Fluid Mechanics Solutions 2nd edition Hibbeler [2 ed.] 1292247304, 9781292247304

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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) mm # MN, (b) Mg>mm, (c) km>ms, (d) kN>(mm)2.

SOLUTION

a) mm # MN = (10 - 3 m)(106 N) = 103 N # m = kN # m 6

b) Mg>mm = (10 g2 >(10

-3

3

Ans.

m2 = 10 g>m = Gg>m

c) km>ms = (103 m)> 110 - 3 s2 = 106 m>s = Mm>s 2

Ans.

9

-3

2

9

2

Ans. 2

Ans.

d) kN> 1mm2 = (10 N2 >(10 m) = 10 N>m = GN>m

Ans: a) kN # m b) Gg>m c) Mm>s d) GN>m2 1

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1–2. Evaluate each of the following to three significant figures, and express each answer in SI units using an appropriate prefix: (a) 34.86(106)4 2 mm, (b) (348 mm)3, (c) (83 700 mN)2.

SOLUTION a) 3 4.86 ( 106 ) 4 2 mm = 34.861106 24 2 110 - 3 m2 = 23.621109 2 m = 23.6 Gm b) (348 mm)3 = 3 348 ( 10 - 3 ) m 4 3 = 42.14 ( 10 - 3 ) m3 = 42.1110 - 3 2 m3 c) 183,700 mN2 2 = 3 83,700 ( 10 - 3 ) N 4 2 = 7.006 ( 103 ) N2 = 7.01(103 2 N2

2

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Ans. Ans. Ans.

Ans: a) 23.6 Gm b) 42.1110 - 3 2 m3 c) 7.01(103 2 N2

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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) c)

(34 mm)(0.0763 Ms)>263 mg = 334110-3 2 m4 3 0.07631106 2s 4 > 3 263110-6 21103 2 g4 = 9.86 ( 106 ) m # s>kg = 9.86 Mm # s>kg

(4.78 mm)(263 Mg) =

3 4.78 ( 10-3 ) m 4 3 263 ( 106 ) g 4

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

Ans.

Ans.

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

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*1–4. Convert the following temperatures: (a) 250 K to degrees Celsius, (b) 322°F to degrees Rankine, (c) 230°F to degrees Celsius, (d) 40°C to degrees Fahrenheit.

SOLUTION a) TK = TC + 273;

250 K = TC + 273

Ans.

TC = -23.0°C

Ans.

b) TR = TF + 460 = 322°F + 460 = 782°R c) TC = d) TC =

5 5 1T - 322 = 1230°F - 322 = 110°C 9 F 9 5 1T - 322; 9 F

40°C =

5 1T - 322 9 F

Ans. Ans.

TF = 104°F

Ans: a) -23.0°C b) 782°R c) 110°C d) 104°F 4

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1–5. The tank contains a liquid having a density of 1.22 slug>ft3. Determine the weight of the liquid when it is at the level shown.

4 ft 1 ft

2 ft 3 ft

SOLUTION The specific weight of the liquid and the volume of the liquid are g = rg = ( 1.22 slug>ft3 )( 32.2 ft>s2 ) = 39.284 lb>ft3 V = ( 4 ft )( 2 ft )( 2 ft ) = 16 ft3 Then the weight of the liquid is W = g V = ( 39.284 lb>ft3 )( 16 ft3 ) = 628.54 lb = 629 lb

Ans.

Ans: W = 629 lb 5

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1–6. 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: W = 91.5 N 6

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1–7. The bottle tank has a volume of 0.35 m3 and contains 40 kg of nitrogen at a temperature of 40°C. Determine the absolute pressure in the tank.

SOLUTION The density of nitrogen in the tank is r =

40 kg m = = 114.29 kg>m3 V 0.35 m3

From the table in Appendix A, the gas constant for nitrogen is R = 296.8 J>kg # K. Applying the ideal gas law, p = rRT

p = ( 114.29 kg>m3 ) (296.8 J>kg # K)(40°C + 273) K = 10.621106 2Pa

Ans.

= 10.6 MPa

Ans: p = 10.6 MPa 7

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*1–8. The bottle tank contains nitrogen having a temperature of 60°C. Plot the variation of the pressure in the tank (vertical axis) versus the density for 0 … r … 5 kg>m3. Report values in increments of ∆p = 50 kPa.

SOLUTION

From the table in Appendix A, the gas constant for nitrogen is R = 296.8 J>kg # K. The constant temperature is T = (60°C + 273) K = 333 K. Applying the ideal gas law, p = rRT

p = r1296.8 J>kg # K)(333 K) p = 198,834r2 Pa p (kPa) 3

r (kg>m )

150 1.52

200 2.02

= 198.8r2 kPa

250 2.53

300 3.04

350 3.54

Ans. p(kPa)

400 4.05

400

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

300 200 100 0

8

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1

2

3 (a)

4

5

r(kg ym3)

Ans: p = 198.8r2 kPa

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1–9. Determine the specific weight of hydrogen when the temperature is 85°C and the absolute pressure is 4 MPa.

SOLUTION

From the table in Appendix A, the gas constant for hydrogen is R = 4124 J>kg # K. Applying the ideal gas law, p = rRT

4 ( 106 ) N>m2 = r(4124 J>kg # K)(85°C + 273) K r = 2.7093 kg>m3 Then the specific weight of hydrogen is g = rg = ( 2.7093 kg>m3 )( 9.81 m>s2 ) = 26.58 N>m3 = 26.6 N>m3

Ans.

Ans: g = 26.6 N>m3 9

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1–10. 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 smaller 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

Ans: TC = 26.9°C 10

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1–11. The tanker carries 900(103) 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>ft3.

SOLUTION The specific weight of the crude oil is go = Sogw = 0.940 ( 62.4 lb>ft3 ) = 58.656 lb>ft3 The volume of the crude oil is Vo = 3900(103) bl4 a

42 gal 1 bl

Then, the weight of the crude oil is

ba

1 ft3 b = 5.0535(106) ft3 7.48 gal

Wo = goVo = 158.656 lb>ft3 235.05351106 2 ft3 4 = 296.41 ( 106 ) lb = 296 ( 106 ) lb

Ans.

Ans: Wo = 296 ( 106 ) lb 11

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*1–12. 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 (rw)1 =

m ; V1

1000.0 kg>m3 =

m 36 m2(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: h = 3.05 m 12

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1–13. Determine the weight of carbon tetrachloride that should be mixed with 15 lb of glycerin so that the combined mixture has a density of 2.85 slug>ft3.

SOLUTION From the table in Appendix A, the densities of glycerin and carbon tetrachloride at s.t.p. are rg = 2.44 slug>ft3 and rct = 3.09 slug>ft3, respectively. Thus, their volumes are given by rg = rct =

mg Vg

; 2.44 slug>ft3 =

115 lb2 > 132.2 ft>s2 2

Wct > 132.2 ft>s2 2 mct ; 3.09 slug>ft3 = Vct Vct

The density of the mixture is rm =

Vg

Vg = 0.1909 ft3 Vct = 10.01005Wct 2 ft3

Wct > 132.2 ft>s2 2 + 115 lb2 > 132.2 ft>s2 2 mm ; 2.85 slug>ft3 = Vm 0.1909 ft3 + 0.01005Wct Wct = 32.5 lb

Ans.

Ans: Wct = 32.5 lb 13

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1–14. The tank contains air at a temperature of 18°C and an absolute pressure of 160 kPa. If the volume of the tank is 3.48 m3 and the temperature rises to 42°C, determine the mass of air that must be removed from the tank to maintain the same pressure.

SOLUTION

For T1 = 118°C + 2732 K = 291 K and R = 286.9 J>kg # K for air (Appendix A), the ideal gas law gives p1 = r1RT1;

160(103) N>m2 = r1(286.9 J>kg # K)(291 K) r1 = 1.9164 kg>m3

Thus, the mass of the air at T1 is m1 = r1V = 11.9164 kg>m3 2 13.48 m3 2 = 6.6692 kg

For T2 = 142°C + 2732 K = 315 K, and R = 286.9 J>kg # K, p2 = r2RT2;

160(103) N>m2 = r2(286.9 J>kg # K)(315 K) r2 = 1.7704 kg>m3

Thus, the mass of air at T2 is m2 = r2V = 11.7704 kg>m3 2 13.48 m3 2 = 6.1611 kg

Finally, the mass of air that must be removed is

∆m = m1 - m2 = 6.6692 kg - 6.1611 kg = 0.508 kg

Ans.

Ans: ∆m = 0.508 kg 14

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1–15. The tank contains 4 kg of air at an absolute pressure of 350 kPa and a temperature of 18°C. If 0.8 kg of air is added to the tank and the temperature rises to 38°C, determine the resulting pressure in the tank.

SOLUTION

For T1 = (18°C + 273) K = 291 K, p1 = 350 kPa and R = 286.9 J>kg # K for air (Appendix A), the ideal gas law gives p1 = r1RT1;

350 ( 103 ) N>m2 = r1 1286.9 J>kg # K)(291 K) r1 = 4.1922 kg>m3

Since the volume is constant, V =

m1 m2 m2 = ; r2 = r r1 r2 m1 1

Here, m1 = 4 kg and m2 = 14 + 0.82 kg = 4.8 kg r2 = a

4.8 kg 4 kg

b 14.1922 kg>m3 2 = 5.0307 kg>m3

Again, applying the ideal gas law with T2 = (38°C + 273) K = 311 K,

p2 = r2RT2; = 15.0307 kg>m3 21286.9 J>kg # K2(311 K) = 448.861103 2 Pa = 449 kPa

Ans.

Ans: p2 = 449 kPa 15

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*1–16. The 8-m-diameter spherical balloon is filled with helium that is at a temperature of 28°C and an absolute pressure of 106 kPa. Determine the weight of the helium contained in the balloon. The volume of a sphere is V = 43pr 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 = rV = ( 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: W = 446 N 16

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1–17. Gasoline is mixed with 8 ft3 of kerosene so that the volume of the mixture in the tank becomes 12 ft3. Determine the specific weight and the specific gravity of the mixture at standard temperature and pressure.

SOLUTION From the table in Appendix A, the densities of gasoline and kerosene at s.t.p. are rg = 1.41 slug>ft3 and rk = 1.58 slug>ft3, respectively. The volume of gasoline is Vg = 12 ft3 - 8 ft3 = 4 ft3 Then the total weight of the mixture is therefore Wm = rggVg + rkgVk = ( 1.41 slug>ft3 )( 32.2 ft>s2 ) 14 ft2 2 + 11.58 slug>ft3 2132.2 ft>s2 218 ft3 2 = 588.62 lb

Thus, the specific weight and specific gravity of the mixture are gm =

Wm 588.62 lb = = 49.05 lb>ft3 = 49.1 lb>ft3 Vm 12 ft3

Ans.

Sm =

49.05 lb>ft3 gm = = 0.786 gw 62.4 lb>ft3

Ans.

Ans: gm = 49.1 lb>ft3 Sm = 0.786 17

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1–18. Determine the change in the density of oxygen when the absolute pressure changes from 345 kPa to 286 kPa, while the temperature remains constant at 25°C. This is called an isothermal process.

SOLUTION Applying the ideal gas law with T1 = (25°C + 273) K = 298 K, p1 = 345 kPa and R = 259.8 J>kg # K for oxygen (table in Appendix A), p1 = r1RT1;

345 ( 103 ) N>m2 = r1(259.8 J>kg # K)(298 K) r1 = 4.4562 kg>m3

For p2 = 286 kPa and T2 = T1 = 298 K, p2 = r2RT2;

286 ( 103 ) N>m2 = r2(259.8 J>kg # K)(298 K) r2 = 3.6941 kg>m3

Thus, the change in density is ∆r = r2 - r1 = 3.6941 kg>m3 - 4.4562 kg>m3 = - 0.7621 kg>m3 = -0.762 kg>m3

Ans.

The negative sign indicates a decrease in density.

Ans: ∆r = -0.762 kg>m3 18

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1–19. 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 W = MT g = (1987.80 kg) ( 9.81 m>s2 ) = 19 500 N = 19.5 kN

Ans.

Ans: W = 19.5 kN 19

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*1–20. 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 = 1.5p ft . Then, the 3

weight of the rain water is Ww = gwVw = ( 62.4 lb>ft3 )( 1.5p ft3 ) = 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 ft3 Thus, gc =

W 93.6p lb = = 0.02971 lb>ft3 Vc ′ 3150p ft3

Then Wc = gcVc = a0.02971

lb 52803 ft3 b c (6.50) a bd 1 ft3

= 28.4 ( 109 ) lb

Ans.

Ans: Wc = 28.4 ( 109 ) lb 20

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1–21. A volume of 8 m3 of oxygen initially at 80 kPa of absolute pressure and 15°C is subjected to an absolute pressure of 25 kPa while the temperature remains constant. Determine the new density and volume of the oxygen.

SOLUTION

From the table in Appendix A, the gas constant for oxygen is R = 259.8 J>kg # K. Applying the ideal gas law, 80 ( 103 ) N>m3 = r1(259.8 J>kg # K)(15°C + 273) K

p1 = r1RT1;

r1 = 1.0692 kg>m3 For T2 = T1 and p2 = 25 kPa , p1 r1RT1 = ; p2 r2RT2

p1 r1 = r2 p2

1.0692 kg>m3 80 kPa = r2 25 kPa r2 = 0.3341 kg>m3 = 0.334 kg>m3

Ans.

The mass of the oxygen is m = r1V1 = ( 1.0692 kg>m3 )( 8 m3 ) = 8.5536 kg Since the mass of the oxygen is constant regardless of the temperature and pressure, m = r2V2; 8.5536 kg = ( 0.3341 kg>m3 ) V2 = 25.6 m3

Ans.

Ans: r2 = 0.334 kg>m3 V2 = 25.6 m3 21

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1–22. When a pressure of 650 psi is applied to a solid, its specific weight increases from 310 lb>ft3 to 312 lb>ft3. Determine the approximate bulk modulus.

SOLUTION Differentiating V =

W with respect to g, we obtain g dV = -

W dg g2

Then EV = -

dp = dV>V

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

=

dp dg>g

Therefore, 650 lb>in2

EV = a

1312 lb>ft3 - 310 lb>ft3 2 310 lb>ft3

b

= 100.75 ( 103 ) psi = 1011103 2 psi Ans.

The more precise answer can be obtained from p

EV =

Lpi

g

dp

dg Lgi g

=

650 lb>in2 p - pi = = 101.07 ( 103 ) psi = 101 ( 103 ) psi Ans. g 312 lb>ft3 lna b ln a b gi 310 lb>ft3

22

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Ans: EV = 1011103 2 psi

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1–23. 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 = -dp > ( dV>V ) .

dp dV = V EV

V2

p

2 dV 1 = dp LV1 V EV Lp1

ln a

V1 1 b = ∆p 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: ∆r = 2.02% r1 23

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*1–24. 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.

Ans: ∆p = 0.958 kip>in2 (ksi) 24

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1–25. At a point deep in the ocean, the specific weight of seawater is 64.2 lb>ft3. Determine the absolute pressure in lb>in2 at this point if at the surface the specific weight is g = 63.6 lb>ft3 and the absolute pressure is pa = 14.7 lb>in2. Take EV = 48.7(106) lb> ft2.

SOLUTION Differentiating V =

W with respect to g, we obtain g dV = -

W dg g2

Then EV = -

dp = dV>V

dp W a - 2 dg b n ( W>g ) g

dp = EV

=

dp dg>g

dg g

Integrate this equation with the initial condition at p = pa, g = g0, then p

Lpa

g

dr = EV

dg Lg0 g

p - pa = EV ln

g g0

p = pa + EV ln

Substitute pa = 14.7 lb>in2, EV = c 48.7(106)

g g0

lb 1 ft 2 da b = 338.19(103) lb>in2 2 12 in ft

g0 = 63.6 lb>ft3 and g = 64.2 lb>ft3 into this equation, p = 14.7 lb>in2 + 3338.19(103) lb>in2 4 c lna = 3.190(103) psi

64.2 lb>ft3 63.6 lb>ft3

bd

= 3.19(103) psi

Ans.

Ans: p = 3.19(103) psi 25

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1–26. A 2-kg mass of oxygen is held at a constant temperature of 50°C 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: EV = 220 kPa 26

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1–27. The viscosity of SAE 10 W30 oil is m = 0.100 N # s>m2. Determine its kinematic viscosity. The specific gravity is So = 0.92. Express the answer in SI and FPS units.

SOLUTION The density of the oil can be determined from ro = Sorw = 0.92 ( 1000 kg>m3 ) = 920 kg>m3

Then,

0.100 N # s>m2 mo no = r = = 108.70 ( 10-6 ) m2 >s = 109 ( 10-6 ) m2 >s o 920 kg>m3

Ans.

In FPS units,

no = c 108.70 ( 10-6 )

2 1 ft m2 da b s 0.3048 m

= 1.170 ( 10-3 ) ft2 >s = 1.17110 - 3 2ft2 >s

27

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Ans.

Ans: no = 109 ( 10-6 ) m2 >s = 1.17110 - 3 2 ft2 >s

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*1–28. If the kinematic viscosity of glycerin is n = 1.15(10-3) m2 >s, determine its viscosity in FPS units. At the temperature considered, glycerin has a specific gravity of Sg = 1.26.

SOLUTION The density of glycerin is rg = Sg rw = 1.26 ( 1000 kg>m3 ) = 1260 kg>m3 Then, ng =

my rg

; 1.15110 - 3 2 m2 >s =

mg 1260 kg>m3

mg = a1.449

N#s 1 lb 0.3048 m 2 ba ba b 2 4.448 N 1 ft m

= 0.03026 lb # s>ft2 = 0.0303 lb # s>ft2

Ans.

Ans: mg = 0.0303 lb # s>ft2 28

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1–29. 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 is 16.8 s -1. Since blood is a non-Newtonian fluid, determine its apparent viscosity at A.

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: ma = 8.93(10-3) N # s>m2 29

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1–30. The plate is moving at 0.6 mm>s when the force applied to the plate is 4 mN. If the surface area of the plate in contact with the liquid is 0.5 m2, determine the approximate viscosity of the liquid, assuming that the velocity distribution is linear.

0.6 mm!s

4 mN

4 mm

SOLUTION The shear stress acting on the fluid contact surface is

t=

4110 - 3 2 N F = = 8.00110 - 3 2 N>m2 A 0.5 m2

Since the velocity distribution is assumed to be linear, the velocity gradient is a constant.

t= m

du ; dy

8.00 110 - 3 2 N>m2 = m c

0.6110 - 3 2m>s 4110 - 3 2m

m = 0.0533 N # s>m2

d

Ans.

Ans: m = 0.0533 N # s>m2 30

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1–31. 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 = (4.23y1>3) mm>s, where y is in mm. Determine the shear stress within the fluid at y = 5 mm. Take m = 0.630(10-3) N # s>m2.

P 10 mm y

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = 14.23y1>3 2 mm>s

du 1 = c 14.232y-2>3 d s - 1 dy 3 = a

At y = 5 mm,

1.41 y2>3

b s-1

du 1.41 = a 2>3 b s - 1 = 0.4822 s - 1 dy 5

The shear stress is t= m

du = 30.630110 - 3 2N # s>m2 410.4822 s - 1 2 dy

= 0.3038110 - 3 2 N>m2

Ans.

= 0.304 mPa

du S ∞ and so t S ∞. Hence, the equation cannot be applied Note: When y = 0, dy at this point.

Ans: t = 0.304 mPa 31

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*1–32. 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 = (4.23y1>3) mm>s, where y is in mm. Determine the minimum shear stress within the fluid. Take m = 0.630(10-3) N # s>m2.

P 10 mm y

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = 14.23y1>3 2 mm>s

du 1 = c 14.232y-2>3 d s - 1 dy 3 = a

1.41 y2>3

b s-1

The velocity gradient is smallest when y = 10 mm and this minimum value is a

du 1.41 b = a 2>3 b s - 1 = 0.3038 s - 1 dy min 10

The minimum shear stress is tmin = m a

du b = dy min

3 0.630 ( 10-3 ) N # s>m2 4 10.3038 s - 1 2

= 0.1914110 - 3 2 N>m2

Ans.

= 0.191 MPa

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

Ans: tmin = 0.191 MPa 32

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1–33. The Newtonian fluid is confined between a plate and a fixed surface. If its velocity profile is defined by u = (8y - 0.3y2) 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 = 0.482 N # s>m2.

P 4 mm y

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y. u = ( 8y - 0.3y2 ) mm>s du = (8 - 0.6y) s -1 dy At the plate and the fixed surface, y = 4 mm and y = 0, respectively. Thus, a

The shear stresses are tp = ma tfs = ma

a

du b = 38 - 0.6(4)4 s -1 = 5.60 s -1 dy p

du b = 38 - 0.6(0)4 s -1 = 8.00 s -1 dy fs

du b = ( 0.482 N # s>m2 ) (5.60 s -1) = 2.699 N>m2 = 2.70 Pa dy p

du b = ( 0.482 N # s>m2 ) (8.00 s -1) = 3.856 N>m2 = 3.86 Pa Ans. dy fs

Ans: tp = 2.70 Pa tfs = 3.86 Pa 33

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1–34. The Newtonian fluid is confined between the plate and a fixed surface. If its velocity profile is defined by u = (8y - 0.3y2) 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 15(103) mm2 in contact with the fluid. Take m = 0.482 N # s>m2.

P 4 mm y

u

SOLUTION Since the velocity distribution is not linear, the velocity gradient varies with y given by u = ( 8y - 0.3y2 ) mm>s du = (8 - 0.6y) s -1 dy At the plate, y = 4 mm. Thus, a

The shear stress is tp = ma

du b = 38 - 0.6(4)4 s -1 = 5.60 s -1 dy p

du b = ( 0.482 N # s>m2 ) 15.60 s -1 2 = 2.6992 N>m2 dy p

Thus, the force P applied to the plate is

P = tpA = (2.6992 N>m2) 3 15 ( 103 ) mm2 4 a = 0.04049 N = 0.0405 N

2 1m b 1000 mm

Ans.

Ans: P = 0.0405 N 34

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1–35. 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 b = ma 32p(0.015 m)4(0.05 m) 0.001 m

m = 0.8498 N # s>m2

Ans.

Thus, 8N v = ( 0.8488 N # s>m2 ) a b 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 y = 2.00 m>s 35

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*1–36. A plastic strip having a width of 0.2 m and a mass of 150 g passes between two layers A and B of paint having a viscosity of 5.24 N # s>m2. Determine the force P required to overcome the viscous friction on each side if the strip moves upwards at a constant speed of 4 mm>s. Neglect any friction at the top and bottom openings, and assume the velocity profile through each layer is linear.

P 8 mm

6 mm

A

B

0.30 m

SOLUTION Since the velocity distribution is assumed to be linear, the velocity gradient will be constant. For layers A and B, a

4 mm>s 4 mm>s du du b = = 0.5 s -1 a b = = 0.66675 s -1 dy A 8 mm dy B 6 mm

The shear stresses acting on the surfaces in contact with layers A and B are tA = ma tB = ma

du b = (5.24 N # s>m2)(0.5 s - 1) = 2.62 N>m2 dy A

P

du b = (5.24 N # s>m2)(0.6667 s - 1) = 3.4933 N>m2 dy B

Thus, the shear forces acting on the contact surfaces are

FA = tAA = (2.62 N>m2)[(0.2 m)(0.3 m)] = 0.1572 N FB = tBA = (3.4933 N>m2)[(0.2 m)(0.3 m)] = 0.2096 N

W 5 0.15(9.81)N FB

FA

Consider the force equilibrium along y axis for the FBD of the strip, Fig. a. + c ΣFy = 0;

(a)

P - 0.15(9.81) N - 0.1572 N - 0.2096 N = 0 Ans.

P = 1.8383 N = 1.84 N

Ans: P = 1.84 N 36

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1–37. A plastic strip having a width of 0.2 m and a mass of 150 g passes between two layers A and B of paint. If force P = 2 N is applied to the strip, causing it to move at a constant speed of 6 mm>s, determine the viscosity of the paint. Neglect any friction at the top and bottom openings, and assume the velocity profile through each layer is linear.

P 8 mm

6 mm

A

B

0.30 m

SOLUTION Since the velocity distribution is assumed to be linear, the velocity gradient will be constant. For layers A and B, a

6 mm>s du b = = 0.75 s -1 dy A 8 mm

a

6 mm>s du b = = 1.00 s -1 dy B 6 mm

The shear stresses acting on the surfaces of the strip in contact with layers A and B are tA = ma tB = ma

P 5 2N

du b = m(0.75 s - 1) = (0.75m) N>m2 dy A

du b = m(1.00 s - 1) = (1.00m) N>m2 dy B

Thus, the shear forces acting on these contact surfaces are

FA = tAA = (0.75m N>m2)[(0.2 m)(0.3 m)] = (0.045m) N FA

FB = tBA = (1.00m N>m2)[(0.2 m)(0.3 m)] = (0.06m) N Consider the force equilibrium along the y axis for the FBD of the strip, Fig. a. + c ΣFy = 0;

FB W 5 0.15(9.81)N

2 N - 0.045m N - 0.06m N - 0.15(9.81) N = 0

(a)

m = 5.0333 N # s>m

2

= 5.03 N # s>m2

Ans.

Ans: m = 5.03 N # s>m2 37

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1–38. The tank containing gasoline has a long crack on its side that has an average opening of 10 μm. The velocity through the crack is approximated by the equation u = 10 1 109 2 3 10 1 10-6 2 y - y2 4 m>s, where y is in meters, measured upward from the bottom of the crack. Find the shear stress at the bottom, y = 0, and the location y within the crack where the shear stress in the gasoline is zero. Take mg = 0.317 1 10-3 2 N # s>m2.

10 mm

SOLUTION

Gasoline is a Newtonian fluid. y(m)

The rate of change of shear strain as a function of y is

10(10–6)

du = 10 ( 109 ) 3 10 ( 10-6 ) - 2y 4 s -1 dy

u

10(109)[10(10–6)y – y2] m s

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

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

u(m s) (a)

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 mm 38

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1–39. The tank containing gasoline has a long crack on its side that has an average opening of 10 μm. If the velocity profile through the crack is approximated by the equation u = 10 1 109 2 3 10 1 10-6 2 y - 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 1 10-3 2 N # s>m2.

10 mm

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)

y(10–6 m)

10.0

10.0

7.50

7.50

5.00

5.00

2.50

2.50

0

0.10

0.20

u(m s)

0.30

–40

–30

–20

–10 0

(a )

10

20

30

40

τ(Ν m2)

(b)

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 Figs. a and b, respectively.

y( 10-6 m ) t( N>m2 )

0

1.25

2.50

3.75

5.00

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

0–6 m)

y(10–6 m) 10.0 7.50 5.00 2.50

0.10

0.20

0.30

u(m s)

–40

(a )

–30

–20

–10 0

10

20

30

40

τ(Ν m2)

(b)

39

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*1–40. 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 = BeC>T

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

ln 3 1.00 ( 10-3 ) 4 = ln ( BeC>293 )

- 6.9078 = ln B + ln eC>293 - 6.9078 = ln B + C>293 ln B = - 6.9078 - C>293

At T = (50 + 273) K = 323 K, m = 0.554 ( 10

-3

(1)

) N # s>m2. Thus,

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

ln 3 0.554 ( 10-3 ) 4 = ln ( BeC>323 )

-7.4983 = ln B + ln eC>323 C -7.4983 = ln B + 323 C ln B = - 7.4983 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.

Ans: C = 1863 K B = 1.73 ( 10-6 ) N # s>m2 40

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1–41. 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 ) e1863>T At T = (10 + 273) K = 283 K,

m = 1.732 ( 10-6 ) e1863 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 ) e1863 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 41

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1–42. 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

42

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

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1–43. 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 43

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*1–44. 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 ⤿

+ 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)

8.688(10–6)

N (b)

Ans: T = 0.218 mN # m 44

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1–45. Determine the torque T required to rotate the disk with a constant angular velocity of v = 30 rad>s. The oil has a thickness of 0.15 mm. Assume the velocity profile is linear, and m = 0.428 N # s>m2.

150 mm

T

SOLUTION Oil is a Newtonian fluid. The speed of the oil on the bottom contact surface of the disk is U = wr = (30r) m>s. Since the velocity profile is assumed to be linear as shown in Fig. a, the velocity gradient will be constant given by (30r) m>s du V = [200(103)r] s -1 = = dy t 0.15(10-3) m

U 5 (30r) mys

t5 0.15(10–3) m

So then, t= m

y

du = (0.428 N # s>m2)[200(103)r] s -1 dy

u (a)

= [85.6(103)r] N>m2 The shaded differential element shown in Fig. b has an area of dA = 2prdr. Thus, dF = tdA = (85.6(103)r)(2prdr) = 171.2p(103)r 2dr. Moment equilibrium about point O in Fig. b requires

&

+ ΣMo = 0 ;

T T -

T =

L

0.15 m

rdF = 0

L0

r 0.15m

L0

r[171.2p(103)r 2dr] = 0

O

0.15m

171.2p(103)r 3dr

= 171.2p(103) a

dF 5 tdA

T

r 4 0.15m b` 4 0

dr

(b)

= 68.07 N # m = 68.1 N # m

Ans.

Ans: T = 68.1 N # m 45

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1–46. Determine the torque T required to rotate the disk with a constant angular velocity of v = 30 rad>s as a function of the oil thickness t. Plot your results of torque (vertical axis) versus the oil thickness for 0 … t … 0.15(10-3) m for values every 0.03 (10-3) m. Assume the velocity profile is linear, and m = 0.428 N # s>m2.

150 mm

T

SOLUTION

y

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

0 ∞

0.03 340

0.06 170

0.09 113

0.12 85.1

0.15 68.1

U 5 (30r) mys

Oil is a Newtonian fluid. The speed of the oil on the bottom contact surface of the disk is U = wr = (30r) m>s. Since the velocity profile is assumed to be linear as shown in Fig. a, the velocity gradient will be constant, given by

t5 0.15(10–3) m u

(30r) m>s du U 30r -1 = = = a bs dy t t t

So then,

(a) 0.15 m

30r -1 12.84r du t= m = (0.428 N # s>m2) a s b = a b N>m2 dy t t

r

The shaded differential element shown in Fig. b has an area of dA = 2prdr. Thus,

O

12.84r 25.68p 2 dF = tdA = a b(2prdr) = r dr. Moment equilibrium about point O t t

&

L

T -

T T = =

(b)

rdF = 0

L0

0.15m

L0

ra

0.15m

dF 5 tdA

T

in Fig. b requires + ΣMo = 0;

dr

T (N·m) 400

25.68p 2 r dr b = 0 t

300

25.68p 3 r dr = 0 t

200

25.68p r 4 0.15m a b` t 4 0

100

0.010211 0.0102 = a b N#m = a b N # m where t is in meters. t t

0

0.03

0.06

0.09 (c)

0.12

0.15

t(10–3)m

Ans.

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

Ans: T = a

0.0102 b N#m t

where t is in meters.

46

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1–47. 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

A

r

L

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

du dy

(vr) = m t

F=

T

Considering the moment equilibrium of the tube, Fig. a, ΣM = 0; T = 2(m) T =

T - 2tAr = 0 (vr) t

2 µ r 2L t

O r

(2prL)r

4pmvr 3L t

Ans.

(a)

Ans: T =

4pmvr 3L t

47

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*1–48. The tube rests on a 1.5-mm thin film of oil having a viscosity of m = 0.0586 N # s>m2. If the tube is rotating at a constant angular velocity of v = 4.5 rad>s, determine the shear stress in the oil at r = 40 mm and r = 80 mm. Assume the velocity profile within the oil is linear.

v 5 4.5 rad !s 40 mm

T

80 mm

SOLUTION Oil is a Newtonian fluid. Since the velocity profile is assumed to be linear, the velocity gradient will be constant. At r = 40 mm and r = 80 mm, (4.5 rad>s)(40 mm) du wr ` = = dy r = 40 mm t 1.5 mm = 120 s -1

(4.5 rad>s)(80 mm) du wr ` = = dy r = 80 mm t 1.5 mm = 240 s -1

Then the shear stresses in the oil at r = 40 mm and 80 mm are t' r = 40 mm = ma

du b` = ( 0.0586 N # s>m2 ) 1120 s - 1 2 = 7.032 N>m2 = 7.03 Pa dy r = 40 mm Ans.

t' r = 80 mm = ma

du b` = ( 0.0586 N # s>m2 ) 1240 s - 1 2 = 14.064 N>m2 = 14.1 Pa dy r = 80 mm Ans.

Ans: t' r = 40 mm = 7.03 Pa t' r = 80 mm = 14.1 Pa 48

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1–49. The tube rests on a 1.5-mm thin film of oil having a viscosity of m = 0.0586 N # s>m2. If the tube is rotating at a constant angular velocity of v = 4.5 rad>s, determine the torque T that must be applied to the tube to maintain the motion. Assume the velocity profile within the oil is linear.

v 5 4.5 rad !s 40 mm

T

80 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 du U wr = = . is U = vr. Thus, dy t t

t= m

mvr du = dy t

Thus, the shear force the oil exerts on the differential element of area dA = 2pr dr shown shaded in Fig. a is dF = tdA = a

mvr 2pmv 2 b(2pr dr) = r dr t t

Considering the moment equilibrium of the tube about point D, Lri

&

+ ΣMO = 0;

ro

rdF - T = 0

T =

= Substituting the numerical values, T =

Lri

ro

rdF =

2pmv ro 3 r dr t Lri

pmv 4 2pmv r 4 ro a b` = 1r o - r i4 2 t 4 ri 2t

p10.0586 N # s>m2 214.5 rad>s2 231.5 110-3 2 m4

= 0.01060 N # m = 0.0106 N # m

10.084 - 0.044 2

Ans.

Ans: T = 0.0106 N # m 49

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1–50. 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

u

SOLUTION

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

(2)

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

dz

dF φ

dr = ds cos u tan u

ds =

dr sin u

z (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

dz

θ ds (b)

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

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

T -

L

rdF = 0 T = =

=

L

rdF =

2pmv R 3 r dr t sin u L0

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

pmvR4 2t sin u

Ans.

Ans: T =

pmvR4 2t sin u

50

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1–51. The city of Denver, Colorado is at an elevation of 1610 m above sea level. Determine how hot one can prepare boiling water to make a cup of tea.

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

Ans.

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

Ans: Tboil = 94.6°C 51

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*1–52. 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>ft2 - a = a659.52

704.4 lb>ft2 - 629.6 lb>ft2 30 000 lb>ft2 - 27 500 lb>ft2

1 ft 2 lb b a b = 4.58 psi ft2 12 in.

b(29 000 ft - 27 500 ft) Ans.

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 52

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1–53. A boat propeller is rotating in water that has a temperature of 15°C. What is the lowest absolute water pressure that can be developed at the blades so that cavitation will not occur?

SOLUTION From the table in Appendix A, the vapor pressure of water at T = 15°C is pv = 1.71 kPa Cavitation (boiling of water) will occur if the water pressure is equal or less than pv. Thus, Ans.

pmin = pv = 1.71 kPa

Ans: pmin = 1.71 kPa 53

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1–54. As water at 20°C flows through the transition, its pressure will begin to decrease. Determine the lowest absolute pressure it can have without causing cavitation.

SOLUTION From the table in Appendix A, the vapor pressure of water at T = 20°C is pv = 2.34 kPa Cavitation (or boiling of water) will occur when the water pressure is equal to or less than pv. Thus, Ans.

p min = 2.34 kPa

Ans: pmin = 2.34 kPa 54

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1–55. 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?

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

Ans.

Ans: pmax = 0.363 lb>in2 55

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*1–56. 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?

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

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

pmax = pv = 3.17 kPa

Ans: pmax = 3.17 kPa 56

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1–57. For water falling out of the tube, there is a difference in pressure ∆p between a point located just inside and a point just outside of the stream due to the effect of surface tension s. Determine the diameter d of the stream at this location.

SOLUTION

d

Consider a length L of the water column. The free-body diagram of half of this column is shown in Fig. a. Consider the force equilibrium along the y-axis, ΣFy = 0;

2sL + po 3d(L)4 - pi 3d(L)4 = 0

However, pi - po = ∆p. Then

2s = ( pi - po ) d

d =

2s ∆p

Ans. z po

d

y

x

pi

s

L

s (a)

Ans: d =

2s ∆p

57

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1–58. 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°. Take gst = 490 lb>ft3 and s = 0.00492 lb>ft. Assume that each particle has the shape of a sphere, where V = 43 pr 3.

SOLUTION The weight of a steel particle is !

4 d 3 245p 3 W = gstV = ( 490 lb>ft3 ) c p a b d = d 3 2 3

W

!

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

d r =2

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

(a)

245p 3 d 3 d = 7.762 ( 10-3 ) ft

0.00492pd =

= 0.0931 in.

Ans.

Ans: d = 0.0931 in. 58

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1–59. Sand grains fall a short distance into a tank of water. Determine the largest average diameter of a grain that will float on the water with a contact angle of 180°. Take gs = 180 lb>ft3 and s = 0.00492 lb>ft. Assume each grain has the shape of a sphere, where V = 43 pr 3.

SOLUTION The weight of a sand grain is 4 d 3 W = gsV = ( 180 lb>ft3 ) c p a b d = 130p d 3 2 lb 3 2

!

W

!

Consider the force equilibrium along the vertical by referring to the FBD of the sand grain, Fig. a. + c ΣFy = 0;

d (0.00492 lb>ft) c 2p a b d - 30p d 3 = 0 2

r=

d 2

(a)

12 in. d = 10.01281 ft2 a b ft

Ans.

= 0.1537 in. = 0.154 in.

Ans: d = 0.154 in. 59

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*1–60. 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. Set D = 0.12 in.

D

h

SOLUTION

508

Using the result, h =

2s cos u rgr

From the table in Appendix A, for mercury, r = 26.3 slug>ft3 and s = 31.9 ( 10-3 )

2c 31.9 ( 10-3 ) h = a26.3 =

slug ft

3

lb . ft

lb d cos (180° - 50°) ft

b a32.2

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

3 - 9.6852 ( 10-3 ) ft 4 a

12 in. b 1 ft

Ans.

= - 0.116 in.

The negative sign indicates that a depression occurs.

Ans: h = -0.116 in. 60

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1–61. 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

508

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>ft3, 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>ft3 )( 32.2 ft>s2 ) 3(d>2)(1 ft>12 in.)4 - 0.01395 b in. d

§a

12 in. b 1 ft

where d is in in.

The negative sign indicates that a depression occurs.

Ans: For d = 0.075 in., h = 0.186 in. 61

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1–62. Water is at a temperature of 30°C. Plot the height h of the water as a function of the gap w between the two glass plates for 0.4 mm … w … 2.4 mm. Use increments of 0.4 mm. Take s = 0.0718 N>m.

w

h

D

SOLUTION When water contacts the glass wall, u = 0°. The weight of the rising column of water is W = gwV = rwg1hwD2 = rwghwD Consider the force equilibrium by referring to the FBD of the water column, Fig. a. + c ΣFy = 0;

21sD2 - rwghwD = 0 h =

sD sD

2s rWgw

W

From the table in Appendix A, rw = 995.7 kg>m3 at T = 30°C. Then h = £ = a

210.0718 N>m2

( 995.7 kg>m )( 9.81 m>s ) 3w110 2 m4 3

14.7 b mm w

-3

2

§a

h

103 mm b 1m

w (a)

Ans. h(mm)

where w is in mm.

40

For 0.4 mm 6 w 6 2.4 mm w (mm) h (mm)

0.4 36.8

30

0.8 18.4

1.2 12.3

1.6 9.19

2.0 7.35

2.4 6.13

20 10

The plot of h vs. w is shown in Fig. b. 0

0.4

0.8

1.2 1.6 (b)

2.0

2.4

w (mm)

Ans: h = a

14.7 b mm w

where w is in mm.

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1–63. Because cohesion resists any increase in the surface area of a liquid, it actually tries to minimize the size of the surface. Separating the molecules and thus breaking the surface tension requires work, and the energy provided by this work is called free-surface energy. To show how it is related to surface tension, consider the small element of the liquid surface subjected to the surface tension force F. If the surface stretches dx, show that the work done by F per increase in area is equal to the surface tension in the liquid.

Dy

Dx

F dx

SOLUTION F = s∆y work = Fdx

work>area increase =

s∆ydx = s (Q.E.D) ∆ydx

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*1–64. The glass tube has an inner diameter d and is immersed in water at an angle u from the horizontal. 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, ΣFx = 0;

"

L

prgd 2L s(pd) sin u = 0 4 L =

4s rgd sin u

Ans.

# d W=

N

!Pgd2 L 4

(a)

Ans: L = 4s>(rgd sin u) 64

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1–65. The glass 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

θ

N

60 0.002m

40

(a )

20 0

W = [9.81(10–3 ) h] N

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 65

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1–66. 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 combined to support itself in water having a temperature of T = 20°C. Take s = 72.7 mN>m, and assume the legs to be thin cylinders with a contact angle of u = 0°.

SOLUTION

P = 3.5316(10 –3) N

The force supported by the legs is

l

P =

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

l

= 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;

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

Ans.

(a)

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.

Ans: l = 24.3 mm 66

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1–67. The triangular glass rod has a weight of 0.3 N and is suspended on the surface of the water, for which s = 0.0728 N>m. Determine the vertical force P needed to pull the rod free from the surface.

P

608

SOLUTION

608 608

80 mm

The free-body diagram of the rod is shown in Fig. a. For water, its surface will be almost parallel to the surface of the rod (u ≈ 0°) at the point of contact. When the rod is on the verge of being pulled free, consider the force equilibrium along the vertical. + c ΣFy = 0;

P - W - 2T = 0

P - 0.3 N - 2{(0.0728 N>m)[3(0.08 m)]} = 0

P

Ans.

P = 0.3349 N = 0.335 N

T 5 sL

T 5 sL W 5 0.3N (a)

Ans: P = 0.335 N 67

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*1–68. The triangular glass rod has a weight of 0.3 N and is suspended on the surface of the water. If it takes a force of P = 0.335 N to lift it free from the surface, determine the surface tension of the water.

P

608

SOLUTION

608 608

80 mm

The free-body diagram of the rod is shown in Fig. a. For water, its surface will be almost parallel to the surface of the rod (u ≃ 0°) at the point of contact. When the rod is on the verge of being pulled free, consider the force equilibrium along the vertical. + c ΣFy = 0; 0.335 N - 0.3 N - 2{s[3(0.08 m)] = 0

P

Ans.

s = 0.07292 N>m = 0.0729 N>m

T 5 sL

T 5 sL W 5 0.3N (a)

Ans: s = 0.0729 N>m 68

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–1. Show that Pascal’s law applies within a fluid that is accelerating, provided there are 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.

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–2. 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 the density of the mud 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)

Ans: rm = 1390 kg>m3 70

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–3. 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 1 ft 12 in. m2 4.4482 N

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 1 ft 12 in. m2 4.4482 N

Ans. Ans.

Ans: ps = 16.0 kPa = 2.31 psi pd = 10.6 kPa = 1.54 psi 71

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–4. Oxygen in a tank has an absolute pressure of 130 kPa. Determine the pressure head in mm of mercury. The atmospheric pressure is 102 kPa.

SOLUTION The gauge pressure of the oxygen in the tank can be determined from pabs = patm + pg 130 kPa = 102 kPa + pg pg = 28 kPa From the table in Appendix A, rHg = 13 550 kg/m3 p = gHghHg 3

28(10 ) N>m2 = 3(13 550 kg>m3)(9.81 m>s2)4hHg hHg = 0.2106 m = 211 mm

Ans.

Ans: hHg = 211 mm 72

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–5. If the piezometer measures a gage pressure of 10 psi at point A, determine the height h of the water in the tube. Compare this height with that using mercury. Take rw = 1.94 slug>ft3 and rHg = 26.3 slug>ft3.

h

0.5 ft

A

SOLUTION

Patm

Here, the absolute pressure to be measured is p = pg + patm = a 10

lb 12 in. 2 lb ba b + patm = 11440 + patm 2 2 2 1 ft in ft

Patm hw

For the water piezometer, Fig. a,

p = patm + pw;

pHg

pw

lb = patm + (1.94 slug>ft3)(32.2 ft>s2)(h + 0.5 ft) ft3

(1440 + patm)

Ans.

hw = 22.55 ft = 22.6 ft

P (a)

hHg

P (b)

For the mercury piezometer, Fig. b, p = patm + pHg; (1440 + patm)

lb = patm + (26.3 slug>ft3)(32.2 ft>s2)(h + 0.5 ft) ft3 Ans.

hHg = 1.20 ft

Ans: hw = 22.6 ft hHg = 1.20 ft 73

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–6. 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 ( 103 ) N>m2 = ( 13 550 kg>m3 )( 9.81 m>s2 ) hHg Ans.

hHg = 0.3009 m = 301 mm

Ans: hHg = 301 mm 74

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

D

2–7. The field storage tank is filled with oil. The 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 in 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 maximum pressure occurs in the tank? What is this value? Take ro = 1.78 slug>ft3.

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>ft3 )( 32.2 ft>s2 ) (4 ft) = 229.26

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

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>ft3 )( 32.2 ft>s2 ) (10 ft) = 573.16 lb>ft2 a

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

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

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

D

*2–8. 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.

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>ft3 )( 32.2 ft>s2 ) (8 ft + 4 ft) = a687.79

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

Ans.

Ans: ( pmax )g = 4.78 psi 76

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. A

2–9. 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. The atmospheric pressure is 14.7 psi.

h

B

SOLUTION From the Appendix, pct = 3.09 slug>ft3. 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>ft3 )( 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

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 77

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 1 ft

2–10. 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>ft3.

3 ft

B

SOLUTION h

The gauge pressures at points A and B are

= a246.5

6 ft

A

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

2 ft

lb 1 ft 2 b = 1.71 psi b a 12 in. ft2

Ans.

pB = gea hB = ( 49.3 lb>ft3 ) (7 ft - 6 ft) = a49.3

lb 1 ft 2 b = 0.342 psi b a 12 in. ft2

Ans.

Ans: pA = 1.71 psi, pB = 0.342 psi 78

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 1 ft

2–11. The soaking bin contains ethyl alcohol used for cleaning automobile parts. If the gage 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>ft3.

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 3 12 in. ft

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

Ans.

h = 7.46 ft

Ans: ( pA)g = 1.87 psi h = 7.46 ft 79

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–12. 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 G B 40 m 1m C D

SOLUTION 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: h = 5.33 m 80

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–13. If the water in the structure in Prob. 2–12 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 G B 40 m 1m C

SOLUTION

D

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

E

(pD)w = (pD)o

5m 1m

10 m

rwghw = rogho

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

Ans.

h = 6.00 m

Ans: h = 6.00 m 81

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–14. The tank is filled with water and gasoline at a temperature of 20°C to the depths shown. If the absolute air pressure at the top of the tank is 200 kPa, determine the gage pressure at the bottom of the tank. Would the results be different if the tank had a flat bottom rather than a curved one? The atmospheric pressure is 101 kPa.

0.8 m 0.2 m

SOLUTION The absolute pressure at the bottom of tank is 1pb 2 abs = pair + pg + pw

Here, pg and pw can be determined with rg = 726 kg>m3 and rw = 998.3 kg>m3 at T = 20°C (table in Appendix A). Then 1pb 2 abs = 2001103 2 N>m2 + 1726 kg>m3 219.81 m>s2 210.8 m2 + 1998.3 kg>m3 219.81 m>s2 210.2 m2 = 207.661103 2 kPa

Thus, the gauge pressure at the bottom of the tank can be determined using 1pb 2 abs = 1pb 2 g + patm

207.661103 2 kPa = 1pb 2 g + 101 kPa

Ans.

(pb)gage = 106.66 kPa = 107 kPa

No, the pressure at the bottom of the tank does not depend on its shape.

Ans: (pb)gage = 107 kPa The pressure at the bottom of the tank does not depend on its shape. 82

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–15. At the bottom of a bottle a bubble within carbonated water has a diameter of 0.2 mm. Determine the bubble’s diameter when it reaches the surface. The temperature of the water and bubbles is 10°C, and the atmospheric pressure is 101 kPa. Assume that the density of the water is the same as that of pure water. 300 mm

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, the mass of the CO2 in the bubble, is also constant, then V p = constant m>V (1)

pV = constant

At T = 10°C, rw = 999.7 kg>m3. The pressure due to the static water can be determined using p = gh. Then the absolute pressure at the bottom of the bottle can be determined from pb = patm + gw hw = 101(103) N>m2 + (999.7 kg>m3)(9.81 m>s2)(0.3 m) = 103.94(103) Pa = 103.94 kPa At the surface of the bottle, the absolute pressure is ps = patm = 101 kPa Using Eq. (1), we can write pbVb = psVs 4 4 db 3 (103.94 kPa)c p(0.1 mm)3 d = (101 kPa) c p a b d 3 3 2 db = 0.201923 mm

Ans.

= 0.202 mm

Ans: db = 0.202 mm 83

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–16. 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 h. The density of the fluid at the surface is r0.

SOLUTION p=0

The fluid is considered compressible. EV = However, V =

dp dV>V

m . Then, r

z=h z

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

p

dp dr>r

(a)

At the surface, where p = 0, r = r0, Fig. a, then p dr dp = Lr0 r L0 r

EV

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

h

gdz p>E L0 L0 r0e EV 1 1 - e - p>EV 2 = gh dp

V

=

r0

1 - e - p>EV =

r0gh EV

p = - EV ln a1 -

r0gh b EV

Ans.

84

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Ans: p = -EV ln a1 -

rogh b EV

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–17. 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 r0 = 1000 kg>m3. Compare this result with water that is assumed 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

3 2.20 ( 10 ) 9

r

dp = dp N>m 4 3 r L0 L1000 kg>m r

2

p = 2.20 ( 109 ) ln a p

r = 1000 e2.20(10 )

r b 1000

(1)

9

Also, dp = rgdz dp = 9.81dz r

(2)

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

= 9.81dz

p

1000e2.20(10 ) 9

Since the pressure p = 0 at z = 0 and p at z = 300 m, r

dp

L0 1000e 2.20(10 ) p

9

p

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

p

9

0

=

L0

300 m

= 9.81z `

9.81dz 300 m 0

85

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2–17. 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: Compressible: p = 2.945 MPa Incompressible: p = 2.943 MPa 86

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–18. As the weather balloon ascends, measurements indicate that the temperature begins to decrease linearly from T0 at z = 0 to Tf at z = h. If the absolute pressure of the air at z = 0 is p0, determine the pressure as a function of z.

SOLUTION We will first determine the absolute temperature as a function of z: T = c T0 - a

T0 - T f h

T0 h - 1T0 - Tf 2z

bz d =

h

Using this result to apply the ideal gas law, p = rRT;

r =

p = RT

= However,

z

Rc

p T0 h - 1T0 - Tf 2z h

d

ph R3T0 h - 1T0 - Tf 2z4

dp = - gdz = -rgdz Substitute the result of r into this equation: dp = -

phgdz R3T0 h - 1T0 - Tf 24z

gh dp dz = - c d p R T0 h - 1T0 - Tf 2z

When z = 0, p = p0. Then p

gh z dp dz = R L0 T0 h - 1T0 - Tf 2z Lp p 0

ln p ` ln

p

p0

= -

z gh 1 e ln 3T0 h - 1T0 - Tf 2z4 f ` R T0 - Tf 0

T0 h - 1T0 - Tf 2z gh p = ln c d p0 R1T0 - Tf 2 T0 h gh 0 - Tf 2

T0 h - 1T0 - Tf 2z R1T p ln = ln e c d p0 T0 h gh 0 - Tf 2

T0 h - 1T0 - Tf 2z R1T p = c d p0 T0 h p = p0 c 1 - a

T0 - Tf T0 h

bz d

f

gh R1T0 - Tf2

Ans. Ans: p = p0 c 1 - a

T0 - T f T0h

bz d

gh R1T0 - Tf2

87

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–19. As the balloon ascends, measurements indicate that the temperature begins to decrease at a constant rate, from T = 60°F at z = 0 to T = 50°F at z = 2500 ft. If the absolute pressure of the air at z = 0 is p = 14.7 psi, plot the variation of pressure (vertical axis) versus altitude for 0 … z … 2500 ft. Give values for increments of ∆z = 500 ft.

SOLUTION We will first determine the absolute temperature as a function of z. T = c 520 - a

520 - 510 bz d °K = 1520 - 0.004z2 °K 2500

Using this result to apply the ideal gas law with R = 1716 ft # lb>slug # °K,

z

p p p r = = = 3 RT 17161520 - 0.004z2 892.32110 2 - 6.864z

p = rRT; However,

dp = - gdz = -rgdz Substitute the result of r into this equation: dp = -

When z = 0, p = a14.7 p

p(psi)

p132.22 892.321103 2 - 6.864z

15.00

dz

14.50

dp 32.2 dz = p 892.321103 2 - 6.864z

14.00

2

lb 12 in. b = 2116.8 lb>ft2. Then ba 2 1 ft in

13.50

z dp dz = - 32.2 3 L2116.8 p L0 892.32110 2 - 6.864z

ln p `

p

2116.8

= - 32.2e -

13.00

z 1 ln 3892.321103 2 - 6.864z4 f ` 6.86 0

0

500

1000

1500 (a)

2000

2500

z(ft)

892.321103 2 - 6.864z p d ln a b = 4.6911 ln c 2116.8 892.321103 2 ln a

892.321103 2 - 6.864z 4.6911 p d b = ln e c f 2116.8 892.321103 2 892.321103 2 - 6.864z 4.6911 p d = c 2116.8 892.321103 2

p = e 2116.8 31 - 7.6923110-6 2z4 4.6911

For 0 … z … 2500 ft,

z(ft) p(psi)

0 14.70

lb 1 ft 2 fa b 2 12 in. ft

p = 514.731 - 7.6923110-6 2z4 4.6911 6 psi where z is in ft.

500 14.44

1000 14.18

1500 13.92

2000 13.67

2500 13.42 Ans: p = 514.731 - 7.69110-6 2z4 4.69 6 psi where z is in ft.

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

88

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–20. Using the data in Appendix A, make a graph showing how the atmospheric pressure p in psia (horizontal axis) varies with elevation h in feet. Plot values of p every 2500 ft for 0 … h … 15 000 ft.

SOLUTION Elevation (ft)

0

2500

5000

7500

10 000

12 500

15 000

r(psf)

2116

1932

1761

1602

1456

1320

1195

p(psia)

14.69

13.42

12.23

11.25

10.11

9.17

8.30

a

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

The plot of elevation vs. psia is shown in Fig. a. Elevation (ft) 15,000 12,500 10,000 7500 5000 2500 0

8

9

10 11 12 13 (a)

14

15

p(psia)

89

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–21. A liquid has a density that varies with depth h, such that r = 10.001h + 1.602 slug>ft3, where h is in feet. Determine the pressure due to the liquid when h = 80 ft.

SOLUTION

z

For the compressible liquid, dp = - gdz = -rgdz

h

However, z = h0 - h where h0 is a constant, Fig. a. Then dz = -dh. Substituting this result into the above equation, it becomes,

p

h0 z

dp = rgdh Integrating this equation using the gage pressure p = 0 at h = 0 and p at h, then L0

p

dp =

L0

h

p = c 10.0161h2 + 51.52h2

lb 1 ft 2 d a b ft2 12 in.

= 30.1118110-3 2h2 + 0.3578h4 psi

When h = 80 ft,

(a)

10.001h + 1.62132.22dh

where h is in feet.

p = 30.1118110-3)(802 2 + 0.357818024 psi = 29.34 psi = 29.3 psi

Ans.

Ans: p = 29.3 psi 90

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–22. A liquid has a density that varies with depth h, such that r = (0.001h + 1.60) slug>ft3, where h is in feet. Plot the variation of the pressure due to the liquid (vertical axis) versus depth for 0 … h … 100 ft. Give values for increments of 20 ft.

SOLUTION For the compressible liquid, dp = - gdz = -rgdz However, z = h0 - h where h0 is a constant, Fig. a. Then dz = -dh. Substituting this result into the above equation, it becomes dp = rgdh Integrating this equation using the gauge pressure p = 0 at h = 0 and p at h, then L0

p

dp =

L0

h

10.001h + 1.602132.22dh

p = c (0.0161h2 + 51.52h)

lb 1 ft 2 d a b ft2 12 in.

= 30.1118110-3 2h2 + 0.3578h4 psi

Ans.

where h is in feet.

For 0 … h … 100 ft,

h(ft) p(psi)

0

20

40

60

80

100

0

7.20

14.5

21.9

29.3

36.9

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

p(psi) z

40 30

h p

h0 z

20 10 0

(a)

20

40

60 (b)

80

100

h(ft)

Ans: p = 30.112110-3 2h2 + 0.358 h4 psi where h is in feet. 91

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–23. 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 law, 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 = p0 a

T0 - Cz g>RC b T0

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–24. Determine the temperature at an elevation of z = 15 km. Also, what is the pressure at this elevation? Assume the stratosphere begins at z = 11 km (see Fig. 2–11).

SOLUTION Within the troposphere, 0 … z 6 11.0(103) m, the temperature TC as a function of z is TC = T0 - Cz (the equation of the straight line shown in Fig. 2–11). From this figure, we obtain 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

The absolute temperature is therefore

- 6.50 ( 10-3 ) z 4 °C

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

3 288

- 6.50 ( 10-3 ) z 4 K

(1)

p Substitute the ideal gas law p = rRT or r = into dp = -gdz = -rgdz, RT dp = -

p gdz RT

g dp = dz p RT

(2)

From the table in Appendix A, the gas constant for air is R = 286.9 J>kg # K. Also, p = 101.3 kPa at z = 0. Then substitute Eq. (1) into (2). p

z 9.81 m>s2 dp dz = -a b # 286.9 J>kg K L0 288 - 6.50 ( 10-3 ) z L101.3(103) Pa p p

ln p `

ln c

101.3(103) Pa

p

101.3 ( 103 )

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

d = 5.2605 ln c

288 - 6.50 ( 10

p = 101.3 ( 103 ) c

288

-3

)z

d

288 - 6.50 ( 10-3 ) z 288

z 0

d

5.2605

93

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*2–24. Continued

At z = 11.0 ( 103 ) m, p = 101.3 ( 10 ) • 3

288 -

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

5.2605

¶

= 22.58 ( 103 ) Pa

At z = 15(103) m, it is into the stratosphere in the region 11(103) m 6 z 6 20.1(103) m, of which the temperature is constant, Fig. 2–11. Thus, Ans.

TC = - 56.5°C 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 L22.58(103) Pa p p

ln p ` ln

22.58(103) Pa

p

22.58 ( 103 )

At z = 15 ( 103 ) m,

= -0.1579 ( 10-3 ) z `

z

11(103) m

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

-3 3 p = 3 22.58(103)e0.1579(10 ) 311(10 ) - z4 4 Pa

p =

3 22.58 ( 103 ) e0.1579(10

-3

) 3 11(103) - 15(103) 4 4

= 12.00 ( 103 ) Pa = 12.0 kPa

Pa Ans.

Ans: TC = -56.5°C p = 12.0 kPa 94

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–25. A heavy cylindrical glass is inverted and then placed down at the bottom of a 12–ft-deep swimming pool. Determine the height ∆h of water within the glass when it is at the bottom. Assume the air in the glass remains at the same temperature as the atmosphere. Hint: Account for the change in volume of air in the glass due to the pressure change. The atmospheric pressure is patm = 14.7 psi.

12 ft F

SOLUTION When submerged, the density and hence the volume of the air change due to pressure changes. Applying the ideal gas law, p = rRT

1 ft

Dh

0.5 ft

m However, r = . Then V m RT V pV = mRT p =

Since the mass and temperature of the air are constant, mRT is also constant. Thus, (1)

p1V1 = p2V2

lb 12 in. 2 lb ba b = 2116.8 2 , V1 = p10.25 ft2 2 11 ft2 = 2 1 ft in ft 0.0625p ft3. When submerged, the water rises to the height of ∆h. Thus,

When p1 = patm = a 14.7

p2 = patm + gwh = 2116.8 = (2928 - 62.4∆h)

lb ft2

lb lb + a62.4 3 b(13 ft - ∆h) ft2 ft

V2 = p10.25 ft2 2 11 ft - ∆h2 = 30.0625p11 - ∆h24 ft3

Substitute these values into Eq. (1).

a 2116.8

lb lb b 10.0625p ft3 2 = c 12928 - 62.4∆h2 2 d 30.0625p11 - ∆h24 ft3 ft2 ft 2116.8 = 12928 - 62.4∆h211 - ∆h2 62.4∆h2 - 2990.4∆h + 811.2 = 0

Choose ∆h 6 1 ft, ∆h = 0.2728 ft Ans.

= 3.27 in.

Ans: ∆h = 3.27 in. 95

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–26. The 150-mm-diameter container is filled to the top with glycerin, and a 50-mm-diameter pipe is inserted within it to a depth of 300 mm. If 0.00075 m3 of kerosene is then poured into the pipe, causing the displaced glycerin to overflow, 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: h = 135 mm 96

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

50 mm

2–27. The 150-mm-diameter container is filled to the top with glycerin, and a 50-mm-diameter pipe is inserted within it to a depth of 300 mm. Determine the maximum volume of kerosene that can be poured into the pipe while causing the displaced glycerin to overflow, so the kerosene does not come out from the bottom end. How high h does the kerosene rise above the glycerin?

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 hke =

rgl rke

hgl

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

Ans.

A kerosene h C

0.3 m

B

glycerin

(a)

Ans: hke = 164 mm, Vke = 0.912(10-3) m3 97

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–28. 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

D

E 120 mm 100 mm

B

Mercury

Referring to Fig. a, the manometer rule gives 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 )

Ans: pB = 17.8 kPa 98

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–29. Determine the level h′ of water in the tube if the depths of oil and water in the tank are 0.6 m and 0.8 m, respectively, and the height of mercury in the tube is h = 0.08 m. Take ro = 900 kg>m3, rw = 1000 kg>m3, and rHg = 13 550 kg>m3.

0.6 m

h

SOLUTION

0.8 m

Referring to Fig. a, hAB = 0.6 m, hBC = 0.8 - h′ and hCD = h = 0.08 m. Then the manometer rule gives

100 mm

h9

pA + poghAB + rwghBC - rHgghCD = pD Here, pA = pD = 0, since points A and D are exposed to the atmosphere. 0 + 1900 kg>m3 21g210.6 m2 + 11000 kg>m3 21g210.8 m - h′2 - 113 550 kg>m3 21g210.08 m2 = 0 h′ = 0.256 m = 256 mm

Ans.

Note: Since 0.1 m 6 h′ 6 0.8 m, the solution is OK!

A

hAB 5 0.6 m B

D hCD 5 0.08 m

hBC 5 0.8 m 2 h9

C

(a)

Ans: h′ = 256 mm 99

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–30. 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>ft3 ) (2ft + 4ft) = a374.4 pB = gBhB = ( 62.4 lb>ft3 ) (3 ft) = a187.2

lb 1 ft 2 b a b = 2.60 psi Ans. ft2 12 in.

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

A

Ans.

Ans: pA = 2.60 psi, pB = 1.30 psi 100

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–31. 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>ft3 ) (2 ft) = a124.8

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

Ans.

Ans: pC = 0.870 psi 101

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

Water

* 2–32. 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. Take 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>m3 )( 9.81 m>s2 ) (0.1 m) + 13.55 ( 1000 kg>m3 )( 9.81 m>s2 ) (0.15 m)

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: pB - pA = 18.5 kPa 102

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–33. 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: pA = 18.2 kPa 103

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–34. 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: h = 893 mm 104

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–35. 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 pipes. Take Sc = 0.953, SHg = 13.55, and Sel = 1.03.

A

B

C D

2 ft h

Mercury

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>ft3 ) (1.5 ft) + (13.55) ( 62.4 lb>ft3 ) (1 ft) - (1.03) ( 62.4 lb>ft3 ) (0.5 ft) = a15 pA = 1257.42

lb 1 ft 2 a b = 8.73 psi ft2 12 in.

Ans.

lb 12 in. 2 b ba 2 1 ft in

A

C

B hBD = 0.5 ft

D

hAC = 1.5 ft hCD = 1 ft

(a)

Ans: pA = 8.73 psi 105

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–36. 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>ft3 ) (2.5 ft - h) + 13.55 ( 62.4 lb>ft3 ) (h) a 1 ft in2 25 lb 12 in. 2 b -(1.03) ( 62.4 lb>ft3 ) (0.5 ft) = a 1 ft in2 h = 1.134 ft = 1.13 ft

3 ft

0.5 ft

hAC = 2.5 ft – h

A

pA + gchAC + gHghCD - gelhBD = pB

2 ft

C B

Ans.

hCD = h

hBD = 0.5 ft

D

(a)

Ans: h = 1.13 ft 106

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 75 mm

2–37. 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 the pressure difference between A and B. rw = 1000 kg>m3.

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, pA = (pw)1 + pa = rwg(hw)1 + pa

And

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: pB - pA = 736 Pa 107

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 75 mm

2–38. Solve Prob. 2–37 if the top segment is filled with an oil for which ro = 800 kg>m3.

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)

Ans: pB - pA = 1.32 kPa 108

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

2–39. 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: pB = 15.1 kPa 109

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

*2–40. 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

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>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)

Ans: pB = 20.8 kPa 110

M02_HIBB9290_01_SE_C02_ANS.indd 110

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–41. Determine the height h of the mercury in the tube if the level of water in the tube is h′ = 0.3 m and the depths of the oil and water in the tank are 0.6 and 0.5 m, respectively. Take ro = 900 kg>m3, rw = 1000 kg>m3, and rHg = 13 550 kg>m3.

A

0.6 m

D B

0.5 m

C

h

h9 5 0.3 m

SOLUTION Referring to Fig. a, hAB = 0.6 m, hBC = 0.8 m - 0.3 m = 0.5 m and hCD = h. Then the manometer rule gives pA + roghAB + rwghBC - rHgghCD = pD Here, pA = pD = 0, since points A and D are exposed to the atmosphere. 0 + 1900 kg>m3 21g210.6 m2 + 11000 kg>m3 21g210.5 m2 - 113 550 kg>m3 21g21h2 = 0

Ans.

h = 0.07675 m = 76.8 mm

A

hAB 5 0.6 m B hBC 5 0.5 m

D hCD 5 h

C

h9 5 0.3 m

(a)

Ans: h = 76.8 mm 111

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. A

2–42. The micro-manometer is used to measure small differences in pressure. The reservoirs R and upper portion of the lower tubes are filled 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 cross-sectional 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

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 - gt e 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 bg - a bg + gt - gR d AR R AR L

e 2

pA - pB = 2gR a

= ec gt - a1 -

At At bgR - a bg d AR AR L

Ans. (a)

Ans: pA - pB = ec gt - a1 -

At At bgR - a bg d AR AR L

112

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–43. 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. Take 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,

Liquid 1 initial level

h2 h AR = a b = At a b 2 2 At h2 = a bh AR

h1

A

B

h2 2

Liquid 1 h Liquid 2 initial level

Substitute this result into Eq. (1).

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

h 2

At bh AR

At br g d AR 1

(2)

h2

Liquid 2

(a)

When r1 = rke = 814 kg>m3, r2 = rHg = 13 550 kg>m3 and pA - pB = 40 Pa, 40 N>m2 = hc ( 13 550 kg>m3 )( 9.81 m>s2 ) - a1 h = 0.3191(10-3 ) m = 0.319 mm

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

Ans.

Ans: h = 0.319 mm 113

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. A

*2–44. Determine the difference in pressure pA - pB between the centers A and B of the closed pipes, which are filled with kerosene. The mercury in the inclined-tube manometer has the level shown. Take SHg = 13.55 and Sk = 0.82.

B 100 mm D

400 mm

C

608

SOLUTION Referring to Fig. a, hAC = 0.4 m, hCD = 0.4 m - 0.1 m = 0.3 m and hBD = 0.1 m. Then the manometer rule gives pA + rkghAC - rHgghCD - rkghBD = pB pA + 0.8211000 kg>m3 219.81 m>s2 210.4 m2 - 13.5511000 kg>m3 219.81 m>s2 210.3 m2 - 0.8211000 kg>m3 219.81 m>s2 210.1 m2 = pB pA - pB = 37.461103 2 Pa = 37.5 kPa A

Ans.

A

B 100 mm

B D

400 mm

hBD 5 0.1 m

hAC 5 0.4 m

hCD 5 0.3 m

Mercury C

608 (a)

Ans: pA - pB = 37.5 kPa 114

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–45. The pipes at A and B contain oil and the inclinedtube manometer is filled with oil and mercury. Determine the pressure difference between A and B. Take ro = 1.70 slug>ft3 and rHg = 26.3 slug>ft3.

A

12 in.

6 in.

608

B 8 in.

10 in. 5 in.

SOLUTION 6 10 sin 60° 8 Referring to Fig. a, hAC = ft, hCD = ft and hBD = ft. Then the 12 12 12 manometer rule gives pA + roghAC + rHgghCD - roghBD = pB pA + rog1hAC - hBD 2 + rHgghCD = pB

pA + 11.70 slug>ft3)(32.2 ft>s2 2a

6 8 10 sin 60° ft ft b + 126.3 slug>ft3)(32.2 ft>s2 2a ft b = pB 12 12 12

pB - pA = a602.05

lb 1 ft 2 b = 4.18 psi ba 2 12 in. ft

Ans.

A 6 hAC 5 ft 12

hCD 5

60° C

B

10 sin 60° ft 12

hBD 5

8 ft 12

D

(a)

Ans: pB - pA = 4.18 psi 115

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

2–46. 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.

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;

100 mm

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)

Ans: FB = 937 N 116

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

2–47. 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 = 61 073 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>in ) a b 3 (p)(2.5 ft)2 4 1 ft

pB = 60 psi

2

= 169 646 lb = 170 kip

Ans.

1.5 ft

FB

1.5 ft (b)

Ans: FB = 61.1 kip, FA = 170 kip 117

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

6 ft

*2–48. Seepage is assumed to occur beneath the concrete wall, producing a linear distribution of hydrostatic pressure as shown. Determine the resultant force on a 1-ft wide portion of the wall and its location, measured to the right and upward from point A. 15 ft

A

SOLUTION Since the width of the dam considered is b = 1 ft, the intensity of the distributed load at the base A of the dam is

6 ft

3

wA = gwhAb = 162.4 lb>ft 2115 ft211 ft2 = 936 lb>ft

Then, the resultant forces of the triangular distributed loads shown in Fig. a are Fh = Fn =

1 1 w h = 1936 lb>ft2115 ft2 = 7020 lb 2 A A 2 1 1 w L = 1936 lb>ft216 ft2 = 2808 lb 2 A 2

u FR

Then, the magnitude of the resultant force is

y A

wA

FR = 2F 2h + F 2n

= 217020 lb2 2 + 12808 lb2 2 = 7560.77 lb = 7.561103 2 lb

Ans.

and its direction is

u = tan-1 a

15 ft

x

Fh

Fn 2808 lb b = tan-1 a b = 21.8° Fh 7020 lb

(a)

Fv

Ans.

The location of the resultant force from A is x =

y =

1 16 ft2 = 2 ft 3

Ans.

1 115 ft2 = 5 ft 3

Ans.

Ans: FR = u = x = y =

7.561103 2 lb 21.8° 2 ft 5 ft

118

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. A

2–49. 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 1.25 m. Also, determine the location of this resultant, measured from the top surface of the oil. Take ro = 900 kg>m3.

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 = F 1 + F 2 + F 3 =

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)

(c)

Ans: FR = 29.3 kN, yP = 1.51 m

119

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–50. 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 container needed to open it. The gate is pinned at B and rests on a rubber seal at A.

h

B

6 ft 308

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>ft3 ) (h - 3 ft)(4 ft) = (249.6h - 748.8) lb>ft wA = gwhAb = ( 62.4 lb>ft3 ) (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. + 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: h = 5.63 ft 120

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–51. Determine the critical height h of the water level that causes the concrete gravity dam to be on the verge of tipping over due to water pressure. The density of concrete is rc = 2.40 Mg>m3. Hint: Work the problem using a 1-m width of the dam. 18 m

h

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

B

A

wB = rwghb = 11000 kg>m3 219.81 m>s2 21h211 m2 = 19810h2 N>m

6m

The resulting triangular distributed load is shown on the FBD of the dam, Fig. a, and its resultant is F =

1 1 w h = 19810h2h = 14905h2 2 N 2 B 2

W 5 1.2714(106) N

The weight of the concrete dam is

F 5 4905h2

1 W = rCgV = 12400 kg>m3 219.81 m>s3 2 c 16 m2118 m211 m2 d 2

h 3

6

= 1.2714110 2 N

The dam will overturn about point A. Write the moment equation of equilibrium about point A by referring to Fig. a. + ΣMA = 0;

Ax wB 2 (6 m) 5 4 m 3

Ay

(a)

h 31.27141106 2 N414 m2 - 4905h2 a b = 0 3 Ans.

h = 14.60 m = 14.6 m

Ans: h = 14.6 m 121

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

* 2–52. A swimming pool has a width of 12 ft and a side profile as shown. Determine the resultant force the water exerts on walls AB and DC, and on the bottom BC.

D 3 ft

C

8 ft B 20 ft

SOLUTION

A

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 wB = ghABb = ( 62.4 lb>ft3 ) (8 ft)(12 ft) = 5990.4 lb>ft

8 ft

FAB

wC = ghDCb = ( 62.4 lb>ft3 ) (3 ft)(12 ft) = 2246.4 lb>ft Using these results, the distributed loads acting on walls AB and CD and bottom BC are shown in Figs. a, b, and c. FAB =

1 1 wBhAB = (5990.4 lb>ft)(8 ft) = 23 962 lb = 24.0 kip 2 2

Ans.

B wB = 5990.4 lb/ft

FDC =

1 1 w h = (2246.4 lb>ft)(3 ft) = 3369.6 lb = 3.37 kip 2 C CD 2

Ans.

(a)

1 1 (w + wC)LBC = 3 5990.4 lb>ft + 2246.4 lb>ft 4 (20 ft) 2 B 2 = 82 368 lb = 82.4 kip

FBC =

D

Ans.

SOLUTION II The same result can also be obtained as follows. For wall AB,

3 ft

FCD

FAB = ghABAAB = ( 62.4 lb>ft3 ) (4 ft) 3 8 ft(12 ft) 4 = 23 962 lb = 24.0 kip

Ans.

For wall CD,

C wC = 2246.4 lb/ft

FCD = ghCDACD = ( 62.4 lb>ft3 ) (1.5 ft) 3 3 ft(12 ft) 4 = 3369.6 lb = 3.37 kip Ans.

For floor BC,

(b)

FBC = ghBCABC = ( 62.4 lb>ft3 ) (5.5 ft) 3 20 ft(12 ft) 4 = 82 368 lb = 82.4 kip Ans.

FBC

w B = 5990.4 lb/ft

w C = 2246.4 lb

C B

20 ft (c)

Ans: FAB = 24.0 kip FDC = 3.37 kip FBC = 82.4 kip 122

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–53. Determine the smallest thickness 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 = 2.40 Mg>m3. Hint: Work the problem using a 1-m width of the dam.

3m

6m

SOLUTION 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 3

2

3

wb = rwghb = 11000 kg>m 219.81 m>s 219 m211 m2 = 88.29110 2 N>m

A b

The resultant force of the triangular distributed load shown on the FBD of the dam, Fig. a, is 1 1 F = wbh = 388.291103 2 N>m419 m2 = 397.3051103 2 N 2 2 The dam can be subdivided into a triangular and a rectangular part, and the weight of each of these parts is 1 Wt = rc gVt = 12400 kg>m3 219.81 m>s2 2 c 1b213 m211 m2 d = 335.321103 2b4 N 2

Wr = rc gVr = 12400 kg>m3 219.81 m>s2 23b16 m211 m24 = 3141.261103 2b4 N

The dam will tip about point O. Writing the moment equation of equilibrium about point O, Fig. a, + ΣMO = 0;

b 2 3141.261103 2b4 a b + 335.321103 2b4 a bb 2 3

1 - 3397.311103 2 N4 c 19 m2 d = 0 3

94,176b2 - 1.1919151106 2 = 0

Ans.

b = 3.5576 m = 3.56 m

Wt 5 [35.32(103)b] N Wr 5 [141.26(103)b] N b 2 F 5 397.31(103) N 1 (9 m) 3 wb5

88.29(103)

Ox Nym 2b 3

Oy

(a)

Ans: b = 3.56 m 123

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–54. The uniform control gate AB is pinned at A and rests on the smooth surface at B. If the gate has a mass of 8.50 Mg, determine the maximum depth of water h in the reservoir that will cause the gate to be on the verge of opening. The gate has a width of 1 m.

3m A h

308 B

2m

SOLUTION The depths of points A and B are hA = h - 2 m - 3 m sin 30° = 1h - 3.5 m2 and hB = 1h - 2 m2. Thus, the intensities of the distributed load at points A and B are wA = rwghAb = 11000 kg>m3 219.81 m>s2 21h - 3.5 m211 m2 = 398101h - 3.524 N>m wB = rwghBb = 11000 kg>m3 219.81 m>s2 21h - 2 m211 m2 = 398101h - 224 N>m

Thus,

F1 = wAlAB = 398101h - 3.52413 m2 = 329.431103 21h - 3.524 N F2 =

1 1 1wB - wA 2lAB = 398101h - 22 - 98101h - 3.52413 m2 = 22.07251103 2 N 2 2

Since the gate is required to be on the verge to open, NB = 0. Write the moment equation of equilibrium about point A by referring to the FBD of the gate, Fig. a. + ΣMA = 0;

329.431103 21h - 3.52411.5 m2 + 322.07251103 2N412 m2 - 3850019.812 N41cos 30°211.5 m2 = 0

Ans.

h = 4.9537 m = 4.95 m

Ay

1.5 m W 5 (8500)(9.81) N

Ax wA

2 (3 m) 5 2 m 3

30°

30°

1.5 m

F1 F2

wB

NB 5 0

(a)

Ans: h = 4.95 m 124

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

2 ft

2–55. Determine the critical height h of the water level before the concrete gravity dam starts to tip over. 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

B

A

SOLUTION

4 ft

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 wB = gwhb = ( 62.4 lb>ft3 ) (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 =

1 1 wBh = (62.4h)h = 31.2h2 2 2

h 3

1 W2 = gCV2 = ( 150 lb>ft3 ) c (2 ft)(12 ft)(1 ft) d = 1800 lb 2

Ax

The dam will overturn about point A. Referring to the free-body diagram of the dam, Fig. a,

h = 10.83 ft = 10.8 ft

3 ft

F = 31.2h2

W1 = gCV1 = ( 150 lb>ft3 )3 2 ft(12 ft)(1 ft) 4 = 3600 lb

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

w 2 = 1800 lb 2 (2 ft) 3

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

+ ΣMA = 0;

w 1 = 3600 lb

w B = 62.4 h Ay (a)

Ans.

Ans: h = 10.8 ft 125

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

2 ft

*2–56. Determine the critical height h of the water level before the concrete gravity dam starts to tip over. Assume water also seeps under the base of the dam and produces a uniform pressure under 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 ) (h)(1) = 62.4h lb>ft 3

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

F2 = wBLB = 62.4h(4 ft) = 249.6h It is convenient to subdivide the dam into two parts. The weight of each part is w1 = gCV1 = ( 150 lb>ft3 ) 3 (2 ft)(12 ft)(1 ft) 4 = 3600 lb

1 w2 = gCV2 = ( 150 lb>ft3 ) 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, + ΣMA = 0;

h 2 31.2h2 a b + 249.6h(2 ft) - (3600 lb)(3 ft) - (1800 lb) c (2 ft)d = 0 3 3 10.4h3 + 499.2h - 13 200 = 0

Solve numerically, Ans.

h = 9.3598 ft = 9.36 ft

w 1 = 3600 lb 3 ft 2 (2 ft) 3 w 2 = 1800 lb

F1 = 31.2h2 h 3

Ax

w B = 62.4h Ay

2 ft

F2 = 249.6h (a)

Ans: h = 9.36 ft 126

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–57. 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 3 ft

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

B 3 ft

wA = gwhAb = ( 62.4 lb>ft3 ) (3 ft)(2 ft) = 374.4 lb>ft wB = gwhBb = ( 62.4 lb>ft3 ) (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,

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

+ ΣMA = 0;

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

Ans.

wA = 374.4 lb yft

F1 5 1123.2!2 lb

Ax

2 (3!2 ft) 3 1 (3!2 ft) 2

Ay

F2 5 561.6!2 lb

3!2 ft

NB

wB = 748.8 lb yft

y

x

Ans: NB = 1.32 kip FA = 1.06 kip

(a)

127

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–58. The uniform rectangular relief gate AB has a weight of 800 lb and a width of 2 ft. Determine the components of reaction at the pin B and the normal reaction at the smooth support A. 9 ft

SOLUTION

B

Here, hB = 9 ft and hA = 9 ft + 6 ft sin 60° = 14.20 ft. Thus, the intensities of the distributed load at B and A are wB = gwhBb = 162.4 lb>ft3 219 ft212 ft2 = 1123.2 lb>ft

Thus,

6 ft A

wA = gwhAb = 162.4 lb>ft3 2114.20 ft212 ft2 = 1771.68 lb>ft

608

1Fp 2 1 = 11123.2 lb>ft216 ft2 = 6739.2 lb

1Fp 2 2 =

1 11771.68 lb>ft - 1123.2 lb>ft216 ft2 = 1945.44 lb 2

Write the moment equation of equilibrium about B by referring to the FBD of the gate, Fig. a. + ΣMB = 0;

1800 lb2cos 60°13 ft2 + 16739.2 lb213 ft2 + 11945.44 lb214 ft2 -NA cos 60°16 ft2 = 0

Ans.

NA = 9733.12 lb = 9.73 kip

Using this result to write the force equations of equilibrium along x and y axes, d+ ΣFx = 0; Bx - 16739.2 lb2sin 60° - 11945.44 lb2sin 60° = 0 Bx = 7521.12 lb = 7.52 kip

Ans.

+ c ΣFy = 0; 9733.12 lb - 800 lb - 16739.2 lb2cos 60° - 11945.44 lb2cos 60° - By = 0 Ans.

By = 4590.80 lb = 4.59 kip

By

800 lb (FP)1 5 6739.2 lb

wB

Bx 1 (6 ft) 5 3 ft 2

60°

(FP)2 5 1945.44 lb 2 (6 ft) 5 4 ft 3 6 ft

wA

60° NA (a)

Ans: NA = 9.73 kip Bx = 7.52 kip By = 4.59 kip 128

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–59. 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 (wC)L = rwghBC (b) = ( 1000 kg>m

3

3.5 m C

)( 9.81 m>s ) (4 m)(2 m) 2

= 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, + 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 +

2 (3.5 m) d - FC(6 m) = 0 3

Ans.

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

6m

FL = 156.96(103) N

Dx

2 2.5 m + —(3.5 m) 3

FR = 120.17(103) N FC

wL = 78.48(103) N/m

wR = 68.67(103) N/m

Ans: FC = 25.3 kN h = 3.5 m

129

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–60. The tide gate 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

SOLUTION

h

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 1 1 (wC)LhBC = 3 19.62 ( 103 ) h 4 h = 9.81 ( 103 ) h2 2 2 1 1 FR = (wC)RhAC = 3 68.67 ( 103 ) N>m 4 (3.5 m) = 120.17 ( 103 ) N 2 2 FL =

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 +

+ ΣMD = 0;

2 2 hb - 120.17 ( 103 ) c2.5 m + (3.5 m)d 3 3 - FC (6 m) = 0

58.86 ( 103 ) h2 - 3.27 ( 103 ) h3 - 580.83 ( 103 ) - 6FC = 0 FC = ( 9.81h2 - 0.545h3 - 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 6 m – h + —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

1

2

(a)

h(m) FC(kN)

3 (b)

4

5

6

3.5

4

4.5

5.0

5.5

6.0

0

25.3

52.2

80.3

109.3

138.6

Ans: FC = ( 9.81 h2 - 0.545 h3 - 96.8 ) kN where h is in meters. 130

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

D

2–61. 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 wB = rghBb = ( 3.09 slug>ft

3

2 ft

A

2 ft

C

2√2 ft

)( 32.2 ft>s ) (2 ft)(6 ft) = 1193.976 lb>ft 2

wA = rghAb = ( 3.09 slug>ft3 )( 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. + MRB = ΣMB;

B

1 2 (6268.37 lb)d′ = (3581.93 lb)c (3 ft) d + (2686.45 lb) c (3 ft) d 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 131

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–62. Determine the resultant force that the water and oil together exert on the wall ABC. The wall has a width of 2 m. Also, determine the location of this resultant measured from the top of the tank. Take ro = 900 kg>m3.

A 1.5 m B

3m

SOLUTION C

Since the wall has a constant width, the intensities of the distributed loading at B and C, Fig. a, are wB = roghABb = 1900 kg>m3 219.81 m>s2 211.5 m212 m2 = 26.4871103 2 N>m = 26.487 kN>m wC = wB + rwghBCb = 26.4871103 2 N>m + 11000 kg>m3 219.81 m>s2 213 m212 m2 = 85.3471103 2 N>m = 85.347 kN>m

A

The resultant force can be determined by adding the areas of the triangles and rectangle, Fig. b. Here, F1 =

1.5 m

1 126.487 kN>m211.5 m2 = 19.87 kN 2

B

F2 = 126.487 kN>m213 m2 = 79.46 kN F3 = Thus, the resultant force is

wB 5 26.487 kNym

3m

1 158.86 kN>m213 m2 = 88.29 kN 2

C

FR = ΣF = F1 + F2 + F3

wC 5 85.347 kNym (a)

= 19.87 kN + 79.46 kN + 88.29 kN Ans.

= 187.62 kN = 188 kN

Each of the force components measured from the top acts through the centroid of its respective area. y1 =

2 1 2 11.5 m2 = 1.00 m y2 = 1.5 m + 13 m2 = 3.00 m y3 = 1.5 m + 13 m2 = 3.50 m 3 2 3

The location of the resultant force can be determined by equating the moment of the resultant force and the force components about A, by referring to Figs. b and c. $

+ 1MR 2 A = ΣMA;

1187.62 kN2yp = 119.87 kN211.00 m2 + 179.46 kN213.00 m2 + 188.29 kN213.50 m2 yp = 3.024 m = 3.02 m

Ans.

y1 5 1.00 m A

A F1 5 19.87 kN

y2 5 3.00 m

5

y3 5 3.50 m F2 5 79.46 kN F3 5 88.29 kN 26.487 kNym (b)

yP

FR 5 187.62 kN

85.347 kNym 2 26.487 kNym 5 58.86 kNym

(c)

Ans: FR = 188 kN yp = 3.02 m

132

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–63. Determine the critical height h of the water level that causes the concrete gravity dam to be on the verge of tipping over. Assume water also seeps under the base of the dam and produces a uniform pressure under the dam. The density of concrete is rc = 2400 kg>m3. Hint: Work the problem using a 1-m width of the dam.

18 m h

B

SOLUTION

A 6m

We will consider the dam having a width of b = 1 m. Then the intensity of the distributed load at the base of the dam is wB = rwghb = 11000 kg>m3 219.81 m>s2 21h211 m2 = 19810h2 N>m

W 5 1.2714 (106) N

The resultant forces of the triangular distributed load and uniform distributed load due to the pressure of seepage water shown on the FBD of the dam, Fig. a, are F1 =

1 1 w h = 19810h21h2 = 14905h2 2 N 2 B 2

2 (6 m) 5 4 m 3 Ax

h 3 wB

F2 = wBLB = 19810h216 m2 = 158 860h2 N

3 m Ay F2 5 58 860h

The weight of the concrete dam is

(a)

1 W = rc gV = 12400 kg>m 219.81 m>s 2 c 16 m2118 m211 m2 d 2 3

F1 5 4905h2

2

= 1.27141106 2 N

The dam will overturn about point A. Write the moment equation of equilibrium about point A by referring to Fig. a. + ΣMA = 0;

Solving numerically,

h 31.2714(106) N4(4 m) - 4905h2 a b - 58 860h(3 m) = 0 3 1635h3 + 17 6581103 2h - 5.08551106 2 = 0 Ans.

h = 12.1583 m = 12.2 m

Ans: h = 12.2 m 133

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–64. 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

SOLUTION

1.5 m

pC = pB = pA + rghAB

D

= 200 ( 103 ) N>m2 + ( 1000 kg>m3 )( 9.81 m>s2 ) (2 m) 1.25 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 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

1.25 m 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 Using these results, the distributed loads acting on plates BC and CD are shown in Figs. a and b, respectively.

w B = 384.335(103) Nym

N 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

FBC = wBLBC = c 384.335 ( 103 )

FCD

= 595.82 ( 103 ) N = 596 kN

FBC (a)

w C = 384.335(103) Nym

Ans.

(FCD)1

1.5 m

(FCD)2

wD – wC w D = 410.086(103) Nym (b)

Ans: FBC = 480 kN, FCD = 596 kN 134

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

2–65. 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 6 m. If the plate has a width of 1.5 m and a mass of 30  Mg, determine the required minimum height h of the water at B so that seepage will not occur at D.

2m C B 4m h

SOLUTION D

Referring to the geometry shown in Fig. a, x h = ; 5 4

x =

3m

5 h 4

The intensities of the distributed load shown in the FBD of the gate, Fig. b, are

C

w1 = rwgh1b = 11000 kg>m3 219.81 m>s2 212 m211.5 m2 = 29.431103 2 N>m w2 = rwgh2b = 11000 kg>m3 219.81 m>s2 216 m211.5 m2 = 88.291103 2 N>m

w3 = rwgh3b = 11000 kg>m3 219.81 m>s2 21h211.5 m2 = 314.7151103 2h4 N>m

and act at

d1 =

1 2 15 m2 = 2.5 m d2 = 15 m2 = 3.3333 m 2 3

[30(103)(9.81)] N Cx

B

4

2.5 m d 3

3

d1 d2

F3 F1

w3

F2

3

3147.15110 2 N412.5 m2 + 3147.15110 2 N413.3333 m2 3 - 3301103 219.812 N4 a b 12.5 m2 5 3

(a)

5

The seepage is on the verge of occurring when the gate is about to open. Thus, it is required that ND = 0. Write the moment equation of equilibrium about point C by referring to Fig. a. + ΣMC = 0;

D

3m

w1

1 5 d3 = 5 m - a hb = 15 - 0.4167h2 m 3 4

3

x

h

Then, the resultant forces of these distributed loads are

F1 = w1lCD = 329.431103 2 N>m415 m2 = 147.151103 2 N 1 1 F2 = 1w2 - w1 2lCD = 388.291103 2 N>m - 29431103 2 N>m415 m2 = 147.151103 2 N 2 2 1 1 5 F3 = w3lBD = 314.7151103 2h4 a hb = 39.1968751103 2h2 4 N 2 2 4

5m

B

4m

w2

2

- 39.196875110 2h 415 - 0.4167h2 = 0

ND 5 0 (b)

3.8320h3 - 45.9844h2 + 416.925 = 0

Solving numerically, Ans.

h = 3.5987 m = 3.60 m

Ans: h = 3.60 m 135

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–66. 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

d

C

3 ft

B

h

wA = gwhAb = ( 62.4 lb>ft3 ) (3 ft)(2 ft) = 374.4 lb>ft wB = gwhBd = ( 62.4 lb>ft3 ) (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 2 (3 ft) = 2 ft — 3

F1 = wALAB = 374.4 lb>ft(3 ft) = 1123.2 lb 1 1 ( wB - wA ) LAB = (748.8 lb>ft - 374.4 lb>ft)(3 ft) = 561.6 lb 2 2 Thus, the resultant force is F2 =

FR = F1 + F2 = 1123.2 lb + 561.6 lb = 1684.8 lb = 1.68 kip

w A = 374.4 lbyft FA = 0

1.5 ft

d

Ans.

d – 1.5 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. + ΣMC = 0;

2 ft – d

F2 = 561.6 lb

(561.61b)(2 ft - d) - (1123.2 lb)(d - 1.5 ft) = 0 Ans.

d = 1.67 ft

Cx

FB = 0

F1 = 1123.2 lb w B = 748.8 lbyft (a)

Ans: FR = 1.68 kip d = 1.67 ft 136

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–67. Determine the placement d of the pin on the 3-ft-diameter 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>ft3 ) (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, d = yp - 3 ft = 4.625 ft - 3 ft = 1.625 ft = 1.62 ft

Ans.

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 137

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. * 2–68. 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 oil surface 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 A = (1 m)(1.5 m) +

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

Ix =

1m

1 (1.5 m)(1.5 m) = 2.625 m2 2

= 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

= 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 138

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–69. The tapered settling tank is filled with oil. Determine the resultant force the oil exerts on the trapezoidal cleanout plate located at its end. How far from the oil surface 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

pdA =

L2.5 m

= 22 072.5y2 - 2943y3 `

= 72.84 ( 10

3

x

8829y(5 - y)dy

1.25 m

A

h=y

x

1.5 m dy

4m 2.5 m

) N = 72.8 kN

Ans.

B 0.5 m 0.5 m y (a)

And it acts at

yP = =

LA

ypdA FR

=

1 72.84 ( 103 )

1 72.84 ( 103 ) N

1 14 715y3

= 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 139

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–70. Ethyl alcohol is pumped into the tank, which has the shape of a four-sided pyramid. When the tank is completely filled, determine the resultant force acting on each side, and its location measured from the top A along the side. Use the formula method. Take 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. 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

2m

2m

B

4m

Ans.

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 140

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A 1m

2–71. The bent plate is 2 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.

3m

3m

B

SOLUTION The horizontal loading on the plate is due to the pressure on the vertical projected area of the plate. Since the plate has a constant width of b = 2 m, the intensities of the horizontal distributed load at A and B are

4m

1 (4 m) (F ) v 2 3

2 (4 m) 3 A

(Fv)1

wA = rwghAb = 11000 kg>m3 219.81 m>s2 211 m212 m2 = 19.621103 2 N>m

wB = rwghBb = 11000 kg>m3 219.81 m>s2 217 m212 m2 = 137.341103 2 N>m

wA O

y

1 (4 m) 2 Ax

The resultant of the distributed load shown in the FBD of the plate, Fig. a, is

1Fh 2 1 = wAlAB = 319.621103 2 N>m416 m2 = 117.721103 2 N = 117.72 kN

1 1 1w - wA 2lAB = 3137.341103 2 N>m - 19.621103 2 N>m416 m2 2 B 2 = 353.161103 2 N = 353.16 kN

1Fh 2 2 =

1 2 (6 m) 2 (6 m) 3 (Fh)1 (Fh)2

F3 and F4 are not in the FBD are equal to the weight of water contained in the respective shaded rectangular and triangular blocks, Fig. a. 1Fv 2 1 = rwgVr = 11000 kg>m3 219.81 m>s2 234 m11 m212 m24 = 78.481103 2 N = 78.48 kN

1 1Fv 2 2 = rwgVt = 11000 kg>m3 219.81 m>s2 2c 14 m213 m)(2 m2 d = 117.721103 2 N = 117.72 kN 2

wB

4m NB

Write the moment equation of equilibrium about points A and O by referring to Fig. a. + ΣMA = 0;

1 2 1 1117.72 kN2 c 16 m2 d + 1353.16 kN2 c 16 m2 d + 178.48 kN2 c 14 m2 d 2 3 2 2 + 1117.72 kN2 c 14 m2 d - NB 14 m2 = 0 3

Ans.

NB = 559.17 kN = 559 kN

+ ΣMo = 0;

1 2 1 1117.72 kN2 c 16 m2 d + 1353.16 kN2 c 16 m2 d - 178.48 kN2 c 14 m2 d 2 3 2 1 - 1117.72 kN2 c 14 m2 d - Ay 14 m2 = 0 3

Ans.

Ay = 362.97 kN = 363 kN

Write the force equation of equilibrium along the x axis. S+ ΣFx = 0; 117.72 kN + 353.16 kN - Ax = 0 Ax = 470.88 kN = 471 kN

Ans.

Ans: NB = 559 kN, Ay = 363 kN, Ax = 471 kN 141

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–72. 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

608

SOLUTION

B A

The resultant force is 1 FR = gee hA = ( 44.5 lb>ft3 ) (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,

5 ft

5 ft

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)

Ans: FR = 18.5 kip d = 3 ft 142

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–73.

Solve Prob. 2–72 using the integration method.

C 12 ft

SOLUTION

608

With respect to x and y axes established, the equation of side AB of the plate, Fig. a, is y - 0 12 - 0 5 = ; x = y x - 0 5 - 0 12

B A

5 ft

5 ft

Thus, the area of the differential element shown shaded in Fig. a is dA = 2xdy = 5 5 2 a ybdy = y dy. The pressure acting on this differential element is p = gh = 12 6 ( 44.5 lb>ft3 )( y sin 60° ) = 38.54y. Thus, the resultant force acting on the entire plate is FR =

L

p dA =

A

= 10.71 y3 `

L0

12 ft

12 ft

5 (38.54y) a y dyb 6

0

= 18.50 ( 10 ) lb = 18.5 kip 3

Ans.

And it acts at

yP =

LA

y p dA =

FR

=

1 18.50 ( 10

3

) L0

1 18.50 ( 10

3

)

= 9.00 ft

12 ft

5 y (38.54y) a y dyb 6

( 8.03y4 ) `

12 ft 0

Thus, Ans.

d = 12 ft - yp = 3.00 ft 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 )

143

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

4m

2–74. If the tank is filled with vegetable oil, determine the resultant force that the oil exerts on plate A, and its location measured from the bottom of the tank. Use the formula method. Take rvo = 932 kg>m3.

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. +(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

w t = 27.429(103) Nym

F1 = 54.858(103) N FR = 73.143(103) N

1m (2 m)

d

w

b=

45.715(103) Nym

O

O

F2 = 18.286(103) N (a)

(b)

Ans: FR = 73.1 kN d = 917 mm 144

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

4m

2–75. If the tank is filled with vegetable oil, determine the resultant force that the oil exerts on plate B, and its location measured from the bottom of the tank. Use the formula method. Take rvo = 932 kg>m3.

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 = gvohA = ( 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 from

h = 2.5 m

yp

(0.75 m)4

yP =

B

p Ix 4 + y = + 2.5 m 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 145

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–76.

4m

Solve Prob. 2–75 using the integration method. 4m

1.5 m 2.5 m 1m

5m

SOLUTION

2m

B

A

With respect to x and y axes established, the equation of the circumference of the circular plate is y

x 2 + y2 = 0.752;

x = 20.752 - y2

Thus, the area of the differential element shown shaded in Fig. a is dA = 2x dy = 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

h = 2.5 – y

0.75 m

L-0.75 m

pdA =

0.75 m

= 18 285.84

9142.92(2.5 - y) c 220.752 - y2 dy d

x

x dy

(2.5 - y) a20.752 - y2 b dy L-0.75 m

y

0.75 m y = 22 857.3c y20.752 - y2 + 0.752 sin-1 d` 0.75 -0.75 m

+ 6095.282 ( 0.752 - y2 ) 3 `

x

0.75 m

-0.75 m

= 40.39 ( 103 ) N = 40.4 kN

Ans.

And it acts at

yP = =

LA

(a)

(2.5 - y)p dA FR 0.75 m

40.39 ( 103 ) L- 0.75 m 1

0.75 m

= 0.4527

(2.5 - y) 3 9142.92(2.5 - y) 4 a220.752 - y2 dyb

(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 m 0.75 m y 0.752 ay20.752 - y2 + a 2 sin-1 + 0.75450 ` bd ` 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

Ans. Ans: FR = 40.4 kN, d = 2.44 m

146

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–77. Determine the resultant force acting on the triangular plate A and the location of the center of pressure, measured from the free water level in the tank. Solve the problem using the formula method.

5m

1.8 m A B

SOLUTION Referring to the dimensions shown in Fig. a, 2 d = 11.8 m2 = 1.20 m y = h = 5 m - 1.20 m = 3.80 m 3 Also,

1m

1m

1 1 bh = (2 m2(1.8 m2 = 1.80 m2 2 2 1 3 1 Ix = bh = (2 m2(1.8 m2 3 = 0.324 m4 36 36

A =

FR = gwhA = (1000 kg>m3 2(9.81 m>s2 2(3.80 m2(1.80 m2 2 = 67.10(103 2 N = 67.1 kN

And it acts at yp =

Ix

yA

+ y =

y5h

yp

The resultant force can now be determined:

0.324 m4 + 3.80 m (3.80 m2(1.80 m2 2

= 3.8474 m = 3.85 m

5m

2m

Ans.

FR

1.8 m

d

(a)

Ans.

Ans: FR = 67.1 kN, yP = 3.85 m 147

M02_HIBB9290_01_SE_C02_ANS.indd 147

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–78.

Solve Prob. 2–77 using the integration method.

5m

1.8 m

SOLUTION

A B

With respect to x and y axes established, the equation of side CD of the plate, Fig. a, is y - 5 3.20 - 5 5 = ; x = 15 - y2 x - 0 1 - 0 9

1m

The area of the differential element shown shaded in Fig. a is dA = 2x dy = 5 10 2c 15 - y2 d dy = 15 - y2dy. The pressure acting on this differential element is 9 9 p = gh = 11000 kg>m3)(9.81 m>s2 2y = 9810y. Thus, the resultant force acting on the entire plate is FR =

LA

5m

p dA =

L3.20 m

19810y2 c 5m

= 10 900

L3.20

10 15 - y2dy d 9

yp =

3.20 m

y

1m 1m x

1.80 m

C

dy D y

(a)

= 67.101103 2 N = 67.1 kN A L

h5y

FR

15y - y 2dy

y3 5 5 = 10 900a y2 b` 2 3 320 m And it acts at

x

x

2

1m

Ans.

yp dA FR

where LA

5m

yp dA =

L3.20 m

y19810y2 c 5m

= 10 900

L3.20 m

10 15 - y2dy d 9

15y2 - y3 2dy

y4 5 m 5 b` = 10 900a y3 3 4 3.20 m = 258.161103 2 N # m

Then yp =

258.161103 2 N # m 67.101103 2 N

= 3.8474 m = 3.85 m

Ans.

Ans: FR = 67.1 kN yP = 3.85 m 148

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2 ft

2–79. 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

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. Σ y∼A y = = ΣA

1 (1.5 ft) 3 3 ft(2 ft) 4 + (1 ft) c (4 ft)(3 ft) d 2 = 1.25 ft 1 3 ft(2 ft) + (4 ft)(3 ft) 2

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 ft4 12 36 2

The area of plate ABCD is

A = 3 ft(2 ft) +

1 (4 ft)(3 ft) = 12 ft2 2

Thus, FR = ghA = ( 50.1 lb>ft3 ) (1.25 ft) ( 12 ft2 ) = 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 149

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y

* 2–80. 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 1 x2 5 1 (0.75 m)2

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 150

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–81.

y

Solve Prob. 2–80 using the integration method.

y2 1 x2 5 1 (0.75 m)2

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

(0.75 - y) a1 b dy (0.75)2 L-0.75 m 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 19 620 1 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 19 620

L-0.75 m

Ans.

- 0 = 17 336 N = 17.3 kN

0.75 m

yP =

1 2

y2

y(0.75 - y) a1 17 336 N

y2 (0.75)2

1 2

b dy

= -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 151

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y y2 1 x2 5 1 (0.75 m)2

2–82. 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 measured from the x axis is y = 4b>3p.

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 Ix = 0.046304 m4

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 152

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–83.

y y2 1 x2 5 1 (0.75 m)2

Solve Prob. 2–82 using the integration method.

0.75 m x 1m

SOLUTION Using a horizontal strip of area dA, dF = p dA 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 0.752

Ans. 1 2

b dy

= 0.4418 m = 442 mm

Ans.

Ans: FR = 3.68 kN yP = 442 mm 153

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y

* 2–84. The trough is filled to its top with carbon disulfide. 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. Solve the problem using the formula method. Take rcd = 2.46 slug>ft3.

1 ft

y 5 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 2 2 bh = (2 ft)(4 ft) = 5.3333 ft2 3 3 8 8 3 Ix = bh = (2 ft)(4 ft)3 = 5.8514 ft4 175 175

A =

With h =

2 2 h = (4 ft) = 1.6 ft, 5 5 FR = ghA = ( 2.46 slug>ft3 )( 32.2 ft>s2 ) (1.6 ft) ( 5.3333 ft2 ) Ans.

= 675.94 lb = 676 lb Since y = h = 1.6 ft, yP = =

Ix + y yA 5.8514 ft4

(1.6 ft) ( 5.3333 ft2 ) = 2.2857 ft = 2.29 ft

+ 1.6 ft Ans.

Ans: FR = 676 lb yP = 2.29 ft 154

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–85.

y

Solve Prob. 2–84 using the integration method. 1 ft

y 5 4x2

4 ft x

SOLUTION Using a horizontal strip of area, FR =

LA

L0

p dA =

= 158.424

L0

= 79.212 a

4 ft

L0

4 ft

( 2.46 slug>ft3 )( 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(p dA) = 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

8 5 2 7 4 ft = 79.212 a y2 - y2 b ` 3 5 0

= 1158.76 lb # ft

1158.76 lb # ft = 1.7143 ft 675.94 lb yP = 4 ft - d d =

= 4 ft - 1.7143 ft Ans.

= 2.2857 ft = 2.29 ft

Ans: FR = 676 lb yP = 2.29 ft 155

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–86. The tank is filled to its top with lubricating oil. Determine the resultant force acting on the semicircular plate ABC, and its location on the plate measured from the base AC of the tank. Use the formula method. Take go = 54.9 lb>ft3.

18 ft

B

SOLUTION Referring to the dimensions shown in Fig. a,

6 ft

608

C

416 ft2 4r 8 = = ft p 3p 3p 8 y = 18 ft - ft = 15.4535 ft p r =

A

h = 115.4535 ft2 sin 60° = 13.3831 ft

Also,

1 1 1pr 2 2 = 316 ft2 2 4 = 18p ft2 2 2 1 p 4 4r 9p2 - 64 9p2 - 64 b = a b 16 ft2 4 = 142.25 ft4 Ix = r - 1pr 2 2 a b = a 8 2 3p 72p 72p

12 ft y 18 ft p

A =

The resultant force can now be determined:

FR = gohA = 154.9 lb>ft3 2113.3831 ft2118p ft2 2 3

= 41.548110 2 lb = 41.5 kip

And it acts at yp =

Ix yA

+ y =

Thus,

h y

FR

60°

d r

Ans.

y (a)

142.25 ft4 + 15.4535 ft = 15.6163 ft 115.4535 ft2118p ft2 2

d = 18 ft - 15.6163 ft = 2.3837 ft = 2.38 ft

Ans.

Ans: FR = 41.5 kip d = 2.38 ft 156

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–87.

Solve Prob. 2–86 using the integration method.

18 ft

B

SOLUTION

6 ft

608

C

With respect to the x and y axes established, the equation of the semicircular plate, Fig. a, is

A

1x - 02 2 + 1y - 182 2 = 62 2

x

2

x + 1y - 182 = 36

x = 236 - 1y - 182 2

The area of the differential element shown shaded in Fig. a is dA = 2x dy = 2236 - 1y - 182 2 dy. The pressure acting on this differential element is p = gh = 154.9 lb>ft3 21y sin 60°2 = 47.5448y. Thus, the resultant force acting on the entire plate is FR =

LA

y dy h 5 y sin60°

12 ft

x

18 ft

L12 ft

p dA =

18 ft

= 95.0896

L12 ft

x

147.5448y232236 - 1y - 182 2 dy4

6 ft

A

C y

2

y236 - 1y - 182 dy

(a)

18 ft

18 ft

y - 18 1 b` = 95.0896e- 2336 - 1y - 182 2 4 3 ` + 9c 1y - 182 236 - 1y - 182 2 + 36 sin-1 a f 3 6 12 ft 12 ft = 41.5481103 2 lb = 41.5 kip

Ans.

And it acts at

yp =

LA

yp dA FR

where LA

18 ft

yp dA =

L12 ft

y147.5448y232236 - 1y - 182 2 dy4 18 ft

= 95.0896

L12 ft

= 95.0896e - a +

y2 236 - 1y - 182 2 dy

18 ft 18 ft y - 18 b 2336 - 1y - 182 2 4 3 ` - 122336 - 1y - 182 2 4 3 ` 4 12 ft 12 ft

18 ft y - 18 333 c 1y - 182 236 - 1y - 182 2 + 36 sin-1 a bd ` r 2 6 12 ft

= 648 829.85 lb # ft Then

yp = Thus,

648 829.85 lb # ft = 15.6163 ft 41.5481103 2 lb Ans.

d = 18 - yp = 2.3837 ft = 2.38 ft

Ans: FR = 41.5 kip d = 2.38 ft

157

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–88. 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.

3.5 m

SOLUTION

1.5 m

Referring to the geometry shown in Fig. a, A =

1 12 m211.5 m2 + 11 m211.5 m2 = 3.00 m2 2

1m1m

1 14.25 m2311 m211.5 m24 + 14.00 m2 c 12 m211.5 m2 d 2 h = y = = 4.125 m 1 11 m211.5 m2 + 12 m211.5 m2 2

Ix =

1 ~ y2 5 3.5 m1 (1.5 m) 3 5 4.00 m

1 11 m211.5 m2 3 + 11 m211.5 m214.25 m - 4.125 m2 2 12 +

yp

1 1 12 m211.5 m2 3 + c 12 m211.5 m2 d 14.125 m - 4.00 m2 2 36 2

FR

= 0.515625 m

The resultant force can now be determined:

1 ~ y1 5 3.5 1 (1.5 m) 2 5 4.25 m

C2

1.5 m C1

2

= 121.401103 2 N = 121 kN yp =

1m 1m 1m

(a) 2

FR = gwhA = 11000 kg>m 219.81 m>s 214.125 m213.00 m 2 And it acts at

3.5 m

C2

4

2

1m

Ix 0.515625 m4 + y = + 4.125 m yA 14125 m213.00 m2 2

= 4.1667 m = 4.17 m

Ans.

Ans.

Ans: FR = 121 kN yp = 4.17 m 158

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–89.

Solve Prob. 2–88 using the integration method.

3.5 m

SOLUTION With respect to the x and y axes established, the equation of side EF of the plate, Fig. a, is y - 3.5 5 - 3.5 = ; x = 3.8333 - 0.6667y x - 1.5 0.5 - 1.5

1.5 m

1m1m

The area of the differential element shown shaded in Fig. a is dA = 2x dy = . 213.8333 - 0.6667y2dy = 17.6667 - 1.3333y2dy. The pressure acting on this differential element is p = gh = 11000 kg>m3)(9.81 m>s2 2y = 9810y. Thus, the resultant force acting on the entire plate is FR =

LA

x

5m

p dA =

L3.5 m

5m

= 9810

L3.5 m

19810y217.6667 - 1.3333y2dy

h5y

17.6667y - 1.3333y2 2dy

= 981013.8333y2 - 0.4444y3 2 ` LA

yp =

3.5 m 1.5 m 1.5 m x

P

5m

x

y E

1.5 m (x, y) F dy

3.5 m

= 121.401103 2 N = 121 kN

And it acts at

1m

Ans.

0.5 m 0.5 m y (a)

yp dA FR

where LA

5m

yp dA =

y19810y217.6667 - 1.3333y2dy L3.5 m 5m

= 9810

L3.5 m

17.6667y2 - 1.3333y3 2dy

= 981012.5556y3 - 0.3333y4 2 `

= 505.831103 2 N # m

Thus, yp =

505.831103 2 N # m 121.401103 2 N

5m

3.5 m

= 4.1667 m = 4.17 m

Ans.

Ans: FR = 121 kN yP = 4.17 m 159

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–90. The control gate ACB is pinned at A and rests 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>ft3 ) (6 ft)(3 ft) = 1123.2 lb>ft wB = gw hBb = ( 62.4 lb>ft3 ) (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. + 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

1 (1.5) = 0.75 ft 2

Ax 2 (1.5) = 1 ft 3

F2 Ay

F1

F3

(a)

wB

Ans: WC = 2.18 kip 160

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–91. The control gate ACB is pinned at A and rests 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 shown in Fig. a. wC = gw hCb = ( 62.4 lb>ft3 ) (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 1 F3 = 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.

3 561.6(h

+ ΣMA = 0;

- 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

F3

F1

1 —(1.5) = 0.75 ft 2 wB (a)

Ans: h = 9.50 ft 161

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. A

*2–92. 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.

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>ft3 ) (4 ft)(8 ft) = 1996.8 lb>ft

10 ft

w2 = gwh2b = ( 62.4 lb>ft3 ) (12 ft)(8 ft) = 5990.4 lb>ft w3 = gwh3b = ( 62.4 lb>ft3 ) (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) = 19 968 lb 1 (5990.4 lb>ft - 1996.8 lb>ft)(10 ft) = 19 968 lb 2 1 5 F3 = (499.2h lb>ft) a hb = ( 312h2 ) lb 2 4 F2 =

and act at

6 ft (a)

10 ft = 5 ft 2 2 d2 = (10 ft) = 6.667 ft 3 1 5 d3 = 10 ft - a hb = (10 - 0.4167h) ft 3 4

d1 =

50 000 lb Cx

For seepage to occur, the reaction at D must be equal to zero. Referring to the FBD of the gate, Fig. b, + ΣMC = 0;

d3

w1

5 ft 4

3 (50 000 lb) a b(5 ft) + ( 312h2 lb ) (10 - 0.4167h) ft 5

Cy

5

3

- (19 968 lb)(5 ft) - (19 968 lb)(6.667 ft) = 0

F3

- 130 h3 + 3120 h2 - 82 960 = 0 Solving numerically,

F1

d2

w3

Ans.

h = 5.945 ft = 5.95 ft 6 8 ft

d1

F2 w2 (b)

Ans: h = 5.95 ft 6 8 ft 162

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 1m

2–93. 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. The fluid is water.

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

3m

2m

B

wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (1 m)(1.5 m) = 14.715 ( 103 ) N>m wA = rwghAb = ( 1000 kg>m3 )( 9.81 m>s2 ) (1 m)(1.5 m) = 14.715 ( 103 ) N>m Using these results, the distributed load acting on the plate is shown on the freebody diagram of the gate, Fig. a.

2.5 m

F1 wA = 14.715(103) Nym

Ax

F1 = wALAB = ( 14.715 ( 103 ) N>m ) (5 m) = 73.575 ( 103 ) N 1 1 F2 = (wB - wA)LBC = (73.575 ( 103 ) N>m - 14.715 ( 103 ) N>m)(4 m) 2 2

F4 Ay

= 117.72 ( 103 ) N

C 2 2 m 3 (4 m) F3

2 m + 2 (3 m) 3

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. 1 F4 = rwg V = ( 1000 kg>m )( 9.81 m>s ) c (3 m)(4 m)(1.5 m) d = 88.29 ( 103 ) N 2 3

wB = 73.575(103) Nym

5m

2

NB (a)

Considering the free-body diagram of the gate, Fig. 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 - 88.29 ( 103 ) N a2 m + (3 m) b = 0 3

NB = 193.748 ( 103 ) N = 194 kN + S ΣFx = 0;

Ans.

Ax - 58.86 ( 103 ) N - 117.72 ( 103 ) N = 0 Ax = 176.58 ( 103 ) N = 177 kN

+ c ΣFy = 0;

Ans.

-Ay - 73.575 ( 103 ) N - 88.29 ( 103 ) N + 193.748 ( 103 ) N = 0 Ay = 31.88 ( 103 ) N = 31.9 kN

Ans.

Ans: NB = 194 kN Ax = 177 kN Ay = 31.9 kN 163

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

2–94. The gate is 1.5 m wide, is pinned at A, and rests on the smooth support at B. Determine the reactions at these supports due to the water pressure.

3m

B

SOLUTION The horizontal loading on the plate is due to the pressure on the vertical projected area of the plate, Fig. a. Since the plate has a constant width of b = 1.5 m, the intensity of the horizontal distributed load at B is given by wB = rwghBb = 11000 kg>m3 219.81 m>s2 13 m211.5 m2 = 44.1451103 2 N>m

Ay

4 Q3 2 pR m

Ax

Fv 4(3 m) 5 4 m 3p p O 2 (3 m) 3

And its resultant force is Fh =

1 1 w l = 344.1451103 2 N>m413 m2 = 66.21751103 2 N = 66.2175 kN 2 B DB 2

The vertical force acting on the plate is equal to the weight of the water contained in the block shown shaded in Fig. a. p Fv = rwgVAOB = 11000 kg>m3 219.81 m>s2 2 c 13 m2 2 11.5 m2 d 4

Fh wB 3m NB (a)

= 104.011103 2 N = 104.01 kN

Write the moment equations of equilibrium about points A and O by referring to the FBD of the plate, Fig. a. 2 4 + ΣMA = 0; NB 13 m2 - 166.2175 kN2 c 13 m2 d - 1104.01 kN2 a3 - b m = 0 p 3 NB = 104.01 kN = 104 kN Ans. + ΣMO = 0;

4 2 mb - 166.2175 kN2 c 13 m2 d = 0 p 3 Ay = 0 Ans.

Ay 13 m2 + 1104.01 kN2 a

Write the force equation of equilibrium along the x axis. S+ ΣFx = 0;

Ax - 66.2175 kN = 0 Ax = 66.2175 kN Ans.

= 66.2 kN

Ans: NB = 104 kN Ay = 0, Ax = 66.2 kN 164

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–95. 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

-1

2m a b = 30° 4m

A

B 4m

2m

2m

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: FR = 442 kN 165

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y

*2–96. The wall is in the form of a parabola. Determine the magnitude and direction of the resultant force on the wall if it is 8 ft wide.

12 ft

SOLUTION

y5

12 ft

1 2 x 12

The horizontal loading on the wall is due to the pressure on the vertical projected area of the wall, Fig. a. Since the wall has a constant width of b = 8 ft, the intensity of the horizontal distributed load at the base of the wall is

Thus,

x

w = gwhb = 162.4 lb>ft3 2112 ft218 ft2 = 5.99041103 2 lb>ft Fv

1 1 Fh = wh = 35.99041103 2 lb>ft4112 ft2 = 35.94241103 2lb 2 2

The vertical force acting on the wall is equal to the weight of the water contained in the block above the wall (shown shaded in Fig. a). From the inside back cover of the text, the volume of this block (parabolic cross-section) is 2 2 V = ahb = 112 ft2112 ft218 ft2 = 768 ft3 3 3

Thus,

Fh w

FR

12 ft u 12 ft (a)

Fv = ywV = 162.4 lb>ft3 21768 ft3 2 = 47.92321103 2 lb

Then the magnitude of the resultant force is

FR = 2F 2h + F 2v = 2335.94241103 2 lb4 2 + 347.92321103 2 lb4 2 And its direction is u = tan-1 a

= 59.9041103 2 lb = 59.9 kip

47.92321103 2 lb Fv b = tan-1 c d = 53.13° = 53.1° Fh 35.94241103 2 lb

Ans.

