption that the density remains constant.

Ans: 275.025 kPa 512

*5–56. Air at 15°C and an absolute pressure of 250 kPa flows through the 200-mm-diameter duct at VA = 20 m>s. Determine the rise in pressure, ∆r = pB - pA, when the air passes through the transition and into the 400-mm-diameter duct. The temperature of the air remains constant.

200 mm

A

SOLUTION Assume the flow is steady and the control volume can be classified as fixed since its volume does not change with time. Applying the continuity equation, 0 rdV + rV # dA = 0 0t L L cv cs 0 - rAVAAA + rBVBAB = 0 But we can assume rA = rB, so that VAAA = VBAB p 4

p 4

( 20 m>s ) c (0.2 m)2 d = VB c (0.4 m)2 d VB = 5 m>s

The density of the air is determined from the ideal gas law 250 ( 103 )

pA = rARTA;

N = rA ( 286.9 J>kg # k ) (15° + 273) K m2

rA = 3.0256 kg>m3 Applying Bernoulli’s equation since the air is assumed incompressible and no temperature change occur, we have pA pB VA2 VB2 + + gzA = + + gzB rA rB 2 2 pA 3.0256 kg>m3

+

(20 m>s)2 2

+ 0 =

pB 3.0256 kg>m3

+

(5 m>s)2 2

+ 0

∆p = pB - pA Ans.

= 567 Pa

513

400 mm

B

5–57. Water flows in a rectangular channel over the 1-m drop. If the width of the channel is 1.5 m, determine the volumetric flow in the channel.

A VA

0.5 m

1m

B 0.2 m

VB

SOLUTION Continuity Equation. Consider the water from A to B as the control volume.

ΣV # A = 0;

0 rdV + rV # dA = 0 0t L L cv cs 0 - VAAA + VBAB = 0 - VA [0.5 m(1.5 m)] + VB [0.2 m(1.5 m)] = 0 (1)

VA = 0.4VB

Bernoulli Equation. Since surfaces A and B are exposed to the open atmosphere, pA = pB = 0. If we set the datum at the lower base of the channel, zA = (1 m + 0.5 m) = 1.5 m and zB = 0.2 m. pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 0 +

VA2 VB2 + ( 9.81 m>s2 ) (1.5 m) = 0 + + ( 9.81 m>s2 ) (0.2 m) 2 2

VB2 - VA2 = 25.506

(2)

Solving Eqs. (1) and (2) yields VA = 2.204 m>s

VB = 5.510 m>s

Discharge. Q = VAAA = ( 2.204 m>s ) [0.5 m(1.5 m)] = 1.65 m3 >s

514

Ans.

Ans: 1.65 m3 >s

5–58. Air at the top A of the water tank has a pressure of 60 psi. If water issues from the nozzle at B, determine the velocity of the water as it exits the hole, and the average distance d from the opening to where it strikes the ground.

A

4 ft B 2 ft

C

d

SOLUTION Bernoulli Equation. Since water is discharged into atmosphere at B, pB = 0. Also, it is discharged from a large source, VA ≅ 0. If we set the datum at B, zA = 4 ft and zB = 0. pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 a60 °

lb 12 in. 2 ba b 1 ft in2

62.4 lb>ft 3 32.2 ft>s2

¢

+ 0 + ( 32.2 ft>s2 ) (4 ft) = 0 +

VB2 + 0 2

VB = 95.78 ft>s

Vertical Motion.

(+T)

(sC)y = (sB)y + (VB)y t + 2 = 0 + 0 +

1 2 at 2 c

1 ( 32.2 ft>s2 ) t 2 2

t = 0.3525 s Horizontal Motion. + 2 1S (sC)x = (sB)x + (VB)xt

d = 0 + ( 95.78 ft>s ) (0.3525 s) = 33.8 ft

Ans.

Ans: 33.8 ft 515

5–59. Air is pumped into the top A of the water tank, and water issues from the small hole at B. Determine the distance d where the water strikes the ground as a function of the gage pressure at A. Plot this distance (vertical axis) versus the pressure pA for 0 … pA … 100 psi. Give values for increments of ∆pA = 20 psi.

A

4 ft B 2 ft

C

d

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Since the water is discharged into the atmosphere at B, pB = 0. Also, it is discharged from a large reservoir, VA = 0. If we set the datum at B, zA = 4 ft and zB = 0. pAa °

12 in 2 b 1 ft

62.4 lb>ft 3 32.2 ft>s2

+ 0 + ( 32.2 ft>s2 ) (4 ft) = 0 + ¢

VB = Vertical Motion,

(+T)

1 2148.62 pA

(sC)y = (sB)y + (VB)yt + 2 ft = 0 + 0 +

1 2 a t 2 C

VB2 + 0 2

+ 257.6 2 ft>s

1 ( 32.2 ft>s2 ) t 2 2

t = 0.3525 s Horizontal Motion, + 2 1S

(sC)x = (sB)x + (VB)xt

516

5–59.

Continued

d = 0 + d =

1 2148.62 pA

1 0.352 2149.6pA

The plot of d vs pA is shown in Fig. a pA(psi) d(ft)

+ 257.6 2 (0.3525 s)

+ 258. 2 ft where pA is in psi

0

20

40

60

80

100

5.66

20.0

27.8

33.8

38.8

43.3

Ans.

d(ft) 50

40

30

20

10

pA(psi) 0

20

40

60

80

100

(a)

Ans: d = 1 0.3522149.6 pA + 258 2 ft, where pA is in psi. 517

*5–60. Determine the height h of the water column and the average velocity at C if the pressure of the water in the 6-in.-diameter pipe at A is 10 psi and water flows past this point at 6 ft>s.

D

h 0.5 in. B A

SOLUTION Bernoulli Equation. Since the water column achieves a maximum height at D, VD = 0. Here, B and D are open to the atmosphere, pB = pD = 0. If the datum is set 3 3 horizontally at A, zA = 0, zB = ft = 0.25 ft, and zD = h + ft = h + 0.25 ft . 12 12 From A to D, pA pD VD2 VD2 + gzA = + gzD + + r r 2 2 lb 12 in. 2 ba b (6 ft>s)2 1 ft in2 + = 0 + 0 + ( 32.2 ft>s2 ) (h + 0.25 ft) 2 62.4 lb>ft 3

a10

32.2 ft>s2 Ans.

h = 23.39 ft = 23.4 ft From A to B, pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 lb 12 in. 2 ba b (6 ft>s)2 VB2 1 ft in2 + + 0 = 0 + + ( 32.2 ft>s2 ) (0.25 ft) 2 2 62.4 lb>ft 3

a10

32.2 ft>s2 Ans.

VB = 38.81 ft>s

Continuity Equation. Consider the water in the pipe from A to B to be the control volume, 0 rdV + rV # dA = 0 0t L L cv cs 0 - VAAA + VBAB + VCAC = 0 - ( 6 ft>s ) £ pa

2 2 2 3 0.25 3 ft b § + ( 38.81 ft>s ) £ pa ft b § + VC £ pa ft b § = 0 12 12 12

Ans.

VC = 5.73 ft>s

518

3 in. C

5–61. If the pressure in the 6-in.-diameter pipe at A is 10 psi, and the water column rises to a height of h = 30 ft, determine the pressure and velocity in the pipe at C.

D

h 0.5 in. B A

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady the Bernoulli’s equation is applicable. Writting this equation between points A and D, realizing that pD = 0 (D is open to atmosphere) and VD = 0 (the water column achives a maximum height at D). If the datum coincides with the 3 Central Stream line, zA = 0, and zD = a ft b + 30 ft = 30.25 ft . 12 pD pA VA2 VD2 + + gzA = + + gzD rw rw 2 2

a10 °

lb 12 in 2 ba b 1 ft in2

62.4 lb>ft 3 32.2 ft>s2

¢

+

VA2 2

+ 0 = 0 + 0 + ( 32.2 ft>s2 ) (30.25 ft)

2

VA = 230.97 2

VA = 21.49 ft>s

Between points B and D where pB = 0 (B is open to atmosphere) and 3 ft = 0.25 ft, zB = 12 pD pB VB2 VD2 + + gzB = + + gzD rw rw 2 2 0 +

VB2 + ( 32.2 ft>s2 ) (0.25 ft) = 0 + 0 + ( 32.2 ft>s2 ) (30.25 ft) 2

VB2 = 966 VB = 43.95 ft>s 2 Consider the fixed control volume that contains the water in the pipe from A to C, continuity requires that 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCAC = 0 ( - 21.49 ft>s) £ pa

2 2 2 3 0.25 3 ft b § + (43.95 ft>s) £ pa ft b § + VC £ pa ft b § = 0 12 12 12

Ans.

VC = 21.188 = 21.2 m>s

519

3 in. C

5–61. Continued

Between points A and C, pC pA VC2 VA2 + + 0 = + = 0 rw rw 2 2 a10 °

lb 12 in. 2 ba b 1 ft in2

62.4 lb>ft 3 32.2 ft>s2

+

( 21.49 ft>s ) 2 2

¢

pC

= °

pC = a1452.62

62.4 lb>ft 3 32.2 ft>s2

+

( 21.188 ft>s ) 2 2

¢

lb 1 ft 2 ba b = 10.1 psi 2 12 in. ft

Ans.

Ans: VC = 21.2 m>s pC = 10.1 psi 520

5–62. Determine the velocity of the flow out of the vertical pipes at A and B, if water flows into the Tee at 8 m>s and under a pressure of 40 kPa.

0.5 m

B 30 mm

5m

8 m/s C

SOLUTION

50 mm

3m

Continuity Equation. Consider the water within the pipe to be the control volume.

A

0 r dV + r V # dA = 0 0t Lcv Lcs

30 mm

0 - VCAC + VBAB + VAAA = 0 - ( 8 m>s ) 3 p(0.025 m)2 4 + VB 3 p(0.015 m)2 4 + VA 3 p(0.015 m)2 4 = 0 VA + VB = 22.22

(1)

Bernoulli Equation. Since the water discharged into the atmosphere at A and B, pA = pB = 0. If we set the datum horizontally through point C, zB = 5 m and zA = -3 m. pB pA VB2 VA2 + gzB = + gzA + + r r 2 2 0 +

VB2 VA2 + ( 9.81 m>s2 ) (5 m) = 0 + + ( 9.81 m>s2 ) ( - 3 m) 2 2 VA2 - VB2 = 156.96

(2)

Solving Eqs. (1) and (2) yields VA = 14.6 m>s

Ans.

VB = 7.58 m>s

Ans.

Note: Treating A and B as if they lie on the same streamline is a harmless shortcut. Officially, the solution process should proceed by considering two streamlines that each run through C.

Ans: VA = 14.6 m>s VB = 7.58 m>s 521

5–63. The open cylindrical tank is filled with linseed oil. A crack having a length of 50 mm and average height of 2 mm occurs at the base of the tank. How many liters of oil will  slowly drain from the tank in eight hours? Take ro = 940 kg>m3.

4m

3m

SOLUTION The linseed oil can be considered as an ideal fluid (incompressible and inviscid). Here, we assume that the flow is steady. Therefore, Bernoulli’s equation is applicable. Applying this equation between A and B Fig. a, where both are on the streamline shown, pA pB VA2 VB2 + gzA = + gzB + + r r 2 2

A

Since the tank is a large reservoir, VA _ 0. Therefore, VA2 is negligible. Since A and B are exposed to the atmosphere, pA = pB = 0. Here, the datum is set through point B. Then

h B

VB2 0 + 0 + gh = 0 + + 0 2

Datum

VB = 22gh = 22 ( 9.81 m>s2 ) h = 219.62 h

(1)

The control volume changes with time. The volume of the control volume at a particular instant is V = pr 2h = p(2 m)2h = (4ph) m3 0V 0h = 4p 0t 0t

(2)

Thus, 0 r dV + ro V # dA = 0 0t Lcv o Lcs ro

0V + roVBAB = 0 0t 0V + VBAB = 0 0t

Using Eq 1 and 2 4p

0h = - ( 219.62 h)[(0.05 m)(0.002 m)] 0t 0h = - 35.2484 ( 10-6 ) 2h 0t

0h = 35.2484 ( 10-6 ) 23 m = 61.1 ( 10-6 ) m>s, which indeed 0t is very small as compare to VB = 219.62 (3 m) = 7.67 m>s. Thus, the assumption When h = 3 m, VA = -

of VA _ 0 is quite reasonable. Integrating the above equation

522

(a)

5–63. Continued

h

L3 m 2h 0h

1

2h 2 `

h

3m

= -

L0

8(3600 s)

35.2484 ( 10-6 ) 0t

= -35.2484 ( 10-6 ) t `

8(3600 s) 0

1

2ah 2 - 1.732b = - 1.0152 h = 1.4993 m Thus, the volume of the leakage is Vle = p(2 m)2(3 m - 1.4993 m) = ( 18.858 m3 ) a

1000 L b = 18.9 ( 103 ) liters 1 m3

523

Ans.

Ans: 18.9 1 103 2 liters

*5–64. At the instant shown, the level of water in the conical funnel is y = 200 mm. If the stem has an inner diameter of 5 mm, determine the rate at which the surface level of the water is dropping.

100 mm

100 mm

A 300 mm y

SOLUTION We will select the vertical streamline containing the points A and B. Here the flow is steady since the stem opening is small in relation to the volume of water in the funnel. In other words, the water level in the funnel will drop at a very slow rate, which can be considered constant for a large value of y. The pressures pA = pB = 0 (the static gauge pressure) and the gravity datum is at B. Bernoulli Equation. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 0 +

VA2 VB2 + ( 9.81 m>s2 ) (0.250 m) = 0 + + 0 2 2

(1)

Continuity Equation. The velocities at A and B can be related by applying the continuity equation. A fixed control volume contains the water within the funnel at the instant shown. When y = 200 mm, the surface area at A has a radius which is determined by proportion. rA 100 mm = 200 mm 300 mm rA = 66.67 mm Thus 0 rdV + rV # dA = 0 0t L L cv cs 0 - VAAA + VBAB = 0 or

- VA 3 p(0.06667 m)2 4 + VB 3 p(0.0025 m)2 4 = 0 VB = 711.1VA

Substituting into Eq. (1) and solving, yields VA = 0.00311 m>s = 3.11 mm>s VB = 2.21 m>s dy = -VA = - 3.11 mm>s dt

Ans.

By comparison, note that the steady flow assumption is appropriate since VA ≈ 0 for these larger values of ya

524

5 mm B

50 mm Datum

5–65. If the stem of the conical funnel has a diameter of 5 mm, determine the rate at which the surface level of water is dropping as a function of the depth y. Assume steady flow. Note: For a cone, V = 13 pr 2h.

100 mm

100 mm

A 300 mm y

SOLUTION The control volume considered is deformable. It contains the water in the funnel and stem. Here, the volume of the water V0 in the stem is constant. Referring to the geometry shown in Fig. a r 0.1 = ; y 0.3

r =

5 mm B

1 y 3 0.1 m

Thus, the volume of the control volume is V =

1 2 1 1 2 1 pr h + V0 = pa yb (y) + V0 = py3 + V0 3 3 3 27 dy dV 1 = py2 dt 9 dt

r

Continuity requires that

0.3 m

0 rdV + rV # dA = 0 0t Lcv Lcs

y

Since r is constant for water, ra

dV + VBAB b = 0 dt

dy 1 py 2 + VB 3 p(0.0025 m)2 4 = 0 9 dt dy dt

= -£

56.25 ( 10-6 ) y2

The Negative sign indicates that y is decreasing.

(1)

§ VB

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Writing this equation dy between points A and B, realizing that VA = - , pA = pB = 0 (A and B are open dt to atmosphere) zB = 0 and zA = y + 0.05 m (datum is set through B), Then pB pA VA2 VB2 + + gzA = + + gzB rw rw 2 2 0 +

50 mm Datum

( dy>dt ) 2 2

+ ( 9.81 m>s2 ) (y + 0.05 m) = 0 +

VB2 = a VB =

dy dt

VB2 + 0 2

2

b + 19.62(y + 0.05)

dy 2 b + 19.62(y + 0.05) A dt

(2)

a

525

5–65. Continued

Substituting this result into Eq. 1, 56.25 ( 10-6 ) dy = -£ § dt y2

dy 2 b + 19.62(y + 0.05) B dt a

19.62(y + 0.05) dy = °¢ m>s dt B 0.3160(109)y4 - 1 19.6(y + 0.05) dy = °¢ m>s where y is in meters Ans. dt B 0.316(109)y4 - 1

Ans: dy 19.6(y + 0.05) = °¢ m>s, dt A 0.316 1 109 2 y4 - 1

where y is in meters. 526

5–66. Water flows from the large container through the nozzle at B. If the absolute vapor pressure for the water is 0.65 psi, determine the maximum height h of the contents so that cavitation will not occur at B.

A

h

B 0.2 ft

0.5 ft C

SOLUTION Bernoulli Equation. Since the water is discharged through B from a large source, VA ≅ 0. Here, A and C are open to the atmosphere, pA = pC = 0. Also, at B, the pressure is required to be equal to the vapor pressure (cavitation). Then,

0.3 ft

( pB ) abs = ( pB ) g + patm 0.65 psi = ( pB ) g + 14.7 psi

( pB ) g = a -14.05

lb 12 in. 2 lb ba b = - 2023.2 2 2 1 ft in ft

If we set the datum at C, zC = 0, zB = 0.5 ft, and zA = 0.5 ft + h. From A to B, pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 - 2023.2

0 + 0 + ( 32.2 ft>s2 ) (0.5 ft + h) = °

lb ft 2

62.4 lb>ft 3 32.2 lb>ft 2

+

VB2 + ( 32.2 ft>s2 ) (0.5 ft) 2

¢

VB2 = 64.4 h + 2088.05

(1)

From A to C, pC pA VC2 VA2 + gzA = + gzC + + r r 2 2 0 + 0 + ( 32.2 lb>ft 2 ) (0.5 ft + h) = 0 +

VC2 + 0 2

VC2 = 64.4 h + 32.2

(2)

Continuity Equation. Consider the water within the container of the control volume. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VBAB + VCAC = 0 -VB 3 p(0.1 ft)2 4 + VC 3 p(0.15 ft)2 4 = 0

(3)

VB = 2.25VC

Solving Eqs. (1) through (3) yields

VB = 50.6 ft>s VC = 22.5 ft>s Ans.

h = 7.36 ft

Ans: 7.36 ft 527

5–67.

Water drains from the fountain cup A to cup B. Determine the depth h of the water in B in order for steady flow to be maintained. Take d = 25 mm.

A

100 mm C

20 mm

B

h D

SOLUTION

d

Bernoulli Equation. Since A, C, B, and D are exposed to the atmosphere, pA = pC = pB = pD = 0. To maintain steady flow, the level of water in cups A and B must be constant. Thus, from A to C with the datum set at C, zC = 0 and zA = 0.1 m, pA pC VC2 VA2 + gzA = + gzC + + r r 2 2 0 + 0 + ( 9.81 m>s2 ) (0.1 m) = 0 +

VC2 + 0 2

VC = 1.401 m>s Continuity Equation. Consider the units within the cup B as the control volume. To meet the continuity requirement at C and D, 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VCAC + VDAD = 0 - ( 1.401 m>s ) 3 p(0.01 m)2 4 + VD 3 p(0.0125 m)2 4 = 0 VD = 0.8964 m>s

Bernoulli Equation. From B to D with datum set at D, zD = 0 and zB = h, pD pB VB2 VD2 + gzB = + gzD + + r r 2 2 0 + 0 + ( 9.81 m>s2 ) h = 0

( 0.8964 m>s ) 2 2

+ 0 Ans.

h = 0.04096 m = 41.0 mm

Ans: 41.0 mm 528

*5–68. Water drains from the fountain cup A to cup B. If the depth in cup B is h = 50 mm, determine the velocity of the water at C and the diameter d of the opening at D so that steady flow occurs.

A

100 mm C

20 mm

B

h D

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is required to be steady, Bernoulli’s equation is applicable. Since A, B, C and D are exposed to the atmosphere, pA = pB = pC = pD = 0. To maintain the steady flow, the level of water in cups A and B must be constant. Thus, VA = VB = 0. Between A and C with the datum at C, zC = 0 and zA = 0.1 m, pC pA VC2 VA2 + + gzA = + + gzC rw rw 2 2 0 + 0 + ( 9.81 m>s2 ) (0.1 m) = 0 +

VC2 + 0 2 Ans.

VC = 1.401 m>s = 1.40 m>s Between B and D with the datum at D, zB = 0.05 m and zD = 0. pB pD VB2 VD2 + + gzB = + + gzD rw rw 2 2 0 + 0 + ( 9.81 m>s2 ) (0.05 m) = 0 +

VD2 + 0 2

VD = 0.9904 m>s = 0.990 m>s The fixed control volume that contains the water in cup B will be considered. Continuity requires that 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VCAC + VDAD = 0 -(1.401 m>s) 3 p(0.01 m)2 4 + (0.9904 m>s)a d D = 0.02378 m = 23.8 mm

p 2 dD b = 0 4

529

Ans.

d

5–69. As air flows downward through the venturi constriction, it creates a low pressure A that causes ethyl alcohol to rise in the tube and be drawn into the air stream. If the air is then discharged to the atmosphere at C, determine the smallest volumetric flow of air required to do this. Take rea = 789 kg>m3 and ra = 1.225 kg>m3.

30 mm

B A

20 mm

200 mm D

40 mm

C

SOLUTION Manometer Equation. Since D is exposed to the atmosphere, pD = 0. Referring to Fig. a, pD + rea ghAD = pA 0 - ( 789 kg>m3 )( 9.81 m>s2 ) (0.2 m) = pA pA = -1548.02 Pa

A

Bernoulli Equation. Since air is discharged into the atmosphere at C, pC = 0. Neglecting the elevation terms, pA pC VC2 VA2 + + gzA = + + gzC ra ra 2 2

hAD = 0.2 m

VC2 VA2 - 1548.02 Pa + + 0 = 0 + + 0 3 2 2 1.225 kg>m

D

VA2 - VC2 = 2527.38

(1)

Continuity Equation. Consider the air within the tube of the control volume. 0 rdV + rV # dA = 0 0t Lcv Lcs

(a)

0 - VAAA + VCAC = 0 - VA 3 p(0.01 m)2 4 + VC 3 p(0.02 m)2 4 = 0

(2)

VA = 4VC

Solving Eqs. (1) and (2),

VA = 51.92 m>s VC = 12.98 m>s Discharge. Q = VCAC = ( 12.98 m>s ) 3 p(0.02 m)2 4 = 0.0163 m3 >s

Ans.

530

Ans: 0.0163 m3 >s

5–70. As air flows downward through the venturi constriction, it creates a low pressure at A that causes ethyl alcohol to rise in the tube and be drawn into the air stream. Determine the velocity of the air as it passes through the tube at B in order to do this. The air is discharged to the atmosphere at C. Take rea = 789 kg>m3 and ra = 1.225 kg>m3.

30 mm

B A

20 mm

200 mm D

C

40 mm

SOLUTION Manometer Equation. Since D is exposed to the atmosphere, pD = 0. Referring to Fig. a, pD + rea ghAD = pA 0 - ( 789 kg>m2 )( 9.81 m>s2 ) (0.2 m) = pA pA = - 1548.02 Pa Bernoulli Equation. Since air is discharged into the atmosphere at C, pC = 0. Neglecting the elevation terms, pA pC VC2 VA2 + + + gzA = + gzC ra ra 2 2 VC2 VA2 - 1548.02 Pa + + 0 = 0 + + 0 2 2 1.225 kg>m3 VA2 - VC2 = 2527.38

(1)

Continuity Equation. Consider the air in the tube of the control volume. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VCAC = 0 - VA 3 p(0.01 m)2 4 + VC 3 p(0.02 m)2 4 = 0 VA = 4VC

(2)

Solving Eqs. (1) and (2),

VA = 51.92 m>s VC = 12.98 m>s Also, consider the air in the tube the control volume. 0 - VBAB + VCAC = 0 - VB 3 p(0.015 m)2 4 + ( 12.98 m>s ) 3 p(0.02 m)2 4 = 0 VB = 23.1 m>s

Ans.

Ans: 23.1 m>s 531

5–71. Water from the large closed tank is to be drained through the lines at A and B. When the valve at B is opened, the initial discharge is QB = 0.8 ft 3 >s. Determine the pressure at C and the initial volumetric discharge at A if this valve is also opened.

C

4 ft 3 in. A

B

2 in.

SOLUTION QB = VBAB 0.8 ft 3 >s = VB £ p a

2 1 ft b § 12

VB = 36.67 ft>s

and QA = VAAA = VA c p a

2 1.5 ft b d = 0.04909 VA 12

(1)

Bernoulli Equation. Since the water is discharged from a large source, VC ≅ 0. Also, the water is discharged into the atmosphere at A and B, pA = pB = 0. If the datum coincides with the horizontal line joining A and B, zA = zB = 0 and zC = 4 ft. From C to B, pC pB VC2 VB2 + gzC = + gzB + + r r 2 2 pC °

62.4 lb>ft 3 32.2 ft>s2

+ 0 + ( 32.2 ft>s2 ) (4 ft) = 0 + ¢

pC = 1053.28 lb>ft 2 a

From C to A,

( 36.67 ft>s ) 2 2

+ 0

1 ft 2 b = 7.31 psi 12 in.

Ans.

pA pC VC2 VA2 + gzC = + gzA + + r r 2 2

1 1053.28 lb>ft2 2 °

62.4 lb>ft 3 32.2 ft>s2

¢

+ 0 +

1 32.2 ft>s2 2 (4 ft)

= 0 +

VA2 + 0 2

VA = 36.67 ft>s

Substituting the results of VA into Eq. (1), QA = 0.04909(36.67) = 1.80 ft 3 >s

Ans.

Ans: pC = 7.31 psi

532

QA = 1.80 ft 3 >s

*5–72. Water from the large closed tank is to be drained through the lines at A and B. When the valve at A is opened, the initial discharge is QA = 1.5 ft 3 >s. Determine the pressure at C and the initial volumetric discharge at B when this valve is also opened.

C

4 ft 3 in. A

SOLUTION QA = VAAA 1.5 ft 3 >s = VA c p a

2 1.5 ft b d 12

VA = 30.558 ft>s

and QB = VBAB = VB c p a

2 1 ft b d = 0.02182 VB 12

(1)

Bernoulli Equation. Since the water is discharged from a large source, VC ≅ 0. Also, the water is discharged into the atmosphere at A and B, pA = pB = 0. If the datum coincides with the horizontal line joining A and B, zA = zB = 0 and zC = 4 ft. From C to A, pC pA VC2 VA2 + gzC = + gzA + + r r 2 2 pC °

62.4 lb>ft 3 32.2 ft>s2

+ 0 + ( 32.2 ft>s2 ) (4 ft) = 0 + ¢ pC = 655.17 lb>ft 2 a

From C to B,

1 30.558 ft>s 2 2 2

1 ft 2 b = 4.55 psi 12 in.

+ 0

Ans.

pC pB VC2 VB2 + gzC = + gzB + + r r 2 2 655.17 lb>ft 2 °

62.4 lb>ft 3 32.2 ft>s2

¢

+ 0 +

1 32.2 ft>s2 2 (4 ft)

= 0 +

VB2 + 0 2

VB = 30.56 ft>s

Substituting the result of VB into Eq. (1), QB = 0.02182(30.56) = 0.667 ft 3 >s

Ans.

533

B

2 in.

5–73. Determine the volumetric flow and the pressure in the pipe at A if the height of the water column in the Pitot tube is 0.3 m and the height in the piezometer is 0.1 m.

0.1 m 0.3 m A

C

B

50 mm 150 mm

SOLUTION Continuity Equation. Consider the water in the pipe from A to B as the control volume. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - VA 3 p(0.025 m)2 4 + VB 3 p(0.075 m)2 4 = 0 VA = 9VB

(1)

Bernoulli Equation. Since point C is a stagnation point, VC = 0. If we set the datum to coincide with the horizontal line connecting points A, B, and C, zA = zB = zC = 0. The pressures at B and C are pB = rwghB = ( 1000 kg>m3 )( 9.81 m>s2 ) (0.1 m + 0.075 m) = 1.71675 ( 103 ) Pa pC = rwghC = ( 1000 kg>m3 )( 9.81 m>s2 ) (0.3 m) = 2.943 ( 103 ) Pa pC pA VC2 VA2 + gzA = + gzC + + r r 2 2 pA 1000 kg>m3

+

2.943 ( 103 ) Pa VA2 + 0 = + 0 + 0 2 1000 kg>m3

pA + 500VA2 = 2.943 ( 103 )

(2)

From C to B, pB pC VC2 VB2 + gzC = + gzB + + r r 2 2 2.943 ( 103 ) Pa 3

1000 kg>m

+ 0 + 0 =

1.71675 ( 103 ) Pa 3

1000 kg>m

+

VB2 + 0 2

VB = 1.566 m>s Using this result to solve Eqs. (1) and (2), VA = 14.094 m>s pA = - 96.38 ( 103 ) Pa = - 96.4 kPa

Ans.

Q = VBAB = ( 1.566 m>s ) 3 p(0.075 m)2 4

Ans.

Thus

Q = 0.0277 m3 >s

534

Ans: pA = -96.4 kPa Q = 0.0277 m3 >s

5–74. The mercury in the manometer has a difference in elevation of h = 0.15 m. Determine the volumetric discharge of gasoline through the pipe. Take rgas = 726 kg>m3.

150 mm 75 mm B

A

h

SOLUTION Since gasoline can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable writing this equation between points A and B on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB rga rga 2 2 With reference to the datum set on the central streamline, zA = zB = 0. From Appendix A, rga = 726 kg>m3. Then pA 3

726 kg>m

+

pB VA2 VB2 + + 0 = + 0 3 2 2 726 kg>m

pB - pA = 363 ( VA2 - VB2 )

(1)

Consider the gasoline in the pipe as the control volume. Continuity requires that 0 rdV + rV # dA = 0 0t Lcv Lcs

A

B

hAC = a

hBD = 0.15 m + a

0 - VAAA + VBAB = 0 C

- VA 3 p(0.0375 m)2 4 + VB 3 p(0.075 m)2 4 = 0

hCD = 0.15 m D

VA = 4VB

(a)

Substitute this result into Eq. (1) pB - pA = 363 3 ( 4VB ) 2 - VB2 4

pB - pA = 5445 VB2

(2)

Write the Manometer equation from point A to B, pA + rga ga + rHg gh - rgas g(a + h) = pB pB - pA = ( rHg -rgas ) gh From Appendix A, rHg = 13550 kg>m3. Thus pB - pA = ( 13550 kg>m3 - 726 kg>m3 )( 9.81 m>s2 ) (0.15 m) pB - pA = 18.871 ( 103 ) Pa Substitute this result into Eq. (2), 18.871 ( 103 ) = 5445 VB2 VB = 1.862 m>s Thus, the discharge through the pipe is Q = VBAB = ( 1.862 m>s ) 3 p(0.075 m)2 4 = 0.0329 m3 >s 535

Ans. Ans: 0.0329 m3 >s

5–75. If water flows into the pipe at a constant rate of 30 kg>s, determine the pressure acting at the inlet A when y = 0.5 m. Also, what is the rate at which the water surface at B is rising when y = 0.5 m? The container is circular.

B 30!

30!

y

A 0.3 m

SOLUTION The volume of the control volume changes with time. Since it contains the water in the container. Referring to the geometry shown in Fig. a, the volume is y 1 V = 2pΣgA = 2pe (0.075 m)(0.15 m)(y) + a0.15 m + tan 30° b c (y tan 30°)(y) d f 3 2 = 2p ( 0.05556y3 + 0.075 tan 30°y2 + 0.01125y ) m3

Then 0y 0y 0y 0V = 2p a0.1667y2 + 0.15 tan 30°y + 0.01125 b 0t 0t 0t 0t = 2p ( 0.1667y2 + 0.15 tan 30°y + 0.01125 )

Thus

0y 0t

0 r dV + rwV # dA = 0 0t Lcv w Lcs rw

0V - rwVAAA = 0 0t

(1000 kg) 3 2p ( 0.1667y2 + 0.15 tan 30°y + 0.01125 ) 4

0y -30 kg>s = 0 0t

0y 3 = 0t 200p ( 0.1667y2 + 0.15 tan 30°y + 0.01125 )

At the instant y = 0.5 m, VB =

0y 3 = 2 0t 200p 3 0.1667 ( 0.5 ) + 0.15 tan 30° (0.5) + 0.01125 4 = 0.04962 m>s = 0.0496 m>s

Ans.

Since water can be considered an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable writing this equation between points A and B on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Here, pB = patm = 0 since point B is exposed to the atmosphere. With reference to the datum set through point A, zA = 0 and zA = 0.5 m. Also, from the mass flow rate # m = rwVAAA; 30 kg>s = ( 1000 kg>m3 )( VA ) 3 p(0.05 m)2 4 VA = 3.8197 m>s

Thus, pA 3

1000 kg>m

+

( 3.8197 m>s ) 2 2

+ 0 = 0 +

( 0.04962 m>s ) 2 2

+ ( 9.81 m>s2 ) (0.5 m)

536

0.1 m

5–75.

Continued

pA = - 2.389 ( 103 ) Pa = - 2.39 kPa VB2 Note: Since VB is very small, then the term can be neglected. 2 pA 1000 kg>m3

+

( 3.8197 m>s ) 2 2

Ans.

+ 0 = 0 + 0 + ( 9.81 m>s2 ) (0.5 m)

pA = - 2.390 ( 103 ) Pa = - 2.39 kPa 0.15 m

y tan 30˚

0.15 m +

y

y tan 30˚ 3

30˚ 0.075 m

(a)

Ans: dy>dt = 0.496 m>s, p = - 2.39 kPa 537

*5–76. Carbon dioxide at 20°C passes through the expansion chamber, which causes mercury in the manometer to settle as shown. Determine the velocity of the gas at A. Take rHg = 13 550 kg>m3.

300 mm

A

C

B 150 mm

150 mm 30!

40 mm

100 mm

SOLUTION Bernoulli Equation. From Appendix A, rCO2 = 1.84 kg>m2 at T = 20° C . If we set the datum to coincide with the horizontal line connecting points A and B, zA = zB = 0. pA pB VA2 VB2 + + gzA = + + gzB rCO2 rCO2 2 2 pA 1.84 kg>m3

+

VA2 2

+ 0 =

pB 1.84 kg>m3

+

VB2 2

A

B

h 0.1 m sin30˚

+ 0

pB - pA = 0.920 ( VA2 - VB2 )

(a)

(1)

Continuity Equation. Consider the gas from A to B to the control volume. 0 rdV + rV # dA = 0 0t L L cv cs 0 - VAAA + VBAB = 0 - VA 3 p(0.075 m)2 4 + VB 3 p(0.15 m)2 4 = 0 VB = 0.25 VA

(2)

Manometer Equation. Referring to Fig. a, h = 0.1 m sin 30° - 0.04 m = 0.01 m. Then, neglecting the weight of the CO 2, pA + rHg gh = pB pA + ( 13 550 kg>m3 )( 9.81 m>s2 ) (0.01 m) = pB (3)

pB - pA = 1329.255 Equating Eqs. (1) and (3), 0.920 ( VA2 - VB2 ) = 1329.255 Substituting Eq. (2) into this equation, 0.9375VA2 = 1444.84 Thus,

Ans.

VA = 39.3 m>s

538

0.04 m

5–77. Determine the kinetic energy coefficient a if the velocity distribution for laminar flow in a smooth pipe has a velocity profile defined by u = Umax 1 1 - (r>R)2 2 .

r R

SOLUTION 1 a = # 2 V 2rV # dA mV Lcs V 3dA Lcs V dA = a = V 3A ( rVA ) V 2 Lcs r

a =

3

V 3A L0 1

2u 3max

R

2u 3max

R

R

u3(2prdr) 3

r2 °1 - 2 ¢ rdr a = 2 3 R R V L0 R V L0

a = a =

8

2u 3max

R8 V 3 L0 a =

V =

R

3

( R2 - r 2 ) 3rdr

( R6r - 3R 4r 3 + 3R 2r 5 - r 7 ) dr

2u 3max 8

R V

3

°

R8 1 u max ¢ = a b 8 4 V

(1)

u max R 1 r2 udA = °1 - 2 ¢(2prdr) 2 A Lcs pR L0 R V =

2u max

R 4 L0 V =

R

( R2r - r 3 ) dr

1 u 2 max

Substitute into Eq. (1), Ans.

a = 2

Ans: 2 539

5–78. Determine the kinetic energy coefficient a if the velocity distribution for turbulent flow in a smooth pipe is assumed to have a velocity profile defined by Prandtl’s oneseventh power law, u = Umax 1 1 - r>R 2 1>7.

r R

SOLUTION 1 a = # 2 V 2r V # dA mV L cs V 3dA Lcs a = V dA = rVAV 2 Lcs V 3A r

3

R

1 u3(2prdr) V 3A L0

a =

2u 3maxp

R hV A L0

a =

3

R

3

(R - r)7rdr

3

Set y = R - r dy = -dr a =

R R V 3A L

y 73 (R - y)( -dy)

3 7

2u 3maxp 7 17 7 17 c R7 R7d 3 3 10 17 R7 V A

a =

49u 3maxR2p

a = V =

0

2u 3maxp

3

85 V A

=

49u 3max

(1)

85 V 3

u max R 1 1 udA = 1 (R - r)7 (2pr dr) AL cs R 7A L0 V =

2pu max 1 7

RA

V =

LR

2pu max 1 7

RA V =

0

1

y 7(R - y)( -dy)

7 15 7 15 c R 7 - R7 d 8 15

pR2u max 49 c d 60 pR2

V =

49 u 60 max

Substitute into Eq. (1), Ans.

a = 1.058 = 1.06

Ans: 1.06 540

5–79. Oil flows through the constant-diameter pipe such that at A the pressure is 50 kPa, and the velocity is 2 m>s. Determine the pressure and velocity at B. Draw the energy and hydraulic grade lines for AB using a datum at B. Take ro = 900 kg>m3.

0.75 m

A

5m

2 m/s

5m C 30° B

SOLUTION Bernoulli Equation. Since the pipe has a constant diameter, Ans.

VB = VA = 2 m>s

With reference to the datum through B, zA = (10 m) sin 30° = 5 m and zB = 0. pB pA VA2 VB2 + zA = + zB + + g g 2g 2g N ( 2 m>s ) 2 ( 2 m>s ) 2 pB m2 + + 5m = + + 0 3 2 2 3 2 ( 900 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s ) ( 900 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s2 ) 50 ( 103 )

pB = 94.145 ( 103 ) Pa = 94.1 kPa

Ans.

El and HGL. EL will have a constant value of H =

pA VA2 + zA + g 2g

N ( 2 m>s ) 2 m2 = + + 5 m = 10.9 m ( 900 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 50 ( 103 )

Here, the velocity head is constant, with a value of (2 m>s)2 VA2 = = 0.204 m 2g 2 ( 9.81 m>s2 ) Thus, the HGL will be 0.204 m below and parallel to the EL. A plot of the EL and HGL is shown in Fig. a.

10.9 m

10.7 m

EGL

Velocity head = 0.204 m

HGL Hydrautic head

Datum A

B (a)

Ans: VB = 2 m>s pB = 94.1 kPa 541

*5–80. Oil flows through the constant-diameter pipe such that at A the pressure is 50 kPa, and the velocity is 2 m>s. Plot the pressure head and the gravitational head for AB using a datum at B. Take ro = 900 kg>m3.

0.75 m

A

5m

2 m/s

5m C 30° B

SOLUTION Bernoulli Equation. Since the pipe has a constant diameter, VB = VA = 2 m>s . With reference to the datum through B, zA = (10 m) sin 30° = 5 m and zB = 0. pA PB VA2 VB2 + zA = + zB + + g g 2g 2g N ( 2 m>s ) 2 m2 + + 5m = ( 900 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 50 ( 103 )

pB

( 900 kg>m3 )( 9.81 m>s2 )

+

(2 m>s)2 2 ( 9.81 m>s2 )

pB = 94.145 ( 103 ) Pa Therefore, the pressure head at A and B is pA = g

N m2 = 5.66 m 3 ( 900 kg>m )( 9.81 m>s2 )

pB = g

N m2 = 10.66 m ( 900 kg>m3 )( 9.81 m>s2 )

50 ( 103 )

94.145 ( 103 )

The gravitational head coincides with the centerline of the pipe. A plot of the pressure head and gravitational head is shown in Fig. a.

10.7 m

5.66 m

Pressure head Gravitational head

5m

A

B

Datum

(a)

542

+ 0

5–81. Water at a pressure of 80 kPa and a velocity of 2 m>s at A flows through the transition. Determine the velocity and the pressure at B. Draw the energy and hydraulic grade lines for the flow from A to B using a datum at B.

100 mm A 50 mm

300 mm

C 50 mm B

150 mm

SOLUTION Continuity Equation. Consider the water in the pipe as the control volume.

0.204 m 8.66 m

0 rdV + rV # dA = 0 0t L L cv

EGL 8.45 m

cs

3.26 m

0 - VAAA + VBAB = 0 - ( 2 m>s ) 3 p(0.05 m)2 4 + VB 3 p(0.025 m)2 4 = 0 VB = 8 m>s

HGL

Ans.

Bernoulli Equation. With reference to the datum through B, zA = 0.3 m and zB = 0. pA pB VA2 VB2 + zA = + zB + + g g 2g 2g

A

pB

( 1000 kg>m3 )( 9.81 m>s2 )

Datum

(a)

N ( 2 m>s ) 2 m2 + + 0.3 m ( 1000 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 80 ( 103 )

=

B

+

(8 m>s)2 2 ( 9.81 m>s2 )

pB = 52.943 ( 103 ) Pa = 52.9 kPa

+ 0 Ans.

EL and HGL. EL will have a constant value of H =

pA VA2 + zA + g 2g

N (2 m>s)2 m2 = + + 0.3 m = 8.66 m 3 2 ( 1000 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s2 ) 80 ( 103 )

The velocity head before A is

( 2 m>s ) 2 VA2 = = 0.204 m 2g 2 ( 9.81 m>s2 ) The velocity head after A is

( 8 m>s ) 2 VA2 = = 3.26 m 2g 2 ( 9.81 m>s2 ) The EL and HGL are plotted as shown in Fig. a.

Ans: VB = 8 m>s pB = 52.9 kPa 543

5–82. Water at a pressure of 80 kPa and a velocity of 2 m>s at A flows through the transition. Determine the velocity and the pressure at C. Plot the pressure head and the gravitational head for AB using a datum at B.

100 mm A 50 mm

300 mm

C 150 mm

50 mm B

SOLUTION Continuity Equation. Consider the water in the pipe as the control volume. 0 rdV + rV # dA = 0 0t L L cv

cs

0 - VAAA + VBAB = 0 - ( 2 m>s ) 3 p(0.05 m)2 4 + VB 3 p(0.025 m)2 4 = 0 VB = 8 m>s

Ans. Ans.

Also, vC = vB = 8 m>s

Bernoulli Equation. With reference to the datum through C, zA = 0.3 m - 0.15 m = 0.15 m and zC = 0. pC pA VC 2 VA2 + zA = + zC + + g g 2g 2g N ( 2 m>s ) 2 m2 + + 0.15 m ( 1000 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 80 ( 103 )

=

pC

( 1000 kg>m3 )( 9.81 m>s2 )

+

( 8 m>s ) 2 + 0 2 ( 9.81 m>s2 )

pC = 51.47 ( 103 ) Pa = 51.5 kPa

Ans.

8.15 m Pressure head

5.40 m 5.10 m

Gravitational head

0.3 m A

B

Datum

(a)

Ans: VC = 8 m>s pC = 51.5 kPa 544

5–83. Water flows through the constant-diameter pipe such that at A it has a velocity of 6 ft>s and a pressure of 30  psi. Draw the energy and hydraulic grade lines for the flow from A to F using a datum through CD.

10 ft

6 ft/s

B 5 ft

A 15 ft

C

10 ft

20 ft

E F 45!

D

SOLUTION EL and HGL. With reference to the datum through CD, zA = 15 ft . Thus, EL will have a constant value of H =

=

pA VA2 + zA + g 2g a30

lb 12 in. 2 ba b ( 6 ft>s ) 2 1 ft in2 + + 15 ft = 84.8 ft 62.4 lb>ft 3 2 ( 32.2 ft>s2 )

Since the pipe has a constant diameter, the velocity head will have a constant value of

( 6 ft>s ) 2 VA2 = 0.559 ft = 2g 2 ( 32.2 ft>s2 ) A plot of the EL and HGL is shown in Fig. a.

0.559 ft

EGL

84.8 ft 84.2 ft

HGL

A

B

C

D

E F

Datum

(a)

Ans: EGL at 84.8 ft, HGL at 84.2 ft 545

*5–84. The hose is used to siphon water from the tank. Determine the smallest pressure in the tube and the volumetric discharge at C. The hose has an inner diameter of 0.75 in. Draw the energy and hydraulic grade lines for the hose using a datum at C.

B 1 ft

A¿

A 0.5 ft

2 ft C

SOLUTION Bernoulli Equation. Since the tank is a large source, VA _ 0. Here, A and C are exposed to the atmosphere, pA = pC = 0. If the datum coincides with the horizontal line through C, zA = 2 ft, zB = 3 ft, and zC = 0. From A to C, pA pC VC2 VA2 + zA = + zC + + g g 2g 2g 0 + 0 + 2 ft = 0 +

VC2 2 ( 32.2 ft>s2 )

+ 0

VC = 11.35 ft>s The smallest pressure occurs at the maximum height of the hose, point B. Since the hose has a constant diameter, VB = VC = 11.35 ft>s . From A to B, pA pB VA2 VB2 + zA = + zB + + g g 2g 2g 0 + 0 + 2 ft = pB = a - 187.2

Discharge.

pB 62.4 lb>ft 3

+

( 11.35 ft>s ) 2 + 3 ft 2 ( 32.2 ft>s2 )

lb 1 ft 2 ba b = - 1.30 psi 2 12 in. ft

Q = VCAC = ( 11.35 ft>s ) £ p a VB2 2g

= 0.0348 ft 3 >s

=

VC2 2g

2 ft

Ans.

2 0.375 ft b § 12

Ans.

= 2 ft

EGL

HGL A

B

C

Datum

546

5–85. The hose is used to siphon water from the tank. Determine the pressure in the hose at points A′ and B. The hose has an inner diameter of 0.75 in. Draw the energy and hydraulic grade lines for the hose using a datum at B.

B 1 ft

A¿

A 0.5 ft

2 ft C

SOLUTION Bernoulli Equation. Since the tank is a large source, VA _ 0. Here, A and C are exposed to the atmosphere, pA = pC = 0. If the datum coincides with the horizontal line through C, zA = 2 ft, zB = 3 ft, and zC = 0. From A to C, pA pC VC2 VA2 + zA = + zC + + g g 2g 2g 0 + 0 + 2 ft = 0 +

VC2 2 ( 32.2 ft>s2 )

+ 0

VC = 11.35 ft>s Since the hose has a constant diameter, VA′ = VB = VC = 11.35 ft>s . From A to A′ and A to B, pA pA′ VA2 VA′2 + zA = + zA′ + + g g 2g 2g 0 + 0 + 2 ft =

and

pA′ = a - 124.8

pA′ 62.4 lb>ft 3

+

( 11.35 ft>s ) 2 + 2 ft 2 ( 32.2 ft>s2 )

lb 1 ft 2 ba b = - 0.867 psi ft 2 12 in.

Ans.

pA pB VA2 VB2 + zA = + zB + + g g 2g 2g 0 + 0 + 2 ft = pB = a - 187.2 A

–1 ft

–3 ft

pB 62.4 lb>ft

3

+

( 11.35 ft>s ) 2 + 3 ft 2 ( 32.2 ft>s2 )

lb 1 ft 2 ba b = - 1.30 psi 2 12 in. ft B

Ans.

C

Datum

EGL

HGL

Ans: pA′ = -0.867 psi pB = - 1.30 psi 547

5–86. Water is siphoned from the open tank. Determine the volumetric discharge from the 20-mm-diameter hose. Draw the energy and hydraulic grade lines for the hose using a datum at B.

A 0.4 m C 0.3 m B

SOLUTION Bernoulli Equation. Since the tank is a large source, VA = 0. Here, A and B are exposed to the atmosphere, pA = pB = 0. If the datum is set to coincide with the horizontal line through B, zA = 0.3 m + 0.4 m = 0.7 m and zB = 0. pB pA VA2 VB2 + zA = + zB + + g g 2g 2g 0 + 0 + 0.7 m = 0 +

VB2 2 ( 9.81 m>s2 )

+ 0

VB = 3.706 m>s Discharge. Q = VBAB = ( 3.706 m>s ) 3 p(0.01 m)2 4 = 0.00116 m3 >s

0.7 m

Ans.

EGL

HGL A

B

Datum

548

Ans: 0.00116 m3 >s

5–87. Gasoline is siphoned from the large open tank. Determine the volumetric discharge from the 0.5-in.diameter hose at B. Draw the energy and hydraulic grade lines for the hose using a datum at B.

A 2 ft 1.25 ft C

2 ft B

SOLUTION Bernoulli Equation. Since the tank is a large source, VA = 0. Here, A and B are exposed to the atmosphere, pA = pB = 0. If the datum is set to coincide with the horizontal line through B, zA = 2 ft + 1.25 ft = 3.25 ft and zB = 0. pB pA VA2 VB2 + zA = + zB + + g g 2g 2g 0 + 0 + 3.25 ft = 0 +

VB2 2 ( 32.2 ft>s2 )

+ 0

VB = 14.47 ft>s Discharge. Q = VBAB = ( 14.47 ft>s ) £ p a = 0.0197 ft 3 >s

3.25 ft

2 0.25 ft b § 12

Ans.

EGL

HGL A

B

Datum

549

Ans: 0.0197 ft 3 >s

*5–88. The pump discharges water at B at 0.05 m3 >s. If the friction head loss between the intake at A and the outlet at B is 0.9 m, and the power input to the pump is 8 kW, determine the difference in pressure between A and B. The efficiency of the pump is e = 0.7.

B

0.5 m A

SOLUTION Discharge. 0.05 m3 >s = VA 3 p(0.25 m)2 4

Q = VAAA;

VA = 0.2546 m>s

0.05 m3 >s = VB 3 p(0.125 m)2 4

Q = VBAB;

VB = 1.019 m>s # Here, ( Ws ) out K Pout = ePin = 0.7(8 kW) = 5.6 kW . Pout = Qghpump

5.6 ( 103 ) W = ( 0.05 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 )( hpump ) hpump = 11.42

Energy Equation. Take the water in the system from A to B as the control volume. If the datum coincides with the horizontal line through A, zA = 0 and zB = 2 m. pA pB VA2 VB2 + zA + hpump = + zB = V + hturb + hL + + g g 2g 2g

( 0.2546 m>s ) 2 + 0 + (11.42 m) = ( 1000 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s2 ) ( 1.019 m>s ) 2 pB + + 2 m + 0.9 m ( 1000 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pA 3

2

+

pB - pA = 83.06 ( 103 ) Pa = 83.1 kPa

550

0.25 m

2m

5–89. The power input of the pump is 10 kW and the friction head loss between A and B is 1.25 m. If the pump has an efficiency of e = 0.8, and the increase in pressure from A to B is 100 kPa, determine the volumetric flow of water through the pump.

B

0.5 m A

0.25 m

2m

SOLUTION From the discharge, Q = VA 3 p(0.25 m)2 4

Q = VAAA;

VA = a

16Q b m>s p

VB = a

64Q b m>s p

Q = VB 3 p(0.125 m)2 4

Q = VBAB;

# # Here, ( Ws ) out = r ( Ws ) in = 0.8(10 kW) = 8 kW . Then # ( Ws ) out = Qghpump 8 ( 103 ) W = Q ( 1000 kg>m3 )( 9.81 m>s2 ) hpump hpump = a

0.8155 bm Q

Consider the fixed control volume that contains water in the system from A to B. Writing the energy equation with the datum set through A, zA = 0 and zB = 2 m, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA

( 1000 kg>m )( 9.81 m>s ) 3

2

pB

( 1000 kg>m3 )( 9.81 m>s2 )

+

+

a

16Q 2 b p

2 ( 9.81 m>s a

64Q 2 b p

2

2 ( 9.81 m>s2 )

)

+ 0 +

0.8155 = Q

+ 2 m + 0 + 1.25 m

pB - pA =

8000 - 194.54 ( 103 ) Q2 - 31.88 ( 103 ) Q

100 ( 103 ) =

8000 - 194.54 ( 103 ) Q2 - 31.88 ( 103 ) Q

100 =

8 - 194.54Q2 - 31.88 Q

194.54Q3 + 131.88Q - 8 = 0 Solving numerically, Q = 0.06024 m3 >s = 0.0603 m3 >s

Ans.

551

Ans: 0.0603 m3 >s

5–90. As air flows through the duct, its absolute pressure changes from 220 kPa at A to 219.98 kPa at B. If the temperature remains constant at T = 60°C, determine the head loss between these points. Assume the air is incompressible.

B A

30 m

SOLUTION Energy Equation. Take the air from A to B to be the control volume. From Appendix A, ra = 1.060 kg>m3 at T = 60° C . The continuity condition requires that VA = VB = V since the duct has a constant diameter. Also, the pipe is level, so that zA = zB = z. pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 220 ( 103 ) N>m2

( 1.060 kg>m3 )( 9.81 m>s2 )

+ +

V2 + z + 0 = 2g

219.98 ( 103 ) N>m2

( 1.060 kg>m3 )( 9.81 m>s2 )

V2 + z + 0 + hL 2g Ans.

hL = 1.92 m

Ans: 1.92 m 552

5–91. Water in the reservoir flows through the 0.2-m-diameter pipe at A into the turbine. If the discharge at B is 0.5 m3 >s, determine the power output of the turbine. Assume the turbine runs with an efficiency of 65%. Neglect frictional losses in the pipe.

C

20 m

0.2 m B

A

SOLUTION 6m

Q = VBAB 0.5 m3 >s = VB 3 p(0.1 m)2 4 VB = 15.92 m>s

Energy Equation. Take the water in the pipe-turbine system from A to B to be the control volume. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the discharge is drawn from a large source, so VC = 0. If we set the datum through B, zC = 20 m and zB = 0. pB pC VC2 VB2 + zC + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 20 m + 0 = 0 +

( 15.92 m>s ) 2 + 0 + hturb + 0 2 ( 9.81 m>s2 )

hturb = 7.090 m # ( Ws ) in = Qgwhturb = ( 0.5 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (7.090 m) = 34.77 ( 103 ) W = 34.77 kW

Thus, Ans.

Pout = ePin = 0.65 (34.77) = 22.6 kW

Ans: 22.6 kW 553

*5–92. Water in the reservoir flows through the 0.2-m-diameter pipe at A into the turbine. If the discharge at B is 0.5 m3 >s, determine the power output of the turbine. Assume the turbine runs with an efficiency of 65%, and there is a head loss of 0.5 m through the pipe.

C

20 m

0.2 m B

6m

SOLUTION Q = VBAB 0.5 m3 >s = VB 3 p(0.1 m)2 4 VB = 15.92 m>s

Energy Equation. Take the water in the pipe-turbine system from A to B to be the control volume. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the discharge is drawn from a large source, so VC = 0. If we set the datum through B, zC = 20 m and zB = 0. pB pC VC2 VB2 + zC + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 20 m + 0 = 0 +

( 15.92 m>s ) 2 + 0 + hturb + 0.5 m 2 ( 9.81 m>s2 )

hturb = 6.590 m # Ws = Qgwhturb = ( 0.5 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (6.590 m) = 32.32 ( 103 ) W = 32.32 kW

# Ws = 32.32 kW(0.65) = 21.0 kW

Ans.

554

A

5–93. A 300-mm-diameter horizontal oil pipeline extends 8 km connecting two large open reservoirs having the same level. If friction in the pipe creates a head loss of 3 m for every 200 m of pipe length, determine the power that must be supplied by a pump to produce a flow of 6 m3 >min through the pipe. The ends of the pipe are submerged in the reservoirs. Take ro = 880 kg>m3.

SOLUTION Energy Equation. Take the oil in the pipes system to be the control volume. Here, pin = pout = 0 since the reservoirs are opened to the atmosphere. Also, Vin = Vout = 0 since the reservoirs are large. Since both reservoirs are at the same level zin = zout = z. pout pin Vout2 Vin2 + zin + hpump = + zout + hturb + hL + + g g 2g 2g 0 + 0 + z + hpump = 0 + 0 + z + 0 + (8000 m)a hpump = 120 m # WS = Qghpump = ( 6 m3 > min. ) a

3m b 200 m

1 min. b ( 880 kg>m3 )( 9.81 m>s2 ) (120 m) 60 s

= 103.59 ( 103 ) W = 104 kW

Ans.

Ans: 104 kW 555

5–94. A pump is used to deliver water from a large reservoir to another large reservoir that is 20 m higher. If the friction head loss in the 200-mm-diameter, 4-km-long pipeline is 2.5 m for every 500 m of pipe length, determine the required power output of the pump so the flow is 0.8 m3 >s. The ends of the pipe are submerged in the reservoirs.

SOLUTION Consider the fixed control volume as the water contained in the piping system. Here, pin = pout = 0, since the reservoirs are opened to the atmosphere. Also, Vin = Vout = 0 since the reservoirs are large. If the datum is set at the surface of the lower reservoir, zin = 0 and zout = 20 m. pin pout Vout2 Vin2 + + zin + hpump = + zout + hturb + hL + gw g 2g 2g 0 + 0 + 0 + hpump = 0 + 0 + 20 m + 0 + a hpump = 40 m

2.5 m b(4000 m) 500 m

Thus, the required power output of the pump is # WS = Qghpump = ( 0.8 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (40 m) = 313.92 ( 103 ) W

Ans.

= 314 kW

Ans: 314 kW 556

5–95. Water is drawn from the well at B through the 3-in.diameter suction pipe and discharged through the pipe of the same size at A. If the pump supplies 1.5 kW of power to the water, determine the velocity of the water when it exits at A. Assume the frictional head loss in the pipe system is 1.5V 2 >2g. Note that 746 W = 1 hp and 1 hp = 550 ft # lb>s.

2 ft A 3 ft

4 ft

3 ft C

3 ft D 5 ft B

SOLUTION P = Qghpump 1500 W a

1 hp 746 W

ba

550 ft # lb>s 1 hp

b = £ V£ pa

2 1.5 ft b § § ( 62.4 lb>ft 3 ) hpump 12

361.0 V Energy Equation. Take the water from D to A to be the control volume. Since A and D are exposed to the atmosphere, pA = pD = 0. Also, VD = 0 since the water is drawn from a large reservoir. Since the pipe has a constant diameter, the continuity condition requires that the water flows with a constant velocity of V into the pipe. With reference to the datum through B, zD = 5 ft and zA = 5 ft + 3 ft + 4 ft = 12 ft . hpump =

pA pD VD2 VA2 + zD + hpump = + zA + hturb + hL + + g g 2g 2g 0 + 0 + 5 ft + a

361.04 V2 1.5 V 2 b = 0 + + 12 ft + 0 + V 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) V 3 + 180.32 V - 9300.50 = 0

Solving numerically, Ans.

V = 18.19 ft>s = 18.2 ft>s

Ans: 18.2 ft>s 557

*5–96. Draw the energy and hydraulic grade lines for the pipe BCA in Prob. 5–95 using a datum at point B. Assume that the head loss is constant along the pipe at 1.5V 2 >2g.

2 ft A 3 ft

4 ft

3 ft C

3 ft D 5 ft B

SOLUTION # Ws = QghS 1500W a

1 hp 746 W

ba

550 ft # lb>s 1 hp

b = £ V £ pa

1.5 2 ft b § § ( 62.4 lb>ft 3 ) hs 12

361.0 V Energy Equation. Take the water from D to A to be the control volume. Since A and D are exposed to the atmosphere, pA = pD = 0. Also, VD = 0 since the water is drawn from a large reservoir. Since the pipe has a constant diameter, the continuity condition requires that the water flows with a constant velocity of V into the pipe. With reference to the datum through B, zD = 5 ft and zA = 5 ft + 3 ft + 4 ft = 12 ft . hpump =

pA pD VD2 VA2 + zD + hpump = + zA + hturb + hL + + g g 2g 2g 0 + 0 + 5 ft + a

2

20.6 ft 2

361.04 V 1.5 V b = 0 + + 12 ft + 0 + V 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s 2 )

EGL

17.1 ft

HGL

15.5 ft 12 ft

V 3 + 180.32V - 9300.50 = 0

Solving,

5 ft

V = 18.19 ft>s

D, B –0.139 ft

EGL and HGL. A + B, HB = HD =

pD VD2 + zD = 0 + 0 + 5 ft = 5 ft + g 2g

–4.38 ft

The total head loss from B to A is (hL)Tot =

EGL

0.761 ft

1.5 ( 18.19 ft>s ) 2 1.5V 2 = = 7.708 ft 2g 2 ( 32.2 ft>s2 )

Thus, the head loss per foot length of the pipe is

7.708 ft (5 + 3 + 3 + 3 + 4 + 2) ft

= 0.3854 ft>ft. And, the total head just before the pump is

( HC - ) = 5 ft - ( 0.3854 ft>ft ) (11 ft) = 0.761 ft Just after the pump, a head of hpump =

361.04 = 19.85 ft is added. Thus, 18.19

( HC + ) = 0.761 ft + 19.85 ft = 20.61 ft

558

C HGL

A

Datum

5–96. Continued

Then, the total head at A is HA = 20.6 ft - ( 0.3854 ft>ft ) (9 ft) = 17.1 ft Since the pipe has a constant diameter, the velocity head has a constant value of

( 18.19 ft>s ) V2 = = 5.14 ft 2g 2 ( 32.2 ft>s2 ) 2

Therefore, the HGL will always be 5.14 ft below and parallel to the EGL. Both are plotted as shown in Fig. a.

559

5–97. Determine the initial volumetric flow of water from tank A into tank B, and the pressure at end C of the pipe when the valve is opened. The pipe has a diameter of 0.25 ft. Assume that friction losses within the pipe, valve, and connections can be expressed as 1.28V 2 >2g, where V is the average velocity of flow through the pipe.

A

B

10 ft C

D

E

4 ft

2 ft

SOLUTION Energy Equation. We will write the energy equation between points A and B since these points are exposed to the atmosphere, pA = pB = patm = 0. Take the water from A to B to be the control volume. Also, the tanks can be considered as large reservoirs, VA = VB ≃ 0 with reference to the datum set through the base of the tank, zA = 10 ft and zB = 4 ft. pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 10 ft + 0 = 0 + 0 + 4 ft + 0 + hL hL = 6 ft Then hL = 1.28

VC2 ; 2g

6 ft = (1.28) £

VC2 2 ( 32.2 ft>s2 )

VC = 17.37 ft>s

§

Thus, the discharge is Q = VC AC = ( 17.37 ft>s ) 3 p(0.125 ft)2 4 = 0.853 ft 3 >s

Ans.

Now take the water from A to C to be the control volume. With reference to the datum set through the base of the tank, zA = 10 ft and zC = 2 ft. Thus, pA pC VC 2 VA2 + + zA + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 10 ft + 0 =

pC =

pC 62.4 lb>ft

3

+

( 17.37 ft>s ) 2 + 2 ft + 0 + 6 ft . 2 ( 32.2 ft>s2 )

( - 167.70 lb>ft 2 ) a

1 ft 2 b = -1.16 psi 12 in

560

Ans.

Ans: Q = 0.853 ft 3 >s, p = -1.16 psi

5–98. Draw the energy and hydraulic grade lines between points A and B using a datum set at the base of both tanks. The valve is opened. Assume that friction losses within the pipe, valve, and connections can be expressed as 1.28V 2 >2g, where V is the average velocity of flow through the 0.25-ft-diameter pipe.

A

B

10 ft C

D

E

4 ft

2 ft

SOLUTION We will write the energy equation between points A and B. So take that water from A to B to be the control volume. Since these points are exposed to the atmosphere, pA = pB = patm = 0. Also, the tank can be considered as large reservoir, VA = VB ≃ 0. With reference to the datum set through the base of the tank, zA = 10 ft and zB = 4 ft. pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 10 ft + 0 = 0 + 0 + 4 ft + 0 + hL hL = 6 ft Then hL = 1.28

VC2 ; 2g

6 ft = (1.28) £

VC2 2 ( 32.2 ft>s2 )

VC = 17.37 ft>s

§

Thus, the velocity head is hv =

( 17.37 ft>s ) 2 VC2 = = 4.6875 ft 2g 2 ( 32.2 ft>s2 )

Based on these results, the energy and hydraulic grade lines are plotted in Fig. a

10 ft EGL hv = 4.6875 ft

hL = 6 ft

5.3125 ft 4 ft HGL A –0.6875 ft

C

B

Datum

Ans: VC = 17.4 ft>s 561

5–99. Water is drawn into the pump, such that the pressure at the inlet A is - 35 kPa and the pressure at B is 120 kPa. If the discharge at B is 0.08 m3 >s, determine the power output of the pump. Neglect friction losses. The pipe has a constant diameter of 100 mm. Take h = 2 m.

C B

h

A

SOLUTION Energy Equation. Take the water from A to B to be the control volume. Since the pipe has a constant diameter, VA = VB = V. If the datum is set through A, zA = 0 and zB = 2 m. With hL = 0, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g - 35 ( 103 ) N>m2

( 1000 kg>m3 )( 9.81 m>s2 )

+

V2 + 0 + hpump = 2g

120 ( 103 ) N>m2

( 1000 kg>m3 )( 9.81 m>s2 ) +

V2 + 2m + 0 + 0 2g

hpump = 17.80 m # Ws = Qghpump = ( 0.08 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (17.80 m) = 13.97 ( 103 ) W = 14.0 kW

Ans.

Ans: 14.0 kW 562

5–100. Draw the energy and hydraulic grade lines for the pipe ACB in Prob. 5–99 using a datum at A.

C B

h

A

SOLUTION Discharge. Since the pipe has a constant diameter, the velocity in the pipe is constant throughout the pipe as required by the continuity condition. Q = VA;

0.08 m3 >s = V 3 p(0.05 m)2 4 V = 10.19 m>s

Energy Equation. Take the water from A to B to be the control volume. With reference to the datum through A, zA = 0 and zB = 2 m. With hL = 0.

5.29 19.5

EGL

14.2

HGL

pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g N V2 m2 = + + 0 + ( 1000 kg>m3 )( 9.81 m>s2 ) 2g - 35 ( 103 )

A –3.57

N V2 m2 = + + 2m + 0 + 0 ( 1000 kg>m3 )( 9.81 m>s2 ) 2g 120 ( 103 )

hpump

hpump = 17.80 m EGL and HGL. Since no losses occur, the total head before the pump is H =

EGL 1.72

pA VA2 + zA + g 2g

N ( 10.19 m>s ) 2 m2 = + + 0 = 1.72 m 3 2 ( 1000 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s2 ) -35 ( 103 )

After the pump, a head of 17 . 80 m is added to the water and becomes H = 1.72 m + 17.80 m = 19.5 m The velocity head has a constant value of

( 10.19 m>s ) V2 = = 5.29 m 2g 2 ( 9.81 m>s2 ) 2

The HGL is always 5.29 m below and parallel to the EL. Both are plotted as shown in Fig. a.

563

C

B Pump

HGL

5.29 m (a)

Datum

5–101. Water is drawn into the pump, such that the pressure at A is - 6 lb>in2 and the pressure at B is 20 lb>in2, If the volumetric flow at B is 4 ft 3 >s, determine the power output of the pump. The pipe has a diameter of 4 in. Take h = 5 ft and rw = 1.94 slug>ft 3.

C B

h

A

SOLUTION Energy Equation. Take the water from A to B to be the control volume. Since the pipe has a constant diameter, VA = VB = V. If we set the datum through A, zA = 0 and zB = 5 ft. With hL = 0, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g lb 12 in. 2 lb 12 in. 2 20 2 a a b b 2 2 ft ft V in in + + 0 + hpump = slug slug 2g ft ft 1.94 3 a32.2 2 b 1.94 3 a32.2 2 b s s ft ft -6

+ hpump = 64.93 # Ws = Qpump ghpump

V2 + 5 ft + h + 0 + 0 2g

= ( 4 ft 3 >s )( 1.94 slug>ft 3 )( 32.2 ft>s2 ) (64.93 ft) = 16225.36 ft .lb>s °

= 29.5 hp

1 hp

550 ft .lb>s

¢

Ans.

Ans: 29.5 hp 564

5–102. Draw the energy and hydraulic grade lines for the pipe ACB in Prob. 5–101 with reference to the datum at A.

C B

h

A

SOLUTION Discharge. Since the pipe has a constant diameter, the water velocity in the pipe is constant throughout the pipe as required by the continuity condition. 2 2 Q = VA; 4ft 3 >s = V £ p a ft b § 12

V = 45.84 ft>s

32.6 ft 83.7 ft

EGL

51.1 ft

HGL

Energy Equation. Take the water from A to B to be the control volume. With reference to the datum through A, zA = 0 and zB = 5 ft. With hL = 0, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 2

lb 12 in. ba b 1 ft V2 in2 = + + 0 + hpump = 3 2 ( 1.94 slug>ft )( 32.2 ft>s ) 2g a -6

+

18.8 ft

2

lb 12 in. ba b 1 ft in2 ( 1.94 slug>ft 3 )( 32.2 ft>s2 ) a20

V2 + 5 ft + 0 + 0 2g

A –13.8 ft

EGL

C HGL

Pump

B

Datum

32.6 ft (a)

hpump = 64.93 ft EGL and HGL. Since no losses occur, the total head before the pump is H =

pA VA2 + zA + g 2g

lb 12 in. 2 ba b (45.84 ft>s)2 1 ft in2 + + 0 = 18.80 ft = 18.8 ft ( 1.94 slug>ft 3 )( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) a -6

After the pump, a head of 59.93 ft is added to the water and becomes H = 18.80 ft + 64.93 ft = 83.7 ft The velocity head has a constant value of (45.84ft>s)2 V2 = = 32.6 ft 2g 2 ( 32.2 ft>s2 ) The HGL is always 32.6 ft below and parallel to the EL. Both are plotted as shown in Fig. a.

Ans: V = 45.8 ft>s, hpump = 66.1 ft 565

5–103. The pump draws water from the large reservoir A and discharges it at 0.2 m3 >s at C. If the diameter of the pipe is 200 mm, determine the power the pump delivers to the water. Neglect friction losses. Construct the energy and hydraulic grade lines for the pipe using a datum at B.

C

8m

A 3m

B

SOLUTION Q = VCAC 2.07 m

0.2 m3 >s = V 3 p(0.1 m)2 4 VC = 6.366 m>s

Energy Equation. Take the water from A to C to be the control volume. Since A and C are exposed to the atmosphere, pA = pC = 0. Also, since the water is drawn from a large reservoir, VA = 0. If we set the datum through B, zA = 3 m and zC = 8 m. With hL = 0,

EGL

8m

HGL

3m

pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g

0.934 m A

( 6.366 m>s ) + 8m + 0 + 0 2 ( 9.81 m>s2 ) 2

0 + 0 + 3 m + hpump = 0 +

10.1 m

EGL HGL

2.07 m C

Datum

Pump (a)

hpump = 7.066 m # Ws = Qpump ghpump = ( 0.2 m3 >s )( 1000 kg>m3 )( 9.81m>s2 ) (7.066 m) = 13.86 ( 103 ) W = 13.9 kW

Ans.

EGL and HGL. Since no loss occurs, the total head before the pump is H =

pA VA2 + zA = 0 + 0 + 3 m = 3 m + gA 2g

After the pump, a head of hpump = 7.066 m is added to the water and becomes H = 3 m + 7.066 m = 10.1 m The velocity head has a constant value of

( 6.366 m>s ) 2 V2 = = 2.07 m 2g 2 ( 9.81 m>s ) The HGL will always be 2.07 m below and parallel with EGL. Both are plotted as shown in Fig. a.

Ans: 13.9 kW 566

*5–104. Solve Prob. 5–103, but include a friction head loss in the pump of 0.5 m, and a friction loss of 1 m for every 5 m length of pipe. The pipe extends 3 m from the reservoir to B, then 12 m from B to C.

C

8m

A 3m

B

SOLUTION From the discharge

12.5 m

0.2 m3 >s = VC 3 p(0.1 m)2 4

Q = VCAC;

2.07 m

VC = 6.366 m>s

Consider the fixed control volume as the water contained in the piping system. Since A and C are exposed to the atmosphere pA = pC = 0. Also, since the water is drawn from a large reservoir VA = 0. If we set the datum through B, zA = 3 m and zC = 8 m.

EGL

10.1 m

HGL

pA pC VC2 VA2 + + zA + hpump = + + zC + hturb + hL gW gW 2g 2g 0 + 0 + 3 m + hpump = 0 +

( 6.366 m>s ) 2 1m + 8 m + 0 + 0.5 m + a b(15 m) 2 5m 2 ( 9.81 m>s )

3.0 m 2.40 m

hpump = 10.57 m

2.07 m HGL

Thus, the power the pump delivers to the water is # Ws = QgWhpump

A

= ( 0.2 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (10.57 m) Ans.

= 20.7 kw The total hand at A is pA VA2 + zA = 0 + 0 + 3 m = 3 m + gA 2g

The total head before the pump is H = 3m - a

1m b(3 m) = 2.40 m 5m

After the pump, a loss in head of 0.5 m and a head of hpump = 10.57 m is added to the water. H = 2.40 m - 0.5 m + 10.57 m = 12.47 m = 12.5 m The total head at C is H = 12.47 m - a

1m b(12 m) = 10.07 m = 10.1 m 5m

The velocity head has a constant value of

( 6.366 m>s ) 2 V2 = = 2.066 m = 2.07 m 2g 2 ( 9.81m>s2 ) The HGL will always be 2.07 m below and parallel with EGL. Both are plotted as shown in Fig. a.

567

C

Pump (a)

= 20.73 ( 103 ) W

H =

EGL

5–105. The turbine removes potential energy from the water in the reservoir such that it has a discharge of 20 ft 3 >s through the 2-ft-diameter pipe. Determine the horsepower delivered to the turbine. Construct the energy and hydraulic grade lines for the pipe using a datum at point C. Neglect friction losses.

A

15 ft B 6 ft C

SOLUTION EGL

Q = VCAC

0.629 ft

21 ft

20 ft 3 >s = VC 3 p(1 ft)2 4

20.4 ft

VC = 6.366 ft>s

HGL

Energy Equation. Take the water from A to C to be the control volume. Since A and C are exposed to the atmosphere, pA = pC = 0. Also, since the water is drawn form a large reservoir, VA = 0. With reference to the datum through C, zA = 15 ft + 6 ft = 21 ft and zC = 0. With hL = 0, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g

0.629 ft

( 6.366ft>s ) 2 0 + 0 + 21 ft + 0 = 0 + + hturb + 0 2 ( 32.2ft>s2 ) hturb = 20.37 ft # Ws = Qghturb = ( 20 ft 3 >s )( 62.4 lb>ft 3 ) (20.37 ft) = 25423 ft .lb>sa

1 hp

550 ft .lb>s

b = 46.2 hp

EGL A Turbine

0.629 ft

C HGL

Datum

(a)

Ans.

EGL and HGL. Since no loss occurs, the total head before the pump is H =

pA VA2 + zA = 0 + 0 + 21 ft = 21 ft + g 2g

After the turbine, a head of hturb = 20.37 ft is drawn from the water and becomes H = 21 ft - 20.37 ft = 0.629 ft The velocity head has a constant value of ( 6.366 ft>s ) 2 V2 = = 0.629 ft 2g 2 ( 32.2 ft>s2 ) The HGL will always be 0.629 ft below and parallel with the EGL. Both are plotted as shown in Fig. a.

Ans: 46.2 hp 568

5–106. The turbine C removes 300 kW of power from the water that passes through it. If the pressure at the intake A is pA = 300 kPa and the velocity is 8 m>s, determine the pressure and velocity of the water at the exit B. Neglect the frictional losses between A and B.

1.5 m

1m 8 m/s

C

A B

SOLUTION The control volume considered contains the water in the turbine case from A to B. Since the flow is steady, the continuity condition become 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - (8 m>s) 3 p ( 0.5 m ) 2 4 + VB 3 p ( 0.75 m ) 2 4 = 0 VB = 3.556 m>s = 3.56 m>s

Ans.

Here, Q = VAAA = ( 8 m>s ) 3 p ( 0.5 m ) 2 4 = 2p m3 >s. The turbine head can be determined from # Ws = Qgwhturb; 300 ( 103 ) W = ( 2p m3 >s )( 9810 N>m3 ) hturb hturb = 4.8671 m

With reference to the datum that contain points A and B, Z A = ZB = 0. The energy equation between points A and B is pB pA VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 300 ( 103 ) N>m2 9810 N>m3

+

( 8 m>s ) 2 ( 3.556 m>s ) 2 pB + 0 + 0 = + + 0 + 4.8671 m + 0 9810 N>m3 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pB = 277.93 ( 103 ) Pa = 278 kPa

Ans.

Ans: V = 3.56 m>s, p = 278 kPa 569

5–107. The pump has a volumetric flow of 0.3 ft 3 >s as it moves water from the pond at A to the one at B. If the hose has a diameter of 0.25 ft, and friction losses within it can be expressed as 5 V 2 >g, where V is the average velocity of the flow, determine the horsepower the pump supplies to the water.

6 ft

B

10 ft

A

SOLUTION Q = VA 0.3 ft 3 >s = V 3 p(0.125 ft)2 4 V = 6.112 ft>s

Energy Equation. Take the water from A to B to be the control volume. Since A and B are both free surfaces, pA = pB = 0. Also, VA = VB = 0 since both ponds are large reservoirs. If the datum passes through A, zA = 0 and zB = 6 ft + 10 ft = 16 ft. pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 + 0 + 16 ft + 0 + 5£

( 6.111 ft>s ) 2 § ( 32.2 ft>s2 )

hpump = 21.80 ft # Ws = Qghpump = ( 0.3 ft 3 >s )( 62.4 lb>ft 3 ) (21.80 ft) 1 hp ft # lb ¢ = 0.742 hp = 408.1 s ° 550ft # lb>s

Ans.

Ans: 0.742 hp 570

*5–108. Water from the reservoir passes through a turbine at the rate of 18 ft 3 >s. If it is discharged at B with a velocity of 15 ft>s, and the turbine withdraws 100 hp, determine the head loss in the system.

A

80 ft B

SOLUTION

# Ws = Qghs

(100 hp)a

550 ft # lb>s 1 hp

b = ( 18 ft 3 >s )( 62.4 lb>ft 3 ) hs

hs = 48.97 ft

Energy Equation. Take the water from A to B to be the control volume. Since A and B are both free surfaces, pA = pB = 0. Also, due to the large source at the reservoir, VA = 0. If the datum passes through B, zA = 80 ft and zB = 0. pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 80 ft + 0 = 0 +

( 15 ft>s ) 2 + 0 + 48.97 ft + hL 2 ( 32.2 ft>s2 ) Ans.

hL = 27.5 ft

571

5–109. The vertical pipe is filled with oil. When the valve at A is closed, the pressure at A is 160 kPa, and at B it is 90 kPa. When the valve is open, the oil flows at 2 m>s, and the pressure at A is 150 kPa and at B it is 70 kPa. Determine the head loss in the pipe between A and B. Take ro = 900 kg>m3.

B

A

2 m/s

SOLUTION Static Pressure. When the valve is closed pA = pB + rogzB 160 ( 103 ) N>m2 = 90 ( 103 ) N>m2 + ( 900 kg>m3 )( 9.81 m>s2 ) (zB) zB = 7.928 m Energy Equation. Take the oil from A to B to be the control volume. Since the pipe has a constant diameter, continuity requires VA = VB = V. If the datum passes through A, zA = 0 and zB = 7.928 m. With hS = 0, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 150 ( 103 ) N>m2

( 900 kg>m3 )( 9.81 m>s2 )

+

V2 + 0 + 0 = 2g

70 ( 103 ) N>m2

( 900 kg>m3 )( 9.81 m>s2 )

+

V2 + 7.928 m + hL 2g Ans.

hL = 1.13 m

Ans: 1.13 m 572

5–110. It is required that a pump be used to discharge water at 80 gal>min from a river to a pond, at B. If the frictional head loss through the hose is 3 ft, and the hose has a diameter of 0.25 ft, determine the required power output of the pump. Note that 7.48 gal = 1 ft 3.

10 ft

A

B 4 ft

SOLUTION Energy Equation. Take the water from A to B to be the control volume. Since A and B are free surfaces, pA = pB = 0. Also, both the river and pond are large reservoirs, VA = VB = 0. If the datum is at the free surface at A, zA = 0 and zB = 4 ft + 10 ft = 14 ft . With hL = 3 ft, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 + 0 + 14 ft + 0 + 3 ft hpump = 17.0 ft # Ws = Qghpump # 80 gal 1 min ft 3 Ws = a ba ba b ( 62.4 lb>ft 3 ) (17 ft) min 60 s 7.48 gal = 189.1 ft # lb>s °

1 hp

550 ft # lb>s

Ans.

¢ = 0.344 hp

Ans: 0.344 hp 573

5–111. A 6-hp pump with a 3-in-diameter hose is used to drain water from a large cavity at B. Determine the discharge at C. Neglect friction losses and the efficiency of the pump. 1 hp = 550 ft # lb>s.

SOLUTION

(6 hp)a

3 ft

B

C

8 ft A

# Ws = Qghpump 550 ft # lb>s 1 hp

b = V£ pa hpump =

2 1.5 ft b § ( 62.4 lb>ft 3 )( hpump ) 12

1077.4 V

Energy Equation. Take the water in the cavity B and in the base to C to be the control volume. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the water is drawn from a large cavity, so that VB = 0. If we set the datum through B, zB = 0 and zC = 3 ft . Since the pump supplies a head of water, hs is a negative quantity. pB pC VC2 VB2 + zB + hpump = + zC + hturb + hL + + g g 2g 2g 0 + 0 + 0 + a

1077.4 V2 b = 0 + + 3 ft + 0 + 0 V 2 ( 32.2 ft>s2 ) V 3 + 193.2V = 69381.76

Solving numerically, V = 39.52 ft>s Discharge. Q = VA = ( 39.52 ft>s ) £ p a = 1.94 ft 3 >s

2 1.5 ft b § 12

Ans.

574

Ans: 1.94 ft 3 >s

*5–112. The pump is used with a 3-in.-diameter hose to draw water from the cavity. If the discharge is 1.5 ft 3 >s, determine the required power developed by the pump. Neglect friction losses.

3 ft 8 ft A

SOLUTION From the discharge, Q = VCAC;

1.5 ft 3 >s = VC £ p a

2 1.5 ft b § 12

VC = 30.56 ft>s

The fixed control volume contains the water in the system. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the water is drawn form a large reservoir, VB = 0. If we set the datum through B, zB = 0 and zC = 3 ft . pC pB VC2 VB2 + + zB + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

( 30.56 ft>s ) 2 + 3 ft + 0 + 0 2 ( 32.2 ft>s2 )

hpump = 17.50 ft The required power output of the pump is # Ws = Qgwhpump = ( 1.5 ft 3 >s )( 62.4 lb>ft 3 ) (17.50 ft) = a1637.97 = 2.98 hp

B

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

575

C

5–113. Solve Prob. 5–112 by including frictional head losses in the hose of 1.5 ft for every 20 ft of hose. The hose has a total length of 130 ft.

3 ft

B

C

8 ft A

SOLUTION From the discharge, Q = VCAC;

1.5 ft 3 >s = VC £ p a

2 1.5 ft b § 12

VC = 30.56 ft>s

The fixed control volume contains the water in the system. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the water is drawn from a large reservoir, VB = 0. If we set the datum through B, zB = 0 and zC = 3 ft . pB pC VC2 VB2 + + zB + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

( 30.56 ft>s ) 2 1.5 ft + 3 ft + 0 + a b(130 ft) 2 20 ft 2 ( 32.2 ft>s )

hpump = 27.25 ft

The required power output of the pump is # Ws = Qgwhpump = ( 1.5 ft 3 >s )( 62.4 lb>ft 3 ) (27.25 ft) = a2550.57

= 4.64 hp

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

Ans: 4.64 hp 576

5–114. The flow of air through a 200-mm-diameter duct has an absolute inlet pressure of 180 kPa, a temperature of 15°C, and a velocity of 10 m>s. Farther downstream a 2-kW exhaust system increases the outlet velocity to 25 m>s. Determine the density of the air at the outlet, and the change in enthalpy of the air. Neglect heat transfer through the pipe.

B A

30 m

SOLUTION

Ideal Gas Law. Referring to Appendix A, R = 286.9 J>kg # K. pin = rinRTin 180 ( 103 ) N>m2 = rin(286.9 J>kg # K)(15° + 273) K rin = 2.178 kg>m3 # m = rinVinAin = ( 2.178 kg>m3 )( 10 m>s ) 3 p(0.1 m)2 4 = 0.6844 kg>s

Energy Equation. With a a

dWturb dQ b = a b = 0 and zin = zout = z, dt in dt

dWpump dWturb Vout2 Vin2 dQ . b - a b + ° ¢ = £ °hout + + gzout ¢ - °hin + + gzin ¢ § m dt in dt dt 2 2 0 - 0 + °

dWpump dt

¢ = £ °hout +

Vout2 Vin2 . + gz¢ - °hin + + gz¢ § m 2 2

∆h = hout - hin = ° = £

( 10 m>s ) 2 2

-

dWpump 1 Vout2 Vin2 ¢ + ° ¢ # m 2 2 dt

( 25 m>s ) 2 2

= 2660 J>kg = 2.66 kJ>kg

§ +

2000 N # m>s 0.6844 kg>s

Ans.

Ans: 2.66 kJ>kg 577

5–115. Nitrogen gas having an enthalpy of 250 J>kg is flowing at 6 m>s into the 10-m-long pipe at A. If the heat loss from the walls of the duct is 60 W, determine the enthalpy of the gas at the exit B. Assume that the gas is incompressible with a density of r = 1.36 kg>m3.

0.15 m 6 m/s A

B 10 m

SOLUTION Mass Flow Rate.

#

m = rVA = ( 1.36 kg>m3 )( 6 m>s ) 3 p(0.15 m)2 4 = 0.5768 kg>s

Since the density and diameter of the duct are constant, continuity requires Vin = Vout = V Energy Equation. The flow is steady.Take the nitrogen on the pipe to be the control volume. Here, a a

dWpump dWturb dQ b = - 60 J>s and ° ¢ = ° ¢ = 0. With zin = zout = z, dt in dt dt

dWpump dWturb Vout2 Vin2 dQ # + gzout ¢ - °hin + + gzin ¢ § m b - a b + ° ¢ = £ °hout + dt in dt dt 2 2

-60 J>s - 0 + 0 = £ °hout +

V2 V2 + gz¢ - °250 J>kg + + gz¢ § ( 0.5768 kg>s ) 2 2 Ans.

hout = 145.98 J>kg = 146 J>kg

Ans: 146 J>kg 578

*5–116. The measured water pressure at the inlet and exit portions of the pipe are indicated for the pump. If the flow is 0.1 m3 >s, determine the power that the pump supplies to the water. Neglect friction losses.

300 kPa 50 mm 200 kPa 75 mm A

SOLUTION Flow: Q = VAAA;

0.1 m3 >s = VA 3 p(0.0375 m)2 4 VA = 22.64 m>s

Q = VBAB;

0.1 m3 >s = VB 3 p(0.025 m)2 4 VB = 50.93 m>s

Energy Equation. Take the water from A to B to be the control volume. If we set the datum through A, zA = 0 and zB = 2 m. With hL = 0, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 200 ( 103 ) N>m2

( 1000 kg>m3 )( 9.81 m>s2 ) =

+

( 22.64 m>s ) 2 + 0 + hpump 2 ( 9.81 m>s2 )

300 ( 103 ) N>m2

( 1000 kg>m3 )( 9.81 m>s2 )

+

( 50.93 m>s ) 2 + 2m + 0 + 0 2 ( 9.81 m>s2 )

hpump = 118.3 m # Ws = Qghpump = ( 0.1 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (118.3 m)

= 116.04 ( 103 ) W = 116 kW

579

Ans.

B 2m

5–117. The wave overtopping device consists of a floating reservoir that is continuously filled by waves, so that the water level in the reservoir is always higher than that of the surrounding ocean. As the water drains out at A, the energy is drawn by the low-head hydroturbine, which then generates electricity. Determine the power that can be produced by this system if the water level in the reservoir is always 1.5 m above that in the ocean, The waves add 0.3 m3 >s to the reservoir, and the diameter of the tunnel containing the turbine is 600 mm. The head loss through the turbine is 0.2 m. Take rw = 1050 kg>m3.

1.5 m 0.3 m

A

SOLUTION From the discharge, Q = Vout Aout;

0.3 m3 >s = Vout 3 p(0.3 m)2 4 Vout = 1.061 m>s

Take the water within the turbine to be the control volume. We will apply the energy equation between the inlet and the outlet. pin pout Vout2 Vin2 + zin + hpump = + zout + hturb + hL + + g g 2g 2g Here, Vin = 0 since the inlet is the surface of a large reservoir. pin = pout = patm = 0 since the inlet and outlet are exposed to the atmosphere. Here, the datum is set at the ocean water level. Then, zin = 1.5 m and zout = 0.3 m. 0 + 0 + 1.5 m + 0 = 0 +

( 1.061 m>s ) 2 + 0.3 m + hturb + 0.2 m 2 ( 9.81 m>s2 )

hturb = 0.9426 m The turbine is # Ws = Qgswhturb = ( 0.3 m3 >s ) 3 ( 1050 kg>m3 )( 9.81 m>s2 ) 4 (0.9426 m) = 2912.83 W

Ans.

= 2.91 kW

Ans: 2.91 kW 580

5–118. Crude oil is pumped from the test separator at A to the stock tank using a galvanized iron pipe that has a diameter of 4 in. If the total pipe length is 180 ft, and the volumetric flow at A is 400 gal>min, determine the required horsepower supplied by the pump. The pressure at A is 4 psi, and the stock tank is opened to the atmosphere. The frictional head loss in the pipe is 0.25 in.>ft, and the head loss for each of the four bends is K 1 V 2 >2g 2 , where K = 0.09 and V is the velocity of the flow in the pipe. Take go = 55 lb>ft 3. Note that 1 ft 3 = 7.48 gal.

B

30 ft A

SOLUTION The discharge is

Thus,

Q = °400

gal min

¢°

1 ft 3 1 min ¢° ¢ = 0.8913 ft 3 >s 7.48 gal 60 s 0.8913 ft 3 >s = VA £ pa

Q = VAAA;

VA = 10.21 ft>s

2 2 ft b § 12

Energy Equation. Take the water from A to B on the control volume. Then pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL go go 2g 2g Since point B is exposed to atmosphere, pB = patm = 0. Also, pA = a4

lb 12 in 2 ba b = 576 lb>ft 2 1 ft in2

Since the stock tank is a large reservoir VB ≃ 0. The frictional head loss is

( hL ) f = c

( 0.25>12 ) ft ft

d (180 ft) = 3.75 ft

There are four bends between point A and B. Thus, the head losses due to the bends is

( hL ) M = 4 °k

( 10.21 ft>s ) 2 V2 ¢ = 4W (0.09) £ § ¶ = 0.5831 ft 2g 2 ( 32.2 ft>s2 )

581

5–118. Continued

With reference to the datum set through point A, zA = 0 and zB = 30 ft . Substituting these results into the energy equation, 576 lb>ft 2 55.0 lb>ft 3

+

( 10.21 ft>s ) 2 + 0 + hpump = 0 + 0 + 30 ft + 0 + (3.75 ft + 0.5831 ft) 2 ( 32.2 ft>s2 ) hpump = 22.24 ft

The required output power can be determined from # Ws = Qgcohpump = ( 0.8913 ft 3 >s )( 55.0 lb>ft 3 ) (22.24 ft) = (1090.23 ft # lb>s)a

= 1.98 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: 1.98 hp 582

5-119. The pump is used to transport water at 90 ft 3 >min from the stream up the 20-ft embankment. If frictional head losses in the 3-in.-diameter pipe are hL = 1.5 ft, determine the power output of the pump.

B

20 ft A

SOLUTION Energy Equation. Take the water from A to B to be the control volume. Then we will apply the energy equation between point A on the surface of water in the stream and point B at the pipe’s exit. Here, points A and B are exposed to the atmosphere, pA = pB = patm = 0. Since the stream can be considered as a large reservoir, VA ≃ 0. Here the discharge is

Then,

Q = °90

Q = VBAB;

ft 3 1 min ¢° ¢ = 1.5 ft 3 >s min 60 s

1.5 ft 3 >s = VB £ pa

2 1.5 ft b § 12

VB = 30.56 ft>s

With reference to the datum set through point A, zA = 0 and zB = 20 ft. pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gW gW 2g 2g 0 + 0 + 0 + hpump = 0 +

( 30.56 ft>s ) 2 + 20 ft + 0 + 1.5 ft 2 ( 32.2 ft>s2 )

hpump = 36.00 ft The power output of the pump can be determined from # Ws = QgWhpump = ( 1.5 ft 3 >s )( 62.4 lb>ft 2 ) (36.00 ft) = (3369.57 ft # lb>s) ° = 6.13 hp

1 hp

550 ft # lb>s

¢

Ans.

Ans: 6.13 hp 583

*5–120. The pump is used to transfer carbon tetrachloride in a processing plant from a storage tank A to the mixing tank C. If the total head loss due to friction and the pipe fittings in the system is 1.8 m, and the diameter of the pipe is 50 mm, determine the power developed by the pump when h = 3 m. The velocity at the pipe exit is 10  m>s, and the storage tank is opened to the atmosphere rct = 1590 kg>m3.

6m

h

C

SOLUTION Take the carbon tetrachloride in the tank and pipe to point B to be the control volume. Then we will apply the energy equation between point A on the surface of carbon tetrachloride in the storage tank and point B at the pipe’s exit. Here points A and B are exposed to atmosphere, pA = pB = patm = 0. Since the storage tank can be considered as a large reservoir, VA ≃ 0. With reference to the datum set through point A, zA = 0 and zB = 6 m - 3 m = 3m. Also, gct = rct g = ( 1590 kg>m3 )( 9.81 m>s2 ) = 15597.9 N>m3. pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + gct gct 2g 2g 0 + 0 + 0 + hpump = 0 +

B

A

( 10 m>s ) 2 + 3 m + 0 + 1.8 m 2 ( 9.81 m>s2 )

hpump = 9.8968 m Here, Q = VBAB = (10 m>s) 3 p(0.025 m)2 4 = 6.25p ( 10-3 ) m3 >s. Then, the power output of the pump can be determined from # WS = Qgct hpump = 3 6.25 p ( 10-3 ) m3 >s 4 ( 15597.9 N>m3 ) (9.8968 m)

Ans.

= 3031.04 W = 3.03 kW

584

5–121. The pump takes in water from the large reservoir at A and discharges it at B at 0.8 ft 3 >s through a 6-in.-diameter pipe. If the frictional head loss is 3 ft, determine the power output of the pump.

B 8 ft C 12 ft A

SOLUTION Q = VBAB 0.8 ft 3 >s = VB £ pa

2 3 ft b § 12

VB = 4.074 ft>s

Energy Equation. Takes the water in the reservoir and pipe system to B at the control volume. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, VC = 0 since the water is drawn from the large reservior. If the datum passes through C, zC = 0 and zB = 8 ft. With hL = 3 ft, =

pC pB VC2 VB2 + zC + hpump = + zB + hturb + hL + + g g 2g 2g

0 + 0 + 0 + hpump = 0 +

( 4.074 ft>s ) 2 + 8 ft + 0 + 3 ft 2 ( 32.2 ft>s2 )

hpump = 11.26 ft # WS = Qghpump = ( 0.8 ft 3 >s )( 62.4 lb>ft 3 ) (11.26 ft) = 562.0 ft # lb>s °

1 hp

550 ft # lb>s

¢ = 1.02 hp

Ans.

Ans: 1.02 hp 585

5–122. Air and fuel enter a turbojet engine (turbine) having an enthalpy of 800 kJ>kg and a relative velocity of 15 m>s. The mixture exits with a relative velocity of 60 m>s and an enthalpy of 650 kJ>kg. If the mass flow is 30 kg>s, determine the power output of the jet. Assume no heat transfer occurs.

SOLUTION Energy Equation. Take the Air and fuel and engine the control volume. Here, dQ a b = 0 and zin = zout = z. dt in a

dWpump dWturb Vout2 Vin2 dQ # + gzout ¢ - °hin + + gzin ¢ § m b - a b + ° ¢ = £ °hout + dt in dt dt 2 2

0 - a

dWturbine b + 0 = £ °650 ( 103 ) J>kg + dt a

( 60 m>s ) 2 2

+ gz¢ - °800 ( 103 ) J>kg +

dWturbine b = 4.449 ( 106 ) W = 4.45 MW dt

( 15 m>s ) 2 2

+ gz¢ § ( 30 kg>s )

Ans.

Ans: 4.45 MW 586

5–123. Water flows into the pump at 600 gal>min and has a pressure of 4 psi. It exits the pump at 18 psi. Determine the power output of the pump. Neglect friction losses. Note that 1 ft 3 = 7.48 gal.

0.75 ft

0.5 ft A

B

SOLUTION Q = (600 gal>min)a

1 min 1 ft 3 b° ¢ = 1.337 ft 3 >s 60 s 7.48 gal

1.337 ft 3 >s = VA 3 p(0.375 ft)2 4

Q = VAAA;

VA = 3.026 ft>s

1.337 ft 3 >s = VB 3 p(0.25 ft)2 4

Q = VBAB;

VB = 6.809 ft>s

Energy Equation. Take the water from A to B as the control volume. Since the centerline of the pipe lies on the same horizontal line, zA = zB = z. With hL = 0, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 4 lb>in2 a

12 in. 2 b 1 ft

62.4 lb>ft

3

( 3.026 ft>s ) + z + hpump = 2 ( 32.2 ft>s2 ) 2

+

18 lb>in2 a

12 in. 2 b 1 ft

62.4 lb>ft

3

hpump = 32.89 ft # WS = Qghpump = ( 1.337 ft 3 >s )( 62.4 lb>ft 3 ) (32.89 ft) = 2743 ft # lb>s a

1 hp

550 ft # lb>s

b = 4.99 hp

+

( 6.809 ft>s ) 2 + z + 0 + 0 2 ( 32.2 ft>s2 )

Ans.

Ans: 4.99 hp 587

*5–124. The 5-hp pump has an efficiency of e = 0.8 and produces a flow velocity of 3 ft>s through the pipe at A. If the frictional head loss within the system is 8 ft, determine the difference in the water pressure between A and B.

0.75 ft

SOLUTION From the discharge, Q = (3 ft>s) 3 p(0.375 ft)2 4 = 0.421875p ft 3 >s

Q = VAAA;

0.421875p ft 3 >s = VB 3 p(0.25 ft)2 4

Q = VBAB;

VB = 6.75 ft>s

The power output of the pump is given by # # Wsout Wsout e = # ; 0.8 = 5hp Wsin Thus, the pump head is # Wsout = Qgwhpump ;

(4hp)a

550ft # lb>s 1 hp

# Wsout = 4hp

b = ( 0.421875p ft 3 >s )( 62.4 lb>ft 3 ) hpump

hpump = 26.60 ft

The fixed control volume contains the water in the system from A to B. Since the centerline of the pipe lies on the same elevation, ZA = ZB = Z. pB pA VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g pA 62.4 lb>ft 3

=

( 3 ft>s ) 2 + z + 26.60 ft 2 ( 32.2 ft>s2 )

+ pB

62.4 lb>ft

pB - pA = ( 1125.30 lb>ft 2 ) a

3

+

0.5 ft A

( 6.75 ft>s ) 2 + z + 0 + 8 ft 2 ( 32.2 ft>s2 )

1 ft 2 b = 7.814 psi = 7.81 psi 12 in

588

Ans.

B

5–125. The water tank is being drained using the 1-in.-diameter hose. If the flow out of the hose is 5 ft 3 >min, determine the head loss in the hose when the water depth is d = 6 ft. h

SOLUTION Energy Equation: Take the water in the tank and fill the hose print B as the central volume. We will apply the energy equation between point A on the surface of water in the tank and point B at the hose’s exit. Here, points A and B are exposed to atmosphere, pA = pB = patm = 0. Since the tank can be considered as a large reservoir, VA _ 0. Here the discharge is

Then

Q = °5

ft 3 1 min ¢a b = 0.08333 ft 3 >s min 60 s 2

Q = VB AB;

0.5 0.08333 ft >s = VB £ p° ft ¢ § 12 3

VB = 15.28 ft>s

With reference to the datum through point B, zA = 6 ft and zB = 0. pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g Then 0 + 0 + 6 ft + 0 = 0 +

( 15.28 ft>s ) 2 + 0 + hL 2 ( 32.2 ft>s2 ) Ans.

hL = 2.38 ft

Ans: 2.38 ft 589

5–126. The pump at C produces a discharge of water at B of 0.035 m3 >s. If the pipe at B has a diameter of 50 mm and the hose at A has a diameter of 30 mm, determine the power output supplied by the pump. Assume frictional head losses within the pipe system are determined from 3V B2 >2g.

B

30 m C A

SOLUTION Q = VBAB 0.035 m3 >s = VB 3 p(0.025 m)2 4 VB = 17.825 m>s

Energy Equation. Take the water in the lower reservoir and in the pipe system to B as the control volume. Since A and B are exposed to the atmosphere, pA = pB = 0. Also, VA = 0 since water is drawn from a large reservoir. If the datum coincides with the free surface A, zA = 0 and zB = 30 m. pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 +

( 17.825 m>s ) 2 3(17.825 m>s)2 + 30 m + 0 + 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) hpump = 94.78 m # WS = Qghpump

= ( 0.035 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (94.78 m) = 32.5 ( 103 ) W = 32.5 kW

Ans.

Ans: 49.2 kW 590

5–127. Determine the power output required to pump sodium coolant at 3 ft 3 >s through the core of a liquid metal fast-breeder reactor if the piping system consists of 23  stainless steel pipes, each having a diameter of 1.25 in. and length of 4.2 ft. The pressure at the inlet A is 47.5 lb>ft 2 and at the outlet B it is 15.5 lb>ft 2. The frictional head loss for each pipe is 0.75 in. gNa = 57.9 lb>ft 3.

B

4.2 ft

A

SOLUTION Take the sodium passing through the water as the control volume. Then we will apply the energy equation between point A (inlet) and point B (outlet). Since the pipes have constant diameter, the continuity condition requires that VA = VB = V. With reference to the datum set through point A, zA = 0 and zB = 4.2 ft . Here, the 1 ft head loss is hL = 23(0.75 in) = (17.75 in)a b = 1.4375 ft 12 in pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gNA gNA 2g 2g

47.5 lb>ft 2 57.9 lb>ft 3

+

15.5 lb>ft 2 V2 V2 + 0 + hpump = + + 4.2 + D + 1.4375 3 2g 2g 57.9 lb>ft hpump = 5.085 ft

Thus, the power output required can be determined from # WS = QgNA hpump = ( 3 ft 3 >s )( 57.9 lb>ft 3 ) (5.085 ft) = (883.234 ft # lb>s) °

1 hp

550 ft # lb>s

= 1.61 hp

¢

Ans.

Ans: 1.61 hp 591

*5–128. If the pressure at A is 60 kPa, and the pressure at B is 180 kPa, determine the power output supplied by the pump if the water flows at 0.02 m3 >s. Neglect friction losses.

75 mm B

0.5 m 50 mm

SOLUTION 0.02 m3 >s = VA 3 p(0.025 m)2 4

Q = VAAA;

VA = 10.186 m>s

0.02 m3 >s = VB 3 p(0.0375 m)2 4

Q = VBAB;

VB = 4.527 m>s

Energy Equation. Take the water from A to B as the control volume. If we set the datum through B, zA = 0 and zC = 0.5 m. With hL = 0, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 60 ( 103 ) N>m2

( 1000 kg>m3 )( 9.81 m>s2 ) +

+

( 10.186 m>s ) 2 180 ( 103 ) N>m2 + 0 + h = pump ( 1000 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 )

( 4.527 m>s ) 2 + 0.5 m + 0 + 0 2 ( 9.81 m>s2 )

hpump = 8.489 m # WS = Qghpump = ( 0.02 m2 >s )( 1000 kg>m3

)( 9.81 m>s2 ) (8.489 m)

= 1.666 ( 103 ) W = 1.67 kW

Ans.

592

A

5–129. The pump supplies a power of 1.5 kW to the water producing a volumetric flow of 0.015 m3 >s. If the total frictional head loss within the system is 1.35 m, determine the pressure difference between the inlet A and outlet B of the pipes.

75 mm B

0.5 m 50 mm

A

SOLUTION From the discharge 0.015 m3 >s = VA 3 p(0.025 m)2 4

Q = VAAA;

VA = 7.639 m>s

0.015 m3 >s = VB 3 p(0.0375 m)2 4

Q = VBAB;

VB = 3.395 m>s

From the power supplied to the water, # ( Ws ) out = Qghpump; 1.5 ( 103 ) W = ( 0.015 m3 >s

)( 1000 kg>m3 )( 9.81 m>s2 ) hpump

hpump = 10.19 m

The fixed control volume contains the water in the system from A to B. If we set the datum through A, zA = 0 and zB = 0.5 m. pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gW gW 2g 2g pA

( 1000 kg>m )( 9.81 m>s ) 3

2

+ +

(7.639 m>s)2 2 ( 9.81 m>s

2

)

(3.395 m>s)2 2 ( 9.81 m>s2 )

+ 0 + 10.19 m =

pB

( 1000 kg>m3 )( 9.81 m>s2 )

+ 0.5 m + 0 + 1.35 m

pB - pA = 105.27 ( 103 ) Pa = 105 kPa

Ans.

Ans: pB - pA = 105 kPa 593

5–130. The circular hovercraft draws in air through the fan A and discharges it through the bottom B near the ground, where it produces a pressure of 1.50 kPa on the ground. Determine the average velocity of the air entering at A that is needed to lift the hovercraft 100 mm off the ground. The open area at A is 0.75 m2. Neglect friction losses. Take ra = 1.22 kg>m3.

A

B C

0.75 m

0.75 m

SOLUTION Between B and C. pin pout Vout2 Vin2 + zin + hpump = + zout + hturb + hL + + g g 2g 2g 1.50 ( 103 ) 1.22(9.81)

+ 0 + 0 + 0 = 0 +

VC2 + 0 + 0 + 0 2(9.81)

VC = 49.59 m>s Between A and C, 0 rVdA + rV # dA = 0 0t L L cv cs 0 + VA ( 0.75 m2 ) - (49.59 m>s)(2p)(0.75 m)(0.1 m) = 0 Ans.

VA = 31.2 m>s

Ans: 31.2 m>s 594

6–1. Determine the linear momentum of a mass of fluid in a 0.2-m length of pipe if the velocity profile for the fluid is a paraboloid as shown. Compare this result with the linear momentum of the fluid using the average velocity of flow. Take r = 800 kg>m3.

u ! 4 (1 " 100 r 2) m/s 0.1 m r

0.2 m

SOLUTION

dr

The shell differential element that has a thickness dr and length 0.2 m shown shaded in Fig. a has a volume of dV = (2prdr)(0.2 m) = 0.4prdr. Thus, the mass of this element is dm = rdV = ( 800 kg>m3 ) (0.4prdr) = 320prdr. The linear momentum of the fluid is L =

L

r

0.1 m (a)

v dm

m

=

L0

0.1 m

4 ( 1 - 100r 2 ) (320pr dr)

= 1280p

L0

= 1280p a

0.1 m

( r - 100r 3 ) dr

0.1 m r2 - 25r 4 b ` 2 0

= 10.05 kg # m>s = 10.1 kg # m>s

Ans.

The ring differential element shown shaded in Fig. a has an area of dA = 2prdr. Therefore Vavg =

=

v dA L A L0

=

4 ( 1 - 100r 2 ) (2pr dr) p(0.1 m)2

8p =

0.1 m

L0

0.1 m

( r - 100r 3 ) dr

p(0.1 m)2 8p a

0.1 m r2 - 25r 4 b ` 2 0

p(0.1 m)2

= 2 m>s The mass of the fluid is mrV = ( 800 kg>m3 ) 3 p(0.1 m)2 4 (0.2 m) = 1.6p kg. Thus, L = mVavg = rVVavg = (1.6p kg) ( 2 m>s ) = 10.05 kg # m>s = 10.1 kg # m>s

Ans.

Ans: L = 10.1 kg # m>s by either method. 595

6–2. Flow through the circular pipe is turbulent, and the velocity profile can be modeled using Prandtl’s one-seventh power law, v = Vmax ( 1 - r>R ) 1>7. If r is the density, show that the momentum of the fluid per unit time passing through the pipe is ( 49>72 ) pR2rV 2max. Then show that Vmax = (60>49)V, where V is the average velocity of the flow. Also, show that the momentum per unit time is ( 50>49 ) pR2rV 2.

r R

SOLUTION

R dr

The amount of mass per unit time passing through a differential ring element of area dA (shown shaded in Fig. a) on the cross-section is # dm = rVdA dA

Then the momentum per unit time passing through this element is # # dL = (dm)V = (rVdA)V = rV 2dA

(a)

Thus, for the entire cross-section, # L =

L

# dL =

A

L

rV 2dA

A

Here dA = 2prdr. Then # L =

R

1

r 7 2 rJVmax a1 - b R (2prdr) R L0 R

2

L0 r Let u = 1 - , then r = R(1 - u) and dr = -Rdu. Also, the integration limits are R r = 0, u = 1 and r = R, u = 0. Thus, = 2prV 2max

# L = 2prV 2max

L1

= 2pR2rV 2max

r a1 -

0

r 7 b dr R

2

R(1 - u)au 7 b( - Rdu)

L1

= 2pR2rV 2max a

0

9

2

au 7 - u7 bdu

7 16 7 9 0 49 u 7 - u7 b ` = pR2rV 2max 16 9 72 1

596

r

(Q.E.D.)

6–3. Oil flows at 0.05 m3 >s through the transition. If the pressure at the transition C is 8 kPa, determine the resultant horizontal shear force acting along the seam AB that holds the cap to the larger pipe. Take ro = 900 kg>m3.

300 mm

A

200 mm D

C B

SOLUTION

p = 8 kPa C

FR 2

We consider steady flow of an ideal fluid. Q = VC AC;

0.05 m >s = VC 3 p(0.15 m) 3

VC = 0.7074 m>s

Q = VD AD;

p =0 D

2

4

C

0.05 m3 >s = VD 3 p(0.1 m)2 4

D

FR 2

VD = 1.592 m>s

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Since D is open to the atmosphere, pD = 0.

(a)

Linear Momentum. Since the flow is steady and incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣF = rQ(VD - VC); S

3 8 ( 103 ) N>m2 4 3 (p)(0.15 m)2 4 F = 526 N

- FR = ( 900 kg>m3 )( 0.05 m3 >s )( 1.592 m>s - 0.7074 m>s ) Ans.

Ans: 526 N 597

*6–4. A small marine ascidian called a styela fixes itself on the sea floor and then allows moving water to pass through it in order to feed. If the opening at A has a diameter of 2 mm, and at the exit B the diameter is 1.5 mm, determine the horizontal force needed to keep this organism attached to the rock at C when the water is moving at 0.2 m>s into the opening at A. Take r = 1050 kg>m3.

A

B

C

SOLUTION

A

The flow is steady and the sea water can be considered as an ideal fluid (incompressible and inviscid) such that average velocities can be used and rsw = 1050 kg>m3. The control volume considered contains the sea water in “styela”, Fig. a. Since the depth of points A and B are almost the same, the pressure forces acting on opened control surfaces A and B can be considered the same and an assumed to cancel each other. Continuity requires

B

0 r dV + rswV # dA = 0 0t Lcv sw Lcs 0 - VAAA + VBAB = 0 - ( 0.2 m>s ) 3 p(0.001 m)

2

4

+ VB 3 p(0.00075 m)

VB = 0.3556 m>s

F 2

4

(a)

= 0

Applying the linear momentum equation, ΣF =

0 Vr dV + VrswV # dA 0t Lcv sw Lcs

Writing the scalar component of this equation along x axis by referring to the FBD of the control volume, Fig. a + 2 ΣFx 1S

-F =

= 0 +

( - VA ) rsw ( - VAAA ) + ( - VB ) rsw ( VBAB )

( - 0.2 m>s )( 1050 kg>m2 ) 5 - ( 0.2 m>s ) 3 p(0.001 m)2 4 6 +

( - 0.3556 m>s )( 1050 kg>m3 ) 5 ( 0.3556 m>s ) 3 p(0.00075 m)2 4 6

F = 0.103 ( 10-3 ) N = 0.103 mN

Ans.

Note: The direction of F implies that if the styela were detached from the rock, it would drift upstream. In reality, it would drift downstream due to forces on its closed surface, which were not considered.

598

6–5. Water exits the 3-in.-diameter pipe at a velocity of 12 ft>s and is split by the wedge diffuser. Determine the force the flow exerts on the diffuser. Take u = 30°.

12 ft/s 3 in.

u

SOLUTION We consider steady flow of an ideal fluid. A

QA = VAAA = ( 12 ft>s ) Jpa = 0.5890 ft 3 >s

2 1.5 ft b R 12

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Since this is free flow, pA = pB = pC = 0. Linear Momentum. Since the change in elevation is negligible and the pressure at A, B, and C is zero gauge, VA = VB = VC = 12 ft>s (Bernoulli equation). The flow is steady and incompressible. ΣF = or

C

B F (a)

0 V rdV + V rV # dA 0t Lcv Lcs

+ c ΣFy = rQB(VB)y + rQC(VC)y - rQA(VA)y F = °

62.4 lb>ft 3 32.2 ft>s2

¢ 3 QB ( - 12 cos 15° ft>s ) + QC ( - 12 cos 15° ft>s ) - ( 0.5890 ft 3 >s )( -12 ft>s ) 4

= 1.9379 3 7.0686 - 12 cos 15° ( QB + QC ) 4

However, QB + QC = QA = 0.5891 ft 3 >s. Then, F = 1.9379[7.0686 - 12 cos 15°(0.5890)]

Ans.

= 0.4668 lb = 0.467 lb

Ans: 0.467 lb 599

6–6. Water exits the 3-in.-diameter pipe at a velocity of 12 ft>s, and is split by the wedge diffuser. Determine the force the flow exerts on the diffuser as a function of the diffuser angle u. Plot this force (vertical axis) versus u for 0 … u … 30°. Give values for increments of ∆u = 5°.

12 ft/s 3 in.

u

SOLUTION

A

The discharge is QA = VAAA = ( 12 ft>s ) c p a

2 1.5 ft b d = 0.1875p ft 3 >s 12

The free-body diagram of the control volume is shown in Fig. a. Since this is a free flow, pA = pB = pC = 0. Also, since the change in elevation is negligible, VA = VB = VC = 12 ft>s. The flow is steady and incompressible. Thus ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

The vertical component of this equation gives + c ΣFy = 0 + F = °

3 - ( VA ) y 4 r ( - VAAA )

624 lb>ft 3 32.2 ft>s2

+

3 - ( VB ) y 4 r ( VBAB )

¢ 3 ( -12 ft>s )( -0.1875p ft 3 >s ) +

F = 23.25 3 0.1875p - ( QB + QC ) cos u>2 4

+

3 - ( VC ) y 4 r ( VCAC )

( - 12 cos u>2 ft>s ) QB + ( -12 cos u>2 ft>s ) QC 4

However, continuity requires that QA = QB + QC. Then F = 23.25 ( 0.1875p - 0.1875p cos u>2 ) F = [13.7 ( 1 - cos u>2 ) ] lb

Ans.

The plot of F vs u is shown in Fig. b.

600

C

B F (a)

6–6. Continued

u(deg.)

0

5

10

15

20

25

30

F(lb)

0

0.0130

0.0521

0.117

0.208

0.325

0.467

5

10

15

20

25

F(lb) 0.5

0.4

0.3

0.2

0.1

(deg.) 0

30

(b)

601

Ans: F = 3 13.7 (1 - cos u>2 ) 4 lb

6–7. Water flows through the hose with a velocity of 4 m>s. Determine the force that the water exerts on the wall. Assume the water does not splash back off the wall.

100 mm 4 m/s

SOLUTION We consider steady flow of an ideal fluid. Q = VA = ( 4 m>s ) 3 p(0.05 m)2 4 = 0.03142 m3 >s

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Since the flow is free, pA = pB = 0. Linear Momentum. The horizontal component of flow velocity is zero when the water jet hits the wall, (Vout)x = 0. Since the flow is steady and incompressible, ΣF =

A

B

F

(a)

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣFx = 0 + ( VA ) (r) ( - QA ) S - F = ( 4 m>s )( 1000 kg>m3 )( - 0.03142 m3 >s )

Ans.

F = 126 N

Ans: 126 N 602

*6–8. The nozzle has a diameter of 40 mm. If it discharges water with a velocity of 20 m>s against the fixed blade, determine the horizontal force exerted by the water on the blade. The blade divides the water evenly at an angle of u = 45°.

40 mm

C A

u

B

SOLUTION

C

We consider steady flow of an ideal fluid. QA = VAAA = ( 20 m>s ) 3 p(0.02 m)2 4 = 0.02513 m3 >s

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Since this is a free flow, pA = pB = pC. Linear Momentum. Since the change in elevation is negligible and the pressure at A, B, and C is zero gauge, VA = VB = VC = 20 m>s (Bernoulli equation). Since the flow is steady and incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣF = d x

( - VB ) x rQB - ( VC ) x rQC + ( VA ) x r ( - QA )

- F = ( 1000 kg>m3 ) 3 QB ( - 20 m>s ) (cos 45°) + QC ( - 20 m>s ) (cos 45°) - ( 20 m>s )( 0.02513 m3 >s ) 4 F = 1000 3 ( QB + QC ) (20 cos 45°) + 0.5027 4

However, QB + QC = QA = 0.02513 m3 >s . Then

F = 1000[0.02513(20 cos 45°) + 0.5027] Ans.

= 858.09 N = 858 N

603

F

A

B (a)

u

6–9. The nozzle has a diameter of 40 mm. If it discharges water with a velocity of 20 m>s against the fixed blade, determine the horizontal force exerted by the water on the blade as a function of the blade angle u. Plot this force (vertical axis) versus u for 0 … u … 75°. Give values for increments of ∆u = 15°. The blade divides the water evenly.

40 mm

C u

A

u

B

SOLUTION

C

The discharge is

The free-body diagram of the control volume is shown in Fig. a. Since this is a free flow, pA = pB = pC = 0. Also, since the change in elevation is negligible, VA = VB = VC = 20 m>s (Bernoulli’s equation). The flow is steady and incompressible. Thus ΣF =

F

A

Q = VAAA = ( 20 m>s ) 3 p(0.02 m)2 4 = 0.008p m3 >s

B (a)

0 VrdV + VrV # dA 0t Lcv Lcs

The horizontal component of this equation gives + ΣF = 0 + d x

3 - (VA)x 4 r( - VAAA)

+ (VB)xr(VBAB) + (VC)xr(VCAC)

F = ( 1000 kg>m3 ) 3 ( 20 m>s )( 0.008p m3 >s ) + ( 20 sin u m>s ) QB + ( 20 sin u m>s ) QC 4 F = 20 ( 103 ) 3 0.008p + (QB + QC) sin u 4

However continuity requires that QA = QB + QC. Then, F = 20 ( 103 ) [0.008p + (0.008p) sin u] Ans.

F = [160p(1 + sin u)] N where u is in deg. The plot of F vs u is shown in Fig. b. u(deg.) F(N)

0

15

30

45

60

75

503

633

754

858

938

988

F(N) 1000 900 800 700 600 500 400 300 200 100 (deg.)

0

15

30

45

60

75

(b)

604

Ans: F = 3 160p (1 + sin u) 4 N

6–10. A speedboat is powered by the jet drive shown. Seawater is drawn into the pump housing at the rate of 20 ft 3 >s through a 6-in.-diameter intake A. An impeller accelerates the water and forces it out horizontally through a 4-in.-diameter nozzle B. Determine the horizontal and vertical components of thrust exerted on the speedboat. The specific weight of seawater is gsw = 64.3 lb>ft 3.

B

45!

A

SOLUTION Consider the control volume to be the jet drive and the water it contains, Fig. a. From the discharge Q = VAAA; Q = VBAB;

20 ft 3 >s = VA c pa

20 ft 3 >s = VB c pa

2 3 ft b d 12

Th

VA = 101.86 ft>s

2 2 ft b d 12

VB = 229.18 ft>s T

Here the flow is steady. Applying the Linear Momentum equation, ΣF =

(a)

0 VrdV + VrV # dA 0t Lcv Lcs

Writing the horizontal and vertical scalar components of this equation by referring to the FBD of the control volume, Fig. a, + ΣFx = 0 + ( VA cos 45° ) r ( - VAAA ) + VBr ( VBAB ) S Th =

3 ( 101.86 ft>s ) cos 45° 4 °

64.3 lb>ft 3 32.2 ft>s2

= 6276.55 lb = 6.28 kip

¢ ( - 20 ft 3 >s ) + ( 229.18 ft>s ) °

64.3 lb>ft 3 32.2 ft>s2 Ans.

¢ ( 20 ft 3 >s )

+ c ΣFy = 0 + ( VA sin 45° ) r ( -VAAA ) - Tv =

3 ( 101.86 ft>s ) sin 45° 4 °

Tv = 2876.54 lb = 2.88 kip

64.3 lb>ft 3 32.2 ft>s2

¢ ( - 20 ft 3 >s )

Ans.

The thrust components on the speedboat are equal and opposite to those exerted on the water.

Ans: Th = 6.28 kip Tv = 2.88 kip 605

6–11. Water flows out of the reducing elbow at 0.4 ft 3 >s. Determine the horizontal and vertical components of force that are necessary to hold the elbow in place at A. Neglect the size and weight of the elbow and the water within it. The water is discharged to the atmosphere at B.

0.5 ft

A

60! B

0.25 ft

SOLUTION

F

y

p

0.4 ft 3 >s = VA 3 p(0.25 ft)2 4

Q = VAAA;

A

F

VA = 2.0372 ft>s

x

A

Continuity equation

0 rdV + V # dA = 0 0t Lcv Lcs

B

0 - VAAA + VBAB = 0 3

p =0 B

(a)

2

- 0.4 ft >s + VB(p)(0.125 ft) = 0 VB = 8.149 ft>s

Bernoulli equation. Neglecting elevation change pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 pA °

62.4 lb>ft 3 32.2 ft>s 2

+

( 2.037 ft>s ) 2 2

¢

+ 0 = 0 +

( 8.149 ft>s ) 2 2

+ 0

pA = 60.3234 lb>ft 2

The free-body diagram is shown in Fig. a. Linear Momentum equation 0 VrdV + VrV # dA 0t Lcv Lcs

ΣF = + ΣFx = 0 + rQ aVB - VA b S x x

-Fx + ( 60.3234 lb>ft 2 ) 3 (p)(0.25 ft)2 4 = ° Fx = 10.3 lb

+ c ΣFy = rQ 3 - VBy + 0 4 - Fy = °

62.4 lb>ft 3 32.2 ft>s2

Fy = 5.47 lb

62.4 lb>ft 3 32.2 ft>s2

¢ ( 0.4 ft 3 >s ) 3 8.149 ft>s(cos 60°) - 2.0372 ft>s 4 Ans.

¢ ( 0.4 ft 3 >s )( - 8.149 ft>s ) (sin 60°)

Ans.

Ans: Fx = 10.3 lb Fy = 5.47 lb 606

*6–12. Oil flows through the 100-mm-diameter pipe with a velocity of 5 m>s. If the pressure in the pipe at A and B is 80 kPa, determine the x and y components of force the flow exerts on the elbow. The flow occurs in the horizontal plane. Take ro = 900 kg>m3.

A

5 m/s

100 mm

60! B

SOLUTION We consider steady flow of an ideal fluid.

p = 80 kPa A

Q = VA = ( 5 m>s ) 3 p(0.05 m) = 0.03927 m3 >s

2

4

A

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Here, pA = pB = 80 kPa.

B

y

p = 80 kPa

0 VrdV + VrV # dA 0t Lcv Lcs

B

(a)

or + ΣFx = 0 + (VA)x r( -Q) + (VB)x rQ S - Fx +

3 80 ( 103 ) N>m2 4 3 p(0.05 m)2 4

+

3 80 ( 103 ) N>m2 4 3 p(0.05 m)2 4 cos 60°

= ( 900 kg>m3 )( 0.03927 m3 >s )( - 5 m>s cos 60° - 5 m>s ) Fx = 1207.55 = 1.21 kN

Ans.

+ c ΣFy = 0 + 0 + (VB)y r( -Q) - Fy +

3 80 ( 103 ) N>m2 4 3 p ( 0.05 m ) 2 4 sin 60° Fy = 697 N

x

F

Linear Momentum. Since the flow is steady incompressible ΣF =

F

60

= ( 5 m>s sin 60° )( 900 kg>m3 )( - 0.03927 m3 >s ) Ans.

607

6–13. The speed of water passing through the elbow on a buried pipe is V = 8 ft>s. Assuming that the pipe connections at A and B do not offer any force resistance on the elbow, determine the resultant horizontal force F that the soil must exert on the elbow in order to hold it in equilibrium. The pressure within the pipe at A and B is 10 psi.

F

B 5 in.

A 8 ft/s

45!

45!

SOLUTION

F

We consider steady flow of an ideal fluid. Q = VA = ( 8 ft>s ) c pa = 1.091 ft 3 >s

2 2.5 ft b d 12

A

p = 10 psi B

Control Volume. The free-body diagram of the control volume is shown in Fig. a.

p = 10 psi A

(a)

Linear Momentum. Since the flow is steady and incompressible, ΣF =

B

0 VrdV + VrV # dA 0t Lcv Lcs

or + c ΣFy = 0 + ( VA ) y(r)( - Q) + 2J ( 10 lb>in2 ) a

( - VB ) y rQ

2 62.4 lb>ft 3 12 in 2 2.5 b cos 45° c p a ft b d R - F = ° ¢ ( 1.091 ft 3 >s ) 3 -8 ft>s cos 45° - 8 ft>s cos 45° 4 1 ft 12 32.2 ft>s2

Ans.

F = 301.60 lb = 302 lb

Ans: 302 lb 608

6–14. Water flows through the 200-mm-diameter pipe at 4 m>s. If it exits into the atmosphere through the nozzle, determine the resultant force the bolts must develop at the connection AB to hold the nozzle onto the pipe.

200 mm

B 100 mm

4 m/s

A

SOLUTION

y

The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3 average velocities will be used. The control volume contains the water in the nozzle as shown in Fig. a. Continuity requires

x Fin

0 rdV + rV # dA = 0 0t Lcv Lcs

F

0 - Vin Ain + Vout Aout = 0 - ( 4 m>s ) 3 p(0.1 m)2 4 + Vout 3 p(0.05 m)2 4 = 0

(a)

Vout = 16 m>s

Applying the Bernoulli’s equation between two points on the control streamline with pout = patm = 0, pout pin Vout2 Vin2 + + + zin = + zout gw gw 2g 2g pin 9810 N>m3

+

( 4 m>s ) 2 ( 16 m>s ) 2 + 0 = 0 + + 0 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pin = 120 ( 103 ) N>m2

Thus, the pressure force acting on the inlet control surface on the FBD of the control volume is Fin = pin Ain =

3 120 ( 103 ) N>m2 4 3 p(0.1 m)2 4

= 3769.91 N

Applying the linear momentum equation, ΣF =

0 VrdV + VrwV # dA 0t Lcv Lcs

Write the scalar component of this equation along x axis, referring to Fig. a + 2 ΣFx 1S

= 0 + Vout rw ( Vout Aout ) + Vin rw ( - Vin Ain )

3769.91 N - F = ( 16 m>s )( 1000 kg>m3 )( 16 m>s ) 3 p(0.05 m)2 4 + ( 4 m>s )( 1000 kg>m3 )( -4 m>s ) 3 p(0.1 m)2 4 Ans.

F = 2261.95 N = 2.26 kN

Ans: 2.26 kN 609

6–15. The apparatus or “jet pump” used in an industrial plant is constructed by placing the tube within the pipe. Determine the increase in pressure (PB - PA) that occurs between the back A and front B of the pipe if the velocity of the flow within the 200-mm-diameter pipe is 2 m>s, and the velocity of the flow through the 20-mm-diameter tube is 40 m>s. The fluid is ethyl alcohol having a density of rea = 790 kg>m3. Assume the pressure at each cross section of the pipe is uniform.

20 mm C

2 m/s

2 m/s

A

200 mm

40 m/s

B

SOLUTION The flow is steady and the ethyl alcohol can be considered an ideal fluid (incompressible and inviscid) Such that rea = 790 kg>m3. Average velocities will be used. The control volume considered is shown in Fig. a. Continuity requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - (VA)t(AA)t - (VA)p(AA)p + (VB)p(AB)p = 0 - ( 40 m>s ) 3 p(0.01 m)2 4 - ( 2 m>s ) e p 3 (0.1 m)2 - (0.01 m)2 4 f + ( VB ) p 3 p(0.1 m)2 4 = 0 (VB)p = 2.38 m>s

Within the tube, zC = zA and VC = VA, so by Bernoulli’s equation, pC = pA. Furthermore, because pC at the tube exit equals pC in the surrounding pipe flow, which again by Bernoulli’s equation equals pA in the pipe, it follows that pA is the same inside and outside the tube. The pressure forces on the inlet and outlet control surfaces are FA = pAAA = pA 3 p(0.1 m)2 4 = 0.01 ppA FB = pBAB = pA 3 p(0.1 m)2 4 = 0.01 ppB

Applying the linear momentum equation, ΣF =

0 Vr dV + VreaV # dA 0t Lcv ea Lcs

Writing the scalar component of this equation along the x axis by referring to Fig. a, + ΣFx = 0 + (VB)prea(VB)p(AB)p + (VA)t rea 3 - (VA)t(AA)t 4 + (VA)prea 3 -(VA)p(AA)p 4 S

0.01ppA - 0.01ppB = ( 2.38 m>s ) 2 ( 790 kg>m3 ) 3 p(0.1 m)2 4 - ( 40 m>s ) 2 ( 790 kg>m3 ) 3 p(0.01 m)2 4 - ( 2 m>s ) 2 ( 790 kg>m3 ) 5 p 3 (0.1 m)2 - (0.01 m)2 4 6 ∆P = pB - pA = 11.29 ( 103 ) pa = 11.3 kPa

FA

Ans.

FB

C

(a)

Ans: 11.3 kPa 610

*6–16. The jet of water flows from the 100-mm-diameter pipe at 4 m>s. If it strikes the fixed vane and is deflected as  shown, determine the normal force the jet exerts on the vane.

A

C

SOLUTION

A

Bernoulli Equation. Since the water jet is a free flow, pA = pB = pC = 0. Also, if we neglect the elevation change in the water jet, the Bernoulli equation gives VA2

VB2

VC2

pA pB pC = = + + + g g g 2g 2g 2g 0 +

B

VA2 VB2 = 0 + = 0 + 2g 2g

( 4 m>s ) 2 Ans.

QC = VC AC = ( 4 m>s ) 3 p(0.05 m)2 4 = 0.03142 m3 >s

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Since the flow is steady incompressible. 0 VrdV + VrV # dA 0t Lcv Lcs

ΣFn = 0 +

Fn

45

C

n

2g

The discharge at C is

ΣF =

Ft

B

VA = VB = 4 m>s

Fn =

4 m/s

t

We consider steady flow of an ideal fluid.

or

45!

100 mm

( - QC ) (r) ( - VC ) n

( - 0.03142 m3 >s )( 1000 kg>m3 )( - 4 m>s sin 45° ) Fn = 88.9 N

611

Ans.

(a)

6–17. The jet of water flows from the 100-mm-diameter pipe at 4 m>s. If it strikes the fixed vane and is deflected as shown, determine the volume flow towards A and towards B if the tangential component of the force that the water exerts on the vane is zero.

A

45!

100 mm

C

SOLUTION

4 m/s

B

t

We consider steady flow of an ideal fluid.

A

Bernoulli Equation. Since the water jet is a free flow, pA = pB = pC = 0. Also, if we neglect the elevation change in the water jet, the Bernoulli equation gives VA2

VC2

VB2

pA pB pC = = + + + g g g 2g 2g 2g 0 +

VA2 VB2 = 0 + = 0 + 2g 2g

Ft

Fn

45

C

B

( 4 m>s ) 2

n

2g

(a)

VA = VB = 4 m>s The discharge at C is QC = VCAC = ( 4 m>s ) 3 p(0.05 m)2 4 = 0.03142 m3 >s

Continuity Equation.

0 rdV + V # dA = 0 0t Lcv Lcs 0 - QC + QA + QB = 0

QA + QB = 0.03142

(1)

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Here, it is required that Ft = 0. Since the flow is steady incompressible, ΣF = or

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣF = r 3 Q ( V ) + Q ( V ) - Q ( V ) 4 ; t A A t B B t C C t a

0 = ( 1000 kg>m3 ) 3 QA ( 4 m>s ) + QB ( - 4 m>s ) - 0.03142 m3 >s ( - 4 m>s cos 45° ) 4 (2)

QA - QB = -0.02221

Solving Eqs. (1) and (2), QA = 0.00460 m3 >s

QB = 0.0268 m3 >s

612

Ans.

Ans: QA = 0.00460 m3>s QB = 0.0268 m3>s

6–18. As water flows through the pipe at a velocity of 5 m>s, it encounters the orifice plate, which has a hole in its center. If the pressure at A is 230 kPa, and at B it is 180 kPa, determine the force the water exerts on the plate.

A

75 mm

5 m/s

B

200 mm

SOLUTION

PA = 230 kPa

PB = 180 kPa

We consider steady flow of an ideal fluid. Take the water from A to B to be the control volume. Continuity Equation. Since the diameters of the pipe at A and B are equal, continuity requires

F

A

B

(a)

VA = VB = 5 m>s The free-body diagram of the control volume is shown in Fig. a. Linear Momentum. The flow is steady and incompressible since points A and B are selected at a sufficient distance from the gate. ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣFx = 0 + ( VA ) r( - Q) + ( VB ) r(Q) S -F +

3 230 ( 103 ) N>m2 4 3 p(0.1 m)2 4

F = 1570.80 N = 1.57 kN

-

3 180 ( 103 ) N>m2 4 3 p(0.1 m)2 4

= rQ(V - V) = 0

Ans.

Ans: 1.57 kN 613

6–19. Water enters A with a velocity of 8 m>s and pressure of 70 kPa. If the velocity at C is 9 m>s, determine the horizontal and vertical components of the resultant force that must act on the transition to hold it in place. Neglect the size of the transition.

20 mm B 40 mm 30!

C

A 50 mm

8 m/s

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. The average velocities will be used. The control volume contains the water in the transition as shown in Fig. a. The continuity condition requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VC AC = 0 - ( 8 m>s ) 3 p(0.025 m)2 4 + VB 3 p(0.01 m)2 4 + ( 9 m>s ) 3 p(0.02 m)2 4 = 0 VB = 14 m>s

Write the Bernoulli’s equation between A and B, and A and C, pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g 70 ( 103 ) N>m2 3

9810 N>m

+

( 8 m>s ) 2 ( 14 m>s ) 2 pB + 0 = + + 0 3 9810 N>m 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pB = 4 ( 103 ) N>m2 pC pA VC2 VA2 + + zA = + + zC gw gw 2g 2g

70 ( 103 ) N>m2 9810 N>m3

+

( 8 m>s ) 2 ( 9 m>s ) 2 pC + 0 = + + 0 9810 N>m3 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pC = 61.5 ( 103 ) N>m3

The pressure forces acting on the inlet and outlet control surfaces indicated on the FBD of the control volume are FA = pAAA = FB = pBAB = FC = pCAC =

3 70 ( 103 ) N>m2 4 3 p(0.025 m)2 4

3 4 ( 10 ) N>m 4 3 p(0.01 m) 4 3

2

2

= 43.75p N

= 0.4p N

3 61.5 ( 103 ) N>m2 4 3 p(0.02 m)2 4

= 24.6p N

614

9 m/s

6–19. Continued

Applying the Linear momentum equation, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

Writing the scalar component of this equation along x and y axes, + 2 ΣFx = 0 + ( - VB cos 30° )( rw )( VB AB ) + VC rw ( VC AC ) 1S

Fx + (0.4p N) cos 30° - 24.6p N = - ( 14 m>s ) (cos 30°) ( 1000 kg>m3 )( 14 m>s ) 3 p(0.01 m) + ( 9 m>s )( 1000 kg>m3 )( 9 m>s ) 3 p(0.02 m)2 4

Fx = 125 N S

4

Ans.

( + c ) ΣFy = 0 + ( VB sin 30° )( rw )( VBAB ) + VArw ( - VAAA )

Fy + 43.75p N - (0.4p N) sin 30° = ( 14 m>s ) (sin 30°) ( 1000 kg>m3 )( 14 m>s ) 3 p(0.01 m)2 4 + ( 8 m>s )( 1000 kg>m3 )( -8 m>s ) 3 p(0.025 m)2 4

Ans.

Fy = + 231.69 N = 232 NT FB = 0.4 N

y

Fx

x FC = 24.6 N

30˚

Fy

FA = 43.75 N (a)

Ans: Fx = 125 N Fy = 232 N 615

*6–20. Crude oil flows through the horizontal tapered 45° elbow at 0.02 m3 >s. If the pressure at A is 300 kPa, determine the horizontal and vertical components of the resultant force the oil exerts on the elbow. Neglect the size of the elbow.

A

50 mm

135!

30 mm B

SOLUTION The flow is steady and crude oil can be considered as an ideal fluid (incompressible and inviscid) such that rco = 880 kg>m3 average velocities will be used. The control volume considered contains the crude oil in the elbow as shown in Fig. a. From the discharge, 0.02 m3 >s = VA 3 p(0.025 m)2 4

Q = VAAA;

0.02 m >s = VB 3 p(0.015 m) 3

Q = VBAB;

2

Applying Bernoulli’s equation between A and B,

4

VA = 10.19 m>s VB = 28.29 m>s

FA

y

pA pB VA2 VB2 + + zA = + + zB gco gco 2g 2g

300 ( 103 ) N>m2

( 880 kg>m )( 9.81 m>s ) 3

2

+

x

( 10.19 m>s ) 2 ( 28.29 m>s ) 2 pB + 0 = + + 0 2 3 2 ( 880 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s ) Fx

pB = - 6.596 ( 103 ) Pa The negative sign indicates that suction occurs at B. The pressure for acting on the inlet and outlet control surfaces indicated on the FBD of the control volume are FA = pAAA = FB = pBAB =

3 300 ( 103 ) N>m2 4 3 p(0.025 m)2 4

45˚

= 589.05 N

3 6.596 ( 103 ) N>m2 4 3 p(0.015 m)2 4

= 4.663 N

FB

Applying the linear momentum equation, ΣF =

Fy

(a)

0 # 0t L VrcodV + L VrcoV dA cv cs

Writing the scalar component of this equation along x and y axis by referring to Fig. a + ΣFx = 0 + S

( - VB cos 45° )( rco )( VBAB )

( - 4.663 N) cos 45° - Fx =

( - 28.29 m>s ) cos 45° ( 880 kg>m3 )( 0.02 m3 >s )

Fx = 349 N d + c ΣFy = 0 +

Ans.

( - VB sin 45° ) rco ( VBAB ) + ( - VA ) rco ( - VAAA )

Fy - (4.663 N) sin 45° - 589.05 N =

( - 28.29 m>s ) sin 45° ( 880 kg>m3 )( 0.02 m3 >s ) + ( - 10.19 m>s )( 880 kg>m3 )( - 0.02 m3 >s )

Fy = 419 N c

616

6–21. The hemispherical bowl of mass m is held in equilibrium by the vertical jet of water discharged through a nozzle of diameter d. If the volumetric flow is Q, determine the height h at which the bowl is suspended. The water density is rw.

h

SOLUTION The flow is steady and water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Average velocities will be used. From the discharge the velocity of the water leaving the nozzle (point A on the control volume shown in Fig. a) is p Q = VAa d 2 b 4

Q = VAAA;

VA =

B

mg

4Q pd 2

h

B

Applying Bernoulli’s equation between points A and B on the central streamline with pA = pB = 0, zA = 0 and zB = h, VA2

C

VB2

pB pA + + zA = + + zB gw gw 2g 2g 0 + a

4Q

pd 2g

2

b

A (a)

2

+ 0 = 0 +

VB2 2g

+ h

16Q2 - 2gh A p2d 4

VB =

C (b)

(1)

By considering the FBD of the control volume shown in Fig. b, where B and C are the inlet and outlet control surfaces, ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of this equation along the y axis realizing that by Bernoulli’s equation VC = VB =

16Q2 - 2gh and Q = VA, A p2d 4

+ c ΣFy = 0 + VBrw ( - VB AB ) +

( - VC ) rw ( VC AC )

-mg = VBrw( - Q) - VBrwQ mg = 2rwQVB Substituting Eg. 1 into this equation mg = 2rwQ h =

8Q2

16Q2 - 2gh A p2d 4

p2d 4g

-

m2g

Ans.

8rw2Q2

Ans: h =

617

8Q2 p2d 4g

-

m2g 8rw2 Q2

6–22. The 500-g hemispherical bowl is held in equilibrium by the vertical jet of water discharged through the 10-mm-diameter nozzle. Determine the height h of the bowl as a function of the volumetric flow Q of the water through the nozzle. Plot the height h (vertical axis) versus Q for 0.5 1 10-3 2 m3 >s … Q … 1 1 10-3 2 m3 >s. Give values for increments of ∆Q = 0.1 1 10-3 2 m3 >s.

h

SOLUTION

B

The flow is steady and water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Average velocities will be used. From the discharge, the velocity of the water leaving the nozzle (point A on the control volume as shown in Fig. a) is Q = VAAA; VA = c

h

Q = VA 3 p(0.005 m)2 4 40 ( 103 ) Q d m>s p

A

Applying Bernoulli’s equation between points A and B on the central streamline with pA = pB = 0, zA = 0 and zB = h, pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g 0 + c

2 40 ( 103 ) Qd p

2 ( 9.81 m>s2 )

+ 0 = 0 +

VB2 2 ( 9.81 m>s2 )

0.5 (9.81) N

+ h B

1.6 ( 10 ) 2 Q - 19.62 h (1) p2 By considering the FBD of the fixed control volume shown in Fig. b, where B and C are the inlet and outlet control surfaces, VB =

ΣF =

9

A

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of the equation along y axis realizing that Vc = VB =

(a)

1.6 ( 109 ) Q2 - 19.62 h and Q = VA, A p2 + c ΣFy = 0 + VBrw ( -VB AB ) +

( -VC ) rw ( VC AC )

618

C

C (b)

6–22. Continued

-0.5(9.81)N = ( 1000 kg>m3 ) c -2 a h = c

A

1.6 ( 109 ) Q2 - 19.62 h bQ d p2

8.26 ( 106 ) Q4 - 0.307 ( 10-6 ) Q2

The plot of h vs. Q is shown in Fig. c Q ( 10-3 m3 >s ) h(m)

d m, where Q is in m3 >s

Ans.

0.5

0.6

0.7

0.8

0.9

1.0

0.439

0.839

2.12

3.42

4.81

6.31

7.96

0

h(m) 8 7 6 5 4 3 2 1

0

(10–3 m3 s( 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

(c)

Ans: h = c 619

8.26 ( 106 ) Q4 - 0.307 ( 10-6 ) Q2

d m

6–23. Water flows into the rectangular tank at the rate of 0.5 ft 3 >s from the 3-in.-diameter pipe at A. If the tank has a width of 2 ft and an empty weight of 150 lb, determine the apparent weight of the tank caused by the flow at the instant h = 3 ft.

A

h ! 3 ft B

3.5 ft

SOLUTION We consider steady flow of an ideal fluid. Q = VAAA;

0.5 ft 3 >s = V £ p a

VA = 10.186 ft>s

2 1.5 ft b § 12

PA = 0

The control volume is the water in the tank. Its free-body diagram is shown in Fig. a. The weight of the water in the control volume is W = gwV = ( 62.4 lb>ft 3 ) 3 (3.5 ft)(2 ft)(3 ft) 4 = 1310.4 lb. Here, A is exposed to the atmosphere, pA = 0. ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

A W = 1310.4 lb

150 lb

N (a)

or + c ΣFy = 0 + N - 150 lb - 1310.4 lb =

( -VA ) r( - Q)

( - 10.186 ft>s ) °

62.4 lb>ft 3 32.2 ft>s2

N = 1470 lb = 1.47 kip

¢ ( - 0.5 ft 3 >s )

Ans.

Ans: 1.47 kip 620

*6–24. The barge is being loaded with an industrial waste liquid having a density of 1.2 Mg>m3. If the average velocity of flow out of the 100-mm-diameter pipe is VA = 3 m>s, determine the force in the tie rope needed to hold the barge stationary.

10 m

VA

2m

B

SOLUTION

A

W

PA = 0 A

We consider steady flow of an ideal fluid.

T

Q = VAAA = ( 3 m>s ) 3 p(0.05 m)2 4 = 0.023562 m3 >s

N

The control volume is the barge and its contents. Its free-body diagram is shown in Fig. a. Since the flow is free, pA = 0. Linear Momentum. Since the flow is steady and incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣFx = 0 + S T =

( - VA ) r( - Q)

( - 3 m>s )( 1.2 ( 103 kg>m3 )( - 0.023562 m3 ) ) Ans.

T = 84.8 N

621

(a)

6–25. The barge is being loaded with an industrial waste liquid having a density of 1.2 Mg>m3. Determine the maximum force in the tie rope needed to hold the barge stationary. The waste can enter the barge at any point within the 10-m region. Also, what is the speed of the waste exiting the pipe at A when this occurs? The pipe has a diameter of 100 mm.

10 m B

VA

A 2m

SOLUTION The maximum force developed in the tie rope occurs when the velocity V of the flow is maximum. This happens when the flow achieves the maximum range, ie, Sx = 10 m. Consider the vertical motion by referring to Fig. a. 1 2

1 2

( + T ) Sy = ( S0 ) y + ( v0 ) y t + ac t 2; 2 m = 0 + 0 + ( 9.81 m>s2 ) t 2 t = 0.6386 s The horizontal motion gives + 2S 1d x

= ( S0 ) x + ( v0 ) x t;

10 m = 0 + VA(0.6386 s) Ans.

VA = 15.66 m>s = 15.7 m>s

The fixed control volume considered is the barge and its contents as shown in Fig. b. Since the flow is free, pA = 0. The flow is steady and incompressible. Then ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

Writing the horizontal scalar component of this equation by referring to the FBD of the control volume, Fig. b, +2 1S

ΣFx = 0 + ( - VA)r ( - VA ) A T = 0 +

( -15.66 m>s )( 1200 kg>m3 ) 3 - ( 15.66 m>s ) p(0.05 m)2 4

= 2311.43 N

Ans.

= 2.31 kN

y

Sx = 10 m

W

pA = 0 A

V

x

T

Sy = 2 m

(a)

N (b)

Ans: VA = 15.7 m>s T = 2.31 kN 622

6–26. A nuclear reactor is cooled with liquid sodium, which is transferred through the reactor core using the electromagnetic pump. The sodium moves through a pipe at A having a diameter of 3 in., with a velocity of 15 ft>s and pressure of 20 psi, and passes through the rectangular duct, where it is pumped by an electromagnetic force giving it a 30-ft pumphead. If it emerges at B through a 2-in.-diameter pipe, determine the restraining force F on each arm, needed to hold the pipe in place. Take gNa = 53.2 lb>ft 3.

A F

B

F 2 in.

SOLUTION

y

The flow is steady and the liquid sodium can be considered as an ideal fluid (incompressible and inviscid) such that gNA = 53.2 lb>ft 3. Average velocities will be  used. The control volume contains the liquid in the pipe and the transition as shown in Fig. a.

0 rdV + rV # dA = 0 0t Lcv Lcs

- ( 15 ft>s ) c p a

2 2 1.5 1 ft b d + VB c p a ft b d = 0 12 12

VB = 33.75 ft>s

Applying the energy equation with hs = - 30 ft (negative sign indicates pump head), lb 12 in 2 ba b = 2880 lb>ft 2 and hl = 0, in 1 ft

53.2 lb>ft 3

pB pA VA2 VB2 + + ZA + ht + hl = + + ZB + ht + hl gNA gNA 2g 2g

+

2F

(a)

0 - VAAA + VBAB = 0

2880 lb>ft 2

x

FB

Continuity requires

pA = a20

15 ft/s 3 in.

( 15 ft>s ) 2 ( 33.75 ft>s ) 2 pB + + 0 + (30 ft ) 0 = + 0 53.2 lb>ft 3 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) pB = 3720.90 lb>ft 2

Thus, the pressure force acting on opened control surfaces at A and B are FA = pA AA = ( 2880 lb>ft 2 ) c p a

2 1.5 ft b d = 141.37 lb 12

FB = pB AB = ( 3720.90 lb>ft 2 ) c p a

Applying the linear momentum equation ΣF =

2 1 ft b d = 81.18 lb 12

0 Vr dV + VrNAV # dA 0t Lcv NA Lcs

623

FA

6–26. Continued

Writing the scalar component of this equation along x axis by referring to Fig. a + 2 ΣFx = 0 + ( - VB ) rNA ( VBAB ) + ( - VA ) rNA ( - VAAA ) 1S 81.18 lb - 141.37 lb + 2 F =

( - 33.75 ft>s ) a

+

( -15 ft>s ) a

53.2 lb>ft 3 2

32.2 ft>s

53.2 lb>ft 3 32.2 ft>s2

F = 18.7 lb

b e ( 33.75 ft>s ) c p a

b e ( -15 ft>s ) c p a

2 1 ft b d f 12

2 1.5 ft b d f 12

Ans.

Note: This solution assumes that the electromagnetic pump is mounted on the outside of the duct, so that the EM force of the pump on the liquid is canceled by the equal and opposite reaction force on the pump, transferred to the pipe. y

x

2F

FB

FA

(a)

Ans: 18.7 lb 624

6–27. Air flows through the closed duct with a uniform velocity of 0.3 m>s. Determine the horizontal force F that the strap must exert on the duct to hold it in place. Neglect any force at the slip joints A and B. Take ra = 1.22 kg>m3.

A 0.3 m/s

C B

3m

2m 1m

SOLUTION

F2 pB = 0.4392 Pa

Assume the air is incompressible and non-viscids. Q = VAAA = ( 0.3 m>s ) (3 m)(1 m) = 0.9 m3 >s

Continuity requires

0 rdV + rV # dA = 0 0t Lcv Lcs

F2

0 - 0.9 m3 >s + VB(1 m)(1 m) = 0 VB = 0.9 m>s

Apply the Bernoulli’s equation between A and B. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 pA 1.22 kg>m3

+

( 0.3 m>s ) 2 2

+ 0 = 0 +

( 0.9 m>s ) 2 2

+ 0

pA = 0.4392 Pa Linear Momentum equation + ΣFx = 0 VrdV + VrV # dA S 0t Lcv Lcs - F + ( 0.4392 N>m2 ) (3 m) ( 1 m ) = 0 + ( 0.3 m>s )( 1.22 kg>m3 )( - 0.9 m3 >s ) + ( 0.9 m>s )( 1.22 kg>m3 )( 0.9 m3 >s ) Ans.

F = 0.659 N

Ans: 0.659 N 625

*6–28. As oil flows through the 20-m-long, 200-mmdiameter pipeline, it has a constant average velocity of 2 m>s. Friction losses along the pipe cause the pressure at B to be 8 kPa less than the pressure at A. Determine the  resultant friction force on this length of pipe. Take ro = 880 kg>m3.

A

SOLUTION Here ∆ p = 8 kPa and so the force developed by the pressure difference is Fp = 8 ( 103 ) N>m2(p)(0.1 m)2 = 251.3 N The free-body diagram is shown in Fig. a. Applying the linear momentum equation ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

251.3 - F = 0 + ( 2 m>s )( 880 kg>m3 )( 2 m>s ) (p) ( 0.1 m2 ) + ( 2 m>s )( 880 kg>m3 )( 2 m>s ) (p) ( 0.1 m2 ) Ans.

F = 251 N F

F

(a)

626

B

6–29. Oil flows through the 50-mm-diameter vertical pipe assembly such that the pressure at A is 240 kPa and the velocity is 3 m>s. Determine the horizontal and vertical components of force the pipe exerts on the U-section AB of the assembly. The assembly and the oil within it have a combined weight of 60 N. Take ro = 900 kg>m3.

3 m/s 50 mm A

0.4 m

B 0.4 m

SOLUTION Bernoulli Equation: Because the diameter is the same at A and B, VA = VB = V. With the datum at B, pA pB V2 V2 + gzA = + gzB + + r r 2 2 240 ( 103 ) Pa 900 Kg>m3

+ ( 9.81 m>s2 ) (0.4 m) =

pB 900 Kg>m3

+ 0

pB = 243.532 ( 103 ) Pa Linear Momentum: ΣF = + 2 Fx 1S

0 VrdV + VrV # dA 0t Lcv Lcs Ans.

= 0 + 0 = 0

( + c ) Fy + 3 243.532 ( 103 ) Pa 4 p(0.025 m)2 - 3 240 ( 103 ) Pa 4 p(0.025 m)2 - 60 N = 0 + ( - V)r( - VA) + ( - V)r(VA)

Ans.

Fy = 53.1 N

Ans: Fx = 0 Fy = 53.1 N 627

6–30. Water flows into the tank at the rate of 0.05 m3 >s from the 100-mm-diameter pipe. If the tank is 500 mm on each side, determine the compression in each of the four springs that support its corners when the water reaches a depth of h = 1 m. Each spring has a stiffness of k = 8 kN>m. When empty, the tank compresses each spring 30 mm. h!1m

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid). Hence Average velocities are used and pw = 1000 kg>m3.The control volume contains the water in the pipe and the tank and it is fixed instantaneously, Fig. a. From the discharge Q = Vin Ain;

- 0.05 m3 >s = Vin 3 p(0.05 m)2 4

Vin = 6.366 m>s

Q = Vout Aout; 0.05m >s = Vout 3 (0.5 m) - p(0.05 m) 3

2

2

Applying the linear momentum equation ΣF =

4

0.05 mm Ain

0.5 m

Aout 0.5 m

Vout = 0.2065 m>s

0 VrdV + VrV # dA 0t Lcv Lcs

Writing the scalar equation along the y axis by refering to the FBD of the control volume, Fig. a, + c F = 0 + Vout rwVout Aout +

F

( - Vin ) rw ( - Vin Ain )

(a)

.5 m

F = ( 0.2065 m>s ) 2 ( 1000 kg>m3 ) 3 (0.5 m)2 - p(0.05 m)2 4 + ( 6.366 m>s ) 2 ( 1000 Kg>m3 ) 3 p(0.05 m)2 4 in

= 328.63 N

The weight of the water in the tank at a depth of 1m is

WT = 3412.5 N

WN = rWgVW = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (0.5 m)2(1 m) 4 = 2452.5 N

The weight of the empty tank is

F = 328.63 N

Wt = 4kx = 4 3 8 ( 103 ) N>m 4 (0.03 m) = 960 N

Thus, the total weight is

WT = WW + Wt = 2452.5 N + 960 N = 3412.5 N

4Fsp

Equilibrium of the FBD of the tank, Fig. b, requires + c ΣFy = 0;

(b)

4Fsp - 328.63 N - 3412.5 N = 0 Fsp = 935.28 N

Thus, the compression of the spring is Fsp = kx;

935.28 N =

3 8 ( 103 ) N>m 4 x

x = 0.1169 m = 117 mm

Ans.

Ans: 117 mm 628

6–31. The 300-kg circular craft is suspended 100 mm from the ground. For this to occur, air is drawn in at 18 m>s through the 200-mm-diameter intake and discharged to the ground as shown. Determine the pressure that the craft exerts on the ground. Take ra = 1.22 kg>m3.

1.5 m

1.5 m 200 mm C

A

B

100 mm

SOLUTION We consider steady flow of an ideal fluid. Q = VCAC = ( 18 m>s ) 3 p(0.1 m)2 4 = 0.5655 m3 >s

Take the control volume to be the craft and the air inside it. Its free-body diagram is shown in Fig. a. Since the flow is open to the atmosphere, pC = 0. Linear Momentum. Since no air escapes from the hovercraft vertically, Vout = 0. Since the flow is steady incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + c ΣFy = 0 + ( - VC)r( - Q) p 3 p(1.5 m)2 - 300 kg ( 9.81 m>s2 ) =

( - 18 m>s )( 1.22 kg>m3 ) - ( 0.5655 m3 >s ) 4

Ans.

p = 418 Pa

(300 kg) (9.81 m s2 ( p =0 C

p

Ans: 418 Pa 629

*6–32. The cylindrical needle valve is used to control the flow of 0.003 m3 >s of water through the 20-mm-diameter tube. Determine the force F required to hold it in place when x = 10 mm.

20 mm

20 mm 20!

x

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocities will be used. The control volume is shown in Fig. a. From the discharge, 0.003 m3 >s = Vin 3 p(0.01 m)2 4

Q = Vin Ain;

Vin = 9.549 m>s

From the geometry shown in Fig. b, r

0.01 m - 0.01 m tan 10°

=

0.01 m ; 0.01 m tan 10°

r = 0.008237 m

Thus,

Then

Aout = p 3 (0.01 m)2 - (0.008237 m)2 4 = 0.1010 ( 10-3 ) m2 Q = Vout Aout;

0.003 m3 >s = Vout 3 0.1010 ( 10-3 ) m2 4 Vout = 29.70 m>s

Applying Bernoulli’s equation between the center points of the inlet and outlet control surfaces where pout = patm = 0 2 pout pin Vout Vin2 + + zin = + + zout gw gw 2g 2g

pin 9810 N>m3

+

( 9.549 m>s ) 2 ( 29.70 m>s ) 2 + 0 = 0 + + 0 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pin = 395.35 ( 103 ) Pa

Thus, the pressure force exerted on the inlet control surface is Fin = pinAin =

3 395.35 ( 103 ) N>m2 4 3 p(0.01m)2 4

= 124.20 N

Applying the linear momentum equation, ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component along x axis by referring to the FBD of the control volume shown in Fig. a + ΣFx = 0 + VoutrwVoutAout + Vinrw( - VinAin) S

630

F

*6–32. Continued

However, Q = VoutAout = VinAin = 0.003 m3 >s

124.20 N - F = ( 29.70 m>s )( 1000 kg>m3 )( 0.003 m3 >s ) + ( 9.549 m>s )( 1000 kg>m3 )( - 0.003 m3 >s ) Ans.

F = 63.76 N = 63.8 N

Note: For simplicity, the effect of the slight deflection of the stream, away from the central axis, has been neglected. If it were accounted for, F would be slightly (6 2,) larger. .0 0.01 m tan tan 10

r

F

Fin

A in

. 0.01 m

10

(a) (b)

631

. 0.01 m

A out

.0 0.01 m

6–33. The cylindrical needle valve is used to control the flow of 0.003 m3 >s of water through the 20-mm-diameter tube. Determine the force F required to hold it in place for any position x of closure of the valve.

20 mm

20 mm 20!

x

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocities will be used. The control volume is shown in Fig. a. From the discharge, 0.003 m3 >s = Vin 3 p(0.01 m)2 4

Q = Vin Ain;

Vin = 9.549 m>s

From the geometry shown in Fig. b,

r 0.01 m = ; 0.01 m 0.01 m - x tan 10° tan 10°

r = 0.01 m - (tan 10°)x

Thus,

Then

Aout = p 3 (0.01 m)2 - (0.01 m - (tan 10°)x)2 4 = 0.01108x - 0.09768x2 Q = VoutAout; 0.003 m3 >s = Vout ( 0.01108x - 0.09768x2 ) Vout = a

1 b m>s 3.693x - 32.559x2

Applying the energy equation between the center points of the inlet and outlet control surfaces, where pout = patm = 0. pin pout Vout2 Vin2 + + zin + hpump = + + zout + hturb + hL gw gw 2g 2g

( 9.549 m>s ) + 0 + 0 = 0 + 2 ( 9.81 m>s2 ) 2

pin 3

9810 N>m

+

pin £

500

( 3.693x - 32.559x2 ) 2

a

2 1 b 3.693x - 32.559x2 + 0 + 0 + 0 2 ( 9.81 m>s2 )

- 45.595 ( 103 ) § Pa

Thus, the pressure force on the inlet control surface is Fin = pinAin = £

500

( 3.693x - 32.559x2 ) 2 =

0.05p

- 45.595 ( 103 ) § 3 p(0.01 m)2 4

( 3.693x - 32.559x2 ) 2

- 14.324

632

F

6–33. Continued

Applying the linear momentum equation, ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component along the x axis by referring to the FBD of the control volume shown in Fig. a, + ΣFx = 0 + VoutrwVoutAout + Vinrw( - VinAin) S However, Q = VoutAout = VinAin = 0.003 m3 >s 0.05p

( 3.693x - 32.559x )

2 2

- 14.324 - F = a

1 b ( 1000 kg>m3 )( 0.003 m3 >s ) 3.693x - 32.559x2

+ ( 9.549 m>s )( 1000 kg>m3 )( - 0.003 m3 >s )

F = £

97.7x2 - 11.1x + 0.157

( 3.69x - 32.6x2 ) 2

Ans.

+ 14.3 § N

Note: As in the preceding problem, the slight effect of the 10° deflection of the stream has be neglected. 0.01 m tan 10

r

F

Fin

A in

0.01 m 0.01 m

10

(a) (b)

A out

x

Ans: F = £ 633

97.7x2 - 11.1x + 0.157

( 3.69x - 32.6x2 ) 2

+ 14.3 § N

6–34. The disk valve is used to control the flow of 0.008 m3 >s of water through the 40-mm-diameter tube. Determine the force F required to hold the valve in place for any position x of closure of the valve.

40 mm

x F

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocities will be used. The control volume is shown in Fig. a. From the discharge, 0.008 m3 >s = Vin 3 p(0.02 m)2 4

Q = VinAin;

Vin = 6.366 m>s

The cross-sectional area of the outlet control surfaces is Aout = 2p(0.02 m)x = (0.04px) m2

Then

0.008 m3 >s = Vout(0.04px)

Q = VoutAout;

Vout = a

0.06366 b m>s x

Applying Bernoulli’s equation between the center points of the inlet and outlet control surfaces, where pout = patm = 0. pin pout Vout2 Vin2 + + zin = + + zout gw gw 2g 2g pin 9810 N>m3

+

( 6.366 m>s )

2

2 ( 9.81 m>s2 )

pin = c

+ 0 = 0 +

a

0.06366 2 b x

2 ( 9.81 m>s2 )

+0

2.026 - 20.264 ( 103 ) d Pa x2

Thus, the pressure force on the inlet control surface is Fin = pinAin = c = c

2.026 - 20.264 ( 103 ) d 3 p(0.02 m)2 4 x2

2.546 ( 10-3 ) x2

- 25.465 d N A out

Fin

F A in

(a)

634

6–34. Continued

Applying the linear momentum equation ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of this equation along x axis by referring to the FBD of the control volume shown in Fig. a S ΣFx = 0 + Vinrw( - VinAin) However, Q = VinAin = 0.008 m3 >s. Thus 2.546 ( 10-3 ) x2

- 25.465 - F = ( 6.366 m>s )( 1000 kg>m3 )( -0.008 m3 >s ) F = £

2.55 ( 10-3 ) x2

Ans.

+ 25.5 § N

Ans: F = £ 635

2.55 ( 10-3 ) x2

+ 25.5 § N

6–35. The toy sprinkler consists of a cap and a rigid tube having a diameter of 20 mm. If water flows through the tube at 0.7 1 10 - 3 2 m3 >s, determine the vertical force the wall of the tube must support at B. Neglect the weight of the sprinkler head and the water within the curved segment of the tube. The weight of the tube and water within the vertical segment AB is 4 N.

A

0.75 m

B

SOLUTION Q = VA 0.7 ( 10

-3

) m3 >s = V(p)(0.01 m)2 V = 2.228 m>s

Since the hose has a constant diameter, continuity requires VA = VB = V = 2.228 m>s Applying Bernoulli’s equation between A and B, with the datum of B, VA2

A

VB2

pA pB + zA = + zB + + g g 2g 2g 0 +

2

V + 0.75 m = 2g

pB

( 1000 kg>m3 )( 9.81 m>s2 )

W

+

2

V + 0 2g

pB = 7357.5 Pa The free-body diagram of the control volume is shown in Fig a. Applying the linear momentum equation in the vertical direction, for steady flow ΣF = + c ΣFy = 0 +

pB AB

Fy (a)

0 VrdV + VrV # dA 0t Lcv Lcs

( -VA ) r(QA) + VB r( - QB)

c ΣFy = -2 Vr Q

( 7357.5 N>m2 ) (p)(0.01 m)2 - Fy - 4 N = -2 ( 2.228 m>s )( 1000 kg>m3 )( 0.7 ( 10-3 ) m3 >s ) Ans.

Fy = 1.43 N

Ans: 1.43 N 636

*6–36. The toy sprinkler consists of a cap and a rigid tube having a diameter of 20 mm. Determine the flow through the tube such that it creates a vertical force of 6 N in the tube at B. Neglect the weight of the sprinkler head and the water within the curved segment of the tube. The weight of the tube and water within the vertical segment AB is 4 N.

A

0.75 m

B

SOLUTION Since the hose has a constant diameter, continuity requires VA = VB = V Applying Bernoulli’s equation between A and B, with the datum at B, pB pA VA2 VB2 + zA = + zB + + g g 2g 2g 0 +

V2 + 0.75 m = 2g

pB

( 1000 kg>m3 )( 9.81 m>s2 )

+

V2 + 0 2g

A

pB = 7357.5 Pa The free-body diagram of the control volume is shown in Fig. a. Applying the linear momentum equation in the vertical direction for steady flow, ΣF =

W

0 VrdV + VrV # dA 0t Lcv Lcs

+ c ΣFy = 0 + ( - VA)rQA + VB r( -QB)

pB AB

Fy (a)

+ c ΣFy = -2VrQ

( 7357.5 N>m2 ) (p)(0.01 m)2 - 4 - 6 = - 2(V) ( 1000 kg>m3 ) (V)(p)(0.01 m)2 V = 3.4981 m>s Q = VA = ( 3.4981 m>s ) (p)(0.01 m)2 = 1.10 ( 10-3 ) m3 >s

637

Ans.

6–37. Air flows through the 1.5 ft-wide rectangular duct at 900 ft 3 >min. Determine the horizontal force acting on the end plate B of the duct. Take ra = 0.00240 slug>ft 3.

A B 3 ft 1 ft

SOLUTION

F2

Q = 900 ft 3 >min ( 1 min.>60 s ) = 15 ft 3 >s VA =

VB =

15 ft 3 >s

(3 ft)(1.5 ft) 15 ft 3 >s

(1 ft)(1.5 ft)

= 3.33 ft>s

pA

= 10 ft>s

F2

Apply Bernoulli’s equation between A and B. pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 pA 0.00240 slug>ft

3

+

( 3.33 ft>s ) 2 2

+ 0 = 0 +

( 10 ft>s ) 2 2

+ 0

pA = 0.10667 lb>ft 2 Using the free-body diagram, Fig. a the linear momentum equation becomes + ΣFx = 0 V rdV + Vx rV # dA S 0t Lcv x Lcs

( 0.10667 lb>ft 2 ) (3 ft)(1.5 ft) - F = 0 + ( 3.33 ft>s )( 0.00240 slug>ft 3 )( - 15 ft 3 >s ) + ( 10 ft>s )( 0.00240 slug>ft 3 )( 15 ft 3 >s ) Ans.

F = 0.24 lb

Ans: 0.24 lb 638

6–38. Air at a temperature of 30°C flows through the expansion fitting such that its velocity at A is 15 m>s and the absolute pressure is 250 kPa. If no heat or frictional loss occurs, determine the resultant force needed to hold the fitting in place.

15 m/s A

100 mm

B

250 mm

SOLUTION

Using the ideal gas law with R = 286.9 J>kg # k for air (Appendix A),

F

250 ( 103 ) N>m2 = rA ( 286.9 J>kg # k ) (273 + 30) k

pA = rARTA;

rA = 2.8759 kg>m3

p = 250 )10

pB = rB ( 286.9 J>kg # k ) (273 + 30) k

pB = rBRTB;

rB =

A

3 11.5034 ( 10 ) pB 4 kg>m -6

3

3

) Pa p = 40 )10

(1)

(a)

Consider the fixed control volume to be the air contained in the expansion fitting as shown in Fig. a. Continuity requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 + rA( -VAAA) + rB(VBAB) = 0

( 2.8759 kg>m ) 5 - ( 15 m>s ) 3 p(0.05 m)2 4 6 + 3 11.5034 ( 10-6 ) pB 4 5 VB 3 p(0.125 m)2 4 6 = 0 3

VB = c

0.6 ( 106 ) pB

(2)

d m>s

Since the fitting remains horizontal, zA = zB = z. The energy equation gives pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gA gB 2g 2g £

0.6 ( 106 )

2

§

pB ( 15 m>s ) 2 pB + + z + 0 + 0 + + z + 0 = ( 2.8759 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 3 11.0534 ( 10-6 ) pB 4 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 250 ( 103 ) N>m2

pB = 40 ( 103 ) Pa

Substituting this result into Eqs. (1) and (2) rB = 0.4601 kg>m3

VB = 15 m>s

Since the flow is steady, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

639

) Pa

3

B

6–38. Continued

Writing the horizontal scalar component of this equation by referring to the FBD of the control volume, Fig. a, + ΣFx = 0 + VArA ( - VAAA ) + VBrB ( VBAB ) S

3 250 ( 103 ) N>m2 4 3 p(0.05 m)2 4

- F -

3 40 ( 103 ) N>m2 4 3 p(0.125 m)2 4

= ( 15 m>s )( 2.8759 kg>m3 ) 5 - ( 15 m>s ) 3 p(0.05 m)2 4 6 + ( 15 m>s )( 0.4601 kg>m3 ) 5 ( 15 m>s ) 3 p(0.125 m)2 4 6 Ans.

F = 0

Ans: F = 0 640

6–39. Air at a temperature of 30°C flows through the expansion fitting such that its velocity at A is 15 m>s and the pressure is 250 kPa. If heat and frictional loss due to the expansion causes the temperature and absolute pressure of the air at B to become 20°C and 7.50 kPa, determine the resultant force needed to hold the fitting in place.