Ans.

Ans: FR = 59.9 kip u = 53.1° 166

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–97. Determine the horizontal and vertical components of reaction at the hinge A and the horizontal normal reaction at B caused by the water pressure. The gate has a width of 3 m.

3m

SOLUTION

B 3m

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, 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

A

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

∼y 1 = 1 (3 m) = 1.5 m 2

∼y 2 = 1 (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

( Fv ) 2 = rwgV2 = ( 1000 kg>m3 )( 9.81 m>s2 ) c p ( 3 m ) 2(3 m) d = 66.2175p ( 103 ) N = 66.2175p kN 4

They act at

∼x 1 = 1 (3 m) = 1.5 m 2

~ x1

∼x 2 =

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)

167

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2–97. Continued

Considering the equilibrium of the FBD of the gate in Fig. 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 S+ ΣFx = 0;

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

Ans: NB = 397 kN, Ax = 0, Ay = 473 kN 168

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. y

2–98. The 5-m-wide overhang is in the form of a parabola. Determine the magnitude and direction of the resultant force on the overhang.

3m

1 y 5 — x2 3

3m

SOLUTION 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 + F v2 = 2(220.725 kN)2 + (294.3 kN)2 = 367.875 kN = 368 kN Ans.

Its direction is

u = tan-1 a y

Fv 294.3 kN b = tan-1 a b = 53.13° = 53.1° Fh 220.725 kN 1 y— = —x2 3

— x

Ans.

B

C

3m — y

u

Fh

x

A Fv

3m

wA FR (a)

Ans: FR = 368 kN u = 53.1° 169

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–99. Determine the resultant force that water exerts on the overhanging sea wall along ABC. The wall is 2 m wide.

1.5 m

B A 2m C 2.5 m

SOLUTION 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: FR = 179 kN 170

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y y 5 4x½

*2–100. Determine the magnitude and direction of the resultant hydrostatic force the water exerts on the parabolic face AB of the wall if it is 3 m wide.

B

8m

A

SOLUTION

x 4m

The horizontal loading on the wall is due to the pressure on the vertical projected area of the wall, Fig. a. Since the wall has a constant width of b = 3 m, the intensity of the horizontal distributed load at the base of the wall is 3

2

Fv

3

w = rwghb = 11000 kg>m 219.81 m>s 218 m213 m2 = 235.44110 2 N>m

B

Thus,

Fh =

1 1 wh = 3235.441103 2 N>m418 m2 = 941.761103 2 N = 941.76 kN 2 2

The vertical force acting on the wall is equal to the weight of water contained in the imaginary block above the wall (shown shaded in Fig. a), but acts vertically upward. From the inside back cover of the text, the volume of this block (parabolic crosssection) is 1 1 V = ahb = 14 m218 m213 m2 = 32.0 m3 3 3

Thus,

3

2

8m

u A

Fh

FR w

4m (a)

3

Fv = rwgV = 11000 kg>m 219.81 m>s 2132.0 m 2 = 313.921103 2 N = 313.92 kN

Then the magnitude of the resultant force is

FR = 2F 2n + F 2v = 21941.76 kN2 2 + 1313.92 kN2 2 = 992.70 kN = 993 kN Ans.

And its direction is

u = tan-1 a

Fv 313.92 kN b = tan-1 a b = 18.43° = 18.4° F 941.76 kN

Ans.

Ans: FR = 993 kN u = 18.4° 171

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y

2–101. 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 5 4x

h

SOLUTION

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 ) (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 = 2Fh 2 + Fv 2 = 2(392.4 kN)2 + (261.6 kN)2 = 471.61 kN = 472 kN

And its direction is

Fv 261.6 kN u = tan a b = tan-1 a b = 33.69° Fh 392.4 kN -1

FR

—

Fv

x

Ans. C

B y2 = 4x

Ans. 4m Fh u

—

y

wA

x

A 4m (a)

Ans: FR = 472 kN, u = 33.7° 172

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

y

2–102. 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 5 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 )( 9.81m>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

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,

Fv

y

h Fh

2

1 1 h 5 3 ahb = a b(h)(5 m) = h 3 3 4 12 Fv = rwg V = ( 1000 kg>m3 )( 9.81 m>s2 ) a

Then the magnitude of the resultant force is FR = 2F h2 + F v2

FR = 2 3 24.525 ( 103 ) h2 4 2 +

3 4087.5h3 4 2

x

wA A h2 4

5 3 h b = 4087.5h3 12 h(m) FR(kN)

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. 173

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–103. Determine the resultant force the water exerts on the quarter-circular wall AB if it is 3 m wide.

A

3m

B

SOLUTION The horizontal loading on the wall is due to the pressure on the vertical projected area of the wall, Fig. a. Since the wall has a constant width of b = 3 m, the intensity of the horizontal distributed load at the base of the wall is

Thus, Fh =

w = rwghb = 11000 kg>m3 219.81 m>s2 213 m213 m2 = 88.291103 2 N>m 1 1 wh = 388.291103 2N>m413 m2 = 132.4351103 2 N = 132.435 kN 2 2

Thus,

u

3m

FR Fw

Ans.

The vertical force acting on the wall is equal to the weight of the water contained in the shaded block above the wall, Fig. a. The cross-sectional area of the block is A = 13 m213 m2 -

Fv

w (a)

1 3p13 m2 2 4 = 19 - 2.25p2 m2 4

Fv = rwgV = 11000 kg>m3)(9.81 m>s2 2319 - 2.25p2 m2 413 m2 = 56.841103 2 N = 56.84 kN

Then, the magnitude of the resultant force is

FR = 2F 2h + F 2v = 21132.435 kN2 2 + 156.84 kN2 2 = 144.12 kN = 144 kN Ans.

And its direction is

u = tan-1 a

Fv 56.84 kN b = tan-1 a b = 23.23° = 23.2° Fh 132.435 kN

Ans.

Ans: FR = 144 kN, u = 23.2° 174

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

C

B

*2–104. 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.

6 ft A

SOLUTION Referring to the geometry shown in Fig. a, p (6 ft)2 = (36 - 9p) ft2 4

AADB = (6 ft)(6 ft) -

x =

(3 ft) 3 (6 ft)(6 ft) 4 - c a6 -

8 p b ft d c (6 ft)2 d p 4

(36 - 9p) ft2

= 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>ft3 ) (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) ft2 4 (4 ft) = 1928.33 lb

Write the moment equation of equilibrium about A by referring to Fig. b. + 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

1.3402 ft

6 ft

Fh =

A

6 ft C1

C2 6 ft

1 (6) = 2 ft — 3

wA

Ax Ay

6 ft (b)

(a)

Ans: TBC = 1.93 kip Ax = 2.56 kip Ay = 1.93 kip

175

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Unless otherwise Unless otherwisestated, stated,take takethe thedensity density of of water water to be to 3 3 be and and its specific its specific weightweight to be to be rw =rw1000 = 1000 kg>m kg>m gw = Also,3. assume Also, assume all pressures all pressures are gage are gage pressures. pressures gw =lb>ft 62.4 62.43.lb>ft unless stated otherwise. 2–105. The parabolic 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.

y 3m 3 y 5 16 (16 – x2 )

B

3m

SOLUTION

A

C

The horizontal loadings on the plates are due to the pressure on the vertical projected areas of the plates, Fig. a and b. Since the plates have a constant width of b = 4 m, the intensities of the horizontal distributed load at points B, C and A are

4m By

wB = rwghBb = 11000 kg>m3 219.81 m>s2 213 m214 m2 = 117.721103 2 N>m

[(Fw)BC]2

wA = wC = rwghCb = 11000 kg>m3 219.81 m>s2 216 m214 m2 = 235.441103 2 N>m

31Fh 2 AB 4 1 = 31Fh 2 BC 4 1 = wBhBC = 3117.721103 2 N>m413 m2 = 353.161103 2 N = 353.16 kN

3m

~ y1 ~ y2

1 1 1w - wB 2hBC = 3235.441103 2 N>m - 117.721103 2 N>m413 m2 2 C 2

wC

Cx Cy

= 176.581103 2 N = 176.58 kN

And they act at 1 ∼y 3 = ∼y 1 = 13 m2 = 1.5 m 2

Bx

wB

[(Fw)BC]1

Thus,

31Fh 2 AB 4 2 = 31Fh 2 BC 4 2 =

x

(a)

1 ∼y 2 = ∼y 4 = 13 m2 = 1 m 3

The vertical force acting on plate AB is equal to the weight of the water contained in the imaginary block above the plate (shown shaded in Fig. b). 31Fv 2 AB 4 1 = rwgV1 = 11000 kg>m3 219.81 m>s2 2314 m213 m214 m24 = 470.881103 2 N = 470.88 kN

1 31Fv 2 AB 4 2 = rwgV2 = 11000 kg>m3 219.81 m>s2 2c 13 m)(4 m214 m2 d = 156.961103 2 N = 156.96 kN 3

[(Fv)AB]1

And they act at

∼x 1 = 1 14 m2 = 2 m 2

∼x 2 = 1 14 m2 = 1 m 4

+ ΣMA = 0;

-By 142 + 1353.16 kN211.5 m2 + 1176.58 kN211 m2

- 1235.44 kN213 m2 + 1470.88 kN212 m2 + 1156.96 kN211 m2 = 0 By = 274.68 kN = 275 kN

~ x2

1

wB

Bx

Ans.

Bx = 235.44 kN = 235 kN

[(Fv)AB]2

3m

Write the moment equations of equilibrium about points C and A by referring to the FBDs of plates BC, Fig. a, and AB, Fig. b, respectively. + ΣMC = 0; Bx(3 m) - (353.16 kN)(1.5 m) - (176.58 kN)(1 m) = 0

~ x1

[(Fh)AB]1 [(Fh)AB]2

2 3m

By

~ y4

Ans.

wA

4m

~ y3

Ax

Ay (b)

Ans: Bx = 235 kN, By = 275 kN 176

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

C

2–106. The semicircular 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 8 Mg with center of mass at G. It is 4 m wide.

3m T

G

A 2m

B

SOLUTION The horizontal loading on the gate is due to the pressure on the vertical projected area of the gate, Fig. a. Since the gate has a constant width of b = 4 m, the intensity of the distributed load at point B is W 5 [8000(9.81) N]

wB = rwghCb = 11000 kg>m3)(9.81 m>s2)(6 m)(4 m2 = 235.441103 2 N>m Fh =

and it acts at

1 1 wChC = 3235.441103 2 N>m416 m2 = 706.321103 2 N 2 2

Fv = rwgVBACEB

And it acts at

4 m p

C Ay

2 y = 16 m2 = 4 m 3 The upward force on BE and downward force on CE is equal to the weight of water contained in blocks BACDEB (imaginary) and CDEC, respectively. Thus, the net upward force on BEC is equal to the weight of water contained in block BACEB shown shaded in Fig. a. Thus, p = 11000 kg>m 219.81 m>s 2 c 13 m2 2 14 m2 d 2 3

Fv

D

Thus,

x5

y54m

Ax

E

3m

T

2m

Fh wB

B

NB

(a)

2

= 176.581103 2p N x =

413 m2 3p

=

4 m p

When the gate is on the verge of opening, NB = 0. Write the moment equation of equilibrium about point A by referring to the FBD of the gate, Fig. a. + ΣMA = 0;

3800019.812 N412 m2 + 3706.321103 2 N414 m - 3 m2 - 3176.581103 2p N4 a

4 mb - T = 0 p

T = 156.961103 2 N # m = 157 kN # m

Ans.

This solution can be simplified if one realizes that the resultant force due to the water pressure on the gate will act perpendicular to the circular surface, thus acting through center A of the semicircular gate and so producing no moment about this point. + ΣMA = 0;

3800019.812N412 m2 - T = 0

T = 156.961103 2 N # m = 157 kN # m

Ans.

Ans: T = 157 kN # m 177

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

C

2–107. The semicircular 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 normal reaction at B. The gate has a weight of 8 Mg with center of mass at G. It is 4 m wide. Take T = 0.

3m T

G

A 2m

B

SOLUTION The horizontal loading on the gate is due to the pressure on the vertical projected area of the gate, Fig. a. Since the gate has a constant width of b = 4 m, the intensity of the distributed load at point B is wB = rwghCb = 11000 kg>m3)(9.81 m>s2)(6 m)(4 m2 = 235.441103 2 N>m

W 5 [8000(9.81) N]

Thus,

1 1 Fh = wChC = 3235.441103 2 N>m416 m2 = 706.321103 2 N 2 2 and it acts at

Fv

D

x5

4 m p

C Ay

2 16 m2 = 4 m 3 The upward force on BE and downward force on CE is equal to the weight of water contained in blocks BACDEB (imaginary) and CDEC, respectively. Thus, the net upward force on BEC is equal to the weight of water contained in block BACEB shown shaded in Fig. a. Thus,

y54m

y =

Ax

E

3m

Fh wB

p Fv = rwgVBACEB = 11000 kg>m3)(9.81 m>s2 2 c 13 m2 2 14 m2 d 2

T

2m

B

NB

(a)

= 176.581103 2p N

And it acts at

x =

413 m2 3p

=

4 m p

Write the moment equation of equilibrium about point A and B with T = 0 by referring to the FBD of the gate, Fig. a. + ΣMA = 0;

3800019.812 N412 m2 + 370 6321103 2 N414 m - 3 m2

- 3176.581103 2p N4 a

4 mb - NB 13 m2 = 0 p

NB = 52.321103 2 N = 52.3 kN

+ ΣMB = 0; Ax 13 m2 + 3800019.812 N412 m2 - 3176.581103 2p N4 a - 3706.321103 2 N416 m - 4 m2 = 0 Ax = 6541103 2 N = 654 kN

Ans. 4 mb p Ans.

Write the force equation of equilibrium along y axis.

+ c ΣFy = 0; 176.581103 2p N - 800019.812 N - Ay = 0 Ay = 476.261103 2 N = 476 kN

Ans.

Ans: NB = 52.3 kN Ax = 654 kN Ay = 476 kN

178

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–108. Plate AB has a width of 1.5 m and a radius of 3 m. Determine the horizontal and vertical components of reaction at the pin A and the vertical reaction at the smooth stop B due to the water pressure.

A

3m

B

SOLUTION The horizontal loading on the gate is due to the pressure on the vertical projected area of the gate, Fig. a. Since the gate has a constant width of b = 1.5 m, the intensity of the horizontal distributed load at point B is wB = rwghBb = 11000 kg>m3)(9.81 m>s2)(3 m)(1.5 m2 = 44.1451103 2 N>m

Thus,

1 1 Fh = wBhB = 344.1451103 2 N>m413 m2 = 66.21751103 2 N = 66.2175 kN 2 2 And it acts at y =

Ay Ax

x2

Fv

x1 O

y 3m

NB

Fh wB (a)

2 13 m2 = 2 m 3

The vertical force acting on the gate is equal to the weight of water contained in the imaginary block (shown shaded in Fig. a) above the gate, but acts upward. 1 Fv = rwgV = 11000 kg>m3)(9.81 m>s2 2 c 1p)(3 m2 2 d 11.5 m2 4 = 104.011103 2 N = 104.01 kN

And it acts at x1 =

413 m2 3p

=

4 m p

x2 = a3 -

4 b m = 1.7268 m p

Write the moment equations of equilibrium about points A and O by referring to the FBD of the gate, Fig. a. + ΣMA = 0;

166.2175 kN)(2 m2 + 1104.01 kN)(1.7268 m2 - NB 13 m2 = 0

Ans.

NB = 104.01 kN = 104 kN

+ ΣMO = 0;

166.2175 kN)(2 m2 - (104.01 kN) a Ay = 0

4 mb - Ay 13 m2 = 0 p

Ans.

Write the force equation of equilibrium along the x axis. + ΣF = 0 ; S x

66.2175 kN - Ax = 0 Ans.

Ax = 66.2175 kN = 66.2 kN

Ans: NB = 104 kN Ay = 0 Ax = 66.2 kN 179

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

C

4m

2–109. The Tainter 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.

T G

A

608 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

9 p 2 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 (FBC)v = gwVBCEB = ( 1000 kg>m

3

= 53.985 ( 10

3

)N

3 5000(9.81) N 4 (4 m) -

+

)( 9.81 m>s ) 3 1.8344 m (3 m) 4

E

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)

60°

B

! 30°

T = 196.2 ( 10

3

- T = 0

A

C2 "

4.5 m 4.5 cos 30° m

(a)

- T = 0

Ans.

A

C1

A

2.25 m

(b)

(c)

5000(9.81) N/m 4m

This solution can be simplified if one realizes that the resultant force will act 2 m) perpendicular to the circular surface. Therefore, FBC will act through point A and so —(4.5 3 produces no moment about this point. Hence,

3 5000(9.81) N 4 (4 m)

2 = — m —(4.5 cos 30° m) 3

2

T = 196.2 ( 103 ) N # m = 196 kN # m

+ ΣMA = 0;

)

D C 2

When the gate is on the verge of opening, NB = 0. Referring to the free-body diagram of the gate in Fig. d, + ΣMA = 0;

)

2 — 3

_ x

Ay T Ax

(FBC)h = 297.98(103) N/m

) N # m = 196 kN # m

Ans.

4.1397 m (FBC)v = 53.985(103) N/m (d)

Ans: T = 196 kN # m 180

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

C

2–110. The Tainter 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 smooth spillway crest B. The gate has a mass of 5 Mg and a center of mass at G. It is 3 m wide. Take T = 0.

4m T G

B

A

608 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), 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 x ΣA = c (4.5 m)2 d - (4.5 m)(4.5 cos 30° m) = 1.8344 m2 and the horizontal 6 2 D C distance measured from its centroid to point A is

x =

ΣxA = ΣA

a

9 1 p 2 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 2

1.8344 m

E

)

A 60°

B

)

2 = — m —(4.5 cos 30° m) 3

2 — 3

C1 ! 30°

A

C2

A

" 4.5 m 4.5 cos 30° m

= 4.1397 m (a)

Thus, the magnitude of the vertical component is

(b)

(c)

(FBC)v = gwVBCEB = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 1.8344 m2(3 m) 4

5000(9.81) N 4m 5000(9.81) N/m = 53.985 ( 10 ) N 4m 2.25 m Ay 2 (4.5 m) 2.25 m — Considering the free-body diagram of the gate in Fig. d, 3 T 2 —(4.5 m) 2 3 3 + ΣMA = 0; 3 5000(9.81) N 4 (4 m) + 3 297.98 ( 10 ) N 4 c (4.5 m) - 2.25 m d 3

3

-

3 53.985 ( 103 ) N 4 (4.1397 m)

- NB(4.5 cos 30° m) = 0 NB = 50.345 ( 103 ) N = 50.3 kN + c ΣFy = 0;

+ ΣF = 0 ; S x

Ax

Ax

(FBC)h = 297.98(103) Nym

Ans.

50.345 ( 103 ) N + 53.985 ( 103 ) N - 5000(9.81) N - Ay = 0 Ay = 55.28 ( 103 ) N = 55.3 kN

4.1397 m

(FBC)h = 297.98(103) N/m

Ay

4.1397 m 4.5 cos 30° m (FBC)v = 53.985(103) N/m (FBC) = 53.985(103) N NB (d) (d)

Ans.

297.98 ( 103 ) N - Ax Ax = 297.98 ( 103 ) N = 298 kN

Ans.

Ans: NB = 50.3 kN Ay = 55.3 kN Ax = 298 kN 181

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

A

2–111. The 6-ft-wide Tainter gate in the form of a quartercircular 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 Tainter gate. What is the moment of this force about the bearing?

458

D

O

458

12 ft

B

SOLUTION 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 ∼ b = 7.2025 ft x1 = a 3 p>4 2 ∼ x 2 = (12 cos 45° ft) = 5.6569 ft 3

p 1 (7.2025 ft) c (12 ft)2 d - (5.6569 ft) c (2)(12 sin 45° ft)(12 cos 45° ft) d 4 2

x =

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,

Fh =

x2 A 12 ft

A

45° O 45° 12 ft

D

45° O 45° C1 12 ft

B

45° O

Fv

45°

C2 B

B (a)

wB = gwhBb = ( 62.4 lb>ft3 ) (16.971 ft)(6 ft) = 6353.78 lb>ft 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>ft3 )( 41.097 ft2 ) (6 ft) = 15.3868 ( 103 ) lb = 15.3868 kip Thus, the magnitude of the resultant force is FR = 2Fh 2 + F v2 = 2(53.9136 kip)2 + (15.3868 kip)2 = 56.07 kip = 56.1 kip Ans.

Its direction is

u = tan-1 a

15.3868 kip Fv b = 15.93° = 15.9° b = tan-1 a Fh 53.9136 kip

Ans.

By referring to Fig. b, the moment of FR about O is

+ (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.

182

M02_HIBB9290_01_SE_C02_ANS.indd 182

Ans: FR = 56.1 kip u = 15.9° 1MR 2 O = 0

24/02/17 4:35 PM

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–112. Determine the horizontal and vertical components of reaction at the hinge A and the horizontal reaction at the smooth surface B caused by the water pressure. The plate has a width of 4 ft.

9 ft

A

SOLUTION The horizontal loading on the gate is due to the pressure on the vertical projected area of the gate, Fig. a. Since the gate has a constant width b = 4 ft, the intensities of the horizontal distributed load at A and B are

6 ft

B

O

wA = gwhAb = 162.4 lb>ft3)(9 ft)(4 ft2 = 2246.4 lb>ft wB = gwhBb = 162.4 lb>ft3)(15 ft)(4 ft2 = 3744 lb>ft

Thus,

F

3

1Fh 2 1 = wAlAD = 12246.4 lb>ft)(6 ft2 = 13.4784110 2 lb 1Fh 2 2 =

1 1 1w - wA 2lAD = 13744 lb >ft - 2246.4 lb>ft)(6 ft2 = 4.49281103 2 lb 2 B 2

Ay wA

and they act at

∼y 1 = 1 16 ft2 = 3 ft 2

∼y 2 = 2 16 ft2 = 4 ft 3

∼y 3 = 1 16 ft2 = 2 ft 3

The vertical force acting on the gate is equal to the weight of the water contained in the imaginary block above the gate (shown shaded in Fig. a), but acts upward. For 1Fv 2 1, 3

3

1Fv 2 1 = gwVADEF = 162.4 lb>ft 2316 ft)(9 ft)(4 ft24 = 13.4784110 2 lb

~ y2 ~ y3

~ y1

~ x1

(Fv)1

E (Fv)2

~ x2 D

Ax (Fh)1 (Fh)2 wB

O (a)

NB

And it acts at

x1 =

1 16 ft2 = 3 ft 2

For 1Fv 2 2, we need to refer to the geometry shown in Fig. b. Here, 1 AADB =AADBO - AABO = 16 ft)(6 ft2 - 3p16 ft2 2 4 = 136 - 9p2 ft2 4 Then, 1Fv 2 2 = gwVADB = 162.4 lb>ft3 23136 - 9p2 ft2 414 ft2 = 1.92831103 2 lb

And it acts at

∼x = 2

13 ft2316 ft)(6 ft24 -

136 - 9p2 ft2

~ x2 A

3 ft

D

5

6 ft

= 4.6598 ft

4(6 ft) 3p

A

D

O

416 ft2 p c 16 ft2 2 d 3p 4

C1

A

6 ft 2

C1 6 ft

B

O

6 ft

B

O

(b)

183

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2–112. Continued

Write the moment equation of equilibrium about points A and B by referring to the FBD of the gate, Fig. a. + ΣMA = 0;

313.47841103 2 lb413 ft2 + 34.492841103 2 lb414 ft2

+ 313.47841103 2 lb413 ft2 + 31.92831103 2 lb414.6598 ft2 - NB 16 ft2 = 0 3

+ ΣMo = 0;

NB = 17.9712110 2 lb = 18.0 kip

Ans.

Ax = 0

Ans.

313.47841103 lb)(3 ft2 + 31.92831103 2 lb414.6598 ft2 - 313.47841103 2 lb413 ft2 - 34.49281103 2 lb412 ft2 - Ax 16 ft2 = 0

Write the force equation of equilibrium along the y axis. + c ΣFy = 0; 13.47841103 2 lb + 1.92831103 2 lb - Ay = 0 Ay = 15.40671103 2 lb = 15.4 kip

Ans.

Ans: NB = 18.0 kip Ax = 0 Ay = 15.4 kip 184

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–113. The sluice gate for a water channel is 2 m wide and in the closed position, as shown. Determine the magnitude of the resultant force of the water acting on the gate. Also, what is the smallest torque T that must be applied to open the gate if its mass is 6 Mg with its center of mass at G?

3m 3m O

T 308 G

SOLUTION

308 2m

The horizontal loading on the gate is due to the pressure on the vertical projected area of the gate, Fig. a. Since the gate has a constant width of b = 2 m, the intensities of the horizontal distributed load at points A and B are wA = rwghAb = 11000 kg>m3)(9.81 m>s2)(3 m)(2 m2 = 58.861103 2 N>m

wB = rwghBb = 11000 kg>m3)(9.81 m>s2)(3m + 3 sin 60° m)(2 m2 = 109.831103 2 N>m

Thus,

1Fh 2 1 = w4hAB = 358.861103 2 N>m413 sin 60° m2 = 152.921103 2 N 1 1 1Fh 2 2 = 1wB - wA 2hAB = 3109.831103 2 N>m - 58.861103 2 N>m413 sin 60° m2 2 2 = 66.221103 2 N Then Fh = 1Fh 2 1 + 1Fh 2 2 = 152.921103 2 N + 66.221103 2 N

Here, 1Fh 2 1 and 1Fh 2 2 act at

∼y 1 = 1 13 sin 60° m2 = 1.2990 m 2

Oy

~ x2

3m 1

T

wA 2

3 sin 60° m

~ y1 (Fh)1

~ y2

(Fh)2

∼y 2 = 2 13 sin 60° m2 = 1.7320 m 3

2 cos 30° m

The vertical force is equal to the weight of the water contained in the imaginary block above the gate (shown shaded in Fig. a), but acts upward. 1Fv 2 1 = rwgV1 =

(Fv)2

Ox

= 219.141103 2 N = 219.14 kN

(Fv)1

~ x1

1 1000 kg>m3)(9.81 m>s2 2313 m)(1.5 m)(2 m2 = 88.291103 2 N

NB

W 5 [6000(9.81) N]

(a)

1 1Fv 2 2 = rwg V2 = 11000 kg>m3)(9.81 m>s2 2 c 13 sin 60° m)(1.5 m)(2 m2 d = 38.231103 2 N 2

Then

1Fv 2 = 1Fv 2 1 + 1Fv 2 2 = 88.291103 2 + 38.231103 2 = 126.521103 2 N = 126.52 kN

Here, 1Fv 2 1 and 1Fv 2 2 act at 1 x1 = 1.5 m + 11.5 m2 = 2.25 m 2 The magnitude of the resultant force is

x2 = 1.5 m +

1 11.5 m2 = 2 m 3

FR = 2F 2h + F 2v = 21219.14 kN2 2 + 1126.52 kN2 2 = 253.04 kN = 253 kN Ans.

When the gate is on the verge of opening, NB = 0. Write the moment equation of equilibrium about point O by referring to the FBD of the gate, Fig. a. + ΣMO = 0; T + 388.291103 2 N412.25 m2 + 338.231103 2 N412 m2 - 3152.921103 2 N411.2990 m2 - 366.221103 2 N411.7320 m2 - 3600019.812 N412 cos 30° m2 = 0

T = 140.181103 2 N # m = 140 kN # m

Ans. Ans: FR = 253 kN T = 140 kN # m

185

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–114. The steel cylinder has a specific weight of 490 lb>ft3 and acts as a plug for the 1-ft-long slot in the tank. Determine the resultant vertical 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

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 p (0.35 ft)2 d (1 ft) = 1.4526 ft3 2

V1 = c 2.35 ft(0.7 ft) -

The volume of the shaded block in Fig. b is V2 = 2 e 0.1 ft(2.35 ft) + c = 0.5037 ft3

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>ft3 )( 1.4526 ft3 - 0.5037 ft3 ) = 59.21 lb T The weight of the cylinder is W = gstVC = ( 490 lb>ft2 ) 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: N = 248 lb 186

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–115. The steel cylinder has a specific weight of 490 lb>ft3 and acts as a plug for the 1-ft-long slot in the tank. Determine the resultant vertical 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 ft3 2

The volume of the shaded block in Fig. b is V2 = 2e 0.35 ft(0.1 ft) + c = 0.10372 ft3

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>ft3 )( 0.10372 ft3 - 0.05258 ft3 ) = 3.192 lb c The weight of the cylinder is W = gstVC = ( 490 lb>ft2 ) 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: N = 185 lb 187

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–116. The Tainter 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. Also, what is 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

O

208 208 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 + 2 m sin 40°)(1.5 m) = 48.347 ( 103 ) N>m Then (Fh)1 = (Fh)2 =

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 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, 2 2 m sin 20° ~ = 1.3064 m x2 = 1.3064 m cos 20° = 1.2276 m ≥ 3£ 20 p 180 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 r =

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

188

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*2–116. Continued

The magnitude of the resultant force is FR = 2Fh2 + F v2 = 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, + 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 water acting on the gate 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

w

y1

x1

1

=

20˚ 20˚

40˚

(Fh)1

2m

(Fh)2 w

x3

x2

2m

2

– 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 189

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–117. Solve the first part of Prob. 2–116 by the integration method using polar coordinates. 2m 2m T G

O

208 208 1.5 m

SOLUTION 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 = [19 620(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 = p dA = 19 620(1 + sin u)(3 du). = 58.86 ( 103 ) (1 + sin u) du 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 1 sin 2u b du 2

(dF)h = 58.86 ( 103 ) acos u +

1 1 - cos 2u b du 2 2

(dF )v = 58.86 ( 103 ) asin u +

The horizontal and vertical components of the resultant force are Fh =

L

(dF )h = 58.86 ( 103 )

= 58.86 ( 103 ) c sin u -

L0

2p 9

1 sin 2u b du 2

acos u + 2p

1 cos 2u d ` 9 4 0

= 49.994 ( 103 ) N = 49.994 kN

Fv =

L

(dF)v = 58.86 ( 103 )

= 58.86 ( 103 ) a - cos u +

L0

2p 9

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 + F v2 = 2(49.994 kN)2 + (19.825 kN)2 = 53.78 kN = 53.8 kN Ans.

Ans: FR = 53.8 kN

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

z

2–118. The cylindrical tank is filled with gasoline and water to the levels shown. Determine the horizontal and vertical components of the resultant force on its hemispherical end. Take rg = 726 kg>m3.

0.5 m 1m 1m

x

SOLUTION z

The vertical component of the resultant force is equal to the total weight of the gasoline and water contained in the hemisphere. Here,

h 5 1.5 m 2 z

3

1 4 1Fv 2 g = rggVg = 31726 kg>m3)(9.81 m>s2 24 c a b(p)(1 mb R = 2.3741103 2p kN 4 3 1Fv 2 w Then

3 1 4 = rwgVw = 311000 kg>m3)(9.81 m>s2 24 c a b(p)(1 mb d = 3.2701103 2p kN 4 3

Fv = 1Fv 2 g + 1Fv 2 w = 2.3741103 2p N + 3.2701103 2p N = 5.6441103 2p N = 17.7 kN

(Fv)g Pg

1.5 m

~ z ~ z

z hw 5 2z Pw

(Fv)w

x

Ans.

The equation of the pressure as a function of z for gasoline, Fig. a, is

(a)

Pg = rgghg = 1726 kg>m3)(9.81 m>s2)(1.5 m - z2 = 7.1221103)(1.5 - z2

z

When hg = 1.5 m1z = 02, pg ' z = 0 = 7.1221103)(1.5 - 02 = 10.6831103 2 N>m2. Then for water realizing that hw = -z, pw = pg ' z = 0 + rwghw 3

x

y2 1 z2 5 1 2

3

2

= 10.683110 2 N>m + 11000 kg>m )(9.81 m>s 21 -z2 = 310.6831103 2 - 9.811103 2z4 N>m2

Here, the differential force dF acting on the differential element of area dA = 2y dz = 221 - z2dz is df = p dA = p3221 - z2dz4. For gasoline,

y

z

y

y

dz

1m

dFg = 7.1231103)(1.5 - z23221 - z2dz4 = 14.2441103 231.521 - z2 - z21 - z2 4dz

Integrating,

1Fg 2 h = 14.2441103 2

L0

= 14.2441103 2 e

(b)

1m

31.521 - z2 - z21 - z2 4dz

1m 1.5 1 3z21 - z2 + sin -1z4 - c - 211 - z2 2 3 d f ` 2 3 0

= 12.0331103 2 N

For water,

dFw = 110.6831103 2 - 9.811103 2z23221 - z2dz4 = 19.621103 231.08921 - z2 - z21 - z2 4dz

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2–118. Continued

Integrating, 0

1Fw 2 h = 19.621103 2

L-1 m

= 19.621103 2 e Then

31.08921 - z2 - z21 - z2 4dz

0 1.089 1 3z21 - z2 + sin -1z4 - c - 211 - x2 3 d f ` 2 3 -1 m

= 23.3211103 2 N

Fh = 1Fg 2 h + 1Fw 2 h = 12.0331103 2 N + 23.3211103 2 N = 35.3541103 2 N = 35.4 kN

Ans.

Ans: Fv = 17.7 kN Fh = 35.4 kN 192

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–119. The hollow spherical float controls the level of water within the tank. If the water is at the level shown, determine the horizontal and vertical components of the force acting on the supporting arm at the pin A, and the normal force on the smooth support B. Neglect the weight of the float.

B A

4 in.

3 in. 18 in.

SOLUTION The vertical buoyant force acting on the spherical float is 3 1 4 3 Fv = gwVsub = 162.4 lb>ft3 2 e c pa ft b d f = 0.65p lb 2 3 12

+ ΣMA = 0; + ΣFx = 0; S

4 in.

10.65p lb)(18 in.2 - NB 14 in.2 = 0 NB = 2.925p lb = 9.19 lb

Ax

2.925p lb - Ax = 0

Ay

Ax = 2.925p lb = 9.19 lb + c ΣFy = 0;

Fv

NB

Consider the equilibrium of the FBD of the floating system shown in Fig. a.

18 in. (a)

0.65p lb - Ay = 0 Ans.

Ay = 0.65p lb = 2.04 lb

Ans: NB = 9.19 lb Ax = 9.19 lb Ay = 2.04 lb 193

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. *2–120. The solid ball is made of plastic having a density of rp = 48 kg>m3. Determine the tension in the cable AB if the ball 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 ball is V = 43pr 3.

0.9 m

0.6 m

SOLUTION The weight of the ball is

A

4 Wb = rpgVb = 148 kg>m )(9.81 m>s 2 c p10.6 m2 3 d = 426.04 N 3 3

2

The submerged volume is equal to the volume of the ball since it is fully submerged. Thus, the buoyant force is 4 Fb = rwgVsub = 11000 kg>m3)(9.81 m>s2 2 c p10.6 m2 3 d = 8875.88 N 3

Wb

Referring to the FBD of the ball, Fig. a, the equilibrium requires + c ΣFy = 0;

B

8875.88 N - 426.04 N - TAB = 0 Ans.

TAB = 8449.84 N = 8.45 kN

The tension in cable AB remains the same since the buoyant force does not change once a body is fully submerged, which means that it is independent of the submerged depth.

Fb TAB (a)

Ans: TAB = 8.45 kN 194

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

h

2–121. 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 = 1 Mg>m3. 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

0.25 m

For equilibrium, the buoyant force on each float is required to be + c ΣFy = 0;

Fb - 6082.2 N = 0

(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 0.3927 m3 6 0.620 m3, 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: h = 136 mm 195

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

25 mm

2–122. 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 at a depth hs. 0.1410 N = ( 1000 kg>m3 )( 9.81 m>s2 ) 3 (0.025 m)2hs 4

Fb = rwgVs;

hs = 0.023 m

Referring to Fig. a, V1 = V2 - V3

3 p(0.025 m) 4 (0.1 m) 2

=

3 p(0.025 m)2 4 h

- (0.02 m)2 (0.023 m) Ans.

h = 0.1073 m = 107 mm

The mass of the ice cube is Mi = riVi = ( 920 kg>m3 )( 0.025 m ) 3 = 0.014375 kg Thus, the rise in water level due to the additional water of the melting ice cubes 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: h = 107 mm 196

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

1 ft

2–123. Water in the container is originally at a height of h = 3 ft. If a block having a specific weight of 50 lb>ft3 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

h

SOLUTION The weight of the block is Wb = gbVb = ( 50 lb>ft3 ) 3 (1 ft)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>ft3 ) VDisp VDisp = 0.8013 ft3

The volume of the water is Vw = 2 ft(2 ft)(3 ft) = 12 ft3 When the level of the water in the container has a height of h, Vw = V′ - VDisp 12 ft3 = 4h ft3 - 0.8013 ft3 Ans.

h = 3.20 ft

Ans: h = 3.20 ft 197

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

3 ft

*2–124. Determine the height at which the oak block will float above the water surface. The specific weight of the wood is gw = 48 lb>ft3.

h 9 ft

6 ft

SOLUTION Referring to the geometry shown in Fig. a, y 3 = ; 9 + y 6

y = 9 ft y

And v h + 9 = ; 3 9 Thus, the volume of the oak block is

r =

1 1h + 92 3

3 ft h 9 ft

1 1 VO p16 ft2 2 118 ft2 - p13 ft2 2 19 ft2 = 189p ft3 3 3 And the submerged volume of the wooden block is Vsub =

2 1 1 1 p 35832 - 1h + 92 3 4 p16 ft2 2 118 ft2 - pc 1h + 92 d 1h + 92 = 3 3 3 27

r 92h 6 ft (a)

Then the weights of the wooden block and the buoyant force are Ww = gOVO = 148 lb>ft3 21189p ft3 2 = 9072p lb Fb = gwVsub = 162.4 lb>ft3 2 e

WO

p 35832 - 1h + 92 3 4 = 2.3111p35832 - 1h + 92 3 4 f 27

Referring to the FBD of oak block, Fig. b, equilibrium requires + c ΣFy = 0;

3

2.3111p35832 - 1h + 92 4 - 9072p lb = O h = 3.4000 ft = 3.40 ft

Ans. Fb (b)

Ans: h = 3.40 ft 198

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–125. The container of water has a mass of 20 kg. Block B has a density of 7840 kg>m3 and a mass of 30 kg. Determine the total compression or elongation of each spring when the block is fully submerged in the water.

kC 5 5 kN!m

C

SOLUTION

B

The volume of block B is determined from

rB =

MB ; VB

7840 kg>m3 =

30 kg VB

VB = 3.8265110-3 2 m3

D

D

kD 5 2 kN!m

Here, block B is fully submerged. Then Vsub = VB. Thus, the buoyant force Fb = rwgVsub = 11000 kg>m3)(9.81 m>s2 233.8265110-3 2 m3 4 = 37.54 N

(Fsp)C

Referring to the FBD of block B, Fig. a, + c ΣFy = 0;

1Fsp 2 C + 37.54 N - 3019.812 N = 0

Referring to the FBD of the container, Fig. b, + c ΣFy = 0;

21Fsp 2 D - 37.54 N - 2019.812 N = 0

Thus, the change in length of springs C and D are dC = dD =

1Fsp 2 C KC

1Fsp 2 D KC

=

Fb

1Fsp 2 C = 256.76 N

30(9.81) N

1Fsp 2 D = 116.87 N

256.76 N = 0.05135 m = 51.4 mm 51103 2 N>m

116.87 N = = 0.05843 m = 58.4 mm 21103 2 N>m

(a)

Ans. (Fsp)C

Ans.

20(9.81) N

(Fsp)D

(Fsp)D (b)

Ans: dC = 51.4 mm dD = 58.4 mm 199

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–126. The container of water has a mass of 20 kg. Block B has a density of 7840 kg>m3 and a mass of 30 kg. Determine the total compression or elongation of each spring when the block is fully submerged in the water.

C

kC 5 5 kN!m

SOLUTION

B

The volume of block B is determined from MB ; VB

rB =

7840 kg>m3 =

30 kg VB

VB = 3.8265110-3 2 m3

Here, block B is fully submerged. Then Vsub = VB. Thus, the buoyant force is Fb = rwg Vsub = 11000 kg>m3)(9.81 m>s2 233.8265110-3 2 m 3 4 = 37.54 N

D

kD 5 2 kN!m

E

kE 5 5 kN!m

Referring to the FBD of block B, Fig. a, + c ΣFy = 0 ;

1Fsp 2 C + 37.54 N - 3019.812 N = 0

Thus, the change in length of spring C is 1Fsp 2 C

dC =

KC

=

(Fsp)C

1Fsp 2 C = 256.76 N

256.76 N = 0.05135 m = 51.4 mm 51103 2 N>m

Fb

Ans. 30(9.81) N

Springs D and E are subjected to the same force, 1Fsp 2 D = 1Fsp 2 E, since both are arranged in series. Referring to the FBD of the container, Fig. b, + c ΣFy = 0 ;

1Fsp 2 D - 37.54 N - 2019.812 N = 0

Thus, the change in length of spring D and E are dD =

1Fsp 2 D

=

dE =

1Fsp 2 D

=

KD KE

(a)

1Fsp 2 D = 233.74 N

233.74 N = 0.1169 m = 117 mm 21103 2 N>m

233.74 N = 0.04675 m = 46.7 mm 51103 2 N>m

(Fsp)C 20(9.81) N

Ans. Ans.

(Fsp)D

(Fsp)D (b)

Ans: dC = 51.4 mm dD = 117 mm dE = 46.7 mm 200

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–127. The hot-air balloon contains air having a temperature of 180°F, while the surrounding air has 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) ft3. 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>ft3 ra ' T = 180°F = 0.00193 slug>ft3

wa

Thus, the weight of the air inside the balloon is Wa ' T = 180° F = ra ' T = 180° FgV = ( 0.00193 slug>ft3 = 7457.52 lb

)( 32.2 ft>s2 ) 3 120 ( 103 ) ft3 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>ft3 )( 32.2 ft>s2 ) 3 120 ( 103 ) ft3 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: WL = 1.50 kip 201

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

600 kN

*2–128. 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 air and water temperature is 12°C? The balloons are at an average depth of 20 m. Neglect the mass of the air and the balloon. The volume of a sphere is V = 43pr 3.

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 13.08pr 3 - 42.125 kN = 0 r = 1.008 m = 1.01 m

+ c ΣFy = 0;

Ans.

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

80(9.81) kN T

(Fb)b = 13.08 !r3

T

+

(Fb)B = 100.55 kN (a)

Ans: r = 1.01 m m = 15.6 kg

T = 42.125 kN (b)

202

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–129. When loaded with gravel, the barge floats in water at the depth shown. If its center of gravity is at G, determine whether the barge will restore itself when a wave causes it to roll slightly at 9°.

G O 2m

1.5 m

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 x = = 0.3168 m 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 y = = 0.7751 m 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

1.5 m

N

1.5 – 3 tan 9˚ = 1.0248 m

y

3m

3m 3m Fb

x (a)

Ans: It will restore itself. 203

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–130. When loaded with gravel, the barge floats in water at the depth shown. If its center of gravity is 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. Ans. 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. 204

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–131. The can of alcohol rests on the floor of a hoist. Determine the maximum pressure developed at the base of the can if the hoist is moving upward with (a) a constant velocity of 10 ft>s, and (b) a constant upward acceleration of 5 ft>s2. Take gal = 49.3 lb>ft3.

SOLUTION a) Since the hoist is travelling with a constant velocity, it is in equilibrium. Thus, 5 ft

p = gah = 149.3 lb>ft3)(5 ft2 = 1246.5 lb>ft2 2 a

b) Since the hoist is accelerating, p = gaha1 +

ac b g

1 ft 2 b = 1.71 psi 12 in.

= 149.3 lb>ft3)(5 ft2 a1 + = (284.78 lb>ft2) a

5 ft>s2 32.2 ft>s2

Ans.

b

1 ft 2 b = 1.98 psi 12 in.

Ans.

Ans: a. p = 1.71 psi b. p = 1.98 psi 205

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 0.5 m

*2–132. The truck carries an open container of water. If it has a constant deceleration 1.5 m>s2, determine the angle of inclination of the surface of the water and the pressure at the bottom corners A and B.

a

1.5 m A

B

4m

SOLUTION The free surface of the water in the decelerated tank is shown in Fig. a. tan u =

1.5 m>s2 ac = g 9.81 m>s2

1.5 m h A

Ans.

u = 8.6935° = 8.69°

2m

Dh

2m

Dh

u 5 8.6935° A

1.5 m

hB

B (a)

From the geometry in Fig. a, ∆h = 12 m2 tan 8.6935° = 0.3058 m

Since ∆h 6 0.5 m, the water will not spill. Thus, hA = 1.5 m - 0.3058 m = 1.1942 m and hB = 1.5 m + 0.3058 m = 1.8058 m. Then pA = rwghA = 11000 kg>m3)(9.81 m>s2)(1.1942 m2

Ans.

pB = rwghB = 11000 kg>m3)(9.81 m>s2)(1.8058 m2

Ans.

= 11.7151103 2 N>m2 = 11.7 kPa

And

= 17.7151103 2 N>m2 = 17.7 kPa

Ans: u = 8.69°, pA = 11.7 kPa, pB = 17.7 kPa 206

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. a

2–133. The closed rail car is 2 m wide and filled with water to the level shown. Determine the pressure that acts at A and B when the car has a constant acceleration of 4 m>s2.

6m

3m 2m A

B

T

SOLUTION tan u =

4 h′ = 9.81 6 62x

h′ = 2.446 m

x u

Also, u = tan

-1

h9

4 = 22.183° 9.81

u 3m

u

Empty volume in tank is 216)(12 = 12 m3. During accelerating we require

B

A 6m

1 12 = 2a xb 1xtan 22.183°2 2 x = 5.425 m

pA = 1100029.8113 - 5.425 tan 22.183°2

Ans.

= 7.73 kPa

pB = 100019.81213 + 16 - 5.42521 tan 22.183°2 Ans.

= 31.7 kPa

Ans: pA = 7.73 kPa, pB = 31.7 kPa 207

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

18 ft

2–134. 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 has 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>ft3 ) (7.5 ft) = 468 lb>ft2

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 ft3 The volume of water after the car accelerates is V′ =

1 (9 ft + 3.4099 ft)(18 ft)(6 ft) = 670.14 ft3 6 810 ft3 2

(OK!)

Thus, the amount of water spilled is ∆V = V - V′ = 810 ft3 - 670.14 ft3 = 139.86 ft3 = 140 ft3

Ans.

The pressure at B when the car accelerates is pB = gwhB = ( 62.4 lb>ft3 ) (9 ft) = 561.6 lb>ft2 = 562 lb>ft2 Ans.

With acceleration:

ac = 10 ftys2

u

9 ft

7.5 ft

B

h

Ans: At rest: pB = 468 lb>ft2 With acceleration: ∆V = 140 ft3 pB = 562 lb>ft2

18 ft (a)

208

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

0.2 m

2–135. 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

Dh Dh u

h9B hA = 0.4 m

h9A

Imaginary free surface

hB = 0.4 m

(a)

Ans: h′A = 0.171 m, h′B = 0.629 m 209

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

0.2 m

*2–136. A large container of benzene is being transported by the truck. Determine its maximum constant acceleration so that no benzene will spill from the vent tubes A or B. When the truck is 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. 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

Ans.

1.5 m

1.5 m

h′B = 0.7 m Imaginary free surface

h′B = 0.4 m

(a)

Ans: ac = 1.96 m>s2 210

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–137. 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 tank.

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 211

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. A

2–138. 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.

1m C

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 =

a

1m B

2m

3m

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 pB = rwghB = ( 1000 kg>m

3

Ans.

)( 9.81 m>s ) (1.3884 m) 2

= 13.62 ( 103 ) Pa = 13.6 kPa

Ans.

C

2m

C

B

2m

3m

Ans: pC = 23.6 kPa, pB = 13.6 kPa 212

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

a

2–139. 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

w a g

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 motion, Fig. c. + ΣFx = max; S

- apx +

N (a)

g(dx dy dz) 0px dxbdy dz + px dy dz = ax 0x g

y

g dx dpx = a g x ax

In y direction, + c Σ Fy = may;

py dx dz - apy +

dpy = - g dy a1 +

0 py 0y ay g

dybdx dz - g dx dy dz =

g dx dy dz ay g

b

x

f ay

a = g sin f

x′

(b)

At the surface, p is constant, so that dpx + dpy = 0, or dpx = -dpy. ay g dx ax = -g dy a1 + b g g

px dy dz

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

(

px +

dpx dx

(

dx dy dz

dx

Since at the surface,

(

py +

dy = - tan u dx then

dpy dy

(

dy dx dz

py dx dz

tan u = tan f

(c)

or Q.E.D.

u = f

213

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

a

*2–140. The cart is given a constant acceleration a up the inclined plane. 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)

x

(a )

Equating Eqs. (1) and (2), we obtain tan u =

ax

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 =

a cos f a sin f + g

Q.E.D

214

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

0.6 m

A

SOLUTION H = =

D

C

2–141. The sealed tube 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 C and D.

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: pD - pC = 28.1 kPa 215

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

0.6 m

A

SOLUTION H = =

D

C

2–142. The sealed tube 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 A and B.

0.5 m

v2R2 2g

B

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)

Ans: pB - pA = 28.1 kPa 216

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–143. A woman stands on a horizontal platform that is rotating at 0.75 rad>s. If she is holding a cup of tea, and the center of the cup is 2 ft from the axis of rotation, determine the slope angle of the liquid’s surface. Neglect the size of the cup.

SOLUTION Since the 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 = ( 0.75 rad>s ) (2 ft) = 1.125 ft>s2 2

Thus, the slope of the tea surface is m = tan u =

1.125 ft>s2 ar = = 0.3494 g 32.2 ft>s2 Ans.

u = 2.00°

Ans: u = 2.00° 217

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

0.1 m

0.1 m

*2–144. Determine the maximum height d the glass can be filled with water so that no water spills out when the glass is rotating at 15 rad>s. 0.2 m d

SOLUTION

0.05 m 0.05 m v

From the geometry shown in Fig. a, d′ 0.05 m = ; d′ = 0.2 m d′ + 0.2 m 0.1 m

0.1 m 0.1 m

Then r d + 0.2 m = ; 0.05 m 0.2 m

r 0.2 m

r = 0.251d + 0.22

Thus, the volume of the empty space in the container shown shaded in Fig. a is 1 1 Ves = p10.1 m2 2 10.4 m2 - p30.251d + 0.224 2 1d + 0.22 3 3 1 = p30.004 - 0.06251d + 0.22 3 4m3 3 For the condition that the water is about to spill, the parabolic profile of the free water surface is shown in Fig. b. h = a

h = c

v2 2 br 2g

115 rad>s2 2

219.81 m>s2 2

r

0.05 m

d

d9

(a) 0.1 m 0.1 m h

d 10.1 m2 2

(b)

(O.K!)

= 0.1147 m 6 0.2 m

Since the empty space in the glass must remain the same, the volume of the paraboloid shown shaded in Fig. b must be equal to this volume. Here, the volume of the paraboloid is equal to one half the volume of the cylinder of the same radius and height. Vparab = Ves 1 1 3p10.1 m2 2 410.1147 m2 = p30.004 - 0.06251d + 0.22 3 4 2 3 d = 0.1316 m = 0.132 m

Ans.

Ans: d = 0.132 m 218

M02_HIBB9290_01_SE_C02_ANS.indd 218

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

0.1 m

0.1 m

2–145. The glass is filled with water to a height of d = 0.1 m. To what height d = d′ does the water rise against the wall of the glass when the platform has an angular velocity of v = 15 rad>s?

0.2 m d

SOLUTION 0.05 m 0.05 m

From the geometry shown in Fig. a,

v

d″ 0.05 m = ; d″ + 0.2 m 0.1m

d″ = 0.2 m

Then, r 0.3 m = ; 0.05 m 0.2 m

0.1 m 0.1 m

r = 0.075 m

r

Thus, the volume of the empty space in the container shown shaded in Fig. a is Ves

0.2 m

1 1 = p10.1 m2 2 10.4 m2 - p10.075 m2 2 10.3 m2 = 0.77083p110-3 2 m 3 3

h = c

v2 2 br 2g

0.05 m

d 5 0.1 m

d0

The parabolic profile of the free water surface is shown in Fig. a. Then h = a

r

(a)

115 rad>s2 2

219.81 m>s2 2

= 11.4679R2

d R2

R

Since the empty space in the glass must remain the same, the shaded volume shown in Fig. b [paraboloid segment (1) and segment (2)] must be equal to this volume. Here, the volume of the paraboloid is equal to one half the volume of the cylinder of the same radius and height.

2

R 1

d9

h

0.05 m d0 5 0.2 m

V1 + V2 = Ves 1 1 1 3pR2 111.4679R2 24 + c p10.1 m2 2 10.4 m2 - pR1 1d′ + 0.22 d = 0.77083p110-3 2 m 2 2 3 1 5.7339R4 - R2 1d′ + 0.22 + 0.5625110-3 2 = 0 (1) 3 From the geometry shown in Fig. b, d′ + 0.2 m R ; = 0.2 0.05

(b)

d′ + 0.2 = 4R

Substitute this result into Eq. (1). 5.7339R4 Solving numerically,

4 3 R + 0.5625110-3 2 = 0 3

R = 0.087852 m Thus, d′ + 0.2 = 410.0878522 Ans.

d′ + 0.1514 m = 0.151 m Also, h = 11.467910.0878522 2 = 0.0885 m 6 d′

(O.K!)

Ans: d′ = 0.151 m

219

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise.

ro

ri

2–146. The drum has a hole in the center of its lid and is filled to a height d with a liquid having a density r. If the drum is then placed on the rotating platform and it attains an angular velocity of v, determine the inner radius ri of the liquid where it contacts the lid.

d

do

SOLUTION The volume of the paraboloid empty space must be the same as the volume of the empty space when the liquid is not spinning. Since the volume of the paraboloid is equal to one half the volume of the cylinder of the same radius and height, then

v

Vparab = Ves 1 1pr 2i h2 = pr 2o 1do - d2 2

ri

r 2oh = 2r 2o 1do - d2

So then,

h =

ro

(1) h

v2 2 r 2g i

(2)

do d

Substitute Eq. (2) into (1). r 2i a

v2 2 br i = 2r 2o 1do - d2 2g r 4i = ri =

4gr 2o 1do - d2 v2

4g r 2o 1do c 2 v

- d2

(a)

d

1>4

Ans.

Ans: ri = c

4gr 2o 1do - d2 2

v

d

1>4

220

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Unless otherwise stated, take the density of water to be rw = 1000 kg>m3 and its specific weight to be gw = 62.4 lb>ft3. Also, assume all pressures are gage pressures unless stated otherwise. 2–147. The tube is filled with oil to the level h = 0.4 m. Determine the angular velocity of the tube so that the pressure at O becomes - 15 kPa. Take So = 0.92.

v

h 5 0.4 m

O

SOLUTION

0.8 m

0.8 m

The level of oil in the tube will not change. Therefore, the imaginary surface of the oil will be that shown in Fig. a. The pressure of point O is negative, since it is located above the imaginary surface. po = - gho;

R 5 0.8 m

- 151103 2 N>m2 = -0.9211000 kg>m3)(9.81 m>s2 2ho ho = 1.6620 m

Thus

0.4 m

h = ho + 0.4 m = 1.6620 m + 0.4 m = 2.0620 m So then, h = a

2.0620 m = c

O

h

v2 2 br 2g

Imaginary surface

ho 5 1.6620

v2 d 10.8 m2 2 219.81 m>s2 2

(a)

Ans.

v = 7.9506 rad>s = 7.95 rad>s

Ans: v = 7.95 rad>s 221

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3–1. A two-dimensional flow field for a fluid can be described by V = {(2x + 1)i - (y + 3x)j} 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.

SOLUTION 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)j6 m>s

x

!

= [2(2) + 1]i - [3 + 3(2)] j

The magnitude of V is

= 55i - 9j6 m>s

V = 2V x2 + V y2 = 2 ( 5 m>s ) 2 +

( - 9 m>s ) 2 = 10.3 m>s

Ans.

Thus,

Vy Vx

b = tan-1 a

9 m> s 5 m>s

V (a)

As indicated in Fig. a, the direction of V is defined by u = 360° - f, where f = tan-1 a

Vy = 9 m/s

b = 60.95° Ans.

u = 360° - 60.95° = 299°

Ans: V = 10.3 m>s, u = 299° 222

M03_HIBB9290_01_SE_C03_ANS.indd 222

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3–2. A two-dimensional flow field for a liquid can be described by V = 5 1 5y2 - x 2 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

(3x + y)j6 m>s

Vy = 13 m/s

+ 33(5) + ( - 2)4j

V

= 515i + 13j6 m>s

V = 2V x2 + V y2 = 2 ( 15 m>s ) 2 + ( 13 m>s ) 2 = 19.8 m>s

Ans.

Vx = 15 m/s

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

!

x

(a)

b = 40.9°

Ans.

Ans: V = 19.8 m>s u = 40.9° 223

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3–3. A two-dimensional flow field for a fluid is defined by V = 314x2i - 12y + 1x2j4 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at 11 m, 2 m2 , and its direction measured counterclockwise from the positive x axis.

SOLUTION The velocity vector for a particle at x = 1 m and y = 2 m is V = 54xi - (2y + x)j6 m>s

u

= 54(1)i - 32(2) + 14j6 m>s

The magnitude of V is

x

f

= 54i - 5j6 m>s

V = 2V x2 + V y2 = 2(4 m>s)2 + ( - 5 m>s)2 = 6.40 m>s

Ans.

f = tan - 1 a

Vy Vx

b = tan - 1 a

5 m>s 4 m>s

Vy 5 5 mys

V (a)

As indicated in Fig. a, the direction of V is defined by u = 360° - f, where

Thus,

Vx 5 4 mys

b = 51.34° Ans.

u = 360° - 51.34° = 308.66° = 309°

Ans: V = 6.40 m>s u = 309° 224

M03_HIBB9290_01_SE_C03_ANS.indd 224

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*3–4. A two-dimensional flow field for a liquid is defined by V = 31x + 2y2 2i + 12x 2 - 3y2j4 m>s, where x and y are in meters. Determine the magnitude of the velocity of a particle located at 1- 2 m, 2 m2, and its direction measured counterclockwise from the positive x axis.

SOLUTION The velocity vector of a particle at x = -2 m and y = 2 m is

y

V = 5(x + 2y2)i + (2x 2 - 3y)j6 m>s

= 5[ - 2 + 2(2)2]i + 32( -2)2 - 3(2)4j6 m>s

The magnitude of V is

V

Vy 5 2 mys

= 56i + 2j6 m>s

u Vx 5 6 mys

V = 2Vx 2 + Vy 2 = 2 ( 6 m>s ) 2 + ( 2 m>s ) 2 = 6.324 m>s = 6.32 m>s

x

Ans.

As indicated in Fig. a, the direction of V is defined by u = tan-1 a

Vy Vx

b = tan-1 a

2 m> s 6 m> s

Ans.

b = 18.43° = 18.4°

Ans: V = 6.32 m>s u = 18.4° 225

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3–5. A flow field is defined by u = 12x 2 + 12 m>s and v = 1xy2 m>s , where x and y are in meters. Determine the equation of the streamline that passes through point (3 m, 1 m), and find the velocity of a particle located at this point. Draw this streamline.

SOLUTION The x and y components of velocity at point (3 m, 1 m) are u = 2(32) + 1 = 19 m>s S = 9 m>s c

v = 3(3)(1) The magnitude of the velocity is

V = 2u 2 + v 2 = 2 ( 19 m>s ) 2 + ( 9 m>s ) 2 = 21.02 m>s = 21.0 m>s

Ans.

And its direction is defined by

9 m>s v b = 25.34° = 25.3° u = tan - 1 a b = tan - 1 a u 19 m>s

Ans.

Using the definition of the slope of the streamline and initial condition y = 1 m at x = 3 m, dy v = ; dx u

dy xy = 2 dx 2x + 1 y

x dy x dx = 2 y L1 m L3 m 2x + 1

ln y `

y 1m

=

ln y =

x 1 ln (2x 2 + 1) ` 4 3m

1 2x 2 + 1 ln a b 4 19

ln y = ln c a y = a

2x 2 + 1 1>4 b d 19

2x 2 + 1 1>4 b m 19

Ans.

The values of x and the corresponding values of y are tabulated below x(m)

0

{1

{2

{3

{4

{5

{6

y(m)

0.479

0.630

0.830

1

1.15

1.28

1.40

The plot of y vs x is shown in Fig. a. y(m) 1.50 1.00 0.50 –6

–4

–2

0

2

4

6

x(m)

(a)

226

M03_HIBB9290_01_SE_C03_ANS.indd 226

Ans: V = 21.0 m>s u = 25.3° 2 y = 1 2x 19+ 1 2 1>4 m

25/02/17 11:58 AM

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3–6. A flow field for a fluid is defined by u = 12 + y2 m>s and v = 12y2 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

L 2y

ln y +

dy =

L

u = (2 ! y) m/s

streamline

y

dy 2y = dx 2 + y 2 + y

V

x x (a)

dx

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° 227

M03_HIBB9290_01_SE_C03_ANS.indd 227

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3–7. Particles travel within a flow field defined by 1 3 V = 3 16 y i + 6j 4 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (3 m, 4 m), and find the velocity of a particle located at this point. Draw this streamline.

SOLUTION

y(m)

The x and y components of velocity at point (3 m, 4 m) are

8 6

1 3 u = (4 ) = 4 m>s S 16

4

v = 6 m>s c

2

The magnitude of the velocity is

0

V = 2u2 + v2 = 2(4 m>s)2 + (6 m>s)2 = 7.211 m>s = 7.21 m>s

Ans.

and its direction is defined by u = tan

-1

4

6

8

x(m)

–2 –4 –6

6 m>s v a b = tan -1 a b = 56.31° = 56.3° u 4 m>s

Ans.

Using the definition of the slope of the streamline and initial condition y = 4 m at x = 3 m, dy v = ; dx u

2

–8 (a)

dy 6 = dx 1 3 y 16 y

x

L4 m

y3dy = 96

dx L3 m

x y4 y ` = 96x ` 4 4m 3m

y4 = 4(96x - 224)

Ans.

The values of x and the corresponding values of y are tabulated below. x(m)

2.33

3

y(m)

0

{4

4

5

6

7

8

{5.03 {5.66 {6.13 {6.51 {6.83

The plot of y vs x is shown in Fig. a.

Ans: V = 7.21 m>s u = 56.3° y4 = 4196x - 2242 228

M03_HIBB9290_01_SE_C03_ANS.indd 228

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*3–8. 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 v 4 = = dx u 2y2 y2 dy =

L

streamline

! u = 2 y2 m/s

dy = tan u dx

L

V

y

2 dx

x x

1 3 y = 2x + C 3

(a)

At x = 1 m, y = 2 m. Then 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° 229

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3–9. A flow field is defined by u = 10 m>s and v = -3 m>s. If metal flakes are released into the flow at point (2 m, 1 m), draw the streamline, streakline, and pathline for these particles.

SOLUTION Using the definition of the slope of the streamline and initial condition y = 1 m at x = 2 m, dy v = ; dx u

dy -3 = dx 10 y

L1 m

x

dy = -

y - 1 = -

3 dx 10 L2 m 3 (x - 2) 10

1 (16 - 3x) 10

y =

Ans.

The plot of y vs x is shown in Fig. a. The flow is steady. Therefore, the streakline and pathline coincide with the streamline and share the same equation.

y(m) 2 1.60 1 0

Streakline and pathline coincide with streamline. 1

2

3

4

5

6 5.33

x(m)

(a)

230

M03_HIBB9290_01_SE_C03_ANS.indd 230

Ans: 1 y = 10 116 - 3x2

25/02/17 11:59 AM

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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 = 10.8 + 0.6y2 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 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 + 0.6 y) 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

x dy dx = L0 1.6 + 1.2y L0

(a)

y 1 ln(1.6 + 1.2y) ` = x 1.2 0

lna

!

1.6 + 1.2y 1.6

lna1 +

y(m)

b = 1.2x

y=

4 1.2 x ( e – 1) m 3

3 yb = 1.2x 4

1 +

3 y = e1.2x 4 4 y = ( e1.2x - 1 ) m 3

streamline

Ans.

Using this result, the streamline is shown in Fig. b.

x(m) (b)

231

M03_HIBB9290_01_SE_C03_ANS.indd 231

Ans: y = 43 1e1.2x - 12

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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 = 10.8x2 m>s, where x is in meters. Also, buoyancy and thermal winds cause the balloon to rise at a rate of v = 11.6 + 0.4y2 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

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 +

! u = (0.8x) m/s

streamline

1 2 yb = ln x 4

1 2 yb = x 4

x

y = 4 ( x 1>2 - 1 )

Ans.

1m (b)

Using this result, the streamline is shown in Fig. b.

Ans: y = 4 ( x 1>2 - 1 ) 232

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*3–12. A flow field for gasoline is defined by u = 1 8y 2 m>s, and v = 12x2 m>s where x and y are in meters. Determine the equation of the streamline that passes through point (2 m, 4 m). Draw this streamline.

SOLUTION Using the definition of the slope of streamline and initial condition y = 4 m at x = 2 m, dy v = ; dx u

dy 2x = dx 8>y y

x dy 1 = x dx 4 L2 m L4 m y

ln y `

y 4m

ln

=

1 2 x x ` 8 2m

y 1 = (x 2 - 4) 4 8 1

2

y = 4e8(x

- 4)

Ans.

The values of x and the corresponding values of y are tabulated below: x(m)

0

{1

{2

{3

{4

4

y(m)

2.43

2.75

4

7.47

17.9

55.2

The plot of y vs x is shown in Fig. a.

y(m) 60 50 40 30 20 10 –5

–4

–3

–2

–1

0

1

2

3

4

5

x(m)

(a)

Ans: 1

2

y = 4e8 (x

- 4)

233

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3–13. A fluid has velocity components of u = 13x 2 + 12 m>s and v = 14txy2 m>s, where x and y are in meters and t is in seconds. Determine the streamlines that pass through point (1 m, 3 m) at times t = 1 s and t = 1.5 s. Plot these streamlines for 0 … x … 5 m.

SOLUTION Since the velocity components are functions of time and position, the flow can be classified as unsteady nonuniform flow. The slope of the streamline is dy v = ; dx u y

dy 4txy x b = = 4tya 2 dx 3x 2 + 1 3x + 1

y

3m

ln ln

=

60 50

t 5 1.5 s

40

x dy x bdx = 4t a 2 L3 m y L1 m 3x + 1

ln y `

y(m)

30 20

x 2t ln (3x 2 + 1) ` 3 1m

t51s

10

y 2t 3x 2 + 1 = ln a b 3 3 4

0

1

2

3

4

5

x(m)

(a)

y 3x 2 + 1 2t>3 = ln c a b d 3 4

For t = 1 s,

y = c 3a

3x 2 + 1 2t>3 b d m 4

For t = 1.5 s,

y = c 3a

3x 2 + 1 2>3 b d m 4

Ans.

3 y = c (3x 2 + 1) d m 4

Ans.

The values of x and the corresponding values of y are tabulated below. For t = 1 s, x(m)

0

1

2

3

4

5

y(m)

1.19

3

6.58

11.0

15.9

21.4

x(m)

0

1

2

3

4

5

y(m)

0.75

3

9.75

21

36.75

57

For t = 1.5 s,

The plot of these streamlines are shown in Fig. a.

Ans: For t = 1 s, y =

3 3 1 3x

For t = 1.5 s, y = 234

M03_HIBB9290_01_SE_C03_ANS.indd 234

2

2 4 m,

+ 1 2>3 4 2

3 34 13x

+ 124 m

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3–14. A particle travels along the streamline defined by y2 + 3 = 2x, where x and y are in meters. If its speed is 3 m>s when it is at x = 3 m, y = 13 m, determine the x and y components of its velocity at this point. Sketch the velocity on the streamline.

SOLUTION The values of x and the corresponding values of y are tabulated below. x(m)

1.50

2

3

4

5

6

y(m)

0

1

1.732

2.24

2.65

3

The plot of the streamline is shown in Fig. a. Taking the derivative of the streamline equation, 2y

dy = 2 dx dy 1 = dx y

At x = 3 m, y2 + 3 = 2(3)

y = 1.732 m

Then dy 1 ` = tan u ` = ; dx x = 5 m 1.732 x=3 m

u = 30°

Therefore the x and y components of velocity are

u = (3 m>s) cos 30° = 2.598 m>s = 2.60 m>s

Ans.

v = {(3 m>s) sin 30° = {1.50 m>s

Ans.

y(m) 3

3 mys v 5 1.50 mys

u 5 30°

2

u 5 2.60 mys

1 0

1

2

3

4

5

6

x(m)

–1 –2 –3 (a)

Ans: u = 2.60 m>s v = {1.50 m>s 235

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3–15. 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 236

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*3–16. A flow field is defined by u = 13x2 ft>s and v = 16y2 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 = (6 y) ft/s

dy = tan u dx

streamline

!

dy 6y v = = dx u 3x dy

L 2y

=

V

u = (3 x) ft/s

y

x

dx 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.

237

M03_HIBB9290_01_SE_C03_ANS.indd 237

Ans: y = x 2 >9

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3–17. The velocity for an oil flow is defined by V = 10.5y2i + 2j2 m>s, where y is in meters. What is the equation of the streamline that passes through point (1 m, 3 m)? If a particle is at this point when t = 0, at what point is it located when t = 2 s?

SOLUTION Since the velocity components are a function of position only, the flow can be classified as steady nonuniform flow. Here u = (0.5y2) m>s and v = 2 m>s. The slope of the streamline is defined by dy v = ; dx u

dy 2 4 = = 2 dx 0.5y2 y y

L3 m

y2dy = 4

x

L1 m

dx

x y3 y ` = 4x ` 3 3m 1m

y3 = 12x + 15

(1)

Ans.

From the definition of velocity, dy = 2 dt y

L3 m

dy =

y`

y 3m

L0

= 2t `

2s

2 dt 2s 0

y - 3 = 4

Ans.

y = 7m Substitute this result into Eq. (1). 73 = 12x + 15

Ans.

x = 27.33 m = 27.3 m

Ans: y3 = 12x + 15 y = 7m x = 27.3 m 238

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3–18. A fluid has velocity components of u = [30>12x + 12] m>s and v = 12ty2 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, dx 30 = u = ; dt 2x + 1

x

t

(2x + 1)dx = 30

L2 m

dt L2 s

x

t

( x2 + x ) `

= 30 t `

2m

2

2s

x + x - 6 = 30(t - 2) t =

dy = v = 2ty; dt

1 2 ( x + x + 54 ) 30

(1)

y

t dy = 2 t dt L6 m y L2 s

ln y `

y 6m

ln

= t2 `

t

y(m)

2s

y = t2 - 4 6

50

y 2 = et - 4 6

40

2

y = 6et

-4

(2)

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

30 20

y = 6e

1 900

(x2 + x + 54)2 - 4

Ans.

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

0

1

2

3

x(m)

4

(a )

Ans: 1

2

y = 6e 900 1x

+ x + 5422 - 4

239

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3–19. A particle travels along a streamline defined by y3 = 8x - 12, where x and y are in meters. 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 = dx 3y2

When x = 1 m, y3 = 8(1) - 12;

y = -1.5874

Then dy 8 ` = tan u ` = ; 2 dx x = 1 m 3( 1.5874) 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)

240

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*3–20. field u meters. for 0 …

The circulation of a fluid is defined by the velocity = 1 3 - 12x 2 m>s and v = 4 m>s, where x is in Plot the streamline that passes through the origin x 6 6 m.

SOLUTION Since the velocity component is a function of position only, the flow can be classified as steady nonuniform flow. Using the definition of the slope of a streamline, dy v = ; dx u

dy 4 = dx 3 - 12 x L0

y

x

dy = 4

L0 3 -

dx 1 2x

y = -8 ln a3 y = c 8 ln a y = c 8 ln a

3

3 -

x 1 xb ` 2 0 1 2x

bd m

6 bd m 6 - x

Ans.

The values of x and the values of corresponding y are tabulated below. x(m) y(m)

0 0

1 1.46

2 3.24

3 5.55

4 8.79

5 14.33

6 ∞

The plot of this streamline is shown in Fig. a. y(m) 20 15 10 5 0

1

2

3

4

5

6

x(m)

(a)

Ans: y = c 8 ln a

6 bd m 6 - x

241

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3–21. The circulation of a fluid is defined by the velocity field u = 16 - 3x2 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 flow. Using the definition of the slope of a streamline, dy v = ; dx u

dy 2 = dx 6 - 3x L0

y

x

dx L0 6 - 3x x 2 y = - ln (6 - 3x) ` 3 0

dy = 2

6 - 3x 2 y = - ln a b 3 6 y=

2 2 ln a b 3 2 - x

Ans.

The plot of this streamline is show in Fig. a.

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=

2 2 ln a b 3 2 - x

242

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3–22. 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.

SOLUTION

y(m)

Using the definition of velocity, for 0 … t 6 10 s, dx = u; dt

B

dx = -2 dt L0

x

dx = - 2

L0

dt

10 A

(1) –20

When t = 10 s, x = - 2(10) = - 20 m.

–15

dy = 3 dt L0

y

dy = 3

C

20

t

x = ( -2t) m

dy = v; dt

30

–10

–5

0

5

x(m)

(a)

L0

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 2

Ans.

For 10 6 t … 15 s, dx = u; dt

dx = 5 dt t

x

L-20 m

dx = 5

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

t

dy = -2

dt L10 s

y - 30 = - 2(t - 10) y = ( -2t + 50) m

(4)

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.

Ans: 3 For 0 … t 6 10 s, y = - x 2 2 For 10 s 6 t … 15 s, y = - x + 22 5

243

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3–23. A two-dimensional flow field for benzene is defined by V = 31y2 - 62i + 12x + 32j4 m>s, where x and y are in meters. Determine a streamline that passes through point 12 m, 4 m2 and the velocity at this point. Sketch the velocity on the streamline.

SOLUTION Since the velocity component is a function of position only, the flow can be classified as steady nonuniform flow. Using the definition of the slope of a streamline, dy v = ; dx u

dy 2x + 3 = 2 dx y - 6

y

L4 m

(y2 - 6) dy =

a

x

(2x + 3)dx L2 m

y x y3 - 6yb ` = (x2 + 3x) ` 3 4m 2m

y3 - 18y = 3x2 + 9x - 38

Ans.

The values of y and the corresponding values of x are tabulated below. y(m)

0

1

2

3

4

5

6

2.45

x(m)

2.36

1.54

0.863

0.932

2

3.66

5.64

0.762

The plot of this streamline is shown in Fig. a. The x and y components of velocity at point (2 m, 4 m) are u = 42 - 6 = 10 m>s

S

v = 2(2) + 3 = 7 m>s c Thus, the magnitude of this velocity is V = 2u2 + v2 = 2(10 m>s)2 + (7 m>s)2 = 12.20 m>s = 12.2 m>s

Ans.

and its direction is defined by

7 m>s v u = tan -1 a b = tan -1 a b = 34.99° = 35.0° u 10 m>s

Ans.

The sketch of this velocity on the streamline is shown in Fig. a.

y(m) 6 5

v 5 7 mys

V 5 12.2 mys

u 5 35.0°

4

u 5 10 mys

3 2

Ans:

1 0

1

2

3

4

5

6

y3 - 18y = 3x2 + 9x - 38

x(m)

V = 12.2 m>s

(a)

u = 35.0° 244

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*3–24. Air flows uniformly through the center of a horizontal duct with a velocity of V = 1 14t 3 + 3 2 m>s, where t is in seconds. Determine the acceleration of the flow when t = 3 s.

SOLUTION Since the flow is along the horizontal (x axis), v = w = 0 and u = V. Also, the velocity is a function of time t only. Therefore, the convective acceleration is zero, 0V so that V = 0. 0x 0V 0V a = + V 0t 0x 3 2 = t + 0 4

When t = 3 s,

3 = a t 2 b m>s2 4 a =

3 2 ( 3 ) = 6.75 m>s2 4

Ans.

Note: The flow is unsteady since its velocity is a function of time.

Ans: a = 6.75 m>s2 245

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3–25. Oil flows through the reducer such that particles along its centerline have a velocity of V = 16x + 4t2 in.>s, where x is in inches and t is in seconds. Determine the acceleration of a particle at x = 8 in. if it arrives when t = 3 s.

24 in.

x

SOLUTION Since the flow is along x axis, v = w = 0 and u = V. 0u 0u + u 0t 0x = 4 + (6x + 4t)(6)

a =

= 4 + 12(3x + 2t) = When t = 3 s, x = 8 in. Then

5431

+ 3(3x + 2t) 4 6

a = 4 5 1 + 3 3 3(8) + 2(3) 4 6 in.>s2 = 364 in.>s2

Ans.

Note: The flow is unsteady since its velocity is a function of time.

Ans: a = 364 in.>s2 246

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1 2 x yt 2 ft>s 3–26. A fluid has velocity components of u = 1 16 and v = 16x - t2 ft>s where x and y are in feet and t is in seconds. Determine the magnitude of acceleration of a particle passing through the point (2 ft, 1 ft) if it arrives when t = 2 s.

SOLUTION 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. 0u 0u 0u ax = + u + v 0t 0x 0y =

1 1 1 2 1 2 xy + x yta xytb + (6x - t) a x 2tb 16 16 8 16

= c

ay =

1 2 1 x ay + y2t 2 + 6 xt - t 2 b d ft>s2 16 8

0v 0v 0v + u + v 0t 0x 0y

= -1 + a

1 2 x ytb(6) + (6x - t)(0) 16

3 = a x 2yt - 1b ft>s2 8

When t = 2 s, x = 2 ft and y = 1 ft. ax = ay =

1 2 1 ( 2 ) c 1 + (2) ( 12 )( 22 ) + 6(2)(2) - 22 d = 5.5 ft>s2 16 8 3 2 ( 2 ) (1)(2) - 1 = 2 ft>s2 8

Thus, the magnitude of acceleration is a = 2a 2x + a 2y = 2 ( 5.5 ft>s2 ) 2 + ( 2 ft>s2 ) 2 = 5.85 ft>s2

Ans.

Ans: a = 5.85 ft>s2 247

M03_HIBB9290_01_SE_C03_ANS.indd 247

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3–27. A fluid flow is defined by u = 16x 2 - 3y2 2 m>s and v = 14xy + y2 m>s, where x and y are in meters. Determine the magnitudes of the velocity and acceleration of a particle at point (2 m, 2 m).

SOLUTION Since the velocity components are a function of position only, the flow can be classified as steady nonuniform flow. The x and y velocity components of the particles at x = 2 m and y = 2 m are u = 6 ( 22 ) - 3 ( 22 ) = 12 m>s v = 4(2)(2) + 2 = 18 m>s The magnitude of the particle’s velocity is V = 2u2 + v2 = 2(12 m>s)2 + (18 m>s)2 = 21.63 m>s = 21.6 m>s

Ans.

The x and y components of the particle’s acceleration, with w = 0, are ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + ( 6x 2 - 3y2 ) (12x) + (4xy + y)( - 6y) = ( 72x 3 - 60xy2 - 6y2 ) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 6x 2 - 3y2 ) (4y) + (4xy + y)(4x + 1) = (40x 2y - 12y3 + 8xy + y) m>s2 The magnitude of the particle’s acceleration is a = 2a x2 + a y2 = 2 ( 72 m>s2 ) + ( 258 m>s2 ) = 267.86 m>s2 = 268 m>s2

Ans.

At x = 2 m and y = 2 m,

ax = 72 ( 23 ) - 60(2) ( 22 ) - 6 ( 22 ) = 72 m>s2 ay = 40 ( 22 ) (2) - 12 ( 23 ) + 8(2)(2) + 2 = 258 m>s2

Ans: V = 21.6 m>s a = 268 m>s2 248

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* 3–28. A fluid flow is defined by u = 13x 2 - 4y2 m>s and v = 16xy2 m>s, where x and y are in meters. Determine the velocity and acceleration of particles passing through point (1 m, 2 m).

SOLUTION Since the velocity components are a function of position only, the flow can be classified as steady nonuniform flow. The x and y velocity components of the particles at point x = 1 m and y = 2 m are u = 3 ( 12 ) - 4(2) = -5 m>s = 5 m>s d v = 6(1)(2) = 12 m>s c The magnitude of the velocity is V = 2u2 + v2 = 2 ( - 5 m>s ) 2 + ( 12 m>s ) 2 = 13.0 m>s

Ans.

Its direction is 12 m>s v uv = tan-1 a b = tan-1 a b = 67.4° u 5 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, 0u 0u 0u ax = + u + v 0t 0x 0y = 0 + (3x 2 - 4y)(6x) + 6xy( - 4) = ay =

3 6x(3x2

- 8y) 4 m>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + (3x 2 - 4y)(6y) + 6xy(6x) =

3 6y(9x2

At point x = 1 m and y = 2 m,

- 4y) 4 m>s2

ax = 6(1) 3 3 ( 12 ) - 8(2) 4 = - 78 m>s2 = 78 m>s2 d ay = 6(2) 3 9 ( 12 ) - 4(2) 4 = 12 m>s2 c

The magnitude of the acceleration is

a = 2a x2 + a y2 = 2 ( - 78 m>s2 ) 2 + ( 12 m>s2 ) 2 = 78.92 m>s2 = 78.9 m>s2 Ans.

Its direction is

ua = tan - 1 a

12 m>s2 ay b = tan-1 a b = 8.75° ax 78 m>s2

Ans.

Ans: V = 13.0 m>s uv = 67.4° a = 78.9 m>s2 ua = 8.75°

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3–29. A fluid flow is defined by u = 10.4x 2 + 2t2 m>s and v = 10.8x + 2y2 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 x = 2 m, y = 1 m if it arrives when t = 3 s.

SOLUTION Since the velocity components are functions of time and position the flow can be classified as unsteady nonuniform flow. The x and y velocity components of the particles at point x = 2 m and y = 1 m, where t = 35, are u = 0.4 ( 22 ) + 2(3) = 7.60 m>s S v = 0.8(2) + 2(1) = 3.60 m>s c The magnitude of the velocity is V = 2u2 + v2 = 2 ( 7.60 m>s ) 2 + ( 3.60 m>s ) 2 = 8.41 m>s

Ans.

Its direction is 3.60 m>s v uv = tan-1 a b = tan-1 a b = 25.3° u 7.60 m>s

Ans.

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. 0u 0u 0u ax = + u + v 0t 0x 0y = 2 + (0.4x 2 + 2t)(0.8x) + (0.8x + 2y)(0) = (0.32x 3 + 1.6xt + 2) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + (0.4x 2 + 2t)(0.8) + (0.8x + 2y)(2) =

3 0.32x2

+ 1.6x + 4y + 1.6t 4 m>s2

When t = 3 s, x = 2 and y = 1 m.

ax = 0.32 ( 23 ) + 1.6(2)(3) + 2 = 14.16 m>s2 S ay = 0.32 ( 22 ) + 1.6(2) + 4(1) + 1.6(3) = 13.28 m>s2 c The magnitude of the acceleration is a = 2a x2 + a y2 = 2 ( 14.16 m>s2 ) 2 + ( 13.28 m>s2 ) 2 = 19.4 m>s2

Ans.

Its direction is

ua = tan-1 a

13.28 m>s2 ax b = tan-1 a b = 43.2° ay 14.16 m>s2

Ans. Ans: V = 8.41 m>s uv = 25.3°

a = 19.4 m>s2

ua = 43.2° 250

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3–30. A fluid flow is defined by u = 14xy2 ft>s and v = 13y2 ft>s, where x and y are in feet. Determine the equation of the streamline passing through point (1 ft, 2 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, dy dy 3y v 3 = ; = = dx u dx 4xy 4x y

L2 ft y`

x

dy = y 2 ft

=

y - 2 =

3 dx 4 L1 ft x x 3 ln x ` 4 1 ft

3 ln x 4

3 y = a ln x + 2b ft 4

Ans.

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. 0u 0u 0u ax = + u + v 0t 0x 0y = 0 + (4xy)(4y) + 3y(4x) = ay =

3 4xy(4y

+ 3) 4 ft>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + (4xy)(0) + 3y(3) = (9y) ft>s2 At x = 1 ft, y = 2 ft, ax = 4(1)(2) 3 4(2) + 3 4 = 88 ft>s2 S

ay = 9(2) = 18 ft>s2 c

The magnitude of the acceleration is

a = 2a x2 + a y2 = 2 ( 88 ft>s2 ) 2 + ( 18 ft>s2 ) 2 = 89.8 ft>s2

Ans.

Its direction is

u = tan-1 a

ay ax

b = tan-1 a

18 ft>s2 88 ft>s2

Ans.

b = 11.6°

Ans: Steady 3 y = a ln x + 2b ft 4 a = 89.8 ft>s2 u = 11.6° 251

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24 in.

3–31. Oil flows through the reducer such that particles along its centerline have a velocity of V = 14xt2 in.>s, where x is in inches and t is in seconds. Determine the acceleration of a particle at x = 16 in. if it arrives when t = 2 s. 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

Ans.

Note: The flow is unsteady since its velocity is a function of time.

Ans: a = 1088 in.>s2 252

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*3–32. A fluid flow is defined by u = 1 14y2 2 m>s and 1 2 v = 1 16 x y 2 m>s, where x and y are in meters. Determine the equation of the streamline passing through point (2 m, 3 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, dy v = ; dx u

dy = dx y

1 2 16 x y 1 2 4y x

=

x2 4y

1 y dy = x 2dx 4 L2 m L3 m y2 y 1 3 x y` = (x ) ` 2 3m 12 2m 1 3 2 y = (x + 46) 6

Ans.

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

1 1 1 = 0 + a y2 b(0) + a x 2yb a yb 4 16 2 1 2 2 = x y m>s 32 0v 0v 0v ay = + u + v 0t 0x 0y 1 1 1 1 = 0 + a y2 b a xyb + a x 2yb a x 2 b 4 8 16 16 xy 2 3 2 (8y + x ) m>s = 256 At point x = 2 m and y = 3 m, ax =

1 2 2 (2 )(3 ) = 1.125 m>s2 S 32 2(3)

[8(32) + 23] = 1.875 m>s2 c 256 The magnitude of the acceleration is ay =

a = 2a x2 + a y2 = 2 ( 1.125 m>s2 ) 2 + ( 1.875 m>s2 ) 2 = 2.19 m>s2

Ans.

Its direction is

u = tan-1 a

ay ax

b = tan-1 a

1.875 m>s2 1.125 m>s2

Ans.

b = 59.0°

Ans:

1 3 (x + 46) 6 a = 2.19 m>s2 u = 59.0° y2 =

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3–33. A fluid flow is defined by V = 52xi + 5j6 m>s, where x is in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 12 m, 1 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

SOLUTION

y(m)

Since the velocity components are a function of position but not time, the flow can be classified as steady nonuniform flow. The x and y velocity components at the point x = 2 m and y = 1 m are

4

S

u = 2(2) = 4 m>s

2

v = 5 m>s c

1

The magnitude of the velocity is 2

2

2

2

V = 2u + v = 2(4 m>s) + (5 m>s) = 6.40 m>s

Ans.

0

u 5 4 mys 1

2

3

4

5

x(m)

–1

For two-dimensional flow, the Eulerian description gives a =

3

V 5 6.40 mys v 5 5 mys

0V 0V 0V + u + v 0t 0x 0y

(a)

Write the scalar components of this equation along the x and y axes. ax =

0u 0u 0u + u + v 0t 0x 0y

y(m) 4

= 0 + 2x(2) + (5)(0) 2

3

= (4x) m>s 0v 0v 0v ay = + u + v 0t 0x 0y

2

At point (2 m, 1 m),

ax = 4(2) = 8 m>s2 S

a 5 8 mys2

1

= 0 + (2x)(0) + 5(0) = 0

0

ay = 0

1

2

3

4

5

x(m)

–1

Thus, the magnitude of acceleration is a = ax = 8 m>s2

Ans.

(b)

Using the definition of the slope of the streamline, dy v = ; dx u

dy 5 = dx 2x y

x

L1 m

dy =

y`

y 1m

=

y - 1 =

5 dx 2 L2 m x x 5 ( ln x) ` 2 2m

5 x ln 2 2

5 x y = a ln + 1b m 2 2

Ans.

The values of x and the corresponding values of y are tabulated below. x(m)

1

2e -2>5 0 y(m) - 0.733

2

3

4

5

1

2.01

2.73

3.29

The sketches of velocity and acceleration on the streamline at point (2 m, 1 m) are shown in Figs. a and b, respectively.

Ans: V = 6.40 m>s a = 8 m>s2 5 x y = a ln + 1b m 2 2

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3–34. A fluid flow is defined by u = 10.5y2 m>s and v = 1 - 0.3x2 m>s, where x and y are in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 14 m, 3 m2. Find the equation of the streamline passing through this point, and show the velocity and acceleration at the point on this streamline.

SOLUTION

y(m)

Since the velocity components are a function of position not time, the flow can be classified as steady nonuniform flow. The x and y velocity components at the point x = 4 m and y = 3 m are

5 4

U 5 1.50 mys

3

u = 0.5(3) = 1.50 m>s S

v 5 1.20 mys

2

v = - 0.3(4) = - 1.20 m>s = 1.20 m>s T

1

The magnitude of the velocity is V = 2u2 + v2 = 2(1.50 m>s)2 + ( -1.20 m>s)2 = 1.92 m>s

Ans.

0

1

2

3

4

5

x(m)

6

(a)

For two-dimensional flow, the Eulerian description gives a =

V 5 1.92 mys

0V 0V 0V + u + v 0t 0x 0y y(m)

Write the scalar components of this equation along the x and y axes.

5

0u 0u 0u + u + v ax = 0t 0x 0y

4

= 0 + (0.5y)(0) + ( -0.3x)(0.5)

ay 5 0.45 mys2

3

= ( - 0.15x) m>s2 0v 0v 0v ay = + u + v 0t 0x 0y

2 1

= 0 + (0.5y)( - 0.3) + ( -0.3x)(0) = ( - 0.15y) m>s

ax 5 0.600 mys2

0

2

a 5 0.75 mys2 1

2

3

4

5

6

x(m)

(b)

At x = 4 m and y = 3 m, ax = - 0.15(4) = - 0.6 m>s2 = 0.600 m>s2 2

ay = - 0.15(3) = - 0.45 m>s = 0.450 m>s

d 2

T

The magnitude of the acceleration is a = 2a x2 + a y2 = 2( - 0.6 m>s2)2 + ( - 0.45 m>s2)2 = 0.750 m>s2

Ans.

Using the definition of the slope of the streamline, dy v = ; dx u

dy -0.3x = dx 0.5y y

5

L3 m

5a

y dy = -3

x

x dx L4 m

y2 y x2 x b` = -3a b ` 2 3m 2 4m

5y2 + 3x 2 = 93

Ans.

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3–34. (Continued)

The values of x and the corresponding values of y are tabulated below. x(m)

0

1

2

3

4

5

5.57

y(m)

4.31

4.24

4.02

3.63

3

1.90

0

The sketches of velocity and acceleration on the streamline at point (4 m, 3 m) are shown in Fig. a and b, respectively.

Ans: V = 1.92 m>s a = 0.750 m>s2 5y2 + 3x 2 = 93 256

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3–35. A velocity field for oil is defined by u = 13y2 m>s, v = 10.6t 2 + 22 m>s, where t is in seconds and y is in meters. If a particle is at the origin when t = 0, determine its acceleration and position when t = 2 s.

SOLUTION Since the velocity components are a function of both position and time, the flow can be classified as unsteady nonuniform flow. Using the definition of velocity, dy = v = 0.6t 2 + 2; dt

L0

y

dy =

L0

t

(0.6t 2 + 2)dt

y = (0.2t 3 + 2t) `

t 0

3

y = (0.2t + 2t) m

dy = u = 3y = 3(0.2t 3 + 2t); dt

L0

x

dx = 3

L0

t

(0.2t 3 + 2t)dt

x = 33(0.05t 4 + t 2)4 m

When t = 2 s,

y = 0.2(23) + 2(2) = 5.60 m 4

Ans.

2

Ans.

x = 330.05(2 ) + 2 4 = 14.4 m

For two-dimensional flow, the Eulerian description gives 0V 0V 0V a= + u + v 0t 0x 0y Write the scalar components of this equations along the x and y axes. 0u 0u 0u ax = + u + v 0t 0x 0y = 0 + 3y(0) + (0.6t 2 + 2)(3) = 33(0.6t 2 + 2)4 m>s2

ay =

0v 0v 0v + u + v 0t 0x 0y

= 1.2t + 3y(0) + (0.6t 2 + 2)(0) = (1.2t) m>s2 When t = 2 s, ax = 330.6(2)2 + 24 = 13.2 m>s2 S ay = 1.2(2) = 2.40 m>s2 c The magnitude of the acceleration is a = 2a x2 + a y2 = 2 ( 13.2 m>s2 ) 2 + ( 2.40 m>s2 ) 2 = 13.4 m>s2

Ans.

And its direction is

u = tan-1 a

ay ax

b = tan-1 a

2.40 m>s2 13.2 m>s

2

Ans.

b = 10.3°

Ans: y = 5.60 m x = 14.4 m a = 13.4 m>s2 u = 10.3°

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*3–36. The velocity for the flow of a gas along the center streamline of the pipe is defined by u = 110x 2 + 200t + 62 m>s, where x is in meters and t is in seconds. Determine the acceleration of the particle at A if it is just leaving the nozzle when t = 0.01 s.

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

+

0u = 20x 0x

+ ( 10x 2 + 200t + 6 ) (20x) 4 m>s2

3 10 ( 0.62 )

= 339 m>s2

0u 0u + u 0t 0x

+ 200(0.01) + 6 4 320(0.6)4 6 m>s2

Ans.

Ans: a = 339 m>s2 258

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3–37. A fluid flow is defined by u = 15y2 2 m>s and v = 14x - 12 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 flow. The slope of the streamline is dy v = ; dx u

4

dy 4x - 1 = dx 5y2 y

3

x

L1 m

5y2dy =

y(m)

ay = 20 m/s2 a = 36.1 m/s2 2

L1 m

(4x - 1)dx

1 ( 6x2 - 3x + 2 ) where x is in m 5 For two-dimensional flow, the Eulerian description is

1

y3 =

ax = 30 m/s2 0

0V 0V 0V a = + u + v 0t 0x 0y

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 = 2a x2 + a y2 = 2 ( 30 m>s2 ) 2 + ( 20 m>s2 ) 2 = 36.1 m>s2

Ans.

Its direction is

u = tan-1 a

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°

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3–38. Air flowing through the center of the duct decreases 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 a particle as a function of time if x = 0 when t = 0.

B A VB 5 2 m!s

VA 5 8 m!s

x 3m

SOLUTION Since the velocity is a function of position only, the flow can be classified as steady nonuniform flow. Since the velocity varies linearly with x, V = VA + a

VB - VA 2 - 8 bx = 8 + a b x = (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

Ans.

Using the definition of velocity, dx = V = 8 - 2x; dt

x

t

dx = dt L0 8 - 2x 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

260

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Ans.

Ans: V = (8 - 2x) m>s a = 4(x - 4) m>s2 x = 411 - e -2t 2 m

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3–39. A fluid flow is defined by u = 18t 2 2 m>s and v = 17y + 3x2 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) if it arrives when t = 2 s.

SOLUTION Since the velocity components are functions of time and position, the flow can be classified as unsteady nonuniform flow. 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 uv = tan-1 a b = tan-1 a b = 17.4° 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 = 2a x2 + a y2 = 2 ( 32 m>s2 ) 2 + ( 166 m>s2 ) 2 = 169 m>s2

Ans.

Its direction is

ua = tan-1 a

ay ax

b = tan-1 a

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° 261

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*3–40. A fluid flow is defined by V = 5 4xi + 2j 6 m>s, where x is in meters. Determine the magnitude of the velocity and acceleration of a particle that passes through point 11 m, 2 m2. Find the equation of the streamline passing through this point, and show 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 flow. At point (1 m, 2 m), u = 4x = 4(1) = 4 m>s v = 2 m>s The magnitude of velocity is

y(m)

V = 2u2 + v2 = 2 ( 4 m>s ) 2 + ( 2 m>s ) 2 = 4.47 m>s

Ans. 3

For two-dimensional flow, the Eulerian description gives 0V 0V 0V a = + u + v 0t 0x 0y

2

Writing the scalar components of this equation along the x and y axes,

u = 4 mys

1

0u 0u 0u ax = + u + v 0t 0x 0y

0

= 0 + 4x(4) + 2(0) = 16x ay =

V = 4.47 mys

v = 2 mys

0v 0v 0v + u + v 0t 0x 0y

1

2

3 (a)

4

5

x(m)

y(m)

= 0 + 4x(0) + 2(0) = 0 3

At point (1 m, 2 m), ax = 16(1) = 16 m>s2

1

Thus, the magnitude of the acceleration is a = ax = 16 m>s2

Ans.

Using the definition of the slope of the streamline, dy v = ; dx u

a = 16 mys2

2

ay = 0

0

1

2

3 (b)

4

5

x(m)

dy 2 1 = = dx 4x 2x 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 = a ln x + 2b 2 262

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3–41. A fluid flow is defined by u = 12x 2 - y2 2 m>s and v = 1 - 4xy2 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 show 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 flow. At point (1 m, 1 m), u = 2x 2 - 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 + ( 2x 2 - y2 ) (4x) + ( -4xy)( - 2y) = 4x ( 2x 2 - y2 ) + 8xy2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x 2 - y2 ) ( - 4y) + ( - 4xy)( - 4x) = -4y ( 2x 2 - y2 ) + 16x 2y 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 = 2a x2 + a y2 = 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 = -4xy dx 2x 2dy + 4xy dx - y2dy = 0 However, d ( 2x 2y ) = 2 ( 2xy dx + x 2dy ) = 2x 2dy + 4xy dx. Then d ( 2x 2y ) - y2dy = 0

263

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3–41. (Continued)

Integrating this equation, 2x 2y -

y3 = C 3

with the condition y = 1 m when x = 1 m,

y(m)

13 2 ( 12 ) (1) = C 3 C =

3.0 2.5

5 3

2.0

Thus, 2x 2y -

1.5

y3 5 = 3 3

1.36 1.0

6x 2y - y3 = 5 x2 =

v = 4 mys

y + 5 6y

Ans. 0

0.5

3.0

dx Set = 0; dy

dy

=

2y3 - 5

x(m)

2.5 a = 17.0 m/s2

2 2.0 ay = 12 m/s

y = 1.357 m 1.5

u = 1 m/s

3 1.0 x 2 =1.36 1.357 + 5

1.5 1.0

v = 4 m/s

ax = 12 m/s2

V = 4.12 m/s

0.5

0.5

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) 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 0

2.0

3.0

2.5

x = 0.960 m

1.5

y(m)

12xy2

2y3 - 52.0 = 0 The corresponding x is

1.0 0.960 (a)

6y ( 3y2 ) - ( y3 + 5 ) (6) 2y3 - 5 dx = = 2 dy 6y2 ( 6y ) y(m) dx

V = 4.12 mys

0.5

3

Taking the derivative of this equation with respect to y, 2x

u = 1 mys

0.5

1.0

1.5

2.0

0

0.5

1.0

1.5

2.0

x(m)

0.960

Ans: V = 4.12 m>s a = 17.0 m>s2 y3 + 5 x2 = 6y 264

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3–42. A fluid flow is defined by u = 12y2 2 m>s and v = 18xy2 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 = (8 xy) m/s

dy 8xy 4x = = 2 dx y xy

dy v = ; dx u y

L2 m

y2 y 2 2

! u = (2 y2 ) m/s

x

y dy = `

2m

L1 m

streamline

4x dx

= 2x 2 `

y

x

x

1m

y - 2 = 2x 2 - 2 2 2

y = 4x

V

x (a)

2

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 + 64x 2y ) m>s2 At point x = 1 m and y = 2 m, ax = 32(1) ( 22 ) = 128 m>s2 ay =

3 16 ( 23 )

+ 64 ( 12 ) (2) 4 = 256 m>s2

The magnitude of the acceleration is

a = 2a x2 + a y2 = 2 ( 128 m>s2 ) 2 + ( 256 m>s2 ) 2 = 286 m>s2

Ans.

Its direction is

u = tan-1 a

ay ax

b = tan-1 a

256 m>s2 128 m>s2

b = 63.4°

Ans.

u

Ans: y = 2x a = 286 m>s2 u = 63.4°

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3–43. The velocity of gasoline, along the centerline of a tapered pipe, is given by u = 14tx2 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 = 4 t dt L2 m x L0.15 ln x `

x 2m

= 2 t2 `

t 0.15

x = 2t 2 - 0.02 2 x 2 e2t - 0.02 = 2 ln

2

x = 2e2t

- 0.02

x = 2e2(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.

Ans: a = 100 m>s2 266

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*3–44. A fluid flow is defined by u = 18x2 ft>s and v = 18y2 ft>s, where x and y are in feet. Determine the equation of the streamline 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 = (8 y) ft/s

y

x dy dx = y L1 ft L2 ft x

ln y `

y 1 ft

= ln x `

ln y = ln y=

V

! u = (8 x) ft/s x

y

x

x

2 ft

x 2

(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 = 2a x2 + a y2 = 2 ( 128 ft>s2 ) 2 + ( 64 ft>s2 ) 2 = 143 ft>s2

Ans.

Its direction is

u = tan-1 a

ay ax

b = tan-1 a

64 ft>s2 128 ft>s2

b = 26.6°

Ans.

u

Ans: y = x>2, a = 143 ft>s2 u = 26.6° u 267

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3–45. A fluid flow is defined by V = 5 4yi + 2xj 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 12 m, 1 m2. Find the equation of the streamline passing through this point, and show 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 = 2a x2 + a y2

= 2 ( 16 m>s ) 2 + ( 8 m>s ) 2 = 17.9 m>s2

Ans.

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 2 268

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3–45. (Continued)

The plot of this streamline is shown is Fig. a. x(m)

22 0

y(m)

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

1

2

3

4

5

6

x(m)

(a)

Ans: V = 5.66 m>s a = 17.9 m>s2 269

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3–46. As the valve is closed, oil flows through the nozzle such that along the center streamline it has a velocity of V = 310.5 + 0.8x 2 216 - 2t2 m>s, where x is in meters and t is in seconds. Determine the acceleration of an oil particle at x = 0.2 m if it arrives when t = 2 s.

A B

x 0.3 m

SOLUTION Here, V has only an x component. Then V = a. Since V is a function of time and position, the flow can be classified as unsteady nonuniform flow. Since v = w = 0, a = Here,

0u 0u + u 0t 0x

0u = 3(0.5 + 0.8x 2)( -2) = 3 - 6(0.5 + 0.8x 2)4 m>s2 0t 0u = 3(6 - 2t)(1.6x) = 4.8x(6 - 2t) 1>s 0x Thus, a = - 6(0.5 + 0.8x 2) + 3(0.5 + 0.8x 2)(6 - 2t)34.8x(6 - 2t)4 = 3 -6(0.5 + 0.8x 2) + 14.4x(0.5 + 0.8x 2)(6 - 2t)2 4 m>s2

When t = 2 s, the particle at x = 0.2 m has an acceleration of

a = -630.5 + 0.8(0.22)4 + 14.4(0.2)30.5 + 0.8(0.22)436 - 2(2)4 2 = -3.192 m>s2 + 6.129 m>s2 = 2.937 m>s2 = 2.94 m>s2

Ans.

Note that the local acceleration ( - 3.192 m>s2) causes the particle to decelerate, since the flow is being decreased due to the closing value. The convective acceleration (6.129 m>s2) causes the particle to accelerate, since the nozzle constricts as x increases. The net result is the particle is accelerating at 2.937 m>s2.

Ans: a = 2.94 m>s2 270

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3–47. Airflow through the duct is defined by the velocity field u = 12x 2 + 82 m>s and v = 1 - 8x2 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

x51m

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 a =

0V 0V 0V + u + v 0t 0x 0y

Writing the scalar components of this equation along the x and y axes: ax = = = ay =

0u 0u 0u + u + v 0t 0x 0y

30

+ ( 2x 2 + 8 ) (4x) + ( -8x)(0) 4

3 4x ( 2x2

+ 8 ) 4 m>s2

0v 0v 0v + u + v 0t 0x 0y

= 0 + ( 2x 2 + 8 ) ( -8) + ( -8x)(0) = At point (0, 0),

3 - 8 ( 2x2

+ 8 ) 4 m>s2

ax = 4(0) 3 2 ( 02 ) + 8 4 = 0

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 = 2a x2 + a y2 = 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°

Ans.

u

Using the definition of the slope of the streamline, dy v - 8x = = ; dx u 2x 2 + 8

L

dy = -8

x dx 2 2x + 8 L

y = - 2 ln ( 2x + 8 ) + C 2

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3–47. (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 2x + 8

Ans.

2

For the streamline passing through point (1 m, 0), 0 = - 2 ln3 2 ( 12 ) + 8 4 + C

y = c 2 ln a

For point (0, 0),

C = 2 ln 10 10 bd m 2x 2 + 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 –1 –5

–4

–3

–2

1 0

2

3

–1

4 y = 2 ln

5

x(m)

10 2x2 + 8

–2 –3 y = –2 ln –4

8 2x2 + 8

Ans: At point (0, 0), a = 64 m>s2 T At point (1 m, 0), a = 89.4 m>s2, u = 63.4° For the streamline passing through point (0, 0),

(a)

8 bd m 2x 2 + 8 For the streamline passing through point (1 m, 0), y = c 2 ln a y = c 2 ln a

10 bd m 2x + 8 2

272

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*3–48. 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

16 m

n

The n - s coordinate system is established with the 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

as s (a)

0V V2 b + 0t n R

= 0 +

( 5 m>s ) 2

16 m Thus, the magnitude of the particle’s acceleration is a = 2a s2 + a n2 = 2 ( 3 m>s2

= 1.5625 m>s2

) 2 + ( 1.5625 m>s2 ) 2

= 3.38 m>s2

Ans.

Ans: a = 3.38 m>s2 273

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3–49. Water flows into the drainpipe such that it only has a radial velocity component V = 1 -3>r2 m>s, where r is in meters. Determine the acceleration of a particle located at point r = 0.5 m, u = 20°. At t = 0, r = 1 m. s

r 5 0.5 m u

SOLUTION Fig. a is based on the initial condition (t = 0) that r0 = 1 m. 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 straight line. s r

r0 = 1 m (a)

Ans: a = 72 m>s2 274

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3–50. 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. r59m

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.

Ans: a = 4 m>s2 275

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3–51. A particle located at a point within a fluid flow has velocity components of u = - 2 m>s and v = 3 m>s, and acceleration components of ax = - 1.5 m>s2 and ay = -2 m>s2. Determine the magnitudes of the streamline and normal components of acceleration of the particle.

SOLUTION The direction of the tangent of the streamline is the same as that of velocity, which is given by v 3 u = tan-1 a b = tan-1 a b = 56.31° u 2 Thus, the unit vector that defines the direction of this tangent is us = -cos 56.31°i + sin 56.31°j = -0.5547i + 0.8321j The acceleration of the particle expressed as a Cartesian vector is a = 5 -1.5i - 2j6 m>s2

The magnitude of the acceleration is

a = 2a x2 + a y2 = 2( - 1.5 m>s2)2 + ( - 2 m>s2)2 = 2.50 m>s2

Thus, the component of the acceleration along the streamline is given by as = a # us = ( - 1.5i - 2j) # ( - 0.5547i + 0.8321j) = ( -1.5)( -0.5547) + ( -2)(0.8321) = - 0.8321 m>s2 = 0.832 m>s2

Ans.

The negative sign indicates that as is directed in the sense opposite to that of us. The normal component of the acceleration is an = 2a 2 - a s 2

= 2(2.50 m>s2)2 - (0.8321 m>s2)2

= 2.357 m>s2 = 2.36 m>s2

Ans.

Ans: as = 0.832 m>s2 an = 2.36 m>s2 276

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*3–52. 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

= 2.25 m>s2

as = 3 m/s2

an = 2.25 m/s2

! = 36.9º

Thus, the magnitude of the acceleration is a = 2a s2 + a n2 = 2 ( 3 m>s2 ) 2 + ( 2.25 m>s2 ) 2 = 3.75 m>s2

Ans.

And its direction is

4m

a = 3.75 m/s2

2

2.25 m>s an u = tan-1 a b = tan-1 a b = 36.9° as 3 m>s2

Ans. (a)

The plot of the acceleration on the streamline is shown in Fig. a.

Ans: a = 3.75 m>s2 u = 36.9° 277

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3–53. Air flows around the front circular surface. If the steady-stream velocity is 4 m>s upstream from the surface, and the velocity along the surface is defined by V = 116 sin u2 m>s, determine the magnitudes of the streamline and normal components of acceleration of a particle located on the surface at u = 30°.

V 4 m!s u

SOLUTION

0.5 m

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 278

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3–54. Fluid particles have velocity components of u = 12x2 m>s and v = 14y2 m>s, where x and y are in meters. Determine the acceleration of a particle located at point (2 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 flow. 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, ax =

0u 0u 0u + u + v 0t 0x 0y

= 0 + 2x(2) + 4y(0) = (4x) m>s2 ay =

0v 0v 0v + u + v 0t 0x 0y

= 0 + 2x(0) + 4y(4) = (16y) m>s2 At point x = 2 m and y = 1 m, ax = 4(2) = 8 m>s2 S ay = 16(1) = 16 m>s2 c The magnitude of the acceleration is a = 2a x2 + a y2 = 2 ( 8 m>s2 ) 2 + ( 16 m>s2 ) 2 = 17.89 m>s2 = 17.9 m>s2

Ans.

Its direction is u = tan-1 a

ay ax

b = tan-1 a

16 m>s2 8 m>s2

The slope of the streamline is

Ans.

b = 63.43° = 63.4° dy 4y 2y = = dx 2x x

dy v = ; dx u y

x dy dx = 2 y L1 m L2 m x

ln y `

y 1m

= 2(ln x) `

x 2m

x ln y = 2 ln 2

x 2 ln y = lnc a b d 2 x 2 y = a b 2

1 y = x2 4

Ans.

Ans: a = 17.9 m>s2 u = 63.4° y = 14x 2

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3–55. A fluid has velocity components of u = 14xy2 m>s and v = 12y2 m>s, where x and y are in meters. Determine the magnitudes of the streamline and normal components of acceleration of a particle located at point (2 m, 1 m).

SOLUTION Since the velocity components are independent of time, but a function of position, the flow can be classified as steady nonuniform flow. The direction of the tangent of the streamline is the same as that of velocity, which is given by 2y v 1 u = tan-1 a b = tan-1 a b = tan-1 a b u 4xy 2x

At point (2 m, 1 m),

u = tan-1 c

1 d = 14.04° 2(2)

Thus, the unit vector that defined the direction of this tangent is us = cos 14.04°i + sin 14.04°j = 0.9701i + 0.2425j For two-dimensional flow where w = 0, the Eulerian description is 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 + (4xy)(4y) + 2y(4x) = 16xy2 + 8xy = 38xy(2y + 1)4 m>s2

ay =

0v 0v 0v + u + v 0t 0x dy

= 0 + 4xy(0) + 2y(2) = (4y) m>s2 At the point (2 m, 1 m), ax = 8(2)(1)32(1) + 14 = 48 m>s2 ay = 4(1) = 4 m>s2

280

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3–55. (Continued)

The acceleration of the particle expressed as a Cartesian vector is a = 548i + 4j6 m>s2

The magnitude of the acceleration is

a = 2a x2 + a y2 = 2(48 m>s2)2 + (4 m>s2)2 = 48.17 m>s2

Thus, the component of the acceleration along the streamline is given by as = a # us = (48i + 4j) # (0.9701i + 0.2425j) = 48(0.9701) + 4(0.2425) = 47.54 m>s2 = 47.5 m>s2

Ans.

The normal component of the acceleration is given by an = 2a 2 - a s 2

= 2(48.17 m>s2)2 - (47.54 m>s2)2 = 7.761 m>s2 = 7.76 m>s2

Ans.

Ans: as = 47.5 m>s2 an = 7.76 m>s2 281

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4–1. Water flows along the triangular channel having the cross section shown. If the average velocity is 3 m>s, determine the volumetric discharge.

0.4 m 0.1 m 0.4 m

3 m!s

308

SOLUTION Referring to the geometry shown in Fig. a, the width b of the water surface is given by b 0.4 m = ; b = 0.32 m 0.4 m 0.5 m

0.4 m b

The vertical cross-sectional area of the flow (shown shaded in Fig. a) is 1 A = (0.32 m)(0.4 m) = 0.064 m2 and its positive direction is indicated on Fig. a. 2 Q = V#A

30° 0.4 m V 5 3 mys

30°

(a)

= 3(3 m>s) cos 30°4(0.064 m2) = 0.1663 m3 >s = 0.166 m3 >s

Ans.

282

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0.5 m

Ans: Q = 0.166 m3 >s

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4–2. Determine the mass flow of nitrogen in an 8-in.diameter duct if it has a velocity of 15 ft>s. The nitrogen has a temperature of 90°F, and the gage pressure is 10 psi. The atmospheric pressure is 14.7 psi.

SOLUTION From the table in Appendix A, R = 1775

8 in.

ft # lb for Nitrogen. Here the absolute slug # °R

pressure is p = patm + pg = 10 psi + 14.7 psi = a24.7

lb 12 in 2 b = 3556.8 lb>ft2 ba 1 ft in2

and absolute temperature is T = 90°F + 460 = 550°R. p = rRT 3556.8 lb>ft2 = ra1775

ft # lb b(550°R) slug # °R

r = 0.003643 slug>ft3 Here, the flow cross-sectional area is A = pa rate is given by

2 4 p ft b = ft2. Then, the mass flow 12 9

# m = rVA = (0.003643 slug>ft3)(15 ft>s) a

p 2 ft b 9

= 0.01908 slug>ft3 = 0.0191 slug>ft3

Ans.

Ans: # m = 0.0191 slug>ft3 283

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4–3. Nitrogen gas flows through the 8-in.-diameter duct. If it has a velocity of 30 ft>s and the gage pressure is maintained at 15 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 = 25°F. The atmospheric pressure is 14.7 psi.

SOLUTION From the table in Appendix A, R

8 in.

ft # lb = 1775 for Nitrogen. Here, the absolute slug # °R

pressure is p = patm + pg = 15 psi + 14.7 psi = a29.7

2

lb 12 in. b = 4276.8 lb>ft2 ba 1 ft in2

and the absolute temperature is T = TF + 460. Applying Eq. 1–11,

0.04

0.01 0

2.4095 b slug>ft3 TF + 460

is given by

0.05

0.02

ft # lb 4276.8 lb>ft = ra1775 b(TF + 460) slug # °R 2

Here, the flow cross-sectional area is A = pa

0.06

0.03

p = rRT

r = a

·

m (slug ys)

25

50

75

100

TF (°F)

(a)

2 4 p ft b = ft2. Then the mass flow rate 12 9

# m = rVA 2.4095 p # m = ca b slug>ft3 d (30 ft>s) a ft2 b TF + 460 9 25.2319 # m = a b slug>s TF + 460

where TF is in °F

# The values of TF and the corresponding values of m are tabulated below. TF (°F) # m (slug>s)

0

25

50

75

100

0.0549

0.0520

0.0495

0.0472

0.0451

# The plot of m vs TF is shown in Fig. a.

Ans.

284

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*4–4. 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 velocity at the exit. The exit has a diameter of 0.3 m.

0.3 m

SOLUTION

ft # lb for CO2 . Here, the absolute pressure slug # R lb 12 in. 2 b = 2980.8 lb>ft2 and + pg = 14.7 psi + 6 psi = a20.7 2 b a 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>ft3 # m = rVA

= ( 0.004977 slug>ft3 )( 20 ft>s ) (p) a = 8.69 ( 10-3 ) slug>s

2 2 ft b 12

Ans.

Ans: # m = 8.69 ( 10 - 3 ) slug>s 285

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4–5. Determine the mass flow of air in the duct if it has a velocity of 45 ft>s. The air has a temperature of 70°F, and the (gage) pressure is 6 psi. The atmospheric pressure is 14.7 psi. 18 in. 15 in.

SOLUTION

ft # lb for air. Here, the absolute slug # °R lb 12 in. 2 + pg = 14.7 psi + 6 psi = a20.7 2 b a b = 2980.8 lb>ft2 1 ft in

From the table in Appendix A, R = 1716 pressure is p = patm

and the absolute temperature is T = 70°F + 460 = 530°R. p = rRT 2980.8 lb>ft2 = ra1716

ft # lb b(530°R) slug # °R

r = 0.003277 slug>ft3 Here, the flow cross-sectional area is A = a mass flow rate is given by

18 15 15 2 ft b a ft b = ft . Then, the 12 12 8

# m = rVA = (0.003277 slug>ft3)(45 ft>s) a

15 2 ft b 8

Ans.

= 0.2765 slug>s = 0.277 slug>s

Ans: # m = 0.277 slug>s 286

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4–6. Air flows through the duct at a velocity of 50 ft>s. If the temperature is maintained at 70°F, plot the variation of the mass flow (vertical axis) versus the (gage) pressure for the range of 0 … p … 10 psi. Give values for increments of ∆ p = 2 psi. The atmospheric pressure is 14.7 psi.

18 in. 15 in.

SOLUTION

ft # lb for air. Here, the absolute slug # °R + pg = (pg + 14.7) psi and the absolute temperature is

From the table in Appendix A, R = 1716 pressure is p = patm

·

m (slug ys)

T = 70°F + 460 = 530°R.

0.4

p = rRT c (pg + 14.7)

0.3

lb 12 in 2 ft # lb b = ra1716 b(530°R) da 2 1 ft slug # °R in

0.2 0.1

-3

r = 30.1583(10 )(pg + 14.7)4 slug>ft

Here, the flow cross-sectional area is A = a mass flow rate is given by # m = rVA

3

0

18 15 15 2 ft b a ft b = ft . Thus, the 12 12 8

2

4

6

8

10

Pg (psi)

(a)

15 2 # m = 30.1583(10-3)(pg + 14.7) slug>ft3 4(50 ft>s) a ft b 8 # m = 30.01484(pg + 14.7)4 slug>s

where pg is in psi.

The values of pg and the corresponding values of in are tabulated below pg (psi) # m (slug>s)

0

2

4

6

8

10

0.218

0.248

0.278

0.307

0.337

0.367 Ans.

The plot of pg vs in is shown in Fig. a.

287

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4–7. 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 V

The velocity profile in Fig. a can be expressed as U max - 0 y - 0 = ; h h y 0 2 2

y = 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

h 2

a1 -

= 2wU max ay =

wU max h 2

Uma x

2 yb d (wdy) h 2 ybdy h h

y2 2 b` h 0

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 =

288

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*4–8. 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

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

V2

5 ft s

V # dA = the volume enclosed by the velocity profile 0.6 ft

0.3 ft 1

V2 = 5 ft>s V2

Then, V1 = 5 ft>s + V2 = 10 ft>s The volume of the cone can be computed by V = Q =

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 ft3 >s

The average velocity is V =

Q 1.05p = = 2.92 ft>s A p(0.6 ft)2

Ans.

Ans: V = 2.92 ft>s 289

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4–9. 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>ft3. 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

5 ft s

V # dA = 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 0.3 ft

1 The volume of the cone can be computed by V = pr 2h. Then, 3

2 (a)

1 1 Q = p(0.6 ft)2 ( 10 ft>s ) - p(0.3 ft)2 ( 5 ft>s ) 3 3

Then the mass flow is

V2

= 1.05p ft3 >s 3

54.7 lb>ft # m = rQ = a b ( 1.05p ft3 >s ) = 5.60 slug>s 32.2 ft>s2

Ans.

Ans: # m = 5.60 slug>s 290

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4–10. A fluid flowing between two plates has a velocity profile that is assumed to be parabolic, where 4umax u = (hy - y2). Determine the average velocity and h2 volumetric discharge in terms of umax. The plates have a width of w.

y

h –– 2

umax

h –– 2

u

SOLUTION y

The differential rectangular element of thickness dy shown shaded in Fig. a having base area of dA = wdy will be considered. Thus, Q =

=

=

=

=

LA L0

h 2

u dA

h

4umax h2

4wumax h2

h 2

(hy - y2)(wdy)

L0

u dy

u

h 2

(hy - y )dy

Umax (a)

y3 h 4wumax hy2 a b` 2 3 0 h2 2 whumax 3

Ans.

Also, since the velocity profile is a parabola, Q =

LA

V # dA = volume under the velocity diagram

2 = c (h)(umax ) d w 3 =

2 whumax 3

Ans.

Therefore 2 whumax Q 3 2 V = = = umax A wh 3

Ans.

Ans: Q =

2 2 whumax, V = umax 3 3

291

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4–11. Determine the average velocity V of the fluid if it has the velocity profile shown. The channel is 2 m wide.

y 0.9 m!s u

0.6 m y

SOLUTION

y

The velocity profile in Fig. a can be expressed as y u = ; 0.9 m>s 0.6 m

0.9 mys

3 u = a yb m>s 2

u

The differential rectangular element of thickness dy shown shaded in Fig. a having base area of dA = (2 m)dy will be considered. Thus, Q =

LA

y

L0 L0

0.6 m

=

=

3 2 2 0.6 m y 2 0

u

3 a yb(2 dy) 2

(a)

3y dy

= 0.540 m3 >s

Also,

dy

u dA

0.6 m

=

0.6 m

Ans.

V # dA = the volume under the velocity diagram L 1 = c (0.6 m)(0.9 m>s) d (2 m) 2

Q =

Therefore

= 0.540 m3 >s V =

0.540 m3 >s Q = = 0.450 m>s A (0.6 m)(2 m)

Ans.

Ans.

Ans: Q = 0.540 m3 >s V = 0.450 m>s 292

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*4–12. Determine the mass flow of the fluid if it has the velocity profile shown. The channel is 2 m wide. Take r = 1600 kg>m3.

y 0.9 m!s u

0.6 m y

SOLUTION

y

The velocity profile in Fig. a can be expressed as y u = ; 0.9 m>s 0.6 m

0.9 mys

2 u = a yb m>s 3

u

The differential rectangular element of thickness dy shown shaded in Fig. a having base area of dA = (2 m)dy will be considered. Thus # m =

LA

rvdA = (1600 kg>m3)

= 4800

L0

= 4800a

L0

0.6 m

0.6 m

0.6 m

dy

y

3 a yb(2 dy) 2

u

ydy

(a)

y2 0.6 m b2 2 0

Ans.

= 864 kg>s

Ans: # m = 864 kg>s 293

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4–13. The liquid in the rectangular channel has a velocity profile defined by u = 13y1>2 2 m>s, where y is in meters. Determine the volumetric discharge if the width of the channel is 2 m.

y

1

u 5 (3y 2 ) m!s

0.9 m

u

SOLUTION

y

The differential rectangular element of thickness dy shown shaded in Fig. a having base area of dA = (2 m)dy will be considered. Thus, Q =

=

LA L0

= 6

u

vdA

0.9 m

0.9 m

L0

2.846 mys

dy

y

(3y1>2)(2dy)

u

0.9 m

y1>2dy

(a)

0.9 m 2 = 6a y3>2 b 2 3 0

= 3.415 m3 >s = 3.42 m3 >s

Ans.

Also, when y = 0.9 m, u = 310.91>2 2 = 2.846 m>s. Since the velocity profile is a half-parabola, V # dA = The volume under the velocity diagram L 2 = c (0.9 m)(2.846 m>s) d (2 m) 3

Q =

= 3.415 m3 >s = 3.42 m3 >s

294

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Ans.

Ans: Q = 3.42 m3 >s

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4–14. The liquid in the rectangular channel has a velocity profile defined by u = (3y1>2) m>s, where y is in meters. Determine the average velocity of the liquid. The channel has a width of 2 m.

y

1

u 5 (3y 2 ) m!s

0.9 m

u

SOLUTION

y

The differential rectangular element of thickness dy shown shaded in Fig. a having base area of dA = (2 m)dy will be considered. Thus, Q =

=

LA L0

= 6

u

vdA

0.9 m

0.9 m

L0

2.846 mys

dy

y

(3y1>2)(2dy)

u

0.9 m

y1>2dy

(a)

0.9 m 2 = 6a y3>2 b 2 3 0

= 3.415 m3 >s

Also, when y = 0.9 m, u = 310.91>2 2 = 2.846 m>s. Since the velocity profile is a half-parabola, V # dA = The volume under the velocity diagram L 2 = c (2.846 m)(0.9 m) d (2 m) = 3.415 m3 >s 3

Q =

Thus, the average velocity is given by V =

3.415 m3 >s Q = = 1.897 m>s = 1.90 m>s A (0.9 m)(2 m)

Ans.

Ans: V = 1.90 m>s 295

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4–15. Water flows along the semicircular trough with an average velocity of 3.6 m>s. Determine the volumetric discharge.

0.1 m 0.4 m 3.6 m!s

SOLUTION

b

Referring to the geometry shown in Fig. a, 2

b = 2(0.4 m) - (0.1 m) = 0.3873 m u = cos -1 a

u A

0.4 m

0.1 m p rad b = (75.52°) a b = 1.3181 rad 0.4 m 180°

30°

V 5 3.6 mys

Thus, the vertical cross-sectional area of the flow (shown shaded in Fig. a) is given by A =

0.1 m

b u

2

308

(a)

1 1 (0.4 m)2 32(1.3181 rad)4 - 32(0.3873 m)4(0.1 m) 2 2

= 0.1722 m2

The positive direction of this area is indicated on Fig. a. Applying Eq. 4–6, Q = V#A = 3(3.60 m>s) cos 30°4(0.1722 m2) = 0.5368 m3 >s = 0.537 m3 >s

Ans.

296

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Ans: Q = 0.537 m3 >s

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*4–16. The 30-mm-diameter nozzle ejects water such that the stream strikes the wall at B, where h = 4 m. Determine the volumetric flow through the nozzle.

B A

h

30 8 1.2 m 12 m

SOLUTION The x-y coordinate system is established as shown in Fig. a. First, we will consider the horizontal motion. ( S+ )

y

(SB)x = (SA)x + (VA)xt 12 = 0 + VA cos 30°t t =

12 VA cos 30°

A

Then, vertical motion gives

30°

(SA)y 5 1.2 m 0

1 (SB)y = (SA)y + (VA)yt + act 2 2

(+ c)

B

VA

(1)

4 = 1.2 + VA sin 30°t +

(SB)x 5 12 m

(SB)y 5 4 m x

1 ( - 9.81)t 2 2

4.905t 2 - 0.5VAt + 2.8 = 0

(2)

Solving Eqs. (1) and (2), VA = 15.10 m>s

t = 0.9174 s

Thus, the volumetric flow rate is given by Q = VAAA = (15.10 m>s)3p(0.015 m)2 4

= 0.01068 m3 >s = 0.0107 m3 >s

Ans.

297

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Ans: Q = 0.0107 m3 >s

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4–17. 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 8 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 = 9.8 1 m s2

Ans.

y

VA 30˚

A

Sy = 3 m Sx = 24 (a)

298

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Ans: Q = 0.0294 m3 >s

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4–18. 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 8 B

A

3m R

ac = 9.81 m s2

SOLUTION The x-y coordinate system with origin at A is established as shown in Fig. a.

VA

Sy = 3 m

30˚

Horizontal Motion.

( d+ )

A

sx = (s0)x + (VA)x t R = 0 + (VA cos 30°)t t =

VA

Vertical Motion.

( + c)

y

Sx = R

R 2R = cos 30° 23VA

sy = (s0)y + (VA)y t + -3 m = 0 + (VA

(1) (a )

1 2 at 2 c

Q(m3 s) 0.03

1 sin 30°)t + ( -9.81 m>s2 ) t 2 2

0.02

4.905t 2 - 0.5VAt - 3 = 0

(2)

Substitute Eq (1) into (2) 4.905 a

2R 23VA

Thus, the discharge is Q = VA;

Q = £

2

b - 0.5VA a

VA = a

0

2R 23VA

1

2 6.54R2 b 0.5774R + 3

Q = a 0.00502R 1

(0.5774R + 3)2

0.01 5

10

15

20

25

R(m)

(b)

b - 3 = 0

1

2 6.54R2 b 3 p(0.025 m)2 4 0.5774R + 3

§ m3 >s where R is in m

Ans.

The plot of Q vs R is shown in Fig. b R(m)

0

5

10

15

20

25

Q(m3 >s)

0

0.0103

0.0170

0.0221

0.0263

0.0301

Ans:

299

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Q = c

0.00502R (0.577R + 3)1>2

d m3 >s where R is in m

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4–19. The human heart has an average discharge of 0.1110-3 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.

300

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Ans: n = 7.071109 2

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*4–20. Rain falls vertically with an average speed of 15 ft>s and accumulates in the bin. It is observed that the level of water in the bin remains constant when the water flows out from the 2-in.-diameter pipe at 3 ft>s. Determine the amount of falling rainwater in a cubic foot of air. Also, if the average radius of a drop of rain is 0.14 in., determine the number of raindrops in a cubic foot of air. Hint: The volume of a drop is V = 43pr 3.

3 ft

2 ft

SOLUTION The discharge Qp of the pipe, which is the volume discharged in 1 s, is equal to the volume of the rain water contained in the air of volume equal to that of the volume shown in Fig. a. Here Qp = VpAp = (3 ft>s) c pa

And the volume of the air is

2 1 p 3 ft b d = ft >s 12 48

Qa = (3 ft)(2 ft)(15 ft>s) = 90 ft3 >s

v 5 15 ftys

p 3 In other words, 90 ft3 of air contains ft of rain water. Therefore, for 1 ft3 of air, it 48 contains p 3 ft ° 48 ¢ V = (1 ft3) = 0.7272(10-3) ft3 = 0.727(10-3) ft3 90 ft3

3 ft

2 ft (a)

Ans.

Then the number of rain drops in 1 ft3 of air is n =

0.7272(10-3) ft3 3 4 0.14 pa ft b 3 12

Ans.

= 109.32 = 109

Ans: V = 0.727(10-3) ft3 n = 109 301

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4–21. Air flows through the gap between the vanes at 0.75 m3 >s. Determine the 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.

208

408

200 mm VB

VA

408

B

208

A

SOLUTION The discharge can be calculated using Va # dA Lcs

Q =

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 m >s = (VB cos 20°) 3 (0.2 m)(0.4 m) 4

Ans.

3

VB = 9.98 m>s

Ans.

140˚ AA

20˚

40˚

AB VB

VA

(a)

Ans: VA = 12.2 m>s, VB = 9.98 m>s 302

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4–22. Kerosene flows through the nozzle at 0.25 m3 >s. Determine the time it takes for a particle on the x axis to travel from x = 0 to x = 0.2 m. Plot the distance-versustime graph for the particle, if x = 0 at t = 0. Use average velocities at the cross section for the solution.

300 mm 25 mm

100 mm

x

x

SOLUTION Since the flow is assumed to be one-dimensional and incompressible, its velocity can be determined using Q u = A

0.075 m

r

0.025 m

0.025 m

r = 0.025 m + a

Then, Q u = = A

0.30 0.25

2

p 1 (10 - 25x) m d = c (10 - 25x)2 d m2 100 10(103)

0.20 0.15 0.10

0.25 m3 >s

p (10 - 25x)2 d m2 10(103)

c

(a) x(m)

0.075 m 1 b 3(0.3 - x) m4 = c (10 - 25x) d m 0.3 m 100

Thus, the cross-sectional area of the nozzle as a function of x is A = pr 2 = pc

(0.3 2 x) m

x

From the geometry shown in Fig. a, the radius r, of the nozzle’s cross-section as a function of x is

2500 = c d m>s p(10 - 25x)2

0.05 0

Using the definition of velocity and the initial condition of x = 0 at t = 0,

0.005 0.010 0.015 0.020

t(s)

(b)

dx 2500 = u = dt p(10 - 25x)2 L0

t

x

dt =

p (10 - 25x)2dx 2500 L0 x

t = t = t =

p (625x 2 - 500x + 100)dx 2500 L0 p 625 3 a x - 250x 2 + 100xb 2500 3

p (25x 3 - 30x 2 + 12x) 300

(1)

When x = 0.2 m, t =

p 325(0.23) - 3.0(0.22) + 12(0.2)4 300

Ans.

= 0.01466 s = 0.0147 s

Using Eq. (1), the following tabulation can be computed and the plot of x vs t is shown in Fig. b. x(m) t(s)

0 0

0.05 0.00553

0.10 0.00969

0.15 0.01266

0.20 0.01466

0.25 0.01587

0.30 0.01649 Ans: t = 0.0147 s

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4–23. Kerosene flows through the nozzle at 0.25 m3 >s. Determine the velocity and acceleration of a particle on the x axis at x = 0.25 m. Use average velocities at the cross section for the solution.

300 mm 25 mm

100 mm

x

x

SOLUTION Since the flow is assumed to be one dimensional and incompressible, its velocity can be determined using u =

0.075 m

Q A

0.025 m

From the geometry shown in Fig. a, the radius r of the nozzle’s cross section as a function of x is r = 0.025 m + a

r 0.025 m (0.3 2 x) m

x (a)

0.075 m 1 b 3(0.3 - x) m4 = c (10 - 25x) d m 0.3 m 100

Thus, the cross-sectional area of the nozzle as a function of x is A = pr 2 = pc Then, u =

Q = A

At x = 0.25 m,

2 p 1 (10 - 25x) m d = c (10 - 25x)2 d m2 100 10(103)

0.25 m3 >s

p c (10 - 25x)2 d m2 10(103)

= c

2500 d m>s p(10 - 25x)2

2500 = 56.59 m>s = 56.6 m>s p310 - 25(0.25)4 2

u =

Ans.

The acceleration can be determined using a= Since Q is constant,

0u 0u + u 0t 0x

0u = 0, that is, no local change takes place 0t

Here

Then

125(103) 0u 2500 1 = 3( - 2)(10 - 25x) -3( - 25)4 = c d p 0x p(10 - 25x)3 s a= 0 + u = 0 + c = c

At x = 0.25 m, a=

0u 0x 125(103) 2500 d c d p(10 - 25x)2 p(10 - 25x)3

0.3125(109)

p2(10 - 25x)5

0.3125(109) 2

p 310 - 25(0.25)4 5

d m>s2

= 42.70(103) m>s2 = 42.7(103) m>s2

Ans.

Ans: u = 56.6 m>s a = 42.7(103) m>s2

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*4–24. At two specific instants during a heartbeat, the velocity profiles of blood passing through a straight portion of the aorta can be modeled by a cosine curve as shown. Determine the volumetric flow through the aorta for each instant.

r

r

R

u 5 a cos(

p r) 2R

u5

7p a [2 cos( r) 1 3 ] 2 6R

SOLUTION The differential ring element of area dA = 2πrdr shown shaded in Fig. a will be considered. The volumetric flow rate can then be determined. For v = a cos a R

dr

p br, 2R

dA r

p Q1 = V # dA = c a cos a r b d (2prdr) 2R L L0 = 2pa

L0

R

c r cos a

= 4Rac r sin a = For v =

p r b d dr 2R

(a)

R p 2R p rb + cos a r b d ` p 2R 2R 0

4(p - 2) 2 aR p

Ans.

a 7p c 2 cos a br + 23 d , 2 6R Q2 =

L

V # dA =

= pa

L0

= pac

R

R

9 7p c 2 cos a r b + 23 d (2prdr) 2 6R L0

c 2r cos a

7p r b + 23r d dr 6R

12R 7p 72R2 7p 23 2 R r sin a rb + cos a r b + r d` 7p 6R 6R 2 49p2 0

= 0.9908aR2 = 0.991aR2

Ans.

Ans:

4(p - 2) 2 aR p Q2 = 0.991aR2 Q1 =

305

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4–25. The radius of the circular duct varies as r = [0.06(1 - 1x)4 m, where x is in meters. The flow of water at A is Q = 0.003 m3 >s at t = 0, and it is increasing at dQ>dt = 0.003 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 = 0.3 m. Use average velocities at the cross section for the solution.

r 0.4 m

60 mm

x B A

x

SOLUTION The discharge as a function of time t is Q = 0.003 m3 >s + (0.003 m3 >s2) t = 30.003 (1 + t)4 m3 >s

The cross-sectional area of the flow as a function of x is A = pr 2 = p30.06(1 - 2x)4 2 = 30.0036p(1 - 2x)2 4 m2

Thus, the velocity of the flow is u =

30.003(1 + t)4 m3 >s 5(1 + t) Q = = c d m>s A 30.0036p(1 - 2x)2 4 m2 6p(1 - 2x)2

Using the definition of velocity and the initial condition when t = 0, x = 0, dx = u; dt 6p

L0

x

5(1 + t) dx = dt 6p(1 - 2x)2

(1 - 2x)2dx = 5

6pax +

6pc 0.3 +

(1 + t) dt

x x2 4 t2 t - x 3>2 b ` = 5at + b ` 2 3 2 0 0

6pax + When x = 0.3 m,

L0

t

x2 4 t2 - x 3>2 b = 5at + b 2 3 2

0.32 4 t2 - (0.33>2) d = 5at + b 2 3 2

t 2 + 2t - 0.9493 = 0 t =

- 2 { 222 - 4(1)( -0.9493) 2(1)

7 0 Ans.

t = 0.3962 s = 0.396 s

Ans: t = 0.396 s 306

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4–26. The radius of the circular duct varies as r = 30.06(1 - 1x)4 m, where x is in meters. If the flow of water at A is Q = 0.003 m3 >s at t = 0, and it is increasing at dQ>dt = 0.003 m3 >s2, determine the velocity and acceleration of a water particle at x = 0.3 m when t = 0.4 s. Use average velocities at the cross section for the solution.

r 0.4 m

60 mm

x B A

x

SOLUTION The discharged as a function of time t is Q = 0.003 m3 >s + 10.003 m3 >s2 2t = 30.003 11 + t24 m3 >s

The cross-sectional area of the flow as a function of x is A = pr 2 = p30.0611 - 1x2 m4 2 = 30.0036p11 - 1x2 2 4 m2

Thus, the velocity of the flow is u =

30.00311 + t24 m3 >s 511 + t2 Q = = c d m>s 2 2 A 30.0036p11 - 2x2 4 m 6p11 - 2x2 2

At x = 0.3 m and t = 0.4 s, u =

511 + 0.42 6p11 - 20.32 2

= 1.815 m>s = 1.82 m>s

Ans.

The acceleration can be determined using a =

0u 0u + u 0t 0x

Here 0u 5 = c d m>s2 0t 6p11 - 2x2 2

511 + t2 1 1 0u c - 211 - 2x2 -3 a - x -2 b d = 0x 6p 2 =

511 + t2

= c

6p

c

1

2x11 - 2x2 3

511 + t2

6p2x11 - 2x2

At x = 0.3 m and t = 0.4 s,

3

d

1 s

d

0u 5 = = 1.2968 m>s2 0t 6p11 - 20.32 2

511 + 0.42 0u 1 = = 7.3286 3 0x s 6p20.311 - 20.32

Also, u = 1.815 m>s. Then

1 a = 1.2968 m>s2 + 11.815 m>s2 a7.3286 b s = 14.60 m>s2 = 14.6 m>s2

Ans. Ans: u = 1.82 m>s a = 14.6 m>s2

307

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4–27. The water in the tank is being drained at A with the average velocity 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

4 m!s

SOLUTION Since the volume of the control volume changes with time, local changes occur. Also, the water flows out of the control volume through the open (outlet) control surface. This causes convective changes to take place.

Water line

Outlet control surface AA

VA 5 4 mys

308

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*4–28. Air flows through the transition, and its temperature decreases as it flows from inlet B to outlet A. 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.

9 m!s

1 .5 m!s

B A

SOLUTION Since the density of the air within the control volume at any given point remains constant, the flow is steady and no local changes occur. However, 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.

VB 5 9 mys AB Inlet control surface

AA

VA 5 1.5 mys

Outlet control surface

309

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4–29. The average velocities of water flowing steadily through the nozzle are indicated. 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

B

8 m!s

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

310

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4–30. Air flows through the tapered duct, and during this time heat is being added at an increasing rate, changing the density of the air within the duct. Select a control volume that contains the air in the duct, and indicate the open control surfaces. Show the positive direction of their areas, and 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

7 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 Aout

Vout = 7 m s

Note: If the heating is constant, then no local changes will occur.

311

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4–31. 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.

1 .5 m!s

B 2.25 m!s C A

SOLUTION

0.75 m!s

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. VB 5 1.5 mys

AB Inlet control surface

VC 5 2.25 mys AC Outlet control surface

AA

VA 5 0.75 mys

312

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*4–32. Air is pumped into the tank, and at the instant shown it has an average velocity of 9 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.

9 m!s

A

SOLUTION Since the density of the air changes with time, local changes occur. Also, the air flows into the control volume through the open (inlet) control surface. This causes convective changes. Inlet control surface VA 5 9 mys AA

313

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4–33. The flat plate is moving to the right at 2 m>s. Water is ejected from the nozzle at A at an average velocity of 4 m>s. Outline a moving control volume that contains the water on the plate. Indicate the open control surfaces, and show the positive direction of their areas through which flow occurs. Also, indicate the directions of the relative velocities through the control surfaces and determine the magnitude of the relative velocity through the inlet control surface. Identify the local and convective changes that occur. Assume water to be incompressible. For the analysis, why is it best to consider the control volume as moving?

4 m!s

2 m!s

A

SOLUTION If the control volume is considered moving with the plate, then the flow can be considered steady as measured relative to the control volume. No local changes occur. Also, the water flows in and out through the open (inlet and outlet) control surfaces. This causes convective changes. (Vwycs)out

Aout

(Vwycs)in 5 2 mys

Outlet control surface

Ain Inlet control surface Aout (Vwycs)out

Ans: A moving control volume allows the flow to be treated as steady. 314

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4–34. 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 within it. Also, select another control volume to be just the water within 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.

3 m!s

B

A 12 m!s

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.

VB 5 3 mys

VA 5 12 mys AB Inlet control surface

AA Outlet control surface

VB 5 3 mys

VA 5 12 mys AB

AA Outlet control surface

315

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4–35. Compressed air is being released from the tank, and at the instant shown it has an average 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

316

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*4–36. The blade on the turbine is moving to the left 6 m>s. Water is ejected from the nozzle at A at an average velocity of 2 m>s. Outline a 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 directions of the relative velocities through the control surfaces and determine the magnitude of the relative velocity through the inlet control surface. Identify the local and convective changes that occur. Assume water to be incompressible. For the analysis, why is it best to consider the control volume as moving?

2 m!s

6 m!s

A

SOLUTION The control volume is considered to be moving with the blade, so that the flow can be considered steady as measured relative to the control volume. No local changes occur. Also, the water flows in and out through the open (inlet and outlet) control surfaces. This causes convective changes.

(Vw cs)out Outlet control surface

Aout

(Vw cs)in = 8 m s Ain

Inlet control surface

317

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4–37. 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 a moving control volume as the jet engine and the air and fuel within it. 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. For an analysis, why is it best to consider this control volume to be moving?

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

Ans: A moving control volume allows the flow to be treated as steady. 318

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4–38. 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

319

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4–39. Water flows through the pipe at A at 60 kg>s, and then out of B with a velocity of 4 m>s. Determine the average velocity at which it flows in through C.

VA A 100 mm B VB 60 mm

SOLUTION

160 mm C

The fixed control volume considered 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. For the flow at A, # mA = rVAAA

VC VA AA

60 kg>s = 11000 kg>m3 21VAAA 2 VAAA = 0.06 m3 >s

Since the fluid is water, which has a constant density, then the continuity equation can be simplified as

AB

0 rdV + rV # dA = 0 0 t Lcv Lcs

AC

0 + VBAB - VAAA - VCAC = 0 2

3

VB

VC (a)

2

14 m>s23p10.08 m2 4 - 0.06 m /s - VC 3p10.03 m2 4 = 0 VC = 7.224 m>s = 7.22 m>s

Ans.

Ans: VC = 7.22 m>s 320

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*4–40. If water flows in the wye through pipes at A and C with a mass flow of 60 kg>s and 20 kg>s, respectively, determine the velocity of the flow out of the wye through the pipe at B.

VA A 100 mm B VB 60 mm

SOLUTION

160 mm C

The fixed control volume considered 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. Since mA = rVA, then the continuity equation can be simplified as

VC

VA

0 rdV + rV # dA = 0 0 t Lcv Lcs # # # 0 + mB - mA - mC = 0

AA

However, # mB = rBVBAB = 11000 kg>m3 21VB 23p10.08 m2 2 4 = 16.4pVB 2 kg>s.

AB

VB

Then

AC

16.4pVB 2 kg>s - 60 kg>s - 20 kg>s = 0

VC

Ans.

VB = 3.979 m>s = 3.98 m>s

(a)

Ans: VB = 3.98 m>s 321

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4–41. The jet engine draws in air at 1800 kg>s and jet fuel at 0.916 kg>s. If the density of the air–fuel mixture at the exhaust is 3.36 kg>m3, determine the velocity of the exhaust relative to the plane. The exhaust nozzle has a diameter of 0.8 m.

SOLUTION The moving control volume shown in Fig. a will be considered. Here, the control volume moves with the plane; thus, the flow is steady relative to the plane. # Therefore, no local changes occur within this control volume. Since m = rVf>cs # A, the continuity equation can be simplified as

Vaycs

Aa

Ae Veycs (a)

0 rdV + rVf>cs # dA 0 t Lcv Lcs # # 0 - ma - mf + rMVe>csAe = 0 0 - 1800 kg>s - 0.916 kg>s + 13.36 kg>m3 21Ve>cs 23p10.4 m2 2 4 = 0 Ve>cs = 1066.31 m>s = 1.071103 2 m>s

Ans.

Ans: Ve>cs = 1.07(103) m>s 322

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4–42. Nitrogen flows into the tank at A at VA = 3 m>s, and helium flows in at B at VB = 8 m>s. Both enter at a gage pressure of 260 kPa and a temperature of 150°C. Determine the steady mass flow of the mixed gas at C. The atmospheric pressure is patm = 101.3 kPa.

VC 0.4 m C

SOLUTION

0.3 m

From the table in Appendix A, the gas constants for Nitrogen and Helium are RN = 296.8 J> kg # K and RHe = 2077 J> kg # K, respectively. Here, the absolute pressure is p = patm + pg = 1101.3 + 2062 kPa = 361.3 kPa and the absolute temperature is T = 150°C + 273 = 4232 K.

B

A

VA

0.2 m VB

p = r NRNT

361.3110 2 N>m2 = rN 1296.8 J>kg # K21423 K2 3

rN = 2.8778 kg>m3

And VC

p = rHeRHeT

361.31103 2 N>m2 = rHe 12077 J>kg # K21423 K2

AC

rHe = 0.4112 kg>m3

The fixed control volume shown in Fig. a will be considered since the flow is steady, there is no change in the volume, and therefore no local changes occur within this control volume. Since the densities of the fluids through the open control surfaces are different but of constant values, the continuity equation gives 0 rdV + rV # dA = 0 0 t Lcv Lcs # 0 - rNVAAA - rHeVBAB + mm = 0

VA 5 3 mys

0 - 12.8778 kg>m3 213 m>s23p10.15 m2 2 4

# - 10.4112 kg>m3 218 m>s23p10.1 m2 2 4 + mm = 0 # mm = 0.7136 kg>s = 0.714 kg>s

AA

AB VB 5 8 mys (a)

Ans.

Ans: # mm = 0.714 kg>s 323

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4–43. Nitrogen flows into the tank at A at VA = 3 m>s, and helium flows in at B at VB = 8 m>s. Both enter at a gage pressure of 260 kPa and a temperature of 150°C. Determine the velocity of the mixed gas leaving the tank at a steady rate at C. The mixture has a density of r = 2.520 kg>m3. The atmospheric pressure is patm = 101.3 kPa.

VC 0.4 m C

0.3 m

SOLUTION

B

A

VA

From the table in Appendix A, the values of gas constant for Nitrogen and Helium are RN = 296.8 J> kg # K and RHe = 2077 J>kg # K, respectively. Here, the absolute pressure is p = patm + pg = 1101.3 + 2602 kPa = 361.3 kPa and the absolute temperature is T = 150°C + 273 = 423 K.

0.2 m VB

p = r NRNT

361.31103 2 N>m2 = rN 1296.8 J>kg # K21423 K2 rN = 2.8778 kg>m3

And p = rHeRHeT 3

2

361.3110 2 N>m

VC

= rHe 12077 J>kg # K21423 K2

AC

rHe = 0.4112 kg>m3

The fixed control volume shown in Fig. a will be considered since the flow is steady, there is no change in the volume, and therefore no local changes occur within this control volume. Since the densities of the fluids through the open control surfaces are different but of constant values, the continuity equation gives 0 rdV + rV # dA = 0 0 t Lcv Lcs 0 - rNVAAA - rHeVBAB + rmVCAC = 0 - 12.8778 kg>m3 213 m>s23p10.15 m2 2 4

VA 5 3 mys

VC = 2.253 m>s = 2.25 m>s

AB VB 5 8 mys (a)

- 10.4112 kg>m3 218 m>s23p10.1 m2 2 4

+ 12.520 kg>m3 21VC 23p10.2 m2 2 4 = 0

AA

Ans.

Ans: VC = 2.25 m>s 324

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*4–44. 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 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 rdV + rV # dA = 0 0 t 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: V = 9.42 m>s 325

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4–45. The flat strip is sprayed with paint using the six nozzles, which are attached to the pipe. The strip is 50 mm wide and the paint is to be 1 mm thick. If the 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 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 rdV + rV # dA = 0 0 t 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

Each nozzle has to cover 0.5 m of length and put

1 ( 10-3 ) m

= 0.1667 ( 10-3 ) m 6 thickness of paint on the strip. Thus, the volume of paint required is 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 V =

21.22 bs D2 0.5 m

( 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

20

25

30

V(m s)

30

20

10

0

5

10

15

D(mm)

Ans: V = (0.0236 D2) m>s where D is in mm

(a)

326

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4–46. 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 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 y = w = 0, we have aA =

0VA 0t

= 1.6t 0 t

= 2s

= 3.20 m>s2

Ans.

Ans: VB = 28.8 m>s, aA = 3.20 m>s2 327

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4–47. 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 speed 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 = Then -V a

p 2 p p d and Aout = (0.2 m)2 - d 2. 4 4 4

p 2 p p d 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 328

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*4–48. 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

Vin

d

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

The plot of Vout vs d is shown in Fig. a.

V

out (m

d m>s where d is in mm

50

75

100

125

150

0.133

0.327

0.667

1.28

2.57

d(mm) Vout(m>s)

p 3 0.04 - (10-6)d 2 4 f = 0 4

Ans.

s)

3

2

1

d(mm) 0

25

50

75

100

125

150

(a)

Ans: Vout = c

2 ( 10-6 ) d 2 0.04 - ( 10-6 ) d 2

d m>s where d is in mm

329

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4–49. 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

SOLUTION

200 mm 200 mm

B 0.15 m!s

0.2 m!s

The control volume considered is fixed as it contains 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

- ( 0.2 m>s ) 3 p(0.15 m)2 4 + ( 0.15 m>s ) 3 p(0.1 m)2 4 +

L0

V = Vma x (1 – 100r2 )

0.1 m

V max ( 1 - 100r ) (2prdr) = 0 2

- 3 ( 10-3 ) pm3 + 5 ( 10-3 ) pV max = 0

r

Ans.

V max = 0.6 m>s

(a )

Note: The integral in the above equation is equal to the volume under the velocity profile, while in this case is a paraboloid. L

Vc dA =

A

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: Vmax = 0.6 m>s 330

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4–50. The unsteady flow of glycerin is such that at A it has a velocity of VA = (0.8t + 2) m>s, where t is in seconds. Determine the acceleration of a fluid particle located at x = 0.3 m when t = 0.5 s. Hint: Determine V = V(x, t), then find the material derivative.

0.4 m

0.1 m

0.2 m B A x

SOLUTION The fixed control volume shown in Fig. a will be considered. Since the volume does not change with time, no local changes occur within this control volume. Referring to the geometry shown in Fig. b, the radius of transition at an arbitrary distance x is r - 0.05 m 0.4 m - x = ; 0.05 m 0.4 m

VA

r = 10.1 - 0.125x2 m

AA

A

Here, since the density of glycerin is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0 t Lcv Lcs

Then,

0.0110.8t + 22

10.1 - 0.125x2

a =

2

0V 0V + V 0t 0x

(a)

0.05 m 2

- 10.8t + 223p10.1 m2 4 + V3p10.1 - 0.125x2 4 = 0 V = c

x

0.05 m

0 - VAAA + VA = 0 2

V

0.05 m

r 0.4m 2 x

x (b)

d m>s

Here 0V 0.008 = c d m>s2 0t 10.1 - 0.125x2 2

0V = 0.0110.8t + 223 - 210.1 - 0.125x2 -3 1 -0.12524 0x = c

0.002510.8t + 22 10.1 - 0.125x2 3

d 1>s

At x = 0.3 m and t = 0.5 s, V =

0.0130.810.52 + 24 30.1 - 0.12510.324 2

= 6.144 m>s

0V 0.008 = = 2.048 m>s2 0t 30.1 - 0.12510.324 2

Then

0.002530.810.52 + 24 0V = = 24.576 1>s 0x 30.1 - 0.12510.324 3 a = 2.048 m>s2 + 16.144 m>s2124.576 1>s2 = 153.04 m>s2 = 153 m>s2

Ans. Ans: a = 153 m>s2

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4–51. The unsteady flow of glycerin is such that at A it has a velocity of VA = (5t 1>2) m>s, where t is in seconds. Determine the acceleration of a fluid particle located at x = 0.3 m when t = 1 s. Hint: Determine V = V(x, t), then find the material derivative.

0.4 m

0.1 m

0.2 m B A x

SOLUTION The fixed control volume shown in Fig. a will be considered. Since the volume does not change with time, no local changes occur within this control volume. Referring to the geometry shown in Fig. b, the radius of the transition at an arbitrary distance x is, r - 0.05 m 0.4 m - x = ; r = 10.1 - 0.125x2 m 0.05 m 0.4 m

VA

AA

A

Here, since the density of the glycerin is constant and the average velocities will be used, the continuity equation can be simplified as

0.05 m

0 - VAAA + VA = 0

0.05 m

- 15t 1>2 23p10.1 m2 2 4 + V3p10.1 - 0.125x2 2 4 = 0

Here

0.05 m

r 0.4m 2 x

x

0.05t 1>2 V = c d m>s 10.1 - 0.125x2 2 a =

x (a)

0 rdV + rV # dA = 0 0 t Lcv Lcs

Then,

V

(b)

0V 0V + V 0t 0x

1 1 0.05a t -2 b 2 0.025 0V = = c 1>2 d m>s 2 0t 10.1 - 0.125x2 t 10.1 - 0.125x2 2

1 0V = 0.05t 2 3 -210.1 - 0.125x2 -3 1 -0.12524 0x

0.0125t 1>2 = c d 1>s 10.1 - 0.125x2 3

At x = 0.3 m and t = 1 s

V = c

0.05111>2 2

3 0.1 - 0.12510.34 2

4 = 12.8 m>s

0V 0.025 = 1>2 = 6.40 m>s2 0t 11 230.1 - 0.12510.324 2 Then

0.0125111>2 2 0V = = 51.2 1>s 0x 30.1 - 0.12510.324 3

a = 6.40 m>s2 + 112.8 m>s2151.2 1>s2 = 661.76 m>s2 = 662 m>s2

Ans. Ans: a = 662 m>s2 332

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*4–52. Water flows through the nozzle at 0.02 m3 >s. Determine the velocity V of a particle as it moves along the centerline as a function of x.

50 mm

20 mm

x 80 mm

SOLUTION The fixed control volume shown in Fig. a will be considered. Since the volume does not change with time, no local changes occur within this control volume. Referring to the geometry shown in Fig. b, the radius of the nozzle at an arbitrary distance x is r - 0.01 m 0.015 m = ; r = 10.025 - 0.1875x2 m 0.08 m - x 0.08 m

Since the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0 t Lcv Lcs 0 - VAAA + VA = 0 Realizing that QA = VAAA = 0.02 m3 >s,

- 0.02 m3 >s + V3p10.025 - 0.1875x2 2 4 V = c

0.015 m

32 d m>s p11 - 7.50x2 2

Ans.

r

0.01 m

0.01 m x

VA

AA

A

V

0.08m 2 x (b)

(a)

Ans: V = c

32 d m>s p11 - 7.50x2 2

333

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4–53. Water flows through the nozzle at 0.02 m3 >s. Determine the acceleration of a particle as it moves along the centerline as a function of x.

50 mm

20 mm

x 80 mm

SOLUTION The fixed control volume shown in Fig. a will be considered. Since the volume does not change, no local changes occur with this control volume. Referring to the geometry shown in Fig. b, the radius of the nozzle at an arbitrary distance x is 0.015 m r - 0.01 m 0.015 m = ; r = 10.025 - 0.1875x2 m 0.08 m - x 0.08 m

r

0.01 m

0.01 m x

0.08m 2 x

Since the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0 t Lcv Lcs

(a)

VA

AA

0 - VAAA + VA = 0

A

V

(b)

Realizing that QA = VAAA = 0.02 m3 >s,

- 0.02 m3 >s + V3p10.025 - 0.1875x2 2 4 V = c

Then

32 d m>s p11 - 7.50x2 2

a = Here, Then

0V = 0 0t

0V 0V + V 0t 0x

0V 32 480 = 3 - 211 - 7.50x2 -3 1 -7.5024 = c d 1>s p 0x p11 - 750x2 3 32 480 a = 0 + c p11 - 7.50x2 2 d c p11 - 7.50x2 3 d = c

15.4(103)

2

p (1 - 7.50x)5

d m>s2

Ans.

Ans: a = c

15.4(103) p2(1 - 7.50x)5

d m>s2

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Vp

4–54. The cylindrical plunger traveling at Vp = 10.004t 1>2 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. The volume of the ball is V = 43pr 3.

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

0.05 m

Since rp is constant, it can be factored out of the integrals. Also, the average velocity will be used. Thus, the above equation becomes rp

dV + rpVAAA = 0 dt

Since QA = VAAA,

y

dV + QA = 0 dt

(1)

The volume of the control volume is

A

V = p(0.025 m)2y + V0 = 0.625 ( 10-3 ) py + V0

V0

(a)

dy dV = 0.625(10-3)p dt dt dy 1 = -Vp = ( - 0.004t 2 ) m>s. The negative sign indicates that Vp is directed dt in the opposite sense to positive y. However,

1 dV = 0.625 ( 10-3 ) p( - 0.004t 2 ) = dt

3 -2.5 ( 10-6 ) pt 4 m3 >s 1 2

The negative sign indicates that the volume is decreasing. 1

- 2.5 ( 10-6 ) pt 2 + QA = 0 1

The volume of the ball is Vs =

QA = ( 2.5 ( 10-6 ) pt 2 ) m3 >s

4 3 4 pr = 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: t = 48.5 s

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Vp

4–55. The cylindrical plunger traveling at Vp = 10.004 t 1>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. What distance must the plunger travel when d = 10 mm? Is this realistic? The volume of the ball is V = 43 pr 3.

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 + VAAA = 0 (1) 0t Since QA = VAAA, the volume of the control volume is V = a

p 2 d by + V0 4

0.05 m

y

0V p 0y = d2 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.

A

V0

(a)

1 1 0V p = d 2( -0.004t 2 ) = ( -0.001pd 2t 2 ) m3 >s 0t 4

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

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 given by Vs =

L0 L0

t

QAdt = Vs or 0.5625 ( 10-3 ) p m3 Vs = 1 QA 0.001pd 2t 2

t

dt = L0

t

1

t 2 dt = 0.5625 d - 2 2 3 t 2 = 0.5625 d - 2 3 4

4

t = 0.8929 d - 3 = 0.893 d -3

Ans.

336

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4–55. Continued

d (mm)

10

20

30

40

50

t (s)

414

164

95.8

65.3

48.5

t (s)

400 350 300 250 200 150 100 50 d (mm) 0

10

20

30

40

50

(b)

To find the distance s the plunger travels when d = 10 mm, we equate the volume of the spherical mold and the cylindrical volume the plunger passes through. 4 p p10.075 m2 3 = 10.010 m2 2 s 3 4

Ans.

s = 22.5 m

This is unrealistic. For the manufacture at an object measuring only 15 cm across.

Ans: t = 0.893d -4>3, s = 22.5 m, unrealistic 337

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*4–56. 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 velocity 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>m3)(400 m>s) 3 p(0.02 m)2 4 rsVs A 0V = = rw 0t 850 kg>m3 = - 0.02070 m3 >s

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 0 V>0t 3600 s 0.02070 m3 >s

Ans.

Ans: t = 2.48 hr 338

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4–57. 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 its 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

V

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 sfa

1 hr b 3600

339

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4–57. 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

993 b m>s where t is in hrs t

340

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4–58. With every breath, air enters the trachea, its flow split equally into two main bronchi, and then passes through about 150 000 bronchial tubes before entering the alveoli. If the air flow into the 18-mm-diameter trachea is 12 liter>min., determine the velocity of the air in the trachea and the main bronchi, which have a diameter of 12 mm. Note: The diameter of the alveoli is about 250 μm, and because there are so many of them, the flow is reduced to practically zero, so gaseous exchange is carried out by diffusion.

SOLUTION The fixed control volume considered is the air contained within the trachea and two main bronchi shown in Fig. a. Since the volume of this control volume does not change with time, no local changes occur within this control volume. Here, the density of the air is considered constant, and so the continuity equation can be simplified as 0 rdV + rV # dA = 0 0 t Lcv Lcs

Vt

(1)

0 - VtAt + 2VbAb = 0

At

liter 1 m3 1 min ba ba b = 0.2110-3 2 m3 >s. The volumetric flow Here, Q = a12 min 1000 liter 60 s rate gives Q = VtAt; 0.2110-3 2 m3 >s = Vt 3p10.009 m2 2 4 Vt = 0.7860 m>s = 0.786 m>s

Ab

Ab

Ans.

Vb

Vb

Then Eq. (1) gives

(a)

- 0.2110-3 2 m3 >s + 2Vb 3p10.006 m2 2 4

Ans.

Vb = 0.8842 m>s = 0.884 m>s

Ans: Vt = 0.786 m>s, Vb = 0.884 m>s 341

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4–59. A liquid flows through the drainpipe such that it has a parabolic velocity profile of u = 10(1 - 400r 2) m>s, where r is in meters. Determine the time needed to drain the cylindrical tank from a depth of h = 2 m at t = 0 to a depth of 1 m. The tank has a diameter of 1.5 m.

1.5 m

h

SOLUTION The control volume considered is the water contained in the tank. Thus, its volume changes with time. Applying the continuity equation, r

0 rdV + rV # dA = 0 0 t Lcv Lcs

100 mm

Since the liquid is considered incompressible, r is a constant which can be factored out of the integrals r

0V + r V # dA = 0 0t Lcs 0V + Qout - Qin = 0 0t

(1)

Here, Qin = 0 and Qout = The integral

LA

LA

vdA =

L0

0.05 m

1011 - 400r 2 212prdr2 = 0.0125p m3 >s

vdA can also be determined by computing the volume under the

velocity profile, which in this case is a paraboloid. LA

vdA =

1 2 1 pr h = p10.05 m2 2 110 m>s2 = 0.0125p m3 >s 2 2

The volume of the control volume at a particular instant is

Then

V = p10.75 m2 2h = 10.5625ph2 m3 dV = 0.5625p m2 dh

Using the chain rule dV dV dh dh = * = 0.5625p dt dh dt dt Substitute these results into Eq. (1) 0.5625p

dh + 0.0125p - 0 = 0 dt

dh = -0.02222 dt 1m

L2m

dh = - 0.02222

L0

t

dt

-1 = - 0.02222 t t = 45 s

Ans. Ans: t = 45 s 342

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*4–60. Oil flows into a mixing tank through pipe A with an average velocity of 6 m>s, and glycerin flows into the tank through pipe B at 3 m>s. Determine the average density at which the mixture flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m3 tank. Take ro = 880 kg>m3 and rg ly = 1260 kg>m3.

80 mm

100 mm A

B

SOLUTION The control volume considered is the fluid mixture contained in the tank and pipes as shown in Fig. a. This is a fixed control volume, so its volume does not change with time (for incompressible fluid), and therefore no local changes take place. Since the densities of oil and glycerin are constant and the average velocity will be used, the continuity equation can be simplified as

C 120 mm

0 rdV + rV # dA = 0 0 t Lcv Lcs 0 - VAAA - VBAB + VCAC = 0 0 - 16 m>s23p10.05 m2 2 4 - 13 m>s23p10.04 m2 2 4 + VC 3p10.06 m2 2 4 = 0

VA

VB

AA

AB

VC = 5.50 m>s

The conservation of mass for steady flow requires 0 rdV + rV # dA = 0 0 t Lcv Lcs 0 - roVAAA - rglyVBAB + rmixVCAC = 0

rmix = =

roVAAA + rglyVBAB VCAC 1880 kg>m3 216 m>s23p10.05 m2 2 4 + 11260 kg>m3 213 m>s23p10.04 m2 2 4 15.50 m>s23p10.06 m2 2 4

= 972.12 kg>m3 = 972 kg>m3

Ans.

AC VC (a)

Ans: rmix = 972 kg>m3 343

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4–61. Oil flows into the mixing tank through pipe A at 0.05 m3 >s, and glycerin flows into the tank through pipe B at 0.015 m3 >s. Determine the average density of the two fluids as the mixture flows out from the tank through the exit pipe at C. Assume uniform mixing of the fluids occurs within the 4@m3 tank. Take ro = 880 kg>m3 and rg ly = 1260 kg>m3.

80 mm

100 mm A

B

SOLUTION The control volume considered is the fluid mixture contained in the tank and pipes as shown in Fig. a. This is a fixed control volume so its volume does not change with time (for incompressible fluid) and therefore no local changes take place. Since the density of oil and glycerin are constant and Q = VA, the continuity equation can be simplified as

C 120 mm

0 rdV + rV # dA = 0 0t Lcv Lcs 0 - QA - QB + QC = 0

VA

VB

0 - 0.05 m3 >s - 0.015 m3 >s + QC = 0

AA

AB

QC = 0.065 m3 >s

The conservation of mass for steady flow requires

0 rdV + rV # dA = 0 0t Lcv Lcs 0 - r0QA - rglyQB + rmixQC = 0 rmix = =

roQA + rglyQB QC (880 kg/m3)(0.05 m3 >s) + (1260 kg>m3)(0.015 m3 >s)

AC

0.065 m3/s

= 967.69 kg>m3 = 968 kg>m3

Ans.

VC (a)

Ans: rmix = 968 kg>m3 344

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4–62. 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>ft3.

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 = a400 QB = a200

gal h gal h

ba ba

3 V0

1.5 ft

+ 2.25py 4 ft3

V # dA and Lcs

y Initial water level

1 ft3 1h ba b = 0.01485 ft3 >s 7.48 gal 3600 s

VA

AB

AA

1 ft3 1h ba b = 0.007427 ft3 >s 7.48 gal 3600 s

VB

(a)

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: 0y = 3.15 ( 10-3 ) ft>s 0t 345

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4–63. Gasoline flows into the tank through two pipes. At A the velocity is 4 m>s. Determine the rate at which the level of gasoline is rising in the tank as a function of the flow through the inlet pipe B. Plot this rate (vertical axis) versus the discharge for 0 … QB … 0.1 m3 >s. Give values for increments of ∆ QB = 0.02 m3 >s.

1.5 m

120 mm B

200 mm A

SOLUTION The control volume considered is the gasoline contained in the tank as shown in Fig. a. The volume of this control volume changes with time as the level of gasoline in the tank rises. Since the density of the gasoline is constant and Q = VA, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs rg c

Initial gasoline level

y

0 dV + V # dA d = 0 0t Lcv Lcs

VA

dV - VAAA - QB = 0 dt

AB

AA

VB

1.5 m (a)

(1)

If the initial volume of the gasoline in the tank is V0, then the volume of the gasoline in the tank for a particular instant is V = V0 + pr 2y = V0 + p(1.5 m)2y = (V0 + 2.25py) m3 Then Eq. (1) becomes d (V + 2.25py) - (4 m>s)3p(0.1 m)2 4 - QB = 0 dt 0 dy 2.25p = QB + 0.04p dt

dy QB = a + 0.0178b m>s dt 2.25p

0y 0t

dy are tabulated below. dt

3

0

0.02

0.04

0.06

0.08

0.10

(m>s)

0.0178

0.0206

0.0234

0.0263

0.0291

0.0319

QB(m >s) The plot of

0.05 0.04 0.03

where QB is in m3 >s

The values of QB and the corresponding values of

dy mys dt

Ans. 0.02 0.01

0

0.02

0.04

0.06

0.08

0.10

QB(m 3 ys)

(b)

0y vs QB is shown in Fig. b. 0t

346

M04_HIBB9290_01_SE_C04_ANS.indd 346

Ans: dy QB = a + 0.0178b m>s dt 2.25p where QB is in m3 >s

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*4–64. Gasoline flows into the cylindrical tank through two pipes. At A the velocity is 4 m>s and at B it is 6 m>s. Determine the rate at which the level of gasoline is rising in the tank.

1.5 m

120 mm B

200 mm A

SOLUTION The control volume considered is the gasoline contained in the tank as shown in Fig. a. The volume of this control volume changes with time as the level of gasoline in the tank rises. Since the density of the gasoline is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs rg c

Initial gasoline level

y

0 dV + V # dA d = 0 0t Lcv Lcs

VA

dV - VAAA - VBAB = 0 dt

(1)

If the initial volume of the gasoline in the tank is V0, then, the volume of the gasoline in tank for a particular instant is

AB

AA

VB

1.5 m (a)

V = V0 + pr 2y = V0 + p(1.5 m)2y = (V0 + 2.25py) m3 Then Eq. (1) becomes d (V + 2.25py) - (4 m>s)3p(0.1 m)2 4 - (6 m>s)3p(0.06 m)2 4 = 0 dt 0 2.25p

dy = 0.0616p dt

dy = 0.02738 m>s = 0.0274 m>s dt

Ans.

Ans: dy = 0.0274 m>s dt 347

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4–65. The cylindrical syringe is actuated by applying a force on the plunger. If this causes the plunger to move forward at 20 mm>s, determine the velocity of the fluid passing out of the needle.

20 mm!s 0.5 mm

10 mm

SOLUTION The deformable control volume considered is shown in Fig. a. If the volume of the control volume is initially V0 then at any instant its volume is V = V0 -

p (0.01 m)2x = 4

3 V0

- 25 ( 10-6 ) px 4 m3

With the fluid assumed to be incompressible, r is constant. Since VA and AA are in p the same sense, QA = VAAA = VA c (0.0005 m)2 d = 362.5 ( 10-9 ) pVA 4 m3 >s. 4

A

x

Then, the continuity equation can be simplified as

(a)

0 rdV + rV # dA = 0 0t Lcv Lcs rc

AA VA

0 (V) + VA AA d = 0 0t

0 3 V - 25 ( 10-6 ) px 4 + 62.5 ( 10-9 ) pVA = 0 0t 0 - 25 ( 10-6 ) p

dx + 62.5 ( 10-9 ) pVA = 0 dt

However, dx = 20 mm>s = 0.02 m>s dt Then

3 -25 ( 10-6 ) p 4 (0.02)

+ 62.5 ( 10-9 ) pVA = 0 Ans.

VA = 8.00 m>s

Ans: VA = 8.00 m>s 348

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4–66. Water enters the conical tank from the bottom with a velocity of 1.5 m>s. Determine the rate at which the water level at the surface is rising in terms of h.

3m

h

100 mm

SOLUTION

1.2 m

The deformable control volume considered is shown in Fig. a. Referring to the geometry shown in Fig. b, r 3m - h = ; r = [0.4(3 - h)] m 1.2 m 3m Then the volume of the control volume at a particular instant is V =

h

1 1 p(1.2 m)2(3 m) - p30.4(3 - h)4 2(3 - h) 3 3

A

= 31.44p - 0.05333p(3 - h)3 4 m3

V

The water is incompressible, so rw is constant. Also, average velocity will be used. Then, the continuity equation can be simplified as

(a)

0 rdV + rV # dA = 0 0t Lcv Lcs rw a

dV + VAb = 0 dt

3m

d 31.44p - 0.05333p(3 - h)3 4 + 3 -(1.5 m>s)3p(0.05 m)2 4 = 0 dt -0.05333pc 3(3 - h)2 a 0.16p(3 - h)2

r h

dh b d - 0.00375p = 0 dt

1.2 m (b)

dh = 0.00375p dt

dh 3 = c d m>s dt 128(3 - h)2

Ans.

Ans: dh 3 = c d m>s dt 128(3 - h)2 349

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4–67. 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 )( 120 m>s ) 3 p(0.0075 m)2 4 = 0 0t dr = - 0.101 kg> ( m3 # s ) dt

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: dr = -0.101 kg> ( m3 # s ) unsteady dt 350

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*4–68. The natural gas (methane) and crude oil mixture enters the separator at A at 6 ft3 >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>ft3, rme = 1.29110-3 2 slug>ft3. Note 1 ft3 = 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 0 - rmixVAAA + roVC AC + rmeVD AD = 0 QC = VCAC = a800

gal min

ba

(1)

1 ft3 1 min ba b = 1.783 ft3 >s and QA = VAAA = 6 ft3 >s. 7.48 gal 60 s

Substituting these results and the given densities into Eq. (1),

+

- rmix(6 ft3 >s) + ( 1.71 slug>ft3 )( 1.783 ft3 >s )

3 1.29 ( 10-3 ) slug>ft3 4 ( 300 ft>s ) c p a rmix = 0.5094 slug>ft3

2 1 ft b d = 0 12

gmix = rmix g = ( 0.5094 slug>ft3 )( 32.2 ft>s2 ) = 16.4 lb>ft3

Ans.

Ans: gmix = 16.4 lb>ft3 351

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4–69. The natural gas (methane) and crude oil mixture having a density of 0.51 slug>ft3 enters the separator at A at 6 ft3 >s, and crude oil flows out through the pipe at C at 800 gal>min. Determine the 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>ft3, rme = 1.29110-3 2 slug>ft3. Note 1 ft3 = 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 0 - rmixVA AA + rCOVC AC + rmeVD AD = 0

(1)

From Appendix A, rCO = 1.71 slug>ft3 and rme = 1.29 ( 10-3 ) slug>ft3 at T = 68°F Also, QC = a800

gal min

ba

1 ft3 1 min ba b = 1.783 ft3 >s and QA = VAAA = 6 ft3 >s. 7.48 gal 60 s

Substituting these results into Eq. 1,

- ( 0.51 slug>ft3 )( 6 ft3 >s ) + ( 1.71 slug>ft3 )( 1.783 ft3 >s ) +

3 1.29 ( 10-3 ) slug>ft3 4 ( VD ) c p a VD = 422 ft>s

2 1 ft b d = 0 12

Ans.

Ans: VD = 422 ft>s 352

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4–70. At the instant considered, the tank contains air at a temperature of 30°C and absolute pressure of 480 kPa. If the air escapes through the nozzle at 0.05 m3 >s, and the volume of the tank is 6 m3, determine the rate of change in the density of the air within the tank at this instant. Is the flow steady or unsteady?

A

SOLUTION

From the table in Appendix A, the gas constant for air is R = 286.9 J>kg # K. Here, T = 30°C + 273 = 303 K and p = 480 kPa. Applying the ideal gas law,

V

(a)

p = rRT 480(103) N>m2 = r(286.9 J>kg # K)(303 K) r = 5.5216 kg>m3 The control volume considered contains air of changing density but constant volume, Fig. a. Therefore, local changes take place. Applying the continuity equation, 0 rdV + rV # dA = 0 0t Lcv Lcs dr (V) + rVA = 0 dt Here, at the instant considered, Q = VA = 0.05 m3 >s. Then

dr (6 m3) + (5.5216 kg>m3)(0.05 m3 >s) = 0 dt

dr = - 0.04601 kg>(m3 # s) = - 0.0460 kg>(m3 # s) dt

Ans.

The negative sign indicates that the density of air is decreasing. Since the air pressure within the tank is decreasing with time the flow rate will be affected hence the flow is unsteady.

Ans: dr = - 0.0460 kg>(m3 # s) dt 353

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4–71. The water in the cylindrical storage tank is being drained using a pump. If the water is pumped out of the tank at 0.2 m3 >s, determine the rate at which the level in the tank is descending.

3m Pumped

h

d

SOLUTION The control volume considered is the volume of water contained in the tank, Fig. a, which changes with time. Here the density of the water is constant and the average velocities will be used. Then, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv L ra

A

V

(a)

dV + VAb = 0 dt

Realizing that Q = VA = 0.2 m3 >s,

dV + 0.2 m3 >s = 0 dt

(1)

The volume of the control volume at a particular instant is V = pr 2h = p(1.5 m)2h = (2.25ph) m3 Then dV dh = 2.25p dt dt Substitute this result into Eq. (1), 2.25p

dh + 0.2 = 0 dt

dh = - 0.02829 m>s = -0.0283 m>s dt

Ans.

The negative sign indicates that the water level is descending.

Ans: dh = - 0.0283 m>s dt 354

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*4–72. The water in the cylindrical storage tank is being pumped through a pipe having a diameter d. Determine the rate at which the level is descending as a function of d if the liquid is pumped out of the tank with an average velocity of 6 m>s. Plot this rate (vertical axis) versus the diameter for 0 … d … 0.3 m. Give values for increments of ∆ d = 0.05 m.

3m Pumped

h

d

SOLUTION The control volume considered is the volume of water contained in the tank, Fig. a, which changes with time. Here the density of water is constant and the average velocity will be used. Then the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t LCV L ra

A (a)

dV + VAb = 0 dt

dh mys dt

dV p + (6 m>s) a d 2 b = 0 dt 4

0

dV 3p 2 + d = 0 dt 2

(1)

Here the volume of the control volume at a particular instant is 2

2

0.05 0.10 0.15 0.20 0.25 0.30

d(m)

– 0.01 – 0.02

3

V = pr h = p(1.5 m) h = (2.25ph) m

– 0.03

Then dV dh = 2.25p dt dt

– 0.04 – 0.05

Substitute this result into Eq. (1) 2.25p

V

– 0.06

dh 3p 2 + d = 0 dt 2

(b)

dh 2 = a - d 2 b m>s dt 3

Ans.

where d is in meters

The negative sign indicates that the water level is descending. The values of d and the corresponding values of d(m) dh dt

(m>s)

0 0

dh are tabulated below dt

0.05 - 0.00167

0.10 - 0.00667

0.15 - 0.015

0.20 - 0.0267

0.25 - 0.0417

0.30 - 0.06

The plot of dh>dt vs. d is given in Fig. (b).

Ans: dh 2 = a - d 2 b m>s dt 3 where d is in meters 355

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4–73. 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 - 83.331103 2y2 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 y

B

( VAC )avg .

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, L

6 mm

D

0.5 m

SOLUTION

# mCD = r

u

C

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: # mAB = 4.41 g>s # mCD = 2.94 g>s 356

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4–74. Kerosene flows into the rectangular tank through pipes A and B, with velocities of 3 ft>s and 2 ft>s, respectively. It exits at C with a velocity 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. Ignore the effect of gravity on the falling kerosene.

3 ft!s

2 ft!s

4 in. A

B 8 in.

6 ft C

12 in.

1 ft!s

y

4 ft

SOLUTION The control volume is the volume of the kerosene in the tank including the two downflows. 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 p 3 = ft >s dt 18

(1)

The volume of the control volume at a particular instant is V = (6 ft)(4 ft)y + Thus

2 2 p 1 p 2 5p 5p a ft b (6 - y) + a ft b (6 - y) = c a24 by + d ft3 4 3 4 3 36 6

0V 5p 0y = a24 b 0t 36 0t

Substituting this result into Eq. (1),

a24 -

5p 0y p b = 6 0t 18

dy = 0.00816 ft>s dt

Ans.

Ans: dy = 0.00816 ft>s dt 357

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4–75. Kerosene flows into the 4-ft-diameter cylindrical tank through pipes A and B with velocities of 3 ft>s and 2 ft>s, respectively. It exits at C with a velocity of 1 ft>s. Determine the time required to fill the tank if y = 0 when t = 0. Ignore the effect of gravity on the falling kerosene.

3 ft!s

2 ft!s

4 in. A

B 8 in.

6 ft C

12 in.

1 ft!s

y

4 ft

SOLUTION The control volume is the volume of the kerosene in the tank including the two downflows. 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 p 3 = ft >s dt 18

(1)

The volume of the control volume at a particular instant is V = Thus,

2 2 p p 1 p 2 139p 5p (4 ft)2y + a ft b (6 - y) + a ft b (6 - y) = a y + b ft3 4 4 3 4 3 36 6

dV 139p dy = dt 36 dt Substituting this result into Eq. (1), 139p dy p = 36 dt 18 L0

6 ft

t

dy = 6 =

t = (417) a

2 dt 139 L0 2 t 139

1 min b = 6.95 min. 60 s

Ans.

Note: The solution assumes that the valves were opened a short time before t = 0, so that at t = 0, both flows have just reached the floor of the tank. Ans: t = 6.95 min 358

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*4–76. The cylindrical tank is filled with oxygen using a hose having an inside diameter of 0.25 in. If, at the instant considered, the oxygen enters the tank with an average velocity of 20 ft>s and has a density of 7.80(10-3) slug>ft3, determine the rate of change in the density of the air within the tank at this instant

6 ft

2 ft

SOLUTION The control volume considered is the oxygen contained in the tank as shown in Fig. a. The volume of this control volume does not change but the density of oxygen changes with time and therefore results in local changes. Here average velocity will be used. Then the continuity equation can be simplified as

A

V

(a)

0 rdV + rV # dA = 0 0t Lcv Lcs dr (V) + r( -VA) = 0 dt The volume of the control volume is V = pr 2h = p(1 ft)2(6 ft) = 6 p ft3. Then dr (6p ft3) dt

3 7.80(10-3) slug>ft3 4 e (20 ft>s) c

2 p 0.25 a ft b d f = 0 4 12

dr = 2.821(10-6) slug>(ft3 # s) dt = 2.82(10-6) slug>(ft3 # s)

Ans.

Ans: dr = 2.82(10-6) slug>(ft3 # s) dt 359

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4–77. Water flows into the cylindrical tank through pipes A and B, with velocities of 5 ft>s and 3 ft>s, respectively. Determine the rate at which the surface of the water is rising. Ignore the effect of gravity on the falling water.

5 ft!s

3 ft!s

4 in. A

B 6 in.

6 ft y

4 ft

SOLUTION The control volume considered contains the water in the tank as shown in Fig. a. It is deformable and therefore its volume changes with time, causing local changes to take place. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified as 0 r dV + rwV # dA = 0 0t Lcv w Lcs

VA

VB

AA

AB

y

dV rw c + ( - VAAA) + ( -VBAB) d = 0 dt

(a)

dV - VAAA - VBAB = 0 dt

2 2 dV 2 3 - (5 ft>s) c pa ft b d - (3 ft>s) c pa ft b d = 0 dt 12 12

dV 47 = a pb ft3 >s dt 144

(1)

The volume of the control volume at a particular instant is V = p(2 ft)2 y + Then

2 2 p 1 p 1 563p 13p a ft b (6 - y) + a ft b (6 - y) = a y + b ft2 4 3 4 2 144 24

dV 563p 0y = dt 144 0t Substitute this result into Eq. (1) 563p dy 47 = p 144 dt 144 dy 47 = a b ft>s = 0.0835 ft>s dt 563

Ans.

Ans: dy = 0.0835 ft>s dt 360

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4–78. Water flows into the cylindrical tank through pipes A and B, with velocities of 5 ft>s and 3 ft>s, respectively. Determine the time required to fill the tank if y = 0 when t = 0. Ignore the effect of gravity on the falling water.

5 ft!s

3 ft!s

4 in. A

B 6 in.

6 ft y

4 ft

SOLUTION The control volume considered contained the water in the tank as shown in Fig. a. It is deformable and therefore its volume changes with time causing the local changes to take place. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified as 0 r dV + r V # dA = 0 0t Lcv w Lcs rw c

VA

VB

AA

AB

y

0V + ( -VAAA) + ( -VBAB) d = 0 0t

(a)

0V - VAAA - VBAB = 0 0t

2 2 0V 2 3 - (5 ft>s) c pa ft b d - (3 ft>s) c pa ft b d = 0 0t 12 12

0V 47 = a pb ft3 >s 0t 144

(1)

The volume of the control volume at a particular instant is V = p(2 ft)2y + Then

2 2 p 1 p 1 563p 13p a ft b (6 - y) + a ft b (6 - y) = y + y 4 3 4 2 144 24

0V 563p 0y = 0t 144 0t Substitute this result into Eq. (1) 563p 0y 47 = p 144 0t 144

L0

0y 47 = a b ft>s 0t 563

6 ft

t

0y = 6 =

47 0t 563 L0 47 t 563

Ans.

t = 71.87 s = 71.9 s

Note: The solution assumes that the valves were opened a short time before t = 0, so that at t = 0, both flows have just reached the floor of the tank. Ans: t = 71.9 s 361

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4–79. The cylinder is pushed down into the tube at a rate of V = 5 m>s. Determine the velocity of the liquid as it rises in the tube. V

150 mm

y

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

200 mm

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 )

0.075 m

Continuity Equation. 0 r dV + rwVf>cs # dA = 0 0t Lcv w Lcs

y2

Since no water enters or leaves the control volume at any instant,

rwVf>cs # dA = 0.

Lcs

y1 y0

Then, y0 –y1

0 r dV = 0 0t Lcv w

(b)

0.1 m (a)

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,

0y1 = Vr = 5 m>s . Then 0t 0y2 = 1.2857(5 m>s) = 6.43 m>s 0t

Ans.

Ans: dy = 6.43 m>s dt 362

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*4–80. 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

SOLUTION

y

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 )

200 mm

= 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 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 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: Vr = 3.11 m>s 363

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4–81. 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 velocity VC of the mixture leaving 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>ft3 and rke = 1.59 slug>ft3.

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

AA VA = 4 ft s

Here,

VC AC

QA = VA AA = (4 ft>s) 3 p(0.15 ft)2 4 = 0.09p ft3 >s

AB V = 6 ft s B

QB = VB AB = (6 ft>s) 3 p(0.1 ft)2 4 = 0.06p ft3 >s

Then rm =

4 ft

( 1.70 slug>ft3 )( 0.09p ft3 >s ) + ( 1.59 slug>ft3 )( 0.06p ft3 >s )

= 1.656 slug>ft3

0.09p ft3 >s + 0.06p ft3 >s

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, 0 rdV + rV # dA = 0 dt Lcv Lcs 0 - rbVAAA - rkeVBAB + rmVcAc = 0 - ( - 1.70 slug>ft3)(0.09 pft3 >s) - (1.59 slug>ft3 )( 0.06p ft3 >s ) + (1.656 slug>ft3)(VC) 3 p(0.2 ft)2 4 = 0

Ans.

VC = 3.75 ft>s

Ans: rm = 1.656 slug>ft3 VC = 3.75 ft>s 364

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4–82. 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 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>ft3 and rke = 1.59 slug>ft3.

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 ft3 >s Then

AB V = 6 ft s B

VC = 6 ft s 4 ft (a)

( 1.70 slug>ft )( 0.09p ft >s ) + ( 1.59 slug>ft )( 0.06p ft >s ) 3

rm =

AA VA = 4 ft s

AC

QB = VB AB = (6 ft>s) 3 p(0.1 ft)2 4 = 0.06p ft3 >s QC = VC AC = (5 ft>s) 3 p(0.2 ft)2 4 = 0.2p ft3 >s

y

= 1.656 slug>ft3

3

3

3

0.09p ft3 >s + 0.06p ft3 >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>ft3 ) a12 b - ( 1.70 slug>ft3 )( 0.09p ft3 >s )

- ( 1.59 slug>ft3 )( 0.06p ft3 >s ) + ( 1.656 slug>ft3 )( 0.2p ft3 >s ) = 0 0y = -0.0131 ft>s 0t

Ans.

The negative sign indicates the level of the mixture is falling.

Ans: dy rm = 1.66 slug>ft3, = -0.0131 ft>s, falling dt 365

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4–83. The block with a square base is pushed down into the cylindrical tank with a velocity of Vb = 1.8 ft>s. Determine the velocity of the liquid as it rises in the tank. V

1 ft

y

SOLUTION The deformable control volume shown in Fig. a will be considered. If the initial liquid level in the cylindrical tank is y0, Fig. b, then the volume of this control volume at a particular instant is

2 ft

V = p11 ft2 2 1y0 - yb 2 + p11 ft2 2 1yl + yb 2 - 11 ft2 2 1yl + yb 2 = 3p1y0 - yb 2 + 1p - 121yl + yb 24 ft3

Applying the continuity equation,

0.5 ft

0 r dV + r Vf>cv # dA = 0 0t Lcv Lcs

y,

r Vf>cs # dA = 0. Lcs Also, the density of the liquid remains constant, so the continuity equation can be simplified as

yb

Since no liquid enters or leaves the control volume at any instant,

0 r dV = 0 0t Lcv

1 ft (a)

dV r = 0 dt d3p1y0 - yb 2 + 1p - 121yl + yb 24 dt

-p

(b)

= 0

dyl dyb dyb + 1p - 12 a + b = 0 dt dt dt 1p - 12

However,

y0

y0 2 yb

dyb = V = 1.8 ft>s. Then dt

dyl dyb = dt dt

dyb dyl 1 = a b dt p - 1 dt

dyl 1 = a b 11.8 ft>s2 dt p - 1

Ans.

= 0.840496 ft>s = 0.840 ft>s

Ans: dyl = 0.840 ft>s dt 366

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*4–84. Determine the speed V at which the square-base block must be pushed down into the cylindrical tank so that the liquid in the tank rises with a velocity of 6 ft>s. V

1 ft

SOLUTION The deformable control volume shown in Fig. a will be considered. If the initial liquid level in the cylindrical tank is y0, Fig. b, then the volume of this control volume at a particular instant is

y

V = p11 ft2 2 1y0 - yb 2 + p11 ft2 2 1yl + yb 2 - 11 ft2 2 1yl + yb 2

2 ft

= 3p1y0 - yb 2 + 1p - 121yl + yb 24 ft3

Applying the continuity equation,

0 r dV + r Vf>cs # dA = 0 0t Lcv Lcs

0.5 ft

r Vf>cs # dA = 0. Lcs Also, the density of the liquid remains constant, so the continuity equation can be simplified as Since no liquid enters or leaves the control volume at any instant,

y, yb

0 r dV = 0 0t Lcv r

dV = 0 dt

d3p(y0 - yb) + (p - 1)(yl + yb 24 dt

-p

1 ft (a)

(b)

= 0

dyl dyb dyb + (p - 1) a + b = 0 dt dt dt (p - 1)

However,

y0

y0 2 yb

dyb dyl = V and = 6 ft>s. Then dt dt V = a

dyl dyb = dt dt

dyb p - 1 dyl = a b dt 1 dt

p - 1 b 16 ft>s2 = 12.8496 ft>s = 12.8 ft>s 1

Ans.

Ans: V = 12.8 ft>s 367

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4–85. Oil flows into the trapezoidal container at a constant rate of 1800 kg/min. Determine the rate at which the oil level is rising when y = 1.5 m. The container has a constant width of 0.5 m. Take ro = 880 kg>m3. y

A 60 8

60 8

3m

SOLUTION The volume of the control volume considered changes with time (deformable) since it contains the oil in the tank for which its level y is rising, Fig. a. Referring to the geometry shown, its volume is 1 223 V = e c 3m + a3m yb d y f 10.5 m2 2 3 = c

Then

1 118y - 223y2 2 d m3 12

y Ï3 5 y tan 60° 3

3m2

2 Ï3 y 3

y Ï3 5 y tan 60° 3

y V

dy dy dy dV 1 1 = a18 - 423y b = 19 - 223y2 dt 12 dt dt 6 dt

A

60°

60° 3m (a)

The density of the oil is constant and the average velocity will be used. Then the continuity equation can be simplified as 0 rdV + r V # dA = 0 0t Lcv Lcs ro

dV + 1 -roVA2 = 0 dt

kg 1 min # Here roVA = mo = a1800 ba b = 30 kg>s. Then min 60 s

dy 1 1880 kg>m3 2 c 19 - 223y2 d - 30 kg>s = 0 6 dt dy 9 = dt 4419 - 223y2

When y = 1.5 m, dy 9 = = 0.05377 m>s = 0.0538 m>s dt 4439 - 22311.524

Ans.

Ans: dy = 0.0538 m>s dt 368

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4–86. Oil flows into the conical frustum at a constant rate of 1800 kg>min. Determine the rate at which the level is rising when y = 1.5 m. Hint: The volume of a cone is V = 13 pr 2h. Take ro = 880 kg>m3. y

A 60 8

60 8

SOLUTION

3m

The volume of the control volume considered changes with time (deformable) since it contained the oil in the tank, Fig. a, for which its level is rising. Referring to the geometry shown, a =

y y 23 23 = a ba b = a yb m tan 60° 3 23 23

h = 1.5 tan 60° = r = 1.5 - a = a

323 m 2

h

3 23 1 yb m = c 19 - 223y2 d m 2 3 6

a r

Then the volume of the control volume is V =

Then

V

2 1 323 1 1 323 p11.5 m2 2 a mb - pc 19 - 223y2 d a m - yb 3 2 3 6 2

A

60°

y 60°

1.5 m

1.5 m (a)

p = c 14y3 - 1823y2 + 81y2 d m3 36

dy dy dy dV p = c 12y2 - 3623y + 81 d dt 36 dt dt dt dy p = 34y2 - 1223y + 274 12 dt

The density of the oil is constant and the average velocity will be used. Then the continuity equation can be simplified as 0 rdV + r V # dA = 0 0t Lcv Lcs r

dV + 1 -roVA2 = 0 dt

kg 1 min # Here roVA = mo = a1800 ba b = 30 kg>s. Then min 60 s 1880 kg>m3 2 c

dy p 14y2 - 1223y + 272 d - 30 kg>s = 0 12 dt

dy 9 d m>s = c 2 dt 22p14y - 1223y + 272

At y = 1.5 m, dy dt

=

9 2

22p3411.5 2 - 122311.52 + 274

= 0.0270 m>s

Ans.

Ans: dy = 0.0270 m>s dt

369

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4–87. 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 = (8.02y1>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 cross-sectional area of 24 in2.

308

308

y 5 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 ) ft3 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: t = 5.99 s 370

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*4–88. 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 = (8.02y1>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.

308

308

y 5 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 ) ft3 2

30˚ 30˚

Continuity Equation. Since water has a constant density. rw c

y tan 30˚

0 dV + Vf>cs # dA d = 0 0t Lcv Lcs

y

A V

0 V + VA = 0 0t

(a)

1 0 24 2 ( 1.1547y2 ) + ( 8.02y2 ) a ft b = 0 0t 144

2.3094y

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: t = 2.73 s 371

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4–89. 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 0r V = Q ( rw - r ) 0t

0r Q t 0t = V L0 Lrs rw - r r

- ln ( rw - r ) `

r

= rs

Q t V

rw - r Q - ln a b = t rw - rs V t =

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° ¢ 3 0.03 m >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: Vw = 0.677 m3 372

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4–90. As nitrogen is pumped into the closed cylindrical # tank, the mass flow through the tube is m = (0.8r-1>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 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 ft3 The density of the nitrogen within the control volume changes with time and therefore contributes to local changes. # Continuity Equation. Realizing that m =

(a )

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>ft3 10p

0.125 slug>ft3. Then, when t = 5 s, p r = c

3(5) 10p

3

+ a

2

0.125 2 3 b d slug>ft3 p

= 0.618 slug>ft3

Ans.

Ans: r = 0.618 slug>ft3 373

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4–91. As nitrogen is pumped into the closed cylindrical # tank, the mass flow through the tube m = (0.8r-1>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 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 ft3 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 0r -1 4p - 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 ft3

r = a

=

3 3 3t + ro2 b slug>ft3 10p

0.125 slug>ft3. Then, when t = 10 s, p r = c

3(10) 10p

3

+ a

2

0.125 2 3 b d slug>ft3 p

= 0.975 slug>ft3

Ans.

Ans: r = 0.975 slug>ft3 374

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*4–92. Hydrogen is pumped into the closed cylindrical tank such that the mass flow through the tube is # m = (0.5r-1>3) kg3 >s. Determine the density of the hydrogen within the tank when t = 4 s from the time the pump is turned on. Assume that initially there is 0.2 kg of hydrogen in the tank.

0.5 m

1.5 m

SOLUTION The fixed control volume considered is shown in Fig. a. This control volume has a constant volume of V = p(0.5 m)2(1.5 m) = 0.375p m3 The density of the hydrogen within the control volume changes with time, and therefore contributes to local change. Applying the continuity equation, realizing # that m =

V

r V # dA,

Lcs

A (a)

0 rdV + r V # dA = 0 0t Lcv Lcs V 0.375p

dr # - m = 0 dt

dr - 0.5r-1>3 = 0 dt dr 4 -1>3 = r dt 3p

Integrating with the initial condition r = ro at t = 0, L0

t

r

dt = t = t =

Here ro =

0.2 kg 0.375p m3

=

3p r1>3dr 4 Lro 3p 3 4>3 r a r b` 4 4 ro

9p 4>3 1r - r4>3 o 2 16

r = a

3>4 16t + r4>3 kg>m3 o b 9p

8 kg>m3. Then when t = 4 s, 15p

r = c

16142 9p

+ a

8 4>3 3>4 b d kg>m3 15p

= 1.903 kg>m3 = 1.90 kg>m3

Ans.

Ans: r = 1.90 kg>m3 375

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4–93. Hydrogen is pumped into the closed cylindrical tank such that the mass flow through the tube is # m = (0.5r1>2) kg>s, where r is the density of the gas in the tank in kg>m3. Determine the density of the hydrogen within the tank when t = 4 s from the time the pump is turned on. Assume that initially there is 0.2 kg of hydrogen in the tank.

0.5 m

1.5 m

SOLUTION The fixed control volume considered is shown in Fig. a. This control volume has a constant volume of V = p(0.5 m)2(1.5 m) = 0.375p m3 The density of the hydrogen within the control volume changes with time, and therefore contributes to local changes. Applying the continuity equation realizing # that m =

r V # dA, Lcs 0 rdV + r V # dA = 0 0t Lcv Lcs

V

A (a)

dr # V - m = 0 dt 0.375p

dr - 0.05r1>2 = 0 dt dr 2 1>2 = r dt 15p

Integrating with the initial condition r = ro at t = 0, t r dr 2 dt = 15p L0 L0 r1>2 r 2 t = 2r1>2 ` 15p ro

2 t = 21r1>2 - r1>2 o 2 15p

Here ro =

0.2 kg 0.375p m3 r = c

r = a

=

2 t 3 + r1>2 o b kg>m 15p

8 kg>m3. Then when t = 4 s, 15p

4 8 1>2 2 + a b d = 0.2469 kg>m3 = 0.247 kg>m3 15p 15p

Ans.

Ans: r = 0.247 kg>m3 376

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1m

4–94. Brine or concentrated salt water in the cylindrical tank has an initial density of rb = 1250 kg>m3. Pure water is pumped into the tank at A at 0.02 m3>s and mixes with the salt water. 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 solution is reduced by 10% of its original value. The tank remains full throughout the mixing process.

3m

B A

SOLUTION The control volume considered is the volume of the tank. It is a fixed control volume, since its volume does not change throughout the mixing. However, the density of the contained salt water varies. Applying the continuity equation, 0 rdV + r V # dA 0t Lcv Lcs dr V + rQ - rwQ dt dr V dt dr rw - r

= 0 = 0 = Q1rw - r2 =

Q dt V

Integrate this equation with the initial condition at t = 0, r = rb gives dr Q t dt = V L0 Lrb rw - r r

r

-ln1rw - r2 `

= rb

Q t V

rw - r Q - lna b = t rw - rb V t =

rw - rb V ln a b rw - r Q

Here, V = p10.5 m2 2 13 m2 = 0.75p m3, and it is required that r = 0.9rb = 0.911250 kg>m3 2 = 1125 kg>m3. Then t = a

1000 kg>m3 - 1250 kg>m3 0.75p m3 b ln a b = 81.659 s 0.02 m3 >s 1000 kg>m3 - 1125 kg>m3

The amount of water to be added is

Vw = Qt = 10.02 m3 >s2181.659 s2 = 1.633 m3 = 1.63 m3

Ans.

Ans: Vw = 1.63 m3 377

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4–95. 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 40 mm>s. Determine the speed at which the plastic rises in the container as a function of yc. The container has a width of 200 mm.

40 mm!s

608

100 mm

300 mm

SOLUTION The control volume shown shaded in Fig. a can be considered fixed at a particular instant. At this instant, no local changes occur since the molten plastic is incompressible and therefore its density is constant. It we use the average velocity, the continuity equation can be simplified as

0.3 m 2

VB AB B

0 rdV + rV # dA = 0 0t Lcv L

B

0.1 m

yc

AA 60°

realizing that VAAA = QA, then (1)

VBAB - QA = 0

2 Ï3 y 3 c

yd

A

0 + VBAB - VAAA = 0

Here, VB =

yc

608

VA

60°

0.3 m (a)

dyc . From the geometry shown in Fig. a dt 223 p yc b 10.2 m2 - 10.1 m2 2 = 10.05214 - 0.230yc 2 m2 3 4

AB = a0.3 m -

The volume of the die submerged in the plastic is p Vd = 10.1 m2 2yd = 10.0025pyd 2 m3 4 The discharge through control surface A at the instant considered is QA = Realizing that

dyd dt

dyd dVd = a0.0025p b m3 >s dt dt

= 0.04 m>s, then QA = 0.0025p10.042 = 0.1110-3 2p m3 >s

Substitute these results into Eq. (1), dyc 10.05214 - 0.2309yc 2 - 0.1110-3 2p = 0 dt 0.1110-3 2p dyc = c d m>s dt 0.0521 - 0.231yc

where yc is in meters

Ans.

Ans: 0.1110-3 2p dyc = c d m>s dt 0.0521 - 0.231yc where yc is in meters 378

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5–1. Determine the required average change in pressure if the water flows from A to B with a deceleration of 3 ft>s2 along a horizontal streamline. Take rw = 1.94 slug>ft3.

A

B

2 ft

SOLUTION 1 dp dv + V + g sin u = 0 rw ds ds However, V

dp ∆p dV = , r = 1.94 slug>ft3 and u = 0° = ab = - 3ft>s2, ds ds 2ft w

Since the streamline is horizontal. Then a

∆p 1 ba b + 1 -3 ft>s2 2 + 0 = 0 3 2 1.94 slug>ft

∆p = 111.64 lb>ft2 2 a

1 ft 2 b = 0.0808 psi 12 in.

Ans.

Positive ∆p indicates the pressure increases as the water flows from A to B.

Ans: ∆p = 0.0808 psi 379

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5–2. If the volumetric flow of water through the pipe is 0.05 m3 >s, determine the pressure difference between points A and B. The flow occurs on the horizontal plane. Assume the velocity is constant over the cross section.

100 mm

B

A 300 mm

SOLUTION

s

Referring to the coordinate system shown in Fig. a, we notice that dR = -dr since r and R axes are opposite in sense. Also, since the pipe is lying on the horizontal plane, dz = 0. Then dp dz rwV 2 - rwg = dr dr R rwV 2 dp = dR R

0.3 m

-

B

n

Integrating, with p = pA and R = RA at A and p = pB and R = RB at B, pB

LpA

dp = rwV 2

RB

LRA

pB - pA = rwV 2 ln

R A

0.4 m

dR R

RB RA

(1)

(a)

Using the water discharge, Q = VA;

0.05 m3 >s = V3p10.05 m2 2 4 V =

20 m>s p

Substitute rw = 1000 kg>m3, RA = 0.3 m, RB = 0.4 m and V = pB - pA = 11000 kg>m3 2 a

20 m>s into Eq. 1: p

2 20 0.4 m m>s b lna b p 0.3 m

= 11.661103 2Pa = 11.7 kPa

Ans.

Positive results indicates that pressure at B is greater than that at A.

Ans: pB - pA = 11.7 kPa 380

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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>ft3. 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>ft3

lb 12 in. 2 da b 2 1 ft in ¥ + 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>ft3 ) (40 ft)

as = 152 ft>s2

Ans.

Ans: as = 152 ft>s2 381

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*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>ft3. 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>ft3 40 ft ∆p = -14.22

lb = -14.2 lb>ft2 ft2

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>ft3 ) (40 ft) ( 150 ft>s2 ) ∆p = -14.2

lb ft2

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.

Ans: ∆p = - 14.2

lb ft2

382

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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. For the calculation, assume the velocity is constant over the cross section.

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 Ans.

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

1.25 ri

1.5 ri

1.75 ri

2ri

383

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5–6. The water in an open channel drainage canal flows with a velocity of VA = 4 m>s into the drainpipe that crosses a highway embankment. Determine the volumetric discharge through the pipe. Neglect any head losses. 4 m!s

3 ft 8 ft

A

24 in. B

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Since point B is exposed to the atmosphere, pB = patm = 0. Here, VA = 4 ft>s with reference to the datum set through point B, zA = 8 ft + 3 ft = 11 ft and zB = 0. The pressure head at A is (3 ft). Applying the Bernoulli’s equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 (3 ft) (32.2 ft>s2) +

(4 ft>s)2 2

+ (32.2 ft>s2)(11 ft) = 0 +

VB2 + 0 2

VB = 30.29 ft>s Thus, the volumetric flow rate is given by Q = VBA = (30.29 ft>s) c pa

2 12 ft b d = 95.16 ft3 >s = 95.2 ft3 >s 12

384

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Ans.

Ans: Q = 95.2 ft3 >s

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5–7. Water flows out of a faucet at A at 6 m>s. Determine the velocity of the water just before it strikes the ground at B. A 6 m!s

1.75 m

B

SOLUTION If the datum is set at B, then zA = 1.75 m and zB = 0. Since the flow from A to B is in the atmosphere, pA = pB = 0. Applying Bernoulli’s equation between A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 0 +

(6 m>s)2 2

+ (9.81 m>s2)(1.75 m) = 0 +

VB2 + 0 2 Ans.

VB = 8.387 m>s = 8.39 m>s

Ans: VB = 8.39 m>s 385

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*5–8. Water flows through the 30-mm-diameter pipe at 0.002 m3 >s and is ejected from the 10-mm-diameter nozzle at B. Determine the velocity and pressure of the water at point A.

A

B

300 mm

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between A and B, pA pB VA2 VB2 + + gzA = + + gzB rw r 2 2 Since point B is exposed to the atmosphere, pB = patm = 0. Here point A and B have the same elevation. Thus, zA = zB = z. pA 3

100 kg>m

VA2 VB2 + gz = 0 + + gz 2 2

+

pA = 500 1VB2 - VA2 2

Continuity requires that Q = VBAB;

(1)

0.002 m3 >s = VB 3p(0.005 m)2 4

VB = 25.46 m>s = 25.5 m>s

Q = VAAA,

3

0.002 m >s = VA 3p(0.015 m)2 4

VA = 2.829 m>s = 2.83 m>s

Ans.

Substitute these results into Eq.(1) pA = 500(25.462 - 2.8292) = 320.22(106) Pa Ans.

= 320 kPa

Ans: VA = 2.83 m>s pA = 3.20 kPa 386

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5–9. Water flows through the 30-mm-diameter pipe and is ejected with a velocity of 25 m>s at B from the 10-mmdiameter nozzle. Determine the pressure and the velocity of the water at A.

A

B

300 mm

SOLUTION Since the flow is steady and density of water is constant, the continuity equation can be simplified as 0 pdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 -VA 3p10.015 m224 + 125 m>s23p10.005 m2 2 4 = 0 VA = 2.7778 m>s = 2.78 ms

Ans.

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B, 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. Thus, zA = zB = z. pA 1000 kg>m3

+

(2.7778 m>s) 2

+ gz = 0 +

(25 m>s)2 2

+ gz

pA = 308.64(103)Pa = 309 kPa

Ans.

Ans: VA = 2.78 m>s pA = 309 kPa 387

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5–10. Drainage under a canal is provided using a 1-ft-diameter drainpipe. Determine the flow through the pipe. Neglect any head losses.

A 8 ft B 1 ft

5.5 ft 0.5 ft

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Here, the pressure at end B of the concrete pipe is pA = gwhB = 162.4 lb>ft3 15.5 ft - 0.5 ft2 = 312 lb>ft2

Also, the water is drawn from a large reservoir where the level of water at A can be assumed constant and so VA _ 0. Since A is exposed to atmosphere, pA = patm = 0. With reference to the datum set through B, zA = 8 ft - 0.5 ft = 7.5 ft and zB = 0. Applying Bernoulli’s equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 312 lb>ft2

0 + 0 + (32.2 ft>s2)(7.5 ft) = a

62.4 lb>ft

32.2 ft>s2

VB = 12.69 ft>s

+

3

b

V B2 + 0 2

Then, the flow rate through the pipe is given by Q = VBA = 112.69 ft>s23p10.5 ft2 2 4 = 9.966 ft3 >s = 9.97 ft3 >s

388

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Ans.

Ans: Q = 9.97 ft3 >s

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5–11. 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 A C 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.

Ans: (pB)abs = 73.0 kPa 389

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*5–12. The jet airplane is flying in still air, A, at an altitude of 4 km. If the air flows past point C near the wing at 90 m>s, measured relative to the plane, determine the difference in pressure between the air near the leading edge B of the wing and point C.

80 m!s A C 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 0.8194 kg>m3

+

( 80 m>s ) 2 2

+ 0 =

pC 0.8194 kg>m3

+

( 90 m>s ) 2 2

+ 0 Ans.

pB - pC = 3.32 kPa

Ans: pB - pC = 3.32 kPa 390

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5–13. A fountain is produced by water that flows up the tube at 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 3

1000 kg>m

+

V02 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;

3

0.08 m >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 391

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5–14. A fountain is produced by water that flows up the tube at 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

5 mm

A

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 1000 kg>m3 p = Continuity requires that Q = VA AA;

Q = VB AB;

+

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

392

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5–14. 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.

r2

d 73.0 kPa,

393

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5–15. Air is drawn into the 200-mm-diameter cylinder through the opening at A. If the piston is moving downward at a constant velocity of 10 m>s, determine the average pressure within the cylinder and the force required to move the piston. Take ra = 1.23 kg>m3.

200 mm

A

B

SOLUTION

F

Assume that air is incompressible and inviscid and the flow is steady then Bernoulli’s equation can be applied between points C and B on the central streamline, where C is an exterior point.

FB 5 PBAB

pC pB VC2 VB2 + + gzC = + + gzC ra ra 2 2 Since the density of air is small, the elevation terms can be neglected. Since the air is taken from the atmosphere, which is a large reservoir, pC = patm = 0 and VC = 0. Then F

pB VB2 0 + 0 = + ra 2 pB =

(a)

(10 m>s)2 VB2 ra = - c d 11.23 kg>m3 2 2 2 = - 61.5 Pa

Ans.

The negative pressure indicates suction or partial vacuum. Referring to FBD of piston, Fig. a, ( + c )ΣFy = 0;

pBAB - F = 0

F = pBAB = 161.5 N>m2 3p10.1 m2 2 4 =1.932 N

Ans.

=1.93 N

Ans: pB = - 61.5 Pa F = 1.93 N 394

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*5–16. The level of mercury in the manometer has the reading shown. Determine the velocity of the water flowing from the nozzle at B. Neglect any head losses. Take gHg = 846 lb>ft3.

A

6 in.

30 in.

3 ft

15 in.

B 2 in.

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Here, VA = 0 since point A is a stagnation point. Also, pB = patm = 0 since the water is discharged into the atmosphere at B. The pressure at A can be determined from the measurement of the manometer shown in Fig. a

A

hAC 5 D

pA + gwhAC - gHgheD = patm

5 15 hCD 5 ft 5 ft 4 12

5 5 pA + a62.4 lb>ft3 a ft b - 1846 lb>ft3 2 a ft b = 0 2 4

30 5 ft 5 ft 12 2

C

pA = 901.51 lb>ft2

with reference to the datum set through point B, zA = 3 ft and zB = 0. Applying Bernoulli’s equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 901.5 lb>ft2 a

62.4 lb>ft3 32.2 ft>s2

b

+ 0 + 132.2 ft>s2 213 ft2 = 0 +

VB2 + 0 2 Ans.

VB = 33.52 ft>s = 33.5 ft>s

Ans: VB = 33.5 ft>s 395

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5–17. A fountain ejects water through the two nozzles A and B, which have inner diameters of 10 mm. If the velocity of the flow at point C in the 50-mm-diameter pipe is 2 m>s, determine the pressure in the pipe at this point and the velocity of the water through each nozzle.

B 2m

A 1.2 m C

SOLUTION

50 mm

Since the water can be considered as an ideal fluid (incompressible and inviscid) the Bernoulli’s equation is applicable. Applying this equation between C and A,

2 m!s

pC pA VC2 VA2 + + gzC = + + gzA rw rw 2 2 with reference to the datum set at c, zC = 0 and zA = 1.2 m. Also, the water is ejected into the atmosphere at A, pA = 0. Then pC 3

1000 kg>m

+

12 m>s2 2 2

Between C and B,

+ 0 = 0 +

VA2 + 19.81 m>s2 211.2 m2 2

pC = 1500 V A2 + 97722Pa

(1)

pC pB VC2 VB2 + + gzC = + + gzB rw rw 2 2 Here, zB = 3.2 m and also pB = 0 (Point B is exposed to atmosphere) pC 1000 kg>m3

+

12 m>s2 2 2

+ 0 = 0 +

VB2 + 19.81 m>s2 213.2 m2 2

pC = (500V B2 + 29392)Pa

(2)

The fixed control volume considered contains the water within the pipe. The density of the water is constant and the average velocities will be used. Then the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VCAC + VAAA + VBAB = 0 - 12 m>s23p10.025 m2 2 4 + VA 3p10.005 m2 2 4 + VB 3p10.005 m2 2 4 = 0 VA + VB = 50

(3)

Solving Eqs. (1), (2), and (3) VA = 25.39 m>s = 25.4 m>s VB = 24.61 m>s = 24.6 m>s

Ans.

3

Ans.

pC - 332.16(10 ) Pa = 332 kPa

Ans: VA = 25.4 m>s, VB = 24.6 m>s, pC = 332 kPa 396

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5–18. A fountain ejects water through the two nozzles A and B, which have inner diameters of 10 mm. Determine the velocity of the water stream passing through each of the nozzles and the pressure at point C if the flow through the 50-mm-diameter pipe is 0.005 m3 >s.

B 2m

A 1.2 m C

SOLUTION

50 mm

Since the water can be considered as an ideal fluid (incompressible and inviscid), the Bernoulli equation is applicable. Applying this equation between C and A,

2 m!s

pC pA VC2 VA2 + + gzC = + + gzA rw rw 2 2 with reference to the datum set at c, zc = 0 and zA = 1.2 m. Also, water is ejected into atmosphere at A, pA = 0 the velocity of the flow at point C is given by VC =

0.005 m3 >s Q 8 = = m>s. p AC p10.035 m2 2 pC 1000 kg>m3

+

a

2 8 m>s b VA2 p + 0 = 0 + + 19.81 m>s2 211.2 m2 2 2

pc = 1500V A2 + 8529.722 Pa

Between C and B

(1)

pc pB VC2 VB2 + + gzC = + + gzB rw rw 3 2 Here, zB = 3.2 m and also pA = 0. (Point B is exposed to atmosphere.) pc 1000 kg>m3

+

a

2 8 m>s b VB2 p + 0 = 0 + + 19.81 m>s2 213.2 m2 2 2

pc = 1500 V B2 + 28149.722 Pa

(2)

The fixed control volume considered contained the water within the pipe. The density of the water is constant and the average velocities will be used. Then the continuity equation can be simplified as 0 0dV + rV # dA = 0 0t Lcv Lcs 0 - VCAC + VAAA + VBAB = 0 Here Q = VCAC = 0.005 m3 >s,

- 0.005 m3 >s + VA 3p10.005 m2 2 4 + VB 3p10.005 m2 2 4 = 0 VA + VB =

200 p

(3)

Solving Eqs. (1), (2), and (3) VA = 32.14 m>s = 32.1 m>s Ans.

VB = 32.52 = 31.5 m>s 3

Ans.

pC = 524.99(10 ) Pa = 525 kPa

Ans: VA = 32.1 m>s, VB = 31.5 m>s, pC = 525 kPa

397

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5–19. Heavy rain has caused reservoir A to reach a height of 6 ft above the pipe at B. Determine the flow through the concrete culvert buried beneath the embankment. Neglect any head losses.

A 6 ft

1.5 ft B

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Here, the pressure at end B of the concrete. Culvert is pB = 0. Also, the water is drawn from a large reservoir where the level of water at A can be assumed constant, and so VA _ 0. Since A is exposed to the atmosphere, pA = patm = 0. With reference to the datum set through B, zA = 6 ft and zB = 0. Applying Bernoulli’s equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 0 + 0 + 132.2 ft>s2 216 ft2 = 0 +

VB2 + 0 2

VB = 2132.22162 = 19.66 ft>s

Then, the flow through the culvert is given by Q = VBA = (19.66 ft>s)3p(0.75 ft)2 4 = 34.74 ft3 >s = 347.7 ft3 >s

398

M05_HIBB9290_01_SE_C05_ANS.indd 398

Ans.

Ans: Q = 34.7 ft3 >s

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*5–20. A fire hydrant supplies water under a pressure of 35 psi. If a 3-in.-diameter hose is connected to it at A and the hose extends 80 ft to the fire truck inlet at B, determine the pressure of the water as it arrives at B. The friction loss is 15 ft for every 100 ft of hose. The inlet at B is 1.5 ft higher than the hydrant outlet.

B A

SOLUTION The fixed control volume considered contains the water within the hose. With reference to the datum set through A, zB = 1.5 ft and zB = 0. Since the diameter is constant throughout the hose, VA = VB = V. Write the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g a35

lb 12 in. 2 ba b 2 pB VA2 VA2 1 ft 15 ft in + + 0 + 0 = + + 1.5 ft + 0 + a b(80 ft) 2g 2g 100 ft 62.4 lb>ft3 62.4 lb>ft3 pB = a4197.6

lb 1 ft 2 ba b = 29.15 psi 2 12 in. ft

Ans.

Ans: pB = 29.15 psi 399

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5–21. 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. 2

3

1000 kg>m

+

B h

BD

= 0.2 m

2

pA pB VA VB + + gzA = + + gzB r r 2 2 pA

= 0.15 m A

C

pB VA2 + 0 = + 0 + 0 2 1000 kg>m3

D

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: VA = 3.51 m>s 400

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5–22. The sewage siphon regulates the level of water in the large holding tank A. Determine the flow through the 3-in.-diameter pipe when the water levels are as shown. Neglect any head losses.

A 4 ft

5 ft

B 3 ft

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Here, the pressure at end B of the pipe is pB = gwhB = 162.4 lb>ft3 2hB = 62.4 hB

Also, the water is drawn from a large reservoir where the level of water at A can be assumed constant and so VA _ 0. Since A is exposed to the atmosphere, pA = patm = 0. With reference to the datum set through B, zA = hB + 19 ft - 3 ft2 = hB + 6 ft and zB = 0. Applying Bernoulli’s equation between points A and B pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 62.4 hB

0 + 0 + (32.2 ft>s2(hB + 6 ft) = a

62.4 lb>ft3

32.2 ft>s2 VB = 19.66 ft>s

+ b

V B2 + 0 2

Thus, the flow through the pipe is given by Q = VBA = (19.66 ft>s) c pa

2 1.5 ft b d = 0.9649 ft3 >s = 0.965 ft3 >s 12

401

M05_HIBB9290_01_SE_C05_ANS.indd 401

Ans.

Ans: Q = 0.965 ft3 >s

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5–23. If the manometer contains mercury, determine the volumetric flow of air through the circular duct. Take rHg = 13 550 kg>m3 and ra = 1.23 kg>m3.

A

V

B

10 mm

2m

SOLUTION

A

B

The air is assumed to be an ideal fluid (incompressible and inviscid) and the flow is steady. Then Bernoulli’s equation is applicable. Applying this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2

hHg 5 0.01 m

Since point B is a stagnation point, VB = 0. Since the density of air is small, the elevation terms can be neglected.

(a)

pB VA2 + = + 0 3 2 1.23 kg>m 1.23 kg>m3 pA

pB - pA = 0.615 V A2

(1)

Referring to Fig. a, write the manometer equation from points A to B realizing that the pressure due to air column is negligible, since the density of air is small. pA + rHgghHg - pB = 0 pA + (13550 kg>m3)(9.81 m>s2)(0.01 m) - pB = 0 (2)

pB - pA = 1329.255 Equating Eq. (1) and (2) 0.615 VA2 = 1329.255 VA = 46.49 m>s Thus, the volumetric flow is Q = VA = (46.49 m>s)[p(1 m)2] = 146.06 m3 >s = 146 m3 >s

Ans.

402

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Ans: Q = 146 m3 >s

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*5–24. One method of producing energy is to use a tapered channel (TAPCHAN), which diverts sea water into a reservoir as shown. 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 hB at the back B of the channel to prevent the water from entering the reservoir.

B A

C

VA 5 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: hB = 3.32 m 403

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5–25. A 6-m-high chimney has a circular opening A at its base. If air flows into it at 3.75 m>s, determine the speed of the air as it exits at the top B. Also, what is the pressure difference between A and B? Take ra = 1.20 kg>m3. B

600 mm

SOLUTION Since the air can be assumed to be an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Consider the fixed control volume to be the air within the chimney; the continuity equation gives

3.75 m!s 6m

A

200 mm

0 rdV + rV # dA = 0 0t L Lcs 0 - VAVA + VBAB = 0 - 13.75 m>s2[p10.2 m2 2] + VB[p10.6 m2 2] = 0 VB = 0.4167 m>s = 0.417 m>s

Ans.

Applying Bernoulli’s equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Setting the datum at A,

= a

1.20 kg>m3 2

pA pB VA2 VB2 + + 0 = + + gh ra ra 2 2 ra ∆p = pA - pB = 1 VB2 - VA2 2 + ra gh 2

b 310.4167 m>s2 2 - 13.75 m>s2 2 4 + 11.20 kg>m3 19.81 m>s2 216 m2

Ans.

= 62.3 Pa

Ans: VB = 0.417 m>s, pA - pB = 62.3 Pa 404

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5–26. When the valve at A is opened, the initial discharge of oil is 0.9 ft3 >s. Determine the depth h of oil in the tank. Take rke = 1.58 slug>ft3 and ro = 1.78 slug>ft3.

C Kerosene

1.5 ft

B

Oil 0.25 ft

h

A

SOLUTION Using the oil discharge, the initial velocity of the oil at A is Q = VAAA;

0.9 ft3 >s = VA 3p(0.25 ft)2 4

VA = a

57.6 b ft>s p

Since the oil can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady at the instant considered, Bernoulli’s equation is applicable. pB pA VB2 VA2 + + gzB = + + gzA ro ro 2 2 Since oil is discharged from a large tank, VA ≈ 0. The pressure at B is pB = rkeghbc = 11.58 slug>ft2 2132.2 ft>s2 2 = 76.314 lb>ft2 since the oil discharges to the atmosphere, pA = 0. If the datum is set at A, zA = 0 zB = h. 76.314 lb>ft 1.78 slug>ft2

+ 0 + 132.2 ft>s2 2 = 0 +

a

2 57.6 ft>s b p + 0 2

Ans.

h = 3.888 ft = 3.89 ft

Ans: h = 3.89 ft 405

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5–27. If the velocity of water changes uniformly along the transition from VA = 10 m>s to VB = 4 m>s, determine the pressure difference over the distance 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. p(x) V 2(x) pA VA2 + + gzA = + + gzx r r 2 2 pA 1000 kg>m3

+

( 10 m>s ) 2 2

+ 0 =

p(x)

( 1000 kg>m3 )

+

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 2

= ( 30x - 4.5x2 ) kPa

+ 0

Ans.

Ans: p(x) - pA = (30x - 4.5 x2) kPa 406

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*5–28. 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. pA pC 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

+ 0 Ans.

6ms VC V A = 10 m s

VB = 4 m s A

x = 1.5 m 2m C

B

0.5 m

(a )

Ans: pC - pA = 34.9 kPa 407

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5–29. Air is pumped into the top of the tank so that the pressure at B is 40 kPa. Determine the discharge of water through the drainpipe when h = 5 m. B

h A 0.15 m

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points B and A, pB pA VB2 VA2 + + gzB = + + gzA rw rw 2 2 Since the tank is a large reservoir, VB ≃ 0. Also, the water is discharged to the atmosphere at A, pA = patm = 0, with reference to the datum set through A, zA = 0 and zB = 5 m 401103 2 N>m2 1000 kg>m3

+ 0 + 19.81 m>s2 215 m2 = 0 +

V A2 + 0 2

VA = 13.345 m>s

Thus, the discharge of water through value at A is Q = VAAA = (13.345 m>s)3p(0.075)2 4

= 0.2358 m3 >s = 0.236 m3 >s

408

M05_HIBB9290_01_SE_C05_ANS.indd 408

Ans.

Ans: Q = 0.236 m3 >s

01/03/17 4:11 PM

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5–30. Air is pumped into the top of the tank so that the pressure at B is 40 kPa. Determine the discharge of the water as a function of the height h. Plot the discharge (vertical axis) versus h for 1 m … h … 6 m. Give values for increments of ∆h = 1 m.

B

h A 0.15 m

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points B and A, pB pA VB2 VA2 + gzB = + + gzA rw rw 2 2 Since the tank is a large reservoir, VB = 0. Also, the water is discharged to the atmosphere at A, pA = patm = 0. With reference to the datum set through A, zA = 0 and zB = h. 40(103) N>m2 1000 kg>m3

+ 0 + 19.81 m>s2 2h = 0 +

VA =

1 280

V A2 + 0 2

+ 19.62h 2 m>s

Thus, the discharge of water through valve at A is Q = VAAA Q =

1 280

+ 19.62h 2 3p10.075 m2 2 4

= 30.005625p280 + 19.62h4 m3 >s

Ans.

The values of h and the corresponding of Q are tabulated: h(m) Q 1 m >s 2 3

1 0.176

2 0.193

3 0.208

4 0.222

5 0.236

6 0.248

The plot of Q vs h is shown in Fig. a. Q(m3ys) 0.30 0.25 0.20 0.15 0.10 0.05 0

1

2

3

4

5

6

h (m)

(a)

409

M05_HIBB9290_01_SE_C05_ANS.indd 409

Ans: Q = 30.005625 p 280 + 19.62 h4 m3 >s

01/03/17 4:11 PM

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5–31. The pipe assembly is mounted vertically. Determine the pressure at A if the velocity of the water ejected from B is 0.75 m>s.

200 mm B

1.2 m

SOLUTION Consider the fixed control volume containing the water in the pipe. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified as

A 80 mm

0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 -VA 3p10.04 m2 2 4 + 10.75 m>s23p10.1 m2 2 4 = 0

VA = 4.6875 m>s

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 with reference to the datum set through point A, zA = 0 and zB = 1.2 m. Since point B is exposed to the atmosphere, pB = patm = 0. pA 1000 kg>m3

+

14.6875 m>s2 2 2

+ 0 = 0 +

10.75 m>s2 2 2

pA = 1066.91103 2 Pa = 1.07 kPa

+ 19.81 m>s2 211.2 m2

Ans.

Ans: pA = 1.07 kPa 410

M05_HIBB9290_01_SE_C05_ANS.indd 410

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*5–32. A river has an average width of 5 m. Just after its flow falls 2 m to the lower elevation, the depth becomes h = 0.8 m. Determine the volumetric discharge.

1.2 m A

B

2m

h

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B on the streamline along the water surface, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Since the surface of water is exposed to the atmosphere, pA = pB = patm = 0 with reference to the datum set through point B, zA = 2 m + 1.2 m - 0.8 m = 2.4 m and zB = 0. 0 +

VA2 VB2 + 19.81 m>s2 212.4 m2 = 0 + + 0 2 2 VB2 - VA 2 = 47.088 m2 >s2

(1)

Consider the fixed control volume that contains the water between the cross sections of the river through points A and B. Since the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - VA 35 m11.2 m24 + VB 35 m10.8 m24 = 0 VA = 0.6667VB

(2)

Solving Eqs. (1) and (2) VB = 9.206 m>s

VA = 6.138 m>s

Thus, the discharge is Q = VAAA = 16.138 m>s235 m11.2 m24 = 36.83 m3 >s = 36.8 m3 >s

411

M05_HIBB9290_01_SE_C05_ANS.indd 411

Ans.

Ans: Q = 36.8 m3 >s

01/03/17 4:12 PM

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5–33. A river has an average width of 5 m and flows with an average velocity of 6 m>s at A. Determine its depth h just after the flow falls 2 m.

1.2 m A

B

2m

h

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B on the streamline along the water surface, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Since the surface of water is exposed to the atmosphere, pA = pB = patm = 0. With reference to the datum set through point B, zA = 2 m + 1.2 m - h = 13.2 - h2 m and zB = 0 0 +

16 m>s2 2 2

+

1 9.81 m>s2 2 13.2

- h2 = 0 +

V B2 + 0 2

V B2 = 98.784 - 19.62h

(1)

Consider the fixed control volume that contains the water within the cross section of the river through points A and B. Since the density of the water is constant and the average velocity will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - 16 m>s235 m11.2 m24 + VB 35 m1h24 VB = a

Substitute Eq. (2) into (1) a

7.2 b m>s h

(2)

7.2 2 b = 98.784 - 19.62h h

19.62h3 - 98.784h2 + 51.84 = 0 Solving numerically, Ans.

h = 0.7888 m = 0.789 m

Ans: h = 0.789 m 412

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5–34. 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 inviscid) 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

Ans: pA = 1.23 kPa 413

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5–35. 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 ft3 >s

Ans.

414

M05_HIBB9290_01_SE_C05_ANS.indd 414

Ans: Q = 66.3 ft3 >s

01/03/17 4:12 PM

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*5–36. Air enters the circular nozzle with a velocity of 30 m>s and then exits to the atmosphere at B. Determine the pressure at A. The density of air is assumed to be constant at ra = 1.20 kg>m3.

1.2 m 0.4 m A

B

SOLUTION Assume that air is an ideal fluid (incompressible and inviscid) and the flow is steady then Bernoulli’s equation is applicable. Applying this equation between points A and B on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Since point B is exposed to the atmosphere, pB = patm = 0. The elevation terms can be neglected since the density of air is considered small. pA

+

1.20 kg>m3 pA =

130 m>s2 2 2

30.61VB2

= 0 +

V B2 2 (1)

- 90024 Pa

The fixed control volume considered contained the air within the nozzle. Since the density of air is assumed to be constant and average velocities will be used, the continuity equation can be simplified as 0 rd V + rV # dV 0t Lcv L 0 - VAAA + VBAB = 0 - 130 m>s23p10.2 m2 2 4 + VB 3p10.6 m2 2 4 = 0 VB = 3.333 m>s

Substituting this result into Eq. (1) pA = 30.613.3332 - 90024

Ans.

= -533.33 Pa = - 533 Pa

The negative sign indicates that the pressure at A is partial vacuum. A difference of less than 1 kPa from atmospheric confirms the reasonableness of the incompressibility assumption.

Ans: pA = - 533 Pa 415

M05_HIBB9290_01_SE_C05_ANS.indd 415

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5–37. Water flows through the transition at 0.3 m3 >s, which causes the water to rise within the piezometer at A to a height of hA = 350 mm. Determine the height hB. hB

hA B

400 mm

A 250 mm

SOLUTION The fixed control volume considered contains the water in the transition between cross section through A and B. From the given discharge, 19.2 Q = VAAA; 0.3 m3 >s = VA 3p10.125 m2 2 4 VA = m>s p 7.5 Q = VBAB; 0.3 m3 >s = VB 3p10.2 m2 2 4 VB = m>s p

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation can be applied. Applying this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 The static pressures at A and B are given by pA = rwghA = rw 19.81 m>s2 210.35 m2 = 13.4335 rw 2 Pa pB = rwghB = rw 19.81 m>s2 21hB 2 = 19.81 rwhB 2 Pa

with reference to the datum set to coincide with the horizontal streamline connecting A and B, zA = zB = 0. 3.4335rw + rw

a

2 2 19.2 7.5 a m>s b m>s b 9.81rwhB p p + 0 = + + 0 rw 2 2

Ans.

hB = 1.963 m = 1.96 m

Ans: hB = 1.96 m 416

M05_HIBB9290_01_SE_C05_ANS.indd 416

03/03/17 2:37 PM

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5–38. If the water in piezometers A and B rises to hA = 250 mm and hB = 950 mm, respectively, as it flows through the transition, determine the volumetric flow. hB

hA B

400 mm

A 250 mm

SOLUTION The fixed control volume considered contains the water in the transition between the cross sections through A and B since the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VBAB + VAAA = 0 -VB 3p10.2 m2 2 4 + VA 3p10.125 m2 2 4 = 0 VA = 2.56 VB

(1)

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation can be applied. Applying this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 The static pressures at A and B are given by pA = rwghA = rw 19.81 m>s2 210.25 m2 = 12.4525rw 2 Pa pB = rwghB = rw 19.81 m>s2 210.95 m2 = 19.3195rw 2 Pa

with reference to the datum set to coincide with the horizontal streamline connecting A and B, zA = zB = 0. 2.4525rw 9.3195rw VA2 VB2 + + 0 = + + 0 rw rw 2 2 V A2 - V B2 = 13.734

(2)

Solving Eqs. (1) and (2) VB = 1.573 m>s

VA = 4.026 m>s

Then, the discharge is given by Q = VBAB = 11.573 m>s23p10.2 m2 2 40.1976 m3 >s = 0.198 m3 >s

417

M05_HIBB9290_01_SE_C05_ANS.indd 417

Ans.

Ans: Q = 0.198 m3 >s

01/03/17 4:12 PM

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5–39. Air at 20°C flows through the circular duct such that the absolute pressure is 100.8 kPa at A, and 101.6 kPa at B. Determine the volumetric discharge through the duct.

B

A

250 mm 150 mm

SOLUTION Assume that air is an ideal fluid (incompressible and inviscid) and the flow is steady. Then, Bernoulli’s equation is applicable. Apply this equation between points A and B on central streamline, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Since the pressures at A and B are very close to atmospheric pressure, them from the table in Appendix A, ra = 1.202 kg>m3 at T = 20°C. This density remains constant, since the air is assumed to be incompressible. The elevation terms can be neglected since the density of air is considered small. 100.81103 2 N>m2 1.202 kg>m3

+

101.61103 2 N>m2 VA2 VB2 = + 2 2 1.202 kg>m3

VA2 - VB2 = 1331.11

(1)

The fixed control volume considered contains the air within the duct. Since the density of the air is constant and average velocities will be used, the continuity equation can be simplified 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - VA 3p10.15 m2 2 4 + VB 3p(0.25 m)2 4 = 0 VA = 2.7778 VB

(2)

Solving Eqs (1) and (2) VB = 14.08 m>s

VA = 39.11 m>s

Then the discharge is Q = VAAA = 139.11 m>s23p10.15 m2 2 4 = 2.764 m3 >s = 2.76 m3 >s

418

M05_HIBB9290_01_SE_C05_ANS.indd 418

Ans.

Ans: Q = 2.76 m3 >s

01/03/17 4:12 PM

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*5–40. Air at 20°C flows through the circular duct such that the absolute pressure at A is 100.8 kPa and the velocity is 40 m>s. Determine the absolute pressure and the velocity of the air at B.

B

A

250 mm 150 mm

SOLUTION Assume that air is an ideal fluid (incompressible and inviscid) and the flow is steady. Then, Bernoulli’s equation is applicable. Applying this equation between points A and B on central streamline, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Since the pressures at A and B (assumed) are very close to atmospheric pressure, then from the table in Appendix A, ra = 1.202 kg>m3 at T = 20°C. This density remain constant, since the air is assumed to be incompressible. The elevation terms can be neglected since the density of air is considered small. 100.8(103) N>m2 1.202 kg>m3

+

140 m>s2 2 2

=

pB 1.202 kg>m3

+

V B2 2

pB = 1101761.6 - 0.601VB2 2 Pa

(1)

The fixed control volume considered contains the air within the duct. Since the density of the air is constant and average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - 140 m>s23p10.15 m2 2 4 + VB 3p10.25 m2 2 4 = 0 VB = 14.4 m>s

Ans.

pB = 3101761.6 - 0.601114.42 24

Ans.

Substitute this result into Eq. (1), = 101636.98 Pa = 102 kPa

Since pB is quite close to atmospheric pressure, the assumption is OK!