15 m/s

100 mm

A B

250 mm

SOLUTION

Using the ideal gas law with R = 286.9 J>kg # k for air (Appendix A), pA = rARTA;

F

250 ( 103 ) N>m2 = rA ( 286.9 J>kg # k ) (273 + 30) k rA = 2.8759 kg>m3

pB = rBRTB;

p = 250 (10

7.50 ( 103 ) N>m2 = rB ( 286.9 J>kg # k ) (273 + 20) k

A

3

( Pa p = 7500 Pa

rB = 0.08922 kg>m3

B

(1)

Consider the fixed control volume to be the water contained in the expansion fitting as shown in Fig. a. The continuity requires

(a)

0 rdV + rV # dA = 0 0t Lcv Lcs 0 + rA( -VAAA) + rB(VBAB) = 0

( 2.8759 kg>m3 ) 5 - ( 15 m>s ) 3 p(0.05 m)2 4 6 + 3 0.08922 kg>m3 4 5 VB 3 p(0.125 m)2 4 6 = 0 VB = 77.36 m>s

Since the flow is steady, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

Writing the horizontal scalar component of this equation by referring to the FBD of the control volume, Fig. a + ΣFx = 0 + VArA ( - VAAA ) + VBrB ( VBAB ) S

3 250 ( 103 ) N>m2 4 3 p(0.05 m)2 4

- F - ( 7.5 ( 103 ) N>m2 ) 3 p(0.125 m)2 4

= ( 15 m>s )( 2.8759 kg>m3 ) 5 - ( 15 m>s ) 3 p(0.05m)2 4 6

+ ( 77.36 m>s )( 0.08922 kg>m3 ) 5 ( 77.36 m>s ) 3 p(0.125 m)2 4 6

Ans.

F = 1.57 kN

Ans: 1.57 kN 641

*6–40. Water flows through the pipe C at 4 m>s. Determine the horizontal and vertical components of force exerted by elbow D necessary to hold the pipe assembly in equilibrium. Neglect the size and weight of the pipe and the water within it. The pipe has a diameter of 60 mm at C, and at A and B the diameters are 20 mm.

4 5 3

B A

4 m/s D C

SOLUTION Assume water is incompressible. We have steady flow. Q = 4 m>s (p)(0.03 m)2 = 0.011310 m3 Continuity requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - 4 m>s(p)(0.03 m)2 + VA (p)(0.01 m)2 + VB(p)(0.01 m)2 = VA + VB = 36

(1)

Fy

Bernoulli Equation. pC pA VC2 VA2 + gzC = + gzA + + r r 2 2 pC 3

1000 kg>m

+

( 4 m>s )

2

2

+ 0 = 0 +

Fx p

VA2 + 0 2

VA2 = 16 + 0.002 pC

C

(a)

(2)

pC pB VC2 VB2 + gzC = + gzB + + r r 2 2 pC 1000 kg>m3

+ VB2

( 4 m>s ) 2 2

+ 0 = 0 +

VB2 + 0 2 (3)

= 16 + 0.002 pC

From Eqs. (2) and (3), VA = VB. From Eq. (1), VA = VB = 18 m>s Thus

( 18 m>s ) 2 = 16 + 0.002 pC pC = 154 kPa The free-body diagram is shown in Fig. a.

642

*6–40. Continued

Linear momentum. ΣF =

0 V rdV + V rV # dA 0t Lcv Lcs

+ ΣFx = 0 + ( VC ) (r) ( - VCAC ) + a -VA 3 brVAAA + 0 S 5

Fx + 154 ( 103 ) (p) ( 0.03 m ) 2 = ( 4 m>s )( 1000 kg>m3 )( - 4 m>s ) (p) ( 0.03m ) 2 3 - ( 18 m>s ) a b ( 1000 kg>m3 )( 18 m>s ) (p)(0.01 m)2 5

Ans.

Fx = - 542 N = 542 N

4 + c ΣFy = 0 + VAa br VAAA + VBrVBAB 5

4 Fy = 18 m>s a b ( 1000 kg>m3 )( 18 m>s ) (p)(0.01 m)2 + 18 m>s ( 1000 kg>m3 )( 18 m>s ) (p)(0.01 m)2 5 Fy = 183 N c

Ans.

643

6–41. The truck dumps water on the ground such that it flows from the truck through a 100-mm-wide opening at an angle of 60°. The length of the opening is 2 m. Determine the friction force that all the wheels of the truck must exert on the ground to keep the truck from moving at the instant the water depth in the truck is 1.75 m.

A 1.75 m

B

60!

SOLUTION We consider steady flow of an ideal fluid. A

Bernoulli Equation. Since A and B are exposed to the atmosphere, pA = pB = 0. Since the water discharges from a large reservoir, VA ≅ 0. If the datum is at B, zA = 1.75 m and zB = 0.

Ww

pA pB VA2 VB2 + zA = + zB + + g g 2g 2g 0 + 0 + 1.75 m = 0 +

VB2 2 ( 9.81 m>s2 )

60

F

+ 0

B N

VB = 5.860 m>s The discharge at B is QB = VBAB = ( 5.860 m>s ) (2 m)(0.1 m) = 1.172 m3 >s

Take the control volume to be the dump truck and its contents. Its free-body diagram is shown in Fig. a. Linear Momentum. Since the flow is steady incompressible, ΣF = or

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣFx = (VB)x r (QB) S F = ( 5.860 m>s cos 60° )( 1000 kg>m3 )( 1.172 m3 >s ) F = 3.43 kN

Ans.

Ans: 3.43 kN 644

6–42. The fireman sprays a 2-in.-diameter jet of water from a hose at the burning building. If the water is discharged at 1.5 ft 3 >s, determine the magnitude of the velocity of the water when it splashes on the wall. Also, find the normal reaction of both the fireman’s feet on the ground. He has a weight of 180 lb. Neglect the weight of the hose, the water within it, and the normal reaction of the hose on the ground.

B A

30° 5 ft 3 ft

C

SOLUTION We consider steady flow of an ideal fluid. Q = VAAA 1.5 ft 3 >s = VA £ p a

2 1 ft b § 12

180 lb A

VA = 68.75 ft>s Bernoulli Equation. Since the water jet from A and B is free flow, pA = pB = 0. If the datum is at A, zA = 0 and zB = 5 ft - 3 ft = 2 ft .

0 +

(68.75 ft>s)

2

2 ( 32.2 ft>s2 )

+ 0 = 0 +

VB2 2 ( 32.2 ft>s2 )

C p

F

C

pB pA VA2 VB2 + zA = + zB + + g g 2g 2g

30˚

N (a)

+ 2 ft Ans.

VB = 67.81 ft>s = 67.8 ft>s

Take the control volume to be the fireman and hose CA and the water within it. Its free-body diagram is shown in Fig. a. Here, the pressure at C, pC, acts horizontally. Linear Momentum. Since the flow is steady incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

+ c ΣFy = 0 + ( VA ) yr(Q) N - 180 lb = 68.75 ft>s sin 30° °

N = 279.93 lb = 280 lb

62.4 lb>ft 3 32.2 ft>s2

¢ ( 1.5 ft 3 >s )

Ans.

Ans: VB = 67.8 ft>s N = 280 lb 645

6–43. The fountain sprays water in the direction shown. If the water is discharged at 30° from the horizontal, and the cross-sectional area of the water stream is approximately 2 in2, determine the normal force the water exerts on the wall at B.

VA A

60! B 30! 2.5 ft 15 ft

SOLUTION

VA

We consider steady flow of an ideal fluid.

B

30˚

A

Motion of Water Jet. Consider the horizontal motion by referring to Fig. a.

VB

15 ft

+ sx = ( so ) x + ( vo ) xt S

30˚

(a)

15 ft = 0 + ( VA cos 30° ) t

(1) t

Referring to Fig. a, vertical motion gives 1 + c sy = ( so ) y + ( vo ) y t + at 2 2

n B

C Ft

30˚

0 = 0 + ( VA sin 30° ) t +

1 ( - 32.2 ft>s2 ) t 2 2

(2)

60˚

D

Fn

Solving Eqs. (1) and (2) yields VA = 23.62 ft>s

(b)

t = 0.7334 s

Bernoulli Equation. Since the water jet from A and B is free flow, pA = pB = 0. If the datum passes through A and B, zA = zB = 0. pB pA VA2 VB2 + zA = + zB + + g g 2g 2g 0 +

( 23.62 ft>s ) 2 2g

+ 0 = 0 +

VB2 + 0 2g

VB = 23.62 ft>s The discharge of the flow is Q = VAAA = ( 23.62 ft>s ) £ ( 2 in2 ) a

1 ft 2 b § = 0.3280 ft 3 >s 12 in.

Take the control volume to be the portion of water striking the wall. Its free-body diagram is shown in Fig. b. Linear Momentum. Here, VB is perpendicular to the wall. Since the flow is steady incompressible, ΣF = ΣFn = 0 +

0 VrdV + VrV # dA 0t Lcv Lcs

( - VB ) r( - Q)

Fn = ( 23.62 ft>s ) ° Fn = 15.0 lb

62.4 lb>ft 3 32.2 ft>s2

¢ ( 0.3280 ft 3 >s )

646

Ans.

Ans: 15.0 lb

*6–44. The 150-lb fireman is holding a hose that has a nozzle diameter of 1 in. If the nozzle velocity of the water is 50 ft>s, determine the resultant normal force acting on both the man’s feet at the ground when u = 30°. Neglect the weight of the hose, the water within it, and the normal reaction of the hose on the ground.

50 ft/s u

A

4 ft B

SOLUTION The discharge of the flow is QA = VAAA QA = ( 50 ft>s ) £ pa

150 lb

2 0.5 ft b § 12

A 30˚ pA = 0

QA = 0.2727 ft 3 >s

B

The free-body diagram of the control volume is shown in Fig. a. F

Linear Momentum. Since the flow is steady incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

N

or + c ΣFy = 0 + ( VA ) yrQ N - 150 lb = 50 ft>s sin 30° ° N = 163 lb

62.4 lb>ft 3 32.2 ft>s2

¢ ( 0.2727 ft 3 >s )

647

Ans.

6–45. The 150-lb fireman is holding a hose that has a nozzle diameter of 1 in. If the velocity of the water is 50 ft>s, determine the resultant normal force acting on both the man’s feet at the ground as a function of u. Plot this normal reaction (vertical axis) versus u for 0° 6 u 6 30°. Give values for increments of ∆u = 5°. Neglect the weight of the hose, the water within it, and the normal reaction of the hose on the ground.

50 ft/s u

A

4 ft B

SOLUTION The discharge of the flow is 150 lb

2 0.5 Q = VAAA = ( 50 ft>s ) £ pa ft b § = 0.2727 ft 3 >s 12

Here the flow is steady. Applying linear momentum equation.

pA = 0

0 ΣF = VrdV + VrV # dA 0t Lcv Lcs Writing the vertical scalar component of this equation by referring to the FBD of the control volume shown in Fig. a. + c ΣFy = 0 + ( VA ) y r ( VAAA ) + 0

3 ( 50 ft>s ) sin u 4 °

N - 150 lb =

62.4 lb>ft 3 32.2 ft>s2

F

N

¢ ( 0.2727 ft 3 >s )

N = (150 + 26.4 sin u) lb where u is in deg.

(a)

Ans.

The plot of N vs u is shown in Fig. a u(deg.) N(lb)

0

5

10

15

20

25

30

150

152.30

154.59

156.84

159.04

161.17

163.21

N(lb) 165

160

155

150

(deg.) 0

5

10

15

20

25

30

(b)

Ans: N = (150 + 26.4 sin u) lb

Note: See solution 6–44 regarding the effects of hose tension. 648

6–46. The 150-lb fireman is holding a hose that has a nozzle diameter of 1 in. If the velocity of the water is 50 ft>s, determine the resultant normal force acting on both the man’s feet at the ground if he holds the hose directly over his head at u = 90°. Neglect the weight of the hose, the water within it, and the normal reaction of the hose on the ground.

50 ft/s u

A

4 ft B

SOLUTION The flow is Q = VAAA = ( 50 ft>s ) £ pa

Linear momentum

ΣF =

2 0.5 ft b § = 0.2727 ft 3 >s 12

150 lb

0 VrdV + VrV # dA 0t Lcv Lcs

N

or + c ΣFy = 0 + ( VA ) yr(Q) N - 150 lb = ( 50 ft>s ) ° N = 176 lb

62.4 lb>ft 3 32.2 ft>s2

¢ ( 0.2727 ft 3 >s )

Ans.

Ans: 176 lb 649

6–47. Water at A flows out of the 1-in.-diameter nozzle at 8 ft>s and strikes the 0.5-lb plate. Determine the height h above the nozzle at which the plate can be supported by the water jet.

B h

A

1 in.

SOLUTION We consider steady flow of an ideal fluid. Discharge.

Wp = 0.5 lb

Q = VAAA = ( 8 ft>s ) £ p a

C

2 0.5 ft b § = 0.04363 ft 3 >s 12

Take the control volume of the plate and portion of water striking it. Its free-body diagram is shown in Fig. a. Since the jet has free flow, the pressure at any point is zero gauge.

B (a)

Linear Momentum. Since the flow is steady incompressible, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + T ΣFy = 0 +

( -VB ) r( - Q) 0.5 lb =

( - VB ) °

62.4 lb>ft 3 32.2 ft>s2

¢ ( - 0.04363 ft 3 >s )

VB = 5.913 ft>s

Bernoulli Equation. If the datum coincides with the horizontal line through A, zB = h and zA = 0. pB pA VA2 VB2 + zA = + zB + + g g 2g 2g 0 +

(8 ft>s)2 2 ( 32.2 ft>s2 )

+0=0+

( 5.913 ft>s ) 2 +h 2 ( 32.2 ft>s2 ) Ans.

h = 0.4508 ft = 0.451 ft

Ans: 0.451 ft 650

*6–48. Water at A flows out of the 1-in.-diameter nozzle at 18 ft>s. Determine the weight of the plate that can be supported by the water jet h = 2 ft above the nozzle.

B h

A

SOLUTION We consider steady flow of an ideal fluid. Bernoulli Equation. Since the jet is free flow, the pressure at any point is zero gauge. If the datum passes through A, zA = 0 and zB = 2 ft. pA pB VA2 VB2 + zA = + zB + + g g 2g 2g 0 +

C

( 18 ft>s ) 2 VB2 + 0 = 0 + + 2 ft 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 )

B (a)

VB = 13.97 ft>s The discharge is Q = VAAA = ( 18 ft>s ) £ pa

2 0.5 ft b § = 0.09817 ft 3 >s 12

Take the control volume as the plate and a portion of water striking it. Its free-body diagram is shown in Fig. a. Linear Momentum. Since the flow is steady incompressible, ΣF =

0 VrVdV + VrVdA 0t Lcv Lcs

or + T ΣFy = 0 +

( - VB ) r( -Q) Wp =

( -13.97 ft>s ) °

F = Wp

62.4 lb ft 3 ¢ ( - 0.09817 ft 3 >s ) 32.2 ft>s2

Wp = 2.658 lb = 2.66 lb

651

Ans.

1 in.

6–49. Water flows through the hose with a velocity of 2 m>s. Determine the force F needed to keep the circular plate moving to the right at 2 m>s.

50 mm

15 mm F

2 m/s

B

A

SOLUTION

C

We consider steady flow of an ideal fluid.

F

B

Take the control volume as the plate and a portion of water striking it.

(a)

Continuity Equation. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - ( 2 m>s ) 3 p(0.025 m)2 4 + VB 3 p(0.0075 m)2 4 = 0 VB = 22.22 m>s

Relative Velocity. Relative to the control volume, the velocity at B is + S

Vf>cs = Vf - Vcv = 22.22 m>s - 2 m>s = 20.22 m>s

Thus, the flow onto the plate is Qf>cs = Vf>cs AB = ( 20.22 m>s ) 3 p(0.0075 m)2 4 = 0.003574 m3 >s

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + ( Vf>cs ) B r ( -Qf>cs ) S x - F = ( 20.22 m>s ) ( 1000 kg>m3 )( - 0.003574 m3 >s ) F = 72.3 N

Ans.

Ans: 72.3 N 652

6–50. Water flows through the hose with a velocity of 2 m>s. Determine the force F needed to keep the circular plate moving to the left at 2 m>s.

50 mm

15 mm F

2 m/s

B

A

SOLUTION

C

We consider steady flow of an ideal fluid.

F

B

Take the control volume as the plate and a portion of water striking it.

(a)

Continuity Equation. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - ( 2 m>s ) 3 p(0.025 m)2 4 + VB 3 p(0.0075 m)2 4 = 0 VB = 22.22 m>s

Relative Velocity. Relative to the control volume, the velocity at B is + ( Vf>cs ) B = Vf - Vcs = 22.22 m>s S

( - 2 m>s ) = 24.22 m>s

Thus, the relative flow onto the plate is Qf>cs = ( Vf>cs ) C AB = ( 24.22 m>s ) 3 p(0.0075 m)2 4 = 0.004280 m3 >s

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or -F = 0 + ( Vf>cs ) Bx r ( - Qf>cs ) - F = ( 24.22 m>s )( 1000 kg>m3 )( - 0.004280 m3 >s ) F = 104 N

Ans.

Ans: 104 N 653

6–51. The large water truck releases water at the rate of 45 ft 3 >min through the 3-in.-diameter pipe. If the depth of the water in the truck is 4 ft, determine the frictional force the road has to exert on the tires to prevent the truck from rolling. How much force does the water exert on the truck if the truck is moving forward at a constant velocity of 4 ft>s and the flow is maintained at 45 ft 3 >min?

4 ft/s

4 ft

SOLUTION We consider steady flow of an ideal fluid. For the case when the truck is required to be stationary, the control volume is the entire truck and its contents. Here the flow is steady. The FBD of the control volume is shown in Fig. a. F

The discharge is Q = a45

Thus, the velocity at the outlet is Q = Vout A out;

(a)

ft 3 1 min ba b = 0.75 ft 3 >s min 60 s

0.75 ft 3 >s = Vout £ p a

2 1.5 ft b § 12

Vout = 15.28 ft>s

Applying the linear momentum equation by referring to Fig. a, 0 Vr dV + VrwV # dA 0t Lcv w Lcs

ΣF =

Writing the scalar component of this equation along x axis, + ΣFx = 0 + Vout rwVout Aout d F = ( 15.28 ft>s ) °

62.4 lb>ft 3 32.2 ft>s2

¢ ( 0.75 ft 3 >s ) = 22.2 lb

Ans.

For the case when the truck is moving with a constant velocity, the same control volume is considered, but it moves with this constant velocity. Then, the flow measured relative to the control volume is steady. From the discharge, the relative velocity at the outlet is Q = ( Vout>cs ) Aout;

0.75 ft 3 >s = Vout>cs £ p a

2 1.5 ft b § 12

Vout>cs = 15.28 ft>s Applying the linear momentum equation by referring to Fig. a, but this time using the relative velocity, ΣF =

0 V r dV + Vw>cs rwVw>cs # dA 0t Lcv w>cs w Lcs

Applying the scalar component of this equation along x axis, + ΣFx = 0 + ( Vout>cs )( rw )( Vout>cs A out ) S F = ( 15.28 ft>s ) °

62.4 lb>ft 3 32.2 ft>s2

¢ ( 0.75 ft 3 >s ) = 22.2 lb

Ans. Ans: 22.2 lb 654

*6–52. A plow located on the front of a truck scoops up a liquid slush at the rate of 12 ft 3 >s and throws it off perpendicular to its motion, u = 90°. If the truck is traveling at a constant speed of 14 ft>s, determine the resistance to motion caused by the shoveling. The specific weight of the slush is gs = 5.5 lb>ft 3.

u

B

A 0.25 m

SOLUTION

W

We consider steady flow of an ideal fluid. F

A

Take the slush in context with the blade as the control volume. Relative Velocity. Since the slush is at rest before it enters control volume, then the velocity at A relative to control volume is + S

( Vf>cs ) A = Vf - Vcs = 0 - 14 ft>s = 14 ft>s d

Linear Momentum. Here, Qf>cs = 12 ft 3 >s and ( Vf>cs ) B = ( Vf>cs ) A = 14 ft>s . Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + S

( - VA ) r( - Q) -F =

( - 14 ft>s ) °

5.5 lb>ft 3 32.2 ft>s2

¢ ( -12 ft 3 >s )

F = -28.69 lb = 28.7 lb d

Ans.

655

(a)

N

6–53. The truck is traveling forward at 5 m>s, shoveling a liquid slush that is 0.25 m deep. If the slush has a density of 125 kg>m3 and is thrown upwards at an angle of u = 60° from the 3-m-wide blade, determine the traction force of the wheels on the road necessary to maintain the motion. Assume that the slush is thrown off the shovel at the same rate as it enters the shovel.

u

B

A 0.25 m

SOLUTION

B

60°

We consider steady flow of an ideal fluid. W

Fx

Take the slush in context with the blade as the control volume. Relative Velocity. Since the slush is at rest before it enters the control volume, then the velocity at A relative to the control volume is + Vf>cs = Vf - Vcs = 0 - 5 m>s = 5 m>s d S

A Fy (a)

Thus, the flow rate of snow onto the shovel is Qf>cs = Vf>cs AA = ( 5 m>s ) 3 0.25 m ( 3 m ) 4 = 3.75 m3 >s

Linear Momentum. Here, ( Vf>cs ) B = ( Vf>cs ) A = 5 ft>s. Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + S

( - Vf>cs ) Ax r ( - Qf>cs ) + ( Vf>cs ) Bx r ( Qf>cs ) Fx = 0 +

( - 5 m>s )( 125 kg>m3 )( - 3.75 m3 >s )

Fx = 3.52 kN

+ ( 5 m>s cos 60° )( 125 kg>m3 )( 3.75 m3 >s )

Ans.

Ans: 3.52 kN 656

6–54. The boat is powered by the fan, which develops a slipstream having a diameter of 1.25 m. If the fan ejects air with an average velocity of 40 m>s, measured relative to the boat, and the boat is traveling with a constant velocity of  8 m>s, determine the force the fan exerts on the boat. Assume that the air has a constant density of ra = 1.22 kg>m3 and that the entering air at A is essentially at rest relative to the ground.

1.25 m A

B

SOLUTION We consider steady flow of an ideal fluid.

out

Relative Velocity. Since the air is at rest before it enters the control volume, then the inlet velocity relative to the control volume is + ( Vf>cs ) A = Vf - Vcs = 0 - 8 m>s = 8 m>s d S

in

F

(a)

The outlet velocity relative to the control volume is ( Vf>cv ) out = 40 m>s . Then, the flow of air in and out of the fan is Qf>cs = ( Vf>cs ) BAB = ( 40 m>s ) 3 p(0.625 m)2 4 = 49.09 m3 >s

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + d ΣFx = 0 + ( Vf>cs ) Ar ( - Qf>cs ) + ( Vf>cs ) Br ( Qf>cs ) = ( 1.22 kg>m3 ) 3 ( 8 m>s )( - 49.09 m3 >s ) + ( 40 m>s )( 49.09 m3 >s ) 4 = 1.92 kN

Ans.

Ans: 1.92 kN 657

6–55. A 25-mm-diameter stream flows at 10 m>s against the blade and is deflected 180° as shown. If the blade is moving to the left at 2 m>s, determine the horizontal force F of the blade on the water.

B 2 m/s F 10 m/s 25 mm

SOLUTION

A

B

We consider steady flow of an ideal fluid. F

Take the control volume as the water on the blade. Relative Velocity. Relative to the control volume, the velocity at A is + ( Vf>cs ) A = Vf - Vcs = 10 m>s - ( - 2 m>s ) = 12 m>s S S

A (a)

Thus, the flow rate onto the vane is Qf>cs = ( Vf>cs ) AAA = ( 12 m>s ) 3 p(0.0125 m)2 4 = 0.005890 m3 >s

Linear Momentum. Here, ( Vf>cs ) B = ( Vf>cs ) A = 12 m>s (Bernoulli equation). Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + ( Vf>cs ) Ar ( - Qf>cs ) + S

( - Vf>cs ) Br ( Qf>cs )

- F = ( 1000 kg>m3 ) 3 ( 12 m>s )( - 0.005890 m3 >s ) + F = 141 N

( - 12 m>s )( 0.005890 m3 >s ) 4

Ans.

Ans: 141 N 658

*6–56. Solve Prob. 6–55 if the blade is moving to the right at 2 m>s. At what speed must the blade be moving to the right to reduce the force F to zero?

B 2 m/s F 10 m/s 25 mm

SOLUTION

A

B

Consider the control volume as the water on the blade. The velocity of the water at A relative to the control volume is

F

+ )(Vf>cs)A = 10 m>s - 2 m>s = 8 m>s S (S To satisfy the Bernoulli’s equation, ( Vf>cs ) B = 8 m>s d for small elevations. The flow is steady relative to control volume. 0 ΣF = V rdV + Vf>cs rVf>cs # dA 0t Lcv f>cv Lcs Writing the horizontal scalar component of this equation by referring to the FBD of the control volume shown in Fig. a + ΣFx = 0 + (Vf>cs)A r 3 - (Vf>cs)A AA 4 + S

3 -(Vf>cs)B 4 r 3 (Vf>cs)BAB 4

However, Q = ( Vf>cs ) A AA = ( Vf>cs ) B AB and

( Vf>cs ) B = ( Vf>cs ) A. Then

-F = - 2r ( Vf>cs ) A 3 ( Vf>cs ) A AA 4 F = 2r ( Vf>cs ) A2 AA

(1)

F = 2(1000 kg>m3) (8 m>s)2 3 p(0.0125 m)2 4 = 62.8 N

Ans.

By inspecting Eq (1), F = 0 if ( Vf>cs ) A = 0. Then

+ ( Vf>cs ) A = Vw - Vb S 0 = 10 m>s - Vb Vb = 10 m>s S

Ans.

659

A (a)

6–57. The vane is moving at 80 ft>s when a jet of water having a velocity of 150 ft>s enters at A. If the crosssectional area of the jet is 1.5 in2, and it is diverted as shown, determine the horsepower developed by the water on the blade. 1 hp = 550 ft # lb>s.

B 45! 80 ft/s

30! A

SOLUTION

B

Fy

We consider steady flow of an ideal fluid. Take the control volume as the water on the blade.

45°

Relative Velocity. Applying the relative velocity equation to determine the velocity relative to the vane, VA>cs, and the angle u, of the jet in a stationary frame,

30°

VA>cs = VA - Vcs + (S

)

(+c)

Fx

A (a)

VA>cs cos 30° = 150 cos u - 80

(1)

VA>cs sin 30° = 150 sin u

(2)

Solving Eqs. (1) and (2), u = 14.53°

VA>cs = 75.29 ft>s

Here, ( Vf>cs ) A = VA>cs = 75.29 ft>s . Thus,the relative flow rate at the vane is Qf>cs = ( Vf>cs ) A A = (75.29 ft>s) c 1.5 in2 a

1 ft 2 b d = 0.7842 ft 3 >s 12 in.

Linear Momentum. Here, ( Vf>cs ) A = VA>cs = 75.29 ft>s (Bernoulli equation). Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + ( VA>cs ) r ( - Qf>cs ) + S x -Fx = °

62.4 lb>ft 3 32.2 ft>s2

Fx = 179.99 lb

( - VB>cs ) x r ( Qf>cs )

¢ 3 ( 75.29 ft>s cos 30° )( - 0.7842 ft 3 >s ) +

( -75.29 ft>s cos 45° )( 0.7842 ft3 >s ) 4

Thus, the power of the water jet can be determined from

#

W = F # V = FxV = (179.99 lb)(80 ft>s) = a14399.40

= 26.2 hp

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

Ans: 26.2 hp 660

6–58. The car is used to scoop up water that is lying in a trough at the tracks. Determine the force needed to pull the car forward at constant velocity v for each of the three cases. The scoop has a cross-sectional area A and the density of water is rw.

v

v

F1

F2

(a)

(b) v

F3

SOLUTION The control volume considered consists of the car and the scoop. This control volume has only inlet control surface (the scoop) but no outlet control surface. Since this same control volume can be used for cases a, b, and c, F1 = F2 = F3 = F. Here, # ma = rwVA

# mf = 0

Ve = 0

(c)

dVcv = 0 (constant velocity) dt

Along x axis, + ΣFx = m S

dVcv # # # + maVcv - ( ma + mf ) Ve dt

F = 0 + rwVAV = rwAV 2 Therefore F1 = F2 = F3 = rwAV 2

Ans.

Ans: F1 = F2 = F3 = rwAV 2 661

6–59. Flow from the water stream strikes the inclined surface of the cart. Determine the power produced by the stream if, due to rolling friction, the cart moves to the right with a constant velocity of 2 m>s. The discharge from the 50-mm-diameter nozzle is 0.04 m3 >s. One-fourth of the discharge flows down the incline, and three-fourths flows up the incline.

B A 2 m/s

60! C

SOLUTION

B 60˚

We consider steady flow of an ideal fluid.

Relative Velocity. The velocity of the jet at A is VA =

Fx

A

Take the control volume as a portion of water striking the cart.

C

0.04 m3 >s Q = = 20.37 m>s AA p(0.025 m)2

Fy

(a)

Thus, the velocity at A relative to the control volume is + S

VA>cs = VA - Vcs = 20.37 m>s - 2 m>s = 18.37 m>s

Here, VB>cs = VC>cs = VA>cs = 18.37 m>s can be determined using the Bernoulli equation and neglecting the elevation change. Thus, the relative flow at A, B, and C are QA>cs = VA>cs AA = (18.37 m>s) 3 p(0.025 m)2 4 = 0.03607m3 >s QB>cs =

QC>cs =

3 3 ( Q ) = ( 0.03607 m3 >s ) = 0.02705 m3 >s 4 A>cs 4

1 1 ( Q ) = ( 0.03607 m3 >s ) = 0.009018 m3 >s 4 A>cs 4

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

or + ΣFx = r 3 QB>cs(VB>cs)x + QC>cs(VC>cs)x - QA>cs(VA>cs)x 4 S

- Fx = (1000 kg>m3) 3 ( 0.02705 m3 >s ) (18.37 m>s cos 60°) + ( 0.009018 m3 >s ) ( -18.37 m>s cos 60°) -

( 0.03607 m3 >s ) (18.37 m>s) 4

Fx = 497.04 N

Thus, the power of the jet stream can be determined from

#

W = F # V = FxV = (497.04 N)(2 m>s) Ans.

= 994.09 W = 994 W

Ans: 994 W 662

*6–60. Water flows at 0.1 m3 >s through the 100-mm-diameter nozzle and strikes the vane on the 150-kg cart, which is originally at rest. Determine the velocity of the cart 3 seconds after the jet strikes the vane.

100 mm A

B

SOLUTION

V

A

We consider steady flow of an ideal fluid. F

Take the control volume as the water on the cart. Relative Velocity. The velocity of the jet at A is VA =

B

0.1 m3 >s Q = = 12.73 m>s AA p(0.05 m)2

(a)

Thus, the velocity at A relative to the control volume is + S

(150 kg)(9.81 m s2 )

VA>cs = VA - Vcs = 12.73 - V S

Here, ( Vf>cs ) A = ( Vf>cs ) B = VA>cv . Thus, the relative flow rate onto the vane is

a= d dt

F

x

Qf>cs = ( Vf>cs ) A AA = (12.73 - V) 3 p(0.05 m)2 4 = 2.5 ( 10-3 ) p(12.73 - V)

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, 0 ΣF = VrdV + VrV # dA 0t Lcv Lcs or + ΣFx = 0 + S

N (b)

( - Vf>cs ) B r ( Qf>cs ) + ( Vf>cs ) A r ( - Qf>cs )

- F = 1000 kg>m3 3 - (12.73 - V)(2.5 ( 10-3 ) p(12.73 - V) + (12.73 - V)( - 2.5 ( 10-3 ) p(12.73 - V) 4 F = 5p(12.73 - V)2

Equation of Motion. Referring to the free-body diagram of the cart in Fig. b, + ΣFx = ma; S

5p(12.73 - V)2 = (150 kg)a L0

3s

V

dt =

t # 03 s = 3 =

dV b dt

30 dV p L0 (12.73 - V)2

V 30 1 a b` p 12.73 - V 0

30 1 1 a b p 12.73 - V 12.73

Ans.

V = 10.19 m>s = 10.2 m>s

663

6–61. Water flows at 0.1 m3 >s through the 100-mmdiameter nozzle and strikes the vane on the 150-kg cart, which is originally at rest. Determine the acceleration of the cart when it attains a velocity of 2 m>s.

100 mm A

B

SOLUTION

V

A

The velocity of the jet at A can be determined from the discharge. Q = VAAA;

0.1 m3 >s = VA 3 p(0.05 m)2 4

F

VA = 12.73 m>s

The velocity at A relative to the control volume is + S

B

( Vf>cs ) A = VA - Vcs = (12.73 - V) m>s S

(a)

To satisfy Bernoulli’s equation ( Vf>cs ) B = (12.73 - V) m>s d for small equations The flow is steady relative to control volume. ΣF =

0 V rdV + Vf>cs rVf>cs # dA 0t Lcv f>cv Lcs

(150 kg)(9.81 m s2 ) a= d dt

F

x

Writing the horizontal scalar component of this equation by referring to the FBD of the control volume shown in Fig. a, + S

ΣFx = 0 + ( Vf>cs ) A r 3 - ( Vf>cs ) A AA 4 +

3 - ( Vf>cs ) B 4 r 3 ( Vf>cs ) B AB 4

N

However, Q = ( Vf>cs ) AAA = ( Vf>cs ) BAB and ( Vf>cs ) B = ( Vf>cs ) A. Then

(b)

- F = -2r ( Vf>cs ) A 3 ( Vf>cs ) AAA 4 F = 2r ( Vf>cs ) 2AAA

F = 2 ( 1000 kg>m3 ) (12.73 - V)2 3 p(0.05 m)2 4 =

3 5p(12.73

- V)2 4 N

Referring to the FBD of the cart Fig. b, + ΣFx = ma; S

When V = 2m>s,

5p(12.73 - V)2 = 150 a a = c a =

p (12.73 - V)2 d m>s2 30

p (12.73 - 2)2 = 12.06 m>s2 = 12.1 m>s2 30

Ans.

Ans: 12.1 m>s2 664

6–62. Determine the rolling resistance on the wheels if the cart moves to the right with a constant velocity of Vc = 4 ft>s when the vane is struck by the water jet. The jet flows from the nozzle at 20 ft>s and has a diameter of 3 in.

B 20 ft/s

V

A 3 in.

SOLUTION

B

We consider steady flow of an ideal fluid.

Fx

Take the control volume as the water on the cart.

A

Relative Velocity. The velocity at A relative to the control volume is + S

(a)

Fy

VA>cs = VA - Vcv = 20 ft>s - 4 ft>s = 16 ft>s

Here, ( Vf>cs ) in = ( Vf>cs ) B = VA>cs = 16 ft>s (Bernoulli equation). Thus, the relative flow rate onto the vane is Qf>cs = ( Vf>cs ) A AA = (16 ft>s) c p a

2 1.5 ft b d = 0.7856 ft 3 >s 12

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, 0 ΣF = VrdV + VrV # dA 0t Lcv Lcs or + ΣFx = 0 + ( VA>cs ) r ( -Qf>cs ) S -Fx = ( 16 ft>s ) a

62.4 lb>ft 3 32.2 ft>s2

Fx = 24.35 lb = 24.4 lb

b ( - 0.7854 ft 3 >s )

Ans.

Ans: 24.4 lb 665

6–63. Determine the velocity of the 50-lb cart in 3 s starting from rest if a stream of water, flowing from the nozzle at 20 ft>s, strikes the vane and is deflected upwards. The stream has a diameter of 3 in. Neglect the rolling resistance of the wheels.

B 20 ft/s

V

A 3 in.

SOLUTION

B

We consider steady flow of an ideal fluid.

Fx

Take the control volume as the water on the cart.

A

Relative Velocity. The velocity of the jet at A relative to the control volume is + S

VA>cs = VA - Vcs = (20 - V) ft>s

Qf>cs = ( Vf>cs ) AAA = (20 - V) c pa

1.5 ft b d = 0.015625p(20 - V) 12

a= d dt

Fx

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, 0 ΣF = VrdV + VrV # dA 0t Lcv Lcs

Fy

Fy 50 lb

Here, ( Vf>cs ) in = ( Vf>cs ) B = VA>cs . Thus, the relative flow rate onto the vane is 2

(a)

x

N (b)

or + ΣFx = 0 + ( VA>cs ) r ( - Qf>cs ) S - Fx = (20 - V)a

62.4 lb>ft 3 32.2 ft>s2

Fx = 0.09513(20 - V)2

b(0.015625 p(20 - V))

Equation of Motion. Referring to the free-body diagram of the cart in Fig. b, + ΣFx = ma; S

0.09513(20 - V)2 = a 0.06126

L0

3s

V

dt =

0.06126(t) # 30 s = a 0.1838 = a

50 lb dV b 32.2 ft>s2 dt

dV 2 L0 (20 - V)

V 1 b` 20 - V 0

1 1 b 20 - V 20

Ans.

V = 15.72 ft>s = 15.7 ft>s

Ans: 15.7 ft>s 666

*6–64. Water flows through the Tee fitting at 0.02 m3 >s. If the water exits the fitting at B to the atmosphere, determine the horizontal and vertical components of force, and the moment that must be exerted on the fixed support at A, in order to hold the fitting in equilibrium. Neglect the weight of the fitting and the water within it.

60 mm B C

100 mm

150 mm 200 mm

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that pw = 1000 kg>m3. Average velocities will be used. The control volume consists of the fitting the fixed support and the water contained. From the discharge, 0.02 m3 >s = VA 3 p(0.05 m)2 4

Q = VCAC;

0.02 m3 >s = VB 3 p(0.03 m)2 4

Q = VBAB;

A

VC = 2.546 m>s y

VB = 7.074 m>s

rB = 0.15 m

Applying Bernoulli’s equation between C and B, with pB = patm = 0. x

pC pB V 2C VB2 + zC = + zB + + gw gw 2g 2g pC

+

3

9810 N>m

(2.546 m>s)2 2 ( 9.81 m>s

2

)

+ 0 = 0 +

(7.074 m>s)2 2 ( 9.81 m>s2 )

+ 0

FC

pC = 21.775 ( 103 ) N>m2

rC = 0.2 m

Then the pressure force on inlet control surface C is FC = pCAC =

3 21.775 ( 103 ) N>m2 4 3 p(0.05 m)2 4

= 171.02 N

Ax MA

Applying the linear momentum equation, ΣF =

0 VrdV + VrwV # dA 0t Lcv Lcs

Ay

writing the scalar component of this equation along the x and y axes by referring to the free-body diagram, Fig. a + ) ΣFx = 0 + VC rw ( - VC AC ) (S Ax + 171.02 N = (2.546 m>s) ( 1000 kg>m3 )( - 0.02 m3 >s ) Ax = - 221.95 N = 222 N d

Ans.

( + c ) ΣFy = 0 + VB rw ( VB AB ) Ay = (7.074 m>s) ( 1000 kg>m3 )( 0.02 m3 >s )

Ans.

= 141.47 N = 141 N

Applying the angular momentum equation, ΣM =

0 ( r * V ) rdV + ( r * V ) rV # dA 0t Lcv Lcs

writing the scalar component of this equation about point A by referring to Fig. a, aT + ΣMA = 0 + rBVB rw(VB AB) - rCVC rw ( - VC AC ) MA - (171.02 N)(0.2 m) = (0.15 m)(7.074 m>s) ( 1000 kg>m3 )( 0.02 m3 >s )

- (0.2 m)(2.546 m>s) ( 1000 kg>m3 )( - 0.02 m3 >s )

MA = 65.61 N # m = 65.6 N # m

667

Ans.

(a)

6–65. Water flows through the Tee fitting at 0.02 m3 >s. If the pipe at B is extended and the pressure in the pipe at B is 75 kPa, determine the horizontal and vertical components of force, and the moment that must be exerted on the fixed support at A, to hold the fitting in equilibrium. Neglect the weight of the fitting and the water within it.

60 mm B C

100 mm

150 mm 200 mm

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocities will be used. The control volume consists of the fitting, fixed support and the contained water. From the discharge, 0.02 m3 >s = VC 3 p(0.05 m)2 4

Q = VCAC;

VC = 2.546 m>s

0.02 m3 >s = VB 3 p(0.03 m)2 4

Q = VBAB;

A

y

VB = 7.074 m>s

rB = 0.15 m F B

Applying Bernoulli’s equation between A and B,

x

pB pC VC2 VB2 + + zC = + + zB gw gw 2g 2g pC 3

9810 N>m

+

(2.546 m>s)2 2 ( 9.81 m>s2 )

+ 0 =

75 ( 103 ) N>m2 3

9810 N>m

(7.074 m>s)2

+

2 ( 9.81 m>s2 )

+ 0

FC

pC = 96.775 ( 103 ) N>m2

rC = 0.2 m

Then the pressure forces on the inlet and outlet control surfaces at C and B are FC = pCAC =

3 96.775 ( 103 ) N>m2 4 3 p(0.05 m)2 4

FB = pBAB =

3 75 ( 103 ) N>m2 4 3 p(0.03 m)2 4

= 760.07 N

Ax MA

= 212.06 N Ay

Applying linear momentum equation, ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

(a)

writing the scalar component of this equation along the x and y axes by referring to the free-body diagram, Fig. a, + ) ΣFx = 0 + VA rw ( - VA AA ) (S Ax + 760.07 N = (2.546 m>s) ( 1000 kg>m3 )( - 0.02 m3 >s ) Ax = -811 N = 811 N d

Ans.

( + c ) ΣFy = 0 + VB rw ( VB AB ) Ay - 212.06 N = (7.074 m>s) ( 1000 kg>m3 )( 0.02 m3 >s ) Ay = 353.53 N = 354 N c

Ans.

Applying the angular momentum equation, 0 ( r * V ) rwdV + ( r * V ) rwV # dA ΣM = 0t Lcv Lcs writing the scalar component of this equation about point A by referring to Fig. a, aT + ΣMA = 0 + rBVBrw(VBAB) - rCVC rw ( - VCAC ) MA - (760.07 N)(0.2 m) - (212.06 N)(0.15 m) = (0.15 m)(7.074 m>s) ( 1000 kg>m3 )( 0.02 m3 >s )

- (0.2 m)(2.546 m>s) ( 1000 kg>m3 )( - 0.02 m3 >s )

MA = 215.22 N # m = 215 N # md

Ans.

668

Ans: Ax = 811 N Ay = 354 N MA = 215 N # m

6–66. Water flows into the bend fitting with a velocity of 3 m>s. If the water exists at B into the atmosphere, determine the horizontal and vertical components of force, and the moment at C, needed to hold the fitting in place. Neglect the weight of the fitting and the water within it.

200 mm

150 mm 3 m/s

150 mm

A

30!

C B

150 mm

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocities will be used. The control volume consists of the bend fitting and the contained water. The discharge is Q = VC AC = (3 m>s) 3 p(0.075 m)2 4 = 0.016875p m3 >s

The water exits at B into the atmosphere. Then pB = patm = 0. Since the diameter of the bend fitting is constant, VB = VC = 3 m>s and the elevation change is small. Therefore pC = pB = 0. As a result, no pressure force acting on the control volume. The FBD of the control volume is shown in Fig. a. Applying the linear momentum equation, 0 Vr dV + VrwV # dA 0t Lcv w Lcs

ΣF =

Writing the scalar components of this equation along the x and y axes by referring to the free-body diagram, Fig. a, + 2 ΣFx 1S Cx =

= 0 + VB cos 30°rw(VBAB) + VCrw( -VCAC)

3 (3 m>s) cos 30° 4 ( 1000 kg>m3 )( 0.016875p m3 >s ) = -21.31 N = 21.3 N d

+ (3 m>s) ( 1000 kg>m3 )( - 0.016875p m3 >s ) Ans.

+ c ΣFy = 0 + ( - VB sin 30°)(rw)(VBAB) - Cy =

3 - (3 m>s) sin 30° 4 ( 1000 kg>m3 )( 0.016875p m3 >s )

Cy = 79.52 N = 79.5 NT

Ans.

Applying the angular momentum equation, ΣM =

0 (r * V)rwdV + (r * V)rwV # dA 0t L L cv

cs

writing the scalar component of this equation about point C by referring to Fig. a, a+ ΣMC = 0 + ( - rBVB sin 30°)rw(VBAB) -MC = - (0.2 m) 3 (3 m>s) sin 30° 4 ( 1000 kg>m3 )( 0.016875p m3 >s ) MC = 15.9 N # mb

Ans.

y

Cy

rB = 0.2 m

x

MC Cx

30˚

Ans: Cx = 21.3 N Cy = 79.5 N MC = 15.9 N # m

(a)

669

6–67. Water flows into the bend fitting with a velocity of 3 m>s. If the water at B exits into a tank having a gage pressure of 10 kPa, determine the horizontal and vertical components of force, and the moment at C, needed to hold the fitting in place. Neglect the weight of the fitting and the water within it.

200 mm

150 mm 3 m/s

150 mm

A

30!

C B

150 mm

SOLUTION The flow is steady and water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocities will be used. The fixed control volume consists of the bend fitting and the contained water. Since the diameter of the pipe is constant, VB = VA = 3 m>s. Also the charge in elevation is negligible, pA = pB = 10 kPa, to satisfy Bernoulli’s equation. Then FA = FB = Also, the discharge is

3 10 ( 103 ) N>m2 4 3 p ( 0.075 m ) 2 4

= 56.25p N

FA

y Cy

x MC

Q = VAAA = VBAB = (3 m>s) 3 p(0.075 m)2 4 = 0.016875p m3 >s

The FBD of the control volume is shown in Fig. a. Applying the linear momentum equation, ΣF =

0.2 m

Cx 30˚

(a)

FB

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of this equation along x and y axes by referring to the FBD, Fig. a + 2 ΣFx = 0 + VB cos 30°rw(VBAB) + VApw( - VAAA) 1S

56.25p N - [(56.25p N) cos 30°] - Cx = [(3 m>s) cos 30°](1000 kg>m3)(0.016875p m3 >s) + (3 m>s)(1000 kg>m3)( - 0.016875p m3 >s)

Cx = 44.98 N = 45.0 N d

Ans.

+ c ΣFy = 0 + ( - VB sin 30°)(rw)(VBAB) (56.25p N) sin 30° - Cy = [ -(3 m>s) sin 30°] ( 1000 kg>m3 )( 0.016875p m3 >s ) Cy = 167.88N = 168 NT

Ans.

Applying the angular momentum equation, ΣM =

0 (r * V)rwdV + (r * V)rwV # dA 0t L L cv

cs

Writing the scalar component of this equation about point C by referring to the FBD, Fig. a, a+ ΣMC = 0 + ( - rBVB sin 30°)rW(VBAB) [(56.25 p N) sin 30°](0.2 m) - MC = - (0.2 m) 3 ( 3 m>s ) sin 30° 4 ( 1000 kg>m3 )( 0.016875p m3 >s ) MC = 33.58 N # m = 33.6 N # m b

Ans.

Ans: Cx = 45.0 N Cy = 168 N MC = 33.6 N # m 670

*6–68. Water flows into the pipe with a velocity of 5 ft>s. Determine the horizontal and vertical components of force, and the moment at A, needed to hold the elbow in place. Neglect the weight of the elbow and the water within it.

8 in. 5 ft/s A

3 in.

1.5 in.

SOLUTION We consider steady flow of an ideal fluid. Take the control volume as the elbow and the water within it. Q = VAAA = (5 ft>s) c p a

= 0.2454 ft 3 >s

Continuity Equation.

2 1.5 ft b d 12

8 ft 12

0 rdV + V # dA = 0 0t L L cv

MA

cs

Ax

0 - VAAA + VBAB = 0 - 0.2454 ft 3 >s + VB Jp a

0.75 2 ft b R = 0 12

Ay

pA

VB = 20 ft>s

pB = 0

Applying the Bernoulli equation between A and B,

(a)

pB pA VA2 VB2 + gzA = + gzB + + r r 2 2 pA

(5 ft>s)2

+

2

62.4 lb ft 3 ° ¢ 32.2 ft>s2

+ 0 = 0 +

(20 ft>s)2 2

+ 0

pA = 363.354 lb>ft 2 = 2.523 lb>in2

The free-body diagram of the control volume is shown in Fig. a. Here, water is discharged into the atmosphere at B. Therefore, pB = 0. Linear Momentum. Referring to Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

+ ΣFx = rQ[(VB)x - (VA)x]; S -Ax + 2.523 lb>in2 3 p(1.5 in.)2 4 = a

62.4 slug>ft 3 b ( 0.2454 ft 3 >s ) (0 - 5 ft>s) 32.2

Ax = 20.2 lb d

Ans.

+ c ΣFy = rQ[(VB)y - (VA)y]; - Ay = a

62.4 slug>ft 3 b ( 0.2454 ft 3 >s ) ( -20 ft>s - 0) 32.2

Ay = 9.51 lb T

671

Ans.

B

*6–68. Continued

Angular Momentum. Referring to Fig. a, ΣM =

0 (r * V)rdV + (r * V)rV # dA 0t L L cv

or a + ΣMA = ΣrQVd;

- MA = a

cs

62.4 8 slug>ft 3 b ( 0.2454 ft 3 >s ) c a - ft b(20 ft>s) - 0 d 32.2 12 Ans.

MA = 6.34 lb.ftb

672

6–69. The bend is connected to the pipe at flanges A and B as shown. If the diameter of the pipe is 1 ft and it carries a volumetric flow of 50 ft 3 >s, determine the horizontal and vertical components of force and the moment exerted at the fixed base D of the support. The total weight of the bend and the water within it is 500 lb, with a mass center at point G. The pressure of the water at A is 15 psi. Assume that no force is transferred to the flanges at A and B.

G A

45! 4 ft D

SOLUTION From the discharge, 50 ft 3 >s = V 3 p(0.5 ft)2 4

Q = VA;

VA = VB = V = 63.66 ft>s The flow is steady and water can be considered as an ideal fluid (incompressible and inviscid), such that gw = 62.4 lb>ft 2. Average velocities will be used. Bernoulli Equation pB = 1983.5 lb>ft 2 FA = pAAA = 15 lb>in.2(p)(6 in.)2 = 1696.46 lb FB = pBAB = 1983.5 lb>ft 2(p)(0.5 ft)2 = 1557.84 lb pA PB VA2 VB2 + gzA = + gzB + + r r 2 2 15(144) lb>ft 2 °

62.4 lb>ft 2 2

32.2 ft>s

+

V2 + 0 = 2

¢

pB °

62.4 lb>ft 2 32.2 ft>s2

Applying the linear momentum equation. ΣF =

+

V2 ( 32.2 ft>s2 ) (4 ft sin 45°) 2

¢

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of this equation along x and y axes by referring to the FBD of the control volume, Fig. a + ΣFx = 0 + VArw( - VAAA) + (VB cos 45°)rw(VBAB) S 1696.46 lb - [(1557.84 lb) cos 45°] - Dx = (63.66 ft>s) ° +

62.4 lb>ft 3 32.2 ft>s2

3 ( 63.66 ft>s ) cos 45° 4 °

¢ ( -50 ft 3 >s )

62.4 lb>ft 3 32.2 ft>s2

¢ ( 50 ft 3 >s )

Ans.

Dx = 2401.6 lb = 2.40 kip + c ΣFy = 0 + (VB sin 45°)rw(VBAB) Dy - 500 - [(1557.84) sin 45°] =

3 ( 63.66 ft>s ) sin45° 4 °

B

1.5 ft

62.4 lb>ft 3 32.2 ft>s2

¢ ( 50 ft 3 >s )

Ans.

Dy = 5963.3 lb = 5.96 kip

673

4 ft

6–69. Continued

Applying the Angular Momentum equation ΣM =

0 (r * V)rw dV + (r * V)rwV # dA 0t L L cv

cs

Writing the scalar component of this equation about D by referring to the FBD a+ ΣMD = 0 + ( -rAVA)rw( - VAAA) + ( - rBVB cos 45°)rw(VBAB) MD + [(1559.84 lb) cos 45°](4 ft) - (1696.46 lb)(4 ft) - (500 lb)[(1.5 ft) cos 45°] = -(4 ft) ( 63.66 ft>s ) °

62.4 lb>ft 3 32.2 ft>s2

¢ ( -50 ft 3 >s ) + ( - 4 ft) 3 ( 63.66 ft>s ) cos45° 4 °

MD = 10136.8 lb # ft = 10.1 kip # ft

62.4 lb>ft 3 32.2 ft>s2

Ans.

¢ ( 50 ft 3 >s )

y FB

500 lb x 1.5 ft

45˚

FA

4 ft

4 ft

Dx MD

Dy

Ans: Dx = 2.40 kip Dy = 5.96 kip MD = 10.1 kip # ft

(a)

674

6–70. The fan blows air at 6000 ft 3 >min. If the fan has a weight of 40 lb and a center of gravity at G, determine the smallest diameter d of its base so that it will not tip over. Assume the airstream through the fan has a diameter of 2 ft. The specific weight of the air is ga = 0.076 lb>ft 3.

2 ft

A G

B 0.75 ft 6 ft

SOLUTION We consider steady flow of an ideal fluid.

Then,

Q = a

6000 ft 3 1 min ba b = 100 ft 3 >s min 60 s

d 2

d 2

100 ft 3 >s = VB 3 p(1 ft)2 4

Q = VBAB;

VB = 31.83 ft>s

Take the control volume as the fan and air passing through it. The free-body diagram of the control volume is shown in Fig. a. Here, tipping will occur about point C. Angular Momentum. Air is sucked into the fan at A from a large source of still air, VA ≅ 0. Referring to Fig. a, ΣMc =

0 (r * V)rdV + (r * V)rV # dA 0t L L cv

a+ 40 lba0.75 ft +

cs

0.076 lb>ft 3 d b = a b ( 100 ft 3 >s ) [6 ft(31.83 ft>s) - 0] 2 32.2 ft>s2 d = 0.7539 ft = 0.754 ft

Ans.

40 lb 0.75 ft

6 ft

C F d 2

N

(a)

Ans: 0.754 ft 675

6–71. When operating, the air-jet fan discharges air with a speed of V = 18 m>s into a slipstream having a diameter of 0.5 m. If the air has a density of 1.22 kg>m3, determine the horizontal and vertical components of reaction at C, and the vertical reaction at each of the two wheels, D. The fan and motor have a mass of 25 kg and a center of mass at G. Neglect the weight of the frame. Due to symmetry, both of the wheels support an equal load. Assume the air entering the fan at A is essentially at rest.

0.5 m A

0.25 m V B G

2m

D C 0.75 m

SOLUTION

0.25 m

We consider steady flow of an ideal fluid. Take the control volume to be the fan and the air passing through it. Q = VBAB = (18 m>s) 3 p(0.25 m)2 4 = 3.5343 m3 >s

The free-body diagram of the control volume is shown in Fig. a. Angular Momentum. Referring to Fig. a, ΣM =

0 (r * V)rdV + (r * V)rV # dA 0t L L cs cv

or 2ND(0.75 m) - 25 kg ( 9.81 m>s2 ) (1m) = 0 + ( 1.22 kg>m3 )( 3.5343 m3 >s ) [( - 2 m)(18 m>s) - 0] Ans.

ND = 60.02 N = 60.0 N

Linear Momentum. Referring to Fig. a, ΣF =

0 VrdV + VrV # dA 0t L L cv cs

or + ΣFx = 0 + VBrQ S Cx = (18 m>s) ( 1.22 kg>m3 )( 3.5343 m3 >s )

Ans.

Cx = 77.6 N

+ c ΣFy = 0 + 0 Cy + 2(60.02 N) - 25 kg ( 9.81 m>s2 ) = ( 1.22 kg>m3 )( 3.5343 m3 >s ) (0 - 0)

Ans.

Cy = 125.22 N = 125 N (25 kg)(9.81 m s2)

2m

Cx 0.75 m Cy

2 ND

0.25 m

Ans: ND = 60.0 N Cx = 77.6 N Cy = 125 N

(a)

676

*6–72. If the air has a density of 1.22 kg>m3, determine the maximum speed V that the air-jet fan can discharge air into the slipstream having a diameter of 0.5 m at B, so that the fan does not topple over. The fan and motor have a mass of 25 kg and a center of mass at G. Neglect the weight of the frame. Due to symmetry, both of the wheels support an equal load. Assume the air entering the fan at A is essentially at rest.

0.5 m A

0.25 m V B G

2m

D C 0.75 m

SOLUTION Consider the control volume to be the fan and the air passing through it, Fig. a. Since the inlet A and outlet B are opened to the atmosphere, pA = pB = 0. The free-body diagram of the control volume is shown in Fig. a. Here, if the fan is about to topple about C, ND = 0. Applying the angular momentum equation ΣM =

0 (r * V)rdV + (r * V)rV # dA 0t L L cs cv

And writing the scalar component of the equation about C by referring to the FBD, a + ΣMC = 0 + ( - rBVB)(ra)(VBAB) - (25 kg) ( 9.81 m>s2 ) (1 m) = - 3 (2m)VB 4 ( 1.22 kg>m3 ) VB 3 p(0.25 m)2 4 VB = 22.63 m>s = 22.6 m>s

(25 kg)(9.81 m s2 )

2m

Cx 0.75 m Cy

2 ND

0.25 m

(a)

677

Ans.

0.25 m

6–73. Water flows through the curved pipe at a speed of 5 m>s. If the diameter of the pipe is 150 mm, determine the horizontal and vertical components of the resultant force, and the moment acting on the coupling at A. The weight of the pipe and the water within it is 450 N, having a center of gravity at G.

B 150 mm G 0.2 m

0.5 m

A 0.45 m

SOLUTION Take the control volume as the pipe and the water within it. QA = VAAA = (5 m>s) 3 p(0.075 m)2 4

0.2 m

3

= 0.08836 m >s

Bernoulli Equation, where VA = VB. Datum at A, the free-body diagram of the control volume is shown in Fig. a. Here, water is discharged into the atmosphere at B. Therefore, pB = 0.

G

pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 pA 3

1000 kg>m

+

2

pB = 0 0.5 m

W pA

2

V V + 0 = 0 + + (9.81 m>s2)(0.5 m) 2 2

Ax

MA

pA = 4905 Pa

Ay

Linear Momentum. Referring to Fig. a,

0.45 m (a)

0 VrdV + VrV # dA ΣF = 0t L L cv cs + ΣFx = 0 + pQ 3 (VB)x - (VA)x 4 ; S

Ans.

Ax = pQ(0 - 0) = 0

+ c ΣFy = 0 + pQ 3 (VB)y - (VA)y 4 ;

-Ay +

3 4905 N>m2 4 3 p(0.075 m)2 4

Ay = 520 N

- 450 N = ( 1000 kg>m3 )( 0.08836 m3 >s )( - 5m>s - 5 m>s ) Ans.

Angular Momentum. Referring to Fig. a, ΣM =

0 (r * V)rdV + (r * V)rV # dA 0t L L cv cs

a + ΣMD = 0 + ΣrQVd; - MA - (450 N)(0.2 m) = ( 1000 kg>m3 )( 0.08836 m3 >s ) [( - 0.45 m)(5 m>s) - 0] MA = 109 N # m

Ans.

Ans: Ax = 0 Ay = 520 N MA = 109 N # m 678

6–74. The chute is used to divert the flow of water. If the flow is 0.4 m3 >s and it has a cross-sectional area of 0.03 m2, determine the horizontal and vertical force components at the pin A, and the horizontal force at the roller B, necessary for equilibrium. Neglect the weight of the chute and the water on it.

B

4m

A

SOLUTION Take the control volume as the chute and the water on it.

3m

0.4 m >s = V ( 0.03 m 3

Q = VA;

2

)

V = 13.33 m>s

The free-body diagram of the control volume is shown in Fig. a. Here, pA = pB = 0 since points A and B are exposed to the atmosphere, Angular Momentum. Referring to Fig. a, 0 r * VrdV + (r * V)rV # dA 0t L L cs cv

ΣM = a + ΣMA = 0 + ΣrQVd;

- Bx(4 m) = ( 1000 kg>m3 )( 0.4 m3 >s ) [0 - 3 m(13.33 m>s)] Bx = 4000 N = 4 kN

Ans.

Linear Momentum. Referring to Fig. a, ΣF =

0 VrdV + VrV # dA 0t L L cv cs

+ ΣFx = 0 + (VA)rQ S 4000 N + Ax = (13.33 m>s) ( 1000 kg>m3 )( 0.4 m3 >s ) Ax = 1.33 kN

Ans.

+ c ΣFy = 0 + VBrQ Ay = (13.33 m>s) ( 1000 kg>m3 )( 0.4 m3 >s )

Ans.

Ay = 5.33 kN

Bx

pB = 0

4m

pA = 0 3m

Ax

Ay

(a)

Ans: Bx = 4 kN Ax = 1.33 kN Ay = 5.33 kN 679

6–75. Water flows through A at 400 gal>min and is discharged to the atmosphere through the reducer at B. Determine the horizontal and vertical components of force, and the moment acting on the coupling at A. The vertical pipe has an inner diameter of 3 in. Assume the assembly and the water within it has a weight of 40 lb and a center of gravity at G. 1 ft 3 = 7.48 gal.

B

2 in. G 18 in.

SOLUTION

A

The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that gw = 62.4 lb>ft 3. Average velocities will be used. The control volume consists of the vertical pipe, reducer and the contained water as shown in Fig. a. The discharge is

Thus,

Q = a400

gal min

ba

1 ft 3 1 min ba b = 0.8913 ft 3 >s 7.48 gal 60 s

G W

rAB = 1.5 ft 2

1.5 ft b R 12 2 1 0.8913 ft 3 >s = VB Jp a ft b R 12

0.8913 ft 3 >s = VA Jp a

Q = VAAA; Q = VBAB;

VA = 18.16 ft>s

FA

VB = 40.85 ft>s

Applying Bernoulli’s equation between points A and B with pB = patm = 0 and zB = 1.5 ft, pB pA VA2 VB2 + zA = + zB + + gw gw 2g 2g pA 62.4 lb>ft

3

+

(18.16 ft>s)2 2 ( 32.2 ft>s

2

)

2 ( 32.2 ft>s2 )

(a)

+ 1.5 ft

pA = 1391.28 lb>ft 2 Then the pressure force acting on the inlet control surface A, indicated in the FBD of the control volume, is FA = pAAA = ( 1391.28 lb>ft 2 ) Jp a

Applying the linear momentum equation, ΣF =

2 1.5 ft b R = 68.28 lb 12

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar components of this equation along the x and y axes by referring to Fig. a + ΣFx = 0 + ( - VB)rw(VBAB) S - Ax = ( - 40.85 ft>s)a

62.4 lb>ft 3 32.2 ft>s2

MA

Ay

(40.85 ft>s)2

+ 0 = 0 +

Ax

b(0.8913 ft 3 >s)

Ax = 70.56 lb = 70.6 lb

680

Ans.

6–75. Continued

+ c ΣFy = 0 + VArw( - VAAA) - 40 lb + 68.29 lb - Ay = (18.16 ft>s)a

62.4 lb>ft 3 32.2 ft>s2

Ay = 59.65 lb = 59.7 lb

b( -0.8913 ft 3 >s)

Ans.

Applying the angular momentum equation, ΣM =

0 (r * V)rwdV + (r * V)rwV # dA 0t Lcv Lcs

Writing the scalar component of this equation about point A, a + ΣMA = 0 + rABVB(rwVBAB) MA = (1.5 ft) ( 40.85 ft>s ) °

62.4 lb>ft 3 32.2 ft>s

¢ ( 0.8913 ft 3 >s )

= 105.84 lb # ft = 106 lb # ft

Ans.

Ans: Ax = 70.6 lb Ay = 59.7 lb MA = 106 lb # ft 681

*6–76. The waterwheel consists of a series of flat plates that have a width b and are subjected to the impact of water to a depth h, from a stream that has an average velocity of V. If the wheel is turning at v, determine the power supplied to the wheel by the water.

R v

V

SOLUTION Using a fixed control volume, with water entering on the left with velocity V and exiting on the right with (x-component) velocity vR (the speed of the plates), we apply the angular momentum equation: 0 (r * V)rdV + (r * V)rV # dA 0t Lcv Lcs -T = 0 + RVrw( -VA) + RvRrw(VA)

a+ ΣMhub =

where T is the torque or moment exerted by the water on the wheel and -T is the torque exerted by the wheel on the water. So then, since A = bh,

# and since W = Tv,

T = rwbhRV(V - vR)

Ans.

P = rwbhvRV(V - vR)

682

h

6–77. Air enters into the hollow propeller tube at A with a mass flow of 3 kg>s and exits at the ends B and C with a velocity of 400 m>s, measured relative to the tube. If the tube rotates at 1500 rev>min, determine the frictional torque m on the tube.

B

0.5 m M A 1500 rev/min

0.5 m

SOLUTION The flow is periodic hence it can be considered steady in the mean. The air is assumed to be an ideal fluid (incompressible and inviscid) such that its density is constant. Average velocities will be used. The control volume consists of the hollow propeller and the contained air. Its FBD is shown in Fig. a

C

The velocity of point B (or C) is VB = vr = Ja1500

B

rev 2p rad 1 min ba ba b R (0.5 m) = 25p m>s S min 1 rev 60 s

Thus, the velocity of the air ejected from B (or C) is

0.5 m

Va = VB + Va>B M

+ 2 Va 1d

=

( -25p m>s ) + ( 400 m>s ) = 321.46 m>s d

A 0.5 m

Applying the angular momentum equation, ΣM =

0 (r * V)rdV + (r * V)rV # dA 0t Lcv Lcs

C

Writing the scalar component about point A,

(a)

a+ ΣMA = 0 + 2 3 rABVBraVBAB 4

# Here raVBAB = mB = 1.5 kg>s . Then

M = 2(0.5 m) ( 321.46 m>s )( 1.5 kg>s ) = 482.19 N # m = 482 N # m

Ans.

Ans: 482 N # m 683

6–78. The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 10  mm, and the water is supplied through the hose at 0.008 m3 >s and is ejected horizontally, through the four arms. Determine the torque required to hold the arms from rotating.

350 mm

SOLUTION We consider steady flow of an ideal fluid relative to the control volume. Take the control volume as the sprinkler and the water within it. Due to symmetry and the continuity condition, the discharge from each nozzle is Q = ( 0.008 m3 >s ) >4 = 0.002 m3 >s . Q = VA;

Z

5m

0.3

0.002 m3 >s = V 3 p ( 0.005 m2 ) 4 V = 25.46 m>s

The free-body diagram of the control volume is shown in Fig. a. Here, water is discharged to the atmosphere at the nozzle, p = 0.

M F

Angular Momentum. Referring to Fig. a, ΣM =

(a)

0 (r * V)rdV (r * V)rV # dA 0t Lcv Lcs or a+ ΣMA = ΣrQVd;

M = 4 3( 1000 kg>m

3

)( 0.002 m3 >s )4 3 0.35 m ( 25.46 m>s ) - 0 4

= 71.30 N # m = 71.3 N # m

Ans.

Ans: 71.3 N # m 684

6–79. The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 10  mm, and the water is supplied through the hose at 0.008 m3 >s and is ejected horizontally, through the four arms. Determine the steady-state angular velocity of the arms. Neglect friction.

350 mm

SOLUTION We consider steady flow of an ideal fluid relative to the control volume. Take the control volume as the sprinkler and the water within it. Due to symmetry and the continuity condition, the discharge from each nozzle is Q = ( 0.008 m3 >s ) >4 = 0.002 m3 >s . Q = Vf>nA;

Z

5m

0.3

0.002 m3 >s = Vf>n 3 p ( 0.005 m2 ) 4 Vf>n = 25.46 m>s

The velocity of the nozzle is

M=0

Vn = vr = v(0.35 m) = 0.35 v

F

Thus, the velocity of the flow can be determined from

(a)

Vf = Vn + Vf>n Vf = - 0.35v + 25.46 The free-body diagram of the control volume is shown in Fig. a. Here, water is discharged to the atmosphere at the nozzle, p = 0. Angular Momentum. Referring to Fig. a, ΣM =

0 (r * V)rdV + (r * V)rV # dA 0t Lcv Lcs or a + ΣMA = ΣrQdV;

0 = 4 3( 1000 kg>m

3

)( 0.002 m3 >s )4 [0.35 m( -0.35v + 25.46)]

v = 72.76 rad>s = 72.8 rad>s

Ans.

Ans: 72.8 rad>s 685

*6–80. The 5-mm-diameter arms of a rotating lawn sprinkler have the dimensions shown. Water flows out relative to the arms at 6 m>s, while the arms are rotating at 10 rad>s. Determine the frictional torsional resistance at the bearing A, and the speed of the water as it emerges from the nozzles, as measured by a fixed observer.

10 rad/s

200 mm

50 mm

A

60!

SOLUTION 150˚

Referring to the geometry shown in Fig. a, the cosine and sine laws give r = 2502 + 2002 - 2(50)(200) cos 150° = 244.6 mm sin a sin 150° = ; 0.05 m 0.2446 m

0.2 m

0.05 m ß

α

r

a = 5.867°

Then

(a)

b = 180° - 150° - 5.867° = 24.133° Thus, the velocity of the tip of the arm is

24.133˚ VW

Vt = vr = ( 10 rad>s ) (0.2446 m) = 2.446 m>s c Referring to the velocity vector diagram shown in Fig. b, the relative velocity equation gives

Vw t = 6 m s

Vw = Vt + Vw>t J + 2 1S

1+c2

(Vw)x 2.446 m>s R + J R = J c (Vw)ydT

(b)

6 m>s R 24.133° 15

(V0w)5x m= 5.476 m>s d

- (Vw)x = - ( 6 m>s ) cos 24.133°

- (Vw)y = 2.446 m>s - ( 6 m>s ) sin 24.133°

(Vw)y = 0.007339 m>s T

The magnitude of Vw is Vw = 2(Vw)x2 + (Vw)y2 = 2 ( 5.476 m>s ) 2 + ( 0.007339 m>s ) 2 = 5.476 m>s = 5.48 m>s

Ans.

The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that rw = 1000 kg>m3. Average velocity will be used. The control volume consists of the entire arm and the contained water as shown in Fig. a. Applying the angular momentum equation, 0 (r * V)rwdV + (r * V)rwV # dA 0t Lcv Lcs Writing the scalar component of this equation about point A, by referring to the 24 control 133 FBD of the volume, Fig. a, ΣM =

a + ΣMA = 0 + r(Vw)yrw(Vw>t A) m s )( 1000 kg>m3 ) 5 ( 6 m>s ) 3 p(0.0025 m)2 4 6 Vw t = m>s M = (0.2446 m) ( 0.007339

= 2.1145 ( 10-4 ) N # m

# = 0.211 mN (b) m

Ans.

M r = 0.2446 m (c)

686

6–81. The airplane is flying at 250 km>h through still air as it discharges 350 m3 >s of air through its 1.5-m-diameter propeller. Determine the thrust on the plane and the ideal efficiency of the propeller. Take ra = 1.007 kg>m3.

3 ft

SOLUTION The average velocity of the air flow through the propeller (control volume) is 350 m3 >s = V 3 p(0.75 m)2 4

Q = VA;

V = 198.06 m>s

Here, V1 = a250

km 1000 m 1h ba ba b = 69.44 m>s h 1 km 3600 s

V =

V1 + V2 ; 2

198.06 m>s =

( 69.44 m>s ) + V2 2

V2 = 326.67 m>s The ideal efficiency is e =

2 ( 69.44 m>s ) 2V1 = = 0.3506 = 0.351 V1 + V2 69.44 m>s + 326.67 m>s

Ans.

The thrust of the propeller is F = =

rpR2 ( V22 - V12 ) 2

( 1.007 kg>m3 ) (p)(0.75 m)2 2

= 90.66 ( 103 ) N = 90.7 kN

3 ( 326.67 m>s ) 2

- ( 69.44 m>s ) 2 4

Ans.

Ans: e = 0.351 F = 90.7 kN 687

6–82. The airplane travels at 400 ft>s through still air. If the air flows through the propeller at 560 ft>s, measured relative to the plane, determine the thrust on the plane and  the ideal efficiency of the propeller. Take ra = 2.15 1 10 - 3 2 slug>ft 3.

3 ft

SOLUTION The propeller and air within it is the control volume. We consider steady flow of an ideal fluid relative to the control volume. Here, V1 = 400 ft>s and V = 560 ft>s. V =

V1 + V2 ; 2

560 ft>s =

400 ft>s + V2 2

V2 = 720 ft>s The ideal efficiency is e =

2 ( 400 ft>s ) 2V1 = = 0.7143 = 0.714 V1 + V2 400 ft>s + 720 ft>s

Ans.

The thrust of the propeller is F = =

ppR2 ( V22 - V12 ) 2

( 2.15 ( 10-3 ) slug>ft 3 ) (p)(1.5 ft)2 2

= 2723.38 lb = 2.72 kip

3 ( 720 ft>s ) 2

- ( 400 ft>s ) 2 4

Ans.

Ans: e = 0.714 F = 2.72 kip 688

6–83. A boat has a 250-mm-diameter propeller that discharges 0.6 m3 >s of water as the boat travels at 35 km>h in still water. Determine the thrust developed by the propeller on the boat.

SOLUTION The propeller and water within is the control volume. The average velocity of the water through the propeller is Q = VA;

0.6 m3 >s = V 3 p(0.125 m)2 4 V = 12.22 m>s

Here, V1 = a35

km 1000 m 1h ba ba b = 9.722 m>s h 1 km 3600 s V =

V1 + V2 ; 2

12.22 m>s =

9.722 m>s + V2 2

V2 = 14.72 m>s The thrust of the propeller is F = =

rpR2 ( V22 - V12 ) 2

( 1000 kg>m3 ) (p)(0.125 m)2 2

= 3.001 ( 103 ) N = 3.00 kN

3 ( 14.72 m>s ) 2

- ( 9.722 m>s ) 2 4

Ans.

Ans: 300 kN 689

*6–84. A ship has a 2.5-m-diameter propeller with an ideal efficiency of 40%. If the thrust developed by the propeller is 1.5 MN, determine the constant speed of the ship in still water and the power that must be supplied to the propeller to operate it.

SOLUTION The propeller and water within it is the control volume. The ideal efficiency is e =

2V1 ; V1 + V2

0.4 =

2V1 V1 + V2

(1)

V2 = 4V1

The thrust of the propeller is F =

rpR2 ( V22 - V 12 ) ; 2

1.5 ( 106 ) N =

( 1000 kg>m3 ) (p)(1.25 m)2 2

( V 22 - V 12 )

V 22 - V 12 = 611.15

(2)

Solving Eqs. (1) and (2) yields V1 = 6.383 m>s = 6.38 m>s Ans.

V2 = 25.53 m>s The power output is

#

Wout = FV1 =

3 1.5 ( 106 ) N 4 ( 6.383 m>s )

= 9.575 ( 106 ) W = 9.575 MW

Thus, the power supply to the propeller is

#

Win =

Pout 9.575 MW = = 23.94 MW = 23.9 MW e 0.4

690

Ans.

6–85. The fan is used to circulate air within a large industrial building. The blade assembly weighs 200 lb and consists of 10 blades, each having a length of 6 ft. Determine the power that must be supplied to the motor to lift the assembly off its bearings and allow it to freely turn without friction. What is the downward air velocity for this to occur? Neglect the size of the hub H. Take ra = 2.36 1 10 - 3 2 slug>ft 3. H

6 ft

SOLUTION The blade and air within it is the control volume. The flow is steady and the air can be considered as an ideal fluid (incompressible and inviscid) such that ra = 2.36 ( 10-3 ) slug>ft 3. Average velocities will be used. To lift the blade assembly off the bearings, the thrust must be equal to the weight of the assembly, i.e, F = 200 lb. Since the air enters the blade assembly from the surroundings which is at rest, V1 = 0. F =

ra pR2 ( V 22 - V 12 ) ; 2

200 lb =

2.36 ( 10-3 ) slug>ft 3 3 p(6 ft)2 4 2

V2 = 38.71 ft>s = 38.7 ft>s V =

( V 22 - 0 ) Ans.

0 + 38.71 ft>s V1 + V2 = = 19.36 ft>s 2 2

The power required by the motor is

#

W = FV = (200 lb) ( 19.36 ft>s ) = a3871.22 = 7.04 hp

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

Ans: V# 2 = 38.7 ft>s W = 7.04 hp 691

6–86. The 12-Mg helicopter is hovering over a lake as the suspended bucket collects 5 m3 of water used to extinguish a fire. Determine the power required by the engine to hold the filled water bucket over the lake. The horizontal blade has a diameter of 14 m. Take ra = 1.23 kg>m3.

SOLUTION The helicopter, bucket, water, and air within the helicopter blade is the control volume. The flow is steady and the air can be considered as an ideal fluid (incompressible and inviscid) such that ra = 1.23 kg>m3. Average velocities will be used. To maintain the hovering, the thrust produced by the rotor blade must be equal to the weight of the helicopter and the water. Thus, F =

3 12 ( 103 ) kg 4 ( 9.81 m>s2 )

+ ( 1000 kg>m3 )( 9.81 m>s2 )( 5 m3 )

= 166.77 ( 103 ) N

Since the air enters the blade from the surroundings, which is at rest, V1 = 0. F =

rapR2 ( V 22 - V 12 ) ; 2

166.77 ( 103 ) N =

( 1.23 kg>m3 ) 3 p(7 m)2 4 2

(V 22 - 0)

V2 = 41.97 m>s V =

0 + 41.97 m>s V1 + V2 = = 20.985 m>s 2 2

Thus, the power required by the engine is # W = FV = 3 166.77 ( 103 ) N 4 ( 20.985 m>s ) = 3.4997 ( 106 ) W

Ans.

= 3.50 MW

Ans: 3.50 MW 692

6–87. The airplane has a constant speed of 250 km>h in still air. If it has a 2.4-m-diameter propeller, determine the force acting on the plane if the speed of the air behind the propeller, measured relative to the plane, is 750 km>h. Also, what is the ideal efficiency of the propeller, and the power produced by the propeller? Take ra = 0.910 kg>m3.

250 km/h

SOLUTION The airplane moves in the still air and the control volume is attached to the airplane, km 1000 m 1h which is travelling with a constant velocity of a250 ba ba b = h 1 km 3600 s 69.44 m>s. Then the inlet velocity is V1 = 69.44 m>s. Relative to the control volume, the flow is steady. The air can be considered as an ideal fluid (incompressible and inviscid) such that ra = 0.910 kg>m3. Average velocities will be used. The outlet velocity is V2 = a750 km ba 1000 m ba 1 h b = 208.33 m>s. The thrust on the h 1 km 3600 s plane is F = =

rapR2 ( V 22 - V 12 ) ; 2

( 0.910 kg>m3 ) 3 p ( 1.2 m ) 2 4 2

= 79.41 ( 103 ) N

3 ( 208.33 m>s ) 2

- ( 69.44 m>s ) 2 4

The power generated by the propeller is

#

W0 = FV1 =

3 79.41 ( 103 ) N 4 ( 69.44 m>s )

= 5.515 ( 106 ) W = 5.51 MW

Ans.

The efficiency of the propeller is e =

2(69.44 m>s) 2V1 = = 0.5 V1 + V2 69.44 m>s + 208.33 m>s

Ans.

Ans: F# = 79.4 kN W = 5.51 MW e = 0.5 693

*6–88. The 12-kg fan develops a breeze of 10 m>s using a 0.8-m-diameter blade. Determine the smallest dimension d for the support so that the fan does not tip over. Take ra = 1.20 kg>m3. G

500 mm

SOLUTION Take the fan and air within it as the control volume. The flow is steady and the air can be considered as an ideal fluid (incompressible and inviscid) such that ra = 1.20 kg>m3. Average velocities can be used. Since the air enters the blade from the surroundings which is at rest, V1 = 0. Here, V2 = 10 m>s. F = =

0.6 d 0.4 d

rapR2 ( V 22 - V 12 ) 2

( 1.20 kg>m3 ) 3 p(0.4 m)2 4

= 9.6p N

2

12(9.81) N

3 ( 10 m>s ) 2

- 04

F

Referring to the FBD of the fan shown in Fig. a, and writing the moment equation of equilibrium about point A, a+ ΣMA = 0;

0.5 m

Ax

[12(9.81) N](0.4 d) - (9.6p N)(0.5 m) = 0 Ans.

d = 0.320 m = 320 mm

0.4d (a)

694

Ay

6–89. The airplane is flying at 160 ft>s in still air at an altitude of 10 000 ft. The 7-ft-diameter propeller moves the air at 10 000 ft 3 >s. Determine the power required by the engine to turn the propeller, and the thrust on the plane.

160 ft/s

SOLUTION Take the propeller and air within it as the control volume. Since the airplane moves in the still air and the control volume is attached to the airplane, which is travelling with a constant velocity of the 160 ft>s, then the inlet velocity is V1 = 160 ft>s. Relative to the control volume, the flow is steady. The air can be considered as an ideal fluid (incompressible and inviscid) such that at an altitude of 10,000 ft, ra = 1.754 ( 10-3 ) slug>ft 3. Average velocity will be used. From the discharge 10 000 ft 3 >s = V 3 p(3.5 ft)2 4

Q = VA;

160 ft>s + V2 V1 + V2 ; 259.84 ft>s = 2 2 The thrust on the plane is V =

F = =

V = 259.84 ft>s V2 = 359.69 ft>s

rapR2 ( V 22 - V 12 ) 2

3 1.754 ( 10-3 ) slug>ft3 4 3 p(3.5 ft)2 4 2

= 3.503 ( 103 ) lb = 3.50 kip

3 ( 359.69 ft>s ) 2

- ( 160 ft>s ) 2 4

Ans.

The power required to turn the propeller is

#

Wi = FV = =

3 3.503 ( 103 ) lb 4 ( 259.84 ft>s )

3 910.22 ( 103 ) ft # lb>s 4 c

= 1655 hp

1 hp

550 ft # lb>s

d

Ans.

Ans: F# = 3.50 kip W = 1655 hp 695

6–90. The airplane is flying at 160 ft>s in still air at an altitude of 10 000 ft. The 7-ft-diameter propeller moves the air at 10 000 ft 3 >s. Determine the propeller’s ideal efficiency, and the pressure difference between the front and back of the blades.

160 ft/s

SOLUTION Take the propeller and air within it as the control volume. Since the airplane moves in the still air and the control volume is attached to the airplane, which is travelling with a constant velocity of 160 ft>s, then the inlet velocity is V1 = 160 ft>s. Relative to the control volume the flow is steady and the air can be considered as an ideal fluid (incompressible and inviscid) such that at an altitude of 10,000 ft, ra = 1.754 ( 10-3 ) slug>ft 3. Average velocities will be used. From the discharge 10 000 ft 3 >s = V 3 p(3.5 ft)2 4

Q = VA; V =

V1 + V2 ; 2

259.84 ft>s =

160 ft>s + V2 2

V = 259.84 ft>s V2 = 359.69 ft>s

The ideal efficiency of the propeller is e =

2 ( 160 ft>s ) 2V1 = = 0.616 V1 + V2 160 ft>s + 359.69 ft>s

Ans.

The pressure difference is ∆p = p4 - p3 = raV(V2 - V1) =

3 1.754 ( 10-3 ) slug>ft3 4 ( 259.84 ft>s ) 3 359.69 ft>s

= a91.01

lb 1 ft 2 ba b = 0.632 psi 2 12 in ft

- ( 160 ft>s ) 4

Ans.

Ans: e = 0.616 ∆p = 0.632 psi 696

6–91. Plot Eq. 6–15 and show that the maximum efficiency of a wind turbine is 59.3% as stated by Betz’s law.

SOLUTION eturb =

V22 V2 1 c1 - a 2b d c1 + a b d 2 V1 V1

0.6 0.5

W W0

0.4 0.3 0.2 0.1 0.0

0

0.2

0.4

0.6

0.8

V2 V1

#

W # = 0.593 = 59.3, W0 when

V2 1 = . V1 3

697

1

*6–92. The wind turbine has a rotor diameter of 40 m and an ideal efficiency of 50% in a 12 m>s wind. If the density of the air is ra = 1.22 kg>m3, determine the thrust on the blade shaft, and the power withdrawn by the blades.

12 m/s

SOLUTION e =

V2 V2 2 1 c1 - a b d c1 + d 2 V1 V1

Solving the cubic equation with e = 0.5, we find V2 >V1 = 0.6180 as the nonzero solution. Then V2 = 0.6180 ( 12 m>s ) = 7.416 m>s and

12 m>s + 7.416 m>s V1 + V2 = = 9.708 m>s 2 2 The thrust on the blades is V =

F = =

rapR2 ( V22 - V12 ) 2

( 1.22 kg>m3 ) p(20 m)2 2

3 ( 12 m>s ) 2

= 68.220 ( 103 ) N = 68.2 kN

- ( 7.416 m>s ) 2 4

Ans.

The power withdrawn by the blades is

#

W = FV =

3 68.220 ( 103 ) N 4 ( 9.708 m>s )

= 662.3 ( 103 ) W

Ans.

= 662 kW

698

40 mm

6–93. The wind turbine has a rotor diameter of 40 m and an efficiency of 50% in a 12 m>s wind. If the density of the air is ra = 1.22 kg>m3, determine the difference between the pressure just in front of and just behind the blades. Also find the mean velocity of the air passing through the blades. 12 m/s

40 mm

SOLUTION e =

V2 V2 2 1 c1 - a b d c1 + d 2 V1 V1

Solving the cubic equation with e = 0.5, we find V2 >V1 = 0.6180 as the nonzero solution. Then V2 = 0.6180 ( 12 m>s ) = 7.416 m>s and V =

12 m>s + 7.416 m>s V1 + V2 = = 9.71 m>s 2 2

Ans.

The thrust on the blades is F = =

rapR2 ( V22 - V12 ) 2

( 1.22 kg>m3 ) p(20 m)2 2

= 68.220 ( 103 ) N

3 ( 12 m>s ) 2

- ( 7.416 m>s ) 2 4

The pressure difference is ∆p =

68.220 ( 103 ) N F = = 54.3 Pa A p(20 m)2

Ans.

Ans: V = 9.71 m>s ∆p = 54.3 Pa 699

6–94. The jet engine on a plane flying at 160 m>s in still air draws in air at standard atmospheric temperature and pressure through a 0.5-m-diameter inlet. If 2 kg>s of fuel is added and the mixture leaves the 0.3-m-diameter nozzle at 600 m>s, measured relative to the engine, determine the thrust provided by the turbojet. 160 m/s

SOLUTION From Appendix A, at standard atmospheric pressure and temperature (15° C), the density of air is ra = 1.23 kg>m3. Thus, . ma = raVA = ( 1.23 kg>m3 )( 160 m>s ) 3 p(0.25 m)2 4 = 38.64 kg>s The thrust of the turbojet is . . . T = (ma + mf)Ve - maVcv

= ( 38.64 kg>s + 2 kg>s )( 600 m>s ) - ( 38.64 kg>s )( 160 m>s ) = 18.20 ( 103 ) N = 18.2 kN

Ans.

Ans: 18.2 kN 700

6–95. The jet engine is mounted on the stand while it is being tested. Determine the horizontal force that the engine exerts on the supports, if the fuel–air mixture has a mass flow of 11 kg>s and the exhaust has a velocity of 2000 m>s.

2000 m/s

SOLUTION

W

Take the control volume as the engine and the fluid within it. We consider steady dVcv = 0, Vcv = 0, and flow of an ideal fluid. Since the turbojet is at rest in still air, dt . ma = 0. Referring to the free-body diagram of the turbojet in Fig. a, . . + 2 ΣF = m dVcv + m. V - ( m 1d x a cv a + mf ) Ve dt

Fh F

- Fh = 0 + 0 - ( 0 + 11 kg>s )( 2000 m>s )

(a)

Ans.

Fh = 22 kN

This is the magnitude of the force the supports exert on the engine, and therefore also the magnitude of the equal and opposite force the engine exerts on the supports.

Ans: 22 kN 701

*6–96. The jet plane has a constant velocity of 750 km>h. Air enters its engine nacelle at A having a cross-sectional area of # 0.8 m2. Fuel is mixed with the air at me = 2.5 kg>s and is exhausted into the ambient air with a velocity of 900 m>s, measured relative to the plane. Determine the force the engine exerts on the wing of the plane. Take ra = 0.850 kg>m3.

A

SOLUTION The control volume is considered to be the entire engine and its contents which move with a constant velocity. The flow, measured relative to the control volume, 1h km 1000 m . is steady. Here, Vcv = a750 ba ba b = 208.33 m>s, mf = 2.5 kg>s h 1 km 3600 s and Ve = 900 m>s. Thus, . ma = raVcvAA = ( 0.850 kg>m3 )( 208.33 m>s )( 0.8 m2 ) = 141.67 kg>s The thrust developed is . . . T = - 3 maVcv - ( ma + mf ) Ve 4

= - 3 ( 141.67 kg>s )( 208.33 m>s ) - ( 141.67 kg>s + 2.5 kg>s )( 900 m>s ) 4 = 100.24 ( 103 ) N = 100 kN

Ans.

702

6–97. The jet engine is mounted on the stand while it is being tested with the braking deflector in place. If the exhaust has a velocity of 800 m>s and the pressure just outside the nozzle is assumed to be atmospheric, determine the horizontal force that the supports exert on the engine. The fuel–air mixture has a flow of 11 kg>s.

B

30!

A 30!

SOLUTION Under test conditions, with the pressure just outside the nozzle assumed to be atmospheric, the deflector is irrelevant since it is not attached to the engine. Since the engine is at rest in still air, dVcv >dt = 0 and Vcv = 0, so that the support reaction force F, which points rightward, is given by + 2 ΣFx 1d

= m

dVcv . . . + maVcv - ( ma + mf)Ve dt

- F = 0 + 0 - ( 11 kg>s )( 800 m>s ) Ans.

F = 8800 N = 8.80 kN

Ans: 8.80 kN 703

6–98. If an engine of the type shown in Prob. 6–97 is attached to a jet plane, and it operates the braking deflector with the conditions stated in that problem, determine the speed of the plane in 5 seconds after it lands with a touchdown velocity of 30 m>s. The plane has a mass of 8 Mg. Neglect rolling friction from the landing gear.

B

30!

A 30!

SOLUTION Assume that the fuel is only a small fraction of the fuel-air mixture, so that . . . ma ≈ ma + mf = 11 kg>s. Then the force equation for the whole plane, of mass mp, is + 2 1d

ΣFx = mp

dV . . . + maV - (ma + mf)Ve dt

0 = (8000 kg)

dV + ( 11 kg>s ) V - ( 11 kg>s )( - 800 m>s - V ) cos 30° dt

-8000

dV = 11[V(1 + cos 30°) + 800 cos 30°] dt

-8000

dV = 11(1.8660V + 692.82) dt V

-

s

8000 dV = 11 dt L0 L30 m>s 1.866 V + 692.82

8000 1.866V + 692.82 ln a b = 55 1.866 748.80

Ans.

V = 24.9 m>s

Ans: 24.9 m>s 704

6–99. The boat has a mass of 180 kg and is traveling forward on a river with a constant velocity of 70 km>h, measured relative to the river. The river is flowing in the opposite direction at 5 km>h. If a tube is placed in the water, as shown, and it collects 40 kg of water in the boat in 80 s, determine the horizontal thrust T on the tube that is required to overcome the resistance due to the water collection.

v ! 5 km/h T

SOLUTION Consider the boat, tube, and water within it as the moving control volume. We consider steady flow of an ideal fluid relative to the control volume. dm 40 = = 0.5 kg>s dt 80 vD>t = (70)a ΣFx = m

1000 b = 19.444 m>s 3600

dmi dV + vD>i dt dt

Ans.

T = 0 + 19.444(0.5) = 9.72 N

Ans: 9.72 N 705

*6–100. The jet is traveling at a constant velocity of 400 m>s in still air, while consuming fuel at the rate of 1.8 kg>s and ejecting it at 1200 m>s relative to the plane. If the engine consumes 1 kg of fuel for every 50 kg of air that passes through the engine, determine the thrust produced by the engine and the efficiency of the engine.

400 m/s

SOLUTION The control volume considered is the entire airplane and its contents which moves with a constant velocity. We consider steady flow of an ideal fluid. The flow measured relative to the control volume is steady. Here, . . Vcv = 400 m>s, mf = 1.8 kg>s, ma = 50 ( 1.8 kg>s ) = 90 kg>s and Ve = 1200 m>s . . . T = - 3maVcv - ( ma + mf)Ve 4

= - 3 ( 90 kg>s )( 400 m>s ) - ( 90 kg>s + 1.8 kg>s )( 1200 m>s ) 4 = 74.16 ( 103 ) N = 74.2 kN

Ans.

The useful power output of the engine is

#

3 74.16 ( 103 ) N 4 ( 400 m>s )

W0 = TV =

= 29.664 ( 106 ) W

Some of the power produces the kinetic energy per unit time of the exhaust fuel-air mixture. Its velocity relative to the ground is Vmix = Ve - Vcv = 1200 m>s 400 m>s = 800 m>s. Thus, the power loss is

#

1 . . (m + mf) Vmix2 2 a 1 = ( 90 kg>s + 1.8 kg>s )( 800 m>s ) 2 2

Wl =

= 29.376 ( 106 ) MW The efficiency of the engine is

#

29.664 ( 106 ) W W0 e = # = W0 + Pl 29.664 ( 106 ) W + 29.376 ( 106 ) W Ans.

= 0.502

706

6–101. The jet boat takes in water through its bow at 0.03 m3 >s, while traveling in still water with a constant velocity of 10 m>s. If the water is ejected from a pump through the stern at 30 m>s, measured relative to the boat, determine the thrust developed by the engine. What would be the thrust if the 0.03 m3 >s of water were taken in along the sides of the boat, perpendicular to the direction of motion? If the efficiency is defined as the work done per unit time divided by the energy supplied per unit time, then determine the efficiency for each case.

10 m/s

SOLUTION The control volume considered is the entire boat and its contents, which moves with a constant velocity. The flow, measured relative to the control volume, is # steady. Water is considered to be incompressible. Here, Vcv = 10 m>s, mf = 0, . 3 3 mw = rQ = ( 1000 kg>m )( 0.03 m >s ) = 30 kg>s. and Ve = 30 m>s. The thrust is # # # T1 = - 3 mwVcv - (mw + mf)Ve 4 = - 3 ( 30 kg>s )( 10 m>s ) - ( 30 kg>s + 0 )( 30 m>s ) 4

= 600 N

Ans.

If the intake of water is perpendicular to the direction of motion, Vcv = 0. Then # # # T2 = 3 mwVcv - ( mw + mf ) Ve 4 = - 3 ( 30 kg>s ) (0) - ( 30 kg>s + 0 )( 30 m>s ) 4

Ans.

= 900 N

The power output for both cases can be determined from

#

(Wo)1 = T1V = (600 N)(10 m>s) = 6000 W

#

(Wo)2 = T2V = (900 N)(10 m>s) = 9000 W Some of the power produces the kinetic energy per unit time of the ejected water. Its velocity relative to ground is V = Ve - Vcv = 30 m>s - 10 m>s = 20 m>s. For both cases, the power loss in the same and is

#

Wl =

1 # 1 # (m + mf)V 2 = ( 30 kg>s + 0 )( 20 m>s ) 2 = 6000 W 2 w 2

Ans.

Thus, the efficiency for each case is

#

(Wo)1

6000 W # # = = 0.5 (Wo)1 + (Wo), 6000 W + 6000 W # (Wo)2 9000 W # = = 0.6 e2 = # (Wo)2 + (Wo), 9000 W + 6000 W

e1 =

Ans. Ans.

Ans: T1 = T2 = e1 = e2 = 707

600 N 900 N 0.5 0.6

6–102. The 10-Mg jet plane has a constant speed of 860 km>h when it is flying horizontally. Air enters the intake I at the rate of 40 m3 >s. If the engine burns fuel at the rate of 2.2 kg>s, and the gas (air and fuel) is exhausted relative to the plane with a speed of 600 m>s, determine the resultant drag force exerted on the plane by air resistance. Assume that the air has a constant density of ra = 1.22 kg>m3.

I

2

2 1

SOLUTION Take the plane and its contents as the control volume. We consider steady flow of an ideal fluid. Vcv = a860

km 1h 1000 m . ba ba b = 238.89 m>s and ma = rQ = ( 1.22 kg>m3 )( 40 m3 >s ) = 48.8 kg>s h 3600 s 1 km

Since the airplane is traveling with constant speed,

dVcv = 0. Referring to the freedt

body diagram of the jet plane in Fig. a, + 2 ΣFx 1d

= m

dVcv # # # + maVcv - ( ma + mf ) Ve dt

- FD = 0 + ( 48.8 kg>s )( 238.89 m>s ) - ( 48.8 kg>s + 2.2 kg>s )( 600 m>s ) FD = 18.94 ( 103 ) N = 18.9 kN

Ans.

W FD

FL (a)

Ans: 18.9 kN 708

6–103. The jet is traveling at a speed of 500 mi>h, 30° above the horizontal. If the fuel is being spent at 10 lb>s, and the engine takes in air at 900 lb>s, whereas the exhaust gas (air and fuel) has a relative speed of 4000 ft>s, determine the acceleration of the plane at this instant. The drag resistance of the air is FD = ( 0.07v2 ) lb, where the speed is measured in ft>s. The jet has a weight of 15 000 lb. Take 1 mi = 5280 ft.

500 mi/h

30!

SOLUTION The control volume considered is the entire jet and its contents as shown in Fig. a which is accelerating. We consider steady flow of an ideal fluid relative to the control volume. Here, Vcv = a500

1h mi 5280 ft ba ba b = 733.33 ft>s h 1 mi 3600 s

FD = 0.07Vcv2 = 0.07 ( 733.332 ) = 37 644.44 lb 900 lb>s # ma = = 27.9503 slug>s 32.2 ft>s2 10 lb>s # = 0.3106 slug>s mf = 32.2 ft>s2 Ve = 4000 ft>s Referring to the FBD of the control volume, Fig. a, + ΣFx = m d

dVcv # # # + maVcv - ( ma + mf ) Ve dt

- (15000 lb) sin 30° - 37644.44 lb = a

15000 lb bacv + ( 27.9503 slug>s )( 733.33 ft>s ) 32.2 ft>s2

- ( 27.9503 slug>s + 0.3106 slug>s )( 4000 ft>s ) acv = 101.76 ft>s2 = 102 ft>s2 X FD

Ans.

W = 15000 lb 30˚

Ful (a)

Ans: 102 ft>s2 709

*6–104. The 12-Mg jet airplane has a constant speed of 950 km>h when it is flying along a horizontal straight line. Air enters the intake scoops S at the rate of 50 m3 >s. If the engine burns fuel at the rate of 0.4 kg>s, and the gas (air and fuel) is exhausted relative to the plane with a speed of 450 m>s, determine the resultant drag force exerted on the plane by air resistance. Assume that air has a constant density of 1.22 kg>m3.

v ! 950 km/h

S

SOLUTION The control volume considered is the entire jet and its contents as shown in Fig. a. We consider steady flow of an ideal fluid relative to the control volume. Here Vcv = a950

1h km 1000 m ba ba b = 263.89 m>s h 1 km 3600 s

# ma = raQ = ( 1.22 kg>m3 )( 50 m3 >s ) = 61 kg>s # mf = 0.4 kg>s

Ans.

Ve = 450 m>s

dVcv = 0, since the jet Referring to the FBD of the control volume, Fig. a, with dt travels with a constant velocity, we have dV # # + ΣFx = m cv + m# aVcv - ( m d a + mf ) Ve dt - FD = 0 + ( 61 kg>s )( 263.89 m>s ) - ( 61 kg>s + 0.4 kg>s )( 450 m>s ) FD = 11.53 ( 103 ) N = 11.5 kN

Ans.

[12(103)(9.81)] N

x

FD

Ful (a)

710

6–105. A commercial jet aircraft has a mass of 150 Mg and is cruising at a constant speed of 850 km>h in level flight (u = 0°). If each of the two engines draws in air at a rate of 1000 kg>s and ejects it with a velocity of 900 m>s relative to the aircraft, determine the maximum angle u at which the aircraft can fly with a constant speed of 750 km>h. Assume that air resistance (drag) is proportional to the square of the speed, that is, FD = cV 2, where c is a constant to be determined. The engines are operating with the same power in both cases. Neglect the amount of fuel consumed.

u

SOLUTION The control volume considered is the entire plane and its contents as shown in Fig. a, which is accelerating. We consider steady flow of an ideal fluid relative to the control volume. Here, Vcv = a850 Vcv = a750

1h km 1000 m ba ba b = 236.11 m>s (u = 0°) h 1 km 3600 s km 1000 m 1h ba ba b = 208.33 m>s h 1 km 3600 s

# ma = 2 ( 1000 kg>s ) = 2000 kg>s # mf = 0 (negligible) Ve = 900 m>s

Referring to the FBD of the control volume, Fig. a, along the x axis with

dVcv = 0 dt

(constant velocity), we have ΣFx = m

5 - 3 150 ( 103 ) (9.81) N 4 6 sin u

dVcv # # + maVcv - ( ma + mf ) Ve dt - c ( 208.33 m>s2 ) 2 = 0 + ( 2000 kg>s )( 208.33 m>s ) - ( 2000 kg>s + 0 )( 900 m>s )

1.4715 ( 106 ) sin u + 43.403 ( 103 ) c = 1.3833 ( 106 )

(1)

For level flight, u = 0°. Then - c ( 236.11 m>s ) 2 = 0 + ( 2000 kg>s )( 236.11 m>s ) - ( 2000 kg>s + 0 )( 900 m>s ) c = 23.817 Substituting this result into Eq. (1), 1.4715 ( 106 ) sin u + 43.403 ( 103 ) (23.817) = 1.3833 ( 106 ) Ans.

u = 13.74° = 13.7° W=

x

[150(103)](9.81)N

FD = CV2

FuL

Ans: 13.7°

(a)

711

6–106. A missile has a mass of 1.5 Mg (without fuel). If it consumes 500 kg of solid fuel at a rate of 20 kg>s and ejects it with a velocity of 2000 m>s relative to the missile, determine the velocity and acceleration of the missile at the instant all the fuel has been consumed. Neglect air resistance and the variation of gravity with altitude. The missile is launched vertically starting from rest.

v

SOLUTION The control volume consists of the missile and its contents as shown in Fig. a, which is accelerating upward. We consider steady flow of an ideal fluid relative to the control volume. The mass of the control volume as a function of time t is # M = Mo - mf t = 3 (1.5 + 0.5) ( 103 ) kg 4 - ( 20 kg>s ) t =

3 2 ( 103 )

- 20t 4 kg

#

Referring to the FBD of the control volume, Fig. a, with ma = 0 and Ve = 2000 m>s, + c ΣFy = m - 3 2 ( 103 ) - 20 t 4 (9.81) N =

W = [2)10 3) – 20t[(9.81) N

dVcv # # # + maVcv - ( ma + mf ) Ve dt

5 3 2 ( 103 )

- 20 t 4 kg 6

dV + 0 - ( 0 + 20 kg>s )( 2000 m>s ) dt

40 ( 103 ) dV = - 9.81 dt 2 ( 103 ) - 20 t

(1) (a)

Integrating this equation with the initial condition V = 0 at t = 0, L0

V

dV = V =

L0

40 ( 103 )

t

°

- 9.81¢dt

2 ( 103 ) - 20 t

3 - 2 ( 103 ) ln 3 2 ( 103 )

V = 2 ( 103 ) ln c

2 ( 10

3

2 ( 10

3

)

- 20 t 4 - 9.81t 4 2

) - 20 t

t 0

(2)

d - 9.81 t

The time required to consume all the fuel is

mf 500 kg = 25 s t = # = 20 kg>s mf Substituting this result into Eqs. (1) and (2) a =

40 ( 103 ) 2 ( 103 ) - 20(25)

V = 2 ( 103 ) ln £

- 9.81 = 16.9 m>s2

2 ( 103 ) 2 ( 103 ) - 20(25)

Ans.

§ - 9.81(25) = 330 m>s

Ans.

Ans: 330 m>s 712

6–107. The rocket has a weight of 65 000 lb, including the solid fuel. Determine the constant rate at which the fuel must be burned, so that its thrust gives the rocket a speed of 200 ft>s in 10 s starting from rest. The fuel is expelled from the rocket at a speed of 3000 ft>s relative to the rocket. Neglect air resistance and the variation of gravity with altitude. v

SOLUTION The control volume considered consists of the rocket and its contents as shown in Fig. a, which is accelerating upwards. We consider steady flow of an ideal fluid relative to the control volume. The mass of the control volume as a function of time t is 65000 lb # # # M = Mo - mf t = - mf t = (2018.63 - mf t) slug 32.2 ft>s2

#

Referring to the FBD of the control volume, Fig. a with ma = 0 and Ve = 3000 ft>s, + c ΣFy = m

dVcv # # # + maVcv - ( ma + mf ) Ve dt

W = (2018.63 – m ˙ f t)(32.2) lb

# dV # # - ( 2018.63 - mf t ) (32.2) = ( 2018.63 - mf t ) + 0 - ( 0 + mf )( 3000 ft>s ) dt # 3000 mf dV = # - 32.2 dt 2018.63 - mf t Integrating this equation with the initial condition V = 0 at t = 0 and the requirement V = 200 ft>s at t = 10 s, # 200 ft>s 10 s 3000 mf dV = ° # - 32.2¢dt 2018.63 - mf t L0 L0 200 =

3 - 3000 ln (2018.63

200 = 3000 ln ° ln °

(a)

10 s # - mf t) - 32.2 t 4 2 0

2018.63 # ¢ - 322 2018.63 - 10 mf

2018.63

# ¢ = 0.174 2018.63 - 10 mf 2018.63

0.174 # = e 2018.63 - 10 mf

# mf = 32.2 slug>s

Ans.

Ans: 32.2 slug>s 713

*6–108. The rocket is traveling upwards at 300 m>s and discharges 50 kg>s of fuel with a velocity of 3000 m>s measured relative to the rocket. If the exhaust nozzle has a cross-sectional area of 0.05 m2, determine the thrust of the rocket.

SOLUTION Take the rocket and its contents as the control volume. The thrust T needed to overcome W, FD, and m

dVcv is dt

# T = mfVe = ( 50 kg>s )( 3000 m>s ) = 150 ( 103 ) N = 150 kN Ans.

714

6–109. The balloon has a mass of 20 g (empty) and it is filled with air having a temperature of 20°C. If it is released, it begins to accelerate upwards at 8 m>s2. Determine the initial mass flow of air from the stem. Assume the balloon is a sphere having a radius of 300 mm.

8 m/s2 5 mm

W = 0.1962 N

SOLUTION The control volume considered is the balloon and the air contained within it, Fig. a. The initial flow measured relative to the accelerated control volume is treated as approximately steady. At T = 20 °C, ra = 1.202 kg>m3. The initial mass and weight of the balloon are m = mb + ma = 0.02 kg + ( 1.202 kg>m3 ) c = 0.1559 kg

4 p (0.3 m)3 d 3

W = mb g = (0.02 kg) ( 9.81 m>s2 ) = 0.1962 N

(a)

We neglect the weight of the air inside because it is counter-acted by buoyancy. Thus, ΣF = m

dVcv 0 + V r dV + Vf>cs(raVf>cs # dA) dt 0t L f>cv a L cv

cs

Writing the scalar components of this equation along the y axis by referring to the FBD of the control volume, Fig. a. + c ΣFy = m

dVcv + 0 + ( -Ve)ra(VeAe) dt

- 0.1962 N = (0.1559 kg) ( 8 m>s2 ) - ( 1.202 kg>m3 ) 3 p(0.0025 m)2 4 Ve2 Ve = 247.33 m>s

Thus, the initial mass flow is . me = raVeAe = ( 1.202 kg>m3 )( 247.33 m>s ) 3 p(0.0025 m)2 4 = 0.00584 kg>s

Ans.

Ans: 0.00584 kg>s 715

6–110. The rocket has an initial total mass m0, including . the fuel. When it is fired, it ejects a mass flow of me with a velocity of ve measured relative to the rocket. As this occurs, the pressure at the nozzle, which has a cross-sectional area Ae, is pe. If the drag force on the rocket is FD = ct, where t is the time and c is a constant, determine the velocity of the rocket if the acceleration due to gravity is assumed to be constant.

a0

SOLUTION The control volume considered is the entire rocket and its contents, which accelerates upward. We consider steady flow of an ideal fluid relative to the control volume. The FBD of the control volume is shown in Fig. a. Here, the mass of the # rocket as a function of time t is m = m0 - met. Thus, the weight of the rocket as # a function of time t is W = mg = (m0 - me t)g. The gage pressure force on the nozzle is Fe = peAe. dVcv 0 + Vf>cs rdV + Vf>cs rVf>cs # dA ΣF = m dt 0t L L cv cv

FD = ct

Writing the scalar component of this equation along the y axis by referring to Fig. a,

#

+ c ΣFy = (m0 - me t)

#

dV + 0 + dt

( -Ve )( reVeAe )

Here, me = reVeAe. Then

#

#

reAe - ct - ( m0 - met ) g = ( m0 - met )

#

dV # - meVe dt

peAe meVe dV ct = # + # # - g dt m0 - met m0 - met m0 - met

W = (m0 – m· e t)g

Integrating this equation with the initial condition V = 0 at t = 0, # V t peAe meVe ct dV = a # + # # - gbdt L L m0 - met m0 - met m0 - met 0

0

t peAe m0 c ct # # # V = e -Ve ln(m0 - met) - # ln(m0 - met) - c - # - # 2 ln ( m0 - met ) d - gt f 2 me me me 0

= Ve lna = aVe +

peAe m0c m0 m0 ct # b + # ln a # b + # - # ln # - gt 2 m0 - met me m0 - met me me m0 - met m0

peAe m0c m0 c # - # 2 blna # b + a # - gbt me m0 - met me me

Ans. F e = pe A e (a)

Ans: V = aVe + 716

pe Ae m0c m0 c - # 2 blna # # b + a # - gbt me me m0 - met me

6–111. The cart has a mass M and is filled with water that has an initial mass m0. If a pump ejects the water through a nozzle having a cross-sectional area A, at a constant rate of v0 relative to the cart, determine the velocity of the cart as a function of time. What is the maximum speed of the cart, assuming all the water can be pumped out? The frictional resistance to forward motion is F. The density of the water is r.

SOLUTION The control volume considered is the entire cart assembly as shown in Fig. a which is accelerating. Here, the mass flow rate of the water is . mf = rVeA Thus, the mass of the control volume as a function of time t is . m = (M + m0) - met = m + m0 - rVeAt . Referring to the FBD of the control volume, Fig. a with ma = 0, + ΣFx = m S

W

x F

dVcv . . . + maVcv - ( ma + mf)Ve dt

-F = (M + m0 - rVeAt)

dV + 0 - (0 + rVeA)Ve dt

N (a)

rVe2A - F dV = dt (M + m0) - rVeAt Integrating this equation with the initial condition V = 0 at t = 0, L0

V

t

dV =

rVe2A - F d dt L0 (M + m0) - rVeAt c

V = =

t rVe2A - F c ln(M + m0 - rVe At) d ` rVeA 0

rVe2A - F M + m0 lna b rVeA M + m0 - rVeAt

Ans.

m0 m0 , so t empty = . = me rVeA Vmax =

rVe2A - F M + m0 lna b rVe A M

Ans.

Ans: Vmax =

717

rVe2A - F M + m0 lna b rVe A M

*6–112. The 10-Mg helicopter carries a bucket containing 500 kg of water, which is used to fight fires. If it hovers over the land in a fixed position and then releases 50 kg>s of water at 10 m>s, measured relative to the helicopter, determine the initial upward acceleration of the helicopter as the water is being released.

a

SOLUTION The control volume considered consists of the helicopter and the bucket containing water as shown in Fig. a, which is accelerating upward. We consider steady flow of an ideal fluid relative to the control volume. The initial mass of the control volume is M0 = 10 ( 103 ) kg + 0.5 ( 103 ) kg = 10.5 ( 103 ) kg Since the helicopter is hovering before the water is released, its weight and the water’s initial weight are balanced by the uplift generated by the rotor blade. Therefore, they are not shown in the FBD of the control volume, Fig. a. Referring to # # the FBD of the control volume with ma = 0, mf = 50 kg>s, Ve = 10 m>s, + c ΣFy = m 0 = a0 =

dVcv . . . + maVcv - (ma + mf)Ve dt

3 10.5 ( 103 ) kg 4

dV + 0 - ( 0 + 50 kg>s )( 10 m>s ) dt

dV = 0.0476 m>s2 c dt

Ans.

(a)

718

6–113. The missile has an initial total weight of 8000 lb. The constant horizontal thrust provided by the jet engine is T = 7500 lb. Additional thrust is provided by two rocket boosters B. The propellant in each booster is burned at a constant rate of 80 lb>s, with a relative exhaust velocity of 3000 ft>s. If the mass of the propellant lost by the jet engine can be neglected, determine the velocity of the missile after the 3-s burn time of the boosters. The initial velocity of the missile is 375 ft>s. Neglect drag resistance.

B T

SOLUTION Take the missile and its contents as the control volume. We consider steady flow of an ideal fluid relative to the control volume. . At any instant t, the total mass of the missile is m = m0 = mf t. Referring to the free-body diagram of the missile in Fig. a. + ΣF = m S

dVcv . - mVe dt

T = 7500 lb

(a)

. dV . T = (m0 - mf t) - mfVe dt . T + mfVe dV = . dt m0 - mf t Integrating gives V

LV0

V`

. T + mfVe . bdt L0 m0 - mf t t

dV =

a

. t T + mfVe . ln(m0 - mf t) ` . mf V0 0 . T + mfVe m0 V = c lna . # b d + V0 m0 - mf t mf V

Here, m0 =

= -

8000 lb = 248.45 slug 32.2 ft>s2

80 lb>s . b = 4.969 slug>s mf = 2a 32.2 ft>s2 Ve = 3000 ft>s t = 3s

T = 7500 lb

V0 = 375 ft>s

Substituting these values into the expression of V, V = a

7500 lb + ( 4.969 slug>s )( 3000 ft>s ) 4.969 slug>s

V = 654.02 ft>s = 654 ft>s

blna

248.45 slug 248.45 slug - 4.969 slug>s(3 s)

b + 375 ft>s

Ans.

Ans: 654 ft>s 719

6–114. The rocket has an initial mass m0, including the fuel. For the comfort of the crew, it must maintain a constant upward acceleration a0. If the fuel is expelled from the rocket at a relative speed ve, determine the rate at which the fuel should be consumed to maintain the motion. Neglect air resistance, and assume that the gravitational acceleration is constant.

a0

SOLUTION The control volume considered is the entire rocket and its contents as shown in Fig. a, which accelerates upward. We consider steady flow of an ideal fluid relative to # the control volume. The FBD of the control volume is shown in Fig. a. Here, mf is a . function of time t. Also, mf is the negative of the rate of change of the rocket’s mass m. dm . . Thus, mf = . Applying Eq. (6–16) with ma = 0, dt dVcv . . . + c ΣFy = m + maVcv - ( ma + mf)Ve dt - mg = ma0 + 0 - a0 m(a0 + g) dm = dt Ve

m

dm Vb dt e

W = mg

(1)

t

a0 + g dm = dt Ve L L m 0

mo

ln

a0 + g m = -a bt m0 Ve

a0 + g m = e - a bt m0 Ve m = m0 e - a

a0 + g bt Ve

(a)

Substitute this result into Eq (1) a0 + g m0 dm . mf = = (a0 + g)e - a bt dt Ve Ve

Ans.

Ans: m0 dm . mf = = (a + g)e - (a0 + g)t>Ve dt Ve 0 720

6–115. The second stage B of the two-stage rocket weighs 2500 lb (empty) and is launched from the first stage with a velocity of 3000 mi>h. The fuel in the second stage weighs 800 lb. If it is consumed at the rate of 75 lb>s, and ejected with a relative velocity of 6000 ft>s, determine the acceleration of the second stage B just after the engine is fired. What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of gravity and air resistance.

B

SOLUTION

A

Take the second stage of the rocket and its contents as the control volume. We consider steady flow of an ideal fluid relative to the control volume. When second 2500 lb + 800 lb stage is fired, the total mass is m = = 102.48 slug. Since the effect 32.2 ft>s2 of gravity and air resistance can be neglected, ΣFy = 0. ΣFy = m

dVcv . - mfVe dt

0 = (102.48 slug) a =

dV 75 slug>s b ( 6000 ft>s ) - a dt 32.2

dV = 136.36 ft>s2 = 136 ft>s2 dt

Just before all the fuel is consumed, m = ΣFy = m

2500 lb = 77.64 slug 32.2 ft>s2

dV . - mfVe dt

0 = (77.64 slug) a =

Ans.

dV 75 - a slug>s b ( 6000 ft>s ) dt 32.2

dV = 180 ft>s2 dt

Ans.

Ans: When second stage is fired, a = 136 ft>s2. Just before all the fuel is consumed, a = 180 ft>s2. 721

7–1. As the top plate is pulled to the right with a constant velocity U, the fluid between the plates has a linear velocity distribution as shown. Determine the rate of rotation of a fluid element and the shear-strain rate of the element located at y.

U

y

u

h

SOLUTION

U

We consider steady flow of an ideal fluid. Referring to the velocity profile shown in Fig. a,

u h

U u = y h

u U = ; y h

y

And (a)

v = 0 The rate of rotation or average angular velocity of the fluid element is vz =

1 0v 0u 1 u U U a b = a0 - b = = b 2 0x 0y 2 h 2h 2h

Ans.

The rate of shear strain is

#

gxy =

0v 0u U + = 0x 0y h

Ans.

Ans: vz =

#

gxy 722

U 2h U = h

7–2. A flow is defined by its velocity components u = 1 4x2 + 4y2 2 m>s and v = ( - 8xy) m>s, where x and y are in meters. Determine if the flow is irrotational. What is the circulation around the rectangular region?

y

B

C

0.4 m

O

0.3 m

A

x

SOLUTION We consider ideal fluid flow. 0v 0 = ( - 8xy) = - 8y 0x 0x 0u 0 ( 4x2 + 4y2 ) = 8y = 0y 0y 1 0v 0u 1 b = ( -8y - 8y) = -8y vz = a 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Along edge OA, y = 0. Then u OA = 4x2 + 4 ( 02 ) = ( 4x2 ) m>s Along edge AB, x = 0.3 m. Then vAB = -8(0.3)y = ( - 2.4y) m>s Along edge BC, y = 0.4 m. Then u BC = 4x2 + 4 ( 0.42 ) = ( 4x2 + 0.64 ) m>s Along edge CO, x = 0. Then vCO = - 8(0)y = 0

Here, uOA and vAB are directed in the positive sense of dx and dy, respectively. uOA # dx and vCO # dy vAB # dy, are positive. However, uBC # dx and C C C C are negative since uBC and vCO are directed in the negative sense of dx and dy, respectively. Thus,

Γ = = =

C

V # ds =

L0

0.3 m

C

4x2dx +

uOA # dx + L0

C

vAB # dy -

0.4 m

( - 2.4y)dy -

L0

C

0.3 m

uBC # dx -

C

vCO # dy

( 4x2 + 0.64 ) dx - 0

0.4 m 0.3 m 4 3 0.3 m 4 x ` - 1.2y2 ` - a x3 + 0.64xb ` 3 0 3 0 0

= -0.384 m2 >s

Ans.

723

Ans: - 0.384 m2 >s

7–3. A uniform flow V is directed at an angle u to the horizontal as shown. Determine the circulation around the rectangular region.

0.3 m C

B 0.5 m

V A

O u

SOLUTION We consider ideal fluid flow. The component of V along edges OA and BC is

30˚

u = V cos u 30˚

and the component of U along edges AB and CO is

30˚

v = V sin u Here, u and v are directed in the same sense as dsOA and d AB , respectively. Thus,

30˚

u # ds BC and v # ds CO are negative since u and v are directed in the opposite C C sense to that of ds BC and ds CO, respectively. Thus, the circulation can be determined as Γ = =

C

V # ds =

L0

0.3 m

C

u # ds OA +

V cos udsOA +

L0

C

v # ds AB -

C

0.5 m

V sin udsAB -

u # ds BC -

L0

C

v # ds CO

0.3 m

V cos udsBC -

L0

0.5 m

V sin udsCO Ans.

= 0

Note: The irrotational flow always produces Γ = 0. In this case, this result is to be expected since the flow is irrotational.

Ans: Γ = 0 724

*7–4. The velocity within the eye of a tornado is defined by vr = 0, vu = (0.2r) m>s, where r is in meters. Determine the circulation at r = 60 m and at r = 80 m.

80 m

SOLUTION We consider ideal fluid flow. Since vu is always tangent to the circle, v # ds = vu ds. For r = 60 m, vu = 0.2(60) m>s = 12 m>s and ds = rdu = 60du. Γr = 60 m =

C

V # ds =

L0

2p

vuds =

L0

2p

2 12(60du) = 720u # 2p 0 = 1440p m >s

Ans.

For r = 80 m, vu = 0.2(80) m>s = 16 m>s, and ds = rdu = 80du. Γr = 80 m =

C

V # ds =

L0

2p

vuds =

L0

2p 2 16(80du) = 1280u # 2p 0 = 2560p m >s Ans.

725

60 m

7–5. Consider the fluid element that has dimensions in polar coordinates as shown and whose boundaries are defined by the streamlines with velocities v and v + dv. Show that the vorticity for the flow is given by z = - 1 v>r + dv>dr 2 .

D !

C v # dv

dr

A

B

r "u

SOLUTION We consider ideal fluid flow. The circulation of the flow around element ABCD can be determined from Γ =

C

V # ds

= vSAB + (v + dv)( - SCD) = v(r∆u) + (v + dv) 3 - (r + dr)∆u 4 = -vdr∆u - rdv∆u - dvdr∆u Neglect the second order terms Γ = - vdr∆u - rdv∆u = - ∆u(vdr + rdv) The area of the element, again, neglecting higher-order terms, is A = (r∆u)dr Thus, the vorticity is z =

- ∆u(vdr + rdv) Γ = A (r∆u)dr

= -a

dv v + b r dr

(Q.E.D)

726

v

7–6. Determine the stream and potential functions for the two-dimensional flow field if V0 and u are known.

y V0

u x

SOLUTION We consider ideal fluid flow. The velocity components are u = V0 cos u0

v = V0 sin u0

0u 0v + = 0 + 0 = 0 is satisfied, the establishment 0x 0y of a stream function is possible using the velocity components, Since the continuity equation

0c = u; 0y

0c = V0 cos u0 0y

Integrating this equation with respect to y, (1)

c = (V0 cos u0)y + f(x) Also, -

0c = v; 0x

-

0 3 ( V0 cos u0 ) y + f(x) 4 = V0 sin u0 0x

Integrating this equation with respect to x,

0 3f(x) 4 = -V0 sin u0 0x

f(x) = ( -V0 sin u0)x + C Setting C = 0 and substituting this result into Eq. (1), c = ( V0 cos u0 ) y - ( V0 sin u0 ) x c = V0 3 ( cos u0 ) y - ( sin u0 ) x 4

Ans.

0u 1 0v 1 a b = (0 - 0) = 0, the flow is indeed irrotational. Thus, the 2 0x 0y 2 potential function exists. Since, vz =

Using the velocity components, 0f = u; 0x

0f = V0 cos u0 0x

Integrating this equation with respect to x, f = ( V0 cos u0 ) x + f(y) Also,

0f = v; 0y

(1)

0 3 V cos u0x + f(y) 4 = V0 sin u0 0y 0

Integrating this equation with respect to y,

0 3f(y) 4 = V0 sin u0 0y

f(y) = ( V0 sin u0 ) y + C Setting C = 0 and substituting this result into Eq. 1, f = ( V0 cos u0 ) x + ( V0 sin u0 ) y f = V0 3 ( cos u0 ) x + ( sin u0 ) y 4

Ans. 727

Ans: c = V0 3(cos u0)y - (sin u0)x4

f = V0 3(cos u0)x + (sin u0)y4

7–7. A two-dimensional flow is described by the stream function c = 1 xy3 - x3y 2 m2 >s, where x and y are in meters. Show that the continuity condition is satisfied and determine if the flow is rotational or irrotational.

SOLUTION We consider ideal fluid flow. u =

0c 0 ( xy3 - x3y ) = ( 3xy2 - x3 ) m>s = 0y 0y

v = -

0c 0 ( xy3 - x3y ) = - ( y3 - 3x2y ) m>s = ( 3x2y - y3 ) m>s = 0x 0x

Then, 0u 0 ( 3xy2 - x3 ) = ( 3y2 - 3x2 ) s-1 = 0x 0x 0v 0 ( 3x2y - y3 ) = ( 3x2 - 3y2 ) s-1 = 0y 0y 0u 0 ( 3xy2 - x3 ) = 6xy s-1 = 0y 0y 0v 0 ( 3x2y - y3 ) = 6xy s-1 = 0x 0x This gives, 0u 0v + = 3y2 - 3x2 + 3x2 - 3y2 = 0 0x 0y Thus, the flow field satisfies the continuity condition, vz = =

1 0v 0u a b 2 0x 0y

1 ( 6xy - 6xy ) = 0 2 Ans.

The flow field is irrotational since vz = 0.

Ans: irrotational 728

*7–8. If the stream function for a flow is c = (3x + 2y), where x and y are in meters, determine the potential function and the magnitude of the velocity of a fluid particle at point (1 m, 2 m).

SOLUTION We consider ideal fluid flow. u =

0c 0 = (3x + 2y) = 2 m>s 0y 0y

v = -

0c 0 = - (3x + 2y) = -3 m>s 0x 0x

Since u and v are constant, the magnitude of the flow velocity at any point in the flow field is the same and is given by V = 2u2 + v2 = 2(2 m>s)2 + ( - 3 m>s)2 = 3.606 m>s = 3.61 m>s

Ans.

Applying, u =

0f ; 0x

2 =

0f 0x

Integrating with respect to x, f = 2x + f(y) Substituting this result into, v =

0f ; 0y

-3 =

0 [2x + f(y)] 0y

-3 = 0 +

0 [f(y)] 0y

Integrating with respect to y, f(y) = -3y + C Setting C = 0, thus Ans.

f = 2x - 3y

729

7–9. The velocity profile of a very thick liquid flowing along the channel of constant width is approximated as u = 13y2 2 mm>s, where y is in millimeters. Determine the stream function for the flow and plot the streamlines for c0 = 0, c1 = 1 mm2 >s, and c2 = 2 mm2 >s.

y u ! (3y2) mm/s

10 mm x

SOLUTION We consider ideal fluid flow. The x and y components of the constant flow velocity are u = ( 3y2 ) mm>s u =

0c ; 0y

v = 0

3y2 =

0c 0y

Integrating with respect to y, c = y3 + f(x) v = -

0c ; 0x

0 = 0 =

0 3 3 y + f(x) 4 0x

0 [f(x)] 0x

Integrating with respect to x f(x) = C Thus, c = y3 + C Setting C = 0, c = y3

Ans.

c0 = 0, y = 0 c1 = 1, y = 1 c2 = 2, y = 1.26 y

1.26 1 0

c2 = 2 c1 = 1 c0 = 0

x

Ans: c = y3 730

7–10. The velocity profile of a very thick liquid flowing along the channel of constant width is approximated as u = 13y2 2 mm>s, where y is in millimeters. Is it possible to determine the potential function for the flow? If so, what is it?

y u ! (3y2) mm/s

10 mm x

SOLUTION We consider ideal fluid flow. The x and y components of flow velocity are u = ( 3y2 ) mm>s

v = 0

Here, 0v = 0 0x 0u 0 = ( 3y2 ) = (6y) rad>s 0y 0y vz =

1 0v 0u 1 a b = (0 - 6y) = - 3y 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Thus, the potential function cannot be established since it requires the flow to be irrotational.

Ans: f cannot be established. 731

7–11. The liquid confined between two plates is assumed to have a linear velocity distribution as shown. Determine the stream function. Does the potential function exist?

1.2 m/s

A 10 mm

B 0.2 m/s

SOLUTION

1.2 m s

We consider ideal fluid flow. From the geometry of Fig. a, the x component of velocity is

u

u - 0.2 1.2 - 0.2 = ; y 0.01

u = (100y + 0.2) m>s

0.01 m y

Also, since the velocity distribution is directed along the x axis, v = 0. u =

0c ; 0y

100y + 0.2 =

0c 0y

0.2 m s (a)

Integrating with respect to y, c = 50y2 + 0.2y + f(x) Substituting this result into, v = -

0c ; 0x

0 = -

0 3 50y2 + 0.2y + f(x) 4 0x

0 [f(x)] = 0 0x Integrating with respect to x, f(x) = C Setting this constant equal to zero,

c = 50y2 + 0.2y

Ans.

Here 0v = 0; 0x

0u 0 = (100y + 0.2) = 100 rad>s 0y 0y

Thus, vz =

1 0v 0u 1 a b = (0 - 100) = -50 rad>s 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Therefore, it is not possible to establish the potential function.

Ans: c = 50y2 + 0.2y, f cannot be established. 732

*7–12. The liquid confined between two plates is assumed to have a linear velocity distribution as shown. If the pressure at the top surface of the bottom plate is 600 N>m2, detemine the pressure at the bottom surface of the top plate. Take r = 1.2 Mg>m3.

1.2 m/s

A 10 mm

B 0.2 m/s

SOLUTION

1.2 m s

We consider ideal fluid flow. From the geometry of Fig. a, the x component of velocity is

u

u - 0.2 1.2 - 0.2 = ; y 0.01

u = (100y + 0.2) m>s

0.01 m y

Also, since the velocity distribution is directed along the x axis, v = 0. Here, 0v = 0; 0x vz =

0u 0 = (100y + 0.2) = 100 rad>s 0y 0y

0.2 m s (a)

1 0v 0u 1 a b = (0 - 100) = -50 rad>s 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Thus, the Bernoulli equation can not be applied at points A and B. Instead, we will first apply the Euler equation along the x axis, 0u 0 0u 0 with = 100y + 0.2 = 0 and = (100y + 0.2) = 100 rad>s, 0x 0x 0y 0y -

1 0p 0u 0u = u + v r 0x 0x 0y 1 0 3 -rgy + f(x) 4 = 0 + 0 = 0 r 0x

0 [f(x)] = 0 0x

Integrating this equation with respect to x, f(x) = C Thus, p = - rgy + C At point B, y = 0 and p = 600 N>m2. Then, 600

N = - 1.2 ( 103 )( 9.81 m>s2 ) (0) + C m2 C = 600

N m2

Thus, p = ( - rgy + 600)

N m2

At point A, y = 0.01 m. Then, pA =

3 -1.2 ( 103 )( 9.81 m>s2 ) (0.01 m)

= 482.28

N = 482 Pa m2

+ 600 4

N m2

733

Ans.

7–13. A two-dimensional flow has a y component of velocity of v = (4y) ft>s, where y is in feet. If the flow is ideal, determine the x component of velocity and find the magnitude of the velocity at the point x = 4 ft, y = 3 ft. The velocity of the flow at the origin is zero.

SOLUTION We consider ideal fluid flow. In order to satisfy the continuity condition, 0u 0v + = 0 0x 0y Here, 0v 0 (4y) = 4 s-1 = 0y 0y Then, 0u + 4 = 0 0x 0u = -4 0x Integrating with respect to x, u = - 4x + f(y) Since ideal flow is irrotational,

and since

0(4y) 0v = = 0, 0x 0x

0v 0u = 0 0x 0y

0u = 0 0y u = g(x)

Thus, Ans.

u = ( -4x) ft>s At x = 4 ft and y = 3 ft, u = -4(4) = - 16 ft>s

v = 4(3) = 12 ft>s

V = 2u2 + v2 = 2 ( - 16 ft>s ) 2 + ( 12 ft>s ) 2 = 20 ft>s

Ans.

Ans: u = ( - 4x) ft>s V = 20 ft>s 734

7–14. A two-dimensional flow field is defined by its components u = (3y) m>s and v = (9x) m>s, where x and y are in meters. Determine if the flow is rotational or irrotational, and show that the continuity condition for the flow is satisfied. Also, find the stream function and the equation of the streamline that passes through point (4 m, 3 m). Plot this streamline.

SOLUTION

y 3

We consider ideal fluid flow. 0v 0x 0u 0y 0u 0x 0v 0y

0 (9x) 0x 0 (3y) = 0y 0 = (3y) 0x 0 = (9x) 0y =

= 9 rad>s = 3 rad>s 3.61 4

x

= 0 = 0

Thus, vz =

1 0v 0u 1 a b = (9 - 3) = 3 rad>s 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Also,

Ans.

0u 0v + = 0 + 0 = 0 0x 0y The flow satisfies the continuity condition. Thus, u =

0c ; 0y

3y =

0c 0y

Integrating with respect to y, c =

3 2 y + f(x) 2

Substituting this result into v = -

0c ; 0x

9x = -

0 3 2 c y + f(x) d 0x 2

-9x = 0 + Integrating with respect to x,

0 3f(x) 4 0x

9 f(x) = - x2 + C 2 Thus, setting C = 0, 3 9 1 c = y2 + a - x2 + C b = ( 3y2 - 9x2 ) 2 2 2 From the slope of the stream function,

Ans.

dy v 9x 3x = = = dx u 3y y y

L3 m

ydy =

x

3xdx L4 m

y2 y 3 x ` = x2 ` 2 3m 2 4m

y2 = 3x2 - 39

y = 23x2 - 39

Ans. 735

Ans: rotational 1 c = 1 3y2 - 9x2 2 2 y = 23x2 - 39

7–15. Water flow through the horizontal channel is defined by the stream function c = 2 1 x2 - y2 2 m2 >s. If the pressure at B is atmospheric, determine the pressure at point (0.5 m, 0) and the flow per unit depth in m2 >s.

y

B

c!0 1.5 m c ! 0.5 m2/s

SOLUTION We consider ideal fluid flow. The velocity components are u =

0c = ( -4y) m>s 0y

The continuity equation At point A (0.5 m, 0),

x

A

v = -

0c = ( - 4x) m>s 0x

1.5 m

0u 0v + = 0 + 0 = 0 is indeed satisfied. 0x 0y vA = -4(0.5) = -2 m>s

uA = 0 Thus,

VA = vA = -2 m>s At point B (1.5 m, 1.5 m), u B = - 4(1.5) = - 6 m>s

vB = -4(1.5) = -6 m>s

Thus, VB = 2uB2 + vB2 = 2 ( - 6 m>s ) 2 +

( - 6 m>s ) 2 = 272 m>s = 8.485 m>s

1 0v 0u 1 a b = 3 -4 - ( -4) 4 = 0, the flow is irrotational. Thus, 2 0x 0y 2 Bernoulli’s equation can be applied between two points on the different streamlines such as points A and B. Since vz =

pA pB VA2 VB2 = + gzA = = + gzB rw rw 2 2 Since the flow occurs in the horizontal plane, zA = zB. Also, pB = patm = 0. pA 1000 kg>m3

+

( 2 m>s ) 2 2

= 0 +

( 8.485 m>s ) 2 2

pA = 34 ( 103 ) Pa = 34 kPa

Ans.

c2 - c1 = 0.5 m2 >s - 0 = 0.5 m2 >s

Ans.

The flow per unit depth is

Ans: pA = 34 kPa

736

c2 - c1 = 0.5 m2 >s

*7–16. A flow field is defined by the stream function c = 2 1 x2 - y2 2 m2 >s, where x and y are in meters. Determine the flow per unit depth in m2 >s that occurs through AB, CB, and AC as shown.

y

B

C

3m

x

A 4m

SOLUTION We consider ideal fluid flow. At point A, x = 0, y = 0. Thus, cA = 2 ( 02 - 02 ) = 0 At point B, x = 4 m, y = 3 m. Thus cB = 2 ( 42 - 32 ) = 14 m2 >s

At point C, x = 0, y = 3 m

cC = 2 ( 02 - 32 ) = - 18 m2 >s

The flow rates per unit depth through AB, BC, and AC are qAB = cB - cA = 14 m2 >s - 0 = 14 m2 >s qBC = cB - cC = 14 m >s qAC

Ans.

( - 18 m >s ) = 32 m >s = cA - cC = 0 - ( - 18 m2 >s ) = 18 m2 >s 2

2

2

Note that the flow satisfies the continuity condition through ABC since

ΣV # A = 0 - qBC + qAB + qAC = - 32 m2 >s + 14 m2 >s + 18 m2 >s = 0

737

Ans. Ans.

7–17. A fluid has the velocity components shown. Determine the stream and potential functions. Plot the streamline for c0 = 0, c1 = 1 m2 >s, and c2 = 2 m2 >s.

y

4 m/s 3 m/s

SOLUTION

x

We consider ideal fluid flow. Here, the flow velocity has constant x and y components. u = 4 m>s

y

v = 3 m>s

ψ2 = 2 m2 s ψ1 = 1 m2 s ψ0 = 0

Applying u = Integrating with respect to y,

0c ; 0y

4 =

0c 0y

0.5 0.25 x

c = 4y + f(x) Substituting this result into v = -

0c ; 0x

3 =

0 34y + f(x) 4 0x

3 = -0 -

0 3f(x) 4 0x

0 3f(x) 4 = - 3 0x

Integrating with respect to x,

f(x) = -3x + C Setting C = 0, we get c = 4y + ( - 3x) Ans.

c = 4y - 3x Applying u = Integrating with respect to x,

0f ; 0x

4 =

0f 0x

f = 4x + f(y) Substituting this result into v =

Integrating with respect to y,

0f ; 0y

0 34x + f(y) 4 0y 0 3 = 0 + 3f(y) 4 0y 3 =

f(y) = 3y + C Setting C = 0, we get Ans.

f = 4x + 3y

Ans: c = 4y - 3x f = 4x + 3y 738

7–18. A two-dimensional flow field is defined by its components u = 1 2x2 2 ft>s and v = 1 - 4xy + x2 2 ft>s, where x and y are in feet. Determine the stream function, and plot the streamline that passes through point (3 ft, 1 ft).

SOLUTION We consider ideal fluid flow. 0u 0v + = 4x + ( - 4x) = 0 is satisfied, then the 0x 0y establishment of stream function is possible. Since the continuity equation

Using the definition of velocity components with respect to the stream function, 0c = u; 0y

0c = 2x2 0y

Integrating this equation with respect to y, c = 2x2y + f(x)

(1)

Also, -

0c = v; 0x

-

0 3 2x2y + f(x) 4 = - 4xy + x2 0x

-4xy -

0 3f(x) 4 = -4xy + x2 0x

0 3f(x) 4 = -x2 0x Integrating this equation with respect to x, 1 f(x) = - x3 + C 3 Substituting this result into Eq. (1), c = 2x2y -

1 3 x + C 3

Here, C is an arbitary constant that we will set equal to zero. The stream function can be expressed as c = 2x2y -

1 3 x 3

Ans.

For the streamline passing through point (3 ft, 1 ft), c = 2(3)2(1) Thus, 2x2y y =

1 3 (3) 3

1 3 x = 9 3

x3 + 27 6x2

739

7–18. Continued

The plot of stream function is shown in Fig. a. 0 ∞

x(m) y(m)

1 4.67

2 1.46

3 1

4 0.948

5 1.01

6 1.125

7 1.26

8 1.40

6x2 ( 3x2 ) - ( x3 + 27 ) (12x) dy = = 0 dx ( 6x2 ) 2 6x4 - 324x = 0 6x ( x3 - 54 ) = 0 x = 3.780 ft The corresponding y =

3.7803 + 27 6 ( 3.7802 )

= 0.945

y (m) 8 7 6 5 4 3 2

(3.78, 0.945)

1 x (m) 1

2

3

4

5

6

7

8

(a)

Ans: 1 c = 2x2y - x3 3 740

7–19. The stream function for a flow field is defined by c = 1 4>r 2 2 sin 2u. Show that continuity of the flow is satisfied, and determine the r and u velocity components of fluid particles at point r = 2 m, u = (p>4) rad. Plot the streamline that passes through this point.

SOLUTION We consider ideal fluid flow. Using the r and u velocity components with respect to stream function 1 0c 1 4 8 = a 2 b(2 cos 2u) = 3 cos 2u r 0u r r r 0c 8 8 = - a - 3 sin 2u b = 3 sin 2u vu = 0r r r vr =

The continuity equation

vr 0vr 1 0vu 8 24 + + = 4 cos 2u + a - 4 cos 2u b r 0r r 0u r r

16 cos 2u = 0 is indeed satisfied. r4 p At point r = 2 m and u = , 4 8 p vr = 3 cos c 2 a b d = 0 4 2 +

vu =

c =

Ans.

8 p sin c 2 a b d = 1 m>s 3 4 2

Ans.

4 p sin c 2 a b d = 1 4 22

Therefore, the stream function that passes through this point is 1 =

4 sin 2u r2

r 2 = 4 sin 2u The plot of this streamline is shown in Fig. a. y

π θ = 4

r=2m

x

π θ = 4

Ans: vr = 0 vu = 1 m>s

(a)

741

*7–20. A flow field has velocity components u = (x - y) ft>s and v = -(x + y) ft>s, where x and y are in feet. Determine the stream function, and plot the streamline that passes through the origin.

SOLUTION We consider ideal fluid flow. u =

0c 0c ; x - y = 0y 0y

Integrating this equation with respect to y, c = xy -

y2 + f(x) 2

Substituting this result into v = -

0c ; 0x

- (x + y) = x + y = y - 0 +

Integrating with respect to x,

y2 0 c xy + f(x) d 0x 2

0 3f(x) 4 0x

0 3f(x) 4 = x 0x f(x) =

x2 + C 2

Thus, setting C = 0, y2 x2 Ans. + 2 2 Evaluate c(x, y) at the origin, x = y = 0. This equation gives c = 0 - 0 + 0 = 0 Then, for c = 0, c = xy -

y2 x2 + xy = 0 2 2 x2 - y2 + 2xy = 0 The plot of this equation is shown in Fig. a. y

c= 0

x

c= 0

(a)

742

7–21. A flow is described by the stream function c = (8x - 4y) m2 >s, where x and y are in meters. Determine the potential function, and show that the continuity condition is satisfied and that the flow is irrotational.

SOLUTION We consider ideal fluid flow. u = v =

0c 0 = (8x - 4y) = -4 m>s 0y 0y

0c 0 = - (8x - 4y) = -8 m>s 0x 0x

Applying u =

0f 0f ; -4 = 0x 0x

Integrating with respect to x, f = -4x + f(y) Substituting this result into v =

0f 0 ; - 8 = [ - 4x + f(y)] 0y 0y -8 = 0 +

Integrating with respect to y,

0 f(y) 0y

0 3f(y) 4 = - 8 0y f(y) = -8y + C

Thus, f = - 4x + ( - 8y + C) Omitting the integration constant, Ans.

f = -4x - 8y Here, 0u 0x 0v 0y 0u 0y 0v 0x

0 ( -4) 0x 0 = ( - 8) 0y 0 = ( - 4) 0y 0 = ( - 8) 0x =

= 0 = 0 = 0 = 0

Then, 0u 0v + = 0 + 0 = 0 0x 0y The flow field satisfies the continuity condition. Also, vz =

1 0v 0u 1 a b = (0 - 0) = 0 2 0x 0y 2

Ans: f = -4x - 8y

The flow field is irrotational since vz = 0.

743

7–22. The stream function for a flow field is defined by c = 2r 3 sin 2u. Determine the magnitude of the velocity of fluid particles at point r = 1 m, u = (p>3) rad, and plot the streamlines for c1 = 1 m2 >s and c2 = 2 m2 >s.

SOLUTION

y (m)

We consider ideal fluid flow. The velocity components are vr =

1 0c ; r 0u

vr =

0c ; 0r

vu =

vu = -

The continuity equation

c = 2 m2 s

1 3 2r 3(2 cos 2u) 4 = ( 4r 2 cos 2u ) m>s r

( - 6r 2 sin 2u ) m>s

vr 0vr 1 0vu + + = 4r cos 2u + 8r cos 2u r 0r r 0u c = 1 m2 s

+ ( - 12r cos 2u) = 0 is indeed satisfied.

x (m)

At point r = 1 m, u = p>3 rad, (a)

p vr = 4 ( 12 ) cos c 2 a b d = - 2 m>s 3

p vu = -6 ( 12 ) sin c 2 a b d = - 5.196 m>s 3

Thus, the magnitude of the velocity is

V = 2vr2 + vu2 = 2 ( - 2 m>s ) 2 +

For c = 1 m2 >s,

1 = 2r 3 sin 2u

u(rad)

0

r(m)



For c = 2 m2 >s

p 12 1.00

p 6 0.833

2 = 2r 3 sin 2u

u(rad)

0

r(m)



p 12 1.260

p 6 1.049

( - 5.196 m>s ) 2 = 5.57 m>s r3 =

p 4 0.794

r3 = p 4 1.00

Ans.

1 2 sin 2u p 3 0.833

5p 12 1.00

p 2 ∞

1 2 sin 2u p 3 1.049

5p 12 1.260

p 2 ∞

Ans: 5.57 m>s 744

7–23. An ideal fluid flows into the corner formed by the two walls. If the stream function for this flow is defined by c = 1 5 r 4 sin 4u 2 m2 >s, show that continuity for the flow is satisfied. Also, plot the streamline that passes through point r = 2 m, u = (p>6) rad, and find the magnitude of the velocity at this point.

y

45!

x

SOLUTION We consider ideal fluid flow. p rad, 6

For the stream f unction passing through point r = 2 m, u = p c = 5 ( 24 ) sin c 4 a b d = 4023 m2 >s 6

y (m)

Thus, the stream function passing through this point is 4023 = 5r 4 sin 4u r 4 sin 4u = 823 The plot of this streamline is shown in Fig. a u(rad)

0

p 24

p 12

p 8

p 6

r(m)



{2.29

{2.0

{1.93

{2.0

5p 24 {2.29

p 4 ∞

45°

x (m)

The radial and transverse components of velocity are vr =

1 0c 1 = 3 5r 4(4 cos 4u) 4 = 20r 3 cos 4u r 0u r vu = -

(a)

0c = -20r 3 sin 4u 0r

The continuity equation vr 0vr 1 0vu + + r 0r r 0u = 20r 2 cos 4u + 60r 2 cos 4u + ( - 80r 2 cos 4u) = 0 is indeed satisfied At the point r = 2 m, u = p>6 rad, p vr = 20 ( 23 ) cos c 4 a b d = -80 m>s 6

p vu = - 20 ( 23 ) sin c 4 a b d = - 138.56 m>s 6 Thus, the magnitude of the velocity is V = 2vr2 + vu2 = 2 ( - 80 m>s ) 2 +

( - 138.56 m>s ) 2 = 160 m>s

Ans.

Ans: 160 m>s 745

*7–24. The horizontal flow confined by the walls is defined 4 by the stream function c = c 4r 4>3 sin a u b d m2 >s, where r 3 is in meters. Determine the magnitude of the velocity at point r = 2 m, u = 45°. Is the flow rotational or irrotational? Can the Bernoulli equation be used to determine the difference in pressure between the two points A and B?

r 45!

We consider ideal fluid flow. 4 4 4 1 0c 1 0 4 1 4 16 1 4 = a4r 3 sin u b = c 4r 3 a cos u b d = r 3 cos u r 0u r 0u 3 r 3 3 3 3

vu = -

0c 4 0 4 16 1 4 = - a4r 3 sin u b = - r 3 sin u 0r 0r 3 3 3

At the point r = 2 m, u = 45°. vr =

16 1 4 ( 23 ) cos c (45°) d = 3.360 m>s 3 3

vu = -

16 1 4 ( 23 ) sin c (45°) d = - 5.819 m>s 3 3

Thus, the magnitude of the velocity is

V = 2vr2 + vu2 = 2 ( 3.360 m>s ) 2 +

( - 5.819 m>s ) 2 Ans.

= 6.72 m>s vr =

0f 16 1 0f 4 ; r 3 cos u = 0r 3 3 0r

Integrating with respect to r, 4

f = 4r 3 cos

4 u + f(u) 3

Substituting this result into, vu =

1 0f ; r 0u -

-

4 4 1 0 4 16 1 r 3 sin u = c 4r 3 cos u + f(u) d 3 3 r 0u 3

1 16 1 4 4 4 1 0 r 3 sin u = ( 4r 3 ) c - sin u d + 3f(0) 4 3 3 3 3 r 0u

Integrating with respect to u,

u O

SOLUTION vr =

B

A

0 3f(u) 4 = 0 0u f(u) = C

Setting the constant equal to zero, 4 4 f = 4r 3 cos u 3

Since the potential function can be established, the flow is irrotational. Therefore, the Bernoulli equation is applicable between any two points in the flow, including points A and B.

746

7–25. The horizontal flow between the walls is defined by 4 the stream function c = c 4r 4>3 sin a u b d m2 >s, where r is 3 in meters. If the pressure at the origin O is 20 kPa, determine the pressure at r = 2 m, u = 45°. Take r = 950 kg>m3.

B

A r 45!

u O

SOLUTION We consider ideal fluid flow. vr =

4 4 4 1 0c 1 0 4 1 4 16 1 4 a4r 3 sin u b = c 4r 3 a cos u b d = = r 3 cos u r 0u r 0u 3 r 3 3 3 3

vu = -

0c 4 0 4 16 1 4 = - a4r 3 sin u b = - r 3 sin u 0r 0r 3 3 3

At point A, where r = 2 m u = 45°, vr =

16 1 4 ( 23 ) cos c (45°) d = 3.360 m>s 3 3

vu = -

16 1 4 ( 23 ) sin c (45°) d = -5.819 m>s 3 3

At the origin O, where r = 0,

vr = v u = 0

Thus, the magnitude of the velocity at these two points is Vo = VA = 2vr2 + vu2 = 2 ( 3.360 m>s ) 2 +

( - 5.819 m>s ) 2

= 6.720 m>s

vr =

0f ; 0r

0f 16 1 4 r 3 cos u = 3 3 0r

Integrating with respect to r, 4 4 f = 4r 3 cos u + f(u), 3

Substituting this result into, vu = -

1 0f ; r 0u

-

4 16 1 4 1 0 4 r 3 sin u = c 4r 3 cos u + f(u) d 3 3 r 0u 3

1 16 1 4 4 4 1 0 r 3 sin u = ( 4r 3 ) c - sin u d + 3f(u) 4 3 3 3 3 r 0u

0 3f(u) 4 = 0 0u

Integrating with respect to u,

f(u) = C Setting this constant equal to zero, 4 4 f = 4r 3 cos u 3 Since the potential function can be established, the flow is irrotational. Therefore, the Bernoulli equation is applicable between any two points in the flow, including points A and O.

p pO VO2 V2 = + + r r 2 2 p 950 kg>m3

+

(6.720 m>s)2 2

N m2 = + 0 950 kg>m3 20 ( 103 )

p = - 1.448 ( 103 ) Pa = - 1.45 kPa

Ans. 747

Ans: -1.45 kPa

7–26. The flat plate is subjected to the flow defined by the stream function c = 3 8r 1>2 sin (u>2) 4 m2 >s. Sketch the streamline that passes through point r = 4 m, u = p rad, and determine the magnitude of the velocity at this point.

r

SOLUTION We consider ideal fluid flow. For the stream function passing point r = 4 m and u = p rad, 1 p c = 8 ( 42 ) sin = 16 2 Thus, the stream function passing through this point is 1 u 16 = 8r 2 sin 2 1 u 2 r sin = 2 2 The plot of this function is shown in Fig. a u(rad) r(m)

0 ∞

7p 4p 3p 2p 5p 5p p 6 3 3 2 3 6 59.71 16.0 8.00 5.33 4.29 4.00 4.29 5.33 8.00 16.0 p 6

p 3

p 2

2p ∞

x (m)

ψ = 16

y (m)

(a)

748

11p 6 59.71

u

7–26. Continued

The radial and transverse components of velocity are 1 1 u 1 1 0c = c 8r 2 a cos b d = vr = r 0u r 2 2

4 cos 1

r2

0c 1 1 u vu = = - a8r - 2 sin b = 0r 2 2 0vr vr 1 0vu + + = The continuity equation r 0r r 0u

4 cos

u 2

3

r2

u 2

4 sin

u 2

1

r2 - 2 cos

+ ±

3

r2

u 2

2 cos ≤ + ±-

3

r2

u 2

≤ = 0

is indeed satisfied. At point r = 4 m, u = p rad, 4 cos vr =

vu = -

p 2

1

42 p 4 sin 2 1

42

= 0

= -2 m>s

Thus, the magnitude of the velocity is Ans.

V = vu = 2 m>s

Ans: 2 m>s 749

7–27. An A-frame house has a window A on its right side. If the stream function that models the flow as this side is defined as c = (2r 1.5sin 1.5 u) ft 2 >s, show that continuity of the flow is satisfied, and then determine the wind speed past the window located at r = 10 ft, u = (p>3) rad. Sketch the streamline that passes through this point.

A r ! 10 ft u ! 120"

SOLUTION We consider ideal fluid flow. For the stream function passing through point r = 10 ft, u = p>3 rad, c = 2 ( 101.5 ) sin 31.5 ( p>3 ) 4 = 2 ( 101.5 )

Thus, the stream function passing through this point is 2(101.5) = 2r 1.5 sin 1.5u r 1.5 sin 1.5u = 101.5 The plot of this stream function is shown in Fig. a u(rad)

0

r(ft)



p 12

p 6

p 4

18.97 12.60 10.54

p 3 10

5p 12

p 2

7p 12

10.54 12.60 18.97

2p 3 ∞

y (ft)

ψ = 2 (101.5)

120° =

x (ft)

2π rad 3

(a)

750

7–27. Continued

The radial and transverse components of velocity are vr =

1 1 0c 1 = 3 2r 1.5(1.5 cos 1.5u) 4 = 3r 2 cos 1.5u r 0u r

vu = -

The continuity equation + a-

4.5 cos 1.5u 1

r2

0c 1 = - 3r 2 sin 1.5u 0r

vr 0vr 1 0vu 3 cos 1.5u 1.5 cos 1.5u + + = + 1 1 r 0r r 0u 2 r r2

b = 0 is indeed satisfied.

At the window, where r = 10 ft, u = 120° = 1

vr = 3 ( 10 2 ) cos c1.5 a 1

2 p rad, 3

2p b d = - 9.4868 ft>s 3

v u = - 3 ( 10 2 ) sin c 1.5a

Thus, the magnitude of the wind velocity is

2p bd = 0 3

Ans.

V = vr = 9.49 ft>s

Ans: 9.49 ft>s 751

*7–28. The stream function for a horizontal flow near the corner is c = (8xy) m2 >s, where x and y are in meters. Determine the x and y components of the velocity and the acceleration of fluid particles passing through point (1 m, 2 m). Show that it is possible to establish the potential function. Plot the streamlines and equipotential lines that pass through point (1 m, 2 m).

y

A

B

x

SOLUTION We consider ideal fluid flow. For the stream function passing through point (1 m, 2 m), c = 8(1)(2) = 16 Thus, 16 = 8xy

y =

Using the velocity components, u =

0c = (8x) m>s 0y

The continuity equation

v = -

2 x

0c = ( -8y) m>s 0x

0u 0v + = 8 + ( - 8) = 0 is indeed satisfied. 0x 0y

The acceleration components are ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 8x(8) + ( - 8y)(0) = (64x) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 8x(0) + ( - 8y)( - 8) = (64y) m>s2 At point (1 m, 2 m), u = 8(1) = 8 m>s S v = - 8(2) = - 16 m>s = 16 m>s T

Ans.

ax = 64(1) = 64 m>s2 S ay = 64(2) = 128 m>s2 c

Ans.

1 0v 0u 1 a b = (0 - 0) = 0, the flow is irrotational. Therefore, it is 2 0x 0y 2 possible to establish potential function. Using the velocity components,

Since vz =

0f = u; 0x

0f = 8x 0x

Integrating this equation with respect to x f = 4x2 + f(y)

(1)

752

7–28. Continued

Also 0f = v; 0y

0 3 4x2 + f(y) 4 = - 8y 0y

0 3 f(y) 4 = - 8y 0y Integrating this equation with respect to y

f(y) = - 4y2 + c

Setting C = 0 and substituting this result into Eq. 1 f = 4x2 - 4y2 f = 4(x2 - y2)

Ans.

The potential function passing through point (1 m, 2 m), Then f = 4 ( 12 - 22 ) = - 12 Thus - 12 = 4 ( x2 - y2 )

y2 = x2 + 3

The plots of the stream and potential functions are shown in Fig. a. y (m)

For the stream function x(m) y(m)

0 ∞

1 2

2 1

3 0.667

4 0.5

5 0.4

6 0.333

6

For the potential function x(m) y(m)

0 1.73

1 2

7

y2 = x2 + 3

2 2.65

3 3.46

4 4.36

5 5.29

6 6.24

(Potential function)

5 4 3 2

y = 2x (Stream function)

1 x (m) 0

1

2

3

4 (a)

753

5

6

7–29. The stream function for horizontal flow near the corner is defined by c = (8xy) m2 >s, where x and y are in meters. Show that the flow is irrotational. If the pressure at point A (1 m, 2 m) is 150 kPa, determine the pressure at point B (2 m, 3 m). Take r = 980 kg>m3.

y

A

B

x

SOLUTION We consider ideal fluid flow. Using the velocity components, u =

0c = (8x) m>s 0y

The continuity equation

v = -

0c = -( - 8y) m>s 0x

0u 0v + = 8 + ( - 8) = 0 is indeed satisfied 0x 0y

At point A(1 m, 2 m), u A = 8(1) = 8 m>s

vA = - 8(2) = -16 m>s

Thus, VA2 = uA2 + vA2 = ( 8 m>s ) 2 + At point B(2 m, 3 m) u B = 8(2) = 16 m>s

( - 16 m>s ) 2 = 320 m2 >s2

vB = -8(3) = -24 m>s

Thus, V B2 = u B2 + vB2 = ( 16 m>s ) 2 +

( - 24 m>s ) 2 = 832 m2 >s2

1 0v 0u 1 a + b = (0 + 0) = 0, the flow is irrotational. Therefore, 2 0x 0y 2 Bernoulli’s equation is applicable to two points located on the different streamlines such as points A and B.

Since vz =

pA pB VA2 VB2 + gzA = + gzB + + r r 2 2 Since the flow is in the horizontal plane, zA = zB = z. pA pB VA2 VB2 + gz = + gz + + r r 2 2 pB = pA +

r ( VA2 - VB2 ) 2

pB = 150 ( 103 ) N>m2 +

980 kg>m3

= - 100.88 ( 103 ) N>m2

2

( 320 m2 >s2 - 832 m2 >s2 ) Ans.

= - 101 kPa

Ans: - 101 kPa 754

7–30. A flow has velocity components u = 1 2x2 2 ft>s and v = ( -4xy + 8) ft>s, where x and y are in feet. Determine the magnitude of the acceleration of a particle located at point (3 ft, 2 ft). Is the flow rotational or irrotational? Also, show that continuity of flow is satisfied.

SOLUTION We consider ideal fluid flow. 0v 0 = ( - 4xy + 8) = -4y 0x 0x 0u 0 = ( 2x2 ) = 0 0y 0y 0u 0 = ( 2x2 ) = 4x 0x 0x 0v 0 = ( - 4xy + 8) = -4x 0y 0y Thus, vz =

1 0v 0u 1 a b = ( -4y - 0) = - 2y 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Also

0u 0v + = 4x + ( - 4x) = 0 0x 0y The flow satisfies the continuity condition. Since

0u 0v = = 0 (steady flow) 0t 0t ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 2x2 )( 4x ) + ( - 4xy + 8)(0) = 8x3 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 2x2( -4y) + ( -4xy + 8)( -4x) = - 8x2y + 16x2y - 32x = 8x2y - 32x Thus, at x = 3 ft, y = 2 ft ax = 8 ( 33 ) = 216 ft>s2 ay = 8(32)(2) - 32(3) = 48 ft>s2 The magnitude of the acceleration is a = 2ax2 + ay2 = 2 ( 216 ft>s2 ) 2 + ( 48 ft>s2 ) 2 = 221.27 ft>s2 = 221 ft>s2

Ans. Ans: rotational a = 221 ft>s2

755

7–31. The potential function for a flow is f = 1 x2 - y2 2 ft 2 >s, where x and y are in feet. Determine the magnitude of the velocity of fluid particles at point A (3  ft, 1 ft). Show that continuity is satisfied, and find the streamline that passes through point A.

SOLUTION We consider ideal fluid flow. From the velocity components u = Since vz =

0f = 2x 0x

v =

0f = -2y 0y

1 0v 0u 1 a b = (0 - 0) = 0, the flow is indeed irrotational. 2 0x 0y 2

At point (3 ft, 1 ft),

u = 2(3) = 6 ft>s

v = -2(1) = -2 ft>s

Then the magnitude of the velocity is V = 2u2 + v2 = 2 ( 6 ft>s ) 2 +

( - 2 ft>s ) 2 = 6.32 ft>s

Ans.

0y 0v 0u 0u = 2 and = - 2. Since + = 2 + ( -2) = 0, the potential function 0x 0y 0x 0y satisfies the continuity condition. Using the velocity components. Here

0c = u; 0y

0c = 2x 0y

Integrating this equation with respect to y, (1)

c = 2xy + f(x) Also, -

0c = v; 0x

-

2y +

0 32xy + f(x) 4 = -2y 0x 0 3f(x) 4 = 2y 0x

0 3f(x) 4 = 0 0x Integrating this equation with respect to x, f(x) = C

Setting C = 0, and substituting this result into Eq. 1 c = 2xy For the streamline passing through point (3 ft, 1 ft), c = 2(3)(1) = 6 Thus, 6 = 2xy Ans.

xy = 3

Ans: V = 6.32 ft>s xy = 3 756

*7–32. The flow around the bend in the horizontal channel can be described as a free vortex for which vr = 0, vu = (8>r) m>s, where r is in meters. Show that the flow is irrotational. If the pressure at point A is 4 kPa, determine the pressure at point B. Take r = 1100 kg>m3.

A B C 2m

r u

0.5 m

SOLUTION We consider ideal fluid flow. vr =

0f ; 0r

0 =

0f 0r

Integrating with respect to r, f = f(u) Substituting this result into vu =

1 0f ; r 0u

8 1 0 = 3f(u)4 r r 0u

0 3f(u)4 = 8 0u

Integrating with respect to u,

f = f(u) = 8u + C Since the potential function can be established, the flow is irrotational and Bernoulli’s equation can be applied between points A and B. The magnitude of velocity at A and B is VA = (vu)A =

8 8 = = 3.2 m>s rA 2.5

VB = (vu)B =

8 8 = = 4 m>s rB 2

Applying the Bernoulli equation, pB pA VB2 VA2 = + + r r 2 2 pB 1100 kg>m3

+

( 4 m>s ) 2 2

N m2 = + 1100 kg>m3 4 ( 103 )

( 3.2 m>s ) 2 2 Ans.

pB = 832 Pa

757

7–33. The velocity components for a two-dimensional flow are u = (8y) ft>s and v = (8x) ft>s, where x and y are in feet. Determine if the flow is rotational or irrotational, and show that continuity of flow is satisfied.

SOLUTION We consider ideal fluid flow. 0v 0 = (8x) = 8 rad>s 0x 0x 0u 0 = (8y) = 8 rad>s 0y 0y 0u 0 = (8y) = 0 0x 0x 0v 0 = (8x) = 0 0y 0y Thus, vz =

1 0v 0u 1 a b = (8 - 8) = 0 2 0x 0y 2

Since vz = 0, the flow is irrotational. Also,

Ans.

0u 0v + = 0 + 0 = 0 0x 0y The flow satisfies the continuity condition.

Ans.

Ans: irrotational 758

7–34. The velocity components for a two-dimensional flow are a u = (8y) ft>s and v = (8x) ft>s where x and y are in feet. Find the stream function and the equation of the streamline that passes through point (4 ft, 3 ft). Plot this streamline.

SOLUTION

y

We consider ideal fluid flow.

3

0c u = ; 0y

0c 8y = 0y

Integrating with respect to y, 4

c = 4y2 + f(x) Then, v = -

0c ; 0x

8x = -

Integrating with respect to x,

2.65

0 3 4y2 + f(x)4 0x

-8x = 0 +

x

0 3f(x)4 0x

f(x) = -4x2 + C Thus, c = 4y2 +

( - 4x2 + C ) = 4 ( y2 - x2 ) + C

Omitting the constant, C, c = 4 ( y2 - x2 )

Ans.

From the slope of the stream function, dy v 8x x = = = dx u 8y y y

L3 ft

x

ydy =

L4 ft

xdx

y2 y x2 2 x 2 = 2 3 ft 2 4 ft y2 = x2 - 7

Also, at (4 ft, 3 ft),

y = { 2x2 - 7

Ans.

c = 4 ( (3)2 - (4)2 ) = - 28 Then, 4 ( y2 - x2 ) = - 28 y = { 2x2 - 7

Ans: c = 4 1y2 - x22 y = { 2x2 - 7 759

7–35. The stream function for the flow field around the 90° corner is c = 8r 2 sin 2u. Show that the continuity of flow is satisfied. Determine the r and u velocity components of a fluid particle located at r = 0.5 m, u = 30°, and plot the streamline that passes through this point. Also, determine the potential function for the flow.

y

r x

SOLUTION We consider ideal fluid flow. From the r and u velocity components to, vr =

1 0c 1 = ( 8r 2 ) (2 cos 2u) = 16r cos 2u r 0u r

0c = -(16r sin 2u) = -16r sin 2u 0r vr 0vr 1 0vu The continuity equation + + = 16 cos 2u + 16 cos 2u + r 0r r 0u ( -32 cos 2u) = 0 is indeed satisfied. vu =

At point r = 0.5 m, u = 30°, Ans.

vr = 16(0.5) cos 32(30°) 4 = 4 m>s

Since sin u = Therefore,

Ans.

vu = -16(0.5) sin 32(30°) 4 = - 6.93 m>s

y x y 2xy x , cos u = , then sin 2u = 2 sin u cos u = 2 a ba b = 2 . r r r r r

At point r = 0.5 m, u = 30°,

c = 8r 2 a

2xy r2

b = 16xy

x = r cos u = (0.5 m) cos 30° = y = r sin u = (0.5 m) sin 30° =

23 m 4 1 m 4

Then c = 16°

23 1 ¢ a b = 23 4 4

Thus, the streamline passing through this point is 23 = 16xy y =

23 16x

760

7–35. Continued

y (m) 1.1 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.25 0.2 0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

x (m)

0.4333 (a)

The plot of this streamline is shown in Fig. a x(m)

0

0.1

0.2

y(m)



1.08

0.541 0.361 0.271 0.217

0.3

0.4

0.5

0.6

0.7

0.8

0.180

0.155

0.135 0.120 0.108

0.9

1.0

The velocity components with respect to stream function are u =

0c = 16x 0y

v = -

0c - 16y 0x

1 0v 0u 1 a b = (0 - 0) = 0, the flow is irrotational. Therefore, it is 2 0x 0y 2 possible to established the potential function using the velocity components, Since vz =

0f 0f = u; = 16x 0x 0x Integrating this equation with respect to x, f = 8x2 + f(y)

(1)

Also, 0f = v; 0y

0 3 8x2 + f(y)4 = -16y 0y

0 3f(y) 4 = - 16y 0y

Integrating this equation with respect to y

f(y) = - 8y2 + C Setting C = 0, and substituting this result in Eq. 1 f = 8x2 - 8y2 + C f = 8 ( x2 - y2 )

Ans.

761

Ans: vr = 4 m>s vu = - 6.93 m>s f = 8 1x2 - y 2 2

*7–36. The stream function for a concentric flow is defined by c = - 4r 2. Determine the velocity components vr and vu, and vx and vy. Can the potential function be established? If so, what is it?

y

r u

SOLUTION We consider ideal fluid flow. Using the r, u velocity components vr =

1 0c 1 = (0) = 0 r 0u r

vu = -

Ans.

0c = - 3 - 4(2r) 4 = 8r 0r

Ans.

Since r 2 = x2 + y2, then c = - 4 ( x2 + y2 ) . Using the velocity components, vx = u =

0c = -4(2y) = - 8y 0y

vy = v = -

Ans.

0c = - 3 -4(2x) 4 = 8x 0x

Ans.

1 0v 0u 1 a b = 38 - ( -8) 4 = 16 ≠ 0, the flow is rotational. 2 0x 0y 2 Therefore, the potential function cannot be established. Since vz =

762

x

7–37. A fluid has velocity components u = (x - y) ft>s and v = -(x + y) ft>s, where x and y are in feet. Determine the stream and potential functions. Show that the flow is irrotational.

SOLUTION We consider ideal fluid flow. 0u 0v + = 1 + ( - 1) = 0 is satisfied, then the 0x 0y establishment of a stream function is possible. Using the velocity components, Since the continuity equation

0c = u; 0y

0c = x - y 0x

Integrating this equation with respect to y, 1 2 y + f(x) 2

c = xy -

(1)

Also, -

0c = v; 0x

-

0 1 = c xy - y2 + f(x) d = - (x + y) 0x 2

y - 0 +

0 3f(x) 4 = x + y 0x

0 3f(x) 4 = x 0x Integrating this equation with respect to x, f(x) =

1 2 x + C 2

Setting C = 0 and substituting this result into Eq. 1, c = xy c =

1 2 1 y + x2 2 2

1 2 ( x - y2 + 2xy ) 2

Ans.

Using the definition of velocity components with respect to the potential function, 0f = u; 0x

0f = x - y 0x

Integrating this equation with respect to x, f =

1 2 x - xy + f(y) 2

(2)

Also, 0f = v; 0y

0 1 2 c x - xy + f(y) d = -(x + y) 0y 2 -x +

0 3f(y) 4 = - x - y 0y

0 3f(y) 4 = - y 0y

763

7–37. Continued

Integrating this equation with respect to y f(y) =

1 2 y + C 2

Setting C = 0 and substituting this result into Eq. 2, f =

1 2 1 x - xy - y2 2 2

f =

1 2 ( x - y2 - 2xy ) 2

Ans.

Here 0v 0u = - 1 and = -1 0x 0y Since vz =

1 0v 0u 1 a b = 3 -1 - ( -1) 4 = 0, the flow is irrotational. 2 0x 0y 2

Ans:

764

c =

1 2 ( x - y2 + 2xy ) 2

f =

1 2 ( x - y2 - 2xy ) 2

7–38. A fluid has velocity components u = (2y) ft>s and v = (2x - 10) ft>s, where x and y are in feet. Determine the stream and potential functions.

SOLUTION We consider ideal fluid flow. 0u 0v + = 0 + 0 = 0 is satisfied, then the 0x 0y establishment of stream function is possible. Since the continuity equation

Using the definition of velocity components with respect to stream function, 0c = u; 0y

0c = 2y 0y

Integrating this equation with respect to y, c = y2 + f(x)

(1)

Also, -

0c = v; 0x

-

0 = 0x

3 y2

+ f(x) 4 = 2x - 10

0 3f(x) 4 = 10 - 2x 0x

Integrating this equation with respect to x,

f(x) = 10x - x2 + C Setting C = 0, and substituting this result into Eq. 1, c = y2 + 10x - x2 c = y2 - x2 + 10x

Ans.