Ans: VB = 14.4 m>s pB = 102 kPa 419

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5–41. Water flows through the pipe transition with a velocity 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. pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA 3

1000 kg>m

+

( 6 m>s ) 2 2

pB

+ 0 =

3

1000 kg>m pB - pA = 16 875 Pa

+

( 1.5 m>s ) 2 2

+ 0 (1)

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)

Ans: h = 137 mm 420

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5–42. Water from a faucet tapers from a diameter of 0.5 in. to 0.3 in. after falling 10 in. Determine the 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 421

M05_HIBB9290_01_SE_C05_ANS.indd 421

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5–43. 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 m = rVAAA = a slug>ft3 b ( 2.827 ft>s ) c a ft b d 32.2 4 12 = 0.00747 slug>s

Ans.

Ans: # m = 0.00747 slug>s 422

M05_HIBB9290_01_SE_C05_ANS.indd 422

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*5–44. Water flows horizontally past the pitot tube such that mercury within the manometer is displaced as shown. Determine the mass flow if the pipe has a diameter of 100 mm. Take rHg = 13 550 kg>m3.

A B 50 mm 100 mm

75 mm

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B on the central streamline,

A

B

hAC 5 0.1 m

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2

hBD 5 0.125 m

C D

Here, VB = 0 since point B is a stagnation point. The datum is set to coincide with the central streamline. Then zA = zB = 0. hCD 5 0.025 m

pB VA2 + + 0 = + 0 + 0 3 2 1000 kg>m 1000 kg>m3 pA

pB - pA = 500 VA2

(a)

(1)

Referring to Fig. a, write the manometer equation from A to B. pA + rwghAC + rHgghCD - rwghBD = pB 3

pA + 11000 kg>m 219.81 m>s2 210.1 m2 + 113550 kg>m3 219.81 m>s2 210.025 m2 - 11000 kg>m3 219.81 m>s2 210.125 m2 = pB pB - pA = 3077.8875

(2)

Equating Eqs. (1) and (2), 3077.8875 = 500VA2 VA = 2.481 m>s Then, the mass flow is # m = rwVAAA = 11000 kg>m3 212.481 m>s23p10.05 m2 2 4 =19.49 kg>s = 19.5 kg>s

Ans.

Ans: # m = 19.5 kg>s 423

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5–45. If the pressure at A is 325 kPa, and the velocity of the water at this point is 2.25 m>s, determine the pressure in the pipe at B if the pressure at C is 175 kPa. The flow occurs in the horizontal plane.

C 20 mm

A 80 mm B 40 mm

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and C, pA pc VC2 VA2 + + gzA = + + gzC rw rw 2 2 Since the flow occurs in the horizontal plane, the elevation terms can be ignored. 3251103 2 N>m2 1000 kg>m3

+

12.25 m>s2 2 2

=

1751103 2 N>m2 1000 kg>m3

+

V C2 2

VC = 17.4 7 m>s

Between A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 3251103 2 N>m2 3

1000 kg>m

+

12.25 m>s2 2 2

=

pB 3

1000 kg>m

+

V B2 2

pB = 310001327.53 - 0.5V B2 24 Pa

(1)

The fixed control volume considered contains the water within the pipe. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCAC = 0 - 12.25 m>s23p10.04 m2 2 4 + VB 3p10.02 m2 2 4 + 117.47 m>s23p10.01 m2 2 4 = VB = 4.633 m>s

Substitute this result into Eq. (1), pB = 10003327.53 - 0.5 1 4.6332 2 4 = 316.80 1 103 2 Pa = 317 kPa

Ans.

Ans: pB = 317 kPa 424

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5–46. If the pressure at A is 215 kPa, and the velocity of the water at this point is 2.25 m>s, determine the pressure in the pipe at B if the water is discharged into the atmosphere at C.

C 20 mm

A 80 mm B 40 mm

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and C, pA pC VC2 VA2 + + gzA = + + gzc rw rw 2 2 Since the water is discharged into the atmosphere at C, pC = patm = 0. Also, the elevation terms can be ignored since the flow occurs in the horizontal plane. 2151103 2 N>m2 3

1000 kg>m

+

12.25 m>s2 2 2

= 0 +

VC2 2

VC = 20.86 m>s

Between A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 215(103) N>m2 3

1000 kg>m

+

12.25 m>s2 2 2

=

pB 3

1000 kg>m

+

V B2 2

pB = 31000(217.53 - 0.5VB2)4 Pa

The fixed control volume considered contained the water within the pipe. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified. 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCAC = 0 - 12.25 m>s23p10.04 m2 2 4 + VB 3p10.02 m2 2 4 + 120.86 m>s23p10.01 m2 2 4 = 0 VB = 3.785 m>s

Substitute this result into Eq. (1), pB = 10003217.53 - 0.5 1 3.7852 2 4 = 210.371102 2 Pa = 210 kPa

Ans.

Ans: pB = 210 kPa 425

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5–47. 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 m11.5 m2] + VB [0.2 m11.5 m2] = 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. pA pB 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

426

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Ans.

Ans: Q = 1.65 m3 >s

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*5–48. If the difference in the level of mercury within the manometer is 80 mm, determine the volumetric flow of the water. Take rHg = 13 550 kg>m3.

100 mm 40 mm B

A

80 mm

SOLUTION B

Referring to Fig. a, the manometer equation written from A to B along the centerline is

A

pA + rwghAC + rHgghCD - rwghBD = pB pA + 11000 kg>m3 219.81 m>s2 21hAC 2 + 113550 kg>m3 219.81 m>s2 210.08 m2 - 11000 kg>m3 219.81 m>s2 21hAC + 0.08 m2 = pB pB - pA = 9849.24

(1)

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B,

C

hBD 5 a 1 0.08 m D

hAC

hCD 5 0.08 m

(a)

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 with reference to the datum set to coincide with the horizontal streamline connecting A and B, zA = zB = 0. pA 3

1000 kg>m

+

pB VA2 VB2 + 0 = + + 0 3 2 2 1000 kg>m

pB - pA = 500(VA2 - VB2)

(2)

The fixed control volume considered contained the water within the transition. Since the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VBAB + VAAA = 0 - VB 3p10.05 m2 2 4 + VA 3p10.02 m2 2 4 = 0

(3)

VA2 - VB2 = 19.69848

(4)

Substitute Eq. (1) into (2)

VA = 6.25VB

Solving Eqs. (3) and (4) Then the discharge is

VB = 0.7194 m>s

VA = 4.496 m>s

Q = VBAB = 10.7194 m>s23p10.05 m2 2 4 = 0.005650 m3 >s

= 0.00565 m3 >s

Ans.

427

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Ans: Q = 0.00565 m3 >s

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5–49. As the air flows through the transition from A to B, its absolute pressure drops from 101.8 kPa to 101.3 kPa. If the temperature of the air remains constant at 20°C, determine the mass flow of the air through the duct.

400 mm 200 mm

A

B

SOLUTION Here, the change in pressure is small. Therefore, air can be assumed incompressible. Also, the flow is steady. Thus, Bernoulli equation is applicable. Applying this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Since the pressure at A and B is very close to atmosphere pressure, from the table in Appendix A, ra = 1.202 kg>m3 at T = 20°C. Since the density of air is considered small, the elevation terms are negligible. 101.81103 2 N>m2 1.202 kg>m3

+

101.31103 2 N>m3 VA2 VB2 = + 2 2 1.202 kg>m3

VB2 - VA2 = 831.95

(1)

The fixed control volume considered contained the air within the transition since the density of the air is assumed to be constant and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 -VA 3p10.2 m2 2 4 + VB 3p10.1 m2 2 4 = 0

(2)

VA = 0.25VB

Solving Eqs. (1) and (2) VB = 29.79 m>s

VA = 7.447 m>s

Then, the mass flow rate is # ma = raVBAB = 11.202 kg>m3 2129.79 m>s23p10.1 m2 2 4 = 1.124 kg>s = 1.12 kg>s

Ans.

Ans: # ma = 1.12 kg>s 428

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5–50. Air flows through the transition with a velocity of 8 m>s at A. If the absolute pressure at A is 101.8 kPa, determine the pressure at B. The temperature of the air remains constant at 20°C.

400 mm 200 mm

A

B

SOLUTION The fixed control volume contains the air within the transition. Since the density of the air is assumed to be constant (incompressible) and the average velocities will be used, the continuity equation can be simplified as 0 rdV + rV # dA = 0. 0t Lcv Lcs 0 - VAAA + VBAB = 0 - (8 m>s)3p10.02 m2 2 4 + VB 3p10.01 m2 2 4 = 0 VB = 32.0 m>s

The change in pressure is assumed to be small. Therefore, air can be assumed incompressible. Also, the fluid is steady. Thus Bernoulli’s equation is applicable. Applying this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB ra ra 2 2 Since the pressure at A is very close to atmospheric pressure from the table in Appendix A, ra = 1.202 kg>m3 at T = 20°C. Since the density of air is considered small, the elevation terms are negligible. 101.81103 2 N>m2 1.202 kg>m3

+

18 m>s2 2 2

=

pB 1.202 kg>m3

+

132.0 m>s2 2 2

pa = 101.2231103 2Pa = 101.2 kPa

Ans.

The very small change in pressure is consistent with the assumption that the air is incompressible and the density remains constant.

Ans: pB = 101.2 kPa 429

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5–51. Determine the height h of the water column and the 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

3 in. C

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 a10

lb 12 in. 2 ba b 2 16 ft>s2 2 1 ft in + = 0 + 0 + 132.2 ft>s2 21h + 0.25 ft2 2 62.4 lb>ft3 32.2 ft>s2

Ans.

h = 23.39 ft = 23.4 ft From A to B, pA pB VA2 VB2 + + gzA = + + gzB r r 2 2 a10

lb 12 in. 2 ba b 2 16 ft>s2 2 VB2 1 ft in + + 0 = 0 + + 132.2 ft>s2 210.25 ft2 2 2 62.4 lb>ft3 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 cs cv 0 - VAAA + VBAB + VCAC = 0 - 16 ft>s2 £ pa

2 2 2 3 0.25 3 ft b § + 138.81 ft>s2 £ pa ft b § + VC £ pa ft b § = 0 12 12 12

Ans.

VC = 5.73 ft>s

Ans: h = 23.4 ft, VB = 38.8 ft>s VC = 5.73 ft>s 430

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*5–52. 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 of the water in the pipe at C.

D

h 0.5 in. B A

3 in. C

SOLUTION Since water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady the Bernoulli’s equation is applicable. Writing this equation between points A and D, realizing that pD = 0 (D is open to atmosphere) and VD = 0 (the water column achieves a maximum height at D). If the datum coincides with the Central Stream line, zA = 0, and zD = a

3 ft b + 30 ft = 30.25 ft . 12

pA pD VA2 VD2 + + gzA = + + gzD rw rw 2 2

a10 °

lb 12 in. 2 ba b 2 1 ft in

62.4 lb>ft

3

32.2 ft>s2

+

¢

VA2 2

+ 0 = 0 + 0 + 132.2 ft>s2 2130.25 ft2

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 zB = ft = 0.25 ft, 12 pB pD VB2 VD2 + + gzB = + + gzD rw rw 2 2 0 +

VB2 + 132.2 ft>s2 210.25 ft2 = 0 + 0 + 132.2 ft>s2 2130.25 ft2 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 1 - 21.49 ft>s2 £ pa

2 2 2 3 0.25 3 ft b § + 143.95 ft>s2 £ pa ft b § + VC £ pa ft b § = 0 12 12 12

Ans.

VC = 21.188 = 21.2 m>s

431

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5–52. Continued

Between points A and C, pA pC VC2 VA2 + + 0 = + = 0 rw rw 2 2 a10 °

12 in. 2 lb b a b 1 ft in2

62.4 lb>ft3 32.2 ft>s2

+

( 21.49 ft>s ) 2

¢

pC = a1452.62

2

pC

= °

62.4 lb>ft3 32.2 ft>s2

+

( 21.188 ft>s ) 2 2

¢

lb 1 ft 2 b = 10.1 psi b a ft2 12 in.

Ans.

Ans: VC = 21.2 m>s pC = 10.1 psi 432

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5–53. Determine the velocity out of the pipes at A and B if water flows into the Tee at 8 m>s and under a pressure of 40 kPa. The system is in the vertical plane.

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 433

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5–54. Air is pumped into the top A of the tank such that it has a pressure of 20 psi. Determine the velocity of the water when it strikes the ground at C. Consider the tank to be a large reservoir.

A

8 ft B C

3 ft

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable applying this equation between points A and B,

(VC)y 5 13.90 ftys

(VC)x 5 59.06 ftys u VC

pA pB V A2 V B2 + + gzA = + + gzB rw rw 2 2

(a)

Since the water is discharged into the atmosphere at B, pB = 0. Also since it is considered discharged from a large reservoir, the level of water at A can be assumed constant and VA ≃ 0 with reference to the datum set through B. zA = 8 ft and zB = 0. lb 12 in. 2 b a b V B2 1 ft in2 2 + 0 + 132.2 ft>s 218 ft2 = 0 + + 0 2 62.4 lb>ft3 a b 32.2 ft>s2

a20

VB = 59.06 ft>s

For horizontal motion, 1VB 2 x = VB = 59.06 ft>s. S Thus

1 S+ 2

1VC 2 x = 1VB 2 x = 59.06 ft>s S

For vertical motion, 1VB 2 y = 0, a = 32.2 ft>s2 T and 1Sc2 y = 3 ft T

(+T)

1VC 2 2y = 1VB 2 2y + 2a1sc 2 y

1VC 2 2y = 02 + 2132.2 ft>s2 213 ft2

1VC 2 y = 13.90 ft>s T 13.89964028

Thus, the magnitude of the velocity at C is

VC = 21VC 2 2x + 1VC 2 2y = 2159.06 ft>s2 2 + 113.90 ft>s2 2 = 60.67 ft>s = 60.7 ft>s Ans. And its direction, Fig. a, is defined by u = tan -1 c

1VC 2 y

d = tan -1 a

1VC 2 x

13.90 ft>s 59.06 ft>s

b = 13.24° = 13.2°

Ans.

Ans: VC = 60.7 ft>s, u = 13.2 ° 434

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5–55. Air is pumped into the top A of the tank. Determine the speed at which the water strikes the ground at C as a function of the pressure at A. Plot this velocity (vertical axis) versus the pressure pA for 0 … pA … 60 psi. Give values for increments of ∆pA = 15 psi. Consider the tank to be a large reservoir.

A

8 ft B C

3 ft

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. applying this equation between points A and B,

VC (ftys) 100

pA pB V A2 V B2 + + gzA = + + gzB rw rw 2 2

80

Since the water is discharged into the atmosphere at B, pB = 0. Also, it is considered discharged from a large reservoir, the level of water at A can be assumed constant and VA ≃ 0. With reference to the datum set through B, zA = 8 ft and zB = 0.

40

12 in. 2 pA a b 1 ft 62.4 lb>ft3 32.2 ft>s2

60

20 0

+ 0 + 132.2 ft>s2 218 ft2 = 0 +

VB =

3 1148.62pA

V B2 + 0 2

15

30

45

60

PA(psi)

(a)

+ 515.2 4 ft>s

For horizontal motion, 1VB 2 x = VB = 1 1148.62 pA + 515.22 ft>s S , Thus

1 S+ 2

1VC 2 x = 1VB 2 x = 1 1148.62 pA + 515.22 ft>s S

For vertical motion, 1VB 2 y = 0, a = 32.2 ft>s2 T and 1Sc2 y = 3 ft T

(+T)

1VC 2 2y = 1VB 2 2y + 2a1Sc2 y

1VC 2 2y = 02 + 2132.2 ft>s2 213 ft2 1VC 2 y = 13.90 ft>s T

Thus, the magnitude of the velocity at c is

VC = 21VC 2 2x + 1VC 2 2y

= 21148.62 pA + 515.22 + 13.902 = 1 2149 pA + 708.42 ft>s

where pA is in psi.

Ans.

The values of pA and the corresponding values of pC and tabulated below. pA 1psi2

VC 1ft>s2

0

15

30

45

60

26.6

54.2

71.9

86.0

98.1

The plot of VC VS pA is shown in Fig. a.

435

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Ans: VC = 1 2149pA + 7082 ft>s

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*5–56. 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 0 + 0 + gh = 0 +

h B Datum

VB2 + 0 2

VB = 22gh = 22 ( 9.81 m>s2 ) h = 219.62 h

(1)

(a)

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 ) 1h 0t

0h = 35.2484 ( 10-6 ) 13 m = 61.1 ( 10-6 ) m>s, which indeed 0t is very small as compared to VB = 119.62 (3 m) = 7.67 m>s. Thus, the assumption of VA _ 0 is quite reasonable. Integrating the above equation

When h = 3 m, VA = -

436

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5–56. 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 Ans. 1 m3

Ans: Vle = 18.9(103) liters 437

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5–57. The 9-in.-diameter bucket is to be used to slowly water the tree. If it is filled to the top with water, determine 1 the time for the bucket to empty if there is a 16 -in.-diameter hole drilled into its bottom. Assume the velocity through the hole is much greater than at the water surface.

9 in. 12 in.

SOLUTION The deformable control volume that contains the water within the bucket at a given instant will be considered. the volume of this control volume is V =

A

p 2 db y 4

y

Take the derivative of V with respect to time t, p dy dV = db2 dt 4 dt

(1)

AB

Applying the continuity equation, realizing that the density of water is constant, VB

0 r dV + r V # dA = 0 0t Lcv Lcs

Here AB =

rw a

(a)

dV + VBAB b = 0 dt

p 2 d h. Since rw ≠ 0, 4 dV p + VB a d 2h b = 0 dt 4

(2)

Substitute Eq. (1) into (2),

p 2 dy p db V a d 2h b = 0 4 dt B 4 dy dh 2 = - a b VB dt db

(3)

Here, points A and B are exposed to the atmosphere, pA = pB = patm = 0 with reference to the datum set through point B, zA = y and zB = 0. Since the water can be considered as an ideal fluid (incompressible and inviscid), Bernoulli’s equation is applicable. Applying this equation between points A, and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 0 +

VA2 VB2 + gy = 0 + + 0 2 2

438

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5–57. Continued

Since, from the continuity condition VB W VA aVB = a V A2 the term is negligible. Then 2

9 in. 1 16

in.

2

b VA = 20736VA b

VB = 12gy

Substitute Eq. (4) into (3),

dy dh 2 = - a b 22gy dt db dt = - a

db 2 dy b dh 22gy

Integrating with y = y0 at the t = 0 and y = 0 at time t, L0

t

dt = t = -

Here db =

db 2 0 dy b 22g dh Ly0 1y 1

a

0 db 2 db 2 2 b 12y1>2 2 ` = a b y10>2 y0 22g dh 22g dh

1

a

1>16 9 3 1 12 ft = ft, dh = ft = ft, and y0 = ft = 1 ft, Then 12 4 12 192 12

t = c

2 22132.2 ft>s2 2

4a

3>4 ft 1>192 ft

2

b 111>2 2 = 5167.885 = 86.1 min

Ans.

Ans: t = 86.1 min 439

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5–58. Determine the velocity of water at B and C if the pressure of the water in the 8-in.-diameter pipe at A is 15 psi and the velocity at this point is 9 ft>s. The water is discharged into the atmosphere at B.

2 in. B

12 ft

8 in.

A

C

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between 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. With reference to the datum set through points A and C, zA = 0 and zB = 12 ft. lb 12 in. 2 b a b 19 ft>s2 2 VB2 1 ft in2 + 0 = 0 + + 132.2 ft>s2 2112 ft2 + 2 2 62.4 l>ft3 a b 32.2 ft>s2 VB = 39.21 ft>s = 39.2 ft>s Ans.

a15

The fixed control volume considered contains the water within the pipe bounded by cross sections at A, B and C. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified as 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCAC = 0 - 19 ft>s2 c pa

2 2 2 4 1 4 ft b d + 139.21 ft>s2 c pa ft b d + VC c pa ft b d = 0 12 12 12

Ans.

VC = 6.549 ft>s = 6.55 ft>s

Ans: VB = 39.2 ft>s VC = 6.55 ft>s 440

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5–59. The solution is ejected from the 20-mm-diameter syringe through a 0.4-mm-diameter needle. If the pressure developed within the syringe is assumed to be constant at 50 kPa, determine the velocity of the solution through the needle and the maximum height the liquid rises in the air. Take r = 1135 kg>m3. Assume the velocity through the needle is much greater than through the syringe.

hmax 50 mm

20 mm

F

SOLUTION The solution can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. The datum is established as shown in Fig. a. Applying this equation between points A and B both on the central streamline, pA pB VA2 VB2 + + gzA = + + gzB r r 2 2 20 mm 2 aVB = a b VA = 2500 VA b, 0.4 mm

C

hmax

Since from the continuity condition VB W VA VA2 the term is negligible. Since the tip B of the needle is exposed to the atmosphere, 2 pB = 0. With reference to the datum, zA = 0 and zB = 0.05 m. Then 501103 2 N>m2 1135 kg>m3

VB2 + 0 + 0 = 0 + + 19.81 m>s2 210.05 m2 2

V = VB = 9.334 m>s = 9.33 m>s

B 0.05 m A

Datum (a)

Ans.

Using the continuity equation, - VAAA + VBAB = 0 - VA 3 p10.01m2

2

4

+ 19.334 m>s2 5 p 3 10.2110-3 2m 4 2 6 = 0 VA = 0.00373 m>s

This result confirms that VA is indeed very small. Again, applying Bernoulli’s equation between points A and C, pA pC VC2 VA2 + + gzA = + + gzC r r 2 2 VA2 will be neglected. Since point C is in the atmosphere, pC = 0. Here, 2 it is required that VC = 0 (a condition when the maximum height is achieved). With reference to the datum, zA = 0 and zC = hmax. Then Again the term

501103 2 N>m2 1135 kg>m3

+ 0 + 0 = 0 + 0 + 19.81 m>s2 2hmax .

hmax = 4.491 m = 4.49 m

Ans. Ans: V = 9.33 m>s hmax = 4.49 m

441

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*5–60. The solution is ejected from the 20-mm-diameter syringe through a 0.4-mm-diameter needle. Determine the velocity of the solution through the needle and the maximum height h max the solution rises in the air, each as a function of the force F applied to the plunger. Assume the pressure within the syringe is constant and the velocity through the needle is much greater than through the syringe. Plot the velocity and maximum height (vertical axis) as a function of the force for 0 … F … 20 N. Give values for increments of ∆F = 5 N. Take r = 1135 kg>m3.

hmax 50 mm

20 mm

F

SOLUTION The solution can be considered as an ideal fluid (incompressible and inviscid). Also, the flow is steady. Therefore, Bernoulli’s equation is applicable. Datum is established as shown in Fig. a. Applying this equation between points A and B of which both points are on the central streamline, 2

pA pB VA VB + + gzA = + + gzB r r 2 2 Since from the continuity condition VB W VA aVB = a

the term

VA2 2

12 10 8 6

2

20 mm b VA = 2500 VA b , 0.4 mm

is negligible. Since the tip B of the needle is exposed to the atmosphere,

pB = 0. Here, pA =

VB (mys)

101103 2 F F = = F. With reference to the datum, 2 p A p10.01 m2

4 2 0

5 10 0.175

15

20

15

20

F(N)

zA = 0 and zB = 0.05 m. Then 101103 2 F p

3

1135 kg>m

hmax (m)

+ 0 + 0 = 0 +

6

VB2 + 19.81 m>s2 210.05 m2 2

VB = 1 25.6090F - 0.9812 m>s VB = 1 25.61F - 0.9812 m>s

where F is in N

5 4

Ans.

3 2

Again, applying Bernoulli’s equation between A and C, when C is the high point of the exit stream,

1 0

pA pC VC2 VA2 + + gzA = + + gzC r r 2 2

5

10

F(N)

VA2 will be neglected. Since point C is in the atmosphere, pC = 0. Here, 2 it is requirement that VC = 0 (a condition when the maximum height is achieved). With reference to the datum, zA = 0 and zC = hmax. Again the term

101103 2 F p

1135 kg>m3

+ 0 + 0 = 0 + 0 + 19.81 m>s2 2h max .

h max = 0.2859 F h max = 0.286 F

Ans.

where F is in N 442

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5–60. Continued

The necessary data to plot VB vs F and hmax vs F are tabulated below and the plots of VB vs F and hmax vs F are shown in Figs. b and c, respectively. 0.175

5

10

15

20

VB 1m>s2

0

5.20

7.42

9.12

10.5

F1N2

0

5

10

15

20

hmax 1m2

0

1.43

2.86

4.29

5.72

F1N2

Ans: VB = 1 25.61F - 0.9812 m>s where F is in N h max = 0.286 F where F is in N 443

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5–61. 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, pB pD 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: h = 41.0 mm 444

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5–62. 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 is maintained.

A

100 mm C

20 mm

B

h D

SOLUTION

d

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. pA pC 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 - VC AC + VD AD = 0 -(1.401 m>s) 3 p(0.01 m)2 4 + (0.9904 m>s) a dD = 0.02378 m = 23.8 mm

p 2 dD b = 0 4

Ans.

Ans: VC = 1.40 m>s dD = 23.8 mm 445

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5–63. 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 ft3 >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 = VB AB 0.8 ft3 >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>ft3 32.2 ft>s2

+ 0 + ( 32.2 ft>s2 ) (4 ft) = 0 + ¢

pC = 1053.28 lb>ft2 a

From C to A,

( 36.67 ft>s ) 2 2

+ 0

1 ft 2 b = 7.31 psi 12 in.

Ans.

pC pA VC2 VA2 + + gzC = + + gzA r r 2 2

1 1053.28 lb>ft2 2 °

62.4 lb>ft3 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 ft3 >s

Ans.

446

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Ans: pC = 7.31 psi QA = 1.80 ft3 >s

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*5–64. 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 ft3 >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

B

2 in.

SOLUTION QA = VAAA 1.5 ft3 >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>ft3 32.2 ft>s2

+ 0 + ( 32.2 ft>s2 ) (4 ft) = 0 + ¢ pC = 655.17 lb>ft2 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>ft2 °

62.4 lb>ft3 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 ft3 >s

Ans.

447

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Ans: pC = 4.55 psi QB = 0.667 ft3 >s

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5–65. 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 308

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

+

A

B

h 0.1 m sin30˚

pB VA2 VB2 + 0 = + + 0 2 2 1.84 kg>m3

pB - pA = 0.920 ( VA2 - VB2 )

0.04 m (a)

(1)

Continuity Equation. Consider the gas from A to B to the control volume. 0 r dV + r V # dA = 0 0t Lcv Lcs 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 CO2, 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

Ans: VA = 39.3 m>s 448

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5–66. Determine the discharge of water from the funnel at the instant y = 50 mm. For the solution, consider the volume of water to be a cone. Note: For a cone, V = 13pr 2h.

50 mm

50 mm

A

200 mm y

B

10 mm

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B on the vertical streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Here, points A and B are exposed to the atmosphere. Then pA = pB = 0. With reference to the datum set through point B, zA = 0.05 m and zB = 0. 0 +

VA2 VB2 + 19.81 m>s2 210.05 m2 = 0 + + 0 2 2 VB2 - VA2 = 0.981

(1)

0.05 m r

0.2 m y

(a)

The deformable control volume that contains the water within the funnel at a given instant will be considered. Referring to the geometry shown in Fig. a r 0.05 m = ; y 0.2 m Thus, the volume of this control volume is V = Then

r = 10.25 y2 m

1 2 1 p pr h = p10.25y2 2 1y2 = a y3 b m3 3 3 48 0V p 2 0y = y 0t 16 0t

(2)

Applying the continuity equation, realizing that the density of water is conduct, 0 r dV + r V # dV = 0 0t Lcv Lcs

Since rw ≠ 0,

rw a

0V + VBAB b = 0 0t

0V + VB 3 p10.005 m2 2 4 = 0 0t

(3)

449

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5–66. Continued

Substitute Eqs. (2) into (3).

However,

p 2 0y y + 325110-6 2p4VB = 0 16 0t

0y = - VA. Then 0t p 2 y 1 - VA 2 + 325110-6 2p4VB = 0 16 VA = c

At y = 0.05 m,

VA = c

Substitute this result into Eq. (1)

0.4110-3 2 y2

0.4110-3 2 0.052

d VB

d VB = 0.16VB.

VB = 1.0034 m>s Thus, the discharge is Q = VBAB = 11.0034 m>s23p(0.005 m)2 4 = 78.81 110-6 2 m3 >s

= 78.8 110-6 2 m3 >s

Ans.

450

M05_HIBB9290_01_SE_C05_ANS.indd 450

Ans: Q = 78.8 m3 >s

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5–67. Determine the rate at which the surface level of water is dropping in the funnel as a function of the depth y. Assume steady flow. Plot this rate (vertical axis) versus the depth y for 20 mm 6 y 6 120 mm. Give values for increments of 20 mm. Consider the volume of water to be a cone. Note: For a cone, V = 13pr 2h.

50 mm

50 mm

A

200 mm y

B

10 mm

SOLUTION The deformable control volume that contained the water within the funnel and its stem at a given instant will be considered. Referring to the geometry shown in Fig. a r 0.05 m = ; y 0.2 m

0.05 m r

r = (0.25 y) m

Thus, the volume of this control volume is

0.2 m

1 1 p V = pr 2h = p10.25y2 2 1y2 = a y3 b m3 3 3 48

Then

0V p 2 0y = y 0t 16 0t

y

(a)

(1)

Applying the continuity equation realizing that the density of water is constant, dy (mys) dt

0 r dV + r V # dA = 0 0t Lcv Lcs

Since rw ≠ 0,

Substitute Eq. (1) into (2)

rw a

0

0V + VBAB b = 0 0t

y(m)

–0.1 –0.2

0V + VB 3p10.005 m2 2 4 = 0 0t

(2) –0.3

p 2 0y y + 325(10-6)p4VB = 0 16 0t

However,

0.02 0.04 0.06 0.08 0.10 0.12

(b)

0y = -VA. Then 0t p 2 y 1 - VA 2 + 325110-6 2p4VB = 0 16 VB = 12500 y2 2VA

(3)

451

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5–67. Continued

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B. On the vertical streamline, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 Here, points A and B are exposed to the atmosphere. Then pA = pB = 0. With reference to the datum set through point B, zA = y and zB = 0. 0 +

VA2 VB2 + 19.81 m>s2 2y = 0 + + 0 2 2 V B2 - V A2 = 19.62y

(4)

Substitute Eq. (3) into (4) 312500y2 2VA 4 2 - V 2A = 19.62y

VA = a Thus,

19.62y b m>s B 6.251106 2y4 - 1

19.6y dy = ab m>s where y is in meters dt B 6.251106 2y4 - 1

Ans.

The negative sign indicates that y is decreasing (the level of water is dropping). The 0y values of y and the corresponding values of are tabulated below. 0t y1m2 dy 1m>s2 dt The plot of

0.02

0.04

0.06

0.08

0.10

0.12

undef.

-0.229

- 0.121

- 0.0785

- 0.0561

- 0.0426

dy vs y is shown in Fig. b. dt

Ans: dy 19.6 y = ¢≤ m>s, dt B6.25(106)y4 - 1 where y is in meters 452

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*5–68. Determine the average velocity and the pressure in the pipe at A if the height of the water column in the pitot tube is 9 in. and the height in the piezometer is 3 in. 9 in. 3 in. A 2 in.

C

B

4 in.

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points C and B. pC pB VC2 VB2 + + gzC = + + gzB rw rw 2 2 Since point C is a stagnation point, VC = 0, with reference to the datum set through the horizontal central streamline, zB = zc = 0. The pressures at points B and C are pB = gw hB = 162.4 lb>ft3 2 a

pC = gw hC = 162.4 lb>ft3 2 a

Thus,

46.8 lb>ft2 a

62.4 lb>ft2 32.2 ft>s

2

9 ft b = 46.8 lb>ft2 12

26 lb>ft2

+ 0 + 0 = b

5 ft b = 26 lb>ft2 12

a

62.4 lb>ft2 32.2 ft>s2

VB = 4.633 ft>s.

+ b

V B2 + 0 2

The fixed control volume considered contains the water within the cross sections through A and B. Since the density of water is constant and the average velocities will be used, the continuity equation can be simplified as 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - VA c pa

2 2 1 2 ft b d + 14.633 ft/s2 c pa ft b d = 0 12 12

VA = 18.53 ft/s = 18.5 ft/s

Ans.

Again, applying the Bernoulli’s equation between points A and C, pA pC V C2 V A2 + + gzA = + + gzC rw rw 2 2 With reference to the same datum, zA = zC = 0 pA a

62.4 lb>ft

+

3

32.2 ft>s2

b

118.53 ft>s2 2 2

pA = 1 - 286 lb>ft2 2 a

46.8 lb>ft2

+ 0 = a

62.4 lb>ft3 32.2 ft>s

1 ft 2 b = -1.99 psi 12 in.

+ 0 + 0 b Ans.

The negative sign indicates that pA is partial vacuum or suction.

Ans: VA = 18.5 ft/s pA = -1.99 psi

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5–69. Water is contained within the bowl, which has a surface defined by y = 83r 2 ft. If the 0.25-in.-diameter drain plug is opened when y = 18 in., determine the time needed to drain the water to the level of the hole 1y = 3 in.2.

y r

y

B 3 in.

SOLUTION The deformable control volume that contains the water within the bowl at a given instant will be considered. The volume of this control volume is V = However, y =

1 2 pr y 2

3y 8 2 r or r 2 = . Then 3 8 V =

3y 1 3p 2 pa by = y 2 8 16

Take the derivative of V with respect to time t,

dy dV 3p 3p dy = a2y b = y dt 16 dt 8 dt

(1)

Applying the continuity equation realizing that the density of water is constant, 0 r dV + r V # dA = 0 0t Lcv Lcs rw a

Since rw ≠ 0,

Substitute Eq. (1) into (2)

dV + VBAB b = 0 dt

2 dV 1 + VB c pa ft b d = 0 dt 96

(2)

3p dy p y + V = 0 8 dt 9216 B y

dy VB = dt 3456

(3)

Here, points A and B are exposed to the atmosphere, pA = pB = patm = 0. With reference to the datum, zA = y and zB = 0.25 ft. Since the water can be considered as an ideal fluid (incompressible and inviscid), Bernoulli’s equation is applicable. Applying this equation between points A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 0 +

VA2 VB2 + 132.2 ft>s2 2y = 0 + + 132.2 ft>s2 210.25 ft2 2 2

454

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5–69. Continued

Since VB W VA, the term

VA2 is negligible. Then 2 (4)

VB = 264.4 1y - 0.252

Substitute Eq. (4) into (3)

y

dy 1 = 264.4 1y - 0.252 dt 3456 dt = -430.66a

Integrating with y = L0

ydy

1y - 0.25

b

18 3 ft = 1.5 ft at time t = 0 and y = ft = 0.25 ft at time t, 12 12 0.25 ft

t

dt = - 430.66 t = 430.66c

y dy

L1.5 ft 1y - 0.25

21 -0.5 - y2 3

= 641.98 s = 10.7 min

1y - 0.25 d `

0.25 ft 1.5 ft

Ans.

Ans: t = 10.7 min 455

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5–70. Water flows through the constant-diameter pipe such that at A the pressure is 80 kPa, and the velocity is 4 m>s. Determine the pressure and velocity at B. Draw the energy and hydraulic grade lines from A to B with reference to the datum set through A.

10 m

B 5

3

4 A

4 m!s

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B,

8.97 m 8.15 m

Velocity head

EGL HGL

pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g

Hydraulic head

Since the pipe has a constant diameter, continuity requires Ans.

VB = VA = 4 m>s

3 With reference to the datum set through A, zA = 0 and zB = 110 m2 a b = 6 m. 5 Then 80(103) N>m2 11000 kg>m3 219.81 m>s2 2

+

14 m>s2 2

219.81 m>s2 2

+ 0 =

pB 11000 kg>m3)(9.81 m>s2 2

pB = 21.141103 2pa = 21.1 kPa

+

14 m>s2 2

8m

0

219.81 m>s2 2

Datum

(a)

+ 6m

Ans.

EGL will have a constant value of H = =

pA VA2 + zA + gw 2g 801103 2 N>m2

11000 kg>m3 219.81 m>s2 2

= 0.89704 m

+

14 m>s2 2

219.81 m>s2 2

+ 0

Here, the velocity head is constant with a value of 14 m>s2 2 VA2 = = 0.8155 m 2g 219.81 m>s2 2

Thus, the HGL will be 0.8155 m below and parallel to the EGL. A plot of the EGL and HGL is shown in Fig. a

Ans: VB = 4 m>s pB = 21.1 kPa 456

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5–71. Water flows through the constant-diameter pipe such that at A the pressure is 80 kPa, and the velocity is 4 m>s. Plot the pressure head and the elevation head from A to B with reference to the datum set through A.

10 m

B 5

3

4 A

4 m!s

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B,

8.15 m 6m

pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g

Gravitational head 2.15 m

Since the pipe has a constant diameter, continuity requires Ans.

VB = VA = 4 m>s

3 With reference to the datum set through A, zA = 0 and zB = 110 m2 a b = 6 m. 5 Then 801103 2 N>m2 3

2

11000 kg>m 219.81 m>s 2

+

14 m>s2 2

2

219.81 m>s 2

+ 0 =

Pressure head

pB 3

2

11000 kg>m 219.81 m>s 2

pB = 21.141103 2 Pa

+

14 m>s2 2

0

219.81 m>s2 2

8m

Datum

(a)

+ 6m

Therefore, the pressure heads at A and B are

801103 2 N>m2 pA = = 8.155 m gw 11000 kg>m3 219.81 m>s2 2 21.141103 2 N>m2 pB = = 2.155 m gw 11000 kg>m3 219.81 m>s2 2

The gravitational head coincides with the centerline of the pipe. A plot of pressure head and gravitational head are shown in Fig. a.

457

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*5–72. Water at a pressure of 12 psi and a velocity of 5 ft>s at A flows through the transition. Determine the velocity and the pressure at B. Draw the energy and hydraulic grade lines with reference to the datum set through A.

3 in. B

12 ft

6 in.

A C 8 ft

SOLUTION

8 ft

2 ft

The fixed control volume considered contains the water within the transition and the pipe between the cross section through A and B since the density of the water is constant and the average velocity will be used, the continuity equation can be simplified as 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - 15 ft>s2 c pa

2 2 3 1.5 ft b d + VB c pa ft b d = 0 12 12

Ans.

VB = 20 ft>s

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B, pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g With reference to the datum set though point A, zA = 0 and zB = 12 ft. a12

lb 12 in. 2 ba b 2 15 ft>s2 2 120 ft>s2 2 pB 1 ft in + + 0 = + + 12 ft 2 3 3 2132.2 ft>s 2 2132.3 ft>s2 2 62.4 lb>ft 62.4 lb>ft pB = a615.8 5

EGL will have a constant value of H =

=

1 ft 2 lb ba b = 4.28 psi 2 12 in. ft

Ans.

pA VA2 + + zA gw 2g a12

lb 12 in. 2 b a b 15 ft>s2 2 1 ft in2 + + 0 2132.2 ft>s2 2 62.4 lb>ft3

= 28.08 ft

The velocity head just before the transition (point A) is 15 ft>s2 2 VA2 = = 0.388 ft 2g 2132.2 ft>s2 2

458

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5–72. Continued

The velocity head after the transition (point C) is 120 ft>s2 2 VB2 = = 6.211 ft 2g 2132.2 ft>s2 2

The EGL and HGL are plotted as shown in Fig. a.

28.08 ft

0.388 ft

EGL

27.69 ft

HGL

6.21 ft

21.87 ft

0

2 ft

18 ft

Datum

(a)

Ans: VB = 20 ft>s pB = 4.28 psi 459

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5–73. Water at a pressure of 12 psi and a velocity of 5 ft>s at A flows through the transition. Plot the pressure head and the elevation head from A to B with reference to the datum set through A.

3 in. B

12 ft

6 in.

A C 8 ft

SOLUTION

8 ft

2 ft

The fixed control volume considered contained the water within the transition and the pipe between the cross section through A and B. Since the density of the water is constant and the average velocity will be used, the continuity equation can be simplified as 0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - 15 ft>s2 c pa

2 2 3 1.5 ft b d + VB c pa ft b d = 0 12 12

VB = 20 ft>s

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between points A and B, pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g with reference to the datum set through point A, zA = 0 and zB = 12. a12

lb 12 in. 2 ba b 2 15 ft>s2 2 120 ft>s2 2 pB 1 ft in + + 0 = + + 12 ft 2 3 3 2132.2 ft>s 2 2132.2 ft>s2 2 62.4 lb>ft 62.4 lb>ft pB = 615.85 lb>ft2

Between A and C pA pC VC2 VA2 + + zA = + + zC gw gw 2g 2g Here VC = VB = 20 ft>s. Also, zC = zA = 0 a12

lb 12 in. 2 b a b 15 ft>s2 2 120 ft>s2 2 pC 1 ft in2 + + 0 = + + 0 2132.2 ft>s2 2 2132.2 ft>s2 2 62.4 lb>ft3 62.4 lb>ft3 pC = 1364.65 lb>ft2

460

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5–73. Continued

Therefore, the pressure head at A, B, and C are

pA = gw

lb 12 in. 2 ba b 2 1 ft in = 27.69 ft 62.4 lb>ft3

a12

615.85 lb>ft2 pB = = 9.87 ft gw 62.4 lb>ft3 1364.65 lb>ft2 pC = = 21.87 ft gw 62.4 lb>ft3 The gravitational head coincides with the centerline of the pipe. Plots of the pressure head and the gravitational head are shown in Fig. a.

27.69 ft

Pressure head

21.87 ft

12 ft 9.87 ft

0

2 ft

10 ft

18 ft

Datum

Gravitational head

461

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5–74. The hose is used to siphon water from the river. Determine the smallest pressure in the hose and the volumetric discharge at C. The hose has an inner diameter of 50 mm. Draw the energy and hydraulic grade lines for the hose with reference to the datum set through C.

B A

2m A9 8m

C

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between A and C, 6m

pA pC VC2 VA2 + + zA = + + zC gw gw 2g 2g Since the river is a large reservoir, VA ≃ 0. Here, A and C are exposed to the atmosphere, pA = pC = 0, and with reference to the datum set through C, zA = 8 m - 2 m = 6 m and zC = 0. 0 + 0 + 6m = 0 +

EGL

6m HGL A A9 B

C

Datum

(a)

V C2 2

219.81 m>s 2

VC = 10.85 m>s

The smallest pressure occurs at the maximum height of the hose, point B, between A and B. pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g Since the hose has a constant diameter, continuity requires VB = VC = 10.85 m>s. Also, with reference to the same datum, zB = 8 m. 0 + 0 + 6m =

pB 11000 kg>m3)(9.81 m>s2 2

+

110.85 m>s2 2

219.81 m>s2 2

+ 8m

pB = - 78.481103 2 Pa = - 78.5 kPa

Ans.

Negative sign indicates pB is partial vacuum or suction. The discharge is given by. Q = VCAC = 110.85 m>s23p10.025 m2 2 4 = 0.0213 m2 >s

Ans.

EGL will have a constant value of H =

pA VA2 + + zA = 0 + 0 + 6 m = 6 m gw 2g

The velocity head will be of a constant value throughout the hose. This value is 110.85 m>s2 2 VC2 = = 6m 2g 219.81 m>s2 2

HGL will be 6 m below and parallel to EGL. A plot of EGL and HGL are shown in Fig. a.

462

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Ans: pB = - 78.5 kPa Q = 0.0213 m3 >s

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5–75. The hose is used to siphon water from the river. Determine the pressure in the hose at point A′. The hose has an inner diameter of 50 mm. Draw the energy and hydraulic grade lines for the hose with reference to the datum set through point B.

B A

2m A9 8m

C

SOLUTION Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Applying this equation between A and C

A A9 B –2 m

pA pC VC2 VA2 + + zA = + + zC gw gw 2g 2g

Since the river is a large reservoir, VA ≃ 0. Here, A and C are exposed to the atmosphere, pA = pC = 0, and with reference in the datum set through point B, –8 m zA = - 2 m and zc = - 8 m. 0 + 0 + 1 - 2 m2 = 0 +

VC2 2

219.81 m>s 2

VC = 10.85 m>s

C

Datum

EGL

6m HGL (a)

+ 1 -8 m2

Between A and A′, this equation gives pA pA VA2 VA2 + zA = + zA + + gw gw 2g 2g Since the hose has a constant diameter, continuity requires VA′ = VC = 10.85 m>s. Also with reference to the same datum, zA′ = -2 m. 0 + 0 + 1 -2 m2 =

pA′ 3

2

11000 kg>m 219.81 m>s 2

+

pA′ = - 58.86 1103 2 Pa = - 58.9 kPa

110.85 m>s2 2

219.81 m>s2 2

+ 1 - 2 m2 Ans.

Negative sign indicates pA′ is partial vacuum or suction. EGL will have a constant value of H =

pA VA2 + + zA = 0 + 0 + 1 -2 m2 = - 2 m gw 2g

The velocity head will be of a constant value throughout the hose. This value is 110.85 m>s2 2 VC2 = = 6m 2g 219.81 m>s2 2

HGL will be 6 m below and parallel to EGL. A plot of EGL and HGL are shown in Fig. a.

Ans: pA′ = - 58.9 kPa 463

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*5–76. The siphon spillway provides an automatic control of the level in the reservoir within a desired range. In the case shown, flow will begin when the water level in the  reservoir rises above the crown C of the conduit. Determine the flow through the siphon if h = 4 ft. Also, draw the energy and hydraulic grade lines for the siphon conduit, with reference to the datum set through B. The siphon has a diameter of 8 in., and the water is at a temperature of 80°F. Neglect any head loss.

A C 10 ft

h

A9

28 ft

SOLUTION

B

2 ft

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Since points A and B are exposed to the atmosphere, pA = pB = patm = 0. Also, the water is drawn from a large reservoir. Thus, the water level in the reservoir can be considered as constant and so VA ≃ 0. The height h of the crown will not affect the flow, provided the cavitation does not occur at C and h 6 10 ft. with reference to the datum set through point B, zA = 38 ft - 2 ft = 36 ft. and zB = 0. Write the Bernoulli equation between A and B, 36 ft

pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g 0 + 0 + 36 ft = 0 +

VB2 2132.2 ft>s2 2

+ 0

EGL

HGL

VB = 48.15 ft>s

Datum (a)

Using the result of VC = VB = 48.15 ft>s to write the Bernoulli equation between A and C with zA = 36 ft and zC = 28 ft + 4 ft - 2 ft = 30 ft. pA pC VC2 VA2 + + zA = + + zC gw gw 2g 2g 0 + 0 + 36 ft =

pC 62.4 lb>ft3

pC = a - 1872

+

148.15 ft>s2 2

2132.2 ft>s2 2

+ 30 ft

1 ft 2 lb b a b = - 13.0 psi ft2 12 in.

From the table in Appendix A, the vapor pressure for the water at 80°F is 1pr 2 ab = 0.507 psia. Then its gage pressure is given by 1pr 2 abs = 1pr 2 g + patm; 0.507 psi = 1pr 2 g + 14.7 psi

1pr 2 g = -14.19 psi

Since pr 7 (pr)g, cavitation will not occur at C. Then the flow rate can be determined from Q = VB A = 148.15 ft>s2 c pa

2 4 ft b d = 16.81 ft3 >s = 16.8 ft3 >s 12

Ans.

Since the diameter of the siphon conduit is constant, the velocity head is constant along the conduit: 148.15 ft>s2 2 V2 = = 36 ft. 2g 2132.2 ft>s2 2

The plot of EGL and HGL with reference to the datum set through B is shown in Fig. a.

464

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Ans: Q = 16.8 ft3 >s

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5–77. What is the maximum permissible height h of the crown in the siphon conduit in Prob. 5–76 before it affects the flow? The water is at a temperature of 80°F.

A C 10 ft A9

h

28 ft

SOLUTION

B

2 ft

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable. Since points A and B are exposed to the atmosphere, pA = pB = patm = 0. Also, the water is drawn from a large reservoir. Thus, the water level in the reservoir can be considered as constant and so VA ≃ 0. The height h of the crown will not affect the flow, provided the cavitation does not occur at C and h 6 10 ft. with reference to the datum set through point zA = 38 ft - 2 ft = 36 ft. and zB = 0. Write the Bernoulli equation between A and B, pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g 0 + 0 + 36 ft = 0 +

VB2 + 0 2g

VB2 = 36 ft 2g For the cavitation to occur at C, pc … pr. For water at T = 80 1pr 2 abs = 0.507 psia. In gage pressure, it is 1pr 2 abs = 1pr 2 g + patm; 0.507 psi = 1pr 2 g + 14.7 psi 1pr 2 g = a -14.193

12 in. 2 lb lb ba b = -2043.79 2 2 1 ft in ft

with reference to the datum set through A′, zA = 10 ft and zC = h. Also, set pC = (pr)g = - 2043.79 lb>ft2 and VC = VB (since the diameter of the conduit is constant). Write Bernoulli’s equation between A and C, pA pC VC2 VA2 + + zA = + + zC gw gw 2g 2g 0 + 0 + 10 ft =

- 2043.79 lb>ft2 62.4 lb>ft3

+ 36 ft + h Ans.

h = 6.753 ft = 6.75 ft

Ans: h = 6.75 ft 465

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5–78. A piezometer and a manometer containing mercury are connected to the venturi meter. If the levels are indicated, determine the volumetric flow of water through the meter. Draw the energy and hydraulic grade lines. Take gHg = 846 lb>ft3.

9 in.

18 in.

SOLUTION Referring to Fig. a, the manometer rule written from A to B gives

B

A

pA + gw hAD - gHg hDE - gwhBE = pB pA + 162.4 lb>ft3 211 ft2 - 1846 lb>ft3 210.5 ft2162.4 lb>ft3 210.5 ft2 = pB pA - pB = 391.8

C 1 in.

4 in.

4 in.

12 in.

(1)

6 in.

Also, from Fig. b, the manometer rule written from C to G gives pC - gw hCF - gHg hFG = patm = 0 pC - 162.4 lb>ft3 211.5 ft2 - 1846 lb>ft3 210.75 ft2 = 0 pC = 728.1 lb>ft2

Since the water can be considered as an ideal fluid (incompressible and inviscid) and the flow is steady, Bernoulli’s equation is applicable with reference to the datum set through points, A and B, zA = zB = 0. Applying Bernoulli’s equation between points A and B, A

pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g pA pB VA2 VB2 + = + gw gw 2g 2g

hAD 5

pA - pB 1 = 1VB2 - V 2A 2 gw 2g

B

12 ft 5 1 ft 12

hBE 5

6 ft 5 0.5 ft 12

hDE 5

6 ft 5 0.5 ft 12

E

D

3

pA - pB =

1 62.4 lb>ft a b 1VB2 - VA2 2 2 32.2 ft>s2

(a)

pA - pB = 0.96891V B2 - V A2 2

(2)

The fixed control volume considered contains the water within the transition. Since the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 r dV + r V # dA 0t Lcv Lcs

hFG 5

9 ft 5 0.75 ft 12

hCF 5

18 ft 5 1.50 ft 12

F

C (b)

0 - VA AA + VB AB = 0 - VA c pa

G

2 2 2 0.5 ft b d + VB c pa ft b d = 0 12 12

VB = 16VA

(3)

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5–78. Continued

Substitute Eq. (1) into (2) 391.8 = 0.96891V B2 - V A2 2

VB2 - VA2 = 404.36

(4)

Solving Eqs. (3) and (4), VA = 1.2593 ft>s

VB = 20.14 ft>s

Then the flow rate is given by Q = VA AA = 11.2593 ft>s2 c pa

The total energy head is H =

2 2 ft b d = 0.1099 ft3 >s = 0.110 ft3 >s 12

Ans.

728.1 lb>ft2 11.2593 ft>s2 2 pC VC 2 + + zC = + + 0 gw 2g 2132.2 ft>s2 2 62.4 lb>ft3 = 11.69 ft.

11.7 ft

The velocity heads at A (or C) and B are

0.0246 ft

EGL

11.2593 ft>s2 2 VC2 = = 0.0246 ft 2g 2132.2 ft>s2 2

6.30 ft

120.14 ft>s2 2 VB2 = 6.30 ft = 2g 2132.2 ft>s2 2

The energy and hydraulic grade lines with reference to the datum set through A, B, and C are plotted as shown in Fig. c

467

M05_HIBB9290_01_SE_C05_ANS.indd 467

A

B

HGL

C

(c)

Ans: Q = 0.110 ft3 >s

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5–79. 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 -351103 2N>m2

11000 kg>m3 219.81 m>s2 2

+

1201103 2N>m2 V2 + 0 + hpump = 2g 11000 kg>m3 219.81 m>s2 2 +

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: # Ws = 14.0 kW 468

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*5–80. Draw the energy and hydraulic grade lines for the pipe ACB in Prob. 5–79 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

5.29 19.5

EGL

14.2

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 = 2 m. With hL = 0. 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 )

1.72 A –3.57

C

B

Datum

Pump

HGL

5.29 m

N V2 m2 = + + 2m + 0 + 0 ( 1000 kg>m3 )( 9.81 m>s2 ) 2g 120 ( 103 )

hpump

EGL

(a)

hpump = 17.80 m EGL and HGL. Since no losses occur, the total head before the pump is H =

pA VA2 + + zA g 2g

N ( 10.19 m>s ) 2 m2 = + + 0 = 1.72 m ( 1000 kg>m3 )( 9.81 m>s2 ) 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.

469

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5–81. Determine the kinetic energy coefficient a if the velocity distribution for laminar flow in a smooth pipe is defined by u = Umax 11 - 1r>R2 2 2.

r R

SOLUTION 1 a = # 2 V 2rV # dA mV Lcs V 3dA Lcs a = V dA = V 3A ( rVA ) V 2 Lcs r

3

V 3A L0 1

a =

2u3max

R

2u3max

R

R

u3(2prdr) 3

r2 a = 2 3 °1 - 2 ¢ rdr R R V L0 R V L0

a = a =

2u3max

R8 V 3 L0 a =

V =

8

R

3

( R2 - r 2 ) 3rdr

( R6r - 3R 4r 3 + 3R 2r 5 - r 7 ) dr

2u3max 8

R V

3

°

R8 1 umax ¢ = a b 8 4 V

(1)

umax R 1 r2 udA = °1 - 2 ¢(2prdr) 2 A Lcs pR L0 R V =

2umax

R 4 L0 V =

R

( R2r - r 3 ) dr

1 u 2 max

Substitute into Eq. (1), Ans.

a = 2

Ans: a = 2 470

M05_HIBB9290_01_SE_C05_ANS.indd 470

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5–82. Determine the kinetic energy coefficient a if the velocity distribution for turbulent flow in a smooth pipe is defined by Prandtl’s one-seventh power law, u = Umax 11 - r>R2 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 =

2u3maxp

R hV A L0

a =

3

R

3

(R - r)7 rdr

3

Set y = R - r dy = -dr a = a =

2u3max p

R R V AL 3 7

2u3maxp

y 37 (R - y)( - dy)

7 17 7 17 R 7 R 7d 10 17 R V A 3 7

a = V =

3

0

3

c

49u3maxR2p 85 V 3A

=

49u3max

(1)

85 V 3

umax R 1 1 udA = 1 (R - r)7 (2pr dr) AL cs R 7A L0 V =

2pumax

V =

1 7

RA

LR

2pumax 1 7

RA V =

0

1

y 7(R - y)( - dy)

7 15 7 15 c R7 - R7 d 8 15

pR2umax 49 c d 60 pR2

V =

49 u 60 max

Substitute into Eq. (1), Ans.

a = 1.058 = 1.06

Ans: a = 1.06 471

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5–83. 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 inlet of the pipe is submerged in the one reservoir and the exit is just above the surface of the other reservoir. Take ro = 880 kg>m3.

SOLUTION Energy Equation. Take the oil in the pipe system plus oil to the surface of the first reservoir to be the control volume. Here, pin = pout = 0 since both ends are opened to the atmosphere. Also, Vin = 0 since the reservoir is large. Since both reservoirs are at the same level, zin = zout = z. Q = Vout A 6 m3 >min = 0.1 m3 >s = Vout 3p10.150 m2 2 4 Vout = 1.4147 m>s

pin pout Vout2 Vin2 + zin + hpump = + zout + hturb + hL + + g g 2g 2g 0 + 0 + z + hpump = 0 +

11.4147 m>s2 2 2

219.81 m>s 2

+ z + 0 + 18000 m2 a

hpump = 120.102 m # WS = Qghpump

= 10.1 m3 >s21880 kg>m3 219.81 m>s2 21120.102 m2 = 103.68 ( 103 ) W = 104 kW

3m b 200 m

Ans.

Ans: # Ws = 104 kW 472

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*5–84. 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 inlet of the pipe is submerged in the one reservoir and the exit is just above the surface of the other reservoir.

SOLUTION Consider the fixed control volume as the water contained in the piping system plus water to the surface of the first reservoir. Here, pin = pout = 0, since both ends opened to the atmosphere. Also, Vin = 0 since the reservoir is large. If the datum is set at the surface of the lower reservoir, zin = 0 and zout = 20 m. Q = Vout A 0.8 m3 >s = Vout 3p10.100 m2 2 4 Vout = 25.465 m>s

pin pout Vout2 Vin2 + zin + hpump = + zout + hturb + hL + + gw g 2g 2g 0 + 0 + 0 + hpump = 0 +

125.465 m>s2 2 219.81 m>s2 2

+ 20 m + 0 + a

hpump = 73.051 m

2.5 m b 14000 m2 500 m

Thus, the required power output of the pump is # WS = Qghpump = ( 0.8 m3 >s )( 1000 kg>m3 )( 9.81 m>s2 ) (73.051 m) = 573.30 ( 103 ) W

Ans.

= 573 kW

Ans: # Ws = 573 kW 473

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5–85. 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

B

3 ft

C

8 ft A

# Ws = Qghpump 550 ft # lb>s 1 hp

b = V£ p a hpump =

2 1.5 ft b § ( 62.4 lb>ft3 )( 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

2 1.5 ft b § 12

= 1.94 ft3 >s

Ans.

474

M05_HIBB9290_01_SE_C05_ANS.indd 474

Ans: Q = 1.94 ft3 >s

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5–86. The pump is used with a 3-in.-diameter hose to draw water from the cavity. If the discharge is 1.5 ft3 >s, determine the required power developed by the pump. Neglect friction losses.

B

3 ft

C

8 ft A

SOLUTION From the discharge, 1.5 ft3 >s = VC £ p a

Q = VCAC;

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 + 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 ft3 >s )( 62.4 lb>ft3 ) (17.50 ft)

= a1637.97 = 2.98 hp

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

Ans: # Ws = 2.98 hp 475

M05_HIBB9290_01_SE_C05_ANS.indd 475

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5–87. Solve Prob. 5–86 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.

B

3 ft

C

8 ft A

SOLUTION From the discharge, Q = VCAC;

1.5 ft3 >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 ft3 >s )( 62.4 lb>ft3 ) (27.25 ft)

= a2550.57 = 4.64 hp

1 hp ft # lb ba b s 550 ft # lb>s

Ans.

Ans: # Ws = 4.64 hp 476

M05_HIBB9290_01_SE_C05_ANS.indd 476

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*5–88. The pump discharges water at B at 0.3 ft3 >s. If the friction head loss between the intake at A and the outlet at B is 2 ft, and the power input to the pump is 2 hp, determine the difference in pressure between A and B. The efficiency of the pump is e = 0.8.

3 in.

A

8 ft

SOLUTION The velocities at A and B can be determined from the discharge and continuity requirements Q = VBAA; Q = VBAB;

2 1.5 0.3 ft >s = VA c pa ft b d 12 3

0.3 ft3 >s = VB c pa

Here, the power output of the pump is

B

2 in.

VA = 6.112 ft>s

2 1 ft b d 12

VB = 13.751 ft>s

550 ft # lb>s # # Wout = eWin = 0.812 hp2 a b = 880 ft # lb>s 1 hp

The pump head can now be determined # 880 ft # lb>s = 10.3 ft3 >s2162.4 lb>ft3 2hpump Wout = Qgw hpump; hpump = 47.01 ft

Consider the fixed control volume that contains water in the pumping system from A to B with reference to the datum set through A, zA = 0 and zB = - 8 ft. Write the energy equation between A and B, pA pB VA2 VB2 + + zA + h pump = + + zB + h turb + hL gw gw 2g 2g pA 62.4 lb>ft3

+

16.112 ft>s2 2

2132.2 ft>s2 2

+ 0 + 47.01 ft =

pB - pA = a3160.71

pB 624 lb>ft2

+

113.751 ft>s2 2 2132.22

lb 1 ft 2 ba b = 21.94 psi = 21.9 psi 2 12 in. ft

+ 1 -82 + 0 + 2 Ans.

Ans: pB - pA = 21.9 psi 477

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5–89. Water is drawn into the pump, such that the pressure at the inlet A is - 6 lb>in2 and the pressure at B is 20 lb>in2. If the discharge at B is 4 ft3 >s determine the power output of the pump. Neglect friction losses. The pipe has a constant diameter of 4 in. Take h = 5 ft and rw = 1.94 slug>ft3.

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

+

V2 + 5 ft + h + 0 + 0 2g

hpump = 64.93 # Ws = Qpump ghpump = ( 4 ft3 >s )( 1.94 slug>ft3 )( 32.2 ft>s2 ) (64.93 ft) = 16225.36 ft .lb>s °

= 29.5 hp

1 hp

550 ft .lb>s

¢

Ans.

Ans: # Ws = 29.5 hp 478

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5–90. Draw the energy and hydraulic grade lines for the pipe ACB in Prob. 5–89 using a 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; 4 ft3 >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, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g 2

12 in. lb 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

12 in. lb ba b 1 ft in2 ( 1.94 slug>ft3 )( 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 b a b (45.84 ft>s)2 1 ft in2 + + 0 = 18.80 ft = 18.8 ft ( 1.94 slug>ft3 )( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) a -6

After the pump, a head of 64.93 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 (45.84 ft>s)2 V2 = = 32.6 ft 2g 2 ( 32.2 ft>s2 ) The HGL is always 32.6 ft below and parallel to the EGL. Both are plotted as shown in Fig. a.

479

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5–91. The turbine removes energy from the water in the reservoir such that it has a discharge of 20 ft3 >s through the 2-ft-diameter pipe. Determine the horsepower delivered to the turbine. Draw 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 Q = VCAC

EGL

20 ft3 >s = VC 3 p(1 ft)2 4

0.629 ft

21 ft

VC = 6.366 ft>s

20.4 ft

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 from 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 A

Turbine

hturb = 20.37 ft

# Ws = Qghturb = ( 20 ft3 >s )( 62.4 lb>ft3 ) (20.37 ft) = 25423 ft # lb>s a

1 hp

550 ft # lb>s

b = 46.2 hp

EGL

0.629 ft

( 6.366ft>s ) 2 + hturb + 0 2 ( 32.2ft>s2 )

0 + 0 + 21 ft + 0 = 0 +

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 ) V2 = = 0.629 ft 2g 2 ( 32.2 ft>s2 ) 2

The HGL will always be 0.629 ft below and parallel with the EGL. Both are plotted as shown in Fig. a.

Ans: # Ws = 46.2 hp 480

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*5–92. 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 ft3 = 7.48 gal.

0.75 ft

0.5 ft A

B

SOLUTION Q = (600 gal>min) a

1 min 1 ft3 b° ¢ = 1.337 ft3 >s 60 s 7.48 gal

1.337 ft3 >s = VA 3 p(0.375 ft)2 4

Q = VAAA;

VA = 3.026 ft>s

1.337 ft3 >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 ft3 >s )( 62.4 lb>ft3 ) (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: # WS = 4.99 hp 481

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5–93. The 5-hp pump has an efficiency of e = 0.8 and produces a 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

0.5 ft A

B

SOLUTION From the discharge, Q = (3 ft>s) 3 p(0.375 ft)2 4 = 0.421875p ft3 >s

Q = VAAA;

0.421875p ft3 >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 0.8 = e = # ; 5hp Wsin Thus, the pump head is # Wsout = Qgwhpump ;

(4hp) a

550ft # lb>s 1 hp

# Wsout = 4hp

b = ( 0.421875p ft3 >s )( 62.4 lb>ft3 ) 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 pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g pA 62.4 lb>ft3 =

+

( 3 ft>s ) 2 + z + 26.60 ft 2 ( 32.2 ft>s2 )

pB 62.4 lb>ft3

pB - pA = ( 1125.30 lb>ft2 ) 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.

Ans.

Ans: pB - pA = 7.81 psi 482

M05_HIBB9290_01_SE_C05_ANS.indd 482

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5–94. Water in the reservoir flows through the 8-in.diameter pipe at A into the turbine. If the discharge at B is 600 ft3 >min, determine the power output of the turbine. Assume the turbine runs with an efficiency of 60%. Neglect frictional losses in the pipe.

C

36 ft A 9 ft

B

8 in.

SOLUTION The

velocity

at

B

can

be

determined

from

ft3 1 min b = 10 ft3 >s. Then Q = a600 ba min 60 s Q = VBAB;

10 ft3 >s = VB c pa VB =

90 ft>s p

the

discharge.

Here,

2 4 ft b d 12

The fixed control volume considered contains the water in the pipe-turbine system and behind the dam. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the discharge is drawn from a large reservoir, VC ≃ 0 with reference to the datum set through B, zB = 0 and zC = 45 ft. pC pB VC2 VB2 + + zC + hpump = + + hturb + hL gw gw 2g 2g

0 + 0 + 45 ft + 0 = 0 +

a

2 90 ft>s b p

2132.2 ft>s2 2

+ hturb + 0

hturb = 32.26 ft

The power input of the turbine is given by # Win = Qgwhturb = 110 ft3 >s2162.4 lb>ft3 2132.26 ft2 = 20.131103 2 ft # lb>s

Thus, the power out put is # # Wout = eWin = 0.6320.131103 2 ft # lb>s4 = 312.08 1103 2 ft # lb>s4 a

= 21.96 hp = 22.0 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Wout = 22.0 hp 483

M05_HIBB9290_01_SE_C05_ANS.indd 483

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5–95. Water in the reservoir flows through the 8-in.diameter pipe at A into the turbine. If the discharge at B is 600 ft3 >min, determine the power output of the turbine. Assume the turbine runs with an efficiency of 60% and there is a head loss of 4.5 ft through the pipe.

C

36 ft A 9 ft

B

8 in.

SOLUTION The velocity at B can be determined from Q = 1600 ft3 >min211 min>60 s2 = 10 ft3 >s. Then Q = VBAB;

10 ft3 >s = VB c pa

the

discharge.

Here,

2 4 ft b d 12

90 ft>s p The fixed control volume considered contains the water in the pipe turbine system and behind the dam. Since B and C are exposed to the atmosphere, pB = pC = 0. Also, the discharge is drawn from a large reservoir, VC ≃ 0 with reference to the datum set through B, zA = 0 and zC = 45 ft. VB =

pC pB VC 2 VB2 + + zC + hpump = + + hturb + hL gw gw 2g 2g 0 + 0 + 45 ft + 0 = 0 +

1 90p ft>s 2 2

2132.2 ft>s2 2

+ hturb + 4.5 ft

hturb = 27.76 ft

The power input of the turbine is given by # Win = Qgwh turb = 110 ft3 >s2162.4 lb>ft3 2127.76 ft2 = 17.32 1103 2 ft # lb>s

Thus, the power output is # # Wout = eWin = 0.6317.321103 2 ft # lb>s4 = 310.39 1103 2 ft # lb>s4 a = 18.89 hp = 18.9 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Wout = 18.9 hp 484

M05_HIBB9290_01_SE_C05_ANS.indd 484

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*5–96. Determine the power delivered to the turbine if the water exits the 400-mm-diameter pipe at 8 m>s. Draw the energy and hydraulic grade lines for the pipe using a datum at point C. Neglect all losses.

A

20 m B 8m C

SOLUTION The fixed control volume considered contains the water within the pipe and the turbine and behind the dam. Since A and C are exposed to the atmosphere, pA = pC = patm = 0. Also, since the water is drawn from a large reservoir, VA ≃ 0. With reference to the datum set through C, zC = 0 and zA = 28 m. Write the energy equation between A and C with hL = 0, pA pC VC2 VA2 + zA + hpump = + zC + hturb + hL + + gw gw 2g 2g 0 + 0 + 28 m + 0 = 0 +

18 m>s2 2

219.81 m>s2 2

+ 0 + h turb + 0

hturb = 24.74 m

With Q = VC AC = 18 m>s23p10.2 m2 2 4 = 0.32p m3 >s, the power delivered to the turbine is given by # Win = Qginhturb = 10.32p m3 >s2311000 kg>m3 219.81 m>s2 24124.74 m2 = 243.97 1103 2 W

Ans.

= 244 kW

Since no losses occur, the total energy head before the turbine is H =

pA VA2 + + zA = 0 + 0 + 28 m = 28 m gw 2g

After the turbine, a head of hturb = 24.74 m is drawn from the water, so that H = 28 m - 24.74 m = 3.26 m The velocity head of the flow in the pipe has a constant value of hv =

18 m>s2 2 VC 2 = = 3.26 m 2g 219.81 m>s2

The HGL is always 3.26 m below and parallel to the EGL. Both are plotted as shown in Fig. a

EGL hV 5 3.26 m HGL

28 m 24.74 m

hturb 5 24.74 m

EGL hV 5 3.26 m HGL

3.26 m Datum

Ans: # Win = 244 kW

485

M05_HIBB9290_01_SE_C05_ANS.indd 485

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5–97. The turbine at C draws a power of 90.5 hp. If the pressure at the intake A is pA = 52 psi and the velocity of the water at that point is 12 ft>s, determine the pressure and velocity of the water at the exit B. Neglect frictional losses between A and B.

24 in. 9 in.

C

B

A

SOLUTION The fixed control volume considered contains the water between the cross sections at A and B. Since the flow is steady, the density of water is constant, and the average velocities will be used, the continuity equation can be simplified as 0 rd V + rV # d V = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - 112 ft>s23p11 ft2 2 4 + VB c pa

2 45 ft b d = 0 12

Ans.

VB = 85.33 ft>s = 85.3 ft>s

With Q = VAAA = 112 ft>s23p11 ft2 2 4 = 12p ft3 >s, the turbine head drawn from water can be determined. # Win = Qgwhturb;

190.5 hp2 a

550 ft # lb>s 1 hp

b = 112p ft3 >s2162.4 lb>ft3 2hturb

hturb = 21.16 ft

Since the transition is horizontal, zA = zB = z. Write the energy equation between A and B, pA pB VC2 VA2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 2 1 52 inlb 2 1 121 in. ft 2 2

63.4 lb>ft3

+

112 ft>s2 2

2132.2 ft>s2 2

+ z + 0 =

pB = a -748.43

pB 62.4 lb>ft2

+

185.33 ft>s2 2

2132.2 ft>s2 2

lb 1 ft 2 b = -5.20 psi b a ft2 12 in.

+ z + 21.16 ft + 0 Ans.

The negative sign indicates that pB is partial vacuum or suction.

Ans: VB = 85.3 ft>s, pB = -5.20 psi 486

M05_HIBB9290_01_SE_C05_ANS.indd 486

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5–98. When the valve at A is closed, the pressure at A is 230 kPa, and at B it is 180 kPa. When the valve is open, the water flows at 6 m>s, and the pressure at A is 210 kPa and at B it is 145 kPa. Determine the head loss in the pipe between A and B.

B

h

SOLUTION When the value is closed, the static pressure at A and B can be related using the manometer equation. pA - rw gh - pB = 0 3

2

3

2

A 3

2

230(10 ) N>m - (1000 kg>m )(9.81 m>s )h - 180(10 ) N>m = 0 h = 5.097 m The fixed control volume considered contains the water within the pipe between cross sections at A and B. Since the pipe has a constant diameter, continuity requires VA = VB = V. With reference to the datum set through A, zA = 0 and zB = 5.097 m. Write the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 210(103) N>m2 (1000 kg>m3)(9.81 m>s2)

+

145(103) N>m2 VA2 VA2 + 0 + 0 = + + 5.097 m + 0 + hL 2g 2g (1000 kg>m3)(9.81 m>s2) Ans.

hL = 1.529 m = 1.53 m

Ans: hL = 1.53 m 487

M05_HIBB9290_01_SE_C05_ANS.indd 487

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5–99. The pump has a volumetric flow of 0.3 ft3 >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 5V 2 >g, where V is the velocity of the flow, determine the horsepower the pump supplies to the water.

6 ft

B

10 ft

A

SOLUTION Q = V 0.3 ft3 >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 = 0 since pond A is large reservoir. If the datum passes through A, zA = 0 and zB = 6 ft + 10 ft = 16 ft. pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL g g 2g 2g 0 + 0 + 0 + hpump = 0 +

16.112 ft>s2 2

2132.2 ft>s2 2

+ 16 ft + 0 + 5£

( 6.112 ft>s ) 2 § ( 32.2 ft>s2 )

hpump = 22.38 ft # Ws = Qghpump = ( 0.3 ft3 >s )( 62.4 lb>ft3 ) (22.38 ft) = 418.95

1 hp ft # lb s ° 550 ft # lb>s ¢ = 0.762 hp

Ans.

Ans: # Ws = 0.762 hp 488

M05_HIBB9290_01_SE_C05_ANS.indd 488

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*5–100. Water from the reservoir passes through a turbine at the rate of 18 ft3 >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

(100 hp) a

# Ws = Qghs 550 ft # lb>s 1 hp

b = ( 18 ft3 >s )( 62.4 lb>ft3 ) 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

Ans: hL = 27.5 ft 489

M05_HIBB9290_01_SE_C05_ANS.indd 489

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5–101. The pump is connected to the 2-in.-diameter hose. If the pump supplies a power of 4.5 hp, determine the discharge at C. Neglect frictional losses. Note that 1 hp = 550 ft # lb>s.

B

4 ft 8 ft

20 ft A C

SOLUTION Realizing that Q = VA = V cpa # Wout = Qgwhpump;

2 1 p ft b d = a V b ft2 >s, then 12 144

14.5 hp2 a

550 ft # lb>s 1 hp

b = a

p V b 162.4 lb>ft3 2 hpump 144

1818.04 V The fixed control volume considered contains the water in the lake and within the hose between the cross sections at B and C. Since B and C are exposed to the atmosphere, pA = pC = patm = 0. Also, the water is drawn from a large reservoir, so that V3 ≃ 0. With reference to the datum set through B, zB = 0 and zC = 4 ft - 20 ft = –16 ft. Write the energy equation between B and C, hpump =

pB pC VC2 VB2 + + zB + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 0 +

1818.04 V2 = 0 + + ( -16 ft) + 0 + 0 V 2(32.2 ft>s2) V 3 - 1030.4 V - 117081.72 = 0

Solving numerically, V = 55.90 ft>s Thus, the discharge is given by Q =

p 155.90 ft>s2 = 1.220 ft3 >s = 1.22 ft3 >s 144

490

M05_HIBB9290_01_SE_C05_ANS.indd 490

Ans.

Ans: Q = 1.22 ft3 >s

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5–102. The pump is connected to the 2-in.-diameter hose. If the average velocity of the water at C is 50 ft>s, determine the required power output of the pump. Neglect friction losses.

B

4 ft 8 ft

20 ft A C

SOLUTION The fixed control volume considered contains the water in the lake and within the hose between the cross sections at B and C. Since B and C are exposed to the atmosphere, pB = pC = patm = 0. Also, the water is drawn from a large reservoir so VB ≃ 0. With reference to the datum set through B, zB = 0 and zC = 4 ft - 20 ft = –16 ft. Write the energy equation between B and C, pB pC VC2 VB2 + + zB + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

150 ft>s2 2

2132.2 ft>s2 2

hpump = 22.82 ft

+ 1 -16 ft2 + 0 + 0

1 ft 2 2 4 = 1.091 ft3 >s, the power output of the With Q = VC AC = 150 ft>s2 3 p 1 12 pump is # Wout = Qgw hpump

= 11.091 ft3 >s2162.4 lb>ft3 2122.82 ft2 = 11553.30 ft # lb>s2 a

= 2.824 hp = 2.82 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Wout = 2.82 hp 491

M05_HIBB9290_01_SE_C05_ANS.indd 491

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5–103. Solve Prob. 5–102 by including frictional head losses in the hose of 1.25 ft for every 15 ft of hose. The hose has a total length of 150 ft.

B

4 ft 8 ft

20 ft A C

SOLUTION The fixed control volume considered contains the water in the lake and within the hose between the cross sections at B and C. Since B and C are exposed to the atmosphere, pB = pC = patm = 0. Also, the water is drawn from a large reservoir so VB ≃ 0. With reference to the datum set through B, zB = 0 and zC = 4 ft - 20 ft = –16 ft. Write the energy equation between B and C, pB pC VC2 VB2 + + zB + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

150 ft>s2 2

2132.2 ft>s2 2

+ 1 - 16 ft2 + 0 + a

hpump = 35.32 ft

1.25 ft b 1150 ft2 15 ft

1 ft 2 2 4 = 1.091 ft3 >s, the power output of the With Q = VC AC = 150 ft>s2 3 p 1 12 pump is # Wout = Qgwhpump

= 11.091 ft3 >s2162.4 lb>ft2135.32 ft2 = 12404 ft # lb>s2 a

1 hp

550 ft # lb>s

= 4.371 hp = 4.37 hp

b

Ans.

Ans: # Wout = 4.37 hp 492

M05_HIBB9290_01_SE_C05_ANS.indd 492

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*5–104. The flow of air at A 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 at B to 25 m>s. Determine 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 . + gzout ¢ - °hin + + gzin ¢ § m b - a b + ° ¢ = £ °hout + dt in dt dt 2 2 0 - 0 + °

dWpump dt

¢ = £ °hout +

∆h = hout - hin = ° = £

( 10 m>s ) 2 2

-

Vout2 Vin2 . + gz¢ - °hin + + gz¢ § m 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: ∆h = 2.66 kj>kg 493

M05_HIBB9290_01_SE_C05_ANS.indd 493

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5–105. The 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

B 2m

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, pA pB 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

Ans.

Ans: # Ws = 116 kW 494

M05_HIBB9290_01_SE_C05_ANS.indd 494

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5–106. Crude oil is pumped from a test separator at A to the stock tank using a 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 horsepower supplied by the pump. The pressure at A is 4 psi, and the stock tank is open to the atmosphere. The frictional head loss in the pipe is 0.25 in.>ft, the head loss at the pipe discharge into the tank is 1.01V 2 >2g2, and for each of the four elbows it is 0.91V 2>2g2, and V is the velocity of the flow in the pipe. Take go = 55 lb>ft3. Note that 1 ft3 = 7.48 gal.

30 ft B

A

SOLUTION The discharge is Q = °400

Thus,

gal min

¢°

1 ft3 1 min ¢° ¢ = 0.8913 ft3 >s 7.48 gal 60 s

0.8913 ft3 >s = V £ pa

Q = VA;

2 2 ft b § 12

V = 10.21 ft>s Energy Equation. Take the water from A to B as the control volume. Then pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL go go 2g 2g It is given that pA = a4

The frictional head loss is

( hL ) f = c

lb 12 in. 2 b = 576 lb>ft2 ba 2 1 ft in

( 0.25>12 ) ft ft

d (180 ft) = 3.75 ft

There are four elbows between points A and B. Thus, the head losses due to the elbows and the discharge into the tank is

( hL ) M = 4 °0.9

( 10.21 ft>s ) ( 10.21 ft>s ) V2 V2 ¢ + 1.0° ¢ = 4• 10.92 £ § ¶1.0£ § = 7.451 ft 2 2g 2g 2 ( 32.2 ft>s ) 2 ( 32.2 ft>s2 ) 2

2

495

M05_HIBB9290_01_SE_C05_ANS.indd 495

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5–106. Continued

With reference to the datum set through point A, zA = 0. On the outflow side, the pressure and elevation heads together add up to 30 ft, relative to the datum through point A. Substituting these results into the energy equation, 576 lb>ft2 55.0 lb>ft3

+

( 10.21 ft>s ) 2 ( 10.21 ft>s ) 2 + 0 + h = + 30 ft + 0 + 13.75 ft + 7.451 ft2 pump 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) hpump = 30.73 ft

The required output power can be determined from # Ws = Qgcohpump = ( 0.8913 ft3 >s )( 55.0 lb>ft3 ) (30.73 ft) = (1506.27 ft # lb>s) a = 2.74 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Ws = 2.74 hp 496

M05_HIBB9290_01_SE_C05_ANS.indd 496

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5–107. The pump is used to transport water at 90 ft3 >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 Q = °90

Then,

Q = VBAB;

ft3 1 min ¢° ¢ = 1.5 ft3 >s min 60 s

1.5 ft3 >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 ft3 >s )( 62.4 lb>ft2 ) (36.00 ft) = (3369.57 ft # lb>s) ° = 6.13 hp

1 hp

550 ft # lb>s

¢

Ans.

Ans: # Ws = 6.13 hp 497

M05_HIBB9290_01_SE_C05_ANS.indd 497

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* 5–108. The pump is used to transfer carbon tetrachloride in a processing plant from a storage tank A to the mixing tank C. If the head loss due to friction in the pipe and in the pipe fittings 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. The storage tank is opened to the atmosphere. Take rct = 1590 kg>m3.

B

A

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, z A = 0 and z B = 6 m - 3 m = 3m. Also, gct = rct g= ( 1590 kg>m3 )( 9.81 m>s2 ) = 15597.9 N>m3. pA pB VA2 VB2 + + z A + hpump = + + z B + hturb + hL gct gct 2g 2g 0 + 0 + 0 + hpump = 0 +

( 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

Ans: # WS = 3.03 kW 498

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5–109. Determine the power that the pump supplies to the water if the velocity of the water at A is 6 m>s and the pressures at A and B are 175 kPa and 350 kPa, respectively. Neglect friction losses.

150 mm B

3m

100 mm A

SOLUTION The fixed control volume considered contains the water within the pipe between cross sections A and B and the pump. Since the flow is steady, the density of the water is constant and the average velocities will be used, the continuity equation can be simplified as 0 rd V + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 2

0 - 16 m>s23p10.05 m2 4 + VB 3p10.075 m2 2 4 = 0 VB = 2.667 m>s

with reference to the datum set through A, zA = 0 and zB = 3 m. Write the energy equation between A, and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 1751103 2 N>m2 3

2

11000 kg>m 219.81 m>s 2

+

16 m>s2 2

2

219.81 m>s 2

+ 0 + hpump =

hpump = 19.37 m

3501103 2 N>m2 3

2

11000 kg>m 219.81 m>s 2

+

12.667 m>s2 2

219.81 m>s2 2

+ 3m + 0 + 0

With Q = VA AA = 16 m>s23p10.05 m2 2 4 = 0.015p m3 >s, the power output of the pump is # Wout = Qgwhpump = 10.015p m3 >s2311000 kg>m3 219.81 m>s2 24119.37 m2

= 8.9531103 2 W

Ans.

= 8.95 kW

Ans: # Wout = 8.95 kW 499

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5–110. The pump delivers water at 120 ft3 >min from the river to the irrigation stream. If the frictional head loss in the 4-in.-diameter hose is hL = 6 ft, determine the power output of the pump.

12 ft B

60 ft

A

SOLUTION The flow velocity in the hose can be determined using the given discharge ft3 1 min Q = a120 ba b = 2 ft3 >s min 60 s Q = VA;

2 ft3 >s = V c pa

2 2 ft b d 12

V =

72 ft>s p

The fixed control volume considered contains the water within the hose and in the pump and river. Since A and B are exposed to the atmosphere, pA = pB = patm = 0. Also, the water is drawn from a river which can be considered as a large reservoir so VA ≃ 0. With reference to the datum set through A, zA = 0 and zB = 60 ft – 12 ft = 48 ft. Write the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g

0 + 0 + 0 + hpump = 0 +

a

2 72 ft>s b p

2132.2 ft>s2 2

+ 48 ft + 0 + 6 ft

hpump = 62.16 ft

The power output of the pump is # Wout = Qgwhpump = 12 ft3 >s2162.4 lb>ft3 2162.16 ft2 = 37.7571103 2 ft # lb>s4 a

1 hp

550 ft # lb>s

= 14.1 hp

b

Ans.

Ans: # Wout = 14.1 hp 500

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5–111. A fire truck supplies 150 gal>min of water to the third story of a building at B. If the friction loss through the 60-ft-long, 2.5-in.-diameter hose is 12 ft for every 100 ft of hose, determine the required pressure developed at the outlet A of the pump located within the truck close to the ground. Also, what is the average velocity of the water as it is ejected through a 1.25-in.-diameter nozzle at B?

B

42 ft

SOLUTION The fixed control volume considered contains the water within the hose and the nozzle between A and B. The flow rate is Q = a150

gal min

ba

A

1 ft3 1 min ba b = 0.3342 ft3 >s 7.48 gal 60 s

From the continuity requirement, the velocity of the flow in the hose and out of the nozzle can be determined from Q = VAAA;

0.3342 ft3 >s = VA c pa

Q = VBAB;

0.3342 ft3 >s = VB c pa

2 1.25 ft b d 12

VA = 9.8046 ft>s

2 0.625 ft b d 12

Ans.

VB = 39.2185 ft>s = 39.2 ft>s

Since the water is discharged into the atmosphere at B, pB = patm = 0. With reference to the datum set through A, zB = 42 ft and zA = 0. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g pA 62.4 lb>ft3

+

19.8046 ft>s2 2 2132.2 ft>s2 2

+ 0 + 0 = 0 +

pA = a4467.26

139.2185 ft>s2 2 2132.2 ft>s2 2

+ 42 ft + 0 + a

1 ft 2 lb ba b = 31.02 psi = 31.0 psi 2 12 in. ft

12 ft b + 160 ft2 100 ft Ans.

Ans: VB = 39.2 ft>s, pA = 31.0 psi 501

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*5–112. 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 3VB2 >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. pA pB 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: # WS = 32.5 kW 502

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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 5 4 (1 2 100 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 = (2pr dr)(0.2 m) = 0.4pr dr. Thus, the mass of this element is dm = rdV = ( 800 kg>m3 ) (0.4pr dr) = 320pr dr. 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 = 2pr dr. 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 503

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6–2. Flow through the circular pipe is turbulent, and the velocity profile can be modeled using Prandtl’s one-seventh power law, u = 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. Also, show that Vmax = (60>49)V, where V is the average velocity of the flow, and 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 = rV dA

r

dA

Then the momentum per unit time passing through this element is # # dL = (dm)V = (rV dA)V = rV 2dA

(a)

Thus, for the entire cross-section, # L =

L

# dL =

A

L

rV 2dA

A

Here dA = 2pr dr. Then # L =

R

1

r 7 2 rJ Vmax a1 - b R (2pr dr) R L0 R

2

L0 r Let u = 1 - , then r = R(1 - u) and dr = - R du. Also, the integration limits are R r = 0, u = 1 and r = R, u = 0. Thus, = 2prV 2max

# L = 2prV 2max

L1

= 2pR 2rV 2max

r a1 -

0

r 7 b dr R

2

R(1 - u) au 7 b( -R du)

L1

= 2pR2rV 2max a

0

9

2

au 7 - u 7 bdu

7 16 7 9 0 49 u 7 - u7 b ` = pR2rV 2max 16 9 72 1

The average velocity is

LA

Vavg =

(Q.E.D.)

V dA A

Here, LA

V dA =

L0

1

R

Vmax a1 R

r 7 b (2pr dr) R 1

r 7 = 2pVmax r a1 - b dr R L0

504

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6–2.

Continued

r , then r = R(1 - u) and dr = - R du. Also, the integration R limits are r = 0, u = 1 and r = R, u = 0. Thus

Again, let u = 1 -

LA

V dA = 2pVmax

L1

= 2pR 2Vmax

0

L1

= 2pR 2Vmax a

49 pR 2Vmax 60

=

1

R(1 - u) au7 b( -R du) 0

8

1

au7 - u7 bdu

7 15 7 8 0 u 7 - u7 b ` 15 8 1

Thus, Vavg = Vmax =

49 60

pR 2Vmax pR 2

60 V 49 avg

(Q.E.D.)

#

Substituting this result into L,

#

L = =

2 49 60 pR 2r a Vavg b 72 49

50 pR 2rV 2avg 49

(Q.E.D.)

505

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6–3. A volumetric discharge of 1.25 m3 >s passes out the pipe. Determine the horizontal force the water exerts on the contraction. 0.9 m

0.6 m

SOLUTION

F 2

The fixed control volume considered contains the water between the cross-sections at A and B of the culvert. Since we consider the steady flow of an ideal fluid (water), continuity requires Q = VAAA; Q = VBAB;

1.25 m3 >s = VA 3p10.45 m2 2 4 1.25 m3 >s = VB 3p10.3 m2 2 4

A

B F 2 (a)

VA = 1.9649 m>s VB = 4.4210 m>s

Since A and B are at the same level, zA = zB = z. Here, pB = patm = 0 since the water is discharged into the atmosphere at B. Write Bernoulli’s equation between points A and B. pA pB VA2 VB2 + + zA = + + zB gw gw 2g 2g pA 3

2

11000 kg>m 219.81 m>s 2

+

11.9649 m>s2 2 2

219.81 m>s 2

+ z = 0 +

pA = 7.8421(103) N>m2

14.4210 m>s2 2 219.81 m>s2 2

+ z

The FBD of the control volume is shown in Fig. a. Since the flow is steady and incompressible, the linear momentum equation can be simplified as ΣF =

0 VrdV + VrV # dA 0t Lcv Lcs

ΣFx = 0 + 1VA 2 xrw 1 -Q2 + 1VB 2 xrwQ

1 S+ 2 ΣFx =

rwQ31VB 2 x - 1VA 2 x4

37.84211103 2 N>m2 43p10.45 m2 2 4 - F = 11000 kg>m3)(1.25 m3 >s)(4.4210 m>s - 1.9649 m>s2 F = 1.91881103 2 N = 1.92 kN

Ans.

Ans: F = 1.92 kN 506

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*6–4. Water flows with a velocity of 6 m>s through the pipe at A. If the pressure at A is 65 kPa, determine the resultant horizontal shear force developed along the seam at C that holds the cap to the larger pipe.

300 mm 6 m!s

200 mm A

B

C

SOLUTION The fixed control volume considered contains the water within the pipe between the cross-sections at A and B. Since we consider the steady flow of an ideal fluid, the continuity equation can be simplified as 0 r dV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 0 - 16 m>s23p10.15 m2 2 4 + VB 3p10.1 m2 2 4 = 0

VB = 13.5 m>s

Since A and B are at the same level, zA = zB = z, write Bernoulli’s equation between A and B. pA pB VA2 VB2 + + gzA = + rw rw 2 + gzB 651103 2 N>m2 1000 kg>m3

+

16 m>s2 2 2

+ gz =

pB 1000 kg>m3

+

113.5 m>s2 2 2

+ gz

pB = - 8.1251103 2 Pa

The negative sign indicates that pB is partial vacuum or suction. The free-body diagram of the control volume is shown in Fig. a. Since the flow is steady and incompressible, the linear momentum equation can be simplified as ΣF =

1 S+ 2 ΣFx =

0 Vr dV + VrV # dA 0t Lcv Lcs 0 + 1VA 2 xr1 - Q2 + 1VB 2 xrQ

ΣFx = rQ31VB 2 x - 1VA 2 x4

Here, Q = VAAA = 16 m>s23p 10.15 m2 2 4 = 0.135p m3 >s. Then by referring to Fig. a, 3651103 2 N>m2 43p10.15 m2 2 4 + 38.1251103 2 N>m2 43p10.1 m2 2 4 - F = 11000 kg>m3)(0.135p m3 >s)(13.5 m>s - 6 m>s2 F = 1668.97 N = 1.67 kN

Ans.

F 2

PA 5 65(103) Pa

A

B

PB 5 8.125(103) Pa

F 2

Ans: F = 1.67 kN

(a)

507

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6–5. Water is ejected from the hose at A with a velocity of 20 ft>s. Determine the force that the water exerts on the floor. Assume the water does not splash back off the floor.

3 in.

A

4 ft

B

SOLUTION

C

The fixed control volume considered contains the water within the water column between A and B. Since A and B are exposed to the atmosphere, pA = pB = patm = 0. With reference to the datum set through B, zA = 4 ft and zB = 0. Write Bernoulli’s equation between A and B. VA2

VB2

pA pB + + gzA = + + gzB rw rw 2 2 0 +

120 ft>s2 2 2

B C F (a)

VB2 + 0 2 VB = 25.64 ft>s

+ 132.2 ft>s2)(4 ft2 = 0 +

The free-body diagram of the control volume considered is shown in Fig. a. We consider steady flow of an ideal fluid. The linear momentum equation along the vertical can be simplified as ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

1 + c 2ΣFy = 0 + 1VB 2 y1r)( -VBAB 2 + 0

The continuity condition requires that Q = VBAB = VAAA = 120 ft>s2 cpa 5 p ft3 >s. Then 16

F = 1 -25.64 ft>s2 a

62.4 lb>ft3 32.2 ft>s2

= 48.79 lb = 48.8 lb

ba-

5 p ft3 >s b 16

2 1.5 ftb d = 12

Ans.

Ans: F = 48.8 lb 508

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6–6. Water flows out of the reducing elbow at 0.6 ft3 >s. Determine the horizontal and vertical components of force that are necessary to hold the elbow in place at A. The pipe and elbow and the water within have a total weight of 250 lb. The water is discharged to the atmosphere at B.

0.25 ft B

3 ft

608

SOLUTION

A

The fixed control volume considered consists of the elbow and the water it contains between cross-sections at A and B. The continuity condition requires

0.5 ft

9.6 Q = VAAA; 0.6 ft >s = VA 3p 10.25 ft2 4 VA = ft>s p 38.4 Q = VBAB; 0.6 ft3 >s = VB 3p 10.125 ft2 2 4 VB = ft>s p Since water discharges into the atmosphere at B, pB = patm = 0. With reference to the datum set through A, zA = 0 and zB = 3 ft. Write the Bernoulli equation between A and B, 3

2

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA a

62.416 lb>ft3 32.2 ft>s2

+ b

a

2 2 9.6 38.4 ft>s b a ft>s b p p + 0 = 0 + + 132.2 ft>s2)(3 ft2 2 2

pA = 322.92 lb>ft2

The free-body diagram of the control volume considered is shown in Fig. a. Since the flow is steady and incompressible, the linear momentum equation, PB 5 0

0 ΣF = Vr dV + VrV # dA 0t Lcv Lcs

250 lb

can be written in scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

Fx - a322.92

0 + 1VA 2 xr1 -VAAA 2 + 1VB 2 xr1VBAB 2 ;

60°

62.4 lb>ft3

lb 9.6 b 3p10.25 ft2 2 4 = a ft>s b a b 1 - 0.6 ft3 >s2 p ft2 32.2 ft>s2 3

+ a-

62.4 lb> ft 38.4 b 10.6 ft3 >s2 cos 60° ft>s b a p 32.2 ft>s2 Fx = 59.85 lb = 59.9 lb

PA 5 322.92 lbyft2

Fx Fy (a)

Ans.

1 + c 2ΣFy = 0 + 0 + 1VB 2 yr1VBAB 2;

Fy - 250 lb = a

62.4 lb>ft3 38.4 sin 60° ft>s b a b 10.6 ft3 >s2 p 32.2 ft>s2

Fy = 262.31 lb = 262 lb

Ans. Ans: Fx = 59.9 lb Fy = 262 lb

509

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6–7. Oil flows through the 100-mm-diameter pipe with a velocity of 8 m>s. If the pressure in the pipe at A and B is assumed to be 60 kPa, determine the xand ycomponents of force the flow exerts on the elbow. The flow occurs in the horizontal plane. Take ro = 900 kg>m3.

8 m!s

A 308 100 mm

B

100 mm

SOLUTION The fixed control volume considered contains oil within the pipe and elbow between cross-sections at A and B. The free-body diagram of this control volume is shown in Fig. a. We consider steady flow of an ideal fluid. Thus, the volumetric flow rate is Q = VA = 18 m>s23p10.05 m2 2 4 = 0.02p m3 >s. The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the xand yaxes, which can be simplified as 3

2

1 S+ 2 ΣFx =

2

0 + 1VA 2 xr1 - VA 2 + 1VB 2 xr1VA 2 ;

360110 2 N>m 43p10.05 m2 4 - 3601103 2 N>m2 43p10.05 m2 2 4 cos 30° - Fx

= (8 m>s)(900 kg>m3)( - 0.02p m3 >s) + (8 cos 30° m>s)(900 kg>m3)(0.02p m3 >s) Fx = 123.74 N = 124 N

3

2

1 + c 2ΣFy = 0 + 0 + 1VB 2 yr1VA 2;

Ans.

360110 2 N>m 43p10.05 m2 2 4 sin 30° - Fy = 1 - 8 sin 30° m>s)(900 kg>m3)(0.02p m3 >s2 Ans.

Fy = 461.81 N = 462 N

Note: Fx and Fy are positive in the usual x- and y- directions. The corresponding components of the equal-and-opposite force the elbow exerts on the flow are negative, as indicated by the subtractive contributions of Fx and Fy in the above equations.

Fy

Fx

PA 5 60 kPa

30°

PB 5 60 kPa (a)

Ans: Fx = 124 N Fy = 462 N 510

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*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 force exerted by the water on the blade. The blade divides the water evenly in the vertical plane at an angle of u = 45°.

40 mm

C A

u

B

SOLUTION

C

We consider steady flow of an ideal fluid.

F

A

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 =

u

B (a )

0 Vr dV + VrV # dA 0t Lcv Lcs

or d+ ΣFx =

( - VB ) xrQB - ( VC ) xrQC + ( VA ) xr ( -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

511

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6–9. The nozzle has a diameter of 40 mm. If it discharges water in the vertical plane with a velocity of 20 m>s against the fixed blade, determine the 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 Vr dV + VrV # dA 0t Lcv Lcs

The horizontal component of this equation gives d+ ΣFx = 0 +

3 - (VA)x4 r( - VAAA)

F = ( 1000 kg>m

3

+ (VB)xr(VBAB) + (VC)xr(VCAC)

) 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)

512

M06_HIBB9290_01_SE_C06_ANS.indd 512

Ans: F = 3 160p (1 + sin u) 4 N

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6–10. Water flows through the elbow with a velocity of 18 ft>s. Determine the horizontal and vertical components of force the support at C exerts on the elbow. The pressure within the pipe at A and B is 28 psi. The flow occurs in the horizontal plane. Assume there is no support at A and B.

18 ft!s 4 in. 4 in. A B 60 8

30 8

C

SOLUTION The fixed control volume considered consists of the elbow and the water contained between the cross-sections at A and B. The free-body diagram of this control volume is shown in Fig. a where the force the support exerts on the elbow equals the force the elbow exerts on the flow. We consider steady flow of an ideal fluid. Thus, the volumetric flow rate is Q = VA = 118 ft>s23p 12>12 ft2 2 4 = 0.5p ft3 >s. The linear momentum equation, 0 Vr dV + VrV # dA 0t Lcv Lcs

ΣF =

can be written in scalar form along the xand yaxes, which can be simplified as

Fx + a28

1 S+ 2 ΣFx =

0 + 1VA 2 xr1 -VA2 + 1VB 2 xr1VA2

lb lb b 3p12 in.2 2 4 cos 60° - a28 2 b 3p12 in.2 2 4 cos 30° in2 in

= 118 cos 60° ft>s2 a

62.4 lb>ft3 32.2 ft>s2

b 1 -0.5p ft3 >s2 + 118 cos 30° ft>s2 a

62.4 lb>ft3 32.2 ft>s2

b 10.5p ft3 >s2 Ans.

Fx = 148.84 lb = 149 lb

1 + c 2ΣFy = 0 + 1VA 2 yr1 -VA2 + 1VB 2 yr1VA2

Fy - 128 lb>in2 23p12 in.2 2 4 sin 60° - 128 lb>in2 23p12 in.2 2 4 sin 30° = 118 sin 60° ft>s2 a

62.4 lb>ft3

b 1 -0.5p ft3 >s2 + 118 sin 30° ft>s2 a

32.2 ft>s2 Fy = 555.495 lb = 555 lb

62.4 lb>ft3 32.2 ft>s

b 10.5p ft3 >s2 Ans.

PA 5 28 psi A B

60°

PB 5 28 psi

30°

Fx Fy

Ans: Fx = 149 lb Fy = 555 lb

(a)

513

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6–11. Oil flows through the 8-in.-diameter pipe with a velocity of 18 ft>s. If it discharges into the atmosphere through the nozzle, determine the total force the bolts must resist at the connection AB to hold the nozzle onto the pipe. Take go = 55 lb>ft3.

B

4 in.

8 in. 18 m!s

A

SOLUTION We will consider the steady flow of an ideal fluid such that the density is constant and the average velocities can be used. The fixed control volume considered contains the oil in the nozzle as shown in Fig. a. The continuity equation can be simplified as 0 r dV + rV # dA = 0 0t Lcv Lcs

F

pin 5 4152.62 lbyft2

(a)

0 - Vin Ain + Vout Aout = 0 - 118 ft>s2 c pa

2 2 4 2 ft b d + Vout c pa ft b d = 0 12 12

Vout = 72 ft>s

Since the exit is exposed to the atmosphere, pout = patm = 0. The central streamline considered is level, zin = zout = z. Applying Bernoulli’s equation between two points on this streamline, pin pout V out2 V in 2 + + gzin = + + gzout r0 r0 2 2 pin a

55 lb>ft

+

3

32.2 ft>s2

b

118 ft>s2 2 2

+ gz = 0 +

172 ft>s2 2 2

+ gz

pin = 4150.62 lb>ft2

The volumetric flow rate is Q = Vin Ain = Vout Aout = 118 ft>s23p14>12 ft2 2 4 = 211 ft3 >s. The free-body diagram of the control volume is shown in Fig. a. The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the xaxis, which can be simplified:

1 S+ 2 ΣFx =

0 + 1Vin 2 xr1 -Q2 + 1Vout 2 xr1Q2

F - 14150.62 lb>ft2 2 c pa

2 55 lb>ft3 4 ft b d = 1 -18 ft>s2 a b 1 - 2p ft3 >s2 12 32.2 ft>s2

+ 1 -72 ft>s2 a

55 lb>ft3

32.2 ft>s2

b 12p ft3 >s2

Ans.

F = 869.30 lb = 869 lb

Ans: F = 869 lb 514

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*6–12. The water jet is ejected from the 4-in.-diameter pipe with a steady velocity of 16 ft>s. If it strikes the surface and is deflected in the vertical plane as shown, determine the normal force the jet exerts on the surface.

A 16 ft!s

C 4 in. 60 8 B

SOLUTION We will consider the steady flow of an ideal fluid such that the density is constant and the average velocities can be used. The fixed control volume considered is the water in the jet and on the wall. Since the water jet is a free flow, pA = pB = pC = patm = 0. Also, if we neglect the elevation change in the water jet, Bernoulli’s equation gives

C Fn

116 ft>s2 2 VA2 VB2 = 0 + = 0 + 2 2 2 VA = VB = 16 ft>s

Ft B

2 2 ftb d = 0.4444p ft3 >s. 12 The free-body diagram of the control volume is shown in Fig. a. The linear momentum equation,

The volumetric flow rate at C is QC = VCA C = 116 ft>s2 c pa ΣF =

n 60°

pA pB pC V C2 VA2 VB2 + = + = + rw rw rw 2 2 2 0 +

A

t (a)

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the n axis, which can be simplified: 1 +Q 2ΣFn = 0 + 1VC 2 nr1 - VCAC 2 Fn = 1 - 16 sin 60° ft>s2 a = 37.49 lb = 37.5 lb

62.4 lb>ft3 32.2 ft>s2

b 1 - 0.4444p ft3 >s2

Ans.

Ans: Fn = 37.5 lb 515

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6–13. The water jet is ejected from the 4-in.-diameter pipe with a steady velocity of 16 ft>s. If it strikes the surface and is deflected in the vertical plane as shown, determine the volumetric flow towards A and towards B if the tangential component of the force that the water exerts on the surface is zero.

A 16 ft!s

C 4 in. 60 8 B

SOLUTION We will consider the steady flow of an ideal fluid such that the density is constant and the average velocities can be used. The fixed control volume considered is the water in the jet and on the wall. Since the water jet is a free flow, pA = pB = pC = patm = 0. Also, if we neglect the elevation change in the water jet, Bernoulli’s equation gives

n 60°

pA pB pC VC2 VA2 VB2 + = + = + rw rw rw 2 2 2 0 +

A

C Fn

116 ft>s2 2 VA2 VB2 = 0 + = 0 + 2 2 2 VA = VB = 16 ft>s

Ft B

2 2 The volumetric flow rate at C is QC = VCA C = 116 ft>s2 cpa ftb d = 0.4444p ft3 >s. 12 The continuity equation can be simplified as

t (a)

0 r dV + rV # dA = 0 0t Lcv Lcs 0 - QC + QA + QB = 0

Q A + Q B = 0.4444p ft3 >s

(1)

The free-body diagram of the control volume is shown in Fig. a. Here, it is required that Ft = 0. The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the t axis, which can be simplified as 1 + R2ΣFt = 1VC 2 t r1 -VCAC 2 + 1VA 2 t r1VAAA 2 + 1VB 2 t r1VBAB 2

0 = 1 -16 cos 60° ft>s2 a + 1 - 16 ft>s2 a Solving Eqs. (1) and (2),

62.4 lb>ft3 32.2 ft>s2

62.4 lb>ft3 32.2 ft>s

2

b 1 -0.4444p ft3 >s2

b 1QA 2 + 116 ft>s2 a

62.4 lb>ft3 32.2 ft>s2

QA - QB = 0.2222p ft3 >s

QA = 1.047 ft3 >s = 1.05 ft3 >s 3

(2)

Ans.

3

Ans.

QB = 0.3491 ft >s = 0.349 ft >s

516

M06_HIBB9290_01_SE_C06_ANS.indd 516

b 1QB 2

Ans: QA = 1.05 ft3 >s QB = 0.349 ft3 >s

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6–14. Water flows through the elbow at 8 ft>s. Assuming the pipe connections at A and B do not offer any resistance on the elbow, determine the resultant horizontal force F that the support 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

SOLUTION We consider steady flow of an ideal fluid.

458

458

Q = VA = ( 8 ft>s ) c pa = 1.091 ft3 >s

2

2.5 ft b d 12 F

Control Volume. The free-body diagram of the control volume is shown in Fig. a. Linear Momentum. Since the flow is steady and incompressible, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

A

B

or

B

+ c ΣFy = 0 + ( VA ) y(r)( -Q) + 2J

p = 10 psi

( 10 lb>in2 ) a

2

A

( -VB ) yrQ 2

12 in. 2.5 b cos 45° c p a ft b d R 1 ft 12

F = 301.60 lb = 302 lb

p = 10 psi (a )

- F = °

62.4 lb>ft

3

32.2 ft>s2

¢ ( 1.091 ft3 >s ) 3 -8 ft>s cos 45° - 8 ft>s cos 45° 4 Ans.

Ans: F = 302 lb 517

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6–15. The steady jet of water flows from the 100-mmdiameter pipe at 4 m>s. If it strikes the fixed vane and is deflected in the vertical plane as shown, determine the normal force the jet exerts on the vane.

A

458

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 pA pB pC VC2 VA2 VB2 = = + + + 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 = 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, ΣF = or ΣFn = 0 + Fn =

0 Vr dV + VrV # dA 0t Lcv Lcs

( -QC ) (r) ( - VC ) n

( -0.03142 m3 >s )( 1000 kg>m3 )( -4 m>s sin 45° )

Fn = 88.9 N

Ans.

Ans: Fn = 88.9 N 518

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*6–16. The steady jet of water flows from the 100-mmdiameter pipe at 4 m>s. If it strikes the fixed vane and is deflected in the vertical plane as shown, determine the volumetric flow towards A and towards B if the tangential component of the force that the water exerts on the vane is zero.

A

458

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 pA pB pC VC2 VA2 VB2 = = + + + 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 r dV + 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 Vr dV + VrV # dA 0t Lcv Lcs

+ ΣF = r 3 Q ( V ) + Q ( V ) - Q ( V ) 4 ; A A t B B t C C t a t

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

519

M06_HIBB9290_01_SE_C06_ANS.indd 519

Ans.

Ans: QA = 0.00460 m3>s QB = 0.0268 m3>s

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6–17. 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

1358

30 mm

SOLUTION

B

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

FA

y

VB = 28.29 m>s x

pA pB VA2 VB2 + + zA = + + zB gco gco 2g 2g 300 ( 103 ) N>m2

+

( 880 kg>m3 )( 9.81 m>s2 )

( 10.19 m>s ) 2 ( 28.29 m>s ) 2 pB + 0 = + + 0 ( 880 kg>m3 )( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 )

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 ( 10 ) N>m 4 3 p(0.025 m) 4 3

2

2

= 589.05 N

3 6.596 ( 103 ) N>m2 4 3 p(0.015 m)2 4

Fy

FB

= 4.663 N

(a)

Applying the linear momentum equation, ΣF =

45˚

0 # 0t L VrcodV + L VrcoV dA cv cs

Writing the scalar component of this equation along the x and y axes by referring to Fig. a. + ΣFx = 0 + S ( -4.663 N) cos 45° - Fx =

( - VB cos 45° )( rco )( VBAB )

( - 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

Ans.

Ans: Fx = 349 N Fy = 419 N 520

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6–18. A speedboat is powered by the jet drive shown. Seawater is drawn into the pump housing at the rate of 20 ft3 >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>ft3. B

45!

A

SOLUTION Consider the control volume to be the jet drive and the water it contains, Fig. a. From the discharge, 2 3 Q = VAAA; 20 ft3 >s = VA c pa ft b d VA = 101.86 ft>s 12 Q = VBAB;

20 ft3 >s = VB c pa

2 2 ft b d 12

Th

VB = 229.18 ft>s T

Here, the flow is steady. Applying the linear momentum equation, (a )

0 Vr dV + VrV # dA ΣF = 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>ft3 32.2 ft>s

2

= 6276.55 lb = 6.28 kip

¢ ( -20 ft3 >s ) + ( 229.18 ft>s ) °

64.3 lb>ft3 32.2 ft>s2 Ans.

¢ ( 20 ft3 >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>ft3 32.2 ft>s2

¢ ( -20 ft3 >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 521

M06_HIBB9290_01_SE_C06_ANS.indd 521

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6–19. The 160-lb man stands on the scale. If the bucket of water weighs 30 lb, and water is flowing into it at 0.09 ft3 >s from a 1-in.-diameter hose, determine the reading on the scale at this instant. Assume the bucket is large so that the rate at which the level of water in the bucket is rising can be neglected.

1 in.

SOLUTION The fixed control volume at the instant considered consists of the man, the bucket and the water it contains as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The average velocity will be used. The velocity of the water entering the bucket at A can be determined using the discharge Q = VAAA;

0.09 ft3 >s = VA c p a

A

30 lb

160 lb

2 0.5 ft b d 12

51.84 ft>s p Here, A is exposed to the atmosphere, pA = patm = 0. The free-body diagram of the control volume is shown in Fig. a. The linear momentum equation VA =

ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the yaxis, which can be simplified as 1 + c 2ΣFy = 0 + 1VA 2 yr1 - VAAA 2

N - 160 lb - 30 lb = a -

N (a) 3

62.4 lb>ft 51.84 b 1 - 0.09 ft2 >s2 ft>s b a p 32.2 ft>s

N = 192.88 lb = 193 lb

Ans.

Ans: N = 193 lb 522

M06_HIBB9290_01_SE_C06_ANS.indd 522

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*6–20. The boat is filled with water using a 50-mm-diameter hose. If the velocity of flow out of the hose is VA = 8 m>s in the direction shown, determine the force in the tie rope needed to hold the boat stationary. Assume the level of water within the boat remains horizontal and the rate at which it rises can be neglected.

5m 8 m!s A

10 m

308 3m

B

SOLUTION The control volume considered consists of the boat and the water it contains as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The average velocities will be used. Here, B is exposed to the atmosphere, pB = 0. The free-body diagram T of the control volume is shown in Fig. a. The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

W B

N (a)

can be written in scalar form along the xaxis, which can be simplified as

1 S+ 2 ΣFx =

0 + 1VB 2 xr1 -VBAB 2

However, 1VB2 x = 1VA 2 x= 18 m>s2 cos 30° = 6.9282 m>s and the continuity condition requires Q = VBAB = VAAA = 18 m>s23p10.025 m2 2 4 = 0.005p m3 >s. Then - T = 16.9282 m>s)(1000 kg>m3)( - 0.005p m3 >s2 T = 108.83 N = 109 N

Ans.

Ans: T = 109 N 523

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6–21. 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>ft3.

A F

15 ft!s

3 in. 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>ft3. Average velocities will be  used. The control volume contains the liquid in the pipe and the transition as shown in Fig. a.

x

2F

FB

Continuity requires 0 r dV + rV # dA = 0 0t Lcv Lcs

FA

(a)

0 - VAAA + VBAB = 0 - ( 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), pA = a20

2880 lb>ft2 53.2 lb>ft3

lb 12 in. 2 ba b = 2880 lb>ft2 and hl = 0, in 1 ft

pA pB VA2 VB2 + + ZA + ht + hl = + + ZB + ht + hl gNa gNa 2g 2g

+

( 15 ft>s ) 2 ( 33.75 ft>s ) 2 pB + + 0 + (30 ft ) 0 = + 0 53.2 lb>ft3 2 ( 32.2 ft>s2 ) 2 ( 32.2 ft>s2 ) pB = 3720.90 lb>ft2

Thus, the pressure force acting on opened control surfaces at A and B are FA = pA AA = ( 2880 lb>ft2 ) c p a

2 1.5 ft b d = 141.37 lb 12

FB = pB AB = ( 3720.90 lb>ft2 ) c p a

2 1 ft b d = 81.18 lb 12

Applying the linear momentum equation, 0 Vr dV + VrNaV # dA ΣF = 0t Lcv Na Lcs

524

M06_HIBB9290_01_SE_C06_ANS.indd 524

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6–21.

Continued

Writing the scalar component of this equation along the x axis by referring to Fig. a,

1 S+ 2 ΣFx =

0 +

( - VB ) rNa ( VBAB ) + ( - VA ) rNa ( -VAAA )

81.18 lb - 141.37 lb + 2F = +

( -33.75 ft>s ) a ( - 15 ft>s ) a

53.2 lb>ft3 32.2 ft>s

53.2 lb>ft3 32.2 ft>s

F = 18.7 lb

2

2

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: F = 18.7 lb 525

M06_HIBB9290_01_SE_C06_ANS.indd 525

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6–22. Water flows out of the reducing elbow at 0.4 ft3 >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 Q = VAAA;

F

y

0.4 ft >s = VA 3 p(0.25 ft) 3

VA = 2.0372 ft>s

Continuity equation.

2

4

p

A

F

x

A

0 r dV + V # dA = 0 0t Lcv Lcs

B

0 - VAAA + VBAB = 0

p =0 B

(a)

- 0.4 ft3 >s + VB(p)(0.125 ft)2 = 0

VB = 8.149 ft>s

Bernoulli equation. Neglecting elevation change, pA pB VA2 VB2 + + gzA = + + gzB r r 2 2 pA °

62.4 lb>ft3 32.2 ft>s 2

+

( 2.037 ft>s ) 2 2

+ 0 = 0 +

¢

( 8.149 ft>s ) 2 2

+ 0

pA = 60.3234 lb>ft2

The free-body diagram is shown in Fig. a. Linear momentum equation. ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

+ ΣFx = 0 + rQ aVB - VA b S x x - Fx + ( 60.3234 lb>ft2 ) 3 (p)(0.25 ft)2 4 = °

62.4 lb>ft3 32.2 ft>s2

Fx = 10.3 lb

¢ ( 0.4 ft3 >s ) 3 8.149 ft>s(cos 60°) - 2.0372 ft>s 4 Ans.

+ c ΣFy = rQ 3 -VBy + 0 4 -Fy = °

62.4 lb>ft3 32.2 ft>s2

Fy = 5.47 lb

¢(0.4 ft3 >s)( -8.149 ft>s)(sin 60°)

Ans.

Ans: Fx = 10.3 lb Fy = 5.47 lb 526

M06_HIBB9290_01_SE_C06_ANS.indd 526

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6–23. Water flows through the 2-in.-diameter pipe assembly such that the velocity at A is 18 ft>s and the pressure is 20 psi. Determine the horizontal and vertical components of force at the support C, assuming there is no support resistance at A and B. The assembly and the water within it have a weight of 102 lb.

B

C 6 ft

2 in. 18 ft!s

SOLUTION

A 6 ft

The fixed control volume considered consists of the pipe assembly and the water it contains between cross-sections at A and B as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The average velocities will be used since the diameter of the pipe assembly is constant, VA = VB = V. With reference to the datum set through A, zA = 0 and zA = 6 ft. Write the energy equation between A and B.

FB 5 17.4 p lb 102 lb

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g

Cx

lb 12 in. 2 a20 2 b a b pB 1 ft V2 V2 in + + 0 + 0 = + + 6 ft + 0 + 0 3 3 2g 2g 62.4 lb>ft 62.4 lb>ft pB = 12505.6 lb>ft2 2 a

MC Cy

FA 5 20p lb

2

1 ft b = 17.4 psi 12 in.

(a)

The pressure forces acting on the inlet A and outlet B of the control surfaces indicated on the FBD of the control volume, Fig. a, are FA = pAA = a20

lb b 3p11 in.2 2 4 = 20p lb in2

FB = pBA = a17.4

The linear momentum equation, ΣF =

lb b 3p11 in.2 2 4 = 17.4p lb in2

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

20p lb - Fx = 118 ft>s2 a

0 + 1VA 2 xr1 -VAAA 2

62.4 lb>ft3 32.2 ft>s

2

b e - 118 ft>s2 c pa

Fx = 76.53 lb = 76.5 lb

2 1 ft b d f 12

1 + c 2ΣFy = 0 + 1VB 2 yr1VBAB 2

Fy - 102 lb - 17.4p lb = 118 ft>s2 a

62.4 lb>ft3 32.2 ft>s

b e 118 ft>s2 c pa

Fy = 170.36 lb = 170 lb

Ans.

2 1 ft b d f 12

Ans.

Ans: Fx = 76.5 lb Fy = 170 lb 527

M06_HIBB9290_01_SE_C06_ANS.indd 527

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*6–24. The pipe AB has a diameter of 40 mm. If water flows through it at a rate of 0.015 m3 >s, determine the horizontal and vertical components of force that must be supported at A to hold the pipe in place. The pipe and water within it have a total mass of 9 kg.

B

2m

SOLUTION The fixed control volume considered consists of the pipe and the water it contains between the cross-sections at A and B, as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Since the diameter of the pipe is constant, VA = VB = V. With reference to the datum set through A, zA = 0 and zB = 2 m. Since water is discharged into the atmosphere at B, pB = patm = 0. Write the Bernoulli equation between A and B.

A

B

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2

9(9.81) N

V2 V2 + + 0 = 0 + + 19.81 m>s2 212 m2 3 2 2 1000 kg>m pA

pA = 19.621103 2 Pa

Thus, the pressure force acting on inlet A of the control surface indicated in the FBD of the control volume, Fig. a, is FA = pAA = 319.621103 2 N>m2 43p10.02 m2 2 4 = 7.848p N

Ay FA (a)

0.015 m3 >s = V3p10.02 m2 2 4 V =

Ax MA

Using the discharge, the velocity of the flow in the pipe is given by Q = VA;

A

37.5 m>s p

The linear momentum equation, ΣF =

0 Vr dV + Vr V # dA 0t Lcv Lcs

can be written in scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

0 + 0

Ans.

Ax = 0

+ c ΣFy = 0 + 1VA 2 yr1 - VAAA 2 + 1VB 2 yr1VBAB 2 - Ay + 7.848p N - 919.812 N

= a

37.5 m>s b 11000 kg>m3 21 - 0.015 m3 >s2 p

+ a-

37.5 m>s b 11000 kg>m3 210.015 m3 >s2 p

Ay = 294.46 N = 294 N

Ans.

Ans: Ax = 0 Ay = 294 N 528

M06_HIBB9290_01_SE_C06_ANS.indd 528

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6–25. Pipe AB has a diameter of 40 mm. Determine the maximum volumetric flow through the pipe if the vertical tensile force developed at A cannot exceed 300 N. The pipe and water within it have a total mass of 9 kg.

B

2m

SOLUTION The fixed control volume considered consists of the pipe and the water it contains between cross-section at A and B, as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Since the diameter of the pipe is constant, VA = VB = V. With reference to the datum set through A, zA = 0 and zB = 2 m. Since water is discharged into the atmosphere at B, pB = patm = 0. Write the Bernoulli equation between A and B. pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA V2 V2 + + 0 = 0 + + (9.81 m>s2)(2 m) 3 2 2 1000 kg>m

A

B

9(9.81) N

pA = 19.621103 2 Pa

Thus, the pressure force acting on inlet A of the control surface indicated in the FBD of the control volume, Fig. a, is

MA

A

Ax

FA = pAA = 319.621103 2 N>m2 43p10.02 m2 2 4 = 7.848p N

Ay 5 300 N

The linear momentum equation, ΣF =

FA 5 7.848p N (a)

0 Vr dV + VrV # dA 0t Lcv L

can be written in scalar form along the yaxis, which can be simplified as 1 + c 2ΣFy = 0 + 1VA 2 yr1 - VAAA 2 + 1VB 2 yr1VBAB 2

Here, it is required that Ay = 300 N, then

7.848p N - 919.812 N - 300 N = V11000 kg>m3 25 - V3p10.02 m2 2 46

+ ( - V)(1000 kg>m3 25V3p10.02 m2 2 46

V = 12.029 m>s

Thus, the volumetric flow rate is Q = VA = 112.029 m>s23p10.02 m2 2 4

= 0.01512 m3 >s = 0.0151 m3 >s

Ans.

529

M06_HIBB9290_01_SE_C06_ANS.indd 529

Ans: Q = 0.0151 m3 >s

02/03/17 12:30 PM

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6–26. A stilling basin is 5 m wide and is used to confine the flow over a spillway. To retard the flow, 12 baffle blocks are used. Determine the horizontal force on each block if the flow is 80 m3 >s.

1.2 m

2.5 m

SOLUTION The fixed control volume considered contains the water between the crosssections upstream and downstream. Since we consider that the flow is steady and incompressible, continuity requires 80 m3 >s = V1 315 m211.2 m24

Q = V1A 1;

3

80 m >s = V2 315 m212.5 m24

Q = V2A 2;

F2

F1 w1

12F

w2

(a)

V1 = 13.33 m>s V2 = 6.40 m>s

The intensities of the distributed pressure loads acting on the upstream and downstream control surfaces are w1 = rwgy1b = 11000 kg>m3 21(9.81 m>s2 211.2 m215 m2 = 58.861103 2 N>m

w2 = rwgy2b = 11000 kg>m3 219.81 m>s2 212.5 m215 m2 = 122.6251103 2 N>m

The resultant forces of the distributed loads are

1 1 w y = 358.861103 2 N>m411.2 m2 = 35.3161103 2 N 2 1 1 2 1 1 F2 = w2y2 = 3122.6251103 2 N>m412.5 m2 = 153.28125 1103 2 N 2 2 F1 =

The FBD of the control volume is shown in Fig. a. Since the flow is steady and incompressible, the linear momentum equation can be simplified as ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

ΣFx = 0 + 1V1 2 xrw 1 -Q2 + 1V2 2 xrwQ

+ ΣFx = rwQ31V2 2 x - 1V1 2 x4 S

35.3161103 2 N - 153.281251103 2 N - 12F

= 11000 kg>m3 2180 m3 >s216.40 m>s - 13.33 m>s2 F = 36.391103 2 N = 36.4 kN

Ans.

Ans: F = 36.4 kN 530

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6–27. Pipe AB has a diameter of 40 mm. If water flows through it at 0.015 m3 >s, determine the horizontal and vertical components of force that must be supported at A to hold the pipe in place. The pipe and the water it contains have a total mass of 12 kg.

0.5 m B

3m

SOLUTION

608

The fixed control volume considered consists of the pipe and the water it contains between cross-sections at A and B as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Since the diameter of the pipe is constant, VA = VB = V. With reference to the datum set through A, zA = 0 and zB = 3 sin 60° m - 0.5 m = 2.0981 m. Since water is discharged in to the atmosphere at B, pb = patm = 0. Write the Bernoulli equation between A and B. pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA 3

1000 kg>m

+

A y

12(9.81) N 60°

V2 V2 + 0 = 0 + + 19.81 m>s2 212.0981 m2 2 2

Ax

pA = 20.581103 2 Pa

Thus, the pressure force acting on inlet A of the control surface indicated in the FBD of the control volume, Fig. a, is

Ay

FA 5 25.86 N

(a)

FA = pAA = 320.581103 2 N>m2 43p10.02 m2 2 4 = 25.86 N

Using the discharge, the velocity of the flow in the pipe is given by 0.015 m3 >s = V3p10.02 m2 2 4

Q = VA;

V = a

The linear momentum equation, ΣF =

37.5 b m>s p

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the xand yaxes. + ΣFx = 0 + 1VB 2 xr1VBAB 2 S 37.5 + - 125.86 N2 cos 60° + Ax = 0 + a m>s b cos 60°11000 kg>m3 21 -0.015 m3 >s2 S p Ans.

Ax = 102.46 N = 102 N 2

+ c 125.86 N2 sin 60° - 112 kg219.81 m>s 2 - A y = 0 + a - a

37.5 m>s b sin 60°11000 kg>m3 21 -0.015 m3 >s2 p

37.5 m>s b 11000 kg>m3 210.015 m3 >s2 p

Ans.

Ay = 238.79 N = 239 N

Ans: Ax = 102 N Ay = 239 N 531

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* 6–28. Pipe AB has a diameter of 40 mm. If the tensile force developed in the pipe at point A cannot exceed 250 N, determine the maximum allowable volumetric flow through the pipe. The pipe and the water it contains have a total mass of 12 kg.

0.5 m B

3m

SOLUTION

608

The fixed control volume considered consists of the pipe and the water it contains between cross-sections at A and B as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Since the diameter of the pipe is constant, VA = VB = V. With reference to the datum set through A, zA = 0 and zB = 3 sin 60° m - 0.5 m = 2.0981 m. Since the water is discharged into atmosphere at B, pB = patm = 0. Write the Bernoulli equation between A and B.

3

1000 kg>m

+

y

12(9.81) N B

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA

A

60°

V2 V2 + 0 = 0 + + 19.81 m>s2 212.0981 m2 2 2 pA = 20.581103 2 Pa

Thus, the pressure force acting on inlet A of the control surface indicated in the FBD of the control volume, Fig. a, is FA = pAA = 320.581103 2 N>m2 43p10.02 m2 2 4 = 25.86 N

MA

NA 5 250 N FA 5 25.86 N

The linear momentum equation, ΣF =

VA

(a)

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the yaxis, which can be simplified as a + ΣFy = 1VA 2 yr1 -VAAA 2 + 1VB 2 yr1VBAB 2

Here, it is required that NA = 250 N. Then

25.86 N - 250N - 1219.812 sin 60° N = V11000 kg>m3 25 -V3p10.02 m2 2 46

+ 1 -V sin 60°211000 kg>m3 25V3p10.02 m2 2 46

V = 11.79 m>s

Thus, the volumetric flow rate is Q = VA = 111.79 m>s23p10.02 m2 2 4

= 0.01482 m3 >s = 0.0148 m3 >s

532

M06_HIBB9290_01_SE_C06_ANS.indd 532

Ans.

Ans: Q = 0.0148 m3 >s

02/03/17 12:30 PM

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6–29. The disk valve is used to control the flow of 0.008 m3 >s of water through the 40-mm-diameter pipe. Determine the force F required to hold the valve in place for any position xof 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

Q = VoutAout;

0.008 m3 >s = Vout(0.04px) 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 0.06366

2

a b ( 6.366 m>s ) 2 x + + 0 = 0 + +0 9810 N>m3 2 ( 9.81 m>s2 ) 2 ( 9.81 m>s2 ) pin

pin = c

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)

533

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6–29. Continued

Applying the linear momentum equation, ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of this equation along the x axis by referring to the FBD of the control volume shown in Fig. a, S ΣFx = 0 + Vin rw( - 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 = £

2.55 ( 10-3 ) x2

+ 25.5 § N

534

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6–30. 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 = VA a 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

pA pB + + zA = + + zB gw gw 2g 2g 0 + a

4Q

pd 2g

b 2

C (b)

A (a)

2

+ 0 = 0 + VB =

VB2 + h 2g

16Q2 B p2d 4

(1)

- 2gh

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 =

8Q2 p2d 4g

-

m2g 8rw2 Q2

535

M06_HIBB9290_01_SE_C06_ANS.indd 535

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6–31. 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

h(m)

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;

8

Q = VA 3 p(0.005 m)2 4

VA = c

6

40 ( 103 ) Q d m>s p

5 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 40 ( 10 ) Qd p 3

0 + c

2 ( 9.81 m>s2 )

4

(a)

3 0.5 (9.81) N 2

2

+ 0 = 0 + VB =

2 ( 9.81 m>s2 )

1.6 ( 109 ) B

1

VB2

p

2

+ h B

2

Q - 19.62 h

(1)

By considering the FBD of the fixed control volume shown in Fig. b, where B and C are the inlet and outlet control surfaces, ΣF =

7

h

0 C

C (b)

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of the equation along the y axis, realizing that VC = VB =

A

1.6 ( 109 ) Q2 - 19.62h and Q = VA, p2 + c ΣFy = 0 + VBrw ( - VB AB ) +

( -VC ) rw ( VC AC )

536

M06_HIBB9290_01_SE_C06_ANS.indd 536

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0.1

0.2

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6–31.

Continued

-0.5(9.81) N = ( 1000 kg>m3 ) c -2 a h = c

A

1.6 ( 109 ) Q2 - 19.62h 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)

B

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

h

6 5 A 4

(a)

3 0.5 (9.81) N 2 1 B

0

C

(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 (c)

(b)

Ans: h = c

8.26 ( 106 ) Q4 - 0.307 ( 10-6 ) Q2

d m

537

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* 6–32. Air flows through the 1.5-ft-wide rectangular duct at 900 ft3 >min. Determine the horizontal force acting on the end plate B of the duct. Take ra = 0.00240 slug>ft3.

A B 3 ft 1 ft

SOLUTION

F2

Q = 900 ft3 >min ( 1 min>60 s ) = 15 ft3 >s 15 ft3 >s

VA =

(3 ft)(1.5 ft) 15 ft3 >s

VB =

(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 3

0.00240 slug>ft

+

( 3.33 ft>s ) 2 2

+ 0 = 0 +

( 10 ft>s ) 2 2

+ 0

pA = 0.10667 lb>ft2 Using the free-body diagram, Fig. a, the linear momentum equation becomes + ΣFx = 0 V r dV + VxrV # dA S 0t Lcv x Lcs

( 0.10667 lb>ft2 ) (3 ft)(1.5 ft) - F = 0 + ( 3.33 ft>s )( 0.00240 slug>ft3 )( -15 ft3 >s ) + ( 10 ft>s )( 0.00240 slug>ft3 )( 15 ft3 >s ) Ans.

F = 0.24 lb

Ans: F = 0.24 lb 538

M06_HIBB9290_01_SE_C06_ANS.indd 538

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6–33. Water flows through the pipe C at 4 m>s. Determine the xand ycomponents of force necessary to hold the horizontal pipe assembly in equilibrium. The pipe has a diameter of 60 mm at C, and at A and B the diameter is 20 mm.

4 5 3

B

A

4 m!s

SOLUTION

D C

Assume water is incompressible. We have steady flow. Q = 4 m>s (p)(0.03 m)2 = 0.011310 m3 Continuity requires 0 r dV + 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 1000 kg>m3

+

( 4 m>s ) 2 2

Fx p

C

VA2 + 0 = 0 + + 0 2

(a)

VA2 = 16 + 0.002pC

(2)

pC pB VC2 VB2 + + gzC = + + gzB r r 2 2 pC 3

1000 kg>m

+

( 4 m>s ) 2 2

+ 0 = 0 +

VB2 + 0 2

VB2 = 16 + 0.002pC

(3)

From Eqs. (2) and (3), VA = VB. From Eq. (1), VA = VB = 18 m>s Thus,

( 18 m>s ) 2 = 16 + 0.002pC pC = 154 kPa The free-body diagram is shown in Fig. a.

539

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6–33.

Continued

Linear momentum. ΣF =

0 V r dV + 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 Fx = -542 N = 542 N d

Ans.

4 + c ΣFy = 0 + VA a brVAAA + 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.

Ans: Fx = 542 N Fy = 183 N 540

M06_HIBB9290_01_SE_C06_ANS.indd 540

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6–34. Water flows through the pipe A at 6 m>s. Determine the x and y components of force the water exerts on the horizontal assembly. The pipe has a diameter of 100 mm at A, and at B and C the diameter is 60 mm. The water is discharged into the atmosphere at B and C.

6 m!s 100 mm A

60 mm

5 4 3

SOLUTION

C

B

The fixed control volume considered contains the water within the pipe system as shown in Fig. a. The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The continuity equation can then be simplified as

60 mm PA 5 16.72(103) Pa

0 r dV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCA C = 0 - 16 m>s23p10.05 m2 2 4 + VB 3p10.03 m2 2 4 + VC 3p10.03 m2 2 4 = 0 VB + VC = 16.67

(1)

Since water is discharged into atmosphere, pB = pC = patm = 0. Applying Bernoulli’s equation between A and B, pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2 pA 1000 kg>m3

+

16 m>s2 2 2

Fx Fy

(a)

VB2 + gz = 0 + + gz 2 VB2 = 0.002pA + 36

(2)

Between A and C, similarly, pA pC VC2 VA2 + + gzA = + + gzC rw rw 2 2 pA 3

1000 kg>m

+

16 m>s2 2 2

+ gz = 0 +

VC2 + gz 2

VC2 = 0.002pA + 36

(3)

Equating Eqs. (2) and (3), VB = VC. Substitute this result into Eq. (1). VB = VC = 8.3333 m>s Thus, 18.3333 m>s2 2 = 0.002pA + 36

pA = 16.721103 2 Pa

541

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6–34.

Continued

The free-body diagram of the control volume is shown in Fig. a. The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in the scalar form along the xand yaxes, which can be simplified as + ΣFx = 0 + 1VB 2 xrw 1VBAB 2 + 1VC 2 xrw 1VCA C 2 S 3 Fx = c - 18.3333 m>s2 a b d 11000 kg>m3 2518.3333 m>s23p10.03 m2 2 46 5 + 18.3333 m>s211000 kg>m3 25 18.3333 m>s23p10.03 m2 2 46

= 78.54 N = 78.5 N

Ans.

+ c ΣFy = 0 + 1VA 2 yrw 1 -VAAA 2 + 1VB 2 yrw 1VBAB 2

Fy - 316.721103 2 N>m2 43p10.05 m2 2 4 = 1 -6 m>s211000 kg>m3 251 -6 m>s23p10.05 m2 2 46 4 + c 1 -8.3333 m>s2 a b d 11000 kg>m3 2518.3333 m>s23p10.03 m2 2 46 5 Fy = 257.00 N = 257 N

Ans.

Ans: Fx = 78.5 N Fy = 257 N 542

M06_HIBB9290_01_SE_C06_ANS.indd 542

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6–35. The 1-in.-diameter pipe ejects water towards the wall as shown. Determine the normal force the water stream exerts on the wall. B 10 ft

458

VA A

2 ft 30 ft

SOLUTION Consider the horizontal motion of water jet by referring to Fig. a.

1 S+ 2 1SB 2 x =

The vertical motion gives

1SA 2 x + 1VA 2 xt;

1 + c 21SB 2 y = 1SA 2 y + 1VA 2 yt +

1 2 at ; 2 c

(1)

30 ft = 0 + VA cos 45°t

8 ft = 0 + VA sin 45°t +

1 1 -32.2 ft>s2 2t 2 2 (2)

Solving Eqs. (1) and (2), VA = 36.29 ft>s

t = 1.1689 s

The discharge of the flow is Q = VBAB = VAAA = 136.29 ft>s2 c p a

2 0.5 ft b d = 0.1980 ft3 >s 12

The fixed control volume considered contains the portion of water striking the wall. Since the water jet is of free flow, the pressure at any point is zero gauge. Its freebody diagram is shown in Fig. b. Since the flow is steady and water can be considered as an ideal fluid (incompressible and inviscid), the linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in the scalar form along the n axis, which can be simplified as ΣFn = 0 + 1VB 2 nr1 -VBAB 2

However, 1VB 2 n = 1VB 2 x = 1VA 2 x = 136.29 cos 45°2 ft>s = 25.66 ft>s. Then

1 S+ 2 Fn

= 1 -25.66 ft>s2 a

62.4 lb>ft3 32.2 ft>s2

Fn = 9.845 lb = 9.85 lb

b 1 - 0.1980 ft3 >s2

Ans.

y C

B

B

Ft VA

y 5 8 ft (SB)x 5 30 ft

n

Fn

45° A

x t

(a)

D (b)

Ans: Fn = 9.85 lb 543

M06_HIBB9290_01_SE_C06_ANS.indd 543

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* 6–36. 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 Vr dV + VrV # dA 0t Lcv Lcs

or + c ΣFy = 0 + ( -VC)r( -Q) pp(1.5 m)2 - 300 kg ( 9.81 m>s2 ) =

( -18 m>s )( 1.22 kg>m3 )( - 0.5655 m3 >s )

Ans.

p = 418 Pa

(300 kg) (9.81 m s2 ( p =0 C

p

Ans: p = 418 Pa 544

M06_HIBB9290_01_SE_C06_ANS.indd 544

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6–37. Water flows through the hose with a velocity of 3 m>s. Determine the force F on the semicylindrical cup if it is moving to the right at 2.5 m>s.

C 50 mm

20 mm

3 m!s

F B

A

D

C

SOLUTION The initial fixed control volume considered contains the water in the nozzle. The flow is steady and water can be considered as an ideal fluid (incompressible and inviscid). The continuity equation can be simplified as

B

0 r dV + rV # dA = 0 0t Lcv Lcs

F D (a)

0 - VAAA + VBAB = 0 - 13 m>s23p10.025 m2 2 4 + VB 3p10.01 m2 2 4 = 0

VB = 18.75 m>s

Subsequently, the fixed control volume that contains a portion of water striking the cup is considered. Its free-body diagram is shown in Fig. a. Relatives to the control volume, the velocity at B is

1 S+ 2 1Vf>cs 2 B

= VB - Vcv = 18.75 m>s - 2.5 m>s = 16.25 m>s S

Relative to the control volume, the flow can be considered steady. Also, the water can be considered as an ideal fluid (incompressible and inviscid). Then, the linear momentum equation, ΣF =

0 V r dV + Vf>cvrV # dA 0t Lcv f>cv Lcs

can be written in the scalar form along the xaxis, which can be simplified as + ΣFx = 0 + 31Vf>cs 2 B 4 xr31 - Vf>cs 2 BAB 4 + 31Vf>cs 2 C 4 xr31Vf>cs 2 CAC 4 S + 31Vf>cs 2 D 4 xr31Vf>cs 2 DAD 4

However, since the change in elevation is negligible, and all pressures are atmospheric, Bernoulli’s equation gives 1Vf>cs 2 C = 1Vf>cs 2 D = 1Vf>cs 2 B = 16.25 m>s

Also, from the relative volumetric flow rate on to the cup and the continuity requirement, Qf>cs = 1Vf>cs 2 BAB = 1Vf>cs 2 C AC + 1Vf>cs 2 D AD = 116.25 m>s23p10.01 m2 2 4 = 1.625110-3 2p

Then

-F = 116.25 m>s211000 kg>m3 23 -1.625110-3 2p m3 >s4

+ 1 -16.25 m>s211000 kg>m3 231.625110-3 2p m3 >s4 F = 165.92 N = 166 N

Ans.

Ans: F = 166 N 545

M06_HIBB9290_01_SE_C06_ANS.indd 545

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6–38. Water flows through the hose with a velocity of 3 m>s. Determine the force F needed to keep the semicylindrical cup moving to the left at 1.5 m>s.

C 50 mm

20 mm

3 m!s

F B

A

D

C

SOLUTION The fixed control volume considered contains the water in the nozzle. The flow is steady and water can be considered as an ideal fluid (incompressible and inviscid). The continuity equation can be simplified as

B

0 r dV + rV # dA = 0 0t Lcv Lcs

F D (a)

0 - VAAA + VBAB = 0 2

- 13 m>s23p10.025 m2 4 + VB 3p10.01 m2 2 4 = 0

VB = 18.75 m>s

Subsequently, the fixed control volume that contains a portion of water striking the cup is considered. Its free-body diagram is shown in Fig. a. Relative to the control volume, the velocity at B is

1 S+ 2 1Vf>cs 2 B

= VB - Vcv = 18.75 m>s - 1 - 1.5 m>s2 = 20.25 m>s S

Relative to the control volume, the flow can be considered steady. Also the water can be considered as an ideal fluid (incompressible and inviscid). Then the linear momentum equation, ΣF =

0 V r dV + Vf>cs rVf>cs # dA 0t Lcv f>cv Lcs

can be written in the scalar form along the xaxis, which can be simplified as + ΣFx = 0 + 31Vf>cs 2 B 4 xr31 -Vf>cs 2 BAB 4 + 31Vf>cs 2 C 4 xr31Vf>cs 2 CAC 4 S + 31Vf>cs 2 D 4 xr31Vf>cs 2 DAD 4

However, since the change in elevation is negligible and all pressures are atmospheric, Bernoulli’s equation gives 1Vf>cs 2 C = 1Vf>cs 2 D = 1Vf>cs 2 B = 20.25 m>s

Also, from the relative volumetric flow rate onto the cup and continuity requirement, Qf>cs = 1Vf>cs 2 BAB = 1Vf>cs 2 CAC + 1Vf>cs 2 DAD = 120.25 m>s23p10.01 m2 2 4 = 2.025110-3 2p m3 >s

Then

-F = 120.25 m>s211000 kg>m3 23 -2.025110-3 2p m3 >s4

+ 1 - 20.25 m>s211000 kg>m3 232.025110-3 2p m3 >s4 F = 257.65 N = 258 N

Ans.

Ans: F = 258 N 546

M06_HIBB9290_01_SE_C06_ANS.indd 546

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6–39. A 20-mm-diameter stream flows at 8 m>s against the blade and is deflected 120° as shown. If the blade is moving to the right at 3 m>s, determine the force F of the blade on the water.

B 308 8 m!s 20 mm

SOLUTION

F

A

B

The fixed control volume that contains a portion of water striking the vane is considered. Its free-body diagram is shown in Fig. a. Relative to the control volume, the velocity at A is

1 S+ 2 1Vf>cs 2 A

30°

= VA - Vcv = 8 m>s - 3 m>s = 5 m>s S

Fx

Since the change in elevation is negligible, Bernoulli’s equation gives A

1Vf>cs 2 B = 1Vf>cs 2 A = 5 m>s

From the relative volumetric flow rate onto the vane and the continuity requirement,

Fy (a)

Qf>cs = 1Vf>cs 2 AAA = 1Vf>cs 2 BAB = 15 m>s23p10.01 m2 2 4 = 0.5110-3 2p m3 >s

Relative to the control volume, the flow can be considered steady. Also, the water can be considered as an ideal fluid (incompressible and inviscid). The linear momentum equation, ΣF =

0 V r dV + Vf>cs rVf>cs # dA 0t Lcv f>cv Lcs

can be written in the scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

0 + 31Vf>cs 2 A 4 xrw 31 - Vf>cs 2 AAA 4 + 31Vf>cs 2 B 4 xrw 31Vf>cs 2 BAB 4

-Fx = 15 m>s211000 kg>m3 23 - 0.5110-3 2p m3 >s4

+ 1 - 5 sin 30° m>s211000 kg>m3 230.5110-3 2p m3 >s4

Fx = 11.78 N

Ans.

1 + c 2ΣFy = 0 + 31Vf>cs 2 B 4 yrw 31Vf>cs 2 BAB 4

Fy = 15 cos 30° m>s211000 kg>m3 230.5110-3 2p m3 >s4 = 6.802 N

Ans.

F = 21Fx2 2 + 1Fy2 2

= 2111.78 N2 2 + 16.802 N2 2

Ans.

= 13.6 N

u = tan -1 a = 30°

6.802 N b 11.78 N

Ans.

Ans: F = 13.6 N, u = 30° 547

M06_HIBB9290_01_SE_C06_ANS.indd 547

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*6–40. Determine the power required to keep the vane moving to the left with a velocity of 6 ft>s. The discharge from the 2-in.-diameter nozzle is 0.3 ft3 >s. Two-thirds of the discharge flows up the incline and one-third flows down and around.

B

2 in.

60 8

6 ft!s

A C

SOLUTION

B

Using the discharge, the velocity of the water jet is Q = VAAA;

0.3 ft3 >s = VA c pa

2 1 ft b d 12

VA = 13.75 ft>s

The fixed control volume that contains a portion of water striking the vane is considered. Its free-body diagram is shown in Fig. a. Relative to the control volume, the velocity at A is

1 S+ 2 1Vf>cs 2 A

= VA - Vcv = 13.75 ft>s - 1 - 6 ft>s2 = 19.75 ft>s

A

Fx

C

Fy (a)

Since the change in elevation is negligible, the Bernoulli’s equation gives 1Vf>cs 2 B = 1Vf>cs 2 C = 1Vf>cs 2 A = 19.75 ft>s

The relative flow rates at A, B, and C are

QA>cs = 1Vf>cs 2 AAA = 119.75 ft>s2 c pa

2 2 Q = 10.13167p ft3 >s2 = 0.2873 ft3 >s 3 A>cs 3 1 1 = QA>cs = 10.13167p ft3 >s2 = 0.1436 ft3 >s 3 3

QB>cs = QC>cs

2 1 ft b d = 0.4309 ft3 >s 12

Relative to the control volume, the flow can be considered steady. Also, the water can be considered as an ideal fluid (incompressible and inviscid). The linear momentum equation, 0 V r dV + Vf>cs rVf>cs # dA 0t Lcv f>cv Lcs can be written in scalar form along the xaxis, which can be simplified as ΣF =

+ S ΣFx = 0 + 31Vf>cs 2 A 4 xrw 1 -QA>cs 2 + 31Vf>cs 2 B 4 xrw 1QB>cs 2 + 31Vf>cs 2 C 4 xrw 1QCC>cs 2

-Fx = 119.75 ft>s2 a

62.4 lb>ft3 32.2 ft>s2

+ 119.75 cos 60° ft>s2 a + 1 - 19.75 ft>s2 a

Fx = 16.49 lb

b 1 - 0.4309 ft3 >s2

62.4 lb>ft3 32.2 ft>s2

62.4 lb>ft3 32.2 ft>s2

b 10.2873 ft3 >s2

b 10.1436 ft3 >s2

Thus, the power of the jet stream can be determined from # W = F # V = FxV = 116.49 lb216 ft>s2 1 hp = 198.96 ft # lb>s2 a b 550 ft # lb>s = 0.180 hp

Ans.

Ans: # W = 0.180 hp 548

M06_HIBB9290_01_SE_C06_ANS.indd 548

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6–41. The large water truck releases water at the rate of 45 ft3 >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.

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 )

3

ft 1 min ba b = 0.75 ft3 >s min 60 s

0.75 ft3 >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, ΣF =

0 Vr dV + VrwV # dA 0t Lcv w Lcs

Writing the scalar component of this equation along the xaxis, d+ ΣFx = 0 + Vout rwVout Aout F = ( 15.28 ft>s ) °

62.4 lb>ft3 32.2 ft>s2

¢ ( 0.75 ft3 >s ) = 22.2 lb

Ans.

Ans: F = 22.2 lb 549

M06_HIBB9290_01_SE_C06_ANS.indd 549

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6–42. The boat is powered by the fan, which develops a slipstream having a diameter of 1.25 m. If the fan ejects air with a velocity of 40 m>s, measured relative to the boat, and the boat is traveling with a 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 air entering at A is essentially at rest relative to the ground.

1.25 m A

B

SOLUTION We consider steady flow of an ideal fluid. 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 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

out

in

F

(a)

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 Vr dV + 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: F = 1.92 kN 550

M06_HIBB9290_01_SE_C06_ANS.indd 550

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6–43. 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 vane. Note that 1 hp = 550 ft # lb>s.

B 458 80 ft!s

308 A

SOLUTION

B

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, VA>cs = VA - Vcs

( S+ ) VA>cs cos 30° = 150 cos u - 80

(1)

( + c ) VA>cs sin 30° = 150 sin u

(2)

30°

Fy

Fx

A (a)

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 ft3 >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 Vr dV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + ( VA>cs ) r ( - Qf>cs ) + S x - Fx = °

62.4 lb>ft3 32.2 ft>s2

Fx = 179.99 lb

( - VB>cs ) x r ( Qf>cs )

¢ 3 ( 75.29 ft>s cos 30° )( -0.7842 ft3 >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: # W = 26.2 hp 551

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*6–44. The car is used to scoop up water that is in a trough between 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

(c)

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

dVcv = 0 (constant velocity) dt

Along the xaxis, + 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 2 552

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6–45. 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. Assume steady flow, all within the vertical plane.

B A 2 m!s

608 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 + VA>cs = VA - Vcs = 20.37 m>s - 2 m>s = 18.37 m>s 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 = 0.03607 m3 >s QB>cs = QC>cs =

3 3 ( QA>cs ) = ( 0.03607 m3 >s ) = 0.02705 m3 >s 4 4

1 1 ( QA>cs ) = ( 0.03607 m3 >s ) = 0.009018 m3 >s 4 4

Linear Momentum. Referring to the free-body diagram of the control volume in Fig. a, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

or + ΣFx = r 3 QB>cs(VB>cs)x + QC>cs(VC>cs)x - QA>cs(VA>cs)x4 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

#

Ws = F # V = FxV = (497.04 N)(2 m>s) Ans.

= 994.09 W = 994 W

Ans: # Ws = 994 W 553

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6–46. A plow located on the front of a truck horizontally scoops up a liquid slush at the rate of 12 ft3 >s and throws it off vertically, perpendicular to its motion, i.e., 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>ft3.

u

B

A

SOLUTION

W

We consider steady flow of an ideal fluid. F

Take the slush in context with the blade as the control volume.

A

Relative Velocity. Since the slush is at rest before it enters the control volume, then the velocity at A relative to control volume is + S

(a)

( Vf>cs ) A = Vf - Vcs = 0 - 14 ft>s = 14 ft>s d

N

Linear Momentum. Here, Qf>cs = 12 ft3 >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 Vr dV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + S -F =

( - VA ) r( - Q)

( - 14 ft>s ) °

5.5 lb>ft3 32.2 ft>s2

F = - 28.69 lb = 28.7 lb d

¢ ( -12 ft3 >s )

Ans.

Ans: F = 28.7 lb 554

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6–47. 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 upward 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

SOLUTION

B

60 °

We consider steady flow of an ideal fluid. 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 Thus, the flow rate of snow onto the shovel is

W

Fx

A Fy (a )

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 Vr dV + VrV # dA 0t Lcv Lcs

or + ΣFx = 0 + S Fx = 0 +

( - Vf>cs ) Axr ( -Qf>cs ) + ( Vf>cs ) Bxr ( Qf>cs ) ( -5 m>s )( 125 kg>m3 )( -3.75 m3 >s )

+ ( 5 m>s cos 60° )( 125 kg>m3 )( 3.75 m3 >s )

Fx = 3.52 kN

Ans.

Ans: F = 3.52 kN 555

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* 6–48. 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 Vr dV + 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: F = 141 N 556

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6–49. Solve Prob. 6–48 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 Bernoulli’s equation, ( Vf>cs ) B = 8 m>s d for small elevations. The flow is steady relative to control volume. ΣF =

A (a )

0 Vf>cv r dV + Vf>cs rVf>cs # dA 0t Lcv 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 + S

( Vf>cs ) A = Vw - Vb 0 = 10 m>s - Vb Vh = 10 m>s S

Ans.

Ans: F = 62.8 N Vh = 10 m>s 557

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6–50. Water flows into the bend fitting with a velocity of 3 m>s. If the water exits 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 is 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 xand yaxes by referring to the free-body diagram, Fig. a,

1 S+ 2 ΣFx = 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, + 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 # m

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)

558

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6–51. Water is discharged into the atmosphere from the pipe at 10 m>s. Determine the horizontal and vertical components of force and the moment that is developed at the fixed support A in order to hold the pipe in equilibrium. Neglect the resistance provided by the pipe at C, and the weight of the pipe and the water within it.

100 mm C

B 300 mm 60 mm

150 mm A

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The control volume considered consists of the fitting, the fixed support and the water contained. From the discharge and continuity requirements,

C FC 5 108.8p N

Q = VBAB = 110 m>s23p10.03 m2 2 4 = 0.009p m3 >s

VC = Q>AC = 10.009p m3 >s2 > 3p10.050 m2 2 4 = 3.6 m>s

Since water is discharged to the atmosphere at B, pB = patm = 0. Also, the charge in elevation is negligible. Applying Bernoulli’s equation between B and C,

1000 kg>m3

+

13.60 m>s2 2 2

+ gt = 0 +

110 m>s2 2 2

rC 5 0.3 m

rB 5 0.15 m

pC pB VC2 VB2 + + gzC = + + gzg rw rw 2 2 pC

B

MA

+ gz

Ax Ay (a)

pC = 43.52 1103 2 Pa

Then the pressure force on inlet control surface C is

FC = pCAC = 343.521103 2 N>m2 43p10.05 m2 2 4 = 108.8p N

The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the x and y axes by referring to the FBD of the control volume shown in Fig a, which can be simplified as

1 S+ 2 ΣFx =

0 + 1VC 2 xrw 1 - VCAC 2

Ax - 108.8p N = 1 - 3.60 m>s211000 kg>m3 21 - 0.009p m3 >s2 Ax = 443.59 N = 444 N

1 + c 2ΣFy = 0 + 1VB 2 yrw 1VBAB 2

- Ay = 1 -10 m>s211000 kg>m3 210.009p m3 >s2 Ay = 282.74 N = 283 N

Ans.

Ans.

The angular momentum equation, ΣM =

0 1r * v2r dV + 1r * v2rV # dA 0t Lcv Lcs

can be written in scalar form about point A by referring to Fig. a, which can be simplified as + ΣMA = 0 + rBVBr1VBAB 2 + rCVCr1 -VCAC 2

1108.8p N210.3 m2 - MA = 10.15 m2110 m>s211000 kg>m3 210.009p m3 >s2

+ 10.3 m213.60 m>s211000 kg>m3 21 -0.009p m3 2

MA = 90.67 N # m = 90.7 N # m

Ans.

Ans: Ax = 444 N Ay = 283 N MA = 90.7 N # m

559

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*6–52. Water flows into the Tee fitting at 3.6 m>s. If a pipe is connected to B 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 pipe in equilibrium. Neglect the resistance provided by the pipe at B and C, and the weight of the pipe and the water within it.

100 mm C

B 300 mm 60 mm

150 mm A

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The control volume considered consists of the fitting, the fixed support and the water contained. From the discharge and continuity requirements, 2

C FC 5 296.3p N

3

Q = VCAC = 13.6 m>s23p10.050 m2 4 = 0.009p m >s 3

2

VB = Q>AB = 10.009p m >s2 > 3p10.030 m2 4 = 10 m>s

Here, the change in elevation is negligible. Applying Bernoulli’s equation between B and C, VC2

VB2

pC pB + + gzC = + + gzB rw rw 2 2 pC 1000 kg>m3

+

13.60 m>s2 2 2

+ gz =

751103 2 N>m2 1000 kg>m3

+

3

pC = 118.52110 2 Pa

B

rC 5 0.3 m

FB 5 67.5p N rB 5 0.15 m

110 m>s2 2 2

+ gz MA

Ax Ay (a)

Then the pressure forces on inlet and outlet control surface C and B are FC = pCAC = 3118.521103 2 N>m2 43p10.05 m2 2 4 = 296.3p N FB = pBAB = 3751103 2 N>m2 43p10.03 m2 2 4 = 67.5p N

The linear momentum equation, ΣF =

0 Vr dV + VrV # dA 0t Lcv Lcs

can be written in scalar form along the x and y axes by referring to the FBD of the control volume shown in Fig. a, which can be simplified as

1 S+ 2 ΣFx =

0 + 1VC 2 xrw 1 - VCAC 2

Ax - 296.3p N = 1 - 3.60 m>s211000 kg>m3 21 -0.009p m3 >s2 Ax = 1032.64 N = 1.03 kN

1 + c 2ΣFy = 0 + 1VB 2 yrw 1VBAB 2

67.5p N - Ay = 1 - 10 m>s211000 kg>m3 210.009p m3 >s2 Ay = 494.80 N = 495 N

Ans.

Ans.

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2rV # dA 0t Lcv Lcs

can be written in scalar form about point A by referring to Fig. a, which can be simplified as + ΣMA = 0 + rBVBr1VBAB 2 + rCVCrw 1 - VCAC 2 1296.3p N210.3 m2 - 167.5p N210.15 m2 - MA

= 10.15 m2110 m>s211000 kg>m3 210.009p m3 >s2

+ 10.3 m213.60 m>s211000 kg>m3 21 - 0.009p m3 >s2 MA = 235.57 N # m = 236 N # m

Ans.

Ans: Ax = 1.03 kN Ay = 495 N MA = 236 N # m

560

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6–53. If water flows into the horizontal bend fitting at A with a velocity of 6 m>s, and exits at B into the atmosphere, determine the xand ycomponents of force and the moment at C needed to hold the fitting in place.

6 m!s

A

150 mm

C

200 mm 400 mm

308

SOLUTION

B

The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The fixed control volume considered consists of the bend fitting and the contained water. The discharge is Q = VCAC = VBAB = 16 m>s23p10.075 m2 2 4 = 0.03375p m3 >s

The water exits at B into the atmosphere. Then pB = patm = 0. Since the diameter of the fitting is constant, continuity requires VB = VC = 6 m>s. Also, Bernoulli’s equation gives pC = pB = 0. As a result, no pressure force is acting on the control volume. The FBD diagram of the control volume is shown in Fig. a. The linear momentum equation, ΣF =

0 Vr dV + Vr V # dA 0t Lcv Lcs

can be written is the scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

0 + 1VC 2 xrw 1 -VCAC 2 + 1VB 2 xrw 1VBAB 2

- Cx = 16 m>s211000 kg>m3 21 -0.03375p m3 >s2 + 1 -6 sin 30° m>s211000 kg>m3 210.03375p m3 >s2 Ans.

Cx = 954.26 N = 954 N

+ c ΣFy = 0 + 1VB 2 yrw 1VBAB 2

- Cy = 1 - 6 cos 30° m>s211000 kg>m3 210.03375p m3 >s2 Cy = 550.94 N = 551 N

Ans.

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2rV # dA 0t Lcv Lcs

can be written in scalar form about point C, which can be simplified as + ΣMC = 0 + 3 - 1rB 2 y1VB 2 x4rw 1VBAB 2 - 31rB 2 x1VB 2 y4rw 1VBAB 2

- MC = 3 - 10.6 m216 sin 30° m>s2411000 kg>m3 210.03375p m3 >s2

- 310.5464 m216 cos 30° m>s2411000 kg>m3 210.03375p m3 >s2

MC = 491.89 N # m = 492 N # m MC

Ans.

Cy

Cx 0.2 m

0.4 m

(rB)y 5 (0.4 1 0.4 sin 30°) m 5 0.6 m

30° 0.4 m

Ans: Cx = 954 N Cy = 551 N MC = 492 N # m

30° (rB)x 5 (0.2 1 0.4 cos 30°) m 5 0.5464 m (a)

561

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6–54. If water flows into the horizontal bend fitting at A with a velocity of 6 m>s, and exits at B into a tank having a gage pressure of 50 kPa, determine the x and y components of force and the moment at C needed to hold the fitting in place.

6 m!s

150 mm

A

C

200 mm 400 mm

308

B

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Average velocities will be used. The fixed control volume considered consists of the bend fitting and the contained water. The discharge is Q = VCAC = VBAB = 16 m>s23p10.075 m2 2 4 = 0.03375p m3 >s

Since the diameter of the fitting is constant, continuity requires VB = VC = 6 m>s. Also, Bernoulli’s equation gives pC = pB = 50 kPa. Then the pressure force on the inlet and outlet control surfaces C and B, respectively, are FB = FC = 3501103 2 N>m2 43p10.075 m2 2 4 = 281.25p N

The FBD of the control volume is shown in Fig. a. The linear momentum equation, ΣF =

0 Vr dV + Vr V # dA 0t Lcv Lcs

can be written in scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

0 + 1VC 2 xrw 1 - VCAC 2 + 1VB 2 xrw 1VBAB 2

281.25p N + 281.25p sin 30° N - Cx

= 16 m>s211000 kg>m3 21 -0.03375p m3 >s2

+ 1 -6 sin 30° m>s211000 kg>m3 210.03375p m3 >s2 Cx = 2279.32 N = 2.28 kN

Ans.

1 + c 2ΣFy = 0 + 1VB 2 yrw 1VBAB 2

281.25p cos 30° N - Cy = 1 - 6 cos 30° m>s211000 kg>m3 210.03375p m3 >s2

Ans.

Cy = 1316.14 N = 1.32 kN

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2rV # dA 0t Lcv Lcs

562

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6–54. Continued

can be written in scalar form about point C, which can be simplified as + ΣMC = 0 + 3 - 1rB 2 y1VB 2 x4rw 1VBAB 2 - 31rB 2 x1VB 2 y4rw 1VBAB 2

1281.25p sin 30° N210.6 m2 + 1281.25p cos 30° N210.5464 m2 - MC = 3 - 10.6 m216 sin 30° m>s2411000 kg>m3 210.03375p m3 >s2

- 310.5464 m216 cos 30° m>s2411000 kg>m3 210.03375p m3 >s2 MC = 1175.07 N # m = 1.18 kN # m

FC

MC

Ans.

Cy

Cx 0.2 m

0.4 m

(rB)y 5 (0.4 1 0.4 sin 30°) m 5 0.6 m

30° 0.4 m FB

30°

(rB)x 5 (0.2 1 0.4 cos 30°) m 5 0.5464 m (a)

Ans: Cx = 2.28 kN Cy = 1.32 kN MC = 1.18 kN # m 563

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6–55. If the velocity through the pipe is 4 m>s, determine the horizontal and vertical components of force and the moment exerted on the fixed support D. The total mass of the bend and the water within it is 20 kg, with a mass center at point G. The pressure of the water at A is 50 kPa. Assume that no force is transferred to the flanges at A and B.

100 mm

B G 608

A 100 mm 600 mm

100 mm

150 mm

SOLUTION

D

The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. Average velocity will be used. The fixed control volume considered consists of the elbow, the support and the contained water. From the discharge and continuity requirement, Q = 14 m>s23p10.05 m2 2 4

Q = VA;

Q = 0.01p m3 >s

FB

Since the change in elevation is negligible, Bernoulli’s equation gives pB = pA = 50 kPa. Thus, the pressure force on inlet and outlet control surfaces A and B, respectively, are FA = FB = pAAA 3501103 2 N>m2 43p10.05 m2 2 4 = 125p N

20(9.81) N

The FBD of the control volume is shown in Fig. a. The linear momentum equation,

60° FA

0 ΣF = Vr dV + Vr V # dA 0t Lcv Lcs can be written in scalar form along the xand yaxes, which can be simplified as

1 S+ 2 ΣFx =

0.1 m

(rA)y 5 (rB)y 5 0.6 m (rB)x 5 0.15 m

0 + 1VA 2 xrw 1 -VAAA 2 + 1VB 2 xrw 1VBAB 2

Dx - 125p N - 125p cos 60° N

= 1 -4 m>s211000 kg>m3 21 - 0.01p m3 >s2 3

MD

Dx = 777.54 N = 778 N

Dy (a)

3

+ 14 cos 60° m>s211000 kg>m 210.01p m >s2

Dx

Ans.

1 + c 2ΣFy = 0 + 1VB 2 yrw 1VBAB 2

Dy - 2019.812 N - 125p sin 60° N = 14 sin 60° m>s211000 kg>m3 210.01p m3 >s2

Ans.

Dy = 645.11 N = 645 N

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2r V # dA 0t Lcv Lcs

can be written in the scalar form about point D, which can be simplified as + ΣMD = 0 + 31rA 2 y1VA 2 x4rw 1 - VAAA 2 - 31rB 2 x1VB 2 y4rw 1VBAB 2 - 31rB 2 y1VB 2 x4rw 1VBAB 2

1125p N210.6 m2 + 1125p sin 60° N210.15 m2 + 1125p cos 60° N210.6 m2 - 312019.812N2410.1 m2 - MD

= 3 - 10.6 m214 m>s2411000 kg>m3 21 -0.01p m3 >s2

- 310.15 m214 sin 60° m>s2411000 kg>m3 210.01p m3 >s2 - 310.6 m214 cos 60° m>s2411000 kg>m3 210.01p m3 >s2 MD = 514.24 N # m = 514 N # m

Ans.

Ans: Dx = 778 N Dy = 645 N MD = 514 N # m

564

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*6–56. Water flows through the 200-mm-diameter pipe bend with a velocity of 8 m>s and discharges into the atmosphere at B. Determine the horizontal and vertical components of force and the moment acting on the coupling at A. The mass of the pipe bend and the water within it is 25 kg, having a center of gravity at G.

600 mm A 608 G B 450 mm 200 mm

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The fixed control volume considered consists of the bend pipe and the contained water. Using the discharge and continuity requirements, Q = VAAA = VBAB = 18 m>s23p10.1 m2 2 4 = 0.08p m3 >s

Since the diameter of the bend pipe is constant, VB = VA = 8 m>s. The water is discharged into atmosphere at B, then pB = patm = 0. Also, the change in elevation is negligible. Then, Bernoulli’s equation gives pA = pB = 0. Thus, no pressure force is acting on the control volume. The FBD of the control volume is shown in Fig. a. The linear momentum equation,

(rB)x 5 0.6 m

MA Ax

60° 0.45 m

Ay

25(9.81) N (a)

0 ΣF = Vr dV + VrV # dA 0t Lcv Lcs can be written in scalar form along the xand yaxes, which can be simplified as ΣFx = 0 + 1VA 2 xrw 1 - VAAA 2 + 1VB 2 xrw 1VBAB 2

Ax = 1 -8 m>s211000 kg>m3 21 -0.08p m3 >s2 + 1 - 8 cos 60° m>s211000 kg>m3 210.08p m3 >s2 Ax = 1005.31 N = 1.01 kN

1 + c 2ΣFy = 0 + 1VB 2 yrw 1VBAB 2

- Ay - 2519.812 N = 1 - 8 sin 60° m>s211000 kg>m3 210.08p m3 >s2 Ay = 1496.00 N = 1.50 kN

Ans.

Ans.

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2r V # dA 0t Lcv Lcs

can be written in scalar form about point A, which can be simplified as + ΣMA = 0 + 31rB 2 x 1VB 2 y4rw 1VBAB 2

MA + 32519.812 N410.45 m2 = 310.6 m218 sin 60° m>s2411000 kg>m3 210.08p m3 >s2 MA = 934.39 N # m = 934 N # m

Ans.

Ans: Ax = 1.01 kN 565

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6–57. Water flows through 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.

B

SOLUTION We consider steady flow of an ideal fluid. Take the control volume as the elbow and the water within it.

8 ft 12

Q = VAAA = (5 ft>s) c p a 3

= 0.2454 ft >s

Continuity Equation.

2 1.5 ft b d 12

MA Ax Ay

pA

0 r dV + V # dA = 0 0t L L cv

pB = 0

cs

0 - VAAA + VBAB = 0 -0.2454 ft3 >s + VB J p a

2 0.75 ft b R 12

(a)

= 0

VB = 20 ft>s

Applying the Bernoulli equation between A and B, pA pB VA2 VB2 + + gzA = + + gzB r r 2 2 pA °

+

(5 ft>s)2

3

62.4 lb ft ¢ 32.2 ft>s2

2

+ 0 = 0 +

(20 ft>s)2 2

+ 0

pA = 363.354 lb>ft2 = 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 Vr dV + 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>ft3 b ( 0.2454 ft3 >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>ft3 b ( 0.2454 ft3 >s ) ( -20 ft>s - 0) 32.2

Ay = 9.51 lb T

Ans.

566

M06_HIBB9290_01_SE_C06_ANS.indd 566

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6–57.

Continued

Angular Momentum. Referring to Fig. a, ΣM =

0 (r * V)r dV + (r * V)rV # dA 0t L L cv

cs

or + ΣMA = ΣrQVd;

- MA = a

62.4 8 slug>ft3 b ( 0.2454 ft3 >s ) c a - ft b(20 ft>s) - 0 d 32.2 12

MA = 6.34 lb # ft

Ans.

Ans: Ax = 20.2 lb, Ay = 9.51 lb MA = 6.34 lb # ft 567

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6–58. The bend is connected to the pipe at flanges A and B as shown. If the diameter of the pipe is 1 ft and the volumetric flow is 50 ft3 >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, and the mass center is 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

B

1.5 ft 458

4 ft

4 ft D

SOLUTION From the discharge,

50 ft3 >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>ft2. Average velocities will be used. Bernoulli Equation. pB = 1983.5 lb>ft2 FA = pAAA = 15 lb>in2(p)(6 in.)2 = 1696.46 lb FB = pBAB = 1983.5 lb>ft2(p)(0.5 ft)2 = 1557.84 lb pA VA2 PB VB2 + + gzA = + + gzB r r 2 2 15(144) lb>ft2 °

62.4 lb>ft2 32.2 ft>s

2

+

V2 + 0 = 2

¢

pB °

62.4 lb>ft2 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 xand yaxes 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>ft3 32.2 ft>s2

3 ( 63.66 ft>s ) cos 45° 4 °

¢ ( -50 ft3 >s )

62.4 lb>ft3 32.2 ft>s2

¢ ( 50 ft3 >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 ) sin 45° 4 °

62.4 lb>ft3 32.2 ft>s2

¢ ( 50 ft3 >s )

Ans.

Dy = 5963.3 lb = 5.96 kip

568

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6–58.

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, + 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>ft3 32.2 ft>s

2

¢ ( - 50 ft3 >s ) + ( - 4 ft) 3 ( 63.66 ft>s ) cos 45° 4 °

MD = 10 136.8 lb # ft = 10.1 kip # ft

62.4 lb>ft3 32.2 ft>s2

Ans.

¢ ( 50 ft3 >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)

569

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6–59. The fan blows air at 6000 ft3 >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>ft3.

2 ft

A G

B 0.75 ft 6 ft

SOLUTION We consider steady flow of an ideal fluid. Q = a

Then,

6000 ft3 1 min ba b = 100 ft3 >s min 60 s

Q = VBAB;

d 2

d 2

100 ft3 >s = VB 3 p(1 ft)2 4 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)r dV + (r * V)rV # dA 0t L L cv

+ 40 lba0.75 ft +

cs

0.076 lb>ft3 d b = a b ( 100 ft3 >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: d = 0.754 ft 570

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* 6–60. The chute has a cross-sectional area of 0.03 m2. If the flow is 0.4 m3 >s, 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, ΣM =

0 r * Vr dV + (r * V)rV # dA 0t L L cs cv

+ 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 Vr dV + 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 571

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6–61. If the propeller tube rotates at a constant rate of 60 rad>s when 1.75 kg>s of air enters at A, determine the frictional torque M acting on the tube. The density of air is ra = 1.23 kg>m3. The diameter of the nozzles at B, C, D, and E is 100 mm.

100 mm

B

0.5 m M

E

A D

60 rad!s

0.5 m

SOLUTION The flow is periodic, hence, for the mean, it can be considered steady. The air is assumed to be an ideal fluid (incompressible and inviscid) such that its density is constant. The control volume considered consists of the propeller and the contained air. Since the air is discharged into atmosphere, pB = pC = pD = pE = patm = 0. Thus, no pressure force acts on these outlet control surfaces. The FBD of the control volume is shown in Fig. a. Due to symmetry, and the continuity requirement, the # # mass flow rate through each nozzle is, m = mA >4 = 11.75 kg>s2 >4 = 0.4375 kg>s. Thus, the velocity of the air relative to the nozzle can be determined from

C 0.5 m

0.5 m

0.5 m

0.5 m B

# m = raVa>nA; 0.4375 kg>s = 11.23 kg>m3 21Va>n 23p10.05 m2 2 4 Va>n = 45.29 m>s

0.5 m

The velocity of the nozzle caused by the rotating propeller is M

Vn = vr = 160 rad>s210.5 m2 = 30 m>s

Consider nozzle B, Fig. a.

E

D

Va = Vn + Va>n

0.5 m

1 d+ 2Va = 1 -30 m>s2 + 45.29 m>s = 15.29 m>s d

The angular momentum equation, ΣM =

C

0 1r * v2r dV + 1r * V2r V # dA 0t Lcv Lcs

(a)

can be written in scalar form about an axis perpendicular to the page passing through A, which can be simplified as + ΣMA = 0 + 431rVa 2ra 1Va>n 21A24

M = 4310.5 m2115.29 m>s210.4375 kg>s24 = 13.38 N # m = 13.4 N # m

Ans.

Ans: M = 13.4 N # m 572

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6–62. The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water is supplied through the hose at 0.006 m3 >s and is ejected through the four nozzles into the atmosphere. Determine the torque required to keep the arms from rotating.

300 mm

308 308

308

308

SOLUTION The flow is steady and the water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The control volume considered consists of the sprinkler and the water contained. It is fixed since the sprinkler is required to be stationary. Since water is discharged into the atmosphere, the pressure on the nozzles pn = patm = 0. Thus, no pressure force acts on the outlet control surfaces. The FBD of the control volume is shown in Fig. a. Due to symmetry and the continuity requirement, the discharge through each nozzle is Q = 10.006 m3 >s2 >4 = 1.5110-3 2 m3 >s. Thus, the velocity of the water exiting from the nozzle is

z

0.3 m

Q = VA; 1.5110-3 2 m3 >s = V3p10.004 m2 2 4 V =

u 5 30°

93.75 m>s p

M

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2r V # dA 0t Lcv Lcs

F (a)

can be written in the scalar form about z axis, which can be simplified as ΣMz = 0 + 431rV cos u2rw 1VA24 M = 4e 10.3 m2 a

93.75 cos 30° m>s b 11000 kg>m3 231.5110-3 2 m3 >s4 f p

= 46.52 N # m = 46.5 N # m

Ans.

Ans: M = 46.5 N # m 573

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6–63. The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water is supplied through the hose at 0.006 m3 >s and is ejected through the four nozzles into the atmosphere. Determine the constant angular velocity of the arms. Neglect friction at the vertical axis.

300 mm

308 308

308

308

SOLUTION The flow is periodic, hence, for the mean, it can be considered steady. The water can be considered as an ideal fluid (incompressible and inviscid) such that its density is constant. The control volume considered consists of the sprinkler and the contained water. Since water is discharged into the atmosphere, the pressure on the nozzles, pn = patm = 0. Thus, no pressure force acts on the outlet control surfaces. The FBD of the control volume is shown in Fig. a. Here, it is required that M = 0. Due to symmetry and the continuity requirement, the discharge through each nozzle is y Q = 10.006 m3 >s2 >4 = 1.5110-3 2 m3 >s. Thus, the velocity of the water exiting from the nozzle relative to the nozzle is

z x

Q = Vw>nA; 1.5110-3 2 m3 >s = Vw>n 3p10.004 m2 2 4

u 5 30°

93.75 Vw>n = m>s p The velocity of the nozzle caused by the rotating propeller is

A M50

Vn = vr = v10.3 m2 = 0.3v Consider at nozzle A, Fig. a.

F

Vw = Vn + Vw>n

(a)

1Vw 2 x = 1Vn 2 x + 1Vw>n 2 x 1Vw 2 x = -0.3v + a

93.75 m>s b cos 30° = 125.84 - 0.3v2 m>s p

The angular momentum equation, ΣM =

0 1r * V2r dV + 1r * V2r V # dA 0t Lcv Lcs

can be written in scalar form about z axis, which can be simplified as ΣMz = 0 + 431rVw 2 xrw 1Vw>n 21A24

0 = 45 10.3 m2125.84 - 0.3v211000 kg>m3 231.5110-3 2 m3 >s46

25.84 - 0.3v = 0 v = 86.145 rad>s = 86.1 rad>s

Ans.

Ans: v = 86.1 rad>s 574

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*6–64. 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 a velocity V. If the wheel is turning at v, determine the power supplied to the wheel by the water.

R v

V

SOLUTION

h

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)r dV + (r * V)rV # dA 0t Lcv Lcs -T = 0 + RVrw( -VA) + RvRrw(VA)

+ 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)

575

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6–65. Air enters into the 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 = J a1500

rev 2p rad 1 min ba ba b R (0.5 m) = 25p m>s S min 1 rev 60 s

B

Thus, the velocity of the air ejected from B (or C) is Va = VB + Va>B

1 d+ 2 Va

=

0.5 m

( - 25p m>s ) + ( 400 m>s ) = 321.46 m>s d M

Applying the angular momentum equation, ΣM =

A

0 (r * V)r dV + (r * V)rV # dA 0t Lcv Lcs

0.5 m

Writing the scalar component about point A,

C

+ ΣMA = 0 + 2 3 rABVBraVBAB 4

(a)

# 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: M = 482 N # m 576

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6–66. 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 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

608

24.133˚ VW Vw

SOLUTION

t

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

(b)

a = 5.867°

150˚ 0.2 m

0.05 m

Then

ß

α

r=

r

b = 180° - 150° - 5.867° = 24.133° Thus, the velocity of the tip of the arm is

(a)

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 = Vt + Vw>t J

1 S+ 2 1+c2

(Vw)x R (Vw)yd T

= J

2.446 m>s

c

R

+ J

Vw t = 6 m s

6 m>s R 24.133°

- (Vw)x = - ( 6 m>s ) cos 24.133°

(Vw)x = 5.476 m>s d

- (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 =

(Vw)y2

(b)

150˚

0.2 m

0.05 m

2(Vw)x2 +

24.133˚ VW

ß = 2 ( 5.476 m>s ) 2 + ( 0.007339 m>s ) 2 r

= 5.476 m>s = 5.48 m>s

M

α r = 0.2446 m

Ans.

(a) The flow is steady and the water can be considered as an ideal fluid (incompressible 3 and inviscid) such that rw = 1000 kg>m . 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,

(c)

0 (r * V)rwdV + (r * V)rwV # dA 0t Lcv Lcs Writing the scalar 24.133˚component of this equation about point A, by referring to the FBD of the V control volume, Fig. a, ΣM =

W

+ ΣMA = 0 + r(Vw)yrw(Vw>t A)

Vw t = 6 m s

M = (0.2446 m) ( 0.007339 m>s )( 1000 kg>m3 ) 5 ( 6 m>s ) 3 p(0.0025 m)2 4 6 = 2.1145 ( 10-4 ) N # m = 0.211 mN # m (b)

Ans.

150˚ 0.2 m ß

M

α r = 0.2446 m

r (a)

Ans: Vw = 5.48 m>s M = 0.211 mN # m

(c)

577

M06_HIBB9290_01_SE_C06_ANS.indd 577

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6–67. 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. Take ra = 1.007 kg>m3.

1.5 m

SOLUTION The average velocity of the air flow through the propeller (control volume) is Q = VA;

350 m3 >s = V 3 p(0.75 m)2 4 V = 198.06 m>s

Here, V1 = a250

1h km 1000 m 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 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: F = 90.7 kN 578

M06_HIBB9290_01_SE_C06_ANS.indd 578

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*6–68. Plot Eq. 6–16 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

1

V2 V1

#

W # = 0.593 = 59.3% W0 when

V2 1 = . V1 3

579

M06_HIBB9290_01_SE_C06_ANS.indd 579

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6–69. The propeller of a boat discharges 67.5 ft3 >s of water as the boat travels at 25 mi>h in still water. If the diameter of the propeller is 15 in., determine the thrust developed by the propeller on the boat.

SOLUTION The flow of water (ideal fluid) can be considered steady relative to the control volume (propeller). Since the boat is travelling in still water 1Vw = 02, the velocity of water entering the control volume relative to the boat can be determined from Vw = VB + Vw>b

1 S+ 2 0

= -25 mi>h + Vw>b

Vw>b = 25 mi>h S

The average velocity of the water through the propeller relative to the control volume (propeller) can be determined from Q = VA; 67.5 ft3 >s = V c pa

Here, V1 = Vw>b = a25 V =

2 7.5 ft b d 12

V = 55.00 ft>s

mi 5280 ft 1h ba ba b = 36.67 ft>s. Then h 1 mi 3600 s

36.67 ft>s + V2 V1 + V2 ; 55.00 ft>s = 2 2

V2 = 73.34 ft>s

The thrust of the propeller is F =

=

rpR2 1V22 - V122 2 a

62.4 lb>ft3 32.2 ft>s

b 1p2 a 2 2

2 7.5 ft b 12

= 4797.3 lb = 4.80 kip

3173.34 ft>s2 2 - 136.67 ft>s2 2 4

Ans.

Ans: F = 4.80 kip 580

M06_HIBB9290_01_SE_C06_ANS.indd 580

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6–70. Determine the largest speed of the breeze that can be generated by the 8-kg fan so that it does not tip over, assuming that slipping will not occur. The blade has a diameter of 400 mm. The mass center is at G. Take ra = 1.20 kg>m3. G

300 mm

SOLUTION The flow is steady and the air can be assumed to be an ideal fluid (incompressible and inviscid) such that its density is constant. Since the air enters the control volume from the surroundings, which is at rest, V1 = 0. Here, V2 = Vb. The thrust on the blade is F = =

rapR2 1V22 - V122 2

11.20 kg>m3 21p210.2 m2 2

2 = 10.024p Vb22 N

100 mm

75 mm

8(9.81) N

1Vb2 - 02

F

Referring to the FBD of the fan shown in Fig. a, and write the moment equation of equilibrium about point A, + ΣMA = 0;

3819.812 N410.075 m2 - 10.024pV b2 210.3 m2 = 0 Vb = 16.13 m>s = 16.1 m>s

0.3 m

Ans. Ax 0.075 m

Ay

(a)

Ans: Vb = 16.1 m>s 581

M06_HIBB9290_01_SE_C06_ANS.indd 581

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6–71. 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 = 2.36110 - 3 2 slug>ft3. 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>ft3. 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>ft3 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: V2 = 38.7 ft>s # W = 7.04 hp 582

M06_HIBB9290_01_SE_C06_ANS.indd 582

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* 6–72. 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 = 3.50 MW

Ans.

Ans: # W = 3.50 MW 583

M06_HIBB9290_01_SE_C06_ANS.indd 583

02/03/17 12:32 PM

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6–73. The wind turbine has an efficiency of 48% in an 8 m>s wind. If the air is at a standard atmospheric pressure with a temperature of 20°C, determine the thrust on the blade shaft and the power withdrawn by the blades.

8 m!s

35 m

SOLUTION The efficiency of the wind turbine can be determined from eturb =

V2 V2 2 1 c1 - a b d c1 + d 2 V1 V1

The above equation if expanded will result in a cubic equation that requires V2 = 0.6454. Then V2 = 0.645418 m>s2 = numerical solution. With e = 0.48, we find V1 5.1633 m>s. The average velocity through the blades is 8 m>s + 5.1633 m>s V1 + V2 = = 6.5817 m>s 2 2 From the table in Appendix A, ra = 1.202 kg>m3 for T = 20°C. The thrust on the propeller is V =

F = =

rapR2 1V 12 - V 22 2 2

11.202 kg>m3 21p2117.5 m2 2

2 = 21.591103 2 N = 21.6 kN

318 m>s2 2 - 15.1633 m>s2 2 4

The power withdrawn by the blades is # Wo = FV = 321.591103 2 N416.5817 m>s2 = 142.101103 2 W = 142 kW

Ans.

Ans.

Ans: F = 21.6 kN # Wo = 142 kW 584

M06_HIBB9290_01_SE_C06_ANS.indd 584

02/03/17 12:33 PM

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6–74. The wind turbine has an efficiency of 48% in an 8 m>s wind. If the air is at a standard atmospheric pressure with a temperature of 20°C, determine the difference between the pressures just in front and just behind the blades. Also find the average velocity of the air passing through the blades. 8 m!s

35 m

SOLUTION The efficiency of the wind turbine can be determined from eturb =

V2 V2 2 1 c1 - a b d c1 + d 2 V1 V1

The above equation if expanded will result in a cubic equation that requires V2 = 0.6454. Then, V2 = 0.645418 m>s2 numerical solution. With e = 0.48, we find V1 = 5.1633 m>s. The average velocity through the blade is 8 m>s + 5.1633 m>s V1 + V2 = = 6.5817 m>s = 6.58 m>s Ans. 2 2 3 From the table in Appendix A, ra = 1.202 kg>m for T = 20°C. The thrust on the propeller is V =

F = =

rapR2 1V 12 - V 22 2 2 11.202 kg>m3 21p2117.5 m2 2

= 21.591103 2 N

2

318 m>s2 2 - 15.1633 m>s2 2 4

Then the pressure difference can be determined from ∆p =

21.591103 2 N F = = 22.44 Pa = 22.4 Pa A p117.5 m2 2

Ans.

Ans: V = 6.58 m>s ∆p = 22.4 Pa 585

M06_HIBB9290_01_SE_C06_ANS.indd 585

02/03/17 12:33 PM

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6–75. Air is drawn into the jet engine at A at 25 kg>s with a velocity of VA = 125 m>s, while the fuel is burned at 0.75 kg>s. Determine the horizontal force that the jet engine exerts on the supports. The exhaust velocity at B is VB = 550 m>s.

A

125 m!s

B 550 m!s

SOLUTION # # With ma = 25 kg>s, Vin = 125 m>s, mf = 0.75 kg>s, and Ve = 550 m>s, # # # T = 1ma + mf 2Ve - maVin = 125 kg>s + 0.75 kg>s21550 m>s2 - 125 kg>s21125 m>s2

= 11 037.5 N

Ans.

= 11.0 kN

Ans: T = 11.0 kN 586

M06_HIBB9290_01_SE_C06_ANS.indd 586

02/03/17 12:33 PM

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*6–76. Air is exhausted from the fixed jet engine at 20 kg>s with a velocity of 480 m>s while it is being tested with an attached brake deflector. Determine the force in each of the two supporting links needed to hold the deflector in the position shown.

158

158

308

308

SOLUTION # # With ma = 20 kg>s, Vin = 480 m>s, mf = 0, and Ve = - 1cos 30°21480 m>s2 = - 415.69 m>s, # # # T = 1ma + mf 2Ve - maVin

= 120 kg>s + 021 - 415.69 m>s2 - 120 kg>s21480 m>s2 = -17 913.84 N d = 17 913.84 N S

The force in each support link is 1 17 913.84 N a b 2 cos 15° = 9272.89 N

F =

Ans.

= 9.27 kN

Ans: F = 9.27 kN 587

M06_HIBB9290_01_SE_C06_ANS.indd 587

02/03/17 12:33 PM

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6–77. The plane has a mass of 6.5 Mg and is propelled using two jet engines. If the plane lands with a horizontal touchdown velocity of 60 m>s when the braking deflector is engaged, determine the velocity of the plane 4 s after it has landed. The exhaust from the engine measured relative to the plane is 900 m>s, and the steady mass flow is 24 kg>s. Neglect rolling resistance of the landing gear and the rate at which the fuel is consumed.

308 A

308

SOLUTION Take the control volume as the entire plane, the two jet engines and deflector, and # # # # # the fluid contained. Assume mf V ma. Then 1ma + mf 2 ≃ ma = 24 kg>s. The horizontal velocity of the exhaust relative to the plane is Ve = 1900 m>s2 cos 30°, leftward. The force equation for the entire plane of mass Mp and velocity V can be written as dV # # # 1 d+ 2ΣFx = Mp + maV + 1ma + mf 2Ve dt dV 0 = 16500 kg2 + 124 kg>s2V + 124 kg>s219002 cos 30° dt dV 6500 = -243V + 900 cos 30°4 dt dV 6500 = -241V + 779.422 dt Integrate this equation with initial condition V = 60 m>s at t = 0. L0

4s

V

24dt = -

6500dV L60 m>s V + 779.42

839.42 V + 779.42 839.42 e0.01477 = V + 779.42 V = 47.69 m>s = 47.7 m>s 96 = 6500 ln

Ans.

Ans: V = 47.7 m>s 588

M06_HIBB9290_01_SE_C06_ANS.indd 588

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6–78. 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 engine. 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 engine 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: T = 18.2 kN 589

M06_HIBB9290_01_SE_C06_ANS.indd 589

02/03/17 12:33 PM

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6–79. The jet is flying at 750 km>h. Still 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 with a velocity of 900 m>s, measured relative to the plane. Determine the force the engine exerts on 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.

Ans: T = 100 kN 590

M06_HIBB9290_01_SE_C06_ANS.indd 590

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*6–80. The jet is traveling at a 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 measured 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

#

Wo = TV =

3 74.16 ( 103 ) N 4 ( 400 m>s )

= 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 . . Wl = (ma + mf) Vmix2 2 1 = ( 90 kg>s + 1.8 kg>s )( 800 m>s ) 2 2 = 29.376 ( 106 ) MW The efficiency of the engine is # 29.664 ( 106 ) W Wo e = # = Wo + Pl 29.664 ( 106 ) W + 29.376 ( 106 ) W

Ans.

= 0.502

591

M06_HIBB9290_01_SE_C06_ANS.indd 591

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6–81. The jet boat takes in water directly through its bow at 0.03 m3 >s while traveling in still water with a 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 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 the ground is V = Ve - Vcv = 30 m>s - 10 m>s = 20 m>s. For both cases, the power loss is 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

#

e1 =

#

(Wo)1

#

(Wo)1 + (Wo),

=

6000 W = 0.5 6000 W + 6000 W

# (Wo)2 9000 W # # e2 = = = 0.6 (Wo)2 + (Wo), 9000 W + 6000 W

Ans. Ans.

Ans: T1 = 600 N T2 = 900 N e1 = 0.5 e2 = 0.6 592

M06_HIBB9290_01_SE_C06_ANS.indd 592

02/03/17 12:33 PM

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6–82. The jet has a speed of 1260 km>h when it is flying horizontally. Air enters the intake scoop S at the rate of 52 m3 >s. If the engine burns fuel at the rate of 1.85 kg>s, and the gas (air and fuel) is exhausted relative to the plane with a speed of 2880 km>h, determine the drag exerted on the plane by air resistance. The plane has a mass of 8 Mg. Assume that the air has a constant density of ra = 1.112 kg>m3.

S

SOLUTION The control volume considered is the entire jet plane and its contents, which moves with a constant velocity. Relative to the control volume, the flow can be considered steady. Air will be assumed as an ideal fluid (incompressible and inviscid). Here, Vcv = a1260

1h km 1000 m ba ba b = 350 m>s h 1 km 3600 s

Ve = a2880

km 1000 m 1h ba ba b = 800 m>s h 1 km 3600 s

# ma = rQ = 11.112 kg>m3 2152 m3 >s2 = 57.824 kg>s

dVcv Since the jet plane is travelling with a constant speed, = 0. Referring to the dt free-body diagram of the jet plane, dVcv # # # + maVcv - 1ma + mf 2Ve d+ ΣFx = m dt -FD = 0 + 157.824 kg>s21350 m>s2 - 157.824 kg>s + 1.85 kg>s21800 m>s2 FD = 27.501103 2 N = 27.5 kN

Ans.

Ans: FD = 27.5 kN 593

M06_HIBB9290_01_SE_C06_ANS.indd 593

02/03/17 12:33 PM

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6–83. The jet has a speed of 750 mi>h when it is flying horizontally. Air enters the intake scoop S at the rate of 2500 ft3 >s. If the engine burns fuel at the rate of 5.50 lb>s, and the gas (air and fuel) is exhausted relative to the plane with a speed of 1800 mi>h, determine the drag exerted on the plane by air resistance. The plane has a weight of 18 000 lb. Assume that air has a constant density of 1.754(10-3) slug>ft3.

S

SOLUTION The control volume considered is the entire jet plane and its contents, which moves with a constant velocity. Relative to the control volume, the flow can be considered steady. Air will be assumed as an ideal fluid (incompressible and inviscid). Here, Vcv = a750

mi 5280 ft 1h ba ba b = 1100 ft>s h 1 mi 3600 s

Ve = a1880

mi 5280 ft 1h ba ba b = 2640 ft>s h 1 mi 3600 s

# ma = rQ = 31.754110-3 2 slug>ft3 412500 ft3 >s2 = 4.385 slug>s lb 1 # mf = a5.5 ba b = 0.1708 slug>s s 32.2 ft>s2

dVcv Since the jet plane is travelling with a constant speed, = 0. Referring to the dt free-body diagram of the jet-plane, dVcv # # # 1 d+ 2ΣFx = m + maVcv - 1ma + mf 2Ve dt - FD = 0 + 14.385 slug>s211100 ft>s2 - 14.385 slug>s + 0.1708 slug>s212640 ft2 FD = 7.2041103 2 lb = 7.20 kip

Ans.

Ans: FD = 7.20 kip 594

M06_HIBB9290_01_SE_C06_ANS.indd 594

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* 6–84. The jet is traveling at 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 speed of 4000 ft>s measured relative to the plane, determine the acceleration of the plane at this instant. The drag 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. Note 1 mi = 5280 ft.

500 mi!h

308

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 # mf = = 0.3106 slug>s 32.2 ft>s2 Ve = 4000 ft>s Referring to the FBD of the control volume, Fig. a, d+ ΣFx = m

dVcv # # # + maVcv - ( ma + mf ) Ve dt

-(15 000 lb) sin 30° - 37 644.44 lb = a

15 000 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: acv = 102 ft>s2 595

M06_HIBB9290_01_SE_C06_ANS.indd 595

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6–85. If the rocket consumes 15 000 lb of solid fuel at a rate of 375 lb>s and ejects it with a velocity of 5(103) ft>s measured relative to the rocket, determine the velocity and acceleration of the rocket at the instant just before all the fuel has been consumed. Neglect air resistance and the variation of gravity with altitude. The rocket has a weight of 48 000 lb at liftoff and starts from rest.

v

SOLUTION The control volume considered consists of the rocket and its contents as shown in Fig. a, which is accelerating upward. We consider flow of an ideal fluid relative to the control volume. The mass of the control volume as a function of time is 375 lb>s 48 000 lb # M = Mo - mf t = a b - a bt 2 32.2 ft>s 32.2 ft>s2 = e

1 3481103 2 - 375t4 f slug 32.2

W 5 [21.5(103) 2 m· f t](9.81) N

Referring to the FBD of the control volume shown in Fig. a, with Ve = 51103 2 ft>s, + c ΣFy = m

- 3481103 2 - 375t4 lb =

dVcv # - mfVe dt

375 lb>s 1 dv 3481103 2 - 375t4 - a b 351103 2 ft>s4 32.2 dt 32.2 ft>s2

1.8751106 2 dv = - 32.2 dt 481103 2 - 375t

(1)

Integrating this equation with the initial condition V = 0 at t = 0, V = e

1.8751106 2 -375

V = 51103 2 ln c

ln 3481103 2 - 375t4 - 32.2t f ` 481103 2

481103 2 - 375t

The time required to consume all the fuel is

(a)

t 0

d - 32.2t

(2)

Mf 15 000 lb t = # = = 40 s 375 lb>s mf Substituting this result into Eqs. (1) and (2), a =

1.8751106 2 dv = - 32.2 = 24.618 ft>s2 = 24.6 ft>s2 dt 481103 2 - 3751402

V = 51103 2 ln c = 585.47 ft>s = 585 ft>s

481103 2

481103 2 - 3751402

Ans.

d - 32.21402

Ans.

Ans: a = 24.6 ft>s2 V = 585 ft>s 596

M06_HIBB9290_01_SE_C06_ANS.indd 596

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6–86. Determine the constant rate at which the fuel in the rocket must be burned, so that its thrust gives the rocket a speed of 500 m>s, 30 s after liftoff. The fuel is expelled from the rocket at a speed of 3000 m>s measured relative to the rocket. The rocket has a total mass of 21.5 Mg, including 6.5 Mg of fuel. 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 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 is # # M = Mo - mf t = 321.51103 2 - mf t4 kg Referring to the FBD of the control volume, Fig. a, and with Ve = 3000 m>s, + c ΣFy = m

W 5 [21.5(103) 2 m· f t](9.81) N

dVcv # - mfVe dt

# # dv # - 321.51103 2 - mf t419.812 = 321.51103 2 - mf t4 - mf 13000 m>s2 dt # 3000mf dv = # - 9.81 dt 21.51103 2 - mf t

Integrating this equation with the initial condition V = 0 at t = 0 and the requirement V = 500 m>s at t = 30 s, L0

500 m>s

dV =

L0

30 s

c

# 3000mf

# - 9.81 d dt 21.51103 2 - mf t

(a)

30 s # 500 = 5 - 3000 ln 321.51103 2 - mf t4 - 9.81t6 ` 0

ln c

21.51103 2

500 = 3000 ln c

# d = 0.2648 21.51103 2 - 30mf

21.51103 2

# d - 294.3 21.51103 2 - 30mf

21.51103 2

0.2648 # = e 21.51103 2 - 30mf

# mf = 166.71 kg>s = 167 kg>s

Ans.

For t = 30 s, the fuel consumed is # 1mf 2 C = mf 1t2 = 1166.71 kg>s2130 s2 = 5.0011103 2 kg

This amount is less than the mass of the fuel carried, which is 6.51103 2 kg. Thus, the solution is OK!

Ans: # mf = 167 kg>s 597

M06_HIBB9290_01_SE_C06_ANS.indd 597

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6–87. 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.

Ans: T = 150 kN 598

M06_HIBB9290_01_SE_C06_ANS.indd 598

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* 6–88. The balloon has a mass of 20 g (empty) and it is filled with air having a temperature of 20°C. When 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. Note V = 43pr 3.

8 m!s2

5 mm

SOLUTION

W = 0.1962 N

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 4 m = mb + ma = 0.02 kg + ( 1.202 kg>m3 ) c p (0.3 m)3 d 3 = 0.1559 kg

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 counteracted 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: # me = 0.00584 kg>s 599

M06_HIBB9290_01_SE_C06_ANS.indd 599

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6–89. 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 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. ΣF = m

dVcv 0 + Vf>cs r dV + Vf>cs rVf>cs # dA dt 0t L L cv cv

FD = ct

Writing the scalar component of this equation along the yaxis 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 m0 ct # b + # ln a # b + # - # 2 ln # - gt m0 - met me m0 - met me me m0 - met peAe m0c m0 c - # 2 blna # # b + a # - gbt me m0 - met me me

Ans. F e = pe A e (a)

Ans: V = aVe +

peAe m0c m0 c # - # 2 blna # b + a # - gbt me m m t m me 0 e e

600

M06_HIBB9290_01_SE_C06_ANS.indd 600

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6–90. The rocket has an initial mass m0, including the fuel. If the fuel is expelled from the rocket at ve measured relative to the rocket, determine the rate at which the fuel should be consumed to maintain a constant acceleration a0. 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 dm . . . Applying Eq. (6–16) with ma = 0, mass m. Thus, mf = dt + c ΣFy = m

dVcv . . . + maVcv - ( ma + mf)Ve dt

W = mg

dm - mg = ma0 + 0 - a0 V b dt e m(a0 + g) dm = dt Ve

m

(1)

t

a0 + g dm = dt Ve L L m 0

mo

ln

a0 + g m = -a bt m0 Ve a +g m = e - 1 V 2t m0 0

(a)

e

m = m0 e - 1

a0 + g Ve

2t

Substitute this result into Eq (1). a +g m0 dm . mf = = (a + g)e - 1 V 2 t dt Ve 0 0

Ans.

e

Ans: m0 # mf = (a + g)e - (a0 + g)t>Ve Ve 0 601

M06_HIBB9290_01_SE_C06_ANS.indd 601

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6–91. The balloon carries a bucket containing 200 kg of water. If it is ascending with a constant velocity of 4 m>s and it then releases 80 kg>s of water through a 100-mmdiameter opening, determine the initial upward acceleration of the balloon as the water is being released. The balloon and empty bucket have a combined a mass of 1.5 Mg.

a

SOLUTION The control volume considered consists of the balloon and the tank containing water as shown in Fig. a, which is ascending with a constant velocity. We consider steady flow of an ideal fluid relative to the control volume. The initial mass of the control volume is M0 = 1.51103 2 kg + 0.21103 2 kg = 1.71103 2 kg

The velocity of the discharged water relative to the control volume can be determined using the mass flow rate of the water. # mw = rwVe A; 80 kg>s = 11000 kg>m3 21Ve 23p10.05 m2 2 4 Ve =

32 m>s p

Fb

Referring to the FBD of the balloon shown in Fig. a, + c ΣFy = m

W

dVcv # - mfVe dt

Fb - W = 31.71103 2 kg4

dV 32 - 180 kg>s2 a m>s b p dt

(a)

Since the balloon is ascending with a constant velocity before the water is released, its weight and the water’s initial weight are balanced by the buoyant force, i.e., Fb - W = 0. Then 0 = 1.71103 2

ai =

dV 2560 p dt

dV = 0.479 m>s2 c dt

Ans.

Ans: ai = 0.479 m>s2 602

M06_HIBB9290_01_SE_C06_ANS.indd 602

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*6–92. The second stage B of the two-stage rocket weighs 2500 lb (empty) and is launched from the first stage with a relative 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 its 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. dVcv . - mfVe dt 75 dV 0 = (102.48 slug) - a slug>s b ( 6000 ft>s ) dt 32.2 dV a = = 136.36 ft>s2 = 136 ft>s2 dt

ΣFy = m

Just before all the fuel is consumed, m =

Ans.

2500 lb = 77.64 slug 32.2 ft>s2

dV . - mfVe dt dV 75 0 = (77.64 slug) - a slug>s b ( 6000 ft>s ) dt 32.2 dV a = = 180 ft>s2 dt

ΣFy = m

Ans.

Ans: When second stage is fired, a = 136 ft>s2. Just before all the fuel is consumed, a = 180 ft>s2. 603

M06_HIBB9290_01_SE_C06_ANS.indd 603

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7–1. As the top plate is pulled to the right with a constant velocity U, the fluid between the plates has the linear velocity distribution 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 We consider steady flow of an ideal fluid. Referring to the velocity profile shown in Fig. a, u U = ; y h

u =

U y h

U

And

u

y = 0

h y

The rate of rotation or average angular velocity of the fluid element is vz =

1 0y 0u 1 u U U a b = a0 - b = = 2 0x 0y 2 h 2h 2h

Ans.

(a)

The rate of shear strain is

#

gxy =

0y 0u U + = 0x 0y h

Ans.

Ans:

U 2h U = h

vz =

#

gxy 604

M07_HIBB9290_01_SE_C07_ANS.indd 604

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7–2. A flow is defined by its velocity components u = 12x 2 + 2y2 2 ft>s and v = 1 - 4xy2 ft>s, where x and y are in feet. Determine if the flow is rotational or irrotational. What is the circulation around the triangular path OABO?

y

B

SOLUTION We consider ideal fluid flow.

4 ft

0u 0 = (2x 2 + 2y2) = 4y 0y 0y 0v 0 = ( -4xy) = -4y 0x 0x vz =

O

1 0v 0u 1 a b = ( -4y-4y) = - 4y 2 0x 0y 2

Since vz ≠ 0, the flow is rotational.

3 ft

A

x

Ans.

Along edge OA, ds = dx i and y = 0. Then C

V # ds = = = =

C

(ui + vj) # dx i

L0

3 ft

L0

3 ft

u dx 32x 2 + 2(0)2 4 dx

2 3 3 ft x ` = 18 ft2 >s 3 0

Along edge AB, ds = dyj and x = 3 ft. Then C

V # ds = = =

C

(ui + vj) # dy j

L0

4 ft

L0

4 ft

v dy - 4(3)y dy

= -6y2 `

4 ft 0

= -96 ft2 >s

Along edge BO, ds = ( - dxi- dyj). Here, y = 43x or x = 34y. C

V # ds = = =

C

(ui + vj) # ( -dxi-dyj)

L0

3 ft

L0

3 ft

= = -

Thus,

-u dx +

L0

L0

4 ft

- v dy

4 2 - c 2x 2 + 2a xb d dx + 3 L0

3 ft

50 2 x dx + 9 L0

4 ft

3 4a yby dy 4

4 ft

3y2dy

4 ft 50 3 3 ft x ` + y3 ` 27 0 0

= 14 ft2 >s

Γ = 18 ft2 >s + ( - 96 ft2 >s) + (14 ft2 >s) = - 64 ft2 >s

M07_HIBB9290_01_SE_C07_ANS.indd 605

Ans. 605

Ans: rotational, Γ = -64 ft2 >s

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7–3. A uniform flow at 6 m>s is directed at an angle of 30° to the horizontal as shown. Determine the circulation around the rectangular path OABCO. 6 m!s

0.8 m

C

B 0.2 m

O

A

30°

SOLUTION We consider ideal fluid flow. Here, V = 56 cos 30°i + 6 sin 30°j6 m>s = 55.196i + 3j6 m>s

Along OA, ds = dxi. Then C

V # ds = =

C

(5.196i + 3j) # dxi

L0

0.8 m

-5.196 dx = 4.157 m2 >s

Along AB, ds = dyi. Then C

V # ds = =

C

(5.196i + 3j) # dyj

L0

0.2 m

Along BC, ds = -dxi. Then C

V # ds =

=

C

(5.196i + 3j) # ( - dxi)

L0

0.8 m

Along CO, ds = - dyj. Then C

V # ds =

= Then

3 dy = 0.6 m2 >s

- 5.196 dx = - 4.157 m2 >s C

(5.196 i + 3j) # ( -dyj)

L0

0.2 m

-3 dy = -0.6 m2 >s

Γ = 4.157 m2 >s + 0.6 m2 >s + ( -4.157 m2 >s) + ( - 0.6 m2 >s) = 0

Ans.

Note: This result is to be expected since Γ = 0 for irrotational flow, which is the type of flow in this problem.

Ans: Γ = 0 606

M07_HIBB9290_01_SE_C07_ANS.indd 606

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*7–4. The velocity within the eye of a tornado is defined by vr = 0, vu = 10.4r2 m>s, where r is in meters. Determine the circulation at r = 100 m and at r = 200 m. 100 m 200 m

SOLUTION We consider ideal fluid flow. Since v u is always tangent to the circle, v # ds = yu ds. For r = 100 m, yu = 0.4(100) m>s = 40 m>s and ds = r du = 100 du. Γr

= 100 m

=

C

V # ds =

L0

2p

yuds =

L0

2p

2p

40(100 du) = 4000 u # 0 = 8000p m2 >s

Ans.

For r = 200 m, yu = 0.4(200) m>s = 80 m>s, and ds = r du = 200 du. Γr

= 200 m

=

C

V # ds =

L0

2p

yuds =

L0

2p 2 80(200 du) = 16 000 u # 2p 0 = 32 000p m >s Ans.

607

M07_HIBB9290_01_SE_C07_ANS.indd 607

Ans: Γr = 100 m = 8000p m2 >s Γr = 200 m = 32 000p m2 >s

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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 = - 1v>r + dv>dr2.

D

C G

y 1 dy

dr

B y

A

r Du

SOLUTION We consider ideal fluid flow. The circulation of the flow around element ABCD can be determined from Γ =

C

V # ds

= ySAB + (y + dy)( -SCD) = y(r∆u) + (y + dy)3 -(r + dr)∆u4 = - y dr∆u - r dy∆u - dy dr∆u Neglect the second order terms. Γ = -y dr∆u - r dy∆u = - ∆u(y dr + r dy) The area of the element, again, neglecting higher-order terms, is A = (r∆u)dr Thus, the vorticity is z =

- ∆u(y dr + r dy) Γ = A (r∆u)dr

= -a

y dy + b r dr

(Q.E.D)

608

M07_HIBB9290_01_SE_C07_ANS.indd 608

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y

7–6. Determine the stream and potential functions for the two-dimensional flow.

SOLUTION

30°

We consider ideal fluid flow. Here, the components of velocity along the x and y axes are v = -5 sin 30° m>s = - 2.5 m>s.

u = 5 cos 30° m>s = 4.330 m>s Since the continuity equation

0u 0v + = 0 + 0 = 0 is satisfied, the 0x 0y

5 m!s

x

establishment of a stream function is possible using the velocity components. 0c = u; 0y

0c = 4.330 0y

Integrating this equation with respect to y, (1)

c = 4.330y + f(x) Also, -

0c = v; 0x

-

0 34.330y + f(x)4 = -2.5 0x

0 3f(x)4 = 2.5 0x

Integrating this equation with respect to x,

f(x) = 2.5 Substituting this result into Eq.(1), c = 4.330y + 2.5x Ans.

= 4.33y + 2.5x Since vz =

1 0v 0u 1 a b = (0 - 0) = 0, the flow is indeed irrotational. Thus, the 2 0x 0y 2

potential function exists. Using the velocity components, 0f = u; 0x

0f = 4.330 0x

Integrating this equation with respect to x, (2)

f = 4.330x + f(y) Also, 0f 0 = v; = 34.330x + f(y)4 = -2.5 0y 0y 0 3f(y)4 = - 2.5 0y

Integrating this equation with respect to y,

f(y) = -2.5y Substituting this result into Eq. (2), f = 4.330x + ( -2.5y) Ans.

= 4.33x - 2.5y

Ans: c = 4.33y + 2.5x, f = 4.33x - 2.5y

609

M07_HIBB9290_01_SE_C07_ANS.indd 609

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7–7. A two-dimensional flow is described by the stream function c = 1 - 2x 3y + 2xy3 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 = ( - 2x 3y + 2xy3) = ( - 2x 3 + 6xy2) m>s 0y 0y

v = -

0c 0 = - ( - 2x 3y + 2xy3) = -( - 6x 2y + 2y3) m>s = (6x 2y - 2y3) m>s 0x 0x

Then 0u 0 = ( - 2x 3 + 6xy2) = ( -6x 2 + 6y2) s -1 0x 0x 0v 0 = (6x 2y - 2y3) = (6x 2 - 6y2) s -1 0y 0y 0u 0 = ( - 2x 3 + 6xy2) = (12xy) s -1 0y 0y 0v 0 = (6x 2y - 2y3) = (12xy) s -1 0y 0y This gives 0u 0v + = ( - 6x 2 + 6y2) + (6x 2 - 6y2) = 0 0x 0y Thus, the flow field satisfied the continuity condition. Also, vz =

1 0v 0u 1 a b = (12xy - 12xy) = 0 2 0x 0y 2

The flow field is irrotational since vz = 0.

Ans.

Ans: irrotational 610

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*7–8. If the stream function for a flow is c = 118x + 12y2ft2 >s, where x and y are in feet, determine the potential function, and the magnitude of the velocity of a fluid particle at point (5 ft, – 4 ft).

SOLUTION We consider ideal fluid flow. The x and y components of velocity are given by u =

0c 0 = (18x + 12y) = 12 ft>s 0y 0y

v = -

0c 0 = (18x + 12y) = -18 ft>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(12 ft >s)2 + ( -18 ft>s)2 = 21.63 ft>s = 21.6 ft>s Ans.

Using the velocity components,

0f = u; 0x

0f = 12 0x

Integrating this equation with respect to x, (1)

f = 12x + f(y) Substitute this result into 0f = v; 0y

Integrating this equation,

0 312x + f(y)4 = -18 0y 0 3f(y)4 = -18 0y f(y) = -18y

Substituting this result into Eq. (1), f = 12x + ( -18y) = (12x - 18y) ft2 >s

Ans.

611

M07_HIBB9290_01_SE_C07_ANS.indd 611

Ans: V = 21.6 ft>s f = (12x - 18y) ft2 >s

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7–9. The velocity profile for a liquid flowing along the channel of constant width is approximated as u = 121y2 m>s, where y is in meters. Determine the stream function for the flow, and plot the streamlines for c = 0, c = 0.5 m2 >s, and c = 1.5 m2 >s.

y u 5 (2 y ) mm!s

1.25 m x

SOLUTION The x and y components of flow velocity are

y

u = (22y) m>s

v = 0

Using the velocity components, 0c 0c u = ; 22y = 0y 0y Integrating this equation with respect to y, 4 c = y3>2 + f(x) 3

1.08

c 5 1.5

0.520

c 5 0.5 x

0

c50 (a)

Also, v = -

0c ; 0x

0 = -

0 = Integrating this equation with respect to x,

0 4 3>2 c y + f(x) R 0x 3 0 3f(x)4 0x

f(x) = C Thus, c =

4 3>2 y + C 3

Setting C = 0, c =

4 3>2 y 3

Ans.

For c = 0, y = 0. For c = 0.5 m2 >s, y = 0.520.

Ans.

2

For c = 1.5 m >s, y = 1.08.

The plot of these streamlines is shown in Fig. a.

Ans:

4 3>2 y 3 For c = 0, y = 0. For c = 0.5 m2 >s, y = 0.520. For c = 1.5 m2 >s, y = 1.08. c =

612

M07_HIBB9290_01_SE_C07_ANS.indd 612

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7–10. The velocity profile of a liquid flowing along the channel of constant width is approximated as u = 121y2 m>s, where y is in meters. Is it possible to determine the potential function for the flow? If so, what is it?

y u 5 (2 y ) mm!s

1.25 m x

SOLUTION We consider ideal fluid flow. The x and y components of flow velocity are u = Here, 0v = 0 0x

1 22y 2 mm>s

v = 0

0u 0 = 1 22y 2 = (y -1>2) rad>s 0y 0y

vz =

1 0y 0u 1 1 b = (0 - y -1>2) = - y -1>2 a 2 0x 0y 2 2

Since vz ≠ 0, the flow is rotational. Thus, the potential function cannot be established since it requires the flow to be irrotational. Ans.

Ans: f cannot be established. 613

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7–11. A two-dimensional flow has a y component of velocity of v = 16y2 m>s, where y is in meters. If the flow is ideal, determine the x component of velocity and find the magnitude of the velocity at the point x = 1 m, y = 2.5 m. 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 0v = (6y) = 6 s -1 0y 0y Then 0u + 6 = 0 0x 0u = -6 0x Integrating this equation with respect to x, (1)

u = - 6x + f(y) Since ideal flow is irrotational, vz = 0. Then 0v 0u = 0 0x 0y and since

0v 0 = (6y) = 0, then 0x 0x 0u = 0 0y

Integrating this equation with respect to y, (2)

u = g(x) Comparing Eq (1) and (2), we notice that f(y) = 0. Then

Ans.

u = ( - 6x) m>s This is consistent with u = v = 0 at the origin. At x = 1m and y = 2.5m, u = -6(1) = - 6 m>s

v = 6(2.5) = 15 m>s

Thus, the magnitude of the velocity at this point is V = 2u2 + v2 = 2( -6 m>s)2 + (15 m>s)2 = 16.16 m>s = 16.2 m>s Ans.

Ans: u = 1 -6x2 m>s, V = 16.2 m>s 614

M07_HIBB9290_01_SE_C07_ANS.indd 614

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*7–12. The liquid confined between the two moving plates has a linear velocity distribution. 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, y = 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 y = -

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, 0y = 0; 0x

0u 0 = (100y + 0.2) = 100 rad>s 0y 0y

Thus, vz =

1 0y 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.

Ans: c = 50y2 + 0.2y, f cannot be established. 615

M07_HIBB9290_01_SE_C07_ANS.indd 615

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7–13. The liquid confined between two moving plates has a linear velocity distribution. If the pressure at the top surface of the bottom plate is 600 N>m2, determine 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, y = 0. Here, 0y = 0; 0x vz =

0u 0 = (100y + 0.2) = 100 rad>s 0y 0y

0.2 m s (a)

1 0y 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 + y 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 Ans.

Ans: pA = 482 Pa 616

M07_HIBB9290_01_SE_C07_ANS.indd 616

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7–14. A two-dimensional flow is defined by its components u = 12y2 m>s and v = 13x2 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 (2 m, 6 m). Plot this streamline.

SOLUTION The necessary partial differentiations are as follows: 0v 0 = 13x2 = 3 rad>s 0x 0x

Thus,

0u 0 = 12y2 = 0 0x 0x

0v 0 = 13x2 = 0 0y 0y vz =

0u 0 = 12y2 = 2 rad>s 0y 0y

1 0v 0u 1 1 a b = 13 - 22 = rad>s 2 0x 0y 2 2

Since vz ≠ 0, the flow is rotational. Also, since

Ans.

0u 0v + = 0 + 0 = 0 0x dy the flow satisfies the continuity condition. Using the x and y components of velocity, u =

0c ; 0y

2y =

0c 0y

Integrating this equation with respect to y, c = y2 + f1x2

(1)

Substituting this result into v = -

0c ; 0x

3x = 3x = -

Integrating this equation with respect to x,

0 2 3y + f1x24 0x

0 3f1x24 0x

3 f1x2 = - x 2 2 Substituting this equation into Eq. 112,

c = y2 -

3 2 x 2

Ans.

From the slope of the streamline, dy v 3x = = dx u 2y y

x

L6 m

2y dy =

y2 2

y 6m

=

y2 - 36 =

3x dx L2 m 3 22x x 2 2m 3 2 1x - 42 2 617

M07_HIBB9290_01_SE_C07_ANS.indd 617

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7–14. Continued

y2 =

y =

3 2 1x + 302 2 3 2 x + 30 A2

Ans.

The plot of this streamline is shown in Fig. a. x1m2 y1m2

0 5.48

1 5.61

2 6

3 6.60

4 7.35

5 8.22

y (m) 10 8 6 4 2 0

1

2

3

4

5

x (m)

(a)

618

M07_HIBB9290_01_SE_C07_ANS.indd 618

Ans: 3 rotational, c = y2 - x 2, 2 3 2 y = x + 30 A2

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7–15. A flow is defined by the stream function c = 10.5x 2 - y2 2 m2 >s, where x and y are in meters. Determine if the flow satisfies the continuity condition through the triangle ABCA.

y

B

C

3m

SOLUTION We consider steady fluid flow. At point A, x = 0 and y = 0. Thus,

x

A 4m

cA = 0.5102 2 - 02 = 0

At point B, x = 4 m and y = 3 m.

cB = 0.5142 2 - 32 = - 1 m2 >s

At point C, x = 0 and y = 3 m.

cC = 0.5102 2 - 32 = - 9 m2 >s

Thus, the flow rates per unit depth through AB, BC and AC are qAB = cA - cB = 0 - 1 - 1 m2 >s2 = 1 m2 >s Then

qBC = cB - cC = -1 m>s2 - 1 -9 m2 >s2 = 8 m2 >s

qAC = cA - cC = 0 - 1 - 9 m2 >s2 = 9 m2 >s qAB + qBC = qAC

Therefore, the flow through triangle ABCA satisfies the continuity condition. Ans.

Ans: yes 619

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*7–16. A two-dimensional flow is defined by its components u = 13x 2 2 m>s and v = 12x 2 - 6xy2 m>s, where x and y are in meters. Determine the stream function, and plot the streamline that passes through point (2 m, 4 m).

SOLUTION We consider steady fluid flow. Since the continuity equation 0u 0v + = 6x + 1 - 6x2 = 0 is satisfied, then the establishment of the stream dx 0y

function is possible. Using the definition of velocity components with respect to the stream function

y (m)

0c + u; 0y

0c = 3x 2 0y

16 14

Integrating this equation with respect to y,

12

c = 3x 2y + f1x2

Also, -

0c = v; 0x

-

10

0 = 33x 2y + f1x24 = 2x 2 - 6xy 0x

8 6

0 3f1x24 = 2x 2 - 6xy -6xy 01x2

4 2

0 3f1x24 = - 2x 2 01x2

0

Integrating this equation with respect to x,

1

2

3

4

5

6

7

8

x (m)

(a)

2 f1x2 = - x 3 3 The stream function can be expressed as 2 3 x 3 For the streamline passing through the point where x = 2 m and y = 4 m, 1 128 c = 39122 2142 - 2123 24 = 3 3 Thus, c = 3x 2y -

128 1 = 19x 2y - 2x 3 2 3 3

2x 3 + 128 9x 2 The plot of this streamline is shown in Fig. a.

Ans.

y =

x1m2 y1m2

0 ∞

1 14.4

2 4

3 2.25

4 1.78

Ans.

5 1.68

6 1.73

7 1.85

8 2

Ans: 2 3 x 3 2x 3 + 128 y = 9x 2

c = 3x 2y -

620

M07_HIBB9290_01_SE_C07_ANS.indd 620

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7–17. A flow has velocity components u = 21y - x2 ft>s and v = 14x + 2y2 ft>s, where x and y are in feet. Determine the stream function, and plot the streamline that passes through the origin.

SOLUTION

y (ft)

We consider steady fluid flow since the continuity equation

0u 0v + = -2 + 2 = 0 dx 0y

10 8

is satisfied, then the establishment of the stream function is possible. Using the definition of velocity components with respect to the stream function, 0c = u; dy

4

0c = 21y - x2 0y

2

Integrating this equation with respect to y, Substituting this result into 0c v = - ; 0x

6

–3

c = y2 - 2xy + f1x2

0 3f1x24 = -4x 0x

–1

2

0

3

x (ft)

–2 –4

0 4x + 2y = 3y2 - 2xy + f1x24 0x 4x + 2y = - e -2y +

–2

–6 –8

0 3f1x24 f 0x

–10 (a)

Integrating this equation with respect to x,

f1x2 = -2x 2 So the stream function can be expressed as c = y2 - 2xy + 1 -2x 2 2

c = y2 - 2x 2 - 2xy

Ans.

Evaluate c1x,y2 at the origin, x = y = 0. This equation gives c = 0 - 0 - 0 = 0. Then, for c = 0, y2 - 2x 2 - 2xy = 0 Solving for y in terms of x by means of the quadratic formula, y = 11 { 232x

Ans.

The plot of this streamline is shown in Fig. a. x1 ft2

-3

-2

-1

0

1

2

3

y1 ft2 2.20 - 8.20 1.46 - 5.46 0.732 - 2.73 0 2.73 - 0.732 5.46 - 1.46 8.20 - 2.20 Note: The convergence of streamlines at the origin is possible because the velocity there is zero.

621

M07_HIBB9290_01_SE_C07_ANS.indd 621

Ans: c = y2 - 2x 2 - 2xy, y = 11 { 232x

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7–18. A flow is described by the stream function c = 12x - 4y2 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 steady fluid flow. The x and y components of velocity are given by u =

0c 0 = 12x - 4y2 = -4 m>s 0y 0y

v = -

0c 0 = - 12x - 4y2 = -2 m>s 0x 0x

The necessary partial derivatives are

0u 0 = 1 -42 = 0 0x 0x

0v 0 = 1 -22 = 0 0y 0y

0u 0 = 1 -42 = 0 0y 0y

0v 0 = 1 -22 = 0 0x 0x

Since 0u 0v = = 0 + 0 = 0 0x 0y the flow field satisfies the continuity condition. Also, vz =

1 0v 0u 1 a b = 10 - 02 = 0 2 0x 0y 2

Since vz = 0, the flow field is irrotational. Thus, the potential function can be established. Using the velocity components, u =

0f ; 0x

-4 =

0f 0x

Integrating this equation with respect to x, f = - 4x + f1y2 Substituting this result into v =

0f ; 0y

-2 = -2 =

0 3 - 4x + f1y24 0y

0 3f1y24 0y

Integrating this equation with respect to y, Thus,

f1y2 = -2y Ans.

f = - 4x - 2y

Ans: f = - 4x - 2y 622

M07_HIBB9290_01_SE_C07_ANS.indd 622

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7–19. Water flow through the horizontal channel is defined by the stream function c = 21x 2 - y2 2 m2 >s. If the pressure at B is atmospheric, determine the pressure at point A (0.5 m, 0) and the flow per unit depth in m2 >s.

y

B

c50 1.5 m c 5 0.5 m2 !s

SOLUTION We consider ideal fluid flow. The velocity components are 0c u = = ( - 4y) m>s 0y The continuity equation At point A (0.5 m, 0),

x

A

0c y = = ( -4x) m>s 0x

1.5 m

0u 0y + = 0 + 0 = 0 is indeed satisfied. 0x 0y yA = - 4(0.5) = -2 m>s

uA = 0 Thus,

VA = yA = -2 m>s At point B (1.5 m, 1.5 m), uB = -4(1.5) = - 6 m>s

yB = - 4(1.5) = -6 m>s

Thus, VB = 2uB2 + yB2 = 2 ( - 6 m>s ) 2 +

( - 6 m>s ) 2 = 272 m>s = 8.485 m>s

1 0y 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 V 2A V 2B = + 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

623

M07_HIBB9290_01_SE_C07_ANS.indd 623

Ans: pA = 34 kPa, c2 - c1 = 0.5 m2 >s

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*7–20. The flat plate is subjected to the flow defined by the stream function c = 3 8r 1>2 sin 1u>22 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

u

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 r 2 sin = 2 2 The plot of this function is shown in Fig. a. u(rad)

0

r(m)

∞

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

11p 6 59.71

2p ∞

x (m)

ψ = 16

y (m)

(a)

624

M07_HIBB9290_01_SE_C07_ANS.indd 624

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7–20. Continued

The radial and transverse components of velocity are 4 cos

1 1 u 1 1 0c = c 8r 2 a cos b d = yr = r 0u r 2 2

1

r2

0c 1 u 1 yu = = - a8r - 2 sin b = 0r 2 2 0y y 0y The continuity equation r + r + 1 u = r 0r r 0u

4 cos

is indeed satisfied.

3

r2

u 2

u 2

4 sin

u 2

1

r2 - 2 cos

+ ±

u 2

3

r2

2 cos ≤ + ±-

u 2

3

r2

≤ = 0

At point r = 4 m, u = p rad, 4 cos yr =

p 2

1

42 4 sin

yu = -

1

42

= 0

p 2

= -2 m>s

Thus, the magnitude of the velocity is Ans.

V = yu = 2 m>s

Ans: V = 2 m>s 625

M07_HIBB9290_01_SE_C07_ANS.indd 625

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7–21. A fluid flows into the corner formed by the two walls. If the stream function for this flow is defined by c = 15 r 4 sin 4u2 m2 >s, show that continuity for the flow is satisfied. Also, plot the streamline that passes through point r = 2 m, u = 1p>62 rad, and find the magnitude of the velocity at this point.

y

458

SOLUTION

x

We consider ideal fluid flow. p rad, 6

For the stream f unction passing through point r = 2 m, u =

y (m)

p c = 5 ( 24 ) sin c 4 a b d = 4023 m2 >s 6

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 °

∞ (a)

The radial and transverse components of velocity are yr =

x (m)

1 0c 1 = 3 5r 4(4 cos 4u) 4 = 20r 3 cos 4u r 0u r

0c = - 20r 3 sin 4u 0r yr 0yr 1 0yu The continuity equation + + = 20r 2 cos 4u + 60r 2 cos 4u + r 0r r 0u ( -80r 2 cos 4u) = 0 is indeed satisfied. yu = -

At the point r = 2 m, u = p>6 rad, p yr = 20 ( 23 ) cos c 4 a b d = - 80 m>s 6

p yu = -20 ( 23 ) sin c 4 a b d = - 138.56 m>s 6

Thus, the magnitude of the velocity is

V = 2y r2 + y u2 = 2 ( - 80 m>s ) 2 +

( - 138.56 m>s ) 2 = 160 m>s

Ans.

Ans: V = 160 m>s 626

M07_HIBB9290_01_SE_C07_ANS.indd 626

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7–22. The air flowing around the inclined plate is defined by its stream function c = 12r 2 sin 3u2 m2 >s. Determine the speed at the point which is located at r = 2 m, u = p>6 rad. The origin is at the corner. Sketch the streamline that passes through this point.

SOLUTION We consider the steady fluid flow. For the streamline passing point p r = 2 m and u = rad, 6 p c = 2122 2 sin c 3a b d = 8 m2 >s 6

60°

Thus, the stream function passing through this point is 8 = 2r 2 sin 3u r = a

2

2 sin 3u The plot of this function is shown in Fig. a.

Ans.

bm

u1rad2

0

p>24

p>12

p>8

p>6

5p>24

p>4

7p>24

p>3

r1 m2

∞

3.23

2.38

2.08

2.00

2.08

2.38

3.23

∞

The radial and transverse components of velocity are

y

1 0c 1 vr = = 3 2r 2 13 cos 3u2 4 = 16r cos 3u2 m>s r 0u r vu = -

c58

0c = 1 - 4r sin 3u2 m>s 0r

The continuity equation

vr 0vr 1 0vu 6r cos 3u 1 + + = + 6 cos 3u + 3 -4r13 cos 3u24 = 0 r 0r r 0u r r

is indeed satisfied. At point A, where r = 2 m, and u = p>6 rad,

p 3

p vr = 6122 cos c 3a b d = 0 6

x (a)

p vu = -4122 sin c 3a b d = - 8 m>s 6

Thus, the magnitude of the velocity is

Ans.

V = vu = 8 m>s

Ans: r = a

2 2sin 3u

b m, V = 8 m>s

627

M07_HIBB9290_01_SE_C07_ANS.indd 627

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7–23. The stream function for a horizontal flow of water near the corner is c = 16xy2 ft2 >s, where x and y are in feet. Determine the x and y components of the velocity and the acceleration of fluid particles passing through point (2 ft, 4 ft). Plot the streamline that passes through this point.

y

SOLUTION

B

A

We consider steady fluid flow for the streamline passing through point 12ft, 4ft2. c = 6122142 = 48 ft2 >s

Thus, the streamline is

48 = 6xy

Ans.

xy = 8

The x and y components of the velocity are u =

0c = 16x2 ft>s 0y

y (ft)

v = -

0c = 1 -6y2 ft>s 0x

8 6

0u 0v The continuity equation + = 6 + 1 -62 = 0 is indeed satisfied. 0x 0y

4 2

The acceleration components are

0u 0u 0u + u + v 0t 0x 0y

ax =

0

1

2

3

4

5

x (ft)

(a)

= 0 + 6x162 + 1 -6y2102 = 136x2 ft>s2 0v 0v 0v + u + v 0t 0x 0y

ay =

At point 12ft, 4 ft2,

x

= 0 + 6x102 + 1 -6y21 -62 = 136y2 ft>s2

u = 6122 = 12 ft>s S 2

ax = 36122 = 72 ft >s S

Ans.

v = -6142 = -24 ft>s = 24 ft>s T 2

ay = 36142 = 144 ft >s c

Ans.

The plot of the streamline is shown in Fig. a. x1ft2

0

1

2

3

4

5

∞

y1ft2

∞

8

4

2.67

2

1.6

0

628

M07_HIBB9290_01_SE_C07_ANS.indd 628

Ans: xy = 8, u = 12 ft>s S , v = 24 ft>s T ax = 72 ft2 >s S , ay = 144 ft2 >s c

07/03/17 2:56 PM

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*7–24. The stream function for the horizontal flow of water near the corner is defined by c = 16xy2 ft2 >s, where x and y are in feet. If the pressure at point A (1 ft, 3 ft) is 10 psi, determine the pressure at point B (2 ft, 4 ft).

y

SOLUTION

A

B

We consider steady fluid flow. The x and y components of velocity are u =

0c = 16x2 ft>s 0y

The continuity equation At point A 11ft, 3 ft2,

v = -

x

0c = 1 -6y2 ft>s 0x

0u 0v + = 6 + 1 -62 = 0 is indeed satisfied. 0x 0y

uA = 6112 = 6 ft>s

vA = - 6132 = - 18 ft>s

Thus, VA2 = uA2 + vA2 = 16 ft>s2 2 + 1 -18 ft>s2 2 = 360 ft2 >s2

At point B 12ft, 4 ft2,

uB = 6122 = 12 ft>s

vB = -6142 = -24 ft>s

Thus,

Since wz =

VB2 = uB2 + vB2 = 112 ft>s2 2 + 1 -24 ft>c2 2 = 720 ft2 >s2

1 0v 0u 1 a b = 10 - 02 = 0, the flow is irrotational. 2 0x 0y 2

Therefore, Bernoulli’s equation is applicable between two points located on the different streamlines, such as points A and B. pA pB VA2 VB2 + + gzA = + + gzB r r 2 2 Since the flow occurs in horizontal plane, zA = zB = z. pA pB VA2 VB2 + + gz = + + gz r r 2 2 pB = pA +

r 1V 2 - VB2 2 2 A

lb 12 in. 2 = a10 2 b a b + 1 ft in = a1091.18

= 7.58 psi

a

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

62.4 lb>ft3 32.2 ft>s2 2

b

1360 ft2 >s - 720 ft2 >s2 Ans.

Ans: pB = 7.58 psi 629

M07_HIBB9290_01_SE_C07_ANS.indd 629

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7–25. The potential function for a flow is f = 61x 2 - y2 2 m2 >s, where x and y are in meters. Determine the magnitude of the velocity of fluid particles at the point (2 m, 3 m). Show that continuity is satisfied, and find the streamline that passes through this point.

SOLUTION We consider steady fluid flow. The x and y components of the velocity are u = Since vz =

0f = 112x2 m>s 0x

v =

0f = 1 -12y2 m>s 0y

1 0v 0u 1 a b = 10 - 02 = 0, the flow is indeed irrotational. 2 0x 0y 2

At point 12 m, 3 m2,

u = 12122 = 24 m>s

v = - 12132 = -36 m>s

Thus, the magnitude of the velocity at this point is V = 2u2 + v2 = 2124 m>s2 2 + 1 - 36 m>s2 2 = 43.27 m>s = 43.3 m>s

Since the continuity equation

Ans.

0u 0v + = 12 + 1 - 122 = 0 is satisfied, the stream 0x 0y

function can be established. Using the velocity components, 0c = u; 0y

0c = 12x 0y

Integrating this equation with respect to y, c = 12xy + f1x2 Substituting this result into -

0c = v; 0x

-

0 312xy + f1x24 = -12y 0x 12y +

Integrating this equation with respect to x,

0 3f1x24 = 12y 0x 0 3f1x24 = 0 0x

f1x2 = C Here, C is an arbitrary constant that can be set equal to zero. Thus, the stream function is c = 12xy For the streamline that passes through point 12m, 3 m2, Then

c = 12(2)(3) = 72 m2 >s 72 = 12xy

Ans.

xy = 6

Ans: V = 43.3 m>s, xy = 6

630

M07_HIBB9290_01_SE_C07_ANS.indd 630

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7–26. The potential function for a flow is f = 1x 2 - y2 2 ft2 >s, where x and y are in feet. Determine the magnitude of the velocity of a fluid particle 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

y =

0f = -2y 0y

1 0y 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

y = - 2(1) = - 2 ft>s

Then the magnitude of the velocity is V = 2u2 + y 2 = 2 ( 6 ft>s ) 2 +

( - 2 ft>s ) 2 = 6.32 ft>s

Ans.

0y 0u 0u 0v = 2 and = - 2. Since + = 2 + ( -2) = 0, the potential function 0y 0x 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 = y; 0x

-

0 32xy + f(x)4 = -2y 0x 2y +

Integrating this equation with respect to x,

0 3f(x)4 = 2y 0x 0 3f(x)4 = 0 0x

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 = 3 631

M07_HIBB9290_01_SE_C07_ANS.indd 631

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7–27. The flow around the bend in the horizontal channel can be described as a free vortex for which vr = 0, vu = (8>r2 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

SOLUTION We consider ideal fluid flow. 0f yr = ; 0r

0.5 m

0f 0 = 0r

Integrating with respect to r, f = f(u) Substituting this result into yu =

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 = (yu)A =

8 8 = = 3.2 m>s rA 2.5

VB = (yu)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

Ans: pB = 832 Pa 632

M07_HIBB9290_01_SE_C07_ANS.indd 632

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*7–28. The velocity components for a two-dimensional flow are u = 18y2 ft>s and v = 18x2 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 We consider ideal fluid flow.

y

u =

0c ; 0y

8y =

0c 0y

3

Integrating with respect to y, c = 4y2 + f(x) Then,

4

y = -

0c ; 0x

8x = -

0 3 4y2 + f(x)4 0x

- 8x = 0 + Integrating with respect to x,

x

2.65

0 3f(x)4 0x

f(x) = -4x 2 + C Thus, c = 4y2 +

( - 4x2 + C ) = 4 ( y2 - x2 ) + C

Omitting the constant, C, c = 4 ( y2 - x 2 )

Ans.