1 0v 0u 1 a b = (2 - 2) = 0, the flow is irrotational. Therefore the 2 0x 0y 2 potential function exists. Since vz =

Using the definition of velocity components with respect to potential function, 0f 0f = u; = 2y 0x 0x Integrating this equation with respect to x, (1)

f = 2xy + f(y) Also, 0f = v; 0y

0 = 32xy + f(y) 4 = 2x - 10 0y

2x +

0 3f(y) 4 = 2x - 10 0y

0 3f(y) 4 = -10 0y

Integrating this equation with respect to y,

f(y) = - 10y + C Setting C = 0, and substituting this result into Eq. 1, f = 2xy - 10y Ans.

f = 2y(x - 5)

Ans: c = y2 - x2 + 10x f = 2y(x - 5)

765

7–39. A fluid has velocity components u = (x - 2y) ft>s and v = -(y + 2x) ft>s, where x and y are in feet. Determine the stream and potential functions.

SOLUTION We consider ideal fluid flow. 0u 0v + = 1 + ( -1) = 0 is satisfied, then the 0x 0y establishment of the stream function is possible using the velocity components, Since the continuity equation

0c = u; 0y

0c = x - 2y 0y

Integrating this equation with respect to y, c = xy - y2 + f(x)

(1)

Also, -

0c = v; 0x

-

0 3 xy - y2 + f(x) 4 = - y - 2x 0x

-y -

0 3f(x) 4 = - y - 2x 0x

0 3f(x) 4 = 2x 0x

Integrating this equation with respect to x,

f(x) = x2 + C Setting C = 0 and substituting this result into Eq. 1, c = x2 - y2 + xy

Ans.

1 0v 0u 1 a b = 3 - 2 - ( - 2) 4 = 0, the flow is irrotational. Therefore, 2 0x 0y 2 the potential function exists. Since vz =

Using the definition of velocity components with respect to potential function, 0f = u; 0x

0f = x - 2y 0x

Integrating this equation with respect to x f =

1 2 x - 2xy + f(y) 2

(1)

Also, 0f = v; 0y

0 1 = c x2 - 2xy + f(y) d = -y - 2x 0y 2

- 2x +

0 3f(y) 4 = - y - 2x 0y

0 3f(y) 4 = - y 0y

Integrating this equation with respect to y

1 f(y) = - y2 + C 2 Setting C = 0 and substituting this result into Eq. 1, 1 2 1 x - 2xy - y2 2 2 1 2 f = ( x - y2 ) - 2xy 2

Ans: c = x2 - y2 + xy

f =

Ans.

766

f =

1 2 ( x - y2 ) - 2xy 2

*7–40. A fluid has velocity components u = 2 1 x2 - y2 2 m>s and v = ( - 4xy) m>s, where x and y are in meters. Determine the stream function. Also show that the potential function exists, and find this function. Plot the streamlines and equipotential lines that pass through point (1 m, 2 m).

SOLUTION 0u 0v + = 4x + ( - 4x) = 0 is satisfied, the stream 0x 0y

Since the continuity equation function can be established. Using the velocity components

0c = u; 0y

0c = 2x2 - 2y2 0y

Integrating this equation with respect to y, c = 2x2y -

2 3 y + f(x) 3

(1)

Also, -

0c = v; 0x

-

0 2 3 2x2y - y3 + f(x) 4 = -4xy 0x 3

4xy +

0 3f(x) 4 = 4xy 0x

0 3f(x) 4 = 0 0x

Integrating this equation with respect to x,

f(x) = C Setting C = 0 and substituting this result into Eq. 1, c = 2x2y c = Since vz =

2 3 y 3

2 y ( 3x2 - y2 ) 3

Ans.

1 0v 0u 1 a b = 3 - 4y - ( - 4y) 4 = 0, the flow is irrotational. Thus, the 2 0x 0y 2

potential function exists. Using the velocity components, 0f = u; 0x

0f = 2x2 - 2y2 0x

Integrating this equation with respect to x f =

2 3 x - 2xy2 + f(y) 3

(2)

767

*7–40. Continued

Also, 0f = v; 0y

0 2 3 c x - 2xy2 + f(y) d = -4xy 0y 3

- 4xy +

0 3f(y) 4 = -4xy 0y

0 3f(y) 4 = 0 0y

Integrating this equation with respect to y,

f(y) = C Setting C = 0 and substituting this result into Eq. 2, f =

2 3 x - 2xy2 3

2 x ( x2 - 3y2 ) Ans. 3 For stream function and potential functions passing through point (1 m, 2 m), f =

c =

2 4 (2) 3 3(1)2 - 22 4 = 3 3

y(m) 7

2 22 f = (1) 3 12 - 3 ( 22 ) 4 = 3 3

Thus, the streamline is

-

6 5

4 2 = y ( 3x2 - y2 ) 3 3 x2 =

4

3

y - 2 3y

3 2

and the equipotential line is -

22 2 = x ( x2 - 3y2 ) 3 3 y2 =

1 x(m)

x3 + 11 3x

–6

–5

–4

–3

–2

–1 0

For the streamline y(m)

1.26

2

3

4

5

6

x(m)

0

{1

{1.67

{2.27

{2.86

{3.45

For the equipotential line x(m)

0

1

1.77

3

4

5

6

0.25

0.50

y(m)



{2

{1.77

{2.05

{2.50

{3.01

{3.55

{3.83

{2.72

The plot of these two functions is shown in Fig. a 768

1

2

3

4

5

6

7–41. If the potential function for a two-dimensional flow is f = (xy) m2 >s, where x and y are in meters, determine the stream function and plot the streamline that passes through the point (1 m, 2 m). What are the velocity and acceleration of fluid particles that pass through this point?

SOLUTION We consider ideal fluid flow. u =

0f 0 = (xy) = y 0x 0x

v =

0f 0 = (xy) = x 0y 0y

u =

0c ; 0y

y =

0c 0y y(m)

Integrating with respect to y, c =

y2 + f(x) 2

ψ = 1.5

Substituting this result into, v = -

0c ; 0x

x = -

0 y2 + f(x) d c 0x 2

x = -0 -

0 3f(x) 4 0x

3

0 3f(x) 4 = -x 0x

Integrating with respect to x,

f(x) = -

x(m)

(a)

x2 + C 2

Setting C = 0, c = c =

y2 x2 + a- b 2 2 1 2 ( y - x2 ) 2

Ans.

When x = 1 m, and y = 2 m. Then, c =

1 2 ( 2 - 12 ) = 1.5 2

For the streamline defined by c = 1.5, its equation is 1 2 ( y - x2 ) = 1.5 2 y2 = x2 + 3 y = 2x2 + 3

Ans:

The plot of this streamline is shown in Fig. a.

c = 769

1 2 ( y - x2 ) 2

7–41. Continued

At c = 1 m, y = 2 m u = y = 2 m>s v = x = 1 m>s V = 2 1 2 m>s 2 2 +

1 1 m>s 2 2 = 2.24 m>s

ax =

0u 0u 0u + u + v = 0 + (2)(0) + 1(1) = 1 m>s2 0t 0x 0y

ay =

0v 0v 0v + u + v = 0 + 2(1) + 1(0) = 2 m>s2 0t 0x 0y

a = 2 1 1 m>s2 2 +

1 2 m>s2 2 2

Ans.

= 2.24 m>s2

Ans.

770

7–42. Determine the potential function for the twodimensional flow field if V0 and u are known.

y

x

V0 u

SOLUTION We consider ideal fluid flow. The velocity components are v = -V0 cos u0

u = V0 sin u0

1 0v 0u 1 a b = (0 - 0) = 0, the flow is indeed irrotational. Thus, the 2 0x 0y 2 potential function exists. Since vz =

Using the velocity components, 0f = u; 0x

0f = V0 sin u0 0x

Integrating this equation with respect to x, (1)

f = V0 sin u0 x + f(y) Also,

0f = v; 0y

0 3 ( V0 sin u0 ) x + f(y) 4 = -V0 cos u0 0y 0 3f(y) 4 = -V0 cos u0 0y

Integrating this equation with respect to y

f(y) = - ( V0 cos u0 ) y + C Setting C = 0, and substituting this result into Eq. 1, f = ( V0 sin u0 ) x - ( V0 cos u0 ) y f = V0 3 ( sin u0 ) x - ( cos u0 ) y 4

Ans.

771

Ans: f = V0 3 ( sin u0 ) x - ( cos u0 ) y 4

7–43. The potential function for a flow is f = 1 x2 - y2 2 ft 2 >s, where x and y are in feet. Determine the magnitude of the velocity of fluid particles at point (3 ft, 1 ft). Show that continuity is satisfied, and find the streamline that passes through this point.

SOLUTION We consider ideal fluid flow. u =

0f 0 = ( x2 - y2 ) = (2x) ft>s 0x 0x

v =

0f 0 = ( x2 - y2 ) = ( -2y) ft>s 0y 0y

Thus, at x = 3 ft, y = 1 ft, u = 2(3) = 6 ft>s v = - 2(1) = - 2 ft>s Then, the magnitude of the flow velocity is

Applying

V = 2u2 + v2 = 2 ( 6 ft>s ) 2 + u =

Integrating with respect to y,

0c ; 0y

( - 2 ft>s ) 2 = 6.32 ft>s

2x =

Ans.

0c 0y

c = 2xy + f(x) Substituting this result into the second of Eq. (8–8), v = -

0c ; 0x

Integrating with respect to x,

0 32xy + f(x) 4 0x 0 2y = 2y + 3f(x) 4 0x 0 3f(x) 4 = 0 0x

- 2y = -

f(x) = C Setting C = 0, then c = 2xy At x = 3 ft, y = 1 ft, c = 2(3)(1) = 6 So the streamline through (3 ft, 1 ft) is 6 = 2xy y =

3 x

Ans.

772

7–43. Continued

Here,

Then,

0u 0 = (2x) = 2 ft>s 0x 0x 0v 0 = ( - 2y) = - 2 ft>s 0y 0y 0u 0v + = 2 + ( -2) = 0 0x 0y

Thus, the flow field satisfies the continuity condition as required.

Ans: V = 6.32 ft>s 3 y = x 773

*7–44. A fluid has velocity components of u = (10xy) m>s and v = 5 1 x2 - y2 2 m>s, where x and y are in meters. Determine the stream function, and show that the continuity condition is satisfied and that the flow is irrotational. Plot the streamlines for c0 = 0, c1 = 1 m2 >s, and c2 = 2 m2 >s.

SOLUTION We consider ideal fluid flow. u = Integrating with respect to y,

0c ; 0y

10xy =

0c 0y

c = 5xy2 + f(x) Substituting this result into 0c ; 0x

0 3 5xy2 + f(x)4 0x 0 5x2 - 5y2 = - 5y2 3f(x)4 0x 0 3f(x)4 = - 5x2 0x Integrating with respect to x, v = -

5 ( x2 - y2 ) = -

5 f(x) = - x3 + C 3 Setting C = 0, then 5 c = 5xy2 + a - x3 b 3 5 2 = x ( 3y - x2 ) 3

Ans.

Here, 0u 0x 0v 0y 0v 0x 0u 0y

Then,

0 (10xy) = (10y) s-1 0x 0 = 3 5 ( x2 - y2 ) 4 = ( -10y) s-1 0y 0 = 3 5 ( x2 - y2 ) 4 = (10x) s-1 0x 0 = (10xy) = (10x) s-1 0y

=

0u 0v + = 10y + ( - 10y) = 0 0x 0y The flow field satisfies the continuity condition. Applying, vz =

1 0v 0u 1 a b = (10x - 10x) = 0 2 0x 0y 2

774

7–44. Continued

The flow field is irrotational since vz = 0. When c = 0,

or

5 x ( 3y2 - x2 ) = 0 3 1 y = { x 23 x = 0

When c = 1, 5 x ( 3y2 - x2 ) = 1 3 y= { When c = 2,

1 x2 + A 5x 3

5 x ( 3y2 - x2 ) = 2 3 y= {

2 x2 + A 5x 3 Ç=2

y

Ç=1

Ç=0

30! x

775

7–45. A fluid has velocity components of u = 1y2 - x22 m>s and v = (2xy) m>s, where x and y are in meters. If the pressure at point A (3 m, 2 m) is 600 kPa, determine the pressure at point B (1 m, 3 m). Also what is the potential function for the flow? Take g = 8 kN>m3.

SOLUTION We consider ideal fluid flow. Applying u =

0f ; 0x

y2 - x2 =

0f 0x

Integrating with respect to x, f = xy2 -

x3 + f(y) 3

Substituting this result into, v =

0f ; 0y

0 x3 c xy2 + f(y) d 0y 3

2xy =

0 3f(y)4 0y

2xy = 2xy - 0 +

Integrating with respect to y,

0 3f(y)4 = 0 0y f(y) = C

Setting C = 0, we have f = xy2 -

x3 3

Ans.

Since the potential function can be established, the flow is irrotational. Thus, the Bernoulli equation can be applied from point A to B. The x and y components of the velocity at these points are u A = ( 22 - 32 ) m>s = - 5 m>s

vA = 32(3)(2) 4 m>s = 12 m>s

u B = ( 3 - 1 ) m>s = 8 m>s 2

2

vB = 32(1)(3) 4 m>s = 6 m>s

Thus, the magnitude of the velocity at these two points is

VA = 2uA2 + vA2 = 2 ( - 5 m>s ) 2 + ( 12 m>s ) 2 = 13 m>s VB = 2uB2 + vB2 = 2 ( 8 m>s2 ) + ( 6 m>s2 ) = 10 m>s

Applying the Bernoulli equation for ideal fluid from A to B, pB pA VB2 VA2 = + + g g 2g 2g pB 8(103) N>m3

+

(10 m>s)2 2 ( 9.81 m>s2 )

N (13 m>s)2 m2 = + 8 ( 103 ) N>m3 2 ( 9.81 m>s2 ) 600 ( 103 )

N pB = 628.13 ( 10 ) 2 = 628 kPa m 3

Ans: Ans.

f = xy2 -

x3 3

pB = 628 kPa 776

7–46. The potential function for a horizontal flow is f = 1 x3 - 5xy2 2 m2 >s, where x and y are in meters. Determine the magnitude of the velocity at point A (5 m, 2 m). What is the difference in pressure between this point and the origin? Take r = 925 kg>m3.

SOLUTION We consider ideal fluid flow. Since the flow is described by the potential function, the flow is definitely irrotational. Therefore, the Bernoulli equation can be applied between any two points. u =

0f 0 = ( x3 - 5xy2 ) = 3x2 - 5y2 0x 0x

v =

0f 0 = ( x3 - 5xy2 ) = -10xy 0y 0y

At point A, x = 5 m, y = 2 m. Thus, u A = 3 ( 52 ) - 5 ( 22 ) = 55 m>s

vA = -10(5)(2) = - 100 m>s

At the origin O, x = 0 and y = 0. Thus, u 0 = 3 ( 02 ) - 5 ( 02 ) = 0

v0 = -10(0)(0) = 0

The magnitude of the velocity at A and O is VA = 2u A2 + vA2 = 2 ( 55 m>s ) 2 +

( - 100 m>s ) 2 = 114.13 m>s = 114 m>s Ans. V0 = 0

Since the flow occurs in the horizontal plane, no change in elevation takes place. Thus, the elevation term can be excluded. Applying the Bernoulli equation for an ideal fluid from O to A, pA pO VO2 VA2 = + + r r 2 2 pO - pA =

r ( VA2 - VO2 ) 2

= °

925 kg>m3 2

¢ 3 ( 114.13 m>s ) 2 - 02 4

= 6.024 ( 106 ) Pa = 6.02 MPa

Ans.

Ans: VA = 114 m>s pO - pA = 6.02 MPa 777

7–47. A fluid has velocity components of u = (10xy) m>s and v = 5 1 x2 - y2 2 m>s, where x and y are in meters. Determine the potential function, and show that the continuity condition is satisfied and that the flow is irrotational.

SOLUTION We consider ideal fluid flow. u =

0f ; 0x

10xy =

0f 0x

Integrating with respect to x, f = 5x2y + f(y) Substituting this result into the second of Eq. (8–12), v =

0f ; 0y

0 3 5x2y + f(y) 4 0y 0 5x2 - 5y2 = 5x2 + 3f(y) 4 0y

5 ( x2 - y2 ) =

Integrating with respect to x,

0 3f(y) 4 = - 5y2 0y

5 f(y) = - y3 + C 3 Setting C = 0, then 5 f = 5x2y + a - y3 b 3 =

5 y ( 3x2 - y2 ) 3

Ans.

Here, 0u 0 = (10xy) = (10y) s-1 0x 0x 0v 0 = 3 5 ( x2 - y2 ) 4 = ( -10y) s-1 0y 0y

0v 0 = 3 5 ( x2 - y2 ) 4 = (10x) s-1 0x 0x 0u 0 = (10xy) = (10x) s-1 0y 0y

Then, 0u 0v + = 10y + ( - 10y) = 0 0x 0y The flow field satisfies the continuity condition. Applying, vz =

Ans:

1 0v 0u 1 a - b = (10x - 10x) = 0 2 0x 0y 2

5 y ( 3x2 - y2 ) 3 1 c = ( y2 - x2 ) 2

f =

The flow field is irrotational since vz = 0.

778

*7–48. A velocity field is defined as u = 2 1 x2 + y2 2 ft>s, v = (- 4xy) ft>s. Determine the stream function and the circulation around the rectangle shown. Plot the streamlines for c0 = 0, c1 = 1 ft 2 >s, and c2 = 2 ft 2 >s.

y

0.6 ft

x 0.5 ft

SOLUTION We consider ideal fluid flow. 0u 0v = = 4x + ( -4x) = 0 is satisfied, the stream 0x 0y function can be established. Using the definition of the velocity components, with respect to stream function,

Since the continuity equation

0c = u; 0y

0c = 2 ( x2 + y2 ) 0y

Integrating this equation with respect to y, c = 2 ax2y +

1 3 y b + f(x) 3

(1)

Also, -

0c = v; 0x

-

0c = - 4xy 0x

0 1 c 2 ax2y + y3 b + f(x) d = -4xy 0x 3 4xy +

0 3f(x) 4 = 4xy 0x

0 3 f(x) 4 = 0 0x

Integrating this equation with respect to x,

f(x) = C Substituting this result into Eq. (1), c = 2 ax2y +

1 3 y b + C 3

c = 2y ax2 +

1 2 y b 3

C is an arbitary constant. If we set it equal to zero then the stream function can be expressed as Ans.

779

*7–48. Continued

0 = 2y°x2 +

For c = 0,

y2 ¢ 3

since x2 +

y2 ≠ 0, 3

then

y = 0 For c = 1 ft 2 >s,

1 = 2y °x2 + x2 =

For c = 2 ft 2 >s,

3 - 2y3 6y

0 6 y 6 1.145

2 = 2y°x2 +

x2 =

y2 ¢ 3

6 - 2y3 6y

y2 ¢ 3 0 6 y 6 1.442

The plot of these streamlines are shown in Fig. a. For c = 1 ft 2 >s y(ft)

0

0.25

0.50

0.75

1.00

x(ft)

{∞

{ 1.407

{ 0.957

{ 0.692

{ 0.408

y(ft) x(ft)

For c = 2 ft 2 >s 0

y(ft)

x(ft)

{∞

1.145

y(ft)

1.442

0

x(ft)

0

0.25

0.50

0.75

1.00

1.25

{ 1.995 { 1.384 { 1.070 { 0.816 { 0.528

y(ft) 1.50

ψ = 2 ft2/s

1.25 1.0 0.75

ψ = 1 ft2/s

0.5 0.25

ψ =0 x(ft)

–2.0

–1.75

–1.5

–1.25

–1.0

–0.75

–0.5

–0.25 0

0.25 (a)

780

0.5

0.75

1.0

1.25

1.5

1.75

2.0

*7–48. Continued

The circulation can be determined using Γ = L0

=

=

L0

0.5 ft

0.5 ft

udx +

L0

2 ( x2 + 0 ) dx +

0.6 ft

L0 +

=

C

vdy +

V # ds

L0

0.5 ft

0.6 ft

- 4(0.5)ydy + L0

L0

u( -dx) + L0

0.5 ft

v( -dy)

0.5 ft

2 ( x2 + 0.62 ) ( -dx)

0.6 ft

- 4(0)y( - dy)

0.6 ft 0.5 ft 2 3 0.5 ft 2 x ` - y2 ` - a x3 + 0.72xb ` + 0 3 0 3 0 0

= - 0.72 ft 2 >s

Ans.

781

7–49. If the potential function for a two-dimensional flow is f = (xy) m2 >s, where x and y are in meters, determine the stream function, and plot the streamline that passes through the point (1 m, 2 m). What are the x and y components of the velocity and acceleration of fluid particles that pass through this point?

SOLUTION We consider ideal fluid flow. Using the velocity components, 0f = y 0x

u =

v =

0f = x 0y

1 0v 0u 1 a b = (1 - 1) = 0, the flow is indeed irrotational. Also, 2 0x 0y 2 0u 0v since the continuity equation + = 0 + 0 = 0 is satisfied, the establishment 0x 0y of a stream function is possible, Since vz =

0c = u; 0y

0c = y 0y

Integrating this equation with respect to y, 1 2 y + f(x) 2

c = Also, -

0c = v; 0x

-

(1)

0 1 2 c y + f(x) d = x 0x 2

Integrating this equation with respect to x

0 3f(x) 4 = -x 0x

1 f(x) = - x2 + C 2 Setting C = 0 and substituting this result into Eq. 1, c =

1 2 1 y - x2 2 2

c =

1 2 ( y - x2 ) 2

Ans.

For the streamline passing through point (1 m, 2 m) c =

1 2 3 ( 2 - 12 ) = 2 2

Thus, 3 1 = ( y2 - x2 ) 2 2 y2 = x2 + 3

782

7–49. Continued

The plot of this streamline is shown in Fig. a 0

x(m)

y(m) 1.73

1

2

3

4

5

6

2

2.65

3.46

4.36

5.29

6.24

y(m) 7 6 5 4 3 2 1 x(m) 0

1

2

3

4

5

6

(a)

At point (1 m, 2 m), the velocity components are u = 2 m>s

Ans.

v = 1 m>s

The acceleration components are ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + y(0) + x(1) = x = 1 m>s2 ay =

Ans.

0v 0v 0v + u + v 0t 0x 0y

= 0 + y(1) + x(0) = y = 2 m>s2

Ans. Ans: u = 2 m>s, v = 1 m>s ax = 1 m>s2 ay = 2 m>s2

783

7–50. A two-dimensional flow is described by the potential function f = 1 8x2 - 8y2 2 m2 >s, where x and y are in meters. Show that the continuity condition is satisfied, and determine if the flow is rotational or irrotational. Also, establish the stream function for this flow, and plot the streamline that passes through point (1 m, 0.5 m).

SOLUTION We consider ideal fluid flow. Here

Since

02f 2

0x condition.

+

02f 0y2

0f = 16x 0x

02f

0f = -16y 0y

02f

0x2 0y2

= 16 = - 16

= 16+ ( - 16) = 0, the potential function f satisfies the continuity

The velocity components can be determined using u =

0f ; 0x

u = (16x) m>s

v =

0f ; 0y

v = ( - 16y) m>s

Then 0v = 0 0x

0u = 0 0y

Thus vz =

1 0v 0u a b = 0 2 0x 0x

Since vz = 0, the flow is indeed irrotational since all flows that can be described by a potential function are irrotational. Using the definition of velocity components with respect to the stream function, 0c = u; 0y

0c = 16x 0y

Integrating this equation with respect to y, (1)

c = 16xy + f(x) Also, -

0c = v; 0x

-

0 316xy + f(x) 4 = - 16y 0x

-16y -

0 3f(x) 4 = - 16y 0x 0 3f(x) 4 = 0 0x

784

7–50. Continued

Integrating this equation with respect to x f(x) = C Setting C = 0 and substituting this result into Eq. 1, Ans.

c = 16xy For the streamline passing through point (1 m, 0.5 m), c = 16(1)(0.5) = 8 Thus, 8 = 16xy;

y =

1 2x

The plot of this stream function is shown in Fig. a x(m)

0

0.5

1

1.5

2

2.5

3

y(m)



1

0.5

0.333

0.25

0.2

0.167

y(m) 1.5

1

0.5

x(m) 0

0.5

1

1.5

2

2.5

3

(a)

Ans: c = 16xy 785

7–51. The y component of velocity of a two-dimensional irrotational flow that satisfies the continuity condition is v = 1 4x + x2 - y2 2 ft>s, where x and y are in feet. Find the x component of velocity if u = 0 at x = y = 0.

SOLUTION We consider ideal fluid flow. In order for the flow to be irrotational, vz = 0. vz = Here,

1 0v 0u a b = 0 2 0x 0y

0v 0 ( 4x + x2 - y2 ) = (4 + 2x) rad>s = 0x 0x Thus, 1 0u c (4 + 2x) d = 0 2 0y 0u = 4 + 2x 0y

Integrating with respect to y, u = 4y + 2xy + f(x) In order to satisfy the continuity condition 0u 0v + = 0 0x 0y Here, 0u 0 0 = 34y + 2xy + f(x) 4 = 2y + 3f(x) 4 0x 0x 0x 0v 0 + ( 4x + x2 - y2 ) = - 2y 0y 0y

Then, 2y +

Integrating with respect to x,

0 3f(x) 4 - 2y = 0 0x 0 3f(x) 4 = 0 0x f(x) = C

Thus, u = 4y + 2xy + C = 2y(2 + x) + C At y = x = 0, u = 0. Then C = 0, and so Ans.

u = 2y(2 + x)

Ans: u = 2y(2 + x) 786

*7–52. The flow has a velocity of V = {(3y + 8)i} ft>s, where y is vertical and is in feet. Determine if the flow is rotational or irrotational. If the pressure at point A is 6 lb>ft2, determine the pressure at the origin. Take g = 70 lb>ft3.

y A 3 ft O

SOLUTION We consider ideal fluid flow. The x and y components of velocity are u = (3y + 8) ft>s

v = 0

Here, 0u = 0; 0x

0u 0 = (3y + 8) = 3 rad>s 0y 0y

Thus, vz =

1 0v 0u 1 a b = (0 - 3) = -1.5 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Thus, the Bernoulli equation can not be applied from O to A. Instead, we will first apply Euler’s equation along the y axis. Here, 0v 0v = = 0. 0x 0y Then, -

1 0p 0v 0v - g = u + v = 0 r 0y 0x 0y

0p = -rg = -g 0y Integrating with respect to y, p = - gy + f(x) Substituting this result into the Euler equation along the x axis, with 0u 0 0u 0 = (3y + 8) = 0 and = (3y + 8) = 3 rad>s, 0x 0x 0y 0y -

1 0p 0u 0u = u + v r 0x 0x 0y

1 0 3 - gy + f(x) 4 = 0 + 0 r 0x 0 3f(x) 4 = 0 0x

787

x

7–52. Continued

Integrating with respect to x, f(x) = C Thus, p = - gy + C At point A, y = 3 ft and p = 6

6

lb . Then, ft 2

lb lb = a - 70 2 b(3ft) + C ft 2 ft C = 216 lb>ft 2

Thus, p = ( - gy + 216) lb>ft 2 At point O, y = 0 Thus, pO = 3 - (70)(0) + 2164 = 216

lb ft 2

lb ft 2

Ans.

788

7–53. A tornado has a measured wind speed of 12 m>s a distance of 50 m from its center. If a building has a flat roof and is located 10 m from the center, determine the uplift pressure on the roof. The building is within the free vortex of the tornado. The density of the air is ra = 1.20 kg>m3.

SOLUTION We consider ideal fluid flow. Since the tornado is a free vortex flow, its velocity components are vr = 0

k r

vu =

Thus V = vu =

k r

It is required that at r = 50 m, V = 12 m>s. Therefore 12 m>s =

k ; 50 m

Then V = a

At r = 10 m,

V =

k = 600 m2 >s

600 b m>s r

600 = 60 m>s 10

Since free vortex flow is irrotational, Bernoulli’s equation can be applied between two points on the different streamlines such as two points on two circular streamlines of radius r = ∞ and r = 10 m. At r = ∞ , V∞ = 0 and pB = 0. Since the flow occurs in the horizontal plane, the gravity term can be excluded. p∞ p V 2 V2 + ∞ = + ra ra 2 2 0 + 0 =

p 1.20 kg>m3

+

( 60 m>s ) 2 2 Ans.

p = 2160 Pa = - 2.16 kPa The negative sign indicates that suction develops.

Ans: -2.16 kPa 789

7–54. Show that the equation that defines a sink will satisfy continuity, which in polar coordinates is written as 0(vr r) 0r

+

0(vu) 0u

= 0.

SOLUTION We consider ideal fluid flow. q For sink flow, vr = and vu = 0. Then, 2pr 0 ( vrr ) 0r

=

q q 0 0 caa- b = 0 b(r) d = 0r 2pr 0r 2p 0vu = 0 0u

Thus, 0(vr r) 0r

+

0vu = 0 + 0 = 0 0u

(Q.E.D)

790

7–55. A source at O creates a flow from point O that is described by the potential function f = (8 ln r) m2 >s, where r is in meters. Determine the stream function, and specify the velocity at point r = 5 m, u = 15°.

O

30! u

SOLUTION We consider ideal fluid flow. The r and u components of velocity are vr =

0f 0 8 = (8 ln r) = a b m>s 0r 0r r

vu =

1 0f 1 0 = (8 ln r) = 0 r 0u r 0u

Applying, vr =

1 0c ; r 0u

8 1 0c = r r 0u

Integrating this equation with respect to u, c = 8u + f(r) Substituting this result, vu = -

0c ; 0r

0 = -

0 38u + f(r) 4 0r

0 = 0Integrating this equation with respect to r,

0 3f(r) 4 0r

f(r) = C Thus, c = 8u + C Setting C = 0, Ans.

c = 8u At r = 5 m, u = 15°, vr =

8 = 1.6 m>s 5

vu = 0 Thus, the magnitude of the velocity is V = 2vr2 + vu2 = 2 ( 1.6 m>s ) 2 + (0)2 = 1.60 m>s2

Ans.

Ans: c = 8u V = 1.60 m>s2 791

*7–56. Combine a source of strength q with a free counterclockwise vortex, and sketch the resultant streamline for c = 0.

SOLUTION We consider ideal fluid flow. Superimposing the streamlines of a source and a free vortex, q c = u - k ln r 2p For c = 0, q u - k ln r 2p q ln r = u 2pk 0 =

q

e ln r = e 2pku q

r = e 2pku This equation represents a logarithmic spiral from the source and its plot is shown in Fig. a.

source

1

(a)

792

7–57. A free vortex is defined by its stream function c = ( - 240 ln r) m2 >s, where r is in meters. Determine the velocity of a particle at r = 4 m and the pressure at points on this streamline. Take r = 1.20 kg>m3. r!4m

SOLUTION We consider ideal fluid flow. The velocity components are vr =

1 0c ; r 0u

vu = -

vr = 0

0c ; 0r

vu = a

Thus, the velocity is

At r = 4 m,

V = vu = a V = a

240 b m>s r

240 b m>s r

240 b m>s = 60.0 m>s 4

Ans.

Since a free vortex flow is irrotational, Bernoulli’s equation can be applied between two points on the different streamlines. In this case, the two points are on the circular streamlines r = ∞ where V0 = 0 and p0 = 0 and r = 4 m where V = 60.0 m>s. Since the flow occurs in the horizontal plane, z0 = z. p p0 V0 2 V2 + gz0 = + gz + + r r 2 2 0 + 0 + gz =

p 3

1.20 kg>m

+

( 60.0 m>s ) 2 2

+ gz Ans.

p = -2160 Pa = -2.16 kPa

Ans: V = 60.0 m>s p = - 2.16 kPa 793

7–58. Determine the location of the stagnation point for a combined uniform flow of 8 m>s and a source having a strength of 3 m2 >s. Plot the streamline passing through the stagnation point.

y

8 m/s

x

SOLUTION We consider ideal fluid flow. This is a case of flow past a half body. The location of the stagnation point P is at Ans.

u = p Using r = r0 =

3 m2 >s q 3 = = m 2pU 2p(8 m>s) 16p

Ans.

The equation of the streamline (boundary of a half body) that passes through the stagnation point P can be determined by applying. r =

r0(p - u) sin u

3 (p - u) 16p r = sin u r =

3(p - u)

y

16p sin u

asymptote

This equation can be written in the form r sin u =

3 (p - u) 16p

3 m 16

Since y = r sin u, this equation becomes

r

P ¨

3 (p - u) y = 16p

source

The half width h of the half body can be determined by setting y = h as u approaches 0 or 2p. Thus, 3 3 (p - u) = m h = 16p 16

3 m 16

x

3 m 16 asymptote (a)

The plot of the half body is shown in Fig. a.

Ans: u = p 3 r = m 16p 794

7–59. As water drains from the large cylindrical tank, its surface forms a free vortex having a circulation of Γ. Assuming water to be an ideal fluid, determine the equation z = f(r) that defines the free surface of the vortex. Hint: Use the Bernoulli equation applied to two points on the surface.

z

r

SOLUTION We consider ideal fluid flow. For a free vortex, the radial and transverse components of velocity are vr = 0

and

vu =

k r

Then V = vu =

A

k r

r

For a circulation Γ, Γ =

C

V # ds = k =

z

2p

k (rdu) = 2p k L0 r

r

Γ 2p

B

Thus, V =

Γ 2pr

Since a free vortex is irrotational flow, Bernoulli’s equation can be applied between two points on different streamlines, such as point A and B shown in Fig. a. Point A is located at (r = ∞ , 0) where pA = patm = 0 and VA = 0, and point B is located Γ at (r, z) where pB = patm = 0 and VB = . Establish the datum through point A, 2pr

z (a)

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2

0 + 0 + g(0) = 0 +

a

Γ 2 b 2pr + g( - z) 2

gz =

Γ2 8p2r 2

z =

Γ2 8p2gr 2

Ans.

Ans: z =

795

Γ2 8p2gr 2

*7–60. Pipe A provides a source flow of 5 m2 >s, whereas the drain, or sink, at B removes 5 m2 >s. Determine the stream function between AB, and show the streamline for c = 0.

y

5 m/s2

5 m/s2 B

A

2m

SOLUTION We consider ideal fluid flow. When the source and sink are superimposed, Fig. a, the resultant stream function is c =

5 m2 >s q q q u2 u1 = (u2 - u1) = (u2 - u1) 2p 2p 2p 2p

Ans.

Here u2 is defined from A and u1 is defined from B. Staying in polar coordinates, c = 0 implies u2 - u1 = 0 or u2 = u1. The graph of the corresponding points is shown in Fig. a, with the direction of flow indicated. Note that “the” steamline has two distinct segments.

y

x

(a)

796

2m

x

7–61. Pipe A provides a source flow of 5 m2 >s, whereas the drain at B removes 5 m2 >s. Determine the potential function between AB, and show the equipotential line for f = 0.

y

5 m/s2

5 m/s2 B

A

2m

x

2m

SOLUTION We consider ideal fluid flow. When the source and sink are superimposed, the resultant potential function is f =

5 m2 >s r2 q q q r2 ln r2 ln r1 = ln = ln 2p 2p 2p r1 2p r1

Ans.

Here r2 is measured from B and r1 is measured from A. Staying in polar coordinates, f = 0 implies ln

r2 = 0 r1 r2 = 1 r1 r1 = r2

Thus, the equipotential line for f = 0 is along the y axis as shown in Fig. a. y

f=0

x

(a)

Ans: 5 m2 >s

r2 2p r1 The equipotential line for f = 0 is along the y axis. f =

797

ln

7–62. A source having a strength of q = 80 ft2 >s is located at point A (4 ft, 2 ft). Determine the magnitudes of the velocity and acceleration of fluid particles at point B (8 ft, !1 ft).

y

A

2 ft 1 ft 4 ft

SOLUTION We consider ideal fluid flow. The radial and transverse components of the velocity are q vr = vu = 0 2pr Thus, the magnitude of the velocity is V = vr =

q 2pr

Here, r = 2(8 ft - 4 ft)2 + ( - 1 ft - 2 ft)2 = 5 ft. Then V =

80 ft 2 >s

2p (5 ft)

Ans.

= 2.546 ft>s = 2.55 ft>s

x any y components of the velocity are u = vr cos u

v = vr sin u

y x Here cos u = and sin u = , r r u =

q x q x a b = a b 2pr r 2p r 2

v =

However, r 2 = x2 + y2. Then u =

q x a 2 b 2p x + y2

x any y components of the acceleration are ax =

y q a 2 b 2p x + y2

0u 0u 0u + u + v 0t 0x 0y

= 0 + = ay =

v =

q y q y a b = a b 2pr r 2p r 2

y2 - x2 y -2xy q q q q x a 2 b W £ § ¶ + a 2 bW £ 2 § ¶ 2 2 2 2 2 2p x + y 2p ( x + y ) 2p x + y 2p ( x + y2 ) 2

q2 4p

2

c

x3 + xy2

( x2 + y2 ) 3

0v 0v 0v + u + v 0t 0x 0y

= 0 + = -

d

- 2xy y x2 - y2 q q q q x a 2 b W £ § ¶ + a b W £ § ¶ 2p x + y2 2p ( x2 + y2 ) 2 2p x2 + y2 2p ( x2 + y2 ) 2

q2

y3 + x2y

4p

( x2 + y2 ) 3

c 2

d 798

4 ft

B

x

7–62. Continued

With respect to point A, the coordinates of point B are B[(8 - 4) ft, ( -1 - 2) ft] = B ( 4 ft, - 3 ft ) . Then ax = -

( 80 ft2 >s ) 2 (4 ft)3 + (4 ft)( -3 ft)2

ay = -

4p2

c

3 (4 ft)2

+ ( - 3 ft)2 4 3

s = -1.038 ft>s2

( 80 ft2 >s ) 2 ( - 3 ft)3 + (4 ft)2( -3 ft) 4p2

c

3 (4 ft)2

Thus, the magnitude of the acceleration is

+ ( -3 ft)2 4 3

s = 0.7781 ft>s2

a = 2ax2 + ay2 = 2 ( - 1.038 ft>s2 ) 2 + ( 0.7781 ft>s2 ) 2 = 1.297 ft>s2 = 1.30 ft>s2

Ans.

As an alternative solution,

a =

0vr 0vr + vr 0t 0r

= 0 + = 2

q q 1 a a- 2b b 2pr 2p r

q2

(2p)2r

2 = 3

(80 ft 2 >s)2

(2p)2(5 ft)3

= 1.30 ft>s2

Ans.

Ans: V = 2.55 ft>s a = 1.30 ft>s2 799

7–63. Two sources, each having a strength of 2 m2 >s, are located as shown. Determine the x and y components of the velocity of fluid particles that pass point (x, y). What is the equation of the streamline that passes through point (0, 8 m) in Cartesian coordinates? Is the flow irrotational?

y

x 4m

SOLUTION We consider ideal fluid flow. When sources (1) and (2) are superimposed, Fig. a, the resultant stream function is q q q ( u + u2 ) u + u = c = 2p 1 2p 2 2p 1 From the geometry shown in Fig. a, u1 = tan-1a

Then,

c =

y x - 4

b

u2 = tan-1a

y x + 4

b

y y q c tan-1a b + tan-1a bd 2p x - 4 x + 4

(1)

The x and y components of velocity are u =

q 0c = ≥ 0y 2p =

v = -

1 1 1 a b + a b¥ 2 x - 4 2 x + 4 y y 1 + a b 1 + a b x - 4 x + 4

q x - 4 x + 4 £ + § 2p (x - 4)2 + y2 (x + 4)2 + y2

q 0c = ≥ 0y 2p =

Here,

1

1 2 y 1 + a b x - 4

£-

y

(x - 4)

2

§ +

y y q £ + § 2p (x - 4)2 + y2 (x + 4)2 + y2

Ans.

1 2 y 1 + a b x + 4

Substituting these results into

vz =

2y(x + 4)

3 (x

y (x + 4)2



Ans.

2y(x - 4) 2y(x + 4) q dv = £ + § dx 2p 3 (x - 4)2 + y2 4 2 3(x + 4)2 + y24 2

2y(x - 4) q du £ + = dy 2p 3 (x - 4)2 + y2 4 2

c-

+ 4)2 + y2 4 2

§

1 0v 0u a b = 0 2 0x 0y

800

4m

7–63. Continued

Since vz = 0, the flow is irrotational. For q = 2 m2 >s, the streamline that passes through point x = 0 and y = 8 m can be determined using Eq. 1, c =

2 8 8 c tan-1a b + tan-1a bd = 0 2p 0 - 4 0 + 4

Thus, the equation of this streamline is tan-1a

y y b + tan-1a b = 0 x - 4 x + 4

tan-1a

y y b = - tan-1a b x - 4 x + 4 y y = x - 4 x + 4

Ans.

x = 0 y x r2

r1

r

1

2

2

1 4m

y x

4m

Ans: u =

v =

q x - 4 £ + 2p 1 x - 42 2 + y2

y q £ + 2p 1 x - 42 2 + y2

The flow is irrotational. x = 0 801

1x 1x

x + 4

+ 42 2 + y2 y

+ 42 2 + y2

§

§

*7–64. The source and sink of equal strength q are located a distance d from the origin as indicated. Determine the stream function for the flow, and draw the streamline that passes through the origin.

y

x

d

SOLUTION We consider ideal fluid flow. When the source and sink are superimposed, Fig. a, the resultant stream function is q q q ( u - u2 ) Ans. u u = c = 2p 1 2p 2 2p 1 Staying in polar coordinates, at the origin u1 = p and u2 = 0. c =

q q (p - 0) = 2p 2

So the streamline satisfies q q ( u - u2 ) = 2 2p 1 u1 = u2 + p Thus, the streamline passing through the origin is the straight segment between the source and the sink, as shown in Fig. a. y

=

q 2

x

(a)

802

d

7–65. Two sources, each having a strength q, are located as shown. Determine the stream function, and show that this is the same as having a single source with a wall along the y axis.

y

x

d

d

SOLUTION We consider ideal fluid flow. When sources (1) and (2) are superimposed, Fig. a, the resultant stream function is c =

q q q ( u + u2 ) u + u = 2p 1 2p 2 2p 1

Ans.

In order for the stream function to be the same as that of a single source and a wall along the y axis, a streamline must exist along the y axis. However, by geometry, along the y axis it is always true that u1 + u2 = {p, so that the value of the stream q q q q function is c = ( {p) { , where + corresponds to the +y axis and 2p 2 2 2 corresponds to the -y axis. y x r2

r1

r

1

2

2

1 d

y x

d

(a)

Ans: c = 803

q ( u + u2 ) 2p 1

7–66. A source q is emitted from the wall while a flow occurs towards the wall. If the stream function is described as c = (4xy + 8u) m2 >s, where x and y are in meters, determine the distance d from the wall where the stagnation point occurs along the y axis. Plot the streamline that passes through this point.

y

d x

SOLUTION We consider ideal fluid flow. Here, x = r cos u and y = r sin u. Then in terms of r and u coordinates, c = 4(r cos u)(r sin u) + 8u c = 2r 2 sin 2u + 8u

(1)

The velocity components are 1 0c 1 1 = 3 2r 2(2 cos 2u) + 8 4 = ( 4r 2 cos 2u + 8 ) r 0u r r 0c vu = = -(4r sin 2u) 0r vr =

At stagnation point p, it is required that these velocity components are equal to zero. vu = - 4r sin 2u = 0 sin 2u = 0

u =

(since r ≠ 0)

2u = 0,

p rad

u = 0,

p rad 2

p rad is chosen and it gives the direction r of the stagnation point. 2 vr =

Since

1 ( 4r 2 cos 2u + 8 ) = 0 r

1 ≠ 0, then r 4r 2 cos 2u + 8 = 0

Substituting u =

p rad and r = d into this equation, 2 p 4d 2 cos c 2 a b d + 8 = 0 2

Ans.

d = 22 m p Substituting u = rad and r = 22 m into Eq 1, 2

p p c = 2 ( 22 ) 2 sin c 2 a b d + 8 a b = 4p 2 2

804

7–66. Continued

Therefore, the streamline passing through the stagnation point is given by r2 =

4p = 2r 2 sin 2u + 8u

4p - 8u 2 sin 2u

The plot of this stream function is shown in Fig. a u(rad)

p 12

p 6

p 4

p 3

5p 12

p 2

r(m)

3.236

2.199

1.772

1.555

1.447

undef.

y (m) 5

12 3 4 6 12

x (m)

(a)

Ans: 22 m 805

7–67. Determine the equation of the boundary of the half body formed by placing a source of 0.5 m2 >s in the uniform flow of 8 m>s. Express the result in Cartesian coordinates.

y 8 m/s

x A

SOLUTION We consider ideal fluid flow. For the flow past a half body, r0 =

0.5 m2 >s q 0.03125 = = m p 2pU 2p(8 m>s)

The equation of the boundary of a half body is given by

r =

r0(p - u) sin u

0.03125 (p - u) p = sin u

0.03125 (p - u) p y Here, y = r sin u and u = tan-1 . Then, this equation becomes x y 32py = p - tan-1 x r sin u =

tan-1

y = p(1 - 32y) x

y = tan [p(1 - 32y)] x

Ans.

Ans:

806

y = x tan 3 p ( 1 - 32y ) 4

*7–68. The leading edge of a wing is approximated by the half body. It is formed from the superposition of the uniform air flow of 300 ft>s and a source. Determine the required strength of the source so that the width of the half body is 0.4 ft.

300 ft/s A r ! 0.3 ft u ! 90" O

SOLUTION We consider ideal fluid flow. For the flow past a half body, r0 =

q q q = = ft 2pU 2p(300 ft>s) 600p

The equation of the boundary of a half body is given by r0(p - u)

r =

sin u q (p - u) q(p - u) 600p r = = sin u 600p sin u r sin u =

q(p - u) 600p

Since y = r sin u, this equation becomes y =

q(p - u) 600p

The half width h of the half body can be determined by setting y = h as u approaches 0 or 2p. Thus, h = Here, h =

q(p - u) 600p

=

q 600

0.4 ft = 0.2 ft. Then 2 0.2 ft =

q 600

q = 120 ft 2 >s

Ans.

807

7–69. The leading edge of a wing is approximated by the half body. It is formed from the superposition of the uniform air flow of 300 ft>s and a source having a strength of 100 ft2 >s. Determine the width of the half body and the difference in pressure between the stagnation point O and point A, where r = 0.3 ft, u = 90". Take r = 2.35(10!3) slug>ft3.

300 ft/s A r ! 0.3 ft u ! 90" O

SOLUTION We consider ideal fluid flow. For the flow past a half body, r0 =

100 ft 2 >s q 1 = = ft 2pU 2p(300 ft>s) 6p

The equation of the boundary of a half body is given by r =

r0(p - u) sin u

1 (p - u) 6p r = sin u r sin u =

1 (p - u) 6p

Since y = r sin u, this equation becomes y =

1 (p - u) 6p

The half width h of the half body can be determined by setting y = h as u approaches 0 or 2p. Thus, h =

1 1 (p - u) = ft 6p 6

1 Here, 2h = 2 a ft b = 0.333 ft 6

Ans.

At the stagnation point O, VO = 0. The r and u components of velocity at point A can be determined using vr =

100 ft 2 >s q + U cos u = + ( 300 ft>s ) cos 90° = 53.05 ft>s 2pr 2p(0.3 ft)

vu = - U sin u = - ( 300 ft>s ) sin 90° = - 300 ft>s Thus, the magnitude of the velocity is V = 2vr2 + vu2 = 2 ( 53.05 ft>s ) 2 +

( - 300 ft>s ) 2 = 304.65 ft>s

The flow past a half body is irrotational. Thus, the Bernoulli equation for an ideal fluid is applicable from point O at A. Neglecting the elevation term, pA pO VO2 VA2 = + + r r 2 2 pO 2.35 ( 10-3 ) slug>ft 3

+ 0 =

pA 2.35 ( 10-3 ) slug>ft 3

∆p = pO - pA = 109.06

+

( 304.65 ft>s ) 2 2

lb 1 ft 2 a b = 0.757 psi ft 2 12 in. 808

Ans.

Ans: 0.333 ft ∆p = 0.757 psi

7–70. The half body is defined by a combined uniform flow having a velocity of U and a point source of strength q. Determine the pressure distribution along the top boundary of the half body as a function of u, if the pressure within the uniform flow is p0. Neglect the effect of gravity. The density of the fluid is r.

y

U r u

x

SOLUTION We consider ideal fluid flow. For the flow past a half body, r0 =

q 2pU

The equation of the boundary of a half body is given by q (p - u) r0(p - u) q(p - u) 2pU r = = = sin u sin u 2pU sin u The r and u components of velocity at any point on the boundary can be determined using vr =

q + U cos u = 2pr

2pc

q q(p - u) 2pU sin u

vu = - U sin u

+ U cos u = d

U sin u + U cos u p - u

Thus, the magnitude of the velocity is V = 2vr2 + vu2 = =

2 U sin u 2 A a p - u + U cos u b + ( -U sin u)

U 2 sin2u + (p - u) sin 2u + (p - u)2 p - u

Since the potential function exists, the flow past a half body is irrotational. The Bernoulli equation is applicable between any two points in the flow. If point A is an arbitrary point on the boundary where VA = V and pA = p, and point O is a point remote from the body where VO = U, then pA p0 VO2 VA2 = + + r r 2 2 p0 U2 p = + + r r 2

U2 3 sin2u + (p - u) sin 2u + (p - u)2 4 (p - u)2

p = p0 -

2

rU

2

2(p - u)2

3 sin2u

+ (p - u) sin 2u 4

Ans.

Ans: p = p0 -

809

rU 2 2(p - u)2

3 sin2u

+ (p - u) sin2u 4

7–71. A fluid flows over a half body for which U = 0.4 m>s and q = 1.0 m2 >s. Plot the half body, and determine the magnitudes of the velocity and pressure in the fluid at the point r = 0.8 m and u = 90o. The pressure within the uniform flow is 300 Pa. Take r = 850 kg>m3.

y

U ! 0.4 m/s r u

x

q ! 1.0 m/s2

SOLUTION We consider ideal fluid flow. The location of the stagnation point p can be determined from r0 =

1.0 m2 >s q = = 0.3979 m = 0.398 m 2pU 2p ( 0.4 m>s )

The half width of the half body is h = pr0 = p(0.3979 m) = 1.25 m The resulting half body is shown in Fig. a. The velocity components of the flow passing a half body at point B where r = 0.8 m and u = 90° are (vr)B =

1.0 m2 >s q + U cos u = + ( 0.4 m>s ) cos 90° = 0.1989 m>s 2pr 2p(0.8 m)

(vu)B = - U sin u = - ( 0.4 m>s ) sin 90° = - 0.4 m>s Thus, the magnitude of the velocity is V = 2(vr)B2 + (vu)B2 = 2 ( 0.1989 m>s ) 2 +

( - 0.4 m>s ) 2 = 0.4467 m>s = 0.447 m>s

Ans.

Since the flow passing a half body is irrotational, Bernoulli’s equation can be applied between two points on the different streamlines such as point A within the uniform flow and point B. Since the flow occurs in the horizontal plane, the gravity term can be excluded. Here VA = U = 0.4 m>s. pB pA VA2 VB2 = + + r r 2 2 r pB = pA + ( VA2 - VB2 ) 2 = 300 Pa + °

850 kg>m3 2

= 283 Pa

¢ 3 ( 0.4 m>s ) 2 - ( 0.4467 m>s ) 2 4

Ans.

y (m)

1.25 m p

x (m) 1.25 m

0.398 m (a)

810

Ans: r0 = 0.398 m h = 1.25 m V = 0.447 m>s p = 283 Pa

*7–72. The half body is defined by a combined uniform flow having a velocity of U and a point source of strength q. Determine the location u on the boundary of the half body where the pressure p is equal to the pressure p0 within the uniform flow. Neglect the effect of gravity.

y U

r

SOLUTION

We consider ideal fluid flow. For the flow past a half body, r0 =

q 2pU

The equation of the boundary of a half body is given by q (p - u) r0(p - u) q(p - u) 2pU r = = = sin u sin u 2pU sin u The r and u components of velocity at any point on the boundary can be determined using q q U sin u vr = + U cos u = + U cos u = + U cos u q(p - u) 2pr p - u 2pc d 2pU sin u vu = - U sin u

Thus, the magnitude of the velocity is V = 2vr2 + vu2 =

2 U sin u 2 A a p - u + U cos u b + ( -U sin u)

U 2sin2u + (p - u) sin 2u + (p - u)2 p - u

=

Since the potential function exists, the flow past a half body is irrotational. The Bernoulli equation is applicable between any two points in the flow. If point A is an arbitrary point on the boundary where VA = V and pA = pO, and point O is a point remote from the body where VO = U, then pO pA VO2 VA2 = + + r r 2 2 2

pA pO U = + + r r 2 pO - pA = Since

rU 2 2(p - u)2

U2 3 sin2u + (p - u) sin 2u + (p - u)2 4 (p - u)2 2

rU 2 2(p - u)2

≠ 0, then

3 sin2 u

+ (p - u) sin 2u 4 = 0

sin2u + (p - u) sin 2u = 0 sin2u + 2(p - u) sin u cos u = 0 sin u[sin u + 2(p - u) cos u] = 0 Since sin u ≠ 0, then sin u + 2(p - u) cos u = 0 tan u + 2(p - u) = 0 Solving by trial and error,

or

u = 1.9760 rad a

180° b = 113.22° = 113° p rad

u = 4.3072 rad a

180° b = 246.8° = 247° p rad

Ans.

Ans. 811

u

x

7–73. The Rankine body is defined by the source and sink, each having a strength of 0.2 m2 >s. If the velocity of the uniform flow is 4 m>s, determine the longest and shortest dimensions of the body.

y

4 m/s

x

0.5 m

0.5 m

SOLUTION We consider ideal fluid flow. The half length of the Rankine oval is 1

2 q b = a a + a2 b Up

b = ca

0.2 m2 >s

(4 m>s)p

1 2

b(0.5 m) + ( 0.5 m ) 2 d = 0.5079 m

Thus, the length of the Rankine oval is

Ans.

L = 2b = 2(0.5079 m) = 1.02 m The half width of the Rankine oval can be determined using h = h =

h2 - a2 2pUh tan a b 2a q

h2 - ( 0.5 m ) 2 2(0.5 m)

tanc

2p(4 m>s)h 0.2 m2 >s

h = ( h2 - 0.25 ) tan (40ph)

d

Solving numerically, and noting that q>2U = 0.2>[2(4)] = 0.025, we start with h = 0.024 and adjust this until we find that h = 0.02423 m Thus, the width of the Rankine oval is Ans.

W = 2h = 2(0.02423) = 0.0485 m

Ans: L = 1.02 m 0.0485 m 812

7–74. The Rankine body is defined by the source and sink, each having a strength of 0.2 m2 >s. If the velocity of the uniform flow is 4 m>s, determine the equation in Cartesian coordinates that defines the boundary of the body.

y

4 m/s

x

0.5 m

0.5 m

SOLUTION We consider ideal fluid flow. The stream function of the flow around the Rankine oval is given by 2ay q c = Uy tan-1a 2 b 2p x + y2 - a2

Since the boundary of the Rankine oval contains the stagnation point where y = 0, then this equation gives c = U(0) -

2(a)(0) q tan-1a 2 b = 0 2p x + 02 - a2

Thus, the equation that describes the boundary of the Rankine oval is 2ay q tan-1a 2 b = 0 2p x + y2 - a2

Uy -

Here, U = 4 m>s, q = 0.2 m2 >s and a = 0.5 m. 4y -

4y -

2(0.5)y 0.2 tan-1 c 2 d = 0 2p x + y2 - 0.52

y 1 tan-1 c 2 d = 0 2 10p x + y - 0.25 y

2

2

x + y - 0.25

Ans.

= tan 40py

Ans: 2

y 2

x + y - 0.25 813

= tan 40py

7–75. A fluid has a uniform velocity of U = 10 m>s. A source q = 15 m2 >s is at x = 2 m, and a sink q = !15 m2 >s is at x = !2 m. Graph the Rankine body that is formed, and determine the magnitudes of the velocity and the pressure at point (0, 2 m). The pressure within the uniform flow is 40 kPa. Take r = 850 kg>m3.

SOLUTION We consider ideal fluid flow. The major and minor axes of the Rankine oval can be determined from 1

2 q b = a a + a2 b Up

= ec

15 m2 >s

(10 m>s)p

= 2.23 m h = h = h =

d (2 m) + (2 m)2 f

1 2

2pUh h2 - a2 tan a b 2a q h2 - (2 m)2 2(2 m)

tan c

2p ( 10 m>s ) h

h2 - 4 4p tan a hb 4 3

15 m2 >s

d

Solving numerically, and noting that q>2U = 15> 32(10) 4 = 0.75, we start with h = 0.74 and adjust this until we find that h = 0.609 m

The resulting Rankine oval is shown in Fig. a Since the flow around a Rankine oval is irrotational, Bernoulli’s equation can be applied between two points on different streamlines where these two points are point A, within the uniform flow and point B (0, 2 m). Since the flow occurs in the horizontal plane, the gravity term can be excluded. Here, VA = U = 10 m>s and the velocity components at B are uB = U +

q x - a x + a £ § 2p (x + a)2 + y2 (x - a)2 + y2

= 10 m>s + a = 11.19 m>s vB =

=

15 m2 >s 2p



0 + 2m 0 - 2m § (0 + 2 m)2 + (2 m)2 (0 - 2 m)2 + (2 m)2

y y q £ § 2 2 2p (x + a) + y (x - a)2 + y2 15 m2 >s

= 0

2p

£

2m 2m § 2 2 (0 + 2 m) + (2 m) (0 - 2 m)2 + (2 m)2

814

7–75. Continued

Thus, Ans.

VB = u B = 11.19 m>s = 11.2 m>s Bernoulli’s equation written between points A and B is pB pA VA2 VB2 = + + r r 2 2 r pB = pA + ( VA2 - VB2 ) 2 = 40 ( 103 ) Pa + °

850 kg>m3 2

= 29.24 ( 103 ) Pa = 29.2 kPa

¢ c ( 10 m>s ) 2 - ( 11.19 m>s ) 2 d

Ans.

y (m)

0.609 m x (m) 0.609 m 2.23 m

2.23 m (a)

Ans: h = 0.609 m V = 11.2 m>s p = 29.2 kPa 815

*7–76. Integrate the pressure distribution, Eq. 7–67, over the surface of the cylinder in Fig. 7–33b, and show that the resultant force is equal to zero.

SOLUTION We consider ideal fluid flow. The pressure distribution around a cylinder is given by p = p0 +

1 2 rU ( 1 - 4 sin2 u ) 2

The force that the pressure exerts on the differential area dA = (adu)L = aLdu is dF = pdA = c p0 +

1 1 rU 2 ( 1 - 4 sin2 u ) d (aLdu) = p0aLdu + raLU 2 ( 1 - 4 sin2 u ) du 2 2

Equating the resultant forces along the x and y axes shown in Fig. a, + S ( FR ) x = ΣFx;

( FR ) x = = = -

L0

2p

L0

2p

L0

2p

dF cos u c p0aLdu +

1 raLU 2 ( 1 - 4 sin2 u ) du d cos u 2

= - p0aL(sin u)& 2p 0 = 0 + c ( FR ) y = ΣFy;

( FR ) y = = = -

2p

p0aL cos udu -

L0

2p

L0

2p

L0

2p

1 raLU 2 ( cos u - 4 sin2 u cos u ) du L0 2

1 4 sin3 u 2p raLU 2 asin u b` 2 3 0

dF sin u c p0aLdu +

1 raLU 2 ( 1 - 4 sin2 u ) du d sin u 2 2p

p0aL sin udu -

= - p0aL ( cos u )& 2p 0 = 0

1 raLU 2 ( sin u - 4 sin3 u ) du L0 2

2p 4 1 raLU 2 c -cos u - c - cos u ( sin2 u + 2 ) d d ` 2 3 0

Therefore, FR = 2 ( FR ) x2 + ( FR ) y2 = 0

(Q.E.D.)

(FR)y ds = ad dF a (FR)x

(a)

816

d

7–77. The tall rotating cylinder is subjected to a uniform horizontal airflow of 3 ft>s. If the radius of the cylinder is 4 ft, determine the location of the stagnation points and the lift per unit length. The circulation around the cylinder is 18 ft 2 >s. Take r = 2.35 1 10-3 2 slug>ft 3.

3 ft/s 4 ft

SOLUTION We consider ideal fluid flow. The lift can be determined by using Fy = rUΓ =

3 2.35 ( 10-3 ) slug>ft3 4 ( 3 ft>s )( 18 ft2 >s )

= 0.127 lb>ft For this case,

18 ft 2 >s Γ = = 0.1194 4pUa 4p ( 3 ft>s ) (4 ft) u = 6.86° and 173°

Ans.

sin u =

Ans.

Since the solution has two roots, there are two stagnation points on the surface of the cylinder.

Ans: Fy = 0.127 lb>ft u = 6.86° and 173° 817

7–78. The 0.5-m-diameter bridge pier is subjected to the uniform flow of water at 4 m>s. Determine the maximum and minimum pressures exerted on the pier at a depth of 2 m.

SOLUTION We consider ideal fluid flow. For the flow around a cylinder, the pressure at any point on the boundary can be determined by using p = p0 +

1 rU 2 ( 1 - 4 sin2 u ) 2

At the depth of h = 2 m, p0 = rgh = ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m) = 19.62 ( 103 ) Pa The pressure extremes occur when

dp = 0. Thus, du

dp 1 = 0 + rU 2(0 - 8 sin u cos u) = 0 du 2 8 sin u cos u = 0 4 sin 2u = 0 Solving, u = 0, 90°, 180°... The maximum pressure occurs when u = 0° or 180°. Thus, pmax = 19.62 ( 103 ) Pa +

1 ( 1000 kg>m3 )( 4 m>s ) 2 3 1 - 4 sin2 0° 4 2

= 27.62 ( 103 ) Pa = 27.6 kPa

Ans.

The minimum pressure occurs when u = 90°. Thus, pmin = 19.62 ( 103 ) Pa +

1 ( 1000 kg>m3 ) (4 m>s)2 31 - 4 sin2 90°4 2

= - 4.38 ( 103 ) Pa = - 4.38 kPa

Ans.

The negative sign indicates that suction occurs.

Ans: pmax = 27.6 kPa pmin = -4.38 kPa 818

7–79. Air flows around the cylinder such that the pressure, measured at A, is pA = - 4 kPa. Determine the velocity U of the flow if r = 1.22 kg>m3. Can this velocity be determined if instead the pressure at B is measured?

A U

45!

B 0.2 m

SOLUTION We consider ideal fluid flow. The pressure at a point removed from the cylinder is atmospheric. Thus, p0 = 0. At point A, u = 45° pA = p0 + -4 ( 103 ) N>m2 = 0 +

1 rU 2 ( 1 - 4 sin2 u ) 2 1 ( 1.22 kg>m2 )( U 2 )( 1 - 4 sin2 45°) 2 Ans.

U = 80.98 m>s = 81.0 m>s

If the pressure is measured at B, the velocity of the uniform flow can be determined using the same equation with u = 180°.

Ans: 81.0 m>s; yes 819

*7–80. The 200-mm-diameter cylinder is subjected to a uniform horizontal flow having a velocity of 6 m>s. At a distance far away from the cylinder, the pressure is 150 kPa. Plot the variations of the velocity and pressure along the radial line r, at u = 90°, and specify their values at r = 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m. Take r = 1.5 Mg>m3.

6 m/s r u 0.1 m

SOLUTION We consider ideal fluid flow. The r and u components of velocity of the uniform flow around a cylinder can be determined using 2

a b cos u = ( 6 m>s ) £ 1 r2

vr = U a1 -

(0.1 m)2 r2

§ cos u = c a6 -

0.06 b cos u d m>s r2

r(m) 0.5 0.4 0.3

2

vu = - U a1 +

When u = 90°,

(0.1 m) a2 0.06 0.2 b sin u = - (6 m>s) c 1 + d sin u = c - a6 + 2 b sin u d m>s 2 r r2 r 0.1 vu = - a6 +

vr = 0

0

0.06 b m>s r2

V =

+

vu2

0.06 2 0.06 = 0 + c - a6 + 2 b = a6 + 2 b m>s r r B 2

(1)

p pO VO2 V2 = + + r r 2 2 N m2 + 1500 kg>m3

( 6 m>s ) 2

2

=

1500 kg>m3

+

a6 +

0.5 0.4 0.3

0.1

0.06 2 b r2 2

0.06 2 b d Pa r2

0

(2)

The values of V and P at r = 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m can be evaluated using Eqs. 1 and 2, respectively, and are tabulated below. r(m) V ( m>s ) Eq. (1) p(kPa) Eq. (2)

v(m s)

0.2

p

p = c 177 ( 103 ) - 750 a6 +

12

r(m)

Flow around a cylinder is irrotational since the potential function exists. Therefore, the Bernoulli equation is applicable. Neglecting the elevation terms,

150 ( 103 )

7.5 6.67

(a)

Thus, the magnitude of the velocity when u = 90° is 2vr2

6.24 6.375

0.1

0.2

0.3

0.4

0.5

12

7.5

6.67

6.375

6.24

69

135

144

147

148

Ans.

The plot of V vs. r and p vs. r are shown in Figs. a and b.

820

69 (b)

135 144

p(kPa) 148 147

7–81. The 200-mm-diameter cylinder is subjected to a uniform flow having a velocity of 6 m>s. At a distance far away from the cylinder, the pressure is 150 kPa. Plot the variation of the velocity and pressure along the radial line r, at u = 0°, and specify their values at r = 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m. Take r = 1.5 Mg>m3.

6 m/s r u 0.1 m

SOLUTION We consider ideal fluid flow. The r and u components of velocity of the uniform flow around a cylinder can be determined using vr = U a1 -

(0.1 m)2 a2 0.06 b cos u = ( 6 m>s ) c 1 d cos u = c a6 - 2 b cos u d m>s 2 r r2 r

vu = -U a1 +

When u = 0°,

v(m s) 5.76 5.625 5.33 4.50

(0.1 m)2 a2 0.06 ( ) bsin u = 6 m>s c 1 + d sin u = c - a6 + 2 b sin u d m>s 2 2 r r r vr = a6 -

0.06 b m>s r2

vu = 0 0

0.1 0.2

Thus, the magnitude of the velocity when u = 0° is V = 2vr2 + vu2 =

B

a6 -

0.06 2 0.06 b + 02 = a6 - 2 b m>s r2 r

( 6 m>s ) 2

P(kPa) 162 177 156 153 152

pO p VO2 V2 = + + r r 2 2 N m2 + 1500 kg>m3

r(m)

(1)

The flow around a cylinder is irrotational since the potential function exists. Therefore, the Bernoulli equation is applicable. Neglecting the elevation terms,

150 ( 103 )

0.3 0.4 0.5 (a)

2

=

p = c 177 ( 103 ) - 750 a6 -

p 1500 kg>m3

+

a6 -

0.06 2 b r2 2

0.06 2 b d Pa r2

(2)

The values of V and P at r = 0.1 m, 0.2 m, 0.3 m, 0.4 m, and 0.5 m can be evaluated using Eqs. 1 and 2, respectively, and are tabulated below. r(m)

0.1

0.2

0.3

0.4

0.5

V(m>s)

0

4.50

5.33

5.625

5.76

0

0.1

0.2 0.3 0.4 0.5 (b)

r(m)

Ans.

p(kPa) 177 162 156 153 152 The plot of V vs. r and p vs. r are shown in Figs. a and b.

Ans: r (m) 0.1 0.2 V (m>s) 0 4.50 p (kPa) 177 162 821

0.3 5.33 156

0.4 0.5 5.625 5.76 153 152

7–82. Air is flowing at U = 30 m>s past the Quonset hut of radius R = 3 m. Find the velocity and absolute pressure distribution along the y axis for 3 m … y … ∞ . The absolute pressure within the uniform flow is p0 = 100 kPa. Take ra = 1.23 kg>m3.

y U A

R u

SOLUTION

x

We consider ideal fluid flow. The velocity components of the flow around the building are vr = U a1 -

a2 bcos u r2

vu = - U a1 +

a2 bsin u r2

Here U = 30 m>s and a = 3 m. Along the y axis u = 90°. Then vr = ( 30 m>s ) c 1 -

(3 m)2 r2

vu = - ( 30 m>s ) c 1 + = -30 a1 +

d cos 90° = 0

(3 m)2 r2

9 bm>s r2

d sin 90°

Thus, the velocity distribution along the y axis is V = vu = 30 a1 +

9 b m>s r2

Ans.

Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines. Here they are point O within the uniform flow and a point along the y axis, p0 p VO2 V2 + + gz0 = + + gz ra ra 2 2 Since the density of air is small, the gravitational terms can be neglected. Here, V0 = U = 30 m>s. Then 100 ( 10

3

) N>m

2

1.23 kg>m3

+

( 30 m>s ) 2

2

+ 0 =

p 1.23 kg>m3

p = c 100 553.5 - 553.5 a1 +

Notice that at r = ∞ ,

+

c 30 a1 + 2

9 2 b d Pa r2

9 2 bd r2

+ 0 Ans.

p = 100 553.5 - 553.5(1 + 0)2 = 100 ( 103 ) Pa = p0

Ans: V = 30 a1 +

9 b m>s r2

p = c 100 553.5 - 553.5 a1 +

822

9 2 b d Pa r2

7–83. The Quonset hut of radius R is subjected to a uniform wind having a velocity U. Determine the resultant vertical force caused by the pressure that acts on the hut if it has a length L. The density of air is r.

y U A

R u

SOLUTION

x

We consider ideal fluid flow. For the flow around a cylinder, the pressure distribution on the boundary is described by 1 rU 2 ( 1 - 4 sin2 u ) 2 Here, p0 is atmospheric pressure. Thus, the net pressure on the boundary is gauge pressure, which is p = p0 +

pg = p - p0 =

1 rU 2 ( 1 - 4 sin2 u ) 2

The force that the gauge pressure exerts on the differential area dA = (Rd u)L = RLd u is dF = pgdA =

1 1 rU 2 ( 1 - 4 sin2 u ) (RLdu) = rRLU 2 ( 1 - 4 sin2 u ) du 2 2

Equating the forces along the y axis shown in Fig. a, + c (FR)y = ΣFy;

(FR)y = -

L0

p

dFd sin u p

= -

1 rRLU 2 ( 1 - 4 sin2 u ) sin u du L 2 0

1 = - rRLU 2 2 L0

p

( sin u - 4 sin3 u ) du

p 4 1 = - rRLU 2 c - cos u - c - cos u ( sin2 u + 2 ) d d 2 3 0

=

5 rRLU 2 3

Ans.

(FR(y ds = Rd¨ dF

(FR(x

R

= (a)

R ¨ d¨

Ans: (FR)y = 823

5 rRLU 2 3

*7–84. The Quonset hut of radius R is subjected to a uniform wind having a velocity U. Determine the speed of the wind and the gage pressure at point A. The density of air is r.

y U A

R u

SOLUTION We consider ideal fluid flow. p At point A, r = R and u = rad. Thus, 2 vr = U a1 -

a2 R2 p bcos u = U a1 bcos = 0 2 r2 R2

p a2 R2 bsin u = -U a1 + bsin = -2U 2 r2 R2

vu = - U a1 +

Thus, the magnitude of the velocity at point A is

VA = 2vr2 + vu2 = 202 + ( - 2U)2 = 2U

Ans.

Here, VA is directed towards the positive x axis. For the flow around a cylinder, the pressure distribution on the boundary is described by p = p0 +

1 rU 2 ( 1 - 4 sin2 u ) 2

Here, p0 is atmospheric pressure. Thus, the net pressure on the boundary is gauge pressure, which is pg = p - p0 = At point A, u =

1 rU 2 ( 1 - 4 sin2 u ) 2

p rad. Then, 2 1 2

p 2

3 2

( pg ) A = rU 2 c 1 - 4 sin2 a b d = - rU 2

The negative sign indicates that suction occurs at A.

824

Ans.

x

7–85. Water flows toward the circular column with a uniform speed of 3 ft>s. If the outer radius of the column is 4 ft, and the pressure within the uniform flow is 6 lb>in2, determine the pressure at point A. Take rw = 1.94 slug>ft 3.

A

6 ft

3 ft/s 4 ft

SOLUTION We consider ideal fluid flow. The velocity components of the flow around the structure are vr = Ua1 -

a2 bcos u r2

vu = -U a1 +

a2 bsin u r2

Here U = 3 ft>s, and a = 4 ft. For point A, r = 6 ft and u = 90°. Then vr = ( 3 ft>s ) c 1 -

(4 ft)2 (6 ft)2

vu = - ( 3 ft>s ) c 1 +

Thus, the magnitude of the velocity at A is

d cos 90° = 0

(4 ft)2 (6 ft)2

d sin 90° = -4.333 ft>s

VA = vu = 4.333 ft>s Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines. Here, they are point O in the uniform flow and point A. p0 pA VO2 VA2 + gz0 = + gzA + + r r 2 2 Since the flow is in the horizontal plane, zD = zA = z. Here, V0 = U = 3 ft>s. Then p0 pA V0 2 VA2 + gz = + gz + + r r 2 2 pA = p0 + = a6

r ( V0 2 - VA2 ) 2

1.94 slug>ft 3 lb 12 in 2 ba b + 3 ( 3 ft>s ) 2 - ( 4.333 ft>s ) 2 4 2 1 ft 2 in

= a854.52 = 5.93 psi

lb 1 ft 2 ba b 2 12 in ft

Ans.

Ans: 5.93 psi 825

7–86. The tall circular building is subjected to a uniform wind having a velocity of 150 ft>s. Determine the location u of the window that is subjected to the smallest pressure. What is this pressure? Take ra = 0.00237 slug>ft 3.

u

85 ft

SOLUTION We consider ideal fluid flow. This is a case of flow around a cylinder where the velocity components are vr = U a1 -

a2 b cos u r2

vu = - U a1 +

On the surface of the building where r = a = 85 ft,

a2 b sin u r2

vu = - 2U sin u

vr = 0

Thus, the velocity of the wind on the surface of the building is V = vu = - 2U sin u The minimum pressure occurs at the point where the magnitude of velocity is maximum, that is when

Then

sin u = 1

or

u = 90°

or

sin u = - 1 Ans.

270°

Therefore Vmax = 2U = 2 ( 150 ft>s ) = 300 ft>s Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as between a point within the uniform flow and a point on the building. Since the flow occurs in the horizontal plane, the gravity term can be excluded. 2 p0 pmin V0 2 Vmax = + + r r 2 2

Here p0 = 0 and V0 = U = 150 ft>s. Then 0 +

( 150 ft>s ) 2 2

=

pmin 0.00237 slug>ft

pmin = a - 79.99

3

+

( 300 ft>s ) 2 2

lb 1 ft 2 ba b = -0.555 psi 2 12 in ft

Ans.

Ans: u = 90° or 270° pmin = -0.555 psi 826

7–87. The tall circular building is subjected to a uniform wind having a velocity of 150 ft>s. Determine the pressure and the velocity of the wind on its walls at u = 0°, 90°, and 150°. Take ra = 0.00237 slug>ft 3.

u

85 ft

SOLUTION We consider ideal fluid flow. This is a case of flow around a cylinder where the velocity components are vr = U a1 -

a2 b cos u r2

vu = -U a1 +

On the surface of the building where r = a = 85 ft, vr = 0

a2 b sin u r2

vu = - 2U sin u

Thus, the velocity of the wind on the surface of the building is V = vu = -2U sin u At u = 0°, 90° and 150°, V ! u = 0° = 2 ( 150 ft>s ) sin 0° = 0

Ans.

V ! u = 90° = 2 ( 150 ft>s ) sin 90° = 300 ft>s

Ans.

V ! u = 150° = 2 ( 150 ft>s ) sin 150° = 150 ft>s

Ans.

Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as between a point within the uniform flow and a point on the building. Since the flow occurs in the horizontal plane, the gravity term can be excluded. p0 p V0 2 V2 = + + r r 2 2 p = p0 +

p 2 ( V0 - V 2 ) 2

Here p0 = 0 and V0 = 150 ft>s. Then P ! u=0° = 0 + °

0.00237 slug>ft 3 2

= a26.66 P ! u=90° = 0 + °

lb 1ft 2 ba b = 0.185 psi 2 12 in ft

0.00237 slug>ft 3 2

= a -79.99 Pu = 150° = 0 + °

¢ 3 ( 150 ft>s ) 2 - 0 4 Ans.

¢ 3 ( 150 ft>s ) 2 - ( 300 ft>s ) 2 4

lb 1ft 2 ba b = - 0.555 psi 2 12 in ft

0.00237 slug>ft 3 2

Ans.

¢ 3 ( 150 ft>s ) 2 - ( 150 ft>s ) 2 4 = 0 827

Ans.

Ans: V ! u = 0° = 0 V ! u = 90° = 300 ft>s V ! u = 150° = 150 ft>s p! u = 0° = 0.185 p! u = 90° = - 0.555 psi p! u = 150° = 0

*7–88. The pipe is built from four quarter segments that are glued together. If it is exposed to a uniform air flow having a velocity of 8 m>s, determine the resultant force the pressure exerts on the quarter segment AB per unit length of the pipe. Take r = 1.22 kg>m3.

8 m/s B

90! A

0.1 m

SOLUTION We consider ideal fluid flow. Here, the pressure at point O removed from the cylinder is atmospheric. Thus, the net pressure on the surface of the cylinder is the gauge pressure given by pg = p - p0 =

1 rU 2 ( 1 - 4 sin2 u ) 2 (FR)y

The force that the gauge pressure exerts on the differential area dA = (adu)(1) = adu is

(FR)x

Equating the forces along the x and y axes shown in Fig. a, + ( FR ) x = ΣFx; d

( FR ) x =

dF cos u Lp>2 1 raU 2 ( 1 - 4 sin2 u ) cos udu Lp>2 2 p

= = = + c ( FR ) y = ΣFy;

1 raU 2 ( cos u - 4 sin2 u cos u ) du Lp>2 2 p 1 4 raU 2 asin u - sin3 u b ` 2 3 p>2

1 raU 2 6

( FR ) y = -

p

dF sinu Lp>2 p

= -

Lp>2

1 raU 2 ( 1 - 4 sin2 u ) sinudu 2

p

= -

1 raU 2 ( sin u - 4 sin3 u ) du Lp>2 2

p 1 4 = - raU 2 c - cos u - c - cos u ( sin2 u + 2 ) d d ` 2 3 p>2

=

= (a)

p

p

=

ds = ad¨ dF

1 1 dF = pgdA = rU 2 ( 1 - 4 sin2 u )( adu ) = raU 2 ( 1 - 4 sin2 u ) du 2 2

5 raU 2 6

Here, U = 8 m>s, r = 1.22 kg>m3, and a = 0.1 m, 1 6

( FR ) x = ( 1.22 kg>m3 )( 0.1 m )( 8 m>s ) 2 = 1.3013 N>m 5 6

( FR ) y = ( 1.22 kg>m3 )( 0.1m )( 8 m>s ) 2 = 6.507 N>m Thus, per unit length the magnitude of the resultant force on the quarter-segments is FR = 2 ( FR ) x2 + ( FR ) y2 = 2 ( 1.3013 N ) 2 + ( 6.507 N ) 2 = 6.64 N>m outward (suction)

828

Ans.

a ¨



7–89. The 1-ft-diameter cylinder is rotating at v = 5 rad>s while it is subjected to a uniform flow having a velocity of 4 ft>s. Determine the lift force on the cylinder per unit length. Take r = 2.38 1 10-3 2 slug>ft 3.

A

4 ft/s

1 ft B u ! 90"

v 0.5 ft

SOLUTION For the corresponding free vortex, at r = 0.5 ft, vu = vr = ( 5 rad>s )( 0.5 ft ) = 2.5 ft>s vr = 0 Thus, the circulation around the cylinder can be determined using Γ =

C

V # ds =

L0

2p

vur du =

L0

2p

2 ( 2.5 ft>s )( 0.5 ft ) du = 1.25u ! 2p 0 = 2.5p ft >s

The uplift force on the cylinder can be determined using Fy = rUΓ =

3 2.38 ( 10-3 ) slug>ft3 4 ( 4 ft>s )( 2.5p ft2 >s )

= 0.0748 lb>ft

Ans.

Ans: 0.0748 lb>ft 829

7–90. The 1-ft-diameter cylinder is rotating at v = 8 rad>s while it is subjected to a flow having a uniform horizontal velocity of 4 ft>s. If the pressure within the uniform flow is 80 lb>ft 2, determine the pressure on the surface B of the cylinder at u = 90°, and at A, where r = 1 ft, u = 90°. Also find the resultant force acting per unit length of the cylinder. Take r = 1.94 slug>ft 3.

A

4 ft/s

1 ft B u ! 90"

v 0.5 ft

SOLUTION We consider ideal fluid flow. For the corresponding free vortex, at r = 0.5 ft, vu = vr = ( 8 rad>s ) (0.5 ft) = 4 ft>s. Thus, the circulation is C

Γ =

V # ds =

L0

2p

L0

v0 ( rdu ) =

2p

4 ( 0.5 du ) = 2 ( 2p ) = 4p ft 2 >s.

The velocity components of the flow around the cylinder are vr = U a1 -

a2 b cos u r2

vu = -U a1 +

a2 Γ b sin u + 2 2pr r

Here, U = 4 ft>s, a = 0.5 ft and u = 90°. For point A, rA = 1 ft. Then

( vr ) A = ( 4 ft>s ) £ 1 ( vu ) A = - ( 4 ft>s ) £ 1 +

Thus, VA = ( vu ) A = 3 ft>s.

( 0.5 ft ) 2 § cos 90° = 0 ( 1 ft ) 2

( 0.5 ft ) 2 4p ft 2 >s § sin 90° + = -3 ft>s ( 1 ft ) 2 2p ( 1 ft )

For point B, on the surface, (vr)B = 0

( vu ) B = - ( 4ft>s ) £ 1 +

Thus, VB = ( vu ) B = 4 ft>s.

( 0.5 ft ) 2 4p ft 2 >s § sin 90° + = - 4ft>s 2 2p (0.5 ft) ( 0.5 ft )

Since the flow is irrotational, Bernoulli’s equation can be applied between two points located on the different streamlines. Here these two points are one within the uniform flow and the other one is point A (or B) p p0 v02 v2 + gz0 = + gz + + r r 2 2 Since the flow occurs in the horizontal plane, z0 = z Here, V0 = U, Thus p = p0 +

r (U 2 - V2) 2

Here, U = 4 ft>s; p0 = 80 lb>ft 2. At point A, pA = 80 lb>ft 2 + = 86.8 lb>ft 2 At point B, pB = 80 lb>ft 2 + = 80 lb>ft 2

1.94 slug>ft 3 2

3 ( 4 ft>s ) 2

1.94 slug>ft 3 2

The resultant force acting on the cylinder is

3 ( 4ft>s ) 2

- ( 3 ft>s ) 2 4 - ( 4ft>s ) 2 4

Fy = -rUΓ = - ( 1.94 slug>ft 3 )( 4 ft>s )( 4p ft 2 >s )

Ans.

Ans. Ans: pA = 86.8 lb>ft 2 pB = 80 lb>ft 2 Fy = 97.5 lb>ft

= - 97.5 lb>ft = 97.5 lb>ft T

830

7–91. The cylinder rotates counterclockwise at 40 rad>s. If the uniform velocity of the air is 10 m>s, and the pressure within the uniform flow is 300 Pa, determine the maximum and minimum pressure on the surface of the cylinder. Also, what is the lift force on the cylinder? Take ra = 1.20 kg>m3.

0.6 m 10 m/s 40 rad/s

SOLUTION We consider ideal fluid flow. For the corresponding free vortex at r = 0.6 m, vu = vr = ( 40 rad>s ) (0.6 m) = 24 m>s. Thus, the circulation is Γ =

C

V # ds =

L0

2p

vu(rdu) =

L0

2p

24(0.6du) = 28.8p m2 >s

Since Γ 7 4pUa = 4p ( 10 m>s ) (0.6 m) = 24p m2 >s, the stagnation point will not be on the surface of the wheel. The pressure at a point on the surface is p = p0 +

2 1 2 Γ rU c 1 - a - 2 sin u + b d 2 2pUa

Since Γ 7 4pUa, the term - 2 sin u +

Γ is the smallest when u = 90°, which 2p va

yields the maximum pressure. Thus, p max = p0 +

2 1 2 Γ rU c 1 - a - 2 + b d 2 2pUa

= 300 Pa +

2 28.8p m2 >s 1 ( 1.20 kg>m3 )( 10 m>s ) 2 e 1 - c -2 + d f 2 2p ( 10 m>s ) (0.6 m)

Ans.

= 350 Pa

Also, we notice that the minimum pressure occurs at a point where u = 90°. Then p min = p0 +

2 1 2 Γ rU c 1 - a2 + b d 2 2pUa

= 300 Pa +

2 28.8p m2 >s 1 ( 1.20 kg>m3 )( 10 m>s ) 2 e 1 - c 2 + d f 2 2p ( 10 m>s ) (0.6 m)

Ans.

= - 802 Pa

Ans: pmax = 350 Pa pmin = - 802 Pa 831

*7–92. A torque T is applied to the cylinder, causing it to rotate counterclockwise with a constant angular velocity of 120 rev>min. If the wind is blowing at a constant speed of 15 m>s, determine the location of the stagnation points on the surface of the cylinder, and find the maximum pressure. The pressure within the uniform flow is 400 Pa. Take ra = 1.20 kg>m3.

120 rev/min 200 mm 15 m/s

SOLUTION We consider ideal fluid flow. For the corresponding free vortex at r = 0.2 m, vu = vr rev 2p rad 1 m ba ba b d (0.2 m) = 0.8p m>s . Thus, the circulation of this min 1 rev 60 s free vortex is = c a120

Γ =

C

V # ds =

L0

2p

vu(rdu) =

L0

2p

(0.8 p)(0.2 du) = 0.32p2 m2 >s

Since Γ 6 4pUa = 4p ( 15 m>s ) (0.2 m) = 12p m2 >s, there exist two stagnation points on the surface. The location of these two points can be found using sin u =

0.32p2 m2 >s Γ = = 0.08378 4pUa 4p ( 15 m>s ) (0.2 m)

u = 4.806° = 4.81° or u = 175.19° = 176°

Ans.

The maximum pressure occurs at the stagnation point where V = 0. Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as one within the uniform flow and the other one at the stagnation point. Here, the gravity term can be neglected since the flow involves air which has a low density. p0 p V0 2 V2 + = + ra ra 2 2 Here V0 = U = 15 m>s, p0 = 400 Pa, p = p max and V = 0. Then 400 N>m2 3

1.20 kg>m

+

( 15 m>s ) 2 2

=

p max 1.20 kg>m3

+ 0 Ans.

p max = 535 Pa

832

T

7–93. A torque T is applied to the cylinder, causing it to rotate counterclockwise with a constant angular velocity of 120 rev>min. If the wind is blowing at a constant speed of 15 m>s, determine the lift per unit length on the cylinder and the minimum pressure on the cylinder. The pressure within the uniform flow is 400 Pa. Take ra = 1.20 kg>m3.

120 rev/min 200 mm 15 m/s

T

SOLUTION We consider ideal fluid flow. For the corresponding free vortex at r = 0.2 m, vu = vr rev 2p rad 1 min ba ba b d (0.2 m) = 0.8p m>s . Thus, the circulation of min 1 rev 60 s this free vortex is = c a120

Γ =

C

V # ds =

L0

2p

vu(r du) =

L0

2p

(0.8p)(0.2 du) = 0.32p2 m2 >s

The “lift” exerted on the cylinder can be determined from

F = rUΓ = ( 1.20 kg>m3 ) (15 m>s) ( 0.32p2 m2 >s ) = 56.8 kN>m

Ans.

The pressure at a point on the surface is p = p0 +

2 1 2 Γ rU c 1 - a - 2 sin u + b d 2 2pUa

We notice from this equation that p will be minimum at a point where u = -90°. Then p = p0 + = p0 +

2 1 Γ ra U 2 c 1 - a - 2 sin ( - 90°) + b d 2 2pUa 2 1 Γ raU 2 c 1 - a2 + b d 2 2pUa

= 400 Pa + = - 99.3 Pa

2 0.32p2 m2 >s 1 2 ( 1.20 kg>m3 )( 15 m>s ) e 1 - c 2 + d f 2 2p ( 15 m>s ) (0.2 m)

Ans.

Ans: F = 5.68 kN>m p = - 99.3 Pa 833

7–94. Liquid is confined between a top plate having an area A and a fixed surface. A force F is applied to the plate and gives the plate a velocity U. If this causes laminar flow, and the pressure does not vary, show that the Navier–Stokes and continuity equations indicate that the velocity distribution for this flow is defined by u = U(y>h), and that the shear stress within the liquid is txy = F>A.

y

F h

Since the flow is steady and is only along the x axis then v = w = 0. Also, the liquid is incompressible. Thus, the continuity equation reduces to 0(ru) 0(rv) 0(rw) 0r + + + = 0 0t 0x 0y 0z 0u + 0 + 0 = 0 0x 0u = 0 0x Integrating this equation with respect to x u = u(y) Using this result, when the pressure p remains constant along x axis, then Navier-Stokes equation along x axis gives ra

0p 0u 02u 0u 0u 0u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z 0 + 0 + 0 + 0 = 0 - 0 + ma0 +

m Since u is a function of y only, y twice.

02u + 0b 0y2

02u = 0 0y2

d 2u 02u = . Integrating this equation with respect to 0y2 dy2 du = C1 dy

(1)

And, (2)

u = C1y + C2 Applying the boundary condition, u = 0 at y = 0 0 = C1(0) + C2

C2 = 0

U = C1(h)

U h

and u = U at y = h, C1 =

Substituting these results into Eq. 1 u = a

u x

SOLUTION

0 + r

U

U by h

(Q.E.D)

834

7–94. Continued

Applying txy = ma Here,

0v 0u + b 0y 0x

0v 0u U = 0 and from Eq. 1 = . Then 0x 0y h txy = ma

U U + 0b = ma b h h

This shows that txy is a constant between the liquid layers. Therefore, its value is equal to that at the bottom surface of the top plate which is txy =

F A

(Q.E.D)

Although not necessary, the Navier-Stokes equation along the y axis gives ra

0p 0v 0v 0v 0v 02v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z r ( 0 + 0 + 0 + 0 ) = r( - g) -

0p + m(0 + 0 + 0 + 0) 0y

0r = - rg 0y Integrating this equation with respect to y by realizing the pressure variation is along y axis only, p = - rgy + C3 Applying the boundary condition p = 0 at y = h, 0 = - rgh + C3

C3 = rgh

Then p = rg(h - y)

835

7–95. The channel for a liquid is formed by two fixed plates. If laminar flow occurs between the plates, show that the Navier–Strokes and continuity equations reduce to 02u>0y2 = (1>m) 0p>0x and 0p>0y = 0. Integrate these equations to show that the velocity profile for the flow is u = (1>(2m)) (dp>dx) 3 y2 - (d>2)2 4 . Neglect the effect of gravity.

y

d/2 d/2

SOLUTION Since the flow is steady and is along the x axis only, then v = w = 0. Also, the liquid is incompressible. Thus, the continuity equation becomes 0(ru) 0(rv) 0(rw) 0r + + + = 0 0t 0x 0y 0z 0 + r

0u + 0 + 0 = 0 0x 0u = 0 0x

Integrating this equation with respect to x, u = u(y) Using this result, the Navier-Stokes equation along the x and y axes gives ra

0p 0u 02u 0u 0u 0u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z r( 0 + 0 + 0 + 0 ) = 0 -

0p 02u + m( 0 + 2 + 0 ) 0x 0y

02u 1 0p = m 0x 0y2 ra

(Q.E.D)

(1)

0p 0v 0v 0v 0v 02v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z r(0 + 0 + 0 + 0) = 0 0p = 0 0y

0p + m(0 + 0 + 0) 0y (Q.E.D)

(2)

Integrating Eq. 2 with respect to y, p = p(x) 02u d 2u Since u is a function of y only and p is a function of x only, then 2 = and 0y dy2 0r dr = . Eq. 1 becomes 0x dx d 2u 1 dp = 2 m dx dy

836

x

*7–95. Continued

Integrating this equation twice with respect to y, du 1 dr = a by + C1 m dx dy u = a Since u is maximum at y = 0. Then 0 = a

Also, u = 0 at y =

(3)

1 dr 2 by + C1y + C2 2m dx

(4)

du = 0 at y = 0. Using Eq. 3 dy

1 dr b(0) + C1 m dx

C1 = 0

d . Using Eq. 4 with C1 = 0, 2

0 = a

1 dr d 2 ba b + 0 + C2 2m dx 2

Substituting these results into Eq. 4, u = a = a

C2 = - a

1 dr d 2 ba b 2m dx 2

1 dr 2 1 dr d 2 by - a ba b 2m dx 2m dx 2 1 dr d 2 b c y2 - a b d 2m dx 2

(Q.E.D)

837

*7–96. Fluid having a density r and viscosity m fills the space between the two cylinders. If the outer cylinder is fixed, and the inner one is rotating at v, apply the Navier–Stokes equations to determine the velocity profile assuming laminar flow.

ro

SOLUTION Since the flow is steady and is along the transverse direction (u axis) only, then vr = vz = 0. Also, the liquid is incompressible. Thus, the continuity equation reduces to 0 ( rvz ) 0r 1 0 ( rrvr ) 1 0 ( rvu ) + + + = 0 0t r 0r 0z rr 0v0u u + 0 = 0 0 + 0 + r 0u 0vu = 0 0u Integrating this equation with respect to u, vu = vu(r) Using this result, the Navier-Stokes equations along the u axis gives ra

= -

0vu vu 0vu vrvu 0vu 0vu + vr + + + vz b 0t 0r r 0u r 0z

vu 02vu 1 0 0vu 1 0p 1 02vu 2 0vr + rgu + mc ar b - 2 + 2 2 + 2 + d r 0u r 0r 0r r r 0u r 0u 0z2

r(0 + 0 + 0 + 0 + 0) = -0 + 0 + m c However, it can be shown that

Thus,

v

ri

vu 1 0 0vu ar b - 2 + 0 + 0 + 0 d r 0r 0r r

vu 1 0 0vu ar b - 2 = 0 r 0r 0r r

vu 0 1 0 1 0 0vu ( rv ) d = c ar b - 2 0r r 0r u r 0r 0r r 0 1 0 ( rv ) d = 0 c 0r r 0r u

838

r

7–96. Continued

Since vu = vu(r), then the above equation can be written in the form of d 1 d c (rv ) d = 0 dr r dr u

Integrating this equation with respect to r,

1 d ( rv ) = C1 r dr u d ( rv ) = C1r dr u Integrating again, rvu = C1a

r2 b + C2 2

C1 C2 r + 2 r

vu =

(1)

At r = ru, vu = 0. Then Eq. 1 gives C1 C (r ) + 2 2 0 r0 At r = ri, vu = vri. Then Eq. 1 gives

(2)

0 =

C1 C2 r + 2 i ri

vri =

(3)

Solving Eq. 2 and 3, C1 = -

2vr i2 r 02

- ri

2

C2 =

vr i2r 02 r 02 - r i2

Substituting these results into Eq. 1, vu = - a vu =

vr i2 r 02

- ri

vr i2

r 02 - r i

a 2

2

br + a

vr i2r 02 r 02 - r i2

r 02 - r 2 b r

1 ba b r Ans.

839

7–97. A horizontal velocity field is defined by u = 2 1 x2 - y2 2 ft>s and v = (- 4xy) ft>s. Show that these expressions satisfy the continuity equation. Using the Navier–Stokes equations, show that the pressure distribution is defined by p = C - rV 2 >2 - rgz.

SOLUTION For the continuity equation, 0(ru) 0(rv) 0(rw) 0r + + + 0t 0x 0y 0z = 0 + 4rx + ( - 4rx) + 0 = 0 (satisfied) The Navier-Stokes equations along the x, y and z axes are ra

0p 0u 02u 02u 02u 0u 0u 0u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z

r 3 0 + ( 2x2 - 2y2 )( 4x ) +

ra

( - 4xy )( - 4y ) + 0 4 = 0 -

0p + m(4 - 4 + 0) 0x

0p = -8r ( x3 + xy2 ) 0x

(1)

0p 0v 0v 0v 0v 02v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2b 0t 0x 0y 0z 0y 0x 0y 0z

r 3 0 + ( 2x2 - 2y2 ) ( - 4y) + ( - 4xy)( -4x) + 0 4 = 0 -

0p + m(0 + 0 + 0) 0y

0r = - 8r ( y3 + x2y ) 0y

ra

(2)

0p 0w 0w 0w 0w 02w 02w 02w + u + v + w b = rgz + ma 2 + 2 + 2 b 0t 0x 0y 0z 0z 0x 0y 0z r(0 + 0 + 0 + 0) = r( -g) -

0p + m(0 + 0 + 0) 0z

0p = -rg 0z

(3)

Integrating Eqs. 1 with respect to x, p = - 8r °

x2y2 x4 + ¢ + f(y) + g(z) 4 2

(4)

Differentiate Eq. 4 with respect to y and equate to Eq. 2 0p = -8r ( x2y ) + f 1(y) = - 8r ( y3 + x2y ) 0y f 1(y) = - 8ry3

840

7–97. Continued

Integrate this equation with respect to y, f(y) = - 2ry4 + C1

(5)

Differentiate Eq. 4 with respect to z and equal to Eq. 3 0p = g1(z) = - rg 0z Integrate this equation with respect to z, (6)

g(z) = - rgz + C2 Substitute Eq. 5 and 6 into 4, p = -8ra

x2y2 x4 + b - 2ry4 - rgz + C 4 2

p = - 2r ( x4 + y4 + 2x2y2 ) - rgz + C

1 1 1 Since V 2 = ( u2 + v2 ) = 3 ( 2x2 - 2y2 ) 2 + ( - 4xy)2 4 = 2 ( x4 + y4 + 2x2y2 ) , 2 2 2 then the above equation becomes p = C -

1 2 rV - rgz 2

(Q.E.D)

841

7–98. The sloped open channel has steady laminar flow at a depth h. Show that the Navier–Stokes equations reduce to 02u>0y2 = - (r g sin u)>m and 0p>0y = -r g cos u. Integrate these equations to show that the velocity profile is u = [(r g sin u)>2m] 1 2hy - y2 2 and the shear-stress distribution is txy = r g sin u (h - y).

y h

u x

SOLUTION Since the flow is steady and is along the x axis only, then v = w = 0. Also, the liquid is incompressible. Thus, the continuity equation reduces to 0(ru) 0(rv) 0(rw) 0r + + + = 0 0t 0x 0y 0z 0 + r

0u + 0 + 0 = 0 0x 0u = 0 0x

Integrating this equation with respect to x, u = u(y) The Navier-Stokes equations along the x and y axes give ra

0p 0u 02u 0u 0u 0u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z r(0 + 0 + 0 + 0) = rg sin u - 0 + m a0 + rg sin u 02u = m 0y2

ra

02u + 0b 0y2

(Q.E.D)

(1)

0r 0v 0v 0v 0v 02v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z r(0 + 0 + 0 + 0) = r( - g cos u) -

0p + m(0 + 0 + 0) 0y

0p = -rg cos u 0y Since u = u(y), then

(Q.E.D)

(2)

02u 02u = 2 . Thus Eq. (1) becomes 2 0y 0y rg sin u d 2u = m dy2

Integrating this equation with respect to y twice yields rg sin u du = y + C1 m dy u = -

(3)

rg sin u 2 y + C1y + C2 2m

(4)

842

7–98. Continued

At y = 0, u = 0. Then, Eq. 4 gives 0 = - 0 + 0 + C2 At y = h, txy = m a

C2 = 0

du b = 0. Then Eq. 3 gives dy 0 = -

rg sin u (h) + C1 m

C1 =

rgh sin u m

Substituting these results into Eq 3 and 4 rg sin u rgh sin u du = y + m m dy rg sin u du = (h - y) m dy u = u =

rg sin u 2 rgh sin u y + y m 2m

rg sin u ( 2hy - y2 ) 2m

(Q.E.D)

The shear stress distribution is txy = m

du = rg sin u(h - y) dy

(Q.E.D)

843

7–99. The laminar flow of a fluid has velocity components u = 6x and v = - 6y, where y is vertical. Use the Navier–Stokes equations to determine the pressure in the fluid, p = p(x, y), if at point (0, 0), p = 0. The density of the fluid is r.

y

A

B

2m

x

0.5 m

SOLUTION Since the flow is steady and the fluid is incompressible, the continuity equation is 0(ru) 0(rv) 0(rw) 0r + + + = 0 + 6r + ( -6r) + 0 = 0 0t 0x 0y 0z is indeed satisfied. Writing the Navier-Stokes equation along the x and y axes gives ra

0r 0u 0u 0u 0u 02u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z r30 + 6x(6) + 0 + 04 = 0 -

0r + 0 0x

0r = - 36rx 0x ra

(1)

0p 0v 0v 0v 0v 02v 02v 02v + u + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z r30 + 0 + ( -6y)( - 6) + 04 = r( -g) -

0p 0y

0p = -rg - 36ry 0y

(2)

Integrating Eq. 1 with respect to x r = - 18rx2 + f(y)

(3)

Takes the partial derivative of Eq. 2 with respect to y and equate it to Eq. 2 0p = f 1(y) = - rg - 36ry 0y Integrate this equation with respect to y, f(y) = - rgy - 18ry2 + C Substitute this result into Eq. 3 p = -18rx2 - pgy - 18ry2 + C p = - r ( 18x2 + 18y2 + gy ) + C

(4)

At point (0, 0), p = 0. Then Eq. 4 gives 0 = r(0 + 0 + 0) + C

C = 0

Thus, the pressure distribution is p(x, y) = - r ( 18x2 + 18y2 + gy )

Ans.

Ans: p = - r (18x2 + 18y2 + gy ) 844

*7–100. The steady laminar flow of an ideal fluid towards  the fixed surface has a velocity of u = 3 10 1 1 + 1> 1 8x3 2 4 m>s along the horizontal streamline AB. Use the Navier–Stokes equations and determine the variation of the pressure along this streamline, and plot it for -2.5 m … x … - 0.5 m. The pressure at A is 5 kPa, and the density of the fluid is r = 1000 kg>m3.

y

A

B

2m

x

0.5 m

SOLUTION We consider ideal fluid flow. Since streamline AB is along the x axis, the velocity of the flow along this streamline will not have components along the y and z axes; ie, v = w = 0. Also, the fluid is ideal and the flow is steady. Writing the Navier-Stoke equation along x axis gives ra

0p 0u 0u 0u 0u 02u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z

rc 0 + 10a1 +

0p 1 30 02u ba - 4 b d = 0 + 0 a 2 + 0 + 0b 3 0x 0x 8x 8x

0p 75r 75r 75r 1 = a1 + b = + 4 3 4 0x 16x7 2x 8x 2x

Integrating this equation with respect to x, p(x) = p(x) = -

25r 2x

3

25r 32x6

-

25r 32x6

+ C

( 16x3 + 1 ) + C

At x = -2.5 m, p = 5 kPa and with r = 1000 kg>m3. Then, 5 = -

25 32 ( - 2.5 ) 6

C = 4.2032r

3 16 ( - 2.5 ) 3

+ 14 + C

p (kPa)

Thus, since r = 1000 kg>m3, p(x) = c -

25 32x6

( 16x3 + 1 ) + 4.2032 d kPa

Ans.

60

The plot of this pressure distribution is shown in Fig. a x(m)

- 2.5

- 2.0

- 1.5

- 1.0

- 0.5

p(kPa)

5.0

5.75

7.84

15.92

54.20

50 40 30 20 10 x(m) –2.5

–2.0

–1.5

–1.0

(a)

845

–0.5

8–1. Investigate if each ratio is dimensionless. a) rV 2 >p, b) Lr>s, c) p>V 2L, d) rL3 >Vm.

SOLUTION 2

a)

rV = p

b)

Lr = s

c)

d)

a

M L 2 ba b L3 T = 1 M a b LT 2

M b T2 L3 = 2 M L a 2b T

rL3 = Vm

Ans.

no

Ans.

no

Ans.

no

Ans.

(L)a

M b M LT 2 = = 4 2 2 V L L L a b L T p

yes

a

M b ( L3 ) L3 = T2 L M a ba b T LT a

Ans: a) yes b) no c) no d) no 846

8–2. Use inspection to arrange each of the following three variables as a dimensionless ratio: a) L, t, V, b)  s, EV, L, c) V, g, L.

SOLUTION L a b(T) T = 1 L

a)

Vt = L

b)

EVL = s

c)

a

Ans.

M b(L) LT 2 = 1 M T2

Ans.

L 2 a b T V2 = = 1 gL L a 2 b(L) T

Ans.

Ans: Vt a) L EVL b) s V2 c) gL 847

8–3. The pressure change that occurs in the aortic artery during a short period of time can be modeled by the equation ∆p = ca ( mV>2R)1>2, where m is the viscosity of blood, V is its velocity, and R is the radius of the artery. Determine the M, L, T, dimensions for the arterial coefficient ca.

SOLUTION 1

The dimensions for the physical variables in the equation ∆p = ca a

Table 8–1 are

mV 2 b given in 2R

M LT 2 M = LT

Pressure change,

∆p

ML-1T -2 =

Viscosity,

m

ML-1T -1

Velocity,

V

Radius,

R

LT -1 =

L T

L

Thus, dimensional homogeneity requires

( M>LT )( L>T ) M = ca £ § 2 L LT

1 2

1

M M 2 = ca a b 2 LT LT 2

1

ca = a

M 2 b LT 2

Ans.

Ans: 1 M 2 a b LT 2 848

*8–4. Determine the Mach number for a jet flying at 800 mi>h at an altitude of 10 000 ft. The velocity of sound in air is determined from c = 2kRT, where the specific heat ratio for air is k = 1.40. Note, 1 mi = 5280 ft.

SOLUTION Referring to the table in Appendix A, at an altitude of 10 000 ft, T = 23.34° F. Also, R = 1716 ft # lb>slug # R. Here, Tabs = (23.34° F + 460) R = 483.34 R. The speed of sound through the air is then c = 2kRTabs

= 21.40 ( 1716 ft # lb>slug # R ) (483.34 R) = 1077.58 ft>s

The velocity of the jet is V =

800 mi 5280 ft 1h a ba b = 1173.3 ft>s 1h 1 mi 3600 s

The Mach number is

M =

1173.3 ft>s V = = 1.089 = 1.09 c 1077.58 ft>s

Ans.

849

8–5. Determine the F, L, T dimensions of the following terms. a) Q>rV, b) rg>p, c) V 2 >2g, d) rgh.

SOLUTION

a)

b)

c)

d)

Q = rV

rg = p

a

L3 b T

2

FT L a 4 ba b T L

a

=

L6 FT 2

Ans.

FT 2 L ba 2 b 4 1 T L = L F a 2b L

Ans.

L 2 a b T V = = L 2g L a 2b T 2

rgh = a

Ans.

FT 2 L F ba 2 b(L) = 2 T L L4

Ans.

Ans: L6 a) FT 2 1 b) L c) L F d) 2 L 850

8–6. Determine the M, L, T dimensions of the following terms. a) Q>rV, b) rg>p, c) V 2 >2g, d) rgh.

SOLUTION

a)

b)

c)

c)

L3 b Q T L5 = = rV M M L a 3 ba b T L a

rg p

=

a

Ans.

M L ba b 1 L3 T 2 = L M a b LT 2

Ans.

L 2 a b T V = = L 2g L a 2b T 2

rgh = a

Ans.

M L M ba 2 b(L) = 3 T LT 2 L

Ans.

Ans: L5 M 1 b) L c) L M d) LT 2 a)

851

8–7. Show that the Weber number is dimensionless using M, L, T dimensions and F, L, T dimensions. Determine its value for water at 70°F flowing at 8 ft>s for a characteristic length of 2 ft. Take sw = 4.98 1 10 - 3 2 lb>ft.

SOLUTION Using the M - L - T base dimensions, 2

We =

rV L = s

a

M L 2 M ba b (L) a 2b 3 T T L = = 1 M M a 2b a 2b T T

Using the F - L - T base dimensions, 2

We =

rV L = s

a

FT 2 L 2 F ba b (L) a b T L L4 = = 1 F F a b a b L L

From the table in Appendix A, r = 1.937 slug>ft 3 at T = 70° F. Substituting numerically, We =

rV 2L = s

( 1.937 slug>ft 3 )( 8 ft>s ) 2(2 ft) = 49.8 ( 103 ) 3 4.98 ( 10-3 ) 4 lb>ft

Ans.

Ans: 49.8 ( 103 ) 852

*8–8. The Womersley number is often used to study blood circulation in biomechanics when there is pulsating flow through a circular tube of diameter d. It is defined as Wo = 12 d22pfr>m, where f is the frequency of the pressure in cycles per second. Like the Reynolds number, Wo is a ratio of inertia and viscous forces. Show that this number is dimensionless.

SOLUTION The dimensions for the physical variables for the Womersley number are Diameter,

d

L

Frequency,

f

T -1

Density,

r

ML-3

Viscosity,

m

ML-1T -1

Then, Wo =

d 2pfr 2A m

= L

= L = L

A

( T -1 )( ML-3 ) ML-1T -1

1 M 1 a ba 3 ba b(L)(T) T M L B 1 = 1 A L2

Thus Wo is a dimensionless number.

853

8–9. The Womersley number is a dimensionless parameter that is used to study transient blood flow through the arteries during heartbeats. It is a ratio of transient to viscous forces and is written as Wo = r 22pfr>m, where r is the vessel radius, f is the frequency of the heartbeat, m the apparent viscosity, and r is the density of blood. Research has shown that the radius r of the aorta of a mammal can be related to its mass m by r = 0.0024m0.34, where r is in meters and m is in kilograms. Determine the Womersley number for a horse having a mass of 350 kg and heartbeat rate of 30 beats per minute (bpm), and compare it to that  of a rabbit having a mass of 2 kg and heartbeat rate of 180 bpm. The viscosities of blood for the horse and rabbit are mh = 0.0052 N # s>m2 and mr = 0.0040 N # s>m2, respectively. The density of blood for both is rb = 1060 kg>m3. Plot this variation of Womersley number (vertical axis) with mass for these two animals. The results should show that transient forces increase as the size of the animal increases. Explain why this happens.

SOLUTION The Womersley number expressed in terms of mass is Wo = 0.0024m0.34 For the horse, fh = a30

2pfr A m

beats 1 min ba b = 0.5 beats>s min 60 s

( Wo ) h = ( 0.0024m0.34 ) ( Wo ) h = 1.9206m0.34

B

32p(0.5) rad>s 4 ( 1060 kg>m3 ) 0.0052 N # s>m2

(1)

For m = 350 kg,

( Wo ) h = 1.9206 ( 3500.34 ) = 14.1 For the rabbit, fr = a180

Ans.

beats 1 min ba b = 3 beats>s min 60 s

( Wo ) r = ( 0.0024m0.34 ) ( Wo ) r = 5.3640m0.34

B

32p(3) rad>s 4 ( 1060 kg>m3 ) 0.0040 N # s>m2

(2)

For m = 2 kg,

( Wo ) r = 6.79

Ans.

The plots of the Womersley number vs mass for the horse and rabbit are shown in Fig. a and b, respectively. From these plots, we notice that Womersley number increases with mass.

854

8–9. Continued

For the horse Eq (1), m(kg)

0

50

100

150

200

250

300

350

400

( Wo ) h

0

7.26

9.19

10.55

11.64

12.55

13.36

14.07

14.73

200

250

300

350

(Wo)h ( 15

10

5

m(kg) 0

50

100

150

400

(a)

For the rabbit Eq (2) m(kg)

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

( Wo(W )

0

4.24

5.36

6.16

6.79

7.32

7.79

8.21

8.59

r

(Wo)r

10

5

m(kg) 0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

(b)

855

Ans: 6.79

8–10. Express the group of variables L, μ, r, V as a dimensionless ratio.

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f (L, m, r, V) = 0. Using the M - L - T system, L

L

m

ML-1T -1

r

ML-3

V

LT -1

Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, m, r, and V are chosen as m = 3 repeating variables and L will become the q variable. Thus, the Π term is Π = marbV cL Π = ( MaL-aT -a )( MbL-3b )( LcT -c ) (L) = Ma + bL-a - 3b + c + 1T -a - c M:

0 = a + b

L:

0 = - a - 3b + c + 1

T:

0 = -a - c

Solving a = -1, b = 1, c = 1 Thus, rVL m

Ans.

m is dimensionless rVL

Ans.

Π = m-1r1V 1L = or This is the Reynolds number. But also Π

-1

=

Ans: m rVL or m rVL 856

8–11. Express the group of variables p, g, D, r as a dimensionless ratio.

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f (p, V, D, r) = 0. Using the M - L - T system, p

ML-1T -2

g

LT -2

D

L

r

ML-3

Here, three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, p, g, and r are chosen as m = 3 repeating variables and D will become the q variable. Thus, the Π term is Π = p ag brcD Π = ( MaL-aT -2a )( L bT -2b )( McL-3c ) (L) = M a + cL-a + b - 3c + 1T -2a - 2b M:

0 = a + c

L:

0 = - a + b - 3c + 1

T:

0 = - 2a - 2b

Solving, a = -1, b = 1, c = 1 Thus, Π = p-1g1r1D =

rgD p

Ans.

Also, Π

-1

=

p is dimensionless rgD

Ans.

Ans: rgD p or p rgD 857

*8–12. The force of buoyancy F is a function of the volume V of a body and the specific weight g of the fluid. Determine how F is related to V and g.

SOLUTION Physical Variables. There are n = 3 variables and the unknown function is f(F, g, V) = 0. Using the F - L - T system, F

F

g

FL-3

V

L3

Here, only two base dimensions are used, so that m = 2. Thus, there is n - m = 3 - 2 = 1 Π term Dimensional Analysis. Here, F and g are chosen as m = 2 repeating variables and V will become the q variable. Thus, the Π term is Π = F ag bV Π = F a ( F bL-3b ) L3 = F a + bL-3b + 3 F:

0 = a + b

L:

0 = -3b + 3

Solving, a = - 1 and b = 1. Thus, Π = F -1g 1V =

gV F

The function can be written as fa

gV b = 0 F

Solving for F using this function Ans.

F = kgV where k is a constant to be determined by experiment.

858

8–13. Show that the hydrostatic pressure p of an incompressible fluid can be established using dimensional analysis by realizing that it depends upon the depth h in the fluid and the fluid’s specific weight g.

SOLUTION Physical Variables. There are n = 3 variables and the unknown function is f (p, g, h) = 0. Using the F - L - T system. p

FL-2

g

FL-3

h

L

Here, two base dimensions are used, so that m = 2. Thus, there is n - m = 3 - 2 = 1 Π term Dimensional Analysis. Here, p and g are chosen as m = 2 repeating variables and h will become the q variable. Thus, the Π term is Π = p ag bh Π = ( F aL-2a )( F bL-3b ) (L) = F a + bL-2a - 3b + 1 0 = a + b 0 = - 2a - 3b + 1 Solving, a = -1 and b = 1. Thus, Π = p-1g 1h =

gh p

The function can be written as fa

gh b = 0 p

Solving for p using this function, (Q.E.D.)

p = kgh where k is a constant to be determined by experiment.

859

8–14. Establish Newton’s law of viscosity using dimensional analysis, realizing that shear stress t is a function of the fluid viscosity μ and the angular deformation du>dy. Hint: Consider the unknown function as f(t, μ, du, dy).

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f(t, m, du, dy) = 0. Here, t has the same dimensions as pressure p. Using the M - L - T system, t

ML-1T -2

m

ML-1T -1

du

LT -1

dy

L

Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Note: Obtaining n 7 m was the reason for treating du and dy as two separate variables. Dimensional Analysis. Here t, m and du are chosen as m = 3 repeating variables and dy will become the q variable. Thus, the Π term is Π = tambducdy Π = ( MaL-aT -2a )( MbL-bT -b )( LcT -c )( L ) = Ma + bL-a + b - c + 1T -2a - b - c M:

0 = a + b

L:

0 = -a - b + c + 1

T:

0 = - 2a - b - c

Solving a = 1, b = - 1, c = -1. Thus, Π = t1m-1du -1dy = The function can be written as fa

t dy a b m du

t dy b = 0 m du

Solving for t using this function, du dy where k is a constant to be determined by experiment.

Ans.

t = km

Ans: t = km 860

du dy

8–15. The period of oscillation t, measured in seconds, of a buoy depends upon its cross-sectional area A, and its mass m, and the specific weight g of the water. Determine the relation between t and these parameters.

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f (A, g, m, t) = 0. The dimensions for t is T. Using the M - L - T system, A = L2 g = ML-2T -2 m = M Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, A, g, and m are chosen as m = 3 repeating variables and t will become the q variable. Thus, the Π term is Π = Aa g b mc t Π = ( L2a )( MbL-2bT -2b ) (Mc)(T) L:

0 = 2a - 2b

M:

0 = b + c

T:

0 = - 2b + 1

Solving, a = 1>2, b = 1>2, c = - 1>2. Thus, 1

1

1

Π = A2 g 2 M - 2 t = t

gA Am

The function can be written as f at

gA b = 0 Am

Solving for t from this function t = k

m A gA

Ans.

where k is a constant to be determined by experiment.

Ans: t = k 861

m A gA

*8–16. Laminar flow through a pipe produces a discharge Q that is a function of the pipe’s diameter D, the change in pressure ∆p per unit length, ∆p> ∆x, and the fluid viscosity, μ. Determine the relation between Q and these parameters.

D

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is ∆p f aQ, D, , mb = 0. Using the M - L - T system, ∆x Q

L3T -1

D

L

∆p ∆x

ML-2T -2

m

ML-1T -1

Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, Q, and D becomes the q variable. Π = Qaa

∆p , and m are chosen as m = 3 repeating variables ∆x

∆p b c b m D = ( L3aT -a )( MbL-2bT -2b )( McL-cT -c )( L ) = Mb + cL3a - 2b - c + 1T -a - 2b - c ∆x

M:

0 = b + c

L:

0 = 3a - 2b - c + 1

T:

0 = - a - 2b - c

Solving, a = - 1>4, b = 1>4, c = -1>4. Thus, ∆p 14 1 ∆p 4 -1 Π = Q a b m 4D = ° ∆x ¢ D ∆x Qm - 14

Therefore, the function can be written as ∆p 14 f ≥ ° ∆x ¢ D ¥ = 0 Qm Solving for Q, Q = kc

D4 ∆p a bd m ∆x

Ans.

where k is a constant to be determined by experiment.

862

8–17. The speed of sound V in air is thought to depend on the viscosity m, the density r, and the pressure p. Determine how V is related to these parameters.

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f(V, m, r, p) = 0. Using the M - L - T system, V

LT -1

m

ML-1T -1

r

ML-3

p

ML-1T -2

Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, V, m, and r are chosen as m = 3 repeating variables and p will become the q variable. Thus, the Π term is Π = V ambrcp Π = ( LaT -a )( MbL-bT -b )( McL-3c )( ML-1T -2 ) = Mb + c + 1La - b - 3c - 1T -a - b - 2 M:

0 = b + c + 1

L:

0 = a - b - 3c - 1

T:

0 = -a - b - 2

Solving, a = -2, b = 0, c = -1. Thus, Π = V -2m0r-1p =

p V 2r

The function can be rewritten as fa

p V 2r

b = 0

Solving for V using this function, V = k

p Ar

Ans.

where k is a constant to be determined by experiment. Notice that V is independent of m.

Ans: V = k 863

p Ar

8–18. The flow Q of gas through the pipe is a function of the density r of the gas, gravity g, and the diameter D of the pipe. Determine the relation between Q and these parameters.

D

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f(Q, r, g, D) = 0. Using the M - L - T system, Q

L3T -1

r

ML-3

g

LT -2

D

L

Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, Q, r, and g are chosen as m = 3 repeating variables and D will become the q variable. Thus the Π term is Π = QarbgcD Π = ( L3aT -a )( MbL-3b )( LcT -2c ) (L) = MbL3a - 3b + c + 1 + T -a - 2c M:

0 = b

L:

0 = 3a - 3b + c + 1

T:

0 = - a - 2c

2 1 Solving, a = - , b = 0, c = . Thus, 5 5 1

2

1

Π = Q -5r0g5D =

g5D 2

Q5

The function can be rewritten as 1



g5D 2

Q5

¢ = 0

Solving for Q using this function, 2

1

Q5 = k′g5D Q = k2gD5

Ans.

where k is a constant to be determined by experiment. Notice that Q is independent of r.

Ans: Q = k2gD5 864

8–19. The velocity V of the stream flowing from the side of the tank is thought to depend upon the liquid’s density r, the depth h, and the acceleration of gravity g. Determine the relation between V and these parameters.

h d

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f (V, r, g, h) = 0. Using the M - L - T system, V = LT -1 r = ML-3 h = L g = LT -2 Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, r, h, and g are chosen as m = 3 repeating variables and V will become the q variable. Thus the Π term is Π = r ah bgcV Π = ( MaL-3a )( Lb )( LcT -2c )( LT -1 ) = MaL-3a + b + c + 1T -2c - 1 M:

0 = a

L:

0 = - 3a + b + c + 1

T:

0 = - 2c - 1

1 1 Solving, a = 0, b = - , c = - . Thus, 2 2 1

V

1

Π = r0h - 2g - 2V =

2gh

The function can be written as f = a

V 2gh

b = 0

Solving for V from this function Ans.

V = k2gh

where k is a constant to be determined by experiment. Notice that V is independent of r.

Ans: V = k2gh 865

*8–20. The pressure p within the soap bubble is a function of the bubble’s radius r and the surface tension s of the liquid film. Determine the relation between p and these parameters.

r

SOLUTION Physical Variables. There are n = 3 variables and the unknown function is f (p, s, r) = 0. Using the F - L - T system, p

FL-2

s

FL-1

r

L

Here, only two base dimensions are used, so that m = 2. Thus, there is n - m = 3 - 2 = 1 Π term Dimensional Analysis. Here, p and s are chosen as m = 2 repeating variables and r will become the q variable. Thus the Π term is Π = p asbr Π = ( F aL-2a )( F bL-b ) (L) = F a + bL-2a - b + 1 F:

0 = a + b

L:

0 = - 2a - b + 1

Solving, a = 1 and b = -1. Thus, Π = p1s -1r =

pr s

The function can be written as fa

pr b = 0 s

Solving for p from this function p = k

s r

Ans.

where k is a constant to be determined by experiment.

866

8–21. The velocity c of a wave on the surface of a liquid depends upon the wave length l, the density r, and the surface tension s of the liquid. Determine the relation between c and these parameters. By what percent will c decrease if the density of the liquid is increased by a factor of 1.5?

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f(l, r, s, c) = 0. Using the F - L - T system, l

L

r

FT 2L-4

s

FL-1

c

LT -1

Here, all three base dimensions are used, so that m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Dimensional Analysis. Here, l, r, and s are chosen as m = 3 repeating variables and c will become the q variable. Thus the Π term is Π = la rb sc c Π = ( La )( F bT 2bL-4b )( F cL-c )( LT -1 ) = F b + cLa - 4b - c + 1T 2b - 1 F:

0 = b + c

L:

0 = a - 4b - c + 1

T:

0 = 2b - 1

1 1 1 , b = , and c = - . Thus, 2 2 2 rl 1 1 1 Π = l2 r 2 s -2 c = c As

Solving, a =

The function can be written as f ac

Solving for c,

rl b = 0 As

c = k

s A rl

Ans.

where k is a constant to be determined by experiment. , of decrease = a1 -

1 b * 100 = 18.4, A 1.5

Ans.

Ans: s A rl 18.4%

c = k

867

8–22. The discharge Q over the weir A depends upon the width b of the weir, the water head H, and the acceleration of gravity g. If Q is known to be proportional to b, determine the relation between Q and these variables. If H is doubled, how does this affect Q?

H A

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f (Q, H, g, b) = 0. Using the M - L - T system, Q

L3T -1

H

L

g

LT -2

b

L

Here, only two base dimensions are used, so that m = 2. Thus, there are n - m = 4 - 2 = 2 Π terms Dimensional Analysis. Here, g and H are chosen as m = 2 repeating variables. Thus the q variables are Q for Π1 and b for Π2. Π1 = gaHbQ = ( LaT -2a )( Lb )( L3T -1 ) = La + b + 3T -2a-1 L:

0 = a + b + 3

T:

0 = -2a - 1

1 5 Solving, a = - and b = - . Thus, 2 2 Q

5

1

Π1 = g - 2H - 2Q =

2gH5

Π2 = gcHdb = ( LcT -2c )( Ld )( L ) = Lc + d + 1T -2c L:

0 = c + d + 1

T:

0 = -2c

Solving, c = 0 and d = - 1. Thus, b H The function can be written as Π = g0H -1b = 2

Q

b ¢ = 0 2gH H Solving for Q, f°

5

,

Q = 2gH5f1a

b b H

Since Q is proportional to b, this function becomes Q = kb2gH3

Ans.

where k is a constant to be determined by experiment. If H is doubled, Q increases by 223 = 2.83 times

868

Ans.

Ans: Q = kb2gH3 increases by 2.83 times

8–23. The capillary rise of a fluid along the walls of the tube causes the fluid to rise a distance h. This effect depends upon the diameter d of the tube, the surface tension s, the density r of the fluid, and the gravitational acceleration g. Determine the relation between h and these parameters.

d

h

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (h, d, s, r, g) = 0. Using the M - L - T system, h

L

d

L

s

MT -2

r

ML-3

g

LT -2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, d, r, and g are chosen as m = 3 repeating variables. Thus the q variables are s for Π1 and h for Π2. Π = d arbges = ( La )( MbL-3b )( LcT -2c )( MT -2 ) = La - 3b + cMb + 1T -2c - 2 1

M:

0 = b + 1

L:

0 = a - 3b + c

T:

0 = -2c - 2

Solving, a = -2, b = - 1, and c = -1. Thus, s Π1 = d -2r-1g -1s = rd 2g Π2 = d erfgih = ( Le )( MfL-3f )( LiT -2i ) (L) = MfLe - 3f + i + 1T -2i M:

0 = f

L:

0 = e - 3f + i + 1

T:

0 = -2i

Solving, e = - 1, f = 0, and i = 0. Thus, h d The function can be written as Π = d -1r0g0h = 2

f1a

s h , b = 0 rd 2g d

Solving for h,

h s = fa 2 b d rd g

h = df a

s b rd 2g

Ans.

Notice that since d appears in the argument of f , we cannot say that h is proportional to d (which in any case we already know from Ch. 1 is not true). We have, however, established that for a given d, h is the same for all scenarios where s>rg is the same. 869

Ans: h = df a

s b rd 2g

*8–24. The torsional resistance T of the thrust bearing depends upon the diameter D of the shaft, the axial force F, the shaft rotation v, and the viscosity m of the lubricating fluid. Determine the relation between T and these parameters.

F

D v

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (T, D, F, v, m) = 0. Using the F - L - T system, T D F

FL L F

v m

T -1 FTL-2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, D, F, and m are chosen as m = 3 repeating variables. Thus the q variables are T for Π1 and v for Π2. Π = DaF bmcT = ( La )( F b )( F cT cL-2c ) (FL) = F b + c + 1La - 2c + 1T c 1

F: L: T:

0 = b + c + 1 0 = a - 2c + 1 0 = c

Solving, a = - 1, b = -1, and c = 0. Thus, T FD

Π = D-1F -1m0T = 1

Π = DeF fmgv = ( Le )( F f )( F gT gL-2g )( T -1 ) = F f + gLe - 2gT g - 1 2

F: L: T:

0 = f + g 0 = e - 2g 0 = g - 1

Solving, e = 2, f = - 1, and g = 1. Thus, Π = D2F -1m1v = 2

mD2v F

The function can be written as f1a

T mD2v , b = 0 FD F

Using this function to solve for T, mD2v T = fa b FD F T = FDf a

mD2v b F

Ans.

Notice that since F and D appear in the argument of f, we cannot say that T is proportional to F or to D. However, we have established that for given F and D, T is the same for all scenarios where mv is the same. 870

8–25. The thickness d of the boundary layer for a fluid passing over a flat plate depends upon the distance x from the plate’s leading edge, the free-stream velocity U of the flow, and the density r and viscosity m of the fluid. Determine the relation between d and these parameters.

U

u y

x

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (d, x, U, r, m) = 0. Using the M - L - T system, d x U

L L LT -1

r

ML-3

m

ML-1T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, x, U, and r are chosen as m = 3 repeating variables. Thus the q variables are d for Π1 and m for Π2. Π = xaU brcd = ( La )( LbT -b )( McL-3c ) (L) = McLa + b - 3c + 1T -b 1

0 = c 0 = a + b - 3c + 1 0 = -b

M: L: T:

Solving, a = -1, b = 0, and c = 0. Thus, Π = x -1V 0r0d = 1

d x

Π = xdU erfm = ( Ld )( LeT -e )( MfL-3f )( ML-1T -1 ) = Mf + 1Ld + e - 3f - 1T -e - 1 2

0 = f + 1 0 = d + e - 3f - 1 0 = -e - 1

M: L: T:

Solving, d = -1, e = - 1, and f = -1. Thus, Π = x -1U -1r-1m = 2

m rUx

or Π = 2

rUx = Re m

The function can be written as d f1a , Reb = 0 x

Solving for d,

d = f (Re) x Ans.

d = xf (Re)

871

Ans: d = xf (Re)

8–26. The discharge Q from a turbine is a function of the generated torque T, the angular rotation v of the turbine, its diameter D, and the liquid density, r. Determine the relations between Q and these parameters. If Q varies linearly with T, how does it vary with the turbine’s diameter D?

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f(Q, T, v, D, r) = 0. Using the F - L - T system, L3T -1 FL T -1 L FT 2L-4

Q T v D r

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Since Q is proportional to T, v, D, and r are chosen as m = 3 repeating variables. Thus, the q variables are Q for Π1 and T for Π2. Π1 = vaDbrcQ = ( T -a )( Lb )( F cT 2cL-4c )( L3T -1 ) = F cLb - 4c + 3T -a + 2c - 1 F:

0 = c

L:

0 = b - 4c + 3

T:

0 = -a + 2c - 1

Solving, a = - 1, b = -3, and c = 0. Thus, Π1 = v-1D-3r0Q =

Q vD3

Π2 = vdDer fT = ( T -d )( Le )( F fT 2fL-4f ) (FL) = F f + 1Le - 4f + 1T -d + 2f F:

0 = f + 1

L:

0 = e - 4f + 1

T:

0 = -d + 2f

Solving, d = - 2, e = -5, and f = -1. Thus, Π2 = v-2D-5r-1T =

T v2D5r

The function can be written as f1a

Q 3

,

T 2

vD v D5r

Solving for Q, Q vD3

= fa

b

T v2D5r

b

T b v2D5r Since Q is proportional to T, Q = vD3f a

Q = vD3k a

T 2

5

vDr

b = k

T vrD2

Ans. 2

where k is a constant to be determined by experiment. Q is inversely proportional to D . 872

Ans: Q = k

T vrD2

8–27. The speed c of a water wave is a function of the wave length l, the acceleration of gravity g, and the average depth of the water h. Determine the relation between c and these parameters.

l c

h

SOLUTION Physical Variables. There are n = 4 variables and the unknown function is f(c, l, g, h) = 0. Using the M - L - T system, LT -1 L LT -2 L

c l g h

Here, only two base dimensions are used, so that m = 2. Thus, there are n - m = 4 - 2 = 2 Π terms Dimensional Analysis. Here, g, and l are chosen as m = 2 repeating variables. Thus the q variables are c for Π1 and h for Π2. Π1 = galbc = ( LaT -2a )( Lb )( LT -1 ) = La + b + 1T -2a - 1 L:

0 = a + b + 1

T:

0 = -2a - 1

1 1 Solving, a = - and b = - . Thus, 2 2 1

c

1

Π1 = g -2l-2c =

2gl

Π1 = gcldh = ( LcT -2c )( Ld ) L = Lc + d + 1T -2c L:

0 = c + d + 1

T:

0 = -2c

Solving, d = -1 and c = 0. Thus, Π2 = g0l-1h =

h l

Therefore, the function is c h f1a , b = 0 2gl l

Solving for c, c

2gl

h = fa b l

l c = 2glf a b h

Note: In fact the equation is C = result just obtained.

Ans.

gl 2ph tan a b , which is consistent with the l B 2p 873

Ans: l c = 2gl f a b h

*8–28. The torque T developed by a turbine depends upon the depth h of water at the entrance, the density of the water r, the discharge Q, and the angular velocity of the turbine v. Determine the relation between T and these parameters.

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f(T, h, r, Q, v) = 0. Using the F - L - T system, T h r Q v

FL L FT 2L-4 L3T -1 T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, r, Q, and v are chosen as m = 3 repeating variables. Thus the q variables are T for Π1 and h for Π2. Π1 = raQbvcT = ( F aT 2aL-4a )( L3bT -b )( T -c ) (FL) = F a + 1L-4a + 3b + 1T 2a - b - c F:

0 = a + 1

L:

0 = -4a + 3b + 1

T:

0 = 2a - b - c

5 1 Solving, a = - 1, b = - , and c = - . Thus, 3 3 5 1 T Π1 = r-1Q -3v - 3T = 5 1 rQ 3v 3 Π2 = rdQevfh = ( F dT 2dL-4d )( L3eT -e )( T -f ) (L) = F dL-4d + 3e + 1T 2d - e - f F:

0 = d

L:

0 = -4d + 3e + 1

T:

0 = 2d - e - f

1 1 Solving, d = 0, e = - , and f = . Thus, 3 3 1 3 1 1 v Π2 = r0Q -3v 3h = a b h Q Therefore, the function is f1 £

1

T 5 3

rQ v

Solving for T,

1 3

,a

v 3 b h§ = 0 Q 1

T = rQ3v3f 3 a 5

1

v 3 b h4 Q

Ans.

874

8–29. The drag force FD on the jet plane is a function of the speed V, the characteristic length L of the plane, and the density r and viscosity m of the air. Determine the relation between FD and these parameters.

V FD L

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f(FD, V, L, r, m) = 0. Using the M - L - T system, MLT -2 LT -1 L ML-3 ML-1T -1

FD V L r m

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, r, V, and L are chosen as m = 3 repeating variables. Thus, the q variables are FD for Π1 and m for Π2. Π1 = raV bLcFD = ( MaL-3a )( LbT -b )( Lc )( MLT -2 ) = Ma + 1L-3a + b + c + 1T -b - 2 M:

0 = a + 1

L:

0 = -3a + b + c + 1

T:

0 = -b - 2

Solving, a = -1, b = - 2, and c = - 2. Thus, Π1 = r-1V -2L-2FD =

FD rV 2L2

Π2 = rdV eLfm = ( MdL-3d )( LeT -e )( Lf )( ML-1T -1 ) = Md + 1L-3d + e + f - 1T -e - 1 M:

0 = d + 1

L:

0 = -3d + e + f - 1

T:

0 = -e - 1

Solving, d = -1, e = - 1, and f = -1. Thus, Π2 = r-1V -1L-1m =

m rVL

or rVL = Re m

Π2 =

Therefore, the function can be written as f1 °

FD rV 2L2

, Re¢ = 0

Solving forFD, FD rV 2L2

= f (Re) FD = rV 2L2f (Re)

Ans. 875

Ans: FD = rV 2L2f (Re)

8–30. The time t needed for ethyl ether to drain from the pipette is thought to be a function of the fluid’s density r and viscosity m, the nozzle’s diameter d, and gravity g. Determine the relation between t and these parameters.

d

SOLUTION t = f(r, m, d, g) or g(t, r, m, d, g) = 0. Thus, n = 5. Using the M - L - T system given in table 8–1 Time,

t

T

Density,

r

ML-3

Viscosity,

m

ML-1 T -1

Diameter,

d,

L

Gravity,

g,

LT -2

Here, m = 3 since three base dimensions M, L and T are involved. Thus, there are n - m = 5 - 3 = 2 Π terms. r, d and g are chosen as m = 3 repeating variables since collectively they contain all three base dimensions as required. The first Π term using t as g the variable is Π1 = rad bgct = ( Ma L-3a )( Lb )( Lc T -2c )( T ) = Ma L-3a + b + c T -2c + 1 Thus, for M:

a=0

L:

-3a + b + c = 0

T:

-2c + 1 = 0

1 1 Solving, a = 0, b = - and c = . Then 2 2 g 1 1 Π1 = r0d -2 g 2t = t Ad

The second Π term using m as g variable is Π2 = rdd egt m = ( Md L-3d )( Le )( Lf T -2f )( ML-1T -1 ) = Md + 1 L-3d + e + f - 1 T -2f - 1

Thus, for M:

d + 1 = 0

L:

-3d + e + f - 1 = 0

T:

-2f - 1 = 0

876

8–30. Continued

3 1 Solving, d = -1, e = - and f = - . Then 2 2 m 3 1 Π2 = r-1d - 2g - 2 m = 3 1 rd 2g2 Then f1 °

g m t, ¢ = 0 A d rd 32 g12

Solving for

g t in this equation, Ad

g m t = f ° 3 1¢ Ad rd 2 g2

t =

m d f ° 3 1¢ Ag rd 2 g2

Ans.

Ans: t =

877

m d f ° 3 1¢ Ag rd 2 g2

8–31. The head loss hL in a pipe depends upon its diameter D, the velocity of flow V, and the density r and viscosity m of the fluid. Determine the relation between hL and these parameters.

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (hL, D, V, r, m) = 0. Using the M – L – T system, hL

L

D

L

V

LT -1

r

ML-3

m

ML-1T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, D, V, and r are chosen as m = 3 repeating variables. Thus, the q variables are hL for Π1 and m for Π2. Π1 = DaV brchL = ( La )( LbT -b )( McL-3c ) (L) = McLa + b - 3c + 1T -b M:

0 = c

L:

0 = a + b - 3c + 1

T:

0 = -b

Solving, a = - 1, b = 0, and c = 0. Thus, Π1 = D-1V 0r0hL =

hL D

Π2 = DdV er fm = ( Ld )( LeT -e )( MfL-3f )( ML-1T -1 ) = Mf + 1Ld + e - 3f - 1T -e - 1 M:

0 = f + 1

L:

0 = d + e - 3f - 1

T:

0 = -e - 1

Solving, d = - 1, e = -1, and f = -1. Thus, Π2 = D-1V -1r-1m =

m rVD

or Π2 =

rVD = Re m

Therefore the function can be written as f1 a

hL , Reb = 0 D

Solving for hL, hL = f (Re) D Ans.

hL = Df (Re)

878

Ans: hL = Df (Re)

*8–32. The pressure difference ∆p of air that flows through a fan is a function of the diameter D of the blade, its angular rotation v, the density r of the air, and the flow Q. Determine the relation between ∆p and these parameters.

D

SOLUTION Physical Variables. There are n = 6 variables and the unknown function is f (∆p, D, v, r, Q) = 0. Using the F – L – T system, ∆p

FL-2

D

L

v

T -1

r

FT 2L-4

Q

L3T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, D, v, and r are chosen as m = 3 repeating variables. Thus, the q variables are ∆p for Π1 and Q for Π2. Π1 = Davbrc ∆p = ( La )( T -b )( F cT 2cL-4c )( FL-2 ) = F c + 1La - 4c - 2T -b + 2c F:

0 = c + 1

L:

0 = a - 4c - 2

T:

0 = -b + 2c

Solving, a = -2, b = - 2, and c = - 1. Thus, Π1 = D-2v-2r-1 ∆p =

∆p rv2D2

Π2 = Ddver fQ = ( Ld )( T -e )( F fT 2fL-4f )( L3T -1 ) = F fLd - 4f + 3T -e + 2f - 1 F:

0 = f

L:

0 = d - 4f + 3

T:

0 = -e + 2f - 1

Solving, d = -3, e = - 1, and f = 0. Thus, Π2 = D-3v-1r0Q =

Q vD3

Therefore, the function can be written as f1 a

∆p Q , b = 0 rv2D2 vD3

Solving for ∆p, ∆p 2

2

rv D

= fa

Q vD3

b

∆p = rv2D2f a

Q vD3

Ans.

b

879

8–33. The period of time t between small water waves is thought to be a function of the wave length l, the water depth h, gravitational acceleration g, and the surface tension s of the water. Determine the relation between t and these parameters.

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (t, l, h, g, s) = 0. Using the M – L – T system, t

T

l

L

h

L

g

LT -2

s

MT -2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, l, g, and s are chosen as m = 3 repeating variables. Thus, the q variables are t for Π1 and h for Π2. Π1 = lagbsct = ( La )( LbT -2b )( McL-2c ) (T) = McLa + bT -2b - 2c + 1 M:

0 = c

L:

0 = a + b

T:

0 = -2b - 2c + 1

1 1 Solving, a = - , b = , and c = 0. Thus, 2 2 1

1

Π1 = l-2g2s0t = t

g Al

Π2 = ldgesfh = ( Ld )( LeT -2e )( MfL-2f ) (L) = MfLd + e + 1T -2e - 2f M:

0 = f

L:

0 = d + e + 1

T:

0 = -2e - 2f

Solving, d = - 1, e = 0, f = 0. Thus, Π2 = l-1g0s0h =

h l

Therefore, the function can be written as f1at

Solving for t, t

g h , b = 0 Al l

g h = fa b Al l

t =

l h fa b Ag l

Ans. Ans: t = 880

l h fa b Ag l

8–34. The drag force FD on the square plate held normal to the wind depends upon the area A of the plate and the air velocity V, density r, and viscosity m. Determine the relation between FD and these parameters. FD

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (FD, V, r, m, A) = 0. Using the M – L – T system, FD

MLT -2

V

LT -1

r

ML-3

m

ML-1T -1

A

L2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, V, r, and A are chosen as m = 3 repeating variables. Thus, the q variables are FD for Π1 and m for Π2. Π1 = V arbAcFD = ( LaT -a )( MbL-3b )( L2c )( MLT -2 ) = Mb + 1La - 3b + 2c + 1T -a - 2 M:

0 = b + 1

L:

0 = a - 3b + 2c + 1

T:

0 = -a - 2

Solving, a = -2, b = - 1, and c = - 1. Thus, FD Π1 = V -2r-1A-1FD = rV 2A Π2 = V dreAfm = ( LdT -d )( MeL-3e )( L2f )( ML-1T -1 ) = Me + 1Ld - 3e + 2f - 1T -d - 1 M:

0 = e + 1

L:

0 = d - 3e + 2f - 1

T:

0 = -d - 1

1 . Thus, 2 m m 1 Π2 = V -1r-1A-2m = = 1 rVL 2 rVA

Solving, d = -1, e = - 1, and f =

or rVL = Re m

Π2 =

Therefore, the function can be written as f1 a

FD rV 2A

, Reb = 0

Solving for FD, FD rV 2A

= f (Re) FD = rV 2Af (Re)

Ans.

881

Ans: FD = rV 2Af (Re)

8–35. The thrust T of the propeller on a boat depends upon the diameter D of the propeller, its angular velocity v, the speed of the boat V, and the density r and viscosity m of the water. Determine the relation between T and these parameters.

V

T

SOLUTION

Physical Variables. There are n = 6 variables and the unknown function is f(T, D, v, V, r, m) = 0. Using the F – L – T system, T

F

D

L T

v

-1

V

LT -1

r

FT 2L-4

m

FTL-2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 6 - 3 = 3 Π terms Dimensional Analysis. Here, r, v, and D are chosen as m = 3 repeating variables. Thus the q variables are T for Π1, m for Π2, and V for Π3. Π1 = ravbDcT = ( F aT 2aL-4a )( T -b )( Lc )( F ) = F a + 1L-4a + cT 2a - b F:

0 = a + 1

L:

0 = -4a + c

T:

0 = 2a - b

Solving, a = - 1, b = -2, and c = - 4. Thus, Π1 = r-1v-2D-4T =

T rv2D4

Π2 = rdveDfm = ( F dT 2dL-4d )( T -e )( Lf )( FTL-2 ) = F d + 1L-4d + f - 2T 2d - e + 1 F:

0 = d + 1

L:

0 = -4d + f - 2

T:

0 = 2d - e + 1

Solving, d = - 1, e = -1, and f = -2. Thus, Π2 = r-1v-1D-2m =

m rvD2

Π3 = rgvhDiV = ( F gT 2gL-4g )( T -k )( Li )( LT -1 ) = F gL-4g + i + 1T 2g - h - 1 F:

0 = g

L:

0 = -4g + i + 1

T:

0 = 2g - h - 1

Solving, g = 0, h = -1, and i = -1. Thus, Π3 = r0v-1D-1V =

V vD

Therefore, the function is f1 a

m T V , , b = 0 2 vD 2 4 rv D rvD

Solving for T,

m T V = fa , b 2 4 rvD2 vD rv D

T = rv2D4f a

Ans: V , b rvD2 vD m

Ans. 882

T = rv2D4f a

V b rvD vD m

2

,

*8–36. The power P of a blower depends upon the impeller diameter D, its angular velocity v, the discharge Q, and the fluid density r and viscosity m. Determine the relation between P and these parameters.

SOLUTION Physical Variables. There are n = 6 variables and the unknown function is f( p, D, v, Q, r, m) = 0. Using the M – L – T system, p

ML2T -3

D

L

v

T -1

Q

L3T -1

r

ML-3

m

ML-1T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 6 - 3 = 3 Π terms Dimensional Analysis. Here, D, v, and m are chosen as m = 3 repeating variables. Thus, the q variables are P for Π1, Q for Π2, and r for Π3. Π1 = DavbmcP = ( La )( T -b )( McL-cT -c )( ML2T -3 ) = Mc + 1La - c + 2T -b - c - 3 M:

0 = c + 1

L:

0 = a - c + 2

T:

0 = -b - c - 3

Solving, a = -3, b = - 2, and c = - 1. Thus, P D3v2m

Π1 = D-3v-2m-1P =

Π2 = DdvemfQ = ( Ld )( T -e )( MfL-fT -f )( L3T -1 ) = M fLd - f + 3T -e - f - 1 M:

0 = f

L:

0 = d - f + 3

T:

0 = -e - f - 1

Solving, d = -3, e = - 1, and f = 0. Thus, Π2 = D-3v-1m0Q =

Q D3v

Π3 = D v m r = ( L )( T -h )( MiL-iT -i )( ML-3 ) = Mi + 1Lg - i - 3T -h - i g h i

g

M:

0 = i + 1

L:

0 = g - i - 3

T:

0 = -h - i

Solving, g = 2, h = 1, and i = - 1. Thus, Π3 = D2v1m-1r =

rD2v m

Therefore, the function can be written as Q rD2v P , , b = 0 D3v2m D3v m Solving for P, f1 a

Q rD2v P = f , a b D3v2m D3v m

P = D3v2mf a

rD2v b D3v m Q

Ans.

,

883

8–37. The discharge Q of a pump is a function of the impeller # diameter D, its angular velocity v, the power output W, and the density r and viscosity m of the fluid. Determine the relation between Q and these parameters.

SOLUTION Physical Variables. There are n = 6 variables and the unknown function is # f(Q, D, v, W, r, m) = 0. Using the M – L – T system, L3T -1 L T -1 ML2T -3 ML-3 ML-1T -1

Q D v # W r m

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 6 - 3 = 3 Π terms Dimensional Analysis. Here, D, v, and P are chosen as m = 3 repeating variables. Thus, the q variables are Q for Π1, r for Π2, and m for Π3.

#

Π1 = DavbW cQ = ( La )( T -b )( McL2cT -3c )( L3T -1 ) = McLa + 2c + 3T -b - 3c - 1 0 = c 0 = a + 2c + 3 0 = -b - 3c - 1

M: L: T:

Solving, a = - 3, b = -1, and c = 0. Thus,

#

Π1 = D-3v-1W 0Q =

Q

D3v # Π2 = Ddve W fr = ( Ld )( T -e )( MfL2fT -3f )( ML-3 ) = Mf + 1Ld + 2f - 3T -e - 3f 0 = f + 1 0 = d + 2f - 3 0 = -e - 3f

M: L: T:

Solving, d = 5, e = 3, and f = - 1. Thus, rD5v3 # W

#

Π2 = D5v3W -1r =

#

Π3 = DgvhW im = ( Lg )( T -h )( MiL2iT -3i )( ML-1T -1 ) = Mi + 1Lg + 2i - 1T -h - 3i - 1 0 = i + 1 0 = g + 2i - 1 0 = -h - 3i - 1

M: L: T:

Solving, g = 3, h = 2, and i = - 1. Thus, D3v2m P Therefore, the function can be written as

#

Π3 = D3v2W -1m =

f1 a

rD5v3 D3v2m # , # b = 0 W Dv W Q 3

,

Solving for Q, Q D3v

= fa

rD5v3 D3v2m # , # b W W

Ans: 5 3

3 2

rD v D v m # b Q = D vf a # , W W 3

Ans. 884

Q = D3vf a

rD5v3 D3v2m # , # b W W

8–38. As the ball falls through a liquid, its velocity V is a function of the diameter D of the ball, its density rb, and the density r and viscosity m of the liquid, and the acceleration due to gravity g. Determine the relation between V and these parameters.

D

V

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f(V, D, rb, r, m) = 0. Using the M – L – T system, V

LT -1

D

L

rb

ML-3

r

ML-3

m

ML-1T -1

g

LT -2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 6 - 3 = 3 Π terms Dimensional Analysis. Here, D, r, and m are chosen as m = 3 repeating variables. Thus, the q variables are V for Π1, rb for Π2, g for Π3. Π1 = DarbmcV = ( La )( MbL-3b )( McL-cT -c )( LT -1 ) = Mb + cLa - 3b - c + 1T -c - 1 M:

0 = b + c

L:

0 = a - 3b - c + 1

T:

0 = -c - 1

Solving, a = 1, b = 1, and c = -1. Thus, Π1 = D1r1m-1V =

rVD m

Π2 = Ddremfrb = ( Ld )( MeL-3e )( MfL-fT -f )( ML-3 ) = Me + f + 1Ld - 3e - f - 3T -f M:

0 = e + f + 1

L:

0 = d - 3e - f - 3

T:

0 = -f

Solving, d = 0, e = - 1, and f = 0. Thus, Π2 = D0r-1m0rb =

rb r

Π3 = Dhrimjg = ( Lh )( MiL-3i )( MjL-jT -j )( LT -2 ) = Mi + jLh - 3i - j + 1T -j - 2 M:

0 = i + j

L:

0 = h - 3i - j + 1

T:

0 = -j - 2

885

8–38. Continued

Solving, h = 3, i = 2, j = - 2. Thus Π3 = D3r2m-2g =

D3e 2g m2

Therefore, the function can be written as fa

rVD rb D3r2g b = 0 , , m r m2

Solving for V,

rb D3r2g rVD b = f1 a , m r m2

V =

m rb D3r2g fa , b r rD m2

Ans.

Ans: V =

886

m rb D3r2g fa , b r rD m2

8–39. The change in pressure ∆p in the pipe is a function of the density r and the viscosity m of the fluid, the pipe diameter D, and the velocity V of the flow. Establish the relation between ∆p and these parameters.

D V

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (∆p, m, D, r, V) = 0. Using the M – L – T system, ∆p

ML-1T -2

m

ML-1T -1

D

L

r

ML-3

V

LT -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, D, r, and V are chosen as m = 3 repeating variables. Thus the q variables are ∆p for Π1 and m for Π2. Π1 = DarbV c ∆p = ( La )( MbT -3b )( LCT -C )( ML-1T -2 ) = Mb + 1La - 3b + c - 1T -c - 2 M:

0 = b + 1

L:

0 = a - 3b + c - 1

T:

0 = -c - 2

Solving, a = 0, b = - 1, and c = - 2. Thus, Π1 = D0r-1V -2 ∆p =

∆p rV 2

Π2 = DdreV fm = ( Ld )( MeL-3e )( LfT -f )( ML-1T -1 ) = Me + 1Ld - 3e + f - 1T -f - 1 M:

0 = e + 1

L:

0 = d - 3e + f - 1

T:

0 = -f - 1

Solving, d = -1, e = - 1, and f = - 1. Thus, Π2 = D-1r-1V -1m =

m rVD

or Π2 =

rVD = Re m

Therefore, the function can be written as f1a

∆p rV 2

, Reb = 0 ∆p = rV 2f (Re)

Ans.

Ans: ∆p = rV 2f (Re) 887

*8–40. The drag force FD on the automobile is a function of its velocity V, its projected area A into the wind, and the density r and viscosity m of the air. Determine the relation between FD and these parameters.

v FD

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (FD, V, A, r, m) = 0. Using the M – L – T system, FD

MLT -2

V

LT -1

A

L2

r

ML-3

m

ML-1T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, V, A, and r are chosen as m = 3 repeating variables. Thus, the q variables are FD for Π1, and m for Π2. Π1 = V aAbrcFD = ( LaT -a )( L2b )( McL-3c )( MLT -2 ) = Mc + 1La + 2b - 3c + 1T -a - 2 M:

0 = c + 1

L:

0 = a + 2b - 3c + 1

T:

0 = -a - 2

Solving, a = - 2, b = -1, and c = - 1. Thus, Π1 = V -2A-1r-1FD =

FD rAV 2

Π2 = V gAhrim = ( LgT -g )( L2h )( MiL-3i )( ML-1T -1 ) = Mi + 1Lg + 2h - 3i - 1T -g - 1 M:

0 = i + 1

L:

0 = g + 2h - 3i - 1

T:

0 = -g - 1

1 Solving, g = - 1, h = - , and i = - 1. Thus, 2 m

1

Π3 = V -1A-2r-1m =

1 2

rVA

=

m rVL

or Π2 =

rVL = Re m

Therefore, the function can be written as f1 a

FD

L , , Reb = 0 rAV 2 2A

Solving for FD in this equation. FD rV 2L2

= f (Re) FD = rV 2L2 3f (Re) 4

Ans. 888

8–41. When an underwater explosion occurs, the pressure p of the shock wave at any instant is a function of the mass of the explosive m, the intial pressure p0 formed by the explosion, the spherical radius r of the shock wave, and the density r and the bulk modulus EV of the water. Determine the relation between p and these parameters. r

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (p, m, p0, r, r, EV) = 0. Using the M - L - T system, p

ML-1T -2

m

M

p0

ML-1T -2

r

L

r

ML-3

EV

ML-1T -2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 6 - 3 = 3 Π terms Dimensional Analysis. Here, p0, m, and r are chosen as m = 3 repeating variables. Thus, the q variables are p for Π1, r for Π2, and EV for Π3. Π1 = p0ambr cr = ( MaL-aT -2a )( Mb )( Lc )( ML-1T -2 ) = Ma + b + 1L-a + c - 1T -2a-2 M:

0 = a + b + 1

L:

0 = -a + c - 1

T:

0 = -2a - 2

Solving, a = - 1, b = 0, and c = 0. Thus, Π1 = p0-1m0r 0p =

p p0

Π2 = p0dmer fr = ( MdL-dT -2d )( Me )( L f )( ML-3 ) = Md + e + 1L-d + f - 3T -2d M:

0 = d + e + 1

L:

0 = -d + f - 3

T:

0 = -2d

Solving, d = 0, e = -1, and f = 3. Thus, Π2 = p00m-1r 3r =

rr 3 m

Π3 = p0gmhr iEV = ( MgL-gT -2g )( Mh )( Li )( ML-1T -2 ) = Mg + h + 1L-g + i - 1T -2g - 2 M:

0 = g + h + 1

L:

0 = -g + i - 1

T:

0 = -2g - 2

889

8–41. Continued

Solving, g = -1, h = 0, and i = 0. Thus, Π3 = p0-1 m0r 0EV =

EV p0

Therefore, the function can be written as p rr 3 EV f1 a , , b =0 p0 m p0

Solving for p using this function, p rr 3 EV = fa , b p0 m p0 p = p0 f a

rr 3 EV , b m p0

Ans.

Ans: p = p0 f a 890

rr 3 EV , b m p0

8–42. The drag force FD acting on a submarine depends upon the characteristic length L of the vessel, the velocity V at which it is traveling, and the density r and viscosity m of the water. Determine the relation between FD and these parameters.

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (F, L,V, r, m) = 0. Using the M - L - T system, FD

MLT -2

L

L

V

LT -1

r

ML-3

m

ML-1T -1

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Dimensional Analysis. Here, V, r, and m are chosen as m = 3 repeating variables. Thus, the q variables are FD for Π1 and L for Π2. Π1 = V arbmcFD = ( LaT - a )( MbL-3b )( McL-cT -c )( MLT -2 ) = Mb + c + 1La - 3b - c + 1T -a - c - 2 M:

0 = b + c + 1

L:

0 = a - 3b - c + 1

T:

0 = -a - c - 2

Solving, a = 0, b = 1, and c = -2. Thus, Π1 = V 0r1m-2FD = Π2 = V dremfL =

FDr m2

( LdT -d )( MeL-3e )( MfL-fT -f )( L ) = Me + fLd - 3e - f + 1T -d - f

M:

0 = e + f

L:

0 = d - 3e - f + 1

T:

0 = -d - f

Solving, d = 1, e = 1, and f = - 1. Thus, Π2 = V 1r1m-1L =

rVL = Re m

The function can be written as f1a

FDr m2

, Reb = 0

Solving for FD, FDr m2

= f (Re) FD =

m2 f (Re) r

Ans.

891

Ans: FD =

m2 f (Re) r

#

8–43. The power W supplied by a pump is thought to be a function of the discharge Q, the change in pressure ∆p between the inlet and outlet, and the density r of the fluid. Use the Buckingham Pi theorem to establish a general relation between these parameters so that an experiment may be performed to determine this relationship.

SOLUTION #

#

W = f (Q, ∆p, r) or g(W, Q, ∆p, r) = 0. Thus, n = 4. Using the M - L - T system given in Table 8–1,

#

Power,

W

ML2T -3

Discharge,

Q

L3T -1

Change in pressure,

∆p

ML-1T -2

Density,

r

ML-3

Here, m = 3 since three base dimensions M, L and T are involved. Thus, there is n - m = 4 - 3 = 1 Π term. r, Q and ∆p are chosen as m = 3 repeating variables since collectively they contain all three base dimensions as required. The only Π # term using W as the q variable is

#

Π = raQb ∆pcW = ( MaL-3a )( L3bT -b )( McL-cT -2c )( ML2T -3 ) = Ma + c + 1L-3a + 3b - c + 2T -b - 2c - 3 Thus, for M:

a + c + 1 = 0

L:

-3a + 3b - c + 2 = 0

T:

-b - 2c - 3 = 0

Solving a = 0, b = - 1 and c = -1. Then

#

Π = r0Q -1 ∆p-1W =

#

W Q∆p

Thus, the general relation between the given physical variables is

#

Ans.

W = CQ∆p where C is a dimensionless constant to be determined from experiment.

Ans: # W = CQ∆p 892

*8–44. The diameter D of oil spots made on a sheet of porous paper depends upon the diameter d of the squirting nozzle, the height h of the nozzle from the surface, the velocity V of the oil, and its density r, viscosity m, and surface tension s. Determine the dimensionless ratios that define this process. d V

h

SOLUTION D = f (d, h, V, r, m, p) or g(D, d, h, V, r, m, s) = 0. Thus, n = 7 using the M - L - T system given in Table 8–1, Diameter of the spot,

D

L

Diameter of the nozzle,

d

L

Height,

h

L

Velocity,

V

LT -1

Density,

r

ML-3

Viscosity,

m

ML-1T -1

Surface tension,

s

MT -2

Here, m = 3 since three base dimensions M, L and T are involved. Thus, there are n - m = 7 - 3 = 4 Π terms. r, V and h are chosen as m = 3 repeating variables since collectively they contain all three base dimensions as required. The first Π term, using m as the q variable, is Π1 = raV bhcm = ( MaL-3a )( LbT -b )( Lc )( ML-1T -1 ) = Ma + 1L-3a + b + c - 1T -b - 1 Thus, for M:

a + 1 = 0

L:

-3a + b + c - 1 = 0

T:

-b - 1 = 0

Solving a = - 1, b = - 1 and c = - 1. Then Π1 = r-1V -1h-1m =

m rVh

rVh 1 = is the Reynolds number. The second Π term, using s as m Π1 the q variable, is

Recognize that

Π2 = rdV ehfs = ( MdL-3d )( LeT -e )( Lf )( MT -2 ) Thus for M:

d + 1 = 0

L:

-3d + e + f = 0

T:

-e - 2 = 0

893

D

8–44. Continued

Solving, d = -1, e = - 2 and f = - 1. Then Π2 = r-1V -2h-1s = Recognize that q variable, is

s rV 2h

rV 2h 1 = is the Weber number. The third Π term, using D as the s Π2

Π3 = r gV hhiD = ( MgL-3g )( LhT -h )( Li )( L ) = MgL-3g + h + i + 1T -h Thus, for M:

g = 0

L:

-3g + h + i + 1 = 0

T:

-h = 0

Solving, g = 0, h = 0 and i = - 1. Then Π3 = r0V 0h-1D =

D h

Since q variable d for fourth Π term has the same dimension as D, Hence Π4 =

d h

Thus, the functional relation is g aRe, We,

D d , b = 0 h h

Ans.

894

8–45. Mist from an aerosol produces droplets having a diameter d, which is thought to depend upon the diameter of the nozzle D, the surface tension s of the droplets, the velocity V at which the droplets are ejected, and the density r and viscosity m of the air. Determine the relation between d and these parameters.

SOLUTION Physical Variables. There are n = 6 variables and the unknown function is f (d, D, V, r, m, s) = 0. Using the M - L - T system, d

L

D

L

V

LT -1

r

ML-3

m

ML-1T -1

s

MT -2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 6 - 3 = 3 Π terms Dimensional Analysis. Here, D, V, and r are chosen as m = 3 repeating variables. Thus, the q variables are d for Π1, m for Π2, and s for Π3. Π1 = DaV brcd = ( La )( LbT -b )( McL-3c ) (L) = McLa + b - 3c + 1T -b M:

0 = c

L:

0 = a + b - 3c + 1

T:

0 = -b

Solving, a = - 1, b = 0, and c = 0. Thus, Π1 = D-1V 0r0d =

d D

Π2 = DdV er fm = ( Ld )( LeT -e )( MfL-3f )( ML-1T -1 ) = Mf + 1Ld + e - 3f - 1T - e - 1 M:

0 = f + 1

L:

0 = d + e - 3f - 1

T:

0 = -e - 1

Solving, d = - 1, e = -1, and f = -1. Thus, Π2 = D-1V -1r-1m =

m rVD

or Π2 =

rVD = Re m

895

8–45. Continued

Π3 = DgV hris = ( Lg )( LhT

-h

M:

0 = i + 1

L:

0 = g + h - 3i

T:

0 = -h - 2

)( MiL-3i )( MT -2 ) = Mi + 1Lg + h - 3iT -h - 2

Solving, g = -1, h = - 2, and i = -1. Thus, Π3 = D-1V -2r-1s =

s rV 2D

or Π3 =

rV 2D = We s

Therefore, the function can be written as d f1a , Re, Web = 0 D

Solving for d,

d = f ( Re, We ) D d = Df ( Re, We )

Ans.

Ans: d = Df ( Re, We ) 896

8–46. Fluid flow depends upon the viscosity m, bulk modulus EV , gravity g, pressure p, velocity V, density r, surface tension s, and a characteristic length L. Determine the dimensionless groupings for these eight variables.

SOLUTION Physical Variables. There are n = 8 variables and the unknown function is f (m, EV, g, p, V, r, s, L) = 0. Using the M - L - T system, m

ML-1T -1

V

LT -1

EV

ML-1T -2

r

ML-3

g

LT -2

s

MT -2

P

ML-1T -2

L

L

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 8 - 3 = 5 Π terms Dimensional Analysis. Here, V, L, and r are chosen as m = 3 repeating variables. Thus, the q variables are m for Π1, EV for Π2, g for Π3, P for Π4, and s for Π5. Π1 = V aLbrcm = ( LaT -a )( Lb )( McL-3c )( ML-1T -1 ) = Mc + 1La + b - 3c - 1T -a - 1 M:

0 = c + 1

L:

0 = a + b - 3c - 1

T:

0 = -a - 1

Solving, a = - 1, b = -1, and c = - 1. Thus, Π1 = V -1L-1r-1m =

m rVL

or Π1 =

rVL = Re m

Π2 = V dLerfEV = ( LdT -d )( Le )( MfL-3f )( ML-1T -2 ) = Mf + 1Ld + e - 3f - 1T -d - 2 M:

0 = f + 1

L:

0 = d + e - 3f - 1

T:

0 = -d - 2

Solving, d = - 2, e = 0, and f = - 1. Thus, Π2 = V -2L0r-1EV =

EV rV 2

or Π2 =

rV 2 = M EV

897

8–46. Continued

Π3 = V gLhrip = ( LgT

-g

)( Lh )( MiL-3i )( ML-1T -2 ) = Mi + 1Lg + h - 3i - 1T -g - 2

M:

0 = i + 1

L:

0 = g + h - 3i - 1

T:

0 = -g - 2

Solving, g = -2, h = 0, and i = - 1. Thus, p

Π3 = V -2L0r-1p =

rV 2

= Eu

Π4 = V jLkrlg = ( L jT -j )( Lk )( MlL-3l )( LT -2 ) = MlL j + k - 3l + 1T -j - 2 M:

0 = l

L:

0 = j + k - 3l + 1

T:

0 = -j - 2

Solving, j = - 2, k = 1, and l = 0. Thus, Π4 = V -2L1r0g =

gL V2

=

2gL V

or Π4 =

V 2gL

= Fr

Π5 = V mLnrps = ( LmT -m )( Ln )( MpL-3p )( MT -2 ) = Mp + 1Lm + n - 3pT -m - 2 M:

0 = p + 1

L:

0 = m + n - 3p

T:

0 = -m - 2

Solving, m = -2, n = - 1, and p = - 1. Thus, Π5 = V -2L-1r-1s =

s rV 2L

or Π5 =

rV 2L = We s

Therefore, the function can be written as Ans.

f (Re, M, Eu, Fr, We) = 0

Ans: f (Re, M, Eu, Fr, We) = 0 898

8–47. The discharge Q over a small weir depends upon the water head H, the width b and height h of the weir, the acceleration of gravity g, and the density r, viscosity m, and surface tension s of the fluid. Determine the relation between Q and these parameters.

H

h

SOLUTION Physical Variables. There are n = 8 variables and the unknown function is f (Q, H, b, h, g, r, m, s) = 0. Using the M - L - T system, Q

L3T -1

g

LT -2

H

L

r

ML-3

b

L

m

ML-1T -1

h

L

s

MT -2

Here, all three base dimensions are used, so that m = 3. Thus, there are n - m = 8 - 3 = 5 Π terms Dimensional Analysis. Here, r, g, and H are chosen as m = 3 repeating variables. Thus, the q variables are Q for Π1, b for Π2, h for Π3, m for Π4, and s for Π5. Π1 = ragbHcQ = ( MaL-3a )( LbT -2b )( Lc )( LbT -1 ) = MaL-3a + b + c + 3T -2b - 1 M:

0 = a

L:

0 = -3a + b + c + 3

T:

0 = -2b - 1

1 5 Solving, a = 0, b = - , and c = - . Thus, 2 2 Q 5 1 Π1 = r0g -2H -2Q = 2gH5

Π2 = rdgeHfb = ( MdL-3d )( LeT -2e )( L f ) (L) = MdL-3d + e + f + 1T -2e

M:

0 = d

L:

0 = -3d + e + f + 1

T:

0 = -2e

Solving, d = 0, e = 0, and f = - 1. Thus, Π2 = r0g0H -1b =

b H

Π3 = rhgiH jh = ( MhL-3h )( LiT -2i )( Lj ) (L) = MhL-3h + i + j + 1T -2i M:

0 = h

L:

0 = -3h + i + j + 1

T:

0 = -2i

Solving, h = 0, i = 0, and j = - 1. Thus, Π3 = r0g0H -1h =

h H

899

8–47. Continued

Π4 = rkglHmm = ( MkL-3k )( LlT -2l )( Lm )( ML-1T -1 ) = Mk + 1L-3k + l + m - 1T -2l - 1 M:

0 = k + 1

L:

0 = -3k + l + m - 1

T:

0 = -2l - 1

1 3 Solving, k = -1, l = - , and m = - . Thus, 2 2 m 3 1 Π4 = r-1g - 2H - 2m = r2gH3 Π5 = rngpHqs = ( MnL-3n )( LpT -2p )( Lq )( MT -2 ) = Mn + 1L-3n + p + qT -2p - 2 M:

0 = n + 1

L:

0 = -3n + p + q

T:

0 = -2p - 2

Solving, n = -1, p = -1, and q = - 2. Thus, Π5 = r-1g -1H -2s =

s rgH2

Therefore, the function can be written as f1 °

Q

m b h s , , , ¢ = 0 2 2gH H H r2gH3 rgH 5

,

Solving for Q, Q

2gH5

= f°

m b h s , , , ¢ H H r2gH3 rgH2

Q = 2gH5f °

m b h s , , , ¢ H H r2gH3 rgH2

Ans.

Ans: Q = 2gH5 f ° 900

m b,h, , s ¢ 3 H H r2gH rgH2

*8–48. If water flows through a 50-mm-diameter pipe at 2 m>s, determine the velocity of carbon tetrachloride flowing through a 60-mm-diameter pipe so that they both have the same dynamic characteristics. The temperature of both liquids is 20°C.

50 mm

SOLUTION Since inertia and viscous forces are predominant, the Reynolds number must be the same for both cases. From Appendix A, rte = 1590 kg>m3, rw = 998.3 kg>m3, mte = 0.958 ( 10-3 ) N # s>m2, and mw = 1.00 ( 10-3 ) N # s>m2 at 20° C . This requires a

rVD rVD b = a b m w m te

( 998.3 kg>m3 )( 2 m>s ) (0.05 m) ( 1590 kg>m3 ) Vte(0.06 m) = 1.00 ( 10-3 ) N # s>m2 0.958 ( 10-3 ) N # s>m2 Ans.

V = 1.00 m>s

901

60 mm

8–49. In order to test the flow over the surface of an airplane wing, a model is built to a scale of 1>15 and is tested in water. If the airplane is designed to fly at 350 mi>h, what should the velocity of the model be in order to maintain the same Reynolds number? Is this test realistic? Take the temperature of both the air and water to be 60°F.

350 mi/h

SOLUTION From Appendix A, for water nm = 12.2 ( 10-6 ) ft 2 >s and for air np = 0.158 ( 10-3 ) ft 2 >s VL VL at 60º F. Since, n = m>r, the Reynolds number can be written as Re = = . n m>r Thus, a

VL VL b = a b n m n p

Vm = a = £

n m Lp ba bV n p Lm p

12.2 ( 10-6 ) ft 2 >s

0.158 ( 10-3 ) ft 2 >s

§a

15 b ( 350 mi>h ) 1 Ans.