From the slope of the stream function, dy y 8x x = = = dx u 8y y y

L3 ft

x

y dy =

L4 ft

x dx

y2 y x2 2 x 2 = 2 3 ft 2 4 ft y2 = x 2 - 7

Also, at (4 ft, 3 ft),

y = { 2x 2 - 7

Ans.

c = 4 ( (3)2 - (4)2 ) = -28 Then, 4 ( y2 - x 2 ) = -28 y = { 2x 2 - 7

Ans: c = 4 ( y2 - x 2 ) y = { 2x 2 - 7 633

M07_HIBB9290_01_SE_C07_ANS.indd 633

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7–29. The stream function for the flow around the 90° corner is c = 8r 2 sin 2u. Show that 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

SOLUTION

r

We consider ideal fluid flow. From the r and u velocity components,

u

yr =

x

1 0c 1 = ( 8r 2 ) (2 cos 2u) = 16r cos 2u r 0u r

0c = -(16r sin 2u) = - 16r sin 2u 0r yr 0yr 1 0yu The continuity equation + + = 16 cos 2u + 16 cos 2u + r 0r r 0u ( -32 cos 2u) = 0 is indeed satisfied. yu =

At point r = 0.5 m, u = 30°, Ans.

yr = 16(0.5) cos 32(30°)4 = 4 m>s

Since sin u = Therefore,

Ans.

yu = - 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 b a 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

634

M07_HIBB9290_01_SE_C07_ANS.indd 634

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7–29. 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

0.3

0.4

0.5

y(m)

∞

1.08

0.541 0.361 0.271 0.217

0.6

0.7

0.8

0.9

1.0

0.180

0.155

0.135 0.120 0.108

The velocity components with respect to stream function are u =

0c = 16x 0y

y = -

0c - 16y 0x

1 0y 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 = 8x 2 + f(y)

(1)

Also, 0f = y; 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 = 8x 2 - 8y2 + C f = 8 ( x 2 - y2 )

Ans. Ans: vr = 4 m>s, vu = - 6.93 m>s f = 8 ( x 2 - y2 ) 635

M07_HIBB9290_01_SE_C07_ANS.indd 635

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y

7–30. The stream function for a circular flow is defined by c = 1 -2r 2 2 m2 >s. Determine the velocity components vr and vu, and vx and vy, at the point r = 2 m, u = - 30°.

2m

30°

x

SOLUTION We consider steady fluid flow. The radial and transverse components of velocity are vr =

1 0c 1 = 102 = 0 r 0u r

vu = The continuity equation

0c = - 3 -212r24 = 4r 0r

vr 0vr 1 0vu + + = 0 + 0 + 0 = 0 r 0r r 0u p is indeed satisfied. At the point where r = 2 m and u = rad, 6 vr = 0

Ans.

vu = 4122 = 8 m>s 2

2

2

2

Ans.

2

Since r = x + y , then c = - 21x + y 2. Then the x and y components of velocity are given by vx = u =

0c = 1 - 4y2 m>s 0y

vy = v = -

0c = 3 - 1 -4x24 m>s = 14x2 m>s 0x

Here, x = 2 cos 30° m = 1.732 m and y = -2 sin 30° = -1 m. Then vx = -41 - 12 = 4 m>s

Ans.

vy = 411.7322 = 6.928 m>s = 6.93 m>s

Ans.

Ans: vr = 0, vu = 8 m>s, vx = 4 m>s, vy = 6.93 m>s 636

M07_HIBB9290_01_SE_C07_ANS.indd 636

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7–31. A fluid has velocity components u = 21 - x + y2 ft>s and v = 13x 2 + 2y2 ft>s, where x and y are in feet. Find the stream function, and determine if the flow is rotational or irrotational. Find the potential function if possible.

SOLUTION We consider steady fluid flow. Since the continuity equation 0u 0v + = -2 + 2 = 0 is satisfied, then the establishment of the stream function is dx 0y possible. Using the definition of velocity components with respect to the stream function, 0c = u; 0y

0c = 21 -x + y2 0y

Integrating this equation with respect to y, c = y2 - 2xy + f1x2 Substituting this result into -

0c = v; 0x

-

0 2 3y - 2xy + f1x24 = 3x 2 + 2y 0x 2y -

Integrate this equation with respect to x,

0 3f1x24 = 3x 2 + 2y 0x

0 3f1x24 = - 3x 2 0x

f1x2 = -x 3 Then

c = y2 - 2xy - x 3

Here, vz =

Ans.

1 0v 0u 1 a b = 16x - 22 2 0x 0y 2

Since vz ≠ 0, the flow is rotational. Thus the establishment of the potential function is not possible. Ans.

Ans: c = y2 - 2xy - x 3 637

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*7–32. A fluid has velocity components u = 14y + 52 ft>s and v = 14x2 ft>s, where x and y are in feet. Determine the stream function and also the potential function if possible.

SOLUTION We consider the steady fluid flow. Since the continuity equation 0u 0v + = 0 + 0 = 0 is satisfied, then the establishment of the stream function is 0x 0y possible. Using the definition of velocity components with respect to stream function, 0c = u; 0y

0c = 4y + 5 0y

Integrating this equation with respect to y, c = 2y2 + 5y + f1x2 Substituting this result into -

0c = v; 0x

-

0 = 32y2 + 5y + f1x24 = 4x 0x

0 3f1x24 = -4x 0x

Integrating this equation with respect to x,

f1x2 = -2x 2 Then

c = 2y2 + 5y - 2x 2 = 21y2 - x 2 2 + 5y

Ans.

1 0v 0u 1 Since vz = a b = 14 - 42 = 0, the flow is irrotational. Therefore, the 2 0x 0y 2

potential function exists. Using the definition of velocity components with respect to potential function, 0f = u; 0x

0f = 4y + 5 0x

Integrating this equation with respect to x, f = 4xy + 5x + f1y2 Substituting this result into 0f = v; 0y

0 = 34xy + 5x + f1y24 = 4x 0y 0 4x + 0 + 3f1y24 = 4x 0y 0 3f1y24 = 0 0y

Integrating this equation with respect to y,

f1y2 = C The arbitrary constant C can be set equal to zero. Then Ans.

f = 4xy + 5x

Ans: c = 21y2 - x 2 2 + 5y f = 4xy + 5x

638

M07_HIBB9290_01_SE_C07_ANS.indd 638

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7–33. A fluid has velocity components u = - 12x + 4y2 m>s and v = 12y - 4x2 m>s, where x and y are in meters. Determine the stream function and also the potential function if possible.

SOLUTION We consider steady fluid flow. Since the continuity equation

0u 0v + = -2 + 2 = 0 0x 0y

is satisfied, then the establishment of the stream function is possible. Using the definition of velocity components with respect to stream function, 0c 0c = u; = - 12x + 4y2 0y 0y Integrating this equation with respect to y, Substituting this result into -

0c = v; 0x

c = - 12xy + 2y2 2 + f1x2 -

0 = 3 - 12xy + 2y2 2 + f1x24 = 2y - 4x 0x 0 2y - 3f1x24 = 2y - 4x 0x 0 3f1x24 = 4x 0x

Integrating this equation with respect to x,

f1x2 = 2x 2 Then

Since vz =

c = - 12xy + 2y2 2 + 2x 2 = 21x 2 - y2 2 - 2xy

Ans.

1 0v 0u 1 a b = 3 - 4 - 1 -424 = 0, the flow is irrotational; therefore, 2 0x 0y 2

the potential function exists using the definition of velocity components with respect to potential function. 0f = u; 0x

0f = - 12x + 4y2 0x

Integrating this equation with respect to x, Substituting this result into 0f = v; 0y

f = - 1x 2 + 4xy2 + f1y2 0 3 - 1x 2 + 4xy2 + f1y24 = 2y - 4x 0y 0 -4x + 3f1y24 = 2y - 4x 0y 0 3f1y24 = 2y 0y

Integrating this equation with respect to y,

f1y2 = y2 Then f = - 1x 2 + 4xy2 + y2 = y2 - x 2 - 4xy

Ans.

Ans: c = 21x 2 - y2 2 - 2xy, f = y2 - x 2 - 4xy

639

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7–34. A fluid has velocity components u = 61x 2 - y2 2 ft>s and v = 1 - 12xy2 ft>s, where x and y are in feet. Determine the stream function and also the potential function if possible. Plot the streamlines and equipotential lines that pass through point (2 ft, 1 ft).

SOLUTION

y (ft)

We consider the steady fluid flow since the continuity equation

6

0u 0v + = 12x + 1 -12x2 = 0 is satisfied, then the establishment of the stream 0x 0y

5

function is possible. Using the definition of velocity components with respect to stream function, 0c = u; 0y

0c = 61x 2 - y2 2 0y

c

4 3

f

2 1

Integrating this equation with respect to y,

0

c = 6x 2y - 2y3 + f1x2 Substituting this result into 0c = v; 0x

1

2

3

4

5

x (ft)

(a)

0c - 36x 2y - 2y3 + f1x24 = -12xy 0x - 12xy -

0 3f1x24 = -12xy 0x 0 3f1x24 = 0 0x

Integrating this equation with respect to x, f1x2 = C

Here, C is an arbitrary constant that can be set equal to zero. Then c = 6x 2y - 2y3

Ans.

For the streamline passing through point 12 ft, 1 ft2,

c = 211233122 2 - 12 4 = 22

Then, the equation of the streamline is

22 = 2y13x 2 - y2 2

y13x 2 - y2 2 = 11 x2 =

Since vz =

y3 + 11 3y

1 0v 0u 1 a b = 3 - 12y - 1 - 12y24 = 0. The flow is irrotational; 2 0x 0y 2

therefore, the potential function exists. Using the definition of velocity components with respect to potential function, 0f = u; 0x

0f = 61x 2 - y2 2 0x

640

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7–34. Continued

Integrating this equation with respect to x, f = 2x 3 - 6xy2 + f1y2 Substituting this result into 0f = v; 0y

0 32x 3 - 6xy2 + f1y24 = -12xy 0y - 12xy +

0 3f1y24 = -12xy 0y 0 3f1y24 = 0 0y

Integrating this equation with respect to y,

f1y2 = C Again, the arbitrary constant C can be set equal to zero. Then f = 2x 3 - 6xy2

Ans.

For the equipotential line passing through point 12 ft, 1 ft2, f = 2122322 - 3112 24 = 4

Then the equation of the equipotential line is

4 = 2x1x 2 - 3y2 2

y2 =

x3 - 2 3x

The plots of the streamline and equipotential line are shown in Fig. a. For the streamline, x1ft2 y1ft2

∞ 0

3.83 0.25

2.72 0.5

2.25 0.75

2 1

179 1.50

1.77 1.75

1.78 2

2.05 3

2.50 3.01 3.55 4 5 6

For the equipotential, x1ft2 y1ft2

1.26 0

2 1

2.5 1.35

3 1.67

3.5 1.97

4 2.27

4.5 2.57

Ans: c = 6x 2y - 2y3, f = 2x 3 - 6xy2 641

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7–35. If the potential function for a two-dimensional flow is f = 1xy2 m2 >s, where x and y are in meters, determine the stream function and plot the streamline that passes through the point 11 m, 2 m2.

SOLUTION

y (m)

We consider ideal fluid flow. u =

0f 0 = (xy) = y 0x 0x

y =

0f 0 = (xy) = x 0y 0y

u =

0c ; 0y

y =

ψ = 1.5

0c 0y

3 x (m)

Integrating with respect to y, c =

y2 + f(x) 2

(a)

Substituting this result into y = -

0c ; 0x

x = -

0 y2 + f(x) d c 0x 2

x = -0 -

Integrating with respect to x,

0 3f(x)4 = -x 0x f(x) = -

0 3f(x)4 0x

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 - x 2 ) = 1.5 2 y2 = x 2 + 3 y = 2x 2 + 3 The plot of this streamline is shown in Fig. a. Ans: c =

1 2 ( y - x2 ) 2

642

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*7–36. The potential function for a flow is f = 81x 2 - y2 2 ft2/s, where x and y are in feet. Determine the streamline that passes through point (2 ft, 4 ft) and the magnitude of the velocity of a fluid particle at this point.

SOLUTION We consider the steady fluid flow. The x and y components of velocity are u = v = Since vz =

0f 0 = 381x 2 - y2 24 = 116x2 ft>s 0x 0x

0f 0 = 381x 2 - y2 24 = 1 - 16y2 ft>s 0y 0y

1 0v 0u 1 a b = 10 - 02 = 0, the flow is irrotational. Therefore, the 2 0x 0y 2

potential function is indeed valid. At point 12 ft, 4 ft2, the x and y components of velocity are u = 16122 = 32 ft>s

v = - 16142 = -64 ft>s

Thus, the magnitude of velocity is V = 2u2 + v2 = 2132 ft>s2 2 + 164 ft>s2 2 = 71.6 ft>s

Ans.

0u 0v + = 16 + 1 -162 = 0 is satisfied, the stream 0x 0y function can be established. Using the definition of components of velocity with respect to the stream function, Since the continuity equation

0c = u; 0y

0c = 16x 0y

Integrating this equation with respect to y, c = 16xy + f1x2 Substituting this result into -

0c = v; 0x

-

0 316xy + f1x24 = -16y 0x

- 16y -

Integrating this equation with respect to x,

0 3f1x24 = -16y 0x 0 3f1x24 = 0 0x

f1x2 = C Here, C is an arbitrary constant that can be set equal to zero. Therefore, c = 16xy For the streamline passing through point 12 ft, 4 ft2,

c = 16122142 = 128 ft2 >s

Thus, the equation of the corresponding streamline is 128 = 16xy

Ans.

xy = 8

Ans: V = 71.6 ft>s xy = 8

643

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7–37. Water has horizontal and vertical velocity components u = 21y2 - x 2 2 m>s and v = 14xy2 m>s, where x and y are in meters. If the pressure at point A (1 m, 4 m) is 450 kPa, determine the pressure at point B (0.5 m, –2 m). Also, what is the potential function for the flow?

SOLUTION We consider the steady fluid flow. Since vz =

1 0v 0u 1 a b = 14y - 4y2 = 0, the 2 0x 0y 2

flow is irrotatinal. Therefore, it is possible to establish the potential function using the definition of the velocity component with respect to potential function, 0f = u; 0x

0f = 21y2 - x 2 2 0x

Integrating this equation with respect to x, f = 2axy2 Substituting the result into 0f = v; 0y

x3 b + f1y2 3

0 x2 c 2axy2 b + f1y2 d = 4xy 0y 3 4xy +

0 3f1y24 = 4xy 0y 0 3f1y24 = 0 0y

Integrating this equation with respect to y, f1y2 = C Here, C is an arbitrary constant that can be set equal to zero. Therefore, 2 f = 2xy2 - x 3 Ans. 3 Again, since the flow is irrotational, Bernoulli’s equation can be applied between points A and B. The x and y components of the velocity at these points are uA = 2 3 42 - 12 4 = 30 m>s uB = 2 3 1 - 22 - 0.5 2

Thus,

2

4

vA = 4112142 = 16 m>s

= 7.5 m>s

vB = 410.521 -22 = -4 m>s

VA2 = uA2 + vA2 = 130 m>s2 2 + 116 m>s2 2 = 1156 m2 >s2

VB2 = uB2 + vB2 = 17.5 m>s2 2 + 1 - 4 m>s2 2 = 72.25 m2 >s2

Applying Bernoulli’s equation between points A and B,

pB pA V B2 VA2 + + qzB = + + gzA r r 2 2

pB 3

1000 kg>m

+

172.25 m2 >s2 2 2

+ 19.81 m>s2 21 - 2 m2 =

4501103 2 N>m2 3

1000 kg>m

pB = 1.05071106 2 Pa = 1.05 MPa

+

1156 m2 >s2 2

+ 19.81 m>s2 214 m2

Ans.

Ans: f = 2xy2 -

2 3 x , pB = 1.05 MPa 3

644

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7–38. The potential function for a horizontal flow of water is f = 14x 2 - 4y2 2 m2 >s, where x and y are in meters. Determine the magnitude of the velocity at point A (2 m, 1 m). What is the difference in pressure between this point and point B (1 m, 3 m)?

SOLUTION Consider the steady fluid flow. The x and y components of velocity are u = v =

0f 0 = 14x 2 - 4y2 2 = 18x2 m>s 0x 0x

0f 0 = 14x 2 - 4y2 2 = 1 - 8x2 m>s 0y 0y

1 0v 0u 1 a b = 10 - 02 = 0, the flow is irrotational. Therefore, the 2 0x 0y 2

Since vz =

potential function is indeed valid. At point A 12 m, 1 m2,

uA = 8122 = 16 m>s vA = -8112 = -8 m>s

VA =

2uA2

+

vA2

= 2116 m>s2 2 + 1 - 8 m>s2 2 = 2320 m>s = 17.9 m>s

Ans.

At point B 11 m, 3 m2,

uB = 8112 = 8 m>s vB = - 8132 = -24 m>s

VB = 2uB2 + vB2 = 218 m>s2 2 + 1 - 24 m>s2 2 = 2640 m>s

Again, since the flow is irrotational, Bernollis’ equation can be applied between point A and B. pA pB V A2 vB2 + + qzA = + + gzB r r 2 2 Neglecting the change in elevation, zA - zB = z. Then pB - pA =

r 1V 2 - V B2 2 2 A

= a

1000 kg>m3 2

b 3 1 2640 m>s2 2 - 1 2320 m>s2 2 4

= 1601103 2 Pa = 160 kPa

Ans.

Ans: VA = 17.9 m>s, pB - pA = 160 kPa 645

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7–39. A fluid has velocity components u = 18xy2 m>s and v = 41x 2 - y2 2 m>s, where x and y are in meters. Determine the potential function and find the circulation around the square path ABCDA.

y C

D

2m A

B 3m

x

SOLUTION 1 0v 0u 1 We consider the steady fluid flow. Since vz = a b = 18x - 8x2 = 0, the 2 0x 0y 2 flow is irrotational. Therefore, the potential function can be established. Using the definition of velocity components with respect to potential function, 0f = u; 0x

1m

2m

0f = 8xy 0x

Integrating this equation with respect to x, f = 4x2y + f1y2

Substituting this result into 0f = v; 0y

0 34x2y + f1y24 = 41x2 - y2 2 0y 4x2 +

0 3f1y24 = 4x2 - 4y2 0y

0 3f1y24 = -4y2 0y

Integrating this equation with respect to y, f1y2 =

-4 3 y 3

Therefore, f = 4x2y -

4 3 y 3

Ans.

Along edge AB, ds = dxi and y = 3 m. Then C

V # ds = =

C

1ui + vj2 # dxi

3m

L1 m

u dx

3m

=

L1 m

8x132 dx = 12x2 2

3m 1m

= 96 m2 >s

646

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7–39. Continued

Along edge BC, ds = dyj and x = 3 m. Then C

V # ds =

C

1ui + vj2 # dyj

5m

=

L3 m

v dy

5m

=

L3 m

4132 - y2 2dy

= 4a9y -

y3 5 m b2 = -58.67 m2 >s 3 3m

Along edge CD, ds = - dxi and y = 5 m. Then C

V # ds =

C

1ui + vj2 # 1 -dxi2 3m

= -

L1 m

u dx

3m

= -

L1 m

8x152 dx

= - 120x 2 2 2

3m 1m

Along edge DA, ds = -dyj and x = 1 m. Then C

V # ds =

C

= -

= - 160 m2 >s

1ui + vj2 # 1 -dyj2 5m

L3 m

v dy

5m

= -

L3 m

411 - y2 2 dy

= -4ay Thus, the circulation is

y3 5 m b2 = 122.67 m2 >s 3 3m

Γ = 96 m2 >s + 1 -58.67 m2 >s2 + 1 - 160 m2 >s2 + 122.67 m2 >s = 0

Ans.

This result is expected since the flow is irrotational.

Ans: f = 4x 2y -

4 3 y ,Γ = 0 3

647

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*7–40. A fluid has horizontal velocity components of u = 1y2 - x 2 2 m>s and v = 12xy2 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 - x 2 =

0f 0x

Integrating with respect to x, f = xy2 -

x3 + f(y) 3

Substituting this result into y =

0f ; 0y

0 x3 c xy2 + f(y) d 0y 3

2xy =

2xy = 2xy - 0 +

Integrating with respect to y,

0 3f(y)4 = 0 0y

0 3f(y)4 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 point B. The x and y components of the velocity at these points are uA = ( 22 - 32 ) m>s = -5 m>s uB = ( 32 - 12 ) m>s = 8 m>s

yA = 32(3)(2)4 m>s = 12 m>s yB = 32(1)(3)4 m>s = 6 m>s

Thus, the magnitude of the velocity at these two points is

VA = 2uA2 + yA2 = 2 ( - 5 m>s ) 2 + ( 12 m>s ) 2 = 13 m>s

VB = 2uB2 + yB2 = 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 )

pB = 628.13 ( 103 )

N (13 m>s)2 m2 = + 8 ( 103 ) N>m3 2 ( 9.81 m>s2 ) 600 ( 103 )

N = 628 kPa m2

Ans.

Ans: f = xy2 -

x3 3

pB = 628 kPa 648

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7–41. A fluid has velocity components of u = 21x 2 + y2 2 ft>s and v = 1 - 4xy2 ft>s. Determine the stream function and the circulation around the rectangle. Plot the streamlines for c = 0, c = 1 ft2 >s, and c = 2 ft2 >s.

y

0.6 ft

x 0.5 ft

SOLUTION 0u 0y = = 4x + ( -4x) = 0 0x 0y is satisfied, the stream function can be established. Using the definition of the velocity components, with respect to stream function, We consider ideal fluid flow. Since the continuity equation

0c = u; 0y

0c = 2 ( x 2 + y2 ) 0y

Integrating this equation with respect to y, c = 2 ax 2y +

1 3 y b + f(x) 3

(1)

Also, -

0c = y; 0x

-

0c = - 4xy 0x

0 1 c 2 ax 2y + y3 b + f(x) d = -4xy 0x 3 4xy +

Integrating this equation with respect to x,

0 3f(x)4 = 4xy 0x

0 3 f(x) 4 = 0 0x

f(x) = C Substituting this result into Eq. (1), c = 2 ax 2y +

1 3 y b + C 3

C is an arbitary constant. If we set it equal to zero, then the stream function can be expressed as c = 2y ax 2 +

1 2 y b 3

Ans.

649

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7–41. Continued

0 = 2y°x 2 +

For c = 0,

y2 ¢ 3

since x 2 +

y2 ≠ 0, 3

then

y = 0 For c = 1 ft2 >s,

1 = 2y °x 2 + x2 =

For c = 2 ft2 >s,

3 - 2y3 6y

0 6 y 6 1.145

2 = 2y°x 2 +

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 ft2 >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 ft2 >s y(ft)

0

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)

0.5

0.75

1.0

1.25

1.5

1.75

Ans: c = 2y(x 2 + 650

M07_HIBB9290_01_SE_C07_ANS.indd 650

2.0

1 2 y ), Γ = -0.72 ft2 >s 3

07/03/17 2:57 PM

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7–42. The x component of velocity of a two-dimensional irrotational flow is u = 4y2 - 2y - 4x 2 ft>s, where x and y are in feet. Find the y component of velocity if v = 0 at x = y = 0.

SOLUTION We consider the steady fluid flow. In order for the flow to be irrotational, it is required that vz = 0. This gives vz = Here,

Thus,

1 0v 0u a b = 0 2 0x 0y

0u 0 = 14y2 - 2y - 4x 2 2 = 18y - 22 rad>s 0y 0y 1 0v c - 18y - 22 d = 0 2 0x

0v = 8y - 2 0x

Integrating this equation with respect to x, v = 8xy - 2x + f1y2 In order to satisfy the continuity condition, 0u 0v + = 0 0x 0y Here, 0v 0 0 = 38xy - 2x + f1y24 = 8x + 3f1y24 0y 0y 0y

Then

0u 0 = 14y2 - 2y - 4x 2 2 = -8x 0x 0x - 8x + 8x +

0 3f1y24 = 0 0y

0 3f1y24 = 0 0y

Integrating this equation with respect to y,

f1y2 = C Therefore, v = 8xy - 2x + C At x = y = 0, v = 0. Then C = 0 and so Ans.

v = 8xy - 2x

Ans: v = 8xy - 2x 651

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7–43. The flow has a velocity of V = 5 13y + 82i6 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

x

SOLUTION We consider ideal fluid flow. The x and y components of velocity are u = (3y + 8) ft>s

y = 0

Here, 0u = 0; 0x

0u 0 = (3y + 8) = 3 rad>s 0y 0y

Thus, vz =

1 0y 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, 0y 0y = = 0. Then, Ans. 0x 0y -

1 0p 0y 0y - g = u + y = 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) = 3 rad>s, = (3y + 8) = 0 and 0y 0y 0x 0x -

1 0p 0u 0u = u + y r 0x 0x 0y

1 0 3 - gy + f(x)4 = 0 + 0 r 0x 0 3f(x)4 = 0 0x

652

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7–43. 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, ft2

lb lb = a - 70 2 b(3 ft) + C 2 ft ft C = 216 lb>ft2

Thus, p = ( - gy + 216) lb>ft2 At point O, y = 0. Thus, pO = 3 - (70)(0) + 2164 = 216

lb ft2

lb ft2

Ans.

Ans: rotational, pO = 216 lb>ft2 653

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*7–44. If a building has a flat roof and is located 20 m from the center of a tornado, determine the uplift pressure on the roof. The building is within the free vortex of the tornado, where the wind speed is 20 m>s at a distance of 40 m from its center. The density of the air is ra = 1.20 kg>m3.

SOLUTION We consider the steady ideal fluid flow. Since the tornado is a free vortex, its velocity components are given by vr = 0

vu =

k r

Thus, V = vu =

k r

It is required that at r = 40 m, V = 20 m>s. Thus, 20 m>s =

k ; 40 m

k = 800 m2 >s

Then

At r = 20 m,

V = a V =

800 b m>s r

800 = 40 m>s 20

Since free vortex is irrotational flow, 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 = 20 m. At r = ∞ , V ∞ = 0 and p ∞ = 0. Since the flow occurs in the horizontal plane, the elevation term can be neglected. p∞ p V2 V2 + ∞ = + ra ra 2 2 0 + 0 =

p 3

1.20 kg>m

+

140 m>s2 2 2

Ans.

p = -960 Pa = - 0.960 kPa

The negative sign indicates that a very modest suction or partial vacuum develops.

Ans: p = -0.960 kPa 654

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7–45. A source at O creates a flow from O that is described by the potential function f = 18 ln r2 m2 >s, where r is in meters. Determine the stream function, and find the velocity at point r = 5 m, u = 15°.

O

308 u

SOLUTION We consider ideal fluid flow. The r and u components of velocity are yr = yu =

0f 8 0 = (8 ln r) = a b m>s 0r 0r r

1 0f 1 0 = (8 ln r) = 0 r 0u r 0u

Applying, yr =

1 0c ; r 0u

8 1 0c = r r 0u

Integrating this equation with respect to u, c = 8u + f(r) Substituting this result, yu = -

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°, yr =

8 = 1.6 m>s 5

yu = 0 Thus, the magnitude of the velocity is V = 2y r2 + y u2 = 2 ( 1.6 m>s ) 2 + (0)2 = 1.60 m>s

Ans.

Ans: c = 8u, V = 1.60 m>s 655

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7–46. 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

eln r = e2pku q

r = e2pku This equation represents a logarithmic spiral from the source and its plot is shown in Fig. a.

source

(a)

656

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7–47. A free vortex is defined by its stream function c = 1 - 240 ln r2 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.

r54m

SOLUTION We consider ideal fluid flow. The velocity components are yr =

1 0c ; r 0u

yu = -

0c ; 0r

Thus, the velocity is V = yu = a

At r = 4 m,

V = a

yr = 0 yu = 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. p0 p V0 2 V2 + + gz0 = + + gz r 2 r 2 0 + 0 + gz =

p 1.20 kg>m3

+

( 60.0 m>s ) 2 2

+ gz Ans.

p = -2160 Pa = -2.16 kPa

Ans: V = 60.0 m>s, p = -2.16 kPa 657

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*7–48. Show that the equation that defines a sink will satisfy continuity, which in polar coordinates is written as 01vr r2 0r

+

0vu = 0. 0u

SOLUTION We consider ideal fluid flow. For sink flow, yr = 0 ( yrr ) 0r

=

q and yu = 0. Then, 2pr

q q 0 0 cab(r) d = a- b = 0 0r 2pr 0r 2p 0yu = 0 0u

Thus, 0(yr r) 0r

+

0yu = 0 + 0 = 0 0u

(Q.E.D)

658

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7–49. 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

y

Using, 3 m2 >s q 3 r = r0 = = = m 2pU 2p(8 m>s) 16p

asymptote

Ans.

The equation of the streamline (boundary of a half body) that passes through the stagnation point P can be determined by applying r =

r

P ¨

r0(p - u)

source

sin u

3 m 16

3 (p - u) 16p r = sin u r =

3 m 16

x

3 m 16 asymptote

3(p - u)

(a)

16p sin u

This equation can be written in the form r sin u =

3 (p - u) 16p

Since y = r sin u, this equation becomes y =

3 (p - u) 16p

The half-width h of the half body can be determined by setting y = h as u approaches 0 or 2p. Thus, h =

3 3 (p - u) = m 16p 16

The plot of the half body is shown in Fig. a.

Ans: u = p, r =

3 m 16p

659

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7–50. 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 yr = 0

and

yu =

k r

Then V = yu =

k r

For a circulation Γ, Γ =

C

V # ds =

k =

2p

k (r du) = 2pk L0 r

A r

Γ 2p

z

r

Thus, V =

Γ 2pr

B

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

(a)

pA pB VA2 VB2 + + gzA = + + gzB rw rw 2 2

0 + 0 + g(0) = 0 +

a

z

Γ 2 b 2pr + g( - z) 2

gz =

Γ2 8p2r 2

z =

Γ2 8p2gr 2

Ans.

Ans: z =

Γ2 8p2gr 2

660

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7–51. Determine the strength q of the source, if the stagnation point P is located at ro = 0.5 m in a combined uniform flow of 4 m> s and the source. Plot the streamline passing through the stagnation point.

y

4 m!s

0.5 m x

q

P

SOLUTION We consider steady ideal fluid flow. This is a case of flow past a half body. Using the location of stagnation point P, ro =

q ; 2pU

0.5 m =

q 2p(4 m>s)

q = 4p m2 >s

Ans.

The equation of the streamline (boundary of a half body) that passes through the stagnation point P can be determined by applying r = r =

r0 1p - u2

sin u 0.51p - u2

y Asymptote

r

0.5p m

sin u

0.25p m P

This equation can be written in the form

Source

r sin u = 0.51p - u2

u

x

0.25p m

0.5p m

Since y = r sin u, this equation becomes y = 0.51p - u2

0.5 m

The half with h of the half body can be determined by setting y = h as u approaches 0 or 2p. Thus,

Asymptote (a)

h = 0.51p - 02 = 0.5p m The plot of the streamline (boundary of the half body) is shown in Fig. a.

661

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Ans: q = 4p m2 >s

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*7–52. Pipe A provides a source flow of 3 m2 >s, whereas the drain, or sink, at B removes 3 m2 >s. Determine the stream function and the equation of the streamline for c = 0.25 m2 >s and c = 0.5 m2 >s.

y

3 m!s2 A

3 m!s2 B

2m

x

2m

SOLUTION We consider the steady ideal fluid flow. When the source and sink are superimposed, the resultant stream function with respect to the x and y axes is c = -

2ay q tan -1 a 2 b 2p x + y2 - a 2

c = -a

For c = 0.25 m2 >s,

3 m2 >s

b tan -1 c

2122y

d x + y2 - 22 4y 3 b f m2 >s c = etan-1 a 2 2p x + y2 - 4 2p

2

0.25 m2 >s =

Ans.

4y -3 b tan-1 a 2 2p x + y2 - 4

4y p tana - b = 2 6 x + y2 - 4 -

4y 23 = 2 3 x + y2 - 4

x 2 + y2 + 413y = 4 x 2 + y2 + 413y + 12132 2 - 12132 2 = 4

x 2 + 1y + 2132 2 = 4 + 12132 2 x 2 + 1y + 2132 2 = 16

For c = 0.5 m2 >s,

0.5 m2 >s = -

Ans.

4y 3 b tan -1 a 2 2p x + y2 - 4

4y p tan a - b = 2 3 x + y2 - 4 - 23 =

4y

2

x + y2 - 4

4

x 2 + y2 +

y = 4 23 4 2 2 2 2 x 2 + y2 + y + a b - a b = 4 23 23 23 x 2 + ay +

2

23

2

b = 4 + a

223 2 16 x + ay + b = 3 3 2

Ans: 2 23

b

c = e-

2

Ans.

4y 3 b f m2 >s tan-1 a 2 2p x + y2 - 4

For c = 0.25 m2 >s,

x 2 + 1y + 2132 2 = 16

For c = 0.5 m2 >s, x 2 + ay +

223 2 16 b = 3 3

662

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7–53. Pipe A provides a source flow of 3 m2 >s, whereas the drain at B removes 3 m2 >s. Determine the potential function and the equations of the equipotential line for f = 1 m2 >s and f = 2 m2 >s.

y

3 m!s2 A

3 m!s2 B

2m

x

2m

SOLUTION We consider the steady ideal fluid flow. When the source and sink are superimposed, the resultant potential function with respect to the x and y axes is f =

21x + a2 2 + y2 q ln 2p 21x - a2 2 + y2

f = a = •

3 m2 >s 2p

b ln £

21x + 22 2 + y2

21x - 22 2 + y2

§

21x + 22 2 + y2 3 ln £ § ¶ m2 >s 2p 21x - 22 2 + y2

(This potential function is for the case of a Rankine oval with u = 0.) For f = 1 m2 >s, 1 m2 >s = 2p

e3 = 4p

e3 = 4p

21x + 22 2 + y2 3 ln £ § 2p 21x - 22 2 + y2 21x + 22 2 + y2 21x - 22 2 + y2

1x + 22 2 + y2 1x - 22 2 + y2

e 3 = 1x 2 - 4x + 4 + y2 2 = x 2 + 4x + 4 + y2 x 2 - 4.123x + y2 = -4

x 2 - 4.123x + a

2

For f = 2 m >s,

4.123 2 4.123 2 b - a b + y2 = -4 2 2 4.123 2 ax b + y2 = 0.2502 2 1x - 2.062 2 + y2 = 0.250 2 m2 >s = 4p

e3 = 8p

e3 = 8p 3

Ans. 21x + 22 2 + y2 3 ln £ § 2p 21x - 22 2 + y2 21x + 22 2 + y2

21x - 22 2 + y2 1x + 22 2 + y2 1x - 22 2 + y2

e 1x 2 - 4x + 4 + y2 2 = x 2 + 4x + 4 + y2 x 2 - 4.0018x + y2 = -4

4.0018 2 4.0018 2 x 2 - 4.0018x + a b - a b + y2 = - 4 2 2 4.0018 2 ax b + y2 = 0.003681 2 1x - 2.002 2 + y2 = 0.00368

M07_HIBB9290_01_SE_C07_ANS.indd 663

Ans: f = b

Ans. 663

2(x + 2)2 + y2 3 ln ¢ ≤ r m2 >s 2p 2(x - 2)2 + y2

For f = 1 m2 >s, (x - 2.06)2 + y2 = 0.250

For f = 2 m2 >s, (x - 2.00)2 + y2 = 0.00368

07/03/17 2:58 PM

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7–54. Two sources, each having a strength of 8 ft2 >s, are located as shown. Determine the x and y components of the velocity of a fluid particle that passes point (8 ft, 12 ft). What is the equation of the streamline that passes through this point?

y

x 12 ft

12 ft

SOLUTION We consider the steady ideal fluid flow. When sources 1 and 2 are superimposed, Fig a, the resultant stream function is q q q u + u = 1u + u2 2 2p 1 2p 2 2p 1

c =

y x

From the geometry shown in Fig. a, u1 = tan-1 a

Then,

y b x - 12

u2 = tan -1 a

y b x + 12

r2 2

8 ft2 >s

c =

c = e

y y c tan-1 a b + tan-1 a bd 2p x - 12 x + 12

u2 12 ft

y y 4 b + tan-1 a b d f ft2 >s c tan-1 a p x - 12 x + 12

r1

r u1

u 12 ft

y x

1

(a)

(1)

The x and y components of velocity are u =

u =

0c 4 = p• £ 0y

1

1 1 1 a b + a b ft>s 2 § x - 12 2 § x + 12 ¶ y y £ 1 + a b 1 + a b x - 12 x + 12

4 x - 12 x + 12 c + d ft>s p 1x - 122 2 + y2 1x + 122 2 + y2

v = -

v = e

0c 4 = p• £ 0x

1

2§ y 1 + a b x - 12

c-

y

1x - 122 2

d +

y y 4 c + d f ft>s p 1x - 122 2 + y2 1x + 122 2 + y2

1 £

2§ y 1 + a b x + 12

c-

y d ft>s 1x + 122 2 ¶

At point (8 ft, 12 ft), u = v =

4 8 - 12 8 + 12 c + d = 0.0150 ft>s p 18 - 122 2 + 122 18 + 122 2 + 122

4 12 12 c + d = 0.124 ft>s p 18 - 122 2 + 122 18 + 122 2 + 122

Ans. Ans.

For point (8 ft, 12 ft), Eq. (1) gives c =

4 12 12 b + tan -1 a b d = - 0.9023 c tan -1 a p 8 - 12 8 + 12

Thus, the equation of the streamline passing through this point is - 0.9023 = tan -1 a

y y 4 b + tan -1 a bd c tan -1 a p x - 12 x + 12

y y b + tan -1 a b = 0.709 x - 12 x + 12

Ans: u = 0.0150 ft>s v = 0.124 ft>s Ans.

tan - 1 a

y y b + tan - 1 a b = - 0.709 x - 12 x + 12

664

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7–55. The source and sink of equal strength 18 ft2 >s are located as shown. Determine the x and y components of velocity of a particle passing through point P and the equation of the streamline passing through this point.

y P

3 ft x

SOLUTION We consider the steady ideal fluid flow. When source and sink are superimposed, the resultant stream function with respect to the x and y axes is c = -

2 ft

2 ft

2ay q tan -1 a 2 b 2p x + y2 - a 2

c = -a

c = e-

18 ft2 >s 2p

b tan -1 c

2122y

2

x + y2 - 22

d

4y 9 b f ft>s2 tan -1 a 2 p x + y2 - 4

The x and y components of velocity are given by u =

0c 9 = p• £ 0y

= v = -

= -

1 1 + a

4y x 2 + y2 - 4

b

2§

x 2 - y2 - 4 36 c 2 d ft>s p 1x + y2 - 42 2 + 16y2 0c 9 = p• £ 0x

1

1 + a

4y 2

2

x + y - 4

b

2§

c

c

1x 2 + y2 - 42142 - 4y12y2 1x 2 + y2 - 42 2

1x 2 + y2 - 42102 - 4y12x2

xy 72 c d ft>s p 1x 2 + y2 - 42 2 + 16y2

At point P 10, 3 ft2,

u = v = -

At point P 10, 2 ft2,

1x 2 + y2 - 42 2

36 36 02 - 32 - 4 d = c 2 ft>s 2 2 2 p 10 + 3 - 42 + 1613 2 13p 102132 72 d = 0 c 2 2 p 10 + 3 - 42 2 + 16132 2

c = -

d

d

¶

¶

Ans. Ans.

4122 9 9 p 9 d = - a b = tan-1 c 2 p p 2 2 0 + 22 - 4

Thus, the equation of the streamline is 4y 9 9 b - = - tan-1 a 2 p 2 x + y2 - 4

tan

4y p b = tan-1 a 2 2 x + y2 - 4 4y p = 2 2 x + y2 - 4

4y 2

x + y2 - 4

= ∞

To satisfy this equation, Ans:

x 2 + y2 - 4 = 0 x 2 + y2 = 4

Ans.

This is an equation of a circle centered at (0, 0) with a radius of 2 ft.

36 ft>s, v = 0 13p x 2 + y2 = 4

u =

665

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*7–56. A source q is emitted from the wall while a flow occurs towards the wall. If the stream function is described as c = 14xy + 8u2 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 yu = = -(4r sin 2u) 0r yr =

At stagnation point P, it is required that these velocity components are equal to zero. yu = -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 yr =

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

666

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*7–56. 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: d = 22 m 667

M07_HIBB9290_01_SE_C07_ANS.indd 667

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7–57. The leading edge of a wing is approximated by the half body. It is formed from the superposition of the uniform airflow of 300 ft > s and a source having a strength of 2880 ft2 >s. Determine the width of the half body and the location of the stagnation point.

1200 ft!s A u 5 608 O

SOLUTION We consider the steady ideal fluid flow. The location of stagnation point O is given by rO =

2880 ft2 >s q 4.8 = = ft = 1.53 ft p 2pU 2p1300 ft>s2

Ans.

q . Thus, the width of the half body is zu w = 2h = q>U = 2880 ft2 >s>300 ft>s

The half-width h of the half body is h =

Ans.

= 9.60 ft

Ans: rO = 1.53 ft, w = 9.60 ft 668

M07_HIBB9290_01_SE_C07_ANS.indd 668

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7–58. The leading edge of a wing is approximated by the half body. It is formed from the superposition of the uniform airflow of 300 ft > s and a source having a strength of 720 ft2 >s. Determine the difference in pressure between the stagnation point O and point A. The air is at a temperature of 60°F under standard atmospheric pressure.

1200 ft!s A u 5 608 O

SOLUTION We consider the steady ideal fluid flow. The location of stagnation point O is given by 720 ft2 >s q 1.2 = = ft p 2pU 2p1300 ft>s2

rO =

Thus, the equation of the boundary of the half body is given by r =

rO 1p - u2 sin u

1.2 a ft b1p - u2 p r = sin u r = For point A, u =

1.21p - u2 p sin u

p rad. Then 3 rA =

1.21p - p>32 p sin 60°

=

813 ft = 0.9238 ft 15

At the stagnation point O, VO = 0. The r and u components of velocity at point A can be determined from vr =

720 ft>s2 q + U cos uA = + 1300 ft>s2 cos 60° = 274.05 ft>s 2prA 2p 10.92382

vu = - U sin uA = - 1300 ft>s2 sin 60° = - 259.81 ft>s

Thus, the magnitude of the velocity at A is

VA = 2vr2 + vu2 = 21274.05 ft>s2 2 + 1 -259.81 ft>s2 2 = 377.63 ft>s

From the table in Appendix A, ra = 0.00237 slug>ft3 at T = 60°F. The flow past a half body is irrotational. Thus, Bernoulli’s equation for an ideal fluid is applicable between points O and A. Neglecting the elevation term, pO pA V o2 V A2 + = + ra ra 2 2 ∆p = pO - pA = = a

ra 1V A2 - VO2 2 2

0.00237 slug>ft3

= a168.98

2

b 3 1377.63 ft>s2 2 - 0 2 4

lb 1 ft 2 ba b 2 12 in. ft

Ans.

= 1.174 psi = 1.17 psi

Ans: pO - pA = 1.17 psi 669

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7–59. 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.

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.

y = x tan [p11 - 32y2]

Ans: y = x tan3p11 - 32y24 670

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*7–60. 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

yu = - U sin u

+ U cos u = d

U sin u + U cos u p - u

Thus, the magnitude of the velocity is V = 2y r2 + y u2 = =

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 p0 pA VO2 VA2 + = + r r 2 2

2

p0 p U + = + r r 2 p = p0 -

U2 3 sin2u + (p - u) sin 2u + (p - u)2 4 (p - u)2 2

rU 2 2(p - u)2

3 sin2u

+ (p - u) sin 2u 4

Ans.

Ans: p = p0 -

671

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rU 2 2(p - u)2

3 sin2u

+ (p - u) sin2u 4

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7–61. A fluid flows horizontally 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 5 0.4 m!s r u

x

q 5 1.0 m!s2

SOLUTION We consider ideal fluid flow. The location of the stagnation point p can be determined from 1.0 m2 >s q = = 0.3979 m = 0.398 m 2pU 2p ( 0.4 m>s )

r0 =

Ans.

The half-width of the half body is h = pr0 = p(0.3979 m) = 1.25 m

Ans.

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 (yr)B =

1.0 m2 >s q + U cos u = + ( 0.4 m>s ) cos 90° = 0.1989 m>s 2pr 2p(0.8 m)

(yu)B = -U sin u = - ( 0.4 m>s ) sin 90° = -0.4 m>s Thus, the magnitude of the velocity is V = 2(yr)B2 + (yu)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. pA pB VA2 VB2 + = + r r 2 2 pB = pA +

r ( V 2 - VB2 ) 2 A

= 300 Pa + °

850 kg>m3

= 283 Pa

2

¢ 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)

Ans: r0 = 0.398 m h = 1.25 m V = 0.447 m>s p = 283 Pa

672

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7–62. The Rankine oval is defined by the source and sink of equal strength. Determine this strength q if the length of the Rankine oval is to be 1.054 m. Also, what is its corresponding width? The velocity of the uniform flow is 4 m >s.

y 4 m!s

x

0.5 m

0.5 m

SOLUTION We consider the steady ideal fluid flow. Using the half-length of the Rankine oval, b = ca

1>2 q ba + a 2 d Up

1>2 q 1.054 m = ec d 10.5 m2 + 10.52 2 f 2 14 m>s2p

q = 0.6969 m2 >s = 0.697 m2 >s

Ans.

The half-width of the Rankine oval can be determined using h = h =

2pUh h2 - a 2 tan a b 2a q h2 - 10.5 m2 2 210.5 m2

tan a

2p14 m>s2h 0.6969 m2 >s

h = 1h2 - 0.252 tan 136.06h2

Solving numerically, and noting that q>2U = h = 0.086 m and adjust until we find

0.6969 m2 >s 214 m>s2

b

= 0.0871, we start with

h = 0.078479 m Thus, the width of the Rankine oval is Ans.

w = 2h = 210.078479 m2 = 0.157 m

Ans: q = 0.697 m2 >s w = 0.157 m 673

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7–63. The Rankine oval 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 oval.

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 - a 2 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 W = 0.0485 m 674

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*7–64. The Rankine oval 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 that defines the boundary of the oval.

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-1 a 2 b 2p x + y2 - a 2 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-1 a 2 b = 0 2p x + 02 - a 2

Thus, the equation that describes the boundary of the Rankine oval is Uy -

2ay q tan-1 a 2 b = 0 2p x + y2 - a 2

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 d = 0 tan-1 c 2 2 10p x + y - 0.25 y

2

2

x + y - 0.25

= tan 40py

Ans.

Ans:

y

x 2 + y2 - 0.25

= tan 40py

675

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7–65. Integrate the pressure distribution, Eq. 7–65, 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 = (a du)L = aL du is dF = p dA = c p0 +

1 1 rU 2 ( 1 - 4 sin2 u ) d (aL du) = p0aL du + raLU 2 ( 1 - 4 sin2 u ) du 2 2

+ S

( FR ) x = -

Equating the resultant forces along the x and y axes shown in Fig. a,

( FR ) x = ΣFx;

= = -

L0

2p

L0

2p

L0

2p

dF cos u c p0aL du +

= 0

( FR ) y = = = -

2p

p0aL cos u du -

= - p0aL(sin u)# 2p 0 -

+ c ( FR ) y = ΣFy;

1 raLU 2 ( 1 - 4 sin2 u ) du d cos u 2

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 p0aL du +

1 raLU 2 ( 1 - 4 sin2 u ) du d sin u 2 2p

p0aL sin u du -

= - p0aL ( cos u )# 2p 0 = 0

1 raLU 2 ( sin u - 4 sin3 u ) du L0 2

2p 1 4 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

d

(a)

676

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7–66. The long rotating cylinder is subjected to a uniform horizontal airflow of 2 m>s. Determine the circulation around the cylinder and the location of the stagnation points if the lift per unit length is 6.56 N>m. Take ra = 1.202 kg>m3.

2 m!s

0.5 m

SOLUTION We consider the steady ideal fluid flow. Using the lift, the circulation can be determined from. Fy = rU Γ;

6.56 N> m = 11.202 kg>m3 212 m>s2Γ Γ = 2.729 m2 >s = 2.73 m2 >s

Ans.

The location of the stagnation points can be determined from sin u =

2.729 m2 >s Γ = = 0.2171 4pUa 4p12 m>s210.5 m2 Ans.

u = 12.5° and 167°

Since we have obtained two solutions, there are two stagnation points on the surface of the cylinder.

Ans: Γ = 2.73 m2 >s u = 12.5° and 167° 677

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7–67. 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 678

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*7–68. Air flows around the cylinder as shown. If r = 1.22 kg>m3, plot the pressure on its surface for 0 … u … p>2 rad for each p>12 rad increment of u.

8 m!s

u 0.3 m

SOLUTION We consider the steady ideal fluid flow. The pressure at a point removed from the cylinder is atmospheric, p0 = patm = 0. The pressure on the surface of the cylinder is given by p = p0 + p = 0 +

When p = 0,

1 2 rU 11 - 4 sin2u2 2

1 11.22 kg>m3 218 m>s2 2 11 - 4 sin2u2 2

p = 339.0411 - 4 sin2u24 Pa

Ans.

0 = 39.0411 - 4 sin2u2 sin2u =

1 4

(a)

1 sin u = { 2 p u = rad 6 2

The plot of pressure distribution on the surface of the cylinder is shown in Fig. a. u1rad.2 p1Pa2

0 39.04

p>12 28.58

p>6 0

p>4 –39.04

p>3 –78.08

5p>12 –106.66

679

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p>2 –117.12

Ans: p = 339.0411 - 4 sin2u24 Pa

07/03/17 2:58 PM

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7–69. 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 variation 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

r(m)

We consider ideal fluid flow. The r and u components of velocity of the uniform flow around a cylinder can be determined using yr = U a1 -

(0.1 m)2 a2 0.06 ( ) b cos u = 6 m>s £ 1 § cos u = c a6 - 2 b cos u d m>s 2 2 r r r

yu = - U a1 +

(0.1 m)2 a2 0.06 b sin u = -(6 m>s) c 1 + d sin u = c - a6 + 2 b sin u d m>s 2 2 r r r

When u = 90°,

0.5 0.4 0.3 0.2 0.1 0

yu = - a6 +

yr = 0

0.06 b m>s r2 (1)

Flow around a cylinder is irrotational since the potential function exists. Therefore, the Bernoulli equation is applicable. Neglecting the elevation terms, pO p V + = + r r 2 2 N 150 ( 103 ) 2 m + 1500 kg>m3

( 6 m>s ) 2 2

v(m s)

r(m)

0.06 2 0.06 = 02 + c - a6 + 2 b = a6 + 2 b m>s r r B VO2

7.5 6.67

12

(a)

Thus, the magnitude of the velocity when u = 90° is V = 2y r2 + y u2

6.24 6.375

2

0.5 0.4 0.3 0.2 0.1

0.06 2 a6 + 2 b p r + = 2 1500 kg>m3

p = c 177 ( 103 ) - 750 a6 +

0.06 2 b d Pa r2

0

69

135 144

p(kPa) 148 147

(b)

(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 ) p(kPa)

0.1

0.2

0.3

0.4

0.5

12 69

7.5 135

6.67 144

6.375 147

6.24 148

Ans.

The plot of V vs. r and p vs. r are shown in Figs. a and b.

Ans: r (m) V (m>s) p (kPa)

0.1 0.2 12 7.5 69 135

0.3 6.67 144

0.4 0.5 6.375 6.24 147 148

680

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7–70. 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 (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

yr = U a1 -

(0.1 m)2 a2 0.06 yu = - U a1 + 2 bsin u = - ( 6 m>s ) c 1 + d sin u = c - a6 + 2 b sin u d m>s 2 r r r

v (m s) 5.76 5.625 5.33 4.50

When u = 0°,

yr = a6 -

0.06 b m>s r2

yu = 0

Thus, the magnitude of the velocity when u = 0° is V = 2y r2 + y u2 =

B

a6 -

0

0.06 2 0.06 b + 02 = a6 - 2 b m>s r2 r

pO p VO2 V2 + = + r r 2 2 N m2 + 1500 kg>m3

( 6 m>s ) 2

0.3 0.4 0.5 (a)

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 )

0.1 0.2

P(kPa) 162 177 156 153 152

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

0.3 5.33 156

0.4 0.5 5.625 5.76 153 152

681

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7–71. Air is flowing at U = 20 m>s past the model of a Quonset hut. Find the magnitude of the velocity and the absolute pressure acting on the windows located on the roof at A, u = 120°, and at B, u = 30°. The absolute pressure within the uniform flow is p0 = 90 kPa. Take ra = 1.23 kg>m3.

y 20 m!s

C

A B 4m u

x

SOLUTION We consider the steady ideal fluid flow. The velocity components of the flow around the building, which we treat as the upper half of a cylinder, are vr = Ua1 -

a2 b cos u r2

a2 b sin u r2

vu = -Ua1 +

Since points A and B are on the boundary surface, r = a = 4 m. vr = 120 m>s2 c 1 -

14 m2 2 14 m2 2

vu = 1 - 20 m>s2 c 1 +

For point A, u = 30°. Then

d cos u = 0

14 m2 2 14 m2 2

d sin u = 1 -40 sin u2 m>s

VA = 1vu 2 A = 1 -40 sin 30°2 m>s = - 20 m>s = 20 m>s

Ans.

30°

For point B, u = 120°. Then

VB = 1vu 2 B = 1 - 40 sin 120°2 m>s = -34.64 m>s = 34.6 m>s

60°

Ans.

Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different stream lines, such as point O within the uniform flow and point A (or B) on the boundary surface. Since the density of air is small, the elevation term can be neglected. Here, VO = U = 20 m>s. For point A, pO pA V O2 V A2 + = + ra ra 2 2 901103 2 N>m2 3

1.23 kg>m

+

120 m>s2 2 2

=

pA 3

1.23 kg>m

+

134.64 m>s2 2 2

pA = 89.511103 2 Pa = 89.5 kPa

For point B,

Ans.

pO pB V O2 V B2 + = + ra ra 2 2 901103 2 N>m2 1.23 kg>m3

+

120 m>s2 2 2

=

pB 1.23 kg>m3

+

120 m>s2 2 2

pB = 901103 2 Pa = 90 kPa

Ans. Ans: VA = VB = pB = pA =

20 m>s 30°60° 34.6 m>s 90 kPa 89.5 kPa

682

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*7–72. The Quonset hut is subjected to a uniform wind having a velocity of 20 m>s. Determine the resultant vertical force caused by the pressure that acts on the roof if it has a length of 10 m. Take ra = 1.23 kg>m3.

y 20 m!s

C

A B 4m u

x

SOLUTION Using symmetry to add an imaginary reflection under the hut, we consider the steady ideal fluid flow. For the flow around a cylinder, the pressure distribution on the boundary is described by p = p0 +

1 2 rU 11 - 4 sin2 u2 2

Here, p0 is atmospheric pressure. Thus, the net pressure on the boundary is gauge pressure, which is 1 2 rU 11 - 4 sin2 u2 2

pg = p - p0 = =

1 11.23 kg>m3 2120 m>s2 2 11 - 4 sin2 u2 2

= 324611 - 4 sin2 u24 N>m2

The force that the gauge pressure exerts on the differential area dA = 1R du2L = 14 du21102 = 40 du is

dF = pgdA = 24611 - 4 sin2 u2140 du2 = 984011 - 4 sin2 u2du

Equating the forces along the y axis + c 1FR 2 y = ΣFy; 1FR 2 y = -

= -

L0

p

L0

p

dF sin u 984011 - 4 sin2 u2 sin u du

= - 9840

L0

p

1 sin u - 4 sin3 u2du

p 4 = - 9840e - cos u - c - cos u1 sin2u + 22 d f ` 3 0

= - 9840a= 32 800 N

10 b 3

Ans.

= 32.8 kN

683

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Ans: 1FR 2 y = 32.8 kN

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7–73. Air is flowing at U = 20 m>s past the model of a Quonset hut. Find the magnitude of the velocity and the gage pressure at point C, where r = 4.5 m and u = 150°. Take ra = 1.23 kg>m3.

y 20 m!s

C

A B 4m u

x

SOLUTION We consider the steady ideal fluid flow. The velocity components of the flow around the (half-cylindrical) building are vr = Ua1 -

a2 b cos u r2

vu = -Ua1 + -

a2 b sin u r2

Here, U = 20 m>s, and a = 4 m. For point C, r = 4.5 m and u = 150°. Then vr = 120 m>s2 c 1 -

14 m2 2

14.5 m2 2

d cos 150° = -3.635 m>s

14 m2 2

vu = 1 - 20 m>s2 c 1 +

14.5 m2 2

Thus, the magnitude of the velocity at C is

d sin 150° = -17.90 m>s

VC = 2vr2 + vu2 = 21 - 3635 m>s2 2 + 1 -17.90 m>s2 2 = 18.27 m>s = 18.3 m>s Ans.

Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as point O with the uniform flow and point C on the streamline around the building. Since Point O is remodeled, its pressure is of atmospheric pressure, pO = patm = 0. Since the density of air is small, the elevation term can be neglected. Here, VO = U = 20 m>s. p0 pC V 02 V C2 + = + ra ra 2 2 pC = p0 +

ra 2 1V 0 - V 2C 2 2

= 0 + a

1.23 kg>m3 2

b 3120 m>s2 2 - 118.27 m>s2 2 4

= 40.79 Pa = 40.8 Pa

Ans.

Ans: VC = 18.3 m>s pC = 40.8 Pa 684

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7–74. Water flows toward the column with a uniform speed of 2 ft>s. Determine the pressure at point A if this point is at a depth of 3 ft below the water surface. Take rw = 1.94 slug>ft3.

2 ft!s

A 2 ft 1 ft

SOLUTION We consider the steady ideal fluid flow. The velocity components of the flow around the column are vr = Ua1 -

a2 b cos u r2

v u = -Ua1 +

a2 b sin u r2

Here, U = 2 ft>s, and a = 1 ft. For point A, r = 2 ft and u = 0°. Then vr = 12 ft>s2 c 1 -

11 ft2 2 12 ft2 2

vu = 1 - 2 ft>s2 c 1 +

d cos u° = 1.50 ft>s

11 ft2 2 12 ft2 2

Thus, the magnitude of the velocity at A is

d sin u° = 0

VA = vr = 1.50 ft>s Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as point O in the uniform flow and point A on a streamline around the column. Here, the pressure at O, 3 ft below the surface, can be determined from pO = rwgh = 11.94 slug>ft3 2132.2 ft>s2 213 ft2 = 187.40 lb>ft2

Also, VO = U = 2 ft>s and the elevation term can be excluded since the flow occurs in the horizontal plane. pO pA VO2 VA2 + = + rw rw 2 2 pA = p0 +

rw 1V0 2 - VA2 2 2

= 187.40 lb>ft2 + a = 1189.10 lb>ft2 2 a

= 1.31 psi

1.94 slug>ft3 2

1 ft 2 b 12 in.

b[12 ft>s2 2 - 11.50 ft>s2 2]

Ans.

Ans: pA = 1.31 psi 685

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7–75. 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>ft3. u 85 ft

150 ft!s

SOLUTION We consider ideal fluid flow. This is a case of flow around a cylinder where the velocity components are a2 b cos u r2

yr = U a1 -

yu = -U a1 +

On the surface of the building where r = a = 85 ft,

a2 b sin u r2

yu = - 2U sin u

yr = 0

Thus, the velocity of the wind on the surface of the building is V = yu = -2U sin u The minimum pressure occurs at the point where the magnitude of velocity is maximum, that is, when sin u = 1

or

u = 90°

or

Then

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>ft3

pmin = a -79.99

+

( 300 ft>s ) 2 2 2

lb 1 ft ba b = -0.555 psi 2 12 in. ft

Ans.

Ans: u = 90° or 270° pmin = - 0.555 psi 686

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*7–76. 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 the building at u = 0°, 90°, and 150°. Take ra = 0.00237 slug>ft3. u 85 ft

150 ft!s

SOLUTION We consider ideal fluid flow. This is a case of flow around a cylinder where the velocity components are yr = U a1 -

a2 b cos u r2

yu = - U a1 +

On the surface of the building where r = a = 85 ft, yr = 0

a2 b sin u r2

yu = - 2U sin u

Thus, the velocity of the wind on the surface of the building is V = yu = - 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>ft3 2

= a26.66 p # u=90° = 0 + °

lb 1ft 2 ba b = 0.185 psi 2 12 in. ft

0.00237 slug>ft3 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>ft3 2

Ans.

¢ 3 ( 150 ft>s ) 2 - ( 150 ft>s ) 2 4 = 0

Ans.

Ans: V # u = 0° = 0 V # u = 90° = 300 ft>s V # u = 150° = 150 ft>s p# u = 0° = 0.185 psi p# u = 90° = -0.555 psi p# u = 150° = 0

687

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7–77. The tube is built from two segments that are glued together at A and B. If it is exposed to a uniform air flow having a velocity of 6 m>s, determine the resultant force the pressure exerts on the top segment AB per unit length of the tube. Take r = 1.22 kg>m3.

6 m!s B

A 308 308 0.2 m

SOLUTION We consider the steady ideal fluid flow. Here, the pressure of a remodeled point O is atmospheric pressure, pO = patm = 0. Thus, the net pressure on the surface of the pipe is the gauge pressure given by 1 pg = p - pO = rU 2 11 - 4 sin2u2 2 The pressure force exerts on the differential element of 1 unit length with area dA = a du112, Fig. a, is 1 1 dF = pgdA = rU 2 11 - 4 sin2 u21a du2 = raU 2 11 - 4 sin 2u2du 2 2 Equating the forces along the x and y axes shown in Fig. a, ( d+ ) ( FR ) x = ΣFx;

5p>6

1FR 2 x = =

Lp>6 Lp6

dF cos u

5p>6

1 raU 2 11 - 4 sin2u2 cosu du 2 p

= =

1 raU 2 1cos u - 4 sin2u cosu2du 2 L0 5p>6 1 4 raU 2 asin u - sin3u b ` 2 3 p>6

= 0

This result is expected, since the pressure distribution is symmetrical about the y axis. 1 + c 21FR 2 y = ΣFy;

1FR 2 y = = -

L6

5p 6

p

dF sin u

5p 6

1 raU 2 11 - 4 sin2u2 sinu du L 2 p 6

5p

6 1 = - raU 2 p 1sin u - 4 sin3u2du 2 L6

5p>6 1 4 = - raU 2 e -cos u - c - cos u1 sin2u + 22 d f ` 2 3 p>6

1 = - raU 2 1 -2232 2 = 23raU 2

y

(FR)y

dF

(FR)x

5 a 30°

ds 5 a du u du

30°

x

(a)

688

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7–77. Continued

Here, U = 6 m>s, r = 1.22 kg>m3, and a = 0.2 m, 1FR 2 x = 0

1FR 2 y = 2311.22 kg>m3 210.2 m216 m>s2 2 = 15.21 N>m

Thus, the resultant force on segment AB per unit length is FR = 1FR 2 y = 15.2 N>m c

Ans.

Ans: FR = 15.2 N>m c 689

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7–78. 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>ft3.

A

6 ft

3 ft!s 4 ft

SOLUTION We consider ideal fluid flow. The velocity components of the flow around the structure are yr = Ua1 -

a2 bcos u r2

yu = - 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 yr = ( 3 ft>s ) c 1 -

(4 ft)2 (6 ft)2

yu = - ( 3 ft>s ) c 1 +

d cos 90° = 0

(4 ft)2 (6 ft)2

Thus, the magnitude of the velocity at A is

d sin 90° = - 4.333 ft>s

VA = yu = 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. pO pA VO2 VA2 + + gzO = + + gzA r r 2 2 Since the flow is in the horizontal plane, zD = zA = z. Here, VO = U = 3 ft>s. Then pO pA VO2 VA2 + + gz = + + gz r r 2 2 pA = pO + = a6

r ( VO2 - VA2 ) 2

1.94 slug>ft3 lb 12 in. 2 b a + 3 ( 3 ft>s ) 2 - ( 4.333 ft>s ) 2 4 b 1 ft 2 in2

= a854.52 = 5.93 psi

lb 1 ft 2 ba b 2 12 in. ft

Ans.

Ans: pA = 5.93 psi 690

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7–79. If the uniform velocity of the air is 15 m>s and the pressure within the uniform flow is 100 kPa, determine the required angular velocity v of the cylinder so that the lift is 96.6 N. Take ra = 1.20 kg>m3.

15 m!s

0.6 m v

SOLUTION We consider the steady ideal fluid flow. Using the “lift”, the circulation can be determined from Fy = rUΓ 96.6 N = 11.20 kg>m3 2115 m>s2Γ Γ = 5.367 m2 >s

For the corresponding free vortex at r = 0.6 m, vu = vr = v10.6 m2 Also, the circulation can be defined by Γ = Γ = 5.367 m2 >s =

C

V # ds

L0

2p

L0

2p

vu 1r du2 v10.6 m210.6 m2du

5.367 = 0.36v12p2 Ans.

v = 2.373 rad>s = 2.37 rad>s

Ans: v = 2.37 rad>s 691

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*7–80. The long cylinder is rotating counterclockwise with a constant angular velocity of 80 rad>s. If the air is blowing at a constant speed of 20 m>s, determine the location of the stagnation points, and find the absolute pressure at point A. The absolute pressure within the uniform flow is 98 kPa. Take ra = 1.20 kg>m3.

80 rad!s 0.3 m 20 m!s

60° 0.5 m A

SOLUTION We consider the steady ideal fluid flow. For the corresponding free vortex at r = 0.3 m, vu = vr = 180 rad>s210.3 m2 = 24 m>s. Thus, the circulation of this free vortex is Γ =

C

V # ds =

L0

2p

vu 1r du2 =

L0

2p

2410.3 du2 = 14.4p m2 >s

Since Γ 6 4pUa = 4p120 m>s210.3 m2 = 24p m2 >s, there exist two stagnation points on the surface. The location of these two points can be found using sin u =

14.4p m2 >s Γ = = 0.6 4pUa 4p120 m>s210.3 m2

u = 36.87° = 36.9° and u = 143.13° = 143°

Ans.

The r and u components of velocity at point A where r = 0.5 m and u = 240° are given by vr = Ua1 -

10.3 m2 2 a2 b cos u = 120 m>s2 c 1 d cos 240° = -6.40 m>s r2 10.5 m2 2

a2 Γ b sin u + 2pr r2 14.4p m2 >s 10.3 m2 2 = 1 - 20 m>s2 c 1 + d sin 240° + 2 2p10.5 m2 10.5 m2 = 37.96 m>s

vu = - Ua1 +

Thus, the magnitude of the velocity at A is VA = 2v2r + v2u = 21 - 6.40 m>s2 2 + 137.96 m>s2 2 = 38.49 m>s

Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as point O in the uniform flow and point A on a streamline around the cylinder. Since the density of the air is small, the elevation terms can be neglected. Here, VO = U = 20 m>s, p0 = 981103 2 Pa. p0 pA vO2 V 2A + = + ra ra 2 2 ra 2 pA = p0 + 1V O - V 2A 2 2 = 981103 2 Pa + a

1.20 kg>m3 2

= 97.351103 2 Pa = 97.4 kPa

b[120 m>s2 2 - 138.49 m>s2 2]

Ans.

Ans: u = 36.9° and 143° pA = 97.4 kPa 692

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7–81. The long cylinder is rotating counterclockwise with a constant angular velocity of 80 rad/s. If the air is blowing at a constant speed of 20 m>s, determine the lift per unit length on the cylinder and the maximum pressure on the cylinder. The absolute pressure within the uniform flow is 98 kPa. Take ra = 1.20 kg>m3.

80 rad!s 0.3 m 20 m!s

60° 0.5 m A

SOLUTION We consider the steady ideal fluid flow. For the corresponding free vortex at r = 0.3 m, vu = vr = 180 rad>s210.3 m2 = 24 m>s. Thus, the circulation of this free vortex is Γ =

C

V # ds =

L0

2p

vu 1r du2 =

L0

2p

2410.3 du2 = 14.4p m2 >s

The “lift” exerted on the cylinder can be determined from

Fy = rUΓ = 11.20 kg>m3 2120 m>s2114.4p m2 >s2 = 1085.73 N>m = 1.09 kN>mT

Ans.

The maximum pressure occurs at a stagnation point, where Vs = 0. Since the flow is irrotational, Bernoulli’s equation can be applied between two points on different streamlines, such as point O within the uniform flow and the stagnation point. Since the density of air is small, the elevation term can be neglected pO ps VO2 V s2 + = + ra ra 2 2 Here, VO = U = 20 m>s, pO = 981103 2 Pa, ps = pmax and Vs = 0. Then 981103 2 Pa

3

1.20 kg>m

+

120 m>s2 2 2

=

pmax

1.20 kg>m2

+ 0

pmax = 98.241103 2 Pa = 98.2 kPa

Ans.

Ans: Fy = 1.09 kN>m T pmax = 98.2 kPa 693

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7–82. The 1-m-long 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, yu = vr = ( 40 rad>s ) (0.6 m) = 24 m>s. Thus, the circulation is Γ =

C

V # ds =

L0

2p

yu(r du) =

L0

Then

2p

24(0.6 du) = 28.8p m2 >s

Fy = rU Γ = 11.20 kg>m3 2110 m>s2128.8p m2 >s2 = 1085.7 N>m For a unit-length cylinder, Ans.

Fy = 1.09 kN

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 cylinder. 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 ya

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 + = - 802 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. Ans: Fy = 1.09 kN pmax = 350 Pa pmin = - 802 Pa

694

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7–83. 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 = U1y>h2, and that the shear stress within the liquid is txy = F>A.

y

U F

h

u x

SOLUTION Since the flow is steady and is only along the x axis then y = w = 0. Also, the liquid is incompressible. Thus, the continuity equation reduces to 0(ru) 0(ry) 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, when the pressure p remains constant along x axis, then the NavierStokes equation along the x axis gives ra

0p 0u 02u 0u 0u 0u 02u 02u + u + y + w b = rgx + ma 2 + 2 + 2b 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

02u d 2u = . 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 by h

(Q.E.D)

695

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*7–84. A horizontal velocity is defined by u = 21x 2 - y2 2 ft>s and v = 1 -4xy2 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(ry) 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 0u 0u 0u 02u 02u 02u + u + y + w b = rgx + ma 2 + 2 + 2b 0t 0x 0y 0z 0x 0x 0y 0z

r 3 0 + ( 2x 2 - 2y2 )( 4x ) +

ra

( -4xy )( -4y ) + 0 4 = 0 -

0p + m(4 - 4 + 0) 0x

0p = -8r ( x 3 + xy2 ) 0x

(1)

0p 0y 0y 0y 0y 02y 02y 02y + u + y + w b = rgy + ma 2 + 2 + 2b 0t 0x 0y 0z 0y 0x 0y 0z

r 3 0 + ( 2x 2 - 2y2 )( - 4y) + ( -4xy)( -4x) + 0 4 = 0 -

0p + m(0 + 0 + 0) 0y

0r = - 8r ( y3 + x 2y ) 0y

ra

(2)

0p 0w 0w 0w 0w 02w 02w 02w + 2b + u + y + w b = rgz + ma 2 + 2 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 Eq. 1 with respect to x, p = - 8r°

x 2y2 x4 + ¢ + f(y) + g(z) 4 2

(4)

Differentiate Eq. 4 with respect to y and equate to Eq. 2. 0p = -8r ( x 2y ) + f 1(y) = - 8r ( y3 + x 2y ) 0y f 1(y) = -8ry3

696

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*7–84. 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 = g 1(z) = - rg 0z Integrate this equation with respect to z. (6)

g(z) = - rgz + C2 Substitute Eq. 5 and 6 into 4. p = - 8ra

x 2y2 x4 + b - 2ry4 - rgz + C 4 2

p = - 2r ( x 4 + y4 + 2x 2y2 ) - rgz + C

1 1 1 Since V 2 = ( u2 + y 2 ) = 3 ( 2x 2 - 2y2 ) 2 + ( -4xy)2 4 = 2 ( x 4 + y4 + 2x 2y2 ) , 2 2 2 then the above equation becomes p = C -

1 2 rV - rgz 2

(Q.E.D)

697

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7–85. The sloped open channel has steady laminar flow within its depth h. Show that the Navier–Stokes equations reduce to 02u>0y2 = - 1r g sin u2 >m and 0p>0y = -r g cos u. Integrate these equations to show that the velocity profile is u = [1r g sin u2 >2m]12hy - y2 2 and the shear-stress distribution is txy = r g sin u 1h - y2.

y h

u x

SOLUTION Since the flow is steady and is along the x axis only, then y = w = 0. Also, the liquid is incompressible. Thus, the continuity equation reduces to 0(ru) 0(ry) 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 + y + w b = rgx + ma 2 + 2 + 2b 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 0y 0y 0y 0y 02y 02y 02y + u + y + w b = rgy + ma 2 + 2 + 2b 0t 0x 0y 0z 0y 0x 0y 0z r(0 + 0 + 0 + 0) = r( -g cos u) 0p = -rg cos u 0y

Since u = u(y), then

0p + m(0 + 0 + 0) 0y (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)

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7–85. 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 Eqs. 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)

699

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7–86. 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.

v

ri ro

r

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 ( ryz ) 0r 1 0 ( rryr ) 1 0 ( ryu ) + + + = 0 0t r 0r r 0u 0z 0 + 0 +

r 0yu + 0 = 0 r 0u 0yu = 0 0u

Integrating this equation with respect to u, yu = yu(r) Using this result, the Navier-Stokes equations along the u axis gives ra

= -

0yu yu 0yu yryu 0yu 0yu + yr + + + yz b 0t 0r r 0u r 0z

02yu yu 1 0 0yu 1 0p 1 02yu 2 0yr + 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 + mc

However, it can be shown that

Thus,

yu 1 0 0yu ar b - 2 + 0 + 0 + 0 d r 0r 0r r

yu 1 0 0yu ar b - 2 = 0 r 0r 0r r

yu 0 1 0 1 0 0yu ( ry ) d = c ar b - 2 0r r 0r u r 0r 0r r 0 1 0 ( ry ) d = 0 c 0r r 0r u

700

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7–86. Continued

Since yu = yu(r), then the above equation can be written in the form of d 1 d c (ry ) d = 0 dr r dr u

Integrating this equation with respect to r,

1 d ( ry ) = C1 r dr u d ( ry ) = C1r dr u Integrating again, ryu = C1 a yu =

r2 b + C2 2

C1 C2 r + 2 r

(1)

At r = ru, yu = 0. Then Eq. 1 gives C1 C (r ) + 2 2 o ro At r = ri, yu = vri. Then Eq. 1 gives

(2)

0 =

C1 C2 ri + 2 ri

vri =

(3)

Solving Eq. 2 and 3, C1 = -

2vr i2 r o2 - r i2

C2 =

vr i2r o2 r o2 - r i2

Substituting these results into Eq. 1, yu = - a yu =

vr i2 r o2

vr i

- ri

2

r o2 - r i

a 2

br + a 2

r o2 - r 2 b r

vr i2r o2 r o2

1 ba b r - ri 2

Ans.

Ans: vu =

vr 2i

a 2

r 2o - r i

r 2o - r 2 b r

701

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7–87. The channel for a liquid is formed by two fixed plates. If laminar flow occurs between the plates, show that the Navier–Stokes and continuity equations reduce to 02u>0y2 = 11>m2 0p>0x and 0p>0y = 0. Integrate these equations to show that the velocity profile for the flow is u = 11>2m2 1dp>dx2 3 y2 - 1d>22 2 4 . Neglect the effect of gravity.

y

d!2

x

d!2

SOLUTION Since the flow is steady and is along the x axis only, then y = w = 0. Also, the liquid is incompressible. Thus, the continuity equation becomes 0(ru) 0(ry) 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 + y + w b = rgx + ma 2 + 2 + 2b 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 = 2 m 0x 0y ra

(Q.E.D)

(1)

0p 0y 0y 0y 0y 02y 02y 02y + u + y + w b = rgy + ma 2 + 2 + 2b 0t 0x 0y 0z 0y 0x 0y 0z r(0 + 0 + 0 + 0) = 0 -

0p + m(0 + 0 + 0) 0y

0p = 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 = m dx dy2

702

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7–87. 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 = 0 = a

(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 1 dr d 2 b a 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)

703

M07_HIBB9290_01_SE_C07_ANS.indd 703

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8–1. Investigate whether each ratio is dimensionless. (a) mV>L3r, (b) m>rVL, (c) 2gL>V, (d) s>rL.

SOLUTION a)

b)

mV L3r

a

=

m = rVL

c)

2gL = V

d)

s = rL

M L ba b LT T 1 = 2 M T L3 a 3 b L

no

Ans.

yes

Ans.

M LT = 1 M L a 3 b a bL T L ca

1>2 L b 1L2 d 2 T = 1 L T

M b L2 T2 = 2 M T a 3 b 1L2 L a

Ans.

yes

Ans.

no

Ans: a) no b) yes c) yes d) no 704

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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 a

a)

Vt = L

b)

EVL = s

c)

L b(T) T = 1 L a

Ans.

M b(L) LT 2 = 1 M T2

Ans.

L 2 b T V = = 1 gL L a 2 b(L) T 2

a

Ans.

Ans: Vt a) L EVL b) s V2 c) gL 705

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8–3. Determine the Mach number for a jet flying at 1500 km>h at an altitude of 3 km. The velocity of sound in air is c = 2kRT, where the specific heat ratio for air is k = 1.40.

SOLUTION Referring to the table in Appendix A, at an altitude of 3 km, T = - 4.491°C. Also, for air, R = 286.9 J>kg # K. Here, Tk = 1 - 4.491 + 273.152 K = 268.659 K. The speed of sound through the air is c = 2kRTK

= 21.401286.9 J>kg # K21268.659 K2 = 328.50 m>s

The velocity of the jet is V = a1500

The Mach number is

M =

km 1000 m 1h ba ba b = 416.67 m>s h 1 km 3600 s 416.67 m>s V = = 1.268 = 1.27 c 328.50 m>s

Ans.

Ans: M = 1.27 706

M08_HIBB9290_01_SE_C08_ANS.indd 706

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*8–4. Determine the F-L-T dimensions of the following terms. (a) Qp>L, (b) rV 2 >m, (c) EV >p, (d) gQL.

SOLUTION a)

QP = L 2

a

L3 F b a 2b T F L = L T

a

FT 2 L 2 ba b T 1 L4 = FT T L2

b)

rV = m

c)

F EV L2 = = 1 p F L2

d)

Ans.

gQL = a

Ans.

Ans.

F L3 FL b a b 1L2 = 3 T T L

Ans.

Ans: F a) T 1 b) T c) 1 FL d) T 707

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8–5. Determine the M-L-T dimensions of the following terms. (a) Qp>L, (b) rV 2 >m, (c) EV >p, (d) gQL.

SOLUTION a)

Qp = L 2

a

L3 M ba b T ML LT 2 = L T3

a

M L 2 b a b T 1 L3 = M T LT

b)

rV = m

c)

M EV LT 2 = 1 = r M LT 2

d)

Ans.

gQL = a

Ans.

Ans.

M L3 ML2 b a b 1L2 = 2 2 T LT T3

Ans.

Ans: ML T3 1 b) T c) 1 a)

d)

ML2 T3

708

M08_HIBB9290_01_SE_C08_ANS.indd 708

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8–6. 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 = 4.98 110 - 3 2 lb>ft.

SOLUTION Using the M - L - T base dimensions, 2

We =

rV L = s

a

M L 2 M b a 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 b a 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>ft3 at T = 70° F. Substituting numerically, We =

rV 2L = s

( 1.937 slug>ft3 )( 8 ft>s ) 2(2 ft) = 49.8 ( 103 ) 3 4.98 ( 10-3 ) 4 lb>ft

Ans.

Ans: We = 49.8(103)

709

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8–7. 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

mV 2 The dimensions for the physical variables in the equation ∆p = caa b given in 2R Table 8–1 are 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 = ca£ LT 2

( M>LT )( L>T ) L 1

M M 2 = caa b 2 LT LT 2

1 2

§

1

ca = a

M 2 b LT 2

Ans.

Ans: a

M 1>2 b LT 2

710

M08_HIBB9290_01_SE_C08_ANS.indd 710

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*8–8. 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 = ( M aL -aT -a)( M bL -3b )( LcT -c ) (L) = M a+ 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: rVL m or m rVL 711

M08_HIBB9290_01_SE_C08_ANS.indd 711

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8–9. 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 = ( M aL -aT -2a)( L bT -2b )( M cL -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-1g 1r1D =

rgD p

Ans.

Also, Π

-1

=

p is dimensionless rgD

Ans.

Ans: Π =

rgD p or p rgD

712

M08_HIBB9290_01_SE_C08_ANS.indd 712

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8–10. 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.

Ans: F = kyV 713

M08_HIBB9290_01_SE_C08_ANS.indd 713

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8–11. Show that the hydrostatic pressure p of an incompressible fluid can be determined 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.

714

M08_HIBB9290_01_SE_C08_ANS.indd 714

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*8–12. A floating body will move up and down with a period of t measured in seconds. The behavior depends upon the mass m of the body, its cross-sectional area A, the acceleration due to gravity, g, and the density r of the water. Determine the relationship between tand these parameters.

SOLUTION The unknown function is t = f1m, A, r, g2 or h1t, m, A, r, g2 = 0. Thus, there are n = 5 variables. Using the M - L - T system, Period, Mass, Area, Density, Acceleration,

t m A r g

T M L2 ML - 3 LT - 2

Here, all three base dimensions are involved, so m = 3. Thus, there are (n - m ) = 5 - 3 = 2 Π terms. Here, m, A, and g are chosen as m = 3 repeating variables. For the first Π term, use q = t: Π1 = maAbg ct = 1M a21L2b 21LcT -2c 21T2 = 1M a21L2b + c 21T 1 - 2c 2 M:

a= 0

T:

1 - 2c = 0

L:

2b +

c =

1 2

1 1 = 0 b = 2 4

Thus, Π1 = A-1>4g 1>2 t =

tg 1>2

A1>4 g 2 1>4 = ta b A

715

M08_HIBB9290_01_SE_C08_ANS.indd 715

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*8–12. Continued

For the second Π term, use q = r: Π2 = mdAeg fr = 1M d 21L2e 21LfT -2f 21ML -3 2 = 1M d + 1 21L2e + f - 3 21T -2f 2

M: T:

d + 1 = 0 -2f = 0

d = –1 f = 0

L:

2e + 0 - 3 = 0

e =

Thus,

3 2

Π2 = m-1A3>2r =

rA3>2 m

Therefore, the function can be written as ha Solving for t, ta

tg 1>2 rA3>2 , b = 0 A1>4 m

g 2 1>4 rA3>2 b = fa b A m t= a

A 1>4 rA3>2 b fa b m g2

Ans.

Ans: t= a

A 1>4 rA3>2 b fa b m g2

716

M08_HIBB9290_01_SE_C08_ANS.indd 716

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8–13. The average velocity V of water flow through a pipe is a function of the pipe’s diameter D, the change in pressure ∆p per unit length, ∆p> ∆x, and the viscosity of the water, m. Determine the relation between V and these parameters.

D V

SOLUTION The unknown function is V = f aD,

∆r ∆r , mb or haV, D, , m b = 0. ∆x ∆x

Thus, there are n = 4 variables using M - L - T, Average Velocity, Diameter, Pressure gradient, Viscosity,

V D ∆r ∆x

LT –1 L

m

ML–1T –1

ML–2T –2

Here, all three base dimensions are involved, so m = 3. Thus, there is n - m = 4 - 3 = 1 Π term Here, V, D, and

∆p are chosen as m = 3 repeating variables. Then q = m. ∆x Π = V aDb a

∆p c b m ∆x

Π = 1LaT -a21Lb 21M cL -2cT -2c 21ML -1T -1 2 L: T: M:

= 1La+ b-2c - 1 21T -a- 2c - 1 21M c + 1 2

a+ b - 2c - 1 = 0 –a- 2c - 1 = 0 c + 1 = 0

Solving, c = –1, a= 1, and b = –2. Thus, Π = VD -2 a Π = Therefore,

or solving for V,

∆p -1 b m ∆x

mV ∆p D2 a b ∆x

mV = C ∆p D2 a b ∆x

V = Cc

D2 ∆p bd a m ∆x

Ans.

where C is a constant to be determined by experiment.

Ans: V = Cc

D2 ∆p ¢ ≤d m ∆x

717

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8–14. 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)( M bL -bT -b )( M cL -3c )( ML -1T -2 ) = M b + 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

p Ar

718

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8–15. 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

V

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 = Qarbg cD Π = ( L3aT -a)( M bL -3b )( LcT -2c ) (L) = M bL3a- 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

- 25 0

1 5

Π = Q rgD =

g 5D 2

Q5

The function can be rewritten as 1

f°

g 5D 2

Q5

¢ = 0

Solving for Q using this function, 2

1

Q5 = k′g 5D Q = k2gD5

Ans.

where k is a constant to be determined by experiment. Notice that Q is independent of r.

Ans: Q = k2gD5 719

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*8–16. 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 bg cV Π = ( M aL -3a)( Lb )( LcT -2c )( LT -1 ) = M aL -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 720

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8–17. 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. Compare the result with that obtained in Sec. 1.10.

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.

Ans: p = k

s r

721

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8–18. 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 = larb 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: c = k

s , percent decrease = 18.4% A rl

722

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8–19. 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 directly 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 = g aH bQ = ( 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 =

2gH 5

Π2 = g cH db = ( 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 = g 0H -1b = 2

f°

Q

2gH Solving for Q,

5

,

b ¢ = 0 H

Q = 2gH 5f1 a

b b H

Since Q is proportional to b, this function becomes Q = kb2gH 3

Ans.

where k is a constant to be determined by experiment. If H is doubled, Q increases by 223 = 2.83 times

Ans.

Ans: Q = kb2gH 3, Q increases by a factor of 2.83.

723

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*8–20. The time t needed for a liquid 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. Show how to obtain the two Π terms 1 1g>d2t and m> 1rd 3>2g 1>2 2 where q = t and m. If the liquid is water at a temperature of 15°C, an experiment produced the data for the time versus the diameter of various tubes. Use this data to plot the relationship between the two Π terms.

d (mm)

t (s)

0.50 1.00 1.50 2.00 2.50 3.00

305 87.6 42.2 25.15 16.9 12.1

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 q, is Π1 = rad bg ct

g t (10 –3) d

= ( M aL -3a)( Lb )( Lc T -2c )( T )

40

= M aL -3a+ b + c T -2c + 1 30

Thus, for M:

a= 0

20

L:

-3a+ b + c = 0

10

T:

-2c + 1 = 0

m 0

1 1 Solving, a= 0, b = - , and c = . Then 2 2 g 1 1 Π1 = r0d -2 g 2t = t Ad

0.01

0.02

0.03

0.04

pd

3y2

g1y2

(a)

The second Π term, using m as q, is Π2 = rdd eg f m

= ( M d L -3d )( Le )( Lf T -2f )( ML -1T -1 ) = M d + 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

724

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8–20. Continued

3 1 Solving, d = -1, e = - , and f = - . Then 2 2 m -1 - 32 - 12 Π2 = r d g m = 3 1 rd 2g 2 Then f1 °

g m t, ¢ = 0 A d rd 32 g 12

Solving for

g t in this equation, Ad

g m t = f ° 3 1¢ Ad rd 2 g 2

From the table in Appendix A, rw = 999.2 kg>m3 and mw = 1.15110-3 2 N # s>m for water at T = 15°C. Together with the given data, the values of Π1 and Π2 are computed and tabulated below. g t Ad m rd 3>2g 1>2

42.7(103)

8.68(103)

3.41(103)

1.76(103)

1.06(103)

0.692(103)

0.0329

0.0116

0.00633

0.00411

0.00294

0.00224

The plot of Π1 vs Π2 is shown in Fig. a

725

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8–21. 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. Take q = d and m.

U

u d

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 = x aU brcd = ( La)( LbT -b )( M cL -3c ) (L) = M cLa+ 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 = x dU erfm = ( Ld )( LeT -e )( M fL -3f )( ML -1T -1 ) = M f + 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 f1 a , Reb = 0 x

Solving for d, d = f (Re) x

Ans.

d = xf (Re)

Ans: d = xf (Re)

726

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8–22. Establish Newton’s law of viscosity using dimensional analysis, realizing that shear stress t is a function of the fluid viscosity m and the angular deformation du>dy. Hint: Consider the unknown function as f(t, m, 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 = ( M aL -aT -2a)( M bL -bT -b )( LcT -c )( L ) = M a+ 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: du , with k determined dy by experiment t = km

727

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8–23. The period of oscillation t, measured in seconds, of a buoy depends upon its cross-sectional area A, 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 = Aag b mc t Π = ( L2a)( M bL -2bT -2b ) (M c)(T) L:

0 = 2a- 2b

M:

0 = b + c

T:

0 = - 2b + 1

Solving, a= 1>2, b = 1>2, c = -1>2. Thus, gA 1 1 1 Π = A2 g 2 M - 2 t = t Am

The function can be written as gA f at 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

m A gA

728

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*8–24. The period of time t between small waves on the surface of a liquid is thought to be a function of the wave length l, the liquid depth h, gravitational acceleration g, and the surface tension s of the liquid. Show how to obtain the two Π terms h>l and t1g>l, where q = t and h. If the liquid is water at a temperature of 15°C, an experiment with the water depth h = 150 mm produced the following results for the period for various measured wavelengths. Use this data to plot the relationship between the two Π terms.

t (s)

l (mm)

0.05 0.10 0.15 0.20 0.25 0.30

25 50 75 100 125 150

SOLUTION The unknown function is t = f1l, h, g, s2, or f′1t, l, h, g, s2 = 0. Thus, there are n = 5 variables. Using the F - L - T system, Period,

t

T

Wave length,

l

L

Liquid depth,

h

L

Gravity,

g

LT –2

s

–1

Surface tension,

t

g l

2.50

FL

2.00 1.50

Here, all three base dimensions are involved, so m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms

1.00 0.50

Here, l, g, and s are chosen as repeating variables. The first Π term, using h as q, is 0

ab c

Π1 = l g s h = 1La21LbT -2b 21F cL -c 21L2

1

2

3

4

5

6

h x

(a)

= La+ b - c + 1T -2bF c

Thus for

L: a+ b - c + 1 = 0 T:

- 2b = 0

F:

c = 0

Solving, a= –1, b = 0, c = 0. Then Π1 = l-1h =

h l

The second Π term, using t as q, is Π2 = ldg esft = 1Ld 21LeT -2e 21F fL -f 21T2 = Ld + e - fT -2e + 1F f Thus for L: T: F:

d + e - f = 0 - 2e + 1 = 0 f = 0

1 1 Solving, d = - , e = , f = 0. Then 2 2 Π2 = l-1>2g 1>2t g = t Al

729

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8–24. Continued

Here, h = 150 mm. Together with the given data, the values of Π1 and Π2 are computed and tabulated below. h l g t Al

6.00

3.00

2.00

1.50

1.20

1.00

0.990

1.40

1.72

1.98

2.21

2.43

The plot of Π1 vs Π2 is shown in Fig. a.

730

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8–25. As the liquid drains from the pipette, the exit velocity V is a function of the liquid’s specific weight g, viscosity m, the nozzle’s diameter d, and the liquid’s level h. Determine the relation between V and these parameters. Take q = h and m. h

SOLUTION d

The unknown function is V = f1g, m, d, h2 or h1V, g, m, d, h2 = 0. Thus, there are n = 5 variables. Using F - L - T system Average velocity,

V

LT –1

Specific weight,

g

FL–3

Viscosity,

m

FTL–2

Diameter,

d,

L

Liquid level,

h

L

V

Here, all three base dimensions are involved, so m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Here, V, g, and d are slected as m = 3 repeating variables. For Π1 term, q = h Π1 = V ag bd ch Π1 = 1LaT -a21F bL -3b 21Lc 21L2 = 1F b 21La- 3b + c + 1 21T -a2

F: b = 0 T: a= 0 L: 0 - 3102 + c + 1 = 0

c = –1

Thus, Π1 = d -1h = For Π2 term, q = m.

h d

Π2 = V dg ed fm Π2 = 1LdT -d 21F eL -3e 21Lf 21FTL -2 2 F: L: T:

= 1F e + 1 21Ld - 3e + f - 2 21T -d + 1 2

e + 1 = 0 d - 3e + f - 2 = 0 -d + 1 = 0

Solving, d = 1, e = –1, and f = –2. Thus. Π2 = Vg -1d -2m = Therefore, the function can be written as

Solving for V,

fa

mV gd 2

mV h , b = 0 gd 2 d

h = fa b d gd gd 2 h V = fa b m d mV

Ans:

2

Ans.

V =

gd 2 h fa b m d

731

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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 relation between Q and these parameters. If Q varies linearly with T, how does it vary with the turbine’s diameter D? Take q = Q and T.

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 f1 a

Q 3

,

T 2

vD v D5r

Solving for Q, Q vD3

= fa

b

T v2D5r

Q = vD3f a

T

b

v2D5r

Since Q is proportional to T,

b

Q = vD3k a

T v2D5r

b = k

T vrD2

Ans.

where k is a constant to be determined by experiment. Q is inversely proportional to D2.

Ans: Q = k

T vrD2

732

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–27. The flow of air Q through the fan is a function of the diameter D of the blade, its angular rotation v, the density r of the air, and the pressure difference ∆p on each side of the blade. Determine the relation between Q and these parameters. Take q = ∆p and Q.

D

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (∆p, D, v, r, Q) = 0. Using the F - L - T system, Pressure change,

∆p

FL -2

Diameter,

D

L

Angular velocity,

v

T -1

Density,

r

FT 2L -4

Flow rate,

Q

L3T -1

Here, all three base dimensions are involved, so 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. For Π1 term, q = ∆p. 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 = For Π2 term, q = Q.

∆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, Q

Π2 = D -3v-1r0Q =

vD3 Therefore, the function can be written as f1 a

∆p 2

2

,

Q

rv D vD3

Solving for Q,

b = 0 Q vD3

= fa

∆p rv2D2

Q = vD3f a

b

∆p rv2D2

Ans.

b

Ans: Q = vD3f ¢

∆p rv2D2

≤

733

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*8–28. The drag 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. Take q = FD and m. 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)( M bL -3b )( L2c )( MLT -2 ) = M b + 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 )( M eL -3e )( L2f )( ML -1T -1 ) = M e + 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 Therefore, the function can be written as Π2 =

f1 a

FD rV 2A

, Reb = 0

Solving for FD, FD rV 2A

= f (Re) FD = rV 2Af (Re)

Ans.

Ans: FD = rV 2Af (Re)

734

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–29. The length l of small water waves is thought to be a function of the period of time t of each wave cycle, the water depth h, gravitational acceleration g, and the surface tension s of the water. Determine the relation between l and these parameters. Take q = l and t.

SOLUTION The unknown function is l = f1t, h, g, s2 or h1l, t, h, g, s2 = 0. Thus, there are n = 5 variables using M - L - T system, Period, Wave length, Water depth, Gravity, Surface tension,

T L L LT -2 MT -2

t l h g s

Here, all three base dimensions are involved so m = 3. Thus, there are n - m = 5 - 3 = 2 Π terms Here, h, g, and s are chosen as m = 3 repeating variables. For Π, term, q = l. Π1 = hag bscl = 1La21LbT -2b 21M cT -2c 21L2 M:

= 1M c 21La+ b + 1 21T -2b - 2c 2

c = 0

T:

-2b - 2102 = 0

b = 0

L:

a+ 0 + 1 = 0

a= -1

Thus,

Π1 = h-1l =

l h

For Π2 term, q = t. Π2 = hdg esft Π2 = 1Ld 21LeT -2e 21M fT -2f 21T2 Π2 = 1M f 21Ld + e 21T -2e - 2f + 1 2

M:

f = 0

T:

- 2e - 2102 + 1 = 0

L: d +

1 = 0 2

e =

1 2

d = -

1 2

Thus, g Π2 = h-1>2g 1>2t = t Ah

Therefore, the function can be written as Solving for l,

g l f1 a , t b = 0 h Ah g l = f at b h Ah g l = h f at b Ah

M08_HIBB9290_01_SE_C08_ANS.indd 735

Ans: Ans.

g l = hf ¢t ≤ Ah

735

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–30. 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. Take q = m, V, and T.

V

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

V

LT -1

D

L

r

FT 2L -4

v

T -1

m

FTL -2

T

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

M08_HIBB9290_01_SE_C08_ANS.indd 736

Ans: V b rvD vD m

Ans.

, 2

T = rv2D4f ¢

m V , ≤ rvD2 vD

736

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–31. 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. Take q = T and h.

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 T -1 - 53 - 13 Π1 = r Q v T = 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. Ans: T = rQ5>3v1>3f c ¢

v 1>3 ≤ hd Q

737

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*8–32. 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. Take q = hL and m.

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 )( M cL -3c ) (L) = M cLa+ 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 )( M fL -3f )( ML -1T -1 ) = M f + 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)

Ans: hL = Df (Re)

738

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8–33. 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. Show how to obtain the two Π terms h>d and s> 1rd 2g2, where q = h and s. If an experiment is performed using water at a temperature of 20°C, and the surface tension is s = 0.0736 N>m, the following data is obtained of the height h versus the diameter d of the tube. Use the data and plot the relationship between the two Π terms.

a

h (mm) d (mm) 30.06 15.03 10.02 7.52 6.01 5.01

h

0.5 1.0 1.5 2.0 2.5 3.0

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 arbg es = ( La)( M bL -3b )( LcT -2c )( MT -2 ) = La- 3b + cM b + 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 erfg ih = ( Le )( M fL -3f )( LiT -2i ) (L) = M fLe - 3f + i + 1T -2i M:

0 = f

L:

0 = e - 3f + i + 1

T:

0 = -2i

h d

Solving, e = - 1, f = 0, and i = 0. Thus,

60

h Π = d -1r0g 0h = 2 d From the table in Appendix A, rw = 998.3 kg>m3, and s = 0.0736 N>m for water at T = 20°C. Together with the given data, the values of Π1 and Π2 are computed and tabulated below. h d s rd 2g

60.12

15.03

6.68

3.76

2.40

1.67

30.06

7.52

3.34

1.88

1.20

0.835

The plot of Π1 vs Π2 is shown in Fig. a

50 40 30 20 10 0

5

10

15

20

25

30

t Pd 2g

(a)

739

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8–34. 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. Take q = ∆p and m.

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)( M bT -3b )( LCT -C )( ML -1T -2 ) = M b + 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 )( M eL -3e )( LfT -f )( ML -1T -1 ) = M e + 1Ld - 3e + f - 1T -f - 1 0 = e + 1

M: 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 f1 a

∆p rV 2

, Reb = 0 ∆p = rV 2f (Re)

Ans.

Ans: ∆p = rV 2f(Re)

740

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8–35. The drag 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. Take q = FD and m.

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 )( M cL -3c )( MLT -2 ) = M c + 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 )( M iL -3i )( ML -1T -1 ) = M i + 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 m 1 Π3 = V -1A-2r-1m = = 1 rVL rVA2 or rVL = Re m

Π2 =

Therefore, the function can be written as f1 a

FD rAV

2

,

L 2A

, Reb = 0

Solving for FD in this equation. FD rV 2L2

= f (Re)

Ans: 2

2

Ans.

FD = rV L 3f (Re)4

FD = rV 2L2 3f (Re)4

741

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*8–36. 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 initial 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. Take q = p, r, and EV.

r

SOLUTION Physical Variables. There are n = 5 variables and the unknown function is f (p, m, p0, r, r, E V) = 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 E V for Π3. Π1 = p0ambr cp = ( M aL -aT -2a)( M b )( Lc )( ML -1T -2 ) = M a+ 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 = ( M dL -dT -2d )( M e )( L f )( ML -3 ) = M d + 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 = ( M gL -gT -2g )( M h )( Li )( ML -1T -2 ) = M g + h + 1L -g + i - 1T -2g - 2 M:

0 = g + h + 1

L:

0 = -g + i - 1

T:

0 = -2g - 2

742

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8–36. Continued

Solving, g = -1, h = 0, and i = 0. Thus, Π3 = p0-1 m0r 0EV =

EV p0

Therefore, the function can be written as f1 a

p rr 3 EV , , 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

rr 3 EV , b m p0

743

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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. Take q = Q, r, and m.

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 )( M cL2cT -3c )( L3T -1 ) = M cLa+ 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 )( M fL2fT -3f )( ML -3 ) = M f + 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,

#

Π2 = D5v3W -1r =

#

rD5v3 # W

Π3 = DgvhW im = ( Lg )( T -h )( M iL2iT -3i )( ML -1T -1 ) = M i + 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 . W Therefore, the function can be written as

#

Π3 = D3v2W -1m =

f1 a

rD5v3 D3v2m # , # b = 0 W D3v W Q

,

Solving for Q, Q 3

Dv

= fa

rD5v3 D3v2m # , # b W W

M08_HIBB9290_01_SE_C08_ANS.indd 744

rD5v3 D3v2m # , # b Q = D vf a W W

Ans:

3

Ans.

Q = D3vf a

rD5v3 D3v2m # , # b W W

744

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8–38. 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. Take q = m, s, D, and d. d V

h

SOLUTION

D

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 = ( M aL -3a)( LbT -b )( Lc )( ML -1T -1 ) = M a+ 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 = ( M dL -3d )( LeT -e )( Lf )( MT -2 ) Thus for M:

d + 1 = 0

L:

-3d + e + f = 0

T:

-e - 2 = 0 745

M08_HIBB9290_01_SE_C08_ANS.indd 745

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8–38. 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 = ( M gL -3g )( LhT -h )( Li )( L ) = M gL -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.

Ans: gaRe, We,

D d , b = 0 h h

746

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# 8–39. 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. # Establish a relation between W and these parameters.

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 = ( M aL -3a)( L3bT -b )( M cL -cT -2c )( ML2T -3 ) = M a+ 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 = r Q ∆p W = 0

-1

-1

# W Q∆p

Thus, the general relation between the given physical variables is # W = CQ∆p

Ans.

where C is a dimensionless constant to be determined from experiment.

Ans: # W = CQ∆p

747

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*8–40. The height H of water as it passes over a small weir depends upon the volumetric discharge Q, 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 H and these parameters. Take q = H, b, and Q.

H

h

SOLUTION The unknown function is H = f1Q, b, h, g, r, m, s2 or h1H, Q, b, h, g, r, m, u2 = 0. Thus, there are n = 8 variables using M – L – T system, Height above weir,

H

L

Density,

r

ML–3

Width of weir,

b

L

Flow rate,

Q

L3T –1

Height of weir

h

L

Viscosity,

m

ML–1T –1

Surface tension,

s

MT –2

Acceleration due to gravity

g

LT

-2

Here, all three base dimensions are involved so m = 3. Thus, there are n - m = 8 - 3 = 5 Π terms Here, r, g, and h are chosen as m = 3 repeating variables. For Π1 term, q = H Π1 = rag bhcH = 1M aL -3a21LbT -2b 21Lc 2L M: T: L:

a= 0 -2b = 0 c + 1 = 0

= 1M a21L -3a+ b + c + 1 21T -2b 2 b = 0 c = –1

Thus, Π1 = h-1H =

H h

For Π2 term, q = b. Π2 = rdg ehfb Π2 = 1M dL -3d 21LeT -2e 21Lf 21L2 M: T: L:

d = 0 -2e = 0 f + 1 = 0

= 1M d 21L -3d + e + f + 1 21T -2e 2 e = 0 f = –1

Thus, Π2 = h-1b = For Π3 term, q = Q.

b h

Π3 = rgg hhiQ = 1M gL -3g 21LhT -2h 21Li 21L3T -1 2 = 1M g 21L -3g + h + i + 3 21T -2h - 1 2

748

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8–40. Continued

M:

g = 0

T:

–2h - 1 = 0

L:

-

1 + i + 3 = 0 2

1 2 5 i = 2 h = -

Thus, Π3 = g -1>2h-5>2Q =

Q g

1>2 5>2

h

=

For Π4 term, q = m

Q 2gh5

Π4 = r jg khlm Π4 = 1M jL -3j 21LkT -2k 21Ll 21ML -1T -1 2 = 1M j + 1 21L -3j + k + l - 1 21T -2k - 1 2

M:

j + 1 = 0

j = –1

T:

-2k - 1 = 0

k = -

L: Thus,

1 2 3 l = 2

1 -31 - 12 + a - b + l - 1 = 0 2 Π4 = r-1 g -1>2 h-3>2m =

For Π5 term, q = s.

m 2r2gh3

Π5 = rmg nhps = 1M mL -3 m 21LnT -2 n 21Lp 21MT -2 2 = 1M m + 1 21L -3 m + n + p 21T -2 n - 2 2

M:

m + 1 = 0

m = –1

T:

–2 n - 2 = 0

n = -1

L:

-3 1–12 + 1–12 + p = 0

p = -2

Thus,

Π5 = r-1g -1h-2s =

s rgh2

Therefore, the function can be written as

Solving for H,

fa

m H b Q s , , , , b = 0 h h 2gh5 2r2gh3 rgh2

m H b Q s = fa , , , b h h 2gh5 2r2gh3 rgh2

m b Q s H = hf a , , , b 5 2 3 h 2gh 2r gh rgh2

Ans.

Ans: m b Q s H = hf a , , , b h 2gh5 2r2gh3 rgh2

749

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8–41. Water and crude oil flow through the pipes with a velocity of 2 m>s and 8 m>s, respectively. Determine the diameter of the oil pipe so that both fluids have the same dynamic characteristics. The temperature of the water and oil is 20°C.

50 mm

SOLUTION

D

Since inertia and viscous forces are predominant, the Reynolds number must be used to perform the dynamic similitude. From Appendix A, rco = 880 kg>m3, rw = 998.3 kg>m3, mco = 30.2110-3 2 N # s>m2 and mw = 1.00110-3 2 N # s>m2 at T = 20°C. The similitude requires

a

rVD rVD b = a b m w m co

1998.3 kg>m3 212 m>s210.05 m2 1.00110-3 2 N # s>m2

=

1880 kg>m3 218 m>s2D 30.2110-3 2N # s>m2

D = 0.4282 m = 428 mm

Ans.

Ans: D = 428 mm 750

M08_HIBB9290_01_SE_C08_ANS.indd 750

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8–42. 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 be the velocity of the model in order to maintain the same Reynolds number? Is this test realistic? Take the temperature of both the air and the water to be 60°F.

350 mi!h

SOLUTION From Appendix A, for water nm = 12.2 ( 10-6 ) ft2 >s and for air np = 0.158 ( 10-3 ) ft2 >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 = £

Lp nm ba bV np Lm p

12.2 ( 10-6 ) ft2 >s

0.158 ( 10-3 ) ft2 >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: Vm = 405 mi>hr 751

M08_HIBB9290_01_SE_C08_ANS.indd 751

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8–43. If the water in a river is flowing at 16 ft>s, and it flows at 4 ft>s in a model, determine the depth of the water in the model if the river has an average depth of 8 ft.

SOLUTION Since the inertia and gravitational forces are predominant in the river flow, the Froude number must be used to perform the dynamic similitude for the model and the prototype.

( Fr ) m = ( Fr ) p a

V 2gd

b

m

= a

dm = a

Vm

Vp

V 2gd 2

b

p

b dp = a

4 ft>s 16 ft>s

2

b 18 ft2 = 0.5 ft = 6 in.

Ans.

Ans: dm = 6 in. 752

M08_HIBB9290_01_SE_C08_ANS.indd 752

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*8–44. If the pressure loss along a measured length of a 50-mm-diameter pipe transporting water is 4 Pa, determine the pressure loss in the same length of a 100-mm-diameter pipe transporting kerosene flowing at a velocity of 4 m>s. The temperature of the water and kerosene is 20°C.

50 mm

SOLUTION

100 mm

The dynamic similitude using Reynolds number must be performed first to relate the flow velocities for water and kerosene. From the table in Appendix A, rke = 814 kg>m3, pw = 998.3 kg>m3, nke = 2.36110-6 2m2 >s and, nw = 1.00110-6 2m2 >s at T = 20° C. a

VD VD b = a b n w n ke

Vw = a

1.00110-6 2 m2 >s 100 mm Vw Dke ba bVke = c da b 14m>s2 nke Dw 50 mm 2.36110-6 2m2 >s = 3.390 m>s.

Ans.

Using this result to perform the dynamic similitude using Euler number, a

∆p rV

2

b = a w

1∆p2 ke = a

∆p rn2

b

ke

814 kg>m3 4 m>s 2 pke Vke 2 ba b ∆pw = a b a b 14 Pa2 = 4.54 Pa rw Vw 998.3 kg>m3 3.39 m>s

Ans.

753

M08_HIBB9290_01_SE_C08_ANS.indd 753

Ans: 1∆p2 ke = 4.54 Pa

06/03/17 12:08 PM

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8–45. 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 on the same model underwater in a channel, what should be the speed of the water 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

nw La ba b ( Va) n a Lw

La b = 1. From Appendix A, Lw ft2 >s at 60° F. Thus,

Since the same model is used for both cases, a nw = 12.2 ( 10-6 ) ft2 >s and na = 0.158 ( 10-3 ) Vw = c

0.0122 ( 10-3 ) ft2 >s 0.158 ( 10-3 ) ft2 >s

d (1)(200 mi>h)

Ans.

= 15.44 mi>hr = 15.4 mi>h

Ans: Vw = 15.4 mi>h 754

M08_HIBB9290_01_SE_C08_ANS.indd 754

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8–46. 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

= a

Vm = =

V 2gL

b

p

Lm Vp B Lp

1 ( 6 ft>s ) = 1.897 ft>s = 1.90 ft>s A 10

Ans.

Ans: Vm = 1.90 ft>s 755

M08_HIBB9290_01_SE_C08_ANS.indd 755

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8–47. 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: vp = 0.755 rad>s 756

M08_HIBB9290_01_SE_C08_ANS.indd 756

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*8–48. If a ship having a length of 100 ft is to be designed to travel at 45 ft>s, determine the speed of a 3-ft-long model in order to maintain the same Froude number.

SOLUTION Performing the dynamic similitude using the Froude number, a

V 2gL

b

m

= a

Vm = a

V 2gL

b

p

Lm 3 ft bVp = a b ( 45 ft>s ) = 7.79 ft>s B Lp B 100 ft

Ans.

Ans: Vm = 7.79 ft>s 757

M08_HIBB9290_01_SE_C08_ANS.indd 757

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8–49. The flow of water around the structural support is 1.2 m>s when the temperature 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

Lp nm ba b np Lm

= 1.2 m>s a

= 14.2 m>s

0.898 ( 10-6 ) m2 >s 1.52 ( 10

-6

) m >s 2

ba

20 b 1

Ans.

Ans: Vm = 14.2 m>s

758

M08_HIBB9290_01_SE_C08_ANS.indd 758

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8–50. If the drag on a 15-m-long airplane is to be determined when the plane is flying at 1200 km>h at an altitude of 10 km, find the speed of the air in a wind tunnel for a 1.5-m-long model if the air has a temperature of 20°C at standard atmospheric pressure.

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

Lp nm ba bV np Lm p

From the table in Appendix A, at air the altitude of 10 km, np = 35.25110-6 2 m2 >s. Also, nm = 15.1110-6 2 m2 >s for air at T = 20°C with standard atmospheric pressure Vm = c

15.1110-6 2 m2 >s

35.25110-6 2 m2 >s

= 5140 km>h

da

15 m b 11200 km>h2 1.5 m

Ans.

Note: The speed of sound at an altitude of 10 km is about 1100 km>h.

Ans: Vm = 5140 km>h 759

M08_HIBB9290_01_SE_C08_ANS.indd 759

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8–51. The flow around the jet plane 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 be the speed of the air 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 bVp np

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: Vm = 4.97 Mm>h, not reasonable 760

M08_HIBB9290_01_SE_C08_ANS.indd 760

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*8–52. A 15-ft-long submarine is intended to travel at 15 mi>h. If it is tested in a wind tunnel using a 1-ft-long model, determine the speed of the air at the same temperature of 40°F. Is this reasonable?

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 Vm = a

Lp Vm ba b 1Vp 2 Vp Lm

From Appendix A for air, nm = 0.147 ( 10-3 ) ft2 >s and for water, np = 16.6 ( 10-6 ) ft2 >s at T = 40° F. Thus, Vm = £

0.147 ( 10-3 ) ft2 >s 16.6 ( 10-6 ) ft2 >s

§a

15 ft b 115 mi>h2 1 ft

Ans.

= 1992.47 mi>h = 1992 mi>h

No, this speed is not reasonable, since it is well above the speed of sound and so the results would not be valid.

Ans: Vm = 1992 mi>h not reasonable 761

M08_HIBB9290_01_SE_C08_ANS.indd 761

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8–53. 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.

SOLUTION Water flow of fluid around the pier causes the inertia and viscous forces to be

0.8 m!s

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

Lp nm ba bV np 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: Vm = 12 m>s 762

M08_HIBB9290_01_SE_C08_ANS.indd 762

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8–54. 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

¢

m

= °

Vm = Here,

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: Vm = 12 km>h 763

M08_HIBB9290_01_SE_C08_ANS.indd 763

06/03/17 12:09 PM

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8–55. The velocity of water waves in a channel is studied in a laboratory using a model of the channel one-twelfth its actual size. Determine the velocity of the 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

Ans: Vp = 20.8 m>s 764

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*8–56. A model of a submarine is used to determine the drag 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 on the model is 20 N, determine the drag 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 b . Thus, = = 1. Here, = a temperature, rm Lm 1 Vm

( FD ) p = (1)(1)2 a

100 2 b ( 20 N ) 1

= 200 ( 103 ) N = 200 kN

Ans.

Ans: ( FD ) p = 200 kN 765

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8–57. 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, a

V V 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: rp = 11.0 kg>m3 766

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8–58. A water pump produces a flow of 8 ft3 >s and a power of 4 hp, using an 8-in.-diameter impeller. What is the flow and power if a 6-in.-diameter impeller is used to maintain dynamic similitude?

SOLUTION Since the viscous force is the dominant force, then Reynolds number similitude must be achieved a

rVD rVD b = a b m 1 m 2

Since water is the fluid involved for both impellers, r1 = r2 and m1 = m2. Then V1D1 = V2D2 V2 D1 = V1 D2

Since V =

(1)

Q where A has the dimension of D2, Eq. (1) becomes A Q2 >D22 Q1 >D12

=

D1 D2

Q2 D2 = Q1 D1

Here, Q1 = 8 ft3 >s and

(2)

D2 6 in. 3 = = . Then D1 8 in. 4 3 Q2 = a b 18 ft3 >s2 = 6 ft3 >s 4

Ans.

For the dynamic similitude involving the change in pressure, the Euler number should be used.

Since r1 = r2,

a

∆p rV 2

b = a 1

1∆p2 1 V12

1∆p2 2 1∆p2 1

=

∆p rV 2

b

2

1∆p2 2

= a

V22

V2 2 b V1

767

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8–58. Continued

Substitute Eq. 1 into this equation, 1∆p2 2 1∆p2 1

As shown in Section 8.1,

# W = C Q∆p

= a

D1 2 b D2

(3)

where C is a constant

Thus, a

# # W W b = a b Q∆p 1 Q∆p 2

# Q2 1∆p2 2 # W2 = a b c d W1 Q1 1∆p2 1

Substitute Eq. (2) and (3) into this equation,

# D2 D1 2 # W2 = a b a b W1 D1 D2 # D1 # W2 = a bW1 D2

# D1 8 in. 4 Here W = 4 hp and = = D2 6 in. 3

# 4 W2 = a b 14 hp2 = 5.33 hp 3

Ans.

Ans: Q2 = 6 ft3 >s # W2 = 5.33 hp 768

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8–59. The jet plane can fly at Mach 2 in 40°F air. What is the comparable speed the plane can fly at in 20°F air? Use Eq. 13–19, c = 2kRT, where k = 1.40 for air.

SOLUTION The Mach number can be expressed as M =

V V = C 2kRT

Peforming the Mach number similitude, V1 2kRT1 V1 2T1

= =

V2 2kRT2 V2

2T2 T2 V2 = a bV B T1 1

For air, the table in Appendix A gives R = 1716 ft # lb>slug # °R. Thus V1 = 2C = 22kRT1 = 221.411716 ft # lb>slug # R2140°F + 4602°R = 2191.99 ft>s Then V2 = a

120°F + 4602°R

B 140°F + 4602°R

b 12191.99 ft>s2

= 2147.70 ft>s = 2.151103 2ft>s

Ans.

Ans: V2 = 2.15(103) ft>s 769

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RES

CUE RES

*8–60. If the jet plane flies at 1500 km > h when the temperature of the air is - 10°C, and the absolute pressure is 60 kPa, what should its speed be so that it has the same Mach number when the air temperature is - 30°C and the absolute pressure is 40 kPa? Assume the air has the same bulk modulus. Use Eq. 13–20, c = 2EV >r.

CUE

51204271

SOLUTION For the Mach number dynamic similitude, a

V V 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, B r c1 A r2 EV A r1 r1 V2 = V A r2 1

From the table in Appendix A, R = 286.9 J>kg # K for air. Applying the ideal gas law, p1 = r1RT1;

601103 2 N>m2 = r1 1286.9 J>kg # K23273 + 1 - 1024K r1 = 0.7952 kg>m3

p2 = r2RT2;

401103 2 N>m2 = r2 1286.9 J>kg # K23273 + 1 - 3024K r2 = 0.5738 kg>m3

Thus V2 = a

0.7452 kg>m3 B 0.5738 kg>m3

b 11500 km>h2 = 1765.88 km>h = 1766 km>h Ans.

Ans: V2 = 1766 km>h 770

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8–61. The drag coefficient on an airplane is defined by CD = 2FD >rV 2L2. If the drag acting on the model of a plane tested at sea level is 0.3 N, determine the drag 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

rV L

2

b = a p

2FD rV 2L2

( FD ) p = ( FD ) m = a

b

rp

rm

m

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: (FD)p = 20.0 kN 771

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8–62. 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

Vm Lm = 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.

Ans: vm = 2.83 (10-9) m2 >s, too low to be practical 772

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8–63. Consider two similar types of hydrofoil boats having a size ratio 3:2. If the maximum lift the larger boat can produce is 8 kN, when it travels with a speed of 60 km>h, determine the maximum lift that the smaller boat can produce and the corresponding speed. Assume the water temperature is the same for both cases. This requires Euler number and Reynolds number similitude.

SOLUTION Assuming Reynolds number dynamic similitude, a

rVL rVL b = a b m s m b

rb ms Vs Lb = a ba ba b mb rs Vb Ls

Lb 3 = and 1Vb 2 = 60 km>h. For the same temperature rb = rs and mb = ms. Here, Ls 2 Then Vs 3 3 3 = 112112 a b = ; Vs = 160 km>h2 = 90 km>h Vb 2 2 2

Ans.

F>L2 F F F = 2 , the Euler number can be written as Eu = = . 2 A L rV rV 2L2 Performing the Euler number dynamic similitude,

Since p =

a

F max rV 2L2

b = a s

1F max 2 s = a

F max rV 2L2

b

b

rs Vs 2 Ls 2 b a b a b 1F max 2 b r b Vb Lb

3 2 2 2 = 112 a b a b 18 kN2 2 3

Ans.

= 8 kN

Ans: Vs = 90 km>h, (Fmax)s = 8 kN 773

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*8–64. 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 ft3 >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

Vm Lm 2 = a b Vp Lp

(1)

Since V = Q>A and A has a dimension of L2, Eq. 1 becomes Qm >Lm2 Qp >Lp 2

1

= a

Lm 2 b Lp

1

Qm Lp 2 Lm 2 a b = a b Qp Lm Lp 1

Qm Lm 2 Lm 2 = a b a b Qp Lp Lp 5

Qm Lm 2 = a b Qp Lp Qm Here

5

Lm 2 = Qp a b Lp

Lm 1 = and Qp = 8000 ft3 >s, then Lp 20 5

Qm

1 2 = ( 8000 ft >s ) a b = 4.47 ft3 >s 20 3

Ans.

The height over the dam is

3

Q = CD 2g LH 2;

3

8000 ft3 >s = 0.71232.2(60 ft) ( H p2 )

Hp = 10.31 ft

Here Hm Lm = ; Hp Lp

Hm 1 = 10.31 ft 20 Ans.

Hm = 0.515 ft

Ans: Qm = 4.47 ft3 >s Hm = 0.515 ft

774

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8–65. 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>1log10Re2 2.58 for 106 6 Re 6 109. Take r = 1000 kg>m3 and n = 1.00110 - 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

Since g is a constant,

V 2gL

b

Vp Vm

Here,

= a

m

= a

Vp = a

Lp

2gL

Lp Lm Lp Lm

= 60 and Vm = 0.5 m>s . Then

Lm

V

b

b

p

1 2

1 2

b Vm

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) V mL m = = 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 log10 ( Re ) m 4 2.58

=

0.455

3 log101.5 ( 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

775

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8–65. 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 ) °

b a b a 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 V pL p

(Re)p =

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 log10 ( Re ) p 4 2.58 Lp

Lm

=

2

0.455

3 log100.6971 ( 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 ) p V p =

3 249.70 ( 103 ) N 4 ( 3.873 m>s )

Ans.

= 967.09 ( 103 ) W = 967 kN

Ans. Ans: Vp = 3.87 m>s (FD)p = 250 kN # W = 967 kN

776

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9–1. Determine the average velocity of the water if the pressure drop from A to B is 100 Pa. The plates are fixed, and the temperature of the water is 20°C.

4 mm

A

B

900 mm

SOLUTION dh = 0 and U = 0. From the table in Appendix A, m = 1.00110-3 2 N # s>m2 dx and r = 998.3 kg>m3 for water at T = 20°C. The average velocity can be determined. Here,

V = = -

a 2 dp 12m dx 10.004 m2 2

1231.00110-3 2 N # s>m2 4

= 0.1481 m>s = 0.148 m>s

a

-100 N>m2 0.9 m

b

Ans.

The Reynolds number is Re =

1998.3 kg>m3 210.1481 m>s210.004 m2 rVa = m 1.00110-3 2 N # s>m2

= 592 6 1400

(OK)

Ans: V = 0.148 m>s 777

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9–2. Determine the flow of the water and the shear stress on each plate if the pressure drop from A to B is 100 Pa. The plates are fixed, and their width is 600 mm, and the temperature of the water is 20°C.

4 mm

A

B

900 mm

SOLUTION dh = 0 and U = 0. From the table in Appendix A, m = 1.00110-3 2 N # s>m2 dx and r = 998.3 kg>m3 for water at T = 20°C. The flow rate can be determined using the equation for volumetric flow between two fixed horizontal plates. Here,

Q = = -

a 3b dp 12m dx 10.004 m2 3 10.6 m2

1231.00110-3 2 N # s>m2 4

a

-100 N>m2 0.9 m

= 0.3556110-3 2 m3 >s = 0.356110-3 2 m3 >s

b

Ans.

To evaluate the Reynolds number, we must first find the average velocity of water flow. a 2 dp V = 12m dx = -

10.004 m2 2

1231.00110-3 2 N # s>m2 4

= 0.1481 m>s

a

-100 N>m2 0.9 m

b

The Reynolds number is Re =

1998.3 kg>m3 210.1481 m>s210.0042 rVa = m 1.00110-3 2 N # s>m2

= 592 6 1400

(OK)

The shear stress on the top and bottom plate, where y = a = 0.004 m and y = 0, respectively, can be determined. tt = tb =

100 N>m2 dp a 0.004 m ay - b = a b a0.004 m b = 0.222 Pa dx 2 0.9 m 2 100 N>m2 dp a 0.004 m ay - b = a b a0 b = 0.222 Pa dx 2 0.9 m 2

Ans. Ans.

Ans: Q = 0.356110-3 2 m3 >s tt = 0.222 Pa, tb = 0.222 Pa 778

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9–3. Glue is applied to the surface of the 200-mm-wide plastic strip 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

( 730 kg>m3 ) c 10 ( 10-3 ) m>s d (0.04 m)

F

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

Ans: F = 0.0258 N 779

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*9–4. The 20-kg uniform plate is released and slides down the inclined plane. If it has a terminal velocity of 2 m>s along the plane, determine the thickness of the oil film under the plate. The plate has a width of 400 mm. Take ro = 900 kg>m3 and mo = 0.0685 N # s>m2.

600 mm

108

SOLUTION When the plate achieves the terminal velocity (constant), it is in equilibrium. Referring to the FBD of the plate shown in Fig. a, R+ ΣFx = 0;

32019.812 N4 sin 10° - Fv = 0

20(9.81) N 10°

Fv = 34.07 N

Thus, the shear stress acting on the plate’s bottom surface is tp =

Fv

Fv 34.07 N = = 141.96 N>m2 Ap 10.6 m210.4 m2

N

We will assume that the steady laminar flow occurs and the oil is incompressible. Here,

(a)

dp dh = 0 and = - sin u and U = ut. Then, dx dx t= t= t=

Umo d a + c 1p + gh2d ay - b a dx 2

Umo a + g1 -sin u2ay - b a 2 Umo a - g ay - b sin u a 2

For this case, u = 10°, t = 141.96 N>m2, y = a, and U = 2 m>s. Then 141.96 N>m2 =

12 m>s210.0685 N # s>m2 2 a

- 1900 kg>m3 219.81 m>s2 2aa -

a b sin 10° 2

766.57a 2 + 141.96a - 0.137 = 0

The Reynolds number is Re =

a = 0.9601110-3 2 m = 0.960 mm

Ans.

1900 kg>m3 212 m>s230.9601110-3 2 m4 rUa = mo 0.0685 N # s>m2 = 25.2 6 1400

(laminar flow OK!)

Ans: a = 0.960 mm 780

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9–5. The plug is pin connected 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(10-3) 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 b (p + gh) = + ga 0x L L 0 - 4 ( 103 ) N>m2

=

+ ( 880 kg>m3 )( 9.81 m>s2 ) a

0.05 m

= -71.367 ( 103 )

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 LA

Q =

L0

=

udA

0.2(10 -3) m

1.1700 ( 106 ) 3 0.2 ( 10-3 ) y - y2 4(0.1p)dy

= 0.36755 ( 106 ) J 0.1 ( 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 43 0.2 ( 10-3 ) m 4 30.5 ( 10-3 ) N # s>m2

= 0.0450 6 1400 (laminar flow)

781

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Ans: Q = 0.490 (10 - 6) m3 >s

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9–6. Plates A and B having a weight of 20 lb and 8 lb, respectively, are connected with a cord and placed on the inclined plane and released. If the oil under the plates has a thickness of 0.02 in., determine the terminal velocity of the plates and the tension in the cord. Both plates have a width of 9 in. Take ro = 1.78 slug>ft3 and mo = 4.18(10-3) lb # s>ft2.

12 in. B

9 in. A 10 8

SOLUTION

8 lb 10°

We will assume that steady laminar flow occurs and the oil is incompressible. Here, dp dh = 0, and = - sin u. dx dx

T

(Fv)B

Umo d a t= + c 1p + gh2 d ay - b a dx 2

NB

Umo a t= + g 1 - sin u2 ay - b a 2

(a)

Umo a - g ay - b sin u a 2

t=

20 lb

1 ft For this case, u = 10° and y = a = 10.02 in.2 a b = 1.6667110-3 2 ft. Then 12 in. t=

10° T

U34.18110-3 2 lb # s>ft2 4 1.6667110-3 2 ft

- 11.78 slug>ft3 2132.2 ft>s2 210.8333110-3 2 ft2 sin 10°

= 12.508u - 0.0082942 lb>ft2

(Fv)A NA (b)

Thus, the shear forces acting on the plate’s bottom surfaces are 1Fv 2 A = tAA = 312.508u - 0.0082942 lb>ft2 4 a 1Fv 2 B = tAB = 312.508u - 0.0082942 lb>ft2 4 a

9 9 ft b a ft b = 11.41075u - 0.00466542 lb 12 12

12 9 ft b a ft b = 11.881u - 0.00622052 lb 12 12

Since the plates achieve the terminal velocity (constant), they are in equilibrium, Referring to the FBD of plates B and A shown in Fig. a and b, respectively, R+ ΣFx = 0; R+ ΣFx = 0; Solving Eqs. (1) and (2),

T + 18 sin 10°2 lb - 11.881u - 0.00622052 lb = 0 T = 11.881u - 1.395412 lb

(1)

20 sin 10° lb - 11.41075u - 0.00466542 lb - T = 0 T = 13.47763 - 1.41075u2 lb

(2)

T = 1.389 lb = 1.39 lb

Ans.

U = 1.480 ft>s = 1.48 ft>s

Ans.

The maximum Reynolds number is Re =

11.78 slug>ft3 211.480 ft>s231.6667110-3 2 ft4 rUa = mo 4.18110-3 2 lb # s>ft2

= 1.05 6 1400

(O.K.)

Note: Normally, the average velocity is used to calculate the Reynolds number.

Ans: T = 1.39 lb, U = 1.48 ft>s

782

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9–7. 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 0 - 19.587 ( 103 ) N>m2 N>m2 dp pout - pin = = = -195.87 ( 103 ) dx L 0.1 m m Q = -

3 0.1 ( 10-3 ) m 4 3(0.1 m) N>m2 a 3b dp = J -195.87 ( 10-3 ) R -3 2 12m dx m 12 3 1.00 ( 10 ) N # s>m 4

= 1.632 ( 10-6 ) m3 >s = 1.63 ( 10-6 ) m3 >s

Ans.

The average velocity is V =

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)

= 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)

783

M09_HIBB9290_01_SE_C09_ANS.indd 783

Ans: Q = 1.63(10-6) m3 >s

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*9–8. A solar water heater mounted on the roof consists of two flat plates. Water enters at A and exits at B. If the gap between the plates is a = 1.5 mm, determine the maximum allowable pressure drop from A to B so that the flow remains laminar. Assume the average temperature of the water is 45°C.

A 2m a

B

58

SOLUTION

From the table in Appendix A, rw = 990.2 kg>m3 and mw = 0.599110-3 2 N # s>m2 at T = 45°C. It is required that steady laminar flow occurs and the water is incompressible. Then, the velocity profile of the flow can be determined. u = Here, U = 0,

U 1 d y c 1 p + gh2 d1ay - y2 2 a 2m dx

dp ∆p dh = and = -sin u. Then this equations becomes dx L dx u = -

1 - ∆p c + g1 - sin u2 d1ay - y2 2 2m L

1 ∆p a + g sin u b1ay - y2 2 2m L

=

The flow rate can be determined from Q = = =

LA

a

U dA =

1 ∆p a 1ay - y2 21b dy2 + g sin u b 2m L L0

y3 a b ∆p 1 a + g sin u b a ay2 b` 2m L 2 3 0 a 3b ∆p a + g sin u b 12m L

Realising that A = ab, then the average velocity is given by V =

Q a 2 ∆p = a + g sin u b A 12m L

Here, a = 1.5 mm = 1.5(10-3) m, L = 2 m and u = 5°. Then V =

31.5110-3 2 m4 2

∆p c + 1990.2 kg>m3 219.81 m>s2 2 sin 5° d 1230.599110-3 2 N # s>m2 4 2 m

= 30.1565110-3 2∆p + 0.26504 m>s

To satisfy the condition of laminar flow, we require Re … 1400 rwVa … 1400 mw 1990.2 kg>m3 230.1565110-3 2∆p + 0.2650431.5110-3 2 m4 0.599110-3 2 N # s>m2

… 1400

0.1565110-3 2∆p + 0.2650 … 0.5646

∆p … 1.9141103 2 Pa

∆p … 1.91 kPa

Ans.

Ans: ∆p … 1.91 kPa

784

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9–9. 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, 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. 2p rad 1 min rev Hence, U = vr = J a180 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 no greater than 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: T = 0.326 N # m 785

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9–10. A 2-mm-wide separation forms between the window frame and the wall of a building. If the difference in pressure between the inside and the outside of the building is 4 Pa, determine the flow of air through the crack. The wall is 150 mm thick and the temperature of the air is 20°C.

0.75 m 2 mm

SOLUTION

From the table in Appendix A, m = 18.1110-6 2 N # s>m2 and r = 1.202 kg>m3

dh for air at T = 20°C. For this case, = 0 and U = 0. Then, the flow rate can be dx determined. Q = = -

a 3b dp 12 m dx 10.002 m2 3 10.75 m2

12318.1110 2 N # s>m 4 -6

2

a

- 4 N>m2 0.15 m

b

= 0.7366110–3 2 m3>s = 0.737110–3 2 m3>s

Ans.

To evaluate the Reynolds number, we must first find the velocity of the air. V = -

- 4 N>m2 10.002 m2 2 a 2 dp = a b = 0.4911 m>s 0.15 m 12 m dx 12318.1110-6 2 N # s>m2 4

Thus, the Reynolds number is Re =

11.202 kg>m3 210.4911 m>s230.002 m4 rVa = m 18.1110-6 2 N # s>m2 = 65.2 6 1400

786

M09_HIBB9290_01_SE_C09_ANS.indd 786

(O.K)

Ans: Q = 0.737110-3 2 m3 >s

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9–11. 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, d+ ΣFx = max ;

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

dp dh = 0 and = - sin 2°. dx dx

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

N (a)

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 Reynolds number is no greater than 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: U = 4.88 m>s 787

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*9–12. 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. U′t m tt = = a′t tb =

( 3 mm>s - Vp ) m 0.8 mm

=

Vp

0.8

V pm V pm Ubm = = ab 0.5 mm 0.5

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

0.8

a´t = 0.8 mm Plate

ab = 0.5 mm

Substituting, =

3 m m s – Vp

ttA - tbA = 0 tt = tb

( 3 - Vp ) m

3 mm s

( 3 - Vp ) m

Vp m 0.5

Vp

Ans.

Vp = 1.154 mm>s = 1.15 mm>s

(a) a=0

Thus, the maximum Reynolds number for the top and bottom layers is 900 kg>m3 3 3 ( 10-3 ) m>s 43 0.8 ( 10-3 ) m 4 rUa Re = = 0.0320 6 1400 = m ( 0.0675 N # s>m2 )

τt A

(OK) τb A (b)

and Re =

x

900 kg>m3 31.154 ( 10-3 ) m>s430.5 ( 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: Vp = 1.15 mm>s 788

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9–13. 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 mean total 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. u = = = -

r

1 0 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

u dA =

A

L 0

-

∆r (ay - y2)(w dy) 2mL a

w∆p = (ay - y2)dy 2mL L 0

= = -

y3 a w∆p ay2 a b` 2mL 2 3 0 wa 3 ∆p 12mL

(1)

The power required can be determined from . W = FV Since F = ∆pA and Q = VA,

#

W = ∆pAV = QAp

(2)

789

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9–13. (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: # W = 0.0153 W 790

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9–14. The flat belts are moving at the speeds shown. Plot both the velocity profile within the oil film and the shearstress distribution. The pressures at A and B are atmospheric. Take mo = 0.45 N # s>m2 and ro = 920 kg>m3.

4 mm

0.6 m!s

A

B 0.2 m!s

SOLUTION Assume that steady laminar flow occurs, and the oil is incompressible. Since the flow dp dh is horizontal a = 0b and no pressure change between A and B a = 0b . Then dx dx d 1p + gh2 = 0 dx

Thus,

u =

and u2 = As a result,

C1 1a2 + u1 mo t = uo

and

1 0

C1 =

(a)

mo 1u2 - u1 2 a

y (10–3) m 5

mo du = C1 = 1u2 - u1 2 dy a

(1)

u = c

4110-3 2 m

2

(2)

d 10.6 m>s - 0.2 m>s2 = 45.0 N>m2

0.6 m>s - 0.2 m>s 4110-3 2 m

u = 1100y + 0.22 m>s

4 3

Here, u1 = 0.2 m>s and u2 = 0.6 m>s. Substitute these values into Eqs. 1 and 2. 0.45 N # s>m2

t (Nym2)

45.0

C2 = u1

u2 - u1 u = a by + u1 a

t= c

4

2

C1 y + C2 mo

C1 102 + C2 mo

5

3

Applying the boundary conditions, u = u1 at y = 0 and u = u2 at y = a, u1 =

y (10–3) m

d y + 0.2 m>s

Ans.

1 0

0.2

0.4

0.6

t (Nym2)

(b)

Ans.

The plots of the shear-stress distribution and the velocity profile are shown in Fig. a and b, respectively. The maximum Reynolds number is Re =

1920 kg>m3 210.6 m>s234(10-3) m4 roVmax a = mo 0.45 N # s>m2 (OK)

= 4.91 6 1400

791

M09_HIBB9290_01_SE_C09_ANS.indd 791

Ans: t = 45.0 N>m2, u = 1100 y + 0.22 m>s

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9–15. A 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>ft3 and mw = 5.46(10-6) lb # s>ft2.

0.5 ft!s

2 ft

1 — in. 16

SOLUTION 3 in.

The Reynolds number is rwVa Re = = mw

( 1.820 slug>ft3 )( 0.5 ft>s ) a

0.0625 ft b 12

5.46 ( 10-6 ) lb # s>ft2

= 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>ft3 )( 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>ft2 4

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

u dA = - 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 -

- y2 4 (0.25 dy)

-3 y3 5.2083(10 ) ft d2 2 0

= -0.5391 ( 10-3 ) (0.5∆p - 58.604)

792

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9–15. (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 b a ft b 12 12

∆p = ( 114.79 lb>ft2 ) 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: ∆p = 0.797 psi 793

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*9–16. 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>(2μ)4(2hy - y2), 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 y = w = 0. Also, the liquid is incompressible. Thus, the countinuity equation reduces to 0(ru) 0(ry) 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 + y + w b = rgx + ma 2 + 2 + 2b 0t 0x 0y 0z 0x 0x 0y 0z r(0 + 0 + 0 + 0) = rg sin u - 0 + ma0 +

d 2u + 0b dy2

rg sin u 02u = m 0y2 ra

0p 0y 0y 0y 0y 02y 02y 02y + m + v + w b = rgy + ma 2 + 2 + 2b 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 = 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 + C 2m

(4)

794

M09_HIBB9290_01_SE_C09_ANS.indd 794

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9–16. (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 2 rgh sin u y + y m 2m

rg sin u ( 2hy - y2 ) 2m

(Q.E.D.)

795

M09_HIBB9290_01_SE_C09_ANS.indd 795

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9–17. A fluid has laminar flow between the two parallel plates, each plate 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

Ut

a

B x Ub

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) Using this result and the requirement that the pressure p remain constant along the x axis, the Navier-Stokes equation along the x axis 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 0 + 0 + 0 + 0 = 0 - 0 + ma0 +

m

02u + 0b 0y2

02u = 0 0y2

02u d 2u Since u is a function of y only, 2 = . Integrate this equation with respect to y 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

C2 = Ub

and u = Ut at y = a, ut = C1(a) + Ub C1 =

Ut - Ub a

Substituting these results into Eq. 2, u = a

Ut - Ub by + Ub a

Ans. 796

M09_HIBB9290_01_SE_C09_ANS.indd 796

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9–17. (Continued)

Applying, txy = ma

Here,

0u 0n + b 0y 0x

Ut - Ub 0v du = 0 and from Eq. (1), . Then = C1 = 0x dy a txy = mc a

m ( Ut - Ub ) Ut - Ub b + 0d = a a

Ans.

The plots 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: Ut - Ub by + Ub a m(Ut - Ub)

u = a txy =

a

797

M09_HIBB9290_01_SE_C09_ANS.indd 797

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9–18. 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 43 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.

Ans: laminar 798

M09_HIBB9290_01_SE_C09_ANS.indd 798

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9–19. Glycerin flows through a 50-mm-diameter horizontal pipe such that the pressure at A is pA = 125 kPa, and at B it is pB = 110 kPa. Determine the shear stress that the glycerin exerts on the wall of the pipe.

A

B 3m

SOLUTION Assume glycerin is incompressible. The shear stress on the pipe’s wall can be determined by applying Eq. 9–16 with r = R regardless of whether the flow is dh laminar or turbulent. For horizontal pipe, = 0. Then dx t= Here, R =

D . Then 2 t=

R ∆p b a 2 L

D ∆p a b 4 L

= a

3 2 3 2 0.05 m 125110 2 N>m - 110110 2 N>m bc d 4 3m

= 62.5 N>m2

Ans.

Ans: t = 62.5 N>m2 799

M09_HIBB9290_01_SE_C09_ANS.indd 799

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*9–20. 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 umax = =

D2 ∆p a b 16m L

(0.1 m)2

16 ( 0.985 N # s>m2 )

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

Ans.

Ans.

The Reynolds number is Re =

ro ( 21 umax ) D roVD = = mo mo

( 960 kg>m3 )1 12 2( 6.345 m>s ) (0.1 m) 0.985 N # s>m2

= 309 6 2300 (laminar flow)

Note: Normally the average velocity is used to calculate the Reynolds number.

Ans: umax = 6.35 m>s tmax = 250 N>m2 800

M09_HIBB9290_01_SE_C09_ANS.indd 800

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9–21. 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

(V)max = 0.3038 m>s = 0.304 m>s

Ans.

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.

Ans: 1V2 max = 0.304 m>s, turbulence will not occur. 801

M09_HIBB9290_01_SE_C09_ANS.indd 801

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9–22. The 3-in.-diameter pipe carries gasoline at a temperature of 68°F. If the pressure drops 3 psi within the 8 ft length, determine the shear stress acting along the wall of the pipe.

8 ft 3 in.

SOLUTION Assume gasoline is incompressible. The shear stress on the pipe’s wall can be determined by applying Eq. 9–16 with r = R regardless of whether the flow is laminar or turbulent. Then this equation becomes t=

R dp dh a + g b 2 dx dx

Since the flow is vertically downward,

dp ∆p dh D = - 1. Also, = , and R = . From dx dx L 2

the table in Appendix A, rg = 1.41 slug>ft3. t= `

D ∆p a - gb ` 4 L

3 lb 12 in. 2 ft b - a3 2 b a b † 12 1 ft §† £ in = - 11.41 slug>ft3 2132.2 ft>s2 2 4 8 ft a

= 6.213 lb>ft2 = 6.21 lb>ft2

Ans.

Ans: t = 6.21 lb>ft2 802

M09_HIBB9290_01_SE_C09_ANS.indd 802

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9–23. The 100-mm-diameter oil pipe is used to transport crude oil at a temperature of 20°C. If the pressure drop over the 8-m length is 500 Pa, determine the average velocity of the flow.

100 mm

A

B

8m

SOLUTION Assume that the flow is steady laminar flow and crude oil is incompressible. Since the pipe is horizontal, where

dh = 0, the average velocity can be determined. dx V =

∆p D2 b a 32mco L

From the table in Appendix A rco = 880 kg>m3 and mco = 30.2110 - 3 2 N # s>m2 for crude oil at T = 20°C. Thus, V =

10.1 m2 2

32330.2110-3 2 N # s>m2 4

= 0.6467 m>s = 0.647 m>s

a

500 N>m2 8m

b

Ans.

The Reynolds number is Re =

1880 kg>m3 210.6467 m>s210.1 m2 rcoVD = mco 30.2110-3 2 N # s>m2 = 1885 6 2300

(OK!)

Ans: V = 0.647 m>s 803

M09_HIBB9290_01_SE_C09_ANS.indd 803

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*9–24. The 100-mm-diameter pipe is used to transport crude oil at a temperature of 20°C. What is the maximum allowable pressure drop if the flow is to be laminar over the 8-m length of pipe?

100 mm

A

B

8m

SOLUTION Assume crude oil is incompressible. From the table in Appendix A, rco = 880 kg>m3 and mo = 30.2110 - 3 2 N # s>m2 for crude oil at T = 20°C. To maintain the steady laminar flow in the pipe, Re … 2300

rcoVD … 2300 mco 1880 kg>m3 21V210.1 m2 30.2110-3 2 N # s>m2

Since the pipe is horizontal, where

32330.2110

-3

V … 0.7893 m>s

dh = 0, the average velocity can be determined. dx

V = (0.1 m)2

… 2300

D2 ∆p a b … 0.7893 m>s 32m L ∆p

a b 2 N # s>m2 4 8 m

… 0.7893 m>s

∆p … 610.24 Pa ∆p … 610 Pa

Ans.

Ans: ∆p … 610 Pa 804

M09_HIBB9290_01_SE_C09_ANS.indd 804

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9–25. SAE 10W-30 oil flows through the 125-mm-diameter pipe with an average velocity of 2 m>s. Determine the drop in pressure caused by viscous friction over the 10-m-long section. Take ro = 920 kg>m3 and mo = 0.2 N # s>m2.

A

B

10 m

SOLUTION The flow is assumed to be steady laminar and the SAE 10W-30 oil is incompressible. Since the pipe is smooth and is along the horizontal, the drop in pressure can be determined. V =

D2 ∆p b; a 32 mo L

2 m>s =

10.125 m2 2

∆p a b 3210.2 N # s>m2 2 10 m

∆p = 8.192(103) Pa = 8.19 kPa

Ans.

The Reynolds number is Re =

1920 kg>m3 212 m>s210.125 m2 roVD = mo 0.2 N # s>m2 (OK!)

= 1150 6 2300

Ans: ∆p = 8.19 kPa 805

M09_HIBB9290_01_SE_C09_ANS.indd 805

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9–26. Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar. The 125-mm-diameter smooth pipe is transporting SAE 10W-30 oil with ro = 920 kg>m3 and mo = 0.2 N # s>m2.

A

B

10 m

SOLUTION SAE 10W-30 oil is incompressible. The flow is required to remain laminar. Thus, Re … 2300 roVD … 2300 mo 1920 kg>m3 21V210.125 m2 0.2 N # s>m2

… 2300

V … 4 m>s Since the pipe is smooth and is along the horizontal, the drop in pressure can be determined. V = 10.125 m2 2

D2 ∆p a b … 4 m>s 32mo L

∆p a b … 4 m>s 3210.2 N # s>m2 2 10 m

∆p … 16.384(103) Pa

∆p … 16. 4 kPa

Ans.

Ans: ∆p … 16.4 kPa 806

M09_HIBB9290_01_SE_C09_ANS.indd 806

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9–27. 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 0.8-mm-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 Assume that steady laminar flow occurs and lymph is incompressible. Here, the flow is vertically upwards, thus, hh - hf ph - pf 0 ( p + gh) = + ga b 0x L L =

( 13 550 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 Applying,

N>m2

1.2 m - 0 b 1.2 m

m

V = -

R2 0 ( p + gh) 8ml 0x

= -

3 0.4 ( 10-3 ) m 4 2

8 ( 0.0016 N # s>m2 )

° -418.97

N>m2 m

¢

= 5.237 ( 10-3 ) m>s = 5.24 mm>s

Ans.

The Reynolds number is Re =

riVD = ml

( 1030 kg>m3 )3 52.37 ( 10-12 ) m>s 43 0.8 ( 10-3 ) m 4 0.0016 N # s>m2

= 27 ( 10-9 ) 6 2300 (laminar flow)

Ans: V = 5.24 mm>s 807

M09_HIBB9290_01_SE_C09_ANS.indd 807

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*9–28. Gasoline at a temperature of 20°C flows through the 3-in.-diameter pipe at the rate of 0.002 ft3 >s. Determine the pressure drop that occurs along a 100-ft length.

SOLUTION The gasoline is incompressible. The average velocity of the flow is given by V =

0.002 ft3 >s Q 0.128 = = ft>s p A 1.5 pa ft b 12

From the table in Appendix A, rg = 1.41 slug>ft3 and mg = 6.62(10-6) lb # s>ft2 for gasoline at 20°C. Thus, the Reynolds number is

Re =

rgVD mg

=

11.41 slug>ft3 2a

0.128 3 ft>s b a ft b p 12

6.62110-6 2 lb # s>ft2

= 2170 6 2300 (laminar)

Since the steady laminar flow occurs, and the flow is horizontal, the pressure drop can be determined. ∆p pD Q = a b; 128mo L 4

0.002 ft3 >s =

pa

4 3 ft b 12

12836.62110

-6

∆p = 0.0138 lb>ft2

∆p

a b 2 lb # ft>s2 4 100 ft

Ans.

Ans: ∆p = 0.0138 lb>ft2 808

M09_HIBB9290_01_SE_C09_ANS.indd 808

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9–29. Determine the shear stress along the wall of a 1-ft-diameter horizontal air duct if the air is at a temperature of 80°F and the average velocity of the flow is 0.35 ft>s.

SOLUTION Assume that air is incompressible. From the table in Appendix A, ra = 0.00228 slug>ft3 and ma = 0.385(10-6) lb # s>ft2. The Reynolds number is Re =

=

raVD ma 10.00228 slug>ft3 210.35 ft>s211 ft2 0.385110-6 2 lb # s>ft2

= 2073 6 2300 1laminar2

Since the steady laminar flow occurs and the duct is horizontal, the drop in pressure can be determined. V =

D2 ∆p a b; 32m L

0.35 ft>s =

11 ft2 2

3230.385110

-6

∆p

a 2 lb # s>ft2 4 L

lb>ft2 ∆p = 4.312(10-6) L ft

b

Subsequently, the shear stress on the duct’s wall, which is also the maximum, can be dh = 0. Also, at the wall, r = R = 0.5 ft determined. Since the duct is horizontal, dx dp ∆p and = . Thus, dx L t=

R ∆p a b 2 L

= a

lb>ft2 0.5 ft b c 4.312110-6 2 d 2 ft

= 1.078110-6 2 lb>ft2 = 1.08110-6 2 lb>ft2

Ans.

809

M09_HIBB9290_01_SE_C09_ANS.indd 809

Ans: t = 1.08110-6 2 lb>ft2

09/03/17 3:16 PM

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9–30. As glycerin flows through a horizontal pipe, the pressure drops 3 psi in a 100-ft length. Determine the shear stress within the glycerin at a distance of 1 in. from the pipe wall. Also find the shear stress along the centerline of the pipe and the volumetric discharge. Take rg = 2.44 slug>ft3 and mg = 31.3(10-3) lb # s>ft2.

6 in.

SOLUTION The glycerin is incompressible. Since the pipe is horizontal, r 0p r ∆p a b = a b 2 0x 2 L

t= Here, ∆p = L

t$ r = 2 in. = a

lb 12 in. 2 b a b lb>ft2 1 ft in2 = 4.32 100 ft ft

a3

For r = 3 in. – 1 in. = 2 in. =

0h = 0. Thus, 0x

1 ft, 6

1>6 ft 2

For r = 0 (centerline),

ba4.32

lb>ft2 ft

b = 0.360 lb>ft2

Ans.

Ans.

t$ r = 0 = 0

The result obtained is valid regardless of whether the flow is laminar or turbulent. However, to determine the average velocity, we assume laminar flow occurs. V =

=

∆p D2 a b 32mgy L a

2 6 ft b 12

32331.3(10-3) lb # s>ft2 4

= 1.078 ft>s

a4.3

lb>ft2 ft

b

Thus, the discharge is Q = VA = 11.078 ft>s2 c p a

The Reynolds number is

Re =

rgyVD mgy

=

2 3 ft b d = 0.2117 ft3 >s = 0.212 ft3 >s 12

12.44 slug>ft3 211.078 ft>s2a 31.3110-3 2 lb # s>ft2

Ans.

6 ft b 12 (O.K!)

= 42.0 6 2300

Ans: t$ r = 2 in. = 0.360 lb>ft2, t0 r = 0 = 0 810

M09_HIBB9290_01_SE_C09_ANS.indd 810

Q = 0.212 ft3 >s

09/03/17 3:16 PM

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9–31. The man blows air through the 3-mm-diameter straw with a velocity of 3 m>s. Determine the force his lips exert on the straw to hold it in place. Assume fully developed flow occurs along the straw. The temperature of the air is 20°C.

225 mm

SOLUTION Assume the air is incompressible. From the table in Appendix A, ra = 1.202 kg>m3 and ma = 18.1(10-6) N # s>m2 at T = 20°C. Thus, the Reynolds number is Re =

11.202 kg>m3 213 m>s233110-3 2 m4 raVD = ma 18.1110-6 2 N # s>m2 = 598 6 2300 (laminar flow)

Since the flow is laminar and the straw is horizontal, the pressure drop can be determined. V =

33110-3 2 m4 2 ∆p D2 ∆p a b a b ; 3 m>s = -6 2 # 32ma L 32318.1110 2 N s>m 4 0.225 m ∆p = 43.44 Pa

0p ∆p 0h = Thus, the shear stress on the straw’s wall can be determined. Here, = 0, 0x L 0x and r = R. Then t=

1.5110-3 2 m 43.44 N>m2 R ∆p a b = c da b = 0.1448 N>m2 2 L 2 0.225 m tA F (a)

Consider the force equilibrium along the horizontal of the FBD of the straw in Fig. a. + ΣFx = 0; S

10.1448 N>m2 252p31.5110-3 2 m410.225 m26 - F = 0

F = 0.3071110-3 2 N = 0.307110-3 2 N

Ans.

Ans:

811

M09_HIBB9290_01_SE_C09_ANS.indd 811

F = 0.307110-3 2 N

09/03/17 3:16 PM

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*9–32. Glycerin at 20°C is ejected through the 50-mmdiameter pipe. Determine the volumetric flow if the pressure drop from A to B is 60.5 kPa. Also plot the shear stress distribution within the glycerin.

50 mm B

4m

A

SOLUTION The glycerin is incompressible. From the table in Appendix A, rgy = 1260 kg>m3 and mgy = 1.50 N # s>m2 at T = 20°C. The positive x axis is in the direction of the 0h flow, which is vertically upward. Thus, = 1. Then 0x t= t=

r 0 1p + gh2 2 0x

x

0.025 m

0.025 m

3 2 r 60.5110 2 N>m c+ 11260 kg>m3 219.81 m>s2 2112 d 2 4m

tmax 5 34.6 Pa

= 1 - 1382.2r2 Pa

The maximum shear stress occurs at the pipe’s wall, where r = R = 0.025 m. Then tmax = -1382.210.0252 = -34.6 Pa The above result is valid regardless of whether the flow is laminar or turbulent. The plot of shear-stress distribution is shown in Fig. a. To determine the flow rate, we will assume the flow is steady laminar. Q = = -

pR4 d 1p + gh2 8mgy dx p10.025 m2 4

811.50 N # s>m2 2

c

- 60.51103 2 N>m2 4m

(a)

+ 11260 kg>m3 219.81 m>s2 2112 d

= 0.2827110-3 2 m3 >s = 0.283110-3 2 m3 >s

Ans.

The average velocity is

V =

0.2827110-3 2 m3 >s Q = = 0.1440 m>s A p10.025 m2 2

The Reynold number is Re =

rgyVD mgy

=

11260 kg>m3 210.1440 m>s210.05 m2 1.50 N # s>m2

(O.K.)

= 6.05 6 2300

Ans:

812

M09_HIBB9290_01_SE_C09_ANS.indd 812

Q = 0.283 ( 10 - 3 ) m3 >s

09/03/17 3:16 PM

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9–33. Glycerin is at a pressure of 15 kPa at A when it enters 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 = = -

The average velocity

+ ( 1260 kg>m3 )( 9.81m>s2 ) a

N>m2 m

4

pR 0 (p + rh) 8mGC 0x p(0.05 m)4

8 ( 1.50 N # s>m2 )

(1)

c -16.111 ( 103 )

N>m2 m

= 0.02636 m3 >s = 0.0264 m3 >s V =

0 - 4m b 4m

d

Ans.

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)

Ans:

813

M09_HIBB9290_01_SE_C09_ANS.indd 813

Q = 0.0264 m3 >s

09/03/17 3:16 PM

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9–34. 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 wall of the pipe, 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. 12 m

B

308 A

SOLUTION Assuming that steady laminar flow occurs, and the castor oil is incompressible. Here, pB - pA hB - hA 0 b (p + gh) = + ga 0x L L =

200 ( 103 ) N>m2 - 550 ( 103 ) N>m2 12 m

= - 24.458 ( 103 )

N>m2

+ ( 960 kg>m3 )( 9.81 m>s2 ) a

12 m sin 30° - 0 b 12 m

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 umax = -

= -

R2 0 (p + gh) 4m 0x (0.015 m)2

4 ( 0.985 N # s>m2 )

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.

814

M09_HIBB9290_01_SE_C09_ANS.indd 814

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9–34.

(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)

Ans: t = 183 N>m2 umax = 1.40 m>s

815

M09_HIBB9290_01_SE_C09_ANS.indd 815

Q = 0.494(10-3) m3 >s

09/03/17 3:16 PM

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9–35. 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 the 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 (p + gh) = + ga b 0x L L =

- 36.5 ( 10-3 ) N>m2 4m

= - 492.2 t=

N>m2

x

+ ( 880 kg>m3 )( 9.81 m>s2 )a

4m - 0 b 4m

0.025 m

0.025 m

m

tmax 5 34.6 Pa

r 0 r (p + gh) = a - 492.2 2 0x 2

N>m2 m

b = ( -246.1r) N>m2

The maximum shear stress occurs at r = 0.025 m. Then

Ans.

tmax = -246.1(0.025) = 6.15 Pa

(a)

The plot of shear-stress distribution is shown in Fig. a.

umax = -

= -

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

= 2.547 m>s = 2.55 m>s umax = 2V;

Ans.

2.547 m>s = 2V

V = 1.273 m>s 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 umax = 2.55 m>s 816

M09_HIBB9290_01_SE_C09_ANS.indd 816

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*9–36. Oil and kerosene are brought together through the wye. 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 r dV + 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

Since the flow is laminar, not turbulent, no mixingwill occur.

Ans.

Ans: no mixing 817

M09_HIBB9290_01_SE_C09_ANS.indd 817

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9–37. 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 volumetric flow. Take ro = 880 kg>m3 and mo = 30.2110-3 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 umax = -

= -

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 Reynolds number is

Re =

r ( 12umax ) D rVD = = m m

( 880 kg>m3 )1 12 2( 0.8651 m>s ) (0.05 m) = 630 6 2300 30.2 ( 10-3 ) N # s>m2

(OK)

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.

Ans:

818

M09_HIBB9290_01_SE_C09_ANS.indd 818

Q = 0.849(10-3) m3 >s

09/03/17 3:16 PM

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9–38. 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 = rogh = ( 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) = + go a b 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) 8mo 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

Ans.

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 =

roVD = mo

( 960 kg>m3 )(0.1120 m>s)(0.015 m) 0.985 N # s>m2

= 1.64 6 2300 (laminar flow)

Ans: t = 3.17 s 819

M09_HIBB9290_01_SE_C09_ANS.indd 819

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9–39. 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

15 mm

300 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 the top and bottom of the stem are pt = rogh = ( 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) = + go a b 0x L L =

0 - 1883.52 N>m2 0.3 m

= -15.696 ( 103 )

Q = -

pR4 0 (p + gh); 8mo 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 8mo

Q = 0.970 N # s>m2

c -15.696 ( 103 )

N>m2 m

d

Ans.

The Reynolds number is

Re =

roVD = mo

( 960 kg>m3 )( 0.1138 m>s )(0.015 m) 0.970 N # s>m2

= 1.69 6 2300 (laminar flow)

Ans:

Q = 0.970 N # s>m2

820

M09_HIBB9290_01_SE_C09_ANS.indd 820

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*9–40. The cylindrical tank is to be filled with crude oil at T = 20°C using the 2-in.-diameter supply pipe. If the flow is to be laminar, determine the shortest time needed to fill the tank from h = 0 to h = 6 ft. Also, what pressure gradient should be maintained to achieve this task? Air escapes through the top of the tank.

6 ft

h 2 in.

SOLUTION The crude oil is incompressible. From the table in Appendix A, rco = 1.71 slug>ft3 and mco = 0.632110 - 3 2 lb # s>ft2 for crude oil at T = 20°C. Since it is required to fill the tank in the shortest time possible, it must be filled with the greatest velocity, yet maintaining laminar flow. This requires Re = 2300 rcoVD = 2300 mco 11.71 slug>ft3 21V2a

2 ft b 12

0.632110-3 2 lb # s>ft2

= 2300

V = 5.1004 ft>s

Subsequently, the flow rate can be determined from Q = VA = 15.1004 ft>s2 c pa

2 1 ft b d = 0.1113 ft3 >s 12

Thus, the shortest time needed to fill the tank is t=

p13 ft2 2 16 ft2 V = = 1524.6 s = 25.4 min Q 0.1113 ft3 >s

Ans.

The required pressure gradient is given by V =

D2 ∆p a b 32mco L

3230.632110-3 2 lb # s>ft2 415.1004 ft>s2 ∆p 32mcoV = = 2 L D2 2 a ft b 12 = 3.713

lb>ft2 ft

= 3.71

lb>ft2

Ans.

ft

Ans: t = 25.4 min lb>ft2 ∆p = 3.71 L ft 821

M09_HIBB9290_01_SE_C09_ANS.indd 821

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9–41. The cylindrical tank is filled with glycerin at 68°F. When the valve at A is opened, glycerin flows out through the 2-in.-diameter pipe. If h = 6 ft at t = 0, determine the time for h = 2 ft. Assume that laminar flow occurs within the pipe.

3 ft

h

6 ft

A

SOLUTION The glycerin is incompressible. From the table in Appendix A, rGy = 2.44 slug>ft3 and mGy = 31.3110-3 2 lb # s>ft2 for glycerin at T = 68°F. The pressures at the inlet and outlet of the pipe are pin = gh = 12.44 slug>ft3 2132.2 ft>s2 2h = 178.568h2

lb ft2

and pout = patm = 0. Thus, the pressure gradient is

lb>ft2 ∆p 78.568h - 0 = = a13.095h b L 6 ft ft

Since the pipe is horizontal, Q =

Q =

pD4 ∆p a b 128mGy L •

pa

4 2 ft b 12

128331.3110-3 2 lb # s>ft2 4

Q = 37.9229110-3 2h4 ft3 >s

¶

113.095h2

Here, the volume of glycerin in the tank is V = p11.5 ft2 2h = 2.25ph. Since the volume is changing over time, d12.25ph2 dV dh = = 2.25p dt dt dt However, dV = Q dt dh - 2.25p = 7.9229110-3 2h dt dh dt = - 892.17 h Integrating this equation using limits at t = 0, h = 6 ft, and at t, h = 2 ft, -

L0

t

dt = -892.17

2 ft

L6 ft

dh h

6 t = 892.17 ln a b = 980.15 s = 16.3 min 2

Ans.

822

M09_HIBB9290_01_SE_C09_ANS.indd 822

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9–41. (Continued)

The maximum average velocity occurs when h = 6 ft. V =

=

∆p D2 a b 32mGy L •

a

2 2 ft b 12

32331.3110 2 lb # s>ft 4 -3

2

= 2.179 ft>s

¶

c 13.095162

lb>ft2 ft

d

The Reynolds number is

Re =

rGyVD mGy

=

12.44 slug>ft3 212.179 ft>s2a 31.3110-3 2 lb # s>ft2

2 ft b 12 (O.K!)

= 28.3 6 2300

Ans: t = 16.3 min 823

M09_HIBB9290_01_SE_C09_ANS.indd 823

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9–42. If the reading of the mercury manometer is h = 0.25 in., determine the volumetric flow of SAE 10W oil through the pipe. Take gHg = 847 lb>ft3 , go = 57.4 lb>ft3 , and mo = 2.09(10-3) lb # s>ft2.

2 ft

3 in. A

B h

SOLUTION We will assume the flow is laminar and the SAE 10W oil is incompressible. The pressure difference between A and B can be determined by writing the manometer equation between these two points by referring to Fig. a. pA + goh′ - gHg h - go 1h′ - h2 = pB

∆p = pB - pA = 1go - gHg 2h

A

h9 2 h h9

B

h

(a)

0.25 = 157.4 lb>ft3 - 847 lb>ft3 2a ft b 12

= -16.45 lb>ft2

The negative sign indicates the pressure drops from A to B since the pipe is horizontal. Q =

=

pD4 ∆p a b 128mo L •

pa

4 3 ft b 12

2 ¶ 16.45 lb>ft a b 2 ft 12832.09110-3 2 lb # s>ft2 4

= 0.3773 ft3 >s = 0.377 ft3 >s

Ans.

Then the average velocity can be determined from V =

The Reynolds number is

0.3773 ft3 >s Q = = 7.686 ft>s 2 A 1.5 pa ft b 12

roVD Re = = mo

a

57.4 lb>ft3 32.2 ft>s2

b17.686 ft>s2a

2.09110-3 2 lb # s>ft2

3 ft b 12 (OK!)

= 1639 6 2300

Ans:

824

M09_HIBB9290_01_SE_C09_ANS.indd 824

Q = 0.377 ft3 >s

09/03/17 3:16 PM

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9–43. Determine the mass flow of SAE 10W oil through the pipe if the mercury manometer reads h = 0.3 in. Take gHg = 847 lb>ft3, go = 57.4 lb>ft3, and mo = 2.09(10-3) lb # s>ft2.

2 ft

3 in. A

B h

SOLUTION We will assume the flow is laminar and the SAE 10W oil is incompressible. The pressure difference between A and B can be determined by writing the manometer equation between these two points. pA + goh′ - gHgh - go 1h′ - h2 = pB

∆p = pB - pA = 1go - gHg 2h

= a57.4 lb>ft2 - 847 lb>ft2 b a

= - 19.74 lb>ft2

0.3 ft b 12

The negative sign indicates the pressure drops from A to B. Since the pipe is horizontal, the volumetric flow rate can be determined by applying Eq. 9–25. Q =

=

pD4 ∆p a b 128mo L •

pa

4 3 ft b 12

2 ¶ 19.74 lb>ft a b 2 ft 12832.09110-3 2 lb # s>ft2 4

= 0.4528 ft3 >s

Then the average velocity can be determined from V =

The Reynolds number is

Re =

0.4528 ft3 >s Q = = 9.224 ft>s 2 A 1.5 pa ft b 12

roVD = mo

a

57.4 lb>ft3 32.2 ft>s

2

b a9.224 ft>s b a

2.09110-3 2 lb # s>ft2

3 ft b 12 (O.K.)

= 1967 6 2300 Therefore, the mass flow is 3

57.4 lb>ft # m = rQ = a b10.4528 ft3 >s2 = 0.807 slug>s 32.2 ft>s2

Ans.

Ans: # m = 0.807 slug>s 825

M09_HIBB9290_01_SE_C09_ANS.indd 825

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*9–44. The Reynolds number Re = rVDh >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

ri

SOLUTION In this case, A = p ( r o2 - r i 2 ) = p 3 (0.06 m)2 - (0.04 m)2 4 = 0.002p m2 Thus,

P = 2p(ro + 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 )

Ans.

Since Re 7 2300, then the flow is not laminar.

Ans.

Ans: Re = 79.14 ( 103 ) , The flow is not laminar. 826

M09_HIBB9290_01_SE_C09_ANS.indd 826

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9–45. 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 vz =

ro

ri

r o2 - r 2i 1 dp 2 r c r - r o2 - a b ln d . 4m dz ln(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, yr = yu = 0. Applying the continuity equation, 0(ryz) 0r 1 0(rryr) 1 0(ryu) + + + = 0 0t r 0r r 0u 0z 0 + 0 + 0 + r

0yz 0z 0yz 0z

= 0 = 0

Since the flow is steady and symmetric about the z axis, it is independent of time t and u. Thus, integration of the above results gives yz = yz(r), which means yz is a function of r only. Applying the Navier-Stokes equation along the z axis, ra

= -

0yz 0t

+ yr

0yz 0r

+

0yz yu 0yz + yz b r 0u 0z

2 0yz 02yz 0p 1 0 1 0 yz + rgz + mc ar b + 2 2 + d 0t r 0r 0r r 0u 0z2

r(0 + 0 + 0 + 0) = -

0p 1 0 0yz + rgz + mc ar b + 0 + 0 d 0z r 0r 0r

Here, gz = 0. Using this result and rearranging the above equation, 0yz 0 r 0p ar b = a b m 0z 0r 0r Integrate the above equation twice. 0yz 0r

=

yz =

C1 r 0p a b + 2m 0z r

(1)

r 2 0p a b + C1 ln r + C2 4m 0z

(2)

Applying the boundary condition at r = ri, yz = 0 and r = ro, yz = 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)

827

M09_HIBB9290_01_SE_C09_ANS.indd 827

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*9–45.

(Continued)

Solving Eqs. 3 and 4, C1 =

r i2 - ro2 0p b r a 4m ln roi 0z

C2 = -

( r i2 ln ro - ro2 ln ri ) 0p 4m ln

a

ro ri

0z

b

Substitute these results into Eqs. 1 and 2. 0yz 0r

=

=

yz =

r i2 - ro2 0p r 0p a b + b r a 2m 0z 4mr ln roi 0z a

0p b ro2 - r i 2 0z a2r b r 4m r ln roi

r i 2 - ro 2 0p r i 2 ln ro - r o2 ln ri 0p r 2 0p a b + a b ln r a b r r 4m 0z 0t 4m ln roi 0z 4m ln roi

=

a

0p b 0z ar 2 + 4m

=

a

ro 0p r i2lnr - ro2lnr - r i 2lnro + ro2lnri + ro2 aln b b ri 0z ≥ r 2 - ro2 + ¥ r 4m o ln ri

=

a

0p b r i2lnr - ro2lnr - r i2lnro + ro2lnri + ro2lnro - ro2lnri 0z £ r 2 - ro2 + § ro 4m ln ri

=

a

0p b 0z £ r 2 - ro2 + 4m

a

0p b ro2 - r i2 0z r £ r 2 - ro2 - a bln § ro 4m ro ln ri

=

yz =

( r i 2 - ro2 ) lnr ln

ro ri

-

r i2ln ro - ro2 ln ri ln

ro ri

b

( ro2 - r i2 ) lnro - ( ro2 - r i2 ) lnr ln

ro r

§

ro2 - r i2 1 0p 2 r c r - ro2 - a bln d ro 4m 0z r ln ri o

(Q.E.D)

828

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9–46. Water flows from a beaker into a 4-mm-diameter tube with an average velocity of 0.45 m>s. Classify the flow as laminar or turbulent if the water temperature is 10°C and if it is 30°C. If the flow is laminar, then find the length of tube 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 flow 1.31 ( 10-6 ) m2 >s

L′ = 0.06 ReD = 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 flow 0.804 ( 10-6 )

L′ = 0.06 ReD = 0.06(2239)(0.004 m) Ans.

= 0.537 m = 537 mm

Ans: At T = 10°C, L′ = 330 mm At T = 30°C, L′ = 537 mm 829

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9–47. The smooth pipe transports water at 60°F. If the pressure at A and B is 35 psi and 33.5 psi, respectively, determine the thickness of the viscous sublayer and the velocity along the centerline of the pipe.

A 4 in.

B 8 ft

SOLUTION dh = 0, and so the dx shear stress at the pipe’s wall, regardless of whether the flow is laminar or turbulent, can be determined from

Assume the water is incompressible. Since the pipe is horizontal,

t0 =

R ∆p a b = 2 L

a

2 lb lb 12 in. 2 ft b a35 2 - 33.5 2 b a b 12 1 ft £ § in in = 2.25 lb>ft2 2 8 ft

From the table in Appendix A, rw = 1.939 slug>ft3 and nw = 12.2110-6 2 ft2 >s Then u* =

2.25 lb>ft2 to = = 1.0772 ft>s A rw B 1.939 slug>ft3

At centerline of the pipe, y =

u*y = vw

2 ft. 12

11.0772 ft>s2a

2 ft b 12

12.2110-6 2 ft2 >s

= 1.4721104 2 7 30 (turbulent flow)

Thus, the velocity at the centerline of the pipe can be determined by u*y u = 2.5 lna b + 5.0 u* vw

u = 2.5 ln31.4721104 24 + 5.0 1.0772 ft>s

Ans.

u = 31.23 ft>s = 31.2 ft>s

Viscous sublayer extends to

u*y = 5. Thus, vw

y =

12.2110-6 2 ft2 >s 5vw = 5c d u* 1.0772 ft>s = 56.63110-6 2 ft

= 0.680110-3 2 in.

Ans.

830

M09_HIBB9290_01_SE_C09_ANS.indd 830

Ans: u = 31.2 ft>s, y = 0.680110-3 2 in.

09/03/17 3:17 PM

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*9–48. Glycerin at 68°F flows through the 6-in.-diameter smooth pipe. Determine its velocity 2 in. from the wall of the pipe. Also, what is the corresponding shear stress? The pressure drop throughout the pipe length is 0.75 psi>ft. Use Eq. 9–33 if the flow is turbulent.

6 in.

SOLUTION dh = 0, and dx so the shear stress at the pipe’s wall, regardless of whether the flow is laminar or turbulent, can be determined from

Assume the glycerin is incompressible. Since the pipe is horizontal,

to =

R ∆p a b = 2 L

a

3 lb 12 in. 2 ft b a0.75 2 b a b 12 1 ft £ § in = 13.5 lb>ft2 2 1 ft

Thus, the shear stress at r = 3 in. - 2 in. = 1 in. can be determined by proportion, 13.5 lb>ft2 t = ; 1 in. 3 in.

t = 4.5 lb>ft2

Ans.

From the table in Appendix A, rGy = 2.44 slug>ft3 and vGy = 12.8110-3 2 ft2 >s Then u* =

At r = 1 in., y = 2 in. = u*y = vGy

13.5 lb>ft2 to = = 2.352 ft>s B rGy B 2.44 slug>ft3 2 ft. 12

12.352 ft>s2a

2 ft b 12

12.8110-3 2 ft2 >s

= 30.62 7 30 (turbulent)

Thus, the velocity at r = 1 in. can be determined by u*y u = 2.5 ln a b + 5.0 u* vGy

u = 2.5 ln 130.622 + 5.0 2.352 ft>s

Ans.

u = 31.88 ft>s = 31.9 ft>s

Ans: t = 4.5 lb>ft2, u = 31.9 ft>s 831

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9–49. Determine the viscous and turbulent shear stress components within the SAE 10W-30 oil 2 in. from the wall of the smooth pipe. The volumetric flow is 1.85 ft3 >s and the pressure drop throughout the pipe length is 0.12 psi>ft. Use the power law velocity profile, Eq. 9–34, to determine the result. Take ro = 1.78 slug>ft3 and mo = 2.09(10-3) lb # s>ft2.

6 in.

SOLUTION Assume the SAE 10W-30 oil is incompressible. The average velocity of the flow is given by V =

The Reynolds number is roVD Re = = mo

1.85 ft3 >s Q = = 9.422 ft>s 2 A 3 pa ft b 12

11.78 slug>ft3 219.422 ft>s2a 2.09110-3 2 lb # s>ft2

6 ft b 12

= 4012 ≈

4000

From Table 9–1, we can use n = 6. Thus, the maximum velocity can be determined by V = Umax c

2n2 d; 1n + 1212n + 12

9.422 ft>s = Umax e

2162 2

16 + 1232162 + 14

Umax = 11.91 ft>s

f

The viscous shear tvisc = m(du>dy). However, y = R - r. Then dy = –dr. Thus, tvisc = -m(du>dr). tvisc = - m • =

dr

mU max r a1 - b nR R

Thus, at r = 3 in. - 2 in. = 1 in. =

tvisc

dc U max a1 1-n n

r 1>n b d R ¶

1 ft, 12

1 ft b 32.09110-3 2 lb # s>ft2 4111.91 ft>s2 12 = ≥1 ¥ 3 3 6a ft b a ft b 12 12 a

1-6 6

= 0.02326 lb>ft2 = 0.0233 lb>ft2

Ans.

1 ft (turbulent or laminar Since the pipe is horizontal, the total shear stress at r = 12 flow) can be determined from

t=

r ∆p a b = 2 L

a

1 lb 12 in. 2 ft b a0.12 2 b a b 12 1 ft £ § in = 0.72 lb>ft2 2 1 ft

The turbulent shear stress is, therefore,

tturb = t - tvisc = 0.720 lb>ft2 - 0.02326 lb>ft2 = 0.697 lb>ft2

Ans.

Ans: tvisc = 0.0233 lb>ft2, tturb = 0.697 lb>ft2

832

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9–50. SAE 10W-30 oil flows through the 30-mm-diameter smooth pipe. If the pressure at A is 200 kPa and at B it is 170 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 if the flow is turbulent. Take ro = 920 kg>m3 and no = 0.1(10-3) m2 >s.

30 mm A

B

4m

SOLUTION Assume the SAE 10W-30 oil is incompressible. The maximum shear stress occurs at the pipe’s wall. Since the pipe is horizontal, it can be determined, regardless of dh whether the flow laminar or turbulent, with = 0 and r = R. dx to =

3 2 3 2 0.015 m 2001102 N>m - 170110 2 N>m R ∆p a b = a bc d 2 L 2 4m

= 56.25 N>m2

Ans.

Then u* =

56.25 N>m2 to = = 0.2473 m>s A ro B 920 kg>m3

At the centerline of the pipe where maximum velocity occurs, y = 0.015 m. Then 10.2473 m>s210.015 m2 u*y = = 37.09 7 30 (turbulent) vo 0.1110-3 2 m2 >s

Thus, the maximum velocity can be determined by

u*y u = 2.5 ln a b + 5.0 u* vo

umax = 2.5 ln 137.092 + 5.0 0.2473 umax = 3.47 m>s The viscous sublayer extends to

y =

Ans.

u*y = 5. Then vo

530.1110-3 2 m2 >s4 5vo = = 2.022110-3 2 m = 2.02 mm u* 0.2473 m>s

Ans.

Ans: t0 = 56.25 N>m2 umax = 3.47 m>s y = 2.02 mm 833

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9–51. SAE 10W-30 oil flows through the 30-mm-diameter smooth pipe at 0.0095 m3 >s. Determine the velocity within the oil 5 mm from the wall of the pipe. Use the power law velocity profile, Eq. 9–34, to determine the result. Take no = 0.1(10-3) m2 >s.

30 mm A

B

4m

SOLUTION Assume the SAE 10W-30 oil is incompressible. The average velocity of the flow is V =

0.0095 m3 >s Q = = 13.44 m>s A p(0.015 m)2

The Reynolds number is Re =

(13.44 m>s)(0.03 m) VD = = 4032 ≈ Vo 0.1(10-3) m2 >s

4000

From Table 9–1, we can use n = 6. Thus the maximum velocity can be determined by V = Umax c

2n2 d; (n + 1)(2n + 1)

13.44 m>s = Umax e

(6 + 1)32(6) + 14

2(62)

Umax = 16.99 m>s

f

Then, the velocity at r = 0.015 m - 0.005 m = 0.01 m can be determined by u u max

= a1 -

r 1>n b R

u 0.01 m 1>6 = a1 b 16.99 m>s 0.015 m

Ans.

u = 14.14 m>s = 14.1 m>s

Ans: u = 14.1 m>s 834

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*9–52. Gasoline at T = 68°F flows through the 3-in.diameter smooth pipe. If the pressure drops from 40 psi to 38.5 psi along the 10-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 gasoline along the centerline of the pipe? The flow is turbulent. Use Eq. 9–33.

3 in. A

B

10 ft

SOLUTION Assume the gasoline is incompressible. Since the pipe is horizontal, the shear stress can be determined, regardless of whether the flow is laminar or turbulent, with dh 1.5 = 0. At the pipe’s wall where the shear stress is maximum, r = R = ft. dx 12 Then lb 12 in. 2 1.5 c 140 - 38.52 2 d a b ft 1 ft R ∆p ¶ in ° 12 ¢ • to = a b = 2 L 2 10 ft = 1.35 lb>ft2

Ans.

At the centerline of the pipe, r = 0. t$ r

= 0

Ans.

= 0

From the table in Appendix A, rg = 1.41 slug>ft3 and vg = 4.70110-6 2 ft2 >s for gasoline at T = 68°F. Then u* =

to 1.35 lb/ft2 = = 0.9785 ft>s. A rg A 1.41 slug>ft3

At the centerline of the pipe where maximum velocity occurs, y =

u*y vg

=

10.9785 ft>s2a

1.5 ft b 12

4.70110-6 2 ft2 >s

1.5 ft. Then 12

= 26,024 7 30

Thus, the maximum velocity can be determined by u*y u b + 5.0 = 2.5 ln a u* vg

u max = 2.5 ln 126,0242 + 5.0 0.9785 ft>s

Ans.

u max = 29.76 ft>s = 29.8 ft>s

Ans: to = 1.35 lb>ft2 t$ r = 0 = 0 u max = 29.8 ft>s 835

M09_HIBB9290_01_SE_C09_ANS.indd 835

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9–53. Gasoline at T = 68°F flows through the 3-in.diameter smooth pipe. If the pressure drops from 40 psi to 38.5 psi along the 10-ft length, determine the shear stress a distance of 1 in. from the wall of the pipe. What is the thickness of the viscous sublayer?

3 in. A

B

10 ft

SOLUTION Assume the gasoline is incompressible. Since the pipe is horizontal, the shear stress can be determined, regardless of whether the flow is laminar or turbulent, with dh 1.5 = 0. At the pipe’s wall where the maximum shear stress occurs, r = R = ft. dx 12 lb 12 in. 2 1.5 c 140 - 38.52 2 d a b ft 1 ft R ∆p ¶ in ° 12 ¢ • to = a b = 2 L 2 10 ft = 1.35 lb>ft2

The shear stress at r = 1.5 in. - 1 in. = 0.5 in. can be determined by proportion. 1.35 lb>ft2 t = 0.5 in. 1.5 in. t = 0.45 lb>ft2

Ans.

From the table in Appendix A, rg = 1.41 slug>ft3 and rg = 4.70110-6 2 ft2 >s for gasoline at T = 68°F. Then the shear velocity is given by u* =

1.35 lb>ft2 to = = 0.9785 ft>s A rg B 1.41 slug>ft3

The laminar sublayer extends to

y =

5vg u*

u*y = 5. Then vg

=

534.70110-6 2 ft2 >s4 0.9785 ft>s

= 324.017110-6 2 ft4 a

= 0.288110-3 2 in.

12 in. b 1 ft

Ans.

Ans: t = 0.45 lb>ft2, y = 0.288 (10 - 3) in. 836

M09_HIBB9290_01_SE_C09_ANS.indd 836

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9–54. The smooth pipe transports water at 120°F with a maximum velocity of 10 ft>s. Determine the pressure drop along the 20-ft length and the shear stress on the wall of the pipe. Also, what is the thickness of the viscous sublayer? Use Eq. 9–33.

6 in.

20 ft

SOLUTION Assume the water is incompressible. From the table in Appendix A, rw = 1.918 slug>ft3 and vw = 6.14110-6 2 ft2 >s for water at 120°F. Then the maximum 3 velocity occurs at the centerline of the pipe, where y = ft. 12 u*y u = 2.5 ln a b + 5.0 u* vw 10 ft>s u*

= 2.5 ln

£

u*a

3 ft b 12

6.14110-6 2 ft2 >s

§

+ 5.0

10 = u*52.5 ln 340 7161103 2u*4 + 5.06

Solving numerically,

u* = 0.34621 ft>s The shear stress on the pipe’s wall can be determined from u* =

to to ; 0.34621 ft>s = A rw A 1.918 slug>ft3

to = 10.2299 lb>ft2 2a

At the centerline of the pipe, u*y = vw

10.34621 ft>s2a

1 ft 2 b = 0.230 lb>ft2 12 in.

Ans.

3 ft b 12

= 14.11103 2 7 30 (turbulent OK) 6.14110-6 2 ft2 >s dh Since the pipe is horizontal where = 0, with r = R, dx 3 ft R ∆p ° 12 ¢ ∆p 2 b; a b to = a 0.2299 lb>ft = 2 L 2 20 ft ∆p = 36.8 lb>ft2

Ans.

The viscous sublayer can be determined using the upper limit for laminar flow. u*y = 5; vw

10.34621 ft>s2y

6.14110-6 2 ft2 >s

y = 388.67110-6 2 ft4 a

= 5

12 in. b = 0.00106 in. 1 ft

Ans.

Ans: t0 = 0.230 lb>ft2, ∆p = 36.8 lb>ft2 y = 0.00106 in. 837

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9–55. Water at 60°F flows through the smooth pipe at 0.480 ft3 >s. If the pressure drop from A to B is 0.02 psi, determine the viscous and turbulent shear stress components within the water at r = 1.5 in. and r = 3 in. from the centerline of the pipe. Use the power law velocity profile, Eq. 9–34.

A

B 6 in.

12 ft

SOLUTION Assume the water is incompressible. The average velocity of the flow is given by V =

0.480 ft3 >s Q = = 2.445 ft>s 2 A 3 pa ft b 12

From the table in Appendix A, rw = 1.939 slug>ft3 and mw = 23.6110-6 2 lb # s>ft2 for water at 60°F. Then the Reynolds number is rwVD Re = = mw

11.939 slug>ft3 212.445 ft>s2a 23.6110-6 2 lb # s>ft2

6 ft b 12

= 1.0041103 2 ≃

11103 2

From Table 9–1, we can use n = 7. Thus, the maximum velocity can be determined by V = u max c

2172 2 2n2 f d ; 2.445 ft>s = u max e 17 + 1232172 + 14 1n + 1212n + 12 umax = 2.993 ft>s

The viscous shear tvisc = ma tvisc = -ma

du b. dv

# du b . However, y = R - r. Then dy = - dr. Thus, dy 1

tvisc = - m = At r = 1.5 in., tvisc =

•

•

dc u max a1 dr

mu max r a1 - b nR R

1-n n

r n b d R ¶

323.6110-6 2 lb # s>ft2 412.993 ft>s2 7a

3 ft b 12

¶

°1 -

1.5 12 ft ¢ 3 12 ft

1-7 7

= 73.125(10-6) lb>ft2 = 73.1(10-6) lb>ft2

Ans.

1.5 Since the pipe is horizontal, the total shear stress at r = ft (turbulent or laminar) 12 can be determined from lb 12 in. 2 1.5 a0.02 2 b a b ft ft r ∆p § in ° 12 ¢ £ t= a b = = 0.0150 lb>ft2 2 L 2 12 ft 838

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*9–55. (Continued)

The turbulent shear stress is, therefore, tturb = t - tvisc = 0.0150 lb>ft2 - 73.125110-6 2 lb>ft2 = 0.0149 lb>ft2

Ans.

3 ft (at pipe’s wall), Notice that tvisc is insignificant compared to tturb. At r = R = 12 the shear stress is due to tvisc only. Thus,

tvisc

lb 12 in. 2 3 a0.02 2 b a b ft ∆p ft R § in ° 12 ¢ £ = a b = 2 L 2 12 ft = 0.03 lb>ft2

Ans.

And Ans.

tturb = 0

Ans: At r = 1.5 in., tvisc = 73.1110-6 2 lb>ft2

tturb = 0.149 lb>ft2 At r = 3 in., tvisc = 0.03 lb>ft2, tturb = 0

839

M09_HIBB9290_01_SE_C09_ANS.indd 839

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*9–56. A 3-in.-diameter smooth horizontal pipe transports kerosene at 68°F. If the pressure drops 17 lb>ft2 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, is D ∆p t0 = = 4 L

a

3 ft b 17 lb>ft2 12 c d = 0.053125 lb>ft2 4 20 ft

From Appendix A, r = 1.58 slug>ft3 and v = 25.4 ( 10-6 ) ft2 >s. Then, u* =

0.053125 lb>ft2 t0 = = 0.1834 ft>s Ar C 1.58 slug>ft3

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 u* v

( 0.1834 ft>s ) (0.125 ft) u = 2.5 lnc d + 5.0 0.1834 ft>s 25.4 ( 10-6 ) ft2 >s u = 4.04 ft>s

Ans.

The maximum Reynolds number is VD Re = = v

( 4.04 ft>s ) a

25.4 ( 10-6 ) ft2 >s

The viscous sublayer extends to a y =

3 ft b 12

u*y v

= 39.76 ( 103 ) 7 2300

(OK)

b = 5. Thus,

5 3 25.4 ( 10-6 ) ft2 >s 4 5v = = 0.6925 ( 10-3 ) ft = 8.31 ( 10-3 ) in. u* 0.1834 ft>s

Ans.

Note: Normally the average velocity is used to calculate the Reynolds number. Here the centerline velocity gives a value almost 20 times the cutoff of 2300, which is sufficient confirmation.

Ans: u = 4.04 ft>s y = 8.31 ( 10-3 ) in. 840

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9–57. 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 log10 Re - 4.6. If Re = 21109 2, plot the velocity profile over the artery wall, and determine the flow at this instant.

3.4 mm

r

SOLUTION The flow rate of blood can be determined from Q = =

LA L0

u dA >

3.4 mm

= 16.72p

L0

3.4 mm

r a1 -

r 1>n b dr 3.4

1

= 16.72p£ a-

= 16.72pc

r 1n b (2pr dr) 3.4

8.36 a1 -

1 2 1 b a + 2b 3.4 n

11.56n r a1 b 2n + 1 3.4

2n + 1 n

-

n = 193.2832npc d (n + 1)(2n + 1) =

1

r n+2 a1 b 3.4

1

1 a-

1 2 1 b a + 1b 3.4 n

11.56n r a1 b n + 1 3.4

n+1 n

d`

r n+1 a1 b §† 3.4

3.4 mm

0

3.4 mm 0

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 =

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

Ans.

r 0.05955 b d mm>s 3.4

Ans:

841

M09_HIBB9290_01_SE_C09_ANS.indd 841

Q = 278 mm3 >s

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10–1. A 3-in.-diameter pipe supplies water with a velocity of 8 ft>s. If the pipe is made of commercial steel and the water temperature is 50°F, determine the friction factor.

SOLUTION We will consider a fully developed steady flow and water is incompressible. From the table in Appendix A, v = 14.1(10-6) ft2 >s for water at 50°F. The Reynolds number is Re =

(8 ft>s) a

3 ft b 12

VD = = 1.42(105) v 14.1(10-6) ft2 >s

For commercial steel, e = 0.00015 ft. Thus, the relative roughness is e 0.00015 ft = = 0.0006 D 3 ft 12 From the Moody diagram, Ans.

f = 0.0200

We can also compare this value with that obtained by applying the Colebrook equation, 1 2f

= - 2 log a

2f

= - 2 log c

1

Solving numerically,

e>D 3.7

+

2.51 Re2f

b

0.0006 2.51 + d 3.7 1.42(105) 2f

f = 0.0199 ≈ 0.02

Ans: f = 0.0200 842

M10_HIBB9290_01_SE_C10_ANS.indd 842

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10–2. Gasoline at T = 68°F flows through an 8-in.-diameter smooth pipe at the rate of 4.15 ft3 >s. Determine the head loss in a 50-ft-long segment of the pipe.

SOLUTION We will consider the fully developed steady flow, and gasoline is incompressible. The average velocity of the flow is V =

4.15 ft3 >s Q = = 11.89 ft>s 2 A 4 pa ft b 12

From the table in Appendix A, vg = 4.70110-6 2 ft2 >s. Thus, the Reynolds number is VD Re = = vg

111.89 ft>s2 a

8 ft b 12

4.70 110-6 2 ft2 >s

For smooth pipe, the Moody diagram gives

= 1.69 1106 2

f = 0.0107 Applying the Darcy–Weisbach equation, hL = f a

L V2 ba b D 2g

= 0.0107

= 1.76 ft

111.89 ft>s2 2 50 ft d c £ 8 § 2132.2 ft>s2 2 a ft b 12

Ans.

Ans: hL = 1.76 ft 843

M10_HIBB9290_01_SE_C10_ANS.indd 843

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10–3. Glycerin flows through a 6-in.-diameter, horizontal smooth pipe with an average velocity of 9 ft>s. Determine the pressure drop in a 12-ft-long segment of the pipe.

SOLUTION We will consider the fully developed steady flow and Glycerin is incompressible. Writing the energy between two points 1, and 2, gives p1 p2 V 12 V 22 + + z1 + hpump = + + z2 + hturb + hL g g 2g 2g p1 p2 V 12 V 22 + + z1 + 0 = + + z2 + 0 + hL g g 2g 2g p1 - p2 V 22 - V 12 = + z2 - z1 + hL g 2g Since the pipe is horizontal, z1 = z2 = z. Also, the average velocity of the flow is constant, V1 = V2 = V due to the constant pipe’s diameter. Then the above equation becomes p1 - p2 = hL g The Darcy–Weisbach equation gives hL = f

L V2 . Then D 2g

p1 - p2 L V2 = f g D 2g p1 - p2 = f

gV 2L 2gD

(1)

From the table in Appendix A, rGy = 2.44 slug>ft3 and mGy = 31.3110-3 2 lb s>ft. The Reynolds number is Re =

rGyVD mGy

=

12.44 slug>ft3 219 ft/s2 a

6 ft b 12

31.3110-3 2 lb # s>ft2

= 350.80 6 2300 (Laminar flow) Thus, f =

64 64 = = 0.1824 Re 350.80

Then Eq. (1) gives p1 - p2 = (0.1824)

= a432.69

£

12.44 slug>ft3)(32.2 ft>s2)(9 ft>s)2(12 ft2 2(32.2 ft>s2) a

6 ft b 12

lb 1 ft 2 ba b = 3.00 psi 2 12 in ft

§ Ans.

Ans: p1 - p2 = 3.00 psi

844

M10_HIBB9290_01_SE_C10_ANS.indd 844

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*10–4. If air at a temperature of 100°F flows through the smooth circular duct at 10.5 ft3 >s, determine the pressure drop that occurs over a 15-ft length of the duct.

15 ft

SOLUTION

8 in.

We will consider the fully developed steady flow and assume air is incompressible. Writing the energy between points 1 and 2, gives p1 p2 V 12 V 22 + z1 + hpump = + z2 + hturb + hL + + g g 2g 2g p1 p2 V 12 V 22 + z1 + 0 = + z2 + 0 + hL + + g g 2g 2g p1 - p2 V 22 - V 12 + z2 - z1 + hL = g 2g Since the density of air is small, the elevation term can be neglected. Also, the average velocity of the flow is constant, V1 = V2 = V due to the constant duct diameter. Then the above equation becomes p1 - p2 = hL g The Darcy–Weisbach equation gives hL = f

L V2 . Then D 2g

p1 - p2 L V2 = f g D 2g p1 - p2 = f

gV 2L 2gD

(1)

From the discharge, the average velocity is 10.5 ft3 >s Q = = 30.08 ft>s 2 A 4 pa in.b 12 From the table in Appendix A, ra= 0.00220 slug>ft3 and mG = 0.396110-6 2 lb # s>ft2. The Reynolds number is V =

raVD Re = = ma

10.00220 slug>ft3)(30.08 ft>s2 a 0.396 110-6 2 lb # s>ft2

8 ft b 12

= 1.11 1105 2 7 2300 turbulent

For the smooth duct, the Moody diagram gives

f = 0.0177 Then Eq. (1) gives p1 - p2 = 10.01772 = a0.3964

£

10.00220 slug>ft3)(32.2 ft>s2)(30.08 ft>s2 2 115 ft2

lb 1 ft 2 b a b ft2 12 in.

= 2.75110-3 2 psi

M10_HIBB9290_01_SE_C10_ANS.indd 845

2(32.2 ft>s2) a

8 ft b 12

§

Ans. 845

Ans: p1 - p2 = 2.75110-3 2 psi

10/03/17 2:36 PM

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10–5. Determine the flow of methane at T = 68°F in ft3 >h through a 1000-ft-long horizontal 1-in.-diameter commercial steel pipe if the pressure drop is 5110-3 2 psi.

1 in.

1,000 ft

SOLUTION We will consider the fully developed steady flow, and treat methane as incompressible. From the table in Appendix A, rme = 1.29(10-3) slug>ft3 and vm = 0.181 (10-3) ft2 >s for methane at T = 68°F. The Reynolds number of the flow is 1 Va ft b 12 VD Re = = = 460.41 V vw 0.18110-3 2 ft2 >s

(1)

Take the methane in the 1000-ft-long pipe as the control volume. Since the pipe has a constant cross section, V1 = V2 = V. Also, z1 = z2 = zsince the pipe is horizontal. lb 12 in. 2 lb Here the pressure drop is p1 - p2 = c 5110-3 2 2 d a b = 0.720 2 per 1 ft in ft 1000 ft of pipe. p1 p2 V 12 V 22 + + z1 + hpump = + + z2 + hturb + hL gme gme 2g 2g p1 p2 V2 V2 + + z+ 0 = + + z + 0 + hL gme gme 2g 2g hL =

p1 - p2 = gme

0.720 lb>ft2

3 1.29110

-3

Using the Darcy–Weisbach equation, hL = f

2 slug>ft3 4 132.2 ft>s2 2

= 17.33 ft

L V2 1000 ft V2 d ; 17.33 ft = f c D 2g ° 1 ¢ 2132.2 ft>s2 2 ft 12 V2 =

0.09302 f

(2)

Assuming the flow is laminar, then f =

64 64 0.1390 = = Re 460.41 V V

Substitute this result into Eq. (2), V2 =

0.09302 0.1390 V

V = 0.6692 ft>s

Then Eq. (1) gives Re = 460.4110.66922 = 308.10 6 2300 (laminar flow), ok. The flow rate is Q = VA = 10.6692 ft>s2 c pa

2 0.5 ft b d = 12

3 3.6499110-3 2 ft3 >s 4 a

= 13.14 ft3 >h = 13.1 ft3 >h 846

M10_HIBB9290_01_SE_C10_ANS.indd 846

3600 s b 1h

Ans.

Ans: Q = 13.1 ft3 >h

10/03/17 2:37 PM

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10–6. Water in the old 15-in.-diameter concrete drain pipe runs full with a flow of 15 ft3 >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 ft3 >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

rV 2 L ba b D 2

62.4 a slug>ft3 b ( 12.22 ft>s ) 2 32.2 12 ft ∆p = (0.07) ≥ ¥≥ ¥ 15 2 ft 12 ∆p = ( 97.28 lb>ft2 ) a

1 ft 2 b = 0.676 psi 12 in.

Ans.

Ans: pA - pB = 0.676 psi 847

M10_HIBB9290_01_SE_C10_ANS.indd 847

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10–7. Water at 20°C flows upwards through the 50-mm-diameter cast iron pipe at 5.88 kg>s. Determine the major head loss that occurs over the 8-m-long vertical segment. Also, what is the pressure at A? The water is discharged into the atmosphere at B.

B

8m

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 998.3 kg>m3 and vw = 1.00110-6 2 m2 >s for water at T = 20°C. For the mass flow rate, the average velocity is # m = rQ; 5.88 kg>s = 1998.3 kg>m3 2 5 V 3 p10.025 m2 2 4 6

A 50 mm

V = 2.9998 m>s

Then, the Reynolds number is Re =

12.9998 m>s210.05 m2 VD = = 1.501105 2 vw 1.00110-6 2 m2 >s

e 0.26 mm = = 0.0052. From the Moody diagram, f = 0.031. Thus, D 50 mm the major head loss can be determined using the Darcy–Weisbach equation. For cast iron,

2

hL = f

12.9998 m>s2 L V2 8m d = 2.2749 m = 2.27 m Ans. = 0.031a bc D 2g 0.05 m 219.81 m>s2 2

Take the water in the pipe as the control volume. Here, pB = patm = 0 since the water is discharged into atmosphere at B. Also, VA = VB = V since the pipe diameter is constant. Applying the energy equation from A to B, with datum set through A, zA = 0 and zB = 8 m, pA pB VA2 VB 2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g pA 3

2

1998.3 kg>m 219.81 m>s 2

+

V2 V2 + 0 + 0 = 0 + + 8 m + 0 + 2.2749 m 2g 2g

pA = 100.621103 2 Pa = 101 kPa

Ans.

Ans: hL = 2.27 m pA = 101 kPa 848

M10_HIBB9290_01_SE_C10_ANS.indd 848

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*10–8. Water at 60°F is delivered horizontally through a 400-ft-long galvanized iron pipe to a sewage treatment plant. If the water is discharged into the atmosphere using a pump that develops a pressure at A of 75 psi, determine the diameter of the pipe if the discharge at B is to be 1500 gal>min.

SOLUTION We will consider the fully developed steady flow, and water is incompressible. The discharge is Q = a1500

gal min

Thus, the average velocity is V =

ba

1 ft3 1 min ba b = 3.3422 ft3 >s 7.48 gal 60 s

3.3422 ft3 >s Q 4.2555 = = p 2 A D2 D 4

From the table in Appendix A, rw = 1.939 slug>ft3 and vw = 12.2110-6 2 ft2 >s for water at 60°F. Thus, the Reynolds number is 4.2555 a b1D2 3.48811105 2 VD D2 Re = = = -6 2 v D 12.2110 2 ft >s

(1)

Since the pipe is horizontal and has constant diameter, hL =

Pin - Pout g

Here, pout = patm = 0. Since the water is discharged into atmosphere. Then lb 12 in. 2 ba b - 0 2 1 ft in hL = = 172.98 ft 11.939 slug>ft3 2132.2 ft>s2 2 a75

Using the Darcy–Weisbach equation, hL = f

2 2 L V2 400 ft 14.2555>D 2 d ; 172.98 ft = f a bc D 2g D 2132.2 ft>s2 2

D5 = 0.6503 f

(2)

For galvanized iron pipe, e = 0.0005 ft. The trial and error iterative procedure is required. The iterations are tabulated as follows: Iteration 1 2

Assumed f 0.021 0.0208

D(ft) (Eq. 2) 0.4237 0.4229

E , D (ft , ft)

Re (Eq. 1)

0.00118

0.00118

f from Moody diagram

3

0.0208

5

0.0208

8.23(10 ) 8.25(10 )

The assumed f in 2nd iteration is almost the same as that given by the Moody diagram. Thus D = 0.4229 ft is an acceptable result. D = (0.4229 ft) a

12 in. b = 5.075 in. 1 ft

1 Use D = 5 @in.@diameter pipe. 8

Ans.

Ans: 1 Use D = 5 @in.@diameter pipe. 8

849

M10_HIBB9290_01_SE_C10_ANS.indd 849

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10–9. A horizontal 40-ft-long galvanized iron pipe having a diameter of 6 in. is used to transport water at a temperature of 50°F. Determine the pressure drop over its length if the velocity is 8 ft>s.

SOLUTION We will consider fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.940 slug>ft3, and vw = 14.1110-6 2 ft2 >s for water at T = 50°F. Thus, the Reynolds number is 6 18 ft>s2 a ft b 12 VD Re = = = 2.84 1105 2 vw 14.1 110-6 2 ft2 >s

For galvanized iron pipe

e 0.0005 ft = = 0.001. From the Moody diagram, D 6 a ft b 12 f = 0.0208

Since the pipe is horizontal and of constant diameter, the energy equation reduces to p1 - p2 L V2 = hL = f gw D 2g Then p1 - p2 = f a

LgwV 2 b 2gD

= 0.0208

£

140 ft211.940 slug>ft3 2132.2ft>s2 218 ft>s2 2

= 1103.30 lb>ft2 2a

= 0.717 psi

2132.2 ft>s2 2a

1 ft 2 b 12 in.

6 ft b 12

§

Ans.

Ans: p1 - p2 = 0.717 psi 850

M10_HIBB9290_01_SE_C10_ANS.indd 850

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10–10. Air at 80°F flows through the commercial steel duct at 300 ft3 >min. Determine the pressure drop along a 10-ft length of the duct. 8 in. 10 ft 6 in.

SOLUTION We will consider the fully developed steady flow and treat the air as incompressible. The hydraulic diameter of the rectangular duct is Dn =

4318 in.216 in.24 4A 1 ft = = 16.8571 in.2 a b = 0.5714 ft P 216 in. + 8 in.2 12 in.

From the table in Appendix A, ra = 0.00228 slug>ft3 and va = 0.169110-3 2 ft2 >s

1 min for air at T = 80°F. Here Q = 1300 ft3 >min2 a b = 5 ft3 >s. Thus the average 60 s velocity of the air is V =

Q = A

Thus, the Reynolds numbers is Re =

a

5 ft3 >s

6 8 ft b a ft b 12 12

= 15 ft>s

115 ft>s210.5714 ft2 VDh = = 5.071104 2 v 0.169110-3 2 ft2 >s

and the relative roughness of the commercial steel duct is e 0.00015 ft = = 0.0002625 Dh 0.5714 ft

From the Moody diagram, f = 0.022. Thus, the major head loss can be determined using the Darcy–Weisbach equation. hL = f

115 ft>s2 2 L V2 10 ft d = 1.345 ft = 0.022a bc Dh 2g 0.5714 ft 2132.2 ft>s2 2

Take air in the 10-ft-long duct as the control volume. Since the duct has a constant cross section, V1 = V2 = V. Also, since the density of air is small, the elevation terms can be neglected. Applying the energy equation, p1 p2 V 12 V 22 + + z1 + hpump = + + z2 + hturb + hL ga 2g ga 2g p1 p2 V2 V2 + + 0 + 0 = + + 0 + 0 + 1.345 ft ga 2g ga 2g p1 - p2 = (0.00228 slug>ft3)(32.2 ft>s2)(1.345 ft) = (0.09875 lb>ft2) a = 0.686(10-3) psi

1 ft 2 b 12 in.

Ans. Ans: p1 - p2 = 0.686(10-3) psi 851

M10_HIBB9290_01_SE_C10_ANS.indd 851

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10–11. If the pressure drop along the 10-ft-long commercial steel duct is 0.68110-3 2 psi, determine the flow Q of 80°F air. 8 in. 10 ft 6 in.

SOLUTION We will consider the fully developed steady flow and treat air as incompressible. From the table in Appendix A, ra = 0.00228 slug>ft3, and va = 0.169110-3 2 ft2 >s for air at T = 80°F. Take air in the 10-ft-long duct as the control volume. Since the duct has a constant cross section, VA = VB = V Also, since the density of air is small, the elevation terms can be neglected Applying the energy equation, pA pB V A2 V B2 + + zA + hpump = + + zB + hturb + hL ga ga 2g 2g pA pB V2 V2 + + 0 + 0 = + + 0 + 0 + hL ga 2g ga 2g

hL =

pA - pB ga

lb 12 in. 2 da b 2 1 ft in = = 1.3338 ft 10.00228 slug>ft3 2132.2 ft>s2 2 c 0.68110-3 2

The hydraulic diameter of the rectangular duct is Dh =

4318 in.216 in.24 4A 1 ft = = 16.8571 in.2 a b = 0.5714 ft P 216 in. + 8 in.2 12 in.

Thus, the Reynolds number is Re =

V10.5714 ft2 VDh = = 3.38121103 2 V va 0.169110-3 2 ft2 >s

(1)

Using the Darcy–Weisbach equation, hL = f

L V2 10 ft V2 d ; 1.3338 ft = f a bc 0.5714 ft 2132.2 ft>s2 2 Dh 2g V =

2.2155

(2)

2f The relative roughness of the commercial steel duct is e 0.00015 ft = = 0.0002625 Dh 0.5714 ft

The trial and error iterative procedure is required. The iterations are tabulated as follows: Iteration 1 2

Assumed f 0.02 0.0218

V(ft , s) (Eq. 2) 15.67

15.00

Re (Eq. 1)

f from Moody diagram

4

0.0218

4

0.022

5.30110 2 5.07110 2

The assumed f in 2nd iteration is very close to that from Moody diagram. Thus, V = 15.00 ft>s is an acceptable result. Then, the flow rate is Q = VA = (15.00 ft>s) c a

6 8 ft b a ft b d = 5.00 ft3 >s 12 12 852

M10_HIBB9290_01_SE_C10_ANS.indd 852

Ans.

Ans: Q = 5.00 ft3 >s

10/03/17 2:37 PM

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*10–12. A horizontal 75-mm-diameter galvanized iron pipe, having a roughness of e = 0.2 mm, is used to transport water at a temperature of 60°C and with a velocity of 3 m>s. Determine the pressure drop over its 12-m length.

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

rV 2 L 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: ∆p = 17.7 kPa 853

M10_HIBB9290_01_SE_C10_ANS.indd 853

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10–13. 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 = 15.1110-6 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 =

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

Ans.

Ans: p1 - p2 = 2.71 Pa 854

M10_HIBB9290_01_SE_C10_ANS.indd 854

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10–14. 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 = 15.1110-6 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. Also, for laminar flow, the friction factor can be determined using f = 64 = 0.0278. Thus, the major head loss can be determined using 2300

64 = Re

( 0.2026 m>s ) L V2 200 m = 0.0278 a b£ § = 0.06791 m 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

855

M10_HIBB9290_01_SE_C10_ANS.indd 855

Ans.

Ans: Q = 0.00608 m3 >s, p1 - p2 = 0.801 Pa

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10–15. Water in the concrete detention pond is to be pumped over the berm and into the lake. If the 8-in.diameter hose has a roughness of e = 0.0006 ft, determine the power output of the pump so that the flow out of the pipe is 2400 gal>min. The hose is 800 ft long. The water temperature is 60°F.

B 60 ft A

SOLUTION We will consider the fully developed steady flow, and the water is incompressible. From the table in Appendix A, rw = 1.939 slug>ft3 and vw = 12.2110-6 2 ft2 >s for water at T = 60°F. Here the flow rate is Q = a2400

gal min

ba

1 ft3 1 min ba b = 5.3476 ft3 >s 7.48 gal 60 s

Then, the flow velocity in the hose is given by Q = VA;

5.3476 ft3 >s = V c pa

2 4 ft b d 12

V = 15.3197 ft>s

The Reynolds number of the flow in the pipe is

VD Re = = vw

115.3197 ft>s2 a

8 ft b 12

= 8.37 1105 2

12.2 110-6 2 ft2 >s

For the hose, the relative roughness is

e 0.0006 ft = = 0.0009. Entering these two D 8 ft 12

values into the Moody diagram, we obtain f = 0.0195. Then the major head loss can be determined using the Darcy–Weisbach equation. 2

hL = f

L V2 800 ft 115.3197 ft>s2 d = 85.2770 ft = 0.0195 c D 2g ° 8 ¢ 2132.2 ft>s2 2 ft 12

Take the water in the detention pond, in the pump, and in the hose as the control volume. Since the water surface at A is exposed to the atmosphere and water is discharged into the atmosphere at B, pA = pB = patm = 0. Also, since the water is drawn from a large reservoir, the water level at A can be considered constant and so VA ≃ 0. With reference to the datum set through A, zB = 60 ft and zA = 0. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

115.3197 ft>s2 2 2132.2 ft>s2 2

+ 60 ft + 0 + 85.2770 ft

hpump = 148.92 ft

Then, the required power output of the pump is # W = gwQhpump = 11.939 slug>ft3 2132.2 ft>s2 215.3476 ft3 >s21148.92 ft2 = 149,722.05 ft # lb>s2 a

1 hp

550 ft # lb>s

b = 90.4 hp

Ans. Ans: # W = 90.4 hp

856

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*10–16. 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

LV2 3m V2 2 = 0.028 a 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,

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL g g 2g 2g V =

-98.7 ( 103 ) N>m2

( 1000 kg>m3 )( 9.81 m>s2 )

+

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: Q = 8.98 liter>s 857

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10–17. The old 15-in.-diameter concrete drain pipe runs full of water with a flow of 15 ft3 >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.

A

B

SOLUTION Water is considered to be incompressible. Since the concrete pipe has a constant diameter. VA = VB = V =

Q = A

15 ft3 >s

2 7.5 pa 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, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g pA pB V2 V2 4 + 12 ft a b + 0 = + 0 + 0 + 1.559 ft + + g g 2g 100 2g pA - pB = ( 62.4 lb>ft2 ) (1.079 ft) = 67.3

lb ft2

Ans.

Ans: pA - pB = 67.3 lb>ft2 858

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10–18. Determine the friction factor of a 2-in.-diameter horizontal tube if natural gas is to flow through it at 20 ft3 >h. The specific weight of the natural gas is gg = 0.051 lb>ft3 and the pressure drop is 0.6110-3 2 psi per 1000 ft.

SOLUTION We will consider the fully developed steady flow and treat natural gas as incompressible. Take the natural gas in the 1000-ft-long tube as the control volume. Since the tube has a constant cross section, V1 = V2 = V. Also, since the tube is horizontal, z1 = z2 = z. lb 12 in. 2 b = 0.0864 lb>ft per Here, the pressure drop is p1 - p2 = c 0.6110-3 2 2 d a 1 ft in 1000 ft of tube. Applying the energy equation, p1 p2 V 12 V 22 + + z1 + hpump = + + z2 + hturb + hL gg gg 2g 2g p1 p2 V2 V2 + z+ 0 = + z + 0 + hL + + gg gg 2g 2g hL =

0.0864 lb>ft2 p1 - p2 = 1.6941 ft = g 0.051 lb>ft3

Here, the flow rate is Q = a20

Thus, velocity of the flow is

ft3 1h ba b = 0.005556 ft3 >s h 3600 s

0.005556 ft3 >s Q = = 0.2546 ft>s 2 A 1 pa ft b 12 Using the Darcy–Weisbach equation, V =

hL = f

L V2 ; D 2g

2

1.6941 ft = f

1000 ft 10.2546 ft>s2 d c 2 2132.2 ft>s2 2 a ft b 12

f = 0.2804 = 0.280

Ans.

Ans: f = 0.280 859

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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 1 min l ba ba b = 0.15 ( 10-3 ) m3 >s 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 loss 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: hL = 76.7 mm 860

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*10–20. Water at T = 70°F flows through the 15-in.diameter concrete pipe from the reservoir at A to the one at B. Determine the flow in gal > min. The length of the concrete pipe is 200 ft. The roughness of the concrete pipe is e = 0.005 ft.

A 30 ft B

SOLUTION We will consider the fully developed steady flow and water is incompressible. From table in Appendix A, rw = 1.937 slug>ft3 and vw = 10.4(10-6) ft2 >s for water at T = 70°F. The Reynolds number of the flow is 15 Va ft b 12 VD Re = = = 120.19 1103 2V vw 10.4110-6 2 ft2 >s

(1)

The major head loss can be determined using the Darcy–Weisbach equation. hL = f

L V2 200 ft V2 = fa bc d = 2.4845 f V 2 D 2g 15>12 ft 2 1 32.2 ft>s2 2

Take the water in reservoir A, and the concrete pipe as the control volume. Since point A is exposed to the atmosphere, pA = patm = 0. The pressure at end B of the pipe is also pB = patm = 0. Also, the water is drawn from a large reservoir, so VA _ 0. With reference to the datum set through B, zA = 30 ft and zB = 0. Applying the energy equation between A and B, pA pB V A2 V B2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 30 ft + 0 = 0 +

V2 + 0 + 0 + 2.4845 f V 2 2132.2 ft>s2 2

V2 + 2.4845f V 2 = 30 64.4 For concrete pipe, the relative roughness is

(2)

e 0.005 ft = = 0.004 D 115>122 ft

The trial and error iterative procedure is required. The iterations are tabulated as follows: Iteration 1

Assumed f 0.028

V(ft , s) Eq. (2) 18.78

Re Eq. (1) 2.26(106)

f from Moody diagram 0.028

The assumed f in the first iteration is almost the same as that from Moody diagram. Thus V = 18.78 ft>s is an acceptable result. Thus the flow rate is Q = VA = 118.78 ft>s2 c pa

2 7.5 ft b d = 23.04 ft3 >s = 23.0 ft3 >s 12

861

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Ans.

Ans: Q = 23.0 ft3 >s

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10–21. Drinking water at 60°F flows at 10 ft>s through a horizontal galvanized iron pipe. If the pressure drop over a 50-ft length is to be no more than 4 psi, determine the smallest allowable diameter D of the pipe.

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.939 slug>ft3 and vw = 12.2110-6 2 ft2 >s for water at T = 60°F. Thus, the Reynolds number is Re =

110 ft>s2D VD = = 8.1967 1105 2D vw 12.2 110-6 2 ft2 >s

(1)

The major head loss can be determined using the Darcy–Weisbach equation. hL = f

110 ft>s2 2 f L V2 50 ft d = 77.64a b = fa bc D 2g D D 2132.2ft>s2 2

Taking the water in the 50-ft-long pipe as the control volume. Since the pipe diameter is constant, VA = VB = V. Also, because it is horizontal, zA = zB = z. Applying the energy equation, pA pB V A2 V B2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g f pA pB V2 V2 + + z+ 0 = + + z + 0 + 77.64a b gw gw 2g 2g D f pA - pB = 77.64a b g D

lb 12 in. 2 ba b 2 f 1 ft in = 77.64a b D 11.939 slug>ft3 2132.2 ft>s2 2 a4

(2)

D = 8.4158 f

For galvanized iron pipe, e = 0.0005 ft. The trial and error iterative procedure is required. The iterations are tabulated as follows: Iteration Assumed f 1 2 3 4

0.015 0.0295 0.0248 0.0258

D(ft) Eq. 2 0.1262 0.2483 0.2087 0.2171

E , D(ft>ft) 0.00396 0.00201 0.00239 0.00230

Re (Eq. 1) 5

1.03(10 ) 2.03(105) 1.71(105) 1.78(105)

f from Moody diagram 0.0295 0.0248 0.0258 0.0255

The assumed f in 4th iteration is very close to that from the Moody diagram. Thus, D = 0.2171 ft is an acceptable result. D = (0.2171 ft) a

12 in. b = 2.606 in. 1 ft

5 Use D = 2 -in.-diameter pipe. 8

Ans.

Ans: 5 Use D = 2 @in.@diameter pipe. 8 862

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10–22. A 450-ft-long horizontal commercial steel pipe having a diameter of 6 in. is used to transport water at T = 70°F. Determine the power output of a pump if the discharge through the pipe is to be 120 ft3 >min and the pressure at the pump inlet is 5 psi. The pipe is open to the atmosphere at its outlet.

SOLUTION We will consider the fully developed steady flow, and water is incompressible. Here, the discharge is ft3 1 min ba b = 2 ft3 >s min 60 s

Q = a120

Thus, the average velocity is

V =

Q = A

2 ft3 >s

pa

2 3 ft b 12

= 10.19 ft>s

From the table in Appendix A, rw = 1.937 slug>ft3 and vw = 10.4110-6 2 ft2 >s for water at T = 70°F. Thus, the Reynolds number is VD Re = = vw

110.19 ft>s2 a

6 ft b 12

10.4 110-6 2 ft2 >s

= 4.90 1105 2

e 0.00015 ft = = 0.0003. Entering these two values D 6 ft 12 into the Moody diagram, we obtain f = 0.0165. Then, the major head loss can be

For commercial steel pipe,

determined using the Darcy–Weisbach equation. hL = f

2 L V2 450 ft 110.19 ft>s2 d = 23.92 ft = 0.0165 c D 2g ° 6 ¢ 2132.2 ft>s2 2 ft 12

Applying the energy equation between inlet and outlet realizing that Vin = Vout (the pipe has constant diameter), pout = patm = 0. (The outlet is open to the atmosphere), and zin = zout = z (the pipe is horizontal). pin pout V out2 V in2 + + zin + hpump = + + zout + hturb + hL gw gw 2g 2g lb 12 in. 2 ba b 2 1 ft V2 V2 in + + z + hpump = 0 + + z + 0 + 23.92 ft 3 2 2g 2g 11.937 slug>ft 2132.2 ft>s 2 a5

hpump = 12.38 ft

Thus, the power output of the pump is # Wout = gwQhpump = (1.937 slug>ft3)(32.2 ft>s2)(2 ft3 >s)(12.38 ft) = (1544.39 ft # lb>s) a = 2.81 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Wout = 2.81 hp 863

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10–23. Air at 60°F is transported by the fan at the rate of 2 ft3 >s through the 12-in.-diameter galvanized iron duct. Determine the head loss over a 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 ft3 >s

2 6 p a ft b 12

= 2.546 ft>s

From Appendix A, r = 0.00237 slug>ft3 and n = 0.158 ( 10-3 ) ft2 >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 ) ft2 >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 ) bc d = 0.028 a D 2g 1 ft 2 ( 32.2 ft>s2 )

= 0.1128 ft = 0.113 ft

Ans.

Ans: hL = 0.113 ft 864

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*10–24. A cast iron pipe having a diameter of 100 mm is used to fill the cylindrical tank with water at T = 25°C. Determine the required power output of a pump necessary to fill the empty tank to a depth of 3 m in 6.5 min. The pipe has a total length of 50 m.

B

2m

12 m

SOLUTION We will consider fully developed steady flow, and water is incompressible. The discharge is Q = c

p11 m2 2 13 m2 6.5 min

Thus, the average velocity is V =

da

A

1 min b = 0.02417 m2 >s 60 s

0.02417 m3 >s Q = = 3.077 m>s A p10.05 m2 2

From the table in Appendix A, rw = 997.1 kg>m3 and vw = 0.898110-6 2 m2 >s for water at T = 25°C. Thus, the Reynolds number is Re =

13.077m>s210.1 m2 VD = = 3.43 1105 2 vw 0.898 110-6 2 m2 >s

e 0.26 mm = = 0.0026. Entering these two values into the Moody D 100 mm diagram, we obtain f = 0.0255. Then, the major head loss can be determined using the Darcy–Weisbach equation. For cast iron pipe,

hL = f

2 L V2 50 m 13.077 m>s2 d = 6.152 m = 0.0255a bc 2 D 2g 0.1 m 219.81 m>s 2

Here pA = pB = patm = 0, Since the water surface at A and pipe outlet at B are exposed to atmosphere. Also VA = 0 since the water is drawn from a large reservoir. With reference to the datum set through A, zA = 0 and zB = 12 m. Applying the energy equation between A and B, pA pA V A2 V B2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

13.077 m>s2 2

219.81 m>s2 2

+ 12 m + 0 + 6.152 m

hpump = 18.635 m Thus, the required power output of the pump is # Wo = Qgw hpump = 10.02417 m3 >s2 3 1997.1 kg>m3 219.81 m>s2 2 4 118.635 m2 = 4404.96 W = 4.40 kW

Ans.

Ans: # Wo = 4.40 kW 865

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10–25. A cast iron pipe having a diameter of 100 mm is used to fill the empty cylindrical tank with water at T = 25°C. If the power output of the pump is 4.5 kW, determine the depth h of the water in the tank 6 minutes after the pump is turned on. The pipe has a total length of 50 m.

B

2m

12 m

A

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 997.1 kg>m3 and vw = 0.898 110-6 2 m2 >s for water at T = 25°C. From the power output of the pump, # Wo = Qgwhpump; 4.51103 2W = V3p10.05 m2 2 431997.1 kg>m3 219.81 m>s2 24hpump hpump =

58.575 V

The major head loss can be determined using the Darcy–Weisbach equation. hL = f

L V2 50 m V2 d = 25.484f V 2 = fa bc D 2g 0.1 m 219.81 m>s2 2

Here, pA = pB = patm = 0, since the water surface at A and pipe outlet at B are exposed to the atmosphere. Also VA = 0 since the water is drawn from a large reservoir. With reference to the datum set through A, zA = 0 and zB = 12 m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 +

58.575 V2 = 0 + + 12 + 0 + 25.484 f V 2 V 219.81 m>s2 2

125.484f + 0.050972V 3 + 12V - 58.575 = 0

(1)

The Reynolds number is Re =

V10.1 m2 VD = = 1.1136 1105 2V vw 0.898 110-6 2 m2 >s

For cast iron pipe,

(2)

e 0.26 mm = = 0.0026. The trial and error iterative procedure is D 100 mm

required. The iterations are tabulated as follows. Iteration

Assumed f

1 2

0.025 0.0255

V(m , s)(Eq. 1) 3.1274 3.1154

Re (Eq. 2)

f from Moody diagram

5

3.48(10 ) 3.47(105)

0.0255 0.0255

The assumed f in 2nd iteration is almost the same as that from Moody diagram. Thus, V = 3.1154 m>s is an acceptable result. Then, the discharge is Q = VA = 13.1154 m>s23p10.05 m2 2 4 = 0.02447 m3 >s

Then, the height h of the tank after 6 min of filling is V = Qt;

3p11 m2 2 4h = 10.02447 m3 >s)(360 s2

Ans.

h = 2.804 m = 2.80 m

Ans: h = 2.80 m

866

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10–26. Water at 70°F flows through the 3-in.-diameter cast iron pipe. If the pressure at A is 6 psi, determine the discharge.

A

308

50 ft

B

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.937 slug>ft3 and vw = 10.4 110-6 2 ft2 >s for water at T = 70°F. The major head loss can be determined using the Darcy–Weisbach equation. hL = f

50 ft V2 L V2 d = 3.1056f V 2 = f c D 2g ° 3 ¢ 2132.2 ft>s2 2 ft 12

Since the water is discharged into the atmosphere at B, pB = patm = 0. Also, VA = VB = V since the pipe has a constant diameter. With reference to the datum set through B, zA = 50 sin 30° ft = 25 ft and zB = 0. Write the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g lb 12 in. 2 ba b 2 1 ft V2 V2 in + + 25 ft + 0 = 0 + + 0 + 0 + 3.1056f V 2 3 2 2g 2g 11.937 slug>ft 2132.2 ft>s 2 a6

V =

The Reynolds number is

3.5370

(1)

2f

3 Va ft b 12 VD Re = = = 2.4038 1104 2V vw 10.4 110-6 2 ft2 >s

(2)

e 0.00085 ft = = 0.0034. The trial and error iterative procedure D 3 ft 12 is required. The iteration is tabulated below: For cast iron pipe

Iteration Assumed f V(ft , s)(Eq. 1) 1 0.027 21.53

Re (Eq. 2)

f from Moody diagram

5.17(105)

0.027

The assumed f is almost the same as that from Moody diagram. Thus, V = 21.53 ft>s is an acceptable result. Then the discharge is given by Q = VA = (21.53 ft>s) Jpa

2 1.5 ft b R = 1.057 ft3 >s = 1.06 ft3 >s 12

867

M10_HIBB9290_01_SE_C10_ANS.indd 867

Ans.

Ans: Q = 1.06 ft3 >s

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10–27. Kerosene at T = 20°C flows through the 100-mmdiameter commercial steel pipe at 22.5 kg>s. Determine the pressure drop that occurs over the 50-m length.

B

50 m

SOLUTION We will consider the fully developed steady flow, and kerosene is incompressible. From the table in Appendix A, rke = 814 kg>m3 and vke = 2.36 110-6 2 m2 >s for kerosene at T = 20°C. The average velocity is given by # 22.5 kg>s = 1814 kg>m3 2 = 5 V 3 p10.05 m2 2 4 6 M = rkeVA The Reynolds number is Re =

608

A

V = 3.519 m>s

13.519 m>s210.1 m2 VD = = 1.49 1105 2 vke 2.36 110-6 2 m2 >s

e 0.045 mm = = 0.00015. Entering these two values D 100 mm into the Moody diagram, we obtain

For commercial steel pipe,

f = 0.0192 Then, the major head loss can be determined using the Darcy–Weisbach equation. 2

hL = f

L V2 50 m 13.519 m>s2 d = 6.061 m = 0.0192a bc D 2g 0.1 m 219.81m>s2 2

Since the pipe has a constant diameter, VA = VB = V with reference to the datum set through A, zA = 0 and zB = 50 sin 60° m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gke gke 2g 2g pA pB V2 V2 + + 0 + 0 = + + 5.0 sin 60°m + 0 + 6.061 m gke gke 2g 2g pA - pB = 49.36 m gke pA - pB = 1814 kg>m3 219.81 m>s2 2149.36 m2 = 394.17(103)Pa = 394 kPa

Ans.

Ans: pA - pB = 394 kPa 868

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*10–28. A 150-mm-diameter galvanized iron pipe is used to transport water at T = 30°C with a velocity of 1.5 m>s. Determine the pressure drop over the 20-m length of the pipe.

B

20 m

SOLUTION

458

We consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 995.7 kg>m3 and vw = 0.804 110-6 2 m2 >s for water at T = 30°C. Thus, the Reynolds number is Re =

A

11.5m>s210.15 m2 VD = = 2.80 1105 2 vw 0.804 110-6 2 m2 >s

e 0.15 mm = = 0.001. D 150 mm Entering these two values into Moody diagram, we obtain For galvanized-iron pipe, the relative roughness is f = 0.0208 Then the major head loss can be determined using the Darcy–Weisbach equation. hL = f

11.5 m>s2 2 L V2 20 m d = 0.3180 m = 0.0208a bc D 2g 0.15 m 219.81 m>s2 2

Since the diameter of the pipe is constant, VA = VB = V. With reference to the datum set through A, zA = 0 and zB = 20 sin 45° m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g pA pB V2 V2 + + 0 + 0 = + + 20 sin 45° m + 0 + 0.3180 m gw gw 2g 2g pA - pB = 14.460 m gw pA - pB = 1995.7 kg>m3 219.81 m>s2 2114.460 m2 = 141.241103 2 Pa = 141 kPa

Ans.

Ans: pA - pB = 141 kPa 869

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10–29. When the valve at A is opened, methane at 20°C flows through the 200-mm-diameter commercial steel pipe at 0.095 m3 >s. Determine the pressure drop over the length AB of the pipe.

A

B

200 m

SOLUTION We will consider the fully developed steady flow, and treat methane as incompressible. From the table in Appendix A, rm = 0.665 kg>m3 and vm = 16.8 110-6 2 m2 >s for methane at T = 20°C. From the discharge, the average velocity is V =

0.095 m3 >s Q = = 12.096 m>s A p10.05 m2 2

Thus, the Reynolds number is Re =

112.096 m>s210.1 m2 VD = = 7.20 1104 2 vm 16.8 110-6 2 m2 >s

For commercial steel pipe, the relative roughness is

e 0.455 mm = = 0.00045. D 100 mm Entering these two values into Moody diagram, we obtain f = 0.0212 Since the pipe is horizontal, the energy equation gives pA - pB L V2 = f g D 2g Since g = rg, then pA - pB = f a = 10.02122 a

rV 2 L ba b D 2

3 2 200 m 10.665 kg>m 2112.096 m>s2 bc d 0.1 m 2

= 2.0631103 2 Pa

Ans.

= 2.06 kPa

Ans: pA - pB = 2.06 kPa 870

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10–30. Water at 80°F is pumped from the well through a 2-in.-diameter pipe having a length of 14 ft. Determine the maximum mass flow at C so that cavitation will not occur. The roughness of the pipe is e = 0.0006 ft, and the gage vapor pressure for water is –14.2 psi.

B

C

12 ft

A

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.934 slug>ft2 and vw = 9.35 110-6 2 ft2 >s. Thus, the Reynolds number is Re =

Va

2 ft b 12

VD = = 1.7825 1104 2V vw 9.35 110-6 2 ft2 >s

(1)

The major head loss can be determined using the Darcy–Weisbach equation. hL = f

14 ft V2 L V2 d = 1.3043f V 2 = f c D 2g ° 2 ¢ 2132.2 ft>s2 2 ft 12

The pressure of B will be the smallest and cavitation will occur here first. For lb 12 in. 2 b = -2044.8 lb>ft2. Since the ba 2 1 ft in water surface at A is open to the atmosphere, pA = patm = 0. Also, VB = V since the hose has a constant diameter. With reference to the datum set through A, zA = 0 and zB = 12 ft Applying the energy equation between A and B, cavitation to occur, pB = a - 14.2

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + 0 =

- 2044.8 lb>ft2 3

2

11.934 slug>ft 2132.2 ft>s 2

+

V2 + 12 ft + 1.304.3f V 2 2132.2 ft>s2 2

11.3043f + 0.015532V 2 - 20.835 = 0

(2)

e 0.0006 ft = = 0.0036. The trial and error D 2 ft 12 iterative procedure is required. The iteration is tabulated below: The relative roughness of the hose is

Iteration Assumed f V(ft>s)(Eq. 2) 1 0.0275 20.13

Re (Eq. 1)

f from Moody diagram

3.59(105)

0.0275

The assumed f is almost the same as that from Moody diagram. Thus, v = 20.13 ft>s is an acceptable result. Then the mass flow rate is # m = rVA =

1 1.934 slug>ft3 2 (20.13 ft>s) Jpa

= 0.8495 slug>s = 0.850 slug>s

2 1 ft b R 12

Ans.

Ans: # m = 0.850 slug>s 871

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10–31. Water from the reservoir at A drains through the 50-mm-diameter pipe assembly. If commercial steel pipe is used, determine the initial discharge into the tank at B when the valve E is closed and F is opened. Take nw = 1.00110-6 2 m2 >s.

A 4m

8m

SOLUTION

B

We will consider the fully developed steady flow, and water is incompressible. The Reynolds number is Re =

V10.05 m2 VD = = 5 1104 2V vw 1.00 110-6 2 m2 >s

F 3m

D

8m

(1)

E 4m

6m C

The major head loss from A to B can be determined using the Darcy–Weisbach equation. hL = f

L V2 16 m V2 d = 16.3099f V 2 = fa bc D 2g 0.05 m 219.81 m>s2 2

Take the control volume as the water contained in the tanks, and pipe between A and B. Here, pA = patm = 0 (water surface exposed to atmosphere), VA ≃ 0 (large tank). With reference to the datum set through B, zB = 0 and zA = 12 m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 12 m + 0 = 3 m + V =

V2 + 0 + 16.3099 f V 2 219.81 m>s2 2

0.7428

2f + 0.003125 For commercial steel pipe, the relative roughness is

(2)

e 0.045 mm = = 0.0009 D 50 mm An iterative procedure is required. The iterations are tabulated as follows: Iteration Assumed f V(m/s)(Eq. 2) Re (Eq. 1) 1 0.022 4.6864 2.343(105) 2 0.0204 4.8436 2.4218(105)

f from Moody diagram 0.0204 0.0204

The assumed f in 2nd iteration is almost the same as that from Moody diagram. Thus, V = 4.8436 m>s is an acceptable result. Therefore the discharge is Q = VA = (4.8436 m>s2) 3 p(0.025 m)2 4 = 0.009510 m3 >s = 0.00951 m3 >s

Ans.

Note: A complete analysis would also include minor losses, as discussed in Sec. 10.2.

872

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Ans: Q = 0.00951 m3 >s

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A

*10–32. Water from the reservoir at A drains through the 50-mm-diameter pipe assembly. If commercial steel pipe is used, determine the initial flow into the pipe from reservoir A when both valves E and F are fully opened. The end C is open to the atmosphere. Take nw = 1.00110-6 2 m2 >s.

4m

8m

B

F 3m

D

8m

SOLUTION

E 4m

6m C

We will consider fully developed steady flow and water as incompressible. Take the control volume as the water contained in the reservoir, tank, and pipe between A, B, and C. Here, the diameter of the pipe is constant. Thus, the continuity equation gives 0 r dV + r V # dA = 0 0 t LCV LCS 0 - VAD a

p 2 p p D b + VCD a D2 b + VBD a D2 b = 0 4 4 4

VAD = VCD + VBD The major head losses for the pipe are 1hL 2 AD = fAD 1hL 2 BD = fBD 1hL 2 CD = fCD

(1)

2 2 LAD VAD VAD 8m 2 d = 8.1549fADVAD = fAD a bc D 2g 0.05 m 219.81 m>s2 2

2 2 LBD VBD VBD 8m 2 d = 8.1549fBDVBD = fBD a bc D 2g 0.05 m 219.81 m>s2 2

2 2 LCD VCD VCD 10 m 2 d = 10.1937fCDVCD = fCD a bc D 2g 0.05 m 219.81 m>s2 2

Here, pA = pC = patm = 0 (exposed to atmosphere), VA = 0 (large reservoir). With reference to the datum set along the horizontal pipe, zA = 12 m, zB = 0 m, and zC = -4 m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 12 m + 0 = 3 m +

2 VBD 2 + 0 m + 0 + 8.1549fADV 2AD + 8.1549fBDVBD 2g

2 fADVAD + afBD +

And between A and C,

1 bV 2 = 1.1036 19.62 BD

(2)

pA pC VC2 VA2 + + zA + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 12 m + 0 = 0 +

2 VCD

219.81 m>s2 2

2 8.1549 fAD VAD + a10.1937 fCD +

The Reynolds number of the flow is Re =

2 2 + 1 -4 m2 + 0 + 8.1549 fADVAD + 10.1937 fCD VCD

1 bV 2 = 16 19.62 CD

(3)

V10.05 m2 VD = = 51104 2V vw 1.00110-6 2 m2 >s

(4)

873

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10–32. Continued

e 0.045 mm = = 0.0009. The D 50 mm trial and error iterative procedure is required. The iterations are tabulated below:

For commercial steel pipe, the relative roughness is

Assumed f fBD Iteration fAD 1 0.02 0.02 2 0.020 0.23

Velocities (m , s)* fCD VAD VBD VCD AD BD 0.02 6.9775 1.3532 5.6236 3.49 6.77 0.020 6.9544 1.3469 5.6078 3.48 6.73

CD 2.81 2.80

The f’s in the 3rd iteration are almost the same as those from the Moody diagram Thus, VAD = 6.9544 m>s is an acceptable result. Thus, the discharge through pipe AD is QAD = VAD AAD = 16.9544 m>s2 3 p10.025 m2 2 4 = 0.0137 m3 >s

Ans.

* Solve Eqs. 1, 2, and 3 to obtain VAD, VBD and VCD. Note: A complete analysis would also include minor losses, as discussed in Sec. 10.2.

874

M10_HIBB9290_01_SE_C10_ANS.indd 874

Ans: QAD = 0.0137 m3 >s

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10–33. 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,

p1 p2 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: p1 - p2 = 133 kPa 875

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10–34. 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 =

rV 2 L ba b D 2

Q , Thus, for Q = Q0, A

∆p1 = f a

L b≥ D

ra

Q0 2 b rQ02 A L ¥ = f a ba b 2 D 2A2

For Q = 2Q0,

∆p2 = f a

L b≥ D

ra

2Q0 2 b 4rQ02 A L ¥ = f a ba b 2 D 2A2

Thus,

% of increase in ∆p =

∆p2 - 1 = ≥ ∆p1

fa

fa

4rQ02 L ba b D 2A2

rQ02 L ba b - 1 D 2A2

¥ * 100% Ans.

= 300%

Ans: percent increase = 300% 876

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10–35. 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 =

( 3 m>s ) (0.075 m) VD = 2.25 ( 105 ) 7 2300 (turbulent flow) = n 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

rV 2 L ba b D 2

= 0.0245 a

( 998.3 kg>m 4m bJ 0.075 m 2

3

)( 3 m>s ) 2

∆p = 5870 Pa = 5.87 kPa

R Ans.