= 405.38 mi>h = 405 mi>hr No, since the velocity is too large for a water tunnel.

Ans: 405 mi>hr 902

8–50. The model of a river is constructed to a scale of 1>60. If the water in the river is flowing at 38 ft>s, how fast must the water flow in the model?

SOLUTION Since the inertia and gravitational forces are predominant in the river flow, the Froude numbers for both the model and the prototype must be same.

( Fr ) p = ( Fr ) m a

V 2gL

b

m

= a

Vm = Vp

V 2gL

b

p

Lm 1 = ( 38 ft>s ) = 4.906 ft>s = 4.91 ft>s A Lp A 60

Ans.

Ans: 4.91 ft>s 903

8–51. Water flowing through a 100-mm-diameter pipe is used to determine the loss in pressure when gasoline flows through a 75-mm diameter pipe at 3 m>s. If the pressure loss in the pipe transporting water is 8 Pa, determine the  pressure loss in the pipe transporting the gasoline. Take ng = 0.465 1 10 - 6 2 m2 >s and nw = 0.890 1 10 - 6 2 m2 >s, rg = 726 kg>m3 , rw = 997 kg>m3.

SOLUTION For the Reynolds number, a

VD VD b = a b n w n g

Vw = Vg a

0.890 ( 10-6 ) m2 >s nw Dg 75 mm ba b = (3 m>s) a ba b -6 2 ng Dw 0.465 ( 10 ) m >s 100 mm

= 4.306 m>s

For the Euler number, a

∆p rv

2

b = a w

∆p rv2

b

∆pg = ∆pw a

g

rg

rw

∆pg = 2.83 Pa

ba

Vg 2 Vw2

b = (8 Pa)a

726 kg>m3 3

997 kg>m

ba

3 m>s 4.306 m>s

b

2

Ans.

Ans: 2.83 Pa 904

*8–52. The effect of drag on a model airplane is to be tested in a wind tunnel with a wind speed of 200 mi>h. If a similar test is performed underwater in a channel, what should the speed of the water be to achieve the same result when the temperature is 60°F?

200 mi/h

SOLUTION Since the viscous and inertia forces are predominant, the Reynolds numbers for both cases must be the same. Since n = m>r, the Reynold’s numbers can be written VL VL as Re = = . Thus, n m>r a

VL VL b = a b n w n a

Vw = a

n w La ba b ( Va ) n a Lw

La b = 1. From Appendix A, Lw ft 2 >s at 60° F. Thus,

Since the same model is used for both cases, a nw = 12.2 ( 10-6 ) ft 2 >s and na = 0.158 ( 10-3 ) Vw = c

0.0122 ( 10-3 ) ft 2 >s 0.158 ( 10-3 ) ft 2 >s

d (1)(200 mi>h)

Ans.

= 15.44 mi>hr = 15.4 mi>h

905

8–53. When a 100-mm-diameter sphere travels at 2 m>s in water having a temperature of 15°C, the drag force is 2.80 N. Determine the velocity and drag force on a 150-mm-diameter sphere traveling through water under similar conditions.

SOLUTION Since the Reynolds number involves V, it will be used to determine the velocity of the 150-mm diameter sphere. Since v = m>r, the Reynold’s numbers can be written VD VD as Re = = . Thus, m>r v a

VD VD b = a b v 2 v 1

V2 = a

v2 D1 ba bV v1 D2 1

Since both spheres move in the same medium (water at 15º C), a V2 = (1)a

v2 b = 1. Thus, v1

100 mm b ( 2 m>s ) = 1.333 m>s = 1.33 m>s 150 mm

Ans.

F F ∝ 2 , the Euler numbers, which involve p, can be used A D to determine the drag force on the 150-mm diameter sphere. This gives

Subsequently, since p =

a

a

p rV 2

2

p rV 2

b

1

F F b = a 2 2b rD2V 2 2 rD V 1

F2 = a

Here,

b = a

r2 D2 2 V2 2 ba b a b F1 r1 D1 V1

r2 = 1 since both spheres move in the same medium (water at 15º C). Thus, r1 F2 = (1)a

150 mm 2 1.333 m>s 2 b a b (2.80 N) 100 mm 2 m>s

Ans.

= 2.80 N

Ans: V2 = 1.33 m>s F2 = 2.80 N 906

8–54. In order to determine the formation of waves around obstructions in a river, a model having a scale of 1>10 is used. If the river flows at 6 ft>s, determine the speed of the water for the model.

SOLUTION For the river, the inertia and gravitational forces are predominant. Thus, the equality of the Froude number will be used. a

V 2gL

b

m

Vm =

=

= a

V 2gL

b

p

Lm Vp B Lp

1 ( 6 ft>s ) = 1.897 ft>s = 1.90 ft>s A 10

Ans.

Ans: 1.90 ft>s 907

8–55. The optimum performance of mixing blades 0.5 m in diameter is to be tested using a model one-fourth the size of the prototype. If the test of the model in water reveals the optimum speed to be 8 rad>s, determine the optimum angular speed of the prototype when it is used to mix ethyl alcohol. Take T = 20°C.

0.25 m 0.25 m

SOLUTION rVD vD and n = m>r, the Reynolds number can be written as Re = = m 2 (vD>2)(D) vD2 = . Thus, m>r 2n

Since V =

a

vD2 vD2 b = a b 2n p 2n m

vp = a

np

nm

ba

Dm 2 b vm Dp

From Appendix A, np = 1.51 ( 10-6 ) m2 >s and nm = 1.00 ( 10-6 ) m2 >s . Thus, vp = c

1.51 ( 10-6 ) m2 >s 1.00 ( 10

= 0.755 rad>s

-6

)

1 2 d a b (8 rad>s) 4

Ans.

Ans: 0.755 rad>s 908

*8–56. The flow of water around the structural support is 1.2 m>s when the temperture is 5°C. If it is to be studied using a model built to a scale of 1>20, and using water at a temperature of 25°C, determine the velocity of the water used with the model.

1.2 m/s

SOLUTION a

VL VL b = a b n m n p

Using Appendix A

Vm = Vp a

n m Lp ba b n p Lm

= 1.2 m>s a

= 14.2 m>s

0.898 ( 10-6 ) m2 >s 1.52 ( 10-6 ) m2 >s

ba

20 b 1

Ans.

909

8–57. A model of a ship is built to a scale of 1>20. If the ship is to be designed to travel at 4 m>s, determine the speed of the model in order to maintain the same Froude number.

SOLUTION a

V 2gL

Vm = =

b

m

= a

Lm Vp B Lp

V 2gL

b

p

1 ( 4 m>s ) = 0.8944 m>s = 0.894 m>s A 20

Ans.

Ans: 0.894 m>s 910

8–58. The flow around the airplane flying at an altitude of 10 km is to be studied using a wind tunnel and a model that is built to a 1>15 scale. If the plane has an air speed of 800  km>h, what should the speed of the air be inside the tunnel? Is this reasonable?

800 km/h

SOLUTION The air flow around the airplane causes the inertia and viscous forces to be predominant. Thus, the Reynolds number will be used. Since n = m>r, the Reynolds VL VL number can be written as Re = = . n m>r a

VL VL b = a b n m n p

Vm = a

Lp

Lm

ba

nm bV np p

From Appendix A, np = 35.25 ( 10-6 ) m2 >s at an altitude of 10 km and nm = 14.61 ( 10-6 ) m2 >s at ground level. Thus, Vm = a

-6 2 15 14.61 ( 10 ) m >s bc d (800 km>h) 1 35.25 ( 10-6 ) m2 >s

Vm = 4973.61 km>h = 4.97 Mm>h

Ans.

No, since a wind speed of Vm = 4.97 Mm>h is extremely difficult to achieve. Also, it is greater than the speed of sound, and so the results would not be valid.

Ans: 4.97 Mm>h 911

8–59. The model of an airplane has a scale of 1>30. If the drag force on the prototype is to be determined when the plane is flying at 600 km>h, find the speed of the air in a wind tunnel for the model if the air has the same temperature and pressure. Is it reasonable to do this test?

SOLUTION Since the air flow around the airplane causes the inertia and viscous forces to be predominant, the Reynolds number will be used. Since n = m>r, the Reynolds VL VL number can be written as Re = = . n m>r a

VL VL b = a b n m n p

Vm = a

n m Lp ba bV n p Lm p

Since the air has the same pressure and temperature in both cases, a Vm = (1)a

30 b(600 km>h) 1

nm b = 1. Thus, np Ans.

= 18 000 km>h = 18 Mm>h

No, Vm = 18 Mm>h is too fast to achieve, and since it is far greater than the speed of sound the results would not be valid even if the test could be done.

Ans: 18 Mm>h 912

8–60. The resistance of waves on a 250-ft-long ship is tested in a channel using a model that is 15 ft long. If the ship travels at 35 mi>h, what should be the speed of the model to resist the waves?

35 mi/h

SOLUTION The wave force on the ship causes the inertia and gravitational forces to be predominant. Thus, the Froude number will be a

V 2gL

Vm = Here,

b

m

= a

Lm V A Lp p

V 2gL

b

p

Lm 15 ft 3 = . Thus, = Lp 250 ft 50 Vm =

3 (35 mi>h) A 50

Ans.

= 8.573 mi>h = 8.57 mi>h

913

8–61. A model of a submarine is built to a scale of 1>25 and tested in a wind tunnel at an airspeed of 150 mi>h. What is the intended speed of the prototype if it is in water at the same temperature of 60°F?

SOLUTION The flow of fluid around the submarine causes the inertia and viscous forces to be predominant. Thus, the Reynolds number will be used. Since n = m>r, the Reynolds VL VL . number can be written as Re = = n m>r a

VL VL b = a b n p n m

Vp = Vma

Lm n p ba b Lp n m

From Appendix A, nm = 0.158 ( 10-3 ) ft 2 >s and np = 12.2 ( 10-6 ) ft 2 >s. Thus, Vp = ( 150 mi>h ) a

12.2 ( 10-6 ) ft 2 >s 1 b£ § 25 0.158 ( 10-3 ) ft 2 >s

Ans.

Vp = 0.4633 mi>h = 0.463 mi>h

Ans: 0.463 mi>h 914

*8–62. The flow of water around the bridge pier is to be studied using a model built to a scale of 1>15. If the river flows at 0.8 m>s, determine the corresponding velocity of the water in the model at the same temperature.

0.8 m/s

SOLUTION Water flow of fluid around the pier causes the inertia and viscous forces to be predominant. Thus, the Reynolds number will be used. Since n = m>r, the Reynolds number can be written as Re = a

VL VL = . n m>r

VL VL b = a b n m n p

Vm = a

n m Lp ba bV n p Lm p

Since water is used for the model and the prototype, a Vm = (1)a

15 b ( 0.8 m>s ) 1

nm b = 1. Thus, np

Ans.

= 12 m>s

Ans: 12 m>s 915

*8–63. The resistance created by waves on a 100-m-long ship is tested in a channel using a model that is 4 m long. If the ship travels at 60 km>h, what should be the speed of the model?

SOLUTION The wave force on the ship causes the inertia and gravitational forces to be predominant. Thus, the Froude number will be °

n 2gL

Vm = Here,

¢

m

= °

n 2gL

¢

p

Lm Vp B Lp

Lm 4m 1 = = . Thus, Lp 100 m 25 Vm =

1 ( 60 km>h ) A 25

Ans.

= 12 km>h

Ans: 12 km>h 916

*8–64. The velocity of water waves in a channel are studied in a laboratory using a model of the channel onetwelfth its actual size. Determine the velocity of waves in the channel if they have a velocity of 6 m>s in the model.

6 m/s

SOLUTION The motion of the wave causes the inertia and gravitational forces to be predominant. Thus, the equality of the Froude numbers will require °

n 2gL

Vp = Here,

Lp Lm

=

¢ = ° p

Lp

B Lm

n 2gL

¢

m

Vm

12 . Thus, 1

Vp =

12 ( 6 m>s ) A1

Ans.

= 20.78 m>s = 20.8 m>s

917

8–65. A model of a submarine is built to determine the drag force acting on its prototype. The length scale is 1>100, and the test is run in water at 20°C, with a speed of 8 m>s. If the drag force on the model is 20 N, determine the drag force on the prototype if it runs in water at the same speed and temperature. This requires that the drag coefficient CD = 2FD >rV 2L2 be the same for both the model and the prototype.

8 m/s

SOLUTION The requirement is a

2FD rV 2L2

b = a p

( FD ) p = a

rp

rm

2FD rV 2L2

ba

Vp Vm

b 2

m

b a

Lp Lm

2

b ( FD ) m

Since the model and prototype run with the same speed in water having the same rp Vp Lp 100 = = 1. Here, = a b . Thus, temperature, rm Lm 1 Vm

( FD ) p = (1)(1)2 a

100 2 b ( 20 N ) 1

= 200 ( 103 ) N = 200 kN

Ans.

Ans: 200 kN 918

*8–66. A model of a plane is built to a scale of 1>15 and is tested in a wind tunnel. If the plane is designed to travel at 800 km>h at an altitude of 5 km, determine the required density of the air in the wind tunnel so that the Reynolds and Mach numbers are the same. Assume the temperature is the same in both cases and the speed of sound in air at this temperature is 340 m>s.

SOLUTION Using the Mach number, V V a b = a b c m c p

The speed of sound in air having the same temperature is the same, cm = cp. Thus, Vm = VP Using the Reynolds number, a

rVL rVL b = a b m m m p

rm = a

mm VP LP ba ba br mP Vm Lm p

For air at the same temperature, mm = mp. From Appendix A, rp = 0.7364 kg>m3 at an altitude of 5 km. Thus, rm = (1)(1)a

15 b ( 0.7364 kg>m3 ) = 11.046 kg>m3 = 11.0 kg>m3 1

Ans.

Note: The result is not reasonable, since the value of rm is not possible with air in realistic conditions.

Ans: 11.0 kg>m3 919

8–67. The motion of water waves in a channel are to be studied in a laboratory using a model one-twelfth the size of the channel. Determine the time for a wave in the channel to travel 10 m if it takes 15 seconds for the wave to travel this distance in the model.

SOLUTION The motion of the wave causes the inertia and gravitational forces to be predominant, so the Froude number will be used. a

V 2gL

Vm = Here, Vm =

b

m

= a

Lm

B Lp

V 2gL

b

p

Vp

sp sm 10 m 10 m = = 0.6667 m>s and Vp = = . Then, tm 15 s tp tp

0.6667 m>s =

1 10 m a b A 12 t p

Ans.

t p = 4.330 s = 4.33 s

Ans: 4.33 s 920

*8–68. It is required that a pump be designed for use in a chemical plant such that it delivers 0.8 m3 >s of benzene with a pressure increase of 320 kPa. What is the expected flow and pressure increase produced by a model one-sixth the size of the prototype? If the model produces a power output of 900 kW, what would be the power output of the prototype?

SOLUTION Since the viscous force is the dominant force, then Reynolds number similitude must be achieved. a

rVL rVL b = a b m m m p

Since benzene will be used for both model and prototype, rm = rp and mm = mp. Then VmLm = VpLp Lp Vm = Vp Lm

(1)

Since Q = VA and A has the dimension of L2, Eq. 1 becomes Qm >Lm2 Qp >Lp2

=

Lp Lm

2 Lp Q m Lp a 2b = a b Q p Lm Lm

Qm = Qp a

Lp

Lm

ba

Here, Qp = 0.8 m3 >s. and

Lm 2 Lm b = Qp a b Lp Lp

Lm 1 = . Then Lp 6

1 Qm = ( 0.8 m3 >s ) a b = 0.133 m3 >s 6

Ans.

For the pressure comparison, Euler number similitude should be used since the pressure change is involved. a

∆P ∆P b = a 2b 2 rV m rV p

Since rm = rp,

(∆P)m Vm2

=

(∆P)p Vp 2

(∆P)m = (∆P)pa

Vm 2 b Vp

921

*8–68. Continued

Substituting Eq 1 into this equation, (∆P)m = (∆P)pa

Lp Lm

Here (∆P)p = 320 kPa and

b

Lp

Lm

2

= 6,

(∆P)m = (320 kPa ) ( 62 ) = 11520 kPa = 11.5 MPa

Ans.

As shown earlier,

#

W = CQ∆p So that

#

#

Wp

Wm = Qp(∆p)p Qm(∆p)m and

#

Here,

#

Wp = Wm a

Qp Qm

ba

(∆p)p (∆p)m

#

b = Wm a

Lp Lm

ba

# Lm Lm 2 b = Wma b Lp Lp

# 1 W = 900 kW a b = 150 kW 6

Ans.

922

8–69. If the jet plane can fly at Mach 2 in air at 35°F, determine the required speed of wind generated in a wind tunnel at 65°F and used on a model built to a scale of 1>25. Hint: Use Eq. 13–24, c = 2kRT, where k = 1.40 for air.

SOLUTION

For air, Appendix A gives R = 1716 ft # lb>slug # R. Thus, Vm 2k RTm Vm

2Tm

=

= Vp

Vp 2k RTp

2Tp

Vp = 2 ( 2kRT ) p

Vp = 221.40 ( 1716 ft # lb>slug # R ) (35° + 460) R = 2181.0 ft>s = 2.18 ( 103 ) ft>s Vm 265° + 460

=

Ans.

2181.0 235° + 460

Vm = 2246.1 ft>s = 2.25 ( 103 ) ft>s

Ans.

Ans: 2.25 ( 103 ) ft>s 923

8–70. The drag coefficient on an airplane is defined by CD = 2FD >rV 2L2. If the drag force acting on the model of a plane tested at sea level is 0.3 N, determine the drag force on the prototype, which is 15 times larger and is flying at 20 times the speed of the model at an altitude of 3 km.

SOLUTION The requirement is a

2FD 2 2

rV L

b = a p

2FD rV 2L2

( FD ) p = ( FD ) m = a

b

m

rp rm

ba

Vp Vm

Lp Lm

b

2

From Appendix A, rp = 0.9092 kg>m at an altitude of 3 km and rm = 1.225 kg>m3 at sea level. Thus

( FD ) p = (0.3 N)a

3

2

b a

0.9092 kg>m3 1.225 kg>m3

ba

= 20.04 ( 103 ) N = 20.0 kN

20Vm 2 15Lm 2 b a b Vm Lm

Ans.

Ans: 20.0 kN 924

8–71. The model of a hydrofoil boat is to be tested in a channel. The model is built to a scale of 1>20. If the lift produced by the model is 7 kN, determine the lift on the prototype. Assume the water temperature is the same in both cases. This requires Euler number and Reynolds number similarity.

SOLUTION For the Reynolds number, a

rVL rVL b = a b m p m m

Vp

Vm

= a

rm mp Lm ba ba b rp mm Lp

At the same temperature, rp = rm and mm = mp. Thus, Vp Vm

=

Lm 1 = Lp 20

F>L2 F F , the Euler number can be written as Eu = = . 2 2 L rV rV 2L2 The Euler number gives Since p = F>A =

a

F F b = a 2 2b 2 2 rV L p rV L m

Fp = a

rp

rm

= (1)a

ba

Vp

Vm

2

b a

Lp

Lm

2

b Fm

1 2 20 2 b a b (7 kN) 20 T

Ans.

= 7 kN

Ans: 7 kN 925

*8–72. The model of a boat is built to a scale of 1>50. Determine the required kinematic viscosity of the water in order to test the model so that the Froude and Reynolds numbers remain the same for the model and the prototype. Is this test practical if  the prototype operates in water at T = 20°C?

SOLUTION For the Froude number a

V 2gL

b

m

= a

Lm Vm = Vp B Lp

V 2gL

b

p

For the Reynolds number, a

VL VL b = a b n p n m

3

nm = a

Vm Lm Lm Lm Lm 2 ba bnp = a bnp = a b np Vp Lp Lp B Lp Lp

nm = a

1 2 b 3 1.00 ( 10-6 ) m2 >s 4 50

From Appendix A, np = 1.00 ( 10-6 ) m2 >s. at T = 20° C . Thus, 3

nm = 2.828 ( 10-9 ) m2 >s = 2.83 ( 10-9 ) m2 >s

Ans.

No, this value is too low to be practical.

926

8–73. If an airplane flies at 800 mi>h at an altitude of 5000  ft, what should its speed be so that it has the same Mach number when it is at 15 000 ft? Assume the air has the same bulk modulus. Use Eq. 13–25, c = 2EV >r.

800 mi/h

2

2

SOLUTION For the Mach number, V V a b = a b c 2 c 1 V2 = a

c2 bV c1 1

EV r1 EV c2 B r2 Since c = and EV is constant, then = = . Thus, c1 A r2 B r EV B r1 V2 =

r1 V A r2 1

From Appendix A, r1 = 2.043 ( 10-3 ) slug>ft 3 at an altitude of 5000 ft and r2 = 1.495 ( 10-3 ) slug>ft 3 at an altitude of 15 000 ft . Thus, V2 = £

2.043 ( 10-3 ) slug>ft 3

B 1.495 ( 10-3 ) slug>ft 3

§ ( 800 mi>h )

Ans.

= 935.20 mi>h = 935 mi>h

Ans: 935 mi>h 927

8–74. A 60-ft-long “check dam” on a river provides a means of collecting debris that flows downstream. If the flow over the dam is 8000 ft 3 >s, and a model of this dam is to be built to a scale of 1>20, determine the flow over the model and the depth of water that flows over its crest. Assume that the water temperature for the prototype and the model is the same. The volumetric flow over the dam can be determined using Q = CD 2gLH 3>2, where CD is the coefficient of discharge, g the acceleration of gravity, L is the length of the dam, and H is the height of the water above the crest. Take CD = 0.71.

60 ft

SOLUTION Since the gravitational force is the dominant force, then Froude number similitude must be achieved. a

V 2gL

b

m

Here gm = gp. Thus Vm 2Lm

=

= a

V 2gL

b

p

Vp 2Lp 1

Lm 2 Vm = a b Vp Lp Since V = Q>A and A has a dimension of L2, Eq 1 becomes Qm >Lm2 Qp >Lp 2

(1)

1

= a

Lm 2 b Lp

1

Q m Lp 2 Lm 2 a b = a b Q p Lm Lp 1

Qm Lm 2 Lm 2 = a ba b Qp Lp Lp 5

Qm Lm 2 = a b Qp Lp Qm = Qp a

Here

5

Lm 2 b Lp

Lm 1 = and Qp = 8000 ft 3 >s, then Lp 20 5

Qm

1 2 = ( 8000 ft >s ) a b = 4.47 ft 3 >s 20 3

Ans.

The height over the dam is

3

Q = CD 2g LH2;

3

8000 ft 3 >s = 0.71232.2(60 ft) ( Hp2 )

Hp = 10.31 ft

Here Hm Lm = ; Hp Lp

Hm 1 = 10.31 ft 20 Ans.

Hm = 0.515 ft

Ans: Qm = 4.47 ft 3 >s Hm = 0.515 ft

928

8–75. A ship has a length of 180 m and travels in the sea where rs = 1030 kg>m3. A model of the ship is built to a 1 >60 scale, and it displaces 0.06 m3 of water such that its hull has a wetted surface area of 3.6 m2. When tested in a towing tank at a speed of 0.5 m>s, the total drag on the model was 2.25 N. Determine the drag on the ship and its corresponding speed. What power is needed to overcome this  drag? The drag due  to  viscous (frictional) forces  can be determined using (FD)f = 1 12 rV 2A 2 CD, where CD is the drag coefficient determined from CD = 1.328> 2Re for Re 6 106 and CD = 0.455> 1log 10Re2 2.58 for 106 6 Re 6 109. Take r = 1000 kg>m3 and n = 1.00 1 10 - 6 2 m2 >s.

SOLUTION Using the scale, Lm 1 = ; Lp 60

Lm 1 = 180 m 60

Lm = 3 m

For Froude number similitude, a

V 2gL

b

m

Since g is a constant,

= a

V 2gL

b

p

1

Lp 2 VP = a b Vm Lm

Here,

Lp Lm

Vp = a

Lp

Lm

1 2

b Vm

= 60 and Vm = 0.5 m>s . Then 1

Vp = ( 602 )( 0.5 m>s ) = 3.873 m>s = 3.87 m>s

Ans.

Next, we will compute the frictional drag force. Here, Vm = 0.5 m>s, Lm = 3 m and nm = 1.00 ( 10-6 ) m2 >s . Then (Re)m =

(0.5 m>s)(3 m) VmLm = = 1.5 ( 10-6 ) nm 1.00 ( 10-6 ) m2 >s

Since 106 6 (Re)m 6 109,

( CD ) m =

3( FD ) f 4 m

0.455

3 log 10 ( Re ) m 4

2.58

=

0.455

3 log 101.5 ( 10 ) 6 4 2.58

= 4.1493 ( 10-3 )

1 1 = a rm Vm2 Am b ( CD ) m = c ( 1000 kg>m3 )( 0.5 m>s ) 2 ( 3.6 m2 ) d 3 4.1493 ( 10-3 ) 4 2 2 = 1.867 N

929

8–75. Continued

Thus, the drag force due to the wave action on the model is

3 ( FD ) g 4 m

= FD -

3 ( FD ) f 4 m

= 2.25 N - 1.867 N = 0.3828 N

Using the equation given in the text

3 ( FD ) g 4 p

= = =

3 ( FD ) g 4 m a

3 ( FD ) g 4 m a 3 ( FD ) g 4 m a

rp rm rp

rm rp rm

= ( 0.3828 N ) °

ba

ba ba

Vp Vm Lp

Lm

Lp

Lm

2

b a

ba

b

Lm

Lp

Lm

3

1030 kg>m3 1000 kg>m3

= 85.17 ( 103 ) N

Lp

b

b

2

2

¢ ( 603 )

The frictional drag force on the prototype must be determined. Here, Vp = 3.873 m>s, Lp = 180 m and np = 1.00 ( 10-6 ) m2 >s . Then (Re)p =

VpLp np

Since 106 6 (Re)p 6 109,

( CD ) p = Here, Ap = Am a

3 ( FD ) f 4 p

=

( 3.873 m>s )( 180 m ) = 0.6971 ( 109 ) 1.00 ( 10-6 ) m2 >s

0.455

3 log 10 ( Re ) p 4 2.58

Lp

Lm

=

2

0.455

3 log 100.6971 ( 109 ) 4 2.58

= 1.6434 ( 10-3 )

b = ( 3.6 m2 )( 602 ) = 12960 m2. Then

1 1 = a rp Vp2 Ap b ( CD ) p = c ( 1030 kg>m3 )( 3.873 m>s ) 2 ( 12960 m2 ) d 3 1.6434 ( 10-3 ) 4 2 2 = 164.53 ( 103 ) N

Thus, the total drag force is

( FD ) p = 3 ( FD ) g 4 p + 3 ( FD ) f 4 p = 85.17 ( 103 ) N + 164.53 ( 103 ) N = 249.70 ( 103 ) N = 250 kN

The power is

#

W = ( FD ) pVp =

3 249.70 ( 103 ) N 4 ( 3.873 m>s )

Ans.

Ans: Vp = 3.87 m>s

= 967.09 ( 103 ) W = 967 kN

Ans.

( FD ) p = 250 kN

#

W = 967 kN 930

9–1. Crude oil flows through the 2-mm gap between the two fixed parallel plates due to a drop in pressure from A to B of 4 kPa. If the plates are 800 mm wide, determine the flow.

2 mm

A

B

0.5 m

SOLUTION dh = 0 and U = 0. From Appendix A, m = 30.2 ( 10-3 ) N # s>m2 and dx r = 880 kg>m3.

Here,

Q = -

= -

a3b dp 12m dx

3 2 ( 10-3 ) m 4 3(0.8 m) 12 3 30.2 ( 10-3 ) N # s>m2 4

£

- 4 ( 103 ) N>m2 0.5 m

= 0.1413 ( 10-3 ) m3 >s = 0.141 ( 10-3 ) m3 >s

§ Ans.

To evaluate the maximum Reynolds number, we must first find the maximum velocity of the oil flow. u max = -

3 2 ( 10-3 ) m 4 2 - 4 ( 103 ) N>m2 a2 dp = £ § = 0.132 m>s 8m dx 0.5 m 8 3 30.2 ( 10-3 ) N # s>m2 4

Thus, the maximum Reynolds number is Re =

ru max a = m

( 880 kg>m3 )( 0.132 m>s ) 3 2 ( 10-3 ) m 4 = 7.719 6 1400 (OK) 30.2 ( 10-3 ) N # s>m2

Note: Normally the average velocity is used to calculate the Reynolds number.

931

Ans: 0.141 ( 10-3 ) m3 >s

9–2. Crude oil flows through the gap between the two fixed parallel plates due to a drop in pressure from A to B of 4 kPa. Determine the maximum velocity of the oil and the shear stress on each plate.

2 mm

A

B

0.5 m

SOLUTION dh = 0 and U = 0. From Appendix A, m = 30.2 ( 10-3 ) N # s>m2 and dx r = 880 kg>m3.

Here,

u max = -

3 2 ( 10-3 ) m 4 2 4 ( 103 ) N>m2 a2 dp = £ § 8m dx 0.5 m 8 3 30.2 ( 10-3 ) m 4

= 0.1324 m>s = 0.132 m>s

Ans.

Thus, the maximum Reynolds number is Re =

rUmax a = m

( 880 kg>m3 )( 0.1324 m>s ) 3 0.2 ( 10-3 ) m 4 = 7.719 6 1400 (OK) 30.2 ( 10-3 ) N # s>m2

At the top and bottom plate, y = a = 2 ( 10-3 ) m, and y = 0, respectively. tt =

4 ( 103 ) N>m2 2 ( 10-3 ) m dp a ay - b = £ § £ 2 ( 10-3 ) m § dx 2 0.5 m 2

= - 8 Pa tb =

4 ( 103 ) N>m2 2 ( 10-3 ) m dp a ay - b = £ § £0 § dx 2 0.5 m 2

= 8 Pa

Ans.

Ans.

Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: u max = 0.132 m>s tt = - 8 Pa tb = 8 Pa 932

9–3. Air at T = 40° F flows with an average velocity of 15 ft>s past the charged plates of the air cleaner. The plates are each 10 in. wide, and the gap between them is 1>8 in. If fully developed laminar flow develops, determine the pressure difference pB - pA between the inlet A and the outlet B.

1 — in. 8 A

B 24 in.

SOLUTION

From Appendix A, at T = 40° F, ra = 0.00247 slug>ft 3 and ma = 0.363 ( 10-6 ) lb # s>ft 2. The Reynolds number of the flow is raVa Re = = ma

( 0.00247 slug>ft 3 )( 15 ft>s ) a

0.125 ft b 12

0.363 ( 10-6 ) lb # s>ft 2

= 1063 6 1400 (laminar flow) Here, the flow can be considered as horizontal flow caused by pressure difference between the inlet and outlet through two fixed parallel plates. Also the air is incompressible. Therefore,

Vavg = -

2

a dp ; 12ma dx

15 ft>s = -

a

2 0.125 ft b 12

12 3 0.363 ( 10

-6

pB - pA

c d ) lb # s>ft 2 4 ( 24>12 ft )

pB - pA = - 1.204 lb>ft 2 = - 1.20 lb>ft 2

Ans.

The negative sign indicates that the pressure drops from the inlet to outlet.

Ans: pB - pA = - 1.20 lb>ft 2 933

*9–4. Glue is applied to the surface of the plastic strip, which has a width of 200 mm, by pulling the strip through  the container. Determine the force F that must be applied to the tape if the tape moves at 10 mm>s. Take rg = 730 kg>m3 and mg = 0.860 N # s>m2.

10 mm/s

40 mm F 40 mm 300 mm

SOLUTION The Reynolds number of the flow is

Re =

rgVa mg

=

τA F

( 730 kg>m3 ) c 10 ( 10-3 ) m>s d (0.04 m)

τA

0.860 N # s>m2

(a)

= 0.3395 Since Re62300, steady laminar flow occurs. Also, the glue is incompressible. Here, the flow is horizontal which is caused by the top moving plastic strip. Therefore, the shear stress acting along the surface of the plastic strip can be determined using Eq. 9–14.

t =

Umg a

=

c 10 ( 10-3 ) m>s d ( 0.860 N # s>m2 ) 0.04 m

= 0.215 N>m2

Consider the horizontal equilibrium of the FBD of the plastic strip in Fig. a, + ΣFx = 0; S

F - 2tA = 0 F = 2tA = 2 ( 0.215 N>m2 ) (0.3 m)(0.2 m) Ans.

= 0.0258 N

934

9–5. The 20-kg uniform plate is released and slides down the inclined plane. If an oil film under its surface is 0.2 mm thick, determine the terminal velocity of the plate along the plane. The plate has a width of 0.5 m. Take ro = 880 kg>m3 and mo = 0.0670 N # s>m2.

0.75 m

15!

SOLUTION Here, the terminal velocity is constant. Thus, the plate is in equilibrium. Referring to the FBD of the plate shown in Fig. a, + ΣFx = 0; S

Fv - [20(9.81) N] sin 15° = 0

20(9.81)N

Fv = 50.78 N

15°

Thus, the shear stress acting on the bottom plate’s surface is tp =

Fr

Fv 50.78 N = 135.41 N>m2 = Ap (0.75 m)(0.5 m)

We will assume that the steady laminar flow occurs and the oil is incompressible. dp dh = 0, = - sin u and U = u t. Then Here dx dx ut u0 a - g ay - b sin u t = a 2

N (a)

For this case u = 15° and a = 0.2 ( 10-3 ) m. At y = a = 0.2 ( 10-3 ) m, t = tp = 135.41 N>m2. Substitute this data into the above equation 135.41 N>m2 =

u t ( 0.0670 N # s>m2 ) 0.2 ( 10

-3

)m

- ( 880 kg>m3 )( 9.81 m>s2 ) c 0.2 ( 10-3 ) m -

0.2 ( 10-3 ) m

Ans.

u t = 0.4049 m>s = 0.405 m>s

2

d (sin 15°)

The Reynolds number is Re =

ru t a = m0

( 880 kg>m3 )( 0.4049 m>s ) 3 0.2 ( 10-3 ) m 4 0.0670 N # s>m2

= 1.06 6 1400 (laminar flow)

Ans: 0.405 m>s 935

9–6. Using pins, the plug is attached to the cylinder such that there is a gap between the plug and the walls of 0.2 mm. If the pressure within the oil contained in the cylinder is 4  kPa, determine the initial volumetric flow of oil up the sides of the plug. Assume the flow is similar to that between parallel plates since the gap size is very much smaller than the radius of the plug. Take ro = 880 kg>m3 and mo = 30.5 1 10-3 2 N # s>m2.

100 mm 50 mm

SOLUTION The flow is assumed to be steady laminar, and the oil is incompressible. Positive x axis is directed in the direction of flow which is vertically upwards. r2 - r1 h2 - h1 0 (p + gh) = + ga b 0x L L =

0 - 4 ( 103 ) N>m2 0.05 m

= - 71.367 ( 103 )

+ ( 880 kg>m3 )( 9.81 m>s2 ) a

N>m2

0.05 m - 0 b 0.05 m

m

Substitute this value and U = 0 into u = 0 -

1

2 3 30.5 ( 10-3 ) N # s>m2 4

J - 71.367 ( 103 )

N>m2 m

u = 1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4

R 3 0.2 ( 10-3 ) y - y2 4

Here b = 2p(0.05 m) = 0.1p m. Then the flow rate is Q =

=

LA L0

udA

0.2(10-3) m

1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4 (0.1p)dy

= 0.36755 ( 106 ) J0.1 ( 10-3 ) y2 -

-3 y3 0.2(10 ) m R` 3 0

= 0.4901 ( 10-6 ) m3 >s = 0.490 ( 10-6 ) m3 >s

Ans.

The average velocity is V =

0.4901 ( 10-6 ) m3 >s Q = = 7.800 ( 10-3 ) m>s A (0.1p m) 3 0.2 ( 10-3 ) m 4

The Reynolds number is Re =

rVa = m0

( 880 kg>m2 ) 3 7.800 ( 10-3 ) m>s 4 3 0.2 ( 10-3 ) m 4 30.5 ( 10-3 ) N # s>m2

= 0.0450 6 1400 (laminar flow)

936

Ans: 0.490 ( 10-6 ) m3 >s

9–7. The boy has a mass of 50 kg and attempts to slide down the inclined plane. If a 0.3-mm-thick oil surface develops between his shoes and the surface, determine his terminal velocity down the incline. Both of his shoes have a total contact area of 0.0165 m2. Take ro = 900 kg>m3 and mo = 0.0638 N # s>m2.

2!

SOLUTION Since the boy is required to move with terminal velocity (constant), a = 0. Referring to the free-body diagram of the boy in Fig. a, + ΣFx = max; d

50(9.81) N a= 0

350(9.81)N4 sin 2° - F = 0

2° x

F = 17.118 N

The shear stress on the oil layer in contact with the boy’s shoes is t =

Here, t =

F 17.118 N = = 1037.47 Pa A 0.0165 m2

F N

dp dh = 0 and = -sin 2°. dx dx

(a)

Um d a + c (p + gh) d ay - b a dx 2

1037.47

U ( 0.0638 N # s>m2 ) N = + 2 m 0.3 ( 10-3 ) m

U = 4.879 m>s = 4.88 m>s

30

+ ( 900 kg>m3 )( 9.81 m>s2 )( -sin 2° ) 4 c 0.3 ( 10-3 ) m Ans.

0.3 ( 10-3 ) m 2

d

Thus, the maximum Reynolds number is Re =

rUa = m

( 900 kg>m3 ) (4.879 m>s) 3 0.3 ( 10-3 ) m 4 0.0638 N # s>m2

= 20.65 6 1400 (OK)

Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: 4.88 m>s 937

*9–8. The 2.5-lb plate, which is 8 in. wide, is placed on the 3° inclined plane and released. If its terminal velocity is 0.2  ft>s, determine the approximate thickness of oil underneath the plate. Take ro = 1.71 slug>ft 3 and mo = 0.632 1 10-3 2 lb # s>ft 2.

12 in. 3!

SOLUTION + ΣF = max; S

(2.5 lb) sin 30° - F = 0

2.5 lb

F = 0.1308 lb

a=0

Thus, the shear stress on the oil layer in contact with the bottom surface of the plate (y = a) is t =

P = A

dp dh = 0 and Here, = - sin 3°. dx dx

0.1963

lb = ft 2

0.1308 lb lb = 0.1963 2 12 8 ft a ft ba ft b 12 12

F N (a)

Um d a + c (p + gh) d ay - b a dx 2

t =

( 0.2 ft>s )( 0.632 ( 10-3 ) lb # s>ft 2 ) a

+

30

+ ( 1.71 slug>ft 3 )( 32.2 ft>s2 ) ( - sin 3°) 4 aa -

1.4409a2 + 0.1963a - 0.1264 ( 10-3 ) = 0 Solving for the positive root,

a = 0.6410 ( 10-3 ) ft = 7.69 ( 10-3 ) in.

Ans.

Thus, the maximum Reynolds number is Re =

rUa = m



( 1.71 slug>ft 3 )( 0.2 ft>s ) 3 0.6410 ( 10-3 ) ft 4 = 0.3469 6 1400 (OK) 0.632 ( 10-3 ) lb # s>ft

Note: Normally the average velocity is used to calculate the Reynolds number.

938

a b 2

x

9–9. The water tank has a rectangular crack on its side having a width of 100 mm and an average opening of 0.1 mm. If laminar flow occurs through the crack, determine the volumetric flow of water through the crack. The water is at a temperature of T = 20° C.

100 mm

2m 100 mm

SOLUTION Assuming that steady laminar flow occurs, and the water is incompressible. The flow can be considered as horizontal flow through two stationary parallel plates driven by a pressure gradient. The pressure of water at the inlet of the crack is pin = rwgh = ( 998.3 kg>m3 )( 9.81 m>s2 ) (2 m) = 19.587 ( 103 ) N>m2 At the outlet, pout = patm = 0. Thus, the pressure gradient is N>m2 0 - 19.587 ( 103 ) N>m2 pout - pin dp = = = -195.87 ( 103 ) dx L 0.1 m m Q = -

3 0.1 ( 10-3 ) m 4 3(0.1 m) N>m2 a3b dp -3 ( ) J 195.87 10 R = 12m dx m 12 3 1.00 ( 10-3 ) N # s>m2 4

The average velocity is V =

= 1.632 ( 10-6 ) m3 >s = 1.63 ( 10-6 ) m3 >s Q = A

The Reynolds number is Re =

rVa = m

1.632 ( 10-6 ) m3 >s

3 0.1 ( 10-3 ) m 4 (0.1 m)

Ans.

= 0.1632 m>s

( 998.3 kg>m3 )( 0.1632 m>s )( 0.1 ( 10-3 ) m ) 1.00 ( 10-3 ) N # s>m2

= 16.29 6 1400 (laminar flow)

939

Ans: 1.63 ( 10-6 ) m3 >s

9–10. A solar water heater consists of two flat plates that rest on the roof. Water enters at A and exits at B. If the pressure drop from A to B is 60 Pa, determine the largest gap a between the plates so that the flow remains laminar. For the calculation, assume the water has an average temperature of 40° C.

A 3m

7.5!

B

a

SOLUTION

From Appendix A at T = 40° C, rw = 992.3 kg>m3 and mw = 0.659 ( 10-3 ) N # s>m2. It is required that steady laminar flow occurs. Also, the water is incompressible. Here 60 N>m2 0 - (3 m) sin 7.5° 0 (p + gh) = + ( 992.3 kg>m3 )( 9.81 m>s2 ) c d 0x 3m 3m = - 1290.60

N>m2 m

u = u =

1

1 0 c (p + gh) d ( ay - y2 ) 2mw 0x

2 3 0.659 ( 10-3 ) N # s>m2 4

u = 0.9792 ( 106 )( ay - y2 )

a -1290.60

N>m2 m

b ( ay - y2 )

The flow rate can be determined from Q =

L A

udA = 0.9792 ( 106 )

L0

a

( ay - y2 ) b dy = 0.1632 ( 106 ) a3b

The average velocity is V =

0.1632 ( 106 ) a3b Q = = 0.1632 ( 106 ) a2 A ab

To satisfy the condition of laminar flow, we require Re = 1400 rwVa = 1400 mw

( 992.3 kg>m3 ) 3 0.1632 ( 106 ) a2 4 a = 1400 0.659 ( 10-3 ) N # s>m2 a = 1.786 ( 10-3 ) m = 1.79 mm

Ans.

Ans: 1.79 mm 940

9–11. The 100-mm-diameter shaft is supported by an oillubricated bearing. If the gap within the bearing is 2 mm, determine the torque T that must be applied to the shaft, so that it rotates at a constant rate of 180 rev>min. Assume no oil leaks out due to sealing, and the flow behavior is similar to that which occurs between parallel plates, since the gap size is very much smaller than the radius of the shaft. Take ro = 840 kg>m3 and mo = 0.22 N # s>m2.

200 mm

180 rev/min T

100 mm

2 mm

SOLUTION The flow can be considered as horizontal flow caused by the top moving plate. Assume that steady laminar flow occurs and the lubricated oil is incompressible. rev 2prad 1 min Hence with U = vr = Ja180 ba ba b R (0.05 m) = 0.3p m>s min 1 rev 605 t =

Um = a

( 0.3p m>s )( 0.22 N # s>m2 ) 0.002 m

= 33p N>m2

The shear force acting on the bearing is F = tA = ( 33p N>m2 ) [2p(0.05 m)(0.2 m)] = 0.66p2 N Equilibrium requires T = Fr = ( 0.66p2 N ) (0.05 m) = 0.326 N # m

Ans.

The Reynolds number is Re =

rUa = m

( 840 kg>m3 )( 0.3p m>s )( 0.002 m ) 0.22 N # s>m2

= 7.20 6 1400 (laminar flow). Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: 0.326 N # m 941

*9–12. The two sections of the building wall have a 10-mm-wide crack between them. If the difference in pressure between the inside and outside of the building is 1.5 Pa, determine the flow of air out of the building through the crack. The air temperature is 30° C.

2m 600 mm

10 mm

SOLUTION For this case,

dh = 0 and U = 0. From dx

Appendix A, m = 18.6 ( 10-6 ) N # s>m2 and r = 1.164 kg>m3 at T = 30° C. Q = -

a3b dp 12m dx =

-(0.01 m)3(2 m)

12 3 18.6 ( 10-6 ) N # s>m2 4

= 0.0224 m3 >s

a

-1.5 N>m2 0.6 m

b

Ans.

To evaluate the maximum Reynolds number, we must first find the maximum velocity of the wind. u max = -

-1.5 N>m2 (0.01 m)2 a2 dp = a b = 1.680 m>s -6 2 8m dx 0.6 m 8 318.6 ( 10 ) N # s>m 4

Thus, the maximum Reynolds number is Re =

ru maxa = m

( 1.164 kg>m3 )( 1.680 m>s ) (0.01m) = 1051.42 6 1400 18.6 ( 10-6 ) N # s>m2

(OK)

Note: Normally the average velocity is used to calculate the Reynolds number.

942

9–13. The belt is moving at a constant rate of 3 mm>s. The 2-kg plate between the belt and surface is resting on a 0.5-mm-thick film of oil, whereas oil between the top of the plate and the belt is 0.8 mm thick. Determine the plate’s terminal velocity as it slides along the surface. Assume the velocity profile is linear. Take ro = 900 kg>m3 and mo = 0.0675 N # s>m2.

3 mm/s 0.8 mm

0.5 mm 300 mm

SOLUTION The velocity profile of the oil flow at the top and bottom layer is shown in Fig. a. The shear stress in each layer is constant.

( 3 mm>s - Vp ) m

tt =

U′t m = a′t

tb =

Vpm Vpm U bm = = ab 0.5 mm 0.5

0.8 mm

=

3 mm s

( 3 - Vp ) m

Vp

0.8 a´t = 0.8 mm

Since the plate is required to move at the terminal velocity Vp (constant), a = 0. Referring to the free-body diagram of the plate in Fig. b, + ΣFx = max; S

3 m m s – Vp

Plate

ttA - tbA = 0 ab = 0.5 mm

tt = tb Substituting,

( 3 - Vp ) m 0.8

=

Vp m

Vp

0.5 Ans.

Vp = 1.154 mm>s = 1.15 mm>s Thus, the maximum Reynolds number for the top and bottom layers is Re =

900 kg>m3 3 3 ( 10-3 ) m>s 4 3 0.8 ( 10-3 ) m 4 rUa = 0.0320 6 1400 = m ( 0.0675 N # s>m2 )

a=0

τt A

(OK)

and Re =

(a) x

τb A (b)

900 kg>m3 31.154 ( 10-3 ) m>s4 30.5 ( 10-3 ) m4 rUa = 0.00769 6 1400 (OK) = m ( 0.0675 N # s>m2 )

Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: 1.15 mm>s 943

9–14. When you inhale, air flows through the turbinate bones of your nasal passages as shown. Assume that for a short length of 15 mm, the flow is passing through parallel plates, the plates having a total mean width of w = 20 mm and spacing of a = 1 mm. If the lungs produce a pressure drop of ∆p = 50 Pa, and the air has a temperature of 20° C, determine the power needed to inhale air.

1 mm

SOLUTION The flow is assumed to be steady laminar and the air is incompressible. Since the density of the air is small, the elevation gradient term is negligible. Also, the “plates” are at rest U = 0. 1 0 u = - c (p + gh) d (ay - y2) 2m 0x = = -

1 ∆r a b(ay - y2) 2m L

∆r (ay - y2) 2mL

Then, the flow can be determined from Q =

L

a

udA =

A

L 0

-

∆r (ay - y2)(wdy) 2mL a

w∆ p = (ay - y2)dy 2mL L 0

= = -

y3 a w∆p ay2 a - b` 2mL 2 3 0 wa3 ∆p 12mL

(1)

The power required can be determined from . W = FV Since F = ∆ pA and Q = VA,

#

W = ∆pAV = QAp

(2)

944

r

9–14. Continued

From Appendix A, at T = 20° C, ra = 1.202 kg>m3 and ma = 18.1 ( 10-6 ) N # s>m2. Substitute the numerical data into Eq. 1 and 2, Q = # W =

(0.02 m)(0.001 m)3 ( -50 N>m2 )

12 3 18.1 ( 10-6 ) N # s>m2 4 (0.015 m)

3 0.3069 ( 10-3 ) m3 >s 4 ( 50 N>m2 )

= 0.3069 ( 10-3 ) m3 >s Ans.

= 0.0153W

The average velocity is

V =

0.3069 ( 10-3 ) m3 >s Q = = 15.35 m>s A (0.02 m) ( 0.001>m )

The Reynolds number is Re =

raVa = ma

( 1.202 kg>m3 )( 15.35 m>s ) (0.001 m) 18.1 ( 10-6 ) N # s>m2

= 1019 6 1400 (laminar flow)

Ans: 0.0153 W 945

9–15. The hydraulic lift consists of a 1-ft-diameter cylinder that fits into the sleeve, such that the gap between them is 0.001 in. If the platform supports a load of 3000 lb, and the  oil is pressurized to 25 psi, determine the terminal velocity  of  the platform. Take ro = 1.70 slug>ft 3 and mo = 0.630 1 10-3 2 lb # s>ft 2. Assume that the flow is laminar, and that it is similar to that which occurs between parallel plates since the gap size is very much smaller than the radius of the cylinder.

3000 lb A

3 ft 1 ft

SOLUTION 3000 lb

Assume the oil is incompressible. The cylinder is moving with a terminal velocity (constant), a = 0. Referring to the free - body diagram of the cylinder in Fig. a, + c ΣFx = max; a25

lb 12 in. 2 ba b 3 p(0.5 ft)2 4 + t[2p(0.5 ft)(3 ft)] - 3000 lb = 0 1 ft in2 t = 18.31 lb>ft

2

Since the gap is small compared to the radius of the cylinder, the oil can be considdh ered as flow between two vertical plates. Here, the oil flows upward, thus, = 1. dx lb 12 in. 2 0 - a25 2 ba b lb>ft 2 dp 1 ft in = = - 1200 and t = 18.31 lb>ft 2 at y = a, Also, dx 3 ft ft =

x

τ

τ

a=0

25 psi (a)

0.001 ft = 83.333 ( 10-6 ) ft 12

Thus, t = 18.31 lb>ft 2 =

dp Um a + c (p + gh) d ay - b a dx 2

U 3 0.630 ( 10-3 ) lb # s>ft 2 4 0.001 ft 12

+ £ ° -1200

lb>ft 2 ft

¢ + ( 1.70 slug>ft 2 )( 32.2 ft>s2 ) (1) § £ 83.333 ( 10-6 ) ft -

83.333 ( 10-6 ) ft 2

Ans.

U = 2.43 ft>s

Ans: 2.43 ft>s 946

§

*9–16. The liquid has laminar flow between the two fixed plates due to a pressure gradient dp>dx. Using the coordinate system shown, determine the shear-stress distribution within the liquid and the velocity profile for the liquid. The viscosity is m.

y

a –– 2 a –– 2

SOLUTION

(τ + dydτ dy (dxdz

Assume the liquid is incompressible. The free-body diagram of the fluid’s differential element of length x, height dy, and width dz, is shown in Fig. a. Applying the momentum equation along the x axis, + ΣF = S x

L cs

Ar # dA;

dp dt pdydz - ap + dxbdydz + at + dybdxdz - tdxdz = 0 dx dy

Dividing by dxdydz, this equation reduces to dp dt = dy dx Integrating with respect to y, t = a

dp by + C1 dx

If the flow is considered laminar, t = m m

du . Then, dy

dp du = a by + C1 dy dx

Again, integrating with respect to y, u =

C1 1 dp 2 y + y + C2 m 2m dx

The integration constant can be evaluated using the boundary condition, u = 0 at a y = { . Substituting, 2 0 =

0 =

C1 a 1 dp a 2 a b + a b + C2 m 2 2m dx 2 a2 dp a + C + C2 8m dx 2m 1

(1)

and 0 =

0 =

x

C1 a 2 a 1 dp a- b + a - b + C2 m 2m dx 2 2 a2 dp a C + C2 8m dx 2m 1

(2)

947

x

( p + dxdp dx (dydz

pdydz

τ dxdz γdxdydz (a)

9–16. Continued

Solving Eqs. 1 and 2, C1 = 0

C2 = -

a2 dp 8m dx

Thus, t =

dp y dx

Ans.

and u =

y 2 a2 dp c4a b - 1d 8m dx a

Ans.

a into the equation for 2 velocity in the text where y′ has its origin at the bottom plate.

This result can also be obtained by substituting y′ = y +

948

9–17. A thin layer of engine oil is trapped between the belts, which are moving in different directions and at different speeds as shown. Plot the velocity profile within the oil film and the shear-stress distribution. The pressures at A and B are atmospheric. Take mo = 0.22 N # s>m2 and ro = 876 kg>m3.

A

0.2 m/s

4 mm

B 0.3 m/s

SOLUTION Assuming that steady laminar flow occurs, and the engine oil is incompressible. Since the flow is horizontal and no pressure different between A and B, then

y(10–3)m

0 (p + rh) = 0 0x C1 y + C2 m0 Applying the boundary conditions at y = 0, u = u 1 and at y = a, u = u 2,

Thus, u =

u1 =

C1 (0) + C2 m0

C2 = u 1 τ (N m2)

–27.5

And u2 =

C1 (a) + u 1 m0

m0 ( u 2 - u 1 )

C1 =

(a)

a

Then t = C1 = u =

m0 ( u 2 - u 1 )

(1)

a

u2 - u1 y + u1 a

y(10–3)m

(2) 4.0

Here u 1 = 0.3 m>s and u 2 = - 0.2 m>s. Substitute this value into Eq. 1 and 2, t =

( 0.22 N # s>m2 )( - 0.2 m>s - 0.3 m>s ) = - 27.5 N>m2 4 ( 10-3 ) m u = c

-0.2 m>s - 0.3 m>s 4 ( 10-3 ) m

3.0 2.4 0.2

Ans.

0.1

d y + 0.3 m>s

–0.2

–0.1

0

u(N m) 0.1

0.2

0.3

Ans.

= (0.3 - 125y) m>s

(b)

The Plot of the shear stress distribution and velocity profile are shown in Fig. a and b respectively. Using V = vmax = 0.5 m>s, Re =

876 kg>m2 ( 0.5 m>s ) (0.004 m) r0Va = = 7.96 6 1400 m0 ( 0.22 N # s>m2 )

Ans: t = - 27.5 N>m2 u = (0.3 - 125y) m>s 949

9–18. The materials testing nuclear reactor has fuel elements in the form of flat plates that allow cooling water to flow between them. The plates are spaced 1>16 in. apart. Determine the pressure drop of the water over the length of the fuel elements if the average velocity of the flow is 0.5 ft>s through the plates. Each fuel element is 2 ft long. Neglect end effects in the calculation. Take rw = 1.820 slug>ft 3 and mw = 5.46 1 10-6 2 lb # s>ft 2.

0.5 ft/s

2 ft

SOLUTION The Reynolds number is rwVa = Re = mw

( 1.820 slug>ft 3 )( 0.5 ft>s ) a

1 — in. 16

0.0625 ft b 12

3 in.

5.46 ( 10-6 ) lb # s>ft 2

= 868 6 1400 (laminar flow) Here, the flow can be considered as vertically downward flow caused by the pressure and elevation gradient between the inlet and outlet through two fixed parallel plates. Also, water is incompressible. Here ∆p hB - hk 0 (p + gh) = + ga b 0x L L = =

∆p 0 - 2 ft + ( 1.820 slug>ft 3 )( 32.2 ft>s2 ) a b 2 ft 2 ft ∆p - 58.604 2

With U = 0, U = -

= -

1 0 c (p + gh) d ( ay - y2 ) 2m 0x 1 2 3 5.46 ( 10

-6

) 16.5>ft 4 2

a

∆p 0.0625 - 58.604b c a by - y2 d 2 12

= - 91.575 ( 103 ) (0.5∆p - 58.604) 3 5.2083 ( 10-3 ) y - y2 4

The flow can be determined from Q =

LA

5.2083(10-3) ft

udA = -91.575 ( 10-3 ) (0.5∆p - 58.604)

L0

3 5.2083 ( 10-3 ) y

= -22.894 ( 103 ) (0.5∆p - 58.604) c 2.6042 ( 10-3 ) y2 = - 0.5391 ( 10-3 ) (0.5∆p - 58.604)

950

- y2 4 (0.25 dy)

-3 y3 5.2083(10 ) ft d2 2 0

9–18. Continued

The average velocity is V =

Q ; A

0.5 ft>s =

- 0.5391 ( 10-3 ) (0.5 ∆ p - 58.604) a

0.0625 3 ft ba ft b 12 12

∆ p = ( 114.79 lb>ft 2 ) a

1 ft 2 b = 0.797 psi 12 in

Ans.

The positive sign indicates that the pressure increases in the direction of flow. The hydrostatic increase in pressure is greater than the decrease due to friction.

Ans: 0.797 psi 951

9–19. The water and oil films have the same thickness a and are subjected to the movement of the top plate. Plot the velocity profile and the shear-stress distribution for each fluid. There is no pressure gradient between A and B. The viscosities of water and oil are mw and mo, respectively.

U a

A

B

a

SOLUTION Assume that steady laminar flow occurs within the water and oil layers, and water and oil are incompressible. Since the flow is horizontal, and no pressure gradient, d (P + gh) = 0 dx For water layer, (1)

tw = C1 uw =

C1 y + C2 mw

(2)

Applying the boundary conditions, at y = 0, u w = 0 and at y = a, u w = U′, Eq. 2 gives C2 = 0

2a Oil layer

C1 (0) + C2 mw

And U′ =

C1 (a) mw

For the oil layer, (4)

t0 = C3 u0 =

C3 y + C4 mO

mw m0 U

(5)

C3 a + C4 mO

(6)

Substituting Eq. 3 into Eq. 6, realizing that C1 = C3, C3 C3 (a) = a + C4 mw mO C4 = C3 a a

τ

0

It is required that at the water and oil interface, y = a, tw = t0. Therefore, from Eq. 1 and 4 C1 = C3. Also, u w = u O = U′. Then Eq. 5 gives U′ =

a

(3) Water layer

0 =

y

1 1 b mw mO

(7)

952

(m0 + mw(a (a)

9–19. Continued

Applying the boundary condition at y = 2a, u o = U, Eq. 5 gives U =

C3 (2a) + C4 mo

(8)

Substitute Eq. 7 into 8

y

2C3a 1 1 U = + C3a a b mo mw mo

2a

ye

Oil layer

mo + mw C3a C3a U = + = C3a a b mw mo mwmo mwmoU (mo +a mw)a

a Water layer

C3 =

l

Substitute this result into Eq. 7,

C4 = a = a = a

mwmoU 1 1 ba b mo mo + mw mw 0

u

0

(m + m (U m0

mwmoU m0 - mw ba b mwmo mo + mw

0

mo - mw( ) bU mo + mw

U

w

(b)

Thus, tw = t0 = uw = uo = uo = From Eq. 3

mwmoU (mo + mw)a

Ans.

moU y (mo + mw)a

Ans.

mwU mo - mw y + a bU (mo + mw)a mo + mw

U 3 m y + a(mo - mw) 4 a(mo + mw) w

U′ = a

Ans.

mo bU mo + mw

Ans:

mwmoU (mo + mw)a moU uw = y (mo + mw)a U uo = [m y + a(mo - mw)] a(mo + mw) w

tw = to =

The plot of the shear distribution and velocity profile are shown in Fig. a and b.

953

*9–20. Use the Navier–Stokes equations to show that the velocity distribution of the steady laminar flow of a fluid flowing down the inclined surface is defined by u = 3rg sin u>(2m) 4 1 2hy - y2 2 , where r is the fluid density and m is its viscosity.

y U

h

u u x

SOLUTION Since the flow is steady and is along the x axis only, then v = w = 0. Also, the liquid is incompressible. Thus, the countinuity equation reduces to 0(ru) 0(rv) 0(rw) 0r + + + = 0 0t 0x 0y 0z 0 + r

0u + 0 + 0 = 0 0x 0u = 0 0x

Integrating this equation with respect to x, u = u(y) Navier-stokes equations along the x and y axes give ra

0p 0u 0u 0u 0u 02u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z r(0 + 0 + 0 + 0) = rg sin u - 0 + m a0 +

d 2u + 0b dy2

rg sin u 02u = 2 m 0y

ra

0p 0v 0v 0v 0v 02v 02v 02v + m + v + w b = rgy + ma 2 + 2 + 2 b 0t 0x 0y 0z 0y 0x 0y 0z r(0 + 0 + 0 + 0) = r( - g cos u) -

0p + m(0 + 0 + 0) 0y

0p = -rg cos u 0y Since u = u(y), then

(2)

02u d 2u = . Thus Eq. 1 becomes 0y2 dy2 rg sin u d 2u = 2 m dy

Integrating this equation with respect to y twice, yields rg sin u du = y + C1 m dy u =

(3)

- rg sin u 2 y + C1y + C 2m

(4)

954

9–20. Continued

At y = 0, u = 0. Thus, Eq. 4 gives 0 = - 0 + 0 + C2 At y = h, txy = m a

C2 = 0

du b = 0. Then Eq. 3 gives dy 0 = -

rg sin u (h) + C1 m

C1 =

rgh sin u m

Substitute these results into Eq. 3 and 4 rg sin u rgh sin u du = y + m m dy rg sin u du = (h - y) m dy

u = -

u =

rg sin u 2m

rg sin u 2m

y2 +

rgh sin u y m

( 2hy - y2 )

(Q.E.D.)

955

9–21. A fluid has laminar flow between the two parallel plates, each moving in the same direction, but with different velocities, as shown. Use the Navier–Stokes equations, and establish an expression that gives the shear-stress distribution and the velocity profile for the fluid. Plot these results. There is no pressure gradient between A and B.

y

A

0(ru) 0(rv) 0(rw) 0r + + + = 0 0t 0x 0y 0z 0u + 0 + 0 = 0 0x 0u = 0. 0x Integrating this equation with respect to x, u = u(y) Using this result and the requirement that the pressure p remain constant along the x axis, the Navier-Stokes equation along x axis gives. 0r 0u 0u 0u 0u 02u 02u 02u + u + v + w b = rgx + ma 2 + 2 + 2 b 0t 0x 0y 0z 0x 0x 0y 0z 0 + 0 + 0 + 0 = 0 - 0 + ma0 +

m

02u + 0b 0y2

02u = 0 0y2

02u d 2u . Integrate this equation with respect to y Since u is a function of y only, 2 = 0y dy2 twice, du = C1 dy

(1)

u = C1y + C2

(2)

And

Applying the boundary condition u = Ub at y = 0, Ub = C1(0) + C2

C 2 = Ub

And u = Ut at y = a, u t = C1(a) + Ub C1 =

Ut - Ub a

Substituting these results into Eq. 2 u = a

B x

Since the flow is steady and is along the x axis only, then v = w = 0. Also, the liquid is incompressible. Thus, the continuity equation reduces to

ra

a

Ub

SOLUTION

0 + r

Ut

Ut - Ub by + Ub a

Ans.

956

9–21. Continued

Applying txy = ma

Here

0n 0u + b 0y 0x

Ut - Ub 0v du = 0 and from Eq. (1), = C1 = . Then 0x dy a txy = mc a

m ( Ut - Ub ) Ut - Ub b + 0d = a a

Ans.

The plot of the shear stress distribution and velocity profile are shown in Fig. a and b respectively

y

y a

a

µ(Ut – Ub( a (a)

u

τ xy Ub

Ut

(b)

Ans: u = a txy =

957

Ut - Ub by + Ub a m(Ut - Ub) a

9–22. The retinal arterioles supply the retina of the eye with blood flow. The inner diameter of an arteriole is 0.08  mm, and the mean velocity of flow is 28 mm>s. Determine if this flow is laminar or turbulent. Blood has a density of 1060 kg>m3 and an apparent viscosity of 0.0036 N # s>m2.

SOLUTION The Reynolds number is given by Re =

rVD = m

( 1060 kg>m3 ) 3 28 ( 10-3 ) m>s 4 3 0.08 ( 10-3 ) m 4 0.0036 N # s>m2

= 0.660

Since Re 6 2300, laminar flow occurs. Here the Reynolds number is much smaller than the limit, therefore there is no danger for the flow to become turbulent.

Ans: laminar 958

9–23. Oil flows through a 3-in.-diameter horizontal pipe such that the pressure drops 1.5 psi within a 10-ft length. Determine the shear stress that the oil exerts on the wall of the pipe.

SOLUTION Assume the oil is incompressible. For the horizontal pipe, we can determine the maximum shear stress at the pipe’s wall regardless of whether the flow is laminar or turbulent using tmax =

=

D ∆p 4 L a

3 lb 12 in. 2 ft b a1.5 2 ba b 12 1 ft £ in § 4 10 ft

= 1.35

lb 1 ft 2 a b = 9.375 ( 10-3 ) psi Ans. ft 2 12 in.

Ans: 9.375 ( 10-3 ) psi 959

*9–24. The 100-mm-diameter horizontal pipe transports castor oil in a processing plant. If the pressure drops 100 kPa in a 10-m length of the pipe, determine the maximum velocity of the oil in the pipe and the maximum shear stress in the oil. Take ro = 960 kg>m3 and mo = 0.985 N # s>m2.

100 mm

10 m

SOLUTION Assuming that steady laminar flow occurs, and the castor oil is incompressible. Since the flow is horizontal, the maximum velocity and the maximum shear stress can be determined by u max = =

D2 ∆p a b 16m L

(0.1 m)2

16 ( 0.985 N # s>m

2

)

c

100 ( 103 ) N>m2 10 m

= 6.345 m>s = 6.35 m>s tmax =

d

3 2 D ∆p 0.1 m 100 ( 10 ) N>m = a ba b 4 L 4 10 m

= 250 N>m2

The maximum Reynolds number is Re =

rou maxD = mo

( 960 kg>m3 )( 6.345 m>s ) (0.1 m) 0.985 N # s>m2

= 618 6 2300 (laminar flow) Note: Normally the average velocity is used to calculate the Reynolds number.

960

Ans.

Ans.

9–25. Most blood flow in humans is laminar, and apart from pathological conditions, turbulence can occur in the descending portion of the aorta at high flow rates as when exercising. If blood has a density of 1060 kg>m3, and the diameter of the aorta is 25 mm, determine the largest average velocity blood can have before the flow becomes transitional. Assume that blood is a Newtonian fluid and has a viscosity of mb = 0.0035 N # s>m2. At this velocity, determine if turbulence occurs in an arteriole of the eye, where the diameter is 0.008 mm.

25 mm

SOLUTION The upper limit of the Reynolds number before laminar flow ceases is Re = 2300 rb(V)maxD = 2300 mb

( 1060 kg>m3 ) (V)max(0.025 m) 0.0035 N # s>m2

= 2300 Ans.

(V)max = 0.3038 m>s = 0.304 m>s with this average velocity and D = 0.008 ( 10-3 ) m, Re =

rbVD = m

( 1060 kg>m3 )( 0.3038 m>s ) 3 0.008 ( 10-3 ) m 4 0.0035 N # s>m2

= 0.736

Since Re 6 2300, laminar flow occurs. Thus, turbulence will not occur in the eye.

Ans: (V)max = 0.304 m>s Turbulence will not occur. 961

9–26. A 50-mm-diameter vertical pipe carries oil having a density of ro = 890 kg>m3. If the pressure drop in a 2-m length of the pipe is 500 Pa, determine the shear stress acting along the wall of the pipe. The flow is downward.

SOLUTION

dh Assume the oil is incompressible. Since the flow is vertically downward, = - 1. dx Here, r = 0.025 m at the pipe wall. t = -

D d (p + gh) = 4 dx 0.050 m 500 Pa c+ ( 890 kg>m3 )( 9.81 m>s2 ) ( -1) d 4 2m

= 112.26 Pa = 112 Pa

Ans.

Ans: 112 Pa 962

9–27. The 1-in.-diameter horizontal pipe is used to transport glycerin. If the pressure at A is 30 psi and at B is 20 psi, determine if the flow is laminar or turbulent.

1 in.

A

B

15 ft

SOLUTION

From Appendix A, r = 2.44 slug>ft 3 and m = 31.3 ( 10-3 ) lb # s>ft 2. The velocity is V = -

= -

D2 ∆p a b 32m L a

2 1 ft b 12

32 3 31.3 ( 10

-3

= 0.6656 ft>s

) lb # s>ft 2 4

£

a20

lb lb 12 in. 2 30 ba b § 1 ft in2 in2 15 ft

Thus, the Reynolds number is rVD Re = = m

( 2.44 slug>ft 3 )( 0.6656 ft>s ) a 31.3 ( 10-3 ) lb # s>ft 2

1 ft b 12

= 4.32 Ans.

Since Re 6 2300, the flow is laminar.

Ans: laminar 963

*9–28. The 1-in.-diameter pipe is used to transport water at 180° F. If the flow is to be laminar, what is the maximum pressure difference between points A and B?

1 in.

A

B

15 ft

SOLUTION Assume the water is incompressible. From Appendix A, r = 1.883 slug>ft 3 and m = 7.20 ( 10-6 ) lb # s>ft 2 at T = 180° F . Thus, Re =

rVD = 2300 m

( 1.883 slug>ft 3 ) V a

1 ft b 12

7.20 ( 10-6 ) lb # s>ft 2

= 2300

V = 0.10553 ft>s For a horizontal pipe, the velocity is V = -

0.10553 ft>s = -

D2 ∆p a b 32m L a

2 1 ft b 12

32 3 7.20 ( 10

∆p = 0.0525 lb>ft 2

-6

∆p

a b ) lb # s>ft 2 4 15

964

Ans.

9–29. Oil having a density of 900 kg>m3 and a viscosity of 0.370 N # s>m2 has a flow of 0.05 m3 >s through the 150-mmdiameter pipe. Determine the drop in pressure caused by viscous friction over the 8-m-long section.

A

B

8m

SOLUTION The flow is assumed to be steady laminar, and the oil is incompressible. Since the pipe is smooth and is horizontal, we can have ∆p =

128m # LQ 4

pD

=

128 ( 0.370 N # s>m2 ) (8 m) ( 0.05 m3 >s ) p(0.15 m)4

= 11.91 ( 103 ) Pa = 11.9 kPa

Ans.

The average velocity is V =

0.05 m3 >s Q = = 2.829 m>s A p(0.075 m)2

The Reynolds number is

Re =

(900 kg>m3) ( 2.829 m>s ) (0.15 m) rVD = m0 0.370 N # s>m2 = 1032.36 6 2300 (laminar flow)

Ans: 11.9 kPa 965

9–30. Oil has a flow of 0.004 m3 >s through the 150-mm-diameter pipe. Determine the drop in pressure caused by viscous friction over the 8-m-long section. Take ro = 900 kg>m3 and mo = 0.370 N # s>m2.

A

B

8m

SOLUTION ∆p = =

128mLQ pD4 128 ( 0.370 N # s>m2 ) (8 m) ( 0.004 m3 >s ) p(0.150 m)4

= 952.89 N>m2 = 953 Pa

Ans.

The maximum velocity can be determined from u max =

=

D2 ∆p a b 16m L (0.15 m)2

16 ( 0.370 N # s>m2 )

°

952.89 8m

N m2 ¢

= 0.4527 m>s Thus, the maximum Reynolds number is Re =

rVD = m

( 900 kg>m3 )( 0.4527 m>s ) (0.15 m) 0.370 N # s>m2

(OK!)

= 165 6 2300

Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: 953 Pa 966

9–31. Lymph is a fluid that is filtered from blood and forms an important part of the immune system. Assuming it is a Newtonian fluid, determine its average velocity if it flows from an artery into a 0.08-µm-diameter precapillary sphincter at a pressure of 120 mm of mercury then passes vertically upwards through the leg for a length of 1200 mm, and emerges at a  pressure of 25 mm of mercury. Take rl = 1030 kg>m3 and ml = 0.0016 N # s>m2.

SOLUTION Assuming that steady laminar flow occurs and lymph is incompressible. Here, the flow is vertically upwards, thus, hh - hf ph - pf 0 b ( p + gh) = + ga 0x L L =

( 13550 kg>m3 )( 9.81 m>s2 ) (0.025 m - 0.12 m) 1.2 m + ( 1030 kg>m3 )( 9.81 m>s ) a

= - 418.97

N>m2

1.2 m - 0 b 1.2 m

m

Applying, V = -

R2 0 ( p + gh) 8ml 0x

= -

3 0.4 ( 10-3 ) m 4 2

8 ( 0.0016 N # s>m

2

)

° -418.97

N>m2

= 5.237 ( 10-3 ) m>s = 5.24 mm>s

m

¢ Ans.

The Reynolds number is Re =

riVD = ml

( 1030 kg>m3 ) 3 52.37 ( 10-12 ) m>s 4 3 0.08 ( 10-6 ) m 4 0.0016 N # s>m2

= 2.69 ( 10-12 ) 6 2300 (laminar flow)

Ans: 5.24 mm>s 967

*9–32. A horizontal pipe is to be used to transport oil. If the pressure loss to be expected over a 500-ft length is 4 psi, determine the largest-diameter pipe that can be used if laminar flow is to be maintained. Take ro = 1.72 slug>ft 3 and mo = 1.40 1 10-3 2 lb # s>ft 2.

SOLUTION Assume the oil is incompressible. Here, the flow is required to be laminar. Thus, Re =

rVD = 2300 m

( 1.72 slug>ft 3 ) VD = 2300 1.40 ( 10-3 ) lb # s>ft 2 V =

1.872 D

(1)

Since the pipe is horizontal, the maximum velocity can be determined using ∆p = L V = V = -

a4

lb 12 in. 2 ba b lb>ft 2 1 ft in2 = - 1.152 . 500 ft ft

D2 ∆p a b 32m L

D2

32 3 1.40 ( 10-3 ) lb # s>ft 2 4

V = 25.71 D2

a - 1.152

lb>ft 2 ft

b

(2)

Solving Eqs. (1) and (2) yields Ans.

D = 0.4176 ft = 5.01 in.

968

9–33. A smooth 100-mm-diameter pipe transfers kerosene at 20° C with an average velocity of 0.05 m>s. Determine the pressure drop that occurs along the 20-m length. Also, what is the shear stress along the pipe wall?

20 m

SOLUTION The kerosene is incompressible. From Appendix A, rke = 814 kg>m3 and mke = 1.92 ( 10-3 ) N # s>m2. The Reynolds number is Re =

rkeVD = mke

( 814 kg>m3 )( 0.05 m>s ) (0.1 m) 1.92 ( 10-3 ) N # s>m2

= 2120 6 2300 (laminar flow) Since the steady laminar flow occurs, and the flow is horizontal, then V =

D2 ∆ p a b; 32mke L

0.05 m>s =

(0.1 m)2 32 3 1.92 ( 10

-3

∆p

a b ) N # s>m2 4 20 m

∆ p = 6.144 Pa = 6.14 Pa

Ans.

Along the pipe’s wall, the shear stress is maximum 2

tmax =

D ∆p 0.1 m 6.144 N>m b = a ba b a 4 L 4 20 m = 7.68 ( 10-3 ) Pa

Ans.

Ans: ∆p = 6.14 Pa tmax = 7.68 ( 10-3 ) Pa 969

9–34. Determine the maximum discharge from an 18-in. -diameter horizontal duct used for air at a temperature of 60° F so that the flow remains laminar.

SOLUTION Assume the oil is incompressible. For laminar flow, the requirement is Re =

rVD = 2300 m and m = 0.374 ( 10-6 ) lb # s>ft 2 at

From Appendix A, r = 0.00237 slug>ft 3 T = 60° F. Thus,

( 0.00237 slug>ft 3 ) V a

18 ft b 12

0.374 ( 10-6 ) lb # s>ft 2

= 2300

Since the duct is horizontal, V = 0.2420 the pressure gradient is V =

0.2420 ft>s =

D2 ∆p a b 32m L a

2 18 ft b 12

32 3 0.374 ( 10

-6

) lb # s>ft 2 4

lb>ft 2 ∆p = 1.2872 ( 10-6 ) L ft

a

∆p b L

Subsequently, the flow rate can be determined from Q =

=

pD4 ∆p a b 128m L pa

4 18 ft b 12

128 3 0.374 ( 10-6 ) lb # s>ft 2 4

£ 1.2872 ( 10-6 )

lb>ft 2 ft

§

= 0.428 ft 3 >s

Ans.

970

Ans: 0.214 ft 3 >s

9–35. Air at 120° F flows through a 3-in.-diameter horizontal duct. Determine the greatest discharge it can have so that the flow remains laminar.

SOLUTION Assume the air is incompressible. For laminar flow, the requirement is Re =

rVD = 2300 m

From Appendix A, r = 0.00213 slug>ft 3 and m = 0.407 ( 10-6 ) lb # s>ft 2 at T = 120° F. Thus,

( 0.00213 slug>ft 3 ) V a

3 ft b 12

0.407 ( 10-6 ) lb # s>ft 2

= 2300

V = 1.7579 ft>s Since the pipe is horizontal, we can determine the pressure gradient from V =

1.7579 ft>s =

D2 ∆p a b 32m L a

2 3 ft b 12

a

∆p

32 3 0.407 ( 10-6 ) lb # s>ft 2 4 L

lb>ft 2 ∆p = 0.3663 ( 10-3 ) L ft

b

Subsequently, the flow rate can be determined from Q =

=

pD4 ∆p a b 128m L pa

4 3 ft b 12

128 3 0.407 ( 10-6 ) lb # s>ft 2 4

£ 0.3663 ( 10-3 )

lb>ft 2 ft

= 0.0863 ft 3 >s

§ Ans.

971

Ans: 0.0431 ft 3 >s

*9–36. Crude oil flows through a 3-in.-diameter horizontal pipe such that the pressure drops 5 psi within a 1000-ft length. Determine the shear stress within the oil at a distance of 0.5 in. from the pipe wall. Also find the shear stress along the centerline of the pipe, and the maximum velocity of the flow. Take ro = 1.71 slug>ft 3 and mo = 0.632 1 10-3 2 lb # s>ft 2.

SOLUTION The flow is assumed to be steady laminar, and the oil is incompressible. Here the 0h = 0. Then pipe is horizontal and smooth. Thus, 0x ∆r 0 ( p + rh) = 0x L Substitute this result into, r ∆p 2 L

t =

Here, since the pressure drops along the flow, ∆p is negative ∆p = a5

lb 12 in. 2 ba b = 720 lb>ft 2 2 1 ft in

For the case r = 1.5 in. - 0.5 in. = 1 =

t> r = >in. =

a

1 ft, 12

1 ft b 720 lb>ft 2 12 lb 1ft 2 £ § = a0.03 2 ba b = 0.208 ( 10-3 ) psi Ans. 2 1000 ft 12 in. ft

For the case r = 0;

Ans.

t> r = o = 0 The maximum velocity is

u max =

a

2 3 ft b 12

2

720 lb>ft D ∆p a b = a b 16m L 16 3 0.632 ( 10-3 ) lb # s>ft 2 4 1000 ft 2

= 4.45 ft>s

the average velocity is V =

a

Ans.

2 3 ft b 12

720 lb>ft 2 D ∆p a b = a b 32m L 1000 ft 32 3 0.632 ( 10-3 ) lb # s>ft 2 4 2

= 2.225 ft>s

The Reynolds number is Re =

roVD = mo

( 1.71 slug>ft 3 )( 2.225 ft>s )( 3>12 ft ) = 1505 6 2300 (laminar flow) 0.632 ( 10-3 ) lb # s>ft 2

972

9–37. Glycerin is at a pressure of 15 kPa at A when it flows through the vertical segment of the 100-mm-diameter pipe. Determine the discharge at B.

A

4m

B 100 mm

SOLUTION The flow is assumed to be steady laminar, and the oil is incompressible. Here the flow is vertically downwards and so the positive x axis must be directed vertically downwards as well. From Appendix A, rGC = 1260 kg>m3 and mGC = 1.50 N # s>m2. rB - rA hB - hA 0 (p + gh) = + ga b 0x LAB LAB =

0 - 15 ( 103 ) N>m2 4m

= - 16.111 ( 103 ) Q = = -

N>m2

+ ( 1260 kg>m3 )( 9.81m>s2 ) a

m

4

pR 0 (p + rh) 8mGC 0x p(0.05 m)4

0 - 4m b 4m

8 ( 1.50 N # s>m2 )

(1)

c - 16.111 ( 103 )

N>m2 m

= 0.02636 m3 >s = 0.0264 m3 >s

d

Ans.

The average velocity

V =

0.02636 m3 >s Q = = 3.356 m>s A p(0.05 m)2

Then, the Reynolds number is Re =

rGCVD = mGC

( 1260 kg>m3 )( 3.356 m>s ) (0.1 m) 1.50 N # s>m2

= 281.93 6 2300 (laminar flow)

973

Ans: 0.0264 m3 >s

9–38. The 50-mm-diameter smooth pipe drains engine oil out of a large tank at the rate of 0.01 m3 >s. Determine the horizontal force the tank must exert on the pipe to hold it in place. Assume fully developed flow occurs along the pipe. Take ro = 876 kg>m3 and mo = 0.22 N # s>m2.

2m

50 mm

SOLUTION

τ maxA

The engine oil is incompressible. The average velocity is

F

0.01 m3 >s Q = = 5.093 m>s V = A p(0.025 m)2

(a)

The Reynolds number is Re =

roVD = mo

( 876 kg>m3 )( 5.093 m>s ) (0.05 m) 0.22 N # s>m2

= 1014 6 2300 (laminar flow) Since the steady laminar flow occurs and the flow is horizontal, then Q =

p(0.05 m)4 ∆p pD4 ∆p a b; 0.01 m3 >s = a b 2 128mo L 128 ( 0.22 N # s>m ) 2 m ∆p = 28.683(10)3 Pa

The maximum shear stress occurs along the pipe’s wall. tmax =

3 2 0.05 m 28.683 ( 10 ) N>m D ∆p = a b£ § = 179.27 N>m2 4 L 4 2m

Consider the horizontal equilibrium of the FBD of the pipe shown in Fig. a, + S ΣFx

= 0;

tmax A - F = 0

F = tmax A = ( 179.27 N>m2 ) [2p (0.025 m)(2 m)] Ans.

= 56.3 N

Ans: 56.3 N 974

9–39. Castor oil is subjected to a pressure of 550 kPa at A and to a pressure of 200 kPa at B. If the pipe has a diameter of 30 mm, determine the shear stress acting on the pipe wall and the maximum velocity of the oil. Also, what is the flow Q? Take rco = 960 kg>m3 and mco = 0.985 N # s>m2. B 12 m 30!

A

SOLUTION Assuming that steady laminar flow occurs, and the castor oil is incompressible. Here pB - pA hB - hA 0 + ga (p + gh) = b 0x L L =

200 ( 103 ) N>m2 - 550 ( 103 ) N>m2 12 m

= -24.458 ( 103 )

+ ( 960 kg>m3 )( 9.81 m>s2 ) a

12 m sin 30° - 0 b 12 m

N>m2 m

The shear stress acting along the pipe’s wall can be determined with r = 0.015 m. t =

N>m2 r 0 0.015 m (p + gh) = J - 24.458 ( 103 ) R 2 0x 2 m = -183.43 N>m2 = 183 N>m2

Ans.

The negative sign indicates that the shear stress is directed in the opposite sense to that of the flow. The maximum velocity and the flow rate are u max = -

= -

R2 0 (p + gh) 4m 0x (0.015 m)2

4 ( 0.985 N # s>m

2

)

J -24.458 ( 103 )

N>m2 m

R Ans.

= 1.397 m>s = 1.40 m>s Q = -

= -

pR4 0 (p + gh) 8m 0x p(0.015 m)4

8 ( 0.985 N # s>m2 )

J -24.458 ( 103 )

N>m2 m

R

= 0.4936 ( 10-3 ) m3 >s = 0.494 ( 10-3 ) m3 >s

Ans.

975

9–39. Continued

The average velocity is V =

1 1 u = ( 1.397 m>s ) = 0.6984 m>s 2 max 2

The Reynolds number is Re =

rcoVD = mco

( 960 kg>m3 )( 0.6984 m>s ) (0.03 m) 0.985 N # s>m2

= 20.42 6 2300 (laminar flow)

976

Ans: t = 183 N>m2 u max = 1.40 m>s Q = 0.494 ( 10-3 ) m3 >s

*9–40. The pressure of crude oil at the bottom of an oil well is determined by lowering a pressure gage down into the well as shown. If the reading gives pA = 459.5 psi, determine the flow of the oil out of the 6-in-diameter pipe at the well head. Assume the pipe is smooth. Take ro = 1.71 slug>ft 3 and mo = 0.632 1 10-3 2 lb # s>ft 2.

1200 ft

SOLUTION The flow is assumed to be steady laminar and crude oil is incompressible. The positive x axis is directed in the direction of flow, which is vertically upward. Here pz - pb hz - hb 0 + ga (r + gh) = b 0x L L

=

lb 12 in. 2 ba b 1200 ft - 0 in2 1 ft 2 + ( 1.71 slug>ft 3 )( 32.2 ft>s2 ) a b 1200 ft 1200 ft

0 - a459.5

lb>ft 2

= - 0.078

ft

The flow rate is Q = -

pR4 0 (p + gh) 8m0 0 x

pa = -

4 3 ft b 12

8 3 0.632 ( 10-3 ) lb # s>ft 2 4

a - 0.078

lb>ft 2 ft

b

= 0.1893 ft 3 >s = 0.189 ft 3 >s

Ans.

The average velocity is

V =

0.1893 ft 3 >s Q = = 0.9642 ft>s 2 A 3 p a ft b 12

The Reynolds number is

Re =

r0VD m0

=

( 1.71 slug>ft 3 )( 0.9642 ft>s ) a 0.632 ( 10-3 ) lb # s>ft 2

6 ft b 12

= 1304 6 2300 (laminar flow)

977

A

9–41. Oil and kerosene are brought together through the wye as shown. Determine if they will mix, that is, create turbulent flow, as they travel along the 60-mm-diameter pipe. Take ro = 880 kg>m3, and rk = 810 kg>m3. The mixture has a viscosity of mm = 0.024 N # m>s2.

40 mm 0.2 m/s

60 mm 0.2 m/s

40 mm

SOLUTION Assume laminar flow and the fluid are incompressible. Continuity Equation. 0 rdV + rV # dA = 0 0 t Lcv Lcs 0 - VoAo - VkAk + VmAm = 0

( - 0.2 m>s ) 3 p(0.02 m)2 4 - ( 0.2 m>s ) 3 p(0.02 m)2 4 + Vm 3 p(0.03 m)2 4 = 0 Vm = 0.1778 m>s

Also, # Σm = 0 # # # - mo - mk + mm = 0 - ( 880 kg>m3 )( 0.2 m>s ) (p)(0.02 m)2 - ( 810 kg>m3 )( 0.2 m>s ) (p)(0.02 m)2 + rm ( 0.1778 m>s ) (p)(0.03 m)2 = 0 rm = 845 kg>m3 Re =

rVD = m

( 845 kg>m3 )( 0.1778 m>s ) (0.06 m) 0.024 N # m>s2

= 375.55 6 2300

Ans.

Since the flow is laminar, not turbulent, no mixing will occur.

Ans: no mixing 978

9–42. Crude oil at 20°C is ejected through the 50-mm-diameter smooth pipe. If the pressure drop from A to B is 36.5 kPa, determine the maximum velocity within the flow, and plot the shear-stress distribution within the oil.

50 mm B

4m

SOLUTION

A

The flow is assumed to be steady laminar, and the oil is incompressible. The positive x axis is in the direction of flow, which is vertically upward. From Appendix A, at T = 20° C, rco = 880 kg>m3 and mco = 30.2 ( 10-3 ) N # s>m2. Here pB - pA hB - hA 0 + ga (p + gh) = b 0x L L =

-36.5 ( 10-3 ) N>m2 4m

= -492.2

x

+ ( 880 kg>m3 )( 9.81 m>s2 ) a

4m - 0 b 4m

0.025 m

N>m2

0.025 m

m τ max = 6.15 Pa

2

t =

N>m r 0 r (p + gh) = a - 492.2 b = ( -246.1r) N>m2 2 0x 2 m

The maximum shear stress occurs at r = 0.025 m. Then Ans.

tmax = -246.1(0.025) = 6.15 Pa The plot of shear stress distribution is shown in Fig. a. u max = -

= -

(a)

R2 0 (r + gh) 4mw 0x (0.025 m)2

4 3 30.2 ( 10-3 ) N # s>m2 4

a -492.2

N>m2 m

b Ans.

= 2.547 m>s = 2.55 m>s u max = 2V;

2.547 m>s = 2V

V = 1.273 m>s

979

9–42. Continued

The Reynolds number is Re =

rwVD = mw

( 880 kg>m3 )( 1.273 m>s ) (0.05 m) 30.2 ( 10-3 ) N # s>m2

= 1855 6 2300 (laminar flow)

Ans: tmax = 6.15 Pa u max = 2.55 m>s 980

9–43. Crude oil is flowing vertically upward through a 50-mm-diameter pipe. If the difference in pressure between two points 3 m apart along the pipe is 26.4 kPa, determine the flow. Take ro = 880 kg>m3 and mo = 30.2 1 10-3 2 N # s>m2.

SOLUTION Assume the oil is incompressible. Assuming that the flow is laminar, the maximum velocity can be determined from 26.4 ( 103 ) Pa dp Pa dh = = -8.8 ( 103 ) and = 1. Thus, dx 3m m dx

u max = -

= -

R2 d (p + gh) 4m dx (0.025 m)2

4 3 30.2 ( 10-3 ) N # s>m2 4

c -8.8 ( 103 )

Pa + ( 880 kg>m3 )( 9.81 m>s2 ) (1) d m

= 0.8651 m>s

Thus, the maximum Reynolds number is Re =

rVD = m

( 880 kg>m3 )( 0.8651 m>s ) (0.05 m) = 1260 6 2300 30.2 ( 10-3 ) N # s>m2

(OK)

Note: Normally the average velocity is used to calculate the Reynolds number.

Therefore, the flow rate is, Q = -

= -

pR4 dp (p + gh) 8m dx p ( 0.025 m ) 4

8 3 30.2 ( 10-3 ) N # s>m2 4

c - 8.8 ( 103 )

Pa + ( 880 kg>m3 )( 9.81 m>s2 ) (1) d m

= 0.8493 ( 10-3 ) m3 >s = 0.849 ( 10-3 ) m3 >s

Ans.

981

Ans: 0.849 ( 10-3 ) m3 >s

*9–44. Castor oil is poured into the funnel so that the level of 200 mm is maintained. It flows through the stem at a steady rate and accumulates in the cylindrical container. Determine the time needed for the level to reach h = 50 mm. Take ro = 960 kg>m3 and mo = 0.985 N # m>s2.

A

200 mm

300 mm

15 mm

SOLUTION h

Assume that steady laminar flow occurs in the stem, and castor oil is incompressible. The pressure at the top and bottom of the stem are 40 mm

pt = r0gh = ( 960 kg>m3 )( 9.81 m>s2 ) (0.2 m) = 1883.52 N>m2 pb = patm = 0 Hence pb - pt hb - ht 0 (p + gh) = b + g0 a 0x L L =

0 - 1883.52 N>m2 0.3 m

= - 15.696 ( 103 )

+ ( 960 kg>m3 )( 9.81 m>s2 ) a

N>m2

0 - 0.3 m b 0.3 m

m

The flow rate is Q = -

= -

pR4 0 (p + gh) 8m0 0x p(0.0075 m)4

8 ( 0.985 N # s>m2 )

J -15.696 ( 103 )

N>m2 m

R

= 19.80 ( 10-6 ) m3 >s

The time required can be determined from Q =

p(0.02 m)2(0.05 m) V ; 19.80 ( 10-6 ) m3 >s = t t t = 3.17 s

The average velocity is V =

19.80 ( 10-6 ) m3 >s Q = = 0.1120 m>s A p(0.0075 m)2

The Reynolds number is Re =

r0VD = m0

( 960 kg>m3 ) (0.1120 m>s)(0.015 m) 0.985 N # s>m2

= 1.64 6 2300 (laminar flow) 982

Ans.

9–45. Castor oil is poured into the funnel so that the level of 200 mm is maintained. It flows through the stem at a steady rate and accumulates in the cylindrical container. If it takes 5 seconds to fill the container to a depth of h = 80 mm, determine the viscosity of the oil. Take ro = 960 kg>m3.

A

200 mm

300 mm

15 mm

SOLUTION h

The flow rate is Q =

p(0.02 m)2(0.08 m) V = = 6.4p ( 10-6 ) m3 >s t 5s

40 mm

The average velocity is V =

6.4p ( 10-6 ) m3 >s Q = = 0.1138 m>s A p(0.0075 m)2

Assume that steady laminar flow occurs, and castor oil is incompressible. Here, the pressure at top and bottom of the stem are pt = r0gh = ( 960 kg>m3 )( 9.81 m>s2 ) (0.2 m) = 1883.52 N>m2 pb = patm = 0 Hence, pb - pt hb - ht 0 (p + gh) = + g0 a b 0x L L =

0 - 1883.52 N>m2 0.3 m

= -15.696 ( 103 )

Q = -

pR4 0 (p + gh); 8m0 0 x

+ ( 960 kg>m3 )( 9.81 m>s2 ) a

0 - 0.3 m b 0.3 m

N>m2 m

6.4p ( 10-6 ) m3 >s = -

p(0.0075 m)4 8m0

m0 = 0.970 N # s>m2

c - 15.696 ( 103 )

N>m2 m

d

Ans.

The Reynolds number is Re =

r0VD = m0

( 960 kg>m3 )( 0.1138 m>s ) (0.015 m) 0.970 N # s>m2

= 1.69 6 2300 (laminar flow) Ans: 0.970 N # s>m2 983

9–46. The resistance to breathing can be measured using a spirometer that measures the time to expire a full volume of air from the lungs. About 20% of the resistance occurs in the medium-size bronchi, where laminar flow exists. The resistance to flow R can be thought of as the driving pressure gradient dp>dx divided by the volumetric flow Q. Determine its value as a function of the diameter D of the bronchi, and plot its values for 2 mm … D … 8 mm. Take ma = 18.9 1 10-6 2 N # s>m2.

SOLUTION Assume that steady laminar flow occurs, the air is incompressible and the bronchi’s wall is smooth. Since the density of air is small, the elevation gradient can be neglected Q = -

= -

pr 4 0 (r + gh) 8ma 0x pr 4 dp 8ma dx

By definition R =

dp>dx Q

= -

8ma pr 4

Substitute r = D>2 into the above equation R =

128 ma

Ans.

pD4

Substitute ma = 18.9 ( 10-6 ) N # s>m2 into above equation, R =

0.770 ( 10-3 ) D4

The plot of R vs D is shown in Fig. a.

984

9–46. Continued

0.002

D(m)

R ( N # s>m

0.003

)

48.13 ( 10

0.001

0.002

6

6

0.004

0.005

0.006

0.007

0.008

) 9.51 ( 10 ) 3.01 ( 10 ) 1.23 ( 10 ) 0.594 ( 10 ) 0.321 ( 10 ) 0.188 ( 106 ) 6

6

6

6

6

R(106( N·S m6

50

40

30

20

10

D(m) 0.003

0.004

0.005

0.006

0.007

0.008

(a)

Ans: R = 985

128ma pD4

9–47. It is thought that the shear stress on the endothelial cells that line the walls of an artery may be important for the development of various vascular disorders. If we assume the velocity profile of the blood flow in an arteriole  (or very small artery) to be parabolic, and the vessel diameter is 80 µm, determine the wall shear stress as a function of the average velocity. Plot the results for  20 mm>s … V … 50 mm>s. Assume here that blood is a Newtonian fluid and take mb = 0.0035 N # s>m2.

80 !m u

SOLUTION Since the velocity profile of the flow is assumed to be parabolic, the flow is steady laminar. Also, blood is incompressible. The average velocity is V = -

R2 0 (r + gh) 8mb 0 x

8 ( 0.0035 N # s>m2 ) 8mb 0 (r + gh) = - a 2 bV = V 0x R 3 40(10-6) m 4 2 = - 17.5 ( 106 ) V

t =

r 3 - 17.5 ( 106 ) 4 V = - 8.75 ( 106 ) rV 2

The negative sign indicates that the shear stress acts in the direction opposite to that of the flow. At the vessel’s wall where t = 40 ( 106 ) m tw = -8.75(106) 3 40(10-6) 4 V

Ans.

tw = - 350V The plot of tw vs V is shown in Fig. a. τw )N m2)

V(m>s) 10

0

20

30

40

50

60

V )10–3 m s)

tw ( N>m2 )

20 ( 10-3 ) 50 ( 10-3 ) -7.00

-17.5

–5

–10

–15

–20

Ans: tw = 350V 986

*9–48. The cylindrical tank is to be filled with glycerin using the 50-mm-diameter pipe. If the flow is to be laminar, determine the shortest time needed to fill the tank to a depth of 2.5 m. Air escapes through the top of the tank.

2m

2.5 m 50 mm

SOLUTION Assume the glycerin is incompressible. Since the requirement is to fill the tank in the shortest time possible, it must be filled with the maximum velocity yet maintain laminar flow. This requires Re =

rVD = 2300 m

From Appendix A, r = 1260 kg>m3 and m = 1.50 N # s>m2. Thus,

( 1260 kg>m3 ) V(0.05 m) 1.50 N # s>m2

= 2300

Since the pipe is horizontal, V = 54.761 m>s the pressure gradient is V =

54.761 m>s =

D2 ∆p a b 32m L (0.05 m)2

a

∆p

32 ( 1.50 N # s>m2 ) L

b

∆p Pa = 1.0514 ( 106 ) L m Subsequently, the flow rate can be determined from Q =

=

pD4 ∆p a b 128m L p(0.05 m)4

Pa

c 1.0514 ( 10 ) d m 128 ( 1.5 N # s>m2 ) 6

= 0.10752 Thus, the shortest time needed to fill the tank is t =

p(1 m)2(2.5 m) V = Q 0.10752 m3 >s

Ans.

t = 73.0 s

987

9–49. The large flask is filled with a liquid having a density r and a viscosity m. When the valve at A is opened, the liquid begins to flow through the horizontal tube, where d 66 D. If h = h1 at t = 0, determine the time when h = h2. Assume that laminar flow occurs within the tube.

D

h

l

SOLUTION The pressure at the inlet and outlet are pin = pgh and pout = patm = 0. Since the 0h = 0. Thus, tube is horizontal 0x

d

A

rgh ∆r 0 (p + gh) = = 0 x L l Q =

prgd 4 pd 4 ∆r pd 4 rgh a b = a b = a bh 128m L 128m l 128ml

(1)

When the fluid level in the flask drops by dh in time dt, then the volume of the fluid in the flask is decreased by p dV = - a D2 bdh 4

However, dV = Qdt. Then

p Qdt = - D2dh 4 dt = -

pD2 dh 4Q

Substitute Eq. 1 into this equation, L

t

dt = -

0

32mlD2

h2

dh rgd 4 L h h1

t =

32mlD2 rgd 4

lna

h1 b h2

Ans.

Ans: 32mlD2 h1 t = ln a b 4 h2 rgd 988

9–50. Oil having a density of ro = 880 kg>m3 and a viscosity of mo = 0.0680 N # s>m2 flows through the 20-mm-diameter pipe at 0.001 m3 >s. Determine the reading h of the mercury manometer. Take rHg = 13 550 kg>m3.

300 mm

20 mm A

B h

SOLUTION We will assume the flow is laminar and the oil is incompressible. The flow is Q = 0.001 m3 >s =

pD4 ∆r a b 128m L p(0.02 m)4

∆p

a b 128 ( 0.0680 N # s>m2 ) 0.3 m

∆p = 5.195 ( 103 ) Pa

(1)

Subsequently, the maximum velocity can be determined from u max =

=

D2 ∆p a b 16m L (0.02 m)2

16 ( 0.0680 N # s>m2

)

c

5.195 ( 103 ) Pa 0.3 m

d

= 6.366 m>s The maximum Reynolds number is Re =

rVD = m

( 880 kg>m3 )( 6.366 m>s ) (0.02 m) 0.0680 N # s>m2

= 1647.72 6 2300 (OK)

Note: Normally the average velocity is used to calculate the Reynolds number. Referring to Fig. a and writing the manometer equation, PA + rogh′ - rHggh - r0g(h′ - h) = PB PB - PA = rogh - rHggh PB - PA = ( 880 kg>m3 - 13550 kg>m3 )( 9.81 N>m2 ) h PB - PA = -124 292.7h Equating Eqs. (1) and this equation, 5.195 ( 103 ) = 124 292.7h Ans.

h = 0.041795 m = 41.8 mm 989

Ans: 41.8 mm

9–51. Oil having a density of ro = 880 kg>m3 and a viscosity of mo = 0.0680 N # s>m2 flows through the 20-mm-diameter pipe. If the mercury manometer reads h = 40 mm, determine the volumetric flow. Take rHg = 13 550 kg>m3.

300 mm

20 mm A

B h

SOLUTION We will assume the flow is laminar and the oil is incompressible. Referring to Fig. a and writing the manometer equation pA + rogh′ - r Hg gh - rog(h′ - h) = pB pB - pA + rogh - r Hg gh pB - pA = ( 880 kg>m3 - 13550 kg>m3 )( 9.81 N>m2 ) (0.04 m) ∆p - pB - pA = -4971.71 Assuming that the flow is laminar and since the pipe is horizontal, the flow is Q =

pD4 ∆p a b 128m L

= -

p(0.02 m)4

128 ( 0.0680 N # s>m2 )

a

-4971.71 Pa b 0.3 m

= 0.957 ( 10-3 ) m3 >s

Ans.

We must check if the flow is laminar, the maximum velocity can be determined from, u max =

D2 ∆p a b 16m L

= -

(0.02 m)2

16 ( 0.0680 N # s>m2 )

a

- 4971.71 Pa b 0.3 m

= 6.093 m>s Thus, the maximum Reynolds number is Re =

rVD = m

( 880 kg>m3 )( 6.093 m>s ) (0.02 m) 0.0680 N # s>m2

= 1576.95 6 2300

(OK)

Note: Normally the average velocity is used to calculate the Reynolds number.

990

Ans: 0.957 ( 10-3 ) m3 >s

*9–52. The Reynolds number Re = rVD>m for an annulus is determined using a hydraulic diameter, which is defined as Dh = 4 A>P, where A is the open cross-sectional area within the annulus and P is the wetted perimeter. Determine the Reynolds number for water at 30°C if the flow is 0.01 m3 >s. Is this flow laminar? Take ri = 40 mm and ro = 60 mm.

ro

SOLUTION In this case, A = p ( r 02 - r i 2 ) = p 3 (0.06 m)2 - (0.04 m)2 4 = 0.002p m2 Thus,

P = 2p(r0 + ri) = 2p 3 (0.06 m) + (0.04 m) 4 = 0.2p m Dh =

4 ( 0.002p m2 ) 0.2p m

= 0.04 m

From Appendix A, at T = 30° C, rw = 995.7 kg>m3 and mw = 0.801 ( 10-3 ) N # s>m2. The average velocity of the flow is V =

0.01 m3 >s Q = = 1.592 m>s A 0.002p m2

The Reynolds number is Re =

rwVDh = mw

( 995.7 kg>m3 )( 1.592 m>s ) (0.04 m) 0.801 ( 10-3 ) N # s>m2

= 79.14 ( 103 ) Since Re 7 2300, then the flow is not laminar.

991

ri

*9–53. A Newtonian fluid has laminar flow as it passes through the annulus. Use the Navier–Stokes equations to  show that the velocity profile for the flow is r 2o - r i2 r 1 dp 2 vr = c r - r 20 - a b ln d . 4m dz ln(ro >ri) ro

ro

SOLUTION Assume that the flow is steady laminar, and the fluid is incompressible. Since the flow is along the z axis of the pipe, then vr = vu = 0. Applying the continuity equation, z 0(rvz) 0r 1 0(rrvr) 1 0(rvu) + + + = 0 0t r 0r r 0u 0z 0 + 0 + 0 + r 0vz 0z

0vz 0z

= 0

= 0

Since the flow is steady and axis symmetric about the z axis, it is independent of time t and u. Thus, integration of the above results gives vz = vz(r), which means vt is a function of r only. Applying the Navier-Stokes equation along the z axis, ra = -

0vz 0t

+ vr

0vz 0r

+

0vz vu 0vz + vz b r 0u 0z

2 0vz 02vz 0p 1 0 1 0 vz + rgz + mc ar b + 2 2 + d 0t r 0r 0r r 0u 0z2

r(0 + 0 + 0 + 0) = -

0p 1 0 0vz + rgz + mc ar b + 0 + 0 d 0z r 0r 0r

Here, gz = 0. Using this result and rearrange the above equation, 0 0vz r 0p ar b = a b m 0z 0r 0r Integrate the above equation twice, 0vz 0r

=

vz =

C1 r 0p a b + 2m 0z r

(1)

r 0p a b + C1 ln + C2 4m 0z

(2)

Applying the boundary condition at y = ri, vz = 0 and y = 10, vz = 0. 0 =

0 =

r 2i 0p a b + C1 ln ri + C2 4m 0z

(3)

r 2o 0p a b + C1 ln ro + C2 4m 0z

(4) 992

ri

*9–53. Continued

Solving Eq. 3 and 4, C1 =

C2 = -

r i2 - r 02 0p b r a 4m ln r0i 0z

( r i2 ln r0 - r 02 ln ri ) 0p 4m ln

a

r0 ri

0z

b

Substitute these results into Eqs. 1 and 2, r i2 - r 02 0p r 0p 0v = a b + b r a 0t 2m 0t 4m ln r0i 0t

=

vt =

a

0p b r 02 - r i 2 0t a2t b r 4m t ln r0i

r i 2 - r 02 0p r i 2 ln r0 - r 02 ln ri 0p t2 0p a b + b ln r a b r0 a r 4m 0t 0t 4m ln ri 0t 4m ln r0i

=

a

0p b 0t at2 + 4m

=

a

r0 0p r i2lnr - r 02lnr - r i 2lnr0 + r 02lnri + r 02 aln b b ri 0t ¥ ≥ r 2 - r 02 + r0 4m ln ri

=

a

0p b r i2lnr - r 02lnr - r i2lnr0 + r 02lnri + r 02lnr0 - r 02lnri 0t § £ r 2 - r 02 + r0 4m ln ri

=

a

0p b 0t £ r 2 - r 02 + 4m

a

0p b r 02 - r i2 0t r £ r 2 - r 02 - a bln § r 4m r0 ln roi

=

vt =

( r i 2 - r 02 ) ln r ln

r0 ri

-

r i2ln r0 - r 02ln ri ln

r0 ri

b

( r 02 - r i2 ) lnro - ( r 02 - r i2 ) lnr ln

r 02 - r 12 1 0p 2 r c r - r 02 - a bln ro 4m 0z r ln ri 0

ro r

§

(Q.E.D) 993

*9–54. A Newtonian fluid has laminar flow as it passes through the annulus. Use the result of Prob. 9–53, and show that the shear-stress distribution for the flow is r o2 - r i 2 1 dp trz = a2r b. 4 dz rln(ro >ri)

ro

SOLUTION Assume that the flow is steady laminar, and the fluid is incompressible. Since the flow is along the z axis of the pipe, then vr = vu = 0. Applying the continuity equation, 0 ( rvz ) 0r 1 0 ( rrvr ) 1 0 ( rvu ) + + + = 0 0t r 0r r 0u 0z 0vz = 0 0 + 0 + 0 + r 0z 0vz = 0 0z Since the flow is steady and axis symmetric about the z axis, it is independent of time t and u. Thus, integration of the above results give vz = vz(r), which means vz is a function of r only. Applying the Navier-Stokes equation along z axis, ra

0rz 0t

+ vr

0vz 0r

+

2 02vz 0vz 0vz vu 0vz 0p 1 0 vz 1 0 ar b + 2 2 + 2 d + vz b = + rgz + m c r 0u 0z 0z r 0r 0r r 0u 0z

0vz 0p 1 0 + rgz + m c ar b + 0 + 0 d 0z r 0r 0r Here, gz = 0. Using this result and rearrange the above equation, r(0 + 0 + 0 + 0) = -

0vz 0 r 0p ar b = a b m 0z 0r 0r

Integrate the above equation twice, 0vz 0r

=

C1 r 0p a b + 2m 0z r

(1)

r 2 0p 2 a b + C1 ln r + C2 4m 0z Applying the boundary condition at y = ri, vz = 0 and y = 10, vz = 0. vz =

0 = 0 =

r i2 0p a b + C1 ln ri + C2 4m 0z

(3)

r02 0p a b + C1 ln r0 + C2 4m 0z

(4)

Solving Eq. 3 and 4, C1 =

C2 =

r i2 - r 02 0p a b r0 0z 4m ln ri

( r i2 ln r0 - r02 ln ri ) 0p r0 4m ln ri

a

0z

(2)

b

994

ri

*9–54. Continued

Substitute these results into Eqs. 1 and 2, 0vz 0r

=

=

vz =

r i2 - r 02 0p r 0p a b + a b r0 0z 2m 0z 4mr ln ri a

0p b r 02 - r i2 0z ± 2r ≤ r0 4m r ln ri

r i 2 - r 02 0p r i2 ln r0 - r 02 ln ri 0p v2 0p a b + a b lnr a b r0 0z r0 4m 0t 0z 4m ln 4m ln ri ri a

0p b 0z ±r2 + 4m

=

a

r0 0p r i2 ln r - r 02 ln r - r i2 ln ro + r i2 ln ri + r 02 aln b b ri 0z ≥ r 2 - r 02 + ¥ r0 4m ln ri

=

a

0p b r i2lnr - r 02lnr - r i 2lnro + r 02lnri + r 02lnro - r 02 ln ri 0z ¥ ≥ r 2 - r 02 + ro 4m ln ri

=

a

0p b 0z ≥ r 2 - r 02 + 4m

=

a

0p b r 0 2 - r i2 0z r ≥ r 2 - r 02 - ± ≤ ln ¥ r0 4m r0 ln ri

=

( r i2 - r 02 ) ln r ln

r0 ri

-

r i2 ln r0 - r 02 ln ri ≤ r0 ln ri

( r 0 2 - r i2 ) ln r0 - ( r 0 2 - r i2 ) lnr ro ln ri

¥

995

*9–54. Continued

The shear stress distribution can be determined from

trz = m

trz =

0vz 0r

=

a

0p 0h + rg b r 02 - r 12 0z 0z ± 2r ≤ r0 4 r ln r1

ro2 - r i2 1 0p °zr ¢ 4 0z r ln (ro >ri)

(Q.E.D.)

Setting 0h>0z = 0 for a horizontal flow gives

996

9–55. As oil flows from A to B, the pressure drop is 40 kPa through the annular channel. Determine the shear stress it exerts on the walls of the channel, and the maximum velocity of the flow. Use the results of Probs. 9–53 and 9–54. Take mo = 0.220 N # s>m2.

60 mm 30 mm

A

B 2m

SOLUTION pz - pA 0r =0z 1 = -

4.0(103)N>m2 2m

= -20 (103)

N>m2

m Substitute this results and numerical values into the equations 1 - for trz - 20 ( 103 ) trz =

N>m2 m

4

= a -10 ( 103 ) r +

At r = ri = 0.03 m, trz `

r = 0.03 m

r = 0.06 m

0.06 m 0.03 m

r ln

¥

19.476 b N>m2 r

= c -10 ( 103 )( 0.03 ) +

19.476 d = 349 N>m3 0.03

= c -10 ( 103 )( 0.06 ) +

19.476 d = - 275 N>m2 0.06

At r = r0 = 0.06 m, trz `

(0.06 m)2 - (0.03 m)2

≥ 2r -

Ans.

Ans.

The negative sign indicates that the shear stress acts in serve that is opposite to that of the flow. Using the equations for Vz. - 20 ( 103 ) vz =

N>m2 m

4 ( 0.220 N # s>m2 )

µ r 2 - ( 0.06 m ) 2 - ≥

( 0.06 m ) 2 - ( 0.03 m ) 2 ln

0.06 m 0.03 m

¥ ln

r ∂ 0.06 m

vz = c -22.727 ( 103 ) r 2 + 88.529 ln ( 16.67r ) + 81.818 d m>s

To determine the maximum velocity, set 0vz 0r

0vz 0r

= 0.

= - 22.727 ( 103 ) r +

88.529 =0 r

Solving, r = 0.04413 m Substitute this result into Eq. 1 (vz)max = - 22.727 ( 103 )( 0.044132 ) + 88.529 ln[16.67(0.04413)] + 81.818 Ans.

= 10.4 m>s 997

Ans: trz % r = 0.03 m = 349 Pa trz % r = 0.06 m = - 275 Pa (vz )max = 10.4 m>s

*9–56. As oil flows from A to B, the pressure drop is 40  kPa through the annular channel. Determine the volumetric flow. Use the result of Prob. 9–53. Take mo = 0.220 N # s>m2.

60 mm 30 mm

A

B 2m

SOLUTION Using the equation for vz vz =

1

4 ( 0.220 N # s>m2 )

vz = Q =

a-

40 ( 103 ) N>m2 2m

b ( r 2 - (0.06 m)2 - £

(0.06 m)2 - (0.03 m)2 0.06 m ln 0.03 m

( - 22.727 ( 103 ) r 2 + 88.529 ln(16.67t) + 81.818 ) m>s L A

0.06

vzdA =

L

0.03

3 - 22.727 ( 103 ) t2

+ 88.529 ln(16.67t) + 81.818 4 2prdr

= 2p 3 - 5.68175 ( 103 ) t4 + 88.529 4 c

r2 r2 d + 40.909r 2 ln(16.67r) 2 4

= 0.0583 m3 >s

Ans.

998

§ ln

r 0.06 m

*9–57. When blood flows through a large artery, it tends to separate into a core consisting of red blood cells and an outer annulus called the plasma skimming layer, which is cell free. This phenomenon can be described using the “cell-free marginal layer model,” where the artery is considered a circular tube of inner radius R, and the cell-free region has a thickness d. The equations that govern ∆p du c 1 d these regions are = cm r d , 0 … r 6 R - d, L r dr c dr

d

R

du p ∆p 1 d = cm r d , R - d … r … R, where mc and L r dr p dr mp are the viscosities (assumed to be Newtonian fluids), and u c and u p are the velocities for each region. Integrate these equations and show that the flow is mp p∆pR4 d 4 Q = c 1 - a1 - b a1 b d. mc 8mpL R

and -

SOLUTION For -

∆p duc 1 d am r b = L r dr c dr

∆p du c d amc r r b = dr dr L

Integrate this equation twice,

∆p du c C1 = r + dr 2mcL mc r uC = -

(1)

∆p 2 C1 r + ln r + C2 mc 4mcL

(2)

For -

du p ∆p 1 d = ampr b L r dr dr

du p ∆p d am r b = r dr p dr L du p dr

= -

up = -

Applying the boundary condition 0 = -

∆p C3 r + 2mpL mpr

(3)

∆p 2 C3 r + ln r + C4 mp 4mpL

(4)

du c = 0 at r = 0, Eq. 1 gives dr

∆p C ( 02 ) + 1 m 2mcL

C1 = 0

u p = 0 at r = R, Eq. 4 gives 0 = C4 =

∆pR2 C3 + ln R + C4 mp 4mpL

∆pR2 C3 ln R mp 4mpL

(5) 999

*9–57. Continued

At the interface where r = R - d, tc = tp. Then using Eqs. 1 and 3 mc mc c -

du p du c = mp dr dr

∆p ∆p C3 (R - d) d = mp c (R - d) + d 2mcL 2mcL mp(R - d) -

∆p ∆p C3 (R - d) = (R - d) + 2L 2L (R - d) C3 = 0

Substitute this result into Eq. 5, C4 =

∆pR2 4mpL

Also, at the interface where r = R - d, u c = u p. Then Eqs. (2) and (4) give -

∆p ∆p ∆pR2 (R - d)2 + C2 = (R - d)2 + 4mcL 4mcL 4mpL C2 =

Then Eqs. 2 and 4 become uc = up =

(R - d)2 ∆p (R - d)2 R2 c + d mc mp mp 4L

(R - d)2 (R - d)2 ∆p r 2 R2 c+ + d mc mp mp 4L mc

∆p ( R2 - r 2 ) 4mpL

The flow rate can be determined from Q =

LA

udA =

L0

R-d

(R - d)2 (R - d)2 ∆p r 2 R2 c+ + d (2prdr) mc mp mp 4L mc R

+ =

LR - d

-

∆p ( R2 - r 2 ) (2prdr) 4mpL

(R - d)2 2 (R - d)2 2 p∆p r4 R2 2 R - d c+ r r + r d` 2L 4mc 2mc 2mp 2mp 0 +

R p∆p R2 2 1 c r - r4 d ` 2mpL 2 4 R-d

1000

*9–57. Continued

Q =

(R - d)4 (R - d)4 R2(R - d)2 p∆p (R - d)4 c+ + d 2L 4mc 2mc 2mp 2mp +

= = = = = =

p∆p R4 R2 1 R4 ca b (R - d)2 + (R - d)4 d 2mpL 2 4 2 4

(R - d)4 R2(R - d)2 R2(R - d)2 (R - d)4 p∆p (R - d)4 R4 c + + + d 2L 4mc 2mp 2mp 4mp 2mp 4mp (R - d)2 (R - d)4 p∆p R4 c + d 2L 4mp 4mc 4mp

p∆pR4 1 1 R - d 4 1 R - d 4 c + a b a b d mp mc mp 8L R R mp p∆pR4 d 4 d 4 c1 + a1 - b - a1 - b d mc 8mpL R R p∆pR4 d 4 mp c 1 + a1 - b a - 1b d mc 8mpL R

mp p∆pR4 d 4 c 1 - a1 - b a1 bd mc 8mpL R

(Q.E.D.)

1001

*9–58. Use the result of Prob. 9–57 to show that if Poiseuille’s equation was used to calculate the apparent  viscosity of blood, it could be written as mp mapp = . mp d 4 c 1 - a1 - b a1 bd mc R

d

R

SOLUTION From Prob. 9–57 Q = Also, Q =

mp p∆pR4 d 4 c 1 - a1 - b a1 bd mc 8mpL R p∆pR4 8mapp L

By comparing these two equations, mapp =

If we take the limit

d S 0, R

mp

mapp =

This result is to be expected.

(Q.E.D.)

mp d 4 1 - a1 - b a1 b mc R

mp 1 - a1 -

mp mc

= mc b

1002

9–59. Glycerin flows from a large tank through the smooth 100-mm-diameter pipe. Determine its maximum volumetric flow if the flow is to remain laminar. How far L from the pipe entrance will fully developed laminar flow begin to occur?

L

100 mm

SOLUTION For laminar flow,

vgl = 1.19 ( 10-3 ) m2 >s Re max =

QD VD = = 2300 ggl Aggl

Q(0.1 m) p(0.05 m)2 ( 1.19 ( 10-3 ) ) m2 >s For fully developed laminar flow

= 2300

Q = 0.215 m3 >s

Ans.

L = 0.06 Re D = 0.06(2300)(0.1 m) Ans.

L = 13.8 m

Ans: Q = 0.215 m3 >s L = 13.8 m 1003

*9–60. Water flows from a beaker into a 4-mm-diameter tube with an average velocity of . Classify the flow 0 45 m>s as laminar or turbulent if the water temperature is and 10°C of if it is 30°C. If the flow is laminar, then find the length pipe for fully developed flow. L

SOLUTION At T = 10° C, n = 1.31 ( 10-6 ) m2 >s

T = 30° C, n = 0.804 ( 10-6 ) m2 >s

At T = 10° C,

Re =

VD = n

( 0.45 m>s ) (0.004 m) laminar = 1374 6 2300 -6 2 flow ( ) 1.31 10 m >s

L′ = 0.06 Re D = 0.06(1374)(0.004 m)

Ans.

= 0.330 m = 330 mm At T = 30° C, Re =

VD = n

( 0.45 m>s ) (0.004 m) laminar = 2239 7 2300 -6 flow 0.804 ( 10 )

L1 = 0.06 Re D = 0.06(2239)(0.004 m) Ans.

= 0.537 m = 537 mm

1004

9–61. A 70-mm-diameter horizontal pipe has a smooth interior surface and transports crude oil at 20°C. If the pressure decrease over a 5-m-long segment is 180 kPa, determine the thickness of the viscous sublayer and the velocity along the centerline of the pipe. The flow is turbulent.

SOLUTION Assume the oil is incompressible. Since the pipe is horizontal, the shear stress that occurs along the wall, regardless of whether the flow is laminar or turbulent is t0 =

3 2 D ∆p 0.07 m 180 ( 10 ) N>m £ § = 630 Pa = 4 L 4 5m

From Appendix A, r = 880 kg>m3 and n = 0.0344 ( 10-3 ) m2 >s. Then, u* =

t0 630 Pa = = 0.8461 m>s Ar A 880 kg>m3

Then the velocity at the centerline of the pipe can be determined by applying u *y u = 2.5 ln a b + 5.0 n u*

( 0.8461 m>s ) (0.035 m) u = 2.5 ln a ≤ + 5.0 0.8461 m>s 0.0344 ( 10-3 )

Ans.

u = 18.5 m>s

The viscous sublayer extends to y =

u *y = 5. Thus, n

5 3 0.0344 ( 10-3 ) m2 >s 4 5n = = 0.2033 ( 10-3 ) m = 0.203 mm u* 0.8461 m>s

Ans.

The maximum Reynolds number is Re =

VD = n

( 18.53 m>s ) (0.07 m) = 37.70 ( 103 ) 7 2300 0.0344 ( 10-3 )

(OK)

Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: u = 18.5 m>s y = 0.203 mm 1005

9–62. Crude oil flows through the 50-mm-diameter smooth pipe. If the pressure at A is 16 kPa and at B is 9 kPa, determine the shear stress and the velocity within the oil 10 mm from the wall of the pipe. Use Eq. 9–33 to determine the result. The flow is turbulent.

50 mm A

B 4m

SOLUTION 3 9 ( 10 (0.05 m) D ∆r = a b†° 4 L 4

3

t0 =

) - 16 ( 103 ) 4 N>m2 4m

= 21.875 Pa

¢†

Total shear stress. By proportion, t 21.875 Pa = 15 mm 25 mm Ans.

t = 13.125 Pa = 13.1 Pa u* = At y = 10 mm = 0.01 m,

21.875 N>m2 t0 = = 0.1577 m>s A r0 C 880 kg>m3

u *y u = 2.5 ln a b + 5.0 n0 u*

( 0.1577 m>s ) (0.01 m) u = 2.5 ln ¢ ≤ + 5.0 0.1577 m>s 0.0344 ( 10-3 ) m2 >s u = 2.30 m>s

Ans.

Ans: t = 13.1 Pa u = 2.30 m>s 1006

9–63. Crude oil flows through the 50-mm-diameter smooth pipe at 0.0054 m3 >s. If the pressure at A is 16 kPa and at B is 9 kPa, determine the viscous and turbulent shearstress components within the oil 10 mm from the wall of the pipe. Use the power-law velocity profile, Eq. 9–34, to determine the result. The flow is turbulent.

50 mm A

B 4m

SOLUTION Q = VA

( 0.0054 m>s2 ) = V ( p(0.025 m)2 ) V = 2.750 m>s Re =

VD = n0

use n = 6 (Table 9–1)

( 2.750 m>s ) (0.05 m) = 3,997 ≈ 4000 0.0344 ( 10-3 ) m2 >s

2n2 ; (n + 1)(2n + 1)

V = u max =

2.750 m>s = u max

2(6)2 (6 + 1)(2(6) + 1)

u max = 3.476 m>s Viscous shear is given by tvise = m ( d u>dy ) = - m ( d u>dr ) , tvise = =

mu max r 1>n - 1 a1 - b nR R

30.2 ( 10-3 ) N # s>m2 ( 3.476 m>s )

6(0.025 m) = 1.502 Pa = 1.50 Pa t0 =

a1 -

0.015 m -5>6 b 0.025 m

Ans.

(0.05 m) 3 9 ( 103 ) - 16 ( 103 ) 4 N>m2 D ∆r = †° ¢† 4 L 4 4

= 21.875 Pa

t 21.875 Pa = , t = 13.125 Pa 15 mm 25 mm t = tvisc + tturb 13.125 Pa = 1.502 Pa + tturb Ans.

tturb = 11.6 Pa

Ans: tvisc = 1.50 Pa tturb = 11.6 Pa 1007

*9–64. Crude oil flows through the 50-mm-diameter smooth pipe. If the pressure at A is 16 kPa and at B is 9 kPa, determine the thickness of the viscous sublayer, and find the maximum shear stress and the maximum velocity of the oil in the pipe. Use Eq. 9–33 to determine the result. The flow is turbulent.

50 mm A

4m

SOLUTION Maximum shear stress t0 =

(0.05 m) 9 ( 103 ) - 16 ( 103 ) N>m2 D ∆r †° ¢† = 4 L 4 4m

= 21.875 Pa = 21.9 Pa u* =

Ans.

21.875 N>m2 t0 = = 0.1577 m>s A r0 C 880 kg>m3

u *y u max = 2.5 ln a b + 5.0 n0 u*

0.1577 m>s (0.025 m) u max = 2.5 ln ¢ ≤ + 5.0 0.1577 0.0344 ( 10-3 ) m2 >s

Ans.

u max = 2.658 m>s y =

B

5 ( 0.0344 ( 10-3 ) ) m2 >s 5n0 = = 0.00109 m u* 0.1577 m>s Ans.

y = 1.09 mm

1008

9–65. Crude oil flows through the 50-mm-diameter smooth pipe at 0.0054 m3 >s. Determine the velocity within the oil 10 mm from the wall of the pipe. Use the power-law velocity profile, Eq. 9–34, to determine the result. The flow is turbulent.

50 mm A

B 4m

SOLUTION Q = VA

( 0.0054 m3 >s ) = V ( p(0.025 m)2 ) V = 2.750 m>s

Re =

VD = n0

Use n = 6 (Table 9–1) V = u max =

( 2.750 m>s ) (0.05 m) = 3997 ≈ 4000 0.0344 ( 10-3 ) m2 >s

2n2 ; (n + 1)(2n + 1)

2.750 m>s = u max =

2(6)2 (6 + 1)(2(6) + 1)

u max = 3.476 m>s u u max

= a1 -

r 1>n b R

u = 3.476 m>s a1 u = 2.98 m>s

0.015 m 1>6 b 0.025 m

Ans.

Ans: 2.98 m>s 1009

9–66. Water flows through the 2-in.-diameter smooth pipe. If the pressure drops 1.5 psi along the 2-ft length, determine the shear stress along the wall of the pipe and at the center of the pipe. What is the velocity of the water along the centerline of the pipe? The flow is turbulent. Use Eq. 9–33. Take gw = 62.4 lb>ft 3 and nw = 16.6 1 10 - 6 2 ft 2 >s

2 in.

2 ft

SOLUTION At the wall

t0 =

D ∆p = a 4 L

( 2>12 ) ft 4



144 in2 b ft 2 ≤ 2 ft

1.5 lb>in2 a

= 4.50 lb>ft 2 = 0.03125 psi.

Ans.

At the center Ans.

t = 0 u * = 2t0 >rw =

4.50 lb>ft 2 62.4 slug>ft 3 S 32.2

= 1.5238 ft>s

u *y u max = 2.5 ln a b + 5.0 nw u*

1.5238 ft>s ( 1>12 ) ft u max ≤ + 5.0 = 2.5 ln¢ 1.5238 ft>s 16.6 ( 10-6 ) ft 2 >s u max = 41.7 ft>s

Ans.

Ans: At the wall t = 0.03125 psi At the center t = 0 u max = 41.7 ft>s 1010

9–67. Water flows through the 2-in.-diameter smooth pipe. If the pressure drops 1.5 psi along the 2-ft length, determine the shear stress at a distance of 0.5 in. from the wall of the pipe. What is the thickness of the viscous sublayer? The flow is turbulent. Take gw = 62.4 lb>ft 3 and nw = 16.6 1 10-6 2 ft 2 >s.

2 in.

2 ft

SOLUTION t0 =

( 2>12 ) ft D ∆p = a b± 4 L 4

144 in2 b ft ≤ 2 ft

1.5 lb>in2 a

= 4.50 lb>ft 2 u * = 2t0 >rw = By proportion,

4.50 lb>ft 2 62.4 a b slug>ft 3 S 32.2

= 1.5238 ft>s

t 4.50 = 0.5 in. 1 in. t = 2.25 lb>ft 2 = 0.0156 psi y =

Ans.

5(16.6) ( 10-6 ) ft 2 >s 5nw = u* 1.5238 ft>s

y = 54.5 ( 10 - 6 ) ft

Ans.

Ans: t = 0.0156 psi y = 54.5(10 - 6) ft 1011

*9–68. The 4-in.-diameter smooth pipe is 12 ft long and transports 70°F water having a maximum velocity of 30 ft>s. Determine the pressure gradient along the pipe and the shear stress on the walls of the pipe. Also, what is the thickness of the viscous sublayer? Use Eq. 9–33.

4 in. 30 ft/s 12 ft

SOLUTION From Appendix A, rw = 1.937 slug>ft 3 and vm = 10.4 ( 10-6 ) ft 2 >s. The maximum 2 velocity occurs at the centerline of the pipe where y = ft = 0.1667 ft . 12 u*y u b + 5.0 = 2.5 ln a u* v u max = 2.5u* ln a 30 ft>s = 2.5u* lnc

u*y v

b + 5.0u*

u* (0.1667 ft)

10.4 ( 10-6 ) ft 2 >s

d + 5.0u*

30 = 2.5u* ln ( 16025.64u* ) + 5.0u* Solving by trial and error, u* = 1.0251 ft>s Using u* =

t0 ; A rw

1.0251 ft>s =

t0 A 1.937 slug>ft 3

t0 = 2.0355 lb>ft 2 = 2.04 lb>ft 2

Ans.

1012

9–68. Continued

Since the shear stress on the wall is caused by viscous laminar flow, then t0 =

D ∆p ; 4 L

2.0355 lb>ft 2 = a

4>12 ft 4

ba

∆p b L

∆p 1 ft 2 b = ( 24.425 lb>ft 2 >ft ) a L 12 in.

Ans.

= 0.170 psi>ft

The viscous sublayer can be determined using the upper limit condition for the laminar flow, u*y = 5; vw

( 1.0251 ft>s ) y = 5 10.4 ( 10-6 ) ft 2 >s

y = 50.727 ( 10-6 ) ft = 0.609 ( 10-3 ) in.

1013

Ans.

9–69. The 100-mm-diameter smooth pipe transports benzene with an average velocity of 7.5 m>s. If the pressure drop from A to B is 400 Pa, determine the viscous and turbulent shear-stress components within the benzene at r = 25 mm and r = 50 mm from the centerline of the pipe. Use a power-law velocity profile, Eq. 9–34. Take rbz = 880 kg>m3 and nbz = 0.75 1 10-6 2 m2 >s.

A

SOLUTION The Reynolds number is Re =

VavgD nbz

=

( 7.5 m>s ) (0.1 m) = 1 ( 106 ) 0.75 ( 10-6 ) m2 >s

From Table 9-1, this Reynolds number corresponds to n = 9. The shear stress distridp d dh bution varies linearly along the radial line with (p + gh) = since = 0 for dx dx dx horizontal pipe. t =

r dp 2 dx

At the wall where t = 0.05 m, Γ & r = 0.05 m = a

2 0.05 m - 400 N>m bc d = -2.50 Pa 2 4m

The shear stress along the wall is due to the viscous effect. Thus at r = 0.05 m, Ans.

tlam = 2.50 Pa tturb = 0

The shear stress at r = 0.025 m is caused by viscous and turbulent stress. It has a value of t & r = 0.025 m = a

2 0.025 m - 400 N>m bc d = -1.25 Pa 2 4m

The shear stress can be determined using Newton’s law of viscosity, which requires the velocity gradient using the power law, u = u max a1 -

Thus,

r 1>n b R

u max du r 1 = a1 - b n dr n R

tlam = msz

-1

a-

u max 1 r 1-n b = a1 - b n R nR R

rbznbzu max u max r 1-n du r 1-n a1 - b n = rbznbz c a1 - b n d = nR R dr nR R

1014

100 mm

B 4m

9–69. Continued

Vavg = u max = c

2n2 d; (n + 1)(2n + 1)

7.5 m>s = u max = c

2 ( 92 ) (9 + 1)(18 + 1)

u max = 8.796 m>s

d

At r = 0.025 m, tlam =

( 880 kg>m3 ) 3 0.75 ( 10-6 ) m2 >s 4 ( 8.796 m>s ) 9(0.05 m)

= 0.02389 N>m2 = 0.0239 Pa

a1 -

0.025 m 1 - 9 b 9 0.05 m

Ans.

The turbulent shear stress component can be determined from tturb = t - tlam = 1.25 N>m2 - 0.02389 N>m2 Ans.

= 1.23 Pa

Ans: At r = 0.05 m, tlam = 2.50 Pa tturb = 0 At r = 0.025 m, tlam = 0.0239 Pa tturb = 1.23 Pa 1015

9–70. A 3-in.-diameter horizontal pipe has a smooth interior surface and transports kerosene at 68°F. If the pressure drops 17 lb>ft 2 in 20 ft, determine the maximum velocity of the flow. What is the thickness of the viscous sublayer? Use Eq. 9–33.

SOLUTION Assume the kerosene is incompressible. The shear stress that occurs along the wall regardless of whether the flow is laminar or turbulent. D ∆p t0 = = 4 L

a

3 ft b 17 lb>ft 2 12 c d = 0.053125 lb>ft 2 4 20 ft

From Appendix A, r = 1.58 slug>ft 3 and n = 25.4 ( 10-6 ) ft 2 >s. Then, u* =

0.053125 lb>ft 2 t0 = = 0.1834 ft>s Ar C 1.58 slug>ft 3

1.5 ft = 0.125 ft . Then the velocity at the centerline of the pipe can be 12 determined by applying

With y =

u *y u = 2.5 ln a b + 5.0 n u*

( 0.1834 ft>s ) (0.125 ft) u = 2.5 lnc d + 5.0 0.1834 ft>s 25.4 ( 10-6 ) ft 2 >s u = 4.04 ft>s

Ans.

The maximum Reynolds number is VD Re = = n

( 4.04 ft>s ) a

25.4 ( 10-6 ) ft 2 >s

The viscous sublayer extends to a y =

3 ft b 12

= 39.76 ( 103 ) 7 2300

(OK)

u *y b = 5. Thus, n

5 3 25.4 ( 10-6 ) ft 2 >s 4 5v = = 0.6925 ( 10-3 ) ft = 8.31 ( 10-3 ) in u* 0.1834 ft>s

Ans.

Ans: u = 4.04 ft>s y = 8.31 ( 10-3 ) in. 1016

9–71. Experimental testing of artificial grafts placed on the inner wall of the carotid artery indicates that blood flow through the artery at a given moment has a velocity profile that can be approximated by u = 8.36(1 - r>3.4)1>n mm>s, where r is in millimeters and n = 2.3 log 10 Re - 4.6. If Re = 2 1 109 ) , plot the velocity profile over the artery wall, and determine the flow at this moment.

3.4 mm

r

SOLUTION The flow rate of blood can be determined from Q = =

LA L0

udA

3.4 mm

= 16.72p

= 16.72p

8.36 a1 -

L0

3.4 mm

r a1 1

£

= 16.72pc

a-

r 1>n b dr 3.4

1 2 1 b a + 2b 3.4 n

a1 -

1

r n+2 b 3.4

a-

1 2 1 b a + 1b 3.4 n 3.4 mm

n d (n + 1)(2n + 1)

193.2832n2p (n + 1)(2n + 1)

For Re = 2 ( 109 ) , n = 2.3 log 3 2 ( 109 ) 4 - 4.6 = 16.79

Substitute this result into the above equation. Q =

1

1

11.56n r 2nn+ 1 11.56n r n n+ 1 2 d a1 b a1 b 2n + 1 3.4 n + 1 3.4 0

= 193.2832npc =

r 1>n b (2prdr) 3.4

193.2832(16.79)2p

(16.79 + 1)[2(16.79) + 1]

The velocity profile is described by u = c 8.36 a1 -

= 278 mm3 >s

r 0.05955 b d mm>s 3.4

The plot of velocity profile is shown in Fig. a

1017

Ans.

a1 -

r n+1 b §3 3.4

3.4 mm

0

9–71. Continued

r(mm)

0

0.5

1.0

1.5

2.0

2.5

3.0

3.4

u(mm>s)

8.36

8.28

8.19

8.08

7.93

7.72

7.36

0

r (mm)

3.4 3.0 2.5 2.0

1.5 1.0 0.5 u (mm s) 0

1

2

3

4

5

6

7

8

9

(a)

1018

Ans: 278 mm3 >s

10–1. Oil flows through a 100-mm-diameter horizontal pipe at 4 m>s. If the pipe is made of cast iron, determine the friction factor. Take no = 0.0344 1 10-3 2 m2 >s.

SOLUTION Oil is considered to be incompressible. The Reynolds number is Re =

VD = n

( 4 m>s ) (0.1 m) = 1.16 ( 104 ) ( 0.0344 ( 10-3 ) m2 >s )

e = 0.26 ( 10-3 ) m for cast iron. Thus, the relative roughness is 0.26 ( 10-3 ) m e = = 0.0026 D 0.1 m From the Moody diagram.

Ans.

f = 0.034 Also, the Colebrook or Haaland equations can be used.

Ans: 0.056 1019

10–2. Oil flows through a 12-in.-diameter horizontal pipe at the rate of 8 ft>s. If the pipe is smooth, determine the head loss in a 20-ft-long horizontal portion of the pipe. Take no = 0.820(10-3) ft 2 >s.

SOLUTION Oil is considered to be incompressible. The Reynolds number is Re =

8 ft>s(1 ft) VD = = 9.76 ( 103 ) n 0.82 ( 10-3 ) ft 2 >s

For a smooth pipe, from the Moody diagram, f = 0.031 Applying, hL = f a

= (0.031)a

L V2 ba b D 2g

( 8 ft>s ) 2 20 ft b° ¢ = 0.616 ft 1 ft 2 ( 32.2 ft>s2 )

Ans.

Ans: 0.616 ft 1020

10–3. Glycerin has a density of 2.46 slug>ft 3 and flows through a 10-in.-diameter pipe at 3 ft>s. If the pressure drops 0.035 psi in an 8-ft-long segment of the pipe, determine the friction factor for the pipe.

SOLUTION Glycerin is considered to be incompressible. Since, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f f = Here, ∆p = a0.035 f =

L rV 2 D 2

2∆pD rV 2L

lb 12 in. 2 ba b = 5.04 lb>ft 2, then 1 ft in2

2 ( 5.04 lb>ft 2 ) a

10 ft b 12

( 2.46 slug>ft 3 )( 3 ft>s ) 2(8 ft)

Ans.

= 0.0474

Ans: 0.0474 1021

*10–4. If air flows through the circular duct at 4 m>s, determine the pressure drop that occurs over a 6-m length of the duct. The friction factor is f = 0.0022. Take ra = 1.092 kg>m3.

6m

4 m/s

0.2 m

SOLUTION Air is considered to be incompressible. Since, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

rV 2 L b° ¢ D 2

∆p = (0.0022)a ∆p = 0.288 Pa

1.092 kg>m3 ( 4 m>s ) 2 6m b£ § 0.4 m 2

1022

Ans.

10–5. Water in the 15-in.-diameter concrete drain pipe runs full with a flow of 15 ft 3 >s. Determine the pressure drop from point A to point B. The pipe is horizontal. Take f = 0.07.

12 ft

15 in.

A

B

SOLUTION Water is considered to be incompressible. The mean velocity of the water flow is 15 ft 3 >s = V(p)a

Q = VA;

2 7.5 ft b 12

V = 12.22 ft>s Since, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

L rV 2 ba b D 2

62.4 a slug>ft 3 b ( 12.22 ft>s ) 2 32.2 12 ft ∆p = (0.07) ≥ ¥≥ ¥ 15 2 ft 12 ∆p = ( 97.28 lb>ft 2 ) a

1 ft 2 b = 0.676 psi 12 in.

Ans.

Ans: 0.676 psi 1023

10–6. Water flows through a horizontal 2-in.-diameter pipe. If the friction factor is f = 0.028 and the flow is 0.006 ft 3 >s, determine the pressure drop that occurs over 3 ft of its length.

SOLUTION Water is considered to be incompressible. The mean velocity of the water flow is Q = VA;

0.006 ft 3 >s = V(p)a

2 1 ft b 12

V = 0.2750 ft>s Since, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

L rV 2 ba b D 2

62.4 slug>ft 3 b ( 0.2750 ft>s ) 2 32.2 ∆p = 0.028a b≥ ¥ 2 ( 2>12 ) ft 3 ft

a

= 0.0369 lb>ft 2

Ans.

Ans: 0.0369 lb>ft 2 1024

10–7. Air is forced through the circular duct. If the flow is 0.3 m3 >s and the pressure drops 0.5 Pa for every 1 m of length, determine the friction factor for the duct. Take ra = 1.202 kg>m3.

200 mm

6m

SOLUTION Air is considered to be incompressible. The mean velocity of the water flow is Q = VA;

0.3 m3 >s = V(p)(0.2 m)2 V = 2.387 m>s

Since, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f

L rV 2 D 2

N b(0.4 m) m2 = f = rV 2L ( 1.202 kg>m3 )( 2.387 m>s ) 2(1 m) 2∆pD

2 a0.5

Ans.

= 0.0584

Ans: 0.0584 1025

*10–8. A 45-mm-diameter commercial steel pipe is used to carry water at T = 20°C. If the head loss in a length of 2 m is 5.60 m, determine the volumetric flow in liters per second.

SOLUTION Water is considered to be incompressible. For commercial steel pipe, e = 0.045 mm. Thus, the relative roughness is e 0.045 mm = = 0.001 D 45 mm From Appendix A, n = 1.00 ( 10-6 ) m2 >s for water at T = 20° C. The Reynolds number is Re = Assuming f = 0.02, then

V(0.045 m) VD = = 45 000 V n 1.00 ( 10-6 ) m2 >s

hL = f

L V2 D 2g

5.60 m = 0.02 a

(1)

(2)

2m V2 bc d 0.045 m 2 ( 9.81 m>s2 )

V = 11.12 m>s Then, Re = 45 000 ( 11.12 m>s ) = 5.00 ( 105 ) Using this value of Re, the Moody diagram gives f = 0.02, which is the assumed value. Thus, V = 11.12 m>s. Q = VA = (11.12 m>s) 3 p(0.0225 m)2 4 = 0.0177 m3 >s = 17.7 liters>s

Ans.

Note: A more direct solution can be obtained from the Colebrook equation. If Eqs. (1) and (2) are substituted into the term 2.51> ( Re1f ) the V will cancel so that the log becomes a constant.

1026

10–9. Water at 60°F flows upwards through the 34 -in. galvanized iron pipe at 2 ft>s. Determine the major head loss that occurs over the 10-ft segment AB. Also, what is the pressure at B if the pressure at A is 40 psi?

B

10 ft

SOLUTION Water is considered to be incompressible. From Appendix A, r = 1.939 slug>ft 3 and n = 12.2 ( 10-6 ) ft 2 >s. Then, the Reynolds number is Re =

( 2 ft>s ) a

A 0.75 in.

0.75 ft b 12

VD = = 1.02 ( 104 ) n 12.2 ( 10-6 ) ft 2 >s

e 0.0005 ft = = 0.008. From the Moody diagram, f = 0.041. D 0.75 a ft b 12 Thus, the head loss can be determined using For galvanized iron,

hL = f

( 2 ft>s ) 2 10 ft L V2 = 0.041≥ § = 0.4074 ft = 0.407 ft ¥£ D 2g 0.75 2 ( 32.2 ft>s2 ) a ft b 12

Ans.

Take the water in the pipe as the control volume. Applying the energy equation from with the datum set at A, pB pA V B2 VA2 + zA + hpump = + + zB + hturb + hL + g g 2g 2g

( 40 lb>in2 ) a

12 in. 2 b 1 ft

( 1.939 slug>ft ) ( 32.2 ft>s ) 3

2

+

pB V2 V2 + 0 + 0 = + 10 ft + 0 + 0.407 ft + g 2g 2g

pB = ( 1.939 slug>ft 3 )( 32.2 ft>s2 ) (81.85 ft) = 5110.20 lb>ft 2 a = 35.5 psi

1 ft 2 b 12 in.

Ans.

Ans: hL = 0.407 ft pB = 35.5 psi 1027

10–10. Determine the diameter of a horizontal 100-m-long PVC pipe that must transport 125 liter>s of  turpentine oil so that the pressure drop does not exceed 500 kPa. Take e = 0.0015 mm, rt = 860 kg>m3, and mt = 1.49 1 10-3 2 N # s>m2.

SOLUTION Assume that fully developed steady flow occurs and water is incompressible. The discharge is l 1 m3 b = 0.125 m3 >s Q = a125 ba s 1000 l

Then the average velocity is

V =

0.125 m3 >s Q 0.1592 = = p 2 A D2 D 4

Thus, the Reynolds number is rVD Re = = mt

0.1592 bD 9.1861(104) D2 = D 1.49(10-3) N # s>m2

( 860 kg>m3 ) a

(1)

Since the pipe is horizontal and has a constant diameter, hL =

∆p = g

500 ( 103 ) N>m2

( 860 kg>m3 )( 9.81 m>s2 )

= 59.27 m

Using the Darcy - Weisbach equation, hL = f

L V2 ; D 2g

100 m ( 0.1592>D ) b£ § D 2 ( 9.81 m>s2 ) 2 2

59.27 m = f a

D5 = 2.1784 ( 10-3 ) f

(2)

Use a trial and error iterative procedure by first choosing the mid-value of f = 0.025. Using Eq (2) then (1), D = 0.1404 m and Re = 6.55 ( 105 ) . For PVC pipe 0.0015 mm P>D = = 0.0000107. Enter the Moody diagram with these values, we 140.35 mm obtain f = 0.0128. Iterate with this value of f, we obtain D = 0.1228 m, P>D = 0.0000122 and Re = 7.48 ( 105 ) . With these values the Moody diagram gives f = 0.0126. Again, with this value of f, D = 0.1224 m, P>D = 0.0000123 and Re = 7.48 ( 105 ) which gives f = 0.0126 that is the same as previous value. Thus, D = 0.1224 m = 122.4 mm Ans.

Use size of D = 123 mm

Ans: 123 mm 1028

10–11. A pipe has a diameter of 60 mm and is 90 m long. When water at 20°C flows through it at 6 m>s, it produces a head loss of 0.3 m when it is smooth. Determine the friction factor of the pipe if, years later, the same flow produces a head loss of 0.8 m.

SOLUTION Water is considered to be incompressible. The head loss of the pipe can be determined for new pipe using, hLn = fn a

L V2 ba b D 2g

hLo = fo a

L V2 ba b D 2g

For the old pipe,

Thus, hLn fn 0.3 m = = fo hLo 0.8 m fo = 2.667 fn From Appendix A, n = 1.00 ( 10-6 ) m2 >s for water at T = 20° C . The Reynolds number for the new smooth pipe is Re = From the Moody diagram,

( 6 m>s ) (0.06 m) VD = = 3.6 ( 105 ) n 1.00 ( 10-6 ) m2 >s fn = 0.0138

Thus Ans.

fo = 2.667(0.0138) = 0.0368

Ans: 0.0368 1029

*10–12. The 15-in.-diameter concrete drain pipe runs full of water with a flow of 15 ft 3 >s. Determine the pressure drop from A to B. The pipe is sloping downward at 4 ft>100 ft. Take f = 0.07.

12 ft

15 in.

SOLUTION Water is considered to be incompressible. Since the concrete pipe has a constant diameter. VA = VB = V =

Q = A

15 ft 3 >s

pa

2 7.5 ft b 12

= 12.22 ft>s

The major head loss from A to B can be determined using

hL = f

( 12.22 ft>s ) 2 L V2 12 ft § = 1.559 ft = 0.07≥ ¥£ D 2g 15 2 ( 32.2 ft>s2 ) a ft b 12

Take the water from A to B as the control volume. Applying the energy equation from with the datum set at B, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pB pA V2 V2 4 + 12 ft a b + 0 = + 0 + 0 + 1.559 ft + + g g 2g 100 2g pA - pB = ( 62.4 lb>ft 2 ) (1.079 ft) = 67.3

lb ft 2

1030

Ans.

A

B

10–13. A processing plant uses water at 70°F that is supplied from a pump at a pressure of 80 psi. Determine the diameter of a horizontal galvanized iron pipe if the flow is to be 1200 gal>min, and the pressure in the pipe after 300 ft is 20 psi.

SOLUTION Assume that fully developed steady flow occurs and water is incompressible. The discharge is Q = a1200

gal min

Thus, the average velocity is V =

ba

1 ft 3 1 min ba b = 2.6738 ft 3 >s 7.48 gal 60 s

2.6738 ft 3 >s Q 3.4044 = = p 2 A D2 D 4

From Appendix A, at T = 70° F, rw = 1.937 slug>ft 3 and nw = 10.4 ( 10-6 ) ft 2 >s. Thus, the Reynolds number is 3.4044 a bD 3.2734 ( 105 ) VD D2 Re = = = n D 10.4 ( 10-6 ) ft 2 >s

(1)

Since the pipe is horizontal and has constant diameter,

hL =

pin - pout = g

a80

lb 12 in 2 lb 12 in 2 ba b a20 ba b 1 ft 1 ft in2 in2 = 138.53 ft ( 1.937 slug>ft 3 )( 32.2 ft>s2 )

Using the Darcy – Weisbach equation, hL = f

L V2 ; D 2g

138.53 ft = f a

2 300 ft ( 3.4044>D ) b£ § 2 D 2 ( 32.2 ft>s )

D5 = 0.38975f

(2)

Using the trial and error iterative procedure by first choosing the mid value of f = 0.025. Using Eq (2) then (1) D = 0.3960 ft and Re = 8.27 ( 105 ) . For galvanized e 0.0005 ft iron pipe, = = 0.00126. Enter the Moody diagram with these values, D 0.3960 ft we obtain f = 0.021. Iterate with this value of f, we obtain D = 0.3825 ft, e = 0.00131 and Re = 8.56 ( 105 ) . With these values, Moody diagram gives D f = 0.0209 which is very close to the previous value. D = (0.3825 ft)a

12 in b = 4.59 in. 1 ft

5 Use 4 in. diameter pipe. 8

Ans.

Ans: 5 Use D = 4 in. 8 1031

10–14. A nail gun operates using pressurized air, which is supplied through the 10-mm-diameter hose. The gun requires 680 kPa to operate with a 0.003 m3 >s airflow. If the air compressor develops 700 kPa, determine the maximum allowable length of hose that can be used for its operation. Assume incompressible flow and a smooth hose. Take ra = 1.202 kg>m3, na = 15.1 1 10-6 2 m2 >s.

A

SOLUTION Take the air in the hose as incompressible. This is the control volume. The mean velocity of the air is V =

0.003 m3 >s Q = = 38.20 m>s A p(0.005 m)2

Thus, the Reynolds number is Re =

VD = n

( 38.20 m>s ) (0.01 m) = 2.53 ( 104 ) 15.1 ( 10-6 ) m2 >s

For smooth hose, the Moody diagram gives f = 0.024. Thus, the head loss can be determined using hL = f

( 38.20 m>s ) 2 L L V2 = 0.024 a b£ § = 178.47 L D 2g 0.01 m 2 ( 9.81 m>s2 )

For a constant diameter hose, VA = VB = V. Also, because there is no change in elevation, zA = zB = z. Applying the energy equation, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g N V2 m2 + + z + 0 = ( 1.202 kg>m3 )( 9.81 m>s2 ) 2g 700 ( 103 )

N V2 m2 + + z + 0 + 178.47 L ( 1.202 kg>m3 )( 9.81 m>s2 ) 2g 680 ( 103 )

Ans.

L = 9.50 m

Ans: 9.50 m 1032

10–15. A 75-mm-diameter galvanized iron pipe, having a roughness of e = 0.2 mm, is used to carry water at a temperature of 60°C and with a velocity of 3 m>s. Determine the pressure drop over its 12-m length if it is horizontal.

SOLUTION Water is considered to be incompressible. From Appendix A, n = 0.478 ( 10-6 ) m2 >s and r = 983.2 kg>m3 for water at T = 60° C. Thus, the Reynolds number is Re = The relative roughness is

( 3 m>s ) (0.075 m) VD = = 4.71 ( 105 ) n 0.478 ( 10-6 ) m2 >s 0.2 ( 10-3 ) m e = = 0.002667 D 0.075 m

From the Moody diagram, f = 0.025 Thus, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

L rV 2 ba b D 2

= 0.025 a

( 983.2 kg>m 12 m b£ 0.075 m 2

= 17.70 ( 103 ) Pa = 17.7 kPa

3

)( 3 m>s ) 2

§

Ans.

Ans: 17.7 kPa 1033

*10–16. Air flows through the galvanized steel duct, with  a  velocity of 4 m>s. Determine the pressure drop along  a  2-m length of the duct. Take ra = 1.202 kg>m3, na = 15.1 1 10-6 2 m2 >s.

200 mm 150 mm

SOLUTION Assume the air is incompressible. The hydraulic diameter of the rectangular duct is Dh =

4(0.15 m)(0.2 m) 4A = = 0.1714 m P 2(0.15 m + 0.2 m)

Then, the Reynolds number is Re =

VDh = n

( 4 m>s ) (0.1714 m) = 4.54 ( 104 ) 15.1 ( 10-6 ) m2 >s

and the relative roughness of the galvanized duct, is

0.15 ( 10-3 ) m e = = 0.000875 Dh 0.1714 m From the Moody diagram, f = 0.0242. Thus the major head loss can be determined using hL = f

( 4 m>s ) 2 L V2 2m = (0.0242)a b£ § = 0.2302 m Dh 2g 0.1714 m 2 ( 9.81 m>s2 )

Take the air in the 2-m long duct as the control volume. Since the duct has a constant cross section, VA = VB = V. Also, it is horizontal, zA = zB = z. Applying the energy equation, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA pB V2 V2 + z + 0 = + z + 0 + 0.2302 m + + g g 2g 2g ∆p = pA - pB = ( 1.202 kg> m3 )( 9.81 m>s2 ) (0.2302 m) = 2.71 Pa

1034

Ans.

2m

10–17. Determine the greatest air flow Q through the galvanized steel duct so that the flow remains laminar. What is the pressure drop along a 200-m-long section of the duct for this case? Take ra = 1.202 kg>m3, na = 15.1 1 10-6 2 m2 >s.

200 mm 150 mm

2m

SOLUTION Assume the air is incompressible. The hydraulic diameter of the rectangular duct is Dh =

4(0.15 m)(0.2 m) 4A = = 0.1714 m P 2(0.15 m) + (0.2 m)

Then, the Reynolds number is Re =

V(0.1714 m) VDh = = 1.1353 ( 104 ) V n 15.1 ( 10-6 ) m2 >s

Since the flow is required to be laminar,

Re = 2300 1.1353 ( 104 ) V = 2300 V = 0.2026 m>s Thus, Q = VA = ( 0.2026 m>s ) (0.15 m) ( 0.2 m ) = 0.006078 m3 >s = 0.00608 m3 >s Ans.

64 Also, for laminar flow, the friction factor can be determined using f = = Re 64 = 0.0278. Thus, the major head loss can be determined using 2300

( 0.2026 m>s ) L V2 200 m b£ § = 0.06791 m = 0.0278 a Dh 2g 0.1714 m 2 ( 9.81 m>s ) 2 2

hL = f

Take the air in the duct as the control volume. Since the duct has a constant cross section, VA = VB = V. Also, it is horizontal, zA = zB = z, Applying the energy equation, pA pB VA2 VB2 + zA + hhump = + zB + hturb + hL + + g g 2g 2g pA pB V2 V2 + z + 0 = + z + 0 + 0.06791 m + + g g 2g 2g ∆p = ( 1.202kg> m3 )( 9.81 m>s2 ) (0.06791 m) = 0.801 Pa

Ans.

Ans: Q = 0.00608 m3 >s ∆p = 0.801 Pa 1035

10–18. Water at 20°C passes through the turbine T using a 150-mm-diameter commercial steel pipe. If the length of the pipe is 50 m and the discharge is 0.02 m3 >s, determine the power extracted from the water by the turbine.

A 4m

T B

SOLUTION Water is considered to be incompressible. The mean velocity of the flow in the pipe is V =

0.02 m3 >s Q = = 1.132 m>s A p(0.075 m)2

From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at 20°C. Thus, the Reynolds number is Re =

VD = n

( 1.132 m>s ) (0.15 m) = 1.70 ( 105 ) 1.00 ( 10-6 ) m2 >s

For commercial steel pipe, the relative roughness is

e 0.045 mm = = 0.0003 D 150 mm From the Moody diagram, f = 0.0182. Thus, the major head loss can be determined using hL = f

( 1.132 m>s ) 2 L V2 50 m = 0.0182 a b£ § = 0.3961 m D 2g 0.15 m 2 ( 9.81 m>s2 )

Take the turbine and water in a portion of the reservoir and pipe as the control volume. Applying the energy equation from A to B with the datum at B, pA pB VA2 VB2 + zA + hhump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 4m + 0 = 0 +

( 1.132 m>s ) 2 + 0 + hturb + 0.3961 m 2 ( 9.81 m>s2 )

hturb = 3.5386 m Applying P = gQ(hturb) = ( 998.3 kg>m3 )( 9.81 m>s2 )( 0.02 m3 >s ) (3.5386 m) = 693.10

N#m = 693 W s

Ans.

Ans: 693 W 1036

10–19. The 20-mm-diameter copper coil is used for a solar hot water heater. If water at an average temperature of T = 50°C passes through the coil at 9 liter>min, determine the major head loss that occurs within the coil. Neglect the length of each bend. Take e = 0.03 mm for the coil.

400 mm

SOLUTION Assume that fully developed steady flow occurs, and water is incompressible. Here the discharge is Q = a9

1 m3 l 1 min ba b = 0.15 ( 10-3 ) m3 >s ba min 1000 l 60 s

Then, the average velocity is V =

0.15 ( 10-3 ) m3 >s Q = = 0.4775 m>s A p(0.01 m)2

Appendix A gives rw = 988 kg>m3 and Nw = 0.561 ( 10-6 ) m2 >s for water at T = 50°C. Thus the Reynolds number is Re =

VD = gw

( 0.4775 m>s ) (0.02 m) = 1.70 ( 104 ) 0.561 ( 10-6 ) m2 >s

e 0.03 mm = = 0.0015, the Moody diagram gives f = 0.030. Here, D 20 mm the total length of the tube is L = 11(0.4 m) = 4.4 m. Thus, the major head losses is Together with

hL = f

L V2 D 2g

( 0.4775 m>s ) 4.4 m b£ § 0.02 m 2 ( 9.81 m>s2 ) 2

= 0.030 a

Ans.

= 76.7 mm

Ans: 76.7 mm 1037

*10–20. Water at 20°C is required to flow through a horizontal commercial steel pipe so that it discharges at 0.013 m3 >s. If the maximum pressure drop over a 5-m length is to be no more than 15 kPa, determine the smallest allowable diameter D of the pipe.

SOLUTION Water is considered to be incompressible. The mean velocity of the flow is V =

0.013 m3 >s Q 0.01655 = = p 2 A D2 D 4

From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at 20° C. Thus, the Reynolds number is 0.01655 a bD 1.655 ( 104 ) VD D2 Re = = = n D 1.00 ( 10-6 ) m2 >s

(1)

The major head loss can be determined using

0.01655 2 b f L V 5m D2 = fa b≥ ¥ = 69.820 ( 10-6 ) 5 hL = f 2 D 2g D D 2 ( 9.81 m>s ) a

2

Take the water in the 5-m-long pipe as the control volume. Since the pipe has a constant diameter, VA = VB = V. Also, because it is horizontal zA = zB = z. Applying the energy equation, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g f pA pB V2 V2 + z + 0 = + z + 0 + 69.820 ( 10-6 ) 5 + + g g 2g 2g D f pA - pB = 69.820 ( 10-6 ) 5 g D N f m2 = 69.820 ( 10-6 ) 5 3 2 ( 998.3 kg>m )( 9.81m>s ) D 15 ( 103 )

D = 0.1354 f 1>5

(2)

For commercial steel pipe, e = 0.045 ( 10 Iteration Assumed f D(m) (Eq. 2) 1 2

0.02 0.0195

0.06194 0.06163

-3

) m. The iterations are tabulated as follows

e m a b Re (Eq. 1) f from Moody diagram D m 0.00073 2.67(105) 0.0195 5 0.00073 2.69(10 ) 0.0195

The assumed f in the 2nd iteration is almost the same as that given by the Moody diagram. Thus, D = 0.06163 m is an acceptable result. Ans.

D = 0.06163 m = 61.6 mm

1038

10–21. Determine the power output required to pump 30 liters>s of crude oil through a 200-m-long horizontal cast iron pipe having a diameter of 100 mm. The pipe is open to the atmosphere at its end. Compare this power requirement with pumping water through the same pipe. The temperature is T = 20°C for both cases.

SOLUTION Assume that fully developed steady flow occurs and crude oil and water are incompressible. Here, the discharge is l 1 m3 Q = a30 ba b = 0.03 m3 >s s 1000l Thus, the average velocity is V =

0.03 m3 >s Q 12 = = m>s 2 p A p(0.05 m)

Applying the Energy Equation between the inlet and outlet, realizing that Vin = Vout = V, and Pin = Pout = P and zin = zout = z, pin pout V out2 V in2 + zin + hpump = + zout + hturb + hL + + g g 2g 2g p V2 V2 + z + 0 = 0 + + z + 0 + hL + g 2g 2g p = hL g For crude oil at 20° C, Appendix A, nco = 0.0344 ( 10-3 ) m2 >s. The Reynolds number is (Re)co

gives

rco = 880 kg>m3

and

12 a m>s b(0.1 m) p VD = = = 1.11 ( 104 ) 7 2300 (Turbulent flow) nco 0.0344 ( 10-3 ) m2 >s

For cast iron pipe,

P 0.26 mm = = 0.0026. Enter these two values into the Moody D 100 mm

Diagram, we obtain f = 0.035. Thus, the head loss can be determined using the DarcyWeisbach equation

(hL)co = f

a

2

2 12 m>s b p

200 m L V = 0.035 a b≥ ¥ = 52.05 m D 2g 0.1 m 2 ( 9.81 m>s2 )

Thus, the power output required is # Wco = QP = gcoQ(hL)co = ( 880 kg>m3 )( 9.81 m>s2 ) ( 0.03 m3 >s ) (52.05) = 13.5 kW

1039

Ans.

10–21. Continued

For water at 20° C, Appendix A gives rw = 998.3 kg>m3 and nw = 1.00 ( 10-6 ) m2 >s. Thus, 12 a m>s b(0.1 m) p VD = (Re)w = = 3.82 ( 105 ) 7 2300 (Turbulent flow) nw 1.00 ( 10-6 ) m2 >s

Together with e>D = 0.0026, the Moody diagram gives f = 0.025. Thus, the head loss is

(hL)w = f

a

2

2 12 m>s b p

L V 200 m = 0.025 a b≥ ¥ = 37.18 m D 2g 0.1 m 2 ( 9.81 m>s2 )

Thus, the power output required is # Ww = gwQ(hL)w = ( 998.3 kg>m3 )( 9.81 m>s2 )( 0.03 m3 >s) (37.18 m) = 10.9 kW

Ans.

Ans: # Wco = 13.5 kW # Ww = 10.9 kW 1040

10–22. Air at 60°F is transported by the fan at the rate of 2 ft 3 >s through the 12-in.-diameter galvanized iron duct. Determine the pressure drop that occurs over a 40-ft-long horizontal section of the duct.

A

B

40 ft

12 in.

SOLUTION Air is considered to be incompressible. The mean velocity of the air in the duct is V =

Q = A

2 ft 3 >s

2 6 pa ft b 12

= 2.546 ft>s

From Appendix A, r = 0.00237 slug>ft 3 and n = 0.158 ( 10-3 ) ft 2 >s for air at T = 60° F. Thus, the Reynolds number is Re =

( 2.546 ft>s ) (1 ft) VD = = 1.61 ( 104 ) n 0.158 ( 10-3 ) ft 2 >s

For the galvanized iron duct, the relative roughness is

e 0.0005 ft = = 0.0005 D 1 ft From the Moody diagram, f = 0.028. Thus, the major head loss along the duct can be determined using hL = f

2 L V2 40 ft ( 2.546 ft>s ) = 0.028 a bc d = 0.1128 ft D 2g 1 ft 2 ( 32.2 ft>s2 )

Since the duct has a constant diameter, VA = VB = V. Also, because it is horizontal, zA = zB = z. Applying the energy equation from A to B, pA pB VA2 VB 2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA pB V2 V2 + z + 0 = + z + 0 + 0.1128 ft + + g g 2g 2g pB - pA = - ( 0.00237 slug>ft 3 )( 32.2 ft>s2 ) (0.1128 ft) = - 0.00861 lb>ft 2

Ans.

Ans: - 0.00861 lb>ft 2 1041

10–23. Air at 60°F is transported by the fan at the rate of 2 ft 3 >s through the 12-in.-diameter galvanized iron duct. Determine the head loss over the 40-ft length.

A

B

40 ft

12 in.

SOLUTION Air is considered to be incompressible. The mean velocity of the air in the duct is V =

Q = A

2 ft 3 >s

pa

2 6 ft b 12

= 2.546 ft>s

From Appendix A, r = 0.00237 slug>ft 3 and n = 0.158 ( 10-3 ) ft 2 >s for air at T = 60° F. Thus, the Reynolds number is Re =

( 2.546 ft>s ) (1 ft) VD = = 1.61 ( 104 ) n 0.158 ( 10-3 ) ft 2 >s

For the galvanized iron duct, the relative roughness is

e 0.0005 ft = = 0.0005 D 1 ft From the Moody diagram, f = 0.028. Thus, the head loss along the duct can be determined using hL = f

2 L V2 40 ft ( 2.546 ft>s ) = 0.028 a bc d 2 D 2g 1 ft 2 ( 32.2 ft>s )

= 0.1128 ft = 0.113 ft

Ans.

Ans: 0.113 ft 1042

*10–24. A 12-in.-diameter hose is used to fill the reservoir with water. If the pressure at the supply A is 38 psi, determine the time needed to raise the depth of the reservoir h = 4 ft. The hose has a length of 100 ft and f = 0.018. The reservoir has a width of 8 ft small. Neglect elevation changes in the hose.

A B h 6 ft

SOLUTION Water is considered to be incompressible. Since the hose has a constant diameter, VA = VB = V. The major head loss can be determined using

hL = f

100 ft L V2 V2 2 = 0.018 C 0.5 £ § = 0.6708 V S ( D 2g 2 32.2 ft>s2 ) a ft b 12

Take the control volume as the water in the hose.

Since the pipe is assumed to be horizontal, zA = zB = z. Applying the energy equation, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g a38

lb 12 in. 2 ba b 1 ft V2 V2 in2 + + z + 0 = 0 + + z + 0 + 0.6708 V 2 3 2g 2g 62.4 lb>ft

V = 11.43 ft>s Thus, Q = VA = ( 11.434 ft>s ) c p a

2 0.25 ft b d = 0.01559 ft 3 >s 12

Then, the time needed to fill the reservoir is t =

(6 ft)(8 ft)(4 ft) 1h V = 12 315.50 s a = b = 3.42 h Q 3600 s 0.01559 ft 3 >s

1043

Ans.

10–25. A 12 -in.-diameter hose is used to fill the reservoir with water. If the pressure at the supply A is 38 psi, determine the depth h two hours after the faucet is turned on. The hose has a length of 100 ft and f = 0.018. The reservoir has a width of 8 ft. Neglect elevation changes in the hose.

A B h 6 ft

SOLUTION Since the hose has constant diameter, VA = VB = V. The major head loss can be determined using

hL = f

100 ft L V2 V2 2 = 0.018 C 0.5 £ § = 0.6708 V S D 2g 2 ( 32.2 ft>s2 ) a ft b 12

Take the control volume as the water in the hose.

Since the pipe is assumed to be horizontal, zA = zB = z. Applying the energy equation, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g a38

lb 12 in. 2 ba b 1 ft V2 V2 in2 + z + 0 = 0 + + z + 0 + 0.6708V 2 + 2g 2g 62.4 lb>ft 3

V = 11.43 ft>s Thus, Q = VA = ( 11.434 ft>s ) c p a

2 0.25 ft b d = 0.01559 ft 3 >s 12

Then, the height h of the reservoir after two hours of filling is V = Qt (6 ft)(8 ft)h = ( 0.01559 ft 3 >s ) (2 h)a h = 2.34 ft

3600 s b 1h

Ans.

Ans: 2.34 ft 1044

10–26. Water flows through the 50-mm-diameter pipe. If the pressures at A and B are the same, determine the flow. Take f = 0.035.

50 mm 20 m A 15! B

SOLUTION Water is considered to be incompressible. Since the pipe has a constant diameter, VA = VB = V. The major head loss can be determined using hL = f

L V2 20 m V2 = 0.035 a bc d = 0.7136V 2 D 2g 0.05 m 2 ( 9.81 m>s2 )

Take the water from A to B as the control volume.

Applying the energy equation with the datum set at B and pA = pB = p, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g p p V2 V2 + (20 m) sin 15° + 0 = + 0 + 0 + 0.7136V 2 + + g g 2g 2g V = 2.693 m>s Thus, Q = VA = ( 2.693 m>s ) 3 p(0.025 m)2 4 = 0.00529 m3 >s

Ans.

1045

Ans: 0.00529 m3 >s

10–27. Oil is ejected from the 3-in.-diameter pipe. If the friction factor for the pipe is f = 0.083, determine the smallest discharge from the pipe that will cause the flange bolts at A to begin to support a tensile force. The pipe weighs 30 lb. Take ro = 1.75 slug>ft 3.

3 in. 5 ft

A

SOLUTION Oil is considered to be incompressible. Take the control volume as the oil in the pipe.

PB = 0

The free-body diagram of the control volume is shown in Fig. a. Here, the weight of oil within the control volume is

30 lb

2 1.5 ft b (5 ft) d = 13.83 lb Ww = gV = ( 1.75 slug>ft )( 32.2 lb>ft ) c p a 12 3

3

Ww = 13.83 lb

Applying the momentum equation, ΣF =

0 VrdV + VrV # dA 0t L L cv

+ c ΣFy = rQ 3 ( Vout ) y - ( Vin ) y 4 ;

cs

pA £ p a

2 1.5 ft b § - 30 lb - 13.83 lb = rQ ( VB - VA ) 12

PA FA = 0 (a)

Since the pipe has a constant diameter, VB = VA = V. Then, pA £ p a

2 1.5 ft b § - 30 lb - 13.83 lb = 0 12

pA = 892.90 lb>ft 2

The head loss between A and B can be determined using

hL = f

L V2 5 ft V2 £ § = 0.02578V 2 = 0.083 ° 3 ¢ 2 ( 32.2 ft>s2 ) D 2g ft 12

Applying the energy equation from A to B with the datum set at A, pB pA VB2 VA2 + zB = + zA - ( hs ) out - hL + + g g 2g 2g 0 +

V2 + 5 ft = 2g

892.90 lb>ft 2

( 1.75 slug>ft ) ( 32.2 lb>ft ) 3

3

+

V2 + 0 - 0 - 0.02578V 2 2g

V = 20.51 ft>s Q = VA = ( 20.51 ft>s ) c p a

2 1.5 ft b d = 1.01 ft 3 >s 12

1046

Ans.

Ans: 1.01 ft 3 >s

*10–28. Oil flows through the 50-mm-diameter pipe at 0.009 m3 >s. If the friction factor is f = 0.026, determine the pressure drop that occurs over the 80-m length. Take ro = 900 kg>m3.

B 80 m A 20!

SOLUTION Oil is considered to be incompressible. Here, the mean velocity is V =

0.009 m3 >s Q = = 4.584 m>s A p (0.025 m)2

The head loss can then be determined using hL = f

( 4.584 m>s ) 2 L V2 80 m = 0.026 a b£ § = 44.55 m D 2g 0.05 m 2 ( 9.81 m>s2 )

Since the pipe has a constant diameter, VA = VB = V. Applying the energy equation from A to B with the datum set at A, pA pB VA2 VB2 + zA + hturb = + zB + hpump + hL + + g g 2g 2g pA pB V2 V2 + 0 + 0 = + zB + 0 + hL + + g g 2g 2g pA - pB = g ( zB + hL ) Here, zB = 80 sin 20° m = 27.36 m. Then, pA - pB = ( 900 kg>m3 )( 9.81 m>s2 ) (27.36 m + 44.55 m) = 634.88 ( 103 ) Pa = 635 k Pa

Ans.

1047

10–29. Oil flows through a 50-mm-diameter cast-iron pipe. If the pressure drop over a 10-m-long horizontal segment is 18 kPa, determine the mass flow through the pipe. Take ro = 900 kg>m3, no = 0.430 1 10-3 2 m2 >s.

SOLUTION Oil is considered to be incompressible. The Reynolds number is Re =

V(0.05 m) VD = = 116.28V n 0.43 ( 10-3 ) m2 >s

From the table, e = 0.26 mm for cast iron. Thus, the relative roughness is e 0.26 mm = = 0.0052 D 50 mm Since,

∆p L V2 = f g D 2g

and g = rg, then ∆p = f a

Assuming f = 0.03, we have

18 ( 103 ) N>m2 = 0.03 a

L rV 2 ba b D 2

( 900 kg>m3 ) V 2 10 m R bJ 0.05 m 2

V = 2.582 m>s

Thus, Re = 116.28(2.582) = 300.23, which is much lower than 2300. Thus, it is reasonable to assume that the flow is laminar. The friction factor can be determined using f =

64 = Re

Thus, ∆p = ∆p = V = =

64 64n = VD VD a b n

rV 2 64n L a ba b VD D 2 32nrVL D2 ∆pD2 32nrL

3 18 ( 103 ) N>m2 4 (0.05 m)2

32 3 0.43 ( 10-3 ) m2 >s ( 900 kg>m3 ) (10 m) 4

= 0.3634 m>s

Re = 116.28 ( 0.3634 m>s ) = 42.3 6 2300

Thus, the mass flow rate is # m = rVA = ( 900 kg>m3 )( 0.3634 m>s ) 3 p ( 0.025 m ) 2 4 = 0.642 kg>s

1048

(OK)

Ans.

Ans: 0.642 kg>s

10–30. A 75-mm-diameter galvanized iron pipe, having a roughness of e = 0.2 mm, is to be used to carry water at a temperature of 60°C and with a velocity of 3 m>s. Determine the pressure drop over its 12-m length if the pipe is vertical and the flow is upward.

SOLUTION Water is considered to be incompressible. From Appendix A, n = 0.478 ( 10-6 ) m2 >s and r = 983.2 kg>m3 for water at T = 60° C. Thus, the Reynolds number is Re = The relative roughness is

( 3 m>s ) (0.075 m) VD = = 4.71 ( 105 ) n 0.478 ( 10-6 ) m2 >s 0.2 ( 10-3 ) m e = = 0.002667 D 0.075 m

From the Moody diagram, f = 0.025 Thus, the head loss can be determined using hL = f

( 3 m>s ) 2 L V2 12 m = 0.025 a b£ § D 2g 0.075 m 2 ( 9.81 m>s2 )

= 1.835 m Applying the energy equation,

p2 p1 V12 V22 + z1 = + z2 + hL + + g g 2g 2g p1 p2 V2 V2 + 0 = + 12 m + 1.835 m + + g g 2g 2g p1 - p2 = 13.835 m g p1 - p2 = (13.835 m) 3 ( 983.2 kg>m3 )( 9.81 m>s2 ) 4 ∆p = 133.44 ( 103 ) Pa = 133 kPa

Ans.

Ans: 133 kPa 1049

10–31. If a pipe has a diameter D and a friction factor f, by what percent will the pressure drop in the pipe increase if the volumetric flow is doubled? Assume that f is constant due to a very large Reynolds number.

SOLUTION Assume the fluid is incompressible. Since ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a Here, V =

L rV 2 ba b D 2

Q , Thus, for Q = Q0, A

∆p1 = f a

L b≥ D

ra

Q0 2 b A L rQ02 ¥ = f a ba 2 b 2 D 2A

For Q = 2Q0,

∆p2 = f a

L b≥ D

ra

2Q0 2 b A L 4rQ02 ¥ = f a ba b 2 D 2A2

Thus,

% of increase in ∆p =

∆p2 - 1 = ≥ ∆p1

fa

fa

L 4rQ02 ba b D 2A2

L rQ02 ba 2 b - 1 D 2A

¥ * 100, Ans.

= 300,

Ans: 300, 1050

*10–32. Methane at 20°C flows through a 30-mm-diameter horizontal pipe at 8 m>s. If the pipe is 200 m long and the roughness is e = 0.4 mm, determine the pressure drop over the length of the pipe.

SOLUTION Methane is considered to be incompressible. From Appendix A, n = 16.8 ( 10-6 ) m2 >s and r = 0.665 kg>m3 for methane at T = 20° C. The relative roughness and Reynolds numbers are 0.0004 m e = = 0.0133 D 0.03 m Re =

( 8 m>s ) (0.03 m) VD = = 1.43 ( 104 ) n 16.8 ( 10-6 ) m2 >s

From the Moody diagram,

f = 0.045 Since ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

L rV 2 ba b D 2

( 0.665 kg>m 200 m b£ 0.03 m 2

3

∆p = 0.045a

)( 8 m>s ) 2

§

∆p = 6.384 ( 103 ) = 6.38 kPa

Ans.

1051

10–33. The galvanized iron pipe is used to carry water at 20°C with a velocity of 3 m>s. Determine the pressure drop that occurs over a 4-m length of the pipe.

75 mm

4m

SOLUTION Water is considered to be incompressible. From the table, e = 0.15 ( 10-3 ) m. Thus, the relative roughness is 0.15 ( 10-3 ) m e = = 0.002 D 0.075 m From Appendix A, n = 1.00 ( 10-6 ) m2 >s and r = 998.3 kg>m3 for water at T = 20° C. Then, the Reynolds number is Re =

VD = n

( 3 m>s ) (0.075 m) = 2.25 ( 105 ) 7 2300 (turbulent flow) 1.00 ( 10-6 ) m2 >s

From the Moody diagram,

f = 0.0245 ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

L rV 2 ba b D 2

( 998.3 kg>m 4m bJ 0.075 m 2

3

= 0.0245 a

)( 3 m>s ) 2

R

∆p = 5870 Pa = 5.87 kPa

Ans.

Ans: 5.87 kPa 1052

10–34. Water at 70°F used for irrigation is to be siphoned from a canal onto a field using a pipe having a roughness of e = 0.00006 ft. If the pipe is 300 ft long, determine its required diameter so that it provides a flow of 0.5 ft 3 >s.

10 ft

SOLUTION Assume that fully developed steady flow occurs and water is incompressible. The average velocity is V =

0.5 ft 3 >s Q 0.6366 = = p 2 A D2 D 4

Appendix A gives rw = 1.937 slug>ft 3 and nw = 10.4 ( 10-6 ) ft 2 >s for water at 70° F. Thus, the Reynolds number is Re =

( 0.6366>D2 ) D 6.1213 ( 104 ) VD = = nw D 10.4 ( 10-6 ) ft 2 >s

(1)

Write the energy equation between the inlet and outlet by realizing that Vin ≃ 0 (large reservoir) and Vout = V, pin = pout = patm = 0, zin = 10 ft and zout = 0, pin pout V out2 Vin2 + zin + hpump = + zout + hturb + hL + + g g 2g 2g 0 + 0 + 10 ft + 0 = 0 +

V2 + 0 + 0 + hL 2(32.2 ft>s2)

hL = 10 -

V2 64.4

Using the Darcy-Weisbach equation, hL = f

L V2 V2 300 ft V2 = fa ; 10 bJ R 64.4 D D 2g 2 ( 32.2 ft>s2 ) V2 a

a

300f + 1b = 644 D

0.6366 2 300f b a + 1b = 644 D D2 f =

1589.01 D5 - D 300

(2)

Assuming D = 0.25 ft for the iteration. Then Eqs (1) and (2) give f = 0.00434, 0.00006 ft Re = 2.45 ( 105 ) and e>D = = 0.00024. Enter the Moody diagram with 0.25 ft the values of Re and e>D, f = 0.017 which is much higher than that computed using Eq. (2). For second iteration, D = 0.321 ft, which gives f = 0.0170. Using Eq. (2), 0.00006 ft this leads to Re = 1.91 ( 105 ) and e>D = = 0.00019 and from Moody 0.321 ft diagram f = 0.0171. This value is very close to that computed using Eq. (2). Thus D = (0.321 ft) a

12 in b = 3.85 in. 1 ft

Ans:

7 Use D = 3 in 8

Ans.

1053

Use D = 3

7 in. 8

10–35. Oil flows at 2 m>s through a horizontal 50-mm-diameter galvanized iron pipe. Determine if the flow is laminar or turbulent. Also, find the pressure drop that occurs over a 10-m length of the pipe. Take ro = 850 kg>m3, mo = 0.0678 N # s>m2.

SOLUTION Oil is considered to be incompressible. The Reynolds number is Re =

rVD = m

( 850 kg>m3 )( 2 m>s ) (0.05 m) 0.0678 N # s>m2

Ans.

= 1253.69 6 2000 (laminar flow) Since the flow is laminar, f =

64 64 = = 0.05105 Re 1253.69

Since, ∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

L rV 2 ba b D 2

( 850 kg>m )( 2 m>s ) 10 m ba b 0.05 m 2 3

= (0.05105)a

2

= 17.36 ( 103 ) Pa = 17.4 kPa

Ans.

Ans: 17.4 kPa 1054

*10–36. For a given volumetric flow, the pressure drop is 5  kPa in a horizontal pipe. Determine the pressure drop if the flow is doubled. The flow remains laminar.

SOLUTION Assume the fluid is incompressible. Q Here, V = . Since the flow is laminar, the friction factor is A f = where Re =

VD . Also, n

64 Re

∆p L V2 = f g D 2g Since g = rg, then ∆p = f a

For Q = Q0,

f1 =

L rV 2 ba b D 2

64 64vA = Q0 Q0D a bD A n

Then,

∆p1 =

64nA L a b≥ Q0D D

ra

Q0 2 b 32rnLQ0 A ¥ = 2 AD2

For Q = 2Q0, f2 =

64 32nA = 2Q0 Q0D a bD A n

Then,

∆p2 = Thus,

32nA L a b≥ Q0D D

ra

2Q0 2 b 64rnLQ0 A ¥ = 2 AD2

64rnLQ0 ∆p2 AD2 = 32rnLQ0 ∆p1 AD2 Here, ∆p1 = 5 kPa. Thus,

∆p2 = 2∆p1

∆p2 = 2(5 kPa) = 10 kPa

Ans. 1055

10–37. Water is pumped from the river through a 40-mm-diameter hose having a length of 3 m. Determine the maximum volumetric discharge from the hose at C so that cavitation will not occur within the hose. The friction factor is f = 0.028 for the hose, and the gage vapor pressure for water is - 98.7 kPa.

B

C

2m A

SOLUTION Water is considered to be incompressible. The pressure at B will be the smallest and cavitation will occur here. For cavitation to occur, pB = - 98.7 kPa. Since the hose has a constant diameter, VB = V. The head loss can be determined from hL = f

L V2 3m V2 2 = 0.028a bJ R = 0.1070V D 2g 0.04 m 2 ( 9.81 m>s2 )

Applying the energy equation from A to B with the datum set at A, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g V =

-98.7 ( 103 ) N>m2

( 1000 kg>m )( 9.81 m>s ) 3

2

+

V2 + 2 m = 0 + 0.1070V 2 2g

V = 7.143 m>s Thus, Q = VA = ( 7.143 m>s ) 3 p(0.02 m)2 4 = 0.00898 m3 >s = 8.98 liter>s

Ans.

Ans: 8.98 liter>s 1056

10–38. The 2-in.-diameter pipe has a roughness of e = 0.0006 ft. If the discharge of water at 60°F through the 1-in.-diameter nozzle at B is 0.15 ft 3 >s, determine the pressure at A.

1 in. B

2 in.

A

30! 8 ft

SOLUTION Water is considered to be incompressible. The mean velocities of the flow at A and B are VA =

VB =

0.15 ft 3 >s Q = = 6.875 ft>s 2 AA 1 pa ft b 12

0.15 ft 3 >s Q = = 27.50 ft>s 2 AB 0.5 pa ft b 12

From Appendix A, r = 1.939 slug>ft 3 and n = 12.2 ( 10-6 ) ft 2 >s for water at T = 60° F . Then, the Reynold number is VAD Re = = n The relative roughness is

( 6.875 ft>s ) a

2 ft b 12

12.2 ( 10-6 ) ft 2 >s

= 9.39 ( 104 )

e 0.0006 ft = = 0.0036 D 2 a ft b 12

From the Moody diagram,

f = 0.0282 Then, the major head loss can be determined using

hL = f

( 6.875 ft>s ) 2 L V2 8 ft = 0.0282 ± ≤£ § = 0.9936 ft D 2g 2 2 ( 32.2 ft>s2 ) a b ft 12

Take the water from A to B as the control volume. Applying the energy equation with the datum set at A, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA

( 1.939 slug>ft 3 )( 32.2 ft>s2 ) pA = a999.24

+

( 6.875 ft>s ) 2 ( 27.50 ft>s ) 2 + 0 = 0 + + 8 ft sin 30° + 0 + 0.9936 ft 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 )

lb 1 ft 2 ba b = 6.94 psi 2 12 in. ft

Ans. Ans: 6.94 psi 1057

10–39. The 2-in.-diameter pipe has a roughness of e = 0.0006 ft. If the water is at T = 60° and the pressure at A is 18 psi, determine the discharge at B.

1 in. B

2 in.

A

30! 8 ft

SOLUTION Water is considered to be incompressible. Take the water in the pipe from A to B as the control volume. To maintain continuity, 0 r0V + rV # dA = 0 0t L L cv

cs

2 2 1 0.5 0 - VA c pa ft b d + VB c p a ft b d = 0 12 12

VB = 4VA

From Appendix A, r = 1.939 slug>ft 3 and n = 12.2 ( 10-6 ) ft 2 >s for water at T = 60° F . Then, the Reynolds number is 2 VAa ft b VAD 12 Re = = = 1.3661 ( 104 ) VA n 12.2 ( 10-6 ) ft 2 >s

(1)

The major head loss from A to B can be determined using hL = f

VA2 L V2 8 ft = f± ≤£ § = 0.7453f VA2 D 2g 2 2 ( 32.2 ft>s2 ) a b ft 12

Applying the energy equation from A to B with the datum set at A,

pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g lb 12 in. 2 ba b ( 4VA ) 2 VA2 1 ft in2 + + 0 + 0 = 0 + + 8 ft sin 30° + 0 + 0.7453f VA2 ( 1.939 slug>ft 3 )( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) a18

VA =

6.125 20.2329 + 0.7453 f

e 0.0006 ft The relative roughness is = = 0.0036. The iterations are tabulated as D 2 follows ft 12 iteration

Assumed f

VA ( ft>s ) Eq. (2)

Re Eq. (1)

f from Moody diagram

1

0.026

12.19

1.67 ( 105 )

0.0278

2

0.0278

12.16

1.66 ( 105 )

0.0278

The assumed f in the 3rd iteration is almost the same as that given by the Moody diagram. Thus, V = 12.16 ft>s is an acceptable result, so that Q = VA = ( 12.16 ft>s ) c pa

2 1 ft b d = 0.265 ft 3 >s 12

Ans.

1058

Ans: 0.265 ft 3 >s

10–40. The section AB of the 100-mm-diameter galvanized iron pipe has a mass of 15 kg. If glycerin is discharged from the pipe at 3 liter>s, determine the pressure at A and the force on the flange bolts at A.

A

3m

SOLUTION Glycerin is considered to be incompressible. Since the pipe has a constant diameter, VA = VB = V. The mean velocity is Q = V = A

( 3 liter>s )( 1 m3 >1000 liter ) p(0.05 m)2

B

= 0.3820 m>s

From Appendix A, r = 1260 kg>m3 and n = 1.19 ( 10-3 ) m2 >s for glycerin. Then, the Reynolds number is Re =

VD = n

FA 3 pA = 31.58(10 ) Pa

( 0.3820 m>s ) (0.1m) = 32.10 6 2300 (laminar flow) 1.19 ( 10-3 ) m2 >s

15(9.81) N

Since the flow is laminar, the friction factor is f =

64 64. = = 1.9938 (round) Re 32.10

Wg = 291.23 N

Then, the head loss can be determined using

FB = 0

( 0.3820 m>s ) L V2 3m = 1.9938 a b£ § = 0.4448 m D 2g 0.1 m 2 ( 9.81 m>s2 ) 2

hL = f

(a)

Take the glycerine in the pipe as the control volume.

Applying the energy equation with the datum set at B, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA V2 V2 + 3m + 0 = 0 + + 0 + 0 + 0.4449 m + g 2g 2g pA = ( 1260 kg>m3 )( 9.81 m>s2 ) (0.4449 m - 3 m) = - 31.58 ( 103 ) Pa = - 31.6 kPa

Ans.

The free-body diagram of the control volume is shown in Fig. a. Here, the weight of glycerin within the control volume is Ww = gV = ( 1260 kg>m3 )( 9.81 m>s2 ) 3 p(0.05 m)2(3 m) 4 = 291.23 N

Writing the momentum equation along the y axis, ΣF =

a31.58 ( 103 )

100 mm

0 VrdV + VrV # dA 0t Lcv Lcs

+ c ΣFy = rQ 3 ( Vout ) y - ( Vin ) y 4

N b 3 p(0.05 m)2 4 - 15(9.81)N - 291.23 N + FA = rQ(V - V) = 0 m2

Ans.

FA = 190.35 N = 190 N

1059

10–41. Oil is ejected from the 3-in.-diameter pipe. If the friction factor for the pipe is f = 0.083, determine the tensile force in the flange bolts at A if the oil is ejected into the air to a height of 12 ft above the end of the pipe at B. The pipe weighs 30 lb. Take ro = 1.75 slug>ft 3.

C B

5 ft

SOLUTION Oil is considered to be incompressible.

A

Take the control volume as the oil in the pipe. It is required that the water jet stop at C. Thus, VC = 0. Applying the energy equation from B to C with the datum set at B, PB = 0

pB pC VC2 VB2 + zB + hpump = + zC + hturb + hL + + g g 2g 2g

30 lb

VB2 0 + + 0 = 0 + 0 + 12 ft + 0 + 0 2g

Ww = 13.83 lb

VB = 22 ( 32.2 ft>s2 ) (12 ft) = 27.80 ft>s

Since the pipe has a constant diameter, VA = VB = V = 27.80 ft>s. The head loss between A and B can be determined using hL = f

( 27.80 ft>s ) 2 L V2 5ft § = 19.92 ft = 0.083 ± ≤£ D 2g 3 2 ( 32.2ft>s2 ) ft 12

PA FA (a)

Applying the energy equation from A to B with the datum set at A, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA V2 V2 + 0 + 0 = 0 + + 5 ft + 19.92 ft + g 2g 2g pA = ( 1.75 slug>ft 3 )( 32.2ft>s2 ) (5 ft + 19.92 ft) = 1404.24

lb ft 2

The free-body diagram of the control volume is shown in Fig. a. Here, the weight of oil within the control volume is Ww = gV = ( 1.75 slug>ft 3 )( 32.2 ft>s2 ) c pa

1.5 2 ft b (5 ft) d = 13.83 lb 12

Applying the momentum equation, using the free body diagram of the control volume, Fig. a, ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

+ c ΣFy = rQ 3 ( Vout ) y - ( Vin ) y 4 ;

+ FA + a1404.24

2 lb 1.5 b c pa ft b d - 30 lb - 13.83 lb = rQ ( VB - VA ) = 0 12 ft 2

Ans.

FA = - 25.1 lb

1060

Ans: 25.1 lb

10–42. The 50-mm-diameter pipe has a roughness of e = 0.01 mm. If the discharge of 20°C water is 0.006 m3 >s, determine the pressure at A.

50 mm B

2m

SOLUTION Water is considered to be incompressible.

A

The velocity of the flow is VA = VB = V =

0.006 m3 >s Q = = 3.056 m>s A p(0.025 m)2

From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at T = 20° C. Then, the Reynolds number is Re =

VD = n

The relative roughness is

( 3.056 m>s ) (0.05 m) = 1.53 ( 105 ) 1.00 ( 10-6 ) m2 >s

e 0.01 mm = = 0.0002 D 50 mm From the Moody diagram, f = 0.018. Then, the head loss can be determined using hL = f

( 3.056 m>s ) 2 L V2 2m = 0.018 a bJ R = 0.3427 m D 2g 0.05 m 2 ( 9.81 m>s2 )

Take the control volume as the water in the pipe.

Applying the energy equation from A to B with the datum set at A, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA V2 V2 + 0 + 0 = 0 + + 2 m + 0 + 0.3427 m + g 2g 2g pA = ( 998.3 kg>m3 )( 9.81 m>s2 ) (2 m + 0.3427 m) = 22.94 ( 103 ) Pa = 22.9 kPa

Ans.

Ans: 22.9 kPa 1061

10–43. The 50-mm-diameter pipe has a roughness of e = 0.01 mm. If the water has a temperature of 20°C and the pressure at A is 50 kPa, determine the discharge at B. 50 mm B

2m

SOLUTION Water is considered to be incompressible.

A

Since the pipe has a constant diameter, VA = VB = V. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at T = 20° C. Then, the Reynolds numbers is Re =

V(0.05 m) VD = = 5 ( 104 ) V n 1.00 ( 10-6 ) m2 >s

(1)

Then, the head loss from A to B can be determined using hL = f

L V2 2m V2 = fa b£ § = 2.0387 f V 2 D 2g 0.05 m 2 ( 9.81 m>s2 )

(2)

Take the control volume as the water in the pipe.

Applying the energy equation from A to B with the datum set at A, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g N V2 V2 m2 + + 0 + 0 = 0 + + 2 m + 0 + 2.0387 f V 2 3 2 2g ( 998.3 kg>m )( 9.81 m>s ) 2g 50 ( 103 )

V =

1.234 2f

e 0.01 mm The relative roughness of the pipe is = = 0.0002. The iterations are D 50 mm tabulated as follows. iteration Assumed f

V(m>s), Eq. (2)

Re, Eq. (1)

f from Moody diagram

1

0.02

8.73

4.36 ( 105 )

0.0155

2

0.0155

9.91

4.96 ( 105 )

0.0152

The assumed f in the second iteration is very close to that given by the Moody diagram. Thus, V = 9.91 m>s is an acceptable result. Thus, Q = VA = ( 9.91 m>s ) 3 p ( 0.025 m ) 2 4 = 0.0195 m3 >s = 19.5 liter>s

Ans.

Note: A more direct solution can be obtained from the Colebrook equation. If Eqs. (1) and (2) are substituted into the term 2.51> ( Re1f ) , the V will cancel so that the log becomes constant.

Ans: 19.5 liter>s 1062

*10–44. A galvanized steel pipe is required to carry water at 20°C with a velocity of 3 m>s. If the pressure drop over its 200-m horizontal length is not to exceed 15 kPa, determine the required diameter of the pipe.

SOLUTION Water is considered to be incompressible. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s. Thus, the Reynolds number is Re =

( 3 m>s ) D VD = = 3 ( 106 ) D n 1.00 ( 10-6 ) m2 >s

(1)

Since the pipe has a constant diameter, VA = VB = V. The major head loss can be determined using hL = f

( 3 m>s ) 2 91.743f L V2 200 m = fa bJ R = 2 D 2g D D 2 ( 9.81 m>s )

Take the control volume to be the water in the pipe.

Also, since the pipe is horizontal, zA = zB = z = VA = VB = V. Applying the energy equation, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 91.743f pA pB V2 V2 + z + 0 = + z + 0 + + + g g 2g 2g D 91.743f pA - pB = g D 91.743f = D

N m2 3 ( 998.3 kg>m )( 9.81 m>s2 ) 15 ( 103 )

(2)

D = 59.898f m For galvanized iron, e = 0.00015 m. The iterations are tabulated as follows iteration Assumed f D(m); Eq. (2) 1 2 3 4

0.02 0.0128 0.0140 0.0138

1.198 0.7667 0.8386 0.8266

Re; Eq. (1)

f from Moody diagram

0.000125

3.59 ( 106 )

0.0128

0.000196

2.30 ( 10

)

0.0140

0.000179

2.52 ( 10

)

0.000181

2.48 ( 10

)

e m a b D m

6 6 6

0.0138

The assumed f in the 4th iteration is almost the same as that given by the Moody diagram. Thus, D = 0.8266 m is an acceptable result. Then, Ans.

D = 0.8266 m = 827 mm

1063

10–45. Water at 70°F is pumped at the rate of 90 gal>min from a river using a 1.5-in.-diameter hose. If the pump must supply a pressure of 30 psi to the water in the hose at C before it enters the sprinkler, determine the required horsepower that must be developed by the pump. The pipe is 120 ft long, and e = 0.05(10-3) ft.

B C 30 ft A

SOLUTION Water is considered to be incompressible. The mean velocity of the flow in the hose is

V =

Q = A

a90

gal min

ba

1 min 1 ft 3 ba b 60 s 7.48 gal

2 0.75 pa ft b 12

= 16.34 ft>s

From Appendix A, r = 1.937 slug>ft 3 and n = 10.4 ( 10-6 ) ft 2 >s. Thus, the Reynolds number is VD Re = = n

( 16.34 ft>s ) a

1.5 ft b 12

10.4 ( 10-6 ) ft 2 >s

= 1.96 ( 105 )

The relative roughness of the pipe is

0.05 ( 10-3 ) ft e = = 0.0004 D 1.5 a ft b 12

From the Moody diagram, f = 0.0185. Thus, the major head loss can be determined using hL = f

( 16.34 ft>s ) 2 L V2 120 ft J = 0.0185 R = 73.64 ft D 2g ° 1.5 ¢ 2 ( 32.2 ft>s2 ) ft 12

Take the control volume to be the water in the pump and hose from A to C. Applying the energy equation with the datum set at A and VC = 16.30 ft>s, pC pA VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g

0 + 0 + 0 + hpump

lb 12 in. 2 ba b ( 16.34 ft>s ) 2 1 ft in2 = + + 30 ft + 0 + 73.64 ft ( 1.937 slug>ft 3 )( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) a30

hpump = 177.05 ft Applying, P = gQhpump = ( 1.937 slug>ft 3 )( 32.2 ft>s2 ) a90 = 2214.47

gal min

ba

1 min 1 ft 3 ba b(177.05 ft) 60 s 7.48 gal

1 hp lb # ft a b = 4.03 hp s 550 ft # lb>s

Ans.

Ans: 4.03 hp 1064

10–46. Sewage, assumed to be water at 20°C, is pumped from the wet well using a pump and a 50-mm-diameter pipe. Determine the maximum discharge from the pump without causing cavitation. The friction factor is f = 0.026. The (gage) vapor pressure for water at 20°C is - 98.7 kPa.

1m C

0.5 m

B 3m

A

SOLUTION Water is considered to be incompressible. The major head loss of the flow from A to B in the pipe can be determined using

( hL ) AB = f

LAB V 2 3m V2 = 0.026 a b£ § = 0.07951V 2 D 2g 0.05 m 2 ( 9.81 m>s2 )

From Appendix A, r = 998.3 kg>m3 for water at 20° C. The cavitation will occur at the juncture, where water is about to enter the pump since the pressure here is the smallest. Thus, pB = - 98.7 kPa. Take the control volume as the water in the pipe from A to B. Applying the energy equation from A to B with the datum at A, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g N V2 m2 0 + 0 + 0 + 0 = + + 3 m + 0 + 0.07951V 2 3 2 ( 998.3 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s2 ) -98.7 ( 103 )

V = 7.365 m>s Thus, Q = VA = ( 7.365 m>s ) 3 p(0.025 m)2 4 = 0.0145 m3 >s

Ans.

1065

Ans: 0.0145 m3 >s

10–47. Sewage, assumed to be water at 20°C, is pumped from the wet well using a pump and a 50-mm-diameter pipe having a friction factor of f = 0.026. If the pump delivers 500 W of power to the water, determine the discharge from the pump.

1m C

0.5 m

B 3m

A

SOLUTION Water is considered to be incompressible. The major head loss for the flow from A to C can be determined using

( hL ) AC = f a

LAC V 2 4.5 m V2 b b£ § = 0.11927V 2 = 0.026 a D 2g 0.05 m 2 ( 9.81 m>s2 )

Take the control volume as the water in the pump and pipe from A to C. From Appendix A, r = 998.3 kg>m3 for water at 20° C. Applying the energy equation from A to C with the datum at A. pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 +

V2 2 ( 9.81 m>s2 )

+ 3.5 m + 0 + 0.11927V 2

hpump = ( 0.17023V 2 + 3.5 ) Here, Q = VA = V 3 p(0.025 m)2 4 = 0.0019635 V.

Applying

500

# Ws = gQhpump N#m = ( 998.3 kg>m3 )( 9.81 m>s2 ) (0.0019635V) ( 0.17023V 2 + 3.5 ) s 3.2734V 3 + 67.302V - 500 = 0

Solving by trial and error, V = 4.0933 m>s Thus, Q = 0.0019635 ( 4.0933 m>s ) = 0.00804 m3 >s

Ans.

1066

Ans: 0.00804 m3 >s

*10–48. Water at 20°C is pumped from the reservoir at A and flows through the 250-mm-diameter smooth pipe. If the discharge at B is 0.3 m3 >s, determine the required power output for the pump that is connected to a 200-m length of the pipe. Draw the energy line and the hydraulic grade line for the pipe. Neglect any elevation differences.

A

B

0.25 m

180 m

20 m

SOLUTION Water is considered to be incompressible.

Head (m)

The mean velocity of the flow in the pipe is

16.7

0.3 m3 >s Q = = 6.112 m>s V = A p(0.125 m)2 From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at 20° C. Thus, the Reynolds number is Re =

VD = n

( 6.112 m>s ) (0.25 m) = 1.53 ( 106 ) 1.00 ( 10-6 ) m2 >s

–1.64

For the smooth pipe, the Moody diagram, gives f = 0.0108. Thus, the major head loss can be determined using. hL = f

( 6.112 m>s ) 2 L V2 200 m = 0.0108 a b£ § = 16.448 m D 2g 0.25 m 2 ( 9.81 m>s2 )

Take the control volume as the water in the pipe and pump from A to B. Since the pipe is assumed to be horizontal, zA = zB = z. Applying the energy equation from A to B, pA pB VA2 V B2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g ( 6.112 m>s ) 2 0 + 0 + z + hpump = 0 + + z + 0 + 16.448 m 2 ( 9.81 m>s2 ) hpump = 18.352 m Applying p = gQ hpump = ( 998.3 kg>m3 )( 9.81 m>s2 )( 0.3 m3 >s ) (18.352 m) = 53.92 ( 103 )

N#m = 53.9 kW s

The velocity head is

( 6.112 m>s ) V2 = = 1.904 m 2g 2 ( 9.81 m>s2 ) 2

The head loss per meter length of pipe is hL 16.448 m = = 0.08224 m L 200 m Using these results, the EL and HGL can be plotted as shown in Fig. a.

1067

1.90

Ans.

1.90

EL HGL

Pump 20 1.90

(a)

200

L (m)

10–49. Water is siphoned from the lake at A and into a river at B through a 150-ft-long cast iron pipe. Determine the smallest diameter D of the pipe so that the discharge is 0.65 ft 3 >s. Take nw = 1.15 1 10-6 2 ft 2 >s.

A 15 ft B

SOLUTION Water is considered to be incompressible. The mean velocity of the flow in the pipe is V =

0.65 ft 3 >s Q 0.8276 = = p 2 A D2 D 4

Thus, the Reynolds number is 0.8276 bD 7.1966 ( 105 ) VD D2 Re = = = n D 1.15 ( 10-6 ) ft 2 >s a

(1)

The major head loss can be determined using

0.8276 2 b f L V 150 ft D2 b = ≥ ¥ = 1.5953 a 5 b = fa hL = f 2 D 2g D ( ) D 2 32.2 ft>s a

2

Take the control volume as the water in the pipe from A to B. Applying the energy equation from A to B with the datum set at B, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0.8276 2 b f D2 0 + 0 + 15 ft + 0 = 0 + + 0 + 0 + 1.5953 a 5 b D 2 ( 32.2 ft>s2 ) a

f = 9.4024 D5 - 0.006667 D

(2)

For cast iron, e = 0.00085 ft . The iterations are tabulated as follows iteration

Assumed D(ft)

1

0.3

2

0.310

e ft a b D ft

0.00283 0.00274

f Eq. (2)

f from the Moody diagram

2.40 ( 106 )

0.0208

0.0253

2.32 ( 10

0.0249

0.025

Re Eq. (1)

6

)

The assumed D in the 2nd iteration yields a value of f using Eq. (2) that is almost the same as that given by the Moody diagram. Thus, D = 0.310 ft is an acceptable result. Then, D = (0.310 ft)a

12 in. b = 3.72 in. 1 ft

Ans.

Ans: 3.72 in. 1068

10–50. Water is delivered into the truck using a pump that creates a flow of 300 liter>min through a 40-mm-diameter hose. If the total length of the hose is 8 m, the friction factor is f = 0.018, and the tank is open to the atmosphere, determine the power that must be supplied by the pump.

A

3m B C

2m

SOLUTION Water is considered to be incompressible. Q = ( 300 liter>min )( 1 min>60 s )( 1 m3 >1000 liter ) = 0.005 m3 >s V =

0.005 m3 >s Q = = 3.979 m>s A p(0.02 m)2

The major head loss from C to A can be determined using hL = f

( 3.979 m>s ) 2 L V2 8m b£ § = 2.905 m = 0.018 a D 2g 0.04 m 2 ( 9.81 m>s2 )

Take the control volume as the water from A to C in the hose and pump. Applying the energy equation from C to A with the datum set at C, pA pC VC2 VA2 + zC + hpump = + zA + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 +

( 3.979 m>s ) 2 + 5 m + 0 + 2.905 m 2 ( 9.81 m>s2 )

hpump = 8.7118 m Applying P = gQh pump = ( 1000 kg>m3 )( 9.81 m>s2 )( 0.005 m3 >s ) (8.7118 m) = 427.31

N#m = 427 W s

Ans.

Ans: 427 W 1069

10–51. Water is delivered at 0.003 m3 >s into the truck using a pump and a 40-mm-diameter hose. If the length of the hose from C to A is 10 m, and the friction factor is f = 0.018, determine the power output of the pump.

A

3m B C

2m

SOLUTION V =

0.003 m3 >s Q = = 2.3873 m>s A p(0.02 m)2

Major head loss

( 2.3873 m>s ) L V2 10 m = 0.018 a ba b D 2g 0.04 m 2 ( 9.81 m>s2 ) 2

hL = f

= 1.3072 m

Applying the energy equation from C to A with the datum at C, pA pC VC2 VA2 + zC + hpump = + zA + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 +

( 2.3873 m>s ) 2 + 5 m + 0 + 1.3072 m 2 ( 9.81 m>s2 )

hpump = 6.5977 m # Ws = gQh pump = ( 1000 kg>m3 )( 9.81 m>s2 )( 0.003 m3 >s ) (6.5977 m) # Ws = 194 W

Ans.

Ans: 194 W 1070

*10–52. The 30-mm-diameter 20-m-long commercial steel pipe transports water at 20°C. If the pressure at A is 200 kPa, determine the volumetric flow through the pipe.

30 mm A

B

20 m

SOLUTION Water is considered to be incompressible. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at 20° C. Thus, the Reynolds number is Re =

V(0.03 m) VD = = 3 ( 104 ) V n 1.00 ( 10-6 ) m2 >s

(1)

The major head loss can be determined using hL = f

L V2 20 m V2 = fa bc d = 33.9789 f V 2 D 2g 0.03 m 2 ( 9.81 m>s2 )

The water in the pipe from A to B. Applying the energy equation from A to B, with zA = zB = z and VA = VB = V. pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g N V2 V2 m2 + + z + 0 = 0 + + z + 0 + 33.9789 f V 2 3 2 2g ( 998.3 kg>m )( 9.81 m>s ) 2g 200 ( 103 )

V =

0.7753

(2)

2f

0.045 mm e = = 0.0015. The D 30 mm

For commercial steel pipe, the relative roughness is interations are as follows: iteration

Assumed f

1

0.023

2

0.0235

V ( m>s ) ; Eq. (2) Re; Eq. (1) f from the Moody diagram 5.112

1.53 ( 105 )

0.0235

5.057

1.52 ( 10

0.0235

5

)

The assumed f of the second iteration is almost the same as that given by the Moody diagram. Thus, the result V = 5.168 m>s is an acceptable result. Then, Q1 = VA = ( 5.057 m>s ) 3 p(0.015 m)2 4 = 0.00357 m3 >s

Ans.

Note: A more direct solution can be obtained from the Colebrook equation. If Eqs. (1) and (2) are substituted into the term 2.51> ( Re1f ) , the V will cancel so that the log becomes constant.

1071

10–53. Determine the power the pump must supply in order to discharge 0.02 m3 >s of water at B from the 100-mm-diameter hose. The friction factor is f = 0.028, and the hose is 95 m long. Neglect minor losses.

B

60 m

A

SOLUTION Water is considered to be incompressible. Since the pipe has a constant diameter, the mean velocity is also constant throughout the pipe and is 0.02 m3 >s Q = = 2.546 m>s V = A p ( 0.05 m ) 2 The head loss from A to B can be determined using hL = f

( 2.546 m>s ) 2 L V2 95 m b£ § = 8.791 m = 0.028 a D 2g 0.1 m 2 ( 9.81 m>s2 )

Take the water in the pipe and pump from A to B as the control volume. Applying the energy equation from A to B with the datum set at A, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 +

( 2.546 m>s ) 2 + 60 m + 0 + 8.791 m 2 ( 9.81 m>s2 )

hpump = 69.122 m Applying, # Ws = gQhpump = ( 1000 kg>m2 )( 9.81 m>s2 )( 0.02 m3 >s ) (69.122 m) = 13.56 ( 103 ) W = 13.6 kW

Ans.

Ans: 13.6 kW 1072

10–54. Determine the power extracted from the water by the turbine at C if the discharge from the pipe at B is 0.02 m3 >s. The pipe is 38 m long, it has a diameter of 100 mm, and the friction factor is f = 0.026. Also, draw the energy line and the hydraulic grade line for the pipe. Neglect minor losses.

A 10 m

C

8m

B

10 m 10 m 10 m

SOLUTION Water is considered to be incompressible. Since the pipe has a constant diameter, the mean velocity is also constant throughout the pipe and is 0.02 m3 >s Q = = 2.546 m>s V = A p(0.05 m)2 The head loss from A to B can be determined using hL = f

2 L V2 38 m ( 2.546 m>s ) = 0.026 a bc d = 3.2654 m D 2g 0.1 m 2 ( 9.81 m>s2 )

Take the control volume as the water in the reservoir and pipe. Applying the energy equation from A to B with the datum set at B, pA pB VA2 V B2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 18 m + 0 = 0 +

( 2.546 m>s ) 2 + 3.2654 m 2 ( 9.81 m>s2 )

(hs)out = 14.40 m (the positive result indicates turbine head) Applying, # Ws = gQ(hs)out = ( 1000 kg>m3 )( 9.81 m>s2 )( 0.02 m3 >s ) (14.40 m) = 2826

N#m = 2.83 kW s

Ans.

The velocity head is

( 2.546 m>s ) V2 = = 0.331 m 2g 2 ( 9.81 m>s2 ) 2

18 m 15.6 m

0.331 m

The head loss per meter length of pipe is hL 3.2654 m = = 0.08593 m>m L 38 m

EL HGL

Using these results the EL and HGL are plotted as shown in Fig. a.

0.331 m

1.19 m 0.859 m 0

turbine

L (m) 28

38

(a)

Ans: 2.83 kW 1073

10–55. The geothermal heat pump works on a closed-loop system. The loop consists of plastic pipe having a diameter of 40 mm, a total length of 40 m, and a roughness factor of e = 0.003 mm. If water at 40°C is to have a flow of 0.002 m3 >s, determine the power output of the pump. Include the minor losses for the 180° bend, KL = 0.6, and for each of the two 90° elbows, KL = 0.4. A

B

SOLUTION Assume fully developed steady flow. Water is assumed incompressible. The average velocity is 0.002 m2 >s Q V = = = 1.592 m>s A p(0.02 m)2 From Appendix A, rw = 992.3 kg>m3, nw = 0.664 ( 10-6 ) m2 >s. The Reynolds number is

( 1.592 m>s ) (0.04 m) VD = = 9.587 ( 104 ) nw 0.66 + ( 10-6 ) m2 >s

Re =

0.003 mm = 0.000075, the Moody diagram gives f = 0.019. 40 mm Here L = 40 m, and there is one 180° bend and two 90° bends. Thus, the total head Together with e>D =

loss is hL = f

L V2 V2 + ΣKL D 2g 2g

= af

L V2 + ΣKL b D 2g

= c 0.019 a

( 1.592 m>s ) 2 40 m b + 0.6 + z(0.4) d ° ¢ 0.04 m 2 ( 9.81 m>s2 )

= 2.6352 m

In a closed circuit, hpump = hL and so # Wpump = gQhL = ( 9.81 m>s2)(992.3 kg>m3)(0.002 m3>s) (2.6352 m) = 51.3 W Ans.

Ans: 51.3 W 1074

*10–56. The horizontal flat-plate solar collector is used for heating water for a swimming pool. It consists of a 1.5-in.-diameter 60-ft long ABS pipe that has a serpentine shape as shown. Water at an average temperature of 120°F is pumped through the pipe at a rate of 0.05 ft 3 >s. If the pressure at A is 32 psi, determine the pressure in the pipe at the exit B. Take the friction factor to be f = 0.019. Include the minor losses of each 180° bend, KL = 0.6, and for each 90° elbow, KL = 0.4.

A B

SOLUTION Assume that fully developed steady flow occurs, and water is incompressible. The average velocity is V =

Q = A

0.05 ft 3 >s

pa

2 0.75 ft b 12

= 4.074 ft>s

The total length of the pipe is 60 ft and there are 11 180° bends and 2 90° bends. Thus, the total lead loss is hL = f

L V2 V2 + ΣKL D 2g 2g

= af

L V2 + ΣKL b D 2g

= £ 0.019 a = 4.258 ft

( 4.074 ft>s ) 2 60 ft b + 11(0.6) + 2(0.4) § £ § 1.5>12 ft 2 ( 32.2 ft>s2 )

Since the pipe is horizontal and has a constant diameter, pA - pB ; hL = g 1b 12 in 2 ba b - pB 1 ft in2 4.258 ft = ( 1.918 slug>ft 3 )( 32.2 ft>s2 ) a32

pB = a4.345

1b 1 ft 2 ba b = 30.2 psi 2 12 in ft

1075

Ans.

10–57. Water is stored in the tank using a pump at A that supplies 300 W of power. If the 50-mm-diameter pipe has a friction factor of f = 0.022, determine the flow into the tank at the instant shown. The tank is open at the top. Include the minor losses of the 90° elbow and the discharge into the reservoir at B.

D

6m 4m C

8m A 1m B

SOLUTION Water is considered to be incompressible. The mean velocity of the flow in the pipe is V =

Q Q = = 509.30Q A p(0.025 m)2

Then, the major head loss from B to D can be determined using

( hL ) BD = f a

(509.30Q)2 LBD V 2 15 m b b£ § = 87 253.96Q2 = 0.022a D 2g 0.05 m 2 ( 9.81 m>s2 )

The minor loss is contributed by the flow discharge, kL = 1.0 and one 90° elbow, kL = 0.9. Thus,

( hL ) mi = ΣkL

(509.30Q)2 V2 = (1.0 + 0.9) £ § = 25 118.56Q2 2g 2 ( 9.81 m>s2 )

Take the control volume as the water in the pipe, and tank, and pump from D to B. Applying the energy equation from B to D with the datum at B, pD pB VB2 VD2 + zB + hpump = + zD + hturb + hL + + g g 2g 2g 0 + 0 + 0 + hpump = 0 + 0 + 5 m + 0 + ( 87 253.96Q2 + 25 118.56Q2 ) hpump = ( 5 + 112 372.53 Q2 ) # W = gQh pump 300

N#m = ( 1000 kg>m3 )( 9.81 m>s2 ) Q ( 5 + 112 372.53Q2 ) s 1 102 374 478Q3 + 49 050Q - 300 = 0

Solving by trial and error, Q = 0.00431 m3 >s

Ans.

1076

Ans: 0.00431 m3 >s

10–58. The pressure of air in a large tank at A is 40 psi. Determine the flow of water at 70°F from the tank after the gate valve at B is fully opened. The 2-in.-diameter pipe is made of galvanized iron. Include the minor losses for the flush entrance, the two elbows, and the gate valve.

6 ft B 16 ft A 4 ft

SOLUTION Assume that fully developed steady flow occurs and water is incompressible. Appendix A gives rw = 1.937 slug>ft 3 and nm = 10.4 ( 10-6 ) ft 2 >s. Then the Reynolds number is Re =

V ( 2>12 ft ) VD = = 16.026 ( 103 ) V nm 10.4 ( 10-6 ) ft 2 >s

10 ft

(1)

The loss coefficient for a flush entrance is 0.5, an elbow is 0.9 and a gate valve is 0.19 the total length of the pipe is L = 10 ft + 16 ft + 6 ft = 32 ft., Thus, the total head loss is hL = f

L V2 V2 + ΣkL D 2g 2g

= af

L V2 + ΣkL b D 2g

= £f a

32 ft V2 b + 0.5 + 2(0.9) + 0.19 § £ § 2>12 ft 2 ( 32.2 ft>s2 )

= (192 f + 2.49) a

V2 b 64.4

Write the energy equation between A and B by realizing that VA ≃ 0 (large reservoir) and VB = V, zA = 4 ft and zB = 16 ft, pA = a40

lb 12 in 2 b = 5760 lb>ft 2 and pB = pamt = 0, hpump = hturb = 0 ba 1 ft in2 pB pA VA2 VB2 + + zA + hpump = + gv gv 2g 2g

5760 lb>ft 2

( 1.937 slug>ft )( 32.2 ft>s ) 2

2

+ 0 + 4 ft + 0 = 0 +

(192f + 3.49)a

+ zB + hturb + hL V2

2 ( 32.2 ft>s

2

)

+ 16 + 0 + (192f + 2.49)a

V2 b = 80.35 64.4

(192f + 3.49)V 2 = 5174.54

V2 b 64.4

(2)

Assume a value of f = 0.025. Eq. (2) then Eq. (1) give V = 24.98 ft>s and 0.0005 ft Re = 4.00 ( 105 ) . For galvanized iron pipe, e>D = = 0.003. Enter these 2>12 ft valves of Re and e>D into the Moody diagram, we obtain f = 0.0265. Repeat the same procedure with this value of f, we obtain V = 24.56 ft>s, Re = 3.94 ( 105 ) and f = 0.0265 which is the same as the previous value. Thus, V = 24.56 ft>s is acceptable. Then the discharge is Q = VA = ( 24.56 ft>s ) £ p a

2 1 ft b § = 0.536 ft 3 >s 12

Ans.

1077

Ans: 0.536 ft 3 >s

10–59. Water at 70°F flows from the tank through the 1-in.diameter galvanized iron pipe. If the faucet (gate valve) is fully opened, determine the volumetric discharge at A. Include the minor losses of the flush entrance, the two elbows, and the gate valve.

B 2 ft

4 ft

A 4 ft

SOLUTION Water is considered to be incompressible. From Appendix A, r = 1.937 slug>ft 3 and n = 10.4 ( 10-6 ) ft 2 >s for water at 70° F . Then, the Reynolds number is Re =

Va

1 ft b 12

VD = = 8012.82V n 10.4 ( 10-6 ) ft 2 >s

2 ft

(1)

The major head loss in the pipe can be determined using

( hL ) ma = f

V2 L V2 8 ft § = 1.4907f V 2 = f± ≤ £ D 2g 1 2 ( 32.2 ft>s2 ) ft 2

The minor head loss is contributed by the flush entrance, kL = 0.5, two 90° elbows, kL = 0.9, and a fully open gate valve, kL = 0.19. Thus,

( hL ) mi = ΣkL

V2 V2 = [0.5 + 2(0.9) + 0.19]£ § = 0.03866V 2 2g 2 ( 32.2 ft>s2 )

Water in the tank and pipe from A to B.

Applying the energy equation from B to A with the datum at B and VB = VA, pA pB VB2 VA2 + zB + hpump = + zA + hturb + hL + + g g 2g 2g 0 + 0 + 4 ft + 0 = 0 +

V2 2 ( 32.2 ft>s2 )

+ 0 + 0 + ( 1.4907 f V 2 + 0.03866V 2 )

1.4907 f V 2 + 0.05419V 2 = 4 V =

2

(2)

21.4907f + 0.05419

0.0005 ft e = = 0.006. The iterations are tabulated as For galvanized iron pipe, D 1 follows: ft 12 iteration

assumed f

V ( m>s ) ; Eq. (2)

1

0.032

6.265

5.02 ( 104 )

0.034

2

0.034

6.176

4.95 ( 104 )

0.034

Re Eq. (1)

f from Moody diagram

The value of f in the second iteration is almost the same as that given by the Moody diagram. Thus, the result V = 6.176 ft>s is acceptable. Then, Q = VA = (6.176 m>s) c pa

2 0.5 ft b d = 0.0337 ft 3 >s 12

Ans.

1078

Ans: 0.0337 ft 3 >s

*10–60. The large tank is filled with water at 70°F to the depth shown. If the gate valve at C is fully opened, determine the power of the water flowing from the end of the nozzle at B. Also, what is the head loss in the system? The galvanizediron pipe is 80 ft long and has a diameter of 2 in. Neglect the minor loss through the nozzle, but include the minor losses at the flush entrance, the two elbows, and gate valve.

A

80 ft

SOLUTION

C

Assume that fully developed steady flow occurs and water is incompressible. Appendix A gives rw = 1.937 slug>ft 3 and nm = 10.4 ( 10-6 ) ft 2 >s . Thus, the Reynolds number is 2 V a ft b 12 VD Re = = (1) = 1.603 ( 104 ) V nm 10.4 ( 10-6 ) ft 2 >s The loss coefficients for a flush entrance, elbow, and gate valve are 0.5, 0.9 and 0.19, respectively. Thus, the total head loss is hL = f = af

L V2 V2 + ΣkL D 2g 2g L V2 + ΣkL b D 2g

= cf a

80 ft V2 b + 0.5 + 2(0.9) + 0.19 d c d 2>12 ft 2 ( 32.2 ft>s2 )

= (480f + 2.49)a

V2 b 64.4

Write the energy equation between A and B, realizing that pA = pB = patm = 0, VA ≃ 0 (large reservoir) and VB = V, zA = 80 ft and zB = 0 (Datum through B), hpump = hturb = 0, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 80 ft + 0 = 0 +

V2 2 ( 32.2 ft>s

2

V2 =

)

+ 0 + 0 + (480f + 2.49)a

5152 480f + 3.49

V2 b 64.4 (2)

0.0005 ft = 0.003. Assume f = 0.026, Eq (2) then 2>12 ft Eq. (1) give V = 17.96 ft>s and Re = 2.88 ( 105 ) . With this value of Re, the Moody diagram gives f = 0.0265. Using this value of f, V = 17.83 ft>s and Re = 2.86 ( 105 ) for which the Moody diagram indicates f = 0.0265, which is the same as the previous value. Thus, V = 17.83 ft>s. Then, the discharge is For galvanized iron pipe, e>D =

Q = VA = ( 17.83 ft>s ) c pa

2 1 ft b d = 0.3890 ft 3 >s 12

1079

B

*10–60. Continued

The energy head at A is HA = 80 ft The head loss is

( hL ) A S B = [480(0.0265) + 2.49]£

( 17.83 ft>s ) 2 § 2 ( 32.2 ft>s2 ) Ans.

= 75.065 ft = 75.1 ft Then at B, HB = HA - ( hL ) A S B = 80 ft - 75.065 ft = 4.935 ft or HB =

( 17.83 ft>s ) 2 pB VB2 + + zB = 0 + + 0 = 4.935 ft gw 2g 2 ( 32.2 ft>s2 )

The power of the water jet is # Ws = gwQHB = ( 1.937 slug>ft 3 )( 32.2 ft>s2 )( 0.3890 ft 3 >s ) (4.935 ft) = a119.73

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

= 0.218 hp

1080

10–61. Water is to be delivered at 0.04 m3 >s to point B, on the ground, 500 m away from the reservoir. Determine the smallest-diameter pipe that can be used if the pump supplies 40 kW of power. Neglect any elevation changes, and take f = 0.02.

A

B 500 m

SOLUTION Water is considered to be incompressible. Since the pipe has a constant diameter, the mean velocity is also constant throughout the pipe and is V =

0.04 m3 >s Q 0.05093 = = p 2 A D2 D 4

The head of the pump can be determined using

hpump

# Ws = = gQ

40 ( 103 )

N#m s

= 101.94 m

( 1000 kg>m2 )( 9.81 m>s2 )( 0.04 m3 >s )

The head loss can be determined using

0.05093 2 b L V 500 m ≥ 0.001322 D2 ¥ = 0.02a b hL = f = D 2g D D5 2 ( 9.81 m>s2 ) a

2

Take the water from A to B as the control volume. Applying the energy equation from A to B noting there is no change in elevation, zA = zB = z, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0.05093 2 b 0.001322 D2 0 + 0 + z + (101.94 m) = 0 + + z + 0 + 2 D5 2 ( 9.81 m>s ) a

101.94D5 - 1.322 ( 10-4 ) D - 0.001322 = 0 Solving by trial and error, Ans.

D = 0.1056 m = 106 mm

Ans: 106 mm 1081

10–62. For an industrial application, water at 70°F flows through the 0.5-in.-diameter commercial steel pipe such that it must exit the gate valve with a discharge of 0.05 ft 3 >s. If the open diameter of the gate valve is also 0.5 in., determine the required pressure a pump must produce at A. Consider only major losses in the pipe. Draw the energy line and hydraulic grade line for the pipe.

D

E

6 ft

B

C

2 ft

A 9 ft

2 ft

SOLUTION Assume that fully developed steady flow occurs and water is incompressible. Appendix A gives rw = 1.937 slug>ft 3 and nw = 10.4 ( 10-6 ) ft 2 >s . Here, the average velocity is Q = A

V =

Thus, the Reynolds number is

Re =

VD = nw

For commercial steel pipe,

0.05 ft 3 >s

2 0.25 pa ft b 12

h (ft)

= 36.67 ft>s

0.5 ( 36.67 ft>s ) a ft b 12 10.4 ( 10-6 ) ft 2 >s

300 272 EGL 143

= 1.47 ( 105 )

Write the energy equation between points A and E , realizing that pE = patm = 0, VA = VE = V, zA = 0 and zE = 8 ft, (datum at A), hpump = hturb = 0, pE pA VA2 VE2 + + zA + hpump = + + zE + hturb + hL gw gw 2g 2g pA

( 1.937 slug>ft )( 32.2 ft>s )

At A, Energy head is hA =

2

+

V2 V2 + 0 + 0 = 0 + + 8 ft + 0 + 271.35 ft 2g 2g

pA = a174 23.42

pA VA2 + + zA = gw 2g

EGL D

0.00015 ft e = = 0.0036. Enter these values of Re and D 0.5>12 ft

( 36.67 ft>s ) 2 L V2 19 ft = 0.0285a b£ § = 271.35 ft D 2g 0.5>12 ft 2 ( 32.2 ft>s2 )

3

V2 = 20.9 ft 2g

57.4 28.9

e>D into the Moody diagram. We obtain f = 0.0285. The total length of the pipe from A to E is L = 2 ft + 9 ft + 6 ft + 2 ft = 19 ft. Thus, the major head loss is hL = f

HGL

lb 1 ft 2 ba b = 121 psi 2 12 in. ft

174 23.42 lb>ft 2

( 1.937 slug>ft 3 )( 32.2 ft>s2 )

+

Ans.

( 36.67 ft>s ) 2 + 0 = 300.23 ft 2 ( 32.2 ft>s2 )

At B, hB = hA - ( hL ) A S B = 300.23 ft - (271.35 ft)a

2 ft b = 271.67 ft 19 ft 1082

B A

E HGL

V2 = 20.9 ft 2g

C Datum (a)

10–62. Continued

At C, hC = hB - (hL)B S C = 271.67 ft - (271.35 ft)a

9 ft b = 143.13 ft 19 ft

At D, hD = hC - (hL)C S D = 143.13 ft - (271.35 ft)a At E, hE = hD - (hL)D S E = 57.44 ft - (271.35 ft)a The velocity head is constant through the pipe which is

6 ft b = 57.44 ft 19 ft

2 ft b = 28.88 ft 19 ft

( 36.67 ft>s ) V2 = = 20.88 ft 2g 2 ( 32.2 ft>s2 ) 2

Using this data, the EGL and HGL are shown in Fig. a.

Ans: 121 psi 1083

10–63. Water at 70°F flows through the 0.5-in.-diameter commercial steel pipe such that it must exit the fully opened gate valve with a discharge of 0.05 ft 3 >s. If the open diameter of the gate valve is also 0.5 in., determine the required pressure a pump must produce in the pipe at A. Include the minor losses of the elbow and gate valve. Draw the energy line and hydraulic grade line for the pipe.

D

E

6 ft

B

C

2 ft

A

SOLUTION

9 ft

2 ft

Assume that fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 1.937 slug>ft 3 and nw = 10.4( 10-6 ) ft 2 >s. Hence, the average velocity is V =

Q = A

Thus, the Reynolds number is

VD = Re = nw

0.05 ft 3 >s

pa

2 0.25 ft b 12

( 36.67 ft>s ) a

= 36.67 ft>s

0.5 ft b 12

10.4 ( 10-6 ) ft 2 >s

h (ft) 361 313 EGL

= 1.47 ( 105 )

0.00015 ft = 0.0036. Enter these values of Re 0.5>12 ft and e>D into Moody diagram. We obtain f = 0.0285. The total length of the pipe from A to E is L = 2 ft + 9 ft + 6 ft = 19 ft. The loss coefficient for the elbow is 0.9, and gate valve it is 0.19. Thus, the total head loss is For commercial steel pipe, e>D =

hL = f

L V2 V2 + ΣkL D 2g 2g

= 331.69 ft

Write the energy equation between points A and E realizing that pE = patm = 0, VA = VE = V, zA = 0 and zE = 8 ft (datum at A), hpump = hturb = 0, pE pA VA2 VE2 + + zA + hpump = + + zE + hturb + hL gw gw 2g 2g

( 1.937 slug>ft )( 32.2 ft>s ) 3

2

+

V2 V2 + 0 + 0 = 0 + + 0 + 8 ft + 0 + 331.69 ft 2g 2g

pA = a21 187.01

At, A the energy head is

pA VA2 + + zA = hA = gw 2g

61.4 28.9

A

( 36.67 ft>s ) 2 19 ft b + 3(0.9) + 0.19 d £ § 0.5>12 ft 2 ( 32.2 ft>s2 )

pA

V2 = 20.9 ft 2g EGL (hL)GV = 3.97 ft

166

D

B

L V2 = af + ΣkL b D 2g = c 0.0285a

(hL)el = 18.8

HGL

lb 1 ft 2 ba b = 147 psi 2 12 in ft

Ans.

lb ( 36.67 ft>s ) 2 ft 2 + + 0 = 360.57 ft 3 2 ( 1.937 slug>ft )( 32.2 ft>s ) 2 ( 32.2 ft>s2 ) 21 187.01

1084

E HGL

V2 = 20.9 ft 2g

C Datum (a)

10–63. Continued

The total major head loss is (hf)L = f

( 36.67 ft>s ) 2 L V2 19 ft = 0.0285° 0.5 ¢ J R = 271.35 ft D 2g 2 ( 32.2 ft>s2 ) 12 ft

The minor head loss for the elbow and gate valve and

( 36.67 ft>s ) V2 = 0.9J R = 18.79 ft 2g 2 ( 32.2 ft>s2 ) 2

(hL)el = (kL)el

( 36.67 ft>s ) V2 = 0.19J R = 3.97 ft 2g 2 ( 32.2 ft>s2 ) 2

(hL)GV = (kL)GV Thus, at B,

hB = hA - (hL)A S B = 360.57 ft - (271.35 ft)a At C, hC = hB - (hL)B S C = 313.22 ft - (271.35 ft)a At D,

2 ft b - 18.79 ft = 313.22 ft 19 ft 9 ft b - 18.79 ft = 165.89 ft 19 ft

hD = hC - (hL)C S D = 165.89 ft - (271.35 ft)a At E, hE = hD - (hL)D S E = 61.41 ft - (271.35 ft)a The velocity head is

6 ft b - 18.79 ft = 61.41 ft 19 ft 2 ft b - 3.97 ft = 28.88 ft 19 ft

( 36.67 ft>s ) V2 = = 20.88 ft 2g 2 ( 32.2 ft>s2 ) 2

Using this data the EGL and HGL are shown in Fig. a.

Ans: 147 psi 1085

*10–64. Water at 200° F enters the radiator at A with an average velocity of 4 ft>s and a pressure of 60 psi. If each 180° bend has a minor loss coefficient of KL = 1.03, determine the pressure at the exit B. The copper pipe has a diameter of 14 in. Take e = 5 ( 10-6 ) ft. The radiator is in the vertical plane.

1.75 ft A

2 ft

SOLUTION

B

Water is considered to be incompressible. From Appendix A, r = 1.869 slug>ft 3 and n = 3.40 ( 10-6 ) s2 >ft. Then, the Reynolds number is 0.25 ( 4 ft>s ) a ft b 12 VD Re = = = 2.45 ( 104 ) n 3.40 ( 10-6 ) ft 2 >s

5 ( 10-6 ) ft e = = 0.00024. From D 0.25 ft 12 the Moody diagram, f = 0.025. Then, the Major head loss can be determined using

The relative roughness of the copper pipe is

(hL)ma = f

15(1.75 ft) ( 4 ft>s ) 2 L V2 § = 7.826 ft = 0.025 £ 0.25 §£ D 2g 2 ( 32.2 ft>s2 ) a ft b 12

The minor head loss due to the 180° bends is (hL)mi = kL

( 4 ft>s ) 2 V2 = 14(1.03) £ § = 3.583 ft 2g 2 ( 32.2 ft>s2 )

Take the control volume as the water in the pipe from A to B. Applying the energy equation from A to B with the datum at B and VA = VB = V, pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g lb 12 in 2 ba b pB 1 ft V2 V2 in2 + + 2 ft + 0 = + + 0 + 0 + 7.826 ft + 3.583 ft ( 1.869 slug>ft 3 )( 32.2 ft>s2 ) 2g ( 1.869 slug>ft 3 )( 32.2 ft>s2 ) 2g a60

pB = a8073.77

lb 1 ft 2 ba b = 56.1 psi 2 12 in. ft

Ans.

1086

*10–65. Water flows at 900 gal>min through the 8-in-diameter pipe. As it passes through the filter the pressure drop is 0.2 psi. Determine the loss coefficient for the filter. There are 7.48 gal>ft 3.

A

B

SOLUTION Water is considered to be incompressible. gal 1 ft 3 1 min ba ba b = 2.005 ft 3 >s. Thus, the mean velocity of Here, Q = a900 min 7.48 gal 60 s

the flow in the pipe is V =

2.005 ft 3 >s Q = = 5.745 ft>s 2 A 4 pa ft b 12

The minor head loss contributed by the filter is (hL)mi = kL

( 5.745 ft>s ) 2 V2 kL £ § = 0.5125 kL 2g 2 ( 32.2 ft>s2 )

Take the control volume as the water in the pipe from A to B.

Applying the energy equation with zA = zB = z and VA = VB = V. pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA 62.4 lb>ft

3

+

pB V2 V2 + + z + 0 = + z + 0 + 0.5125 kL 3 2g 2g 62.4 lb>ft

pA - pB = 31.978 kL 0.2 lb 144 in2 a b = 31.978 kL in2 ft 2

Ans.

kL = 0.901

Ans: 0.901 1087

10–66. Gasoline from the tank of an automobile A is pumped at B through the fuel filter C, and then to the fuel injectors on the engine. The fuel line is stainless-steel 4-mm-diameter tubing. Each fuel injector has a nozzle diameter of 0.5 mm. If the loss coefficient is KL = 0.5 at the flush entrance of the fuel tank, KL = 1.5 through the filter, and KL = 4.0 at each nozzle, find the required power output of the pump to deliver fuel at the rate of 0.15 liter>min to four of the cylinders that are under an average pressure of 300 kPa. The fuel line is 2.5 m long. Take e = 0.006 mm. Assume that the nozzles and the tank are on the same level.

4 mm A

SOLUTION Assume that fully developed steady flow occurs, and gasoline is incompressible. Appendix A gives rg = 726 kg>m3 and ng = 0.465 ( 10-6 ) m2 >s. The discharge is Q = a0.15

l 1 m3 1 min ba ba b = 2.5 ( 10-6 ) m3 >s min 1000 l 60 s

Then the average velocity in the tube is Vt =

2.5 ( 10-6 ) m3 >s Q = = 0.1989 m>s At p(0.002 m)2

Since there are four nozzles in parallel, the discharge for each nozzle is Qn =

1 1 Q = 3 2.5 ( 10-6 ) m3 >s 4 = 0.625 ( 10-6 ) m3 >s 4 4

Then the average velocity in the nozzle is Vn =

0.625 ( 10-6 ) m3 >s Qn = = 3.183 m>s An p(0.00025 m)2

The Reynolds number of the flow in the tube is Re =

(0.1989 m>s)(0.004 m) Vt Dt = = 1711.34 6 2300 ng 0.465 ( 10-6 ) m3 >s

(laminar flow)

Thus, the friction factor can be determined using f =

64 64 = = 0.03740 Re 1711.34

Then the total head loss is hL = f

Vt 2 Vn 2 L Vt 2 + ΣkL + kL D 2g 2g 2g

= af

Vt 2 Vn 2 L + ΣkL b + kL D 2g 2g

= c 0.03740 a = 2.1169 m

( 0.1989 m>s ) ( 3.183 m>s ) 2.5 m b + 0.5 + 1.5 d £ § + 4.0£ § 2 0.004 m 2 ( 9.81 m>s ) 2 ( 9.81 m>s2 ) 2

2

1088

B

C

10–66. Continued

Write the energy equation between the tank and nozzle, realizing that VT ≃ 0 (large reservoir), pT = patm = 0, zT = zn = z (tank and nozzle are in same level), pT pn Vn 2 VT + + zT + hpump = + + zn + hturb + hL gg gg 2g 2g 0 + 0 + z + hpump =

300 ( 103 ) N>m2

( 726 kg>m3 )( 9.81 m>s2 )

+

( 3.183 m>s ) 2 + z + 0 + 2.1169 m 2 ( 9.81 m>s2 )

hpump = 44.756 m Thus, the required power output of the pump is # Ws = ggQhpump = ( 726 kg>m3 )( 9.81 m>s2 ) 3 2.5 ( 10-6 ) m3 >s 4 (44.756 m) = 0.797 W

Ans.

Ans: 0.797 W 1089

10–67. Air at a temperature of 40°C flows through the duct at A with a velocity of 2 m>s. Determine the change in pressure between A and B. Account for the minor loss caused by the sudden change in diameter of the duct.

300 mm 200 mm A

B

SOLUTION Air is considered to be incompressible. Take the air from A to B to the control volume. Use average velocities. The continuity equation gives 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - ( 2 m>s ) 3 p(0.1 m)2 4 + VB 3 p(0.15 m)2 4 = 0 VB = 0.8889 m>s

The head loss is contributed by the minor head loss of the sudden change in diameter.

( hL ) mi = kL

DA 2 2 V 2A V2 = c1 - a b d 2g DB 2g 2

( 2 m>s ) 2 0.2 m 2 = £1 - a b § £ § = 0.06292 m 0.3 m 2 ( 9.81 m>s2 ) From Appendix A, r = 1.127 kg>m3 for air at 40° C. Since the duct is horizontal, zA = zB = z. Applying the energy equation from A to B. pB pA VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA

( 1.127 kg>m3 )( 9.81 m>s2 ) =

pB

( 1.127 kg>m )( 9.81 m>s ) 3

2

+

+

( 2 m>s ) 2 + z + 0 2 ( 9.81 m>s2 )

( 0.8889 m>s ) 2 + z + 0 + 0.06292 m 2 ( 9.81 m>s2 ) Ans.

pB - pA = 1.11 Pa

Ans: pB - pA = 1.11 Pa 1090

*10–68. Water at 20° C flows through the 20-mm-diameter galvanized iron pipe such that it discharges at C from the fully opened gate valve B at 0.003 m3 >s. Determine the required pressure at A. Include the minor losses of the  elbows and gate valve.

3m

A 1m

1m

1m C B

SOLUTION 0.5 m

Water is considered to be incompressible. The mean velocity of the flow in the pipe is V =

0.003 m3 >s Q = = 9.549 m>s A p(0.01 m)2

From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at 20° C. Thus, the Reynolds number is Re =

VD = n

( 9.549 m>s ) (0.02 m) = 1.91 ( 105 ) 1.00 ( 10-6 ) m2 >s

e 0.15 mm = = 0.0075. From D 20 mm the Moody diagram, f = 0.0345. Thus, the major head loss along the pipe can be

For galvanized iron pipe, the relative roughness is determined using

( hL ) ma = f

( 9.549 m>s2 ) L V2 7.5 m = 0.0345 a b£ § = 60.13 m D 2g 0.02 m 2 ( 9.81 m>s2 )

The minor head loss is contributed by four 90° elbows, kL = 0.9 and a gate valve, kL = 0.19. Thus,

( hL ) mi = Σ kL

( 9.549 m>s ) V2 = [4 (0.9) + 0.19] £ § = 17.62 m 2g 2 ( 9.81 m>s2 ) 2

Water in the pipe from A to C.

Applying the energy equation from A to C, with the datum at C and VA = VC = V because the diameter of the pipe is constant, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g pA

( 998.3 kg>m )( 9.81 m>s ) 3

2

+

V2 V2 + 2m + 0 = 0 + + 0 + 17.62 m + 60.13 m 2g 2g

pA = 741.80 ( 103 ) Pa = 742 kPa

Ans.

1091

20 mm 1m

10–69. The pump A and pipe system are used to transport oil to the tank. If the pressure developed by the pump is 400 kPa, and the filter at B has a loss coefficient of KL = 2.30, determine the discharge from the pipe at C. The 50-mm-diameter pipe is made of cast iron. Include the minor losses of the filter and three elbows. Take ro = 890 kg>m3 and no = 52.0 1 10 - 6 2 m2 >s.

0.5 m 2m 4m A

C

B

6.5 m

SOLUTION Water is considered to be incompressible. The Reynolds number for the flow in the pipe is Re =

V(0.05 m) VD = = 961.54V n 52.0 ( 10-6 ) m2 >s

(1)

The major head loss from A to C can be determined using

( hL ) ma = f

L V2 13 m V2 = fa bc d = 13.2518 f V 2 D 2g 0.05 m 2 ( 9.81 m>s2 )

The minor head loss is contributed by the filter, kL = 2.30 and three 90° elbows, kL = 0.9. Thus,

( hL ) mi = ΣkL

V2 V2 = [2.30 + 3(0.9)] c d = 0.2548V 2 2g 2 ( 9.81 m>s2 )

Take the control number as the water in the pipe from A to C. Since the pipe has a constant diameter, VA = VC = V. Applying the energy equation from A to C with the datum at A, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g N V2 V2 m2 + + 0 + 0 = 0 + + 2 m + 0 + (0.2548 + 13.2518f)V 2 3 2 2g ( 890 kg>m )( 9.81 m>s ) 2g 400 ( 103 )

V =

6.6192

(2)

20.2548 + 13.2518f

For cast iron pipe, the relative roughness is

e 0.26 mm = = 0.0052 D 50 mm The iterations are as follows: iteration

Assumed f

V ( m>s ) ; Eq. (2)

Re Eq. (1)

f from Moody diagram

1

0.03

8.195

7.88 ( 103 )

0.039

2

0.039

7.535

7.25 ( 103 )

0.0395

The value of f assumed in iteration (2) is almost the same as that given by the Moody diagram. Thus, V = 7.535 m>s is an acceptable result. Then, Q = VA = ( 7.535 m>s ) 3 p(0.025 m)2 4 = 0.0148 m3 >s

Ans.

1092

Ans: 0.0148 m3 >s

10–70. The pump A and pipe system are used to transport oil to the tank. Determine the required pressure developed by the pump in order to provide a discharge into the tank of 0.003 m3 >s. The filter at B has a loss coefficient of KL = 2.30. The 50-mm-diameter pipe is made of cast iron. Include the minor losses of the filter and three elbows. Take ro = 890 kg>m3 and no = 52.0 1 10 - 6 2 m2 >s.

0.5 m 2m 4m A

C

B

6.5 m

SOLUTION Water is considered to be incompressible. The mean velocity of the oil flow in the pipe is V =

0.003 m3 >s Q = = 1.528 m>s A p(0.025 m)2

Then, the Reynolds number is Re =

VD = n

( 1.528 m>s ) (0.05 m) = 1469 52.0 ( 10-6 ) m2 >s

Since Re 6 2300, the flow is laminar. Thus, the friction factor can be determined using f =

64 64 = = 0.04356 Re 1469

The major head loss from A to C can be determined using

( hL ) ma = f

( 1.528 m>s ) L V2 13 m = 0.04356 a bc d = 1.3477 m D 2g 0.05 m 2 ( 9.81 m>s2 ) 2

The minor head loss is contributed by the filter, kL = 2.30 and three 90° elbows, kL = 0.9. Thus,

( hL ) mi = ΣkL

( 1.528 m>s ) V2 = [2.30 + 3(0.9)] £ § = 0.5949 m 2g 2 ( 9.81 m>s2 ) 2

Take the control volume as the water in the pipe from A to C. Since the pipe has a constant diameter, VA = VC = V. Applying the energy equation from A to C with the datum at A, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g pA

( 890 kg>m )( 9.81 m>s ) 3

2

+

V2 V2 + 0 + 0 = 0 + + 2 m + 0 + 1.3477 m + 0.5949 m 2g 2g

pA = 34.42 ( 103 ) Pa = 34.4 kPa

Ans.

Ans: 34.4 kPa 1093

10–71. Water at 80°F flows at 5 ft>s through the 3 1 4 -in.-diameter pipe at A. If the 2 -in.-diameter gate valve at B is fully opened, determine the pressure in the water at A. The minor loss coefficient is KL = 0.6 for the spout at B. Also account for the minor loss in the elbow, the tee, and the fully opened gate valve. For the pipe take f = 0.016.

2 ft C

0.5 ft 6 ft

B

SOLUTION

D 0.25 ft

Assume that the fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 1.934 slug> ft 3 and nw = 9.35 ( 10-6 ) ft 2 >s at T = 80° F.

3 ft A

Continuity requires

QA = QB VA c pa

2 0.375 2 0.25 ft b d = VB c pa ft b d 12 12

VB = 2.25 VA = 2.25 ( 5 ft>s ) = ( 11.25 ft>s ) The loss coefficients for the elbow, tee along the branch, tee and the fully opened gate valve, are 0.9, 1.8, and 0.19, respectively. Thus the head losses from A to D and D to B are

(hL)AD = afAD

( 5ft>s ) 2 LAD VA2 3 ft + ΣkL b = ≥ 0.016 ± ≤ + 0.9 ¥ D 2g 0.75 2 ( 32.2 ft>s2 ) ft 12 = 0.6475 ft

( hL ) DB = ΣkL

( 11.25 ft>s ) 2 VB2 = (1.8 + 0.6 + 0.19) £ § = 5.090 ft 2g 2 ( 32.2 ft>s2 )

Applying the energy equation between A and B, where pB = 0, zA = 0, zB = 2.75 ft, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + ( hL ) AB gw gw 2g 2g pA 1.934 slug>ft 3 ( 32.2 ft>s2 ) 0 +

+

( 5 ft>s ) 2 + 0 + 0 = 2 ( 32.2 ft>s2 )

( 11.25 ft>s ) 2 + 2.75 ft + 0 + 0.6475 + 5.090 2 ( 32.2 ft>s2 ) pA = 626.77 lb>ft 2 = 4.35 psi

Ans.

Ans: 4.35 psi 1094

*10–72. Water at 80°F flows at 5 ft>s through the 3 4 -in.-diameter copper pipe at A. As the water is flowing, it emerges from the showerhead that consists of 100 holes, 1 each having a diameter of 16 in. Determine the pressure of the water at A if the minor loss coefficient is KL = 0.45 for the showerhead. Also account for the minor loss in the two elbows, the fully opened gate valve, and the tee. For the copper pipe take f = 0.016.

2 ft C

0.5 ft 6 ft

B

SOLUTION

0.25 ft

Assume that fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 1.934 slug>ft 3 and nw = 9.35 ( 10-6 ) ft 2 >s at T = 80° F.

Continuity require

QA = QC

( 5 ft>s ) c p a

2 0.03125 0.375 ft b d = VC £ (100)ap ft b § 12 12

VC = 7.20 ft>s The loss coefficients for the elbow, tee along the pipe run, and the fully opened gate valve, are 0.9, 0.4 and 0.19, respectively. Thus, the head losses from A to D, and D to C are (hL)AD = afADa

D

( 5 ft>s ) 2 LAD V 2A 3 ft b + Σ kL b ≤ + 0.9 ¥ ° ¢ = ≥ 0.016 ± D 2g 0.75 2 ( 32.2 ft>s2 ) ft 12 = 0.6475 ft

( hL ) DC = afDC a

LDC VA2 VL2 b + ΣKL b + kL D 2g 2g 2

( 5 ft>s ) 8.5 ft § = a0.016 ° 0.75 ¢ + 0.4 + 0.19 + 2(0.9) b £ 2 ( 32.2 ft>s2 ) 12 ft + 0.45 °

( 7.20 ft>s ) 2 ¢ 2 ( 32.2 ft>s2 )

= 2.1348 ft Applying the energy equation between A and C, where pC = 0, zA = 0, zC = 8.5 ft, pC pA VC 2 VA2 + zA + hpump = + zC + hturb + ( hL ) AC + + gw gw 2g 2g pA 1.934 slug>ft 3 ( 32.2 ft>s2 ) 0 +

+

( 5 ft>s ) 2 + 0 + 0 = 2 ( 32.2 ft>s2 )

( 7.20 ft>s ) 2 + 8.5 ft + 0 + 0.6475 ft + 2.1348 ft 2 ( 32.2 ft>s2 ) pA = 728.55 lb>ft 2 = 5.06 psi

Ans.

1095

3 ft A

10–73. Water at 20°C is pumped from the reservoir A using a pump that supplies 3 kW of power. Determine the discharge at C if the pipe is made of galvanized iron and has a diameter of 50 mm. Neglect minor losses.

50 mm C 4m

B

1m 4m

SOLUTION

2m

3m

A

Water is considered to be incompressible. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s. Thus, the Reynolds number is Re =

V(0.05 m) VD = = 5(10)4V n 1.00 ( 10-6 ) m2 >s

(1)

The major head loss can be determined using. hL = f

L V2 14 m V2 = fa b£ § = 14.271f V 2 D 2g 0.05 m 2 ( 9.81 m>s ) 2

Take the control volume as the water in the pump and pipe from A to C. Applying the energy equation from A to C, with the datum at A PC VC2 PA VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g

0 + 0 + 0 + 0 = 0 +

V2 2 ( 9.81 m>s2 )

+ 8 m + 0 + 14.271f V 2

hpump = ( 14.271f V 2 + 0.05097V 2 + 8 ) Applying

3 ( 103 )

# Ws = gQhpump

N#m = ( 998.3 kg>m3 )( 9.81 m>s2 ) 3 V(p)(0.025 m)2 4 ( 14.271 f V 2 + 0.05097V 2 + 8 ) s 14.271 f V 3 + 0.05097V 3 + 8V = 156.01

(2)

0.15 mm e = = 0.003. The For galvanized iron pipe, the relative roughness is D 50 mm iterations are as follows iteration 1

Assumed f V ( m>s ) ; Eq. (2) Re; Eq. (1) f from the Moody diagram 0.026

6.3015

3.15 ( 105 )

0.026

The assumed f of this iteration is almost the same as that given by the Moody diagram. Thus, V = 6.3015 m>s is an acceptable result. Then, Q = VA = ( 6.3015 m>s ) 3 (p)(0.025 m)2 4 = 0.0124 m3 >s 1096

Ans. Ans: 0.0124 m3 >s

10–74. Water at 20°C is pumped from the reservoir A using a pump that supplies 3 kW of power. Determine the discharge at C if the pipe is made of galvanized iron and has a diameter of 50 mm. Include the minor losses of the four elbows.

50 mm C 4m

B

1m 4m

SOLUTION

2m

3m

A

Water is considered to be incompressible. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s. Thus, the Reynolds number is Re =

V(0.05 m) VD = = 5 ( 104 ) V n 1.00 ( 10-6 ) m2 >s

(1)

The major head loss along the pipe can be determined using (hL)ma = f

L V2 14 m V2 = fa b£ § = 14.271 f V 2 D 2g 0.05 m 2 ( 9.81 m>s 2 )

The minor head loss is contributed by four 90° elbows, kL = 0.9. Thus,

( hL ) mi = ΣkL

V2 V2 = 4(0.9) £ § = 0.18349V 2 2g 2 ( 9.81 m>s2 )

Take the control volume as the water in the pump and pipe from A to C. Applying the energy equation from A to C,with the datum at A, PC VC2 PA VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g 0 + 0 + 0 + 0 = 0 +

V2 2 ( 9.81 m>s2 )

+ 8 m + 0 + 0.18349V 2 + 14.271f V 2

hpump = ( 14.271fV 2 + 0.23445V 2 + 8 ) Applying

# Ws = gQhpump 3 ( 103 )

N#m = ( 998.3 kg>m3 )( 9.81m>s2 ) 3 V(p)(0.025 m)2 4 s

(14.271 fV 2 + 0.23445V 2 + 8)

14.271f V 3 + 0.23445V 3 + 8V = 156.01

(2)

0.15 mm e = = 0.003. The For galvanized iron pipe, the relative roughness is D 50 mm iterations are as follows: Iteration Assumed f V ( m>s ) ; Eq. (2) Re; Eq. (1) f from the Moody diagram 1

0.026

5.6742

2.84 ( 105 )

0.026

The assumed f of this iteration is almost the same as that given by the Moody diagram. Thus, V = 5.6742 m>s is an acceptable result. Then, Q = VA = ( 5.6742 m>s ) 3 p(0.025 m)2 4 = 0.0111 m3 >s 1097

Ans.

Ans: 0.0111 m3 >s

10–75. If the faucet (gate valve) at E is fully opened and the pump produces a pressure of 350 kPa at A, determine the pressure just to the right of the tee connection C. The valve at B remains closed. The pipe and the faucet both have an inner diameter of 30 mm, and f = 0.04. Include the minor losses of the tee, the two elbows, and the gate valve.

0.65 m D

E

1m

3m

A

C B

5m

SOLUTION Water is considered to be incompressible. Since the pipe has a constant diameter, the mean velocity of the flow throughout the entire pipe is also constant. The major head loss from A to E is

( hL ) ma = f

L V2 9.65m V2 = 0.04 a b£ § = 0.6558V 2 D 2g 0.03m 2 ( 9.81m>s2 )

The minor head loss from A to E, is contributed by the tee along the pipe run, kL = 0.4, two 90° elbows, kL = 0.9, and a gate valve, kL = 0.19. Thus,

( hL ) mi = ΣkL

V2 V2 = [0.4 + 2(0.9) + 0.19] £ § = 0.1218V 2 2g 2 ( 9.81 m>s2 )

Take the control volume as the water in the pipe from A to E in the pipe. Applying the energy equation from A to E with the datum at A, pA pE VA2 VE2 + zA + hpump = + zE + hturb + hL + + g g 2g 2g N V2 V2 m2 + + 0 + 0 = 0 + + 1 m + 0 + 0.1218V 2 + 0.6558V 2 3 2 2g ( 1000kg>m )( 9.81m>s ) 2g 350 ( 103 )

V = 6.678 m>s No minor loss occurs from A to just right of the tee. The major head loss is (hL)ma = f

( 6.678 m>s ) 2 L V2 5m = 0.04 a b£ § = 15.1531 m D 2g 0.03 m 2 ( 9.81 m>s2 )

Take the control volume as the water in the pipe from A to C. Applying the energy equation from A to C, with zA = zC = z, pC pA VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g N V2 m2 + + z + 0 = 3 2 ( 1000kg>m )( 9.81m>s ) 2g 350 ( 103 )

PC

( 1000kg>m )( 9.81m>s ) 3

2

pC = 201.34 ( 103 ) Pa = 201 kPa

+

V2 + z + 15.1531 m 2g Ans.

Ans: 201 kPa 1098

*10–76. Water from the reservoir A is pumped into the large tank B. If the top of the tank is open, and the power output of the pump is 500 W, determine the volumetric flow into the tank when h = 2 m. The cast-iron pipe has a total length of 6 m and a diameter of 50 mm. Include the minor losses for the elbow and sudden expansion. The water has a temperature of 20°C.

B h!2m A 1m

SOLUTION Assume that fully developed steady flow occurs, and water incompressible. Appendix A gives rw = 998.3 kg>m3 and nm = 1.00 ( 10-6 ) m2 >s. The flow is Q = VA = V 3 p(0.025 m)2 4 = 0.625p ( 10-3 ) V

And the Reynolds number is Re =

V(0.05 m) VD = = 5(10)4V nm 1.00 ( 10-6 ) m2 >s

(1)

From the power output of the pump, # Ws = gQhpump; 500 W = ( 998.3 kg>m3 )( 9.81 m>s2 )3 0.625p ( 10-3 ) V 4 hpump hpump =

26.00 V

The loss coefficients for the elbow and sudden expansion are 0.9 and 1.0, respectively. Thus, the total head loss is hL = f

= af

L V2 V2 + ΣKL D 2g 2g

L V2 + ΣKL b D 2g

= £f a

6m V2 b + 0.9 + 1.0 § £ § 0.05m 2 ( 9.81 m>s2 )

= (120f + 1.9) a

V2 b 19.62

Write the energy equation between A and B, realizing that VA = VB ≃ 0 (large reservoir), pA = pB = patm = 0, zA = 0 and zB = 3 m (Datum through A), pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 +

26.00 V2 = 0 + 0 + 3m + 0 + (120f + 1.9) a b V 19.62

(120f + 1.9)V 3 + 58.86V - 510.16 = 0

(2)

1099

10–76. Continued

For the first iteration, assume V = 10 m>s. Then Eq.(1) gives Re = 5 ( 105 ) . For e 0.26 mm cast iron pipe, = = 0.0052. For Re = 5 ( 105 ) the Moody diagram gives D 50 mm f = 0.0305. Using this value of f, solving for V in Eq. 2, getting Re from Eq. 1, and checking on the Moody diagram, we obtain after a few iterations, V = 3.72 m>s. The discharge is then Q = 0.625p ( 10-3 )( 3.72 m>s ) = 0.007304 m3 >s

Ans.

= 0.438 liter>min.

1100

10–77. Water flows down through the vertical pipe at a rate of 3 m>s. If the differential elevation of the mercury manometer is 30 mm as shown, determine the loss coefficient KL for the filter C contained within the pipe. rHg = 13 550 kg>m3.

3 m/s

A

0.5 m

C

SOLUTION B

Water is considered to be incompressible.

20 mm

Writing the manometer equation by referring to Fig. a,

30 mm

pA + rwghAC - rHgghCD - rwghBD = pB pA + ( 1000 kg>m3 )( 9.81 m>s2 ) (0.55 m) - ( 13550 kg>m3 )( 9.81 m>s2 ) (0.03 m) - ( 1000 kg>m3 )( 9.81 m>s2 ) (0.02 m) = pB

A

pB - pA = 1211.54 Pa The minor head loss contributed by the filter is

( hL ) mi = kL

hAC = 0.55 m

2

( 3 m>s) V § = 0.4587kL = kL £ 2g 2 ( 9.81 m>s2 ) 2

hBD = 0.02 m

B C

D

Take the control volume as the water in the pipe from A to B. hCD = 0.03 m

Applying the energy equation with VA = VB = V and datum at B,

(a)

pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA

( 1000 kg>m )( 9.81 m>s ) 3

2

+

V2 + 0.5 m + 0 = 2g

pB - pA 9810 N>m3

pB

( 1000 kg>m )( 9.81 m>s ) 3

2

+

V2 + 0 + 0 + 0.4587kL 2g

= 0.5 - 0.4587kL

1211.54 Pa = 0.5 - 0.4587kL 9810 N>m3 Ans.

kL = 0.821

Ans: 0.821 1101

10–78. Water at T = 20°C flows from the open tank through the 50-mm-diameter galvanized iron pipe. Determine the discharge at the end B if the globe valve is fully opened. The length of the pipe is 50 m. Include the minor losses of the flush entrance, the four elbows, and the globe valve.

A 5m

B 1m

SOLUTION Water is considered to be incompressible. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s for water at 20° C. Then, the Reynolds number is V(0.05 m) VD = = 5 ( 104 ) V n 1.00 ( 10-6 ) m2 >s

Re =

(1)

The “major” head loss from A to B can be determined using

( hL ) f = f

L V2 50 m V2 = fa b£ § = 50.968f V 2 D 2g 0.05 m 2 ( 9.81 m>s2 )

The minor head loss is contributed by the flush entrance ( kL = 0.5 ) , four 90° elbows ( kL = 0.9 ) , and one fully - opened globe valve ( kL = 10.0 ) . Thus,

( hL ) m = ΣkL

V2 V2 § = 0.7187V 2 = [0.5 + 4(0.9) + 10.0] £ 2g 2 ( 9.81 m>s2 )

Take the control volume as the water in the tank and pipe from A to B. Applying the energy equation from A to B with the datum set at B, pA pB VA2 VB 2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g

0 + 0 + 4m + 0 = 0 +

V2 2 ( 9.81 m>s2 ) V =

+ 0 + 0 + ( 50.968 f V 2 + 0.7187V 2 ) 2

(2)

250.968f + 0.7696

0.15 mm e = = 0.003. The iterations For galvanized iron, the relative roughness is D 50 mm are tabulated as follows iteration 1 2 3

Assumed f 0.026 0.0278 0.0280

V ( m>s ) Eq. (2)

Re Eq. (1)

f from Moody diagram

1.382

6.91 ( 10

)

0.278

1.353

6.76 ( 10

)

0.0280

1.349

6.75 ( 10

)

0.0280

4 4 4

The assumed f in the 3rd iteration is almost the same as that given by the Moody diagram. Thus, V = 1.349 m>s is an acceptable result, so that Q = VA = ( 1.349 m>s ) 3 p(0.025 m)2 4 = 0.00265 m3 >s

1102

Ans.

Ans: 0.00265 m3 >s

10–79. An automatic sprinkler system for a yard is made from 12 -in.-diameter PVC pipe, for which e = 5 1 10 - 6 2 ft. If the system has the dimensions shown, determine the volumetric flow delivered to each sprinkler head C and D. The faucet at A delivers water at 70°F with 32 psi pressure. Neglect elevation changes, but include the minor losses at the two elbows and the tee. Also, the loss coefficient of the 1 8 -in.-diameter sprinkler nozzles at C and D is KL = 0.05.

C A

50 ft 50 ft

Assume fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 1.937 slug>ft 3 and nw = 10.4 ( 10-6 ) ft 2 >s at 70° F. The Reynolds number for the flow is 0.5 Va ft b 12 VD Re = = = 4.006 ( 103 ) V nw 10.4 ( 10-6 ) ft 2 >s

(1)

The continuity condition requires

Qp = Qnz Vp c pa

2 2 0.25 0.0625 ft b d = Vnz c pa ft b d 12 12

(2)

Vnz = 16 Vp And QA = QC + QD VA c pa

2 2 2 0.25 0.25 0.25 ft b d = ( Vp ) C c pa ft b d + ( Vp ) D c pa ft b d 12 12 12

VA = ( Vp ) C + ( Vp ) D

(3)

The loss coefficients for the elbow, the tee along the branch and the tee along the pipe run are 0.9, 1.8 and 0.4, respectively. Thus, the head losses from A to B, B to C and B to D are

( hL ) AB = afAB

LAB VA2 + ΣkL b D 2g

= c fABa

( hL ) BC = afBC

VA2 VA2 50 ft ( ) b + 0.9 d ° ¢ = 1200f + 0.9 a b AB 0.5>12 ft 64.4 2 ( 32.2 ft>s2 )

( Vp )C ( Vnz )C2 LBC + ΣkL b + kL D 2g 2g

= c fBC a

2

2

( Vp ) C2 3 16 ( Vp ) C 4 50 ft b + 1.8 d ° ¢ + 0.05W ¶ 2 0.5>12 ft 2 ( 32.2 ft>s ) 2 ( 32.2 ft>s2 ) = ( 1200fBC + 14.6 ) £

( Vp ) C2 64.4

30 ft

B 40 ft

SOLUTION

§

1103

D

10–79. Continued

( hL ) BD = afBD

( Vp )D ( Vnz ) D LBD + ΣkL b + kL D 2g 2g

= c fBDa

2

2

( Vp )D2 3 16 ( Vp ) D 4 2 70 ft b + 0.4 + 0.9 d ° ¢ + 0.05W ¶ 0.5>12 ft 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) = c 1680fBD + 14.1 d £

( Vp )D2 64.4

§

write the energy equations between A and C and A and D realizing that lb - 12 in 2 lb pA = a32 2 ba b = 4608 2 , pC = pD = patm = 0 and zA = zC = zD = z 1 ft in ft (same elevation). pC pA VA2 + + zA + hpump = + gw gw 2g 4608 lb>ft 2

( 1.937 slug>ft )( 32.2 ft>s ) 3

= 0 +

3 16 ( Vp ) C 4 2

2 ( 32.2 ft>s2 )

2

( Vnz ) C2 2g +

+ zC + hturb + (hL)AC VA2

2 ( 32.2 ft>s2 )

+ z + 0 + ( 1200fAB + 0.9 ) a

+ z + 0

( Vp )C VA2 b + ( 1200fBC + 14.6 ) ° ¢ 64.4 64.4 2

( 1200fAB - 0.1 ) VA2 + ( 1200fBC + 270.6 )( Vp ) C2 = 4757.87 pD pA VA2 + + zA + hpump = + gw gw 2g 4608 lb>ft 2

( 1.937 slug>ft 3 )( 32.2 ft>s2 ) = 0 +

3 16 ( Vp ) D 4 2

2 ( 32.2 ft>s2 )

( Vnz ) 2D 2g

+

(4)

+ zD + hturb + ( hL ) AD V 2A

2 ( 32.2 ft>s2 )

+ z + 0 + ( 1200fAB + 0.9 ) a

+ z + 0

( Vp )D VA2 b + ( 1680fBD + 14.1 ) £ § 64.4 64.4 2

( 1200fAB - 0.1 ) VA2 + ( 1680fBD + 270.1 )( Vp )D2 = 4757.87

(5)

Equating Eqs. (4) and (5)

( 1200fBC + 270.6 )( Vp )C2 = ( 1680fBD + 270.1 )( Vp )D2

(6)

0.000005 ft e = = 0.00012. Assuming that fAB = 0.025, fBC = 0.030 D 0.5>12 ft and fBD = 0.0305, and solve Eqs. (3), (5) and (6). We obtain VA = 6.633 ft>s, ( Vp ) C = 3.355 ft>s and ( Vp ) D = 3.277 ft>s. Then Eq. (1) gives (Re)AB = 2.66 ( 104 ) , (Re)BC = 1.34 ( 104 ) and (Re)BD = 1.31 ( 104 ) . The Moody diagram gives fAB = 0.02475, fBC = 0.0285 and fBD = 0.0285. Iterate with these values of f, For PVC pipe,

1104

10–79. Continued

we obtain VA = 6.655 ft>s, ( Vp ) C = 3.363 ft>s and ( Vp ) D = 3.292 ft>s . Then (Re)AB = 2.67 ( 104 ) , (Re)BC = 1.35 ( 104 ) and (Re)BD = 1.32 ( 104 ) with which the Moody diagram gives fAB = 0.02475 and fBC = fBD = 0.0285 which are the same as the previous values. Thus, VA = 6.655 ft>s,

( Vp ) C = 3.363 ft>s ( Vp ) D = 3.292 ft>s

Then the discharges at C and D are QC = ( Vp ) CA = ( 3.363 ft>s ) c p a

2 0.25 ft b d = 0.00459 ft 3 >s 12

QD = ( Vp ) DA = ( 3.292 ft>s ) c p a

2 0.25 ft b d = 0.00449 ft 3 >s 12

1105

Ans.

Ans.

Ans: QC = 0.00459 ft 3 >s QD = 0.00449 ft 3 >s

*10–80. When the globe valve is fully opened, water at 20°C is discharged at 0.003 m3 >s from C. Determine the pressure at A. The galvanized iron pipes AB and BC have diameters of 60 mm and 30 mm, respectively. Consider the minor losses only from the elbows and the globe valve.

5m A 4m B 2m

3m

3m 7m

SOLUTION C

Water is considered to be incompressible. The mean velocity of segments AB and BC of the pipe are VAB =

0.003 m3 >s Q = = 1.061 m>s AAB p(0.03 m)2

VBC =

0.003 m3 >s Q = = 4.244 m>s ABC p(0.015 m)2

From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s. Thus, the Reynolds numbers for the flow segments AB and BC are (Re)AB =

VABDAB = n

(Re)AB =

VBCDBC = v

( 1.061 m>s ) (0.06 m) = 6.37 ( 104 ) 1.00 ( 10-6 ) m2 >s

( 4.244 m>s ) (0.03 m) = 1.27 ( 105 ) 1.00 ( 10-6 ) m2 >s

The relative roughness segments AB and BC, which are made from galvanized iron, are a

e 0.15 mm b = = 0.0025 D AB 60 mm

a

e 0.15 mm b = = 0.005 D BC 30 mm

From the Moody diagram, fAB = 0.027 and fBC = 0.031. Since the pipe segments are in a series, the major head loss from A to C is the sum of the losses in segments AB and BC. Thus,

( hL ) ma = fAB

LBC VBC2 LAB VAB 2 + fBC DAB 2g DBC 2g

( 1.061 m>s ) ( 4.244 m>s ) 11 m 13 m b£ § + 0.031a b£ § 0.06 m 2 ( 9.81 m>s2 ) 0.03 m 2 ( 9.81 m>s2 ) 2

= 0.027a

2

= 12.62 m The minor head loss is contributed by the 90° elbows, kL = 0.9 and the fully-opened globe valve, kL = 10.0. Thus,

( hL ) mi = ΣkL

( 1.061 m>s ) 2 ( 4.244 m>s ) 2 VBC2 VAB2 + ΣkL = 2 ( 0.9 ) £ § + (10.0 + 0.9) £ § 2g 2g 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 )

= 10.11 m

1106

*10–80. Continued

Take the control volume as the water in the pipe from A to C. Applying the energy equation from A to C with the datum set at C, pA pC VC 2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g pA

( 998.3 kg>m3 )( 9.81 m>s2 )

+

( 1.061 m>s ) 2 ( 4.244 m>s ) 2 + 11 m + 0 = 0 + + 0 2 ( 9.81 m>s2 ) 2 ( 9.81m>s2 ) + 0 + 12.62 m + 10.11 m

pA = 123.8 ( 103 ) Pa = 124 kPa

Ans.

1107

10–81. The two water tanks are connected together using the 100-mm-diameter pipes. If the friction factor for each pipe is f = 0.024, determine the flow out of tank C when the valve at A is opened, while the valve at B remains closed. Neglect any minor losses.

C 7m

1m A 10 m

B

D 3m

E 8m

SOLUTION Water is considered to be incompressible. Using V the mean velocity of the flow along the pipe. hL = f

18 m V2 L V2 = 0.024 a b£ § = 0.22018V 2 D 2g 0.1 m 2 ( 9.81 m>s2 )

Take the control volume as the water in the reservoir and pipe from from C to D. Applying the energy equation from C to D with the datum along the pipe, pC pD VC 2 VD2 + zC + hpumb = + zD + hturb + hL + + g g 2g 2g 0 + 0 + 7 m + 0 = 0 + 0 + 3 m + 0 + 0.22018V 2 V = 4.262 m>s Thus, Q = VA = ( 4.262 m>s ) 3 p(0.05 m)2 4 = 0.0335 m3 >s

Ans.

1108

Ans: 0.0335 m3 >s

10–82. The two water tanks are connected together using the 100-mm-diameter galvanized iron pipes. Determine the flow out of tank C when both valves A and B are opened. Neglect any minor losses. Take nw = 1.00 1 10-6 2 m2 >s.

C 7m

1m A 10 m

SOLUTION Water is considered to be incompressible. Water in the pipes A, B, and E. Here, the diameter of the pipes is constant. Thus, 0 r dV + rV # dA = 0 0t L L cv

cs

p p p O - VA c D2 d + VB c D2 d + VE c D2 d = 0 4 4 4 (1)

VA = VB + VE The major head losses for the pipes are

( hL ) A = fA

( hL ) B = fB

( hL ) E = fE

LA VA2 VA2 10 m = fAa b£ § = 5.0968 fAVA2 D 2g 0.1 m 2 ( 9.81 m>s2 ) LB VB2 VB2 9m = fB a b£ § = 4.5872 fBVB2 D 2g 0.1 m 2 ( 9.81 m>s2 ) VE 2 LE VE2 8m = fE a b£ § = 4.0775 fEVE2 D 2g 0.1 m 2 ( 9.81 m>s2 )

Since B and E are parallel, the head loss in these two pipes must be the same.

( hL ) B = ( hL ) E 4.5872fBVB2 = 4.0775fEVE2 VB = 0.9428

fE V B fB E

(2)

Substituting Eq. (2) into (1), VA = a0.9428

fE + 1bVE B fB

(3)

Water in the reservoir and pipes from C to D.

Applying the energy equation from C to D with the datum along pipes A and E, pC pD VC2 VD2 + zC + hpump = + zD + hturb + hL + + g g 2g 2g 0 + 0 + 7 m + 0 = 0 + 0 + 3 m + 0 + 5.0968fAVA2 + 4.0775fEVE2 5.0968fAVA2 + 4.0775fEVE2 = 4

(4) 1109

B E 8m

D 3m

10–82. Continued

The Reynolds numbers for the flow in pipes A, B, and E are (Re)A =

(Re)B =

(Re)E =

VA(0.1 m) VAD = = 1 ( 105 ) VA n 1.00 ( 10-6 ) m2 >s

(5)

VB(0.1 m) VBD = = 1 ( 105 ) VB n 1.00 ( 10-6 ) m2 >s

(6)

VE(0.1 m) VED = = 1 ( 105 ) VE n 1.00 ( 10-6 ) m2 >s

(7)

For the galvanized iron pipe, e = 0.15 mm. Thus, the relative roughness for all the e 0.15 mm pipes is = = 0.0015. The iterations can be begun by assuming the D 100 mm values of fA, fB, and fE and solving for VA, VB, and VE using Eqs. (2), (3), and (4). The iterations are tabulated as follows: Pipe A: iteration

Assumed fA

VA ( m>s )

1

0.021

5.519

( Re ) A 5.52 ( 105 )

0.0215

5.477

5.48 ( 10

iteration

Assumed fB

VB ( m>s )

1

0.023

2.648

( Re ) B 2.65 ( 105 )

0.022

2.658

2.66 ( 10

iteration

Assumed fE

VE ( m>s )

1

0.022

2.871

( Re ) E 2.87 ( 105 )

2.819

2.82 ( 10

2

5

)

fA from Moody diagram 0.0215 0.0215

Pipe B:

2

5

)

fB from Moody diagram 0.022 0.022

Pipe E:

2

0.022

5

)

fE from Moody diagram 0.022 0.022

The values of f assumed in iteration (2) are almost the same as that given by the Moody diagram. Thus, the result VA = 5.477 m>s is acceptable. Then, QA = VA AA = (5.477 m>s) 3 p(0.05 m)2 4 = 0.0430 m3 >s

1110

Ans.

Ans: 0.0430 m3 >s

10–83. Water from the reservoir at A drains through the 30-mm-diameter pipe assembly. If commercial steel pipe is used, determine the initial discharge into B when the valve E is closed and F is opened. Neglect any minor losses. Take vw = 1.00 1 10-6 2 m2 >s.

3m

4m

A 1m

E

D F

3m

2m C

B

SOLUTION Water is considered to be incompressible. The Reynolds number of the flow in the pipe is Re =

V(0.03 m) VD = = 3 ( 104 ) V n 1.00 ( 10-6 ) m2 >s

(1)

The head loss from A to B can be determined using hL = f

6m V2 L V2 = fa b£ § = 10.194fV 2 D 2g 0.03 m 2 ( 9.81 m>s2 )

Take the control volume as the water from B to A. Applying the energy equation from A to B with the datum at, B, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 4 m + 0 = 0 + 0 + 0 + 0 + 10.194 fV 2 V =

0.6264

(2)

2f

For commercial steel pipe, the relative roughness is e 0.045 mm = = 0.0015 D 30 mm The iterations are as follows: iteration

Assumed f

V ( m>s ) ; Eq. (2)

Re Eq. (1)

f from Moody diagram

1

0.025

3.962

1.19 ( 10

)

0.0233

2

0.0233

4.104

1.23 ( 105 )

0.0233

5

The values of f assumed in iteration (2) is almost the same as that given by the Moody diagram. Thus, V = 4.104 m>s is an acceptable result. Then, Q = VA = ( 4.104 m>s ) 3 p(0.015m)2 4 = 0.00290 m3 >s

Ans.

Note: A more direct solution can be obtained from the Colebrook equation. If Eqs. (1) and (2) are substituted into the term 2.51> ( Re1f ) , the V will cancel so the log becomes constant.

1111

Ans: 0.00290 m3 >s

*10–84. Water from the reservoir at A drains through the 30-mm-diameter pipe assembly. If commercial steel pipe is used, determine the initial flow into pipe D from reservoir A when both valves E and F are fully opened. Neglect any minor losses. Take nw = 1.00 1 10-6 2 m2 >s.

3m

4m

A 1m

E

D F

3m B

SOLUTION Water is considered to be incompressible. Take the control volume as the water in the pipes D, E, and F. Here, the diameter of the pipes is constant. Thus, the continuity equation gives 0 r dV + rV # dA = 0 0t L L cv cs p p p O - VD c D2 d + VE c D2 d + VF c D2 d = 0 4 4 4 VD = VE + VF

(1)

The major head losses for the pipes are (hL)D = fD

(hL)E = fE

(hL)F = fF

LD VD2 VD2 3m = fDa b£ § = 5.0968 fDVD2 D 2g 0.03 m 2 ( 9.81 m>s2 ) LE VE2 VE2 6m = fE a b£ § = 10.194 fEVE2 D 2g 0.03 m 2 ( 9.81 m>s2 ) LF VF2 VF2 3m = fF a b£ § = 5.0968 fFVF2 D 2g 0.03 m 2 ( 9.81 m>s2 )

Take the control volume as the water in A and pipes DFB. Applying the energy equation from A to B with the datum at B, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 4 m + 0 = 0 + 0 + 0 + 0 + 5.0968 ( fDVD2 + fFVF2) fDVD2 + fFVF2 = 0.7848

(2)

Take the control volume as the water in system from A to C. Applying the energy equation from A to C with the datum at C, pA pC V C2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g 0 + 0 + 3m + 0 = 0 + 0 + 0 + 5.0968fDVD2 + 10.194fEVE2 fDV D2 + 2fEVE2 = 0.5886

(3)

1112

2m C

*10–84. Continued

The Reynolds number for the flow in pipes D, E, and F are (Re)D =

(Re)E =

(Re)F =

VD(0.03 m) VDD = = 3 ( 104 ) VD n 1.00 ( 10-6 ) m2 >s VE(0.03 m) VED = = 3 ( 104 ) VE n 1.00 ( 10-6 ) m2 >s VF(0.03 m) VFD = = 3 ( 104 ) VF n 1.00 ( 10-6 ) m2 >s

For commercial steel pipe, e = 0.045 mm. Thus, the relative roughness for all the e 0.045 mm pipes is = = 0.0015. The iterations can be begun by assuming the D 30 mm values of fD, fE, and fF and solving for VD, VE, and VF using Eqs. (1), (2), and (3). The iterations are tabulated as follows: Pipe D: iteration

Assumed fD

VD ( m>s )

( Re ) D

fD from Moody diagram

1

0.022

4.8022

1.44 ( 105 )

0.0228

0.0228

4.7082

1.41 ( 10

0.023

iteration

Assumed fE

VE ( m>s )

( Re ) E

fE from Moody diagram

1

0.023

1.3290

3.98 ( 105 )

0.0262

0.0262

1.2601

3.78 ( 10

0.0265

iteration

Assumed fF

VF ( m>s )

( Re ) F

fF from Moody diagram

1

0.023

3.4732

1.04 ( 105 )

0.0235

3.4481

1.03 ( 10

0.0235

2

5

)

Pipe E:

2

5

)

Pipe F:

2

0.0235

5

)

The values of f assumed in iteration (2) are almost the same as that given by the Moody diagram. Thus, the result VD = 4.7082 m>s is acceptable. Then, QD = VD AD = ( 4.7082 m>s ) 3 p(0.015 m)2 4 = 0.00333 m3 >s

1113

Ans.

10–85. Water at 20°C is pumped through the two commercial steel pipes having the lengths and diameters shown. If the pressure developed at A is 230 kPa, determine the discharge at C. Neglect minor losses.

120 mm A

10 m

B

60 mm

C

12 m

SOLUTION Water is considered to be incompressible. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s. Thus, the Reynolds numbers for the flow in segments AB and BC are (Re)AB =

(Re)BC =

VA(0.12 m) VADAB = = 1.2 ( 105 ) VA n 1.00 ( 10-6 ) m2 >s

(1)

VC(0.06 m) VCD = = 6 ( 104 ) VC n 1.00 ( 10-6 ) m2 >s

(2)

Take the control volume as the water in ABC. The continuity equation gives

0 r dV + rV # dA = 0 0t L L cv cs - VAAA + VCAC = 0 - VA c p(0.06 m)2 d + VC 3 p(0.03 m)2 4 = 0 VA = 0.25VC

(3)

Since segments AB and BC of the pipes are in a series, the major head loss from A to C is the sum of the losses for these for two segments. Thus, hL = fAB

LBC VC2 LAB VA2 = fBC DAB 2g DBC 2g

= fAB a

( 0.25VC ) 2 VC2 10 m 12 m b£ § + f a b £ § BC 0.12 m 2 ( 9.81 m>s2 ) 0.06 m 2 ( 9.81 m>s2 )

= ( 0.2655fAB + 10.1937fBC ) VC2 Since the pipe is horizontal, zA = zC = z. Applying the energy equation from A to C, pA pC V C2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g N ( 0.25VC ) 2 VC2 m2 + + z + 0 = 0 + + z + 0 + ( 0.2655fAB + 10.1937fBC ) VC2 3 2 2 ( 998.3 kg>m )( 9.81 m>s ) 2 ( 9.81 m>s ) 2 ( 9.81 m>s2 ) 230 ( 103 )

VC =

4.8462

(4)

20.2655fAB + 10.1937fBC + 0.04778 1114

10–85. Continued

For commercial steel pipe, the relative roughnesses of segments AB and BC are e e 0.045 mm 0.045 mm = = 0.00075. The iteration a b = = 0.000375 and a b D BC 60 mm D AB 120 mm are tabulated as follows: For section AB: iteration

Assumed fAB VC Eq. (4)

( Re ) AB Eq. (1) fAB from Moody diagram

1

0.019

9.565

2.87 ( 105 )

0.0175

2

0.0175

9.809

2.94 ( 105 )

0.0175

For section BC: iteration 1 2

Assumed fBC VA Eq. (4) 0.02 0.0188

( Re ) BC Eq. (1) fBC from Moody diagram

2.391

5.74 ( 105 )

0.0188

2.452

5.88 ( 10

0.0188

5

)

The values of f in the second iteration are almost the same as those given by the Moody diagram. Thus, the result VC = 9.809 m>s and VA = 2.452 m>s are acceptable. Thus, QC = VCAC = ( 9.809 m>s ) 3 p(0.03 m)2 4 = 0.0277 m3 >s

1115

Ans.

Ans: 0.0277 m3 >s

10–86. The horizontal galvanized iron pipe system is used for irrigation purposes and delivers water to two different outlets. If the pump delivers a flow of 0.01 m3 >s in the pipe at A, determine the discharge at each outlet, C and D. Neglect minor losses. Each pipe has a diameter of 30 mm. Also, what is the pressure at A?

A

40 m D

Assume that fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 998.3 kg>m3 and nw = 1.00 ( 10-6 ) m2 >s at T = 20° C. The average velocity in the pipe AB is 0.01 m3 >s Q = = 14.15 m>s A p(0.015 m)2

The Reynolds number for the pipes is Re = For pipe AB,

V(0.03 m) VD = = 30 ( 103 ) V nw 1.00 ( 10-6 ) m2 >s

(1)

(Re)AB = 30 ( 103 )( 14.15 m>s ) = 4.24 ( 105 ) 0.15 mm = 0.005. From the Moody diagram, 30 mm = 0.0305, Thus, the head loss from A to B is

For galvanized iron pipe e>D = fAB

(hL)A S B = fAB

( 14.15 m>s ) 2 LAB VA2 20 m b£ § = 207.42 m = 0.0305a D 2g 0.03 m 2 ( 9.81 m>s2 )

The head loss from B to C and B to D are (hL)B S C = fBC

(hL)B S D = fBD

VC2 VC2 LBC VC2 30 m = fBC a b£ § = ( 1000 fBC ) a b 2 D 2g 0.03 m 2 ( 9.81 m>s ) 19.62 LBC VD2 VD2 VD2 60 m = fBDa b£ § = ( 2000 fBD ) a b 2 D 2g 0.03 m 2 ( 9.81 m>s ) 19.62

write the energy equation between A and C and A and D, realizing that pC = pD = patm = 0, zA = zC = zD = z (same elevation) pA pC VC2 VA2 + zA + hpump = + zC + hturb + ( hL ) A S c + + gw gw 2g 2g pA

( 998.3 kg>m3 )( 9.81 m>s2 ) = 0 +

VC2 2 ( 9.81 m>s

2

)

+

(14.15 m>s)2 2 ( 9.81 m>s2 )

20 m

30 m

C

SOLUTION

VA =

20 m B

+ z + 0

+ z + 0 + 207.42 m + 1000 fBC a

2.0034 ( 10-3 ) pA = ( 1000 fBC + 1 ) VC2 + 3869.39

VC2 b 19.62

1116

(2)

10–86. Continued

SOLUTION pA pD VA 2 VD2 + + zA + hpump = + zD + hturb + (hL)A S D + gw g 2g 2g pA

( 998.3 kg>m )( 9.81 m>s ) 3

2

+

( 14.15 m>s ) 2 + z = 0 + 2 ( 9.81 m>s2 )

VD2 2 ( 9.81 m>s

2

2.0034 ( 10-3 ) pA = ( 2000fBD + 1 ) VD2 + 3869.39

)

+ z + 0 + 207.42 m + (2000fBD)a

VD2 b 19.62

(3)

The continuity condition requires that QA = QC + QD 0.01 m3 >s = VC 3 p(0.015 m)2 4 + VD 3 p(0.015 m)2 4 VC + VD = 14.147

(4)

Assume that fBC = 0.031, fBD = 0.032, and solve Eqs. (2), (3) and (4) We obtain, VC = 8.314 m>s, VD = 5.833 m>s and pA = 3035 kPa. Then Eq. (1) gives (Re)BC = 2.49 ( 105 ) and (Re)BD = 1.75 ( 105 ) . The Moody diagram gives fBC = 0.031 and fBD = 0.032, which are the same as the previous values. Thus, VC = 8.314 m>s

VD = 5.833 m>s Ans.

pA = 3035 kPa The discharges at outlets C and D are QC = VCA = ( 8.314 m>s ) 3 p(0.015 m)2 4 = 0.00588 m3 >s

QD = VDA = ( 5.833 m>s ) 3 p(0.015 m)2 4 = 0.00412 m3 >s

1117

Ans. Ans.

Ans: QC = 0.00588 m3 >s QD = 0.00412 m3 >s

10–87. Determine the discharge at each outlet, C and D, for the pipe network in Prob. 10–86 by considering the minor losses of the elbow and tee.

A

20 m B

20 m

30 m

C

40 m D

SOLUTION Assume that fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 998.3 kg>m3 and nw = 1.00 ( 10-6 ) m2 >s at T = 20° C. The average velocity in pipe AB is VA =

0.01 m3 >s Q = = 14.15 m>s A p(0.015 m)2

The Reynolds number for the pipes are Re = For pipe AB

V(0.03 m) VD = = 30 ( 103 ) V nw 1.00 ( 10-6 ) m2 >s

(1)

(Re)AB = 30 ( 103 )( 14.15 m>s ) = 4.24 ( 105 ) 0.15 mm e = = 0.005. From the Moody diagram, D 30 mm = 0.0305. Thus, the head loss from A to B is

For galvanized iron pipe, fAB

(hL)A S B = fAB

( 14.15 m>s ) 2 LAB VA2 20 m = 0.0305 a b£ § = 207.42 m D 2g 0.03 m 2 ( 9.81 m>s2 )

The loss coefficients for the tee along the branch, tee along the pipe run and the elbow are 1.8, 0.4, and 0.9, respectively. Thus, the head losses from B to C and B to D are (hL)B S C = afBC

( hL ) B S D = afBD

VC2 VC2 LBC 30 m + ΣkL b = c fBC a b + 1.8 d c d D 2g 0.03 m 2 ( 9.81 m>s2 ) = ( 1000fBC + 1.8 ) a

VC2 b 19.62

= ( 2000fBD + 1.3 ) a

VD2 b 19.62

LBD VD2 VD2 60 m + ΣkL b = c fBDa b + 0.4 + 0.9 d c d D 2g 0.03 m 2 ( 9.81 m>s2 )

Write the energy equation between A and C, A and D realizing that pC = pD = patm = 0, zA = zC = zD = z (same elevation) pC pA VC2 VA2 + + zA + hpump = + + zC + hturb + (hL)A S C gw gw 2g 2g pA

( 998.3 kg>m3 )( 9.81 m>s2 ) = 0 +

VC2 2 ( 9.81 m>s2 )

+

(14.15 m>s)2 2 ( 9.81 m>s2 )

+ z + 0

+ z + 207.42 m + (1000fBC + 1.8)a

2.0034 ( 10-3 ) pA = ( 1000fBC + 2.8 ) VC2 + 3869.39

1118

VC2 b 19.62

(2)

10–87. Continued

pA pD VA2 VD2 + + + zA + hpump = + zD + hturb + (hL)A S D gw gw 2g 2g pA

( 998.3 kg>m3 )( 9.81 m>s2 ) = 0 +

VD2 2 ( 9.81 m>s

2

)

+

( 14.15 m>s ) 2 + z 2 ( 9.81 m>s2 )

+ z + 0 + 207.42 m + (2000fBD + 1.3)a

2.0034 ( 10-3 ) pA = ( 2000fBD + 2.3 ) VD2 + 3869.39

VD2 b 19.62

(3)

The continuity condition requires that QA = QC + QD 0.01 m3 >s = VC 3 p(0.015 m)2 4 + VD 3 p(0.015 m)2 4 VC + VD = 14.147

(4)

Assume that fBC = 0.031, fBD = 0.032, and solve Eqs. (2), (3) and (4). We obtain VC = 8.254 m>s, VD = 5.893 m>s and pA = 3080.77 kPa. Then Eq. (1) gives (Re)BC = 2.48 ( 105 ) and (Re)BD = 1.77 ( 105 ) . The Moody diagram gives fBC = 0.031 and fBD = 0.032, which are the same as the previous values. Thus VC = 8.254 m>s

VD = 5.893 m>s Ans.

pA = 3081 kPa The discharge at outlet C and D are QC = VCA = ( 8.254 m>s ) 3 p(0.015 m)2 4 = 0.00583 m3 >s

QD = VDA = ( 5.893 m>s ) 3 p(0.015 m)2 4 = 0.00417 m3 >s

1119

Ans. Ans.

Ans: QC = 0.00583 m3 >s QD = 0.00417 m3 >s

*10–88. Water at 70°F in the container at A is dispensed into the buckets at B and C using the 0.5-in.-diameter commercial steel pipe assembly. Determine the initial discharge at B and C. Neglect any minor losses.

A

3 ft D 1 ft

1 ft

1 ft

2 ft

SOLUTION

C

From Appendix A, r = 1.937 slug>ft and n = 10.4 ( 10 ) ft >s for water at 70° F. Then, the Reynolds numbers for the flow in pipes B and C are -6

3

2

0.5 VB a ft b VBD 12 (Re)B = = = 4006.41VB n 10.4 ( 10-6 ) ft 2 >s 0.5 VC a ft b VCD 12 (Re)C = = = 4006.41VC n 10.4 ( 10-6 ) ft 2 >s

(1)

(2)

The major head loss in pipes B and C

3 ( hL ) ma 4 B

= fB

3 ( hL ) ma 4 C

= fC

VB2 LB VB2 2 ft = fB ± ≤£ § = 0.7453fBVB2 D 2g 0.5 2 ( 32.2 ft>s2 ) ft 12 VC2 LC VC2 4 ft § = 1.4907fCVC2 ≤£ = fC ± D 2g 0.5 2 ( 32.2 ft>s2 ) ft 12

Take the control volume as the water in the reservoir at A and pipe to B. Applying the energy equation from A to B with the datum at B, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 4 ft + 0 = 0 + VB =

VB2 2 ( 32.2 ft>s2 )

+ 0 + 0 + 0.7453fBVB2

2

(3)

20.7453fB + 0.01553

Take the control volume as the water in the reservoir at A and pipe to C. Applying the energy equation from A to B with the datum at C, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g 0 + 0 + 6 ft + 0 = 0 + VC =

VC2 2 ( 32.2 ft>s2 )

+ 0 + 0 + 1.4907fCVC2

6 A 1.4907fC + 0.01553

(4)

1120

0.5 in.

0.5 in. B

10–88. Continued

For commercial steel pipe,

e 0.00015 ft = = 0.0036. The iterations are tabulated D 0.5 ft 12

as follows:

For pipe B: iteration assumed fB VB ( m>s ) ; Eq. (3)

( Re ) B Eq. (1) fB from Moody diagram

1

0.032

10.079

4.04 ( 104 )

0.03

2

0.03

10.275

4.12 ( 104 )

0.03

For pipe C: iteration assumed fC VC ( m>s ) ; Eq. (4) 1 2

0.03 0.03

( Re ) C Eq. (2) fC from Moody diagram

9.979

4.00 ( 104 )

0.03

9.979

4.00 ( 10

0.03

4

)

The assumed values of f in the second iteration are almost the same as those given by the Moody diagram. Thus, the results VB = 10.275 ft>s and VC = 9.979 ft>s are acceptable. Then, QB = VBAB = ( 10.275 ft>s ) c p a

QC = VCAC = ( 9.979 ft>s ) c p a

2 0.25 ft b d = 0.0140 ft 3 >s 12

2 0.25 ft b d = 0.0136 ft 3 >s 12

1121

Ans. Ans.

10–89. Water at 70°F in the container at A is dispensed into the buckets at B and C using the 0.5-in.-diameter commercial steel pipe assembly. Determine the initial discharge at B and C. Include the minor losses at the tee and elbows and at the flush entrance to the pipe at D.

A

3 ft D 1 ft

1 ft

1 ft

2 ft

SOLUTION

C

Water is considered to be incompressible. From Appendix A, r = 1.937 slug>ft 3 and n = 10.4 ( 10-6 ) ft 2 >s for water at 70° F. Then, the Reynolds number for the flow in pipes B and C are (Re)B =

VB a

0.5 ft b 12

VBD = = 4006.41VB n 10.4 ( 10-6 ) ft 2 >s

0.5 VC a ft b VCD 12 (Re)C = = = 4006.41VC n 10.4 ( 10-6 ) ft 2 >s

(1)

(2)

The major head losses in pipes B and C are

3 ( hL ) ma 4 B

= fB

3 ( hL ) ma 4 C

= fC

VB2 LB VB2 2 ft = fB ± ≤£ § = 0.7453fBVB2 D 2g 0.5 2 ( 32.2 ft>s2 ) ft 12 VC2 LC VC2 4 ft ≤£ § = 1.4907fCVC2 = fC ± D 2g 0.5 2 ( 32.2 ft>s2 ) ft 12

The minor head loss for both pipes is contributed by the flush entrance, KL = 0.5, the tee along the branch, KL = 1.8, and one 90° elbow, KL = 0.9. Thus,

3 ( hL ) mi 4 B

= ΣKL

3 ( hL ) mi 4 C

= ΣKL

VB2 VB2 § = 0.04969VB2 = (0.5 + 1.8 + 0.9) £ 2g 2 ( 32.2 ft>s2 ) VC2 VC2 § = 0.04969VC2 = (0.5 + 1.8 + 0.9) £ 2g 2 ( 32.2 ft>s2 )

Take the control volume as the water in reservoir and pipe to B.

Applying the energy equation from A to B with the datum at B, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 0 + 0 + 4 ft + 0 = 0 +

VB2 2 ( 32.2 ft>s2 )

+ 0 + ( 0.7453 fBVB2 + 0.04969VB2 )

( 0.7453 fB + 0.06522 ) VB2 = 4 ft VB =

2

(3)

20.7453fB + 0.06522

1122

0.5 in.

0.5 in. B

10–89. Continued

Take the control volume as the water in reservoir and pipe to C. Applying the energy equation from A to B with the datum at C, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + g g 2g 2g 0 + 0 + 6 ft + 0 = 0 +

VC2 2 ( 32.2 ft>s2 )

+ 0 + ( 1.4907 fCVC2 + 0.04969VC2 )

( 1.4907fC + 0.06522 ) VC2 = 6 ft VC = For commercial steel pipe,

6 A 1.4907fC + 0.06522

(4)

0.00015 ft e = = 0.0036. The iterations are tabulated D 0.5 ft 12

as follows: For pipe B: iteration assumed fB VB ( m>s ) ; Eq. (3)

( Re ) B Eq. (1) fB from Moody diagram

1

0.03

6.758

2.71 ( 104 )

0.0315

2

0.0315

6.715

2.69 ( 104 )

0.0315

For pipe C: iteration assumed fC VC ( m>s ) ; Eq. (4) 1 2

0.026 0.031

( Re ) C Eq. (2) fC from Moody diagram

7.596

3.04 ( 104 )

0.031

7.338

2.94 ( 10

0.031

4

)

The assumed values of f in the second iteration are almost the same as those given by the Moody diagram. Thus, the results VB = 6.715 ft>s and VC = 7.338 ft>s are acceptable. Then, QB = VBAB = ( 6.715 ft>s ) c pa

QC = VCAC = ( 7.338 ft>s ) c pa

2 0.25 ft b d = 0.00916 ft 3 >s 12 2 0.25 ft b d = 0.0100 ft 3 >s 12

1123

Ans. Ans.

Ans: QB = 0.00916 ft 3 >s QC = 0.0100 ft 3 >s

10–90. The two galvanized iron pipes branch to form the loop. Branch CAD is 200 ft long, and branch CBD is 100 ft long. What horsepower pump should be used in branch CAD if an equal flow of 70°F water occurs through each branch? All pipes have a diameter of 3 in. The lines are all horizontal. Neglect minor losses.

A

10 ft/s 3 in.

C

SOLUTION

D

Water is considered to be incompressible. The discharge at pipe C is

B

QC = VCAC = ( 10 ft>s ) £ pa

2

1.5 ft b § = 0.4909 ft 2 >s 12

Since the discharge in pipes A and B is required to be the same, the continuity condition gives QA = QB =

0.4909 ft 2 >s QC = = 0.2454 ft 3 >s 2 2

Thus, the mean velocity in these two pipes is VA = VB = V =

0.2454 ft 3 >s QB = 5 ft>s = 2 A 1.5 pa ft b 12

From Appendix A, r = 1.937 slug>ft 3 and n = 10.4 ( 10-6 ) ft 2 >s . Then, the Reynolds number of the flow in these two pipes is 3 ( 5 ft>s ) a ft b 12 VD (Re)A = (Re)B = = = 1.20 ( 105 ) n 10.4 ( 10-6 ) ft 2 >s

For the galvanized iron pipe,

e 0.0005 ft = = 0.002. D 3 ft 12

From the Moody diagram fA = fB = 0.0245. Then, the major head losses in pipes A and B are

( hL ) A = fA

( hL ) B = fB

( 5 ft>s ) 2 LA V 2 200 ft = 0.0245≥ ¥£ § = 7.6087 ft D 2g 3 2 ( 32.2 ft>s2 ) a ft b 12 (5 ft>s)2 LB V 2 100 ft § = 3.8043 ft = 0.0245≥ ¥£ D 2g 3 2 ( 32.2 ft>s2 ) a ft b 12

Since gravity and minus losses do not play a role, the head loss in these two pipes must be the same. Then, the excessive head loss in pipe A must be overcome by the pump. Thus, hp = ( hL ) A - ( hL ) B = 7.6087 ft - 3.8043 ft = 3.8043 ft

1124

3 in.

10–90. Continued

Applying

# ws = gQhp = ( 1.937 slug>ft 3 )( 32.2 ft>s2 )( 0.2454 ft 3 >s ) (3.8043 ft) = a58.24

lb # ft b± s

1 hp

lb # ft 550 s

≤ = 0.106 hp

Ans.

Ans: 0.106 hp 1125

10–91. Water at 60°F flows into the 2-in.-diameter galvanized iron pipe at 0.3 ft 3 >s. If the pipe branches into two horizontal 1-in.-diameter pipes ABD and ACD, which are 4 ft long and 6 ft long, respectively, determine the flow through each pipe in gal>min. Neglect minor losses.

1 in. B

SOLUTION

1 in.

Water is considered to be incompressible. Take the control volume as the water in pipes A, B, and C. The continuity equation gives 0 r dV + rV # dA = 0 0t L L cv cs 0 - VAAA + VBAB + VCAC = 0 -0.3 ft 3 >s + VB £ pa

2 2 0.5 0.5 ft b § + VC £ pa ft b § = 0 12 12

(1)

VB + VC = 55.0039

From Appendix A, r = 1.939 slug>ft 3 and n = 12.2 ( 10-6 ) ft 2 >s . Thus, the Reynolds numbers for the flow in pipes B and C are 1 VB a ft b VBDB 12 (Re)B = = = 6830.60VB n 12.2 ( 10-6 ) ft 2 >s 1 VC a ft b VCDC 12 (Re)C = = = 6830.60VC n 12.2 ( 10-6 ) ft 2 >s

(2)

(3)

The major head losses in pipes B and C are

( hL ) B = fB

( hL ) C = fC

LB VB2 = fB ≥ D 2g

VB2 4 ft ¥£ § = 0.7453fBVB2 1 2 ( 32.2 ft>s2 ) a ft b 12

LC VC2 = fC ≥ D 2g

VC2 6 ft § = 1.1180fCVC2 ¥£ 1 2 ( 32.2 ft>s2 ) a ft b 12

Since pipes A and B are parallel, their head loss must be the same.

( hL ) B = ( hL ) C 0.7453 fBVB2 = 1.1180 fCVC2 VB = 1.2247

D

A

fC V B fB C

(4)

1126

C

10–91. Continued

For galvanized iron pipe, the relative roughnesses of pipes B and C are a =

e e b = a b D B D C

0.0005 ft = 0.006. First, we need to assume fB and fC and then solve for VB and VC 1 ft 12

using Eqs. (1) and (4). The iterations are tabulated as follows: For pipe B: iteration 1

assumed fB

VB ( ft>s )

0.032

30.28

assumed fB

VC ( ft>s )

( Re ) B Eq. (2) 2.07 ( 10

5

)

fB from Moody diagram 0.032

For pipe C: iteration 1

0.032

24.72

( Re ) C Eq. (2) 1.69 ( 10

5

)

fC from Moody diagram 0.032

The assumed values of f are almost the same as those given by the Moody diagram. Thus, the result VB = 30.28 ft>s and VC = 24.72 ft>s are acceptable. Thus,

QB = VBAB = (30.28 ft>s) £ p a

2 0.5 ft b § = 0.165 ft 3 >s = 74.1 gal>min. 12

QC = VCAC = (24.72 ft>s) £ p a

2 0.5 ft b § = 0.135 ft 3 >s = 60.6 gal>min. 12

Ans.

Ans.

Ans: QB = 74.1 gal>min QC = 60.6 gal>min 1127

*10–92. Water at 60°F flows into the 2-in.-diameter galvanized iron pipe at 0.3 ft 3 >s. If the pipe branches into two horizontal 1-in.-diameter pipes ABD and ACD, which are 4 ft long and 6 ft long, respectively, determine the pressure drop that occurs across each branch from A to D. Neglect minor losses.

1 in. B

SOLUTION

1 in.

Water is considered to be incompressible. Take the control volume as the water in pipe A, B, C. The continuity equation gives 0 r dV + rV # dA = 0 0t L L cv

cs

0 - VAAA + VBAB + VCAC = 0 - 0.3 ft 3 >s + VB £ pa

2 2 0.5 0.5 ft b § + VC £ pa ft b § = 0 12 12

(1)

VB + VC = 55.0039

From Appendix A, r = 1.939 slug>ft 3 and n = 12.2 ( 10-6 ) ft 2 >s . Thus, the Reynolds number for the flow in pipes B and C are 1 VB a ft b VBDB 12 (Re)B = = = 6830.60VB n 12.2 ( 10-6 ) ft 2 >s

1 VC a ft b VCDC 12 (Re)C = = = 6830.60VC n 12.2 ( 10-6 ) ft 2 >s

(2)

(3)

The major head loss in pipes B and C are

( hL ) B = fB

( hL ) C = fC

LB VB2 VB2 4 ft = fB £ §≥ ¥ = 0.7453fBVB2 D 2g ( 1>12 ft ) 2 ( 32.2 ft>s2 ) LC VC2 VC2 6 ft = fC £ §≥ ¥ = 1.1180fCVC2 D 2g ( 1>12 ft ) 2 ( 32.2 ft>s2 )

Since pipes A and B are parallel, their head loss must be the same.

( hL ) B = ( hL ) C 0.7453 fBVB2 = 1.1180 fCVC2 VB = 1.2247

D

A

fC V A fB C

(4)

1128

C

*10–92. Continued

For galvanized iron pipe, the relative roughness of pipes B and C are a =

e e b = a b D B D C

0.0005 ft = 0.006. First, we need to assume fB and fC and then solve for VB and VC 1 ft 12

using Eqs. (1) and (4). The iterations are tabulated as follows: For pipe B: iteration

VB ( ft>s )

assumed fB

1

0.032

30.28

assumed fB

VC ( ft>s )

( Re ) B Eq. (2) 2.07 ( 10

5

fB from Moody diagram

)

0.032

For pipe C: iteration 1

0.032

( Re ) C Eq. (2) 1.69 ( 10

5

24.72

fC from Moody diagram

)

0.032

The assumed values of f are almost the same as those given by the Moody diagram. Thus, the results VB = 30.28 ft>s and VC = 24.72 ft>s are acceptable. Thus, the head loss across either branch (pipe B or C) is

( hL ) B = 0.7453fBVB2 = 0.7453(0.032) ( 30.28 ft>s ) 2 = 21.87 ft Applying the energy equation from A to D with zA = zD = z and VA = VD = V, pA pD VD2 VA2 + zD = + zA - ( hs ) out - hL + + g g 2g 2g pD

( 1.939 slug>ft )( 32.2 ft>s ) 3

2

+

V2 + z = 2g

pA

( 1.939 slug>ft )( 32.2 ft>s ) 3

2

+

V2 + z - 0 - 21.87 ft 2g

pA - pD = ( 1.939 slug>ft 3 )( 32.2 ft>s2 ) (21.87 ft) = a1365.39

lb 1 ft 2 ba b = 9.48 psi ft 2 12 in.

Ans.

1129

10–93. The copper pipe system, which transports water at 70°F, consists of two branches. Branch ABC has a diameter of 0.5 in. and length of 8 ft, whereas branch ADC has a diameter of 1 in. and length of 30 ft. If a pump provides an  inlet flow at A of 67.3 gal>min, determine the flow in gal>min through each branch. Take e = 80 1 10 - 6 2 ft. The system is in the horizontal plane. Include minor losses of the elbows and tees. The diameters at A and C are the same.

B A

SOLUTION Assume that fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 1.937 slug>ft 3 and nw = 10.4 ( 10-6 ) ft 2 >s at T = 70° F. The Reynolds numbers for the flow in the pipes are (Re)ABC

(Re)ADC

0.5 VABC a ft b VABCDABC 12 = = = 4.006 ( 103 ) VABC nw 10.4 ( 10-6 ) ft 2 >s

1 VADC a ft b VADCDADC 12 = = = 8.013 ( 103 ) VADC nw 10.4 ( 10-6 ) ft 2 >s

(1)

(2)

The loss coefficients for the elbow, and the tee along the branch, and the are 0.9 and 1.8 respectively. Thus, the head losses along pipes ABC and ADC are

( hL ) ABC = afABC

LABC V 2ABC + ΣkL b DABC 2g

= ≥ fABC ±

V 2ABC 8 ft § ≤ + 2(1.8) + 2(0.9) ¥ £ 0.5 2 ( 32.2 ft>s2 ) ft 12

= ( 192fABC + 5.4 ) a

( hL ) ADC = afADC

V 2ABC b 64.4

LADC V 2ADC + ΣkL b DADC 2g

= ≥ fADC ±

VADC 2 30 ft § ≤ + 2(1.8) + 4(0.9) ¥ £ 1 2 ( 32.2 ft>s2 ) ft 12 = ( 360fADC + 7.2 ) a

It is required that

VADC2 b 64.4

( hL ) ABC = ( hL ) ADC ( 192fABC + 5.4 ) V 2ABC = ( 360fADC + 7.2 )

(3)

The continuity condition at junction A require that QABC + QADC = ( 67.3 gal>min. )( 1 min.>605 )( 1 ft 3 >7.48 gal ) VABC £ pa

2 2 0.25 0.5 ft b § + VADC £ pa ft b § = 0.15 12 12

VABC + 4VADC =

0.5 in.

345.5 p

(4) 1130

C

1 in.

D

10–93. Continued

For copper pipe a

e 0.00008 ft e 0.00008 ft b = = 0.00192 and a b = D ABC 0.5>12 ft D ADC 1>12 ft

= 0.00096. Assuming that fABC = 0.025 and fADC = 0.0215, Solve Eqs. (3) and (4), we obtain VABC = 25.553 ft>s and VADC = 21.114 ft>s. Then Eqs. (1) and (2) gives (Re)ABC = 1.02 ( 105 ) and (Re)ADC = 1.69 ( 105 ) . The Moody Diagram gives fABC = 0.025 and fADC = 0.0215, which are the same as previous values. Thus VABC = 25.553 ft>s

VADC = 21.114 ft>s

The discharges are QABC = VABC AABC = ( 25.553 ft>s ) £ pa

2 0.25 ft b § = 0.0348 ft 3 >s 12

= 15.6 gal>min.

QADC = VADC AADC = ( 21.114 ft>s ) £ pa = 51.7 gal>min.

2 0.5 ft b § = 0.115 ft 3 >s 12

Ans.

Ans.

Ans: QABC = 15.6 gal>min QADC = 51.7 gal>min 1131

10–94. If the pressure at A is 60 psi and at C it is 15 psi, determine the flow in gal>min through each branch of the pipe system described in Prob. 10–93. Include minor losses of the elbow and tees. The diameter at A and C are the same.

B A

SOLUTION Assume that fully developed steady flow occurs, and water is incompressible. Appendix A give rw = 1.937 slug>ft 3 and nw = 10.4 ( 10-6 ) ft 2 >s at T = 70° F. The Reynolds numbers for the flow in the pipes are (Re)ABC

(Re)ADC

0.5 VABC a ft b VABCDABC 12 = = = 4.006 ( 103 ) VABC nw 10.4 ( 10-6 ) ft 2 >s

1 VADC a ft b VADCDADC 12 = = = 8.013 ( 103 ) VADC nw 10.4 ( 10-6 ) ft 2 >s

(1)

(2)

The loss coefficient for the elbow and the tee along the branch, are 0.9 and 1.8 respectively. Thus, the head losses along pipes ABC and ADC are

( hL ) ABC = afABC

V 2ABC LABC + ΣkL b DABC 2g

= ≥ fABC ±

V 2ABC 8 ft § ≤ + 2(1.8) + 2(0.9) ¥ £ 0.5 2 ( 32.2 ft>s2 ) ft 12

= ( 192fABC + 5.4 ) a

( hL ) ADC = afADC

V 2ABC b 64.4

LADC V 2ADC + ΣkL b DADC 2g

= ≥ fADC ±

V 2ADC 30 ft § ≤ + 2(1.8) + 4(0.9) ¥ £ 1 2 ( 32.2 ft>s2 ) ft 12

= ( 360fADC + 7.2 ) a

V 2ADC b 64.4

Write the energy between A and C realizing that pA = a60 pB = a15

lb 12 in 2 lb ba b = 8640 2 , 2 1 ft in ft

lb 12 in 2 lb ba b = 2160 2 , VA = VC (Same discharge and diameter), 1 ft in2 ft

zA = zC (same elevation),

pA pC VC2 VA2 + + zA + hpump = + + zC + hturb + hL gw gw 2g 2g pA - pC hL = = gw

0.5 in.

lb lb - 2160 2 2 ft ft = 103.89 ft ( 1.937 slug>ft 3 )( 32.2 ft>s2 ) 8640

1132

C

1 in.

D

10–94. Continued

It is required that

( hL ) ABC = ( hL ) ADC = hL Thus

( 192fABC + 5.4 ) a

( 360fADC + 7.2 ) a

For copper pipe a

V 2ABC b = 103.89 ft 64.4

(3)

V 2ADC b = 103.89 ft 64.4

(4)

e 0.00008 ft e 0.00008 ft b = = 0.00192 and a b = D ABC 0.5>12 ft D ADC 1>12 ft

= 0.00096. Assuming that fABC = 0.025 and fADC = 0.0215, then Eqs. (3) and (4) give VABC = 25.612 ft>s and VADC = 21.162 ft>s. It follows that Eqs. (1) and (2) give (Re)ABC = 1.03 ( 105 ) and (Re)ADC = 1.70 ( 105 ) . The Moody diagram gives fABC = 0.025 and fADC = 0.0215, which are the same as the previous values. Thus VABC = 25.612 ft>s

VADC = 21.162 ft>s

The discharges are QABC = VABC AABC = (25.612 ft>s) £ pa

2 0.25 ft b § = 0.0349 ft 3 >s 12

= 15.7 gal>min.

QADC = VADC AADC = (21.162 ft>s) £ pa = 51.8 gal>min.

2 0.5 ft b § = 0.115 ft 3 >s 12

Ans.

Ans.

Ans: QABC = 15.7 gal>min QADC = 51.8 gal>min 1133

11–1. Oil flows with a free-stream velocity of U = 3 ft>s over the flat plate. Determine the distance xcr to where the  boundary layer begins to transition from laminar to  turbulent flow. Take mo = 1.40 1 10-3 2 lb # s>ft and go = 55.1 lb>ft 3.

U

x

SOLUTION Oil is considered to be incompressible. The flow is steady. The transition from a laminar boundary layer occurs at a critical Reynolds number of (Re x) cr = 5 (105) .

(Re x) cr = 5 ( 105 ) =

gUxcr Uxcr Uxcr = = n mg>g mg

( 55.1 lb>ft 3 )( 3 ft>s ) xcr 3 1.40 ( 10-3 ) lb # s>ft 4 ( 32.2 ft>s2 )

Ans.

xcr = 136.36 ft = 136 ft

Ans: 136 ft 1134

11–2. Water at 15°C flows with a free-stream velocity of U = 2 m>s over the flat plate. Determine the shear stress on the surface of the plate at point A.

U

A 200 mm

SOLUTION Water is considered to be incompressible. The flow is steady.

From Appendix A, m = 1.15 ( 10-3 ) N # s>m2 and n = 1.15 ( 10-6 ) m2 >s. Thus, Re x =

( 2 m>s ) (0.2 m) Ux = 3.478 ( 105 ) = n 1.15 ( 10-6 ) m2 >s

Since Re x 6 ( Re x ) cr = 5 ( 105 ) , the boundary layer is still laminar. Thus, t0 = 0.332m a

U b 2Re x x

= 0.332 3 1.15 ( 10-3 ) N>m2 4 a

= 2.25 N>m2

2 m>s 0.2 m

b 23.478 ( 105 )

Ans.

Ans: 2.25 Pa 1135

11–3. The boundary layer for wind blowing over rough  terrain can be approximated by the equation u>U = 1 y>( y + 0.01) 2 , where y is in meters. If the freestream velocity of the wind is 15 m>s, determine the velocity at an elevation y  = 0.1 m and at y = 0.3 m from the ground surface.

15 m/s

y

SOLUTION Air is considered to be incompressible. The flow is steady. Here, y u = U y + 0.01 u = Ua

At y = 0.1 m,

y 15y b = a b m>s y + 0.01 y + 0.01

u " y = 0.1 m =

15(0.1) 0.1 + 0.01

= 13.6 m>s

Ans.

= 14.5 m>s

Ans.

At y = 0.3 m, u " y = 0.3 m =

15(0.3) 0.3 + 0.01

Ans: u " y = 0.1 m = 13.6 m>s u " y = 0.3 m = 14.5 m>s 1136

*11–4. An oil–gas mixture flows over the top surface of the plate that is contained in a separator used to process these two fluids. If the free-stream velocity is 0.8 m>s, determine the maximum boundary layer thickness over the plate’s surface. Take n = 42 1 10-6 2 m2 >s.

0.75 m

1.5 m

0.8 m/s

SOLUTION The mixture is considered to be incompressible. The flow is steady. The transition from a laminar boundary layer occurs at a critical Reynolds number of ( Re x ) cr = 5 ( 105 ) .

( Re x ) cr = 5 ( 105 ) =

Uxcr n

( 0.8 m>s ) xcr 42 ( 10-6 ) m2 >s

xcr = 26.25 m

Since xcr 7 L = 1.5 m, the boundary layer for the entire length of the plate is laminar. The maximum thickness occurs at the end of the plate where x = L = 1.5 m. The Reynolds number at this point is Re x = Then,

Ux = n

( 0.8 m>s ) (1.5 m) = 2.857 ( 104 ) 42 ( 10-6 ) m2 >s

dmax = d " x = 1.5 m =

5.0x 2Re x

=

5.0(1.5 m) 22.857 ( 104 )

= 0.04437 m = 44.4 mm

1137

Ans.

11–5. An oil-gas mixture flows over the top surface of the  plate that is contained in a separator used to process these two fluids. If the free-stream velocity is 0.8 m>s, determine the friction drag acting on the surface of the plate. Take n = 42 1 10-6 2 m2 >s and r = 910 kg>m3.

0.75 m

1.5 m

0.8 m/s

SOLUTION The mixture is considered to be incompressible. The flow is steady. The transition from a laminar boundary layer occurs at a critical Reynolds number of ( Re x ) cr = 5 ( 105 ) .

( Re x ) cr = 5 ( 105 ) =

Uxcr n

( 0.8 m>s ) xcr 42 ( 10-6 ) m2 >s

xcr = 26.25 m

Since xcr 7 L = 1.5 m, the boundary layer for the entire length of the plate is laminar. Here, the Reynolds number at x = L = 1.5 m is Re L =

UL = n

Then, FD =

( 0.8 m>s ) (1.5 m) = 2.857 ( 104 ) 42 ( 10-6 ) m2 >s

0.664brU 2L 2Re L

= 2.57 N

=

0.664(0.75 m) ( 910 kg>m3 )( 0.8 m>s ) 2(1.5 m) 22.857 ( 104 )

Ans.

Ans: 2.57 N 1138

11–6. Wind flows along the side of the rectangular sign. If the air is at a temperature of 60°F and has a free-stream velocity of 6 ft>s, determine the friction drag on the front surface of the sign.

12 ft

6 ft/s

6 ft

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 0.00237 slug>ft 2 and n = 0.158 ( 10-3 ) ft 2 >s. The transition from a laminar boundary layer occurs at a critical Reynolds number of ( Re x ) cr = 5 ( 105 ) .

( Re x ) cr = 5 ( 105 ) =

Uxcr n

( 6 ft>s ) x cr 0.158 ( 10-3 ) ft 2 >s

xcr = 13.17 ft

Since xcr 7 L = 12 ft, the boundary layer for the entire length of the signboard is laminar. Here, the Reynolds number at x = L = 12 ft is Re L = Then, FD =

( 6 ft>s ) (12 ft) UL = = 4.557 ( 105 ) n 0.158 ( 10-3 ) ft 2 >s

0.664brU 2L 2Re L

= 0.00604 lb

=

0.664(6 ft) ( 0.00237 slug>ft 3 )( 6 ft>s ) 2(12 ft) 24.557 ( 105 )

Ans.

Ans: 0.00604 lb 1139

11–7. A flat plate is to be coated with a polymer. If the thickness of the laminar boundary layer that occurs during the coating process at a distance of 0.5 m from the plate’s front edge is 10 mm, determine the free-stream velocity of this fluid. Take n = 4.68 1 10-6 2 m2 >s.

SOLUTION The polymer is considered to be incompressible. The flow is steady. The Reynolds number at x = 0.5 m is Re x = Then, d =

U(0.5 m) Ux = = 1.0684 ( 105 ) U n 4.68 ( 10-6 ) m2 >s

5.0x 2Re x

0.01 m =

5.0(0.5 m)

21.0684 ( 105 ) U

Ans.

U = 0.585 m>s

Ans: 0.585 m>s 1140

*11–8. Compare the thickness of the boundary layer of water with air at the end of the 0.4-m-long flat plate. Both fluids are at 20°C and have a free-stream velocity of U = 0.8 m>s.

0.8 m/s

0.8 m/s

y

0.4 m

SOLUTION Both water and air are to be incompressible. The flow is steady. From Appendix A, nw = 1.00 ( 10 - 6 ) m2 > s and na = 15.1 ( 10 - 6 ) m2 >s. Thus, the Reynolds numbers for water and air at the end of the plate x = L = 0.4 m are (Re L)w = (Re L)a =

( 0.8 m>s ) (0.4 m) UL = = 3.2 ( 105 ) nw 1.00 ( 10 - 6 ) m2 >s

( 0.8 m>s ) (0.4 m) UL = = 2.1192 ( 104 ) na 15.1 ( 10 - 6 ) m2 >s

Since (Re L)w 6 (Re x)cr and (Re L)a 6 (Re x)cr, where (Re x)cr = 5 ( 105 ) , the boundary layers for both water and air are laminar. For x = L = 0.4 m, dw = da =

5.0x 2(Re x)w 5.0x

2(Re x)a

= =

5.0(0.4 m) 23.2 ( 105 )

= 0.003536 m = 3.54 mm

5.0(0.4 m)

22.1192 ( 104 )

= 0.01374 m = 13.7 mm

1141

Ans.

Ans.

11–9. A liquid having a viscosity m, a density r, and a freestream velocity U flows over the plate. Determine the distance x where the boundary layer has a disturbance thickness that is one-half the depth a of the liquid. Assume laminar flow.

U a

x

SOLUTION The liquid is considered to be incompressible. The flow is steady. rUL a The Reynolds number at the x = L is Re L = and at the exit d = . Thus, m 2 d =

5.0x 2Re x

;

a = 2

x =

5.0x rUx B m rUa2 100m

Ans.

Ans: x = 1142

rUa2 100m

11–10. A fluid has laminar flow and passes over the flat plate. If the thickness of the boundary layer at a distance of 0.5 m from the plate’s edge is 10 mm, determine the boundary layer thickness at a distance of 1 m.

6 m/s

10 mm

0.5 m

SOLUTION The fluid is considered to be incompressible. The flow is steady. The Reynolds number at x = 0.5 m and 1 m can be determined using Re x " x = 0.5 m =

U(0.5 m) Ux 0.5U = = n n n

and Re x " x = 1 m =

U(1 m) Ux U = = n n n

At x = 0.5 m, d = 0.01 m. Thus, d =

5.0x 2Re x

;

0.01 m =

5.0(0.5 m) 0.5U A n

U = 125 000 n Thus, at x = 1 m, Re x = d =

U = 125 000. Then, n

5.0x 2Re x

=

5.0(1 m) 2125 000

= 0.01414 m = 14.1 mm

Ans.

Ans: 14.1 mm 1143

11–11. Air at 60°C flows through the very wide duct. Determine the required dimension a of the duct at x = 4 m so that the central 200-mm core flow velocity maintains the constant free-stream velocity of 0.5 m>s.

0.5 m/s

200 mm

a

0.5 m/s

x!4m

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, n = 18.9 ( 10 - 6 ) m2 >s. Thus, the Reynolds number at x = 4 m is Re x =

( 0.5 m>s ) (4 m) Ux = 1.0582 ( 105 ) = n 18.9 ( 10 - 6 ) m2 >s

Since Re x 6 (Re x)cr = 5 ( 105 ) , the boundary layer is laminar throughout the entire length of the duct. Thus, the displacement thickness is d* =

1.721x 2Re x

=

1.721(4 m) 21.0582 ( 105 )

= 0.02116 m = 21.16 mm

The dimension of the square duct at x = 4 m is a = 200 mm + 2d * = 200 mm + 2(21.16 mm) Ans.

= 242 mm

Ans: 242 mm 1144

*11–12. Oil confined in a channel flows past the diverter fin at U = 6 m>s. Determine the friction drag acting on both sides of the fin. Take no = 40 1 10-6 2 m2 >s and ro = 900 kg>m3. Neglect end effects.

U

200 mm

50 mm

SOLUTION Oil is considered to be incompressible. The flow is steady. The transition from a laminar boundary layer occurs at a critical Reynolds number of (Re x)cr = 5 ( 105 ) . (Re x)cr = 5 ( 105 ) =

Uxcr n ( 6 m>s ) xcr 40 ( 10 - 6 ) m2 >s

xcr = 3.33 m

Since xcr 7 L = 0.05 m, the boundary layer for the entire length of the fin is laminar. Here, the Reynolds number at x = L = 0.05 m is

( 6 m>s ) (0.05 m) UL = = 7500 n 40 ( 10 - 6 ) m2 >s The drag force of both sides of the fin is Re L =

FD = 2£

0.664brU 2L 2Re L

= 4.97 N

§ = 2£

0.664(0.2 m) ( 900 kg>m3 )( 6 m>s ) 2(0.05 m) 27500

1145

§

Ans.

11–13. Air at 80°F and atmospheric pressure has a freestream velocity of 4 ft>s. If it passes along the surface of a smooth glass window of a building, determine the thickness of the boundary layer at a distance of 0.2 ft from the leading edge of the window. Also, what is the velocity of air 0.003 ft away from the window’s surface at this point?

SOLUTION Air is considered to be incompressible. The flow is steady. From Appendix A, n = 0.169 ( 10 - 3 ) ft 2 >s for air at T = 80° F. Thus, the Reynolds number at x = 0.2 ft is Re x =

( 4 ft>s ) (0.2 ft) Ux = = 4733.73 n 0.169 ( 10 - 3 ) ft 2 >s

Since Re x 6 (Re x)cr = 5 ( 105 ) , the boundary layer up to x = 0.2 ft is still laminar. Thus, its thickness at this point is d =

5.0x 2Re x

=

5.0(0.2 ft) 24733.73

= (0.01453 ft)a

12 in. b = 0.174 in. 1 ft

Ans.

Also, the velocity of a particle at the point x = 0.2 ft and y = 3 ( 10 - 3 ) ft can be determined using Blasius solution. Here, y x

2Re x =

3 ( 10 - 3 ) ft 0.2 ft

24733.73 = 1.032

Interpolating the values in the table gives u ≈ 0.3396 U

u = 0.3396 ( 4 ft>s ) = 1.36 ft>s (approx.)

Ans.

Ans: d = 0.174 in. u = 1.36 ft>s (approx.) 1146

11–14. Water at 40°C has a free-stream velocity of 0.3 m>s. Determine the boundary layer thickness at x = 0.2 m and at x = 0.4 m on the flat plate.

0.3 m/s

x 0.4 m

SOLUTION Water is considered to be incompressible. The flow is steady. From Appendix A, n = 0.664 ( 10 - 6 ) m2 >s for water at T = 40° C. Thus, the Reynolds number in terms of x is Re x =

( 0.3 m>s ) x Ux = = 4.5181 ( 105 ) x n 0.664 ( 10 - 6 ) m2 >s

At x = 0.2 m and 0.4 m,

Re x " x = 0.2 m = c 4.5181 ( 105 ) d (0.2 m) = 9.0361 ( 104 ) Re x " x = 0.4 m = c 4.5181 ( 105 ) d (0.4 m) = 1.8072 ( 105 )

Since Re x " x = 0.4 m 6 (Re x)cr = 5 ( 105 ) , the boundary layer up to x = 0.4 is still laminar. Thus, its thickness is d " x = 0.2 m = d " x = 0.4 m =

5.0x 2Re x 5.0x

2Re x

= =

5.0(0.2 m) 29.036 ( 104 ) 5.0(0.4 m)

218.072 ( 104 )

= 0.003327 m = 3.33 mm = 0.004705 m = 4.70 mm

Ans. Ans.

Ans: d " x = 0.2 m = 3.33 mm d " x = 0.4 m = 4.70 mm 1147

11–15. Water at 40°C has a free-stream velocity of 0.3 m>s. Determine the shear stress on the plate’s surface at x = 0.2 m and at x = 0.4 m.

0.3 m/s

x 0.4 m

SOLUTION Water is considered to be incompressible. The flow is steady.

From Appendix A, m = 0.659 ( 10 - 3 ) N # m>s and n = 0.664 ( 10 - 6 ) m2 >s for water at T = 40° C. Thus, the Reynolds number in terms of x is Re x =

( 0.3 m>s ) x Ux = = 4.5181 ( 105 ) x n 0.664 ( 10 - 6 ) m2 >s

At x = 0.2 m and 0.4 m,

Re x " x = 0.2 m = c 4.5181 ( 105 ) d (0.2 m) = 9.0361 ( 104 )

Re x " x = 0.4 m = c 4.5181 ( 105 ) d (0.4 m) = 1.8072 ( 105 )

Since Re x " x = 0.4 m 6 (Re x)cr = 5 ( 105 ) , the boundary layer up to x = 0.4 is still laminar. Thus, the shear stress on the plate’s surface is t0 " x = 0.2 m = 0.332m a

U b 2Re x x

= 0.332c 0.659 ( 10 - 3 ) N # s>m2 d a

0.3 m>s 0.2 m

= 0.0987 N>m2 t0 " x = 0.4 m = 0.332m a

U b 2Re x x

= 0.332c 0.659 ( 10 - 3 ) N # s>m2 d a = 0.0698 N>m2

0.3 m>s 0.4 m

b 29.0361 ( 104 )

b 21.8072 ( 105 )

Ans.

Ans.

Ans: t0 " x = 0.2 m = 0.0987 Pa t0 " x = 0.4 m = 0.0698 Pa 1148

*11–16. The boat is traveling at 0.7 ft>s through still water having a temperature of 60°F. If the rudder can be assumed to be a flat plate, determine the boundary layer thickness at the trailing edge A. Also, what is the displacement thickness of the boundary layer at this point? 2 ft

A

1.75 ft

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, n = 12.2 ( 10 - 6 ) ft 2 >s for water at T = 60° F. Thus, the Reynolds number at the trailing edge of the rudder (x = L = 1.75 ft) is Re x =

( 0.7 ft>s ) (1.75 ft) Ux = = 1.0041 ( 105 ) n 12.2 ( 10 - 6 ) ft 2 >s

Since Re x 6 (Re x)cr = 5 ( 105 ) , the boundary layer is laminar for the entire length of the rudder. Thus, its thickness and displacement thickness at the trailing edge are d = and d* =

5.0x 2Re x 1.721x 2Re x

=

=

5.0(1.75 ft)

= (0.02761 ft) a

21.0041 ( 105 ) 1.721(1.75 ft) 21.0041 ( 10

5

)

12 in. b = 0.331 in. 1 ft

= (0.00951 ft) a

12 in. b = 0.114 in. 1 ft

1149

Ans.

Ans.

11–17. The boat is traveling at 0.7 ft>s through water having a temperature of 60°F. If the rudder can be assumed to be a flat plate having a height of 2 ft and a length of 1.75 ft, determine the friction drag acting on both sides of the rudder. 2 ft

A

1.75 ft

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, n = 12.2 ( 10 - 6 ) ft 2 >s for water at T = 60° F. Thus, the Reynolds number at the trailing edge of the rudder (x = L = 1.75 ft) is Re x =

( 0.7 ft>s ) (1.75 ft) Ux = = 1.0041 ( 105 ) n 12.2 ( 10 - 6 ) ft 2 >s

Since Re x 6 (Re x)cr = 5 ( 105 ) , the boundary layer is laminar for the entire length of the rudder. Therefore, the frictional drag force on both surfaces of the rudder is F = 2£

= 2£

0.664brU 2L 2Re L

§

0.664(2 ft) ( 1.939 slug>ft 3 )( 0.7 ft>s ) 2(1.75 ft)

= 0.0139 lb

21.004 ( 105 )

§

Ans.

Ans: 0.0139 lb 1150

11–18. Air at a temperature of 40°F flows at 0.6 ft>s over the plate. Determine the distance x where the disturbance thickness of the boundary layer becomes 1.5 in.

0.6 ft/s

3 in.

x

SOLUTION Air is considered to be incompressible. The flow is steady. We will assume the boundary layer is laminar. From Appendix A, n = 0.147 ( 10 - 3 ) ft 2 >s at T = 40° F. Thus, the Reynolds number in terms of x is Re x =

( 0.6 ft>s ) x Ux = = 4081.63x n 0.147 ( 10 - 3 ) ft 2 >s

Here, the requirement is d = d =

5.0x 2Re x

;

1.5 ft. Thus, 12

0.125 ft =

5.0x 24081.63x

Ans.

x = 2.5510 ft = 2.55 ft

Using this result, Re x = 4081.63(2.5510 ft) = 1.041 ( 104 ) Since Re x 6 ( Re x ) cr = 5 ( 103 ) , the boundary layer is laminar as assumed.

Ans: 2.55 ft 1151

11–19. Determine the friction drag on the bar required to overcome the resistance of the paint if the force F lifts the bar at 3 m>s. Take r = 920 kg>m3 and n = 42 ( 10-6 ) m2 >s.

F 3 m/s

500 mm

2 mm

SOLUTION

50 mm

The point is considered to be incompressible. The relative flow is steady. The Reynolds number at x = L = 0.5 m is Re x =

( 3 m>s ) (0.5 m) Ux = = 0.35714 ( 105 ) n 42 ( 10 - 6 ) m2 >s

Since Re x 6 ( Re x ) cr = 5 ( 105 ) , the boundary layer is laminar throughout the entire length of the bar. Thus, the total frictional force on the bar is FD = Σ =

0.664brU 2L 2Re L

0.664 ( 920 kg>m3 )( 3 m>s ) 2(0.5 m)

= 1.51 N

20.35714 ( 105 )

[2(0.05 m) + 2(0.002 m)] Ans.

Ans: 1.51 N 1152

*11–20. The diverter fin extends 2 ft within the air duct to partition the flow through two separate conduits. Determine the friction drag on the fin if it is 0.3 ft wide and the velocity of the air is 25 ft>s. Take ra = 0.00257 slug>ft 3 and ma = 0.351(10-6) lb # s>ft 2.

2 ft

SOLUTION We will assume that steady flow occurs, and the air is incompressible. The Reynolds number of the flow at the trailing edge (x = 2 ft) is

( 0.00257 slug>ft 3 )( 25 ft>s ) (2 ft) raUL = = 3.661 ( 105 ) Re L = ma 0.351 ( 10-6 ) lb # s>ft 2 Since Re L 6 5 ( 105 ) , laminar flow persist within the boundary layer. Thus, the frictional drag force on the fin surface can be determined using Eq. 11–11. FDf =

0.664braU 2L 2Re L

=

0.664(0.3 ft) ( 0.00257 slug>ft 3 )( 25 ft>s ) 2(2 ft)

= 0.001058 lb

Since this force acts on two surfaces, Fig. a, the total drag force is

1153

FDf (a)

23.661 ( 105 )

( FDf ) T = 2 FDf = 2(0.001058 lb) = 0.00212 lb

FDf

Ans.

11–21. Crude oil at 20°C flows over the surface of the flat plate that has a width of 0.7 m. If the free-stream velocity is U = 10 m>s, plot the boundary layer thickness and the shearstress distribution along the plate. What is the friction drag on the plate?

10 m/s

1.5 m

SOLUTION We will assume that steady flow occurs, and crude oil is incompressible. Appendix A gives rco = 880 kg>m3 and mco = 30.2 ( 10-3 ) N # s>m2. The Reynold number as a function of x is Re x =

rcoUx = mco

( 880 kg>m3 )( 10 m>s ) x = 2.914 ( 105 ) x 30.2 ( 10-3 ) N # s>m2

At x = L = 1.5 m, Re L = 2.914 ( 105 ) (1.5) = 4.371 ( 105 ) 6 5 ( 105 ) . Thus, laminar flow persist within the boundary layer. For the boundary thickness d =

5.0 2Re x

x =

5.0

1

22.914 ( 10 ) x 5

x = c 9.2626 ( 10-3 ) x 2 d m

x(m)

0

0.25

0.5

0.75

1.0

1.25

1.5

d(mm)

0

4.63

6.55

8.02

9.26

10.36

11.34

0.75

1.0

1.25

1.50

The plot of d vs x is show in Fig. a. δ (m)

15

10

5

x (mm)

0

0.25

0.50

(a)

1154

11–21. Continued

For the shear stress, t0 = 0.332m a

U b 2Re x x

= 0.332 3 30.2 ( 10-3 ) N # s>m2 4 a

= a x(m) t0 ( N>m

2

)

54.12 1

x2

10 m>s x

b 22.914 ( 105 ) x

b N>m2

0

0.125

0.25

0.50

0.75

1.0

1.25

1.5



153.08 108.25 76.54

62.50

54.12

48.41

44.19

The plot of t0 vs x is shown in Fig. b. For the frictional drag force, 0.664brcoU 2L

FDf =

2Re L

0.664(0.7 m) ( 880 kg>m3 )( 10 m>s ) 2(1.5 m)

=

24.371 ( 105 )

= 92.8 N 0

Ans.

(N m2(

150

10

50

x (mm) 0

0.25

0.50

0.75

1.0

1.25

1.50

Ans: 92.8 N

(b)

1155

11–22. Castor oil flows over the surface of the flat plate at a free-stream speed of 2 m>s. The plate is 0.5 m wide and 1 m long. Plot the boundary layer and the shear stress versus x. Give values for every 0.5 m. Also calculate the friction drag on the plate. Take rco = 960 kg>m3 and mco = 985 1 10-3 2 N # s>m2.

2 m/s

2m

SOLUTION We will assume that steady flow occurs and castor oil is incompressible. The Reynolds number as a function of x is Re x =

rcoUx = mco

( 960 kg>m3 )( 2 m>s ) x = 1949.24x 985 ( 10-3 ) N # s>m2

At x = L = 2 m, Re L = 1949.24(2) = 3.898 ( 103 ) 6 5 ( 103 ) . Thus, laminar flow persist within the boundary layer. For the boundary thickness, d =

5.0 2Re x

x =

x(m)

0

d(mm)

0

5.0

1

21949.24x 0.5

x = ( 0.1132x2 ) m

1.0

1.5

2.0

80.08 113.25 138.70 160.16

The plot of d vs x is show in Fig. a. δ (mm) 200

150

100

50

x (m)

0

0.50

1.0

1.5

2.0

(a)

1156

11–22. Continued

For the shear stress t0 = 0.332m a

U b 2Re x x

= 0.332 3 985 ( 10-3 ) N # s>m2 4 a

= a

28.876 1

x2

x(m)

t0 ( N>m

2

)

b N>m2

2 m>s x

b 21949.24x

0

0.5

1.0

1.5

2.0



40.84

28.88

23.58

20.42

The plot of t0 vs x is shown in Fig. b. For the frictional drag force, FDf =

=

0.664brcoU 2L 2Re L

0.664(0.5 m) ( 960 kg>m3 )( 2 m>s ) 2(2 m) 23.898 ( 103 )

= 40.8 N

0

Ans.

(N m2(

60

50 40 30 20 10 x (m) 0

0.5

1.0

1.5

2.0

(b)

Ans: 40.8 N 1157

11–23. Assume the boundary layer has a velocity profile that is linear and defined by u = U 1 y>d 2 . Use the momentum integral equation to determine t0 for the fluid passing over the flat plate.

U

U

d

y

u ! U(y/d)

SOLUTION The fluid is considered to be incompressible. The flow is steady. y u Here, = . Substituting this result into the momentum integral equation U d d

t0 = rU 2 t0 = rU 2 t0 = rU 2 t0 = rU 2 t0 =

d u u a1 - bdy dx L0 U U d y y d a1 - bdy dx L0 d d d y2 y d a - 2 bdy dx L0 d d

d d a b dx 6

rU 2 dd 6 dx

(1)

For a laminar boundary layer, Newton’s law of viscosity applies. Thus, t0 = u

y mU du d ` = u cUa b d = dy y = 0 dy d d

(2)

Equating Eqs. (1) and (2), mU rU 2 dd = d 6 dx ddd =

6m dx rU

At the leading edge of the plate, x = 0 and d = 0. Therefore, L0

d

ddd =

6m x dx rU L0

6m x d2 d ` = x` 2 0 rU 0 d =

12m x B rU

Substituting this result into Eq. (2), t0 =

mU 12m x B rU

However, Re x =

= mU

rU rUx U = 0.289m a b B 12mx x B m

rUx . Then, B m

t0 = 0.289m a

U b 2Re x x

Ans.

Note: Compare this result with the one obtained using Blasius’s solution. 1158

Ans: t0 = 0.289m a

U b 2Re x x

*11–24. The wind tunnel operates using air at a temperature of 20°C with a free-stream velocity of 40 m>s. If this velocity is to be maintained at the central 1-m core throughout the tunnel, determine the increased dimension a at the exit in  order to accommodate the growing boundary layer. Show  that the boundary layer is turbulent, and use d * = 0.0463x>(Re x)1>5 to calculate the displacement thickness.

a

1m

a

1m

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, n = 15.1 ( 10-6 ) m2 >s. Thus, the Reynolds number at x = 6 m is Re x =

( 40 m>s ) (6 m) Ux = = 15.894 ( 106 ) n 15.1 ( 10-6 ) m2 >s

Since Re x 7 (Re x)cr = 5 ( 105 ) , the boundary layer is turbulent. Thus the displacement thickness is d* =

0.0463x (Re x)

1 5

=

0.0463(6 m)

3 15.894 ( 106 ) 4

1 5

= 0.01008 m

Thus, the dimension of the square tunnel at exit is a = 1 m + 2d * = 1 m + 2(0.01008 m) = 1.02 m

1159

Ans.

6m

11–25. Assume the turbulent boundary layer for a fluid has a velocity profile that can be approximated by u = U(y>d)1>6. Use the momentum integral equation to determine the boundary layer thickness as a function of x. Use the empirical formula Eq. 11–19, developed by Prandtl and Blasius.

SOLUTION The fluid is considered to be incompressible. The flow is steady. 1

Here,

y 6 u = a b . Substituting this result into the momentum integral equation, U d d d u u t0 = rU 2 a1 - bdy dx L0 U U 1

t0 = rU 2

1

d y 6 y 6 d a b J1 - a b R dy dx L0 d d 1

1

d y 6 y 3 d Ja b - a b R dy t0 = rU dx L0 d d 2

t0 =

3rU 2 dd 28 dx

(1)

Using the empirical formula developed by Prandtl and Blasius, 1

t0 = 0.0225rU 2 a

1

0.0225rU 2 a 1

n 4 b Ud

(2)

3rU 2 dd n 4 b dx = Ud 28 dx 1

v 4 b dx U

d 4dd = 0.21a

Assuming that the boundary layer is initially turbulent, then d = 0 at x = 0. Thus, d

1

x

n 4 d dd = 0.21a b dx U L0 L0 1 4

1

n 4 x 4 5 d d 4 ` = 0.21a b x ` 5 0 U 0 1

5

d 4 = 0.2625 a d = 0.343 a = 0.343 a

1

n 5 4 b x5 U 1

n5 1 5

1

U x5

bx

x

= 0.343≥

However, Re x =

n 4 bx U

1

a

Ux 5 b n

¥

Ux . Then this equation becomes n 0.343x d = 1 (Re x)5

Ans: Ans. 1160

d =

0.343x 1

(Re x)5

11–26. Air enters the square plenum of an air-handling system with a velocity of 6 m>s and a temperature of 10°C. Determine the thickness of the boundary layer and the momentum thickness of the boundary layer at x = 1 m.

300 mm

x

300 mm

SOLUTION We assume steady flow and the air to be incompressible. Appendix A gives ra = 1.247 kg>m3, and ma = 17.6 ( 10-6 ) N # s>m2 at T = 10° C. The Reynolds number as a function of x is Re x =

raUx = ma

( 1.247 kg>m3 )( 6 m>s ) x = 4.251 ( 105 ) x 17.6 ( 10-6 ) N # s>m2

at x = 1 m, Re x = 4.251 ( 105 ) 6 5 ( 105 ) , thus, laminar flow within the boundary layer. d =

5.0 2Re x

x =

ϴ =

5.0 24.251 ( 105 ) (1) 0.664

2Re x

x =

(1 m) = 7.67 mm 0.664

Ans.

(1)

24.251 ( 105 ) (1)

ϴ = 0.00102 m = 1.02 mm

Ans.

Ans: d = 7.67 mm ϴ = 1.02 mm 1161

11–27. Air enters the square plenum of an air-handling system with a velocity of 6 m>s and a temperature of 10°C. Determine the displacement thickness d * of the boundary layer at a point x = 1 m downstream. Also, what is the uniform speed of the air at this location?

300 mm

300 mm

x

SOLUTION We will assume that steady flow occurs and air is incompressible. Appendix A gives ra = 1.247 kg>m3 and ma = 17.6 ( 10-6 ) N # s>m2 at T = 10° C. The Reynolds number as a function of x is Re x =

raUx = ma

( 1.247 kg>m3 )( 6 m>s ) x = 4.251 ( 105 ) x 17.6 ( 10-6 ) N # s>m2

At x = 1 m, Re x = 4.251 ( 105 ) 6 5 ( 105 ) . Thus, laminar flow persist within the boundary layer. Then, Eq. 11–5 can be used to determine the displacement thickness. d* = At x = 1 m

1.721 2Re x

x = £

1.721

1

24.251 ( 10 ) x 5

§ x = 2.6395 ( 10-3 ) x2 Ans.

d * = 0.0026395 m = 2.64 mm Thus, the imaginary cross-sectional area of the duct at x = 1 m is A′ = 30.3 m - 2(0.0026395 m) 4 2 = 0.08686 m2

The continuity condition requires that the discharge through entrance is the same as that through the imaginary cross-section. VinAin = V′A′

( 6 m>s ) (0.3 m)2 = U ( 0.08686 m2 ) Ans.

U = 6.22 m>s

Ans: 6.22 m>s 1162

*11–28. Assume the turbulent boundary layer for a fluid has a velocity profile that can be approximated by u = U1y>d2 1>6. Use the momentum integral equation to determine the displacement thickness as a function of x and Rex. Use the empirical formula, Eq. 11–19, developed by Prandtl and Blasius.

U

d y

SOLUTION The fluid is considered to be incompressible. The flow is steady. 1 y 6 u Here, = a b . Substituting this result into the momentum integral equation, U d d

t0 = rU 2

d u u a1 - bdy dx L0 U U 1

t0 = rU 2

1

t0 = rU 2 t0 =

1

d y 6 y 6 d a b J1 - a b R dy dx L0 d d 1

d y 6 y 3 d Ja b - a b R dy dx L0 d d

3rU 2 dd 28 dx

(1)

Using the empirical formula developed by Prandtl and Blasius, 1

t0 = 0.0225rU 2 a

1

0.0225rU 2 a

n 4 b Ud

(2)

3rU 2 n 4 b dx = dd Ud 28 1

n 4 b dx U Assuming that the boundary layer is initially turbulent, then d = 0 at x = 0. Thus, 1

d 4dd = 0.21a

d

1

x

n 4 d dd = 0.21 a b dx U L0 L0 1 4

1

n 4 x 4 5 d d 4 ` = 0.21a b x ` 5 0 U 0 1

5

d 4 = 0.2625 a d = 0.343 a = 0.343 a

n 4 bx U 1

n 5 4 b x5 U 1

n5 1 5

1

U x5

bx

x

= 0.343≥

1

a

Ux 5 b n

¥ 1163

*11–28. Continued

Ux . Then this equation becomes n 0.343x

However, Re x = d =

(3)

1

( Re x ) 5

The displacement thickness is d* =

L0 d

=

d

a1 -

u bdy U 1

y 6 c 1 - a b d dy d L0 6

6 y7 d b` = ay 7 d 16 0 =

d 7

Substituting Eq. 3 into this result, d* =

0.0490x 1 0.343x £ § = 1 7 ( Re x ) 15 ( Re x ) 5

Ans.

1164

11–29. The laminar boundary layer for a fluid is assumed to be parabolic, such that u>U = C1 + C2 1y>d2 + C3 1y>d2 2. If the free-stream velocity U starts at y = d, determine the constants C1, C2, and C3.

U

d y

SOLUTION We will assume that steady flow occurs, and the fluid is incompressible. Applying the boundary conditions at y = 0, u = 0. Then 0 0 2 0 = C1 + C2 a b + C3 a b d d

Ans.

C1 = 0

And at y = d, u = U. Then

d d 2 1 = 0 + C2 a b + C3 a b d d

(1)

C2 + C3 = 1

Since laminar flow persist within the boundary layer, Newton’s law of viscosity du t = m can be applied. Here dy 2C3 2C3 C2 C2 1 du du = + 2 y = Ua + 2 yb U dy d dy d d d At y = d, t = 0. Then 0 = m 0 = Since

2C3 C2 du ` = mJU a + bR dy y = d d d

mU ( C2 + 2C3 ) d

mU ≠ 0, then d C2 + 2C3 = 0

(2)

Solving Eqs. (2) and (3), C2 = 2

Ans.

C3 = - 1

Ans: C1 = 0 C2 = 2 C3 = - 1 1165

11–30. The laminar boundary layer for a fluid is assumed to be cubic, such that u>U = C1 + C2 1y>d2 + C3 1y>d2 3. If the free-stream velocity U starts at y = d, determine the constants C1, C2, and C3.

U

d y

SOLUTION We will assume that steady flow occurs and the fluid is incompressible. Applying the boundary condition at y = 0, u = 0. Then 0 0 3 0 = C1 + C2 a b + C3 a b d d

C1 = 0

Ans.

And at y = d, u = U. Then

d d 3 1 = 0 + C2 a b + C3 a b d d

(1)

C2 + C3 = 1

Since laminar flow persist within the boundary layer, Newton’s Law of viscosity du t = m can be applied. Here dy 3C3 C2 1 du = + 3 y2 U dy d d At y = d, t = 0. Then 0 = m 0 = Since

3C3 C2 du = Ua + 3 y2 b dy d d

3C3 C2 du ` = mJU a + bR dy y = d d d

mU ( C2 + 3C3 ) d

mU ≠ 0, then d C2 + 3C3 = 0

(2)

Solving Eqs. (1) and (2), C2 =

3 2

C3 = -

1 2

Ans.

Ans: C1 = 0 1166

C2 =

3 2

C3 = -

1 2

11–31. Assume a laminar boundary layer for a fluid can be approximated by u>U = y>d. Determine the thickness of the boundary layer as a function of x and Rex.

U

x

SOLUTION d

t0 = rU 2 t0 = rU 2 t0 =

u d u a1 - bdy dx L0 U U

d y y d a ba1 - bdy dx L0 d d

rU 2 dd 6 dx

Newton’s law of viscosity, t0 = m

U d

Thus mU rU 2 dd = = d 6 dx L0

x

d

d dd =

6m dx L0 rU

6m 1 2 d = x 2 ru d = 3.46

mx A rU

Since Re x = rU * 1m then d =

3.46x

Ans.

2Re x

Ans: d =

1167

3.46x 2Re x

*11–32. Assume a laminar boundary layer for a fluid can be  approximated by u>U = sin 1py>2d2 . Determine the thickness of the boundary layer as a function of x and Rex.

U

x

SOLUTION The fluid is considered to be incompressible. The flow is steady. Substituting

u py = sina b into the moment integral equation, U 2 d d

t0 = rU 2 t0 = rU 2 t0 = rU 2

d u u a1 - bdy dx L0 U U

d d py py sin a b c 1 - sin a b d dy dx L0 2 d 2 d d d py py c sin a b - sin2 a b d dy dx L0 2 d 2 d

From the trigonometric identity,

Then,

sin2 a

py py 1 b = c 1 - cos a b d 2 d 2 d

t0 = rU 2 t0 = rU 2

d py d py 1 1 c sin a b + cos a b - d dy dx L0 2 d 2 d 2

d py d 2d py d 1 c - cos a b + sin a b - y d ` p dx 2 d 2p d 2 0

t0 = 0.1366rU 2

dd dx

(1)

For a laminar boundary layer, Newton’s law of viscosity applies. Thus, t0 = m

du d py ` = m c U sin a bd ` dy y = 0 dy 2 d y=0

t0 = mUc t0 =

p py cos a bd ` 2d 2 d y=0

pmU 2d

(2)

Equating Eqs. (1) and (2), pmU dd = 0.1366rU 2 2d dx ddd =

11.498m dx rU

1168

*11–32. Continued

At the leading edge of the plate, x = 0 and d = 0. Thus, L0

d

d dd =

11.498m x dx rU L0

11.498m x d2 d ` = x` 2 0 rU 0 d2 =

22.995m x rU 1

d = d =

1

1

r2U 2 4.7953x 1

1

1

r2U 2x2 1 2

m Since Re x =

1

4.7953m2x2

=

4.7953x rUx B m

rUx , this equation becomes m d =

4.80x

Ans.

2Re x

1169

11–33. Assume a laminar boundary layer for a fluid can be  approximated by u>U = sin 1py>2d2 . Determine the displacement thickness d * for the boundary layer as a function of x and Rex.

U

x

SOLUTION The displacement thickness is d* = =

L0

d

L0

d

a1 -

a1 - sin a

= a1 -

From solution 11–32, d = So,

u bdy U

2 bd p

py b bdy 2d

4.7953x 2Re x

d* = a1 -

2 4.7953x 1.74x b = p 2Re 2Re x x

Ans: d* =

1170

1.74x 2Re x

11–34. The, velocity profile for a laminar boundary layer of a fluid is represented by u>U = 1.51y>d2 - 0.51y>d2 3. Determine the thickness of the boundary layer as a function of x and Rex.

y U

SOLUTION The fluid is considered to be incompressible. The flow is steady. Substituting

y y 3 u = 1.5 a b - 0.5 a b into the moment integral equation, U d d d

t0 = rU 2

d u u a1 - bdy dx L0 U U

d u u 2 J - a b R dy dx L0 U U d

t0 = rU 2 Here,

y y 3 y y 3 2 u 2 u - a b = J1.5 a b - 0.5 a b R - J1.5 a b - 0.5 a b R U U d d d d

y 6 y 4 y 3 y 2 y = -0.25 a b + 1.5 a b - 0.5 a b - 2.25 a b + 1.5 a b d d d d d

Then, t0 = rU 2

d y 6 y 4 y 3 y 2 y d J -0.25 a b + 1.5 a b - 0.5 a b - 2.25 a b + 1.5 a b R dy dx L0 d d d d d

t0 = 0.1393rU 2

dd dx

(1)

For laminar boundary layer, Newton’s law of viscosity applies. Thus, t0 = m

t0 =

y y 3 du d ` = u £ UJ1.5a b - 0.5a b R § † dy y = 0 dy d d

y=0

1.5mU d

(2)

Equating Eqs. (1) and (2), 1.5mU dd = 0.1393rU 2 d dx ddd =

10.769m dx rU

1171

11–34. Continued

At the leading edge of the plate, x = 0 and d = 0. Thus, L0

d

d dd =

10.769m x dx rU L0

10.769m x d2 d ` = x` 2 0 rU 0 d2 =

21.538m x rU 1

d =

1

4.6410m2x2 1 2

rU

1 2

=

4.6410x 1 2

1 2

rU x

1 2

=

4.6410x

1 2

rUx B m

m rUx Since Re x = , this equation becomes m d =

4.64x

Ans.

2Re x

Ans: d =

1172

4.64x 2Re x

11–35. The velocity profile for a laminar boundary layer of a fluid is represented by u>U = 1.51y>d2 - 0.51y>d2 3. Determine the shear-stress distribution acting on the surface as a function of x and Rex.

y U

SOLUTION The fluid is considered to be incompressible. The flow is steady. Substituting

y y 3 u = 1.5 a b - 0.5 a b into the moment integral equation, U d d d

t0 = rU 2

d u u a1 - bdy dx L0 U U

d u u 2 J - a b R dy dx L0 U U d

t0 = rU 2 Here,

y y 3 y y 3 2 u 2 u - a b = J1.5 a b - 0.5 a b R - J1.5 a b - 0.5 a b R U U d d d d

y 6 y 4 y 3 y 2 y = -0.25 a b + 1.5 a b - 0.5 a b - 2.25 a b + 1.5 a b d d d d d

Then, t0 = rU 2

d y 6 y 4 y 3 y 2 y d J -0.25 a b + 1.5 a b - 0.5 a b - 2.25 a b + 1.5 a b R dy dx L0 d d d d d

t0 = 0.1393rU 2

dd dx

(1)

For a laminar boundary layer, Newton’s law of viscosity applies. Thus, t0 = m

y y 3 du d ` = u £ Uc 1.5a b - 0.5a b d § † dy y = 0 dy d d

y=0

1.5mU t0 = d

(2)

Equation Eqs. (1) and (2), 1.5mU dd = 0.1393U 2 d dx ddd =

10.769m dx rU

1173

11–35. Continued

At the leading edge of the plate, x = 0 and d = 0. Thus, L0

d

d dd =

10.769m x dx rU L0

10.769m x d2 d ` = x` 2 0 rU 0 d2 =

21.538m x rU 1

d =

1

4.6410m2x2 1 2

rU

1 2

=

4.6410x 1 2

1 2

rU x

1 2

1 2

m Since Re x = d =

=

4.6410x rUx B m

rUx , this equation becomes m 4.6410x

(3)

2Re x

Substituting Eq. (3) into Eq. (2), t0 =

1.5mU U = 0.323m a b 2Re x 4.6410x x

Ans.

2Re x

Ans: t0 = 0.323ma 1174

U b 2Re x x

*11–36. A boundary layer for laminar flow of a fluid over the plate is to be approximated by the equation u>U = C1 1y>d2 + C2 1y>d2 2 + C3 1y>d2 3. Determine the constants C1, C2, and C3 using the boundary conditions when y = d, u = U; when y = d, du>dy = 0; and when y = 0, d 2u>dy2 = 0. Find the thickness of the boundary layer as a function of x and Rex using the momentum integral equation.

U

x

SOLUTION The fluid is considered to be incompressible. The flow is steady. Here, y y 2 y 3 u = UJC1 a b + C2 a b + C3 a b R d d d

For the boundary condition u = U at y = d, U = U 3 C1(1) + C2(1) + C3(1) 4

(1)

C1 + C2 + C3 = 1 Subsequently,

3C3y2 2C2y C1 du + R = UJ + dy d d2 d3 For the boundary condition 0 = Uc

du = 0 at y = d, dy

3C3 2C2 C1 + + d d d d

(2)

C1 + 2C2 + 3C3 = 0 For the boundary condition

2

d u = 0 at y = 0, dy2 U ( 2C2 + 0 ) d2

0 =

Ans.

C2 = 0

Substituting this result into Eqs. (1) and (2) and solving, C1 =

3 2

C3 = -

1 2

Ans.

Thus, u 3 y 1 y 3 = a b - a b U 2 d 2 d

Substituting this result into the momentum integral equation, d

t0 = rU 2

d u u a1 - bdy dx L0 U U

d u u 2 J - a b R dy dx L0 U U d

t0 = rU 2 Here,

u 1 y 3 1 y 3 2 u 2 3 y 3 y - a b = J a b - a b R - J a b - a b R U U 2 d 2 d 2 d 2 d

1 y 6 3 y 4 1 y 3 9 y 2 3 y = - a b + a b - a b - a b + a b 4 d 2 d 2 d 4 d 2 d

1175

*11–36. Continued

Then, t0 = rU 2 t0 =

d d 1 y 6 3 y 4 1 y 3 9 y 2 3 y J- a b + a b - a b - a b + a bR dy dx L0 4 d 2 d 2 d 4 d 2 d

39rU 2 dd 280 dx

(1)

For a laminar boundary layer, Newton’s law of viscosity applies. Thus, t0 = m

1 y 3 du d 3 y ` = u £U J a b - a b R § † dy y = 0 dy 2 d 2 d

y=0

3mU 2d Equating Eqs. (1) and (2),

(2)

t0 =

3mU 39rU 2 dd = 2d 280 dx ddd =

140m dx 13rU

At the leading edge of the plate, x = 0 and d = 0. Thus, L0

d

d dd =

140m x dx 13rU L0

140m x d2 d ` = x` 2 0 13rU 0 d2 =

280m x 13rU

1

d =

1

4.6410m2x2 1 2

rU

1 2

=

4.6410x 1 2

1 2

rU x 1 2

m Since Re x =

1 2

=

4.6410x rUx B m

rUx , this equation becomes B m 4.64x d = 2Re x

Ans.

1176

11–37. The train travels at 30 m>s and consists of an engine and a series of cars. Determine the approximate thickness of the boundary layer at the top of a car, x = 18 m from the front of the train. The air is still and has a temperature of 20°C. Assume the surfaces are smooth and flat, and the boundary layer is completely turbulent.

30 m/s

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, n = 15.1 ( 10-6 ) m2 >s for air at T = 20° C . Thus, the Reynolds number at x = 18 m is Re x =

( 30 m>s ) (18 m) Ux = = 3.576 ( 107 ) n 15.1 ( 10-6 ) m2 >s

Since Re x 7 ( Re x ) cr = 5 ( 105 ) , the boundary layer is turbulent. Assuming that the boundary layer is turbulent from x = 0, d =

0.371x

( Re x )

1 5

=

0.371(18 m)

3 3.576 ( 107 ) 4

1 5

= 0.2060 m = 206 mm

Ans.

Ans: 206 mm 1177

11–38. The train travels at 30 m>s and consists of an engine and a series of cars. Determine the approximate shear stress acting on the top of a car at x = 18 m from the front of the train. The air is still and has a temperature of 20°C. Assume the surfaces are smooth and flat, and the boundary layer is completely turbulent.

30 m/s

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 1.202 kg>m3 and n = 15.1 ( 10-6 ) m2 >s for air at T = 20° C . Thus, the Reynolds number at x = 18 m is Re x =

( 30 m>s ) (18 m) Ux = = 3.576 ( 107 ) n 15.1 ( 10-6 ) m2 >s

Since Re x 7 ( Re x ) cr = 5 ( 105 ) , the boundary layer is turbulent. Assuming that the boundary layer is turbulent from x = 0, the shear stress at any point on the top surface of the train is t0 =

0.0288rU 2

( Re x )

1 5

= 0.961 N>m2

=

0.0288 ( 1.202 kg>m3 )( 30 m>s ) 2

3 3.576 ( 107 ) 4

1 5

Ans.

Ans: 0.961 Pa 1178

11–39. A ship is traveling forward at 10 m>s on a lake. If it is 100 m long and the side of the ship can be assumed to be a flat plate, determine the drag force on a 1-m-wide strip along the entire length of the ship. The water is still and has a temperature of 15°C. Assume the boundary layer is completely turbulent.

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, r = 999.2 kg>m3 and n = 1.15 ( 10-6 ) m2 >s for water at T = 15° C . Thus, the Reynolds number at x = L = 100 m is Re L =

( 10 m>s ) (100 m) UL = = 8.6957 ( 108 ) n 1.15 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 107 … Re L 6 109, the frictional drag coefficient is CD =

0.455

( log 10 Re L )

2.58

=

0.455 c log 10 3 869.57 ( 106 ) 4 d

2.58

= 0.0015983

Thus, the frictional drag force can be determined from 1 F = CD a rU 2 bbL 2 = 0.0015983c

1 ( 999.2 kg>m3 )( 10 m>s ) 2 d (1 m)(100 m) 2

= 7.9849 ( 103 ) N = 7.98 kN

Ans.

Ans: 7.98 kN 1179

*11–40. An airplane has wings that are, on average, each 5 m long and 3 m wide. Determine the friction drag on the wings when the plane is flying at 600 km>h in still air at an altitude of 2 km. Assume the wings are flat plates and the boundary layer is completely turbulent.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 1.007 kg>m3 and n = 17.15 ( 10-6 ) m2 >s for air at an altitude

of 2 km. Here, U = a600

1h 1000 m km ba ba b = 166.67 m>s. Thus, the Reynolds h 3600 s 1 km

number at x = L = 3 m is Re L =

( 166.67 m>s ) (3 m) UL = = 2.915 ( 107 ) n 17.15 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 107 … Re L 6 109, the frictional drag coefficient is CD =

0.455

( log 10 Re L )

2.58

=

0.455 c log 10 3 2.915 ( 107 ) 4 d

2.58

= 0.0025447

Since each of the wings has top and bottom surfaces (4 surfaces altogether), the total drag force on the two wings can be determined from 1 F = ΣCD a rU 2 bbL 2 = 4c 0.0025447c

1 ( 1.007 kg>m3 )( 166.67 m>s ) 2 d (5 m)(3 m) d 2

= 2135.44 N = 2.14 kN

1180

Ans.

11–41. The oil tanker has a smooth surface area of 4.5(103) m2 in contact with the sea. Determine the friction drag on its hull and the power required to overcome this force if the velocity of the ship is 2 m>s. Take r = 1030 kg>m3 and m = 1.14 1 10-3 2 N # s>m2.

2 m/s

300 m

SOLUTION We will assume that steady flow occurs and sea water is incompressible. The Reynolds number at the trailing edge of the hull is Re L =

rUL = m

( 1030 kg>m3 )( 2 m>s ) (300 m) = 5.42 ( 108 ) 1.14 ( 10-3 ) N # s>m2

Since 5 ( 105 ) 6 Re L 6 109, the boundary layer on the hull will be laminar and turbulent along the length. Thus, CDf = =

0.455 (log 10 Re L)2.58

-

1700 Re L

0.455 e log 10 c 5.421 ( 108 ) d f

2.58

-

1700 5.421 ( 108 )

= 0.001694

The frictional drag force acting on each side of the hull can be determined from FDf = CDf Aa

rU 2 b 2

= 0.001694 3 4.5 ( 103 ) m2 4 £

( 1030 kg>m3 )( 2 m>s ) 2 2

= 15.70 ( 103 ) N = 15.7 kN

§ Ans.

The power required is P = ( FDf ) T (V) = = 31.40 ( 103 ) W

3 15.70 ( 103 ) N 4 ( 2 m>s ) Ans.

= 31.4 kW

Ans: FDf = 15.7 kN # W = 31.4 kW 1181

11–42. Wind is blowing at 2 m>s as the truck moves forward into the wind at 8 m>s. If the air has a temperature of 20°C, determine the friction drag acting on the flat side ABCD of the truck. Assume the boundary layer is completely turbulent.

8m

C

E 8 m/s 2 m/s

3m F

B

4m

D

A

SOLUTION The air is considered to be incompressible. The relative flow is steady. Here, the velocity of the truck relative to still air is 8 m>s + 2 m>s = 10 m>s. Thus, U = 10 m>s. From Appendix A, r = 1.202 kg>m3 and n = 15.1 ( 10-6 ) m2 >s. Thus, the Reynolds number at x = L = 8 m is Re L =

( 10 m>s ) (8 m) UL = = 5.298 ( 106 ) n 15.1 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 5 ( 105 ) … Re L 6 107, the frictional drag coefficient is CD =

0.0740 (Re L)

1 5

=

0.0740

3 5.298 ( 106 )4

1 5

= 0.0033451

Thus, the frictional drag force on surface ABCD can be determined from 1 F = CDa rU 2 bbL 2 = 0.0033451c

= 6.43 N

1 ( 1.202 kg>m3 )( 10 m>s ) 2 d (4 m)(8 m) 2

Ans.

Ans: 6.43 N 1182

11–43. The wind is blowing at 2 m>s as the truck moves forward into the wind at 8 m>s. If the air has a temperature of 20°C, determine the friction drag acting on the top surface BCFE of the truck. Assume the boundary layer is completely turbulent.

8m

C

E 8 m/s 2 m/s

3m F

B

4m

D

A

SOLUTION The air is considered to be incompressible. The relative flow is steady. Here, the velocity of the truck relative to still air is 8 m>s + 2 m>s = 10 m>s . Thus, U = 10 m>s. From Appendix A, r = 1.202 kg>m3 and n = 15.1 ( 10-6 ) m2 >s. Thus, the Reynolds number at x = L = 8 m is Re L =

( 10 m>s ) (8 m) UL = = 5.298 ( 106 ) n 15.1 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 5 ( 105 ) … Re L 6 107, the frictional drag coefficient is CD =

0.0740 (Re L)

1 5

=

0.0740

3 5.298 ( 106 ) 4

1 5

= 0.0033451

Thus, the frictional drag force on surface BCFE can be determined from 1 F = CDa rU 2 bbL 2 = 0.0033451c = 4.82 N

1 ( 1.202 kg>m3 )( 10 m>s ) 2 d (3 m)(8 m) 2

Ans.

Ans: 4.82 N 1183

*11–44. The flat-bottom boat is traveling at 4 m>s on a lake for which the water temperature is 15°C. Determine the approximate drag force acting on the bottom of the boat if the length of the boat is 10 m and its width is 2.5 m. Assume the boundary layer is completely turbulent.

4 m/s

10 m

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, r = 999.2 kg>m3 and n = 1.15 ( 10-6 ) m2 >s for water at T = 15° C. Thus, the Reynolds number at x = L = 10 m is Re L =

( 4 m>s ) (10 m) UL = = 3.4783 ( 107 ) n 1.15 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 107 … Re L 6 109, the frictional drag coefficient is CD =

0.455 (log 10 Re L)2.58

=

0.455

3 log 10 3 3.4783 ( 107 )44 2.58

= 0.0024785

Thus, the frictional drag force on the bottom surface can be determined from 1 F = CDa rU 2 bbL 2 = 0.0024785c = 495 N

1 ( 999.2 kg>m3 )( 4 m>s ) 2d (2.5 m)(10 m) 2

1184

Ans.

11–45. An airplane is flying at 170 ft>s through still air at an altitude of 5000 ft. If the wings can be assumed to be flat plates, each having a width of 7 ft, determine the boundary layer thickness at their trailing or back edge if the boundary layer is considered to be fully turbulent.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, n = 0.1779 ( 10 - 3 ) ft 2 >s for air at an altitude of 5000 ft. Thus, the Reynolds number at x = L = 7 ft is Re L =

( 170 ft>s ) (7 ft) UL = = 6.689 ( 106 ) n 0.1779 ( 10 - 3 ) ft 2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Thus, the thickness of the boundary layer. d =

0.371x (Re x)

1 5

=

0.371(7 ft)

3 6.689 ( 10 ) 4 6

1 5

= 0.1120 ft a

12 in. b = 1.34 in. 1 ft

Ans.

Ans: 1.34 in. 1185

11–46. An airplane is flying at 170 ft>s though still air at an altitude of 5000 ft. If the wings can be assumed to be flat plates, each having a width of 7 ft and a length of 15 ft, determine the friction drag on each wing if the boundary layer is considered fully turbulent.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 2.043 ( 10-3 ) slug>ft 3 and n = 0.1779 ( 10-3 ) ft 2 >s for air at an altitude of 5000 ft. Thus, the Reynolds number at x = L = 7 ft is Re L =

( 170 ft>s ) (7 ft) UL = = 6.689 ( 106 ) n 0.1779 ( 10-3 ) ft 2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 5 ( 105 ) … Re L 6 107, the frictional drag coefficient is CD =

0.0740

( Re L )

1 5

=

0.0740

3 6.689 ( 106 ) 4

1 5

= 0.0031927

Since each wing has two surfaces, top and bottom, the total frictional drag force can be determined from 1 F = ΣCDa rU 2 bbL 2

1 = 2c 0.0031927c (2.043) ( 10-3 ) slug>ft 3 ( 170 ft>s ) 2 d d (15 ft)(7 ft) 2 Ans.

= 19.8 lb

Ans: 19.8 lb 1186

11–47. An airplane is flying at a speed of 90 m>s. If the wings are assumed to have a flat surface of width 2.5 m, determine the boundary layer thickness d and the shear stress at the trailing or back edge. Assume the boundary layer is fully turbulent. The airplane flies at an altitude of 1 km.

90 m/s B

A 2.5 m

SOLUTION The air is considered to be incompressible. The relative flow is steady. Here, the velocity of the airplane is, U = 90 m>s. From Appendix A, r = 1.112 kg>m3 and n = 15.81 ( 10-6 ) m2 >s for air at an altitude of 1 km. Thus, the Reynolds number at x = L = 2.5 m is Re L =

( 90 m>s ) (2.5 m) UL = = 1.423 ( 107 ) n 15.81 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Therefore, the thickness of the boundary layer and the shear stress at x = L = 2.5 m are d = t0 =

0.371x

( Re L )

1 5

=

0.0288rU 2

( Re L )

1 5

= 9.62 N>m2

0.371(2.5 m)

3 1.423 ( 107 )4 =

1 5

= 0.03441 m = 34.4 mm

Ans.

0.0288 ( 1.112 kg>m3 )( 90 m>s ) 2

3 1.423 ( 107 )4

1 5

Ans.

Ans: d = 34.4 mm t0 = 9.62 Pa 1187

*11–48. An airplane is flying at an altitude of 1 km and a speed of 90 m>s. If the wings are assumed to have a flat surface of width 2.5 m and length 7 m, determine the friction drag on each wing. Assume the boundary layer is fully turbulent.

90 m/s

2.5 m

SOLUTION The air is considered to be incompressible. The relative flow is steady. Here, the velocity of the airplane is 90 m>s . From Appendix A, r = 1.112 kg>m3 and n = 15.81 ( 10-6 ) m2 >s for air at an altitude of 1 km. Thus, the Reynolds number at x = L = 2.5 m is Re L =

( 90 m>s ) (2.5 m) UL = = 1.423 ( 107 ) n 15.81 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Since 107 … Re L 6 109, the frictional drag coefficient can be determined using. CD =

0.455

( log 10Re L ) 2.58

=

0.455

3 log 10 3 1.423 ( 107 )44 2.58

= 0.0028405

Since each wing has two surfaces, top and bottom, the total frictional drag force can be determined from 1 F = ΣCD a rU 2 bbL 2 = 2c0.0028405c

B

A

1 ( 1.112 kg>m3 )( 90 m>s ) 2 d (7 m)(2.5 m)d 2 Ans.

= 447.73 N = 448 N

1188

11–49. The tail of the airplane has an approximate width of 1.5 ft and a length of 4.5 ft. Assuming the air flow onto the tail is uniform, plot the boundary layer thickness d. Give values for every increment of 0.05 ft for the laminar boundary layer, and every 0.25 ft for the turbulent boundary layer. Also, calculate the friction drag on the rudder. The plane is flying in still air at an altitude of 5000 ft with a speed of 500 ft>s.

500 ft/s

4.5 ft

1.5 ft

SOLUTION We will assume that steady flow occurs and the air is incompressible. Appendix A gives ra = 2.043 ( 10-3 ) slug>ft 3 and ma = 0.3637 ( 10-6 ) lb # s>ft 2 for air at an altitude of 5000 ft. The Reynolds number as a function of x is raUx = ma

Re x =

3 2.043 ( 10-3 ) slug>ft3 4 ( 500 ft>s ) x 0.3637 ( 10-6 ) lb # s>ft 2

= 2.809 ( 106 ) x

At the trailing edge where x = L = 1.5 ft, Re L = 2.809 ( 106 ) (1.5 ft) = 4.213 ( 106 ) . Since 5 ( 105 ) 6 Re L 6 109, the boundary layer will be laminar and turbulent. First, we will determine the critical distance xcr where the transition to turbulent flow occurs.

( Re x ) cr = 5 ( 105 ) ;

2.809 ( 106 ) xcr = 5 ( 105 ) xcr = 0.1780 ft

For the laminar boundary layer where x 6 xcr, d =

5.0 2Re x

x(ft) d(in.)

x =

0 0

5.0

1

22.809 ( 10 ) x 6

0.05 0.00801

x = c 2.9835 ( 10-3 ) x2 d ft 0.10 0.0113

0.15 0.0139

0.178 0.0151

1189

11–49. Continued

For the turbulent boundary layer where x 7 xcr, 0.371

d = x(ft) d(in.)

( Re x )

0.178 0.0574

1 5

x =

0.371

4

3 2.809 ( 10 ) x 4

0.25 0.0754

6

0.50 0.1312

1 5

x = ( 0.01904x5 ) ft 0.75 0.1815

1.0 0.2284

1.25 0.2731

1.50 0.3160

The plot of the boundary layer is shown in Fig. a. For the laminar and turbulent boundary layer the frictional drag coefficient can be determined from C Df = =

0.455

( log 10Re L ) 2.58

-

1700 Re L

0.455

3 log 104.213 ( 106 )4 2.58

-

1700 4.213 ( 106 )

= 0.003059

Thus, the frictional drag force can be determined by applying FDf = CDf bLa

raU 2 b 2

= 0.003059(4.5 ft)(1.5 ft) W

3 2.043 ( 10-3 ) slug>ft3 4 ( 500 ft>s ) 2 2

= 5.27 lb



Ans.

δ (in.) 0.35 0.30 0.25

0.20 0.15

0.10 0.05 x (ft) 0

0.10

0.25 xcr = 0.178

0.50

0.75

1.0

1.25

1.50

Ans: 5.27 lb

(a)

1190

11–50. Two hydrofoils are used on the boat that is traveling at 20 m>s. If the water is at 15°C, and if each blade can be considered as a flat plate, 4 m long and 0.25 m wide, determine the thickness of the boundary layer at the trailing or back edge of each blade. What is the drag on each blade? Assume the flow is completely turbulent.

20 m/s

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, r = 999.2 kg>m3 and n = 1.15 ( 10-6 ) m2 >s for water at T = 15° C . Thus, the Reynolds number at x = L = 0.25 m is Re =

( 20 m>s ) (0.25 m) UL = = 4.348 ( 106 ) n 1.15 ( 10-6 ) m2 >s

Here, we assume that the boundary layer is turbulent from x = 0. Thus, the thickness of the boundary layer at x = L = 0.25 m is d =

0.371x

( Re x )

1 5

=

0.371 ( 0.25 m )

3( 4.348 ( 106 ) )4

1 5

= 0.00436 m = 4.36 mm

Ans.

Since 5 ( 105 ) … Re L 6 107, the frictional drag coefficient can be determined using CD =

0.0740

( Re L )

1 5

=

0.0740

3 4.348 ( 106 )4

1 5

= 0.003480

Since each wing has two surfaces, top and bottom, the total frictional drag force can be determined from 1 F = ΣCD a rU 2 bbL 2 = 2c0.003480c

1 ( 999.2 kg>m3 )( 20 m>s ) 2 d (4 m)(0.25 m)d 2 Ans.

= 1390.88 N = 1.39 kN

Ans: d = 4.36 mm F = 1.39 kN 1191

11–51. Two hydrofoils are used on the boat that is traveling at 20 m>s. If the water is at 15°C and if each blade can be considered as a flat plate, 4 m long and 0.25 m wide, determine the drag on each blade. Consider both laminar and turbulent flow boundary layers.

20 m/s

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, r = 999.2 kg>m3 and n = 1.15 ( 10-6 ) m2 >s for water at T = 15° C . Thus, the Reynolds number at x = L = 0.25 m is Re =

( 20 m>s ) (0.25 m) UL = = 4.348 ( 106 ) n 1.15 ( 10-6 ) m2 >s

Since 5 ( 105 ) … Re L 6 109, the frictional drag coefficient considering both laminar and turbulent boundary layers is CD =

0.455

( log 10Re L ) 2.58

-

1700 0.455 1700 = = 0.0030534 Re L 3 log 10 3 4.348 ( 106 )44 2.58 4.348 ( 106 )

Since each wing has two surfaces, top and bottom, the total frictional drag force can be determined from 1 F = ΣCD a rU 2 bbL 2 = 2c0.0030533c

1 ( 999.2 kg>m3 )( 20 m>s2 ) d (4 m)(0.25 m)d 2 Ans.

= 1220.37 N = 1.22 kN

Ans: 1.22 kN 1192

*11–52. An airplane is flying at an altitude of 3 km and a speed of 700 km>h. If each wing is assumed to have a flat surface of width 2 m and length 6 m, determine the friction drag acting on each wing. Consider both laminar and turbulent boundary layers.

120 ft

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 0.9092 kg>m3 and n = 18.63 ( 10-6 ) m2 >s for air at an altitude of 3 km. Here,

1h km 1000 m ba ba b = 194.44 m>s . Thus, the Reynolds number at h 1 km 3600 s x = L = 2 m is

U = a700

Re L =

( 194.44 m>s ) (2 m) UL = = 2.087 ( 107 ) n 18.63 ( 10-6 ) m2 >s

Since 5 ( 105 ) … Re L 6 109, the frictional drag coefficient considering both laminar and turbulent boundary layers is CD = =

0.455

( log 10Re L )

2.58

-

1700 Re L

0.455

3 log 102.087 ( 107 )4 2.58

-

1700 2.087 ( 107 )

= 0.002595

Since each wing has two surfaces, top and bottom, the total frictional drag force can be determined from 1 F = ΣCDa rU 2 bbL 2 = 2c0.002595c

1 ( 0.9092 kg>m3 )( 194.44 m>s ) 2 d (6 m)(2 m)d 2 Ans.

= 1070.65 N = 1.07 kN

1193

11–53. The barge is traveling forward at 15 ft>s in still water having a temperature of 60°F. If the bottom of the barge can be assumed to be a flat plate of length 120 ft and width 25 ft, determine the power of the engine required to overcome the frictional resistance of the water on the bottom of the barge. Consider both laminar and turbulent boundary layers.

120 ft

SOLUTION Water is considered to be incompressible. The relative flow is steady. From Appendix A, r = 1.939 slug>ft 3 and n = 12.2 ( 10-6 ) ft 2 >s for water at T = 60° F . Thus, the Reynolds number at x = L = 120 ft is Re L =

( 15 ft>s ) (120 ft) UL = = 1.475 ( 108 ) n 12.2 ( 10-6 ) ft 2 >s

Since 5 ( 105 ) … Re L 6 109, the frictional drag coefficient considering both laminar and turbulent boundary layers is CD =

0.455

( log 10Re L ) 2.58

= 0.0020051

-

1700 = Re L

0.455

3 log 10 3 1.475 ( 108 ) 4 4 2.58

-

1700 1.475 ( 108 )

Thus, the frictional drag force on the bottom surface can be determined from 1 FD = CD a rU 2 bbL 2 = 0.0020051c

1 ( 1.939 slug>ft 3 )( 15 ft>s ) 2 d (25 ft)(120 ft) 2

= 1312.17 lb Thus, the power required to overcome FD is P = FD # V = (1312.17 lb) ( 15 ft>s ) = a19 683 = 35.8 hp

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

Ans: 35.8 hp 1194

11–54. The plate is 2 m wide and is held at an angle of 12° with the wind as shown. If the average pressure under the plate is 40 kPa, and on the top it is 60 kPa, determine the pressure drag on the plate.

3m 12!

SOLUTION

(FR)y

The air is considered to be incompressible. The flow is steady. The resultant forces of the pressure on the top and bottom surfaces of the plate are Ft = ptA = c 60 ( 103 ) Fn = pnA = c 40 ( 103 )

N d 3 2 m(3 m)4 = 360 ( 103 ) N = 360 kN m2

Fpd = (FR)x

Ft = 360 kN

12°

N d 3 2 m(3 m)4 = 240 ( 103 ) N = 240 kN m2

The pressure drag is equal to the component of the resultant force along the direction of the free-stream flow, which in this case is horizontal. Referring to Fig. a, + ( ) S FR x = ΣFx ; FPD = (360 kN) sin 12° - (240 kN) sin 12° = 24.9 kN Ans.

60 kPa

40 kPa 12° Fb = 240 kN (a)

Ans: 24.9 kN 1195

11–55. Wind blows over the inclined surface and produces the approximate pressure distribution shown. Determine the pressure drag acting over the surface if the surface is 3 m wide.

4m

80 Pa

10 m/s

30!

SOLUTION

(FR)y

The air is considered to be incompressible. The flow is steady. FPD = (FR)x

The resultant force of the pressure on the inclined surface is F =

1 N a80 2 b 3 3 m(4 m)4 = 480 N 2 m

F = 480 N 30°

The pressure drag is equal to the component of the resultant force along the direction of the free-stream flow which in this case is horizontal. Referring to Fig. a, + S (FR)x

= ΣFx;

Ans.

FD = (480 N) sin 30° = 240 N

80 Pa (a)

Ans: 240 N 1196

*11–56. The sign is subjected to a wind profile that produce a pressure distribution that can be approximated by r = 1 112.5 ey0.6 2 Pa, where y is in meters. Determine the resultant pressure force on the sign due to the wind. The air is at a temperature of 20°C, and the sign is 0.5 m wide.

p

3m

y 3m

SOLUTION The air is considered to be incompressible. The flow is steady. The dynamic pressure can be determined from p = ( 112.5ry0.6 ) Pa The force of this pressure on a differential area dA = bdy = (0.5 m)dy is dF = pdA = ( 112.5ry0.6 ) (0.5dy) = 56.25ry0.6dy. Thus, the resultant force on the entire surface of the sign is FR =

LA

dF =

6m

L3 m

56.25ry0.6dy =

56.25ry1.6dy 6 m ` = (414.19r) N 1.6 3m

From Appendix A, r = 1.202 kg>m2 for air at T = 20° C. Thus, FR = 414.19 ( 1.202 kg>m2 ) = 498 N

Ans.

1197

11–57. The air pressure acting from A to B on the surface of a curved body can be approximated as p = (5 - 1.5u) kPa, where u is in radians. Determine the pressure drag acting on the body from 0 … u … 90°. The body has a width of 300 mm.

B 100 mm u A

SOLUTION

(FR)y

The air is considered to be incompressible. The flow is steady. The force of the pressure on a differential area dA = bds = brdu = (0.3 m)(0.1 m)du = 0.03du is dF = pdA = (5 - 1.5u) ( 103 ) (0.03du) = 30(5 - 1.5u)du. The pressure drag is equal to the component of the force along the direction of the free-stream flow, which in this case is horizontal. Referring to Fig. a, + S (FR)x

= ΣFx ;

FPD =

LA

(dF)x =

L0

p 2

dF cos u =

L0

p 2

FPD = (FR)x

dF

30(5 - 1.5u) cos u du

d

p

= 30 3 5 sin u - 1.5(cos u + u sin u) 4 # 02

= 124.31 N = 124 N

Ans.

(a)

Ans: 124 N 1198

11–58. The air pressure acting on the inclined surfaces is approximated by the linear distributions shown. Determine the horizontal force resultant acting on the surface if it is 3 m wide.

3 kPa 3 kPa 45!

5 kPa

6m

20! 6m

SOLUTION

(FR)y

The air is considered to be incompressible. The flow is steady. FPD = (FR)x

The resultant force of the trapezoidal and triangular pressure prism are Ftrap = Ftri =

Ftri = 27 kN

1 N c (5 + 3) ( 103 ) 2 d 36 m(3 m)4 = 72 ( 103 ) N = 72 kN 2 m

45° Ftrap = 72 kN 20° 3 kPa

1 N c 3 ( 103 ) 2 d 36 m(3 m)4 = 27 ( 103 ) N = 27 kN 2 m

5 kPa (a)

Referring to Fig. a, + S (FR)x

= ΣFx ;

FPD = (72 kN) sin 20° + (27 kN) sin 45° = 43.7 kN

Ans.

Ans: 43.7 kN 1199

11–59. The front of the building is subjected to wind that exerts a pressure of p = 1 0.25y1>2 2 lb>ft 2, where y is in feet, measured from the ground. Determine the resultant pressure force on the windward face of the building due to this loading.

y 80 ft

30 ft

SOLUTION The air is considered to be . The flow is steady. The force of the wind pressure on a differential area dA = bdy = (80 ft)dy is 1

1

dF = pdA = ( 0.25y2 ) (80 dy) = 20y2dy. Thus, the resultant force on the entire windward surface is FR =

LA

dF =

L0

30 ft

1 2 3 30 ft 20y2dy = 20 a b y2 ` 3 0

Ans.

= 2190.89 lb = 2.19 kip

Ans: 2.19 kip 1200

*11–60. The building is subjected to a uniform wind having a speed of 80 ft>s. If the temperature of the air is 40°F, determine the resultant pressure force on the front of the building if the drag coefficient is 1.43.

y 80 ft

30 ft

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 0.00247 slug>ft 3 for air at T = 40° F . FD = CDApr

U2 2

= 1.43 380 ft(30 ft)4 ( 0.00247 slug>ft 3 ) £

( 80 ft>s ) 2 2

§

= 27.13 ( 103 ) lb = 27.1 kip

Ans.

1201

11–61. Determine the moment developed at the base A of the square sign due to wind drag if the front of the signboard is subjected to a 16 m>s wind. The air is at 20°C. Neglect the drag on the pole.

2m

2m

3m

SOLUTION

A

The air is considered to be incompressible. The flow is steady. From Appendix A, r = 1.202 kg>m3 and n = 15.1 ( 10-6 ) m2 >s for air at T = 20° C . Thus, the Reynolds number of the flow is Re =

( 16 m>s ) (2 m) UL = = 2.12 ( 106 ) n 15.1 ( 10-6 ) m2 >s

Since Re 7 104, the value of CD for the plate in Table 11–3 can be used. For b 2m = = 1, CD = 1.10. Here, AP = 2 m(2 m) = 4 m2 h 2m

FD = 1415.47 N

4m

Ax

FD = CDAP r

( 16 m>s ) U = 1.10 ( 4 m2 )( 1.202 kg>m3 ) £ § 2 2 2

2

MA Ay (a)

= 676.97 N Here, FD will act through the center of the signboard, as shown on its free-body diagram, Fig. a. Consider the moment equilibrium about point A. a + ΣMA = 0;

MA - (676.97)(4 m) = 0 MA - 2.707 ( 103 ) N # m = 2.71 kN # m

Ans.

Ans: 2.71 kN # m 1202

11–62. The mast on the boat is held in place by the rigging, which consists of rope having a diameter of 0.75 in. and a total length of 130 ft. Assuming the rope to be cylindrical, determine the drag it exerts on the boat if the boat is moving forward at a speed of 30 ft>s. The air has a temperature of 60°F.

SOLUTION We will assume that steady laminar flow occurs, and the air is incompressible. Appendix A gives ra = 0.00237 slug>ft 3 and ma = 0.374 ( 10-6 ) lb # s>ft 2 at T = 60° F . The Reynolds number is raUD = Re = ma

( 0.00237 slug>ft 3 )( 30 ft>s ) a

0.374 ( 10-6 ) lb # s>ft 2

0.75 ft b 12

= 1.188 ( 104 )

With this Reynolds number, the drag coefficient for the cylinder can be obtained using Fig. 11–31, for which CD = 1.3 (approximately). Then the drag force on the rope can be determined by applying. FD = CDAP a = 1.3c a

raV 2 b 2

( 0.00237 slug>ft 0.75 ft b(130 ft) d £ 12 2

3

)( 30 ft>s ) 2

§ Ans.

= 11.3 lb

Ans: 11.3 lb 1203

11–63. Wind at 20°C blows against the 100-mm-diameter flagpole with a speed of 1.20 m>s. Determine the drag on the pole if it has a height of 8 m. Consider the pole to be a smooth cylinder. Would you consider this a significant force?

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 1.202 kg>m3, n = 15.1 ( 10-6 ) m2 >s for air at T = 20° C . Here, it is required that Re = Re =

UD n

1.20 m>s(0.1 m) 15.1 ( 10-6 ) m2 >s

= 7947

With Re = 7947, CD for the cylindrical pole can be determined from the graph. CD ≈ 1.2 (Approx.) Thus,

( 1.20 m>s ) u2 = 1.2(8 m)(0.1 m) ( 1.202 kg>m3 ) ° ¢ 2 2 2

FD = CDAP r

Ans.

FD = 0.831 N This is a very small force.

Ans: 0.831 N 1204

*11–64. Each of the smooth bridge piers (cylinders) has a diameter of 0.75 m. If the river maintains an average speed of 0.08 m>s, determine the drag the water exerts on each pier. The water temperature is 20°C. 0.08 m/s 0.75 m

SOLUTION Water is considered to be incompressible. The flow is steady. From Appendix A, r = 998.3 kg>m3 and n = 1.00 ( 10-6 ) m2 >s . Thus, the Reynolds number of the flow is Re =

UD = n

( 0.08 m>s ) (0.75 m) = 6 ( 104 ) 1.00 ( 10-6 ) m2 >s

Since the pier is a cylinder (rough), the drag coefficient can be determined by entering this Re on the graph which gives CD = 1.4 (approximately). Also, AP = 0.75 m(6 m) = 4.5 m2.

( 0.08 m>s ) U2 ¢ = 1.4 ( 4.5 m2 )( 998.3 kg>m3 ) ° 2 2 2

FD = CDAP r

Ans.

= 20.1 N

1205

6m

11–65. A 60-mi>h wind blows on the side of the truss. If the members are each 4 in. wide, determine the drag acting on the truss. The air is at 60°F, and CD = 1.2. Note that 1 mi = 5280 ft.

F E 10 ft A

B 10 ft

10 ft

C

D

10 ft

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 0.00237 slug>ft 3 for air at T = 60° F. Here, mi 5280 ft 1h U = a60 ba ba b = 88 ft>s . The projected area of the truss’s h 1 mi 3600 s members perpendicular to the air stream is AP = c 6(10 ft) + 32 ( 10 ft ) 2 + ( 10 ft ) 2 d a

4 ft b = 34.14 ft 2 12

( 88 ft>s ) U2 § = 1.2 ( 34.14 ft 2 )( 0.00237 slug>ft 3 ) £ 2 2 2

FD = CDAp r

Ans.

= 375.97 lb = 376 lb

Ans: 376 lb 1206

11–66. A periscope on a submarine has a submerged length of 2.5 m and a diameter of 50 mm. If the submarine is traveling at 8 m>s, determine the moment developed at the base of the periscope. The water is at a temperature of 15°C. Consider the periscope to be a smooth cylinder.

SOLUTION

W

Water is considered to be incompressible. The relative flow is steady. From Appendix A, r = 999.2 kg>m3 and n = 1.15 ( 10-6 ) m2 >s for water at T = 15° C . Thus, the Reynolds number of the flow is Re =

( 8 m>s ) (0.05 m) UD = = 3.48 ( 105 ) n 1.15 ( 10-6 ) m2 >s

1.25 m

Since the periscope is a cylinder (smooth), the drag coefficient can be determined by entering this Re into Fig. 11–31 which gives CD ≅ 0.85 (approx.). Also, AP = 0.05 m(2.5 m) = 0.125 m2.

Ax MA Ay (a)

( 8 m>s ) U2 = 0.85 ( 0.125 m2 )( 999.2 kg>m3 ) £ § 2 2 2

FD = CDAp r

FD = 3197.44 N

= 3397.28 N

Here, FD acts through the mid-length of the periscope’s submerged length as shown in its free-body diagram in Fig. a, a + ΣMA = 0;

MA - 3397.28 N(1.25 m) = 0

MA = 4246.6 N # m = 4.25 kN # m

Ans.

Ans: 4.25 kN # m 1207

11–67. The antenna on the building is 20 ft high and has a diameter of 12 in. Determine the restraining moment at its base to hold it in equilibrium if it is subjected to a wind having an average speed of 80 ft>s. The air is at a temperature of 60°F. Consider the antenna to be a smooth cylinder.

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 0.00237 slug>ft 3 and n = 0.158 ( 10-3 ) ft 2 >s for air at T = 60° F. Thus, the Reynolds number for the flow is Re =

( 80 ft>s ) (1 ft) UD = = 5.06 ( 105 ) n 0.158 ( 10-3 ) ft 2 >s

Since the antenna is a cylinder (smooth), the drag coefficient can be determined by entering this Re into the graph, which gives CD ≅ 0.32 (approx.). Also, AP = 1 ft(20 ft) = 20 ft 2.

( 80 ft>s ) U2 = 0.32 ( 20 ft 2 )( 0.00237 slug>ft 3 ) c d 2 2 2

FD = CDAp r

10 ft

= 48.54 lb

Here, FD acts through the mid-height of the antenna as shown in its free-body diagram in Fig. a, a + ΣMA = 0;

FD = 48.54 lb

Ax

M

(a)

MA - (48.54 lb)(10 ft) = 0 MA = 485 lb # ft

Ans.

Ans: 485 lb # ft 1208

*11–68. The truck has a drag coefficient of CD = 1.12 when it is moving with a constant velocity of 80 km>h. Determine the power needed to drive the truck at this speed if the average front projected area of the truck is 10.5 m2. The air is at a temperature of 10°C.

SOLUTION The air is considered to be incompressible. The flow is steady. Appendix A, r = 1.247 kg>m3 for 1h km 1000 m U = a80 ba ba b = 22.22 m>s . h 1 km 3600 s

From

FD = CDAp r

air

at

T = 10° C .

Here,

( 22.2 m>s ) U2 = 1.12 ( 10.5 m2 )( 1.247 kg>m3 ) c d 2 2 2

= 3620.92 N Thus, the power needed to overcome the drag is

#

W = FD # V = (3620.92 N) ( 22.2 m>s ) = 80.46 ( 103 ) W Ans.

= 80.5 kW

1209

11–69. The truck has a drag coefficient of CD = 0.86 when it is moving with a constant velocity of 60 km>h. Determine the power needed to drive the truck at this speed if the average front projected area of the truck is 10.5 m2. The air is at a temperature of 10°C.

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 1.247 kg>m3 for air at T = 10° C . Here, U = a60

km 1000 m 1h ba ba b = 16.67 m>s . h 1 km 3600 s

FD = CDAP r = 1564 N

( 16.667 m>s ) U2 = 0.86 ( 10.5 m2 )( 1.247 kg>m3 ) c d 2 2 2

Thus, the power needed to overcome the drag is

#

W = FD # V = (1564 N) ( 16.667 m>s ) = 26.07 ( 103 ) W Ans.

= 26.1 kW

Ans: 26.1 kW 1210

11–70. Wind at 10°C blows against the 30-m-high chimney at 2.5 m>s. If the diameter of the chimney is 2 m, determine the moment that must be developed at its base to hold it in place. Consider the chimney to be a rough cylinder.

2m

30 m

SOLUTION W

The air is considered to be incompressible. The flow is steady. From Appendix A, r = 1.247 kg>m and n = 14.2 ( 10 Thus, the Reynolds number of the flow is 3

Re =

( 2.5 m>s ) (2 m) UD = = 3.52 ( 105 ) n 14.2 ( 10-6 ) m2 >s

-6

) m >s for air at T = 10° C . 2

15 m

Since the chimney is a cylinder (rough), the drag coefficient can be determined by entering this Re into the graph, which gives C D ≅ 0.5 (approx.). Also, AP = 2 m ( 30 m ) = 60 m2. FD = CDAp r

( 2.5 m>s ) U = 0.5 ( 60 m2 )( 1.247 kg>m3 ) c d 2 2 2

FD = 116.90 N

2

Ax MA Ay (a)

= 116.90 N

Here, FD acts through the mid-height of the chimney as shown in its free-body diagram in Fig. a, + ΣMA = 0;

MA - (116.90 N)(15 m) = 0

MA = 1753.59 N # m = 1.75 kN # m

Ans.

Ans: 1.75 kN # m 1211

11–71. A rectangular plate is immersed in a stream of oil flowing at 0.5 m>s. Compare the drag acting on the plate if it is oriented so that AB is the leading edge and then when it is rotated 90° counterclockwise so that BC is the leading edge. The plate is 0.8 m wide. Take ro = 880 kg>m3.

C

B 8 m/s

0.2 m

A

0.4 m

SOLUTION Oil is considered to be incompressible. The flow is steady. b 0.8 m = = 4. From the table, CD = 1.19 for a h 0.2 m rectangular plate. Also, AP = (0.8 m)(0.2 m) = 0.16 m2.

When AB is the leading edge,

( FD ) AB = CDAP r = 20.9 N

( 0.5 m>s ) U2 = 1.19 ( 0.16 m2 )( 880 kg>m3 ) c d 2 2 2

Ans.

b 0.8 m = = 2. From the table, CD = 1.15 for a h 0.4 m rectangular plate. Also, AP = (0.8 m)(0.4 m) = 0.32 m2.

When BC is the leading edge,

( FD ) BC = CDAP r = 40.5 N

( 0.5 m>s ) U2 = 1.15 ( 0.32 m2 )( 880 kg>m3 ) c d 2h 2 2

1212

Ans.

Ans: 1FD 2 AB = 20.9 N 1FD 2 BC = 40.5 N

*11–72. The parachute has a drag coefficient of CD = 1.36 and an open diameter of 4  m. Determine the terminal velocity as the man parachutes downward. The air is at 20°C. The total mass of the parachute and man is 90 kg. Neglect the drag on the man.

V

SOLUTION The air is considered to be incompressible. The relative flow is steady. Since the parachutist descends with a constant terminal velocity, the acceleration is zero. Referring to the free-body diagram shown in Fig. a, + c ΣFy = may;

FD - 90(9.81) N = 90(0)

FD = 882.9 N

From Appendix A, r = 1.202 kg>m3 for air at T = 20° C . Here, the projected area of the parachute perpendicular to the air stream is AP = p(2 m)2 = 4pm2. FD = CDAP r

U2 2

882.9 N = 1.36 ( 4p m2 )( 1.202 kg>m3 ) a U = 9.27 m>s

U2 b 2

Ans.

FD

a=0

90(9.81) N (a)

1213

11–73. The parachute has a drag coefficient of CD = 1.36. Determine the required open diameter of the parachute so the man attains a terminal velocity of 10 m>s. The air is at 20°C. The total mass of the parachute and man is 90 kg. Neglect the drag on the man.

V

SOLUTION The air is considered to be incompressible. The relative flow is steady. Since the parachutist descends with a constant terminal velocity, the acceleration is zero. Referring to the free-body diagram shown in Fig. a, + c ΣFy = may;

FD - 90(9.81) N = 90(0)

FD = 882.9 N

From Appendix A, r = 1.202 kg>m3 for air at T = 20° C . Here, the projected area d 2 pd 2 of the parachute perpendicular to the air stream is AP = p a b = . 2 4 FD = CDAP r

U2 2

( 10 m>s ) pd 2 b ( 1.202 kg>m3 ) c d 4 2 2

882.9 N = 1.36a d = 3.71 m

Ans.

FD

a=0

90(9.81) N (a)

Ans: 3.71 m 1214

11–74. The man and the parachute have a total mass of 90 kg. If the parachute has an open diameter of 6 m and the man attains a terminal velocity of 5 m>s, determine the drag coefficient of the parachute. The air is at 20°C. Neglect the drag on the man.

V

SOLUTION The air is considered to be incompressible. The relative flow is steady. Since the parachutist descends with a constant terminal velocity, the acceleration is zero. Referring to the free-body diagram shown in Fig. a, + c ΣFy = may;

FD - 90(9.81) N = 90(0)

FD = 882.9 N

3

From Appendix A, r = 1.202 kg>m for air at T = 20° C . Here, the projected area 6m 2 of the parachute perpendicular to the air stream is AP = p a b = 9p m2. 2 FD = CDAP r

U2 2

882.9 N = CD ( 9p m2 )( 1.202 kg>m3 ) c

( 5 m>s ) 2 2

CD = 2.08

d

Ans.

FD

a=0

90(9.81) N (a)

Ans: 2.08 1215

11–75. The car has a projected front area of 14.5 ft 2. Determine the power required to drive at a constant velocity of 60 mi>h if the drag coefficient is CD = 0.83 and the air is at 60°F. Note that 1 mi = 5280 ft.

60 mi/h

SOLUTION The air is considered to be incompressible. The relative flow is steady. From

Appendix

U = a60

r = 0.00237 slug>ft 3

A,

for

air

at

T = 60° F.

1h mi 5280 ft ba ba b = 88 ft>s . h 1 mi 3600 s FD = CDAP r

Here,

( 88 ft>s ) U2 = 0.83 ( 14.5 ft 2 )( 0.00237 slug>ft 3 ) c d 2 2 2

= 110.44 lb

Referring to the free-body diagram shown in Fig. a, + d ΣFx

= ma;

F - 110.44 lb = 0

F = 110.44 lb

Subsequently, the power needed to produce this drive force is P = F # U = 110.44 lb ( 88 ft>s ) = a9718.80 a=0 x

1 hp ft # lb ba # b = 17.7 hp s 550 ft s lb

Ans.

W

FD = 110.44 lb F (a)

N

Ans: 17.7 hp 1216

*11–76. A 5-m-diameter balloon is at an altitude of 2 km. If it is moving with a terminal velocity of 12 km>h, determine the drag on the balloon.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 1.007 kg>m3 and n = 17.26 ( 10-6 ) m2 >s for air at an altitude

of 2 km. Here, U = a12 number is

Re =

km 1000 m 1h ba ba b = 3.333 m>s. Thus, the Reynolds h 1 km 3600 s

( 3.333 m>s ) (5 m) UD = = 9.656 ( 105 ) n 17.26 ( 10-6 ) m2 >s

Entering this Re into the graph for a sphere, CD ≅ 0.16 (approx.). Here, 5m 2 AP = p a b = 6.25p m2. 2

( 3.333 m>s ) U2 0.16 ( 6.25p m2 )( 1.007 kg>m3 ) c d 2 2 2

FD = CDAP r = 17.6 N

1217

Ans.

11–77. The drag coefficient for the car is CD = 0.28, and the projected area into the 20°C airstream is 2.5 m2. Determine the power the engine must supply to maintain a constant speed of 160 km>h.

160 km/h

SOLUTION The air is considered to be incompressible. The relative flow is steady. Appendix A, r = 1.202 kg>m3 for 1h km 1000 m U = a160 ba ba b = 44.44 m>s. h 1 km 3600 s

air

From

FD = CDAP r

at

T = 20° C.

Here,

( 44.44 m>s ) U2 = 0.28 ( 2.5 m2 )( 1.202 kg>m3 ) £ § 2 2 2

= 831.01 N

Referring to the free-body diagram shown in Fig. a, + d ΣFx

= ma;

F - 831.01 N = 0

F = 831.01 N

Subsequently, the power that must be supplied by the engine to produce this drive force is P = F # U = (831.01 N) ( 44.44 m>s ) = 36.93 ( 103 ) W

Ans.

= 36.9 kW a=0

W x

FD = 831.01 N F (a)

N

Ans: 36.9 kW 1218

11–78. The rocket has a nose cone that is 60° and a base diameter of 1.25 m. Determine the drag of the air on the cone when the rocket is traveling at 60 m>s in air having a temperature of 10°C. Use Table 11–3 for the cone, but explain why this may not be an accurate assumption.

60! 1.25 m

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 1.247 kg>m3 and n = 14.2 ( 10-6 ) m2 >s for air at T = 10° C. Thus, the Reynolds number of the air flow is Re =

( 60 m>s ) (1.25 m) UD = = 5.28 ( 106 ) n 14.2 ( 10-6 ) m2 >s

Since Re 7 104, the value of CD for the cone in the table can be used. For u = 60°, 1.25 m 2 CD = 0.8. Here, AP = p a b = 0.390625p m2. 2

( 60 m>s ) U2 § = 0.8 ( 0.39025p m2 )( 1.247 kg>m3 ) £ 2 2 2

FD = CDAP r

= 2.204 ( 103 ) N = 2.20 kN

Ans.

Ans: 2.20 kN 1219

11–79. A boat traveling with a constant velocity of 2 m>s tows a half submerged log having an approximate diameter of 0.35 m. If  the  drag coefficient is CD = 0.85, determine the tension in the tow rope if it is horizontal. The log is oriented so that the flow is along the length of the log.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Since the flow is along the length of the log and the log has an approximate diameter of 0.35 m, AP = p a

0.35 m 2 b = 0.030625p m2 2

The drag on the log is F = CDAP r

U2 2

( 2 m>s ) 1 = 0.85a b ( 0.030625p m2 )( 1000 kg>m3 ) c d 2 2 2

= 81.78 N

Referring to the free-body diagram of the log in Fig. a, + d ΣFx

= max ;

T - 81.78 N = 0 Ans.

T = 81.78 N = 81.8 N

a=0 x

W FD = 81.77 N

T

N (a)

Ans: 81.8 N 1220

*11–80. A 0.25-lb ball has a diameter of 3 in. Determine the initial acceleration of the ball when it is thrown vertically downward with an initial speed of 20 ft>s. The air is at a temperature of 60°F.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 0.00237 slug>ft 3 and n = 0.158 ( 10-3 ) ft 2 >s for air at T = 60° F. Thus, the Reynolds number is Re =

( 20 ft>s ) a

3 ft b 12

UD = = 3.165 ( 104 ) n 0.158 ( 10-3 ) ft 2 >s

Entering this Re into the graph for a sphere, CD ≅ 0.5 (approx.). Here 2 1.5 AP = p a ft b = 0.015625p ft 2. 12

( 20 ft>s ) U2 = 0.5 ( 0.015625p ft 2 )( 0.00237 slug>ft 3 ) c d 2 2 2

FD = CDAD r

Ans.

= 0.01163 lb

Writing the equation of motion along the y axis by referring to the free-body diagram of the ball in Fig. a, + T ΣFy = may:

0.25 lb - 0.01163 lb = a

0.25 lb ba 32.2 ft>s2

a = 30.7 ft>s2

Ans.

0.25 lb y a

FD = 0.01163 lb (a)

1221

11–81. A 1 ft by 1 ft square plate is held in air at 60°F, which is blowing at 50 ft>s. Compare the drag on the plate when it is held normal and then parallel to the air flow.

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 0.00237 slug>ft 3 and n = 0.158 ( 10-3 ) ft 2 >s for air at T = 60° F. When the plate is held normal to the air flow, the drag is contributed b by pressure drag only. For this case, CD = 1.1 for a square plate where = 1 and h AP = 1 ft(1 ft) = 1 ft 2. Normal:

( 50 ft>s ) U2 = 1.1 ( 1 ft 2 )( 0.00237 slug>ft 3 ) c d 2 2 2

FD = CD AP r = 3.26 lb

Ans.

When the plate is held parallel to the air flow, the drag is contributed by frictional drag only. Here, the Reynolds number for the flow at x = L = 1 ft is Re L =

( 50 ft>s ) (1 ft) UL = = 3.165 ( 105 ) n 0.158 ( 10-3 ) ft 2 >s

Since Re L 6 (Re x)cr = 5 ( 105 ) , the boundary layer throughout the length of the plate is laminar. Since there are two surfaces subjected to flow Parallel: FD = Σ

0.664brU 2L 2Re L

= 0.0140 lb

= 2c

0.664(1 ft) ( 0.00237 slug>ft 3 )( 50 ft>s ) 2(1 ft) 23.165 ( 105 )

d

Ans.

Ans: Normal: FD = 3.26 lb Parallel: FD = 0.0140 lb 1222

11–82. The smooth empty drum has a mass of 8 kg and rests on a surface having a coefficient of static friction of ms = 0.3. Determine the speed of the wind needed to cause it to either tip over or slide. The air temperature is 30°C.

0.3 m

1.25 m

SOLUTION The air is considered to be incompressible. The flow is steady. Here, FD acts through the mid-height of the drum as shown in its free-body diagram in Fig. a, + c ΣFy = 0;

N - 8(9.81)N = 0

+ S ΣFx

FD - F = 0

(1)

[8(9.81) N]x - FD(0.625 m) = 0

(2)

= 0;

a+ ΣMO = 0;

N = 78.48 N

Assuming sliding occurs first,

O

Using this result to solve Eqs. (1) and (2),

x

x = 0.1875 m

Since x 6 0.3 m, the drum will slide before it tips as assumed. From Appendix A, r = 1.164 kg>m3 and n = 16.0 ( 10-6 ) m2 >s for air at T = 30° C. Thus, the Reynolds number of the flow is Re =

FD 0.625 m

F = msN = 0.3(78.48 N) = 23.544 N FD = 23.544 N

8(9.81) N

U(0.6 m) UD = = 37 500U n 16.0 ( 10-6 ) m2 >s

F

N (a)

(3)

Also, AP = (0.6 m)(1.25 m) = 0.75 m2. FD = CDAP r

U2 2

23.544 N = CD ( 0.75 m2 )( 1.164 kg>m3 ) a U2 =

53.938 CD

U2 b 2

(4)

The iterations carried out are tabulated as follows: Iteration 1 2

Assumed CD 1.0 1.2

U ( m>s ) : Eq. (4) Re: Eq. (3) 7.344 2.75 ( 105 ) 6.704 2.51 ( 105 )

CD from the graph gives 1.2 1.2

Since the assumed CD is almost the same as that obtained from the graph in iteration 2, the result of U in the iteration is acceptable. Thus, Ans.

U = 6.70 m>s

Ans: 6.70 m>s 1223

11–83. The smooth empty drum has a mass of 8 kg and rests on a surface having a coefficient of static friction of ms = 0.6. Determine the speed of the wind needed to cause it to either tip over or slide. The air temperature is 30°C.

0.3 m

1.25 m

SOLUTION The air is considered to be incompressible. The flow is steady. Here, FD acts through the mid-height of the drum as shown in its free-body diagram in Fig. a. Considering the equilibrium of the drum, + c ΣFy = 0;

N - 8(9.81) N = 0

+ S ΣFx

FD - F = 0

(1)

[8(9.81) N]x - FD(0.625 m) = 0

(2)

= 0;

a+ ΣMO = 0;

N = 78.48 N

Assuming sliding occurs first,

O

Using this result to solve Eqs. (1) and (2),

x

x = 0.375 m

From Appendix A, r = 1.164 kg>m3 and n = 16.0 ( 10-6 ) m2 >s for air at T = 30° C. Thus, the Reynolds number of the flow is U(0.6 m) UD = = 37 500U n 16.0 ( 10-6 ) m2 >s

F

N

Tipping will occur first. Setting x equal to 0.3 m in Eq. (2), FD = 37.6704 N.

Re =

FD 0.625 m

F = msN = 0.6(78.48 N) = 47.088 N FD = 47.088 N

8(9.81) N

(a)

(3)

Also, AP = (0.6 m)(1.25 m) = 0.75 m2. FD = CDAP r

U2 2

37.67 N = CD ( 0.75 m2 )( 1.164 kg>m3 ) a U2 =

86.301 CD

U2 b 2

The iterations carried out are tabulated as follows: Iteration Assumed CD U ( m>s ) : Eq. (4) Re: Eq. (3) 1 1.0 9.290 3.48 ( 105 ) 2 0.8 10.386 3.89 ( 105 ) 3 0.64 11.61 4.35 ( 105 ) 4 0.50 13.137 4.92 ( 105 ) 5 0.35 15.703 5.89 ( 105 )

(4)

CD from the graph 0.8 0.64 0.50 0.33 0.34

Since the assumed CD is almost the same as that obtained from the graph in iteration 5, the result of U in the iteration is acceptable. Thus, Ans.

U = 15.7 m>s

Ans: 15.7 m>s 1224

*11–84. The blades of a mixer are used to stir a liquid having a density r and viscosity m. If each blade has a length L and width w, determine the torque T needed to rotate the blades at a constant angular rate v. Take the drag coefficient of the blade’s cross section to be CD.

T v

L L

w

SOLUTION The liquid is considered to be incompressible. The relative flow is steady.

z

The drag on the differential area dA = wdr is shown on the free-body diagram of the blade, Fig. a. dFD = CDAP r

T

dFD = CD(wdr)rc

(vr)2 2

d =

1 C rwv2r 2dr 2 D

r

Since the blade rotates with a constant angular velocity, moment equilibrium exists about the z axis. Thus, L

2 dFD(r) - T = 0 L0 L

L

1 dFD(r) = 2 CD rwv2r 3dr T = 2 L0 L0 2 L0

L

r 3dr

r4 2 b 4 0

L

= CDrwv2 a =

dr

dFD

Here, AP = dA = wdr and U = vr. Thus,

= CDrwv2

dr

U2 2

1 CD rwv2L4 4

Ans.

1225

r

dFD

11–85. A ball has a diameter of 60 mm and falls in oil with a terminal velocity of 0.8 m>s. Determine the density of the ball. For oil, take ro = 880 kg>m3 and n0 = 40 ( 10-6 ) m2 >s. Note: The volume of a sphere is V = 43pr 3.

SOLUTION

FB = 0.97635 N

W = 1.109 )10 –3) y

Oil is considered to be incompressible. The relative flow is steady. The initial Reynolds number is Re =

UD = n

( 0.08 m>s ) (0.06 m) = 1200 40 ( 10-6 ) m2 >s

b

a=0

Using this Re, CD ≅ 0.44. Also, AP = p(0.03 m)2 = 0.9 ( 10-3 ) p m2. 2

FD = CDAP r0

(0.8 m) U2 d = 0.44 3 0.9 ( 10-3 ) p m2 4 ( 880 kg>m3 ) c 2 2

FD = 0.36625 N

(a)

= 0.35033 N

4 3 4 pr = p(0.03 m)3 = 36 ( 10-6 ) p m3. Thus, the 3 3 weight of the ball and the bouyant force are

The volume of the ball is V =

W = mg = rbVg = rb 3 36 ( 10-6 ) p m3 4 ( 9.81 m>s2 ) = 1.109 ( 10-3 ) rb

FB = r0Vg = ( 880 kg>m3 ) 3 36 ( 10-6 ) p m3 4 ( 9.81 m>s2 ) = 0.97635 N

Referring to the free-body diagram in Fig. a, + c ΣFy = may;

0.35033 N + 0.97635 N = 1.109 ( 10-3 ) rb rb = 1196 kg>m3 = 1.20 mg>m3

Ans.

Ans: 1.20 mg>m3 1226

11–86. A ball has a diameter of 8 in. If it is kicked with a speed of 18 ft>s, determine the initial drag acting on the ball. Does this force remain constant? The air is at a temperature of 60°F.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 0.00237 slug>ft 3 and n = 0.158 ( 10-3 ) ft 2 >s for air at T = 60° F . Thus, the initial Reynolds number is 8 ( 18 ft>s ) a ft b 12 UD Re = = = 7.59 ( 104 ) n 0.158 ( 10-3 ) ft 2 >s

Entering this Re into the graph for a sphere, CD ≅ 0.5 (approx.). Also, 2 4 AP = p a ft b = 0.1111p ft 2. 12

( 18 ft>s ) U2 § = 0.5 ( 0.1111p ft 2 )( 0.00237 slug>ft 3 ) £ 2 2 2

FD = CDAP r

= 0.0670 lb

Ans.

The drag force on the ball will not remain constant since the velocity of the ball changes. Furthermore, it also depends on the drag coefficient, which is a function of velocity.

Ans: 0.0670 lb 1227

11–87. Particulate matter at an altitude of 8 km in the upper atmosphere has an average diameter of 3 µm. If a particle has a mass of 42.5 ( 10-12 ) g, determine the time needed for it to settle to the earth. Assume gravity is constant, and for air, r = 1.202 kg>m3 and m = 18.1 ( 10-6 ) N # s>m2.

8 km

]

SOLUTION The air is considered to be incompressible. The relative flow is steady.

]

42.5 (10–12( (9.81) N 1000 y

Here, we will assume that Re 6 1. Therefore, the drag is

FD = 3pmUD = 3p 3 18.1 ( 10-6 ) N # s>m2 4 (U) 3 3 ( 10-6 ) m 4

a=0

= 0.1629 ( 10-9 ) pU

Referring to the free-body diagram in Fig. a, 0.1629 ( 10-9 ) pU - £

+ c ΣFy = 0;

42.5 ( 10-12 ) 1000

U = 0.8147 ( 10-3 ) m>s

FD = 0.1629 (10–9( U

§ (9.81) N = 0

(a)

Therefore, the Reynolds number is Re =

rUD = m

( 1.202 kg>m3 ) 3 0.8147 ( 10 - 3 ) m>s 4 3 3 ( 10-6 ) m 4 = 1.623 ( 10-4 ) 6 1 (O.K.) 18.1 ( 10-6 ) N # s>m2

Thus, the time needed to settle t =

8 ( 103 ) m s = = U 0.8147 ( 10-3 ) m>s

= 113.66 days = 114 days

3 9.820 ( 106 ) s 4 a

1 day 1 hr ba b 3600 s 24 hr

Ans.

Ans: 114 days 1228

*11–88. A solid ball has a diameter of 20 mm and a density of 3.00 Mg>m3. Determine its terminal velocity if it is dropped into a liquid having a density of r = 2.30 Mg>m3 and a viscosity of n = 0.052 m2 >s. Note: The volume of a sphere is V = 43pr 3.

SOLUTION

FB =

6

The liquid is considered to be incompressible. The relative flow is steady.

3 L gD

W=

Here, we will assume that Re 6 1. Realizing that m = rLn, then

6

3 b gD

y

FD = 3pmUD = 3prLnUD The volume of the ball is V = ball and the bouyant force are

p p W = mg = rbVg = rb c D3 d g = rb gD3 6 6

FD = 3

Referring to the free-body diagram in Fig. a, p p + c ΣFy = may; 3prLnUD + rLgD3 - rbgD3 = 0 6 6 gD2 ( rb - rL ) 18rLn

Substituting the data, U =

( 9.81 m>s2 ) (0.02 m)2 ( 3000 kg>m3 - 2300 kg>m3 ) 18 ( 2300 kg>m3 )( 0.052 m2 >s )

= 0.001276 m>s = 0.00128 m>s

Ans.

Thus, the Reynolds number is Re =

UD = n

( 0.001276 m>s ) (0.02 m) 0.052 m2 >s

L

(a)

p p FB = rLVg = rLa D3 bg = rL gD3 6 6

U =

a=0

4 3 4 D 3 p pr = p a b = D3. Thus, the weight of the 3 3 2 6

= 0.4907 ( 10-3 ) 6 1

1229

(O.K.)

v UD

11–89. Determine the velocity of the aerosol solid particles when t = 10 s, if when t = 0 they leave the can with a horizontal velocity of 30 m>s. Assume the average diameter of the droplets is 0.4 µm and each has a mass of 0.4 ( 10-12 ) g. The air is at 20°C. Neglect the vertical component of the velocity. Note: The volume of a sphere is V = 43pr 3.

30 m/s

SOLUTION The air is considered to be incompressible. The relative flow is steady.

From Appendix A, ra = 1.202 kg>m3 and ma = 18.1 ( 10-6 ) N # s>m2 for air at T = 20° C . Thus, the maximum Reynolds number is

( Re ) max =

rmaxUmaxD = ma

a=

mg

( 1.202 kg>m3 )( 30 m>s ) 3 0.4 ( 10-6 ) m 4 = 0.7969 18.1 ( 10-6 ) N # s>m2

du dt x

(

(

FD = 21.72 10–12

U

(a)

Since ( Re ) max 6 1, the drag on the droplet is

FD = 3pmaUD = 3p 3 18.1 ( 10-6 ) N # s>m2 4 (U) 3 0.4 ( 10-6 ) m 4 = 21.72 ( 10-12 ) pU

Referring to the free-body diagram of the droplet in Fig. a, + S ΣFx

= max ;

-21.72 ( 10-12 ) pU = c -54300p

L0

0.4 ( 10-12 )

10(10-6) s

dt =

1000 V

kg d a

dU L30 m>s U

dU b dt

-0.543p = ln U # V 30 m>s -0.543p = ln e -0.543p =

V 30

V 30 Ans.

V = 5.45 m>s

Ans: 5.45 m>s 1230

11–90. Impure water at 20°C enters the retention tank and rises to a level of 2 m when it stops flowing in. Determine the shortest time needed for all sediment particles having a diameter of 0.05 mm or greater to settle to the bottom. Assume the density of the particles is r = 1.6 Mg>m3 or greater. Note: The volume of a sphere is V = 43pr 3.

2m 5m 2m

SOLUTION Water is considered to be incompressible. The relative flow is steady. FB =

Here, we will assume that Re 6 1. Realizing that m = rLn, then

6

FD = 3pmUD = 3prLnUD

3 L gD

W=

4 3 4 D 3 p pr = p a b = D3. Thus, the weight of 3 3 2 6 the particles and the bouyant force are

6

3 b gD

y

The volume of the particles is V =

p p W = mg = rVg = rba D3 bg = rbgD3 6 6

p p FB = rLVg = rLa D3 bg = rLgD3 6 6

FD = 3

L

v UD

(a)

Referring to the free-body diagram in Fig. a, p p + c ΣFy = may ; 3prLnUD + rLgD3 - rbgD3 = 0 6 6 U =

a=0

gD3 ( rb - rL ) 18rLn

From Appendix A, rL = 998.3 kg>m and n = 1.00 ( 10-6 ) m2 >s for water at T = 20° C . Substituting the data, 3

( 9.81 m>s2 ) 3 50 ( 10-6 ) m 4 2 ( 1600 kg>m3 - 998.3 kg>m3 ) 18 ( 998.3 kg>m3 ) 3 1.00 ( 10-6 ) m2 >s 4

U =

= 0.8212 ( 10-3 ) m>s Thus, the Reynolds number is Re =

UD = n

( 0.8212 ( 10-3 ) m>s ) 3 50 ( 10-6 ) m 4 = 0.0411 6 1 1.00 ( 10-6 ) m2 >s

(O.K.)

Thus, the time required for the particles to settle is t =

s 2m 1 min = b = 2435.42 s a U 60 s 0.8212 ( 10-3 ) m>s

Ans.

= 40.6 min

Ans: 40.6 min 1231

11–91. A ball having a diameter of 0.6 m and a mass of 0.35 kg is falling in the atmosphere at 10°C. Determine its terminal velocity. Note: The volume of a sphere is V = 43pr 3.

SOLUTION

FB = 1.3835 N

The air is considered to be incompressible. The relative flow is steady.

0.35(9.81) N

From Appendix A, ra = 1.247 kg>m3 and na = 14.2 ( 10-6 ) m2 >s for air at T = 10° C . Thus, the Reynolds number is Re =

U(0.6 m) UD = = 4.225 ( 104 ) U na 14.2 ( 10-6 ) m2 >s

(1)

y

a=0

The projected area perpendicular to the air stream is AP = p(0.3 m)2 = 0.09p m2. FD = CDAP r

U2 U2 = CD ( 0.09p m2 )( 1.247 kg>m3 ) a b 2 2

FD = 0.056115 U 2

= 0.056115pCDU 2

4 4 The volume of the ball is V = pr 3 = p(0.3 m)3 = 0.036p m3. Thus, the bouyant 3 3 force is FB = raVg = ( 1.247 kg>m3 )( 0.36p m3 )( 9.81 m>s2 ) = 1.3835 N Referring to the free-body diagram in Fig. a, 0.056115pCDU 2 + 1.3835 N - 0.35(9.81) N = 0

+ c ΣFy = may;

The iterations carried out are tabulated as follows: Iteration

Asumed CD

U ( m>s ) ; Eq. (2)

Re; Eq. (1)

CD from the graph

1

0.5

4.823

2.04 ( 10

)

0.46

2

0.46

5.028

2.12 ( 105 )

0.45

5

Since the assumed CD is almost the same as that obtained from the graph in iteration 2, the result of U in this iteration is acceptable. Thus, Ans.

U = 5.03 m>s

Ans: 5.03 m>s 1232

*11–92. A raindrop has a diameter of 1 mm. Determine its approximate terminal velocity as it falls. Assume that the air has a constant ra = 1.247 kg>m3 and na = 14.2 ( 10-6 ) m2 >s. Neglect buoyancy. Note: The volume of a sphere is V = 16pD3.

SOLUTION

W y

The air is considered to be incompressible. The relative flow is steady. The Reynolds number is Re =

a=0

U(0.001 m) UD = = 70.422U na 14.2 ( 10-6 ) m2 >s

The volume of the raindrop is V =

(1)

p 3 p D = (0.001 m)3 = 1.667 ( 10-10 ) p m3. Thus, 6 6

its weight is W = mg = rwVg = ( 1000 kg>m3 ) 3 1.667 ( 10-10 ) p m3 4 ( 9.81 m>s2 ) = 5.1365 ( 10-6 ) N 0.001 m 2 b = 2.5 ( 10-7 ) p m2. 2 U2 U2 = CD 3 2.5 ( 10-7 ) p m2 4 ( 1.247 kg>m3 ) a b FD = CDAP r 2 2

Here, AP = p a

= 4.897 ( 10-7 ) CDU 2

Referring to the free-body diagram in Fig. a, 4.897 ( 10-7 ) CDU 2 - 5.1365 ( 10-6 ) N = 0

+ c ΣFy = may; U2 =

10.489 CD

(2)

The iterations carried out are tabulated as follows: Iteration

Assumed CD U ( m>s ) ; Eq. (2)

Re; Eq. (1)

CD from the graph

1

0.5

4.58

323

0.66

2

0.66

3.98

280

0.7

Use CD = 0.7, and Ans.

U = 3.87 m>s

1233

FD (a)

11–93. The 2-Mg race car has a projected front area of 1.35 m2 and a drag coefficient of 1CD 2 C = 0.28. If the car is traveling at 60 m>s, determine the diameter of the parachute needed to reduce the car’s speed to 20 m>s in 4 s. Take 1CD 2 p = 1.15 for the parachute. The air is at 20°C. The wheels are free to roll.

SOLUTION

a=

The air is considered to be incompressible. The relative flow is steady.

du dt x

From Appendix A, r = 1.202 kg>m3 and n = 15.1 ( 10-6 ) m2 >s for art at T = 20° C . U2 U2 = 0.28 ( 1.35 m2 )( 1.202 kg>m3 ) a b = 0.2272U 2 2 2 2 U2 p 2 U = ( CD ) P ( AP ) P r = 1.15 a d b ( 1.202 kg>m3 ) a b = 0.5428d 2U 2 2 4 2

(FD (c =

(FD (P = 0.5428d2U2

0.2272U2

( FD ) C = ( CD ) C ( AP ) Cr

( FD ) P

(a)

Writing the equation of motion along the x axis by referring to the free-body diagram shown in Fig. a, dU + -0.2272U 2 - 0.5428d 2U 2 = 2 ( 103 ) d ΣFx = ma; dt - ( 0.2272 + 0.5428d 2 ) U 2 = 2 ( 103 ) J

- ( 0.2272 + 0.5428d 2 ) 2 ( 103 )

R

L0

4s

0.002 ( 0.2272 + 0.5428d 2 ) =

20 m>s

dt =

-0.002 ( 0.2272 + 0.5428d 2 ) =

dU dt dU 2 L60 m>s U

1 20 m>s ` U 60 m>s

1 1 20 60

Ans.

d = 5.50 m

Ans: 5.50 m 1234

11–94. A 2-mm-diameter sand particle having a density of 2.40 Mg>m3 is released from rest at the surface of oil that is contained in the tube. As the particle falls downward, “creeping flow” will be established around it. Determine the velocity of the particle and the time at which Stokes’ law becomes invalid, at about Re = 1. The oil has a density of ro = 900 kg>m3 and a viscosity of mo = 30.2 1 10-3 2 N # s>m2. Assume the particle is a sphere, where its volume is V = 43pr 3.

SOLUTION

FD

4 W = rVg = ( 2400 kg>m3 ) c p ( 0.001 m ) 3 d ( 9.81 m>s2 ) = 9.8621 ( 10-5 ) N 3 4 Fb = roVg = ( 900 kg>m3 ) c p ( 0.001 m ) 3 d ( 9.81 m>s2 ) = 3.6983 ( 10-5 ) N 3

Fb

FD = 3pm0VD = 3p 3 30.2 ( 10 - 3 ) N # s>m2 4 V(0.002 m) = 5.6926 ( 10-4 ) V

We solve Re = 1 to find V: roVD = 1 m0 V =

m0 = r oD

W (a)

30.2 ( 10 - 3 ) N # s>m2

( 900 kg>m3 ) (0.002 m)

= 0.016778 m>s

Now we integrate, starting with Newton’s Second Law: + T ΣFy = ma; W - Fb - FD = m

dV dt

9.8621 ( 10 - 5 ) - 3.6983 ( 10-5 ) - 5.6926 ( 10-4 ) V 4 dV = (2400) c p ( 0.001 m3 ) d 3 dt

6.1638 ( 10-5 ) - 5.6926 ( 10-4 ) V = 1.00531 ( 10-5 ) 0 L

t

dt =

0.016778

1.00531 ( 10-5 )

L0 6.1638 ( 10-5 ) - 5.6926 ( 10-4 ) V

dV dt

dV Ans.

t = 0.002973 s = 2.97 ms

Ans: V = 16.8 mm>s t = 2.97 ms 1235

11–95. Dust particles having an average diameter of 0.05 mm and an average density of 450 kg>m3 are stirred up by an airstream and blown off the edge of the 600-mm-high desk into a horizontal steady wind of 0.5 m>s. Determine the distance d from the edge of the desk where most of them will strike the ground. The air is at a temperature of 20°C. Note: The volume of a sphere is V = 43 pr 3.

0.5 m/s

600 mm

d

SOLUTION

FD

Due to the smallness of a dust particle, the flow can be assumed steady and often referred to as creeping flow. Also, the air will be assumed incompressible. Appendix  A gives ra = 1.202 kg>m3 and ma = 18.1 ( 10-6 ) N # s>m2. For creeping flow we assume that Re 6 1 so that stokes equation FD = 3pmaVD can be used. The bouyant force is Fb = raVg, and the weight of the dust W = mg = rdVg. Since the dust is creeping in the vertical direction with its terminal velocity (constant), then referring to the FBD in Fig. a, + T ΣFy = 0;

Fb

rdVg - raVg - 3pmaVD = 0

V =

(rd - ra)Vg

W

3pmaD

(a)

4 D 3 p since V = p a b = D3, the above equation becomes 3 2 6 p 6 3pmaD

( rd - ra ) a D3 bg

V =

=

( rd - ra ) gD2 18ma

Substitute the numerical data into this equation to find the terminal downward velocity, V =

( 450 kg>m3 - 1.202 kg>m3 )( 9.81 m>s2 ) 3 0.05 ( 10-6 ) m 4 2 18 3 18.1 ( 10-6 ) N # s>m2 4

= 0.03378 m>s

Then the Reynolds number is Re =

raVD = ma

( 1.202 kg>m3 )( 0.03378 m>s ) 3 0.05 ( 10-3 ) m 4 18.1 ( 10-6 ) N # s>m2

= 0.1122 6 1

1236

(O.K.)

*11–95. Continued

The time for the dust to strike the ground can therefore be determined from t =

h 0.6 m = = 17.76 s v 0.03378 m>s

Thus, the horizontal distance d is d = Ut = ( 0.5 m>s ) (17.76 s) = 8.88 m

Ans.

Ans: 8.88 m 1237

*11–96. A rock is released from rest at the surface of the lake, where the average water temperature is 15°C. If CD = 0.5, determine its speed when it reaches a depth of 600 mm. The rock can be considered a sphere having a diameter of 50 mm and a density of rp = 2400 kg>m3. Note: The volume of a sphere is V = 43pr 3.

600 mm

SOLUTION + T ΣFy =

d ( mV ) dt

Fb

dV VdV = m = m dt ds

Referring to the FBD shown in Fig. a, and realizing that Fb = rw g V, rwV 2 b and W = rsg V, FD = CDAp a 2 rwV 2 VdV b = rsV rsg V - rwg V - CDAr a 2 ds 2(rs - rw)Vg - CDAprwV 2 = 2rsV

VdV ds

W

with the initial condition at s = 0, V = 0, L

s

(a)

V

ds = 2rsV

VdV 2 L 2(rs - rw)Vg - CpAprwV 0

0

Let a = 2(rs - rw)V g and b = CDAprw. Then L

s

V

ds = 2rsV

0

s = s =

VdV 2 L a - bV 0

2rsV 2( - b )

c ln ( a - bV 2 ) d `

V 0

rsV a ln a b b a - bV 2

bs a = ln a b rsV a - bV 2 a bs = e 2 r a - bV sV bV 2 = a - ae V =

FD

bs rsV

a ( 1 - e -bs>rsV ) Ab

(1)

1238

*11–96. Continued

Substituting the numerical data, rs = 2400 kg>m3 , rw = 999.2 kg>m3 , s = 0.6 m 4 V = p(0.025 m)3 = 65.4498 ( 10-6 ) m3 and Ap = p(0.025 m)2 = 0.625 ( 10-3 ) p m2, 3 a = 2 ( 2400 kg>m3 - 999.2 kg>m3 ) 3 65.4498 ( 10-6 ) m3 4 ( 9.81 m>s2 ) = 1.7988 b = CD 3 0.625 ( 10-3 ) p m2 4 ( 999.2 kg>m3 ) = 1.9619CD

rsV = ( 2400 kg>m3 ) 3 65.4498 ( 10-6 ) m3 4 = 0.15708 kg

Then Eq. (1) becomes V = c

0.9169 ( 1 - e -7.4940CD ) d m>s A CD

(2)

Using C D = 0.5, at s = 0.6 m V = 1.34 m>s Ans. The terminal velocity can be obtained by setting s S ∞ . Then Eq. (1) becomes Vt =

0.9169 A CD

Again using C D = 0.5,

Ans.

Vt = 1.35 m>s

1239

11–97. The smooth cylinder is suspended from the rail and is partially submerged in the water. If the wind blows at 8 m>s, determine the terminal velocity of the cylinder. The water and air are both at 20°C. 1m

8 m/s

The fluids are considered incompressible. The relative fluid is steady. From Appendix A, ra = 1.202 kg>m3 and va = 15.1 ( 10-6 ) m2 >s for air and rw = 998.3 kg>m3 and nw = 1.00 ( 10-6 ) m2 >s for water at T = 20° C . If the terminal velocity of the cylinder is V0, Ua = 8 m>s - V0 and Uw - V0. Thus, the Reynolds number for air and water are

( Re ) a ( Re ) w

(8 - V0)(0.25 m) UaD = = = 1.6556 ( 104 ) (8 - V0) na 15.1 ( 10-6 ) m2 >s V0(0.25 m) UwD = = = 2.5 ( 105 ) V0 nw 1.00 ( 10-6 ) m2 >s

N2

N1

a=0 x

(1)

( FD( a

W

(2)

The projected areas perpendicular to the stream for air and water are ( AP ) a = (0.25 m)(1 m) = 0.25 m2 and ( AP ) w = (0.25 m)(0.5 m) = 0.125 m2.

( FD ) a

0.5 m

0.25 m

SOLUTION

( FD ( k

( 8 - V0 ) 2 Ua2 = ( CD ) a ( AP )a ra = ( CD ) a(0.25 m) ( 1.202 kg>m3 ) J R 2 2

(a)

= 0.15025 ( CD ) a ( V02 - 16n0 + 64 )

( FD ) w = ( CD ) w ( AP ) wrw

Uw2 V 02 = ( CD ) w(0.125 m) ( 998.3 kg>m3 ) c d 2 2

= 62.394 ( CD ) wV 02

Writing the equation of motion along the x axis by referring to the free-body diagram in Fig. a, + ΣFx = max; S

( FD ) a - ( FD ) w = 0 ( FD ) a = ( FD ) w 0.15025 ( CD ) a ( V0 2 - 16n0 + 64 ) = 62.394 ( CD ) wV0 2

( CD ) a ( V 02 - 16v0 + 64 ) - 415.27 ( CD ) wV 02 = 0

(3)

The iterations carried out are tabulated as follows: Assumed Iteration 1

Value from the graph

( CD ) a ( CD ) w V0 ( m>s ) ; Eq. (3) (Re)a; Eq. (1) (Re)w; Eq. (2) 1.3

1.3

0.3742

1.26 ( 105 )

9.36 ( 104 )

( CD ) a

( CD ) w

1.4

1.4

Since the assumed CD is almost the same as that obtained from the graph in iteration 1, the result of v0 in the iteration is acceptable. Thus, Ans.

V0 = 0.374 m>s

Ans: 0.374 m>s 1240

11–98. A 5-m-diameter balloon and the gas within it have a mass of 80 kg. Determine its terminal velocity of descent. Assume the air temperature is at 20°C. Note: The volume of a sphere is V = 43pr 3.

SOLUTION

FB = 771.76 N

The air is considered to be incompressible. The relative flow is steady. From Appendix A, ra = 1.202 kg>m3 and va = 15.1 ( 10-6 ) m2 >s for air at T = 20° C . Thus, the Reynolds number is Re =

y

W = 100 (9.81) N

a=0

U(5 m) UD = = 3.3113 ( 105 ) U va 15.1 ( 10-6 ) m2 >s

4 4 5m 3 b = 65.45 m3. Thus, the bouyThe volume of the balloon is V = pr 3 = p a 3 3 2 ant force is FB = raVg = ( 1.202 kg>m3 )( 65.45 m3 )( 9.81 m>s2 ) = 771.76 N

(a)

5m 2 Here, AP = p a b = 6.25 p m2 2 FD = CDAP r

FD = 3.75625 CDU2

U2 U2 CD ( 6.25p m2 )( 1.202 kg>m3 ) a b 2 2

= 3.75625 pCDU 2

Referring to the free-body diagram in Fig. a, 3.75625pCDU 2 + 771.76 N - 80(9.81) N = 0

+ c ΣFy = may; U2 =

1.1051 CD

The iterations carried out are tabulated as follows: Iteration

Assumed CD

U ( m>s ) ; Eq. (2)

1

0.2

2.351

7.78 ( 105 )

0.14

2.809

9.30 ( 10

)

0.155

2.670

8.84 ( 10

)

0.15

2 3

0.14 0.155

Re; Eq. (1) 5 5

CD from the graph

Since the assumed CD is almost the same as that obtained from the graph in iteration 3, the result of U in this iteration is acceptable. Thus, Ans.

U = 2.67 m>s

Ans: 2.67 m>s 1241

11–99. A smooth ball has a diameter of 43 mm and a mass of 45 g. When it is thrown vertically upward with a speed of 20  m>s, determine the initial deceleration of the ball. The temperature is 20°C.

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 1.202 kg>m3 and v = 15.1 ( 10-6 ) m2 >s for air at T = 20° C . Thus, the initial Reynolds number is Re =

FD = 0.1746 N y

(20 m>s)(0.043 m) UD = = 5.695 ( 104 ) v 15.1 ( 10-6 ) m2 >s

a

Entering this Re into the graph for a sphere, CD ≅ 0.5 (aprox.). Here, AP = p a

0.043 m 2 b = 0.46225 ( 10-3 ) p m2. 2

FD = CDADr

0.045(9.81) N

( 20 m>s ) U2 = 0.5 3 0.46225 ( 10-3 ) p m2 4 ( 1.202 kg>m3 ) J R 2 2 2

(a)

Ans.

= 0.1746 N

Referring to the free-body diagram of the ball in Fig. a, + c ΣFy = may;

- 30.045(9.81) N4 - 0.1746 N = 0.045a a = 13.7 m>s2

Ans.

Ans: 13.7 m>s2 1242

*11–100. The parachutist has a total mass of 90 kg and is in free fall at 6 m>s when she opens her 3-m-diameter parachute. Determine the time for her speed to be increased to 10 m>s. Also, what is her terminal velocity? For the calculation, assume the parachute to be similar to a hollow hemisphere. The air has a density of ra = 1.25 kg>m3.

V

SOLUTION Relative to the parachutist, the flow is unsteady and uniform since he is decelerating. Here, the air is assumed to be incompressible. Applying the momentum equation + T ΣFy =

0 VrdV + VrVdA 0t Lcv Lcs

The control volume considered is the parachute and the parachutist. Since there is VrVdA = 0. Also, Vr can be factored out from the Lcs integral since it is independent of V . Also dV = V since the volume of the Lcv control volume is fixed. Realizing that rV = m, the above equation reduces to

FD

no opened control surface,

+ T ΣFy =

d(mv) dt

= m

dv dt

Referring to the FBD shown in Fig. a, and realizing that FD = CDAP a mg - CDAPa

raV 2 dv b = m 2 dt

raV 2 b, 2

2mg - CDAP raV 2 dv = m 2 dt

with the initial condition at t = 0, V = Vo, L0

t

t =

t =

V

dV 2 LV0 2 mg - cDAP raV

dt = 2 m

2m 2( 22mg) 2CDAp ra m 22mgCDAP ra

£ ln °

ln°

22mg + 2CDAP raV

22mg - 2CDAP raV

22mg + 2CDAP raV

22mg - 2CDAP raV

V

¢†

mg Vo

¢ - ln°

(a)

22mg + 2CDAP raVo

22mg - 2CDAP raVo

Substituting the numerical data, m = 90 kg, CD = 1.4 (table 11–3), AP = p ( 1.5 m ) 2 = 2.25 p m2, ra = 1.25 kg>m3, Vo = 6 m>s and V = 12 m>s, we have 22mg = 22 ( 90 kg )( 9.81 m>s2

) = 42.02

2CDAP ra = 21.4 ( 2.25p m2 )( 1.25 kg>m3

) = 3.5171

1243

¢§

(1)

*11–100. Continued

Then t =

42.02 + 3.5171 ( 10) 42.02 + 3.5171 ( 6) 90 £ ln ° ¢ - ln ° ¢§ (42.02)(3.5045) 42.02 - 3.5171(10) 42.02 - 3.5171(6) Ans.

= 0.805 s

Terminal velocity occurs when t = ∞ . By inspecting Eq. (1), this condition can be satisfied if 22mg + 2CDAP raVt = 0 Vt =

2mg 42.02 = = 11.95 m>s = 12.0 m>s A CDAP ra 3.5171

1244

Ans.

11–101. A 3-Mg airplane is flying at a speed of 70 m>s. If each wing can be assumed rectangular of length 5 m and width 1.75 m, determine the smallest angle of attack a to provide lift assuming the wing is a NACA 2409 section. The density of air is r = 1.225 kg>m3.

5m

5m

SOLUTION The air is considered to be incompressible. The relative flow is steady. For two wings, A = 2(5 m)(1.75 m) = 17.5 m2. Thus, the lift is

( 70 m>s ) U2 d = 52521.875 CL = CL ( 17.5 m2 )( 1.225 kg>m3 ) c 2 2 2

FL = CL Ar

The equilibrium along a vertical requires + c ΣFy = 0;

FL - W = 0 52521.875 CL - 3000(9.81) N = 0 CL = 0.560

Entering this value of CL into the graph Ans.

a = 5° (approx.)

Ans: 5° (approx.) 1245

11–102. The 5-Mg airplane has wings that are each 5 m long and 1.75 m wide. It is flying horizontally at an altitude of 3 km with a speed of 150 m>s. Determine the lift coefficient. 5m

5m

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r = 0.9092 kg>m3 for air at an altitude of 3 km. For two wings, A = 2(5 m)(1.75 m) = 17.5 m2. Thus, the lift is

( 150 m>s ) U2 = CL ( 17.5 m2 )( 0.9092 kg>m3 ) c d = 178998.75 CL 2 2 2

FL = CL Ar Equilibrium requires + c ΣFy = 0;

FL - W = 0 178998.75 CL - 5000(9.81) N = 0 Ans.

CL = 0.274

Ans: 0.274 1246

11–103. The 5-Mg airplane has wings that are each 5 m long and 1.75 m wide. Determine its speed in order to generate the same lift when flying horizontally at an altitude of 5 km as it does when flying horizontally at 3 km with a speed of 150 m>s.

5m

5m

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r1 = 0.9092 kg>m3 and r2 = 0.7364 kg>m3 for air at an altitude of 3 km and 5 km, respectively. Here, it is required that (FL)1 = (FL)2 (CL)1A1 r1 U2 = °

U 12 U 22 = (CL)2A2 r2 2 2

( CL ) 1A1r1 ¢U1 A (C ) A r L 2 2 2

Since A1 = A2 and the angle of attack is the same for both cases, (CL)1 = (CL)2. Thus, U2 =

r1 0.9092 kg>m3 U1 = ° ¢ ( 150 m>s ) = 167 m>s A r2 A 0.7364 kg>m3

Ans.

Ans: 167 m>s 1247

*11–104. A 4-Mg airplane is flying at a speed of 70 m>s. If each wing can be assumed rectangular of length 5 m and width 1.75 m, determine the drag on each wing when it is flying at the proper angle of attack a. Assume each wing is a NACA 2409 section. The density of air is ra = 1.225 kg>m3.

SOLUTION The air is considered to be incompressible. The relative flow is steady. For two wings, A = 2(5 m)(1.75 m) = 17.5 m2. Thus, the lift is

( 70 m>s ) U2 = CL ( 17.5 m2 )( 1.225 kg>m3 ) c d = 52521.875CL 2 2 2

FL = CL Ar

The equilibrium along a vertical requires + c ΣFy = 0;

FL - W = 0 52521.875CL - 4000(9.81) N = 0 CL = 0.747

Entering this value of CL into the graph a = 8.20° (approx.) Using this result, the graph gives C D ≅ 0.04 (approx.). For each wing, A = 5 m(1.75 m) = 8.75 m2.

( 70 m>s ) U2 = 0.04 ( 8.75 m2 )( 1.225 kg>m3 ) c d 2 2 2

FD = CDAr

= 1.05 kN

1248

Ans.

11–105. The plane can take off at 250 km>h when it is at an airport located at an elevation of 2 km. Determine the takeoff speed from an airport at sea level.

V

SOLUTION The air is considered to be incompressible. The relative flow is steady. From Appendix A, r1 = 1.007 kg>m3 and r2 = 1.225 kg>m3 for air at an altitude of 2 km and 0 km, respectively. Here, it is required that

( FL ) 1 = ( FL ) 2 ( CL ) 1A1r1 U2 = °

U 21 U22 = ( CL ) 2A2r2 2 2

( CL ) 1A1r1 ¢U1 A (C ) A r L 2 2 2

Since A1 = A2 and the angle of attack is the same for both cases, ( CL ) 1 = ( CL ) 2. Thus, U2 =

r1 1.007 kg>m3 U = ° ¢ ( 250 km>h ) = 227 km>h A r2 1 A 1.225 kg>m3

Ans.

Ans: 227 km>h 1249

11–106. The glider has a weight of 350 lb. If the drag coefficient is CD = 0.456, the lift coefficient is CL = 1.20, and the total area of the wings is A = 80 ft 2, determine the angle u at which it is descending with a constant speed. u

SOLUTION The air is considered to be incompressible. The relative flow is steady. ΣFx′ = max′;

FD - W sin u = 0

FD = W sin u

(1)

ΣFy′ = may′;

FL - W cos u = 0

FL = W cos u

(2)

Dividing Eq. (1) by Eq. (2), FD W sin u = = tan u FL W cos u

(3)

The drag and lift are FD = CDAr FL = CLAr

U2 2

U2 2

Substituting these results into Eq. (3), U2 2 = tan u U2 CLAr 2

CDAr

tan u =

CD 0.456 = CL 1.2 Ans.

u = 20.8°

y´ W a=0

FD x´ FL (a)

Ans: 20.8° 1250

11–107. The glider has a weight of 350 lb. If the drag coefficient is CD = 0.316, the lift coefficient is CL = 1.20, and the total area of the wings is A = 80 ft 2, determine if it can land on a landing strip that is 1.5 km long and located 5 km away from where its altitude is 1.5 km. Assume the density of the air remains constant.

u

SOLUTION The air is considered to be incompressible. The relative flow is steady. ΣFx′ = max′;

FD - W sin u = 0

FD = W sin u

(1)

ΣFy′ = may′;

FL - W cos u = 0

FL = W cos u

(2)

Dividing Eq. (1) by Eq. (2),

= 14.75′

1.5 Km

d

FD W sin u = = tan u FL W cos u

(3)

(a)

The drag and lift are FD = CDAr

U2 2

FL = CLAr

U2 2

Substituting these results into Eq. (3), U2 2 = tan u U2 CLAr 2

CDAr

tan u =

CD 0.316 = CL 1.2

u = 14.75° Referring to the geometry shown in Fig. a, tan 14.75° =

1.5 km d

d = 5.7 km Since 5 km 6 d 6 (5 + 1.5) km = 6.5 km, the glider can land on the landing strip.

Ans: The glider can land. 1251

*11–108. Each of the two wings on a 20 000-lb airplane is to have a span of 25 ft and an average cord distance of 5 ft. When a 1>15 scale model of the wing section (assumed to be infinite) is tested in a wind tunnel at 1500 ft>s, using a gas for which rg = 7.80 1 10-3 2 slug>ft 3, the total drag is 160 lb. Determine the total drag on the wing when the plane is  flying at a constant altitude with a speed of 400 ft>s,  where  ra = 1.75 1 10-3 2 slug>ft 3. Assume an elliptical lift distribution.

SOLUTION We will assume that the flow is steady relative to airplane and the air and the gas is incompressible. For the model plane,

( AP ) m = 2c

1 1 (25 ft) d c (5 ft) d = 1.1111 ft 2 15 15

The drag coefficient for the infinite span can be determined using the model plane.

( FD ) m

( CD ) ∞ =

( AP ) ma

rgVm2 2

= 0.01641

160 lb

= b

( 1.1111 ft2 ) W

3 7.80 ( 10-3 ) slug>ft3 4 ( 1500 ft>s ) 2 2

Since the plane is flying at a constant altitude, it is in vertical equilibrium. This means that the lift is equal to its weight; ie, FL = 20000 lb. Then the lift coefficient is FL

CL =

2

AP a

raV b 2

20000

= [2(25)(5 ft)] W

= 0.5714

3 1.75 ( 10-3 ) slug>ft3 4 ( 400 ft>s ) 2 2



The total drag coefficient can be determined by applying CD = ( CD ) ∞ + = 0.01641 + = 0.03720

CL2 pb2 >AP

0.57142

p ( 25 ft ) 2 >(25 ft)(5 ft)

Thus, the total drag force on the wing is FD = CDAP a

raV 2 b 2

= 0.03720[2(25 ft)(5 ft)] W = 1302 lb

3 1.75 ( 10-3 ) slug>ft3 4 ( 400 ft>s ) 2 2

1252



Ans.



11–109. The glider has a constant speed of 8 m>s through still air. Determine the angle of descent u if it has a lift coefficient of CL = 0.70 and a wing drag coefficient of CD = 0.04. The drag on the fuselage is considered negligible compared to that on the wings, since the glider has a very long wingspan.

8 m/s

u

SOLUTION Since the glider is gliding with a constant velocity, it is in equilibrium. Referring to the FBD of the glider in Fig. a, + ΣFx = 0; R

W sin u - FD = 0 W sin u - CDAP a

+ c ΣFy = 0;

y

raV 2 b = 0 2

W

raV 2 b W sin u = CDAP a 2

(1)

FL

FD

FL - W cos u = 0

CLAP a

raV 2 b - W cos u = 0 2

W cos u = CLAP a

Divided Eq. (1) by Eq. (2) W sin u = W cos u

tan u = u = tan-1a

CDAP a CLAP a

CD CL

(a)

raV 2 b 2

x

(2)

raV 2 b 2

raV 2 b 2

CD 0.04 b = tan-1a b = 3.27° CL 0.7

Ans.

Ans: 3.27° 1253

11–110. The 2000-lb airplane is flying at an altitude of 5000 ft. Each wing has a span of 16 ft and a cord length of 3.5 ft. If each wing can be classified as a NACA 2409 section, determine the lift coefficient and the angle of attack when the plane is flying at 225 ft>s.

16 ft

SOLUTION

3.5 ft

Relative to the airplane, the flow is steady. Also, air is assumed to be incompressible. Appendix A gives ra = 2.043 ( 10-3 ) slug>ft 3. Since the air plane is flying at a constant altitude, equilibrium exists along the vertical. Thus + c ΣFy = 0;

FL - W = 0 CLAP a

raV 2 b - W = 0 2

CL 32(16 ft)(3.5 ft) 4 W

3 2.043 ( 10-3 ) slug>ft3 4 ( 225 ft>s ) 2 2

CL = 0.345

¶ - 2000 lb = 0

Ans.

with this value of CL, Ans.

a = 3° (Approx.)

Ans: CL = 0.345 a = 3° (Approx.) 1254

11–111. The 2000-lb airplane is flying at an altitude of 5000 ft. Each wing has a span of 16 ft and a cord length of  3.5 ft, and it can be classified as a NACA 2409 section. If  the plane is flying at 225 ft>s, determine the total drag on  the wings. Also, what is the angle of attack and the  corresponding velocity at which the condition of stall occurs? 16 ft

SOLUTION

3.5 ft

Relative to the airplane, the flow is steady. Also, air is assumed to be incompressible. Appendix A gives ra = 2.043 ( 10-3 ) slug>ft 3. Since the air plane is flying at a constant altitude, equilibrium exists along the vertical. Thus, + c ΣFy = 0;

FL - W = 0

CLApa

raV 2 b - W = 0 2

CL[2(16 ft)(3.5 ft)] W

(1)

3 2.043 ( 10-3 ) slug>ft3 4 ( 225 ft>s ) 2 2

CL = 0.3453

¶ - 2000 lb = 0

with this value of CL, a = 2.75 with this angle of attack,

( CD ) ∞ = 0.015 The total drag coefficient can be determined using CD = ( CD ) ∞ + = 0.015 + = 0.0233

C L2 pb2 >A

0.34532 p(16 ft) >(16 ft)(3.5 ft) 2

Thus, the drag force on the airplane caused by the wing is FD = CD APa

raV 2 b 2

= 0.023332(16 ft)(3.5 ft) 4 W

3 2.043 ( 10-3 ) slug>ft3 4 ( 225 ft>s ) 2 2

= 135 lb



Ans.

From the text, the condition of stall occurs when the angle of attack is Ans.

a = 20° And the corresponding lift coefficient is CL = 1.50 Again, applying Eq. (1) CL APa

raV 2 b - W = 0 2

1.5[2(16 ft)(3.5 ft)] W

3 2.043 ( 10-3 ) slug>ft3 4 V 2s

Vs = 107.95 ft>s = 108 ft>s

2

¶ - 2000 lb = 0

1255

Ans.

Ans: FD = 135 lb a = 20° Vs = 108 ft>s

*11–112. If it takes 80 kW of power to fly an airplane at 20 m>s, how much power does it take to fly the plane at 25 m>s at the same altitude? Assume CD remains constant.

SOLUTION We will assume that the flow is steady relative to the airplane and the air is incompressible. When the speed is 20 m>s, 80 ( 103 ) W = ( FD ) 1 ( 20 m>s )

P = FDV;

( FD ) = 4000 N Using the drag force equation,

( FD ) 1 = CDApa

raV 12 b 2

4000 N = CD AP c CD =

ra ( 20 m>s ) 2

20 ra AP

2

d

when the speed is 25 m>s,

( FD ) 2 = CD AP a = a

raV22 b 2

ra ( 25 m>s ) 20 b ( Ap ) c d ra Ap 2 2

= 6250 N

Thus, the power regained is

#

W2 = ( FD ) 2V2 = (6250 N) ( 25 m>s ) = 156.25 ( 103 ) W Ans.

= 156 kW

1256

11–113. The plane weighs 9000 lb and can take off from an airport when it attains an airspeed of 125 mi>h. If it carries an additional load of 750 lb, what must be its airspeed before takeoff at the same angle of attack?

125 mi/h

SOLUTION The air is considered to be incompressible. The relative flow is steady. Equilibrium along the vertical requires + c ΣFy = 0;

FL - W = 0 (1)

FL = W The lift is FL = CL Ar

2

U . Thus, using Eq. (1), 2

( FL ) 1 = ( CL ) 1A1r1

U 12 = W1 2

(2)

( FL ) 2 = ( CL ) 2A2r2

U 22 = W2 2

(3)

Dividing Eq. (3) by Eq. (2), U 22 W2 2 = 2 W1 U ( CL ) 1A1r1 1 2

( CL ) 2A2r2

U2 = °

( CL ) 1A1r1W2 ¢U1 A (C ) A r W L 2 2 2

1

Here, A1 = A2 and ( CL ) 1 = ( CL ) 2. Thus, r1 = r2, U2 =

W2 9750 lb U = ° ¢ ( 125 mi>h ) = 130 mi>h A W1 1 A 9000 lb

Ans.

Ans: 130 mi>h 1257

11–114. A baseball has a diameter of 73 mm. If it is thrown with a speed of 5 m>s and an angular velocity of 60 rad>s, determine the lift on the ball. Assume the surface of the ball is smooth. Take ra = 1.20 kg>m3 and na = 15.0 1 10-6 2 m2 >s, and use Fig. 11–50.

60 rad/s

5 m/s

SOLUTION For the given data, vD = 2V Re =

( 60 rad>s ) (0.073 m) = 0.438 2 ( 5 m>s ) ( 5 m>s ) (0.073 m) VD = = 2.43 ( 104 ) v 15.0 ( 10-6 ) m2 >s

Since Re is in the range of 104, the figure in the text can be used to determine the lift coefficient. Here CL ≈ 0.13. Thus, 2

FL = CL APa

raV b 2

= 0.13 3 p ( 0.0365 m ) 2 4 £ = 0.00816 N

( 1.20 kg>m3 )( 5 m>s ) 2 2

§

Ans.

Ans: 0.00816 N 1258

11–115. A 0.5-kg ball having a diameter of 50 mm is thrown with a speed of 10 m>s and has an angular velocity of 400 rad>s. Determine its horizontal deviation d from striking a target a distance of 10 m away. Take ra = 1.20 kg>m3 and na = 15.0 1 10-6 2 m2 >s, and use Fig. 11–50.

y

v

d

x

10 m Top View

SOLUTION From the given data, vD = 2V Re =

( 400 rad>s ) (0.05 m) = 1.0 2 ( 10 m>s ) ( 10 m>s ) (0.05 m) VD = = 3.33 ( 104 ) va 15.0 ( 10-6 ) m2 >s

Since Re is in the range of 104, the figure in the text can be used to determine the lift coefficient. Here, CL ≃ 0.270. Thus, FL = CLApa

raV 2 b 2

= 0.27 3 p(0.025 m)2 4 £

( 1.20 kg>m3 )( 10 m>s ) 2 2

= 0.03181 N

§

The acceleration of the ball in the y-direction is 0.03181 N = ( 0.5 kg ) ay

+ c ΣFy = may;

ay = 0.06362 m>s2 The ball travels with a constant velocity V = 10 m>s in the x- direction. Thus, the time for the ball to strike the wall is t =

Sx 10 m = = 1s V 10 m>s

The displacement d in the y direction for this same time interval is + c sy = ( sy ) 0 + ( vy ) 0t + d = 0 + 0 +

1 2 at ; 2 y

1 ( 0.06362 m>s2 ) (1 s)2 2 Ans.

= 0.03181 m = 31.8 mm

Ans: 31.8 mm 1259

12–1. A large tank contains water that has a depth of 4 m. If the tank is on an elevator that is descending, determine the speed of a wave created on its surface if the rate of descent is (a) constant at 8 m>s, (b) accelerated at 4 m>s2, (c) accelerated at 9.81 m>s2.

SOLUTION

mg

Water is considered to be incompressible. Referring to the free-body diagram of the tank in Fig. a, + T ΣFy = may;

a

mg - N = ma N = m(g - a)

Here, g′ = g - a and the speed of the wave can be determined using c = 2g′y 2

a.) For a = 0, g′ = g - 0 = 9.81 m>s . Then,

c = 2 ( 9.81 m>s2 ) (4 m) = 6.26 m>s

Ans.

c = 2 ( 5.81 m>s2 ) (4 m) = 4.82 m>s

Ans.

c = 20(4 m) = 0

Ans.

N (a)

b.) For a = 4 m>s2, g′ = g - 4 m>s2 = 5.81 m>s2. Then,

c.) For a = 9.81 m>s2, g′ = g - 9.81 m>s2 = 0. Then,

Ans: a) 6.26 m>s b) 4.82 m>s c) 0 1260

12–2. A river is 4 m deep and flows at an average speed of 2 m>s. If a stone is thrown into it, determine how fast the waves will travel upstream and downstream.

SOLUTION Water is considered to be incompressible. The flow is steady. The speed of the wave in still water is c = 2gy

c = 2 ( 9.81 m>s2 ) (4 m) = 6.264 m>s

Thus, applying the relative velocity equation with vw>r = c the speed of a wave traveling upstream is (vu)u = vr + vw>r + (vu)u = - 2 m>s + 6.264 m>s = 4.26 m>s S

Ans.

and the speed of a wave traveling downstream is (vu)d = vr + vw>r Ans.

+ (vu)d = 2 m>s + 6.264 m>s = 8.26 m>s d

Ans: (vu)u = 4.26 m>s (vu)d = 8.26 m>s 1261

12–3. A rectangular channel has a width of 2 m. If the flow is 5 m3 >s, determine the Froude number when the water depth is 0.5 m. At this depth, is the flow subcritical or supercritical? Also, what is the critical speed of the flow?

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow is V =

5 m3 >s Q = = 5 m>s A (2 m)(0.5 m)

Thus, the Froude number is Fr =

V 2gy

=

5 m>s 29.81 m>s2(0.5 m)

Ans.

= 2.26

Since the channel is rectangular and Fr 7 1, the flow is supercritical. The critical depth is 1 3

1 3

( 5 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.8605 m gb ( 9.81 m>s2 ) (2 m)2 Q2

At critical flow, Fr = 1. Then, Fr =

V 2gy

;

1 =

Vc 29.81 m>s2(0.8605 m)

Ans.

Vc = 2.91 m>s

Ans: Fr = 2.26 supercritical Vc = 2.91 m>s 1262

*12–4. A rectangular channel has a width of 2 m. If the flow is 5 m3 >s, determine the Froude number when the water depth is 1.5 m. At this depth, is the flow subcritical or supercritical? Also, what is the critical speed of the flow?

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow is V =

5 m3 >s Q = = 1.667 m>s A (2 m)(1.5 m)

Thus, the Froude number is Fr =

V 2gy

=

1.667 m>s 29.81 m>s2(1.5 m)

Ans.

= 0.434

Since the channel is rectangular and Fr 6 1, the flow is subcritical. The critical depth is 1 3

1 3

( 5 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.8605 m gb ( 9.81 m>s2 ) (2 m)2 Q2

At critical flow, Fr = 1. Then, Fr =

V 2gy

;

1 =

Vc 29.81 m>s2(0.8605 m)

Ans.

Vc = 2.91 m>s

1263

12–5. A rectangular channel transports water at 8 m3 >s. The channel width is 3 m and the water depth is 2 m. Is the flow subcritical or supercritical?

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow is V =

8 m3 >s Q = = 1.3333 m>s A (3 m)(2 m)

Thus, the Froude number is Fr =

V 2gy

=

1.3333 m>s 29.81 m>s2(2 m)

= 0.3010

Since Fr 6 1, the flow is subcritical. Also, since the channel is rectangular, the critical depth is 1 3

1 3

( 8 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.898 m gb ( 9.81 m>s2 ) (3 m)2 Q2

Since y >yc, the flow is subcritical.

Ans: subcritical 1264

12–6. Water flows with an average speed of 6 ft>s in a rectangular channel having a width of 5 ft. If the depth of the water is 2 ft, determine the specific energy and the alternate depth that provides the same flow.

SOLUTION Water is considered to be incompressible. The flow is steady. The flow is Q = VA = ( 6 ft>s ) (5 ft)(2 ft) = 60 ft 3 >s

Since the channel is rectangular, the critical depth is 1 3

1 3

( 60 ft3 >s ) 2 yc = ° 2 ¢ = £ § = 1.648 ft gb ( 32.2 ft>s2 ) (5 ft)2 Q2

and the specific energy is E =

E =

Q2 2gb2y2

+ y

( 60 ft3 >s ) 2 2.2360 + y = + y y2 2 ( 32.2 ft>s2 ) (5 ft)2y2

(1)

For a depth of 2 ft, Eq. (1) gives E =

2.2360 + 2 ft = 2.5590 ft = 2.56 ft (2 ft)2

Ans.

Substituting this result into Eq. (1), 2.5590 ft =

2.2360 + y y2

y3 - 2.5590y2 + 2.2360 = 0 Solving by trial and error, y = 2.00 ft 7 yc

(subcritical)

y = 1.37 ft 6 yc

(supercritical)

y = - 0.814 ft

(unrealistic)

Ans.

Ans: E = 2.56 ft y = 1.37 ft 1265

12–7. Water flows in a rectangular channel with a speed of 3 m>s and depth of 1.25 m. What other possible depth of flow provides the same specific energy?

SOLUTION Water is considered to be incompressible. The flow is steady. Using the continuity equation, 0 rd V + rV # dA = 0 0t L L cv

cs

0 - V1A1 + V2A2 = 0 - (3 m>s)(b)(1.25 m) = V(b)(y) V = a

3.75 b m>s y

(1)

The specific energy is E =

V2 + y 2g

(2)

For a depth of y = 1.25 m E =

( 3 m>s ) 2 + 1.25 m = 1.7087 m 2 ( 9.81 m>s2 )

Substituting this result and Eq. (1) into Eq. (2),

1.7087 m =

a

3.75 2 b y2

2 ( 9.81 m>s2 )

+ y2

y3 - 1.7087y2 + 0.7167 = 0 Solving by trial and error, Ans.

y = 1.021 m = 1.02 m y = 1.25 m y = - 0.562 m

(unrealistic)

At y = 1.25 m, Fr =

V 2gy

=

3 m>s 2 ( 9.81 m>s2 ) (1.25 m)

= 0.8567 6 1

(subcritical)

At y = 1.021 m, Eq. (1) gives V = Fr =

3.75 = 3.674 m>s 1.021 V 2gy

=

3.674 m>s 2 ( 9.81 m>s2 ) (1.021 m)

= 1.161 7 1

(supercritical) Ans: 1.02 m 1266

*12–8. Water flows in a rectangular channel with a mean speed of 6 m>s and depth of 4 m. What other possible average velocity of flow provides the same specific energy?

SOLUTION Water is considered to be incompressible. The flow is steady. Using the continuity equation, 0 rd V + rV # dA = 0 0t L L cv

cs

0 - V1A1 + V2A2 = 0 - ( 6 m>s ) (b)(4 m) = V(b)(y) V = a

24 b m>s y

(1)

The specific energy is E =

V2 + y 2g

(2)

For a depth of y = 4 m E =

( 6 m>s ) 2 + 4 m = 5.8349 m 2 ( 9.81 m>s2 )

Substituting this result and Eq. (1) into Eq. (2),

5.8349 m =

a

24 2 b y

2 ( 9.81 m>s2 )

+ y

y3 - 5.8349y2 + 29.3578 = 0 Solving by trial and error, y = 4m y = 3.7777 m y = -1.9428 m

(unrealistic)

At y = 3.7777 m, Eq. (1) gives

and

V = a Fr =

At y = 4 m, Fr =

24 b m>s = 6.353 m>s = 6.35 m>s 3.7777 V

2gy V 2gy

=

=

6.353 m>s 2 ( 9.81 m>s2 ) (3.7777 m) 6 m>s 2 ( 9.81 m>s2 ) (4 m)

= 1.04 7 1

= 0.958 6 1

Ans.

(supercritical)

(subcritical)

1267

12–9. A rectangular channel having a width of 3 m is required to transport 40 m3 >s of water. Determine the critical depth and critical velocity of the flow. What is the specific energy at the critical depth, and also when the depth is 2 m?

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is 1 3

1 3

( 40 m3 >s ) 2 yc = ° 2 ¢ = £ § = 2.6267 m = 2.63 m gb ( 9.81 m>s2 ) (3 m)2 Q2

Ans.

At critical flow, Fr =

Vc 2gyc

= 1

Vc

2 ( 9.81 m>s2 ) (2.6267 m)

= 1 Ans.

Vc = 5.076 m>s = 5.08 m>s

Also, the specific energy is E =

Q2 2gb2y2

+ y =

( 40 m3 >s ) 2 9.0610 + y = + y 2 2 2 y2 2 ( 9.81 m>s ) (3 m) y

(1)

At y = yc = 2.6267 m, Eq. (1) gives E min =

9.0610 + 2.6267 m = 3.940 m = 3.94 m 2.62672

Also, Emin =

3 3 y = (2.627 m) = 3.94 m 2 c 2

At y = 2 m, Eq. (1) gives E =

9.0610 + 2 m = 4.2652 m = 4.27 m 22

Ans.

Ans: yc = Vc = E min At y 1268

2.63 m 5.08 m>s = 3.94 m = 2 m, E = 4.27 m.

12–10. The channel transports water at 8 m3 >s. If the depth of flow is y = 1.5 m, determine if the flow is subcritical or supercritical. What is the critical depth of flow? Compare the specific energy of the flow with its minimum specific energy. y 2.5 m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangule, the critical depth is 1 3

1 3

( 8 m3 >s ) 2 yc = ° 2 ¢ = £ § = 1.0144 m = 1.01 m gb ( 9.81 m>s2 ) (2.5 m)2 Q2

Ans.

Since y = 1.5 m 7 yc, the flow is subcritical. Also, the minimum specific energy is E min =

3 3 yc = (1.0144 m) = 1.52 m 2 2

Ans.

The specific energy at y = 1.5 m is E =

Q2 2gb2y2

+ y =

( 8 m3 >s ) 2 + 1.5 m = 1.73 m 2 ( 9.81 m>s2 ) (2.5 m)2(1.5 m)2

Ans.

Ans: yc = 1.01 m E min = 1.52 m At y = 1.5 m, E = 1.73 m. 1269

12–11. Water flows within the rectangular channel with a flow of 8 m3 >s. Determine the two possible flow depths, and identify the flow as supercritical or subcritical, if the specific energy is 2 m. Also, plot the specific energy diagram.

y

3m

SOLUTION

y (m)

E=y

Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is yc = °

Q2 2

gb

1 3

Q = 8 m3 s

1 3

¢ = £

( 8 m3 >s ) 2 § = 0.8983 m ( 9.81 m>s2 ) (3 m)2

1.90 yc = 0.898 0.490

2 Emin = 1.35

Also, the minimum specific energy is E min =

subcritical

3 3 y = ( 0.8983 m ) = 1.347 m 2 c 2

supercritical E (m)

(a)

The specific energy is E =

Q2 2 2

2gb y

+ y =

( 8 m3 >s ) 2 + y = 2m 2 ( 9.81 m>s2 ) (3 m)2y2

y3 - 2y2 + 0.3624 = 0 Solving by trial and error, y = 1.90 m > yc

(subcritical)

Ans.

y = 0.490 m< yc

(supercritical)

Ans.

y = - 0.389 m

(unrealistic)

The specific energy diagram is shown in Fig. a.

Ans: y = 1.90 m (subcritical) y = 0.490 m (supercritical) 1270

*12–12. The rectangular channel transports water at 4 m3 >s. Determine the critical depth yc and plot the specific energy diagram for the flow. Indicate y for E = 1.25 m.

y

2m

SOLUTION

y (m)

E=y

Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is yc = °

Q2 2

gb

1 3

Q = 4 m3 s

1 3

¢ = £

( 4 m3 >s ) 2 § = 0.7415 m = 0.742 m ( 9.81 m>s2 ) (2 m)2

Ans.

0.533

Also, the minimum specific energy is Emin

Q2 2gb2y2

1.25 m =

subcritical supercritical E (m)

1.25 Emin = 1.11

3 3 = yc = ( 0.7415 m ) = 1.112 m 2 2

(a)

The specific energy is E =

1.07 yc = 0.742

+ y

( 4 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (2 m)2y2

y3 - 1.25y2 + 0.2039 = 0 Solving by trial and error, y = 1.07 m 7 yc

(subcritical)

Ans.

y = 0.533 m 6 yc

(supercritical)

Ans.

y = - 0.356 m

(unrealistic)

The specific energy diagram is shown in Fig. a.

1271

12–13. The rectangular channel transports water at 8 m3 >s. Determine the critical depth yc and plot the specific energy diagram for the flow. Indicate y for E = 2 m. y

2m

SOLUTION

y (m)

E=y

Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is 1 3

Q = 8 m3 s

1 3

( 8 m3 >s ) 2 yc = ° 2 ¢ = £ § = 1.1771 m = 1.18 m gb ( 9.81 m>s2 ) (2 m)2 Q2

Ans.

subcritical supercritical

0.838

Also, the minimum specific energy is E min =

1.73 yc = 1.18

E (m)

2

Emin = 1.77

3 3 y = (0.1771 m) = 1.766 m 2 c 2

(a)

The specific energy is E =

Q2 2gb2y2

2m =

+ y

( 8 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (2 m)2y2

y3 - 2y2 + 0.8155 = 0 Solving by trial and error, y = 1.73 m 7 yc

(subcritical)

Ans.

y = 0.838 m 6 yc

(supercritical)

Ans.

y = - 0.564 m

(unrealistic)

The specific energy diagram is shown in Fig. a.

Ans: yc = 1.18 m y = 1.73 m (subcritical) y = 0.838 m (supercritical) 1272

12–14. Water flows within the rectangular channel such that the flow is 4 m3 >s. Determine the critical depth of flow and the minimum specific energy. If the specific energy is 8 m, what are the two possible flow depths?

y

2m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is 1 3

1 3

( 4 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.7415 m = 0.742 m gb ( 9.81 m>s2 ) (2 m)2 Q2

Ans.

Also, the minimum specific energy is E min =

3 3 y = (0.7415 m) = 1.112 m = 1.11 m 2 c 2

Ans.

The specific energy is E =

Q2 2gb2y2

8m =

+ y

( 4 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (2 m)2y2

y3 - 8y2 + 0.2039 = 0 Solving by trial and error, y = 8.00 m 7 yc

(subcritical)

Ans.

y = 0.161 m 6 yc

(supercritical)

Ans.

y = - 0.158 m

(unrealistic)

Ans: Emin = 1.11 m yc = 0.742 m y = 8.00 m (subcritical) y = 0.161 m (supercritical) 1273

12–15. The rectangular channel transports water at a flow of 8 m3 >s. Plot the specific energy diagram for the flow and indicate y for E = 3 m.

y

1.5 m

y (m)

SOLUTION

E=y

Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is 1 3

Q = 8 m3 s 2.82

1 3

( 8 m3 >s ) 2 yc = ° 2 ¢ = £ § = 1.4260 m gb ( 9.81 m>s2 ) (1.5 m)2 Q2

0.814

Also, the minimum specific energy is Emin

subcritical

yc = 1.43

supercritical E (m)

3

Emin = 2.14

3 3 = yc = ( 1.4260 m ) = 2.14 m 2 2

(a)

The specific energy is E =

Q2 2gb2y2

3m =

+ y

( 8 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (1.5 m)2y2

y3 - 3y2 + 1.4498 = 0 Solving by trial and error, y = 2.82 m 7 yc

(subcritical)

Ans.

y = 0.814 m 6 yc

(supercritical)

Ans.

y = - 0.632 m

(unrealistic)

The specific energy diagram is shown in Fig. a.

Ans: y = 2.82 m (subcritical) y = 0.814 m (supercritical) 1274

*12–16. The rectangular channel passes through a transition that causes its width to narrow to 1.5 m. If the flow is 5 m3 >s and yA = 3 m, determine the depth of flow at B.

3m 1.5 m A B

yA yB

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energy is E =

Q2 2gb2y2

+ y

At section A, y = 3 m and b = 3 m. Then, E =

( 5 m3 >s ) 2 + 3 m = 3.0157 m 2 ( 9.81 m>s 2 ) (3 m)2(3 m)2

At section B, y = yB and b = 1.5 m. Then, 3.0157 m =

( 5 m3 >s ) 2 + yB 2 ( 9.81 m>s 2 ) (1.5 m)2yB2

yB3 - 3.0157 yB2 + 0.5663 = 0 Solving by trial and error, yB = 2.95 m yB = 0.472 m yB = -0.407 m

(unrealistic)

Also, the critical depths at sections A and B are 1 3

1 3

( 5 m3 >s ) 2 (yc)A = ° 2 ¢ = £ § = 0.6567 m gbA ( 9.81 m>s2 ) (3 m)2 Q2

1 3

1 3

( 5 m3 >s ) 2 (yc)B = ° 2 ¢ = £ § = 1.042 m gbB ( 9.81 m>s2 ) (1.5 m)2 Q2

Since yA = 3 m 7 (yC)A, the flow at section A is subcritical. Since the type of flow must remain the same throughout the channel, the flow at section B is also subcritical. This requires yB 7 (yB)C. Ans.

yB = 2.95 m

1275

12–17. The rectangular channel has a transition that causes its width to narrow to 1.5 m. If the flow is 5 m3 >s and yA = 5 m, determine the depth of flow at B.

3m 1.5 m A B

yA yB

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energy is E =

Q2 2gb2y2

+ y

At section A, y = 5 m and b = 3 m. Then, E =

( 5 m3 >s ) 2 + 5 m = 5.0057 m 2 ( 9.81 m>s 2 ) (3 m)2(5 m)2

At section B, y = yB and b = 1.5 m. Then, 5.0057 m =

( 5 m3 >s ) 2 + yB 2 ( 9.81 m>s 2 )( 1.5 m ) 2yB2

yB3 - 5.0057 yB2 + 0.5663 = 0 Solving by trial and error, yB = 4.98 m yB = 0.349 m yB = - 0.326 m

(unrealistic)

Also, the critical depths at sections A and B are 1 3

1 3

( 5 m3 >s ) 2 (yc)A = ° 2 ¢ = £ § = 0.6567 m gbA ( 9.81 m>s2 ) (3 m)2 Q2

1 3

1 3

( 5 m3 >s ) 2 (yc)B = ° ¢ = £ § = 1.042 m gbB2 ( 9.81 m>s2 ) (1.5 m)2 Q2

Since yA = 5 m 7 (yC)A, the flow at section A is subcritical. Since the type of flow must remain the same throughout the channel, the flow at section B is also subcritical. This requires yB 7 (yC)B. Ans.

yB = 4.98 m

Ans: 4.98 m 1276

12–18. The venturi is placed in the channel in order to measure the volumetric flow. If the depth of flow at A is yA = 2.50 m and at the throat B is yB = 2.35 m, determine the flow through the channel.

1.75 m

1m A B

yA ! 2.50 m yB ! 2.35 m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energy at sections A and B is EA = = EB = =

Q2 2

yA2

2gb

+ yA =

Q2 2 ( 9.81 m>s2 ) (1.75 m)2(2.50 m)2

+ 2.50 m

Q2 + 2.50 375.54 Q2 2

2gb

yB2

+ yB =

Q2 2 ( 9.81 m>s ) (1 m)2(2.35 m)2 2

+ 2.35 m

Q2 + 2.35 108.35

Since the head losses are neglected, EA = EB Q2 Q2 + 2.50 = + 2.35 375.54 108.35 Q = 4.779 m3 >s = 4.78 m3 >s

Ans.

1277

Ans: 4.78 m3 >s

12–19. Water flows at 25 ft 3 >s in the channel, which is originally 6 ft wide and then gradually narrows to b2 = 4 ft. If the original depth of water is 3 ft, determine the depth y2 after it passes through the constriction. What width b2 will produce critical flow y2 = yc?

6 ft b2

3 ft

y2

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energy is E =

Q2

+ y

2gb2y2

At section (1), y = 3 ft and b = 6 ft . Then, E =

( 25ft 3 >s ) 2 + 3 ft 2 ( 32.2 ft>s2 ) (6 ft)2(3 ft)2

= 3.0300 ft At section (2), y = y2 and b = 4 ft . Then 3.0300 ft =

( 25ft 3 >s ) 2 + y2 2 ( 32.2 ft>s2 ) (4 ft)2y22

y23 - 3.0300y22 + 0.6066 = 0 Solving by trial and error, For critical flow y2 = 2.96 ft y2 = 0.489 ft y2 = - 0.419 ft Also, the critical depths at sections (1) and (2) are 1 3

1 3

1 3

1 3

( 25 ft3 >s ) 2 (yc)1 = ° 2 ¢ = £ § = 0.8139 ft gb1 ( 32.2 ft>s2 ) (6 ft)2 Q2

( 25 ft3 >s ) 2 (yc)2 = ° 2 ¢ = £ § = 1.067 ft gb2 ( 32.2 ft>s2 ) (4 ft)2 Q2

Since y1 = 3 ft > ( yc ) 1, the flow at section (1) is subcritical. Since the type of flow must remain the same throughout the channel, the flow at section (2) is also subcritical. This requires y2 7 ( yc ) 2. Ans.

y2 = 2.96 ft When the critical flow occurs, E min = E =

3 y 2 c

3 y 2 c yc = 2.0200 ft

3.0300 ft =

1278

12–19.

Continued

Also, it requires that Fr =

Vc 2gyc

= 1

Vc = 2gyc = 2 ( 32.2 ft>s2 ) (2.0200 ft) Vc = 8.0649 ft>s

Thus, Q = VA;

Q = Vcbcyc 25 ft 3 >s = ( 8.0649 ft>s ) (bc)(2.0200 ft)

Ans.

bc = 1.53 ft

Ans: y2 = 2.96 ft bc = 1.53 ft 1279

*12–20. The rectangular channel has a width of 8 ft and transports water at 30 ft 3 >s. If the depth of flow at A is 6 ft, determine how high h the channel bed should rise in order to produce critical flow at B.

A B 6 ft

yC h

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the critical depth is 1 3

1 3

( 30 ft3 >s ) 2 yc = ° 2 ¢ = £ § = 0.7587 ft gb ( 32.2 ft>s2 ) (8 ft)2 Q2

Also, the specific energies at sections A and B are EA =

Q2 2gb2yA2

+ yA =

( 30 ft3 >s ) 2 + 6 ft 2 ( 32.2 ft>s2 ) (8 ft)2(6 ft)2

= 6.0061 ft EB =

Q2 2gb2yB2

+ yB =

( 30 ft3 >s ) 2 + 0.7587 ft 2 ( 32.2 ft>s2 ) (8 ft)2(0.7587 ft)2

= 1.1380 ft Thus, EA = EB + h 6.0061 ft = 1.1380 ft + h Ans.

h = 4.868 ft = 4.87 ft

1280

12–21. The channel is 2 m wide and transports water at 18 m3 >s. If the elevation of the bed is raised 0.25 m, determine the new depth y2 of the water and the speed of the flow. Is the new flow subcritical or supercritical?

1.5 m

y2

0.25 m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energies at sections (1) and (2) are E1 = = E2 = =

Q2 2gb2 y12

+ y1

( 18 m3>s ) 2 + 1.5 m = 3.3349 m 2 ( 9.81 m>s2 ) (2 m)2(1.5 m)2 Q2 2gb2y22

+ y2 =

( 18 m3>s ) 2 + y2 2 ( 9.81 m>s2 ) (2 m)2y22

4.1284 + y2 y22

Thus, E1 = E2 + h 3.3349 m =

4.1284 + y2 + 0.25 m y22

y23 - 3.0849y22 + 4.1284 = 0 Solving by trial and error, y2 = 2.3136 m y2 = 1.7760 m y2 = - 1.0047 m

(unrealistic)

Also, the critical depth is 1 3

1 3

( 18 m3 >s ) 2 yc = ° 2 ¢ = £ § = 2.021 m gb ( 9.81 m>s2 ) (2 m)2 Q2

Since y1 = 1.5 m 6 yc, the flow at section (1) is supercritical. Since the type of flow must remain the same throughout the channel, the flow at section (2) is also supercritical. This requires y2 6 yc. Ans.

y2 = 1.7760 m = 1.78 m The mean velocity at section (2) is V2 =

18 m3 >s Q ; = 5.07 m>s A2 (2 m)(1.7760 m)

Ans.

Ans: supercritical y2 = 1.78 m V2 = 5.07 m>s 1281

12–22. The channel is 2 m wide and transports water at a flow of 18 m3 >s. If the elevation of the bed is lowered by 0.1 m, determine the new depth y2 of the water.

1.25 m

y2

0.1 m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energies at sections (1) and (2) are E1 = = E2 = =

Q2 2gb2y12

+ y1

( 18 m3 >s ) 2 + 1.25 m = 3.8922 m 2 ( 9.81 m>s2 ) (2 m)2(1.25 m)2 Q2 2gb2y22

+ y2 =

( 18 m3 >s ) 2 + y2 2 ( 9.81 m>s2 ) (2 m)2y22

4.1284 + y2 y22

With h = -0.1 m E1 = E2 + h 3.8922 m =

4.1284 + y2 + ( - 0.1 m) y22

y23 - 3.9922y22 + 4.1284 = 0 Solving by trial and error, y2 = 1.22 m y2 = 3.69 m y2 = - 0.917 m

(unrealistic)

Also, the critical depth is 1 3

1 3

( 18 m3 >s ) 2 yc = ° 2 ¢ = £ § = 2.021 m gb ( 9.81 m>s2 ) (2 m)2 Q2

Since y1 = 1.25 m < yc, the flow at section (1) is supercritical. Since the type of flow must remain the same throughout the channel, the flow at section (2) is also supercritical. This requires y2 6 yc. Ans.

y2 = 1.22 m

Ans: 1.22 m 1282

12–23. Water flows at 18 ft 3 >s through the rectangular channel having a width of 4 ft. If the depth of flow yA = 4 ft, determine if the depth increases or decreases after the water flows over the 0.25-ft rise. What is yB?

A

B yA

yB 0.25 ft

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energies at sections A and B are EA = =

Q2 2gb2yA2

+ yA

( 18 ft3 >s ) 2 + 4 ft 2 ( 32.2 ft>s2 ) (4 ft)2(4 ft)2

= 4.0197 ft EB =

Q2 2gb2yB2

+ yB

=

( 18 ft3 >s ) 2 + yB 2 ( 32.2 ft>s2 ) (4 ft)2yB2

=

0.3144 + yB yB2

Thus, EA = EB + h 4.0197 ft =

0.3144 + yB + 0.25 ft yB2

yB3 - 3.7697yB2 + 0.3144 = 0 Solving by trial and error, yB = 0.301 ft yB = 3.75 ft yB = -0.279 ft

(unrealistic)

Also, the critical depth is 1 3

1 3

( 18 ft3 >s ) 2 yc = ° 2 ¢ = £ § = 0.8568 ft gb ( 32.2 ft>s2 ) (4 ft)2 Q2

Since yA = 4 ft 7 yc, the flow at section A is subcritical. Since the type of flow must remain the same throughout the channel, the flow at section B is also subcritical. This requires yB 7 yc. Ans.

yB = 3.75 ft Since yB 6 yA, the depth of the flow decreases after the rise.

Ans: 3.75 ft decreases 1283

*12–24. Water flows at 18 ft 3 >s through the rectangular channel having a width of 4 ft. If the depth yA = 0.5 ft, determine if the depth yB increases or decreases after the water flows over the 0.25-ft rise. What is yB?

A yA

0.25 ft

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energies at sections A and B are EA = =

Q2 2gb2yA2

+ yA

( 18 ft3 >s ) 2 + 0.5 ft 2 ( 32.2 ft>s2 ) (4 ft)2(0.5 ft)2

= 1.7578 ft EB =

Q2 2gb2yB2

+ yB

=

( 18 ft3 >s ) 2 + yB 2 ( 32.2 ft>s2 ) (4 ft)2yB2

=

0.3144 + yB yB2

Thus,

EA = EB + h 0.3144 + yB + 0.25 ft yB2

1.7578 ft =

yB3 - 1.5078yB2 + 0.3144 = 0 Solving by trial and error,

yB = 0.583 ft yB = 1.33 ft yB = -0.405 ft

(unrealistic)

Also, the critical depth is 1 3

1 3

( 18 ft3 >s ) 2 yc = ° 2 ¢ = £ § = 0.8568 ft gb ( 32.2 ft>s2 ) (4 ft)2 Q2

B yB

Since yA = 0.5 ft 6 yc, the flow at section A is supercritical. Since the type of flow must remain the same throughout the channel, the flow at section B is also supercritical. This requires yB 6 yc.

yB = 0.583 ft

Ans.

Since yB 7 yA, the depth of the flow increases after the rise.

1284

12–25. Water flows within the 4-m-wide rectangular channel at 20 m3 >s. Determine the depth of flow yB at the downstream end and the velocity of flow at A and B. Take yA = 5 m.

A

B yA

yB 0.2 m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energy at section A is EA = =

Q2

+ yA

2gb2yA2

( 20 m3 >s ) 2 + 5m 2 ( 9.81 m>s2 ) (4 m)2(5 m)2

= 5.0510 m The specific energy at section B is EA = EB + h 5.0510 m = EB + 0.2 m EB = 4.8510 m Then, EB =

Q2 2gb2yB2

4.8510 m =

+ yB

( 20 m3 >s ) 2 + yB 2 ( 9.81 m>s2 ) (4 m)2yB2

yB3 - 4.8510yB2 + 1.2742 = 0 Solving by trial and error, yB = 4.7956 m yB = 0.5439 m yB = -0.4885 m

(unrealistic)

Also, the critical depth is 1 3

1 3

( 20 m3 >s ) 2 yc = ° 2 ¢ = £ § = 1.366 m gb ( 9.81 m>s2 ) (4 m)2 Q2

Since yA = 5 m 7 yc, the flow at section A is subcritical. Since the type of flow must remain the same throughout the channel, the flow at section B is also subcritical. This requires yB 7 yc. Ans.

yB = 4.7956 m = 4.80 m The mean velocity at sections A and B is VA =

20 m3 >s Q = = 1 m>s AA 4 m(5 m)

Ans.

VB =

20 m3 >s Q = = 1.04 m>s AB 4 m(4.7956 m)

Ans. Ans: yB = 4.80 m VA = 1 m>s VB = 1.04 m>s 1285

12–26. Water flows within the 4-m-wide rectangular channel at 20 m3 >s. Determine the depth of flow yB at the downstream end and the velocity of flow at A and B. Take yA = 0.5 m.

A

B yA

yB 0.2 m

SOLUTION Water is considered to be incompressible. The flow is steady. Since the channel is rectangular, the specific energy at section A is EA = =

Q2 2gb2yA2

+ yA

( 20 m3 >s ) 2 + 0.5 m 2 ( 9.81 m>s2 ) (4 m)2(0.5 m)2

= 5.5968 m The specific energy at section B is EA = EB + h 5.5968 m = EB + 0.2 m EB = 5.3968 m Then, EB =

Q2 2gb2yB2

5.3968 m =

+ yB

( 20 m3 >s ) 2 + yB 2 ( 9.81 m>s2 ) (4 m)2yB2

yB3 - 5.3968 myB2 + 1.2742 = 0 Solving by trial and error, yB = 0.5107 m yB = 5.3524 m yB = - 0.4662 m

(unrealistic)

Also, the critical depth is 1 3

1 3

( 20 m3 >s ) 2 yc = ° 2 ¢ = £ § = 1.366 m gb ( 9.81 m>s2 ) (4 m)2 Q2

Since yA = 0.5 m 6 yc, the flow at section A is supercritical. Since the type of flow must remain the same throughout the channel, the flow at section B is also supercritical. This requires yB 7 yc. Ans.

yB = 0.5107 m = 0.511 m The mean velocity at sections A and B is VA =

20 m3 >s Q = = 10 m>s AA 4 m(0.5 m)

Ans.

VB =

20 m3 >s Q = = 9.79 m>s AB 4 m(0.5107 m)

Ans. Ans: yB = 0.511 m VA = 10 m>s VB = 9.79 m>s 1286

0.5 m/s

12–27. The rectangular channel is 2 m wide, and the depth of the water is 1.5 m as it flows with an average velocity of 0.5 m>s. Show that the flow is tranquil, and determine the required height h of the bump so that the flow can change to rapid flow after it passes over the bump. What is the new depth y2 for rapid flow?

1.5 m

y2 h

SOLUTION Q = VA = ( 0.5 m>s ) (2 m)(1.5 m) = 1.5 m3 >s 1 3

1 3

( 1.5 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.386 m 6 1.5 m gb ( 9.81 m>s2 ) (2 m)2 Q2

Flow is tranquil. E min =

3 3 y = (0.386 m) = 0.578 m 2 c 2

In general, E = E =

Q2 2gb2y2

+ y =

( 1.5 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (2 m)2y2

0.02867 + y y2

At y = 1.5 m, E = 1.5127 m 1.5127 m =

0.02867 + y y2

y3 - 1.5127y2 + 0.02867 = 0 y2 = 1.5 m 7 0.386 m

(Tranquil)

y2 = 0.145 m 6 0.386

(rapid)

y2 = -0.132 m

(unrealistic)

Ans.

Bump must remove 1.5127 m - 0.578 m = 0.935 m of the specific energy. Thus, Ans.

h = 0.935 m

Ans: y2 = 0.145 m h = 0.935 m 1287

4 m/s

*12–28. The rectangular channel is 2 m wide, and the depth of the water is 0.75 m as it flows with an average velocity of 4 m>s. Show that the flow is rapid, and determine the required height h of the bump so that the flow can change to tranquil flow after it passes over the bump. What is the new depth y2 for tranquil flow?

0.75 m

SOLUTION Q = VA = ( 4 m>s ) (2 m)(0.75 m) = 6 m3 >s 1 3

1 3

( 6 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.97168 m 7 0.75 m gb ( 9.81 m>s2 ) (2 m)2 Q2

Flow is rapid

E min =

3 3 y = (0.97168 m) = 1.4575 m 2 c 2

In general, E = E =

Q2 2 2

2gb y

+ y =

( 6 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (2 m)2y2

0.45872 + y y2

At y = 0.75 m E = 1.5655 m 1.5655 m =

0.45872 + y y2

y3 - 1.5655y2 + 0.45872 = 0 y2 = 0.75 m 6 0.97168 m

(rapid)

y2 = 1.29 m 7 0.97168 m

(tranquil)

y2 = - 0.4742 m

(unrealistic)

Ans.

Bump must remove 1.5655 m - 1.4575 m = 0.108 m of specific energy. Thus, Ans.

h = 108 mm

1288

y2 h

12–29. The rectangular channel has a width of 3 ft, and the depth of the water is originally 4 ft. If the flow is 50 ft3>s, show that the upstream flow is tranquil, and determine the required depth of the water, y′, at the top of the bump so that the downstream flow can be transformed to rapid flow. What is the downstream depth?

4 ft

y¿

y2

SOLUTION 1 3

1 3

( 50 ft3 >s ) 2 yc = ° 2 ¢ = £ § = 2.0509 gb ( 32.2 ft>s2 ) (3 ft)2 Q2

Since 4 ft 7 2.0509 ft The flow is tranquil. At the top of the bump the flow has the critical depth. Ans.

yc = 2.05 ft In general E = E =

Q2 2gb2y2

+ y =

( 50 ft3 >s ) 2 + y 2 ( 32.2 ft>s2 ) (3 ft)2y2

4.313 + y y2

At depth 4 ft, E = 4.270 ft Thus, 4.270 ft =

4.313 + y y2

y3 - 4.270y2 + 4.313 = 0 y2 = 4 ft 7 2.0509 ft

(Tranquil flow) Ans.

y2 = 1.18 ft 6 2.0509 ft (rapid flow) y2 = - 0.912 ft

(unrealistic)

The water surface profiles are

4 ft

4 ft

2.05 ft

4 ft

Tranquil

2.05 ft Rapid

Ans: yc = 2.05 ft y2 = 1.18 ft 1289

12–30. The sluice gate is 5 ft wide. If the flow is 200 ft3>s, determine the depth of flow y2 and the type of flow downstream from the gate. What is the maximum volumetric flow of water that can pass through the gate?

5 ft

y1 ! 10 ft y2

SOLUTION Water is considered to be incompressible. The flow is steady. The critical depth of the flow further downstream from the gate is (y2)c =

2 2 y = (10 ft) = 6.667 ft 3 1 3

Thus, y1 =

Q2 2gb2y22

10 ft =

+ y2

( 200ft3 >s ) 2 + y2 2 ( 32.2ft>s2 ) (5ft)2y22

y23 - 10y22 + 24.845 = 0 Solving by trial and error, y2 = 9.74 ft 7 (y2)c

(subcritical)

y2 = 1.73 ft 6 (y2)c

(supercritical)

y2 = - 1.47 ft The first root is the flow depth just before the gate, and the second root is the flow depth further downstream from the gate. Thus, Ans.

y2 = 1.73 ft The maximum flow rate is Q max = =

8 2 3 gb y1 A 27

8 ( 32.2 ft>s2 ) (5 ft)2(10 ft)2 A 27

= 488 ft 3 >s

Ans.

1290

Ans: y2 = 1.73 ft Qmax = 488 ft 3 >s

12–31. The sluice gate is 5 ft wide. Determine the volumetric flow of water through the gate if y2 = 2 ft. What type of flow occurs?

5 ft

y1 ! 10 ft y2

SOLUTION Water is considered to be incompressible. The flow is steady. y1 = 10 ft =

Q2 2gb2 y22

+ y2 Q2

2 ( 32.2 ft>s ) (5 ft)2(2 ft)2 2

+ 2 ft

Q = 227 ft 3 >s

Ans.

The critical depth of the flow further downstream from the gate is y2 =

2 2 y = (10 ft) = 6.667 ft 3 1 3

Since y2 = 2 ft 6 y2, the flow is supercritical.

Ans: Q = 227 ft 3 >s supercritical 1291

*12–32. The 2-m-wide sluice gate is used to control the flow of water from a reservoir. If the depths y1 = 4 m and y2 = 0.75 m, determine the volumetric flow through the gate and the depth y3 just before the gate.

2m

y1

y2

SOLUTION Water is considered to be incompressible. The flow is steady. y1 = 4m =

Q2 2gb2y22

+ y2 Q2

2 ( 9.81 m>s2 ) (2 m)2(0.75 m)2

+ 0.75 m

Q = 11.97 m3 >s = 12.0 m3 >s

Ans.

Using this result, y1 =

Q2 2gb2y32

4m =

y3

+ y3

( 11.978 m3 >s ) 2 + y3 2 ( 9.81 m>s2 ) (2 m)2y32

y33 - 4y32 + 1.828125 = 0 Solving by trial and error, y3 = 0.750 m y3 = 3.88 m y3 = - 0.628 The first root is the flow depth further downstream from the gate, and the second root is the flow depth just before the gate. Thus, Ans.

y3 = 3.88 m

1292

12–33. The 2-m-wide sluice gate is used to control the flow of water from a reservoir. If the flow is 10 m3>s and y1 = 4 m, determine the depth y2, and depth y3 just before the gate.

2m

y1

y3 y2

SOLUTION Water is considered to be incompressible. The flow is steady. y1 =

Q2 2gb2y2

4m =

+ y

( 10 m3 >s ) 2 + y 2 ( 9.81 m>s2 ) (2 m)2y2

y3 - 4y2 + 1.2742 = 0 Solving by trial and error, y = 3.917 m y = 0.6134 m y = - 0.5303 m

(unrealistic)

The first root is the flow depth just before the gate, and the second root is the flow depth further downstream from the gate. Thus, y2 = 0.613 m

Ans.

y3 = 3.92 m

Ans.

Ans: y2 = 0.613 m y3 = 3.92 m 1293

12–34. The sluice gate and channel both have a width of 2  m. If the depth of flow at A is y1 = 3 m, determine the volumetric flow through the channel as a function of depth y2 and specify Q when the depth y2 is (a) 1 m, (b) 1.5 m.

2m

y1

y3 y2

SOLUTION Water is considered to be incompressible. The flow is steady. y1 =

Q2 2gb2y22

3m =

+ y2 Q2

2 ( 9.81 m>s2 ) (2 m)2y22

+ y2 Ans.

a.) When y2 = 1 m,

Q = c 278.48y22 ( 3 - y2 ) d m3 >s

Ans.

b.) When y2 = 1.5 m,

Q = c 278.48 ( 12 ) (3 - 1) d = 12.5 m3 >s Q = c 278.48 ( 1.52 ) (3 - 1.5) d = 16.3 m3 >s

Ans.

Ans: Q = c 278.48y22 ( 3 - y2 ) d m3 >s

1294

a) 12.5 m3 >s b) 16.3 m3 >s

12–35. Determine the hydraulic radius for each channel cross section.

30! 30! h

l

b (a)

(b)

SOLUTION 30!

a.) For the rectangular section, A = bh

p = 2h + b

30!

l

bh Rh = 2h + b

l

Ans. b

b.) For the triangular section,

(c)

1 1 23 2 1 23 A = 3 (2 l sin 30°)(l cos 30°) 4 = £ 2l a b ° l¢ § = l 2 2 2 2 4

P = 2l

23 2 l 4 23 Rh = = l 2l 8

Ans.

c.) For the trapezoidal section A = 2c

1 3 (l sin 30°)(l cos 30°) 4 d + b(l cos 30°) 2

1 23 23 23 2 23 23 = l2 a b ° ¢ + bl = l + bl = l ( l + 2b ) 2 2 2 4 2 4

P = 2l + b

23 l (l + 2b) 23l (l + 2b) 4 Rh = = 2l + b 4 (2l + b)

Ans.

Ans: a) Rh = b) Rh = c) Rh = 1295

bh 2h + b 23 l 8 23l (l + 2b) 4 (2l + b)

*12–36. The channel has a triangular cross section. Determine the critical depth y = yc in terms of u and the flow Q. y u

SOLUTION For a non-rectangular section, the condition for critical flow is gAc3 Q2btop

(1)

= 1

From the geometry shown in Fig. a,

Then,

btop = 2 a Ac =

yc 2yc b = tan u tan u

yc2 1 1 2yc btop yc = a byc = 2 2 tan u tan u

Substituting these results into Eq. (1), yc2 3 b tan u = 1 2yc 2 Q a b tan u ga

1

yc = a

2Q2 tan2 u 5 b g

Ans.

btop

yc

(a)

1296

u

12–37. A rectangular channel has a width of 2 m, is made of unfinished concrete, and is inclined at a slope of 0.0014. Determine the volumetric flow when the depth of flow of the water is 1.5 m.

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Here, A = (2 m)(1.5 m) = 3 m2 P = 2(1.5 m) + 2 m = 5 m For SI units, 5

Q =

5

1

A3S02 2

nP 3

=

1

( 3 m2 ) 3 ( 0.0014 2 ) 1

2

( 0.014 s>m3 ) (5 m) 3

= 5.70 m3 >s

1297

Ans.

Ans: 5.70 m3 >s

12–38. The channel is made of wood and has a downward slope of 0.0015. Determine the volumetric flow of water if y = 2 ft. y

h 3 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.012 s>m 3 for a wood surface. Here, A = (3 ft)(2 ft) = 6 ft 2 P = 2(2 ft) + 3 ft = 7 ft For FPS units, 5

Q =

2

nP 3

5

1

1.486A3S02

=

1

1.486 ( 6 ft 2 ) 3 ( 0.0015 2 ) 1

2

( 0.012 s>m3 ) (7 ft)3

= 26.0 ft 3 >s

1298

Ans.

Ans: 26.0 ft 3 >s

12–39. The channel is made of wood and has a downward slope of 0.0015. Determine the depth of flow y that will produce the maximum volumetric flow using the least amount of wood when it is flowing full of water, that is, when y = h. What is this volumetric flow?

y

h 3 ft

SOLUTION Water is considered to be incompressible. The flow is steady. The optimum cross-section for a rectangular channel is when y =

b 3 ft = = 1.5 ft . 2 2 Ans.

y = 1.5 ft A = (3 ft)(1.5 ft) = 4.5 ft

2

P = 2(1.5 ft) + 3 ft = 6 ft 1

From the table, n = 0.012 s>m 3 for a wood surface. For FPS units, 5

Q =

2

nP 3

5

1

1.486A3S02

=

1

1.486 ( 4.5 ft 2 ) 3 ( 0.0015 2 ) 1

2

( 0.012 s>m3 ) (6 ft)3

= 17.8 ft 3 >s

1299

Ans.

Ans: y = 1.5 ft Q = 17.8 ft 3 >s

*12–40. The drainage canal has a downward slope of 0.002. If its bottom and sides have weed growth, determine the volumetric flow of water when the depth of flow is 2.5 m.

4m

2.5 m

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.030 s>m 3 for a weedy surface. Here, A = (4 m)(2.5 m) = 10 m2 P = 2(2.5 m) + 4 m = 9 m For SI units, 5

Q =

5

1

A3S02 2

nP 3

=

1

( 10 m2 ) 3 ( 0.002 2 ) 1

2

( 0.030 s>m3 ) (9 m) 3

= 16.0 m3 >s

1300

Ans.

12–41. The sewer pipe, made of unfinished concrete, is required to carry water at 60 ft3>s when it is half full. If the downward slope of the pipe is 0.0015, determine the required internal radius of the pipe. R

SOLUTION 1

From the table, n = 0.014 s>ft 3 for an unfinished concrete surface. Here, p A = R2 P = pR 2 For FPS units, 5

Q =

1

1.486A3S02 2

nP 3 5

60 ft 3 >s =

3 1 p 1.486 a R2 b a0.00152 b 2 1

2

( 0.014 s>ft 3 ) (pR)3

Ans.

R = 2.74 ft

Ans: 2.74 ft 1301

12–42. A 3-ft-diameter unfinished-concrete drain pipe has been designed to transport 80 ft3>s of water at a depth of 2.25 ft. Determine the required downward slope of the pipe.

SOLUTION Water is considered to be incompressible. The flow is steady. From the geometry shown in Fig. a,

Then,

u = cos-1a

0.75 ft b = 60° 1.5 ft

a = 360° - 2u = 360° - 2(60°) = 240° a

p rad 4 b = p rad 180° 3

The cross-sectional area and the wetted perimeter are A =

(1.5 ft)2 4 4 R2 (a - sin a) = c p rad - sin p d = 5.6867 ft 2 2 2 3 3

P = aR =

4 p(1.5 ft) = 2p ft 3 1

From the table, n = 0.014 s>ft 3 for an unfinished concrete surface. For FPS units, 5

Q =

1

1.486A3S02 2

nP 3 5

80 ft 3 >s =

1

1.486 ( 5.6867 ft 2 ) 3S02 1

2

( 0.014 s>ft 3 ) (2p ft)3 Ans.

S0 = 0.0201

0.75 ft

1.5 ft

1.5 ft (a)

Ans: 0.0201 1302

12–43. Water flows uniformly down the triangular channel having a downward slope of 0.0083. If the walls are made of  finished concrete, determine the volumetric flow when y = 1.5 m.

20!

20! y

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.012 s>m 3 for a finished concrete surface. Also, with a = 90° - 20° = 70° and y = 1.5 m for the triangular channel, A =

(1.5 m)2 y2 = = 0.8189 m2 tan a tan 70°

P =

2(1.5 m) 2y = = 3.1925 m sin a sin 70°

For SI units, 5

Q =

5

1

A3S02 2

nP 3

=

1

( 0.8189 m2 ) 3 ( 0.0083 2 ) 1

2

( 0.012 s>m3 ) (3.1925 m)3

= 2.51 m3 >s

1303

Ans.

Ans: 2.51 m3 >s

*12–44. A rectangular channel has a downward slope of 0.006 and a width of 3 m. The depth of the water is 4 m. If the volumetric flow through the channel is 30 m3>s, determine the value of n in the Manning formula.

SOLUTION Water is considered to be incompressible. The flow is steady. The cross-sectional area and wetted perimeter of the rectangular channel are A = (3 m)(4 m) = 12 m2 P = 2(4 m) + 3 m = 11 m For SI units, 5

Q =

1

A3S02 2

nP 3 5

3

30 m >s =

1

( 12 m2 ) 3 ( 0.0062 ) 2

n(11 m) 3 1

Ans.

n = 0.0328 s>m 3

1304

12–45. The channel is made of unfinished concrete and has the cross section shown. If the downward slope is 0.0008, determine the flow of water through the channel when y = 4 m.

3m y 3m

SOLUTION 3m

Water is considered to be incompressible. The flow is steady.

2m

3m

1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Here, A = (8 m)(1 m) + (2 m)(3 m) = 14 m2 P = 2(1 m + 3 m + 3 m) + 2 m = 16 m For SI units, 5

Q =

5

1

A3S02 2

nP 3

=

1

( 14 m2 ) 3 ( 0.0008 2 ) 1

2

( 0.014 s>m3 ) (16 m)3

= 25.9 m3 >s

1305

Ans.

Ans: 25.9 m3 >s

12–46. The channel is made of unfinished concrete and has the cross section shown. If the downward slope is 0.0008, determine the flow of water through the channel when y = 6 m.

3m y 3m

3m

SOLUTION

2m

3m

Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Here, A = (8 m)(3 m) + (2 m)(3 m) = 30 m2 P = 2(3 m + 3 m + 3 m) + 2 m = 20 m For SI units, 5

Q =

5

1

A3S02 2

nP 3

=

1

( 30 m2 ) 3 ( 0.0008 2 ) 1

2

( 0.014 s>m3 ) (20 m)3

= 79.4 m3 >s

1306

Ans.

Ans: 79.4 m3 >s

12–47. The channel has a triangular cross section and is flowing full of water. Determine the volumetric flow if the sides are made of wood and the downward slope is 0.002. 3 ft 50!

50!

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.012 s>m 3 for a wood surface. Also, with a = 50° and y = 3 ft for a triangular channel, A = P =

y2 tan a

=

(3 ft)2 tan 50°

= 7.5519 ft 2

2(3 ft) 2y = = 7.8324 ft sin a sin 50°

For FPS units, 5

Q =

2

nP 3

5

1

1.486 A3S02

=

1

1.486 ( 7.5519 ft 2 ) 3 ( 0.002 2 ) 1

2

( 0.012 s>m3 ) (7.8324 ft)3

= 40.8 ft 3 >s

1307

Ans.

Ans: 40.8 ft 3 >s

*12–48. The drainage pipe is made of unfinished concrete and is sloped downward at 0.0025. Determine the volumetric flow from the pipe if the center depth is y = 3 ft. 2 ft

y

SOLUTION Water is considered to be incompressible. The flow is steady.

2 ft

From the geometry shown in Fig. a,

Then,

1 ft

1 u = cos-1a b = 60° 2

2 ft

a = 360° - 2u = 360° - 2(60°) = 240° a

Therefore,

A =

R2 (a - sin a) = 2

P = aR =

(a)

p rad 4 b = p rad 180° 3

( 2 ft ) 2 4 2

4 8 p (2 ft) = p ft 3 3

4 a p - sin pb = 10.110 ft 2 3 3

1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. For FPS units, 5

Q =

2

nP 3

5

1

1.486A3S02

=

1

1.486 ( 10.110 ft 2 ) 3 ( 0.00252 ) 2

3 8 ( 0.014 s>m ) a p ft b 3 1 3

= 60.8 ft 3 >s

1308

Ans.

12–49. The drainage pipe is made of unfinished concrete and is sloped downward at 0.0025. Determine the volumetric flow from the pipe if the center depth is y = 1 ft. 2 ft

y

SOLUTION Water is considered to be incompressible. The flow is steady. From the geometry shown in Fig. a,

Then,

2 ft

1 u = cos-1a b = 60° 2

1 ft 1 ft

a = 2u = 2(60°) = 120° a

Therefore,

A =

(a)

p rad 2 b = p rad 180° 3

(2 ft)2 2 R2 2 (a - sin a) = a p - sin pb = 2.4567 ft 2 2 2 3 3

2 4 P = aR = a pb(2 ft) = p ft 3 3 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. For FPS units, Q =

1.486A5>2S01>2 nP

2>3

=

1.486 ( 2.4567 ft 2 ) 5>3 ( 0.00251>2 ) 2>3 4 ( 0.014 s>m ) a pb 3 1 3

= 9.14 ft 3 >s Ans.

1309

Ans: 9.14 ft 3 >s

12–50. The culvert carries water and is at a downward slope S0. Determine the depth y that will produce the maximum volumetric flow. R

SOLUTION Assume that uniform steady flow occurs in the channel and water is incompressible. The geometry of the cross-section is shown in Fig. a. From this geometry, the area and wetted perimeter are A =

u

y

1 2 1 R (2u) - (2R sin u)(R cos u) 2 2

A = R2u A = R2u -

1 2 3 R (2 sin u cos u) 4 2 1 2 R sin 2u 2

(1) (2)

P = R(2u) = 2Ru Manning’s equation, in the form of writing Q =

kA5>3S01>2 nP 2>3

= a

R

A5 1>3 kS01>2 b a b n P2

A5 Since k, S0 and n are constant, Q will be maximum if 2 is maximum. This requires P A5 d a 2b P = 0. du Here, A5 dA dP b P2 1 5A4 2 - A5(2P) du du P2 = = 0 2 4 du P dA dP - 2PA5 = 0 5P2A4 du du da

PA4 a5P

Since PA4 ≠ 0; 5P

dA dP - 2A b = 0 du du

dA dP - 2A = 0 du du

(3)

Here dA 1 = R2 - R2 (cos 2u)(2) = R2 (1 - cos 2u) du 2 dP = 2R du Substituting these results, Eqs. (1) and (2) into Eq. (3), 5(2Ru) 3 R2(1 - cos 2u) 4 - 2aR2u -

1 2 R sin 2ub ( 2R ) = 0 2

2R3(3u - 5u cos 2u + sin 2u) = 0 Since 2R3 ≠ 0, then 3u - 5u cos 2u + sin 2u = 0 Solving numerically, u = 2.6391 rad = 151.21°

1310

y

(a)

12–50. Continued

Finally y = R - R cos u = R - R cos 151.21° = 1.8764 R Ans.

= 1.88 R

Ans: 1.88R 1311

12–51. The culvert carries water and is at a downward slope S0. Determine the depth y that will produce maximum velocity for the flow. R u

y

SOLUTION Assume that uniform steady flow occurs in the channel and water is incompressible. The geometry of the flow cross-section is shown in Fig. a. From this geometry, the flow area and wetted perimeter are A =

1 2 1 R (2u) - (2R sin u) (R cos u) 2 2

= R2u = R2u -

1 2 3 R (2 sin u cos u) 4 2

R

1 2 R sin 2u 2

y

P = R(2u) = 2Ru Referring to the Manning equation, 2

(a)

1

kRh 3S0 2 V = n Since k, S0 and n are constants, then V will be a maximum if Rn is maximum. This dRn requires = 0. Here, du A = Rn = P

R2u -

1 2 R sin 2u 2 R 1 sin 2u = - R 2Ru 2 4 u

Then u(cos 2u)(2) - sin 2u(1) dRn 1 = 0 - RJ R = 0 du 4 u2 Since

R (2u cos 2u - sin 2u) = 0 4u 2

R ≠ 0, 4u 2 2u cos 2u - sin 2u = 0 2u - tan 2u = 0

Solving by trial and error, u = 2.2467 rad = 128.73° Finally, y = R - R cos u = R - R cos 128.73° = 1.6256 R Ans.

= 1.63 R

Ans: 1.63R 1312

*12–52. The channel is made of unfinished concrete and has a downward slope of 0.003. Determine the volumetric flow if the depth is y = 2 m. Is the flow subcritical or supercritical?

y

5

3

5

4

3m

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Also, with 3 tan a = , 4 3 sin a = , y = 2 m, and b = 3 m, then 5 A = ya P =

y 2m + bb = (2 m) c 3 + 3 m d = 11.333 m2 tan a 4

2(2 m) 2y + b = + 3 m = 9.6667 m 3 sin a 5

For SI units, 5

Q =

nP

5

1

A3S02 2 3

=

1

( 11.333 m2 ) 3(0.003) 2 1

2

( 0.014 s>m3 ) (9.6667 m)3

The mean velocity of the flow is V =

= 49.30 m3 >s = 49.3 m3 >s Ans.

49.30 m3 >s Q = = 4.350 m>s A 11.333 m2

Thus, the Froude number is Fr =

V 2gy

=

4.350 m>s 29.81 m>s2(2 m)

= 0.9821

Since Fr 6 1, the flow is subcritical.

1313

3 4

12–53. The channel is made of unfinished concrete and has a downward slope of 0.003. Determine the volumetric flow if the depth is y = 3 m. Is the flow subcritical or supercritical?

y

5

3

5

4

3 4

3m

SOLUTION Water is considered to be incompressible. The flow is steady. From the table, n 3 tan a = , sin a = 4 A = ya P =

1

= 0.014 s>m 3 for an unfinished concrete surface. Also, with 3 , y = 3 m, and b = 3 m, then 5

y 3m + bb = (3 m) £ 3 + 3 m § = 21 m2 tan a 4

2(3 m) 2y + 3 m = 13 m + b = 3 sin a 5

For SI units, Q =

A5>3S01>2 nP 2>3

=

( 21 m2 ) 5>3(0.003)1>2 1

( 0.014 s>m3 ) (13 m)2>3

The mean velocity of the flow is V =

= 113.11 m3 >s = 113 m3 >s

Ans.

113.11 m3 >s Q = = 5.386 m>s A 21 m2

Thus, the Froude number is Fr =

V 2gy

=

5.386 m>s 29.81 m>s2(3 m)

= 0.9929

Since Fr 6 1, the flow is subcritical.

Ans: 113 m3 >s subcritical 1314

12–54. The channel is made of finished concrete and has a trapezoidal cross section. If the average velocity of the flow is to be 6 m>s when the water depth is 2 m, determine the required slope.

2 m 40!

40!

3m

SOLUTION Water is considered to be incompressible. The flow is steady. 1 From the table, n = 0.012 s>m 3 for a finished concrete surface. Also, with a = 40°, y = 2 m, and b = 3 m, then A = ya

y tan a

+ bb = (2 m) c

2m + 3 m d = 10.767 m2 tan 40°

P =

2(2 m) 2y + b = + 3 m = 9.2229 m sin a sin 40°

Rh =

A 10.767 m2 = = 1.1674 m P 9.2229

Then,

V =

Rh2>3S01>2 n

6 m>s =

(1.1674 m)2>3S01>2 1

0.012 s>m 3 Ans.

S0 = 0.00422

Ans: 0.00422 1315

12–55. The channel is made of unfinished concrete. Its bed has a drop in elevation of 2 ft for 1000 ft of horizontal length. Determine the volumetric flow when the depth y = 8 ft. Is the flow subcritical or supercritical?

45!

45!

y

10 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Also with a = 45°, y = 8 ft, and b = 10 ft, A = ya P =

y tan a

8 ft + 10 ft d = 144 ft 2 tan 45°

2(8 ft) 2y + b = + 10 ft = 32.627 ft sin a sin 45°

For FPS units, with S0 = Q =

+ bb = (8 ft) c

2 ft = 0.002, 1000 ft

1.486A5>3S01>2 nP 2>3

=

1.486 ( 144 ft 2 ) 5>3 1 0.002 1>2 2 1

( 0.014 s>m3 ) (32.627 ft)2>3

= 1839.17 ft 3 >s = 1839 ft 3 >s

Ans.

The mean velocity of the flow is V =

1839.17 ft 3 >s Q = = 12.772 ft>s A 144 ft 2

Thus, the Froude number is Fr =

V 2gy

=

12.772 ft>s 232.2 ft>s2(8 ft)

= 0.7958

Since Fr 6 1, the flow is subcritical.

Ans: 1839 ft 3 >s subcritical 1316

*12–56. The channel is made of unfinished concrete. Its bed has a drop in elevation of 2 ft for 1000 ft of horizontal length. Determine the volumetric flow when the depth y = 4 ft. Is the flow subcritical or supercritical?

45!

45! 10 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Also with a = 45°, y = 4 ft, and b = 10 ft, then A = ya P =

y tan a

4 ft + 10 ft d = 56 ft 2 tan 45°

2(4 ft) 2y + b = + 10 ft = 21.3137 ft sin a sin 45°

For FPS units with S0 =

Q =

+ bb = (4 ft) c

2 ft = 0.002, 1000 ft

1.486A5>3S01>2 nP 2>3

=

1.486 ( 56 ft 2 ) 5>3 1 0.002 1>2 2 1

( 0.014 s>m3 ) (21.3137 ft)2>3

= 506.15 ft 3 >s = 506 ft 3 >s

Ans.

The mean velocity of the flow is V =

506.15 ft 3 >s Q = = 9.038 ft>s A 56 ft 2

Thus, the Froude number is Fr =

V 2gy

=

9.038 ft>s 232.2 ft>s2(4 ft)

= 0.7964

Since Fr 6 1, the flow is subcritical.

1317

y

12–57. The bed of the channel drops 2 ft per 1000 ft of horizontal length. If it is made of unfinished concrete, and the depth of the water is y = 6 ft, determine the volumetric flow.

y

60!

60!

10 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Also with a = 60°, y = 6 ft, and b = 10 ft, then A = ya P =

y tan a

6 ft + 10 ft d = 80.7846 ft 2 tan 60°

2(6 ft) 2y + b = + 10 ft = 23.8564 ft sin a sin 60°

For FPS units with S0 =

Q =

+ bb = (6 ft) c

2 ft = 0.002, 1000 ft

1.486A5>3S01>2 nP 2>3

=

1.486(80.7846)5>3 1 0.002 1>2 2 1

( 0.014 s>m3 ) (23.8564 ft)2>3

= 864.74 ft 3 >s = 865 ft 3 >s

Ans.

1318

Ans: 865 ft 3 >s

12–58. The bed of the channel drops 2 ft per 1000 ft of horizontal length. If it is made of unfinished concrete, determine the depth of the water if the flow is Q = 800 ft3>s.

y

60!

60!

10 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Also with a = 60°, and b = 10 ft, then A = ya P =

y tan a

y tan 60°

+ 10 ft b = ( 0.5774y2 + 10y ) ft 2

2y 2y + b = + 10 ft = (2.3094y + 10) ft sin a sin 60°

For FPS units with S0 =

Q =

+ bb = y a

2 ft = 0.002, 1000 ft

1.486A5>3S01>2 nP 2>3

800 ft 3 >s =

1.486 ( 0.5774y2 + 10y ) 5>3(0.002)1>2 1

( 0.014 s>m3 ) (2.3094y + 10)2>3 ( 0.5774y2 + 10y ) 5>3 (2.3094y + 10)2>3

= 168.533

Solving by trial and error, Ans.

y = 5.732 m = 5.73 m

Ans: 5.73 m 1319

12–59. Determine the volumetric flow of water through the channel if the depth of flow is y = 1.25 m and the downward slope of the channel is 0.005. The sides of the channel are finished concrete.

2

2 1

y

1 2m

SOLUTION Assume that uniform steady flow occurs in the channel, and water is incompressible. The geometry of the cross-section is shown in Fig. a. For this geometry a 1 = ; 1.25 m 2

a =

2

1.25 m = 0.625 m 2

1.25 m

1

Thus, the area and the wetted perimeter are A =

a

1 3 2 m + 2 m + 2(0.625 m) 4 (1.25 m) = 3.28125 m2 2

2m

a

(a)

P = 2 m + 22(1.25 m)2 + (0.625 m)2 = 4.7951 m 1

From the table, n = 0.012 s>m 3 for finished concrete. Applying Manning’s equation, 5

Q =

1

A3S02 2

nP 3

( 3.28125 m2 ) 3 1 0.0052 2 5

=

1

1

2

( 0.012 s>m3 ) (4.7951 m)3

= 15.0 m3 >s

Ans.

1320

Ans: 15.0 m3 >s

*12–60. Determine the normal depth of water in the channel if the flow is Q = 15 m3>s. The sides of the channel are finished concrete, and the downward slope is 0.005.

2

2 1

y

1 2m

SOLUTION Assume that the uniform steady flow occurs in the channel and the water is incompressible. The geometry of the cross-section is shown in the Fig. a. For this geometry a 1 = ; y 2

1

1 y 2

a =

a

Thus, the area and the wetted perimeter are A =

1 2 y2 + a yb = ( 2 + 15y ) m 2 B 1

From the table, n = 0.012 s>m 3 for finished concrete. Applying Manning’s equation, 5

Q =

5 3

A S0 2

nP 3

1 2

;

15 m3 >s =

1

2

( 0.012 s>m3 )( 2 + 15y ) 3 5

3 1 a y2 + 2yb 2 2

Solving by trial and error,

3 1 1 a y2 + 2yb 1 0.0052 2 2

( 2 + 15y ) 3

2m (a)

1 1 1 c 2 m + 2 m + 2 a yb d y = a y2 + 2yb m2 2 2 2

P = 2m + 2

2

= 2.5456

Ans.

y = 1.249 m = 1.25 m

1321

a

y

12–61. The channel is made of unfinished concrete. If it is required to transport water at 400 ft3>s, determine the critical depth y = yc and the critical slope.

2

2 1

y

1

10 ft

SOLUTION Water is considered to be incompressible. The flow is steady. Referring to the geometry shown in Fig. a, yc a = ; 1 2 b Thus,

25

=

btop

yc a = 2

yc 2

btop = 2 a

b =

yc 2

a

25 y 2 c

2

a

√5 1

√5 b

b

2

yc

1

10 ft

b + 10 ft = (yc + 10) ft

(a)

1 yc 1 Ac = 2c a byc d + 10yc = a yc2 + 10yc b ft 2 2 2 2

Pc = 2 °

25 yc ¢ + 10 = 25yc + 10 2

The critical depth can be determined by satisfying, gAc3 Q2btop

= 1 1 2

( 32.2 ft>s2 ) a yc2 + 10yc b ( 400 ft3 >s ) 2(yc + 10)

3

= 1

3 1 a yc2 + 10yc b 2 = 4968.94 (yc + 10) Solving by trial and error,

Ans.

yc = 3.4605 ft = 3.46 ft Using this result, Ac =

1 (3.4605 ft)2 + 10(3.4605 ft) = 40.59 ft 2 2

Pc = 25(3.4605 ft) + 10 = 17.74 ft 1

From the table, n = 0.014 s>ft 3 for an unfinished concrete surface. For FPS units, 5

Q =

1

1.486A3S02 2

nP 3 5

3

400 ft >s =

1

1.486 ( 40.59 ft 2 ) 3Sc 2 1

2

( 0.014 s>ft 3 ) (17.74 ft)3 Ans.

Sc = 0.002858 = 0.00286 1322

Ans: yc = 3.46 ft Sc = 0.00286

12–62. The channel is made of unfinished concrete. If it is required to transport water at 400 ft3>s, determine its downward slope when the depth of the water is 4 ft. Also, what is the critical slope for this depth and what is the corresponding critical flow?

2

2 1

y

1

10 ft

SOLUTION Water is considered to be incompressible. The flow is steady. Referring to the geometry shown in Fig. a,

Thus,

a 4 ft = ; 1 2 b 4 ft = ; 2 25

btop

a = 2 ft

a

b = 225 ft 2

√5 1

btop = 2(2 ft) + 10 ft = 14 ft

a √5 b

b

2

y = 4 ft

1

10 ft (a)

1 A = 2c (2 ft)(4 ft) d + (10 ft)(4 ft) = 48 ft 2 2

P = 2 1 225 ft 2 + 10 = 18.94 ft 1

From the table, n = 0.014 s>ft 3 for an unfinished concrete surface. For FPS units, 5

Q =

1

1.486A3S02 2

nP 3 5

400 ft 3 >s =

1

1.486 ( 48 ft 2 ) 3S02 1

2

( 0.014 s>ft 3 ) (18.94 ft)3 Ans.

S0 = 0.001784 = 0.00178

For yc = 4 ft, the corresponding flow rate can be determined by satisfying, gAc3 Qc2btop

= 1

( 32.2 ft>s2 )( 48 ft2 ) 3 Qc2(14 ft)

Qc = 504.34 ft 3 >s = 504 ft 3 >s

For FPS units, 5

Q =

= 1 Ans.

1

1.486A3S02 2

nPc 3

5

504.34 ft 3 >s =

1

1.486 ( 48 ft 2 ) 3Sc 2 1

2

( 0.014 s>ft 3 ) (18.94 ft)3 Ans.

Sc = 0.002841 = 0.00284

Ans: S0 = 0.00178 Qc = 504 ft 3 >s Sc = 0.00284 1323

12–63. The unfinished concrete channel is intended to have a downward slope of 0.002 and sloping sides at 60°. If the flow is estimated to be 100 m3>s, determine the base dimension b of the channel bottom.

4m 60!

60!

b

SOLUTION 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. Also, with a = 60° and y = 4 m, then A = ya P =

y 4m + bb = (4 m)a + bb = (9.2376 + 4b) m2 tan a tan 60°

2(4 m) 2y + b = + b = (9.2376 + b) m sin a sin 60°

For SI units, 5

Q =

2

nP 3

(9.2376 + 4b)3 1 0.002 2 2 5

1

A3S02

=

1

1

2

( 0.014 s>m3 ) (9.2376 + b)3 5

(9.2376 + 4b)3 2

(9.2376 + b)3

= 100 m3 >s

= 31.305

Solving by trial and error, Ans.

b = 3.08 m

Ans: 3.08 m 1324

*12–64. A rectangular channel has a width of 2.5 m and is made of unfinished concrete. If it is inclined downward at a  slope of 0.0014, what depth of water will produce a discharge of 12 m3>s?

SOLUTION Water is considered to be incompressible. The flow is steady. The cross-sectional area wetted perimeter of the rectangular channel are A = (2.5 m)y = (2.5y) m2 P = (2y + 2.5) m 1

From the table, n = 0.014 s>m 3 for an unfinished concrete surface. For SI units, 5

1

A3S02

Q =

2

nP 3 (2.5y)3 1 0.00142 2 5

3

12 m >s =

1

1

2

( 0.014 s>m3 ) (2y + 2.5)3

5

y3

2

(2y + 2.5)3

= 0.9750

Solving by trial and error, Ans.

y = 2.11 m

1325

12–65. Determine the angle u of the channel so that it has the best hydraulic triangular cross section that uses the minimum amount of material for a given discharge. u l

l

SOLUTION The cross-sectional area wetted perimeter of the triangular channel are 1 u u 1 A = 2c al sin bal cos b d = l 2 sin u 2 2 2 2

P = 2l

Expressing A in terms of P by eliminating l from the above two equations, A =

P2 sin u 8

(1)

Taking Q is fixed, we rewrite 5

Q =

1

A3S02 2

nP 3

as 2

A = kP5 Where k is constant. Substituing from Eq. (1), 2 P2 sin u = kP 5 8

8

P5 = Setting

8k sin u

dP = 0 obtain the minimum value of P, du sin u(0) - cos u 8 3 dP d P5 = 8kc 5 du sin2 u

0 = cos u

Ans.

u = 90°

Ans: 90° 1326

12–66. Determine the length of the sides a of the channel in terms of its base b, so that the flow at full depth it provides the best hydraulic cross section that uses the minimum amount of material for a given discharge.

120! 120! a

a

b

SOLUTION The cross-sectional area and wetted perimeter of the trapezoidal channel written in terms of flow depth y with a = 60° are A = ya

y2 y y + bb = y a + bb = + by tan a tan 60° 23

2y 2y 4y + b = + b = + b sin a sin 60° 23

P =

Expressing A in terms of P by eliminating b from the above two equation, A = Py - 23y2

(1) 5

Taking Q as fixed, we rewrite Q =

1 2

A3S 0 2

nP 3

as 2

A = kP5 Where K is constant. Substituting from Eq. (1), 2

Py - 23y2 = kP5

dP Taking the derivative of both sides with respect to y and setting = 0 to obtain dy the minimum P, P + y

3 dP dP 2 - 223y = kP - 5 dy 5 dy

P - 223y = 0 However, P = 4y 23

4y 23

+ b. Then,

+ b - 223y = 0

223y -

4y 23

y =

23 b 2

a =

y sin 60°

= b

But,

Then, a =

23 b 2 23 2

= b

Ans.

a = b 1327

Ans: a = b

12–67. Determine the angle u and the length l of its sides so that the channel has the best hydraulic trapezoidal cross section of base b.

u

u

l

l b

SOLUTION The cross-sectional area and wetted perimeter of the trapezoidal channel written in terms of flow depth y with a = u are A = ya

y2 y y + bb = ya + bb = + by tan a tan u tan u

2y 2y + b = + b sin a sin u

P =

Expressing A in terms of P by eliminating b from the above two equations, 2y2 y2 + Py tan u sin u

A =

Taking Q as fixed, we rewrite 1 2

5

A3S 0

Q =

2

nP 3

as 2

A = kP5 and so y2 2y2 2 + Py = kP5 tan u sin u

(1)

We now minimize P with respect to both u and Y , beginning with u. Taking the partial derivative of both sides with respect to u and setting obtain the minimum P, -

y2 2

sin u

+

2

-

y

sin2 u

cos u =

+

2y2cos u 2

sin u 2y2 cos u sin2 u

+ y

0P = 0 to 0u

3 0P 0P 2 = kP- 5 0u 5 0u

= 0

1 2 Ans.

u = 60° Substituting this result into Eq. (1), 2

Py - 23y2 = kP5

(2)

1328

12–67. Continued

Taking the partial derivative of both sides with respect to y and setting obtain the minimum P, P + y

0P = 0 to 0y

0P 2 -3 0P - 223y = kP 5 0y 5 0y

P - 223y = 0 However, P =

2y 423 + b = y + b. Then, sin 60° 3

423 y + b - 223y = 0 3 y =

23 b 2

However,

y = l = sin 60°

¢

13 ≤b 2

¢

13 ≤ 2

Ans.

= b

Ans: u = 60° l = b 1329

*12–68. Show that the width b = 2h(csc u - cot u) in order to minimize the water perimeter for a given crosssectional area and angle u. At what angle u will the wetted perimeter be the smallest for a given cross-sectional area and depth h?

h u

u b

SOLUTION Assume that the uniform steady flow occurs in the channel and water is incompressible. The geometry of the cross-section is shown in Fig. a. For this geometry h = tan u a

l

h

l

a = h cot u

Thus, the area and the wetted perimeter are a

1 A = (b + b + 2h cot u)h = bh + h2 cot u 2

(1)

P = b + 2h csc u

(2)

Using Eq. (1) to express b in term of h and A b =

A - h2 cot u h

Substituting this result into Eq. (2), P =

A - h2 cot u + 2h csc u h

P =

A + h(2 csc u - cot u) h

To have minimum P for a given area (A = constant) and angle u (u = constant), dP we must set = 0. dh dP A = - 2 + 2 csc u - cot u = 0 dh h Substituing Eq. (1) into this equation -a

bh + h2 cot u b + 2 csc u - cot u = 0 h2

b = 2h(csc u - cot u)

(Q .E .D)

(3)

To have minimum P, for a given area (A = constant) and depth (h = constant), we dP must set = 0 du dP = 0 + h 3 2( -csc u cot u) - ( -csc 2u ) 4 = 0 du h ( csc 2u - 2 csc u cot u ) = 0

1330

b (a)

a

*12–68. Continued

Since h ≠ 0. Then csc 2u - 2 csc u cot u = 0 csc u(csc u - 2 cot u) = 0 Since csc u ≠ 0. Then csc u - 2 cot u = 0 1 2 cos u = 0 sin u sin u 1 - 2 cos u = 0 sin u 1 - 2 cos u = 0 cos u =

1 2 Ans.

u = 60° Substituting this result into Eq. (3) b = 2h(csc 60° - cot 60°) = 2h a

The length of the sides is

l = h csc u = h csc 60° = h a

2 1 2 b = h 13 13 13

2 2 b = h 13 13

This shows that the most efficient of all trapozoidal sections would be one with three equal sides and 120° interior angles. Actually, it is a half hexagon which can be inscribed in a semicricle, which is the most efficient among all the shapes.

1331

12–69. Show that when the depth of flow y = R, the semicircular channel provides the best hydraulic cross section.

R

y

SOLUTION Referring to Fig. a, a = 2u. Then, A =

y

P = aR = (2u)R = 2Ru

(a)

Expressing A in terms of P by eliminating R from the above two equations, A = Then,

sin 2u P2 2 a b 8 u u2 2

A = kP5 2 P2 2 sin 2u a b = kP5 8 u u2 -8 2 sin 2u = 8kP 5 2 u u

dP Taking the derivative of both sides with respect to u and setting = 0 to obtain du the minimum P, -

R

R

R2 R2 (a - sin a) = (2u - sin 2u) 2 2

2 2 cos 2u 2 sin 2u 64 -13 dP + = - kP 5 2 2 3 5 du u u u

1 (2 sin 2u - 2u cos 2u - 2u) = 0 u3 2 sin 2u - 2u cos 2u - 2u = 0 Solving by trial and error u = 90° when u = 90°, (Q.E.D)

y = R

1332

R

12–70. A rectangular channel is made of unfinished concrete, and it has a width of 1.25 m and an upward slope of  0.01. Determine the surface profile for the flow if it is 0.8 m3 >s and the depth of the water at a specific location is 0.5 m. Sketch this profile.

SOLUTION Water is considered to be incompressible. The flow is steady. The critical depth is yc = °

Q2 2

gb

1 3

¢ = £

( 0.8 m3 >s ) 2 2

9.81m>s (1.25 m)

Type A2 profile

1 3

2

yc = 0.347 m

§ = 0.3469 m

y = 0.5 m

Since y = 0.5 m > yc the flow is subcritical. Also, the channel has an adverse slope. Referring to the table with these two conditions, the surface profile is of type A2. The plot of this profile is shown in Fig. a.

100 Adverse slope (a)

Ans: A2 1333

12–71. A rectangular channel is made of finished concrete, and it has a width of 1.25 m and an upward slope of 0.01. Determine the surface profile for the flow if it is 0.8 m3 >s and the depth of the water at a specific location is 0.2 m. Sketch this profile.

SOLUTION Water is considered to be incompressible. The flow is steady. The critical depth is yc = °

Q2 2

gb

1 3

¢ = £

type A3 yc = 0.347 m

1 3

( 0.8 m3 >s ) 2 § = 0.3469 m 9.81 m>s2 ( 1.25 m ) 2

100

Since y = 0.2 m < yc the flow is supercritical. Also, the channel has an adverse slope. Referring to the table with these two conditions, the surface profile is of type A3. The plot of this profile is shown in Fig. a.

1 y = 0.2 m

Adverse slope (a)

Ans: A3 1334

*12–72. A rectangular channel is made of finished concrete, and it has a width of 1.25 m and a downward slope of 0.01. Determine the surface profile for the flow if it is 0.8 m3 >s and the depth of the water at a specific location is 0.6 m. Sketch this profile.

SOLUTION Type S1

Water is considered to be incompressible. The flow is steady. The critical depth is yc = °

Q2 gb2

1>3

¢

( 0.8 m3 >s ) 2 = £ § 9.81 m>s2 ( 1.25 m ) 2

1>3

= 0.3469 m

Since y = 0.6 m 7 yc the flow is subcritical. For normal flow, the cross-sectional area and wetted perimeter of the rectangular channel are A = (1.25 m)yn = (1.25yn) m P = ( 2yn + 1.25 ) m From the table, n = 0.012 for a finished concrete surface. For SI units to determine the normal depth, A5>3S01>2 nP 2>3

0.8 m3 >s =

( 1.25yn ) 5>3 ( 0.01 ) 1>2 0.012 ( 2yn + 1.25 ) 2>3

yn5>3

( 2yn + 1.25 ) 2>3

= 0.06618

Solving by trial and error, yn = 0.2447 m Using the result of yc, Ac = (1.25 m)(0.3469 m) = 0.4337 m2 Pc = 2 ( 0.3469 m ) + 1.25 m = 1.9438 m Thus,

( Rh ) c =

Ac 0.4337 m2 = = 0.2231 m Pc 1.9438 m

The critical channel slope can be determined using Sc =

n2gAc btop ( Rh ) c4>3

=

0.0122 ( 9.81 m>s2 )( 0.4337 m2 ) 1.25 m ( 0.2231 m ) 4>3

= 0.00362

Here, y 7 yc 7 yn and S0 7 Sc (steep slope). Referring to the table with these two conditions, the surface profile is of type S1. The plot of this profile is shown in Fig. a.

1335

100

yc = 0.347 m Steep slope (a)

2

Q =

y = 0.6 m 1

yn = 0.245 m

12–73. A rectangular channel is made of finished concrete, and it has a width of 1.25 m and a downward slope of 0.01. Determine the surface profile for the flow if it is 0.8 m3 >s and the depth of the water at a specific location is 0.3 m. Sketch this profile.

SOLUTION Water is considered to be incompressible. The flow is steady. The critical depth is 1 3

1 3

( 0.8 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.3469 m gb 9.81 m>s2 ( 1.25 m ) 2 Q2

type S2

yc = 0.347 m y = 0.3 m

1

100

yn = 0.245 m

Steep slope

Since y = 0.3 m 6 yc the flow is supercritical. For normal flow, the cross-sectional area and wetted perimeter of the rectangular channel are

(a)

A = (1.25 m)yn = (1.25yn) m2 P = ( 2yn + 1.25 ) m From the table, n = 0.012 for a finished concrete surface. For SI units to determine the normal depth, 5

Q =

1

A3S02 2

nP 3 5

0.8 m >s =

1

( 1.25yn ) 3 ( 0.01 ) 2

3

2

0.012 ( 2yn + 1.25 ) 3

5

yn3

2

( 2yn + 1.25 ) 3

= 0.06618

Solving by trial and error, yn = 0.2447 m Using the result of yc, Ac = (1.25 m)(0.3469 m) = 0.4337 m2 Pc = 2 ( 0.3469 m ) + 1.25 m = 1.9438 m Thus,

( Rh ) c =

Ac 0.4337 m2 = = 0.2231 m Pc 1.9438 m

The critical channel slope can be determined using Sc =

n2gAc btop ( Rh ) c

4 3

=

0.0122 ( 9.81 m>s2 )( 0.4337 m2 ) 4

1.25 m ( 0.2231 m ) 3

= 0.00362

Here, yc 7 y 7 yn and S0 7 Sc (steep slope). Referring to the table with these two conditions, the surface profile is of type S2. The plot of this profile is shown in Fig. a.

1336

Ans: S2

12–74. Water flows at 4 m3 >s along a horizontal channel made of unfinished concrete. If the channel has a width of 2 m, and the water depth at the control section A is 0.9 m, approximate the depth at the section where x = 2 m from the control section. Use increments of ∆y = 0.004 m and plot the profile for 0.884 m … y … 0.9 m.

SOLUTION

yB = 0.896 m A

Water is considered to be incompressible. The flow is steady. The critical depth is 1 3

1 3

x (m)

( 4 m3 >s ) 2 yc = ° 2 ¢ = £ § = 0.7415 m gb ( 9.81 m>s2 )( 2 m ) 2 Q2

1

yA = 0.9 m

yC = 0.892 m yD = 0.888 m B y = 0.884 m C D E E 2 3 4

0

0.663

1.306

1.930

(a)

Since yA = 0.9 m 7 yc the flow is subcritical. Also, the channel bed is horizontal. Referring to the table with these two conditions, the surface profile is of type H 2 as shown in Fig. a. Thus, the flow depth decreases as x increases. At the control section A, yA = 0.9 m. Then, VA =

4 m3 >s Q = = 2.2222 m>s AA (2 m)(0.9 m)

( Rh ) A =

(2 m)(0.9 m) AA = = 0.4737 m PA 2(0.9 m) + 2 m

Here, ∆y = 0.004 m is chosen. Then yB = 0.9 m - 0.004 m = 0.896 m. Thus, at section B, VB =

4 m3 >s Q = = 2.2321 m>s AB (2 m)(0.896 m)

( Rh ) B =

(2 m)(0.896 m) AB = = 0.4726 m PB 2(0.896 m) + 2 m

Then, the mean values for segment (1) between sections A and B are Vm =

( Rh ) m = Sm = ∆x =

2.2222 m>s + 2.2321 m>s VA + VB = = 2.2272 m>s 2 2

( Rh ) A + ( Rh ) B 2 n2Vm2 4 3

( Rh )m

=

=

0.4737 m + 0.4726 m = 0.4731 m 2

0.0142 ( 2.2272 m>s ) 2 4

( 0.4731 m ) 3 ( yB - yA ) + ( VB2 - VA2 ) >2g

= 0.6627 m

S0 - Sm

=

= 0.002637 -0.004 m +

3 ( 2.2321 m>s ) 2

- ( 2.2222 m>s ) 2 4 > 3 2 ( 9.81 m>s2 ) 4

0 - 0.002637

1337

2.536

12–74. Continued

Using the same procedure, the computation for the sections that follow are tabulated below: Section A B C D E

Segment 1 2 3 4

y(m)

V ( m>s )

0.9

2.2222

0.896

2.2321

0.892

2.2422

0.888

2.2523

0.884

2.2624

Vm ( m>s ) 2.2272 2.2371 2.2472 2.2573

Rh(m)

(Rh)m(m)

Sm

∆x(m)

0.4731

0.002637

0.6627

0.4720

0.002669

0.6434

0.4709

0.002702

0.6242

0.4698

0.002735

0.6051

0.4737 0.4726 0.4715 0.4703 0.4692

x(m) 0 0.6627 1.3061 1.9303 2.5355

Since the section x = 2 m is between sections D and E, its flow depth can be determined by interpolation of the flow depth between these two sections. y = 0.888 m + a

0.884 m - 0.888 m b(2 m - 1.9303 m) 2.5355 m - 1.9303 m

= 0.88745 m = 0.888 m

Ans.

Ans: 0.888 m 1338

12–75. Water flows at 4 m3 >s along a horizontal channel made of unfinished concrete. If the channel has a width of 2 m, and the water depth at the control section A is 0.9 m, determine the approximate distance x from A to where the depth is 0.8 m. Use increment of ∆y = 0.025 m and plot the profile for 0.8 m … y … 0.9 m

SOLUTION The critical depth is yc = °

1 3

yB = 0.875 m yC = 0.85 m

1 3

( 4 m >s ) ¢ = £ § = 0.7415 m 2 gb ( 9.81 m>s2 )( 2 m ) 2 Q

2

3

2

A

B

C

yD = 0.825 m D yE = 0.8 m E

Since yA = 0.9 m 7 yc the flow is subcritical. Also, the channel bed is horizontal. yA = 0.9 m 1 2 3 4 Referring to the table with these two conditions, the surface profile is of type H 2 x (m) as shown in Fig. a. Thus, the flow depth decreases as x increases. At the control 0 3.823 6.917 9.313 11.041 section A, yA = 0.9 m. Then, (a)

4 m3 >s Q VA = = = 2.2222 m>s AA (2 m)(0.9 m)

( Rh ) A =

(2 m)(0.9 m) AA = = 0.4737 m PA 2(0.9 m) + 2 m

Here, ∆y = 0.025 m is chosen. Then yB = 0.9 m - 0.025 m = 0.875 m. Thus, VB =

4 m3 >s Q = = 2.2857 m>s AB (2 m)(0.875 m)

( Rh ) B =

(2 m)(0.875 m) AB = = 0.4667 m PB 2(0.875 m) + 2 m

Then, the mean values for segment (1) between sections A and B are Vm =

2.2222 m>s + 2.2857 m>s VA + VB = = 2.2540 m>s 2 2

( Rh ) m = Sm =

∆x =

( Rh ) A + ( Rh ) B 2

n2Vm2 4

( Rh ) m3

=

=

0.4737 m + 0.4667 m = 0.4702 m 2

0.0142 ( 2.2540 m>s ) 2 4

( 0.4702 m ) 3

( yB - yA ) + ( VB2 + VA2 ) >2g

= 3.8229 m

S0 - Sm

= 0.002724

=

-0.025 m +

3 ( 2.2857 m>s ) 2

- ( 2.2222 m>s ) 2 4 > 3 2 ( 9.81 m>s2 ) 4

0 - 0.002724

1339

12–75. Continued

Using the same procedure, the computation for the sections that follow are tabulated below: Section A B C D E

Segment 1 2 3 4

y(m)

V ( m>s )

0.9

2.2222

0.875

2.2857

0.850

2.3529

0.825

2.4242

0.800

2.5

Vm ( m>s ) 2.2540 2.3193 2.3886 2.4621

Rh(m)

(Rh)m (m)

Sm

∆x(m)

0.4702

0.002724

3.8229

0.4631

0.002943

3.0941

0.4558

0.003188

2.3960

0.4482

0.003463

1.7284

0.4737 0.4667 0.4595 0.4521 0.4444

x(m) 0 3.8229 6.9170 9.3129 11.0413

Thus, the location of the section having a depth y = 0.8 m is Ans.

x = 11.0 m

Ans: 11.0 m 1340

*12–76. Water flows at 12 m3 >s down a rectangular channel made of unfinished concrete. The channel has a width of 4 m and a downward slope of 0.008, and the water depth is 2 m at the control section A. Determine the distance x from A to where the depth is 2.4 m. Use increment of ∆y = 0.1 m and plot the profile for 2 m … y … 2.4 m.

A 2m

2.4 m B

x

SOLUTION yB = 2.1 m y = 2.2 m y = 2.3 m C D

Water is considered to be incompressible. The flow is steady. The critical depth is yc = °

Q2 gb2

1 3

( 12 m3 >s ) 2

¢ = £

9.81 m>s2(4 m)2

1 3

yA = 2 m

§ = 0.9717 m

Since yA = 2 m 7 yc, the flow is subcritical. For normal flow, the cross-sectional area and the wetted perimeter of the rectangular channel are A = (4 m)yn = ( 4yn ) m

2

P = ( 2yn + 4 ) m From the table, n = 0.014 for an unfinished concrete surface. For SI units to determine the normal depth, 5

1

A3S02

Q =

2

nP 3 5

1

( 4yn ) 3 ( 0.008 ) 2

3

12 m >s =

2

0.014 ( 2yn + 4 ) 3

5

yn3 2

( 2yn + 4 ) 3

= 0.18635

Solving by trial and error, yn = 0.7183 m Using the result of yc, Ac = (4 m)(0.9717 m) = 3.8867 m2 Pc = 2(0.9717 m) + 4 m = 5.9434 m Thus,

( Rh ) c =

Ac 3.8867 m2 = = 0.6540 m Pc 5.9434 m

The critical channel slope can be determined using Sc =

n2gAc 4 3

btop ( Rh ) c

=

E

D

C

B

A

0.0142 ( 9.81 m>s2 )( 3.8867 m2 ) 4

4 m ( 0.6540 m ) 3

= 0.003291

1341

1 0

2

3

4

yE = 2.4 m

11.776 23.662 35.638 47.688 (a)

x (m)

12–76. Continued

Here, yA 7 yc 7 yn and S0 7 Sc (steep slope). Referring to the table with these two conditions, the surface profile is of type S1. Thus, the flow depth increases with x. At the control section A, yA = 2 m. Then, VA =

12 m3 >s Q = = 1.5 m>s AA (4 m)(2 m)

( Rh ) A =

(4 m)(2 m) AA = = 1m PA 2(2 m) + 4 m

Here, ∆y = 0.1 m is chosen. Then yB = 2 m + 0.1 m = 2.1 m. Thus, VB =

12 m3 >s Q = = 1.4286 m>s AB (4 m)(2.1 m)

( Rh ) B =

(4 m)(2.1 m) AB = = 1.0244 m PB 2(2.1 m) + 4 m

Then, the mean values for segment (1) between sections A and B are Vm =

1.5 m>s + 1.4286 m>s VA + VB = = 1.4643 m>s 2 2

( Rh ) m = Sm =

∆x =

=

( Rh ) A + ( Rh ) B 2

n2V 2m

( Rh ) m

4 3

1 m + 1.0244 m = 1.0122 m 2

=

( 0.0142 )( 1.4643 m>s ) 2

=

4

( 1.0122 m ) 3 = 0.0004135

( yB - yA ) + ( VB2 - VA2 ) >2g S0 - Sm

0.1 m +

3 ( 1.4286 m>s ) 2

- ( 1.5 m>s ) 2 4 > 3 2 ( 9.81 m>s2 ) 4

0.008 - 0.0004135

= 11.7760 m

Using the same procedure, the computation for the sections that follow are tabulated below: Section A B C D E

Segment 1 2 3 4

y(m)

V ( m>s )

2

1.5

2.1

1.4286

2.2

1.3636

2.3

1.3043

2.4

1.25

Vm ( m>s ) 1.4643 1.3961 1.3340 1.2772

Rh(m)

(Rh)m (m)

Sm

∆x(m)

1.0122

0.0004135

11.7760

1.0360

0.0003644

11.8863

1.0587

0.0003232

11.9761

1.0803

0.0002884

12.0500

1 1.0244 1.0476 1.0698 1.0909

Thus, the location of the section having a depth of y = 2.4 m is Ans.

x = 47.7 m 1342

x(m) 0 11.7760 23.6623 35.6384 47.6884

12–77. The unfinished concrete channel has a width of 4 ft and slopes upward at 0.0025. At control section A, the water depth is 4 ft and the average velocity is 40 ft>s. Determine the depth of the water y, 40 ft downstream from A. Use increment of ∆y = 0.1 ft and plot the profile for 4 ft … y … 4.4 ft.

40 ft/s A

y

4 ft

B 40 ft

SOLUTION Water is considered to be incompressible. The flow is steady. The flow rate is Q = VAA = ( 40 ft>s ) (4 ft)(4 ft) = 640 ft 3 >s . Thus, the critical depth is yc = a

Q

1 3

2

2

gb

b = £

1 3

2

B

A 1

yA = 4 ft

( 640 ft >s ) § = 9.2639 ft ( 32.2 ft>s2 )( 4 ft ) 2 3

yC = 4.2 ft D C

yE = 4.1 ft

2

yD = 4.3 ft

3

E yE = 4.4 ft

4

x (ft) 0

11.38

Since yA = 4 ft 6 yc, the flow is supercritical. Also, the channel bed has an adverse slope since it slopes upward. Referring to the table with these two conditions, the surface profile is of type A3 as shown in Fig. a. Thus, the flow depth increases with x. At the control section A, yA = 4 ft . Then,

22.51

33.39

44.03

(a)

(4 ft)(4 ft) AA = = 1.3333 ft PA 2(4 ft) + 4 ft

( Rh ) A =

Here, ∆y = 0.1 ft is chosen. Then yB = 4 ft + 0.1 ft = 4.1 ft . Thus, VB =

640 ft 3 >s Q = = 39.0244 ft>s AB (4 ft)(4.1ft) (4 ft)(4.1ft) AB = = 1.3443 ft PB 2(4.1ft) + 4 ft

( Rh ) B =

Then, the mean values for segment (1) between sections A and B are Vm =

40 ft>s + 39.0244 ft>s VA + VB = = 39.5122 ft>s 2 2

( Rh ) m = Sm =

∆x =

( Rh ) A + ( Rh ) B 2

n2Vm2 4 3

k 2 ( Rh ) m

=

=

1.3333 ft + 1.3443 ft = 1.3388 ft 2

0.0142 ( 39.5122 ft>s ) 2 4

( 1.486 ) 2 ( 1.3388 ft ) 3

(yB - yA) + ( VB2 - VA2 ) >2g So - Sm

=

= 5.2276 ft

= 0.09391

0.1 ft +

3 ( 39.0244 ft>s ) 2

- ( 40 ft>s ) 2 4 > 3 2 ( 32.2 ft>s2 ) 4

- 0.0025 - 0.207379

Using the same procedure, the computation for the sections that follow are tabulated below: Section A B C D E

Segment 1 2 3 4

y(ft)

V ( ft>s )

4

40

4.1

39.0244

4.2

38.0952

4.3

37.2093

4.4

36.7865

Vm ( ft>s ) 39.5122 38.5598 37.6523 36.7865

Rh(ft) 1.3333 1.3443 1.3548 1.3651 1.375 1343

(Rh)m(ft)

Sm

1.3388

0.09391

1.3496

0.08849

1.3600

0.08351

1.3700

0.07894

∆x(ft)

x(ft)

11.3797

0

11.1292

11.3797

10.8812

22.5090

10.6354

33.3901 44.0255

12–77. Continued

Since x = 20 ft is between sections D and E, its flow depth can be determined by interpolation of the flow depth between these two sections. y = 4.3 ft + a

4.4 ft - 4.3 ft b(40 ft - 33.3901 ft) 44.0255 ft - 33.3901 ft

= 4.3622 ft = 4.36 ft

Ans.

Ans: 4.36 ft 1344

12–78. Water flows under the partially opened sluice gate, which is in a rectangular channel. If the water has the depth shown, determine if a hydraulic jump forms, and if so, find the depth yC at the downstream end of the jump.

A

6m

B 2m

C yC

SOLUTION Water is considered to be incompressible. The flow is steady. Applying the energy equation between points A and B by neglecting the losses and setting the datum at the channel bed, pA pB VA2 VB2 + zA = + zB + + g g 2g 2g 0 + 0 + 6m = 0 +

VB2 2 ( 9.81 m>s2 )

+ 2m

VB = 8.8589 m>s The Froude number at B is (Fr)B =

VB 2gyB

=

8.8589 m>s 2 ( 9.81 m>s2 ) (2 m)

= 2.00

Since (Fr)B 7 1, the flow at section B is supercritical, which means that a hydraulic jump can occur. yC 1 = c 21 + 8(Fr)B2 - 1 d yB 2

yC 1 = c 21 + 8(2.00)2 - 1 d 2m 2

Ans.

yC = 4.74 m

Ans: 4.74 m 1345

12–79. The sill at A causes a hydraulic jump to form in the channel. If the channel width is 1.5 m, determine the average upstream speed and downstream speed of the water. What amount of energy head is lost in the jump?

2.5 m

4m A

SOLUTION Water is considered to be incompressible. The flow is steady. Here, the flow depths at sections (1) and (2) are y1 = 2.5 m and y2 = 4 m, respectively. y2 y2 2 2V12 = a b + gy1 y1 y1 2V12

( 9.81 m>s2 ) (2.5m)

= a

4m 2 4m b + 2.5 m 2.5 m

Ans.

V1 = 7.1423 m>s = 7.14 m>s

Applying the continuity equation, 0 r dV + r V # dA = 0 0f Lcv Lcs 0 - V1A1 + V2A2 = 0

( - 7.1423 m>s ) [1.5 m(2.5 m)] + V2[(1.5 m)(4 m)] Ans.

V2 = 4.4639 m>s = 4.46 m>s The energy loss during the jump can be determined from hL =

( y2 - y1 ) 3 4y1y2

=

(4 m - 2.5 m)3 4(2.5 m)(4 m) Ans.

= 0.084375 m = 0.0844 m

Ans: V1 = 7.14 m>s V2 = 4.46 m>s hL = 0.0844 m 1346

*12–80. Water flows at 18 m3>s over the 4-m-wide spillway of the dam. If the depth of the water at the bottom apron is 0.5 m, determine the depth y2 of the water after the hydraulic jump.

y2 0.5 m

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow at section (1) of depth y1 = 0.5 m is V1 =

18 m3 >s Q = 9 m>s = A1 (4 m)(0.5 m)

Then, the Froude number of the flow at section (1) is

( Fr ) 1 =

V1 2gy1

=

9 m>s 2 ( 9.81 m>s2 ) (0.5 m)

= 4.0637

Since ( Fr ) 1 7 1, the flow at section (1) is supercritical, which means that a hydraulic jump can occur. y2 1 = c 21 + 8Fr 12 - 1 d y1 2

y2 1 = c 21 + 8 ( 4.0637 ) 2 - 1 d 0.5 m 2 y2 = 2.63 m

1347

Ans.

12–81. Water runs from a sloping channel with a flow of 8 m3>s onto a horizontal channel, forming a hydraulic jump. If the channel is 2 m wide, and the water is 0.25 m deep before the jump, determine the depth of water after the jump. What energy is lost during the jump?

0.25 m y2

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow at section (1) of depth y1 = 0.25 m is 8 m3 >s Q = 16 m>s = A1 (2 m)(0.25 m)

V1 =

Then, the Froude number of the flow at section (1) is

( Fr ) 1 =

V1 2gy1

=

16 m>s 2 ( 9.81 m>s2 ) (0.25 m)

= 10.2168

Since (Fr)1 7 1, the flow at section (1) is supercritical, which means that a hydraulic jump can occur. y2 1 = c 21 + 8Fr 12 - 1 d y1 2

y2 1 = c 21 + 8 ( 10.2168 ) 2 - 1 d 0.25 m 2

Ans.

y2 = 3.4894 m = 3.49 m

hL =

( y2 - y1 ) 3 4y1y2

=

( 3.4894 m - 0.25 m ) 3 4(0.25 m)(3.4894 m) Ans.

= 9.7416 m = 9.74 m

Ans: y2 = 3.49 m hL = 9.74 m 1348

12–82. The hydraulic jump has a depth of 5 m at the downstream end, and the velocity is 1.25 m>s. If the channel is 2 m wide, determine the depth y1 of the water before the jump and the energy head lost during the jump.

1.25 m/s

y1

5m

SOLUTION Water is considered to be incompressible. The flow is steady. Applying the continuity equation, 0 r dV + rV # dA = 0 0t L cv L cs 0 - V1A1 + V2A2 = 0 - V1 3 2 m ( y1 ) 4 + ( 1.25 m>s ) 3 (2 m)(5 m) 4 = 0 6.25 y1

V1 = Thus,

y2 2 y2 2V12 = a b + gy1 y1 y1 2a

6.25 2 b y1

( 9.81 m>s2 ) y1

= a

5m 2 5m b + y1 y1

5y12 + 25y1 - 7.9638 = 0 Solving for the positive root,

Ans.

y1 = 0.30049 m = 0.300 m Using this result, the energy head loss during the jump is hL =

(y2 - y1)3 4y1y2

=

(5 m - 0.30049 m)3 4(0.30049 m)(5 m) Ans.

= 17.27 m = 17.3 m

Ans: y1 = 0.300 m hL = 17.3 m 1349

12–83. Water flows at 30 ft3>s through the 4-ft-wide rectangular channel. Determine if a hydraulic jump will form, and if so, determine the depth of flow y2 after the jump and the energy head that is lost due to the jump.

1 ft

y2

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow at section (1) of depth y1 = 1 ft is V1 =

30 ft 3 >s Q = 7.5 ft>s = A1 (4 ft)(1 ft)

Then, the Froude number of the flow at section (1) is (Fr)1 =

V1 2gy1

=

7.5 ft>s 2 ( 32.2 ft>s2 ) (1 ft)

= 1.3217

Since (Fr)1 7 1 the flow at section (1) is supercritical, which means that a hydraulic jump can occur. y2 1 = c 21 + 8(Fr)12 - 1 d y1 2

y2 1 = c 21 + 8 ( 1.3217 ) 2 - 1 d 1 ft 2

Ans.

y2 = 1.4349 ft = 1.43 ft

Using this result, the energy head loss during the jump is hL =

(y2 - y1)3 4y1y2

=

(1.4349 ft - 1 ft)3 4(1 ft)(1.4349 ft) Ans.

= 0.01433 ft = 0.0143 ft

Ans: y2 = 1.43 ft hL = 0.0143 ft 1350

*12–84. Water flows down the 3-m-wide spillway, and at the bottom it has a depth of 0.4 m and attains a speed of 8 m>s. Determine if a hydraulic jump will form, and if so, determine the speed of the flow and its depth after the jump.

0.4 m

SOLUTION Water is considered to be incompressible. The flow is steady. The Froude number of the flow at section (1) where y1 = 0.4 m is (Fr)1 =

V1 2gy1

=

8 m>s 2 ( 9.81 m>s2 ) (0.4 m)

= 4.039 7 1

Since (Fr)1 7 1, the flow at section (1) is supercritical, which means that a hydraulic jump can occur. y2 1 = c 21 + 8(Fr)12 - 1 d y1 2

y2 1 = c 21 + 8 ( 4.039 ) 2 - 1 d 0.4 m 2

Ans.

y2 = 2.0933 m = 2.09 m

Applying the continuity equation, 0 rdV + r V # dA = 0 0t Lcv Lcs 0 - V1A1 + V2A2 = 0 - ( 8 m>s ) [(3 m)(0.4 m)] + V2[(3 m)(2.0933 m)] = 0 Ans.

V2 = 1.5287 m>s = 1.53 m

1351

y2

12–85. The rectangular channel has a width of 3 m and the depth of flow is 1.5 m. Determine the volumetric flow of water over the rectangular sharp-crested weir. Take Cd = 0.83.

3m 1m 1.5 m

SOLUTION Water is considered to be incompressible. The flow is steady. Qactual = =

3 2 C 22g bH2 3 d

3 2 (0.83) 22 ( 9.81m>s2 ) (3m)(1m)2 3

= 7.35 m3 >s

Ans.

1352

Ans: 7.35 m3 >s

12–86. The rectangular channel is fitted with a 90° triangular weir plate. If the upstream depth of the water within the channel is 2 m, and the bottom of the weir plate is 1.5 m from the bottom of the channel, determine the volumetric flow of water over the weir. Take Cd = 0.61.

90! 2m

1.5 m

SOLUTION Water is considered to be incompressible. The flow is steady. Qactual =

5 8 u C 22gH2 tan 15 d 2

= a

5 8 90° b(0.61) 22 ( 9.81m>s2 ) (0.5m)2 tana b 15 2

= 0.2547 m3 >s

= 0.255 m3 >s

Ans.

1353

Ans: 0.255 m3 >s

12–87. The flow of water over the broad-crested weir is 15 m3>s. If the weir and the channel have a width of 3 m, determine the depth of water y within the channel. Take Cw = 0.80.

y 1m

SOLUTION Water is considered to be incompressible. The flow is steady. 3

2 2 Qactual = Cwb1g a Hb 3

3

2 2 15 m3 >s = 0.8(3m) 2 ( 9.81m>s2 ) c (y - 1) d 3

Ans.

y = 3.3775 m = 3.38 m

Ans: 3.38 m 1354

*12–88. Determine the volumetric flow of water over the broad-crested weir if it is in a channel having a width of 5 ft. Take Cw = 0.95.

6 ft 3 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 3

2 2 Q = Cwb2g a Hb 3

3

2 2 = 0.95(5ft) 232.2ft>s2 c (3ft) d 3

= 76.24 ft 3 >s = 76.2 ft 3 >s

Ans.

1355

13–1. Oxygen is decompressed from an absolute pressure of 600 kPa to 100 kPa, with no change in temperature. Determine the changes in the entropy and enthalpy.

SOLUTION Oxygen is considered to be compressible. From Appendix A, R = 259.8 J>kg # K and k = 1.40. Since there is no change in temperature, ∆T = 0 and T1 = T2. The change in enthalpy is ∆h = cp ∆T = 0

Ans.

The change in entropy is s2 - s1 = cp ln

p2 T2 - R ln T1 p1

∆s = cp ln 1 - ( 259.8 J>kg # K ) ln a = 465 J>kg # K

100 kPa b 600 kPa

Ans.

Ans: ∆h = 0 ∆s = 465 J>(kg # K) 1356

13–2. If a pipe contains helium at a gage pressure of 100 kPa and a temperature of 20°C, determine the density of the helium. Also, determine the temperature if the helium is compressed isentropically to a gage pressure of 250 kPa. The atmospheric pressure is 101.3 kPa.

SOLUTION Helium is considered to be compressible. From Appendix A, R = 2077 J>kg # K and k = 1.66 for helium. Using the universal gas law, P1 = r1R1T1 (100 + 101.3) ( 103 )

N = r1 ( 2077 J>kg # K ) (273 + 20°C) K m2

r1 = 0.3308 kg>m3 = 0.331 kg>m3

Ans.

Since the process is isentropic, k

p2 T2 k - 1 = a b p1 T1

(250 + 101.3) kPa (100 + 101.3) kPa

1.66

= J

1.66 - 1 T2 R (273 + 20°C) K

Ans.

T2 = 365.61 K = 366 K

Ans: r1 = 0.331 kg>m3 T2 = 366 K 1357

13–3. Helium is contained in a closed vessel under an absolute pressure of 400 kPa. If the temperature increases from 20°C to 85°C, determine the changes in pressure and entropy.

SOLUTION Helium is considered to be compressible. From Appendix A, R = 2077 J>kg # K and k = 1.66 for helium. Since the helium is contained in a closed rigid vessel, the mass and volume remain constant throughout the process. Therefore, the density r is a constant. Applying the universal gas law, rRT1 p1 = p2 rRT2 (273 + 20°C) K 400 kPa = p2 (273 + 85°C) K p2 = 488.7 kPa The change in pressure is ∆p = p2 - p1 = 488.74 kPa - 400 kPa = 88.74 kPa = 88.7 kPa Also, cp =

Ans.

1.66 ( 2077 J>kg # K ) kR = = 5223.97 J>kg # K k - 1 1.66 - 1

The change in entropy is s1 - s2 = cp ln

p2 T2 - R ln T1 p1

∆s = ( 5223.97 J>kg # K ) ln

(273 + 85°C) K (273 + 20°C) K

-

∆s = 630.53 J>kg # K = 631 J>kg # K

1 2077 J>kg # K 2 ln a

488.7 kPa b 400 kPa Ans.

The positive result indicates that the entropy increases.

Ans: ∆p = 88.7 kPa ∆s = 631 J>(kg # K) 1358

*13–4. The oxygen in section A of the pipe is at a temperature of 60°C and an absolute pressure of 280 kPa, whereas when it is at B, its temperature is 80°C and the absolute pressure is 200 kPa. Determine the change per unit mass in the internal energy, enthalpy, and entropy between the two sections.

A

SOLUTION Oxygen is considered to be compressible. The flow is steady. From Appendix A, R = 259.8 J>kg # K and k = 1.40. cv = cp =

259.8 J>kg # K R = = 649.5 J>kg # K k - 1 1.40 - 1

1.40 ( 259.8 J>kg # K ) kR = = 909.3 J>kg # K k - 1 1.40 - 1

Then, the changes in internal energy, enthalpy, and entropy are ∆u = cv ∆T = ( 649.5 J>kg # K ) 3 (273 + 80°C) K - (273 + 60°C) K 4 = 12.99 ( 103 ) J>kg = 13.0 kJ>kg

Ans.

∆h = cp ∆T = ( 909.3 J>kg # K ) 3 (273 + 80°C) K - (273 + 60°C) K 4

Ans.

= 18.186 J>kg = 18.2 kJ>kg

sB - sA = cp ln

pB TB - R ln TA pA

∆s = ( 909.3 J>kg # K ) Jln a

273 + 80°C 200 kPa b R - ( 259.8 J>kg # K ) Jlna bR 273 + 60°C 280 kPa

= 140.45 J>kg # K = 140 J>kg # K

Ans.

1359

B

13–5. The hydrogen in section A of the pipe is at a temperature of 60°F and an absolute pressure of 30 lb>in2, whereas when it is at B, its temperature is 100°F and the absolute pressure is 20 lb>in2. Determine the change per unit mass in the internal energy, enthalpy, and entropy between the two sections.

B

A

SOLUTION Hydrogen is considered to be compressible. The flow is steady. From Appendix A, R = 24.66 ( 103 ) ft # lb>slug # R and k = 1.41. cv =

24.66 ( 103 ) ft # lb>slug # R R = = 60.15 ( 103 ) ft # lb>slug # R k - 1 (1.41 - 1)

cp =

1.41 3 24.66 ( 103 ) ft # lb>slug # R 4 kR = = 84.81 ( 103 ) ft # lb>slug # R k - 1 1.41 - 1

Then, the changes in internal energy, enthalpy, and entropy are ∆u = cv ∆T =

3 60.15 ( 103 ) ft # lb>slug # R 4 3 (460

+ 100°F) R - (460 + 60°F) 4

= 2.406 ( 106 ) ft # lb>slug = 2.41 ( 106 ) ft # lb>slug ∆h = cp ∆T =

3 84.81 ( 10 ) ft # lb>slug # R 4 3 (460 3

Ans.

+ 100°F) R - (460 + 60°F) 4

= 3.392 ( 106 ) ft # lb>slug = 3.39 ( 106 ) ft # lb>slug sB - sA = cp ln ∆s =

Ans.

pB TB - R ln TA pA

3 84.81 ft # lb>slug # R 4 c ln a

460 + 100°F bd 460 + 60°F

3 24.66 ( 103 ) ft # lb>slug # R 4 c ln a

= 16.28 ( 103 ) ft # lb>slug # R = 16.3 ( 103 ) ft # lb>slug # R

1360

Ans.

20 lb>in2 30 lb>in2

bd

Ans: ∆u = 2.41 1 106 2 ft # lb>slug ∆h = 3.39 1 106 2 ft # lb>slug ∆s = 16.3 1 106 2 ft # lb>(slug # R)

13–6. The closed tank contains helium at 200°C and under an absolute pressure of 530 kPa. If the temperature is increased to 250°C, determine the changes in density and pressure, and the changes per unit mass in the internal energy and enthalpy of the helium.

SOLUTION The helium is considered to be compressible. From Appendix A, R = 2077 J>kg # K and k = 1.66 for helium. Since the helium is contained in a closed rigid tank, the mass and volume remain constant throughout the process. Therefore, the density r will remain constant. Applying the universal gas law, rRT1 p1 = p2 rRT2 (273 + 200°C) K 530 kPa = p2 (273 + 250°C) K p2 = 586.03 kPa The change in pressure is ∆p = p2 - p1 = 586.03 kPa - 530 kPa Ans.

= 56.03 kPa = 56.0 kPa Also, cv =

2077 J>kg # K R = = 3146.97 J>kg # K k - 1 (1.66 - 1)

cp =

1.66 ( 2077 J>kg # K ) kR = = 5223.97 J>kg # K k - 1 1.66 - 1

The changes in internal energy and enthalpy are ∆u = cv ∆T = ( 3146.97 J>kg # K ) 3 (273 + 250°C) K - (273 + 200°C) K 4 = 157.35 ( 103 ) J>kg = 157 kJ>kg

Ans.

∆h = cp ∆T = ( 5223.97 J>kg # K ) 3 (273 + 250°C) K - (273 + 200°C) K 4 = 261.20 ( 103 ) J>kg = 261 kJ>kg

Ans.

Ans: The density r will remain constant. ∆p = 56.0 kPa ∆u = 157 kJ>kg ∆h = 261 kJ>kg 1361

13–7. The closed tank contains oxygen at 400°F and under an absolute pressure of 30 lb>in. If the temperature decreases to 300°F, determine the changes in density and pressure, and the change per unit mass in the internal energy and enthalpy of the oxygen.

SOLUTION The oxygen is considered to be compressible. From Appendix A, R = 1554 ft # lb>slug # R and k = 1.40 for oxygen. Since the oxygen is contained in a closed rigid tank, the mass and volume remain constant throughout the process. Therefore, the density r will remain constant. Thus, ∆r = 0 Applying the universal gas law, rRT1 p1 T1 = = p2 rRT2 T2 30 lb>in2 p2

=

(460 + 400°F) R (460 + 300°F) R

p2 = 26.51 lb>in2 The change in pressure is ∆p = p2 - p1 = 26.51 lb>in2 - 30 lb>in2 = - 3.488 lb>in2 = - 3.49 lb>in2 Also, cv = cp =

Ans.

1554 ft # lb>slug # R R = = 3885 ft # lb>slug # R k - 1 1.40 - 1

1.40 ( 1554 ft # lb>slug # R ) kR = = 5439 ft # lb>slug # R k - 1 1.40 - 1

The changes in internal energy and enthalpy are ∆u = cv ∆T = ( 3885 ft # lb>slug # R ) 3 (460 + 300°F) R - (460 + 400°F) R 4 = - 388.5 ( 103 ) ft # lb>slug

Ans.

∆h = cp ∆T = ( 5439 ft # lb>slug # R ) 3 (460 + 300°F) R - (460 + 400°F) R 4 = - 543.9 ( 103 ) ft # lb>slug = - 544 ( 103 ) ft # lb>slug

1362

Ans.

Ans: The density r will remain constant. ∆p = - 3.49 lb>in2 ∆u = - 388.5 1 103 2 ft # lb>slug ∆h = - 544 1 103 2 ft # lb>slug

*13–8. A gas has a specific heat that varies with the absolute temperature, such that cp = 1 1256 + 36 728>T 2 2 J>kg # K. If the temperature rises from 300 K to 400 K, determine the change in enthalpy per unit mass.

SOLUTION The gas is considered to be compressible. Since cp is a function of T, the change in enthalpy is cp =

dh dT

h2

Lh1

T2

dh =

LT1

cp dT 400 K

h2 - h1 =

L300 K

a1256 +

= a1256T -

36 728 bdT T2

36 728 400 K b` T 300 K

= 125.63 ( 103 ) J>kg

Ans.

= 126 kJ>kg

1363

13–9. Air has a temperature of 600°R and absolute pressure of 100 psi at A. As it passes through the transition, its temperature becomes 500°R and the absolute pressure becomes 40 psi at B. Determine the changes in the density and the entropy per unit mass of the air.

6 in. 2 in.

B A

SOLUTION The air is considered to be compressible. The flow is steady. From Appendix A, R = 1716 ft # lb>slug # R and k = 1.40 for air. Applying the universal gas law, pA = rARTA;

(100 lb>in2 ) a

12 in. 2 b = rA ( 1716 ft # lb>slug # R ) (600 R) 1 ft

rA = 0.013986 slug>ft 3 pB = rBRTB;

( 40 lb>in2 ) a

12 in. 2 b = rB ( 1716 ft # lb>slug # R ) (500 R) 1 ft

rB = 0.006713 slug>ft 3

Thus, the change in density is ∆r = rB - rA = 0.006713 slug>ft 3 - 0.013986 slug>ft 3 = - 0.00727 slug>ft 3 Also, cv =

Ans.

1716 ft # lb>slug # R R = = 4290 ft # lb>slug # R k - 1 1.40 - 1

The change in entropy is sB - sA = cv ln ∆s =

rB TB - R ln rA TA

1 4290 ft # lb>slug # R 2 ln a

500 R b 600 R

1 1716 ft # lb>slug # R 2 °ln

= 477.3 ft # lb>slug # R = 477 ft # lb>slug # R

0.006713 slug>ft 3 0.013986 slug>ft 3 Ans.

¢

Ans: ∆r = - 0.00727 slug>ft 3 ∆s = 477 ft # lb>(slug # R) 1364

13–10. Air has a temperature of 600°R and absolute pressure of 100 psi at A. As it passes through the transition, its temperature becomes 500°R and the absolute pressure becomes 40 psi at B. Determine the change per unit mass in the internal energy and the enthalpy of the air.

6 in. 2 in.

B A

SOLUTION The air is considered to be compressible. The flow is steady. From Appendix A, R = 1716 ft # lb>slug # R and k = 1.40 for air. cv = cp =

1716 ft # lb>slug # R R = = 4290 ft # lb>slug # R k - 1 1.40 - 1

1.40 1 1716 ft # lb>slug # R 2 kR = = 6006 ft # lb>slug # R k - 1 1.40 - 1

Then, the changes in internal energy and enthalpy are

∆u = cv ∆T = ( 4290 ft # lb>slug # R ) (500 R - 600 R) = - 429 ( 103 ) ft # lb>slug

Ans.

= - 600.6 ( 103 ) ft # lb>slug = - 601 ( 103 ) ft # lb>slug

Ans.

∆h = cp ∆T = ( 6006 ft # lb>slug # R ) (500 R - 600 R)

Ans: ∆u = - 429 1 103 2 ft # lb>slug 1365

∆h = - 601 1 103 2 ft # lb>slug

13–11. Air flows in a horizontal duct at 20°C with a velocity of 180 m>s. If the velocity increases to 250 m>s, determine the corresponding temperature of the air. Hint: Use the energy equation to find ∆h .

SOLUTION The air is considered to be compressible. The flow is steady. Applying the energy equation with a hout +

Vout 2 V in2 = hin + 2 2

∆h = hout - hin = =

dWs dQ b = a b = 0 and zin = zout dt in dt out

1 1 V 2 - V out2 2 2 in

1 31 180 m>s 2 2 2

1 250 m>s 2 2 4

= -15.05 1 103 2 J>kg

From Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. cp =

1.40 1 286.9 J>kg # K 2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

The change in enthalpy is ∆h = cp ∆T;

- 15.05 1 103 2 J>kg =

1 1004.15 J>kg # K 2 3 (273

Tout = 5.01°C

+ Tout) K - (273 + 20°C) K 4 Ans.

Ans: 5.01°C 1366

*13–12. The half-angle a on the Mach cone of a rocket is 20°. If the air temperature is 65°F, determine the speed of the rocket.

SOLUTION The air is considered to be compressible. From Appendix A, R = 1716 ft # lb>slug # R and k = 1.40 for air. c = 1kRT = 21.40 1 1716 ft # lb>slug # R 2 (460 + 65°F) = 1123.06 ft>s

We have sin a =

c ; V

sin 20° =

1123.06 ft>s V

V = 3283.60 ft>s = 3.28 1 103 2 ft>s

Ans.

1367

13–13. Determine the speed of a jet plane that flies at Mach 2.3 and at an altitude of 10 000 ft. Use the Standard Atmospheric Table in Appendix A.

SOLUTION The air is considered to be compressible. From Appendix A, T = 23.34°F at an altitude of 10 000 ft. Also, R = 1716 ft # lb>slug # R and k = 1.40. V = M 1kRT

= 2.321.4 1 1716 ft # lb>slug>R 2 1 460 + 23.34°F 2 R = 2478.43 ft>s = 2.48 1 103 2 ft>s

1368

Ans.

Ans: 2.48 1 103 2 ft>s

13–14. Compare the speed of sound in water and air at a temperature of 20°C. The bulk modulus of water at T = 20°C is EV = 2.2 GPa.

SOLUTION Table in Appendix A gives k = 1.4 and R = 286.9 J>kg # K. Here, T = (273 + 20) K = 293 K. Then cair = 1kRT = 21.4 1 286.9 J>kg # K 2 (293 K) = 343 m>s

Ans.

Table in Appendix A gives rw = 998.3 kg>m3. Then cw =

2.20 1 109 2 N>m2 EV = = 1485 m>s B r C 998.3 kg>m3

Ans.

Ans: cair = 343 m>s cw = 1485 m>s 1369

13–15. Determine the speed of sound in water and in air, both at a temperature of 60°F. Take EV = 311 1 103 2 psi for water.

SOLUTION The water is considered to be compressible. From Appendix A, R = 1716 ft # lb>slug # R and k = 1.40 for air. cair = 1kRT = 21.40 1 1716 ft # lb>slug # R 2 (460 + 60°F) = 1117.70 ft>s = 1.12 1 103 2 ft>s

Ans.

From Appendix A, rw = 1.939 slug>ft 3 at T = 60°F.

cw =

EV = B r S

3 311 1 103 2 lb>in2 4 a

12 in. 2 b 1 ft

1.939 slug>ft 3

cw = 4805.88 ft>s = 4.81 1 103 2 ft>s

Ans.

1370

Ans: cair = 1.12 1 103 2 ft>s cw = 4.81 1 103 2 ft>s

*13–16. A ship is located where the depth of the ocean is 3  km. Determine the time needed for a sonar signal to bounce off the bottom and return to the ship. Assume the  water temperature is 10°C. Take r = 1030 kg>m3 and EV = 2.11 1 109 2 Pa for sea water.

SOLUTION c =

2.11 1 109 2 N>m2 EV = = 1431.27 m>s B r C 1 1030 kg>m3 2

The distance traveled by the sonic wave is s = 2 3 3 1 103 2 m 4 = 6 1 103 2 m. Then, s = ct;

6 1 103 2 m = t = 4.19 s

1 1431.27 m>s 2 t

Ans.

1371

13–17. Determine how fast a race car must travel in 20°C weather in order for M = 0.3.

SOLUTION V V = . For air, table in Appendix A gives k = 1.4, R = 286.9 J>kg # K. c 1kRT Here, T = (273 + 20) K = 293. M =

Then V

31.4 1 286.9 J>kg # K 2 (293 K)

= 0.3 Ans.

V = 102.92 m>s = 103 m>s

Ans: 103 m>s 1372

13–18. A jet plane is flying at Mach 2.2. Determine its speed in kilometers per hour. The air is at 10°C.

SOLUTION The air is considered to be compressible. From Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. V = M 1kRT = 2.231.4 1 286.9 J>kg # K 2 (273 + 10°C) K = a741.72

1 km 3600 s m ba ba b s 1000 m h

= 2.67 1 103 2 km>h

Ans.

1373

Ans: 2.67 1 103 2 km>h

13–19. Water is at a temperature of 40°F. If a sonar signal takes 3 s to detect a large whale, determine the distance from the whale to the ship. Take r = 1.990 slug>ft 3 and EV = 311 1 103 2 psi.

SOLUTION The water is considered to be compressible.

c =

EV = B r S

3 311 ( 103 ) lb>in2 4 a

12 in. 2 b 1ft

( 1.990 slug>ft 3 )

= 4743.89

The distance traveled by the sonic wave is 2s, where s is the distance the whale is from the ship. Thus, 2s = ct s =

ct = 2

1 4743.89 ft>s 2 (3 s)

= 7115.86 ft a = 1.35 mi

2

1 mi b 5280 ft

Ans.

Ans: 1.35 mi 1374

*13–20. Determine the Mach number of a cyclist peddling at 15 mi>h. The air has a temperature of 70°F. 1 mi = 5280 ft.

SOLUTION The air is considered to be compressible. The speed of the cyclist is V = a15

1h 5280 ft mi ba ba b = 22 ft>s h 3600 s 1 mi

From Appendix A, R = 1716 ft # lb>slug # R and k = 1.40 for air. The speed of sound is c = 1kRT

= 21.40 1 1716 ft # lb>slug # R 2 1 460 + 70°F 2 R

= 1128.39 ft>s The Mach number is M =

22 ft>s V = c 1128.39 ft>s Ans.

= 0.0195

1375

13–21. A jet plane has a speed of 600 mi>h when flying at  an altitude of 10 000 ft. Determine the Mach number. 1  mi = 5280 ft. Use the Standard Atmospheric Table in Appendix A.

SOLUTION The air is considered to be compressible. The speed of the jet is V = a600

1h 5280 ft mi ba ba b = 880 ft>s h 3600 s 1 mi

From Appendix A, T = 23.34°F for air at an altitude of 10 000 ft. Also, R = 1716 ft # lb>slug # R and k = 1.40 for air. c = 2kRT

= 21.40 ( 1716 ft # lb>slug # R ) (460 + 23.34°F) R = 1077.6 ft>s

The Mach number is M =

880 ft>s V = c 1077.6 ft>s Ans.

= 0.817

Ans: 0.817 1376

13–22. Determine the half-angle a of the Mach cone at the nose of a jet if it is flying at 1125 m>s in air at 5°C.

SOLUTION The air is considered to be compressible. From Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. c = 2kRT

= 21.40 ( 286.9 J>kg # K ) (273 + 5°C) K = 334.16 m>s

Thus, sin a =

334.16 m>s c = V 1125 m>s Ans.

a = 17.3°

Ans: 17.3° 1377

13–23. A jet plane has a speed of 600 m>s. If the air has a temperature of 10°C, determine the Mach number and the half-angle a of the Mach cone.

600 m/s a a

SOLUTION The air is considered to be compressible. From Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. c = 2kRT

= 21.40 ( 286.9 J>kg # K ) (273 + 10°C) K = 337.15 m>s

Thus, the Mach number is V M = c M =

600 m>s 337.15 m>s

Ans.

= 1.78

The half angle of the Mach cone is sin a =

337.15 m>s c = V 600 m>s Ans.

a = 34.2°

Ans: M = 1.78 a = 34.2° 1378

*13–24. A jet plane passes 5 km directly overhead. If the sound of the plane is heard 6 s later, determine the speed of the plane. The average air temperature is 10°C.

SOLUTION The air is considered to be compressible. From Appendix A, R = 286.9 J>kg # K and k = 1.40. c = 2kRT

= 21.40 ( 286.9 J>kg # K ) (273 + 10°C) K = 337.15 m>s

The half angle of the Mach cone is sin a =

337.15 m>s c = V V

(1)

Referring to the geometry of the half cone shown in Fig. a, sin a =

5000

(2)

2(6V)2 + 50002

S = Vt = V(6)

Equating Eqs. (1) and (2), a

337.15 5000 = V 2(6V)2 + 50002

5000 m (6V)2 + 50002

Ans.

V = 368.67 m>s = 369 m>s

1379

(a)

13–25. The Mach number of the air flow in the wind tunnel at B is to be M = 2.0 with an air temperature of 10°C and absolute pressure of 25 kPa. Determine the required absolute pressure and temperature within the large reservoir at A.

B

A

SOLUTION The air is considered to be compressible. The flow is steady. The air in reservoir A is at rest. Thus, the temperature and pressure here are T0 (stagnation temperature) and p0 (stagnation pressure), respectively. Since k = 1.40 for air, for M = 2.0, T = 0.5556; T0

T0 =

p = 0.1278; p0

p0 =

(273 + 10°C) K 0.5556

Ans.

= 509.4 K = 509 K

25 kPa = 195.61 kPa = 196 kPa 0.1278

Ans.

Ans: T0 = 509 K p0 = 196 kPa 1380

13–26. The absolute stagnation pressure for air is 875 kPa when the stagnation temperature is 25°C. If the absolute pressure for the flow is 630 kPa, determine the velocity of the flow.

SOLUTION The air is considered to be compressible. The flow is steady. Since k = 1.40 for air, here, p 630 kPa = = 0.72 p0 875 kPa Interpolating the values given in the table, we obtain M = 0.7014. Using this result, T = 0.9104; T0

T = 0.9104(273 + 25°C) K = 271.30 K

For air, R = 286.9 J>kg # K. V = M 2kRT = 0.7014 21.40 ( 286.9 J>kg # K ) (271.30) V = 231.55 m>s = 232 m>s

Ans.

Ans: 232 m>s 1381

13–27. The flow at a point in a wind tunnel has a speed of M = 2.5 when the absolute pressure of the air is 16 kPa and the temperature is 200 K. Determine the speed of the air at the point, and also find the temperature and pressure of the air in the supply reservoir.

SOLUTION The air is considered to be compressible. The flow is steady. For air, R = 286.9 J>kg # K and k = 1.40.

V = M 2kRT = 2.521.40 ( 286.9 J>kg # K ) (200 K)

Ans.

= 708.57 m>s = 709 m>s

The air in the supply reservoir is at rest. Thus, the temperature and pressure here are T0 (stagnation temperature) and p0 (stagnation pressure), respectively. Since k = 1.40 for air, for M = 2.5, T = 0.4444; T0 p = 0.05853; p0

T0 =

200 K = 450 K 0.4444

Ans.

p0 =

16 kPa = 273.38 kPa = 273 kPa 0.05853

Ans.

Ans: V = 709 m>s T0 = 450 K p0 = 273 kPa 1382

*13–28. The 4-in.-diameter pipe carries air that is flowing at M = 1.36. Measurements show that the absolute pressure is 60 psi and the temperature is 95°F. Determine the mass flow through the pipe. 4 in.

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 1716 ft # lb>slug # R. V = M 2kRT

= 1.3621.40 ( 1716 ft # lb>slug # R ) (460 + 95°F) R = 1570.39 ft>s

Using the ideal gas law, p = rRT;

a60

lb 12 in. 2 ba b = r ( 1716 ft # lb>slug # R ) (460 + 95°F) R 1 ft in2 r = 0.009072 slug>ft 3

Thus, the mass flow can be determined from 2 2 # m = rVA = ( 0.009072 slug>ft 3 )( 1570.39 ft>s ) £ pa ft b § 12

= 1.243 slug>s = 1.24 slug>s

1383

Ans.

13–29. The 4-in.-diameter pipe carries air that is flowing at M = 0.83. If the stagnation temperature is 85°F and the absolute stagnation pressure is 14.7 psi, determine the mass flow through the pipe. 4 in.

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 1716 ft # lb>slug # R. Since k = 1.40 for air, for M = 0.83, the interpolation of the values in the tables gives T = 0.8789; T = 0.8789(460 + 85°F) R = 479.00 R T0 p = 0.6365; p = 0.6365(14.7 psi) = 9.3565 psi p0 Using the universal gas law, p = rRT; a9.3565

lb 12 in. 2 b = r ( 1716 ft # lb>slug # R ) (479.00 R) ba 1 ft in2 r = 0.001639 slug >ft 3

The velocity of the air is V = M 2kRT

= 0.8321.40 ( 1716 ft # lb>slug # R ) (479.00 R)

= 890.37 ft>s Thus, the mass flow can be determined from 2 2 # m = rVA = ( 0.001639 slug>ft 3 )( 890.37 ft>s ) £ p a ft b § 12

= 0.1274 slug>s = 0.127 slug>s

Ans.

Ans: 0.127 slug>s 1384

13–30. The absolute stagnation pressure for methane is 110 lb>in2 when the stagnation temperature is 70°F. If the  pressure in the flow is 80 lb>in2, determine the corresponding velocity of the flow.

SOLUTION The methane is considered to be compressible. The flow is steady. For methane, k = 1.31 and R = 3099 ft # lb>slug # R (Appendix A). Applying k

p0 = pa1 +

k - 1 2 k-1 M b 2 1.31

110 lb>in2 = M = 0.7106

1 80 lb>in2 2 c 1

+ a

1.31 - 1 1.31 - 1 bM 2 d 2

Using this result, T0 = T a1 + a

k - 1 bM 2 b 2

(460 + 70°F) R = T c 1 + a T = 491.53 R

1.31 - 1 b ( 0.71062 ) d 2

Then, the velocity of the flow is V = M 2kRT = 0.7106 21.31 ( 3099 ft # lb>slug # R ) (491.53 R) = 1003.82 ft >s = 1.00 ( 103 ) ft>s

Ans.

1385

Ans: 1.00 1 103 2 ft>s

13–31. The temperature and absolute pressure of air within the circular duct are 40°C and 800 kPa, respectively. If the mass flow is 30 kg>s, determine the Mach number.

200 mm

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K (table in Appendix A). Using the universal gas law, 800 ( 103 )

p = rRT;

N = r ( 286.9 J>kg # K ) (273 + 40°C) m2 r = 8.9087 kg>m3

Using this result, the velocity of the flow can be determined from

#

m = rVA 30 kg>s = ( 8.9087 kg>m3 ) (V) 3 p(0.1 m)2 4 V = 107.19 m>s The speed of sound is c = 2kRT = 21.40 ( 286.9 J>kg # K ) (273 + 40°C) K = 354.57 m>s

Finally, the Mach number of the flow is M =

107.19 m>s V = c 354.57 m>s Ans.

= 0.302

Ans: 0.302 1386

*13–32. Determine the pressure of air if it is flowing at 1600  km>h. When the air is still, the temperature is 20°C and the absolute pressure is 101.3 kPa.

SOLUTION The air is considered to be compressible. The flow is steady. The velocity of the air is V = a1600

1h km 1000 m ba ba b = 444.44 m>s hr 1 km 3600 s

For air, k = 1.40 and R = 286.9 J>kg # K.

c = 21.40 ( 286.9 J>kg # K ) T = 20.042T

The Mach number of the flow is M =

444.44 m>s V 22.1763 = = c 20.04 2T 2T

(1)

Using this result,

T0 = T a1 +

k - 1 2 M b 2

(273 + 20°C) K = T c 1 + a T = 194.64 K

1.4 - 1 22.1763 2 ba b d 2 2T

Substituting this result into Eq. (1), M =

22.1763 2194.64

= 1.5895

Using this result,

k

k - 1 2 k-1 p0 = pa1 + M b 2

1.4

1.4 - 1 1.4 - 1 101.3 kPa = pc 1 + a b ( 1.58952 ) d 2

Ans.

p = 24.21 kPa = 24.2 kPa

1387

13–33. What are the ratios of the critical pressure, temperature, and density to the stagnation pressure, temperature, and density for methane?

SOLUTION The methane is considered to be compressible. For methane, k = 1.31 (Appendix A). The critical temperature, pressure, and density occur when M = 1. T0 = T a1 + T0 = T *a1 +

k - 1 2 M b 2

1.31 - 1 2 (1) b 2

T* = 0.8658 = 0.866 T0

Ans. k

k - 1 2 (k - 1) p0 = pa1 + M b 2 p0 = p*a1 +

1.31

1.31 - 1 2 (1.31 - 1) (1) b 2

p* = 0.5439 = 0.544 p0

Ans. 1

k - 1 2 ak - 1 b M b r0 = r a1 + 2 r0 = r*a1 +

1

1.31 - 1 2 a1.31 - 1 b (1) b 2

r* = 0.6282 = 0.628 r0

Ans.

Ans: T*>T0 = 0.866 p*>p0 = 0.544 r*>r0 = 0.628 1388

13–34. The nozzle is fitted to a large chamber of air in which the absolute pressure is 175 psi and the absolute temperature is 500°R. Determine the greatest possible mass flow through the nozzle. The throat has a diameter of 2 in.

2 in.

SOLUTION We assume isentropic flow. The greatest mass flow will occur when the nozzle becomes choked, that is, at the throat M = 1. The stagnation pressure and temperature are p0 = 175 psi and T0 = 500°R. The temperature and pressure of the air stream is (M = 1). T* = 0.8333 T0 T * = 0.8333(500°R) = 458.3°R p* = 0.5283 p0 p* = 0.5283(175 psi) = 92.45 psi Rather than using Appendix B, the density can also be found using the ideal gas law with R = 1716 ft # lb>slug # R and k = 1.4 (from table in Appendix A). The mass flow is therefore

#

m = r*V *A = =

p* RT *

M 2kRT *A

( 92.45 lb>in2 ) (12 in.>ft)2 ( 1716 ft # lb>slug # R ) (458.3 R)

(1) 2(1.4) ( 1716 ft # lb>slug # R ) (458.3 R) c p a

# m = 0.388 slug>s

2 1 ft b d 12

Ans.

Ans: 0.388 slug>s 1389

13–35. Nitrogen in the reservoir is at a temperature of 20°C and an absolute pressure of 300 kPa. Determine the mass flow through the nozzle. The atmospheric pressure is 100 kPa.

10 mm

SOLUTION The flow can be considered steady and isentropic and the nitrogen is compressible. The nitrogen in the reservoir is stagnant. Thus, T0 = (273 + 20) K = 293 K and p0 = 300 kPa. Here p = patm = 100 kPa and for nitrogen k = 1.4 (from table in Appendix A). k

p0 = pa1 +

k - 1 2 k-1 M b 2

300 kPa = (100 kPa)c 1 + a

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

M = 1.3578 7 1 (Nozzle choked)

Since the nozzle is choked, the nitrogen will undergo expansion so that its pressure will drop abruptly to patm just to the right of the exit plane. The flow properties just to the left of the exit plane can be determined using M = 1. T0 = T a1 +

k - 1 2 M b 2

293 K = T * c 1 + a

1.4 - 1 b ( 12 ) d 2

T * = 244.17 K

And k

p0 = pa1 +

k - 1 2 k-1 M b 2

300 kPa = p* c 1 +

1.4

1.4 - 1 2 1.4 - 1 (1 ) d 2

p* = 158.48 kPa

Using the ideal gas law with R = 296.8 J>kg # K for nitrogen (from table in Appendix A) p* =

p* RT

*

=

158.48 ( 103 ) N>m2

( 296.8 J>kg # K ) (244.17 K)

= 2.1869 kg>m3

V * = M 2kRT * = (1) 21.4 ( 296.8 J>kg # K ) (244.17 K) = 318.52 m>s

The mass flow rate under the choked condition is the greatest possible for the given stagnation condition and nozzle.

#

m = r*V *A = ( 2.1869 kg>m3 )( 318.52 m>s ) c p(0.005 m)2 d = 0.0547 kg>m3

1390

Ans.

Ans: 0.0547 kg>m3

13–36. The large tank contains air at an absolute pressure of 150 kPa and temperature of 20°C. The 5-mm-diameter nozzle at A is opened to let air out of the tank. Determine the mass flow and the horizontal force that must be applied to the tank to prevent it from moving. The atmospheric pressure is 100 kPa.

A

F

SOLUTION The flow can be considered steady and isentropic and the air is compressible. Since the tank is a large reservoir, the air contained is stagnant. Thus, T0 = (273 + 20) K = 293 K and p0 = 150 kPa. Here, p = patm = 100 kPa and for air k = 1.4 and using ideal gas law. r0 =

p0 = RT0

150 ( 103 ) N>m2

( 286.9 J>kg # K ) (293 K)

= 1.7844 kg>m3 k

p0 = p a1 +

k - 1 2 k-1 M b 2

150 kPa = (100 kPa)c 1 + a

1

1.4 - 1 1.4 - 1 bM 2 d 2

M = 0.7837 6 1 (Nozzle will not choke) Also, for of M,

p 100 kPa = = 0.6667. Applying Eq. 13–31 and Eq. 13–33 with the result p0 150 kPa

T0 = T a1 +

k - 1 2 M b 2

293 K = T c 1 + a T = 260.95 K

1.4 - 1 b ( 0.78372 ) d 2

And 1

r0 = ra1 +

k - 1 2 k-1 M b 2

1.7844 kg>m3 = rc 1 + a

1

1.4 - 1 1.4 - 1 b ( 0.78372 ) d 2

r = 1.3357 kg>m3

With R = 286.9 J>kg # k (from table in Appendix A),

V = M 2kRT = 0.783721.4 ( 286.9 J>kg # K ) (260.95 K) = 253.71 m>s

Thus, the mass flow rate is

#

m = rVA = ( 1.3357 kg>m3 )( 253.71 m>s ) 3 p(0.0025 m)2 4 = 0.00665 kg>s

1391

Ans.

13–36. Continued

Applying the linear momentum equation by referring to the FBD of the control volume shown in Fig. a ΣF =

0 V rdV + Vf>cs rVf>cs # dA 0t Lcv f>cs Lcs

+ ) F = 0 + ( 253.71 m>s )( 1.3357 kg>m3 )( 253.71 m>s ) 3 p(0.0025 m)2 4 (S

Ans.

= 1.69 N

1392

F

(a)

13–37. Nitrogen, at an absolute pressure of 600 kPa and temperature of 800 K, is contained in the large tank. Determine the backpressure in the hose to choke the nozzle and yet maintain isentropic supersonic flow through the divergent portion of the nozzle. The nozzle has an outer diameter of 40 mm, and the throat has a diameter of 20 mm.

20 mm

40 mm

SOLUTION Nitrogen is considered to be compressible. The ratio of the cross-sectional area of the exit plane to that of the throat is p(0.02 m)2 A = = 4 A* p(0.01 m)2 For isentropic supersonic flow to occur through the divergent portion of the nozzle, M 7 1 at the exit plane and the back pressure is p4. Since k = 1.40 for nitrogen, A with * = 4, (choosing M 7 1) gives M = 2.9402. Then using this result, A p4 = 0.02979; p0

p4 = 0.02979(600 kPa) = 17.9 kPa

Ans.

Ans: 17.9 kPa 1393

13–38. Nitrogen, at an absolute pressure of 600 kPa and temperature of 800 K, is contained in the large tank. Determine the backpressure in the hose to choke the nozzle and maintain isentropic subsonic flow through the divergent portion of the nozzle. The nozzle has an outer diameter of 40 mm, and the throat has a diameter of 20 mm.

20 mm

40 mm

SOLUTION Nitrogen is considered to be compressible. The ratio of the cross-sectional area of the exit plane to that of the throat is p(0.02 m)2 A = = 4 A* p(0.01 m)2 For isentropic subsonic flow to occur through the divergent portion of the nozzle, M 6 1 at the exit plane and the back pressure is p3. Since k = 1.40 for nitrogen, A with * = 4, (choosing M < 1) gives M = 0.1465. Then using this result, we obtain A p4 = 0.9851; p4 = 0.9851(600 kPa) = 591 kPa Ans. p0

Ans: 591 kPa 1394

13–39. The large tank contains air at an absolute pressure of 700 kPa and temperature of 400 K. Determine the mass flow from the tank into the pipe if the converging nozzle has an exit diameter of 40 mm and the absolute pressure in the pipe is 150 kPa.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 400 K and p0 = 700 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

k - 1 2 (k - 1) M b P0 = pa1 + 2 700 kPa = 150 kPac 1 + a M = 1.663

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

The flow at the exit plane with M 7 1 (supersonic) is not possible since the nozzle will be choked at the exit plane and an expansion shock wave forms thereafter. Thus, at the exit plane M = 1. T0 = T a1 +

k - 1 2 M b 2

400 K = T *a1 + T * = 333.33 K

1.4 - 1 2 (1) b 2 k

k - 1 2 (k - 1) P0 = pa1 + M b 2

1.4

1.4 - 1 2 1.4 - 1 (1) b 2

700 kPa = p*a1 + p* = 369.80 kPa

Using the universal gas law, p* = p*RT *;

369.80(103)

N = r* ( 286.9 J>kg # K ) (333.33 K) m2

r* = 3.8668 kg>m3 The velocity of the flow at the exit plane is V * = M * 2kRT * = (1) 2(1.40) ( 286.9 J>kg # K ) (333.33 K) = 365.91 m>s

Finally,

#

m = r*V *A* = ( 3.8668 kg>m3 )( 365.91 m>s ) 3 p ( 0.02 m>s ) 2 4 = 1.778 kg>s = 1.78 kg>s

1395

Ans.

Ans: 1.78 kg>s

*13–40. The large tank contains air at an absolute pressure of 700 kPa and temperature of 400 K. Determine the mass flow from the tank into the pipe if the converging nozzle has an exit diameter of 40 mm and the absolute pressure in the pipe is 400 kPa.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 400 K and p0 = 700 kPa. For air k = 1.40 and R = 286.9 J>kg # K. k

k - 1 2 (k - 1) M b P0 = pa1 + 2 700 kPa = 400 kPac 1 + a M = 0.9311

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 6 1 (subsonic) is possible. Using this result, T0 = T a1 +

k - 1 2 M b 2

400 K = T a1 + a T = 340.89 K

1.4 - 1 b(0.9311)2 b 2

Using the universal gas law, 400 ( 103 )

p = rRT;

N = r ( 286.9 J>kg # K ) (340.89 K) m2

r = 4.0899 kg>m3 The velocity of the flow at the exit plane is V = M 2kRT = (0.9311) 21.40 ( 286.9 J>kg # K ) (340.89 K) = 344.53 m>s

Finally,

#

m = rVA = ( 4.0899 kg>m3 )( 344.53 m>s ) 3 p(0.02 m)2 4 = 1.771 kg>s = 1.77 kg>s

1396

Ans.

13–41. The large tank contains air at an absolute pressure of 600 kPa and temperature of 70°C. The Laval nozzle has a throat diameter of 20 mm and an exit diameter of 50 mm. Determine the absolute pressure within the connected pipe so that the nozzle chokes but also maintains isentropic subsonic flow within the divergent portion of the nozzle. Also, what is the mass flow from the tank if the absolute pressure within the pipe is 150 kPa?

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 70°) K = 343 K and p0 = 600 kPa. Since the nozzle is required to choke, M = 1 at the throat section. Here, p(0.025 m)2 A = a b = 6.25 A* p(0.01 m)2

For air, k = 1.40 and R = 286.9 J>kg # K. With

A = 6.25, choose M 6 1 since the A* flow at the exit plane is required to be subsonic. We get M = 0.09307

Using this result, Appendix B gives r = 0.9940; p = (0.9940)(600 kPa) = 596.38 = 596 kPa P0

Ans.

Since pb = 150 kPa 6 596 kPa, the nozzle will be choked. Thus, M = 1 at the throat section. Thus, T* = 0.8333, T0

T * = 0.8333(343 K) = 285.83 K

p* = 0.5283, p* = 0.5283(600 kPa) = 316.97 kPa p0 Using the universal gas law, N = r* 1 286.9 J>kg # K 2 (285.83 K) m2 r* = 3.8652 kg>m3 The velocity of the flow at the throat section is p* = r*RT *;

316.97 1 103 2

V * = M * 2kRT * = (1) 2(1.40) ( 286.9 J>kg # K ) (285.83 K)

Finally,

= 338.83 m>s

# m = r* V *A =

1 3.8652 kg>m3 2 ( 338.83 m>s ) 3 p(0.01 m)2 4

= 0.411 kg>s

Ans.

Ans: For isentropic flow, p = 596 kPa # When p = 150 kPa, m = 0.411 kg>s 1397

13–42. The large tank contains air at an absolute pressure of 600 kPa and temperature of 70°C. The nozzle has a throat diameter of 20 mm and an exit diameter of 50 mm. Determine the absolute pressure within the connected pipe, and the corresponding mass flow through the pipe, when the nozzle chokes and maintains isentropic supersonic flow within the divergent portion of the nozzle.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 70°) K = 343 K and p0 = 600 kPa. Since the nozzle is required to choke, M = 1 at the throat section. Here, p(0.025 m)2 A = ¢ ≤ = 6.25 * A p(0.01 m)2 A For air, k = 1.40 and R = 286.9 J>kg # K. With * = 6.25 and choosing M 7 1 A since the flow at the exit plane is required to be supersonic, we get M = 3.4114 Using this result, p = 0.01488; p0

p = (0.01488)(600 kPa) = 8.9276 kPa = 8.93 kPa

T = 0.3005; T0

T = 0.3005(343 K) = 103.08 K

Ans.

Using the universal gas law, p = rRT;

8.9276 ( 103 )

N = r ( 286.9 J>kg # K ) (103.08 K) m2 r = 0.3019 kg>m3

The velocity of the flow at the exit plane is V = M 2kRT = 3.41142(1.40) ( 286.9 J>kg # K ) (103.08 K) Finally,

= 694.14 m>s

# m = rVA = ( 0.3019 kg>m3 )( 694.14 m>s ) 3 p(0.025 m)2 4 = 0.411 kg>s

Ans.

Ans: p = 8.93 kPa # m = 0.411 kg>s 1398

13–43. The absolute pressure is 400 kPa and the temperature is 20°C in the large tank. If the pressure at the entrance A of the nozzle is 300 kPa, determine the mass flow out of the tank through the exit of the nozzle.

A

40 mm

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 20°) K = 293 K and p0 = 400 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

p0 = p¢1 +

k - 1 2 (k - 1) M ≤ 2 1.4

400 kPa = 300 kPac 1 + a M = 0.6545

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 6 1 (subsonic) at A is possible. Using this result, T0 = T ¢1 +

k - 1 2 M ≤ 2

293 K = T c 1 + a T = 269.88

1.4 - 1 b(0.6545)2 d 2

Using the universal gas law, p = rRT;

300 ( 103 )

N = r ( 286.9 J>kg # K ) (269.88 K) m2 r = 3.8745 kg>m3

The velocity of the flow at A is V = M 2kRT = (0.6545) 21.40 ( 286.9 J>kg # K ) (269.88 K) Finally,

= 215.48 m>s

# m = rVA = ( 3.8745 kg>m3 )( 215.48 m>s ) 3 p(0.02 m)2 4 = 1.05 kg>s

Ans.

Ans: 1.05 kg>s 1399

*13–44. Atmospheric air at an absolute pressure of 103 kPa and temperature of 20°C flows through the converging nozzle into the tank where the absolute pressure at A is 30  kPa. Determine the mass flow into the tank.

A

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the atmosphere; i.e., T0 = (273 + 20°) K = 293 K and p0 = 103 kPa. For air, k = 1.40 and R = 286.9 J>kg # K (Appendix A). k

p0 = p¢1 +

k - 1 2 (k - 1) M ≤ 2 1.4

103 kPa = 30 kPac 1 + a M = 1.4535

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 7 1 (supersonic) is not possible at the exit plane. Since the nozzle will be choked at the exit plane, an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. T0 = T ¢1 +

k - 1 2 M ≤ 2

293 K = T * c 1 + a T * = 244.17 K

1.4 - 1 b(1)2 d 2 k

p0 = p a1 +

k - 1 2 (k - 1) M b 2

103 kPa = p* c 1 + a p* = 54.4130 kPa

1.4

1.4 - 1 1.4 - 1 b 1 12 2d 2

Using the universal gas law, p* = r*RT *;

54.4130 ( 103 )

N = r* ( 286.9 J>kg # K ) (244.17 K) m2

r* = 0.7768 kg>m3 The velocity of the flow at the exit plane can be determined from V * = M* 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (244.17 K) = 313.16 m>s

Finally, # m = r*V *A = ( 0.7768 kg>m3 )( 313.16 m>s ) 3 p(0.02 m)2 4 = 0.306 kg>s

1400

Ans.

40 mm

13–45. Air exits a large tank through a converging nozzle having an exit diameter of 20 mm. If the temperature of the air in the tank is 35°C and the absolute pressure in the tank is 600 kPa, determine the velocity of the air as it exits the nozzle. The absolute pressure outside the tank is 101.3 kPa.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 35°) K = 308 K and p0 = 600 kPa. For air k = 1.40 and R = 286.9 J>kg # K (Appendix A). k

p0 = p a1 +

k - 1 2 (k - 1) M b 2

600 kPa = 101.3 kPac 1 + a M = 1.8198

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 7 1 (supersonic) is not possible at the exit plane. Since the nozzle will be choked at the exit plane, an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. T0 = T ¢1 +

k - 1 2 M ≤ 2

308 K = T * c 1 + a T * = 256.67 K

1.4 - 1 b(1)2 d 2

The velocity of the flow at the exit plane is V * = M 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (256.67 K) = 321 m>s

Ans.

Ans: 321 m>s 1401

13–46. Air exits a large tank through a converging nozzle having an exit diameter of 20 mm. If the temperature of the air in the tank is 35°C and the absolute pressure in the tank is 150 kPa, determine the mass flow of the air as it exits the nozzle. The absolute pressure outside the tank is 101.3 kPa.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those of the tank; i.e., T0 = (273 + 35°) K = 308 K and p0 = 150 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

p0 = p¢1 +

k - 1 2 (k - 1) M ≤ 2 1.4

1.4 - 1 1.4 - 1 150 kPa = 101.3 kPac 1 + a bM 2 d 2

M = 0.7704

The flow with M 6 1 (subsonic) at the exit plane is possible. Using this result, T0 = T ¢1 +

k - 1 2 M ≤ 2

308 K = T c 1 + a T = 275.32 K

1.4 - 1 b(0.7704)2 d 2

Using the universal gas law, p = rRT;

101.3 ( 103 )

N = r ( 286.9 J>kg # K ) (275.32 K) m2 r = 1.2824 kg>m3

The velocity of the flow at the exit plane is V = M 2kRT = (0.7704) 21.40 ( 286.9 J>kg # K ) (275.32 K) = 256.18 m>s

Finally,

# m = rVA = ( 1.2824 kg>m3 )( 256.18 m>s ) 3 p(0.01 m)2 4 = 0.103 kg>s

Ans.

Ans: 0.103 kg>s 1402

13–47. Nitrogen is contained in the large tank under an absolute pressure of 20 psi and at a temperature of 25°F. If the absolute outside pressure is 14.7 psi, determine the mass flow from the nozzle. The throat has a diameter of 0.25 in.

0.25 in.

SOLUTION Nitrogen is considered to be compressible. The stagnation temperature and pressure are of those of the tank; i.e., T0 = (460 + 25°F) R = 485 R and p0 = 20 psi. For nitrogen, k = 1.40 and R = 1775 ft # lb>slug # R.

k

p0 = p¢1 +

k - 1 2 (k - 1) M ≤ 2 1.4

20 psi = (14.7 psi) c 1 + a M = 0.6781

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 6 1 (subsonic) at the exit plane is possible. Using this result, T0 = T ¢1 +

k - 1 2 M ≤ 2

485 R = T c 1 + a T = 444.16 R

1.4 - 1 b(0.6781)2 d 2

Using the universal gas law, p = rRT;

a14.7

lb 12 in 2 ba b = r ( 1775 ft # lb>slug # R ) (444.16 R) 1 ft in2 r = 0.002685 slug>ft 3

The velocity of the flow at the exit plane is V = M 2kRT = (0.6781) 21.40 ( 1775 ft # lb>slug # R ) (444.16 R) = 712.36 ft>s

Finally, 2 0.125 # m = rVA = ( 0.002685 slug>ft 3 )( 712.36 ft>s ) £ pa ft b § 12

= 0.652 ( 10-3 ) slug>s

1403

Ans.

Ans: 0.652 1 10-3 2 slug>s

*13–48. Nitrogen is contained in the large tank under an absolute pressure of 80 psi and at a temperature of 25°F. If the absolute outside pressure is 14.7 psi, determine the mass flow from the nozzle. The throat has a diameter of 0.25 in.

0.25 in.

SOLUTION Nitrogen is considered to be compressible. The stagnation temperature and pressure are of the tank; i.e., T0 = (460 + 25°F) R = 485 R and p0 = 80 psi. For nitrogen, k = 1.40 and R = 1775 ft # lb>slug # R. k

p0 = p¢1 +

k - 1 2 (k - 1) M ≤ 2 1.4

80 psi = (14.7 psi) c 1 + a M = 1.7644

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 7 1 (supersonic) is not possible at the exit plane since the nozzle will be choked and an expansion shock wave forms thereafter. Thus, at the exit plane, M = 1. T0 = T ¢1 +

k - 1 2 M ≤ 2

485 R = T * c 1 + a T * = 404.17 R

1.4 - 1 b(1)2 d 2 k

p0 = pa1 +

k - 1 2 (k - 1) M b 2

1.4

1.4 - 1 1.4 - 1 80 psi = p c 1 + a b ( 12 ) d 2

*

p* = 42.2625 psi

Using the universal gas law, p* = r*RT *;

a42.2625

lb 12 in 2 ba b = r* ( 1775 ft # lb>slug # R ) (404.17 R) 1 ft in2

r* = 0.008483 slug>ft 3

The velocity of the flow at the exit plane is V * = M 2kRT * = (1) 21.40 ( 1775 ft # lb>slug # R ) (404.17 R) = 1002.17 ft>s

Finally, 2 0.125 # m = r*V *A = ( 0.008483 slug>ft 3 )( 1002.17 ft>s ) £ pa ft b § 12

= 0.00290 slug>s

1404

Ans.

13–49. If the fuel mixture within the chamber of the rocket is under an absolute pressure of 1.30 MPa, determine the Mach number of the exhaust if the area ratio of the exit to the throat is 2.5. Assume that fully expanded supersonic flow occurs. Take k = 1.40 for the fuel mixture. The atmosphere has a pressure of 101.3 kPa.

SOLUTION The mixture is compressible. Steady flow occurs relative to the rocket. The stagnation pressure that is in the chamber, i.e., p0 = 1.30 MPa. When the nozzle is choked, M = 1 at the throat section. k

k - 1 2 (k - 1) p0 = p¢1 + M ≤ 2 1.4

1.30 ( 10

6

)

1.4 - 1 1.4 - 1 Pa = p c 1 + a b ( 12 ) d 2

*

p* = 686.8 kPa

Since the back pressure p = 101.3 kPa 6 p*, the nozzle will choke. Therefore to A determine M with * = 2.5. (M 7 1) since the flow is required to be supersonic at A the exit plane. Ans.

M = 2.44

Ans: 2.44 1405

13–50. The large tank contains air at a gage pressure of 170  lb > in2 and temperature of 120°F. The throat of the nozzle has a diameter of 0.35 in., and the exit diameter is 1 in. Determine the absolute pressure in the pipe required to produce a jet that has isentropic supersonic flow through the pipe. Also, what is the Mach number of this flow? The atmospheric pressure is 14.7 psi.

SOLUTION The air is considered to be compressible. The ratio of the cross-sectional area of the exit plane to that of the throat is p (1 in.)2 4 A = = 8.1633 p A* (0.35 in.)2 4 For isentropic supersonic flow to occur through the divergent portion of the nozzle, M 7 1 at the exit plane and the back pressure is at p3. Since k = 1.40 for air, we can A solve this problem using Appendix B. With * = 8.163, (choosing M 7 1) gives A Ans.

M = 3.6992 = 3.70 Then using this result, p4 = 0.009914; p4 = 0.009914(170 psi + 14.7 psi) p0 = 1.83 psi

Ans.

Ans: M = 3.70 p4 = 1.83 psi 1406

13–51. The large tank contains air at a gage pressure of 170 lb>in2 and temperature of 120°F. The throat of the nozzle has a diameter of 0.35 in., and the exit diameter is 1  in. Determine the absolute pressure in the pipe required to choke the nozzle and also maintain isentropic subsonic flow through the pipe. Also, what is the velocity of the flow through the pipe for this condition? The atmospheric pressure is 14.7 psi.

SOLUTION The air is considered to be compressible. The ratio of the cross-sectional area of the exit plane to that of the throat is p (1 in.)2 A 4 = = 8.1633 p A* (0.35 in.)2 4 For isentropic supersonic flow to occur through the divergent portion of the nozzle, M 6 1 at the exit plane and the back pressure is at p3. Since k = 1.40 for air, A = 8.163, for M 6 1, A* M = 0.07111 Then using this result, p3 = 0.9965; p0

p3 = 0.9965(170 psi + 14.7 psi) Ans.

= 184.05 psi = 184 psi T3 = 0.9990; T0

T3 = 0.9990(460 + 120 F) R = 579.41 R

Thus, the velocity of flow at the exit plane can be determined with R = 1716 ft # lb>slug # R for air V3 = M 1kRT = 0.0711121.40 1 1716 ft # lb>slug # R 2 (579.41 R) = 83.9 ft>s

Ans.

Ans: p3 = 184 psi V3 = 83.9 ft>s 1407

*13–52. The diameter of the exit of a converging nozzle is 50 mm. If its entrance is connected to a large tank containing air at an absolute pressure of 500 kPa and temperature of 125°C, determine the mass flow through the nozzle. The ambient air is at an absolute pressure of 101.3 kPa.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 125°C) = 398 K and p0 = 500 kPa. For air, k = 1.40 and R = 286.9 J>kg # K (Appendix A).

k

k - 1 2 k-1 M b p0 = p a1 + 2

1.4

1.4 - 1 1.4 - 1 500 kPa = (101.3 kPa)c 1 + a bM 2 d 2

M = 1.7000

The flow with M 7 1 (supersonic) is not possible at the exit plane. Since the nozzle will be choked at the exit plane, an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. T0 = T a1 +

k - 1 2 M b 2

398 K = T * c 1 + a T * = 331.67 K

1.4 - 1 b(1)2 d 2 k

p0 = p a1 +

k - 1 2 (k - 1) M b 2

500 kPa = p* c 1 + a p* = 264.14 kPa

1.4

1.4 - 1 1.4 - 1 b 1 12 2 d 2

Using the universal gas law, 264.14 1 103 2

p* = r* RT *;

N = r* 1 286.9 J>kg # K 2 (331.67 K) m2

r* = 2.7759 kg>m3

The velocity of the flow at the exit plane is V * = M 2kRT * = (1) 21.40 1 286.9 J>kg # K 2 (331.67 K) = 364.99 m>s

Finally, # m = r* V * A =

1 2.7759 kg>m3 2 1 364.99 m>s 2 3 p(0.025 m)2 4

= 1.99 kg>s

1408

Ans.

13–53. The diameter of the exit of a converging nozzle is 50 mm. If its entrance is connected to a large tank containing air at an absolute pressure of 180 kPa and temperature of 125°C, determine the mass flow from the tank. The ambient air is at an absolute pressure of 101.3 kPa.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 125°) K = 398 K and p0 = 180 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

p0 = p a1 +

k - 1 2 (k - 1) M b 2

180 kPa = 101.3 kPac 1 + a M = 0.9447

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 6 1 (subsonic) at the exit plane is possible. Using this result, T0 = T a1 +

k - 1 2 M b 2

398 K = T c 1 + a T = 337.72 K

1.4 - 1 b(0.9447)2 d 2

Using the universal gas law, p = rRT:

101.3 1 103 2

N = r 1 286.9 J>kg # K 2 (337.72 K) m2

r = 1.0455 kg>m3

The velocity of the flow at the exit plane is V = M 1kRT = (0.9447) 21.40 1 286.9 J>kg # K 2 (337.72 K) = 347.95 m>s

Finally, # m = rVA =

1 1.0455 kg>m3 2 1 347.95 m>s 2 3 p(0.025 m)2 4

= 0.714 kg>s

Ans.

Ans: 0.714 kg>s 1409

13–54. Air flows at VA = 100 m>s at 1200 K and has an absolute pressure of pA = 6.25 MPa. Determine the diameter d of the pipe at B so that M = 1 at B.

VB

100 m/s B 100 mm

A

d

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K. The Mach number at the entrance plane A is MA =

100 m>s V = = 0.1440 1kRT 21.40 1 286.9 J>kg # K 2 (1200 K)

Using the universal gas law, pA = rA RA TA; At the entrance plane A, T0 = TAa1 +

N = rA 1 286.9 J>kg # K 2 (1200 K) m2 = 18.1538 kg>m3

6.25 1 106 2 rA

k - 1 M A2b 2

T0 = (1200 K) c 1 + a T0 = 1204.98 K

1.4 - 1 b(0.1440)2 d 2 1

r0 = rAa1 + r0 =

(k - 1) k - 1 M A2 b 2

1 18.1538 kg>m3 2 c 1

r0 = 18.3427 kg>m3

+ a

1

1.4 - 1 1.4 - 1 b(0.1440)2 d 2

At the exit plane B with MB = 1 since the transition is required to choke, T0 = TB a1 +

k - 1 MB2 b 2

1204.98 K = TB c 1 + a TB = 1004.15 K

1.4 - 1 b(1)2 d 2 1

r0 = rB a1 +

(k - 1) k - 1 M B2 b 2

18.3427 kg>m3 = rB c 1 + a rB = 11.6282 kg>m3

1

1.4 - 1 1.4 - 1 b(1)2 d 2

The velocity of the flow at exit plane B can be determined using VB = M B 1kRTB = (1) 21.40 1 286.9 J>kg # K 2 (1004.15 K) = 635.08 m>s

Finally, the continuity equation requires 0 rdV + V # dA = 0 0t Lcv Lcs

0 + rA VA AA + rB VB AB = 0 - 1 18.1538 kg>m3 2 1 100 m>s 2 3 p(0.05 m)2 4 + 1 11.6282 kg>m3 2 1 635.08 m>s 2 a Ans.

d = 0.04958 m = 49.6 mm

1410

p 2 d b = 0 4

Ans: 49.6 mm

13–55. Air flows at VA = 100 m>s at 1200 K and has an absolute pressure of pA = 6.25 MPa. Determine the diameter d of the pipe at B so that M = 0.8 at B.

VB

100 m/s B 100 mm

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K. The Mach number at the entrance plane A is MA =

100 m>s V = = 0.1440 1kRT 21.40 1 286.9 J>kg # K 2 (1200 K)

Using the universal gas law, pA = rA RA TA;

6.25 1 106 2

N = rA 1 286.9 J>kg # K 2 (1200 K) m2

rA = 18.1538 kg>m3

At the entrance plane A, T0 = TAa1 +

k - 1 M A2b 2

T0 = (1200 K) c 1 + a T0 = 1204.98 K

1.4 - 1 b(0.1440)2 d 2 1

(k - 1) k - 1 r0 = rAa1 + M A2 b 2

r0 =

1 18.1538 kg>m3 2 c 1

r0 = 18.3427 kg>m3

+ a

1

1.4 - 1 1.4 - 1 b(0.1440)2 d 2

At the exit plane B with M B = 0.8, T0 = TB a1 +

k - 1 MB2 b 2

1204.98 K = TB c 1 + a TB = 1068.24 K

1.4 - 1 b(0.8)2 d 2 1

r0 = rB a1 +

(k - 1) k - 1 M B2 b 2

18.3427 kg>m3 = rB c 1 + a rB = 13.5735 kg>m3

1

1.4 - 1 1.4 - 1 b(0.8)2 d 2

The velocity of the flow at exit plane B is VB = M B 1 kRTB = 0.821.40 1 286.9 J>kg # K 2 (1068.24 K) = 524.03 m>s

1411

A

d

13–55. Continued

Finally, the continuity equation requires 0 r dV + rV # dA = 0 0t Lcv Lcs 0 + rA VA AA + rB VB AB = 0 - 1 18.1538 kg>m3 2 1 100 m>s 2 3 p(0.05 m)2 4 + 1 13.5735 kg>m3 2 1 524.03 m>s 2 a Ans.

d = 0.05052 m = 50.5 mm

p 2 d b = 0 4

Ans: 50.5 mm 1412

*13–56. Air flows at 200 m>s through the pipe. Determine the Mach number of the flow and the mass flow if the temperature is 500 K and the absolute stagnation pressure is 200 kPa. Assume isentropic flow.

0.3 m

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K. The Mach number is M =

200 m>s V = 1kRT 21.40 1 286.9 J>kg # K 2 (500 K)

Ans.

= 0.4463 = 0.446 With p0 = 200 kPa,

k

p0 = p0 a1 +

k - 1 2 (k - 1) M b 2

200 kPa = p0 c 1 + a p0 = 174.44 kPa

1.4

14. - 1 1.4 - 1 b(0.4463)2 d 2

Using the universal gas law, p = rRT;

174.44 1 103 2

N = r 1 286.9 J>kg # K 2 (500 K) m2 r = 1.2161 kg>m3

Finally, # m = rAV =

1 1.2161 kg>m3 2 1 200 m>s 2 3 p(0.15 m)2 4

= 17.2 kg>s

1413

Ans.

13–57. Air flows at 200 m> s through the pipe. Determine the pressure within the flow if the temperature is 400 K and  the absolute stagnation pressure is 280 kPa. Assume isentropic flow.

0.3 m

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K. The Mach number is M =

200 m>s V = 1kRT 21.40 1 286.9 J>kg # K 2 (400 K)

= 0.4990 With p0 = 280 kPa,

k

k - 1 2 (k - 1) p0 = p a1 + M b 2 280 kPa = pc 1 + a

1.4

1.4 - 1 1.4 - 1 b(0.4990)2 d 2

Ans.

p = 236.21 kPa = 236 kPa

Ans: 236 kPa 1414

13–58. The converging–diverging nozzle at the end of a supersonic jet engine is to be designed to operate efficiently when the absolute outside air pressure is 25 kPa. If the absolute stagnation pressure within the engine is 400 kPa and the stagnation temperature is 1200 K, determine the exit plane diameter and the throat diameter for the nozzle if the mass flow is 15 kg>s. Take k = 1.40 and R = 256 J>kg # K.

SOLUTION The air is considered to be compressible. There is steady relative flow. The absolute stagnation temperature and pressure are T0 = 1200 K and p0 = 400 kPa. Here, the nozzle is required to operate under choking conditions yet maintain the isentropic flow at the exit plane to have the maximum efficiency. When the nozzle chokes, M = 1 at the throat. T0 = T a1 +

k - 1 2 M b 2

1200 K = T * c 1 + a T * = 1000 K

1.4 - 1 b(1)2 d 2 k

p0 = p a1 +

k - 1 2 (k - 1) M b 2

400 kPa = p* c 1 + a p* = 211.31 kPa

1.4

1.4 - 1 1.4 - 1 b 1 12 2 d 2

Using the universal gas law, 211.31 1 103 2

p* = r* RT *;

N = r* 1 256 J>kg # K 2 (1000 K) m2

r* = 0.8254 kg>m3

The velocity of the flow at the throat is V * = M * 2kRT * = (1) 21.40 1 256 J>kg # K 2 (1000 K) = 598.67 m>s

Thus, the mass flow rate is # m = r*V *A* 15 kg>s =

1 0.8254 kg>m3 2 1 598.67 m>s 2 a

d t = 0.1966 m = 197 mm

p 2 d b 4 t

Ans.

Also, the pressure at the exit plane must be equal to the back pressure, i.e., pe = 25 kPa. k

p0 = pe a1 +

(k - 1) k - 1 M e2 b 2

400 kPa = (25 kPa)c 1 + a M e = 2.4578

1.4

1.4 - 1 1.4 - 1 bM e2 d 2

1415

13–58. Continued

Since M e 7 1, the flow at the exit plane is isentropic supersonic flow. Using the result of Me, k+1

Ae A*

=

1 ≥ Me

1 +

p 2 de 4 p (0.1966 m)2 4

2(k - 1) k - 1 Me2 2 ¥ k + 1 2 1.4 + 1

=

1 ≥ 2.4578

1 + a

d e = 0.3130 m = 313 mm

2(1.4 - 1) 1.4 - 1 b(2.4578)2 2 ¥ 1.4 + 1 b a 2

Ans.

Ans: d t = 197 mm d e = 313 mm 1416

13–59. Natural gas (methane) has an absolute pressure of 400 kPa and flows through the pipe at A at M = 0.1. Determine the diameter of the throat of the nozzle so that M = 1 at the throat. Also, what are the stagnation pressure, the pressure at the throat, and the subsonic and supersonic Mach numbers of the flow through pipe B?

120 mm

200 mm dt B

A C

SOLUTION Methane is considered to be compressible. The flow is steady. For methane, k = 1.31. At the entrance plane A with M A = 0.1 and pA = 400 kPa, k

k-1 k - 1 p0 = pa1 + M A2 b 2

p0 = (400 kPa) c 1 + a

1.31

1.31 - 1 1.31 - 1 b 1 0.12 2 d 2

Ans.

p0 = 402.63 kPa = 403 kPa Since the nozzle chokes,

k+1

AA A*

=

1 ≥ MA

1 + a

2(k - 1) k - 1 bM A2 2 ¥ k + 1 2 1.31 + 1

2(1.31 - 1) 1.31 - 1 p 1 + a b(0.1)2 (0.12 m)2 2 1 4 = ≥ ¥ p 2 0.1 1.31 + 1 dt 4 2

Ans.

d t = 0.04949 m = 49.5 mm Using the result of p0 at the throat where M t = 1, k

p0 = pt a1 +

k-1 k - 1 M t2b 2

402.63 kPa = pt c 1 + a

1.31

1.31 - 1 1.31 - 1 b 1 12 2 d 2

Ans.

pt = 219.00 kPa = 219 kPa

Using the result of d t, k+1

2(k + 1) k - 1 1 + M B2 AB 1 2 ¥ = ≥ MB k + 1 A* 2 1.31 + 1

2(1.31 - 1) 1.31 - 1 1 + a bMB2 2 1 = ≥ ¥ p MB 1.31 + 1 (0.04949 m)2 a b 2 2

p (0.2 m)2 4

2.4443M B0.2684 - 0.155M B2 - 1 Solving by trial and error, M B = 0.0358 6 1 (subsonic)

Ans.

M B = 4.07 7 1 (supersonic)

Ans. 1417

Ans: p0 = 403 kPa d t = 49.5 mm pt = 219 kPa M B = 0.0358 6 1 (subsonic) M B = 4.07 7 1 (supersonic)

*13–60. Air has an absolute pressure of 400 kPa and flows through the pipe at A at M = 0.5. Determine the Mach number at the throat of the nozzle where dt = 110 mm, and the Mach number in the pipe at B. Also, what are the stagnation pressure and the pressure in the pipe at B?

120 mm

200 mm dt B

A C

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40. Applying the equation or Appendix B, with M A = 0.5 and pA = 400 kPa at the entrance plane A, pA = 0.8430; p0

p0 =

AA

AC

A*

= 1.3398;

400 kPa = 474.49 kPa = 474 kPa 0.8430

Ans.

p (0.11 m)2 AA AC 4 = * a b = 1.3398≥ ¥ = 1.1258 p A* A AA 2 (0.12 m) 4

AC From Appendix B with * = 1.1258, choose M C 6 1 (subsonic). The interpolation A of the values gives Ans.

M C = 0.6608 = 0.661 Also, p (0.2 m)2 AA AB 4 = * a b = 1.3398≥ ¥ = 3.7218 p A* A AA 2 (0.12 m) 4

AB

Since the nozzle will not choke, M B 6 1 (subsonic). From Appendix B and interpolating between the values in the table gives Ans.

M B = 0.1578 = 0.158 Using this result or Appendix B, pB = 0.9828; p0

pB = 0.9828(474.49 kPa) Ans.

= 466.30 kPa = 466 kPa

1418

13–61. The tank contains oxygen at a temperature of 70°C and absolute pressure of 800 kPa. If the converging nozzle at the exit has a diameter of 6 mm, determine the initial mass flow out of the tank if the outside absolute pressure is 100 kPa.

6 mm

SOLUTION Oxygen is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 70°C) = 343 K and p0 = 800 kPa. For oxygen, k = 1.40 and R = 259.8 J>kg # K. With p = 100 kPa, k

k - 1 2 k-1 p0 = pa1 + M b 2

1.4

1.4 - 1 1.4 - 1 800 kPa = (100 kPa)c 1 + a bM 2 d 2

M = 2.0143

The flow with M 7 1 (supersonic) is not possible at the exit plane. Since the nozzle will be choked at the exit plane, an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. T0 = T a1 +

k - 1 2 M b 2

343 = T * c 1 + a

T * = 285.83 K

1.4 - 1 b(1)2 d 2 k

k - 1 2 (k - 1) p0 = p a1 + M b 2 *

800 kPa = p* c 1 + a p* = 422.63 kPa

1.4

1.4 - 1 1.4 - 1 b ( 12 ) d 2

Using the universal gas law, p* = r*RT *;

422.63 ( 103 )

N = r* ( 259.8 J>kg.K ) (285.83 K) m2

p* = 5.6912 kg>m3 The velocity of the flow at the exit plane is V * = M* 2kRT * = (1) 21.40 ( 259.8 J>kg # K ) (285.83 K) = 322.43 m>s

Finally, # m = r* V * A* = ( 5.6912 kg>m3 )( 322.43 m>s ) 3 p(0.003 m)2 4 = 0.0519 kg>s

Ans.

Ans: 0.0519 kg>s 1419

13–62. The tank contains helium at a temperature of 80°C and absolute pressure of 175 kPa. If the converging nozzle at the exit has a diameter of 6 mm, determine the initial mass flow out of the tank if the outside absolute pressure is 98 kPa.

6 mm

SOLUTION Helium is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 80°C) = 353 K and p0 = 175 kPa. For helium, k = 1.66 and R = 2077 J>kg # K. With p = 98 kPa, k

p0 = pa1 +

k - 1 2 k-1 M b 2

175 kPa = (98 kPa)c 1 + a M = 0.8864

1.66

1.66 - 1 1.66 - 1 bM 2 d 2

The flow with M < 1 (subsonic) at the exit plane is possible. Using this result, T0 = T a1 +

k - 1 2 M b 2

353 K = T c 1 + a T = 280.32 K

1.66 - 1 b(0.8864)2 d 2

Using the universal gas law, p = rRT;

98 ( 103 )

N = r ( 2077 J>kg # K ) (280.32 K) m2 r = 0.1683 kg>m3

The velocity of the flow at the exit plane is V = M 2kRT = (0.8864) 21.66 ( 2077 J>kg # K ) (280.32 K) = 871.40 m>s

Finally, # m = rVA = ( 0.1683 kg>m3 )( 871.40 m>s ) 3 p(0.003 m)2 4 = 4.15 ( 10 - 3 ) kg>s

1420

Ans.

Ans: 4.15 1 10-3 2 kg>s

13–63. The large tank contains air at 250 K under an absolute pressure of 1.20 MPa. When the valve is opened, the nozzle chokes. The outside absolute atmospheric pressure is 101.3 kPa. Determine the mass flow from the tank. The nozzle has an exit diameter of 40 mm and a throat diameter of 20 mm.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 250 K and p0 = 1.20 MPa. When the nozzle chokes M = 1 at the throat. For air, k = 1.40 and R = 286.9 J>kg # K. At the throat, k

p0 = pa1 +

k - 1 2 k-1 M b 2

1.20 MPa = p* c 1 + a

1.4

1.4 - 1 1.4 - 1 b ( 12 ) d 2

p* = 0.6339 MPa = 633.94 kPa

Since p = 101.3 6 p*, the nozzle will indeed choke. T0 = T a1 +

k - 1 2 M b 2

250 K = T * c 1 + a T * = 208.33 K

1.4 - 1 b(1)2 d 2

Using the universal gas law, p* = r*RT *;

633.94 ( 103 )

N = r* ( 286.9 J>kg # K)(208.33 K) m2

r* = 10.6061 kg>m3 The velocity of the flow at the throat is V * = M* 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (208.33 K) = 289.27 m>s

Finally, # m = r*V *A* = ( 10.6061 kg>m3 )( 289.27 m>s ) 3 p(0.01 m)2 4

Ans.

= 0.964 kg>s

Ans: 0.964 kg>s 1421

*13–64. The large tank contains air at 250 K under an absolute pressure of 150 kPa. When the valve is opened, determine if the nozzle is choked. The outside atmospheric pressure is 90 kPa. Determine the mass flow from the tank. Assume the flow is isentropic. The nozzle has an exit diameter of 40 mm and a throat diameter of 20 mm.

SOLUTION The air is considered to be compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 250 K and p0 = 150 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. Assuming that the p(0.02 m)2 A nozzle chokes, * = = 4. Then, using the equations or Appendix B A p(0.01 m)2 gives M = 0.1465 (subsonic) pB = 0.9851; r0

pB = 0.9851(150 kPa) = 147.77 kPa

Since p = 90 6 pB, the nozzle will choke.

Ans.

Therefore, M = 1 at the throat. Using Appendix B, T* = 0.8333; T0

T * = 0.8333(250 kPa) = 208.33 K

p* = 0.5283 p* = 0.5283(150 kPa) = 79.24233 kPa p0 Using the universal gas law, p* = r*RT *;

79.24233 ( 103 )

N = r* ( 286.9 J>kg # K ) (208.33 K) m2

r* = 1.3258 kg>m3 The velocity of the flow at the throat is V * = M * 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (208.33 K) = 289.27 m>s

Finally,

# m = r*V *A* = ( 1.3258 kg>m3 )( 289.27 m>s ) 3 p(0.01 m)2 4 = 0.120 kg>s

1422

Ans.

13–65. Air at A flows into the nozzle at M = 0.4. Determine the Mach number at C and at B.

50 mm

70 mm

40 mm

A

C

B

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40. Enter Appendix B, with M A = 0.4, the interpolated value is AA

= 1.5901 A* Using this result, p (0.04 m)2 AA AC 4 = * a b = 1.5901≥ ¥ = 1.0177 7 1 (choking will not occur) p A* A AA (0.05 m)2 4

AC

AC

= 1.0177 and choosing M C 6 1 (subsonic), the A* interpolated value in the table gives From Appendix B with

Ans.

M C = 0.8607 = 0.861 Also, p (0.07 m)2 AA AB 4 = * a b = 1.5901 ≥ ¥ = 3.1167 p A* A AA (0.05 m)2 4

AB

Since the nozzle will not choke, M B 6 1 (subsonic). From Appendix B with AB = 3.1167 and interpolating between the values in the table gives A* Ans.

M B = 0.1897 = 0.190

Ans: M C = 0.861 M B = 0.190 1423

13–66. Air at A flows into the nozzle at M = 0.4. If pA = 125 kPa and TA = 300 K, determine the pressure at B and the velocity at B.

50 mm

70 mm

40 mm

A

C

B

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K. Enter Appendix B, with M A = 0.4, the interpolated values are AA A*

= 1.5901

Using the result of

PA = 0.8956 p0 AA A*

TA = 0.9690 T0

,

p (0.04 m)2 AA AC 4 = * a b = 1.5901≥ ¥ = 1.0177 p A* A AA (0.05 m)2 4 AC

Since

AC A*

7 1, the nozzle will not choke. Also, p (0.07 m)2 AA AB 4 = * a b = 1.5901≥ ¥ = 3.1167 p A* A AA (0.05 m)2 4 AB

Since the nozzle will not choke, M B 6 1 (subsonic). From Appendix B with AB = 3.1167 and interpolating between the values in the table gives A* M B = 0.1897 Using this result, Appendix B gives TB = 0.9929 T0

pB = 0.9752 p0

Therefore, TB >T0 TB 0.9929 = = = 1.0246; TB = 1.0246(300 K) = 307.39 K TA TA >T0 0.9690 pB >p0 pB 0.9752 = = = 1.0889; pB = 1.0889(125 kPa) pA pA >p0 0.8956

= 136.11 kPa = 136 kPa Ans.

The velocity of the flow at B is

VB = M B 2kRTB = 0.189721.40 ( 286.9 J>kg # K ) (307.39 K) = 66.7 m>s

Ans.

Ans: pB = 136 kPa VB = 66.7 m>s 1424

13–67. Air flows isentropically into the nozzle at MA = 0.2 and out at MB = 2. If the diameter of the nozzle at  A is 30  mm, determine the diameter of the throat and the diameter at B. Also, if the absolute pressure at A is 300 kPa, determine the stagnation pressure and the pressure at B.

d

30 mm

dt

A

B

SOLUTION The air is considered to be compressible. The flow is steady. The nozzle is required to choke. For air, k = 1.40. k+1

AA *

A

=

1 = ≥ MA

1 +

2(k - 1) k - 1 M A2 2 ¥ k + 1 2 1.4 + 1

2(1.4 - 1) 1.4 - 1 p 1 + a b(0.2)2 (0.03 m)2 2 4 1 ¥ = ≥ p 2 0.2 1.4 + 1 dt 4 2

Ans.

d t = 0.01743 m = 17.4 mm Using this result, k+1

AB A*

=

1 = ≥ MB

1 +

2(k - 1) k - 1 M B2 2 ¥ k + 1 2

1.4 + 1

p 2 dB 4 p (0.01743 m) 2 4

=

1 ≥ 2

2(1.4 - 1) 1.4 - 1 b(2)2 2 ¥ 1.4 + 1 2

1 + a

Ans.

d B = 0.02264 m = 22.6 mm At plane A, k

k-1 k - 1 p0 = pAa1 + M A2 b 2

p0 = ( 300 kPa ) c 1 + a

1.4

1.4 - 1 1.4 - 1 b ( 0.22 ) d 2

Ans.

= 308.48 kPa = 308 kPa

Again, at plane B using this result, k

p0 = pBa1 +

k-1 k - 1 M B2 b 2

308.48 kPa = pB c 1 + a

1.4

1.4 - 1 1.4 - 1 b ( 22 ) d 2

Ans.

pB = 39.43 kPa = 39.4 kPa

1425

Ans: d t = 17.4 mm d B = 22.6 mm p0 = 308 kPa pB = 39.4 kPa

13–68. Air flows isentropically into the nozzle at MA = 0.2 and out at MB = 2. If the diameter of the nozzle at A is 30 mm, determine the diameter of the throat and the diameter at B. Also, if the temperature at A is 300 K, determine the stagnation temperature and the temperature at B.

The air is considered to be compressible. The flow is steady. The nozzle is required to choke. For air, k = 1.40. k+1

AA *

A

1 ≥ MA

=

2(k - 1) k - 1 M A2 2 ¥ k + 1 2

1.4 + 1

2(1.4 - 1) 1.4 - 1 p 1 + a b(0.2)2 (0.03 m)2 2 4 1 = ≥ ¥ p 2 2 1.4 + 1 dt 4 2

Ans.

d t = 0.01743 m = 17.4 mm Using this result, k+1

AB A*

=

1 = ≥ MB

1 +

2(k - 1) k - 1 M B2 2 ¥ k + 1 2 1.4 + 1

p 2 dB 4 p (0.01743 m) 2 4

=

1 ≥ 2

2(1.4 - 1) 1.4 - 1 b(2)2 2 ¥ 1.4 + 1 2

1 + a

Ans.

d B = 0.02264 m = 22.6 mm At plane A, T0 = TAa1 +

k - 1 M A2 b 2

T0 = ( 300 K ) c 1 + a

1.4 - 1 b ( 0.2 ) 2 d 2

Ans.

= 302.4 K = 302 K

Again, at plane B using this result, T0 = TB a1 +

k - 1 M B2 b 2

302.4 = TB c 1 + a TB = 168 K

dt

A

SOLUTION

1 +

d

30 mm

1.4 - 1 b ( 2 )2 d 2

Ans.

1426

B

13–69. Air flows through a pipe having a diameter of 50  mm. Determine the mass flow if the stagnation temperature of the air is 20°C, the absolute pressure is 300 kPa, and the stagnation pressure is 375 kPa.

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are T0 = (273 + 20°) K = 293 K and p0 = 375 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

p0 = pa1 +

k - 1 2 (k - 1) M b 2

375 kPa = 300 kPa c 1 + a M = 0.5737

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

Using this result, T0 = T a1 +

k - 1 2 M b 2

293 K = T c 1 + a T = 274.90 K

1.4 - 1 b(0.5737)2 d 2

Using the universal gas law, p = rRT;

300 ( 103 )

N = r ( 286.9 J>kg # K )( 274.90 K ) m2 r = 3.8037 kg>m3

The velocity of the flow is V = M 2kRT = (0.5737) 21.40 ( 286.9 J>kg # K ) (274.90 K) = 190.64 m>s

Finally, # m = rVA = ( 3.8037 kg>m3 )( 190.64 m>s ) 3 p(0.025 m2) 4 = 1.42 kg>s

Ans.

Ans: 1.42 kg>s 1427

13–70. Air at a temperature of 25°C and standard atmospheric pressure of 101.3 kPa flows through the nozzle into the pipe where the absolute internal pressure is 80 kPa. Determine the mass flow into the pipe. The nozzle has a throat diameter of 10 mm.

10 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of the atmosphere; i.e., T0 = (273 + 25°C) = 298 K and p0 = 101.3 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. At the exit plane, k

k - 1 2 k-1 M b p0 = pa1 + 2

1.4

1.4 - 1 1.4 - 1 101.3 kPa = (80 kPa)c 1 + a bM 2 d 2

M = 0.5906

The flow with M 6 1 (subsonic) at the exit plane is possible. At the exit plane, T0 = T a1 +

k - 1 2 M b 2

298 K = T c 1 + a T = 278.56 K

1.4 - 1 b(0.5906)2 d 2

Using the universal gas law, p = rRT;

80 ( 103 )

N = r ( 286.9 J>kg # K ) (278.56 K) m2 r = 1.001 kg>m3

The velocity of the flow is V = M 2kRT = (0.5906) 21.40 ( 286.9 J>kg # K ) (278.56 K) = 197.57 m>s

Finally, the mass flow rate is # m = rVA = ( 1.001 kg>m3 )( 197.57 m>s ) 3 p(0.005 m)2 4 = 0.01553 kg>s = 0.0155 kg>s

Ans.

Ans: 0.0155 kg>s 1428

13–71. Air at a temperature of 25°C and standard atmospheric pressure of 101.3 kPa flows through the nozzle into the pipe where the absolute internal pressure is 30 kPa. Determine the mass flow into the pipe. The nozzle has a throat diameter of 10 mm.

10 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of that in the atmosphere; i.e., T0 = (273 + 25°C) = 298 K and p0 = 101.3 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

p0 = pa1 +

k - 1 2 k-1 M b 2

101.3 kPa = (30 kPa)c 1 + a M = 1.4418

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 7 1 (supersonic) is not possible since the nozzle will be choked at the exit plane and an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. T0 = T a1 +

k - 1 2 M b 2

298 K = T * c 1 + a T * = 248.33 K

1.4 - 1 b(1)2 d 2 k

p0 = pa1 +

k - 1 2 k-1 M b 2

101.3 kPa = p* c 1 + a p* = 53.5149 kPa

1.4

1.4 - 1 1.4 - 1 b ( 12 ) d 2

Using the universal gas law, p* = r*RT *;

53.5149 ( 103 )

N = r* ( 286.9 J>kg # K ) (248.33 K) m2

r* = 0.7511 kg>m3 The velocity of the flow at the exit plane is V * = M * 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (248.33 K) = 315.83 m>s

Finally, the mass flow rate is . m = r*V *A* = ( 0.7511 kg>m3 )( 315.83 m>s ) 3 p(0.005 m)2 4 = 0.0186 kg>s

Ans. Ans: 0.0186 kg>s

1429

13–72. The large tank contains air at an absolute pressure of 800 kPa and a temperature of 150°C. If the diameter at the end of the converging nozzle is 20 mm, determine the mass flow out of the tank where the standard atmospheric pressure is 101.3 kPa.

20 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of those in the tank; i.e., T0 = (273 + 150°C) = 423 K and p0 = 800 kPa. For air, k = 1.40 and R = 286.9 J>kg # K. k

p0 = pa1 +

k - 1 2 k-1 M b 2

800 kPa = (101.3 kPa) c 1 + a M = 2.006

1.4

1.4 - 1 1.4 - 1 bM 2 d 2

The flow with M 7 1 (supersonic) is not possible at the exit plane since the nozzle will be choked at the exit plane and an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. k - 1 2 M b 2

T0 = T a1 +

423 K = T * c 1 + a T * = 352.5 K

1.4 - 1 b(1)2 d 2 k

k - 1 2 k-1 p0 = p a1 + M b 2 *

1.4

1.4 - 1 2 1.4 - 1 b(1 ) d 800 kPa = p c 1 + a 2 *

p* = 422.63 kPa

Using the universal gas law, p* = r*RT *;

422.63 ( 103 )

N = r* ( 286.9 J>kg # K ) (352.5 K) m2

r* = 4.1789 kg>m3 The velocity of the flow is V * = M * 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (352.5 K) = 376.28 m>s

Finally, the mass flow rate is # m = r*V *A* = ( 4.1789 kg>m3 )( 376.28 m>s ) 3 p(0.01 m)2 4 = 0.4940 kg>s = 0.494 kg>s

1430

Ans.

13–73. A large reservoir contains air at a temperature of T = 20°C and absolute pressure of p = 300 kPa. The air flows isentropically through the nozzle and then through the 1.5-m-long, 50-mm-diameter pipe having an average friction factor of 0.03. Determine the mass flow through the pipe and the corresponding velocity, pressure, and temperature at the inlet 1 and outlet 2 if the flow is choked at section 2.

1

50 mm

2

1.5 m

SOLUTION The flow through the nozzle is considered isentropic and the flow through the pipe is considered Fanno flow since friction is involved, so that M = 1 at outlet 2. Hence Lmax = 1.5 m. f 0.03 Lmax = a b(1.5 m) = 0.9 D 0.05 m

Using this result and interpolating the values tabulated in the Fanno flow table in Appendix B, select M 1 6 1, M 1 = 0.5226

T1 T

*

= 1.1378

V1 V

*

= 0.55742

p1 p*

= 2.0414

For the isentropic flow at inlet 1 with M 1 = 0.5226, the interpolated values tabulated in the isentropic flow tables in Appendix B gives T1

( T0 ) 1

= 0.94820

p1

( p0 ) 1

= 0.83016

Here ( T0 ) 1 = (273 + 20) K = 293 K and ( p0 ) 1 = 300 kPa. T1 = 0.94820(293 K) = 277.82 K = 278 K

Ans.

p1 = 0.83016(300 kPa) = 249.05 kPa = 249 kPa

Ans.

Using the universal gas law with R r1 =

= 286.9 J>kg # K,

249.05 ( 103 ) N>m2 p1 = = 3.1246 kg>m3 = 3.12 kg>m Ans. RT1 ( 286.9 J>kg # K ) (277.82 K)

Applying, V1 = M 1 2kRT1 = (0.5226) 21.4 ( 286.9 J>kg # K ) (277.82 K) = 174.57 m>s = 175 m>s

Thus, the mass flow is . m = r1V1A1 = ( 3.1246 kg>m3 )( 174.57 m>s) 3p(0.025 m)2 4 = 1.07 kg>s

Ans.

Since

T1

T* V1 V* p1 p*

= 1.1378;

T* =

= 0.55742;

V* =

= 2.0414;

p* =

277.82 K = 244 K 1.1378

174.57 m>s 0.55742

Ans. Ans.

= 313 m>s

249.05 kPa = 122 kPa 2.0414

Ans.

1431

Ans: T1 = 278 K p1 = 249 kPa r1 = 3.12 kg>m3 # m = 1.07 kg>s T* = 244 K V* = 313 m>s p* = 122 kPa

13–74. A large reservoir contains air at a temperature of T = 20°C and absolute pressure of p = 300 kPa. The air flows isentropically through the nozzle and then through the 1.5-m-long, 50-mm-diameter pipe having an average friction factor of 0.03. Determine the stagnation temperature and pressure at outlet 2 and the change in entropy between the inlet 1 and outlet 2 if the pipe is choked at section 2.

1

50 mm

1.5 m

SOLUTION The flow through the nozzle is considered isentropic and the flow through the pipe is considered Fanno flow since friction is involved under the choked condition M = 1 at outlet 2. Hence Lmax = 1.5 m f 0.03 Lmax = a b(1.5 m) = 0.9 D 0.05 m

Using this result and performing the interpolation of the values tabulated in the Fanno flow tables in Appendix B, and M 1 6 1, ( p0 ) 1 p1 T1 M 1 = 0.5226 = 1.1378 = 2.0414 = 1.2990 p0* T* p* Since ( p0 ) , = 300 kPa, then

( p0 ) * =

300 kPa = 230.95 kPa = 231 kPa 1.2990

Ans.

Since the process is adiabatic,

( T0 ) * = ( T0 ) 1 = (273 + 20) K = 293 K

Ans.

For isentropic flow at inlet 1 with M 1 = 0.5226, T1

( T0 ) 1

= 0.94820

p1

( p0 ) 1

= 0.83016

Here ( T0 ) 1 = 293 K and ( p0 ) 1 = 300 kPa. T1 = 0.94820(293 K) = 277.82 K p1 = 0.83016(300 kPa) = 249.05 kPa

Since T1 T* p1 p*

= 1.1378;

T* =

277.82 K = 244.17 K 1.1378

= 2.0414;

p* =

249.05 kPa = 122.00 kPa 2.0414

Using the ideal gas law with R = 286.9 J>kg # K (From table in Appendix A), p* = r1 =

p* RT *

=

p1 = RT1

122.00 ( 103 ) N>m2

( 286.9 J>kg # K ) (244.17 K) 249.05 ( 103 ) N>m2

( 286.9 J>kg # K ) (277.82 K)

= 1.7416 kg>m3 = 3.1246 kg>m3

1432

2

13–74. Continued

Applying, cv =

286.9 J>kg # K R = = 717.25 J>kg # K k - 1 1.4 - 1

The change in entropy is ∆s = s* - s1 = cv ln

r1 T* + R ln * T1 r

= ( 717.25 J>kg # K ) ln a = 75.1 J>kg # K

3.1246 kg>m3 1 b + ( 286.9 J>kg # K ) ln ° ¢ 1.1378 1.7416 kg>m3 Ans.

Ans: p0* = 231 kPa T0* = 293 K 75.1 J>(kg # K) 1433

13–75. The duct has a diameter of 200 mm. If the average friction factor is f  =   0.003, and air is drawn into the duct with an inlet velocity of 200 m>s, temperature of 300 K, and absolute pressure of 180 kPa, determine these properties at the exit.

200 mm 1

2

30 m

SOLUTION The flow through the duct is considered Fanno flow since friction is involved. For air, k = 1.4 and R = 286.9 J>kg # K. The Mach number at the inlet is Mi =

Vi 2kRTi

200 m>s

=

21.4 ( 286.9 J>kg # K ) (300 K)

= 0.5762

with this Mach number, interpolated values tabulated in the Fanno flow tables in 0.59361(0.2 m) f = 39.57 m 7 L Appendix B gives Lmax = 0.59361. Thus, Lmax = D 0.003 = 30 m (O.K!) Also this table gives Ti T

*

= 1.1253

Vi V

*

= 0.61122

pi

= 1.8412

p*

Then T* =

T* 300 K (T ) = = 266.60 K Ti i 1.1253

V* =

200 m>s V* ( Vi ) = = 327.21 m>s Vi 0.61122

p* =

p* 180 kPa ( pi ) = = 97.76 kPa pi 1.8412

For the flow properties at the exit, f f f 0.003 L = Lmax - Li - e = 0.59361 - a b(30 m) = 0.14361 D D D 0.2 m

with this value

M e = 0.7383

Te T

*

= 1.08202

Ve V

*

= 0.76802

pe p*

= 1.40890

Then Te = 1.08202(266.60 K) = 288.47 K = 288 K

Ans.

Ve = 0.76802 ( 327.21 m>s ) = 251.30 m>s = 251 m>s

Ans.

pe = 1.40890 (97.76 kPa) = 137.73 kPa = 138 kPa

Ans.

Ans: Te = 288 K Ve = 251 m>s pe = 138 kPa 1434

*13–76. The duct has a diameter of 200 mm. If the average friction factor is f = 0.003, and air is drawn into the duct with an inlet velocity of 200 m>s, temperature of 300 K, and absolute pressure of 180 kPa, determine the mass flow through the duct and the resultant friction force acting on the 30-m length of duct.

200 mm 1

2

30 m

SOLUTION

Pi A

Ff

The flow through the duct is considered Fanno flow since friction is involved. For air, k = 1.4 and R = 286.9 J>kg # K. Using the ideal gas law, ri =

180 ( 103 ) N>m2

Pi = RTi

( 286.9 J>kg # K ) (300 K)

Thus, the mass flow is # m = riViA = ( 2.0913 kg>m3 )( 200 m>s ) 3 p(0.1 m)2 4

Ans.

= 13.14 kg>s = 13.1 kg>s

The Mach number at the inlet is Mi =

Vi 2kRTi

=

200 m>s

21.4 ( 286.9 J>kg # K ) (300 K)

= 0.5762

0.59361(0.2 m) f = Lmax = 0.59361. Thus, Lmax = 0.003 D 39.57 m 7 L = 30 m (O.K!)

with this Mach number, Also, this table gives Ti T*

= 1.1253

Vi V*

= 0.6112

pi p*

= 1.8412

Then T* =

T* 300 K ( Ti ) = = 266.60 K Ti 1.1253

V* =

200 m>s V* ( Vi ) = = 327.21 m>s Vi 0.6112

p* =

p* 180 kPa ( pi ) = = 97.76 kPa pi 1.8412

For the flow properties at the exit f f f 0.003 L = Lmax - Li - e = 0.59361 - a b(30 m) = 0.14361 D D D 0.2 m

with this value,

Me = 0.7383

Te T

*

= 1.08202

Ve V

*

(a)

= 2.0913 kg>m3

= 0.76802

pe p*

= 1.40890

Then Te = 1.08202(266.60 K) = 288.47 K Ve = 0.76802 ( 327.21 m>s ) = 251.30 m>s pe = 1.40890(97.76 kPa) = 137.73 kPa

1435

Pe A

13–76. Continued

Using the ideal gas law, re =

pe = RTe

137.73 ( 103 ) N>m2

( 286.9 J>kg # K ) (288.47 K)

= 1.6642 kg>m3

Applying the linear momentum equation by referring to the FBD of the control volume shown in Fig. a ΣF =

0 VrdV + VrV # dA 0t L L cv

1S +2

cs

pi A - pe A - Ff = 0 + Vi pi( - Vi A) + Ve pe (Ve A) c 180 1 103 2

N N d 3 p(0.1 m)2 4 - c 137.73 1 103 2 2 d 3 p(0.1 m)2 4 - Ff 2 m m

=

1 200 m>s 2 1 2.0913 kg>m3 2 1- 200 m>s 2 3 p(0.1 m)2 4

+ 1 251.30 m>s 2 1 1.6642 kg>m3 2 1 251.30 m>s 2 3 p(0.1 m)2 4

Ans.

Ff = 654.24 N = 654 N

1436

13–77. Air in a large room has a temperature of 24°C and absolute pressure of 101 kPa. If it is drawn into the 200-mm-diameter duct isentropically such that the absolute pressure at section 1 is 90 kPa, determine the critical length of duct Lmax where the flow becomes choked, and the Mach number, temperature, and pressure at section 2. Take the average friction factor to be f = 0.002.

200 mm 1

2 150 m Lmax

SOLUTION Since friction is involved, the flow can be considered as Fanno flow. The air is drawn in isentropically, the isentropic flow table in Appendix B can be used. Here p1 = 90 kPa, p0 = 101 kPa and T0 = (273 + 24) K = 297 K. Enter the table with p1 90 kPa = = 0.89109, and after performing the interpolation, p0 101 kPa M 1 = 0.4092

T1 = 0.96759; T0

T1 = 0.96759(297 K) = 287.37 K

Using this Mach number, f L = 2.1483; D max

Lmax =

2.1483(0.2 m) 0.002

= 214.83 m = 215 m

Ans.

Also, p1 *

p

T1 T*

p* 90 kPa = 34.18 kPa (p ) = p1 1 2.6334

= 2.6334;

p* =

= 1.1611;

T* =

T* 287.37 K = 247.49 K (T ) = T1 1 1.1611

At section 2, f f f 0.002 = 2.1483 - a L = L L b(150 m) = 0.6483 D D max D 1-2 0.2 m M 2 = 0.5650 = 0.565

p2 p* T2 T*

Ans.

= 1.8800;

p2 = 1.8800(34.18 kPa) = 64.26 kPa = 64.3 kPa

Ans.

= 1.1280;

T2 = 1.1280(247.49 K) = 279.17 K = 279 K

Ans.

Ans: Lmax = 215 m M 2 = 0.565 p2 = 64.3 kPa T2 = 279 K 1437

13–78. Air in a large room has a temperature of 24°C and absolute pressure of 101 kPa. If it is drawn into the 200-mm-diameter duct isentropically such that the absolute pressure at section 1 is 90 kPa, determine the mass flow through the duct and the resultant friction force acting on the duct. Also, what is the required length Lmax to choke the flow? Take the average friction factor to be f = 0.002.

200 mm 1 150 m Lmax

SOLUTION

P1A

Since friction is involved, the flow can be considered as Fanno flow. The air is drawn in isentropically, thus the isentropic flow table in Appendix B can be used. Here, p1 = 90 kPa p0 = 101 kPa and T0 = (273 + 24) K = 297 K. Enter the table with p1 90 kPa = = 0.89109 and after performing the interpolation, p0 101 kPa T1 = 0.96759; T0

M 1 = 0.4092

T1 = 0.96759(297 K) = 287.37 K

For air, k = 1.4 and R = 286.9 J>kg # K. The velocity of the flow at section 1 is V1 = M 1 1kRT1 = 0.4092 a 21.4 1 286.9 J>kg # K 2 (287.37 K) b = 139.02 m>s

The density of the air at section 1 can be determined using the ideal gas law. p1 =

p1 = RT1

90 1 103 2 N>m2

1 286.9 J>kg # K 2 (287.37 K)

Thus, the mass flow is # m = r1 V1 A1 =

= 1.0916 kg>m3

1 1.0916 kg>m3 2 1 139.02 m>s 2 3 p(0.1 m)2 4

= 4.7675 kg>s = 4.77 kg>s

Ans.

The duct is choked when the length is equal to the critical length. With M 1 = 0.4092, f L = 2.1483; D max

Lmax =

2.1483(0.2 m) 0.002

= 214.83 m = 215 m

Ans.

Also, p1 *

p

T1 T* V1 V*

= 2.6334;

p* =

p* 90 kPa (p ) = = 34.18 kPa p1 1 2.6334

= 1.1611;

T* =

T* 287.37 K (T ) = = 247.49 K T1 1 1.1611

= 0.44093;

V* =

139.02 m>s V* (V ) = = 315.29 m>s V1 1 0.44093

Using the ideal gas law r* =

p* RT *

=

34.18 1 103 2 N>m2

1 286.9 J>kg # K 2 (247.49 K)

= 0.4814 kg>m3

Applying the linear momentum equation by referring to the FBD of the control volume shown in Fig. a, ΣF = + 2 1S

2

0 VrdV + VrV # dA 0t Lcv Lcs

p1 A - p*A - Ff = 0 + V1r1( - V1A) + V *r* (V *A)

1438

Ff

(a)

P *A

13–78. Continued

c 90 1 103 2 =

N N d 3 p(0.1 m)2 4 - c 34.18 1 103 2 2 d 3 p(0.1 m)2 4 - Ff 2 m m

1 139.02 m>s 2 1 1.0916 kg>m3 2 1 - 139.02 m>s 2 3 p(0.1 m)2 4

+ 1 315.29 m>s 2 1 0.4814 kg>m3 2 1 315.29 m>s 2 3 p(0.1 m)2 4

Ff = 913 N

Ans.

Ans: # m = 4.77 kg>s Lmax = 215 m Ff = 913 N 1439

13–79. The 40-mm-diameter pipe has a friction factor of f = 0.015. A nozzle on the large tank A delivers nitrogen isentropically to the pipe at section 1 with a velocity of 1200  m>s, temperature of 460 K, and absolute pressure of 750 kPa. Determine the mass flow. Show that a normal shock forms within the pipe. Take L′ = 1.35 m.

A

40 mm

1

B 2

L L¿

SOLUTION The flow through the pipe is considered Fanno flow since friction is involved. For nitrogen k = 1.4 and R = 296.8 J>kg # K. The Mach number at the entrance is M1 =

V1 1kRT1

=

1200 m>s

21.4 1 296.8 J>kg # K 2 (460 K)

= 2.7448

With this Mach number, interpolated values tabulated in Fanno flow tables gives 0.48001(0.04 m) f Lmax = 0.48001. Thus, Lmax = = 1.28 m. D 0.015 Since Lmax 6 L = 1.35 m and M 1 7 1, a normal shock forms within the pipe. The flow out of the pipe is choked at M = 1. # m = r VA = a = 8.28 kg>s

750 1 103 2 Pa

1 296.8 J>kg # K 2 (460 K)

b 1 1200 m>s 2 1 p (0.020 m)2 2

Ans.

Ans: 8.28 kg>s 1440

*13–80. The 40-mm-diameter pipe has a friction factor of f = 0.015. The nozzle on the large tank A delivers nitrogen isentropically to the pipe at section 1 with a velocity of 200 m>s and temperature of 460 K. Determine the velocity and temperature of the nitrogen at L = 2 m if L′ = 3 m.

A

40 mm

1

L L¿

SOLUTION M1 =

V1

=

1kRT1

Subsonic flow. For this M 1, f L = 1.4794 D max

200 m>s

21.4 1 296.8 J>kg # k 2 (460 K)

= 0.4575 m>s

0.015 L = 1.4794 0.04 m max Lmax = 3.945 m 7 3 m T = 1.151795 T* T* =

460 K = 399.38 K 1.151795

Therefore at L = 3.945 m - 2 m = 1.945 m a

0.015 b(1.945 m) = 0.7294 0.04 m

M = 0.54977

T = 1.131597, T*

T = 1.131597(399.38 K) = 451.94 K = 452 K

V = M 1kRT = 0.5497721.4 1 296.8 J>kg # K 2 (451.94 K) = 238 m>s

1441

Ans.

Ans.

B 2

13–81. The 100-mm-diameter pipe is connected by a nozzle to a large reservoir of air that is at a temperature of 40°C and absolute pressure of 450 kPa. If the absolute pressure at section 1 is 30 kPa, determine the mass flow through the pipe and the length L of the pipe so that a backpressure of 90 kPa in the tank maintains supersonic flow through the pipe. Assume a constant friction factor of 0.0085 throughout the pipe.

100 mm 1 L

SOLUTION Since friction is involved, the flow can be considered as Fanno flow. The air is drawn in isentropically. Here p1 = 30 kPa, p0 = 450 kPa and T0 = (273 + 40) K = 313 K. p1 30 kPa Enter the table with = = 0.06667 and after performing the interpolation, p0 450 kPa T1 = 0.46129; T0

M 1 = 2.4164

T1 = 0.46129(313 K) = 144.38 K

For air, k = 1.4 and R = 286.9 J>kg # K. The velocity of the flow at section 1 is V1 = M 1 2kRT1 = 2.4164c 21.4 ( 286.9 J>kg # K ) (144.38 K) d = 581.90 m>s

The density of the air at section 1 can be determined using the ideal gas law. r1 =

p1 = RT1

30 ( 103 ) N>m2

( 286.9 J>kg # K ) (144.38 K)

= 0.72424 kg>m3

Thus, the mass flow is # m = r1V1A1 = ( 0.72424 kg>m3 )( 581.90 m>s ) 3 p(0.05 m)2 4 = 3.3100 kg>s = 3.31 kg>s

Ans.

For M 1 = 2.4164 f L = 0.41361 D max

p1 p*

= 0.30791

Then p* =

p* 30 kPa (p ) = = 97.43 kPa p1 1 0.30791

It is required that no compression or expansion wave forms at the exit. Then at section 2, p2 = 90 kPa, so that p2 p*

=

2

90 kPa = 0.92374 97.43 kPa

1442

13–81. Continued

M 2 = 1.0698 7 1

(The flow is not choked)

f L = 0.0050990 D Then f f f L = Lmax L = 0.41361 - 0.0050990 = 0.408511 D 1-2 D D 0.408511(0.1 m) = 4.806 m = 4.81 m L1 - 2 = L = 0.0085

Ans.

Ans: # m = 3.31 kg>s L = 4.81 m 1443

13–82. The 100-mm-diameter pipe is connected by a nozzle to a large reservoir of air that is at a temperature of 40°C and absolute pressure of 450 kPa. If the absolute pressure at section 1 is 30 kPa, determine the mass flow through the pipe and the length L of the pipe so that sonic flow occurs into the tank. What is the required backpressure in the tank for this to occur? Assume a constant friction factor of 0.0085 throughout the pipe.

100 mm 1

2 L

SOLUTION Since friction is involved, the flow can be considered as Fanno flow. The air is drawn in isentropically. Here, p1 = 30 kPa, p0 = 450 kPa and T0 = (273 + 40) K = 313 K. p1 30 kPa Enter the table with = = 0.06667 and after performing the interpolation, p0 450 kPa M 1 = 2.4164

T1 = 0.46129; T1 = 0.46129(313 K) = 144.38 K T0

For air, k = 1.4 and R = 286.9 J>kg # K. The velocity of the flow at section 1 is V1 = M 1 2kRT1 = 2.4164c 21.4 ( 286.9 J>kg # K ) (144.38 K) d = 581.90 m>s

The density of the air at section 1 can be determined using the ideal gas law. r1 =

p1 = RT1

30 ( 103 ) N>m2

( 286.9 J>kg # K )( 144.38 K )

= 0.72424 kg>m3

Thus, the mass flow is

#

m = r1V1A1 = ( 0.72424 kg>m3 )( 581.90 m>s ) 3 p(0.05 m)2 4 = 3.3100 kg>s = 3.31 kg>s

Ans.

M 1 = 2.4164

f L = 0.41361 D max

p1 p*

= 0.30791

Then V* =

581.90 m>s V* ( V1 ) = = 323.67 m>s V1 1.7978

p* =

p* 30 kPa (p ) = = 97.4 kPa p1 1 0.30791

It is required that no shock forms within the pipe, so the pipe must be choked. Therefore f f L1 - 2 = Lmax = 0.41361 D D L = Lmax =

0.41361(0.1 m) 0.0085

= 4.866 m = 4.87 m Ans.

p2 = p* = 97.43 kPa

Ans: # m = 3.31 kg>s L = 4.87 m p2 = 97.4 kPa 1444

13–83. The 100-mm-diameter pipe is connected by a nozzle to a large reservoir of air that is at a temperature of 40°C and absolute pressure of 450 kPa. If the backpressure causes M 1 7 1 and the flow is choked at the exit, section 2, when L = 5 m, determine the mass flow through the pipe. Assume a constant friction factor of 0.0085 throughout the pipe.

100 mm 1

2 L

SOLUTION Since friction is involved, the flow can be considered as Fanno flow. It is required that the mass flow is at its greatest, thus the pipe must be choked, that is, M 2 = 1 at exit. As a result, Lmax = 5 m. Then f 0.0085 b(5 m) = 0.425 L = a D max 0.1 m

Enter this value into the Fanno flow tables in Appendix B and select M 1 7 1. Interpolation gives M 1 = 2.4677

p1 = 0.29809 p*

Since the air is drawn isentropically into section 1, here, T0 = (273 + 40) K = 313 K and p0 = 450 kPa. Enter M 1 = 2.4677 into the tables, interpolation gives T1 = 0.45087 T0

p1 = 0.06155 p0

Then T1 = 0.45087(313 K) = 141.12 K p1 = 0.06155(450 kPa) = 27.70 kPa

For air, k = 1.4 and R = 286.9 J>kg # K. The velocity of the flow at section 1 is V1 = M 1 2kRT1 = 2.4677c 21.4 ( 286.9 J>kg # K )( 141.12 K ) d = 587.51 m>s

Using the ideal gas law, r1 =

27.70 ( 103 ) N>m2 p1 = = 0.6842 kg>m3 RT1 ( 286.9 J>kg # K ) (141.12 K)

Thus, the mass flow is ( m# ) max = r1V1A1 = ( 0.6842 kg>m3 )( 587.51 m>s ) 3 p(0.05 m)2 4 = 3.1571 kg>m3 = 3.16 kg>m3

Ans.

Ans: 3.16 kg>s 1445

13–84. The 100-mm-diameter pipe is connected by a nozzle to a large reservoir of air that is at a temperature of 40°C and absolute pressure of 450 kPa. If the backpressure causes M 1 6 1, and the flow is choked at the exit, section 2, when L = 5 m, determine the mass flow through the pipe. Assume a constant friction factor of 0.0085 throughout the pipe.

100 mm 1 L

SOLUTION Here M 2 = 1 at the exit. As a result Lmax = 5 m. Then, f 0.0085 Lmax = a b(5 m) = 0.425 D 0.1 m

For M 1 6 1, we get

M 1 = 0.61774 With T0 = (273 + 40) K = 313 K, and p0 = 450 kPa, with M 1 = 0.61774, we get T1 = 0.92909 T0

p1 = 0.77304 p0

Then T1 = 0.92909(313 K) = 290.81 K p1 = 0.77304(450 kPa) = 347.87 kPa Therefore, V1 = M 1 2kRT1 = 0.6177421.4 ( 286.9 J>(kg # K) ) (290.81 K) = 211.13 m>s

So that # m = r1V1A1 = °

. m = 6.91 kg>m3

347.87 ( 103 ) Pa

( 286.9 J>(kg # K) ) (290.81 K)

2

¢ ( 211.13 m>s ) p(0.05 m)2 Ans.

1446

13–85. Outside air at a temperature of 60°F is drawn isentropically into the duct and then heated along the duct at  200(103) ft # lb>slug. At section l the temperature is T = 30°F and the pressure is 13.9 psia. Determine the Mach number, temperature, and pressure at section 2. Neglect friction.

3 in. 1

2 4 ft

SOLUTION Air is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered Rayleigh flow. Since the air is drawn in isentropically, ( T0 ) 1 = (460 + 60)°R = 520°R, T1 = (460 + 30)°R = 490°R and p1 = 13.9 psi. T1 490°R = 0.94231 = ( T0 ) 1 520°R M 1 = 0.5532 Enter this value into the Rayleigh flow tables in Appendix B, and after performing the interpolating, p1 *

p

= 1.6802

T1 T

*

= 0.86384

( T0 ) 1

Then

T0*

= 0.76393

13.9 psi p* ( p1 ) = = 8.2728 psi p1 1.6802 T* 490°R (T ) = = 567.23°R T* = T1 1 0.86384 p* =

T0* =

T0*

( T0 ) 1

( T0 ) 1 =

520°R = 680.69°R 0.76393

For air, k = 1.4 and R = 1716 ft # lb>slug # °R. Then, CP = Applying,

1.4 ( 1716 ft # lb>slug °R) kR = = 6006 ft # lb>slug # °R k - 1 1.4 - 1

∆Q = Cp 3 ( T0 ) 2 - ( T0 ) 1 4 ; 200 ( 103 ) ft # lb>slug = ( 6006 ft # lb>slug # °R ) 3 ( T0 ) 2 - 520°R 4 ∆m

( T0 ) 2

553.30°R = = 0.81285 into the Rayleigh flow tables in Appendix B. 680.69°R T0* Interpolation gives

Then enter

Ans.

M 2 = 0.5945 = 0.594 T2 T* p2 p*

= 0.91102; T2 = 0.91102(567.23°R) = 516.76°R = 517°R = 1.6056; p2 = 1.6056(8.2728 psi) = 13.28 psi = 13.3 psi

Ans. Ans.

Since M 2 6 1, the flow is not choked.

Ans: M 2 = 0.594 T2 = 517°R p2 = 13.3 psi 1447

13–86. Outside air at a temperature of 60°F is drawn isentropically into the duct and then heated along the duct at 200 ( 103 ) ft # lb>slug. At section 1 the temperature is T = 30°F and the pressure is 13.9 psia. Determine the mass flow and the change in the entropy per unit mass that occurs between sections 1 and 2.

3 in. 1

2 4 ft

SOLUTION Air is assumed to be inviscid and heat energy is being added. Thus the flow can be considered Rayleigh flow. Since the air is drawn in isentropically. Here, ( T0 ) 1 = (460 + 60)°R = 520°R, T1 = (460 + 30)°R = 490°R and p1 = 13.9 psi. T1 490°R Enter the table with = 0.94231 and after performing the = ( T0 ) 1 520°R interpolation, M 1 = 0.5532 p1 p*

= 1.6802

T1 T*

= 0.86384

( T0 ) 1

Then

T *0

p* =

13.9 psi p* (p ) = = 8.2728 psi p1 1 1.6802

T* =

T* 490°R (T ) = = 567.23°R T1 1 0.86384 T0*

T0* =

( T0 ) 1

( T0 ) 1 =

= 0.76393

520°R = 680.69°R 0.76393

For air, k = 1.4 and R = 1716 ft # lb>slug # °R. Then, cP = Applying,

1.4 ( 1716 ft # lb>slug # °R ) kR = = 6006 ft # lb>slug # °R k - 1 1.4 - 1

∆Q = cp 3 ( T0 ) 2 - ( T0 ) 1 4 ; 200 ( 103 ) ft # lb>slug = ( 6006 ft # lb>slug # °R ) 3 ( T0 ) 2 - 520°R 4 ∆m (T0)2 = 553.30 °R

( T0 ) 2 T *0

=

553.30°R = 0.81285 680.69°R

M 2 = 0.5945 T2 T* p2 p*

= 0.91102; T2 = 0.91102(567.23°R) = 516.76°R = 1.6056; p2 = 1.6056(8.2728 psi) = 13.28 psi

Since M 2 6 1, the flow is not choked.

1448

13–86. Continued

The velocity of the flow at section 1 is V1 = M 1 2kRT1 = 0.5532 a 21.4 ( 1716 ft # lb>slug # °R ) (490°R) b = 600.21 ft>s

The density of air at section 1 is

lb 12 in 2 ba b p1 1 ft in2 r1 = = = 0.0023805 slug>ft 3 RT1 ( 1716 ft # lb>slug # °R ) (490°R) a13.9

Thus, the mass flow is 2 1.5 # m = r1V1A1 = ( 0.0023805 slug>ft 3 )( 600.21 ft>s ) c p a ft b d 12

= 0.07014 slug>s = 0.0701 slug>s

Ans.

The change in entropy from section 1 to 2 can be determined by applying. ∆s = s2 - s1 = cp ln

p2 T2 - R ln T1 p1

= ( 6006 ft # lb>slug # °R ) ln a

13.28 psi 516.76°R b - ( 1716 ft # lb>slug # °R ) ln a b 490°R 13.9 psi

= 397.66 ft # lb>slug # °R = 398 ft # lb>(slug # °R)

Ans.

Ans: # m = 0.0701 slug>s ∆s = 398 ft # lb>(slug # °R) 1449

13–87. Nitrogen having a temperature of T1 = 270 K and absolute pressure of p1 = 330 kPa flows into the smooth pipe at M 1 = 0.3. If it is heated at 100 kJ>kg # m, determine the velocity and pressure of the nitrogen when it exits the pipe at section 2.

1

2

4m

SOLUTION Nitrogen is assumed to be inviscid and the heat energy is being added. Thus, the flow can be considered Rayleigh flow. Here, T1 = 270 K and p1 = 330 kPa. For nitrogen, k = 1.4. M 1 = 0.3 ( T0 ) 1 p1 T1 V1 = 2.1314 = 0.40887 = 0.19183 = 0.34686 * * * p T V T *0 Then p* =

p* 330 kPa ( p1 ) = = 154.83 kPa p1 2.1314

T* =

T* 270 K (T ) = = 660.36 K T1 1 0.40887

For nitrogen, R = 296.8 J>kg # K. Then the velocity of the flow at section 1 is V1 = M 1 2kRT1 = 0.3c 21.4 ( 296.8 J>kg # K)(270 K) d = 100.48 m>s 100.48 m>s V* (V ) = = 523.80 m>s V1 1 0.19183

V* =

Since nitrogen is drawn in to section 1 isentropically, M 1 = 0.3 T1

= 0.98232; ( T0 ) 1 =

( T0 ) 1

( T0 ) 1 T1

( T1 ) =

270 K = 274.86 K 0.98232

Then T0*

T0* =

cp =

( T0 ) 1

( T0 ) 1 =

274.86 K = 792.42 K 0.34686

1.4 ( 296.8 J>kg # K ) kR = = 1038.8 J>kg # K k - 1 1.4 - 1

Applying ∆Q J = cp 3 ( T0 ) 2 - ( T0 ) 1 4 ; c 100 ( 103 ) # d (4 m) = ( 1038.8 J>kg # K ) 3 ( T0 ) 2 - 274.86 K 4 ∆m kg m

( T0 ) 2 = 659.92 K

Then enter

( T0 ) 2 T *0

=

659.92 K = 0.83279 into the Rayleigh flow tables in 792.42 K

M 2 = 0.6131 6 1 (pipe is not choked) p2 p* V2 V*

= 1.5724; p2 = 1.5724(154.83 kPa) = 242.41 kPa = 242 kPa

Ans.

= 0.59113; V2 = 0.59113 ( 523.80 m>s ) = 309.63 m>s = 310 m>s Ans.

1450

Ans: p2 = 242 kPa V2 = 310 m>s

*13–88. Nitrogen having a temperature of T1 = 270 K and absolute pressure of p1 = 330 kPa flows into the smooth pipe at M 1 = 0.3. If it is heated at 100 kJ>kg # m, determine the stagnation temperatures at sections 1 and 2, and the change in entropy per unit mass between these two sections.

1

4m

SOLUTION Nitrogen is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered Rayleigh flow. Here, T1 = 270 K and p1 = 330 kPa. For nitrogen, k = 1.4. M 1 = 0.3 p1 *

p

= 2.1314

T1 T

*

= 0.40887

( T0 ) 1 T0*

= 0.34686

Then p* =

p* 330 kPa ( p1 ) = = 154.83 kPa p1 2.1314

T* =

T* 270 K (T ) = = 660.36 K T1 1 0.40887

Since nitrogen is drawn into section 1 isentropically, M 1 = 0.3 T1

( T0 ) 1

( T0 ) 1 =

= 0.98232;

( T0 ) 1 T1

( T1 ) =

270 K = 274.86 K = 275 K 0.98232

Ans.

Then T0* =

T0*

( T0)1

( T0 ) 1 =

274.86 K = 792.42 K 0.34686

For nitrogen, k = 1.4 and R = 296.8 J>kg # K Then cp =

1.4 ( 296.8 J>kg # K ) kR = = 1038.8 J>kg # K k - 1 1.4 - 1

Applying ∆Q J = cp 3 ( T0 ) 2 - ( T0 ) 1 4 ; c 100 ( 103 ) # d (4 m) = ( 1038.8 J>kg # K ) 3 ( T0 ) 2 - 274.86 K 4 ∆M kg m

( T0 ) 2 = 659.92 K = 660 K

( T0 ) 2 T *0

=

659.92 K = 0.83279 792.42 K

M 2 = 0.6131 6 1 (pipe is not choked) p2 p* T2 T*

2

= 1.5724; p2 = 1.5724(154.83 kPa) = 243.46 kPa = 0.92946; T2 = 0.92946(660.36 K) = 613.78 K

1451

Ans.

13–88. Continued

The change in entropy from section 1 to 2 can be determined by applying. ∆s = s2 - s1 = cp ln

T2 p2 - R ln T1 p1

= ( 1038.8 J>kg # K ) ln a

613.78 K 243.46 kPa b - ( 296.8 J>kg # K ) ln a b 270 K 330 kPa

= 943.35 J>kg # K = 943 J>kg # K

Ans.

1452

13–89. Air is drawn isentropically into the pipe at V1 = 640 m>s, T1 = 80°C, and absolute pressure of p1 = 250 kPa. If it exits the pipe having a speed of 470 m > s, determine the amount of heat per unit mass that the pipe supplies to the air.

1

2

SOLUTION Air is assumed to be inviscid and heat energy is being added. Thus the flow can be considered as Rayleigh flow. Here, T1 = (273 + 80) K = 353 K, p1 = 250 kPa and V1 = 640 m>s. For air, k = 1.4 and R = 286.9 J>kg # K. The Mach number at section 1 can be determined from M1 =

V1 2kRT1

=

640 m>s

21.4(286.9 J>kg # K) (353 K)

= 1.6997

The air is drawn into section 1 isentropically. T1 = 0.63380; (T0)1

(T0)1 =

(T0)1 T1

(T1) =

353 K = 556.96 K 0.63380

Also, enter the value of M 1 into the Rayleigh flow tables in Appendix B, V1 V*

= 1.3744 ;

(T0)1 T0*

V* =

= 0.85977;

640 m>s V* (V ) = = 456.66 m>s V1 1 1.3744

T0* =

Then enter the same table with

V2 V*

T0* 556.96 K (T ) = = 647.80 K (T0)1 0 1 0.85977 =

470 m>s 465.66 m>s

= 1.00932,

M 2 = 1.0113 > 1 (pipe is not choked) (T0)2 T0* cp =

= 0.99990;

(T0)2 = 0.99990(647.80 K) = 647.74 K

1.4(286.9 J>kg # K) kR = = 1004.15 J>kg # K k - 1 1.4 - 1

Then, ∆Q = cp 3(T0)2 - (T0)1 4 ∆m

= (1004.15 J>kg # K) 3647.74 K- 556.96 K4 = 91.157 ( 103 ) J>kg = 91.2 kJ>kg

Ans.

Ans: 91.2 kJ>kg 1453

13–90. Air is drawn into the 100-mm-diameter pipe at V1 = 640 m>s, T1 = 80°C, and absolute pressure of p1 = 250 kPa. If it exits the pipe having a speed of 470 m>s, determine the stagnation temperatures at sections 1 and 2 and the change in entropy per unit mass between these sections.

1

2

SOLUTION Air is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered as Rayleigh flow. Here, T1 = (273 + 80) K = 353 K, p1 = 250 kPa and V1 = 640 m>s. For air, k = 1.4 and R = 286.9 J>kg # K. The Mach number at section 1 is M1 =

V1 1kRT1

=

640 m>s

21.4 1 286.9 J>kg # K 2 (353 K)

= 1.6997

Since the air is drawn into section 1 isentropically, enter the value of M 1 into the isentropic flow tables in Appendix B and after performing the interpolation, T1 = 0.63380; (T0)1 V1 V

*

p1 *

p

T1 T

*

V

*

=

T1

= 0.47577;

p* =

= 0.65391;

T* =

465.66 m>s

353 K = 556.96 K = 557 K 0.63380

Ans.

p* 250 kPa (p ) = = 525.47 kPa p1 1 0.47577 T* 353 K (T1) = = 539.83 K T1 0.65391

T0* =

= 0.85977;

470 m>s

(T1) =

640 m>s V* (V ) = = 465.66 m>s V1 1 1.3744

V* =

T0* V2

(T0)1

= 1.3744;

(T0)1

With

(T0)1 =

T0* 556.96 K (T ) = = 647.80 K (T0)1 0 1 0.85977

= 1.00932

M 2 = 1.0113 7 1 (pipe is not choked)

( T0 ) 2 T0* p2 p* T2 T*

= 0.98695;

p2 = 0.98695(525.47 kPa) = 518.61 kPa

= 0.99614;

T2 = 0.99614(539.83 K) = 537.75 K

cp = Then

= 0.99990; (T0)2 = 0.99990 (647.80 K) = 647.74 K = 648 K Ans.

1.4 1 286.9 J>kg # K 2 kR = = 1004.15 J>kg # K k - 1 1.4 - 1

∆s = s2 - s1 = cp ln =

p2 T2 - R ln T1 p1

1 1004.15 J>kg # K 2 ln a

537.75 K b 353 K

= 213.32 J>kg # K = 213 J>(kg # K)

1 286.9 J>kg # K 2 ln a

1454

518.61 kPa b 250 kPa

Ans.

Ans: 1T0 2 1 = 557 K 1T0 2 2 = 648 K ∆s = 213 J>(kg # K)

13–91. Air from a large reservoir is at a temperature of 275 K and absolute pressure of 101 kPa. It isentropically enters the duct at section 1. If 80 kJ>kg of heat is added to the flow, determine the greatest possible velocity it can have at section 1. The backpressure at 2 causes M 1 6 1.

1

2

SOLUTION Air is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered as Rayleigh flow. Here, (T0)1 = 275 K and (p0)1 = 101 kPa. To obtain the greatest possible V1, the duct must be choked, that is M 2 = 1. For air k = 1.4 and R = 286.9 J>kg # K. Thus, cp = Then

1.4 1 286.9 J>kg # K 2 kR = = 1004.15 J>(kg # K) k - 1 1.4 - 1

∆Q = cp 3 (T0)2 - (T0)1 4 ; ∆m

80 1 103 2 J>kg =

(T0)2 = 354.67 K

1 1004.15 J>(kg # K) 2 3 (T0)2

- 275 K 4

Under choked conditions, (T0)2 (T0)1

T0*

= 1;

T0* = (T0)2 = 354.67 K

275 K = 0.77537 for M 1 6 1, 354.67 K p1 M 1 = 0.5624 = 1.6635 p* Since the air is drawn isentropically into section 1, M 1 = 0.5624

With

T0*

=

T1 = 0.94050; (T0)1

T1 = 0.94050(275 K) = 258.64 K = 259 K

The velocity of the flow at section 1 can be determined from V1 = M 1 1kRT1 = 0.5624c 21.4 1 286.9 J>kg # K 2 (258.64 K) d = 181.27 m>s = 181 m>s

Ans.

Ans: T1 = 259 K V1 = 181 m>s 1455

*13–92. Air from a large reservoir is at a temperature of 275 K and absolute pressure of 101 kPa. It isentropically enters the duct at section 1. If 80 kJ>kg of heat is added to the flow, determine the temperature and pressure at the entrance of the duct. The backpressure at 2 causes M 2 = 1.

1

2

SOLUTION Air is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered as Rayleigh flow. Here, (T0)1 = 275 K and (p0)1 = 101 kPa. For air, k = 1.4 and R = 286.9 J>kg # K. Thus, cp = Then

1.4 1 286.9 J>kg # K 2 kR = = 1004.15 J>kg # K k - 1 1.4 - 1

∆Q = cp 3 (T0)2 - (T0)1 4 ; ∆m

80 1 103 2 J>kg =

(T0)2 = 354.67 K

1 1004.15 J>kg # K 2 3 (T0)2

- 275 K 4

Under choked conditions (T0)2 With

(T0)1 T 0*

T0* =

= 1;

T0* = (T0)2 = 354.64 K

275 K = 0.77537 for M 1 7 1, 354.64 K

M 1 = 2.0930 Since the air is drawn isentropically into section 1, using M 1 = 2.0930 T1 = 0.53302; (T0)1

T1 = 0.53302(275 K) = 146.58 K = 147 K

p1 = 0.11055; (p0)1

p1 = 0.11055(101 kPa) = 11.17 kPa = 11.2 kPa Ans.

1456

Ans.

*13–93. Nitrogen having a temperature of 300 K and absolute pressure of 450 kPa flows from a large reservoir into a 100-mm-diameter duct. As it flows, 100 kJ>kg of heat is added. Determine the temperature, pressure, and density at section 1 if the backpressure causes M 1 7 1 and the flow becomes choked at section 2.

100 mm 1

2

SOLUTION Nitrogen is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered as Rayleigh flow. Here, (T0)1 = 300 K and (p0)1 = 450 kPa. For nitrogen, k = 1.4 and R = 296.8 J>kg # K. cp = Then,

1.4 1 296.8 J>(kg # K) 2 kR = = 1038.8 J>(kg # K) k - 1 1.4 - 1

∆Q = cp 3 (T0)2 - (T0)1 4 ; ∆m

100 1 103 2 J>kg = (T0)2 = 396.26 K

1 1038.8 J> (kg # K) 2 3 (T0)2

- 300 K 4

Assume that the duct will be choked, that is M 2 = 1, and the mass flow is at its greatest. Under this condition (T0)2 T0*

T0* = (T0)2 = 396.26 K

= 1;

Enter the Rayleigh flow tables in Appendix B with Since M 1 7 1, interpolation gives

(T0)1 T0*

=

300 K = 0.75708 396.26 K

M 1 = 2.1946 Since the air is drawn isentropically, M 1 = 2.1946 T1 = 0.50936; (T0)1 p1

( p0 ) 1

= 0.09432;

T1 = 0.50936(300 K) = 152.81 K = 153 K p1 = 0.09432(450 kPa) = 42.44 kPa = 42.4 kPa

Ans. Ans.

Using the ideal gas law, r1 =

p1 = RT1

42.44 1 103 2 N>m2

1 296.8 J>kg # K 2 (152.81 K)

= 0.936 kg>m3

Ans.

Ans: T1 = 153 K p1 = 42.4 kPa r1 = 0.936 kg>m3 1457

13–94. Nitrogen having a temperature of 300 K and absolute pressure of 450 kPa flows from a large reservoir into a 100-mm-diameter duct. As it flows, 100 kJ>kg of heat is added. Determine the mass flow if the backpressure causes M 1 6 1 and the flow becomes choked at section 2.

100 mm 1

2

SOLUTION Nitrogen is assumed to be inviscid and heat energy is being added. Thus, the flow can be considered as Rayleigh flow. Here, (T0)1 = 300 K and (p0)1 = 450 kPa. For nitrogen, k = 1.4 and R = 296.8 J>kg # K. cp = Then,

1.4 1 296.8 J>(kg # K) 2 kR = = 1038.8 J>(kg # K) k - 1 1.4 - 1

∆Q = cp 3 (T0)2 - (T0)1 4 ; ∆m

(T0)2 = 396.26 K

100 1 103 2 J>kg =

1 1038.8 J>kg # K 2 3 (T0)2

- 300 K 4

To have the greatest mass flow, the duct must be choked that is M 2 = 1. Under this condition (T0)2 T0*

T0* = (T0)2 = 396.26 K

= 1;

Enter the Rayleigh flow table with interpolation gives

(T0)1 T0*

=

300 K = 0.75708 since M 1 6 1, 396.26 K

M 1 = 0.5478 Since the air is drawn isentropically into section 1, M 1 = 0.5478 T1 = 0.94338; (T0)1 p1 = 0.81546; ( p0 ) 1

T1 = 0.94338(300 K) = 283.01 K P1 = 0.81546(450 kPa) = 366.96 kPa

The velocity of the flow at section 1 can be determined from V1 = M 1 1kRT1 = 0.5478c 21.4 1 296.8 J>(kg # K) 2 (283.01 K) d = 187.85 m>s

Using the ideal gas law, r1 =

p1 = RT1

366.96 1 103 2 N>m2

1 296.8 J> (kg # K) 2 (283.01 K)

Thus, the mass flow is # m = r1 V1 A1 =

= 4.3687 kg>m3

1 4.3687 kg>m3 2 1 187.85 m>s 2 3 p(0.05 m)2 4

= 6.445 kg>s = 6.45 kg>s

Ans.

Ans: 6.45 kg>s 1458

13–95. The converging nozzle has an exit diameter of 0.25 m. If the fuel-oxidizer mixture within the large tank has an absolute pressure of 4 MPa and temperature of 1800 K, determine the mass flow from the nozzle when the backpressure is a vacuum. The mixture has k = 1.38 and R = 296 J>kg # K.

0.25 m

SOLUTION The mixture is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 1800 K and p0 = 4 MPa. Since the back pressure is a vacuum, p = 0. k

p0 = p a1 +

k - 1 2 k-1 M b 2

1.38

1.38 - 1 1.38 4 MPa = 0 c 1 + a b M2 d 2

M = ∞

The flow with M 7 1 (supersonic) is not possible at the exit plane since the nozzle will be choked at the exit plane and an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. T0 = T a1 +

k - 1 2 M b 2

1800 K = T * c 1 + a T * = 1512.61 K

1.38 - 1 b(1)2 d 2 k

p0 = p a1 +

k - 1 2 (k - 1) M b 2

4 MPa = p* c 1 + a

1.38

1.38 - 1 1.38 - 1 b 1 12 2 d 2

p* = 2.1267 MPa

Using the universal gas law, 2.1267 1 103 2

p* = r* RT *;

N = r* 1 296 J>kg # K 2 (1512.61 K) m2

r* = 4.7499 kg>m3

The velocity of the flow at the exit plane is

V * = M * 2kRT * = (1) 21.38 1 296 J>kg # K 2 (1512.61 K) = 786.05 m>s

Finally,

# m = r* V * A* =

1 4.7499 kg>m3 2 1 786.05 m>s 2 3 p(0.125 m)2 4

= 183.28 kg>s = 183 kg>s

Ans.

Ans: 183 kg>s 1459

*13–96. The converging nozzle has an exit diameter of 0.25 m. If the fuel-oxidizer mixture within the large tank has an absolute pressure of 4 MPa and temperature of 1800 K, determine the mass flow from the nozzle if the atmospheric pressure is 100  kPa. The mixture has k = 1.38 and R = 296 J>kg # K.

0.25 m

SOLUTION The gas is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 1800 K and p0 = 4 MPa. Since the back pressure is a vacuum, p = 0. k

p0 = p a1 +

k - 1 2 k-1 M b 2

1.38

1.38 - 1 1.38 - 1 bM 2 d 4 MPa = 0.1 MPac 1 + a 2

M = 3.0448

The flow with M 7 1 (supersonic) is not possible at the exit plane since the nozzle will be choked at the exit plane and an expansion shock wave will form thereafter. Thus, at the exit plane, M = 1. k - 1 2 M b 2

T0 = T a1 +

1800 K = T * c 1 + a T * = 1512.61 K

1.38 - 1 b(1)2 d 2 k

p0 = p a1 +

k - 1 2 k-1 M b 2

4 MPa = p* c 1 + a p* = 2.1267 MPa

1.38

1.38 - 1 1.38 - 1 b 1 12 2d 2

Using the universal gas law, 2.1267 1 103 2

p* = r* RT *;

N = r* 1 296 J>kg # K 2 (1512.61 K) m2

r* = 4.7499 kg>m3

The velocity of the flow at the exit plane can be determined by applying V * = M * 2kRT * = (1) 21.38 1 296 J>kg # K 2 (1512.61 K) = 786.05 m>s

Finally, # m = r* V * A* =

1 4.7499 kg>m3 21 786.05 m>s 23 p(0.125 m)2 4

= 183.28 kg>s = 183 kg>s

1460

Ans.

13–97. The bottle tank contains 0.13 m3 of oxygen at an absolute pressure of 900 kPa and temperature of 20°C. If the nozzle has an exit diameter of 15 mm, determine the time needed to drop the absolute pressure in the tank to 300 kPa once the valve is opened. Assume the temperature remains constant in the tank during the flow and the ambient air is at an absolute pressure of 101.3 kPa.

15 mm

SOLUTION The oxygen is considered to be compressible. The flow is unsteady. The stagnation pressure decreases from (p0)1 = 900 kPa to (p0)2 = 300 kPa, while the stagnation temperature is assumed to remain constant at T0 = 1 273 + 20°C 2 = 293 K. For oxygen k = 1.40 and R = 259.8 J>kg # K. k

k - 1 2 k-1 M b 2

( p0)2 = pa1 +

1.4

1.4 - 1 1.4 - 1 300 kPa = (101.3 kPa)c 1 + a bM 2 d 2

M = 1.3485

The flow with M 7 1 (supersonic) is not possible at the exit plane since the nozzle will be choked at the exit plane and an expansion shock wave will form thereafter. This condition remains throughout the flow. At the exit plane, M = 1. T0 = T a1 +

k - 1 2 M b 2

293 K = T * c 1 + a T * = 244.17 K

1.4 - 1 b(1)2 d 2 k

p0 = p* a1 +

k - 1 2 k-1 M b 2

p0 = p* c 1 + a p* = 0.5283p0

1.4

1.4 - 1 1.4 - 1 b 1 12 2 d 2

Using the universal gas law, p* = r* RT *;

0.5283 p0 = r* 1 259.8 J>kg # K 2 (244.17 K) r* = 8.3280 1 10-6 2 p0

The velocity of the flow at the exit plane is V * = M * 2kRT * = (1) 21.40 1 259.8 J>kg # K 2 (244.17 K) = 298.01 m>s

Thus, the mass flow rate is # m = r* V* A* = 3 8.3280 1 10-6 2 p0 41 298.01 m>s 23 p(0.0075 m)2 4 = 0.4386 1 10-6 2 p0 The density of air in the tank is determined from the universal gas law p0 = rRT0;

p0 = r 1 259.8 J>kg # K 2 (293 K)

r = 13.1369 1 10-6 2 p0

1461

13–97. Continued

Also, r =

m m = = 7.6923 m V 0.13 m3

Substituting this result into the previous equation, p0 = 0.5855 1 106 2 m

Taking the time derivative

dp0 dm = 0.5855 1 106 2 dt dt

However,

dm # = - m = -0.4386 1 10-6 2 p0, then this equation becomes dt

dp0 = -0.2568 p0 dt 300 kPa

L900 kPa ln p0 `

t dp0 = - 0.2568 dt p0 L0

300 kPa 900 kPa

= -0.2568t Ans.

t = 4.278 s = 4.28 s

Ans: 4.28 s 1462

13–98. The rocket has a converging–diverging exhaust nozzle. The absolute pressure at the inlet to the nozzle is 200 lb>in2, and the temperature of the fuel mixture is 3250 °R. If the mixture flows at 300 ft>s into the nozzle and exits with isentropic supersonic flow, determine the areas of the throat and exit plane. The inlet has a diameter of 18 in. The outside absolute pressure is 14.7 psi. The fuel mixture has k = 1.4 and R = 1600 ft # lb>slug # R.

SOLUTION The mixture is considered compressible. There is steady flow relative to the rocket. The Mach number of the inlet flow is Vin

M in =

1kRTin

=

300 ft>s

21.40 1 1600 ft # lb>slug # R 2 (3250 R)

= 0.1112

At the inlet plane using this result,

k

p0 = pin a1 +

k-1 k - 1 M in2 b 2

p0 = (200 psi) c 1 + a = 201.74 psi

1.4

1.4 - 1 1.4 - 1 b(0.1112)2 d 2

k+1

Ain *

A

=

1 ≥ M in

1 +

2(k - 1) k - 1 M in2 2 ¥ k + 1 2 1.4 + 1

2(1.4 - 1) 1.4 - 1 1 + a b(0.1112)2 p(9 in)2 2 1 = ≥ ¥ 0.1112 1.4 + 1 A* 2

A* = 48.53 in2 = 48.5 in2

Ans.

At the exit plane, k

k-1 k - 1 M out2b p0 = pout a1 + 2

201.74 psi = (14.7 psi) c 1 + a M out = 2.3595

1.4

1.4 - 1 1.4 - 1 bM out2 d 2

k+1

Aout *

A

=

1 ≥ M out

1 +

2(k - 1) k - 1 M out2 2 ¥ k + 1 2

1.4 + 1

Aout 48.53 in2

=

1 ≥ 2.3595

1 + a

2(1.4 - 1) 1.4 - 1 b(2.3595)2 2 ¥ 1.4 + 1 2

Aout = 112.36 in2 = 112 in2

Ans.

1463

Ans: A* = 48.5 in2 Aout = 112 in2

13–99. The rocket has a converging–diverging exhaust nozzle. The absolute pressure at the inlet to the nozzle is 200 lb>in2, and the temperature of the fuel mixture is 3250°R. If the mixture flows at 500 ft>s into the nozzle and exits with isentropic supersonic flow, determine the areas of the throat and exit plane and the mass flow through the nozzle. The inlet has a diameter of 18 in. The outside absolute pressure is 7 psi. The fuel mixture has k = 1.4 and R = 1600 ft # lb>slug # R.

SOLUTION The air is considered to be compressible. There is steady relative flow. Water is considered to be compressible. The Mach number of the inlet flow is M in =

Vin

=

2kRTin

500 ft>s

21.40 ( 1600 ft # lb>slug # R ) (3250 R)

= 0.1853

At the inlet plane using this result,

k

p0 = pina1 +

k-1 k - 1 M in 2 b 2

p0 = (200 psi) c 1 + a = 204.85 psi

1.4

1.4 - 1 1.4 - 1 b(0.1853)2 d 2 k+1

Ain *

A

1 ± M in

=

1 +

2(k - 1) k - 1 M in 2 2 ≤ k + 1 2 1.4 + 1

p (9 in)2 A*

=

1 ≥ 0.1853

1 + a

2(1.4 - 1) 1.4 - 1 b(0.1853)2 2 ¥ 1.4 + 1 2

A* = 79.83 in2 = 79.8 in2

Ans.

At the exit plane, k

k-1 k - 1 M out 2 b 2

p0 = pout a1 +

204.85 psi = (7 psi) c 1 + a M out = 2.8495

1.4

1.4 - 1 1.4 - 1 bM out 2 d 2 k+1

Aout *

A

=

1 ± M out

1 +

2(k - 1) k - 1 M out 2 2 ≤ k + 1 2 1.4 + 1

Aout 79.83 in2

=

1 ≥ 2.8495

1 + a

2(1.4 - 1) 1.4 - 1 b(2.8495)2 2 ¥ 1.4 + 1 2

Aout = 292.89 in2 = 293 in2

Ans.

1464

13–99. Continued

Applying the universal gas law at the inlet plane pin = rin RTin;

a200

lb 12 in. 2 b = rin ( 1600 ft # lb>slug # R ) (3250 R) ba 2 1 ft in

rin = 0.005538 slug>ft 3 Then, the mass flow rate is

2 9 # m = rinVin Ain = ( 0.005538 slug>ft 3 )( 500 ft>s ) Jp a ft b R 12

= 4.894 slug>ft 3 = 4.89 slug>ft 3

Ans.

Ans: A* = 79.8 in2 Aout = 293 in2 # m = 4.89 slug>ft 3 1465

*13–100. The jet engine is tested on the ground at standard atmospheric pressure of 101.3 kPa. If the fuel–air mixture enters the inlet of the 300-mm-diameter nozzle at 250 m >s, with an absolute pressure of 300 kPa and temperature of 800 K, and exits with supersonic flow, determine the velocity of the exhaust developed by the engine. Take k  = 1.4 and R = 249 J>kg # K. Assume isentropic flow.

300 mm dt

SOLUTION The mixture is considered to be compressible. The flow is steady. The Mach number of the inlet flow is M in =

Vin 2kRTin

At the inlet plane,

=

21.40 ( 249 J>kg # K ) (800 K)

= 0.4734

k - 1 M in 2 b 2

T0 = Tin a1 +

T0 = (800 K) c 1 + a p0 = pina1 +

250 m>s

1.4 - 1 b(0.4734)2 d = 835.86 K 2 k

k-1 k - 1 M in 2 b 2

p0 = (300 kPa) c 1 + a = 349.76 kPa

1.4

1.4 - 1 1.4 - 1 b(0.4734)2 d 2

At the exit plane, k

k-1 k - 1 p0 = pe a1 + Me2 b 2

1.4

1.4 - 1 1.4 - 1 349.76 kPa = (101.3 kPa)c 1 + a bM e 2 d 2

M e = 1.4574 T0 = Te a1 +

k - 1 Me2 b 2

835.86 K = Te c 1 + a Te = 586.64 K

1.4 - 1 b(1.4574)2 d 2

The velocity of the flow at the exit plane is Ve = M e 2kRTe = (1.4574) 21.40 ( 249 J>kg # K ) (586.64 K) = 659.08 m>s = 659 m>s

1466

Ans.

de

13–101. The jet engine is tested on the ground at standard atmospheric pressure of 101.3 kPa. If the fuel–air mixture enters the inlet of the 300-mm-diameter nozzle at 250 m>s, with an absolute pressure of 300 kPa and temperature of 800 K, determine the required diameter of the throat dt, and the exit diameter de, so that the flow exits with isentropic supersonic flow. Take k = 1.4 and R = 249 J>kg # K.

300 mm dt

de

SOLUTION The mixture is considered to be compressible. The flow is steady. The Mach number of the inlet flow is M in =

Vin 2kRTin

p0 = pina1 +

=

250 m>s

21.40 ( 249 J>kg # K ) (800 K)

= 0.4734

k

k-1 k - 1 M in2 b 2

p0 = (300 kPa) c 1 + a = 349.76 kPa

1.4

1.4 - 1 1.4 - 1 b(0.4734)2 d 2

k+1

Ain A*

=

1 ± M in

2(k - 1) k - 1 M in 2 2 ≤ k + 1 2

1 +

1.4 + 1

2(1.4 - 1) 1.4 - 1 p 1 + a b(0.4734)2 (0.3 m)2 2 4 1 = ≥ ¥ p 2 0.4734 1.4 + 1 d 4 t 2

Ans.

d t = 0.2541 m = 254 mm At the exit plane, k

k-1 k - 1 Me2 b p0 = pe a1 + 2

1.4

349.76 kPa = (101.3 kPa)c 1 + a M e = 1.4574

1.4 - 1 1.4 - 1 bM e 2 d 2

k+1

Ae A*

=

1 ± Me

1 +

p 2 d 4 e p (0.2541 m)2 4

2(k - 1) k - 1 Me2 2 ≤ k + 1 2 1.4 + 1

=

1 ≥ 1.4574

1 + a

2(1.4 - 1) 1.4 - 1 b(1.4574)2 2 ¥ 1.4 + 1 2

Ans.

d e = 0.2723 m = 272 mm

Ans: d t = 254 mm d e = 272 mm 1467

13–102. The nozzle is attached onto the end of the pipe. The air supplied from the pipe is at a stagnation temperature of 120°C and an absolute stagnation pressure of 800 kPa. Determine the mass flow from the nozzle if the backpressure is 60 kPa.

60 mm 20 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are T0 = (273 + 120°C) = 393 K and p0 = 800 kPa. For air k = 1.40 and R = 286.9 J>kg # K. Assuming that the nozzle chokes, Ae A*

=

p(0.03 m)2 p(0.01 m)2

= 9

Choose M 6 1 (subsonic) by interpolating the values in the table, M e = 0.06446 Using this result, p3 = 0.9971; p3 = 0.9971(800 kPa) = 797.68 kPa p0 Since the back pressure 60 kPa is less than p3, the nozzle chokes. Thus, M = 1 at the throat. T* = 0.8333; T0

T * = 0.8333(393 K) = 327.5 K

p* = 0.5283; p0

p* = 0.5283(800 K) kPa = 422.63 kPa

Using the universal gas law, p* = r*RT *;

422.63 ( 103 )

N = r* ( 286.9 J>kg # K ) (327.5 K) m2

r* = 4.4979 kg>m3 The velocity of the flow at the throat is V * = M * 2kRT * = (1) 21.40 ( 286.9 J>kg # K ) (327.5 K) = 362.69 m>s

Finally, the mass flow is # m = r*V *A* = ( 4.4979 kg>m3 )( 362.69 m>s ) 3 p(0.01 m)2 4 = 0.5125 kg>s = 0.513 kg>s

Ans.

Ans: 0.513 kg>s 1468

13–103. The nozzle is attached onto the end of the pipe. The air in the pipe is at a stagnation temperature of 120°C and an absolute stagnation pressure of 800 kPa. Determine the two values of the backpressure that will choke the nozzle yet produce isentropic flow. Also, what is the maximum velocity of the isentropic flow?

60 mm 20 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are T0 = (273 + 120°C) = 393 K and p0 = 800 kPa. For air k = 1.40 and R = 286.9 J>kg # K. Enter the table in Appendix B with Ae *

A

=

p(0.03 m)2 p(0.01 m)2

= 9

The two values of M are M = 0.06446 6 1

(subsonic)

M = 3.8061 7 1

(supersonic)

For subsonic flow, the table in Appendix B gives pe = 0.9971; pe = 0.9971(800 kPa) = 797.68 kPa = 798 kPa p0

Ans.

For supersonic flow, pe = 0.008553; p0

pe = 0.008553(800 kPa) = 6.846 kPa = 6.85 kPa Ans.

T = 0.2566; T0

T = 0.2566(393 K) = 100.84 K

The maximum velocity of the isentropic flow is V = M 2kRT = (3.8061) 21.40 ( 286.9 J>kg # K ) (100.84 K) = 765.99 m>s = 766 m>s

Ans.

Ans: For subsonic flow, pe = 798 kPa For supersonic flow, pe = 6.85 kPa V = 766 m>s 1469

*13–104. The jet plane creates a shock that forms in air having a temperature of 20°C and absolute pressure of 80 kPa. If it travels at 1200 m>s, determine the pressure and temperature just behind the shock wave.

SOLUTION The air is considered to be compressible. There is steady relative flow. Relative to the plane, the static temperature, pressure, and velocity of the air before the shock are T1 = (273 + 20°C) K = 293 K, p1 = 80 kPa, and V1 = 1200 m>s. For air k = 1.40 and R = 286.9 J>kg # K. The Mach number of the flow before the shock wave is M1 =

V1 2kRT1

=

1200 m>s

21.40 ( 286.9 J>kg # K ) (293 K)

= 3.4980

The static pressure after the shock can be determined using the equation or the table. p2 = 14.1085; p1

p2 = 14.1085(80 kPa) = 1128.68 kPa = 1.13 MPa Ans.

The Mach number after the shock wave can be determined using the equation or the table. M 2 = 0.4512 Using this result and the table the static temperature after the shock is T2 = 3.3123; T1

T2 = 3.3123(293 K) = 970.50 K = 971 K

1470

Ans.

13–105. The jet plane flies at M = 1.8 in still air at an altitude of 15 000 ft. If a shock forms at the air inlet for the engine, determine the stagnation pressure within the engine just behind the shock and the stagnation pressure a short distance within the chamber.

SOLUTION The air is considered to be compressible. There is steady relative flow. The stagnation pressure before the shock is the air pressure at an altitude of 15 000 ft which is ( p0 ) 1 = 1195 lb>ft 2. Ans. Applying the equation or the table with k = 1.40 for air, the stagnation pressure after the shock is

( p0 ) 2 = 0.8127; ( p0 ) 1

( p0 ) 2 = 0.8127 ( 1195 lb>ft 2 ) = 971.16

lb = 971 lb>ft 2 Ans. ft 2

Since no shock wave forms in the engine chamber, the stagnation pressure is constant throughout the chamber after the shock.

1471

Ans: 1p0 2 1 = 1195 lb>ft 2 1p0 2 2 = 971 lb>ft 2

13–106. A shock is produced by a jet plane flying at a speed of 2600 ft>s. If the air is at a temperature of 60°F and an absolute pressure of 12 lb>in2, determine the velocity of the air relative to the plane and its temperature just behind the shock.

SOLUTION The air is considered to be compressible. There is steady relative flow. Relative to the plane, the static temperature, pressure, and velocity of the air before lb the shock are T1 = (460 + 60°F) R = 520 R, p1 = 12 2 , and V1 = 2600 ft>s. For in air k = 1.40 and R = 1716 ft # lb>slug # R. The Mach number of the flow before the shock is M1 =

V1 2kRT1

=

2600 ft>s

21.40 ( 1716 ft # lb>slug # R ) (520 R)

= 2.3262

Using the equation or the table, the Mach number after the shock is M 2 = 0.5313 The temperature and velocity of the flow after the shock can be determined by applying the equations or the table. T2 = 1.9710; T1

T2 = 1.9710(520 R) = 1024.90 R = 1025°R

Ans.

V2 = 0.3207; V1

V2 = 0.3207 ( 2600 ft>s ) = 833.73 ft>s = 834 ft>s

Ans.

Ans: T2 = 1025°R V2 = 834 ft>s 1472

13–107. A shock is produced by a jet plane flying at a speed of 2600 ft>s. If the air is at a temperature of 60°F and an absolute pressure of 12 lb>in2, determine the stagnation pressure and the pressure just behind the shock.

SOLUTION The air is considered to be compressible. There is steady relative flow. Relative to the plane, the static temperature, pressure, and velocity of the air before the shock are T1 = (273 + 60°F) R = 520 R, p1 = 12 psi, and V1 = 2600 ft>s. For air k = 1.40 and R = 1716 ft # lb>slug # R. The Mach number of the flow before the shock is M1 =

V1 2kRT1

=

2600 ft>s

21.40 ( 1716 ft # lb>slug # R ) (520 R)

= 2.3262

Applying the equation or using the table, the Mach number after the shock is M 2 = 0.5313 The stagnation pressure before the shock can be determined using the equation or the table. p1

( p0 ) 1

= 0.07676;

( p0 ) 1 =

12 psi 0.07676

= 156.33 psi

The static stagnation pressure after the shock can be determined using the equations or the table. p2 = 6.1465; p1

( p0 ) 2 = 0.5718; ( p0 ) 1

p2 = 6.1465(12 psi) = 73.76 psi = 73.8 psi

( p0 ) 2 = 0.5718(156.33 psi) = 89.39 psi = 89.4 psi

1473

Ans. Ans.

Ans: p2 = 73.8 psi 1p0 2 2 = 89.4 psi

*13–108. A standing shock occurs in the pipe when the  upstream conditions for air have an absolute pressure of p1 = 80 kPa, temperature T1 = 75°C, and velocity V1 = 700 m>s. Determine the downstream pressure, temperature, and velocity of the air. Also, what are the upstream and downstream Mach numbers?

1 2

SOLUTION The air is considered to be compressible. The flow is steady. For air k = 1.40 and R = 286.9 J>kg # K. The Mach number before the shock wave is M1 =

V1 2kRT1

=

700 m>s

21.40 ( 286.9 J>kg # K ) (273 + 75°C) K

= 1.8723 = 1.87 Ans.

Using the result, the Mach number after the shock can be determined by applying the equation or using the table. Ans.

M 2 = 0.6011 = 0.601

The temperature, pressure, and velocity of the flow after the shock can be determined by using the equations, T2 = 1.5864; T1

T2 = 1.5864(273 + 75°C) = 552.09 K = 552 K

Ans.

p2 = 3.9232; p1

p2 = 3.9232(80 kPa) = 313.85 kPa = 0.314 MPa

Ans.

V2 = 0.4044; V1

V2 = 0.4044 ( 700 m>s ) = 283.07 m>s = 283 m>s

Ans.

1474

13–109. The large tank supplies air at a temperature of 350 K and an absolute pressure of 600 kPa to the nozzle. If the throat diameter is 0.3 m and the exit diameter is 0.5 m, determine the range of backpressures that will cause expansion shock waves to form at the exit.

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of those in the tank, i.e., T0 = 350 K and p0 = 600 kPa. For an expansion shock wave to form at the exit plane, the back pressure pb 6 p4, where p4 is the back pressure at which isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.5 m)2 A 4 = = 2.7778 p A* (0.3 m)2 4 Choose M 7 1, M 1 = 2.5557 Using this result to apply the equation or using the table, p4 = 0.05368; p4 = 0.05368(600 kPa) = 32.21 kPa = 32.2 kPa p0 Thus, Ans.

pb 6 32.2 kPa

Ans: pb 6 32.2 kPa 1475

13–110. The large tank supplies air at a temperature of 350 K and an absolute pressure of 600 kPa to the nozzle. If the throat diameter is 0.3 m and the exit diameter is 0.5 m, determine the range of backpressures that will cause oblique shock waves to form at the exit.

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of those in the tank, i.e., T0 = 350 K and p0 = 600 kPa. For an oblique shock wave to form at the exit plane, the back pressure must be in between pb and p4, which are the back pressures at which the standing normal shock wave is at the exit plane and the back pressure at which isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.5 m)2 A 4 = = 2.7778 p A* (0.3 m)2 4 Choose M 7 1 (supersonic), M 1 = 2.5557 Using this result to apply the equation or using the table, p4 = 0.05368; p4 = 0.05368(600 kPa) = 32.21 kPa p0 Using the result of M1 and p4 to apply the equation, p6 = 7.4534; p6 = 7.4534(32.21 kPa) = 240.05 kPa p4 Thus, Ans.

32.2 kPa 6 pb 6 240 kPa

Ans: 32.2 kPa 6 pb 6 240 kPa 1476

13–111. The nozzle is attached to the end of the pipe that carries air having an absolute stagnation pressure of 60 psi and stagnation temperature of 400°R. Determine the range of backpressures that will cause a standing shock to form within the nozzle.

1 in.

1.5 in.

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are T1 = 400 R and p0 = 60 psi. For a standing shock wave to form between the throat and exit plane of the nozzle, the back pressure must be between p6 and p3, which are the back pressures at which a standing shock wave is at the exit plane and isentropic subsonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (1.5 in.)2 A 4 = = 2.25 p A* (1 in.)2 4 And we obtain, M 1 = 0.2685 (subsonic) M 1 = 2.3282 (supersonic) Using the result of M 1 = 0.2685 to apply the equation or using the table, p3 = 0.9511; p3 = 0.9511(60 psi) = 57.066 psi p0 Similarly, with the result of M 1 = 2.3282, p4 = 0.07652; p4 = 0.07652(60 psi) = 4.5915 psi p0 Using the result of M 1 = 2.3282 and p4 to apply the equation or using the table, p6 = 6.1571; p6 = 6.1571(4.5915 psi) = 28.27 psi p4 Thus, Ans.

28.3 psi 6 pb 6 57.1 psi

Ans: 28.3 psi 6 pb 6 57.1 psi 1477

*13–112. The nozzle is attached to the end of the pipe that carries air having an absolute stagnation pressure of 60 lb>in2 and stagnation temperature of 400°R. Determine the range of backpressures that will cause oblique shocks to form at the nozzle exit.

1 in.

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are T0 = 400 R and p0 = 60 psi. For an oblique shock wave to form at the exit, the back pressure must be between p6 and p4, which are the back pressures at which a standing shock wave is at the exit plane and isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (1.5 in.)2 A 4 = = 2.25 p A* (1 in.)2 4 Choose M 1 7 1 for supersonic flow, M 1 = 2.3282 Using the result of M 1 to apply the equation or using the table, p4 = 0.07652; p4 = 0.07652(60 psi) = 4.5915 psi p0 Using the result of M 1 and p4 to apply the equation or using the table, p6 = 6.1571; p6 = 6.1571(4.5915 psi) = 28.27 psi p4 Thus, Ans.

4.59 psi 6 pb 6 28.3 psi

1478

1.5 in.

13–113. A 200-mm-diameter pipe contains air at a temperature of 10°C and an absolute pressure of 100 kPa. If a shock is formed in the pipe and the speed of the air in front of the shock is 1000 m>s, determine the speed of the air behind the shock.

SOLUTION The air is considered to be compressible. The flow is steady. For air, k = 1.40 and R = 286.9 J>kg # K. The Mach number before the shock is M1 =

V1 2kRT1

=

1000 m>s

21.40 ( 286.9 J>kg # K ) (273 + 10°C) K

= 2.9660

Using this result, the Mach number after the shock can be determined by applying the equation or using the table. M 2 = 0.4772 Using the results of M 1 and M 2 to apply the equation or using the table, V2 = 0.2614; V2 = 0.2614 ( 1000 m>s ) = 261.39 m>s = 261 m>s V1

Ans.

Ans: 261 m>s 1479

13–114. The jet is flying at M = 1.3, where the absolute air pressure is 50 kPa. If a shock is formed at the inlet of the engine, determine the Mach number of the air flow just within the engine where the diameter is 0.6 m. Also, what are the pressure and the stagnation pressure in this region? Assume isentropic flow within the engine.

M ! 1.3

0.4 m 0.6 m

SOLUTION The air is considered to be compressible. There is steady relative flow. Relative to the plane, the static air pressure and Mach number before the shock are p1 = 50 kPa and M 1 = 1.3. For air k = 1.40. The stagnation pressure before the shock can be determined by applying the equation or using the table, p1 50 kPa ( p0 ) 1 = = 0.3609; = 138.54 kPa 0.3609 ( p0 ) 1 The Mach number after the shock can be determined by applying the equation or using the table. M 2 = 0.7860 Using this result to apply the equation, the stagnation pressure after the shock is

( p0 ) 2 = 0.9794; ( p0 ) 1

( p0 ) 2 = 0.9794(138.54 kPa) = 135.68 kPa = 136 kPa

Ans.

At the entrance plane with M 2 = 0.7860, the equation or table gives A2 = 1.0443 A* Here, the area of the exit plane is A3. Then A3 A*

= a

p(0.3 m)2 A2 A3 ba b = 1.0443 J R A* A2 p(0.2 m)2

= 2.3497

Applying the equation or using the table and choosing M 3 6 1 (the flow must be subsonic since the nozzle does not choke), Ans.

M 3 = 0.2561 Using this result to apply the equation at the exit plane, p3 = 0.9554; p3 = 0.9554(135.68 kPa) = 129.63 kPa = 130 kPa ( p0 ) 2

Ans.

Ans: p0 = 136 kPa M = 0.256 p = 130 kPa 1480

13–115. The large tank supplies air at a temperature of 350 K and an absolute pressure of 600 kPa to the nozzle. If the throat diameter is 30 mm and the exit diameter is 60 mm, determine the range of backpressures that will cause oblique shock waves to form at the exit.

30 mm

60 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 350 K and p0 = 600 kPa. For an oblique shock wave to form at the exit plane, the back pressure must be between p6 and p4, which are the back pressures at which a standing shock wave is at the exit plane and isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.06 m)2 A 4 = = 4 p A* (0.03 m)2 4 Choose M 1 7 1 (supersonic), M 1 = 2.9402 Using the result of M 1 to apply the equation or using the table, p4 = 0.02979; p4 = 0.02979(600 kPa) = 17.87 kPa p0 Using the result of M 1 and p4 to apply the equation, p6 = 9.9188; p6 = 9.9188(17.87 kPa) = 177.27 kPa p4 Thus, Ans.

17.9 kPa 6 pb 6 177 kPa

Ans: 17.9 kPa 6 pb 6 177 kPa 1481

*13–116. The large tank supplies air at a temperature of 350 K and an absolute pressure of 600 kPa to the nozzle. If the throat diameter is 30 mm and the exit diameter is 60  mm, determine the range of backpressures that will cause a standing shock to form in the nozzle.

30 mm

60 mm

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation temperature and pressure are T0 = 350 K and p0 = 600 kPa. For a standing normal shock wave to form between the throat and the exit plane of the nozzle, the back pressure must be between p6 and p3, which are the back pressures at which a standing shock wave is at the exit plane and isentropic subsonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.06 m)2 A 4 = = 4 p A* (0.03 m)2 4 and we obtain M 1 = 0.1465 (subsonic) M 2 = 2.9402 (supersonic), Using the result of M 1 = 0.1465 to apply the equation or using the table, p3 = 0.9851; p0

p3 = 0.9851(600 kPa) = 591.06 kPa

Similarly, with the result of M 2 = 2.9402, p4 = 0.02979; p0

p4 = 0.02979(600 kPa) = 17.87 kPa

Using the result of M 2 = 2.9402 and p4 to apply the equation or using the table, p6 = 9.9188; p6 = 9.9188(17.87 kPa) = 177.27 kPa p4 Thus, Ans.

177 kPa 6 pb 6 591 kPa

1482

13–117. A shock is formed in the nozzle at C, where the diameter is 100 mm. If the air flows through the pipe at A at M A = 3.0 and the absolute pressure is pA = 15 kPa, determine the pressure in the pipe at B.

150 mm 100 mm

A

C B

SOLUTION The flow across the shock (at C) is unisentropic but from section C to B is isentropic. First, we need to determine the stagnation pressure at section A. Enter M A = 3.0 into the isentropic flow tables, pA

( p0 ) A

= 0.02722;

( p0 ) A =

( p0 ) A pA

( pA ) =

15 kPa = 551.07 kPa 0.02722

Next, we will consider the flow across the shock. Enter M 1 = 3.0 into the normal shock tables, M 2 = 0.47519

( p0 ) B = 0.32834; ( p0 ) A

( p0 ) B = 0.32834(551.07 kPa) = 180.94 kPa

Next we will consider the isentropic flow from section C to B. Here, an imaginary throat exists based on M 2 = 0.47519. Enter this value into the isentropic flow tables. Interpolation gives AC = 1.3905 ( A* ) ′ Then AB

(A ) ′ *

=

p(0.075 m)2 AB b = 1.3905 £ § = 3.1286 p(0.05 m)2 ( A ) ′ AC AC *

a

Again enter this value into the isentropic flow tables, realizing that M B 6 1 and performing the interpolation, pB M B = 0.1890 = 0.97539 ( p0 ) B Then pB = 0.97539 (180.94 kPa) = 176.49 kPa = 176 kPa

Ans.

Ans: 176 kPa 1483

13–118. Air at a temperature of 20°C and an absolute pressure of 180 kPa flows from a large tank through the nozzle. Determine the backpressure at the exit that causes a shock wave to form at the location where the nozzle diameter is 50 mm.

20 mm

50 mm

80 mm

SOLUTION Since a normal shock develops within the divergent portion of the nozzle, the nozzle is choked, that is, M = 1 at the throat. The flow from the inlet to just to the left of normal shock, section 1, and from just to the right of the shock, section 2, to the exit are isentropic but the flow across the normal shock is unisentropic. First, we will consider the isentropic flow from the inlet to section 1. Here, ( p0 ) 1 = ( p0 ) i = 180 kPa, A1 p(0.025 m)2 = = 6.25. Enter this value into isentropic flow tables, and after A* p(0.01 m)2 interpolation, select M 1 7 1, M 1 = 3.4114 Next consider the flow across the normal shock, Enter M 1 = 3.4109 into normal shock tables. Interpolation gives M 2 = 0.4547

( p0 ) 2 = 0.22995; ( p0 ) 1

( p0 ) 2 = 0.22995(180 kPa) = 41.39 kPa

Finally consider the isentropic flow from section 2 to the exit, which has a different throat (imaginary). Enter M 2 = 0.4547 into the isentropic flow tables, A2 Then

( A* ) ′ Ae

( A* ) ′

= 1.4372

=

A2

( A* ) ′

*

p(0.04 m)2 Ae = 1.4372 £ § = 3.6791 A2 p(0.025 m)2

Again enter this value into the isentropic flow tables, realizing that M e 6 1. Performing the interpolation, M e = 0.1597

pe

( p0 ) e

= 0.98235

Here, ( p0 ) e = ( p0 ) 2 = 41.39 kPa. Then Ans.

pe = 0.98235(41.39 kPa) = 40.66 kPa = 40.7 kPa

Ans: 40.7 kPa 1484

13–119. The nozzle is attached to the large tank A containing air. If the absolute pressure within the tank is 14.7 psi, determine the range of pressures at B that will cause expansion shock waves to form at the exit plane.

1.75 in. 1 in.

A

B

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation pressure is that in the tank; i.e., p0 = 14.7 psi. For an expansion shock wave to form at the exit plane, the back pressure pb 6 p4, where p4 is the back pressure at which isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation at the exit plane with p (1.75 in.)2 A 4 = = 3.0625 p A* (1 in.)2 4 Choose M 7 1 (supersonic), M 1 = 2.6592 Using this result to apply the equation at the exit plane or using the table, p4 = 0.04573; p0

p4 = 0.04573(14.7 psi) = 0.6723 psi

Thus, Ans.

pb 6 0.672 psi

Ans: pb 6 0.672 psi 1485

*13–120. The converging–diverging nozzle is attached to the large tank A containing air. If the absolute pressure within the tank is 14.7 psi, determine the range of pressures at B that will cause a standing normal shock wave to form within the nozzle.

1.75 in. 1 in.

A

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation pressure is that in the tank; i.e., p0 = 14.7 psi. For a standing normal shock wave to form between the throat and the exit plane of the nozzle, the back pressure pb must be between p6 and p3, which are the back pressures at which a standing normal shock wave is at the exit plane and isentropic subsonic flow occurs. For air k = 1.40. Enter the table or apply the equation at the exit plane with p (1.75 in.)2 A 4 = = 3.0625 p A* (1 in.)2 4 We obtain, M 1 = 0.1932 (subsonic) M 2 = 2.6592 (supersonic) Using the result of M 1 = 0.1932 to apply the equation or using the table, p3 = 0.9743; p3 = 0.9743(14.7 psi) = 14.32 psi p0 Similarly, with the result of M 2 = 2.6592, p4 = 0.04573; p0

p4 = 0.04573(14.7 psi) = 0.6723 psi

Using the result of M 2 = 2.6592 and p4 to apply the equation or using the table, p6 = 8.0834; p6 = 8.0834(0.6723 psi) = 5.434 psi p4 Thus, Ans.

5.43 psi 6 pb 6 14.3 psi

1486

B

13–121. The nozzle is attached to the large tank A containing air. If the absolute pressure within the tank is 14.7 psi, determine the range of pressures at B that will cause oblique shock waves to form at the exit plane.

1.75 in. 1 in.

A

B

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation pressure is that in the tank; i.e., p0 = 14.7 psi. For oblique shock waves to form at the exit plane, the back pressure must be between p6 and p4, which are the back pressures at which a standing normal shock wave is at the exit plane  and isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (1.75 in.)2 A 4 = = 3.0625 p A* (1 in.)2 4 Choose M 7 1 (supersonic), M 1 = 2.6592 Using this result to apply the equation at the exit plane or using the table, p4 = 0.04573; p0

p4 = 0.04573(14.7 psi) = 0.6723 psi

Using the result of M 1 and p4, p6 = 8.0834; p6 = 8.0834(0.6723 psi) = 5.434 psi p4 Thus, Ans.

0.672 psi 6 pb 6 5.43 psi

Ans: 0.672 psi 6 pb 6 5.43 psi 1487

13–122. Air in the large reservoir A has an absolute pressure of 70 lb>in2. Determine the range of backpressures at B so that a standing normal shock wave will form within the nozzle.

0.2 m

0.150 m

B A

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation pressure is that in the tank; i.e., p0 = 70 psi. For a standing normal shock wave to form between the throat and the exit plane of the nozzle, the back pressure pb must be between p6 and p3, which are the back pressures at which a standing normal shock wave is at the exit plane and isentropic subsonic flow occurs. For air k = 1.40. Enter the table or apply the equation at the exit plane with p (0.2 m)2 4 A = = 1.7778 p A* 2 (0.15 m) 4 We obtain, M 1 = 0.3500 (subsonic) M 2 = 2.0618 (supersonic) Using the result of M 1 = 0.3500 to apply the equation or using the table, p3 = 0.9188; p0

p3 = 0.9188(70 psi) = 64.31 psi

Similarly, with the result of M 1 = 2.0618, p4 = 0.1161; p0

p4 = 0.1161(70 psi) = 8.1248 psi

Using the result of M 1 = 2.0618 and p4 to apply the equation or using the table, p6 = 4.7930; p4

p6 = 4.7930(8.1248 psi) = 38.94 psi

Thus, Ans.

38.9 psi 6 pb 6 64.3 psi

Ans: 38.9 psi 6 pb 6 64.3 psi 1488

13–123. Air in the large reservoir A has an absolute pressure of 70 lb >in2. Determine the range of backpressures at B so that oblique shock waves appear at the exit.

0.2 m

0.150 m

B A

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation pressure is that in the tank; i.e., p0 = 70 psi. For oblique shock waves to form at the exit plane, the back pressure must be between p6 and p3, which are the back pressures at which a standing normal shock wave is at the exit plane and isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.2 m)2 4 A = = 1.7778 p A* 2 (0.15 m) 4 Choose M 7 1 (supersonic), M 1 = 2.0618 Using this result to apply the equation or using the table, p4 = 0.1161; p0

p4 = 0.1161(70 psi) = 8.1248 psi

Using the result of M 1 and p4 to apply the equation or using the table, p6 = 4.7930; p4

p6 = 4.7930(8.1248 psi) = 38.94 psi

Thus, Ans.

8.12 psi 6 pb 6 38.9 psi

Ans: 8.12 psi 6 pb 6 38.9 psi 1489

*13–124. Air in the large reservoir A has an absolute pressure of 70 lb>in2. Determine the range of backpressures at B so that expansion shock waves appear at the exit.

0.2 m

0.150 m

B A

SOLUTION The air is considered to be compressible. The flow is steady. The stagnation pressure is that in the tank; i.e., p0 = 70 psi. For an expansion shock wave to form at the exit plane, the back pressure p6 6 p4, where p4 is the back pressure at which isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.2 m)2 4 A = = 1.7778 p A* (0.15 m)2 4 Choose M 7 1 (supersonic), M 1 = 2.0618 Using this result to apply the equation or using the table, p4 = 0.1161; p0

p4 = 0.1161(70 psi) = 8.12 psi

Thus, Ans.

pb 6 8.12 psi

1490

*13–125. The cylindrical plug is fired with a speed of 150  m> s in the pipe that contains still air at 20°C and an absolute pressure of 100 kPa. This causes a shock wave to move down the pipe as shown. Determine its speed and the pressure acting on the plug.

150 m/s

SOLUTION We will consider a control volume that contains the shock wave that moves to the right with a speed of Vs. The flow of air can be considered steady with respect to the control volume. The velocity of the air to the left and to the right of the control volume are (Va)1 = Vp (the speed of piston) and (Va)2 = 0, respectively. Therefore, the velocity of the air relative to the control volume can be determined using the relative equation.

Vs

)Va) 2 = Vp

)Va) 2 = 0

2

Va = Vcv + Va>cv

1

To the left of control volume (section 2), (Va>cv)2 = Vs - Vp d

+ ) - Vp = - VS + (Va>cv)2 (d

(a)

To the right of the control volume (section 1) ( d ) 0 = - VS + (Va>cv)1

(Va>cv)1 = Vs d

Since the flow is steady with respect to the moving control volume, continuity requires 0 rdV + rVf>cs # dA = 0 0t Lcv Lcs 0 + p2(Va>cv)2 A + p1[ -(Va>cv)1 A] = 0 p2(Vs - Vp) - p1 VS = 0 r2 Vs = r1 Vs - Vp Since Vs = M1C1, then the above equation becomes r2 M1 = r1 M 1 - Vp >c1

(1)

If we combine Eq. 13–77 and Eq. 13–78 and after going through quite a bit of algebra manipulation, (k + 1) M 12 p2 = p1 (k - 1)M 12 + 2

(2)

Equating Eqs. (1) and (2), (k + 1) M 12 M1 = M 1 - Vp >c1 (k - 1)M 12 + 2

(k - 1)M 12 + 2 = (k + 1)M 12 - (k + 1)(Vp >c1) M 1

M 12 - a

k + 1 Vp ba b M 1 - 1 = 0 2 c1

1491

13–125. Continued

For air, k = 1.4 and R = 286.9 J>kg # K. At T1 = (273 + 20) K = 293 K, c1 = 2kRT1 = 21.4(286.9 J>kg # K)(293 K) = 343.05 m>s. Thus, the above equation becomes M 12 - a

150 m>s 1.4 + 1 ba b M1 - 1 = 0 2 343.05 m>s

M 12 - 0.5247 M 1 - 1 = 0 Solving for the positive root, M 1 = 1.2962 Thus,

Vs = M 1c 1 = 1.2962 ( 343.05 m>s ) = 444.66 m>s = 445 m>s

Ans.

Applying Eq. 13–81, 2(1.4) p2 p2 2k k - 1 1.4 - 1 = M2 ; = c d ( 1.2962 2 ) p1 k + 1 1 k + 1 100 kPa 1.4 + 1 1.4 + 1 p2 = 179.35 kPa = 179 kPa

Ans.

Ans: Vs = 445 m>s p2 = 179 kPa 1492

13–126. Air flows at 800 m>s through a long duct in a wind tunnel, where the temperature is 20°C and the absolute pressure is 90 kPa. The leading edge of a wing in the tunnel is represented by the 7° wedge. Determine the pressure created on its top surface if the angle of attack is set at a = 2°.

3.5! 800 m/s

a 3.5!

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then V1

M1 =

2kRT1

=

800 m>s

21.4(286.9 J>kg # K)(293 K)

3.5° a = 2° 3.5° l = 3.5° + 2 = 5.5°

2 cot bu ( M 12 sin2 bu - 1 ) M12(k + cos 2 bu) + 2

(a)

2 cot bu ( 2.3320 sin bu - 1 ) 2

tan 1.5° =

= 3.5° – 2° = 1.5°

= 2.3320

For the upper surface, uu = 1.5° (Geometry shown in Fig. a). Applying tan uu =

u

2

2.33202(1.4 + cos 2 bu) + 2

Solving by trial and error to find the weak shock angle, bu = 26.5193° The normal component of M 1 is

Then,

3 (M 1)n 4 u

(p2)u p1

=

(p2)u 90 kPa

= M 1 sin bu = 2.3320 sin 26.5193° = 1.0412

2k k - 1 2 3 (M 1)n 4 u k + 1 k + 1

= J

2(1.4)

1.4 + 1

R ( 1.04122 ) -

1.4 - 1 1.4 + 1 Ans.

(p2)u = 98.83 kPa = 98.8 kPa

Ans: 98.8 kPa 1493

13–127. Air flows at 800 m>s through a long duct in a wind tunnel, where the temperature is 20°C and the pressure is 90  kPa. The leading edge of a wing in the tunnel is represented by the 7° wedge. Determine the pressure created on its bottom surface if the angle of attack is set at  a = 2°.

3.5! 800 m/s

a 3.5!

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 =

V1 2kRT1

=

800 m>s

21.4(286.9 J>kg # K)(293 k)

3.5° a = 2° 3.5° l = 3.5° + 2 = 5.5°

2 cot bl ( M 12 sin2 bl - 1 )

(a)

M12(k + cos 2bl) + 2

tan 5.5° =

= 3.5° – 2° = 1.5°

= 2.3320

For the lower surface, ul = 5.5° (Fig. a). Applying tan ul =

u

2 cot bl ( 2.33202 sin2 bl - 1 ) 2.33202(1.4 + cos 2bt) + 2

Solving by trial and error to find the weak shock angle, bl = 29.7697° Then,

Then

3 (M 1)n 4

(p2)l p1

=

(p2)l 90 kPa

l

= M 1 sin bl = 2.3320 sin 29.7697° = 1.1579

2 2k k - 1 3 (M 1)n 4 l k + 1 k + 1

= J

2(1.4)

1.4 + 1

R ( 1.15792 ) -

1.4 - 1 1.4 + 1 Ans.

(p2)l = 125.77 kPa = 126 kPa

Ans: 126 kPa 1494

*13–128. Air flows at 800 m>s through a long duct in a wind tunnel, where the temperature is 20°C and the absolute pressure is 90 kPa. The leading edge of a wing in the tunnel can be represented by the 7° wedge. Determine the pressure created on its top surface if the angle of attack is set at a = 5°.

3.5! 800 m/s

a 3.5!

SOLUTION

u

= 5° – 3.5° = 1.5°

First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 =

V1 2kRT1

=

800 m>s

21.4 ( 286.9 J>kg # K )( 293K )

3.5°

= 2.3320 l

= 3.5 + 5° = 8.5°

For the upper surface, an expansion wave will be formed. For M 1 = 2.3320, its corresponding deflection angle with respect to the reference state can be determined using the Prandtl–Meyer expansion function. k + 1 k - 1 ( M 2 - 1 ) d - tan - 1 ( 2M 2 - 1 ) v = tan - 1 c Ak - 1 Ak + 1 v( M 1 ) =

1.4 + 1 1.4 - 1 tan - 1 c (2.33202 - 1) d - tan - 1 ( 22.33202 - 1 ) = 35.0806° A 1.4 - 1 A 1.4 + 1

The deflection angle of the surface is u = 1.5° (Geometry shown in Fig. a). Then u = v(M 2) - v(M 1);

1.5° = v(M 2) - 35.0806°

v(M 2) = 36.5806°

Enter this value into the table and after performing the interpolation, M 2 = 2.3932 Since the expansion is an isentropic process, the isentropic flow table can be used. For M 1 = 2.3320, the interpolation gives ( p1)u ( p0)u

= 0.076074

For M 2 = 2.3932, ( p2)u ( p0)u

= 0.069134

Using these ratios, ( p2)u ( p1)u Then

=

( p2)u ( p0)u 1 c d = 0.069134 a b = 0.90878 ( p0)u ( p1)u 0.076074

( p2)u = 0.90878 (90 kPa) = 81.79 kPa = 81.8 kPa

1495

Ans.

3.5°

(a)

a = 5°

13–129. Air flows at 800 m>s through a long duct in a wind tunnel, where the temperature is 20°C and the absolute pressure is 90 kPa. The leading edge of a wing in the tunnel can be represented by the 7° wedge. Determine the pressure created on its bottom surface if the angle of attack is set at  a = 5°.

3.5! 800 m/s

a 3.5!

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 =

V1 2kRT1

=

800 m>s

21.4 ( 286.9 J>kg # K )( 293K )

= 2.3320

For the lower surface, ul = 8.5° (Fig. a). tan ul =

2 cot bl ( M12 sin2bl - 1 ) M 12(k + cos 2bt) + 2

tan 8.5° =

2 cot bl (2.33202 sin2bl - 1) 2.33202 (1.4 + cos 2bt) + 2

Solving by trial and error, bl = 32.4599° Then [(M 1)n]l = M 1 sin bl = 2.3320 sin 32.4599° = 1.2516 Then ( p2)l p1

=

( p2)l 90 kPa

2k k - 1 [(M 1)n]2l k + 1 k + 1 = c

2(1.4) 1.4 + 1

d ( 1.25162 ) -

1.4 - 1 1.4 + 1 Ans.

( p2)l = 149.48 kPa = 149 kPa

Ans: 149 kPa 1496

13–130. A jet plane is flying in air that has a temperature of 8°C and absolute pressure of 90 kPa. The leading edge of the wing has the wedge shape shown. If the plane has a speed of 800  m>s and the angle of attack is 2°, determine the pressure and temperature of the air at the upper surface A just in front or to the right of the oblique shock wave that forms at the leading edge.

A

3!

a " 2!

3!

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 =

V1 2kRT1

=

800 m>s

21.4 ( 286.9 J>kg # K )( 273 + 8 ) K

= 2.3813

A

= 3° – 2° = 1° 3°

For the upper surface A, uA = 1° (Geometry shown in Fig. a). tan uA = tan 1° =

a = 2°

2 cot bA ( M 1 2 sin2 bA - 1 ) M 12 (k + cos 2bB) + 2



2 cot bA (2.38132 sin2 bA - 1)

(a)

2.38132 (1.4 + cos 2bB) + 2

Solving by trial and error to find the weak shock angle, bA = 25.5698° The normal component of M 1 is (M 1)n = M 1 sin bA = 2.3813 sin 25.5698° = 1.0278 Enter this value into the normal shock table and performing the interpolation, p2 = 1.06579; pA = p2 = 1.06579 (90 kPa) = 95.92 kPa = 95.9 kPa Ans. p1 T2 = 1.01837; TA = T2 = 1.01837 (281 K) = 286.16 K = 286 K T1

Ans.

Ans: pA = 95.9 kPa TA = 286 K 1497

13–131. A jet plane is flying in air that has a temperature of 8°C and absolute pressure of 90 kPa. The leading edge of the wing has the wedge shape shown. If the plane has a speed of 800  m>s and the angle of attack is 2°, determine the pressure and temperature of the air at the lower surface B just in front or to the right of the oblique shock wave that forms at the leading edge.

A

3!

a " 2!

3!

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 =

V1

=

2kRT1

800 m>s

21.4 ( 286.9 J>kg # K )( 273 + 8 ) K

= 2.3813



For the lower surface b, uB = 5° (Geometry shown in Fig. a). tan uB =

2 cot bB ( M 1 2 sin2 bB - 1 ) M 1 ( k + cos 2bB ) + 2 2

tan 5° =

a = 2°

2



2

B = 3° + 2° = 5°

2 cot bB (2.3813 sin bB - 1) 2.38132 (1.4 + cos 2bB) + 2

(a)

Solving by trial and error to find the weak shock angle, bB = 28.7466° The normal component of M 1 is (M 1)n = M 1 sin bB = 2.3813 sin 28.7466° = 1.1453 Enter this value into the normal shock table and performing the interpolation, p2 = 1.36365; pB = p2 = 1.36365 (90 kPa) = 122.73 kPa = 123 kPa Ans. p1 T2 = 1.09361; TB = T2 = 1.09361 (281 K) = 307.30 K = 307 K T1

Ans.

Ans: pB = 123 kPa TB = 307 K 1498

*13–132. The leading edge on the wing of the aircraft has the shape shown. If the plane is flying at 900 m>s in air that has a temperature of 5°C and absolute pressure of 60 kPa, determine the angle b of an oblique shock wave that forms on the wing. Also, determine the pressure and temperature on the wing just in front or to the right of the shock.

b 5! 5! b

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 = Here, u =

V1 2kRT1

=

d 10° = = 5°. 2 2

tan u = tan 5° =

900 m>s

21.4 ( 286.9 J>kg # K )( 273 + 5 ) K

= 2.6933

2 cot b ( M 1 2 sin2 b - 1 ) M 1 2 ( k + cos 2bB ) + 2 2 cot b (2.69332 sin2 b - 1) 2.69332 (1.4 + cos 2bB) + 2

Solving by trial and error to find the weak shock angle, Ans.

b = 25.5511° = 25.6° The normal component of M 1 is (M 1)n = M 1 sin b = 2.6933 sin 25.5511° = 1.1617

Enter this value into the normal shock table, and after performing the interpolation, p2 = 1.40782; p2 = 1.40782 (60 kPa) = 84.47 kPa = 84.5 kPa p1

Ans.

T2 = 1.10394; T2 = 1.10394 (278 K) = 306.90 K = 307 K T1

Ans.

1499

900 m/s

13–133. A jet plane is flying at M = 2.4, in air having a temperature of 2°C and absolute pressure of 80 kPa. If the leading edge of the wing has an angle of d = 16°, determine the velocity, pressure, and temperature of the air just in front or to the right of the oblique shock that forms on the wing. What is the angle d of the leading edge that will cause the shock wave to separate from the front of the wing?

M ! 2.4 d

SOLUTION Here u =

d 16° = = 8°. 2 2

tan u = tan 8° =

2 cot b ( M 1 2 sin2 b - 1 ) M 12 (K + cos 2b) + 2 2 cot b (2.42 sin2 b - 1) 2.42 (1.4 + cos 2b) + 2

Solving by trial and error to find the weak shock angle, b = 31.1489° The normal component of M 1 is (M 1)n = M 1 sin b = 2.4 sin 31.1489° = 1.2414 Thus, (M 2)n = 0.81751 p2 = 1.63126; p2 = 1.63126 (80 kPa) = 130.50 kPa = 131 kPa p1

Ans.

T2 = 1.15398; T2 = 1.15398 (275 K) = 317.34 K = 317 K T1

Ans.

(M 2)n = M 2 sin (b - u);

0.81751 = M 2 sin (31.1489° - 8°) M 2 = 2.0795

The velocity of the air in front of the shock is V2 = M 2 2kRT2 = 2.0795 c 21.4 ( 286.9 J>kg # K)(317.34 K) d = 742.42 m>s = 742 m>s

1500

Ans.

13–133. Continued

The separation of shock wave occurs when d = 2umax. Here, umax can be determined by plotting u vs b. For M 1 = 2.4, this yields tan u =

2 cot b (2.42 sin2 b - 1) 2.42(1.4 + cos 2b) + 2

b 24.62° 30° 40° 50° 60° 70° 80° u 0° 6.72° 16.56° 23.78° 28.09° 27.76 19.17°

90° 0

62.5°

65°

67.5°

28.54° 28.68° 28.44

(deg.)

max

30 = 28.7

20

10

0

ß (deg.) 10

20

30

40

50

60

70

80

90

Thus, Ans.

d = 2umax = 2(28.7°) = 57.4°

Ans: p2 = 131 kPa T2 = 317 K V2 = 742 m>s d = 57.4° 1501

13–134. The jet plane is flying upward such that its wings make an angle of attack of 15° with the horizontal. The plane is traveling at 700 m>s, in air having a temperature of 8°C and absolute pressure of 90 kPa. If the leading edge of the wing has an angle of 8°, determine the pressure and temperature of the air just in front or to the right of the expansion waves.

4! 4!

700 m/s

SOLUTION

= 11°

Since the surface bends downward from the direction of flow, the flow will undergo isentropic expansion. First, we must determine the Mach number of the flow before the expansion. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 =

V1 2kRT1

=

700 m>s

21.4 ( 286.9 J>kg # K )( 273 + 8 ) K

a = 15° 4° 4°

= 2.0836

For this Mach number, its corresponding deflection angle with respect to the reference state can be determined using the Prandtl-Meyer expansion function. v =

15!

(a)

k + 1 k - 1 ( M 2 - 1) d - tan - 1 ( 2M 2 - 1 ) tan - 1 c Ak - 1 Ak + 1

v(M 1) =

1.4 + 1 1.4 - 1 ( 2.08362 - 1 ) d - tan-1 1 22.08362 - 1 2 = 28.6569° tan - 1 c A 1.4 - 1 A 1.4 + 1

Referring to the geometry in Fig. a, the deflection angle of the surface is u = 11°. Then u = v(M 2) - v(M 1);

11° = v(M 2) - 28.6569° v(M 2) = 39.6569°

Thus, M 2 = 2.5230 Since the expansion is an isentropic process, the isentropic flow table can be used. For M 1 = 2.0836, the interpolation gives T1 = 0.53525 T0

p1 = 0.11219 p0

For M 2 = 2.5230, the interpolation gives T2 = 0.43993 T0

p2 = 0.056476 p0

Using these ratios, T2 T2 T0 1 = a b a b = (0.43993) a b = 0.82191 T1 T0 T1 0.53525

Then,

p2 p0 p2 1 = a b a b = (0.056476) a b = 0.50340 p1 p0 p1 0.11219

T2 = 0.82191 (281 K) = 230.96 K = 231 K

Ans.

p2 = 0.50340 (90 kPa) = 45.31 kPa = 45.3 kPa

Ans.

Ans: T2 = 231 K p2 = 45.3 kPa 1502

13–135. Nitrogen gas at a temperature of 30°C and an absolute pressure of 150 kPa flows through the large rectangular duct at 1200 m>s. When it comes to the transition, it is redirected as shown. Determine the angle b of the oblique shock that forms at A, and the temperature and pressure of the nitrogen just in front or to the right of the wave.

20!

B

b 20! A

SOLUTION First, we must determine the Mach number behind the shock. For nitrogen, k = 1.4 and R = 296.8 J>kg # K. Then M1 = Here, u = 20°.

V1 2kRT1

tan u =

=

1200 m>s

21.4 ( 296.8 J>kg # K )( 273 + 30 ) K

= 3.3819

2 cot b (M 1 2 sin2 b - 1)

tan 20° =

M 1 2(k + cos 2b) + 2 2 cot b (3.38192 sin2 b - 1) 3.38192 (1.4 + cos 2b) + 2

Solving by trial and error to find the weak shock angle, Ans.

b = 35.2345° = 35.2° The normal component of M 1 is (M 1)n = M 1 sin b = 3.3819 sin 35.2345° = 1.9511 p2 = 4.2746; p2 = 4.2746 (150 kPa) = 641.19 kPa = 641 kPa p1

Ans.

T2 = 1.64818; T2 = 1.64818 (303 K) = 499.40 K = 499 K T1

Ans.

Ans: b = 35.2° p2 = 641 kPa T2 = 499 K 1503

*13–136. Nitrogen gas at a temperature of 30°C and an absolute pressure of 150 kPa flows in the large rectangular duct at 1200 m>s. When it comes to the transition, it is redirected as shown. Determine the temperature and pressure just in front or to the right of the expansion waves that form on the duct at B.

20!

B

b 20! A

SOLUTION First, we must determine the Mach number of the flow before the expansion. For nitrogen, k = 1.4 and R = 296.8 J>kg # K. Then M1 =

V1 2kRT1

=

1200 m>s

21.4 ( 296.8 J>kg # K )( 273 + 30 ) K

= 3.3819

For this Mach number, its corresponding deflection angle with respect to the reference state can be determined using the Prandtl–Meyer expansion function or the Prandtl–Meyer expansion table in Appendix B. v =

k + 1 k - 1 ( M 2 - 1) d - tan - 1a 2M 2 - 1b tan - 1 c Ak - 1 Ak + 1

v(M 1) =

1.4 + 1 1.4 - 1 ( 3.38192 - 1 ) d - tan - 1 a 23.38192 - 1b tan - 1 c A 1.4 - 1 A 1.4 + 1

= 56.6072°

The deflection angle of the surface is u = 20°. Then u = v(M 2) - v(M 1);

20° = v(M 2) - 56.6072° v(M 2) = 76.6072°

Enter this value into the Prandtl–Meyer expansion table and after performing the interpolation, M 2 = 4.9667 Since the expansion is an isentropic process, the isentropic flow table can be used. For M 1 = 3.3819, T1 = 0.30423 T0

p1 = 0.015547 p0

For M 2 = 4.9667, the interpolation gives T2 = 0.16856 T0

p2 = 0.001970 p0

Using these ratios, T2 T2 T0 1 = a b = (0.16856) a b = 0.55405 T1 T0 T1 0.30423 Then

p2 p2 p0 1 = a b = 0.001970 a b = 0.12671 p1 p0 p1 0.015597

T2 = 0.55405 (303 K) = 167.88 K = 168 K

Ans.

p2 = 0.12671 (150 kPa) = 19.01 kPa = 19.0 kPa

Ans.

1504

13–137. The wing of a jet plane is assumed to have the profile shown. It is traveling horizontally at 900 m>s, in air having a temperature of 8°C and absolute pressure of 85 kPa. Determine the pressure that acts on the top surface in front or to the right of the oblique shock at A and in front or to the right of the expansion waves at B.

A

SOLUTION First, we must determine the Mach number behind the shock. For air, k = 1.4 and R = 286.9 J>kg # K. Then M1 = Here, u = 3°.

V1 2kRT1

tan u = tan 3° =

=

900 m>s

21.4 ( 286.9 J>kg # K )( 273 + 8 ) K

= 2.6789

2 cot b (M 1 2 sin2 b - 1) M 1 2(k + cos 2b) + 2 2 cot b (2.67892 sin2 b - 1) 2.67892 (1.4 + cos 2b) + 2

Solving by trial and error to find the weak shock angle, b = 24.1083° The normal component of M 1 is (M 1)n = M 1 sin b = 2.6789 sin 24.1083° = 1.0942 p2 = 1.2302; pA = p2 = 1.2302 (85 kPa) = 104.57 kPa = 105 kPa p1

Ans.

(M 2)n = 0.91634 (M 2)n = M 2 sin (b - u);

0.91634 = M 2 sin (24.1083° - 3°) M 2 = 2.5445

The expansion waves will occur at the corner of surface A and B. For M 2 = 2.5445, its corresponding deflection angle with respect to the reference state can be determined using the Prandt–Meyer expansion table. v(M 2) = 40.1534° The deflection angle of the surface is u = 3° + 3° = 6°. Then u = v( M 3 ) - v( M 2 );

6° = v(M 3) - 40.1534° v(M 3) = 46.1534°

Thus, M 3 = 2.8196 Since the expansion is an isentropic process, for M 2 = 2.5445, p2 = 0.054618 p0

1505

3! 3!

B 900 m/s

3!

13–137. Continued

For M 3 = 2.8196, p3 = 0.035762 p0 Using this ratio

Then

p3 p3 p0 1 = a b = 0.035762 a b = 0.65477 p2 p0 p2 0.054618 pB = p3 = 0.65477 (104.57 kPa) = 68.47 kPa = 68.5 kPa

Ans.

Ans: pA = 105 kPa pB = 68.5 kPa 1506

14–1. Water flows at 5 m>s towards the impeller of the axial-flow pump. If the impeller is rotating at 60 rad>s, and it has a mean radius of 200 mm, determine the initial blade angle b1 so that a1 = 90°. Also, find the relative velocity of the water as it flows onto the blades of the impeller.

b1

35!

5 m/s

200 mm 60 rad/s

SOLUTION tan b1 =

5 12

(Vrel(1 Ans.

b1 = 22.62° = 22.6° sin 22.62° =

b1

5

V = Va = 5 m/s

( Vrel ) 1

( Vrel ) 1 = 13.0 m>s

Ans.

b1

(Vt(1 = U = 12 m/s

Ans: b1 = 22.6° ( Vrel ) 1 = 13.0 m>s 1507

14–2. Water flows at 5 m>s towards the impeller of the axial-flow pump. If the impeller is rotating at 60 rad>s, and it has a mean radius of 200 mm, determine the velocity of the water as it exits the blades, and the relative velocity of the water as it flows off the blades of the impeller.

b1 5 m/s

35!

200 mm 60 rad/s

SOLUTION Water is considered to be incompressible. The relative flow is steady. The speed at the midpoint on the impeller is

(Vrel (2

U = vrm = ( 60 rad>s )( 0.2 m ) = 12 m>s

b2 = 35°

Here, Va = 5 m>s and b2 = 35°. Using these results, the velocity diagram of the water at the tail of the impeller blade is shown in Fig. a. From the geometry, Va = ( Vrel ) 2 sin b2;

5 m>s = ( Vrel ) 2 sin35°

( Vrel ) 2 = 8.717 m>s = 8.72 m>s

Va = 5 m/s

(Vt (2 a2 V2

Ans. b2 = 35°

Also, Va =

3U

- ( Vt ) 2 4 tan b2 ;

5 m>s =

3 12 m>s

- ( Vt ) 2 4 tan 35°

U = 12 m/s

( Vt ) 2 = 4.859 m>s

(a)

Using the result of (Vt)2, V2 = 2Va2 + (Vt)2 = 2 ( 5 m>s ) 2 + ( 4.859 m>s ) 2

Ans.

= 6.972 m>s = 6.97 m>s

Ans: ( Vrel ) 2 = 8.72 m>s V2 = 6.97 m>s 1508

14–3. Water flows through the axial-flow pump at 4 ( 10-3 ) m3 >s, while the impeller has an angular velocity of 30 rad>s. If the blade tail angle is 35°, determine the velocity and tangential velocity component of the water when it leaves the blade. rw = 1000 kg>m3. 35!

30 rad/s

75 mm 150 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The radius of the midpoint of the impeller blade is rm = 0.0375 m +

0.0375 m = 0.05625 m 2

Thus, the speed at the midpoint on the impeller is U = vrm = ( 30 rad>s )( 0.05625 m ) = 1.6875 m>s Here, the cross-sectional area within the open region of the impeller is

b2 = 35°

A = p 3 (0.075 m)2 - (0.0375 m)2 4 = 4.21875 ( 10 - 3 ) pm2

U = 1.6875 m/s

Thus, the axial flow velocity can be determined from

Va = 0.3018 m/s

V2

(Vrel(2

a2

(Vt( 2

b2 = 35°

(a)

Q = VaA; 4 ( 10 - 3 ) m3 >s = Va 3 4.21875 ( 10 - 3 ) p 4 m2 Va = 0.3018 m>s

Here, b2 = 35°. Then the velocity diagram of the water at the blade tail is shown in Fig. a. From the geometry of this figure, Va =

3U

- ( Vt ) 2 4 tan b2;

0.3018 m>s =

3 1.6875 m>s

- ( Vt ) 2 4 tan 35°

( Vt ) 2 = 1.2565 m>s = 1.26 m>s

V2 = 2 ( 0.3018 m>s ) 2 + ( 1.2565 m>s ) 2

Ans.

Ans.

= 1.29 m>s

Ans: ( Vt ) 2 = 1.26 m>s V2 = 1.29 m>s 1509

*14–4. Water flows at 6 m>s through an axial-flow pump. If the impeller blade has an angular velocity of v = 100 rad>s, determine the velocity of the water when it is delivered to the stator blades. The impeller blades have a mean radius of 100 mm and the angles shown.

Impeller

Stator 100 mm

45!

6 m/s

30!

v

SOLUTION Water is considered to be incompressible. The relative flow is steady. The velocity of the impeller is U = vrm;

U = 10 m/s 30°

U = ( 100 rad>s ) (0.1 m) = 10 m>s

Using the result of U, the velocity diagram of the water at the blade tail is shown in Fig. a. From the geometry of this figure, U - ( Vt ) 2 = Va tan 30°;

10 m>s - ( Vt ) 2 = ( 6 m>s ) tan 30°

V2

(Vt(2

Va 30°

( Vt ) 2 = 6.536 m>s

(a)

Thus, the magnitude of the water when it leaves the blade tail is V2 = 2Va2 + ( Vt ) 22 = 2 ( 6 m>s ) 2 + ( 6.536 m>s ) 2

Ans.

= 8.87 m>s

1510

(Vrel ( 2

14–5. Water flows at 6 m>s through an axial-flow pump. If the impeller blade angles are 45° and 30° as shown, determine the power supplied to the water by the pump when v = 100 rad>s. The impeller blades have a mean radius of 100 mm. The volumetric flow is 0.4 m3 >s.

Impeller

Stator 100 mm

45!

6 m/s

30!

v

SOLUTION Water is considered to be incompressible. The relative flow is steady.

U = 10 m/s 30°

The velocity of the impeller is U = vrm;

U = ( 100 rad>s ) (0.1 m) = 10 m>s

Va

Using the result of U, the velocity diagram of the water at the blade tail is shown in Fig. a. From the geometry of this figure, U - ( Vt ) 2 = Va tan 30°;

V2

(Vt (2

(Vrel(2

30°

10 m>s - ( Vt ) 2 = ( 6 m>s ) tan 30°

(a)

( Vt ) 2 = 6.536 m>s

U = 10 m/s

From Fig. b, U - ( Vt ) 1 = Va tan 45°

45°

45°

10 - ( Vt ) 1 = 6 m>s tan 45°

V1

( Vt ) 1 = 4 m>s # Wpump = rQU ( Vt2 - Vt1 )

(Vt (1 Va = 6 m/s

= ( 1000 kg>m3 ) ( 0.4 m3 >s ) ( 10 m>s ) ( 6.536 m>s - 4 m>s ) = 10.1 kW

45°

(Vrel(1

Ans. (b)

Ans: 10.1 kW 1511

14–6. If the water velocity onto the impeller is 3 m>s, determine the required initial blade angle b1. Also, what is the power supplied to the water by the pump? The impeller blades have a mean radius of 50 mm and v = 180 rad>s. The volumetric flow is 0.9 m3 >s.

v 60!

SOLUTION

b1

The air is considered to be compressible.

50 mm

U = vr = ( 180 rad>s )( 0.05 m ) = 9 m>s From the kinematic diagram Fig. a, tan b1 =

3 m>s 9 m>s

3 m/s

Ans.

b1 = 18.4° Vt1 = 0 From the kinematic diagram Fig. b, Va = 3 m>s =

3U

- ( Vt ) 2 4 tan b2

3 9 m>s

- ( Vt ) 2 4 tan 60°

Vt2 = 7.268 m>s . Wpump = PQU ( Vt2 - Vt1 ) . Wpump = ( 1000 kg>m3 )( 0.9 m3 >s )( 9 m>s )( 7.268 m>s - 0 ) = 58.9 kW

Ans.

V = Va = 3 m/s

b1

b1 U = 9 m/s (a)

V2 b2 = 60°

U = 9 m/s

Va = 3 m/s

(Vrel(2

b2 = 60°

a2

(Vt (2 (b)

Ans: b#1 = 18.4° W = 58.9 kW 1512

14–7. An axial-flow pump has an impeller with a mean radius of 100 mm that rotates at 1200 rev>min. At the exit the stator blade angle a2 = 70°. If the velocity of the water leaving the impeller is 8  m>s, determine the tangential component of the velocity and the relative velocity of the water at this instant.

SOLUTION Water is considered to be incompressible. The relative flow is steady. The angular velocity of the impeller is v = a1200

rev 1 min 2p rad ba ba b = 40p rad min 60 s 1 rev

Then, the velocity at the midpoint of the impeller is

U = vrm = ( 40p rad>s )( 0.1 m ) = 12.5664 m>s Using this result, a2 = 70°, and V2 = 8 m>s, the velocity diagram of the water at the blade tail is shown in Fig. a. From the geometry of this figure,

V2 = 8 m/s

( Vt ) 2 = V2 cos a2 = ( 8 m>s ) cos 70° = 2.7362 m>s Va = ( Vt ) 2 tan a2 = (2.7362 m>s ) tan 70° = 7.5175 m>s tan b2 = =

b2

Va

a2 = 70°

U = 12.566 m/s

U - ( Vt ) 2 7.5175 m>s

Va

(Vrel(2

b2

(Vt (2 (a)

12.5664 m>s - 2.7362 m>s

b2 = 37.407° U - ( Vt ) 2 = ( Vrel ) 2 cos b2

( Vrel ) =

U - ( Vt ) 2 cos b2

=

12.5664 m>s - 2.7362 m>s cos 37.407° Ans.

= 12.4 m>s

Ans: 12.4 m>s 1513

*14–8. The radial ventilation fan is used to force air into the ducts of a building. If the air is at a temperature of 20°C, and the shaft is rotating at 60 rad>s, determine the power output of the motor. The blades have a width of 30 mm. Air enters the blades in the radial direction and is discharged with a velocity of 50 m>s at the angle shown.

30!

50 m/s

60 rad/s

SOLUTION 125 mm

The air is considered to be incompressible. The relative flow is steady. The velocity diagram of the water on the blade tail is shown in Fig. a. From the geometry of this figure,

( Vt ) 2 = ( 50 m>s ) sin 30° = 25 m>s ( Vr ) 2 = ( 50 m>s ) cos 30° = 43.301 m>s The flow rate in the pump is

(Vr(2

Q = ( Vr ) 2 A2 = ( 43.301 m>s ) [2p(0.125 m)(0.03 m)]

V2 = 50 m/s

= 1.0203 m3 >s

30°

For air at T = 20° C, r = 1.202 kg>m3 (Appendix A). Since the air enters the impeller blade radially, ( Vt ) 1 = 0. The torque applied by the motor is therefore T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1.202 kg>m3 )( 1.0203 m3 >s ) 3 (0.125 m)(25 m>s) - 0 4 = 3.8324 N # m

Thus, the power output of the motor is . Ws = Tv = (3.8324 N # m) ( 60 rad>s ) = 229.94 W

Ans.

= 230 W

1514

(Vt (1 (a)

14–9. The blades of the centrifugal pump are 30 mm wide and are rotating at 60 rad>s. Water enters the blades in the radial direction and flows off the blades with a velocity of 20 m>s as shown. If the discharge is 0.4 m3 >s, determine the torque that must be applied to the shaft of the pump.

20 m/s

100 mm

60 rad/s 250 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The radial component of the water velocity at the tail of the blade is Q = ( Vr ) 2A2

0.4 m3 >s = ( Vr ) 2[2p(0.25 m)(0.03 m)]

( Vr ) 2 = 8.4883 m>s

Here, V2 = 20 m>s. Thus, the velocity diagram of the water on the blade tail is shown in Fig. a. From the geometry of this figure,

V2 = 20 m/s

( Vt ) 2 = 2V 22 - ( Vr ) 22 = 2 ( 20 m>s ) 2 - (8.4883 m>s ) 2 = 18.1094 m>s

a1

(Vt (2

Since the flow enters the impeller blade radially, ( Vt ) 1 = 0. The torque applied to the pump shaft is

(Vr(2 = 8.4883 m/s

T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 0.4 m3 >s ) c ( 0.25 m )( 18.1094 m>s ) - 0 d = 1810.94 N # m = 1.81 kN # m

(a)

Ans.

Ans: 1.81 kN # m 1515

14–10. Air enters the 3-in.-wide blades of the blower in the radial direction and is discharged from the blades at a tail angle of b2 = 38°. Determine the power required to turn the blades at 100 rad>s, producing a discharge of 120 ft3 >s. Take ra = 2.36 1 10-3 2 slug>ft 3.

b2 ! 38"

100 rad/s

10 in. 14 in.

SOLUTION The air is considered to be incompressible. The relative flow is steady. The speed of the point on the blade tail is U2 = vr2 = ( 100 rad>s ) a

14 ft b = 116.67 ft>s 12

Since the air enters the impeller blade radially ∆H =

U2Q cot b2 U22 = g 2pr2bg

( 116.67 ft>s ) 2 32.2 ft>s2

-

(116.67 ft>s) ( 120 ft 3 >s ) cot 38°

= 119.04 ft

2pa

14 3 ft ba ft b ( 32.2 ft>s2 ) 12 12

Using this result the power required is . Ws = ∆HrQg = (119.04 ft) 3 2.36 ( 10 - 3 ) slug>ft 3 4 ( 120 ft 3 >s )( 32.2 ft>s2 )

= a1085.5

1 hp ft # lb ba b = 1.97 hp s 550 ft # lb>s

Ans.

Ans: 1.97 hp 1516

14–11. The radial-flow pump impeller rotates at 600 rev>min. If the width of the blades is 2.5 in., and the blade head and tail angles are as shown, determine the ideal pump head developed by the pump. Water is initially guided radially onto the impeller blades.

b2 ! 20"

b1 ! 35" 3.75 in. 600 rev/min 2 in.

SOLUTION Water is considered to be incompressible. The relative flow is steady. The angular velocity of the impeller is v = a

(Vrel(1

600 rev 1 min 2p rad ba ba b = 20p rad>s min 60 s 1 rev

V1

Thus, the speed of points on the head and tail of the blades are U1 = vr1 = ( 20p rad>s ) a U2 = vr2 = ( 20p rad>s ) a

b1 = 35°

2 ft b = 10.4720 ft>s 12

b1 = 35°

U1 = 10.4720 ft/s

(a)

3.75 ft b = 19.6350 ft>s 12

Since the water is initially guided radially onto the impeller blades V1 = ( Vr ) 1 and ( Vt ) 1 = 0. The velocity diagram of the water on the blade head is shown in Fig. a. From the geometry of this figure V1 = U1 tan b1 = ( 10.472 ft>s ) tan 35° = 7.3326 ft>s Thus, the flow rate is Q = V1A1 = ( 7.3326 ft>s ) c 2p a

The ideal pump head is ∆H =

2 2.5 ft ba ft b d = 1.5997 ft 3 >s 12 12

19.6350 ft>s U2Q cot b2 U 22 = g 2pr2bg 32.2 ft>s2

= 5.4212 ft = 5.42 ft

( 19.6350 ft>s )( 1.5997 ft3 >s ) cot 20° 2pa

3.75 2.5 ft ba ft b ( 32.2 ft>s2 ) 12 12 Ans.

Ans: 5.42 ft 1517

*14–12. The radial-flow pump impeller rotates at 600 rev>min. If the width of the blades is 2.5 in., and the blade head and tail angles are as shown, determine the flow and the ideal power the pump supplies to the water. Water is initially guided radially onto the impeller blades.

b2 ! 20"

b1 ! 35" 3.75 in. 600 rev/min 2 in.

SOLUTION Water is considered to be incompressible. The relative flow is steady. The angular velocity of the impeller is

V2 = 20 m/s

600 rev 1 min 2p rad ba ba b = 20p rad>s v = a min 60 s 1 rev

a1

Thus, the speed of points on the head and tail of the blades are U1 = vr1 = ( 20p rad>s ) a U2 = vr2 = ( 20p rad>s ) a

2 ft b = 10.4720 ft>s 12

(Vr(2 = 8.4883 m/s

3.75 ft b = 19.6350 ft>s 12

(a)

Since the water is initially guided radially onto the impeller blades V1 = ( Vr ) 1 and ( Vt ) 1 = 0. The velocity diagram of the water on the blade head is shown in Fig. a. From the geometry of this figure V1 = U1 tan b1 = ( 10.472 ft>s ) tan 35° = 7.3326 ft>s Thus, the flow rate is Q = V1A1 = ( 7.3326 ft>s ) c 2p a

The ideal pump head is ∆H =

2 2.5 ft ba ft b d = 1.5997 ft 3 >s = 1.60 ft 3 >s 12 12

19.6350 ft>s U2Q cot b2 U 22 = g 2pr2bg 32.2 ft>s2

= 5.4212 ft

Ans.

( 19.6350 ft>s )( 1.5997 ft3 >s ) cot 20° 2pa

3.75 2.5 ft ba ft b ( 32.2 ft>s2 ) 12 12

Finally, the power supplied by the pump is # Ws = ∆HrQg = (5.4212 ft) ( 62.4 lb>ft 3 )( 1.5997 ft 3 >s ) = a541.16

(Vt(2

1 hp ft # lb ba b = 0.9839 hp = 0.984 hp s 550 ft # lb>s

1518

Ans.

14–13. The radial-flow pump has a 60-mm-wide impeller with the dimensions shown. If the blades rotate at 160 rad>s, determine the discharge if the water enters each blade in the radial direction. 40! 60! 75 mm 160 rad/s 200 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady.

(Vrel(1

The speed of a point on the head of the blade is U1 = vr1 = ( 160 rad>s ) (0.075 m) = 12 m>s Here, the water is required to enter the impeller blade radially, so that V1 = ( Vr ) 1 and ( Vt ) 1 = 0. Also, b1 = 60°. The velocity diagram of the water on the blade head is shown in Fig. a. From the geometry of this figure

V1

b1 = 60°

V1 = U1 tan b1 = ( 12 m>s ) tan 60° = 20.7846 m>s Thus, the flow rate in the pump is

b1 = 60°

Q = V1A1 = ( 20.7846 m>s ) [2p(0.075 m)(0.06 m)]

U1 = 12 m/s

= 0.5877 m3 >s = 0.588 m3 >s

Ans.

1519

(a)

Ans: 0.588 m3 >s

14–14. The radial-flow pump has a 60-mm-wide impeller with the radial dimensions shown. If the blades rotate at 160 rad>s and the discharge is 0.3 m3 >s, determine the power supplied to the water. 40! 60! 75 mm 160 rad/s 200 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The speed of points on the head and tail of the blade are

(Vrel(1

U1 = vr1 = ( 160 rad>s ) (0.075 m) = 12 m>s

From the flow rate, the radial components of the water velocity at the head and tail of the blade are Q = ( Vr ) 1A1

0.3 m >s = ( Vr ) 1[2p(0.075 m)(0.06 m)]

V1 a1

3

U1 = 12 m/s

0.3 m3 >s = ( Vr ) 2[2p(0.2 m)(0.06 m)]

(a)

( Vr ) 2 = 3.979 m>s

Here, b1 = 60° and b2 = 40°. Thus, the velocity diagram of the water on the blade head and tail is shown in Figs. a and b, respectively. From the geometry of Fig. a,

( Vr ) 1 = 3 U1 - ( Vt ) 1 4 tan b1;

10.61 m>s =

3 12 m>s

- ( Vt ) 1 4 tan 60°

3 32 m>s

- ( Vt ) 2 4 tan 40°

( Vt ) 1 = 5.874 m>s

From the geometry of Fig. b,

( Vr ) 2 = 3 U2 - ( Vt ) 2 4 tan b2;

3.979 m>s =

(Vt (1

b1 = 60°

( Vr ) 1 = 10.61 m>s

Q = ( Vr ) 2A2

b1 = 60°

(Vr(1 = 10.61 m/s

U2 = vr2 = ( 160 rad>s ) (0.2 m) = 32 m>s

b2 = 40°

(Vr(2 = 3.979 m/s

V2

U2 = 32 m/s

a2

(Vt (2

b2 = 40°

(b)

( Vt ) 2 = 27.258 m>s

The torque applied to the water is T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 0.3 m3 >s ) 3 0.2 m ( 27.258 m>s ) - 0.075 m ( 5.874 m>s ) 4 = 1503.32 N # m

Then, the power supplied to the water is # Ws = Tv = (1503.32 N # m) ( 160 rad>s ) = 240.53 ( 103 ) W = 241 kW Ans.

Ans: 241 kW 1520

14–15. The radial-flow pump has a 60-mm-wide impeller with the radial dimensions shown. If the blades rotate at 160 rad>s and the discharge is 0.3 m3 >s, determine the ideal head developed by the pump. 40! 60! 75 mm 160 rad/s 200 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The speed of points on the head and tail of the blade are

(Vrel(1

U1 = vr1 = ( 160 rad>s ) (0.075 m) = 12 m>s

From the flow rate, the radial components of the water velocity at the head and tail of the blade are Q = ( Vr ) 1A1;

0.3 m3 >s = ( Vr ) 1[2p(0.075 m)(0.06 m)]

V1 a1

U1 = 12 m/s

0.3 m3 >s = ( Vr ) 2[2p(0.2 m)(0.06 m)]

(a)

( Vr ) 2 = 3.979 m>s

Here, b1 = 60° and b2 = 40°. Thus, the velocity diagram of the water on the blade head and tail is shown in Figs. a and b, respectively. From the geometry of Fig. a,

( Vr ) 1 = 3 U1 - ( Vt ) 1 4 tan b1;

10.61 m>s =

3 12 m>s

- ( Vt ) 1 4 tan 60°

3 32 m>s

- ( Vt ) 2 4 tan 40°

b2 = 40°

( Vt ) 1 = 5.874 m>s

(Vr(2 = 3.979 m/s

V2

U2 = 32 m/s

From the geometry of Fig. b,

( Vr ) 2 = 3 U2 - ( Vt ) 2 4 tan b2;

(Vt (1

b1 = 60°

( Vr ) 1 = 10.61 m>s Q = ( Vr ) 2A2;

b1 = 60°

(Vr(1 = 10.61 m/s

U2 = vr2 = ( 160 rad>s ) (0.2 m) = 32 m>s

a2

(Vt (2

b2 = 40°

(b)

3.979 m>s =

( Vt ) 2 = 27.258 m>s

The ideal pump head is ∆H = =

U2 ( Vt ) 2 - U1 ( Vt ) 1 g

( 32 m>s )( 27.258 m>s ) - ( 12 m>s )( 5.874 m>s ) 9.81 m>s2 Ans.

= 81.73 m = 81.7 m

Ans: 81.7 m 1521

*14–16. The velocity of water flowing onto the 40-mm-wide impeller blades of the radial-flow pump is directed at 20° as shown. If the flow leaves the blades at the blade angle of 40°, determine the torque the pump must exert on the impeller.

40!

50!

250 mm

100 mm

V1 20! 60 rad/s

SOLUTION Water is considered to be incompressible. The relative flow is steady.

(Vrel(1

The speeds of points on the head and tail of the blade are

(Vr(1

U1 = vr1 = ( 60 rad>s ) (0.1 m) = 6 m>s U2 = vr2 = ( 60 rad>s ) (0.25 m) = 15 m>s

(Vt (1

Using the result of U1, a1 = 20°, and b1 = 50°, the velocity diagram of the water is shown in Fig. a. Using the law of sines, 6 m>s V1 = ; sin 50° sin 110°

V1

b1 = 50°

110° b1 = 50°

a1 = 20° (a)

V1 = 4.8912 m>s

U1 = 6 m/s

Then,

(Vr(2 = 0.6692 m/s

( Vr ) 1 = ( 4.8912 m>s ) sin 20° = 1.6729 m>s

b2 = 40°

( Vt ) 1 = ( 4.8912 m>s ) cos 20° = 4.5963 m>s

(Vrel(2

Thus, the flow rate of pump is

The radial component of the water velocity on the blade is Q = ( Vr ) 2A2;

0.04204 m3 >s = ( Vr ) 2[2p(0.25 m)(0.04 m)]

( Vr ) 2 = 0.6692 m>s

Using the results of U2 and 1 Vr 2 2 and b2 = 40°, the velocity diagram of the water on the blade tail is shown in Fig. b. From the geometry of this figure,

1 Vr 2 2

=

3 U2

-

1 Vt 2 2 4 tan b2;

0.6692 m>s =

3 15 m>s

1 Vt 2 2

-

1 Vt 2 2 4 tan 40°

= 14.2025 m>s

The torque the pump shaft exerts on the impeller is T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 0.04204 m3 >s ) 3 (0.25 m) ( 14.2025 m>s ) - (0.1 m) ( 4.5963 m>s ) 4

= 129.96 N # m = 130 N # m

Ans.

1522

(Vt (2 (b)

Q = ( Vr ) 1A1 = ( 1.6729 m>s ) [2p(0.1 m)(0.04 m)] = 0.04204 m3 >s

V2 b2 = 40° U2 = 15 m/s

14–17. The velocity of water flowing onto the 40-mm-wide impeller blades of the radial-flow pump is directed at 20° as shown. If the flow leaves the blades at the blade angle of 40°, determine the total head developed by the pump.

40!

50!

250 mm

100 mm

V1 20! 60 rad/s

SOLUTION Water is considered to be incompressible. The relative flow is steady.

(Vrel(1

The speed of points on the head and tail of the blade are

(Vr(1

U1 = vr1 = ( 60 rad>s ) (0.1 m) = 6 m>s U2 = vr2 = ( 60 rad>s ) (0.25 m) = 15 m>s

(Vt (1

Using the result of U1, a1 = 20°, and b2 = 50°, the velocity diagram of the water is shown in Fig. a. Using the law of sines, 6 m>s V1 = ; sin 50° sin 110°

V1

b1 = 50°

110° b1 = 50°

a1 = 20° (a)

V1 = 4.8912 m>s

U1 = 6 m/s

Then,

( Vr ) 1 = ( 4.8912 m>s ) sin 20° = 1.6729 m>s

(Vr(2 = 0.6692 m/s

( Vt ) 1 = ( 4.8912 m>s ) cos 20° = 4.5963 m>s

b2 = 40°

(Vrel(2

Continuity 0 rdV + rV # dA = 0 0t Lcv Lcs

V2

(Vt (2

b2 = 40° U2 = 15 m/s

(b)

0 - ( Vr ) 1A1 + ( Vr ) 2A2 = 0 - ( 1.6729 m>s ) [2p(0.1 m)(0.04 m)] + ( Vr ) 2[2p(0.1 m)(0.04 m)] = 0

( Vr ) 2 = 0.6692 m>s Using the results of U2 and ( Vr ) 2 and b2 = 40°, the velocity diagram of the water on the blade tail is shown in Fig. b. From the geometry of this figure,

( Vr ) 2 = 3 U2 - ( Vt ) 2 4 tan b2;

0.6692 m>s =

3 15 m>s

- ( Vt ) 2 4 tan 40°

( Vt ) 2 = 14.2025 m>s

The ideal pump head ∆H = =

U2 ( Vt ) 2 - U1 ( Vt ) 1 g

( 15 m>s )( 14.2025 m>s ) - ( 6 m>s )( 4.5963 m>s ) 9.81 m>s2 Ans.

= 18.91 m = 18.9 m

Ans: 18.9 m 1523

14–18. Water flows through the centrifugal pump impeller such that the entrance velocity is V1 = 6 m>s and the exit velocity is V2 = 10 m>s. If the discharge is 0.04 m3 >s and the width of each b1ade is 20 mm, determine the torque that must be applied to the pump shaft.

V2

a2

175 mm V 1 v

a1 80 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The radial components of the water velocity at the head and tail of the blade can be determined using the flow rate Q - ( Vr ) 1A1

0.04 m3 >s = ( Vr ) 1[2p(0.08 m)(0.02 m)]

( Vr ) 1 = 3.9789 m>s Q - ( Vr ) 2A2

(Vr(1 = 3.9789 m/s V1 = 6 m/s

a1

0.04 m3 >s = ( Vr ) 2[2p(0.175 m)(0.02 m)]

( Vr ) 2 = 1.8189 m>s

(Vt (1

Here, V1 = 6 m>s and V2 = 10 m>s. Thus, the velocity diagram of the water on the blade head and tail is shown in Figs. a and b, respectively. From the geometry of Fig. a,

( Vt ) 1 = 2V12 - ( Vr )12 = 2 ( 6 m>s ) 2 - ( 3.9789 m>s ) 2 = 4.4909 m>s

(a)

(Vr(2 = 1.8189 m/s

From the geometry of Fig. b,

V2 = 10 m/s

( Vt ) 2 = 2V22 - ( Vr ) 22 = 2 ( 10 m>s ) 2 - ( 1.8189 m>s ) 2 = 9.8332 m>s

a2

The torque supplied by the pump shaft is

(Vt (2

(b)

T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 0.04 m3 >s ) 3 (0.175 m) ( 9.8332 m>s ) - (0.08 m) ( 4.4909 m>s ) 4 = 54.46 N # m = 54.5 N # m

Ans.

Ans: 54.5 N # m 1524

14–19. Water flows through the centrifugal pump impeller at the rate of 0.04 m3 >s. If the blades are 20 mm wide and the velocities at the entrance and exit are directed at the angles a1 = 45° and a2 = 10°, respectively, determine the torque that must be applied to the pump shaft.

V2

a2

175 mm V 1 v a

1

80 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The radial components of the water velocity at the head and tail of the blade can be determined using the flow rate Q = ( Vr ) 1A1

V1

0.04 m3 >s = ( Vr ) 1[2p(0.08 m)(0.02 m)]

a1 = 45°

( Vr ) 1 = 3.9789 m>s Q = ( Vr ) 2A2

(Vr(1 = 3.9789 m/s

0.04 m >s = ( Vr ) 2[2p(0.175 m)(0.02 m)] 3

(Vt (1 (a)

( Vr ) 2 = 1.8189 m>s

Here, a1 = 45° and a2 = 10°. Thus, the velocity diagram of the water on the blade head and tail can be constructed as shown in Figs. a and b, respectively. From the geometry of Fig. a,

(Vr(2 = 1.8189 m/s V2 a2 = 10°

( Vt ) 1 = ( Vr ) 1 cot a1 = ( 3.9789 m>s ) (cot 45°) = 3.9789 m>s

(Vt (2

(b)

From the geometry of Fig. b,

( Vt ) 2 = ( Vr ) 2 cot a2 = ( 1.8189 m>s ) (cot 10°) = 10.3156 m>s The torque supplied by the pump shaft is T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 0.04 m3 >s ) 3 (0.175 m) ( 10.3156 m>s ) - (0.08 m) ( 3.9789 m>s ) 4

= 59.48 N # m = 59.5 N # m

Ans.

Ans: 59.5 N # m 1525

*14–20. Show that the ideal head for a radial-flow pump can be determined from ∆H = (U2V2 cos a2)>g, where V2 is the velocity of the water leaving the impeller blades. Water enters the impeller blades in the radial direction.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Since water enters the impeller blade radially, then ( Vt ) 1 = 0. From the geometry of the velocity diagram of the water at the blade tail shown in Fig. a,

( Vt ) 2 = V2 cos a2

∆H = =

(Vr(2 b2

V2 a2

(Vt (2

Thus,

3 U2 ( Vt ) 2

(Vrel(2

- U1 ( Vt ) 1 4 g

=

(a)

U2 ( V2 cos a2 ) - 0 g

U2V2 cos a2 g

(Q.E.D)

1526

b2 U2

14–21. Water flows radially onto the 40-mm-wide blades of the centrifugal pump impeller and exits with a velocity V2 at an angle of 20°, as shown. If the impeller is turning at 10 rev>s and the flow is 0.04 m3>s, determine the ideal head supplied to the water, the torque required to turn the impeller, and the power supplied to the pump if it has an efficiency of h = 0.65.

V2

40 mm

20! 120 mm

40 mm

10 rev/s

40 mm

120 mm

SOLUTION Assume that the flow is steady and water is incompressible. From the discharge Q = ( Vr ) 2 ( 2pr2b ) ;

0.04 m3 >s = ( Vr ) 2 [2p(0.12 m) (0.04 m)]

( Vr ) 2 = 1.326 m>s

(Vt (2

b2

U2 = 2.4 m/s

The angular speed of the impeller is rev 2p rad v = a10 ba b = 20p rad>s s 1 rev

b2

Then the velocity of the tail of the blade is

U2 = vr2 = ( 20p rad>s ) (0.12 m) = 2.4 p m>s

(Vrel(2 V2

(Vr(2 = 1.326 m/s

a2 = 20° (a) Tail

From the kinematics diagram at the tail, Fig. a,

( Vt ) 2 = ( Vr ) 2 cot a2 = ( 1.326 m>s ) cot 20° = 3.644 m>s b2 = tan - 1 £

( Vr ) 2 1.326 m>s § = tan - 1 a b = 18.80° 2.4 p m>s - 3.644 m>s U2 - ( Vt ) 2

Since the flow enters the blade radially (a1 = 90°), the torque supplied to the impeller’s shaft, is T = rQr2 ( Vr ) 2 cot a2 = ( 1000 kg>m3 )( 0.04 m3 >s ) (0.12 m) ( 1.326 m>s ) cot 20°

= 17.49 N # m = 17.5 N # m

Ans.

The power output of the pump is

#

Wout = PQU2(Vr)2 cot a2 = ( 1000 kg>m3 )( 0.04 m3 >s )( 2.4p m>s )( 1.326 m>s ) cot 20° = 1098.99 W

Thus, the power input to the pump is

# Wout # 1098.99 W Win = = 1690.76 W = 1.69 kW = h 0.65

Ans.

The head supplied by the pump is hpump = =

U2Q cot b2 U 22 g 2pr2bg

( 2.4p m>s ) 2 2

9.81 m>s

-

( 2.4p m>s )( 0.04 m3 >s ) cot 18.80° 2p(0.12 m)(0.04 m) ( 9.81 m>s2 ) Ans.

= 2.80 m

Ans: # T = # 17.5 N m 1Ws 2 pump = 1.69 kW hpump = 2.80 m 1527

14–22. The impeller of the centrifugal pump is rotating at 1200 rev>min and produces a flow of 0.03 m3>s. Determine the speed of the water as it exits the blade, and the ideal power and the ideal head produced by the pump.

30!

60!

50 mm

150 mm

1200 rev/min

SOLUTION 20 mm

Assume that the flow is steady and water is incompressible. From the discharge, Q = ( Vr ) 1 ( 2pr1b ) ;

0.03 m3 >s = ( Vr ) 1 [2p(0.05 m)(0.02 m)]

( Vr ) 1 = 4.775 m>s

Q = ( Vr ) 2 ( 2pr2b ) ;

0.03 m3 >s = ( Vr ) 2 [2p(0.15 m)(0.02 m)]

(Vrel(1

( Vr ) 2 = 1.592 m>s

(Vr(1 = 4.775 m/s

The angular speed of the impeller is v = a1200

rev 2p rad 1 min ba ba b = 40p rad>s min 1 rev 60 s

V1

Then, the velocity of the lead and tail of the blade is

U1 = vr1 = ( 40p rad>s ) (0.05 m) = 2p m>s U2 = vr2 = ( 40p rad>s ) (0.15 m) = 6p m>s From the kinematics diagram at the lead, Fig. a,

4.775 m/s

a1

b1 = 60°

(Vt (1

( Vt ) 1 = U1 - ( Vr ) 1 cot b1 = 2p m>s - ( 4.775 m>s ) cot 60° = 3.527 m>s And at the tail, Fig. b

60°

(a) Lead

( Vt ) 2 = U2 - ( Vr ) 2 cot b2 = 6p m>s - ( 1.592 m>s ) cot 30° = 16.093 m>s

U1 = 2 m/s

(

Thus, the speed of the water leaving the blade is V2 = 3 ( Vr ) 22 + ( Vt ) 22 = 3 ( 1.592 m>s ) 2 + ( 16.093 m>s ) 2 = 16.17 m>s = 16.2 m>s Ans.

The ideal head supplied by the pump is hpump = =

U2 ( Vt ) 2 - U1 ( Vt ) 1

(Vr(2 = 1.592 m/s

g

( t ()1 - ( 2p m>s )( 3.527 m>s ) ( 6p m>s )( 16.093 m>s 60 9.81 m>s2

b2 = 30°

Ans.

= 28.66 m = 28.7 m

(Vt (2

a2

V2

30°

U2 = 6 m/s

(b) Tail

The ideal power supplied by the pump is # Ws = gQhpump = ( 1000 kg>m3 )( 9.81 m>s2 )( 0.03 m3 >s ) (28.66 m) = 8.436 ( 103 ) W

Ans.

= 8.44 kW

Ans: V#2 = 16.2 m>s Ws = 8.44 kW hpump = 28.7 m 1528

14–23. The buckets of the Pelton wheel deflect the 100-mm-diameter water jet 140° as shown. If the velocity of the water from the nozzle is 30 m>s, determine the torque needed to hold the wheel in a fixed position and the torque that maintains an angular velocity of 10 rad>s.

v 2m

30 m/s 140!

SOLUTION Water is considered to be incompressible. The relative flow is steady. The discharge from the nozzle is Q = VA = ( 30 m>s ) 3 p(0.05 m)2 4 = 0.075p m3 >s

Since the bucket is required to be rest, U = 0. Then, Vj>cv = V = 30 m>s.  Referring to the free-body diagram of the control volume shown in Fig. a, + ΣF = rQ 3 ( Vout ) j>cv - ( Vin ) j>cv 4 ; d

40°

Applying the moment equation of equilibrium about its axial by referring to its freebody diagram in Fig. b,

40°

F = ( 1000 kg>m3 )( 0.075p m3 >s ) 3 ( 30 m>s ) cos 40° - ( - 30 m>s) 4

F

F = 12483.43 N = 12.483 kN

a+ ΣMa = 0;

(12483.43 N)(2 m) - T = 0

(a)

Tw = 0 = 24.97 kN # m = 25.0 kN # m

Ans.

When the wheel rotates with a constant angular velocity v = 10 rad>s, the bucket speed is U = vr = ( 10 rad>s ) (2 m) = 20 m>s. Thus, Vj>cv = V - U = 30 m>s - 20 m>s = 10 m>s. Again, Tv = 10 rad>s = rQVj>cv(1 + cos u)r = ( 1000 kg>m3 )( 0.075pm3 >s ) (10 m>s)(1 + cos 40°)(2m) = 8322.29 N # m = 8.32 kN # m

Ans.

1529

Rx 2m F

T Ry (b)

Ans: T 0 v = 0 = 25.0 kN # m T 0 v = 10 rad>s = 8.32 kN # m

*14–24. The buckets of the Pelton wheel deflect the 100-mm-diameter water jet 140° as shown. If the velocity of the water from the nozzle is 30 m>s, determine the power that is delivered to the shaft when the wheel is rotating at a constant angular velocity of 2 rad>s. How fast must the wheel be turning to maximize the power developed by the wheel?

v 2m

30 m/s 140!

SOLUTION Water is considered to be incompressible. The relative flow is steady. The discharge from the nozzle is Q = VA = ( 30 m>s ) 3 p(0.05 m)2 4 = 0.075p m3 >s

When the wheel rotates with a constant angular velocity v = 2 rad>s,  the bucket speed is U = vr = ( 2 rad>s ) (2 m) = 4 m>s. Thus, Vj>cv = V - U = 30 m>s - 4 m>s = 26 m>s. Thus, the power supply to the shaft can be determined with u = 180° - 140° = 40°. . Ws = rQVj>cvU(1 + cos u) = ( 1000 kg>m3 )( 0.075pm3 >s )( 26 m>s )( 4 m>s )( 1 + cos 40° ) = 43.28 ( 103 ) W = 43.3 kW

Ans.

From the text, the maximum power developed by the wheel occurs when the bucket has a speed of 30 m>s V U = = = 15 m>s 2 2 Thus, the correspondence wheel’s angular velocity is 15 m>s U v = = = 7.5 rad>s r 2m

Ans.

1530

14–25. Water flows from a lake through a 300-m-long pipe having a diameter of 300 mm and a friction factor of f = 0.015. The flow from the pipe passes through a 60-mm-diameter nozzle and is used to drive the Pelton wheel, where the bucket deflection angles are 160°. Determine the power and torque produced when the wheel is turning under optimum conditions. Neglect minor losses.

A 20 m

2m

B

SOLUTION Assume that the flow is steady and the water is incompressible. Applying the energy equation between A and B pA pB VA2 VB2 + ZA + hpump = + ZB + hturb + hL + + g g 2g 2g Here pA = pB = patm = 0, VA = 0 (large reservoir), ZA = 20 m and ZB = 0 2 L Vp (Datum through B), hpump = hturb = 0 (no shaft head), VB = Vj and hL = f . D 2g The above equation becomes 0 + 0 + 20 m + 0 = 0 +

Vj 2

+ 0 + 0 + 0.015 a

2 ( 9.81 m>s2 )

Vj 2 + 15 V p2 = 392.4

Vp 2 300 m b£ § 0.3 m 2 ( 9.81 m>s2 )

(1)

The continuity requires that Vj Aj - Vp Ap = 0 Vj 3 p (0.03 m)2 4 - Vp 3 p (0.15 m)2 4 = 0

(2)

Vp = 0.04 Vj

Solving Eqs (1) and (2)

Vj = 19.576 m>s

Vp = 0.7830 m>s

Thus, the flow of the jet is Q = Vj Aj = ( 19.576 m>s ) 3 p(0.03 m)2 4 = 0.05535 m3 >s

When the wheel rotates at optimum condition, the bucket’s speed is U =

Vj 2

=

19.576 m>s 2

= 9.788 m>s

Then Vj>cv = Vj - U = 19.576 m>s - 9.788 m>s = 9.788 m>s The torque produced is T = PQVj>cv (1 + cos u) r = ( 1000 kg>m3 )( 0.05535 m3 >s )( 9.788 m>s ) (1 + cos 20°) (2 m) = 2101.62 N # m = 2.10 kN # m

Ans.

The power produced is

#

Wturb = PQVj>cvU (1 + cos u) = ( 1000 kg>m3 )( 0.05535 m3 >s )( 9.788 m>s )( 9.788 m>s ) (1 + cos 20°)

= 10.285 ( 103 ) W = 10.3 kW

Ans. Ans: T# = 2.10 kN # m Wturb = 10.3 kW

1531

14–26. Water flows through the 400-mm-diameter delivery pipe at 2 m>s. Each of the four 50-mm-diameter nozzles is aimed tangentially at the Pelton wheel, which has bucket deflection angles of 150°. Determine the torque and power developed by the wheel when it is rotating at 10 rad>s.

2 m/s

400 mm

2.5 m 10 rad/s

SOLUTION Steady flow U = vr = ( 10 rad>s ) (2.5 m) = 25 m>s Q = VA = ( 2 m>s ) (p)(0.2 m)2 = 0.2513 m3 >s

At each nozzle

Q = Vn An ; 4

0.2513 m3 >s 4

= Vn p(0.025 m)2

Vn = 32.0 m>s Vj>cv = 32 m>s - 25 m>s = 7 m>s 180° - 150° = 30° T = r Q Vf>cs(1 + cos u) r = ( 1000 kg>m3 )( 0.2513 m3 >s )( 7 m>s ) (1 + cos 30°)(2.5 m)

Ans.

= ( 1000 kg>m3 )( 0.2513 m3 >s )( 7 m>s )( 25 m>s ) (1 + cos 30°)

Ans.

= 8207.2 N # m = 8.21 kN # m # Wturb = r Q Vf>cs U(1 + cos u)

= 82,072 W = 82.1 kW

Ans: T# = 8.21 kN # m Wturb = 82.1 kW 1532

14–27. Water flowing at 4 m>s is directed from the stator onto the blades of an axial-flow turbine at an angle of a1 = 28° and exits at an angle of a2 = 43°. If the blades are rotating at 80 rad>s, determine the required angles b1 and b2 of the turbine blades so that they properly accept, and then deliver the flow to the adjacent stator. The turbine has a mean radius of 600 mm.

b2 b1

a2 ! 43" 4 m/s a1 ! 28"

SOLUTION Water is considered to be incompressible. The relative flow is steady. The speed at the midpoint of the turbine’s blade is U = vrm =

1 80 rad>s 2 (0.6 m)

Stator

= 48 m>s

With this result, Va = 4 m>s, a1 = 28°, and a2 = 43°, the velocity diagram of the water at the head and tail of the impeller blade is shown in Figs. a and b, respectively. From the geometry of Fig. a, Va = ( Vt ) 1 tan a1;

4 m>s = ( Vt ) 1 tan 28°

( Vt ) 1 = 7.5229 m>s Va U - ( Vt ) 1

=

4 m>s 48 m>s - 7.5229 m>s

;

b1 = 5.644° = 5.64°

Ans.

Stator

(Vrel(1 b1

(Vt (1

Then, tan b1 =

Turbine

Va = 4 m/s

a1 = 28°

V1

b1

From the geometry of Fig. b, Va = ( Vt ) 2 tan a2;

4 m>s = ( Vt ) 2 tan 43°

U = 48 m/s

( Vt ) 2 = 4.2895 m>s (a)

Then, tan b2 =

Va U - ( Vt ) 2

=

4 m>s 48 m>s - 4.2895 m>s

;

b2 = 5.229° = 5.23°

Ans.

(Vrel(2 b2

(V 2 (t

Va = 4 m/s

a2 = 43°

V2

b2 U = 48 m/s

(b)

Ans: b1 = 5.64° b2 = 5.23° 1533

*14–28. Water flowing at 4 m>s is directed from the stator onto the blades of the axial-flow turbine, where the mean radius of the blades is 0.75 m. If the blades are rotating at 80  rad>s and the flow is 7 m3>s, determine the torque produced by the water.

b2 b1

a2 ! 43" 4 m/s a1 ! 28"

SOLUTION

Stator

Turbine

Stator

Water is considered to be incompressible. The relative flow is steady. With Va = 4 m>s, a1 = 28°, and a2 = 43°, the velocity diagram of the water at the head and tail of the impeller blade is shown in Figs. a and b, respectively. From the geometry of Fig. a, Va = ( Vt ) 1 tan a1;

4 m>s = ( Vt ) 1 tan 28°

Va = 4 m/s

( Vt ) 1 = 7.5229 m>s From the geometry of Fig. b, Va = ( Vt ) 2 tan a2;

a1 = 28°

4 m>s = ( Vt ) 2 tan 43°

a2 = 43°

( Vt ) 2 = 4.2895 m>s

(Vt (2

The torque produced by the water and exerted on the turbine’s shaft is

= ( 1000 kg>m = - 16.98 ( 10

3

)( 7 m >s ) (0.75 m) 3 4.2895 m>s - 7.5229 m>s 4 3

) N # m = - 17.0 kN # m

1534

V2 (b)

(Vt (1

T = rQrm 3 ( Vt ) 2 - ( Vt ) 1 4 3

Va = 4 m/s

Ans.

(a)

V1

14–29. A stator directs 8 kg>s of gas onto the blades of a gas turbine that is rotating at 20 rad>s. If the mean radius of the turbine blades is 0.8 m, and the velocity of the flow entering the blades is 12 m>s as shown, determine the exit velocity of the gas from the blades. Also, what is the required angle b1 at the entrance of the blades?

20! 12 m/s

15!

SOLUTION

Stator

Turbine

The gas is considered to be incompressible. The relative flow is steady. The speed at the midpoint of the rotor blade is U = vrm = ( 20 rad>s ) (0.8 m) = 16 m>s With the result of U, V1 = 12 m>s, and a1 = 20°, the velocity diagram of the water at the head of the blade is shown in Fig. a. From the geometry of this figure,

U = 16 m/s

( Vt ) 1 = V1 cos a1 = ( 12 m>s ) cos 20° = 11.2763 m>s

b1

V1 = 12 m/s

Va = V1 sin a1 = ( 12 m>s ) sin 20° = 4.1042 m>s

(Vt (1

Then, tan b1 =

Va U - ( Vt ) 1

=

4.1042 m>s

a1 = 20°

16 m>s - 11.2763 m>s

Va

Ans.

b1

Using the result of U, Va, and b2 = 15°, the velocity diagram of the water at the tail of the blade is shown in Fig. b. From the geometry of this figure,

(a)

b1 = 40.99° = 41.0°

Va = Then,

3U

- ( Vt ) 2 4 tan b2;

4.1042 m>s =

(Vrel(1

3 16 m>s - ( Vt ) 2 4 tan 15°

( Vt ) 2 = 0.6828 m>s

V2 = 2Va 2 + ( Vt ) 22 = 2 ( 4.1042 m>s ) 2 + ( 0.6828 m>s ) 2 = 4.1606 m>s = 4.16 m>s

U = 16 m/s b2 = 15°

V1 Ans.

(Vt (2

V2 Va = 4.1042m/ s

b2 = 15° (b)

(Vrel(2

Ans: b1 = 41.0° V2 = 4.16 m>s 1535

14–30. The dimensions of the blades on the axial-flow water turbine are shown. Water passes through the guide vanes at an angle of 60°. If the flow is 0.85 m3>s, determine the velocity of the water as it strikes the mean radius of the blades. Hint: Within the free passage from the guide vanes to the turbine, free-vortex flow occurs; that is, Vtr = constant.

580 mm 200 mm 200 mm

500 mm

60!

SOLUTION Water is considered to be incompressible. The relative flow is steady.

580 mm

The radial component of the water velocity leaving the stator vanes can be determined from the flow rate. Q = ( Vr ) sAs;

0.85 m3 >s = ( Va ) S[2p(0.58 m)(0.2 m)]

( Vr ) S = 1.1662 m>s

Using this result, the velocity diagram of the water leaving the stator vane is shown in Fig. a. From the geometry of this figure,

( Vr ) S = ( Vt ) S tan 60°;

1.1662 m3 >s = ( Vt ) S tan 60°

( Vt ) S = 0.6733 m>s

Since free vortex flow occurs between the stator vane and the blade, Vtr = constant. At the midpoint of the blade, ( rm ) b = (0.2 m + 0.5 m)>2 = 0.35 m. Thus,

( Vt ) srs = ( Vt ) b ( rm ) b;

( 0.6733 m>s ) (0.58 m) = ( Vt ) b(0.35 m)

(Vt (s 60°

( Vt ) b = 1.1158 m>s Due to the continuity condition, the axial component of the water’s velocity that strikes the blade is Q = ( Va ) bAb;

(Vr(s = 1.1662 m/s

Vs (a)

0.85 m3 >s = ( Va ) b 3 p 3 (0.5 m)2 - (0.2 m)2 44

( Va ) b = 1.2884 m>s

Thus, the magnitude of the water’s velocity striking the midpoint of the blade is Vb = 3 ( Va ) b2 + ( Vt ) b2 = 3 ( 1.2884 m>s ) 2 + ( 1.1158 m>s ) 2 = 1.704 m>s = 1.70 m>s

Ans.

Ans: 1.70 m>s 1536

14–31. The velocities on and off the 90-mm-wide blades of the turbine are directed as shown. If V1 = 18 m>s and the  blades are rotating at 80 rad>s, determine the relative velocity of the flow off the blades. Also, determine the blade angles, b1 and b2.

V1

V1 30! 30!

b1 b1

b2

b2 V2

150 mm m

80 rad/s

V2 275 5 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady.

U1 = 22 m/s

The speed of points on the head and tail of the impeller blade are

(Vt (1

a1 = 30°

U1 = vr1 = ( 80 rad>s ) (0.275 m) = 22 m>s

b1 V1 = 18 m/s

U2 = vr2 = ( 80 rad>s ) (0.15 m) = 12 m>s Using the result of U1, a1 = 30°, and V1 = 18 m>s, the velocity diagram of the water at the head of the impeller is shown in Fig. a. From the geometry of this figure,

( Vr ) 1 = V1 sin a1 = ( 18 m>s ) sin 30° = 9 m>s

(Vr(1

(Vrel(1

(a)

U2 = 12 m/s

( Vt ) 1 = V1 cos a1 = ( 18 m>s ) cos 30° = 15.5885 m>s

b1

b2 b2

Then,

(Vrel(2

( Vr ) 1 9 m>s = tan b1 = 22 m>s - 15.5885 m>s U1 - ( Vt ) 1

V2 = 16.5 m/s (b)

Ans.

b1 = 54.53° = 54.5° Also,

( Vr ) 1 = ( Vrel ) 1 sin b1;

9 m>s = ( Vrel ) 1 sin 54.43°

( Vrel ) 1 = 11.05 m>s = 11.1 m>s The discharge in the pump is Q = ( Vr ) 1A1 = ( 9 m>s ) [2p(0.275 m)(0.09 m)] = 0.4455p m3 >s

Since the water exits the blade radially, ( Vt ) 2 = 0 and V2 = ( Vr ) 2. Due to the continuity condition, Q = V2A2;

0.4455p m3 >s = V2 [2p(0.15 m)(0.09 m)]

V2 = 16.5 m>s

Using the results of U2 and V2, the velocity diagram of the blade tail is shown in Fig. b. From the geometry of this figure, tan b2 =

16.5 m>s V2 = ; U2 12 m>s

Ans.

b2 = 53.97° = 54.0°

( Vrel ) 2 = 2V22 + U 22 = 2 ( 16.5 m>s ) 2 + ( 12 m>s ) 2 = 20.40 m>s = 20.4 m>s

Ans.

Ans: b1 = 54.5° b2 = 54.0° ( Vrel ) 2 = 20.4 m>s 1537

*14–32. The velocities on and off the 90-mm-wide blades of a turbine are directed as shown. If the blades are rotating at 80 rad>s and the discharge is 1.40 m3>s, determine the power that the turbine withdraws from the water.

V1

V1 30! 30!

b1 b1

b2

b2 V2

150 mm m

V2 275 5 mm

SOLUTION Water is considered to be incompressible. The relative flow is steady. The radial component of the water’s velocity at the blade head can be determined from the discharge. Q = ( Vr ) 1A1;

1.40 m3 >s = ( Vr ) 1 3 2p(0.275 m)(0.09 m) 4

(Vt (1

( Vr ) 1 = 9.0027 m>s

a1 = 30°

Using this result and a1 = 30°, the velocity diagram of the water at the blade’s head is shown in Fig. a. From the geometry of this figure,

( Vr ) 1 = ( Vt ) 1 tan a1;

9.0027 m>s = ( Vt ) 1 tan 30°

(a)

( Vt ) 1 = 15.5931 m>s Since the water exits the blade radially, ( Vt ) 2 = 0. Thus, the torque the water applies to the impeller is T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 1.40 m3 >s ) 3 0 - (0.275 m) ( 15.5931 m>s ) 4 = - 6.0034 ( 103 ) N # m

Then, the power the turbine withdraws from the water is

#

Ws = Tv =

(Vr(1 = 9.0027 m/s

V1

3 - 6.0034 ( 103 ) N # m 4 ( 80 rad>s )

= - 480.27 ( 103 ) W = - 480 kW

Ans.

1538

80 rad/s

14–33. If the angles for an axial-flow turbine blade are a1 = 30°, b1 = 60° and b2 = 30°, determine the velocity of the water entering and exiting the turbine blades if the mean radius of the blades is 1.5 ft. The turbine is rotating at 70 rad>s.

SOLUTION

U = 105 ft/s

Water is considered to be incompressible. The relative flow is steady.

b1 = 60°

The speed of the midpoint of the blade is U = vrm = ( 70 rad>s ) (1.5 ft) = 105 ft>s Using the result of U, a1 = 30°, and b1 = 60°, the velocity diagram of the water at the head of the blade is shown in Fig. a. Applying the law of sines, V1 U = ; sin b1 sin (180° - a1 - b1)

V1

90°

(Vt (1 a1 = 30°

Va

105 ft>s V1 = sin 60° sin 90°

b1 = 60°

V1 = 90.9327 ft>s = 90.9 ft>s

(a)

Ans.

Then,

(Vrel(1

U = 105 ft/s

Va = V1 sin a1 = ( 90.9327 ft>s ) sin 30° = 45.4663 ft>s 0

Using this result, U, and b2 = 30°, the velocity diagram of the water at(Vthe t (1 blade’s tail is shown in Fig. b. From the geometry of this figure, Va =

3U

- ( Vt ) 2 4 tan b2;

45.4663 ft>s =

3 105 ft>s

( Vt ) 2 = 26.25 ft>s

- ( Vt ) 2 4 tan 30°

= 52.5 ft>s

a2

(Vt (2

Va = 45.4663

b2 = 30°

Thus, the magnitude of V2 is V2 = 2 ( Vt ) 22 + Va 2 = 2 ( 26.25 ft>s ) 2 + ( 45.4663 ft>s ) 2

b2 = 30°

(Vrel(2 Ans.

(b)

Ans: V1 = 90.9 ft>s V2 = 52.5 ft>s 1539

14–34. If the angles for an axial-flow turbine blade are a1 = 30°, b1 = 60° and b2 = 30°, determine the relative velocity of the water at the head and tail of the turbine blades if the mean radius of the blades is 1.5 ft. The turbine is rotating at 70 rad>s.

SOLUTION Water is considered to be incompressible. The relative flow is steady. The speed of the midpoint of the blade is U = vrm = ( 70 rad>s ) (1.5 ft) = 105 ft>s Using this result, a1 = 30°, and b1 = 60°, the velocity diagram of the water at the head of the blade is shown in Fig. a. Applying the law of sines, 105 ft>s V1 V1 U = ; = sin 60° sin 90° sin b1 sin (180° - a1 - b1) V1 = 90.9327 ft>s

( Vrel ) 1 sin a1

=

U ; sin (180° - a1 - b1)

( Vrel ) 1 sin 30°

=

105 ft>s sin 90°

( Vrel ) 1 = 52.5 ft>s

Ans.

Then, Va = V1 sin a1 = ( 90.9327 ft>s ) sin 30° = 45.4663 ft>s Using this result, U, and b2 = 30°, the velocity diagram of the water at the blade’s tail is shown in Fig. b. From the geometry of this figure, Va = ( Vrel ) 2 sin b2;

45.4663 ft>s = ( Vrel ) 2 sin 30°

( Vrel ) 2 = 90.93 ft>s = 90.9 ft>s

Ans.

Ans: ( Vrel ) 1 = 52.5 ft>s ( Vrel ) 2 = 90.9 ft>s 1540

14–35. The blades of an axial-flow turbine have a mean radius of 1.5 ft and are rotating at 70 rad>s. If the angles for a turbine blade are a1 = 30°, b1 = 60° and b2 = 30°, and the flow is 900 ft3>s, determine the ideal power supplied to the turbine.

SOLUTION Water is considered to be incompressible. The relative flow is steady. U = 105 ft/s

The speed of the midpoint of the blade is U = vrm = ( 70 rad>s ) (1.5 ft) = 105 ft>s

b1 = 60°

Using this result a1 = 30°, and b1 = 60°, the velocity diagram of the water at the head of the blade is shown in Fig. a. Applying the law of sines, 105 ft>s V1 = sin 60° sin 90°

V1 U = ; sin b1 sin (180° - a1 - b1)

V1

90°

(Vt (1 a1 = 30°

Va

(Vrel(1

V1 = 90.9327 ft>s

b1 = 60°

Then,

(a)

Va = V1 sin a1 = ( 90.9327 ft>s ) sin 30° = 45.4663 ft>s

( Vt ) 1 = V1 cos a1 = ( 90.9327 ft>s ) cos 30° = 78.75 ft>s

U = 105 ft/s

Using this result, U, and b2 = 30°, the velocity diagram of the water at the blade’s 6 tail is shown in Fig. b. From the geometry of this figure, Va =

3U

- ( Vt ) 2 4 tan b2;

45.4663 ft>s =

3 105 ft>s

( Vt ) 2 = 26.25 ft>s

- ( Vt ) 2 4 tan 30° 1

b2 = 30° a2

(Vt (2

Va = 45.4663

The power of the water supplied to the turbine is therefore

#

Ws = rQU 3 ( Vt ) 2 - ( Vt ) 1 4 = °

62.4 lb>ft 2 32.2 ft>s2

b2 = 30°

(Vrel(2

¢ ( 900 ft >s )( 105 ft>s )( 26.25 ft>s - 78.75 ft>s )

= c - 9.6143 ( 106 ) = - 17.5 ( 103 ) hp

V2

3

1 hp ft # lb da b s 550 ft # lb>s

(b)

Ans.

The negative sign indicates that the power is withdrawn from the water.

Ans: 17.5(103) hp 1541

*14–36. Water is directed at a1 = 50° onto the blades of the Kaplan turbine and leaves the blades in the axial direction. Each blade has an inner radius of 200 mm and outer radius of 600 mm. If the blades are rotating at v = 28 rad>s, and the flow is 8 m3 >s, determine the power the water supplies to the turbine.

v

SOLUTION Water is considered to be incompressible. The relative flow is steady. The axial component of the water at the blade head can be determined from the discharge. Q = ( Va ) 1A1;

8 m3 >s = ( Va ) 1 3 p 3 (0.6 m)2 - (0.2 m)2 4 4

( Va ) 1 = 7.9577 m>s

(Vt (1

Using the result of ( Va ) 1 and a1 = 50°, the velocity diagram of the water at the blade’s head is shown in Fig. a. From the geometry of this figure, (Va)1 = ( Vt ) 1 tan a1;

7.9577 m>s = ( Vt ) 1 tan 50°

Va = 7.9577 m/s

V1

( Vt ) 1 = 6.6773 m>s

(a)

The mean radius of the impeller blade is rm =

a1 = 50°

0.6 m + 0.2 m = 0.4 m 2

The speed of the midpoint of the blade is U = vrm = ( 28 m>s )( 0.4 m ) = 11.2 m>s Since water exits the blade axially, ( Vt ) 2 = 0.

#

W = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 8 m3 >s )( 11.2 m>s )( 0 - 6.6773 m>s ) = 598.29 ( 103 ) W = 598 kW

1542

Ans.

14–37. The blades of the Francis turbine rotate at 40 rad>s as they discharge water at 0.5 m3>s. Water enters the blades at an angle of a1 = 30° and leaves in the radial direction. If the blades have a width of 0.3 m, determine the torque and power the water supplies to the turbine shaft.

1.2 m 40 rad/s 0.4 m V1 30!

SOLUTION

0.3 m

Water is considered to be incompressible. The relative flow is steady. The radial component of the water’s velocity at the blade head is Q = ( Vr ) 1 A1;

0.5 m3 >s = ( Vr ) 1[2p(1.2 m)(0.3 m)]

( Vr ) 1 = 0.2210 m>s

Using this result and a1 = 30°, the velocity diagram of the water at the blade’s head is shown in Fig. a. From the geometry of this figure, ( Vr ) 1 = ( Vt ) 1 tan a1; 0.2210 m>s = ( Vt ) 1 tan 30°

( Vt ) 1 = 0.3829 m>s Since the water exits the blade’s tail radially, ( Vt ) 2 = 0. Thus, the torque the water exerts on the turbine’s shaft is T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4

= ( 1000 kg>m3 )( 0.5 m3 >s ) 3 0 - (1.2 m) ( 0.3829 m>s ) 4 = - 229.72 N # m = -230 N # m

(Vr(1 = 0.2210 m/s

V1

Ans.

Then, the power the water supplies to the turbine is

a1 = 30°

# Ws = Tv = ( - 229.72 N # m )( 40 rad>s )

(Vt (1

(a)

= - 9.1888 ( 103 ) W = - 9.19 kW

Ans.

Ans: T# = 230 N # m Ws = 9.19 kW 1543

14–38. The blades of the Francis turbine rotate at 40 rad>s as they discharge water at 0.5 m3>s. Water enters the blades at an angle of a1 = 30° and leaves in the radial direction. If the blades have a width of 0.3 m and the turbine operates under a total head of 3 m, determine the hydraulic efficiency.

1.2 m 40 rad/s 0.4 m V1 30!

SOLUTION

0.3 m

Water is considered to be incompressible. The relative flow is steady. The radial component of the water’s velocity at the blade head is Q = ( Vr ) 1A1;

0.5 m3 >s = ( Vr ) 1 3 2p(1.2 m)(0.3 m) 4

( Vr ) 1 = 0.2210 m>s

Using this result and a1 = 30°, the velocity diagram of the water at the blade’s head is shown in Fig. a. From the geometry of this figure,

( Vr ) 1 = ( Vt ) 1 tan a1;

0.2210 m>s = ( Vt ) 1 tan 30°

( Vt ) 1 = 0.3829 m>s Since the water exits the blade radially, ( Vt ) 2 = 0. Thus, the torque the water exerts on the turbine’s shaft is T = rQ 3 r2 ( Vt ) 2 - r1 ( Vt ) 1 4 = ( 1000 kg>m

3

(Vr(1 = 0.2210 m/s

)( 0.5 m >s ) 3 0 - (1.2 m) ( 0.3829 m>s ) 4 3

= -229.72 N # m

a1 = 30°

Then, the power the water supplies to the turbine is

#

( - 229.72 N # m )( 40 rad>s )

Ws = Tv =

V1

(Vt (1

(a)

= - 9.1888 ( 103 ) W

#

Using the result of Ws,

#

Ws = ∆HrQg;

9.1888 ( 103 ) W = ∆H ( 1000 kg>m3 )( 0.5 m3 >s )( 9.81 m>s2 ) ∆H = 1.8734 m

The efficiency of the turbine is ∆H 1.8734 m nt = = = 0.6244 = 0.624 ∆Ht 3m

Ans.

Ans: 0.624 1544

14–39. Water enters the 50-mm-wide blades of the turbine with a velocity of 20 m>s as shown. If the blades are rotating at 75 rev>min and the flow off the blades is radial, determine the power the water supplies to the turbine.

20 m/s 35!

75 rev/min

0.6 m

SOLUTION Water is considered to be incompressible. The relative flow is steady. Using V1 = 20 m>s and a1 = 35°, the velocity diagram of the water at the head of the blade is shown in Fig. a. From the geometry of this figure,

( Vr ) 1 = V1 sin a1 = ( 20 m>s ) sin 35° = 11.4715 m>s

(Vt (1 a1 = 35°

( Vt ) 1 = V1 cos a1 = ( 20 m>s ) cos 35° = 16.3830 m>s

(Vr(1

Then, the discharge is

V1 = 20 m/s

Q = (Vr)1 A1 = ( 11.4715 m>s ) 3 2p(0.6 m)(0.05 m) 4 = 2.1623 m3 >s

(a)

Since the water exits the blade radially, ( Vt ) 2 = 0. The torque the water exerts on the turbine’s shaft is T = rQ 3 r2(Vt)2 - r1(Vt)1 4

= ( 1000 kg>m3 )( 2.1623 m3 >s ) 3 0 - (0.6 m) ( 16.3830 m>s ) 4 = - 21.26 1 103 2 N # m

Here, v = a75

rev 1 min 2p rad ba ba b2p rad>s min 60 s 1 rev

Then, the power the water supplies to the turbine is # Ws = Tv = 3 - 21.26 1 103 2 N # m 4 ( 2p rad>s ) = - 166.94 1 103 2 W = - 167 kW

Ans.

Ans: 167 kW 1545

*14–40. Water enters the 50-mm-wide blades of the turbine with a velocity of 20 m>s as shown. If the blades are rotating at 75 rev>min and the flow off the blades is radial, determine the ideal head the turbine draws from the water.

20 m/s 35!

75 rev/min

0.6 m

SOLUTION Water is considered to be incompressible. The relative flow is steady. Using V1 = 20 m>s and a1 = 35°, the velocity diagram of the water at the head of the blade is shown in Fig. a. From the geometry of this figure,

h(m)

( Vr ) 1 = V1 sin a1 = ( 20 m>s ) sin 35° = 11.4715 m>s

800 700

( Vt ) 1 = V1 cos a1 = ( 20 m>s ) cos 35° = 16.3830 m>s 480

Q = ( Vr ) 1 A1 = ( 11.4715 m>s ) 3 2p(0.6 m)(0.05 m) 4 = 2.1623 m3 >s

Since the water exits the blade radially, (Vt)2 = 0. The torque the water exerts on the turbine’s shaft is T = rQ 3 r2(Vt)2 - r1(Vt)1 4

Then, the power the water supplies to the turbine is # Ws = Tv = 3 - 21.26 1 103 2 N # m 4 ( 2p rad>s ) = - 166.94 1 103 2 W

1546

200 100 Q(m3/s(

(a)

rev 1 min 2p rad ba ba b2p rad>s min 60 s 1 rev

= 7.869 m = 7.87 m

300

0.0

= - 21.26 1 103 2 N # m

The ideal head the turbine draws from the water is # 166.94 ( 103 ) W Ws ∆H = = rQg ( 1000 kg>s )( 2.1623 m3 >s )( 9.81 m>s2 )

400

0

= ( 1000 kg>m3 )( 2.1623 m3 >s ) 3 0 - (0.6 m) ( 16.3830 m>s )4

Pump performance curve

500

05 0.0 10 0.0 15 0.0 20 0.0 25 0.0 30 0.0 35 0.0 40

Then, the discharge is

Here, v = a75

hreg

600

Ans.

0.03375

14–41. The pipe system consists of a 2-in.-diameter, 50-ft-long galvanized iron pipe, a fully opened gate valve, two elbows, a flush entrance, and a pump with the pump head curve shown. If the friction factor is 0.025, estimate the flow and corresponding pump head generated by the pump.

B

15 ft A

hpump (ft) 300 250 200 150

SOLUTION

100

Assume that the flow is steady and the water is incompressible, the loss coefficient for flush entrance, fully opened gate valve and elbow are 0.5, 0.19 and 0.9, respectively. Thus, the head loss is hL = af

50 Q gal/min 50 100 150 200 250 300 350

VB2 VB2 L 50 ft + ΣkL b = c 0.025 a b + 0.5 + 0.19 + 2(0.9) d £ § D 2g 2>12 ft 2 ( 32.2 ft>s2 ) = 0.1551 VB2

(1)

However, using the discharge, Q = VBAB;

Q = VB c pa

2 1 ft b d 12

hpump(ft) 250

(2)

VB = 45.8366Q

200

Substitute Eq. (2) into (1), hL = 325.92Q2

0 + 0 + 0 + hpump = 0 +

2 ( 32.2 ft>s2 )

Operating point

150

Write the energy equation between A and B realizing that pA = pB = patm = 0, zA = 0 and zB = 15 ft (Datum through A), VA ≃ 0 (large reservoir) and hturb = 0, 100 pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL 50 gw gw 2g 2g (45.8366Q)2

hpump= 200

+ 15 ft + 0 + 325.92Q

2

0

Q = 325 50 100 150 200 250 300 350

Q ( gal/min(

hpump = ( 358.54Q2 + 15 ) ft Q ( gal>min )

50

100

150

200

250

300

350

Q ( ft 3 >s )

0.111

0.223

0.334

0.446

0.557

0.668

0.780

19.45

32.80

55.05

86.20

126.25

175.20

233.06

hpump(ft)

a

gal min

ba

1 ft 3 1 min ba b = ft 3 >s 7.48 gal 60 s

The plot of these data on the pump performance curve is shown in Fig. a. By locating the operating point, Ans.

Q = 325 gal>min and hpump = 200 ft

Ans: Q = 325 gal>min hpump = 200 ft 1547

14–42. Water at 20°C is pumped from a lake into the tank on the truck using a 50-mm-diameter galvanized iron pipe. If the pump performance curve is as shown, determine the maximum flow the pump will generate. The total length of the pipe is 50 m. Include the minor losses of the five elbows.

hpump (m) 700 600 500 400 300 200 100 0.005 0.015 0.025 0.035

Q (m3/s) B

SOLUTION

8m

Assume that fully developed steady flow occurs, and the water is incompressible. Appendix A gives rw = 998.3 kg>m3 and vw = 1.00 ( 10-6 ) m2 >s at 20° C. The average velocity is

A

Q Q 1600 = = Q 2 p A p(0.025 m)

V = Thus, the Reynolds number is

1600 Qb(0.05 m) p 1.00 ( 10-6 ) m2 >s

=

80 ( 106 ) Q p

(1)

The loss coefficient for an elbow is 0.9. Thus, the total head loss is

h(m)

L V2 V2 hL = f + ΣkL D 2g 2g = af

800 700

L V2 + ΣkL b D 2g

300 200

= (1000f + 4.5) ( 13220.30Q2 )

Write the energy equation between A and B realizing that pA = pB = patm = 0, VA ≃ 0 (large reservoir), zA = 0 and zB = 8 m (Datum through A),

0 + 0 + 0 + hreq = 0 +

2 ( 9.81 m>s2 )

100 0

Q (m3/s( 0

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 2 1600 Qb p

Pump performance curve

500 480 400

2

1600 a Qb p 50 m b + 5(0.9) d = cf a 0.05 m 2 ( 9.81 m>s2 )

a

hreg

600

00 5 01 0 0 01 5 0 02 0 0 02 0 5 03 0 0 03 5 0 04 0

a

0

VD = Re = vw

(a)

0.03375

+ 8 m + 0 + (1000f + 4.5) ( 13220.30Q2 )

hreq = (1000f + 5.5) ( 13220.30Q2 ) + 8

(2)

0.15 mm e = = 0.003. When Q = 0.01 m3 >s,  D 50 mm Re = 2.55 ( 105 ) , the moody diagram gives f = 0.02675. Then Eq. (2) gives hreq = 50.64 m. The other values of hreq for an increment of 0.005 m3 >s for Q are tabulated as follows, For

galvanized

Q ( m3 >s ) hreq(m)

iron

pipe,

0

0.005 0.010

0.015

0.020

0.025

0.03

0.035

0.04

8

18.82

103.19

175.90

270.34

382.79

518.14

674.30

50.64

Plotting this data on the pump performance curve, Fig. a, the intersection point gives Q = 0.03375 m3 >s

Ans.

1548

Ans: 0.03375 m3 >s

14–43. The pump is used to transfer water at 60°F from the river up onto the field for irrigation. The friction factor for the 3-in.-diameter, 30-ft-long hose is f = 0.015, and h = 15 ft. Determine if cavitation occurs when the average velocity through the hose is 18 ft>s. Use the pump performance curves in Fig. 14–16. Neglect minor losses.

B

18 ft/s

h

A

SOLUTION pA pB VA2 VB2 L V2 V2 + zA + hpump = + zB + hturb + f + ΣkL + + g g 2g 2g D 2g 2g

( 14.7 lb>in2 )( 12 in.>ft ) 2

+ 0 - 15 ft + 0 =

62.4 lb>ft 3

pB 62.4 lb>ft 3

+

(18 ft>s)2 2 ( 32.2 ft>s2 )

+ 0 + 0

( 18 ft>s ) 30 ft ≤ + 0 3 2 ( 32.2 ft>s2 ) ft 12 2

+ (0.015 ft) ± pB = 301.77 lb>ft 2 Available suction head at the pump inlet is therefore

( 18 ft>s ) 2 301.77 lb>ft 2 pB VB2 = + = 9.867 ft + g 2g 62.4 lb>ft 3 2 ( 32.2 ft>s2 ) From Appendix A, at T = 60° F, pV = 0.256 psia (NSPH)Avail = 9.867 ft Q = VA = 18 ft>s (p)a From Fig. 14–16,

( 0.256 lb>in2 )( 12 in.>ft ) 2 62.4 lb>ft 3

= 9.28 ft

2 7.48 gal 1.5 60 s ft b a ba b = 397 gal>min 12 min ft 3

(NSPH)req>d ≈ 14 ft 7 9.28 ft

Ans.

Cavitation will occur.

Ans: Cavitation will occur. 1549

*14–44. Water at 80°F is being pumped up h = l2 ft from the river onto the irrigation field using a 3-in.-diameter, 25-ft-long hose that has a friction factor of f = 0.030. If the average velocity through the hose is 18 ft>s, use the pump performance curves in Fig. 14–16, and determine if cavitation occurs in the pump. Neglect minor losses.

B

h

A

SOLUTION pB pA VA 2 VB 2 L V2 V + zA + hpump = + zB + hturb + f + ΣkL + + g g 2g 2g D 2D 2g 14.7 lb>in ( 12 in>ft ) 2 2

62.4 lb>ft 3

+ 0 - 12 ft + 0 =

pB 62.4 lb>ft 3

+

(18 ft>s)

(Vrel(1

2

2 ( 32.2 ft>s2 )

+ 0 + 0

b1

( 18 ft>s ) 25 ft ≤ + 0 3 2 ( 32.2 ft>s2 ) a ft b 12

V = Va = 5 m/s

2

+ (0.030) ± pB = 112.2 lb>ft 2 Available suction head at pump inlet is therefore

( 18 ft>s ) 2 112.2 lb>ft 2 pB VB 2 = + = 6.830 ft + g 2g 62.4 lb>ft 3 2 ( 32.2 ft>s2 ) From Appendix A, at T = 80° F, pV = 0.507 psia (NSPH)Avil = 6.830 ft -

( 0.507 lb>in2 )( 12 in.>ft ) 2 62.4 lb>ft 3

Q = VA = ( 18 ft>s ) (p)a From Fig. 14–16,

= 5.66 ft

2 7.48 gal 1.5 60 s ft b a ba b = 397 gal>min 3 12 min ft

(NSPH)Req>d ≈ 14 ft 7 5.66 ft

Ans.

Cavitation will occur. Ans.

b1 = 22.62° = 22.6° Sin 22.62° =

5

( Vrel ) 1

( Vrel ) 1 = 13.0 m>s

Ans.

1550

b1

(Vt (1 = U = 12 m/s

18 ft/s

14–45. The radial-flow pump having an impeller diameter of 5 in. and the performance curves shown in Fig. 14–16 is to be used to pump water from the reservoir into the tank. Determine the efficiency of the pump if the flow is 400 gal > min. Also, what is the maximum height h to which the tank can be filled? Neglect any losses.

h

SOLUTION For 400 gal>min. From Fig. 14–16, for a 5-in. impeller, Ans.

h = 76% Also, hpump ≈ 80 ft. From the energy equation with datum at reservoir level, pin pout Vout2 Vin2 + zin + hpump = + zout + hturb + h + + g g 2g 2g 0 + 0 + 0 + 80 ft = 0 + 0 + h + 0 + 0

Ans.

h = 80 ft

Ans: h = 76% h = 80 ft 1551

14–46. The radial-flow pump has a 5.5-in.-diameter impeller and the performance curves shown in Fig. 14–16. If it is used to pump water from the reservoir tank to the fill tank, determine the approximate flow when the water is at the elevations shown, when h = 80 ft. Neglect friction losses in  the 3-in.-diameter pipe, but consider the minor losses to be KL = 3.5.

h

5 ft

SOLUTION pA pB VA2 VB2 L V2 V2 + zA + hpump = + zB + hturb + f a b + KL + + g g 2g 2g D 2g 2g 0 + 0 + 5 ft + hpump = 0 + 0 + 80 ft + 0 + 3.5

V2

2 ( 32.2 ft>s2 )

hpump = 75 + 0.05434 V 2 Q = VA = Vp a

Thus,

2 1.5 ft b = 0.04909 V 12

hpump = 75 + 22.55 Q2 This equation is solved along with the curve in Fig. 14–16 for the 5.5-in. impeller. Assume Q = 500 gal>min a

1 min 1 ft 3 ba b = 1.114 ft>s 60 s 7.48 gal

Then hpump = 75 + 22.55(1.114)2 = 103 ft

This is approximately the value in Fig. 14–16. Thus, Ans.

Q = 500 gal>min.

Ans: 500 gal>min 1552

14–47. A radial-flow pump has a 6-in.-diameter impeller, and the performance curves for it are shown in Fig. 14–16. Determine the approximate flow it provides to pump water from the reservoir tank to the fill tank, where h = 115 ft. Neglect minor losses, and use a friction factor of f = 0.02 for the 100-ft-long, 3-in.-diameter hose.

h

5 ft

SOLUTION pA pB VA2 VB2 L V2 V2 + zA + hpump = + zB + hturb + f a b + KL + + g g 2g 2g D 2g 2g 100 ft V2 ≥ + 0 0 + 0 + 5 ft + hpump = 0 + 0 + 115 ft + 0 + 0.02£ 3 ft 2(32.2 ft>s) 12 hpump = 110 + 0.1242 V 2 Q = VA = Vp a

Thus,

2 1.5 ft b = 0.04909 V 12

hpump = 110 + 51.554 Q2 This equation must be solved along with the curve in Fig. 14–16 for the 6-in. impeller. 1 min 1 ft 3 Assume Q = 300 gal>min a ba b = 0.6684 ft 3 >s 60 s 7.48 gal hpump = 110 + 51.554(0.6684)2 = 133 ft

Try

Q = 350 gal>min

get

hpump = 141 ft

Try

Q = 400 gal>min

get

hpump = 151 ft

This is the approximate value in Fig. 14–16. Thus, Ans.

Q = 400 gal>min.

Ans: 400 gal>min 1553

*14–48. A 200-mm-diameter impeller of a radial-flow water pump rotates at 150 rad>s and produces a change in ideal head of 0.3 m. Determine the change in head for a geometrically similar pump that has an impeller diameter of 100 mm and operates at 80 rad>s.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Dynamic similitude of the head coefficient gives ∆H1 v12D12

=

∆H2 v22D22

(0.3 m)

( 150 rad>s ) (0.2 m) 2

2

=

∆H2

( 80 rad>s ) 2(0.1 m)2

∆H2 = 0.0213 m

Ans.

1554

14–49. A 200-mm-diameter impeller of a radial-flow water pump rotates at 150 rad>s and has a discharge of 0.3  m3>s. Determine the discharge for a similar pump that has an impeller diameter of 100 mm and operates at 80 rad>s.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Dynamic similitude of the flow coefficient gives Q1 v1D13

=

Q2 v2D23

0.3 m3 >s

( 150 rad>s ) (0.2 m)

3

=

Q2

( 80 rad>s ) (0.1 m)3

Q2 = 0.02 m3 >s

Ans.

1555

Ans: 0.02 m3 >s

14–50. The temperature of benzene in a processing tank is maintained by recycling this liquid through a heat exchanger, using a pump that has an impeller speed of 1750 rpm and produces a flow of 900 gal>min. If it is found that the heat exchanger can maintain the temperature only when the flow is 650 gal>min, determine the required angular speed of the impeller.

SOLUTION Since Q1, v1 and Q2 are given and D1 = D2, we can use flow coefficient similitude to determine v2. Q1 v1D13

=

Q2 v2D23

v2 = a = a

Q2 D1 3 ba b v1 Q1 D2 650 gal>min 900 gal>min

= 1264 rpm

b ( 13 ) (1750 rpm)

Ans.

Ans: 1264 rpm 1556

14–51. The temperature of benzene in a processing tank is maintained by recycling this liquid through a heat exchanger, using a pump that has an impeller diameter of 6  in. and produces a flow of 900 gal>min. If it is found that the heat exchanger can maintain the temperature only when the flow is 650 gal>min, determine the required diameter of the impeller if it maintains the same angular speed.

SOLUTION Since Q1, D1 and Q2 are given and v1 = v2, we can use flow coefficient similitude to determine D2. Q1 v1D13

=

Q2 v2D23 1

1

Q2 3 v1 3 D2 = a b a b D1 v2 Q1 = a

650 gal>min 900 gal>min

= 5.38 in.

1 3

1

b (1)3 (6 in.)

Ans.

Ans: 5.38 in. 1557

*14–52. A water pump has an impeller that has a diameter of 8 in. and rotates at 1750 rev>min. If the pump provides a discharge of 500 gal>min when operating at a head of 35 ft, determine the discharge and head if a similar pump is used with an impeller that has a diameter of 12 in. and turns at the same rate.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Dynamic similitude of the flow coefficient gives Q1 v1D13

=

Q2 v2D23

500 gal>min v(8 in.)

3

=

Q2 v(12 in.)3 Ans.

Q2 = 1687.5 gal>min Dynamic similitude of the head coefficient gives ∆H1 v12D12

=

(35 ft) 2

v (8 in.)

2

∆H2 v22D22 =

∆H2 2

v (12 in.)2

∆H2 = 78.75 ft

Ans.

1558

14–53. A variable-speed pump requires 28 hp to run at an impeller speed of 1750 rpm. Determine the power required if the impeller speed is reduced to 630 rpm.

SOLUTION #

Since W1, v1 and v2 are given and D1 = D2, power coefficient similitude can be # used to determine W2.

#

#

W1 v13D15

=

W2 v 23D25

#

W2 = a = a

v2 3 D2 5 # b a b W1 v1 D1 630 rpm

1750 rpm

= 1.31 hp

3

b ( 15 ) (28 hp)

Ans.

Ans: 1.31 hp 1559

14–54. The model of a water pump has an impeller with a diameter of 4 in. that discharges 80 gal>min. If the power required is 1.5 hp, determine the power required for the prototype having an impeller diameter of 12 in. that will discharge 600 gal>min.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Dynamic similitude of the power coefficient gives

#

#

Wp

=

vp3Dp5

Wm vm3Dm5

#

Wp = a

vp vm

3

b a

Dp Dm

5

#

(1)

b Wm

Dynamic similitude of the flow coefficient gives Qp vpDp

3

vp vm

=

Qm vmDm3

= a

Dm 3 Qp b a b Dp Qm

(2)

Substituting Eq. (2) into Eq. (1),

#

Wp = a = a

Dm 4 Qp 3 # b a b Wm Dp Qm

4 in. 4 600 gal>min 3 b a b (1.5 hp) 12 in. 80 gal>min

Ans.

= 7.81 hp

Ans: 7.81 hp 1560

14–55. The model of a water pump has an impeller with a diameter of 4 in. that discharges 80 gal>min with a pressure head of 4 ft. Determine the diameter of the impeller of the  prototype that will discharge 600 gal>min with a pressure head of 24 ft.

SOLUTION Water is considered to be incompressible. The relative flow is steady. Dynamic similitude of the head coefficient gives (∆H)p vp2Dp2 a

Dp

=

(∆H)m vm2Dm2 (∆H)p

2

b = c

Dm

(∆H)m

da

vm 2 b vp

(1)

Dynamic similitude of the flow coefficient gives Qp

=

vp Dp3

Qm vmDm3

vm Qm Dp 3 = a ba b vp Qp Dm

(2)

Substituting Eq. (2) into Eq. (1), and rearranging, a

Dp

4

Dm

b = c

Dp = c = a

(∆H)m (∆H)p (∆H)m (∆H)p 1

da 1 4

Qp Qm

d a

b

Qp

Qm

2

1 2

b (Dm)

1

4 ft 4 600 gal>min 2 b a b (4 in.) 24 ft 80 gal>min

Ans.

= 6.9993 in. = 7.00 in.

Ans: 7.00 in. 1561
Hibbeler, R.C. 2015. Solutions Manual, Fluid Mechanics. Pearson

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