Ans: p1 - p2 = 5.87 kPa 877

M10_HIBB9290_01_SE_C10_ANS.indd 877

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*10–36. 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 = 0.000 06 ft. If the pipe is 300 ft long, determine its required diameter so that it provides a flow of 0.5 ft3 >s.

10 ft

SOLUTION Assume that fully developed steady flow occurs and water is incompressible. The average velocity is V =

0.5 ft3 >s Q 0.6366 = = p 2 A D2 D 4

Appendix A gives rw = 1.937 slug>ft3 and nw = 10.4 ( 10-6 ) ft2 >s for water at 70° F. Thus, the Reynolds number is Re =

( 0.6366>D2 ) D 6.1213 ( 104 ) VD = = -6 2 nw D 10.4 ( 10 ) ft >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 V in2 + 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.

7 Use D = 3 in. 8

878

M10_HIBB9290_01_SE_C10_ANS.indd 878

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10–37. A commercial steel pipe is required to carry glycerin at T = 20°C with a mass flow of 14.8 kg>s. If the pressure drop over its 100-m horizontal length is not to exceed 350 kPa, determine the required diameter of the pipe.

SOLUTION We consider the fully developed steady flow, and glycerin is incompressible. From the table in Appendix A, rgy = 1260 kg>m3 and vgy = 1.19 110-3 2 m2 >s for glycerin at 20°C. Using this mass flow rate, the average velocity is given by # m = rgyVA; 14.8 kg>s = 11260 kg>m3)(V23p>41D2 24 V =

0.01496 D2

Thus, the Reynolds number is 0.01496 a b 1D2 VD 12.5677 D2 Re = = = vgy D 1.19110-3 2 m3 >s

Assuming that laminar flow occurs. Thus the friction factor can be determined from f =

64 64 = = 5.0924D Re 12.5677>D

Then the major head loss can be determined using the Darcy–Weisbach equation. hL = f

2 2 L V2 100 m 10.01496>D 2 0.005805 d = = 15.0924D2 a bc 2 D 2g D 219.81 m>s 2 D4

Since the pipe has a constant diameter and is horizontal, the energy equation gives p1 - p2 = hL; g

350 1103 2 N>m2

112.60 kg>m3 219.81 m>s2 2

=

0.005805 D4 Ans.

D = 0.1197 m = 120 mm The value of the Reynolds number is Re =

12.5677 = 105 6 2300 (laminar flow, ok!) 0.1197 m

Ans: D = 120 mm 879

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10–38. Water at 80°F is pumped from the large underground tank at A and flows through the 6-in.-diameter smooth pipe. If the discharge at B is 150 ft3 >min, determine the required power output for the pump that is connected to a 90-ft length of the pipe. Draw the energy grade line and the hydraulic grade line for the pipe with reference to the datum set through A.

10 ft B 50 ft

308

C

30 ft

SOLUTION A

We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.934 slug>ft3 and vw = 9.35110-6 2 ft2 >s for water at 80°F. From the discharge, the average velocity is V =

Q = A

1150 ft3 >min2 a

Then the Reynolds number is Re =

VD = vw

1 min b 60 s

EGL HGL

61.3 ft

= 12.73 ft>s

2 3 pa ft b 12

58.8 ft 57.5 ft

6 112.73 ft>s2 a ft b 12 9.35110-6 2 ft2 >s

2.52 ft

= 6.811105 2

hpump 5 63.2 ft

For the smooth pipe, the Moody diagram gives f = 0.0125. Thus the major head loss can be determined using the Darcy–Weisbach equation,

A

C Pump –1.89 ft

2

112.73 ft>s2 L V 90 ft d = 5.6639 ft hL = f = 0.0125 c D 2g ° 6 ¢ 2132.2 ft>s2 2 ft 12 2

–2.52 ft

Take the control volume as the water contained in the pipe and pump from A to B. Since the water surface at A is open to the atmosphere and water is discharged into the atmosphere at B, pA = pB = patm = 0. Also, the water is drawn from a large tank, VA = 0. With reference to the datum through A, zA = 0 and zB = 30 ft + 50 sin 30° ft = 55 ft. Applying the energy equation between A and B,

55.0 ft

2.52 ft

B

Datum

–4.41 ft (a)

pA pB VA2 VB 2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

112.73 ft>s2 2

2132.2 ft>s2 2

+ 55 ft + 0 + 5.6639 ft

hpump = 63.1812 ft Thus, the required power ouput of this pump is # 1 min Wout = gwQhpump = 11.934 slug>ft2)(32.2 ft>s2 2 c 1150 ft3 >min2 a b d 163.1812 ft2 60 s = 19836.49 ft # lb>s2 a

1 hp

550 ft # lb>s

= 17.9 hp

b

The EGL and HGL will be plotted with reference to the datum set through A. Here, the velocity head is 112.73 ft>s2 2 V2 = = 2.517 ft 2g 2132.2 ft>s2 2

And the head loss per foot length of the pipe is

hL 5.6639 ft = = 0.06293 ft>ft L 90 ft The HGL is 2.517 ft below and parallel to the EGL. The plot is shown in Fig. a

Ans: # Wout = 17.9 hp

880

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10–39. 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 VD = n

Re =

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

64 64. = = 1.9938 (round) Re 32.10

f =

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

100 mm

(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 yaxis, ΣF =

a31.58 ( 103 )

0 VrdV + VrV # dA 0t Lcv Lcs

+ c ΣFy = rQ 3 ( Vout ) y - ( Vin ) y4

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

Ans: pA = - 31.6 kPa, FA = 190 N 881

M10_HIBB9290_01_SE_C10_ANS.indd 881

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*10–40. 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 = 0.018a 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: pA = 22.9 kPa 882

M10_HIBB9290_01_SE_C10_ANS.indd 882

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10–41. 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. 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

A

(1)

Then, the head loss from A to B can be determined using hL = f

L V2 2m V2 = fa b£ § = 2.0387f 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 + 2.0387f 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 1

0.02

2

0.0155

V(m , s), Eq. (2)

Re, Eq. (1)

f from Moody diagram

8.73

4.36 ( 105 )

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 Eq. (2) is solved for f, and the result and Eq. (1) are substituted into the 2.51> ( Re1f ) term, the V will cancel out and the logarithm will become a constant.

Ans: Q = 19.5 liter>s 883

M10_HIBB9290_01_SE_C10_ANS.indd 883

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10–42. Sewage, assumed to be water where r = 998.3 kg>m3, is pumped from the wet well using 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 is -98.7 kPa. Neglect the friction losses in the submerged segment of the pipe.

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 )

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, pA pB VA2 VB2 + zA + hpump = + zB + hturb + hL + + g g 2g 2g N V2 m2 0 + 0 + 0 + 0 = + + 3 m + 0 + 0.07951V 2 ( 998.3 kg>m3 )( 9.81 m>s2 ) 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

884

M10_HIBB9290_01_SE_C10_ANS.indd 884

Ans.

Ans: Q = 0.0145 m3 >s

10/03/17 2:38 PM

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10–43. Sewage, assumed to be water where r = 998.3 kg>m3, is pumped from the wet well using 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. Neglect the friction losses in the submerged segment of the pipe.

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 = 0.026 a b£ § = 0.11927V 2 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. Applying the energy equation from A to C with the datum at A, pA pC VC 2 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

# Ws = gQhpump

500

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.

885

M10_HIBB9290_01_SE_C10_ANS.indd 885

Ans: Q = 0.00804 m3 >s

10/03/17 2:38 PM

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*10–44. Determine the power the pump must supply in order to discharge 45 ft3 >min of water at T = 60°F at B from the 3-in.-diameter hose. The 200-ft-long hose is made of a material having a roughness of e = 0.0005 ft.

B

80 ft

A

SOLUTION We will consider the fully developed steady flow, and water is incompressible From the table in Appendix A, rw = 1.939 slug>ft3 and vw = 12.2 110-6 2 ft2 >s for water

1 min at T = 60°F. From the discharge Q = 145 ft3 >min2 a b = 0.75 ft2 >s, the 60 s average velocity is given by V =

The Reynolds number is

0.75 ft3 >s Q = = 15.28 ft>s 2 A 1.5 pa ft b 12

VD Re = = vw For the hose,

115.28 ft>s2 a

3 ft b 12

12.2110-6 2 ft2 >s

= 3.13 1105 2

e 0.0005 ft = = 0.002. Entering these two values into Moody diagram, D 3 a ft b 12

we obtain f = 0.0242. Then, the major head loss be determined using the Darcy– Weisbach equation. hL = f

2 L V2 200 ft 115.28 ft>s2 d = 70.18 ft = 0.0242 c 2 D 2g ° 3 ¢ 2132.2 ft>s 2 ft 12

Take the water contained in the hose and pump contained between A and B since water surface at A is opened to atmosphere and the water is discharged into atmosphere, pA = pB = patm = 0. Also water is drawn from a large reservoir, VA ≃ 0. With reference to the datum set through A, zA = 0 and zB = 80 ft. Applying the energy equation between A and B, pA pB V A2 V B2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

115.28 ft>s2 2

2132.2 ft>s2 2

+ 80 ft + 0 + 70.18 ft

hpump = 153.80 ft The required power supplied by the pump is # Wo = gwQhpump = 11.939 slug>ft3)(32.2 ft>s2)(0.75 ft3 >s21153.80 ft2 = 17.2021103 2 ft # lb>s2 a = 13.1 hp

1 hp

550 ft # lb>s

b

Ans. Ans: # Wo = 13.1 hp 886

M10_HIBB9290_01_SE_C10_ANS.indd 886

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10–45. Determine the power extracted from the water at 15°C by the turbine at C if the discharge is 0.07 m3 >s. The galvanized iron pipe has a diameter of 150 mm. Also draw the energy grade line and the hydraulic grade line for the pipe.

20 m A B

10 m

30 m

SOLUTION

C

We will consider the fully developed steady flow and water is incompressible. From the table in Appendix A, rw = 999.2 kg>m3 and vw = 1.15 110-6 2 m2 >s for water at T = 15°C. From the discharge, the average velocity is given by V =

0.07 m3 >s Q = = 3.961 m>s A p10.075 m2 2

Thus, the Reynolds number is Re =

13.961 m>s210.15 m2 VD = = 5.17 1105 2 vw 1.15110-6 2 m2 >s

e 0.15 mm = = 0.001. Entering these two values into D 150 mm Moody diagram, we obtain f = 0.0202. Then the major head loss can be determined using the Darcy–Weisbach equation. For galvanized iron pipe,

hL = f

13.961 m>s2 2 L V2 60 m d = 6.462 m = 0.0202a bc D 2g 0.15 m 219.81 m>s2 2

Since the water surface at A is opened to atmosphere and the water is discharged into atmosphere at C, pA = pC = patm = 0. Also, the water is drawn from a large reservoir, VA ≃ 0 with reference to the datum set through C, zA = 40 m and zC = 0. Take the water in the reservoir, the pipe, and turbine from A to C as the control volume and write the energy equation from A to C, pA pC VC2 VA2 + + zA + hpump = + + zC + hturb + hL gw gw 2g 2g 0 + 0 + 40 m + 0 = 0 +

13.961 m>s2 2

219.81 m>s2 2

+ 0 + hturb + 6.462

hturb = 32.7383 m

The power drawn by the turbine can be determined from # W = gQhturb = 1999.2 kg>m3)(9.81 m>s2)(0.07 m3 >s)(32.7383 m2 = 22.461103 2W

Ans.

= 22.5 kW

887

M10_HIBB9290_01_SE_C10_ANS.indd 887

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10–45. Continued

The EGL and HGL will be plotted with reference to the datum set through C. Here, the velocity head is 13.961 m>s2 2 V2 = = 0.7997 m 2g 219.81 m>s2 2

and the head loss per meter length of pipe is

hL 6.462 m = = 0.1077 m>n L 60 m The HGL is 0.7997 m below and parallel to EGL. The plot is shown in Fig. a.

40 m 39.20 m

EGL HGL

0.80 m

Turbine 33.54 m

0.80 m A

B

C

Datum

(a)

Ans: # W = 22.5 kW 888

M10_HIBB9290_01_SE_C10_ANS.indd 888

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10–46. The spillway from a reservoir consists of a 3-ft-diameter concrete pipe having a total length of 750 ft. Determine the flow through the pipe. The roughness of the concrete is e = 0.006 ft, and the temperature of the water is at 60°F.

A

3 ft

450 ft B

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.939 slug>ft3 and vw = 12.2(10-6) ft2 >s for water at T = 60°F. The Reynolds number of the flow in the pipe is Re =

V(3 ft) VD = = 2.45901105 2V vw 12.2110-6 2 ft2 >s

(1)

The major head loss can be determined using Darcy–Weisbach equation hL = f

L V2 750 ft V2 d = 3.8820f V 2 = fa bc D 2g 3 ft 2132.2 ft>s2 2

Take the water contained in the dam and pipe between A and B as the control volume. Since water is drawn from a large reservoir the water level at A can be considered constant and so VA ≃ 0. The water surface at A is exposed to the atmosphere and the water is discharged into the atmosphere at B. Then pA = pB = patm = 0. With reference to the datum set through B, zA = 450 ft and zB = 0. Write the energy equation between A and B, pA pB V A2 V B2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 450 ft + 0 = 0 +

V2 + 0 + 0 + 3.8820f V 2 2132.2 ft>s2 2

V2 + 3.8820f V 2 = 450 64.4

(2)

e 0.006 ft = = 0.002. The trial and D 3 ft error iterative procedure is required. The iterations are tabulated below. For concrete pipe, the relative roughness is

Iteration Assumed f 1 2

0.024 0.0238

V(m , s), [Eq. (2)]

Re [Eq. (1)]

f from Moody diagram

64.34

1.58 ( 10

)

0.0238

64.57

1.59 ( 10

)

0.0238

7 7

The assumed f of 2nd iteration is almost the same as that from the Moody diagram. Thus, V = 64.57 ft>s is an acceptable result. The flow rate is given by Q = VA = 164.57 ft>s2 3 p11.5 ft2 2 4 = 456.45 ft3 >s = 456 ft3 >s

889

M10_HIBB9290_01_SE_C10_ANS.indd 889

Ans.

Ans: Q = 456 ft3 >s

10/03/17 2:39 PM

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10–47. If the length of the 6-in.-diameter commercial steel pipe from A to the pump and from the pump to B is 20 ft and 150 ft, respectively, determine the discharge generated by the 30-hp pump. Also, draw the EGL and HGL for the pipe from A to B. The water temperature is at 50°F.

B 6 ft

A

SOLUTION We will consider the fully developed steady flow, and the water is incompressible. From the table in Appendix A, rw = 1.940 slug>ft3 and vw = 14.1(10-6) ft2 >s for the water at T = 50°F. Then Reynolds number of the flow in the pipe is 6 V a ft b 12 VD Re = = = 3.54611104 2V vw 14.1110-6 2 ft2 >s

(1)

The major head loss can be determined using Darcy–Weisbach equation hL = f

L V2 170 ft V2 d = 5.2795f V 2 = fa bc D 2g 6>12 ft 2132.2 ft>s2 2

Take the water in the river, pump, and pipe between A and B as the control volume. Since the water surface at A is exposed to the atmosphere and the water is discharged into the atmosphere at B, pA = pB = patm = 0. Since the water is drawn from a large reservoir, the water level at A can be considered constant and so VA ≃ 0. With reference to the datum set through A, zB = 6 ft and zA = 0. Write the energy equation between A and B, pA pB VA2 VB 2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g V2 + 6 ft + 5.2795 f V 2 2132.2 ft>s2 2

0 + 0 + 0 + hpump = 0 + hpump =

V2 + 5.2795f V 2 + 6 64.4

The power supplied by the pump is # W = gwQhpump;

130 hp2 a

= 11.940 slug>ft3 2132.2 ft>s2 2 e V c pa a

550 ft # lb>s 1 hp

b

2 3 V2 ft b d f a + 5.2795 f V 2 + 6b 12 64.4

1 + 5.2795f bV 3 + 6V - 1345.23 = 0 64.4

(2)

e 0.0015 ft = = 0.0003. D 6 ft 12 The trial and error iterative procedure is required. The iterations are tabulated below. For commercial steel pipe, the relative roughness is

Iteration Assumed f 1

0.017

2

0.0158

V(ft , s), (Eq. 2)

Re (Eq. 1)

f from Moody diagram

22.57

8.00 ( 105 )

0.0158

23.02

8.16 ( 105 )

0.0158

890

M10_HIBB9290_01_SE_C10_ANS.indd 890

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10–47. Continued

The assumed f in 2nd iteration is almost the same as that from the Moody diagram. Thus, V = 23.02 ft>s is an acceptable result. Then, the discharge is Q = VA = 123.02 ft>s2 c pa

And the pump head is hpump =

2 3 ft b d = 4.52 ft3 >s 12

Ans.

23.022 + 5.279510.01582123.022 2 + 6 = 58.43 ft 64.4

and the velocity head is

123.02 ft>s2 2 V2 = 8.229 ft = 2g 2132.2 ft>s2 2

The head losses from A to the pump and pump to B are 1hL 2 A9pump = 0.0158

2 20 ft 123.02 ft>s2 d = 5.200 ft c ° 6 ¢ 2132.2 ft>s2 2 ft 12

1hL 2 pump9B = 0.0158

2 150 ft 123.02 ft>s2 d = 39.003 ft c 2 ° 6 ¢ 2132.2 ft>s 2 ft 12

Based on these results, the EGL and HGL with reference to the datum set through A shown in Fig. acan be plotted. EGL HGL 53.23 ft

39.00 ft hpump 5 58.43 ft 8.229 ft 6 ft

A A9 B

Datum

5.200 ft (a)

891

M10_HIBB9290_01_SE_C10_ANS.indd 891

Ans: Q = 4.52 ft3 >s

10/03/17 2:39 PM

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*10–48. Water is pumped into the truck at 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 = 0.018 a b£ § = 2.905 m 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, pC pA VC 2 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: P = 427 W 892

M10_HIBB9290_01_SE_C10_ANS.indd 892

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10–49. Water is pumped at 0.003 m3 >s into the top of the truck through 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 = Major head loss

0.003 m3 >s Q = = 2.3873 m>s A p(0.02 m)2

( 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, pC pA 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 = gQhpump = ( 1000 kg>m3 )( 9.81 m>s2 )( 0.003 m3 >s ) (6.5977 m) # Ws = 194 W

Ans.

Ans: # Ws = 194 W 893

M10_HIBB9290_01_SE_C10_ANS.indd 893

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10–50. The sump pump is required to lift water from the pit to a reservoir. If a 2-in.-diameter commercial steel pipe is used, determine the required power supplied by the pump so that the flow is 250 gal>min. The pipe is 175 ft long. The water temperature is at 60°F.

2 in. B 30 ft A

SOLUTION We will consider the fully developed steady flow and the water is incompressible. From the table in Appendix A, rw = 1.939 slug>ft3 and vw = 12.2110-6 2 ft2 >s for water at T = 60°F. Here, the flow rate is Q = a250

gal

min

ba

1 ft3 1 min ba b = 0.5570 ft3 >3 7.48 gal 60 s

Then the flow velocity in the pipe is given by

Q = VA; 0.5570 ft3 >s = V c pa

2 1 ft b d 12

V = 25.53 ft>s

The Reynolds number of the flow in the pipe is VD Re = = vw

125.53 ft>s2 a

2 ft b 12

12.2110-6 2 ft2 >s

= 3.488 1105 2

For commercial steel pipe, the relative roughness is,

e 0.00015 ft = = 0.0005. D 2 ft 12

Enter these two values into the Moody diagram, we obtain f = 0.02. Then the major head loss can be determined using the Darcy–Weisbach equation. (hL)ma = f

2 L V2 175 ft 125.53 ft>s2 d = 212.59 ft = 0.02 c D 2g ° 2 ¢ 2132.2 ft>s2 2 ft 12

Take the water in the pit, the pump, and the pipe between A and B as the control volume. Since water surface at A is exposed to the atmosphere and the water is discharged into the atmosphere at B, pA = pB = patm = 0. Also, the water is drawn from a large reservoir, and so VA ≈ 0. With reference to the datum set through A, zB = 30 ft and zA = 0. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

125.53 ft>s2 2

2132.2 ft>s2 2

+ 30 ft + 0 + 212.59 ft

hpump = 252.71 ft

Then, the required power supplied by the pump is # W = gwQhpump = 11.939 slug>ft3 2132.2 ft>s2 210.5570 ft3 >s21252.71 ft2 = a8789.02

ft # lb b s °

1 hp

550

ft # lb ¢ s

Ans.

= 15.98 hp = 16.0 hp

Ans: # W = 16.0 hp

894

M10_HIBB9290_01_SE_C10_ANS.indd 894

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10–51. The submersible pump is used to fill the cylindrical tank. If the level of water in the tank rises at 0.1 ft>s, when the height h = 3 ft, determine the pressure developed at the exit A of the pump at this instant. The 2-in.-diameter cast iron pipe is 80 ft long. The temperature of the water is 70°F.

2.5 ft

B h

75 ft

SOLUTION We will consider the fully developed steady flow and water is incompressible. Take the water in the tank and the pipe between A and B as the control volume. The continuity requires

A

0 r dV + r V # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 -V c pa

2 1 ft b d + 10.1 ft>s23p11.25 ft2 2 4 = 0 12

V = 22.5 ft>s

From the table in Appendix A, rw = 1.937 slug>ft3 and vw = 10.4110-6 2 ft2 >s for water at T = 70°F. The Reynolds number of the flow in the pipe is VD Re = = vw For cast iron pipe,

122.5 ft>s2 a

2 ft b 12

10.4110-6 2 ft2 >s

= 3.611105 2

e 0.00085 ft = = 0.0051. Entering these two values into the Moody D 2 ft 12

diagram, we obtain f = 0.03. Then, the major head loss can be determined using the Darcy–Weisbach equation. hL = f

122.5 ft>s2 2 L V2 80 ft d = 113.20 ft = 0.03 c D 2g ° 2 ¢ 2132.2 ft>s2 2 ft 12

Since point B is exposed to the atmosphere, pB = patm = 0. With reference to the datum set through A, zB = 75 ft + 3 ft = 78 ft and zA = 0. Write the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g pA 11.937 slug>ft3 2132.2 ft>s2 2

+

122.5 ft>s2 2

2132.2 ft>s2 2

pA = a11.43501103 2

+ 0 + 0 = 0 +

10.1 ft>s2 2

2132.2 ft>s2 2

lb 1 ft 2 b a b = 79.41 psi = 79.4 psi ft2 12 in.

+ 78 ft + 0 + 113.20 ft

Ans.

Ans: pA = 79.4 psi 895

M10_HIBB9290_01_SE_C10_ANS.indd 895

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*10–52. 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 VA2 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, pA pB VA 2 VB 2 + + zA + hpump = + + zB + hturb + hL g g 2g 2g pA

( 1.127 kg>m3 )( 9.81 m>s2 ) =

pB

( 1.127 kg>m3 )( 9.81 m>s2 )

+

+

( 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 896

M10_HIBB9290_01_SE_C10_ANS.indd 896

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10–53. Air at a temperature of 80°F flows through the duct at A with a velocity of 30 ft>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.

18 in.

6 in. B

A

SOLUTION We will consider fully developed steady flow and treat air as incompressible. Take the air contained in the transition between A and B as the control volume. Using the average velocities, the continuity equations give 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB = 0 - 130 ft>s2 c pa

2 2 3 9 ft b d + VB c pa ft b d = 0 12 12

VB = 3.3333 ft>s

The head loss is the minor head loss of sudden change in diameter. Here,

Thus,

KL = c 1 - a 1hL 2 mi = KL

dA 2 2 6 in. 2 2 b d = c1 - a b d = 0.7901 dB 18 in.

130 ft>s2 2 V A2 d = 11.042 ft = 10.79012 c 2g 2132.2 ft>s2

From the table in Appendix A, ra = 0.00228 slug>ft3 for air at 80°F. Since the density of air is small, the elevation terms will be neglected. Applying the energy equation between A and B, pA pB VA2 VB2 + + hpump = + + hturb + hL gw ga 2g 2g pA 10.00228 slug>ft3 2132.2 ft>s2 2

+

130 ft>s2 2

2132.2 ft>s2 2

pB - pA = 10.2027 lb>ft2 2 a

+ 0 =

pB 10.00228 slug>ft3 2132.2 ft>s2 2

1 ft 2 b = 1.41110-3 2 psi 12 in.

897

M10_HIBB9290_01_SE_C10_ANS.indd 897

+

13.3333 ft>s2 2 2132.2 ft>s2 2

+ 0 + 11.042 ft

Ans.

Ans: pB - pA = 1.41110 - 3 2 psi

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10–54. 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

10 ft

Assume that fully developed steady flow occurs and water is incompressible. Appendix A gives rw = 1.937 slug>ft3 and nm = 10.4 ( 10-6 ) ft2 >s. Then the Reynolds number is Re =

V ( 2>12 ft ) VD = = 16.026 ( 103 ) V nm 10.4 ( 10-6 ) ft2 >s

(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>ft2 and pB = patm = 0, hpump = hturb = 0 ba 2 1 ft in

pA pB VA 2 VB2 + + zA + hpump = + + zB + hturb + hL gv 2g gv 2g

5760 lb>ft2

( 1.937 slug>ft )( 32.2 ft>s ) 2

2

+ 0 + 4 ft + 0 = 0 +

(192f + 3.49) a

V2 2 ( 32.2 ft>s

V2 b = 80.35 64.4

2

)

+ 16 + 0 + (192f + 2.49) a

(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. Entering these 2>12 ft values of Re and e>D into the Moody diagram, we obtain f = 0.0265. Repeating 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 ft3 >s 12

Ans.

898

M10_HIBB9290_01_SE_C10_ANS.indd 898

Ans: Q = 0.536 ft3 >s

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10–55. The 100-mm-diameter commercial steel pipe discharges 0.025 m3 >s of water at 25°C through the 40-mm-diameter nozzle at B. Determine the pressure at A. The minor head loss coefficient for the nozzle is KL = 0.15.

40 mm B

3m

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 997.1 kg>m3 and vw = 0.898110-6 2 m2 >s for water at 25°C. To satisfy the continuity condition, the average velocities at A and B are VA =

0.025 m3 >s Q 10 = = m>s p AA p10.05 m2 2

VB =

0.025 m3 >s Q 62.5 = = m>s 2 p AB p10.02 m2

608 A 100 mm

The Reynolds number along the pipe is

10 a m>s b 10.1 m2 p VD Re = = = 3.541105 2 vw 0.898110-6 2 m2 >s

The relative roughness of commercial steel pipe is

e 0.045 mm = = 0.00045 D 100 mm Entering these two values into Moody diagram, we obtain f = 0.018 Then, the major head loss can be determined using the Darcy–Weisbach equation, 2 10 m>s b § 3m £ p L = 0.018a 1hL 2 ma = f b = 0.2789 m 2 D 2g 0.1 m 219.81 m>s 2

a

VA2

The minor head loss at the nozzle is

1hL 2 mi

2 62.5 m>s b § £ p = 10.152 = KL = 3.0259 m 2g 219.81 m>s2 2

a

VB2

Since water is discharged into the atmosphere at B, pB = patm = 0. With reference to the datum set through A, zA = 0 and zB = 3 sin 60° m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g a

2 10 m>s b p

a

2 62.5 m>s b p

pA + + 0 + 0 = 0 + + 3 sin 60° m + 0 + 10.2789 m + 3.0259 m2 gw 219.81 m>s2 2 219.81 m>s2 2 pA = 1997.1 kg>m3 219.81 m>s2 2125.5589 m2 = 250.011103 2 Pa = 250 kPa

Ans. Ans: pA = 250 kPa

899

M10_HIBB9290_01_SE_C10_ANS.indd 899

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*10–56. The 100-mm-diameter commercial steel pipe discharges water at 25°C through the 40-mm-diameter nozzle. If the pressure at A is 250 kPa, determine the discharge. The minor loss coefficient of the nozzle is KL = 0.15.

40 mm B

3m

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 997.1 kg>m3 and vw = 0.898110-6 2 m2 >s for water at 25°C. Take the water contained in the pipe from A to B as the control volume. The continuity condition requires

608 A 100 mm

0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VA 3 p10.05 m2 2 4 + VB 3 p10.02 m2 2 4 = 0 VB = 6.25 VA

The Reynolds number along the pipe is VA 10.1 m2 VAD = = 1.11361105 2VA vw 0.898110-6 2 m2 >s

Re =

(1)

The major head loss can be determined using the Darcy–Weisbach equation 1hL 2 ma = f

VA2 L VA2 3m d = 1.5291f VA2 = fa bc D 2g 0.1 m 219.81 m>s2 2

The minor head loss at the nozzle is 1hL 2 mi = KL

16.25VA 2 2 VB2 d = 0.2986 VA2 = 10.152 c 2g 219.81 m>s2 2

Since water is discharged into the atmosphere at B, pB = patm = 0. With reference to the datum set through A, zA = 0 and zB = 3 sin 60° m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 2501103 2 N>m2

1997.1 kg>m3 219.81 m>s2 2 = 0 +

+

VA2 219.81 m>s2 2

16.25VA 2 2

219.81 m>s2 2

+ 0 + 0

+ 3 sin 60° m + 0 + 1.5291f VA2 + 0.2986 VA2

12.2386 + 1.5291 f2VA2 - 22.9602 = 0

(2)

900

M10_HIBB9290_01_SE_C10_ANS.indd 900

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10–56. Continued

e 0.045 mm = = 0.00045. D 100 mm

For commercial steel pipe, the relative roughness is

The trial and error iterative procedure is required. The iterations are tabulated below. Iteration 1 2

Assumed f 0.02 0.018

VA(m , s) (Eq. 2)

Re (Eq. 1)

f from Moody diagram

3.1810

3.54 ( 10

)

0.018

3.1831

3.54 ( 10

)

0.018

5 5

The assumed in 2nd iteration is almost the same as that from Moody diagram. Thus, VA = 3.1831 m>s is an acceptable result. Then the discharge is given by Q = VAAA = 13.1831 m>s2 3 p10.05 m2 2 4 = 0.0250 m3 >s

901

M10_HIBB9290_01_SE_C10_ANS.indd 901

Ans.

Ans: Q = 0.0250 m3 >s

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10–57. Water at 200°F enters the radiator at A with a 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 within the pipe at the exit B. The copper pipe has a diameter of 14 in. Take e = 5110-6 2 ft. The radiator is in the vertical plane.

1.75 ft A

2 ft

SOLUTION Water is considered to be incompressible. From Appendix A, r = 1.869 slug>ft3 and n = 3.40 ( 10-6 ) s2 >ft. Then, the Reynolds number is

B

0.25 ( 4 ft>s ) a ft b 12 VD Re = = 2.45 ( 104 ) = n 3.40 ( 10-6 ) ft2 >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, pA pB VA 2 VB 2 + + zA + hpump = + + zB + hturb + hL g g 2g 2g lb 12 in. 2 ba b 2 1 ft V2 in + + 2 ft + 0 = 3 2 ( 1.869 slug>ft )( 32.2 ft>s ) 2g a60

pB = a8073.77

pB

( 1.869 slug>ft )( 32.2 ft>s ) 3

1 ft 2 lb ba b = 56.1 psi 2 12 in. ft

2

+

V2 + 0 + 0 + 7.826 ft + 3.583 ft 2g

Ans.

Ans: pB = 56.1 psi 902

M10_HIBB9290_01_SE_C10_ANS.indd 902

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10–58. 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>ft3.

A

B

SOLUTION Water is considered to be incompressible. Here, Q = a900

gal min

ba

the flow in the pipe is

1 ft3 1 min ba b = 2.005 ft3 >s. Thus, the mean velocity of 7.48 gal 60 s V =

2.005 ft3 >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, pA pB VA 2 VB 2 + + 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 ft2

Ans.

kL = 0.901

Ans: KL = 0.901 903

M10_HIBB9290_01_SE_C10_ANS.indd 903

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10–59. 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 four elbows and gate valve.

3m

A 1m

1m

1m C

SOLUTION

B

20 mm

Water is considered to be incompressible. The mean velocity of the flow in the pipe is V =

0.5 m

1m

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 determined using

For galvanized iron pipe, the relative roughness is

( 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.

Ans: pA = 742 kPa 904

M10_HIBB9290_01_SE_C10_ANS.indd 904

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*10–60. Water at 80°F flows at 5 ft>s through the 34 -in.diameter pipe at A. If the 12 -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 gate valve. The gate valve at E is closed. For the pipe take f = 0.016.

2 ft C

0.5 ft 6 ft E 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> ft3 and nw = 9.35 ( 10-6 ) ft2 >s at T = 80° F.

3 ft A

Continuity requires

QA = QB VA c pa

2 2 0.375 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

( 5 ft>s ) 2 LAD VA 2 3 ft + ΣKL b = ≥ 0.016 ± ≤ + 0.9 ¥ D 2g 0.75 2 ( 32.2 ft>s2 ) ft 12 = 0.6475 ft

( 11.25 ft>s ) 2 VB2 = (1.8 + 0.19 + 0.6) £ § = 5.0900 ft 2g 2 ( 32.2 ft>s2 )

( hL ) DB = ΣKL

Applying the energy equation between A and B, where pB = 0, zA = 0, zB = 2.75 ft, pA pB VA 2 VB 2 + + zA + hpump = + + zB + hturb + ( hL ) AB gw gw 2g 2g pA 1.934 slug>ft ( 32.2 ft>s 3

0 +

2

)

+

( 5 ft>s ) 2 + 0 + 0 = 2 ( 32.2 ft>s2 )

( 11.25 ft>s ) 2 + 2.75 ft + 0 + 0.6475 + 5.0900 2 ( 32.2 ft>s2 ) pA = 626.77 lb>ft2 = 4.35 psi

Ans.

Ans: pA = 4.35 psi 905

M10_HIBB9290_01_SE_C10_ANS.indd 905

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10–61. Water at 80°F flows at 5 ft>s through the 34 -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 three elbows, the fully opened gate valve at E, and the tee. The gate valve at B is closed. For the copper pipe take f = 0.016.

2 ft C

0.5 ft 6 ft E B

D 0.25 ft 3 ft

SOLUTION

A

Assume that fully developed steady flow occurs, and water is incompressible. Appendix A gives rw = 1.934 slug>ft3 and nw = 9.35 ( 10-6 ) ft2 >s at T = 80° F.

Continuity requires

QA = QC

( 5 ft>s ) c p a

2 0.375 0.03125 ft b d = VC £ (100) ap ft b § 12 12

VC = 7.20 ft>s

The loss coefficients for the elbows, 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 = afAD a

( 5 ft>s ) 2 LAD V 2A 3 ft b + Σ KL b = ≥ 0.016 ± ≤ + 0.9 ¥ ° ¢ D 2g 0.75 2 ( 32.2 ft>s2 ) ft 12 = 0.6475 ft

( hL ) DC = afDC a

LDC VL2 VA2 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, pA pC V C2 V A2 + + zA + hpump = + + zC + hturb + ( hL ) AC gw gw 2g 2g pA 1.934 slug>ft3 ( 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 = 728.56 lb>ft2 = 5.06 psi

Ans.

Ans: pA = 5.06 psi

906

M10_HIBB9290_01_SE_C10_ANS.indd 906

10/03/17 2:39 PM

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10–62. Water at 15°C is pumped from the reservoir A into the large tank B. If the power output of the pump is 3.70 kW, determine the volumetric flow into the tank when h = 8 m. The commercial steel pipe has a total length of 30 m and a diameter of 75 mm. Include the minor losses for the elbow and sudden expansion into the tank.

B

h

A

SOLUTION We will consider fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 999.2 kg>m3 and vw = 1.15110-6 2 m2 >s for water at T = 15°C. The Reynolds number is Re =

V10.075 m2

1.15110-6 2 m2 >s

= 6.52171104 2V

(1)

The major head loss can be determined using the Darcy–Weisbach equation 1hL2 major = f

L V2 30 m V2 d = 20.3874f V 2 = fa bc D 2g 0.075 m 219.81 m>s2 2

The minor head loss is contributed by one 90° elbow 1KL = 0.92 and the sudden expansion into the tank 1KL = 1.02 V2 V2 d = 0.09684 V 2 = 10.9 + 1.02 c 1hL 2 minor = a KL 2g 219.81 m>s2 2

Take the control volume as the water contained in the reservoir, tank and pipe between A and B Here, VA = VB = 0 (large reservoir) pA = pB = patm = 0 (surfaces at A and B are opened to the atmosphere) with reference to the datum set through A, zA = 0 and zB = 8 m. Applying the energy equation between A and B, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 + 0 + 8 m + 0 + 20.3874f V 2 + 0.09684V 2 hpump = 120.3874f + 0.096842V 2 + 8

From the output power of the pump, # W = gwQhpump ; 3.701103 2W = 1999.2 kg>m219.81m>s2 21V23p10.0375 m2 2 4320.3874f + 0.09684)V 2 + 84 120.3874f + 0.096842V 3 + 8V - 85.4413 = 0

For commercial steel pipe the relative roughness is

(2)

e 0.045 mm = = 0.0006. D 75 mm

The trial and error iterative procedure is required. The iterations are tabulated as follows: Iteration Assumed f 1

0.019

V(m , s) (Eq. 2) 4.6388

Re (Eq. 1) 3.03 ( 10

5

f from Moody diagram

)

0.0187

The assumed f in this iteration is very close to that from Moody diagram. Thus, the result of V = 4.6388 m>s is acceptable. Then, the flow rate is Q = VA = ( 4.6388 m>s ) 3 p ( 0.0375 m ) 2 4 = 0.0205 m3 >s

907

M10_HIBB9290_01_SE_C10_ANS.indd 907

Ans. Ans: Q = 0.0205 m3 >s

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10–63. Water at 80°F flows from the large reservoir through the 3-in.-diameter galvanized iron pipe. Determine the discharge if the globe valve is fully opened. The length of the pipe is 150 ft. Include the minor losses of the flush entrance, the four elbows, and the globe valve.

A

15 ft B 3 ft

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.934 slug>ft3 and vw = 9.35110-6 2 ft2 >s for water at T = 80°F. Then the Reynolds number is 3 Va ft b 12 VD Re = = = 2.67381104 2V vw 9.35110-6 2 ft2 >s

(1)

The major head loss can be determined using the Darcy–Weisbach equation. 1hL 2 major = f

150 ft V2 L V2 d = 9.3168f V 2 = f c D 2g ° 3 ¢ 2132.2 ft>s2 2 ft 12

The minor head loss is contributed by the flush entrance 1KL = 0.52, four 90° elbows 1KL = 0.92, and one fully opened globe valve 1KL = 10.02. Thus 1hL 2 minor a K

V2 V2 d = 0.2189V 2 = 30.5 + 410.92 + 10.04 c 2g 2132.2 ft>s2 2

Take the control volume as the water contained in the pipes and large reservoir from A to B. Since the water surface at A is opened to the atmosphere and the water is discharged into atmosphere at B, pA = pB = patm = 0. Also, water is drawn from a large reservoir, VA ≃ 0. With reference to the datum set through B, zB = 0 and zA = 15 ft - 3 ft = 12 ft. Applying the energy equation between A and B, pA pB VA2 V B2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 12 ft + 0 = 0 +

V2 + 0 + 0 + 9.3168f V 2 + 0.2189 V 2 2132.2 ft>s2 2

19.3168f + 0.23452V 2 - 12 = 0

(2)

e 0.0005 ft = = 0.002. The trial D 3 ft 12 and error iterative procedure is required. The iterations are tabulated as follows:

For galvanized iron pipe, the relative roughness is

Iteration 1 2

Assumed f 0.024 0.025

V(ft , s) (Eq. 2)

Re (Eq. 1)

f from Moody diagram

5.1183

1.37 ( 10

)

0.025

5.0670

1.35 ( 10

)

0.025

5 5

The assumed f in 2nd iteration is almost the same as that from Moody diagram. Thus the result of V = 5.0670 ft>s is acceptable. Thus, the flow rate is given by Q = VA = 15.0670 ft>s2 c pa

2 1.5 ft b d = 0.249 ft3 >s 12

908

M10_HIBB9290_01_SE_C10_ANS.indd 908

Ans. Ans: Q = 0.249 ft3 >s

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*10–64. Water at 50°F flows through the 2-in.-diameter commercial steel pipe such that it exits the gate valve at 120 gal>min. If the open diameter of the gate valve is also 2 in. and the pressure at A is 4.50 psi, determine the required power the pump supplies at B. Consider the minor losses at the two 90° elbows and fully opened gate valve. Draw the energy grade line and hydraulic grade line for the pipe with reference to the datum set through A.

4 ft D

E

20 ft

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.94 slug>ft3 and vw = 14.1110-6 2 ft2 >s for water at T = 50°F. The discharge is Q = a120

gal min

ba

1 min 1 ft3 ba b = 0.2674 ft3 >s 60 s 7.48 gal

C

B

A 5 ft

5 ft

Then, the average velocity is given by

0.2674 ft3 >s Q = = 12.26 ft>s 2 A 1 pa ft b 12 Thus, the Reynolds number is V =

Re =

VD = vw

112.26 ft>s2 a

2 ft b 12

14.1110-6 2 ft2 >s

= 1.451105 2

e 0.00015 ft = = 0.0009. D 2 a ft b 12 Entering these two values into Moody diagram we obtain f = 0.0212. Then the major head loss can be determined using Darcy–Weisbach equation. For commercial steel pipe, the relative roughness is

1hL 2 major = f

112.26 ft>s2 2 L V2 34 ft d = 10.087 ft = 0.0212 c D 2g ° 2 ¢ 2132.2 ft>s2 2 ft 12

The minor head loss caused by two 90° elbows 1KL = 0.92 and the fully opened gate valve 1KL = 0.192 is 112.26 ft>s2 2 V2 d = 4.641 ft 1hL 2 minor = a KL = 3210.92 + 0.194 c 2g 2132.2 ft>s2 2

Take the water contained in the pipe and pump between A and E as the control volume. Since the diameter of the pipe is constant, VA = VE = V. The water is discharged into the atmosphere at E, so pE = patm = 0. With reference to the datum set through A, zA = 0 and zE = 20 ft. Applying the energy equation between A and E, pA pE VA2 V E2 + + zA + hpump = + + zE + hturb + hL gw g 2g 2g lb 12 in. 2 ba b 2 1 ft V2 V2 in + + 0 + hpump = 0 + + 20 ft + 0 + 110.087 ft + 4.641 ft2 3 2 2g 2g 11.94 slug>ft 2132.2 ft>s 2 a4.50

hpump = 24.355 ft

909

M10_HIBB9290_01_SE_C10_ANS.indd 909

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10–64. Continued

Thus, the power supplied by the pump can be determined from # WO = gwQhpump = 11.94 slug>ft3 2132.2 ft>s2 210.2674 ft3 >s2124.355 ft2 = 1406.79 ft # lb>s2 a

1 hp

550 ft # lb>s

= 0.740 hp

b

Ans.

Here, the velocity head is 112.26 ft>s2 2 V2 = = 2.332 ft 2g 2132.2 ft>s2 2

The major head loss per foot length of pipe is

hL 10.087 ft = = 0.2967 ft>ft L 34 ft The minor head losses at C (and D) and E 1KL 2 C 1KL 2 E

112.26 ft>s2 2 V2 d = 2.099 ft = 0.9c 2g 2132.2 ft>s2 2

112.26 ft>s2 2 V2 d = 0.443 ft = 0.19c 2g 2132.2 ft>s2 2

The HGL is always 2.332 ft below and parallel to the EGL. Here,

hA =

pA gw

lb 12 in. 2 b a b 112.26 ft>s2 2 1 ft in2 + + zA = + + 0 = 12.71 ft 2g 2132.2 ft>s2 2 11.94 slug>ft3 2132.2 ft>s2 2 VA2

a4.50

hB- = hA - 1hL 2 A–B = 12.71 ft - 10.2967 ft>ft215 ft2 = 11.22 ft hB+ = hB- + hpump = 11.22 ft + 24.355 ft = 35.58 ft

hC- = hB+ - 1hL 2 B–C = 35.58 ft - 10.2967 ft>ft215 ft2 = 34.09 ft hC+ = hC- - 1hL 2 minor = 34.09 ft - 2.099 ft = 31.99 ft

hD - = hC+ - 1hL 2 C–D = 31.99 ft - 10.2967 ft>ft2120 ft2 = 26.06 ft hD + = hD - - 1hL 2 minor = 26.06 ft - 2.099 ft = 23.96 ft

hE- = hD + - 1hL 2 D–E = 23.96 ft - 10.2967 ft>ft214 ft2 = 22.78 ft

hE + = hE- + 1hL 2 minor = 22.78 ft - 0.443 ft = 22.33 ft

910

M10_HIBB9290_01_SE_C10_ANS.indd 910

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10–64. Continued

Based on the above results, EGL and HGL is plotted as shown in Fig. a H(ft) 35.58

EGL HGL 34.09 31.99 26.06 23.96

2.33 ft

22.78 22.33

11.22

12.71

A

B

C

D

E

Datum

(a)

Ans: # WO = 0.740 hp 911

M10_HIBB9290_01_SE_C10_ANS.indd 911

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10–65. An automatic sprinkler system for a yard is made from 34 -in.-diameter PVC pipe, for which e = 9110 - 6 2 ft. If the system has the dimensions shown, determine the volumetric flow through each sprinkler head C and D. The faucet at A delivers water at 70°F with 40 psi pressure. Neglect elevation changes, but include the minor losses at the two elbows and the tee. Also, the loss coefficient of the 14 -in.-diameter sprinkler nozzles at C and D is KL = 0.05.

A

D B

40 ft

60 ft 50 ft 60 ft C

SOLUTION We will consider fully developed flow, and water is incompressible. From the table in Appendix A, rw = 1.937 slug>ft3 and vw = 10.4110-6 2 ft2 >s at T = 70°C. The Reynolds number for the flow is 0.75 Va ft b VD 12 Re = = = 6.00961 103 2 V vw 10.41 10-6 2 ft2 > s

(1)

Continuity requires

QP = Qnz VP c pa

2 2 0.375 0.125 ft b d = Vnzc pa ft b d 12 12

(2)

Vnz = 9VP

And QA = QC + QD VA c pa

2 2 2 0.375 0.375 0.375 ft b d = 1VC 2 P c pa ft b d + 1VD 2 P c pa ft b d 12 12 12

VA = 1VC 2 P + 1VD 2 P

(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 1hL 2 AB = afAB =

1hL 2 BC

£

fAB

LAB VA2 + ΣKb D 2g VA2 VA2 60 ft d + 0.9 c = 1960f + 0.92 a b AB ° 0.75 ¢ § 2132.2 ft>s2 2 64.4 ft 12

1VC 2p2 1VC 2 2nz LBC = afBC + ΣKb + KL D 2g 2g =

£

fBC

1VC 2 2p 391VC 2 p 4 2 60 ft d ¶ + 1.8 c + 0.05 • ° 0.75 ¢ § 2132.2 ft>s2 2 2132.2 ft>s2 2 ft 12

= 1960fBC + 5.852 c

1VC 2p2 64.4

d

912

M10_HIBB9290_01_SE_C10_ANS.indd 912

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10–65. Continued

2

1hL 2 BD = afBC =

£

fBD

1VD 2 p 1VD 2 2nz LBD + ΣKL b + KL D 2g 2g

1VD 2p2 391VD 2 p 4 2 90 ft d f + 10.4 + 0.92 c + 0.05e ° 0.75 ¢ § 2132.2 ft>s2 2 2132.2 ft>s2 2 ft 12

= 11440fBD + 5.352 c

1VD 2 p2 64.4

d

Applying the energy equation between A and C, and A and D, realizing that lb 12 in. 2 lb ba b = 5760 2 , pC = pD = patm = 0 and zA = zC = zD = z 2 1 ft in ft (same elevation) pA = a40

1VC 2 2nz pA pC VA2 + + zA + hpump = + + zC + hturbine + 1hL 2 AC gw gw 2g 2g 5760 lb>ft2 11.937 slug>ft3 2132.2 ft>s2 2

= 0 +

391VC 2 p 4 2

2132.2 ft>s2 2

+

VA2 2132.2 ft>s2 2

+ z + 0 + 1960fAB + 0.92 a

+ z+ 0 2

1VC 2p VA2 b + 1960fBC + 5.852 c d 64.4 64.4

1960fAB - 0.1)VA2 + 1960fBC + 86.8521VC 2p2 = 5947.34

(4)

2 1VD 2 nz pA pD VA2 + + zA + hpump = + + zD + hturb + 1hL 2 AD gw gw 2g 2g

5760 lb>ft2 2

11.937 slug2132.2 ft>s 2

= 0 +

391VD 2 p 4 2

2132.2 ft>s2 2

+

VA2 2132.2 ft>s2 2

+ z + 0 + 1960fAB + 0.92 a

+ z+ 0

1VD 2p2 VA2 b + 11440fBD + 5.352 c d 64.4 64.4

1960fAB - 0.1)VA2 + 11440fBD + 86.3521VD 2p2 = 5947.34

(5)

1960fBC + 86.8521VC 2p2 = 11440fBD + 86.3521VD 2p2

(6)

Equating Eq. 4 and 5,

913

M10_HIBB9290_01_SE_C10_ANS.indd 913

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10–65. Continued

9110-6 2 ft e = 0.000144. The trial and = D 0.75 ft 12 error iterative procedure is required. The iterations are tabulated below: For PVC pipe, the relative roughness is

Iteration 1 2

fAB 0.024

Assumed f fBC fBD 0.028

0.0205 0.0238

VA

0.028 0.024

Velocities (ft , s)*

10.6006 11.0902

(VC)p

5.4431 5.6800

Re

(VD)p

5.1576 5.4102

AB

f from Moody diagram

BC 4

4

4

6.37(10 ) 3.27(10 ) 6.66(10 )

fAB

fBC

fBD

4

0.0205

0.0238

0.024

4

0.0203

0.0238

0.024

BD

4

3.41(10 )

3.10(10 ) 3.25(10 )

The assumed f in 2nd iteration are very close to those from Moody diagram, the results VA = 11.0902 ft>s, 1VC 2 p = 5.6800 ft>s and 5.4102 ft>s are acceptable. Then the discharges at C and D are QC = AC 1VC 2 p = pa

2 0.375 ft b 15.6800 ft>s2 = 0.0174 ft3 >s 12

QD = AD 1VD 2 p = pa

2 0.375 ft b 15.4102 ft>s2 = 0.0166 ft3 >s 12

Ans. Ans.

*Solve Eq. 3, 5 and 6 to obtain VA, 1VC 2 p and 1VD 2 p

914

M10_HIBB9290_01_SE_C10_ANS.indd 914

Ans: QC = 0.0174 ft3 >s, QD = 0.0166 ft3 >s

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10–66. Water at 70°F is pumped through the two commercial steel pipes having the lengths and diameters shown. If the pressure developed at A is 30 psi, determine the discharge. Neglect minor losses.

6 ft

10 ft B

10 ft A

1.5 in.

C

3 in.

2 ft

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 1.937 slug>ft3 and vw = 10.4110-6 2 ft2 >s. Take the control volume as the water contained in the pipe between A and C. The continuity condition requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VABAAB + VBCABC = 0 - VAB c pa

2 2 1.5 0.75 ft b d + VBC c pa ft b d = 0 12 12

VAB = 0.25 VBC

The Reynolds numbers of the flow along segments AB and BC are

1Re2 AB

VABDAB = = yw

1Re2 BC

10.25VBC 2 a

3 ft b 12

10.4110-6 2 ft2 >s

= 6.00961103 2VBC

1.5 VBC a ft b VBCDBC 12 = = = 1.20191104 2VBC yw 10.4110-6 2 ft2 >s

(1)

(2)

The major head loss from A to C is the sum of the losses in segments AB and BC. Thus, 1hL 2 ma = fAB = fAB

2 2 LBC V BC LAB VAB + fBC DAB 2g DBC 2g 2 10.25VBC 2 2 V BC 18 ft 10 ft d + fBC d c c 2 ° 3 ¢ 2132.2 ft>s 2 ° 1.5 ¢ 2132.2 ft>s2 2 ft ft 12 12

2 = 10.06988fAB + 1.24224fBC 2V BC

Since the water is discharged into the atmosphere at C, pC = patm = 0. With reference to the datum set through A, zA = 0 and zC = 10 ft. Applying the energy equation between A and C, 2 2 pA pC V BC V AB + + zA + hpump = + + zC + hturbine + hL gw gw 2g 2g

12 in. 2 lb ba b 2 2 10.25 VBC 2 2 V BC 1 ft in + + 0 + 0 = 0 + + 10 ft + 0 2 3 2 2132.2 ft>s 2 2132.2 ft>s2 2 11.937 slug>ft 2132.2 ft >s2 a30

2 + 10.06988fAB + 1.24224fBC 2V BC

2 10.06988 fAB + 1.24224 fBC + 0.014562V BC = 59.2625

(3)

915

M10_HIBB9290_01_SE_C10_ANS.indd 915

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10–66. Continued

For commercial steel pipe, the relative roughnesses, for segments AB and BC of the pipe are a

e 0.00015 ft = 0.0006 b = D AB 3 ft 12

a

e 0.00015 ft = 0.0012 b = D BC 1.5 ft 12

The trial and error iterative procedure is required, the iterations are tabulated below. Assumed Iteration 1 2

fAB

fBC

0.02

0.022

0.0193

0.021

f from Moody diagram VBC (ft , s)(Eq. 3)

(Re)AB(Eq. 1)

(Re)BC (Eq. 2)

fAB

fBC

37.00

2.22 ( 10

)

4.45 ( 10

)

0.0193

0.021

37.57

2.26 ( 10

)

4.52 ( 10

)

0.0193

0.021

5 5

5 5

The assumed fAB and fBC in 2nd iteration is almost the same as those from Moody diagram. Thus, the result of VBC = 37.57 ft>s is acceptable. Thus, the discharge through the pipe is Q = VBC ABC = 137.57 ft>s2 c pa = 0.461 ft3 >s

2 0.75 ft b d 12

916

M10_HIBB9290_01_SE_C10_ANS.indd 916

Ans.

Ans: Q = 0.461 ft3 >s

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10–67. Two galvanized iron pipes branch to form a loop. Branch CAD is 120 m long and branch CBD is 60 m long. Determine the power of the pump used in branch CAD if an equal flow of water at 20°C occurs through each branch. All pipes have a diameter of 75 mm. The loop is in the horizontal plane. Neglect minor losses.

D

B

A C 3 m!s

SOLUTION We will consider fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 998.3 kg>m3 and vw = 1.00110-6 2 m2 >s for water at T = 20°C. The flow at pipe C is QC = 13 m>s23p10.0375 m2 2 4 = 0.01325 m3 >s

Since the discharge in pipes CBD and CAD is required to be the same, the continuity condition gives QCBD = QCAD =

0.01325 m3 >s QC = = 0.006627 m3 >s 2 2

Thus, the average velocity in these two pipes is VCAD = VCBD = V =

0.006627 m3 >s QCBD = = 1.50 m>s A p10.0375 m2 2

Then the Reynolds number of the flow in these two pipes is 1Re2 CBD = 1Re2 CAD =

11.50 m>s210.075 m2 VD = = 1.1251105 2 vw 1.00110-6 2 m2 >s

e 0.15 mm = = 0.002. From the D 75 mm Moody diagram f = 0.0255. Then the major head losses in pipes CBD and CAD are For galvanized iron pipe, the relative roughness is

1hL 2 CBD = f 1hL 2 CAD = f

11.50 m>s2 2 LCBD V 2 60 m d = 2.3394 m = 0.0255a bc D 2g 0.075 m 219.81 m>s2 2

11.50 m>s2 2 LCAD V 2 120 m d = 4.6789 m = 0.0255a bc D 2g 0.075 m 219.81 m>s2 2

Since the loop in the horizontal plane, no change in elevation occurs. Also, the minor head loss will be neglected. Then, the excess head loss in pipe CAD must be overcome by the pump to maintain the equal flow in these two pipes, or otherwise they will have different flow as a result of different head loss. Thus, hpump = 1hL 2 CAD - 1hL 2 CBD = 4.6789 m - 2.3394 m = 2.3394 m

The required power output for the pump is # W = gwQhpump = 1998.3 kg>m219.81 m>s2 2 3 0.006627 m2 >s 4 12.3394 m2 = 151.83 W = 152 W

Ans.

Ans: # W = 152 W 917

M10_HIBB9290_01_SE_C10_ANS.indd 917

10/03/17 2:40 PM

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*10–68. Water at 25°C is pumped into the galvanized iron pipe branch consisting of two 50-mm-diameter pipes ABD and ACD, which are 8 m and 12 m long, respectively. Determine the flow through each pipe if the pressure drop from A to D is 75 kPa. Neglect minor losses. The pipe branch is in the horizontal plane.

75 mm

50 mm B A

75 mm D

C 50 mm

SOLUTION We will consider the fully developed steady flow, and water is incompressible. From the table in Appendix A, rw = 997.1 kg>m3 and yw = 0.898110-6 2 m2 >s for water at T = 25°C. Take the control volume as the water contained in pipes B and C. The continuity condition requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - VAAA + VBAB + VCAC = 0 (1)

QA = QB + QC The Reynolds numbers of the flow in pipes B and C are Re =

V10.05 m2 VD = = 5.56791104 2V vw 0.898110-6 2 m2 >s

(2)

The major head losses in pipes B and C are 1hL 2 B = fB 1hL 2 C = fC

LB VB2 VB2 8m d = 8.1549fBVB2 = fB a bc D 2g 0.05 m 219.81 m>s2 2

LC VC2 VC2 12 m d = 12.2324fCVC2 = fC a bc D 2g 0.05 m 219.81 m>s2 2

Since pipes B and C are parallel, their head losses must be the same. 1hL 2 B = 1hL 2 C

8.1549fBVB2 = 12.2324fcVC2 VB = 1.2247a

fC bV B fB C

(3)

Since the diameters of the pipes at A and D are the same, VA = VD = V. Also, the branch is on horizontal plane, then zA = zD = z. Applying the energy equation between A and D, pA pD VA2 VD2 + + zA + hpump = + + zD + hturb + hL gw gw 2g 2g pA pD V2 V2 + + z + hpump = + + z + 0 + 12.2324fCVC2 gw gw 2g 2g pA - pD 1997.1 kg>m3 219.81 m>s2 2

= 12.2324fCVC2

pA - pD = 119.6521103 2fCVC2

751103 2 Pa = 119.6521103 2fCVC2 VC =

0.7917

(4)

2fc 918

M10_HIBB9290_01_SE_C10_ANS.indd 918

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10–68. Continued

e 0.15 mm = = 0.003. Trial D 50 mm and error iterative process is required. The iterations are tabulated below.

For galvanized iron pipes, the relative roughness is

Assumed Iteration

fC

fB

Velocities (m , s) VC (Eq. 4) VB (Eq. 3)

f from Moody diagram (Re)C

(Re)B

fC

fB

1

0.026

0.027

4.9100

5.9011

2.73(105)

3.41(105)

0.0265

0.0265

2

0.0265

0.0265

4.8635

5.9565

2.71(105)

3.32(105)

0.0265

0.0265

The assumed fC and fB in 2nd iteration are almost the same as those from Moody diagram. Thus, VC = 4.8635 m>s and VB = 5.9565 m>s are the acceptable results. Thus, QB = VBAB = 15.9565 m>s23p10.025 m2 2 4 = 0.011695 m3 >s = 0.0117 m3 >s Ans. QC = VCAC = 14.8635 m>s23p10.025 m2 2 4 = 0.009549 m3 >s = 0.00955 m3 >s Ans.

From Eq. (1)

QA = QD = 0.011695 m3 >s + 0.009549 m3 >s = 0.0212 m3 >s

Ans.

Ans: QA = QB = QC = QD = 919

M10_HIBB9290_01_SE_C10_ANS.indd 919

0.0212 m3 >s 0.0117 m3 >s 0.00955 m3 >s 0.0212 m3 >s

10/03/17 2:40 PM

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10–69. Water at 25°C is pumped into the galvanized iron pipe branch consisting of two 50-mm-diameter pipes ABD and ACD, which are 8 m and 12 m long, respectively. Determine the pressure drop from A to D if the discharge through A is 0.0215 m3 > s. The pipe branch is in the horizontal plane. Neglect minor losses.

75 mm

50 mm B A

75 mm D

C 50 mm

SOLUTION We will consider the fully developed steady flow, and water is incompressible from the table in Appendix A, rw = 997.1 kg>m3 and vw = 0.898110-6 2 m2 >s for water at T = 25°C. The Reynolds number of the flow in the pipes is Re =

V10.05 m2 VD = = 5.56791104 2V yw 0.898110-6 2 m2 >s

(1)

Take the control volume as the water contained in pipes B and C. The continuity condition requires 0 rdV + rV # dA = 0 0t Lcv Lcs 0 - 0.215 m3 >s + VB 3p10.025 m2 2 4 + VC 3p10.025 m2 2 4 = 0 VB + VC =

34.4 p

(2)

The major head loss in pipes B and C are 1hL 2B = fB 1hL 2C = fC

LB VB 2 VB 2 8m d = 8.1549fBVB 2 = fB a bc D 2g 0.05 m 219.81 m>s2 2

LC VC 2 VC 2 12 m d = 12.2324fCVC 2 = fC a bc D 2g 0.05 m 219.81 m>s2 2

Since pipes B and C are parallel, their head losses must be the same 1hL 2 B = 1hL 2 C

8.1549fBVB 2 = 12.2324fCVC 2 VB = 21.5a

fC bVC A fB

(3)

e 0.15 mm = = 0.003. D 50 mm The trial and error iterative procedure is required. The iterations are tabulated below.

For galvanized iron pipe, the relative roughness of pipe is

Assumed Iteration 1 2

fC 0.026 0.0265

fB 0.025 0.0265

Velocities (m , s)* VC

4.8688 4.9218

VB

6.0811 6.0280

From Moody diagram (Re)C 5

2.71(10 ) 5

2.74(10 )

fC

fB

5

0.0265

0.0265

5

0.0265

0.0265

(Re)B 3.39(10 ) 3.36(10 )

920

M10_HIBB9290_01_SE_C10_ANS.indd 920

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10–69. Continued

The assumed fC and fB in 2nd iteration are almost the same as those from Moody diagram. Thus, the result of VC = 4.9218 m>s and VB = 6.0280 m>s are acceptable. Since the diameters of the pipes at A and D are the same, VA = VD = V. Also, the branch is in the horizontal plane, then zA = zD = z. Applying the energy equation between A and D, pA pD VA2 VD2 + + + zA + hpump = + zD + hturbine + hL gw gw 2g 2g pA pD V2 V2 + + z+ 0 = + + z + 0 + 12.232410.0265214.928 m>s2 2 gw gw 2g 2g pA - pD 1997.1 kg>m3 219.81 m>s2 2

= 7.8526 m

pA - pD = 76.811103 2 Pa = 76.8 kPa

Ans.

Ans: pA - pD = 76.8 kPa 921

M10_HIBB9290_01_SE_C10_ANS.indd 921

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10–70. 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 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

L V2 18 m 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 VD 2 + + 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.

922

M10_HIBB9290_01_SE_C10_ANS.indd 922

Ans: Q = 0.0335 m3 >s

10/03/17 2:40 PM

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10–71. 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 minor losses. Take nw = 1.00110-6 2 m2 >s.

C 7m

1m A 10 m

B

D 3m

E 8m

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 D 2 d + VB c D 2 d + VE c D 2 d = 0 4 4 4 VA = VB + VE

(1)

The major head losses for the pipes are

( hL ) A = fA ( hL ) B = fB ( hL ) E = fE

LA VA 2 VA 2 10 m = fA a b£ § = 5.0968 fAVA 2 D 2g 0.1 m 2 ( 9.81 m>s2 ) LB VB 2 VB 2 9m = fB a b£ § = 4.5872 fBVB 2 D 2g 0.1 m 2 ( 9.81 m>s2 ) VE 2 LE VE 2 8m = fE a b£ § = 4.0775 fEVE 2 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 4.5872fBVB 2 = 4.0775fEVE 2

VB = 0.9428 Substituting Eq. (2) into (1), VA = a0.9428

fE V B fB E

(2)

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 VC 2 VD 2 + + zC + hpump = + + zD + hturb + hL g g 2g 2g 0 + 0 + 7 m + 0 = 0 + 0 + 3 m + 0 + 5.0968fAVA2 + 4.0775fEVE 2 5.0968fAVA 2 + 4.0775fEVE 2 = 4

(4)

923

M10_HIBB9290_01_SE_C10_ANS.indd 923

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10–71. 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 VB(0.1 m) VBD = 1 ( 105 ) VB = n 1.00 ( 10-6 ) m2 >s

VE(0.1 m) VED = 1 ( 105 ) VE = n 1.00 ( 10-6 ) m2 >s

(5)

(6)

(7)

For the galvanized iron pipe, e = 0.15 mm. Thus, the relative roughness for all the pipes is

e 0.15 mm = = 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

1

0.021

2

VA (m , s)

(Re)A

fA from Moody diagram

5.519

5.52 ( 105 )

0.0215

0.0215

5.477

5.48 ( 10

0.0215

Assumed fB

VB (m , s)

5

)

Pipe B: Iteration 1 2

(Re)B

fB from Moody diagram

2.648

2.65 ( 10

)

0.022

0.022

2.658

2.66 ( 10

)

0.022

Assumed fE

VE (m , s)

0.023

5 5

Pipe E: Iteration 1 2

0.022 0.022

(Re)E

fE from Moody diagram

2.871

2.87 ( 10

)

0.022

2.819

2.82 ( 10

)

0.022

5 5

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

924

M10_HIBB9290_01_SE_C10_ANS.indd 924

Ans.

Ans: Q = 0.0430 m3 >s

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*10–72. The copper pipe system, which transports water at 70°F, consists of two branches. The pipe for branch ABC has a diameter of 0.5 in. and length of 8 ft, whereas the pipe for 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 = 80110 - 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>ft3 and nw = 10.4 ( 10-6 ) ft2 >s at T = 70° F. The Reynolds numbers for the flow in the pipes are (Re)ABC

(Re)ADC

0.5 in.

0.5 VABC a ft b VABCDABC 12 = = = 4.006 ( 103 ) VABC nw 10.4 ( 10-6 ) ft2 >s

1 VADC a ft b VADCDADC 12 = = = 8.013 ( 103 ) VADC nw 10.4 ( 10-6 ) ft2 >s

C

1 in.

D

(1)

(2)

The loss coefficients 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

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 = ≥ fADC ±

It is required that

V 2ABC b 64.4

LADC V 2ADC + ΣkL b DADC 2g

V ADC2 30 ft ≤ + 2(1.8) + 4(0.9) ¥ £ § 1 2 ( 32.2 ft>s2 ) ft 12 = ( 360fADC + 7.2 ) a

V ADC2 b 64.4

( hL ) ABC = ( hL ) ADC ( 192fABC + 5.4 ) V 2ABC = ( 360fADC + 7.2 )

(3)

The continuity condition at junction A requires that QABC + QADC = ( 67.3 gal>min )( 1 min>605 )( 1 ft3 >7.48 gal ) VABC £ pa

2 2 0.25 0.5 ft b § + VADC £ pa ft b § = 0.15 12 12

VABC + 4VADC =

345.5 p

(4) 925

M10_HIBB9290_01_SE_C10_ANS.indd 925

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10–72. 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 ft3 >s 12

QADC = VADC AADC = ( 21.114 ft>s ) £ pa

2 0.5 ft b § = 0.115 ft3 >s 12

= 15.6 gal>min

= 51.7 gal>min

Ans.

Ans.

Ans: QABC = 15.6 gal>min QADC = 51.7 gal>min 926

M10_HIBB9290_01_SE_C10_ANS.indd 926

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10–73. 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. 10–72. 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 give rw = 1.937 slug>ft3 and nw = 10.4 ( 10-6 ) ft2 >s at T = 70° F. The Reynolds numbers for the flow in the pipes are (Re)ABC

(Re)ADC

0.5 in.

0.5 VABC a ft b VABCDABC 12 = = = 4.006 ( 103 ) VABC nw 10.4 ( 10-6 ) ft2 >s

1 VADC a ft b VADCDADC 12 = = = 8.013 ( 103 ) VADC nw 10.4 ( 10-6 ) ft2 >s

C

1 in.

D

(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

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 ±

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), 2 1 ft in ft

zA = zC (same elevation),

pA pC VC2 VA2 + + zA + hpump = + + zC + hturb + hL gw gw 2g 2g pA - pC hL = = gw

lb lb - 2160 2 2 ft ft = 103.89 ft ( 1.937 slug>ft3 )( 32.2 ft>s2 ) 8640

927

M10_HIBB9290_01_SE_C10_ANS.indd 927

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10–73. 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 = 15.7 gal>min

QADC = VADC AADC = (21.162 ft>s) £ pa = 51.8 gal>min

2 0.25 ft b § = 0.0349 ft3 >s 12

2 0.5 ft b § = 0.115 ft3 >s 12

Ans.

Ans.

Ans: QABC = 15.7 gal>min QADC = 51.8 gal>min 928

M10_HIBB9290_01_SE_C10_ANS.indd 928

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11–1. Determine the distance x = xcr over the flat plate to where the boundary layer for kerosene at T = 68°F begins to transition from laminar to turbulent flow.

4 ft!s

x

SOLUTION We will consider the flow as steady, and the kerosene as incompressible. From the table in Appendix A, rke = 1.58 slug>ft3 and vke = 25.4(10-6) ft2 >s for kerosene at T = 68°F. The transition from laminar to turbulent boundary layers occurs at critical Reynolds number of 1Rex2 cr = 51105 2. 1Rex2 cr =

Uxcr ; vke

51105 2 =

14 ft>s2xcr

25.4110-6 2 ft2 >s

xcr = 3.175 ft

Ans.

Ans: xcr = 3.175 ft 929

M11_HIBB9290_01_SE_C11_ANS.indd 929

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11–2. The boundary layer for wind blowing over rough terrain can be approximated by the equation u>U = 3y> 1y + 0.0224, where y is in meters. If the free-stream velocity of the wind is 10 m>s, determine the velocity at y = 0.2 m and at y = 0.4 m from the ground.

10 m!s

y

SOLUTION We will consider the flow steady, and air as incompressible. Here, y u = U y + 0.02 u = a

At y = 0.2 m, u "y

= 0.2 m

At y = 0.4 m, u "y

= 0.4 m

= c = c

y 10y bU = a b m>s y + 0.02 y + 0.02

1010.22 0.2 + 0.02 1010.42 0.4 + 0.02

d m>s = 9.09 m>s

Ans.

d m>s = 9.52 m>s

Ans.

Ans: u " y = 0.2 m = 9.09 m>s u " y = 0.4 m = 9.52 m>s 930

M11_HIBB9290_01_SE_C11_ANS.indd 930

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11–3. Water at 60°F flows over the top surface of the plate. If the free-stream velocity is 1.5 ft>s, determine the boundary layer disturbance thickness and momentum thickness at the back end of the plate.

2 ft

4 ft

1.5 ft!s

SOLUTION The flow is steady, and the water is incompressible. The transition from laminar to turbulent boundary layer occurs at the critical Reynolds number of 1Rex2 cr = 51105 2. From the table in Appendix A, vke = 12.2110-6 2 ft2 >s for water at 60°F. 1Rex2 cr =

Uxcr ; vw

51105 2 =

11.5 ft>s2xcr

12.2110-6 2 ft2 >s

xcr = 4.067 ft

Since xcr 7 L = 4 ft, the boundary layer is laminar over the entire length of the plate. At the end of the plate, x = L = 4 ft. The Reynolds number at this point is Rex =

11.5 ft>s2(4 ft) Ux = = 4.9181105 2 v 12.2110 - 6 2 ft2 >s

Then the boundary layer disturbance thickness at the end of the plate can be determined. d =

5.0x 2Rex

=

5.014 ft2 5

4.918110 2

= 10.02852 ft2a

12 in. b = 0.342 in. 1 ft

Ans.

And the momentum thickness of the boundary layer can be determined. Θ =

0.664x 2Rex

=

0.66414 ft2 5

24.918110 2

= 33.787110-3 2 ft4 a

12 in. b = 0.0454 in. Ans. 1 ft

Ans: d = 0.342 in. Θ = 0.0454 in. 931

M11_HIBB9290_01_SE_C11_ANS.indd 931

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*11–4. If the free-stream velocity of water at 60°F is 1.5 ft>s, determine the friction drag acting on the surface of the plate.

2 ft

4 ft

1.5 ft!s

SOLUTION The flow is steady and the water is in compression. The transition from laminar to turbulent boundary layer occurs at the critical Reynolds number of 1Rex2 cr = 51105 2. From the table in Appendix A, rw = 1.939 slug>ft3. and vx = 12.2110-6 2 ft2 >s for water at 60°F. 1Rex2 cr =

Uxcr ; vx

51105 2 =

11.5 ft>s2xcr

12.2110-6 2 ft2 >s

xcr = 4.067 ft.

Since xcr 7 L = 4 ft, the boundary layer is laminar over entire length of the plate. At the end of the plate, x = L = 4 ft, and the Reynolds number at this point is ReL =

11.5 ft>s214 ft2 UL = = 4.9181105 2 vw 12.2110-6 2 ft2 >s

Then, the friction drag on the plate can be determined. FDf =

=

0.664brU 2L 2ReL

0.66412 ft211.939 slug>ft3 211.5 ft>s2 2 14 ft2

= 0.0330 lb

24.9181105 2

Ans.

Ans: FDf = 0.0330 lb 932

M11_HIBB9290_01_SE_C11_ANS.indd 932

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11–5. 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>ft2 and n = 0.158 ( 10-3 ) ft2 >s. The transition from a laminar boundary layer occurs at a critical Reynolds number of ( Rex ) cr = 5 ( 105 ) .

( Rex) cr = 5 ( 105 ) =

Uxcr n

( 6 ft>s ) x cr 0.158 ( 10-3 ) ft2 >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 ReL = Then, FD =

( 6 ft>s ) (12 ft) UL = 4.557 ( 105 ) = n 0.158 ( 10-3 ) ft2 >s

0.664brU 2L 2Re L

= 0.00604 lb

=

0.664(6 ft) ( 0.00237 slug>ft3 )( 6 ft>s ) 2(12 ft) 24.557 ( 105 )

Ans.

Ans: FD = 0.00604 lb 933

M11_HIBB9290_01_SE_C11_ANS.indd 933

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11–6. If the disturbance thickness of a laminar boundary layer of oil at a distance of 0.75 m from a plate’s front edge is 10 mm, determine the free-stream velocity of the oil. Take no = 40110-6 2 m2 >s.

SOLUTION The flow is steady and the oil is incompressible. The Reynolds number at x = 0.75 m is Rex =

U10.75 m2 Ux = = 18.751103 2U no 40110-6 2 m2 >s

Using the laminar disturbance thickness gives d =

5.0x 2Rex 3

;

5.010.75 m2

0.01 m =

218.751103 2U

Ans.

U = 7.50 m>s 5

5

In this case, Rex = 18.75110 217.502 = 1.41110 2 6 5110 2. Therefore, at x = 0.75 m, the flow is indeed laminar.

Ans: U = 7.50 m>s 934

M11_HIBB9290_01_SE_C11_ANS.indd 934

16/03/17 10:36 AM

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11–7. Determine the maximum free-stream velocity of oil so that if it flows over a plate having a length of 3 m it maintains a laminar boundary layer for the entire length. Take no = 40110-6 2 m2 >s.

SOLUTION The flow is steady and the oil is incompressible. The transition from laminar to turbulent boundary layer occurs at the critical Reynolds number of 1Rex2 cr = 51105 2. Here, xcr = L = 3 m. 1Rex2 cr =

Uxcr ; no

51105 2 =

U max 13 m2

40110-6 2 m2 >s

U max = 6.67 m>s

Ans.

Ans: Umax = 6.67 m>s 935

M11_HIBB9290_01_SE_C11_ANS.indd 935

16/03/17 10:36 AM

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*11–8. Oil has laminar flow and passes over the flat plate. Plot the velocity profile a distance of x = 2 m from the leading edge of the plate. Using Table 11–1, give values of u for every 0.8 increment of 1y>x2 1Rex until u = 0.99U. Take no = 40110-6 2 m2 >s.

5 m!s

y

x

SOLUTION

y(mm)

The flow is steady and the oil is incompressible. The transition from laminar to turbulent boundary layer occurs at the critical Reynolds number of 1Rex2 cr = 51105 2. 1Rex2 cr =

Uxcr ; no

51105 2 =

15 m>s2xcr

40110-6 2 m2 >s

15 10

xcr = 4 m

Since xcr 7 x = 2 m, the boundary layer is laminar at this location. Then, the Reynolds number of the flow at this location is Rex =

20

15 m>s212 m2 Ux = = 2.501105 2 no 40110-6 2 m2 >s

5

0

22.501105 2 2Rex Hence, = = 0.25 mm-1. Using the Blasius solution (Table 11–1), x 2000 mm we can establish the relation between u and y tabulated below. y 2Rex x

0

0.8

1.6

2.4

3.2

4.0

4.8

y1mm2

0

3.20

6.40

9.60

12.8

16.0

19.2

u>U

0

0.26471

0.51676

0.72899

0.87609

0.95552

0.98779

0.99

u1m>s2

0

1.32

2.58

3.64

4.38

4.78

4.94

4.95

1

2

3

4

5

u(mys)

(a)

5.0 20

The plot of u vs y is shown in Fig. a.

936

M11_HIBB9290_01_SE_C11_ANS.indd 936

16/03/17 10:37 AM

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11–9. Water at 15°C, confined in a channel, flows past the diverter fin at U = 2 m>s. Determine the friction drag acting on both sides of the fin, which has a width of 900 mm. Neglect end effects. U 5 2 m!s 200 mm

SOLUTION The flow is steady and the water is incompressible. From the table in Appendix A, rw = 999.2 kg>m3 and vw = 1.15110-6 2 m2 >s for water at T = 15°C. The transition from laminar to turbulent boundary layer occurs at a critical Reynolds number of 1Rex2 cr = 51105 2. 1Rex2 cr =

Uxcr ; vw

51105 2 =

12 m>s2xcr

1.15110-6 2 m2 >s

xcr = 0.2875 m

Since xcr 7 L = 0.2 m, the boundary layer for the entire length of the fin is laminar. Here, the Reynolds number at x = L = 0.2 m is ReL =

12 m>s210.2 m2 UL = = 3.4781105 2 vw 1.15110-6 2 m2 >s

The friction drag on both sides of the fin is FDf = 2c

0.664brU 2L 2ReL

d = 2c

0.66410.9 m21999.2 kg>m3 212 m>s2 2 10.2 m2

= 1.62 N

23.4781105 2

d Ans.

Ans: FDf = 1.62 N 937

M11_HIBB9290_01_SE_C11_ANS.indd 937

16/03/17 10:37 AM

© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6 m!s

11–10. A fluid has laminar flow and passes over the flat plate. If the disturbance thickness of the boundary layer at a distance of x = 0.5 m from the plate’s edge is 10 mm, determine the disturbance thickness at a distance of x = 1 m.

10 mm

x

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 Rex " x = 0.5 m =

U(0.5 m) Ux 0.5U = = n n n

and Rex " x = 1 m =

U(1 m) Ux U = = n n n

At x = 0.5 m, d = 0.01 m. Thus, d =

5.0x 2Rex

;

0.01 m =

5.0(0.5 m) 0.5U A n

U = 125 000 n Thus, at x = 1 m, Rex = d =

U = 125 000. Then, n

5.0x 2Rex

=

5.0(1 m) 2125 000

= 0.01414 m = 14.1 mm

Ans.

Ans: d = 14.1 mm 938

M11_HIBB9290_01_SE_C11_ANS.indd 938

16/03/17 10:37 AM

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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

x54m

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 Rex =

( 0.5 m>s ) (4 m) Ux = 1.0582 ( 105 ) = n 18.9 ( 10 - 6 ) m2 >s

Since Rex 6 (Rex)cr = 5 ( 105 ) , the boundary layer is laminar throughout the entire length of the duct. Thus, the displacement thickness is d* =

1.721x 2Rex

=

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: a = 242 mm 939

M11_HIBB9290_01_SE_C11_ANS.indd 939

16/03/17 10:37 AM

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*11–12. Air at 20°C is blowing at 2 m>s as it passes over the flat surface. Determine the disturbance and displacement thickness of the boundary layer at a distance of x = 0.5 m from the leading edge. What is the velocity of the flow at half the disturbance thickness?

2 m!s

x

SOLUTION We will consider the flow as steady and the air as incompressible. From the table in Appendix A, ra = 1.202 kg>m3 and na = 15.1110-6 2 m2 >s for air at T = 20°C. Thus, the Reynolds number at x = 0.5 m is Rex =

12 m>s210.5 m2 Ux = = 6.62251104 2 va 15.1110-6 2 m2 >s

Since Rex 6 1Rex2 cr = 51105 2, the boundary layer up to x = 0.5 m is still laminar. The disturbance thickness of the laminar boundary layer at this location is d =

5.0

2Rex

x= a

5.0

26.62251104 2

b10.5 m2 = 9.7146110-3 2 m = 9.71 mm Ans.

And the displacement thickness at this location is dA =

1.721 2Rex

x= a

1.721 26.62251104 2

b10.5 m2 = 3.3438110-3 2 m = 3.34 mm

Ans.

d The velocity of the flow at x = 0.5 m and y = = 4.8573110-3 2 m can be 2 determined using the Blasius solution (Table 11-1). 4.8573110-3 2 m y 2Rex = 26.62251104 2 = 2.50 x 0.5 m

Interpolating the values in the table gives u ≃ 0.7496 U

u = 0.749612 m>s2 = 1.4992 m>s Ans.

= 1.50 m>s

Ans: d = 9.71 mm d A = 3.34 mm u = 1.50 m>s 940

M11_HIBB9290_01_SE_C11_ANS.indd 940

16/03/17 10:37 AM

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11–13. Water at 20°C has a free-stream velocity of 500 mm>s. Plot the boundary layer disturbance thickness along the distance x, in increments of 0.1 m for 0 … x … 0.5 m.

500 mm!s

x 0.5 m

SOLUTION The flow is steady and the water is incompressible. From the table in Appendix A, vw = 1.00110-6 2 m2 >s for water at T = 20°C. Thus, the Reynolds number in terms of x is Rex = Then x1m2 Rex

0 0

10.5 m>s2x Ux = = 5.01105 2x vw 1.00110-6 2 m2 >s

0.1

0.2

4

0.3

5

5110 2

1110 2

0.4

5

1.5110 2

5

2110 2

0.5 2.51105 2

Since Rex " x = 0.5 m 6 1Rex2 cr = 51105 2, the boundary layer up to x = 0.5 m is still laminar. Thus, its thickness can be determined. d =

5.0

2Rex

x

The values of d for various x are tabulated below. x1m2 d1mm2

0 0

0.1 2.24

0.2 3.16

0.3 3.87

0.4 4.47

0.5 5.00

The plot of d vs x is shown in Fig. a. d(mm) 5 4 3 2 1 0

0.1

0.2

0.3

0.4

0.5

x(m)

(a)

941

M11_HIBB9290_01_SE_C11_ANS.indd 941

16/03/17 10:37 AM

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11–14. Water at 20°C has a free-stream velocity of 500 mm>s. Determine the drag on the plate’s surface if it has a width of 0.2 m.

500 mm!s

x 0.5 m

SOLUTION The flow is steady and the water is incompressible. From the table in Appendix A, rw = 998.3 kg>m3 and vw = 1.00110-6 2 m2 >s for water at T = 20°C. Thus, the Reynolds number at x = L = 0.5 m is ReL =

10.5 m>s210.5 m2 UL = = 2.501105 2 vw 1.00110-6 2 m2 >s

Since ReL 6 1Rex2 cr = 51105 2, The boundary layer for the entire length of the plate is laminar. The friction drag on the plate’s side can be determined, FDf =

=

0.664brU 2L 2ReL

0.66410.2 m21998.3 kg>m3 210.5 m>s2 2 10.5 m2

= 0.0331 N

22.501105 2

Ans.

Ans: FDf = 0.0331 N 942

M11_HIBB9290_01_SE_C11_ANS.indd 942

16/03/17 10:37 AM

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11–15. Air at a temperature of 60°F flows at 1.2 ft>s over the plate. Determine the distance x where the disturbance thickness of the boundary layer becomes 0.75 in.

1.2 ft!s

3 in.

x

SOLUTION We will consider the flow steady and air as incompressible. From the table in Appendix A, va = 0.158110-3 2 ft2 >s for air at T = 60°F. Thus, the Reynolds number in terms of x is Rex =

11.2 ft>s2x Ux = = 7.59491103 2x va 0.158110-3 2 ft2 >s

We will assume that the boundary layer is laminar. With the requirement 0.75 d = ft = 0.0625 ft, 12 d =

5.0 2Rex

x;

0.0625 ft =

5.0x 27.59491103 2x

x = 1.1867 ft = 1.19 ft

Ans.

Using this result, Rex = 37.59491103 2411.1867 ft2 = 9.0131103 2

Since Rex 6 1Rex2 cr = 51103 2 the boundary layer is laminar as assumed.

Ans: x = 1.19 ft 943

M11_HIBB9290_01_SE_C11_ANS.indd 943

16/03/17 10:37 AM

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*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

SOLUTION

A

1.75 ft

Water is considered to be incompressible. The relative flow is steady. From Appendix A, n = 12.2 ( 10 - 6 ) ft2 >s for water at T = 60°F. Thus, the Reynolds number at the trailing edge of the rudder (x = L = 1.75 ft) is Rex =

( 0.7 ft>s )(1.75 ft) Ux = 1.0041 ( 105 ) = n 12.2 ( 10 - 6 ) ft2 >s

Since Rex 6 (Rex)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 2Rex

=

1.721x 2Rex

=

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

Ans.

Ans.

Ans: d = 0.331 in. d * = 0.114 in. 944

M11_HIBB9290_01_SE_C11_ANS.indd 944

16/03/17 10:37 AM

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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

SOLUTION

A

1.75 ft

Water is considered to be incompressible. The relative flow is steady. From Appendix A, n = 12.2 ( 10 - 6 ) ft2 >s for water at T = 60°F. Thus, the Reynolds number at the trailing edge of the rudder (x = L = 1.75 ft) is Rex =

( 0.7 ft>s ) (1.75 ft) Ux = 1.0041 ( 105 ) = n 12.2 ( 10 - 6 ) ft2 >s

Since Rex 6 (Rex)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 2ReL

§

0.664(2 ft) ( 1.939 slug>ft3 )( 0.7 ft>s ) 2(1.75 ft)

= 0.0139 lb

21.004 ( 105 )

§

Ans.

Ans: F = 0.0139 lb 945

M11_HIBB9290_01_SE_C11_ANS.indd 945

16/03/17 10:37 AM

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11–18. Determine the force F that must be applied to the cable to lift the fully submerged 50-kg plate up to the surface of the glycerin at T = 20°C at a constant speed of 2 m>s. Include the effect of buoyancy.

F

10 mm

2m

SOLUTION The flow is steady and the glycerin is incompressible. From the table in Appendix A, rgl = 1260 kg>m3 and vgl = 1.19110-3 2 m2 >s. Thus, the Reynolds number at x = L = 2 m is ReL =

3m

(2 m>s)(2 m) UL = = 3.3613(103) vgl 1.19(10-3) m2 >s

F

5

Since ReL 6 1Rex2 cr = 5110 2, the boundary layer is laminar throughout the entire length of the plate. Thus, the total friction drag on the plate can be determined. FDf = Σ = c

0.664 brU 2L 2ReL

0.664(2 m)(1260 kg>m3)(2 m>s)2

= 694.97 N

23.3613(103)

d 32(3 m) + 2(0.01 m)4

Fb 5 741.636 N

Consider the force equilibrium along the vertical by referring to the FBD of the plate shown in Fig. a, where Fb = 1260(9.81)(2 * 3 * 0.1) = 741.636 N is the buoyant force. + c ΣFy = 0 ;

F + 741.636 N - 5019.812 N - 694.97 N = 0 Ans.

F = 443.834 N = 444 N

W 5 50(9.81) N

FDf (a)

Ans: F = 444 N 946

M11_HIBB9290_01_SE_C11_ANS.indd 946

16/03/17 10:37 AM

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11–19. Castor oil flows over the surface of the 2-m-long flat plate at a free-stream speed of 2 m>s. 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. The plate is 0.5 m wide. Take rco = 960 kg>m3 and mco = 985110-3 2 N # s>m2.

2 m!s

x 2m

SOLUTION We will assume that steady flow occurs and castor oil is incompressible. The Reynolds number as a function of x is Rex =

rcoUx = mco

( 960 kg>m3 )( 2 m>s ) x = 1949.24x 985 ( 10-3 ) N # s>m2

At x = L = 2 m, ReL = 1949.24(2) = 3.898 ( 103 ) 6 5 ( 103 ) . Thus, laminar flow persists within the boundary layer. For the boundary thickness, d =

5.0 2Rex

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. τ0 (N m2(

δ (mm)

60

200

50 150 40 100

30 20

50 10 x (m)

0

0.50

1.0

1.5

2.0

(a)

x (m) 0

0.5

1.0

1.5

2.0

(b)

947

M11_HIBB9290_01_SE_C11_ANS.indd 947

17/03/17 10:01 AM

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11–19. Continued

For the shear stress, t0 = 0.332m a

U b 2Rex 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 2ReL

0.664(0.5 m)( 960 kg>m3 )( 2 m>s ) 2(2 m) 23.898 ( 103 )

= 40.8 N

Ans.

τ0 (N m2(

60

50 40 30 20 10 x (m) 1.5

2.0

x (m) 0

0.5

1.0

1.5

2.0

(b)

Ans: FDf = 40.8 N 948

M11_HIBB9290_01_SE_C11_ANS.indd 948

17/03/17 9:06 AM

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*11–20. 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.

y

U

d y x

SOLUTION d

t0 = rU 2 t0 = rU 2 t0 =

d u u a1 - bdy dx L0 U U

d y y d a b a1 - bdy dx L0 d d

rU 2 dd d 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 rU L0

6m 1 2 d = x 2 ru d = 3.46

mx A rU

Since Rex = rU * 1m, then

d =

3.46x

Ans.

2Rex

Ans: d =

3.46x 2Rex

949

M11_HIBB9290_01_SE_C11_ANS.indd 949

16/03/17 10:37 AM

© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

11–21. The velocity profile for a laminar boundary layer of a fluid is approximated by u>U = 1.5(y>d) - 0.5(y>d)3. Determine the thickness of the boundary layer as a function of x and Rex.

y

U

d y x

SOLUTION The fluid is considered to be incompressible. The flow is steady. y y 3 u Substituting = 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 u u 2 - a b = J 1.5 a b - 0.5 a b R U U d d

y y 3 - J 1.5 a b - 0.5 a b R d d

2

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

y y 3 du d ` = u £ UJ 1.5a b - 0.5a b R § † dy y = 0 dy d d

y=0

1.5mU t0 = d

(2)

Equating Eqs. (1) and (2), 1.5mU dd = 0.1393rU 2 d dx d dd =

10.769m dx rU

950

M11_HIBB9290_01_SE_C11_ANS.indd 950

16/03/17 10:37 AM

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11–21. 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

1

r2U 2

=

4.6410x 1

1

1 2

m Since Rex =

1

r2U 2x2

rUx , this equation becomes m d =

=

4.6410x rUx B m

4.64x

Ans.

2Rex

Ans: d =

4.64x 2Rex

951

M11_HIBB9290_01_SE_C11_ANS.indd 951

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11–22. The velocity profile for a laminar boundary layer of a fluid is approximated by u>U = 1.5(y>d) - 0.5(y>d)3. Determine the shear-stress distribution acting on the surface as a function of x and Rex.

y

U

d y x

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 u u 2 - a b = J 1.5 a b - 0.5 a b R U U d d

y y 3 - J 1.5 a b - 0.5 a b R d d

2

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 d dd =

10.769m dx rU

952

M11_HIBB9290_01_SE_C11_ANS.indd 952

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11–22. 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

1

r2U 2

=

4.6410x 1

1

1 2

m Since Rex = d =

1

r2U 2x2

rUx , this equation becomes m

=

4.6410x rUx B m

4.6410x

(3)

2Rex

Substituting Eq. (3) into Eq. (2), t0 =

1.5mU U = 0.323m a b 2Rex 4.6410x x

Ans.

2Rex

Ans: t0 = 0.323ma

U b 2Rex x

953

M11_HIBB9290_01_SE_C11_ANS.indd 953

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11–23. The laminar boundary layer for a fluid is assumed to be parabolic, such that u>U = C1 + C2(y>d) + C3(y>d)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 persists 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 ` = mJ U 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 = -1 954

M11_HIBB9290_01_SE_C11_ANS.indd 954

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*11–24. The laminar boundary layer for a fluid is assumed to be cubic, such that u>U = C1 + C2(y>d) + C3(y>d)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

Ans.

C1 = 0

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 persists 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 ` = mJ U 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

C2 =

3 2

C3 = -

1 2

955

M11_HIBB9290_01_SE_C11_ANS.indd 955

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11–25. A laminar boundary layer for a fluid is approximated by u>U = sin (py>2d). 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 1 d b + c - cos a 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 d dd =

11.498m dx rU

956

M11_HIBB9290_01_SE_C11_ANS.indd 956

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11–25. 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 2

1 2

1 2

rU x 1 2

m Since Rex =

1

4.7953m2x2

=

4.7953x rUx B m

rUx , this equation becomes m d =

4.80x

Ans.

2Rex

Ans: d =

4.80x 2Rex

957

M11_HIBB9290_01_SE_C11_ANS.indd 957

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11–26. A laminar boundary layer for a fluid is approximated by u>U = sin (py>2d). 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 11–25, d = So,

u bdy U

2 bd p

py b bdy 2d

4.7953x 2Rex

d* = a1 -

2 4.7953x 1.74x b = p 2Re 2Rex x

Ans: d* =

1.74x 2Rex

958

M11_HIBB9290_01_SE_C11_ANS.indd 958

16/03/17 10:37 AM

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11–27. A boundary layer for laminar flow of a fluid over the plate is approximated by u>U = C1(y>d) + C2(y>d)2 + C3(y>d)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 disturbance 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 = UJ C1 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,

2C2y 3C3y2 C1 du = UJ + + R 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

0 =

U ( 2C2 + 0 ) d2

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 - a b R dy J dx L0 U U d

t0 = rU 2 Here,

u u 2 3 y 1 y 3 - a b = J a b - a b R U U 2 d 2 d

- J

3 y 1 y 3 a b - a b R 2 d 2 d

2

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 959

M11_HIBB9290_01_SE_C11_ANS.indd 959

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11–27. 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 d dd =

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 Rex =

1 2

=

4.6410x rUx B m

rUx , this equation becomes B m 4.64x d = 2Rex

Ans.

Ans: C1 =

3 1 4.64x , C = 0, C3 = - , d = 2 2 2 2Rex

960

M11_HIBB9290_01_SE_C11_ANS.indd 960

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*11–28. 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 throughout the central 1-m core of the tunnel, determine the dimension a at the exit in order to accommodate the growing boundary layer. Show that the boundary layer is turbulent at the exit, and use d * = 0.0463x>(Rex)1>5 to calculate the displacement thickness.

a

1m

a

1m

6m

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 Rex =

( 40 m>s ) (6 m) Ux = = 15.894 ( 106 ) n 15.1 ( 10-6 ) m2 >s

Since Rex 7 (Rex)cr = 5 ( 105 ) , the boundary layer is turbulent. Thus the displacement thickness is d* =

0.0463x (Rex)

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

Ans.

Ans: a = 1.02 m 961

M11_HIBB9290_01_SE_C11_ANS.indd 961

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11–29. 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 disturbance thickness as a function of x and Rex. Use the empirical formula developed by Prandtl and Blasius.

y

U

d y x

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

1

d y 6 y 6 d a b J 1 - a b R dy dx L0 d d 1

d y 6 y 3 d J a b - a b R dy dx L0 d d

3rU 2 dd 28 dx

t0 =

(1)

Using the empirical formula developed by Prandtl and Blasius, 1

1

t0 = 0.0225rU 2 a

3rU 2 n 4 0.0225rU a b dx = dd Ud 28 2

n 4 b Ud

(2)

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

x

1

n 4 d dd = 0.21 a b dx U L0 L0 1 4

1

4 5 d n 4 x d 4 ` = 0.21a b x ` 5 0 U 0 1

5

d 4 = 0.2625 a

n 4 bx U 1

n 5 4 d = 0.343 a b x5 U 1

= 0.343 a

n5 1 5

bx

x

= 0.343≥

1

a

Ux 5 b n

¥

Ux . Then this equation becomes n 0.343x

However, Rex = d =

1

U x5

( Rex)

Ans.

1 5

Ans: d =

0.343x (Rex)1>5

962

M11_HIBB9290_01_SE_C11_ANS.indd 962

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11–30. 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 displacement thickness as a function of x and Rex. Use the empirical formula developed by Prandtl and Blasius.

y

U

d y x

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

1

d y 6 y 6 d a b J 1 - a b R dy dx L0 d d 1

d y 6 y 3 d J a b - a b R dy dx L0 d d

3rU 2 dd 28 dx

t0 =

(1)

Using the empirical formula developed by Prandtl and Blasius, 1

1

0.0225rU 2 a

t0 = 0.0225rU 2 a

3rU 2 n 4 b dx = dd Ud 28

n 4 b Ud

(2)

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

x

1

n 4 d dd = 0.21 a b dx U L0 L0 1 4

1

4 5 d n 4 x 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

¥

963

M11_HIBB9290_01_SE_C11_ANS.indd 963

16/03/17 10:37 AM

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11–30. Continued

Ux . Then this equation becomes n 0.343x

However, Rex = d =

(3)

1

( Rex) 5

The displacement thickness is d* =

L0

d

d

=

u bdy U

a1 -

1

y 6 c 1 - a b d dy d L0 6

= ay =

d 7

6 y7 d b` 7 d 16 0

Substituting Eq. 3 into this result, d* =

1 0.343x 0.0490x £ § = 1 7 ( Rex ) 15 ( Rex) 5

Ans.

Ans: d* =

0.0490x (Rex)1>5

964

M11_HIBB9290_01_SE_C11_ANS.indd 964

16/03/17 10:37 AM

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11–31. A barge is being towed at 3 m>s. Determine the total friction drag of the water on its sides and bottom. The water is still and has a temperature of 15°C. The submerged depth of the barge is 1.5 m. Assume the boundary layer is completely turbulent.

F 3 m!s

20 m

6m 1.5 m

SOLUTION We will consider the relative flow steady, and the water as incompressible. From the table in Appendix A, rw = 999.2 kg>m3 and vw = 1.15110-6 2 m2 >s for water at T = 15°C. Thus, the Reynolds member at x = L = 20 m is ReL =

(3 m>s)(20 m) UL = = 5.2174(107) vw 1.15(10-6) m2 >s

Here, we assume that the boundary layer is turbulent at x = 0. Since 107 … ReL 6 109, CDf =

0.455 1log10ReL 2 2.58

=

0.455 5 log10 35.21741107 246 2.58

= 0.002335

Thus, with b = 211.5 m2 + 6 m = 9 m, the total friction drag on the sides and bottom of the barge is 1 FDf = CDf bL a rU 2 b 2

1 = 0.002335(9 m)(20 m) c (999.2 kg>m3)(3 m>s)2 d 2 = 1890.02 N = 1.89 kN

Ans.

Ans: FDf = 1.89 kN 965

M11_HIBB9290_01_SE_C11_ANS.indd 965

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*11–32. 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 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 ReL =

( 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 … ReL 6 109, the frictional drag coefficient is CD =

0.455 (log10 ReL)2.58

=

0.455

3 log10 3 3.4783 ( 107 )44 2.58

= 0.0024785

Thus, the frictional drag force on the bottom surface can be determined from 1 F = CD a rU 2 bbL 2 = 0.0024785c = 495 N

1 ( 999.2 kg>m3 )( 4 m>s ) 2d(2.5 m)(10 m) 2

Ans.

Ans: F = 495 N 966

M11_HIBB9290_01_SE_C11_ANS.indd 966

16/03/17 10:37 AM

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11–33. An airplane has wings that each, on average, are 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 km 1h 1000 m of 2 km. Here, U = a600 ba ba b = 166.67 m>s. Thus, the Reynolds h 3600 s 1 km number at x = L = 3 m is ReL =

( 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 … ReL 6 109, the frictional drag coefficient is CD =

0.455

( log10 ReL ) 2.58

=

0.455 c log10 3 2.915 ( 10

7

)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

Ans.

Ans: F = 2.14 kN 967

M11_HIBB9290_01_SE_C11_ANS.indd 967

16/03/17 10:37 AM

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11–34. 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. Consider both laminar and turbulent boundary layers. Take r = 1030 kg>m3 and m = 1.14110-3 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 ReL =

rUL = m

( 1030 kg>m3 )( 2 m>s )(300 m) = 5.42 ( 108 ) 1.14 ( 10-3 ) N # s>m2

Since 5 ( 105 ) 6 ReL 6 109, the boundary layer on the hull will be laminar and turbulent along the length. Thus, CDf = =

0.455 (log10 ReL)2.58

-

1700 ReL

0.455 e log10 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 The power required is # W = ( FDf ) T (V) = = 31.40 ( 103 ) W

§ Ans.

3 15.70 ( 103 ) N 4( 2 m>s ) Ans.

= 31.4 kW

Ans: FDf = 15.7 kN # W = 31.4 kW 968

M11_HIBB9290_01_SE_C11_ANS.indd 968

16/03/17 10:37 AM

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11–35. The plane is flying at 450 km>h in still air at an altitude of 3 km. Determine the friction drag on the wing. Assume the wing is flat having the dimensions shown, and the boundary layer is completely turbulent.

12 m 2m

SOLUTION We will consider the relative flow as steady and air as incompressible. From the table in Appendix A, ra = 0.9092 kg>m3 and va = 18.63110-6 2 m2 >s for air at an altitude of km 1000 m 1h 3 km. Here, the free stream velocity is U = a450 ba ba b = 125 m>s h 1 km 3600 s ReL =

1125 m>s212 m2 UL = = 1.34191107 2 va 18.63110-6 2 m2 >s

Here, we assume that the boundary layer is turbulent at x = 0. Since 107 … ReL 6 109 CDf =

0.455 1log10 ReL 2 2.58

=

0.455 5log10 31.34191107 246 2.58

= 0.002867

Since the wing has top and bottom surfaces, the total friction drag on the wing is 1 FDf = 2c CDf bL a rU 2 b d 2

1 = 2e 0.002867112 m212 m2 c 10.9092 kg>m3 21125 m>s2 2 d f 2 = 977.44 N = 977 N

Ans.

Ans: FDf = 977 N 969

M11_HIBB9290_01_SE_C11_ANS.indd 969

16/03/17 10:38 AM

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*11–36. An airplane is flying at 300 km>h in still air at an altitude of 2 km. If the wings can be assumed to be flat plates, each having a width of 1.5 m, determine the disturbance boundary layer thickness at their midpoint and also at their trailing or back edge if the boundary layer is considered to be fully turbulent.

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, va = 17.15110-6 2 m2 >s for air at an altitude of 2 km. Here, the km 1000 m 1h free-stream velocity is U = a300 ba ba b = 83.33 m>s. Thus, h 1 km 3600 s the Reynolds number at x = L = 1.5 m and x = 0.75 m ReL = Rex =

183.33 m>s211.5 m2 UL = = 7.28861106) 7 51105 2 (turbulent) va 17.15110-6 2 m2 >s 183.33 m>s210.75 m2 Ux = = 3.64431106 2 va 17.15110-6 2 m2 >s

Here, we assume that the turbulent flow starts from x = 0. Thus, d" x = 1.5 m = d 0 x = 0.75 m =

0.371L 1ReL 2

1>5

0.371x

1Rex2

1>5

= =

0.37111.5 m2 37.28861106 24 1>5 0.37110.75 m2

33.64431106 24 1>5

= 0.02360 m = 23.6 mm Ans. = 0.01356 m = 13.6 mm

970

M11_HIBB9290_01_SE_C11_ANS.indd 970

Ans.

Ans: d 0 x = 1.5 m = 23.6 mm d 0 x = 0.75 m = 13.6 mm

16/03/17 10:38 AM

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11–37. An airplane is flying at 300 km>h in still air at an altitude of 2 km. If the wings can be assumed to be flat plates, each having a width of 1.5 m and a length of 5 m, determine the friction drag on each wing if the boundary layer is considered to be fully turbulent.

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 1.007 kg>m3 and va = 17.15110-6 2 m2 >s for air at an altitude of 2 km. Here, the free-stream velocity is km 1000 m 1h U = a300 ba ba b = 83.33 m>s. Thus, the Reynolds number at h 1 km 3600 s x = L = 1.5 m is ReL =

183.33 m>s211.5 m2 UL = = 7.28861106 2 v 17.15110-6 2 m2 >s

Here, we assume that the boundary layer is turbulent right from the leading edge, that is, from x = 0. Since 51105 2 6 ReL 6 107 CDf =

0.0740 ReL1>5

=

0.0740

37.28861106 24 1>5

= 0.003138

Since each using has two surfaces, top and bottom, the total friction drag can be determined from 1 FDf = ΣCDf bL a rU 2 b 2

1 = 2e 0.00313815 m211.5 m2 c 11.007 kg>m3 2183.33 m>s2 d f 2 = 164.60 N = 165 N

Ans.

Ans: FDf = 165 N 971

M11_HIBB9290_01_SE_C11_ANS.indd 971

16/03/17 10:38 AM

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11–38. If the wings of an airplane flying at a speed of 200 mi>h are assumed to have a flat surface of width 5 ft, determine the disturbance thickness of the boundary layer and the shear stress at the midpoint of the wing. Assume the boundary layer is fully turbulent. The airplane flies at an altitude of 5000 ft.

200 mi!s B

A 5 ft

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, rA = 2.043110-3 2 slug>ft3 and va = 0.1779(10-3) ft2 >s for air at an altitude of 5000 ft. Here, the free-stream velocity is mi 5280 ft 1h U = a200 ba ba b = 293.33 ft>s. Thus, the Reynolds number at h 1 mile 3600 s x = 2.5 ft is Rex =

1293.33 ft>s212.5 ft2 Ux = = 4.12221106 2 va 0.1779110-3 2 ft2 >s

Here, we assume that boundary layer is turbulent right from the leading edge, that is, from x = 0. Therefore, the disturbance thickness of the boundary layer and the shear stress at x = 2.5 ft are, respectively, d = t0 =

0.371x 1Rex2

1>5

0.0288rU 2 1Rex2

1>5

= =

0.37112.5 ft2 34.12221106 24 1>5

= 10.04408 ft2112 in.>1 ft2 = 0.529 in. Ans.

0.0288 32.043110-3 2 slug>ft3 41293.33 ft>s2 2 34.12221106 24 1>5

= 0.2406 lb>ft2 = 0.241 lb>ft2

Ans.

Ans: d = 0.529 in. t0 = 0.241 lb>ft2 972

M11_HIBB9290_01_SE_C11_ANS.indd 972

16/03/17 10:38 AM

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11–39. The wings of an airplane flying at an altitude of 3 km and a speed of 450 km>h can be assumed to have a flat surface of width 2 m and length 8 m. Determine the friction drag on each wing. Assume the boundary layer is fully turbulent.

450 km!h B

A 2m

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 0.9092 kg>m3 and va = 18.63110-6 2 m2 >s for the air at an altitude of 3 km. Here the free-stream velocity is km 1000 m 1h U = a450 ba ba b = 125 m>s. Thus, the Reynolds number at h 1 km 3600 s x = L = 2 m is ReL =

1125 m>s212 m2 UL = = 1.34191107 2 va 18.63110-6 2 m2 >s

Here, we assume that the boundary layer is turbulent right from the leading edge, that is, from x = 0. Since 107 … ReL 6 109, CDf =

0.455 1log10 ReL 2

2.58

=

0.455 3 log10 1.34191107 24 2.58

= 0.002867

Since each wing has two surfaces, top and bottom, the total friction drag can be determined from 1 FDf = ΣCDf bL a rU 2 b 2

1 = 2e 0.00286718 m212 m2 c 10.9092 kg>m3 21125 m>s2 2 d f 2 = 651.62 N

Ans.

= 652 N

Ans: FDf = 652 N 973

M11_HIBB9290_01_SE_C11_ANS.indd 973

16/03/17 10:38 AM

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*11–40. Determine the disturbance thickness of the boundary layer along the side of the trailer truck at its midlength x = 7.5 ft. The trailer travels at 60 mi>h. The air is still and has a temperature of 60°F. Assume the surface is smooth and flat.

15 ft 60 mi!h

x

8 ft

SOLUTION We will consider the flow as steady and the air as incompressible. From the table in Appendix A, va = 0.158110-3 2 ft2 >s for air at T = 60°F. Here, the free-stream mi 5280 ft 1h velocity is U = a60 ba ba b = 88 ft>s. Thus, the Reynolds number h 1 mi 3600 s at x = 7.5 ft is Rex =

188 ft>s217.5 ft2 Ux = = 4.17721106 2 va 0.158110-3 2 ft2 >s

Since Rex 7 1Rex2 cr = 51105 2, the boundary layer is turbulent at x = 7.5 ft. d =

0.371x

1Rex2 1>5

=

0.37117.5 ft2

34.17721106 24 1>5

= 10.1319 ft2a

12 in. b = 1.58 in. 1 ft

Ans.

Ans: d = 1.58 in. 974

M11_HIBB9290_01_SE_C11_ANS.indd 974

16/03/17 10:38 AM

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11–41. Determine the drag acting on each side of the trailer truck when it is traveling at 60 mi>h. The air is still and has a temperature of 60°F. Assume the surfaces are smooth and flat. Consider both laminar and turbulent boundary layers.

15 ft 60 mi!h

x

8 ft

SOLUTION We will consider the flow as steady and the air as incompressible. From the table in Appendix A, ra = 0.00237 slug>ft3 and va = 0.158110-3 2 ft2 >s for air at T = 60°F. mi 5280 ft 1h Here, the free-stream velocity is U = a60 ba ba b = 88 ft>s. Thus, h 1 mi 3600 s the Reynolds number at x = L = 15 ft is ReL =

188 ft>s2115 ft2 UL = 8.35441106 2 va 0.158110-3 2 ft2 >s

Since 51105 2 6 ReL 6 109, and the boundary layer is not completely turbulent, CDf = =

0.455

1log10 ReL2 2.58 0.455

-

1700 ReL

6

5log10 38.3544110 246

2.58

= 0.0028884

-

1700 8.35441106 2

Thus, the friction drag on one side of the truck is 1 FDf = CDf bL a rU 2 b 2

1 = 0.002888418 ft2115 ft2 c 10.00237 slug>ft3 2188 ft>s2 2d 2 = 3.1807 lb = 3.18 lb

Ans.

Ans: FDf = 3.18 lb 975

M11_HIBB9290_01_SE_C11_ANS.indd 975

16/03/17 10:38 AM

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11–42. The tail of the airplane has an approximate width of 1.5 ft and a length of 4.5 ft. Assuming the airflow onto the tail is uniform, plot the disturbance 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 tail. The plane is flying in still air at an altitude of 5000 ft with a speed of 500 ft>s. Consider both laminar and turbulent boundary layers.

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>ft3 and ma = 0.3637 ( 10-6 ) lb # s>ft2 for air at an altitude of 5000 ft. The Reynolds number as a function of x is raUx = ma

Rex =

3 2.043 ( 10-3 ) slug>ft3 4( 500 ft>s ) x 0.3637 ( 10-6 ) lb # s>ft2

= 2.809 ( 106 ) x

At the trailing edge where x = L = 1.5 ft, ReL = 2.809 ( 106 ) (1.5 ft) = 4.213 ( 106 ) . Since 5 ( 105 ) 6 ReL 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.

( Rex) 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 2Rex

x=

x(ft) d(in.)

0 0

5.0 22.809 ( 106 ) x 0.05 0.00801

1

x = c 2.9835 ( 10-3 ) x2 d ft 0.10 0.0113

0.15 0.0139

0.178 0.0151

976

M11_HIBB9290_01_SE_C11_ANS.indd 976

16/03/17 10:38 AM

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11–42. Continued

For the turbulent boundary layer where x 7 xcr, 0.371

d = x(ft) d(in.)

( Rex)

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

( log10ReL ) 2.58

-

1700 ReL

0.455

3 log104.213 ( 106 )4 2.58

-

1700 4.213 ( 106 )

= 0.003059

Thus, the frictional drag force can be determined by applying FDf = CDf bL a

raU 2 b 2

= 0.003059(4.5 ft)(1.5 ft) •

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: FDf = 5.27 lb

(a)

977

M11_HIBB9290_01_SE_C11_ANS.indd 977

16/03/17 10:38 AM

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11–43. An airplane is flying at an altitude of 10,000 ft and a speed of 450 mi>h in still air. If each wing is assumed to have a flat surface of width 6 ft and length 20 ft, determine the friction drag acting on each wing. Consider both laminar and turbulent boundary layers.

SOLUTION We will consider the relative flow as steady and air as incompressible. From the table in Appendix A, ra = 1.754110-32 slug>ft3 and va = 0.2015110-3 2 for air at an altitude mi 5280 ft 1h of 10 000 ft. Here, the free-stream velocity is U = a450 ba ba b = h 1 mi 3600 s 660 ft>s. Thus, the Reynolds number at x = L = 6 ft is ReL =

1660 ft>s216 ft2 UL = = 1.96531107 2 va 0.2015110-3 2 ft2 >s

Since 51105 2 … ReL 6 109, the friction drag coefficient, considering both laminar and turbulent layers, is CDf = =

0.455 1log10 ReL2 2.58

-

1700 ReL

7

2.58

0.455

3log10 1.9653110 24

= 0.002615

-

1700 1.96531107 2

Since each using has two surfaces, top and bottom, the total friction drag can be determined from 1 FDf = ΣCDf bL a rU 2 b 2

1 = 2e 0.002615120 ft216 ft2a b 311.754110-3 2 slug>ft3 41660 ft>s2 r 2

= 239.78 lb

Ans.

= 240 lb

Ans: FDf = 240 lb 978

M11_HIBB9290_01_SE_C11_ANS.indd 978

16/03/17 10:38 AM

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*11–44. The cargo ship is traveling forward at 30 mi>h in still water having a temperature of 50°F. If the bottom of the ship can be assumed to be a flat plate of length 250 ft and width 50 ft, determine the friction drag of the water on the bottom of the ship. Consider both laminar and turbulent boundary layers.

30 mi!h

250 ft

SOLUTION The relative flow is steady, and the water is incompressible. From the table in Appendix A, rw = 1.94 slug>ft3 and vw = 14.1110-6 2 ft2 >s for water at 50°F. Here, mi 5280 ft 1h the free-stream velocity is U = a30 ba ba b = 44 ft>s. Thus, the h 1 mi 3600 s Reynolds number at x = L = 250 ft is ReL =

144 ft>s21250 ft2 UL = = 7.8014(108) vw 14.1110-6 2 ft2 >s

Since 51105 2 … Re 6 109, the friction drag coefficient, considering both laminar and turbulent boundary layers, is CDf = =

0.455

1log10 ReL2

2.58

-

1700 ReL

0.455

3log10 7.8014(108)4 2.58

-

1700 7.8014(108)

= 0.001618

The friction drag on the flat botton of the tanker can be determined from 1 FDf = CDf bL a rU 2 b 2

1 = 0.001618150 ft21250 ft2 c 11.94 slug>ft3 2144 ft>s2 2 d 2 = 37.9811103 2 lb = 38.0 kip

Ans.

Ans: FDf = 38.0 kip 979

M11_HIBB9290_01_SE_C11_ANS.indd 979

16/03/17 10:38 AM

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11–45. The wall of the building is subjected to a wind that produces a pressure distribution that can be approximated by p = 1215.5ry1>2 2 Pa, where y is in meters. Determine the resultant pressure force on the wall. The air is at a temperature of 20°C, and the wall is 10 m wide. p 6m y

SOLUTION We will consider the flow as steady and the air as incompressible. The force of the pressure on a differential area dA = b dy = (10 m)dy is dF = p dA = (215.5ry1>2)(10 dy) = 2155ry1>2 dy. Thus, the resultant pressure force on the entire wall is FR =

LA

dF =

L0

6m

6m 2 2155ry1>2 dy = 2155ra y3>2 b ` = 21.1151103 2r N 3 0

From the table in Appendix A, ra = 1.202 kg>m3 for air at T = 20°C. Thus,

FR = 321.1151103 2411.202 kg>m3 2 = 25.381103 2 N = 25.4 kN

Ans.

Ans: FR = 25.4 kN 980

M11_HIBB9290_01_SE_C11_ANS.indd 980

16/03/17 10:38 AM

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11–46. The pressure distribution of air as it passes over a cylinder is p = 36 - 16>p2u4 kPa. Determine the pressure drag acting on the cylinder for 0 … u … 180°. The cylinder has a length of 4 m.

p 600 mm u

SOLUTION We will consider the flow as steady and the air as incompressible. The force of the pressure on a differential area dA = b ds = br du = 14 m210.6 m2 du = 2.4 du 6 u is dF = p dA = a6 - u b1103 212.4 du2 = 14.41103 2a1 - b du. The pressure p p 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 ;

1FD 2 p = = =

L A

1dF2 x

L0

p

L0

p

dF cos u 14.41103 2a1 -

= 14.41103 2

L0

p

u b cos u du p

1 u cos u b du p

a cos u -

= 14.41103 2 c sin u -

2 = 14.41103 2a b N p

p 1 (u sin u + cos u) d ` p u

= 9.167(103) N

Ans.

= 9.17 kN

(FR)y dF (FD)p 5 (FR)x

0.6 m

ds du

5

u

(a)

981

M11_HIBB9290_01_SE_C11_ANS.indd 981

Ans: 1FD 2 p = 9.17 kN

16/03/17 10:38 AM

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11–47. The windward roof on the house is 40 ft wide and is subjected to the wind as shown. If the average absolute pressure on the top of the roof is 36 psf, and under the roof, that is, within the attic, it is 24 psf, determine the pressure drag on the windward side of the roof.

25 ft 30°

SOLUTION

(FR)y

We will consider the flow as steady and the air as incompressible. Here, the net pressure on the top of the roof is pnet = 136 - 242 psf = 12 psf. The resultant force of this net pressure is

(FD)p 5 (FR)x

5

FR = 112 lb>ft2 2140 ft2125 ft2 = 121103 2 lb

The pressure drag is equal to the component of resultant force along the direction of the free-stream flow, which in this case is horizontal. Referring to Fig. a, + S

(FR)x = ΣFx ;

1FD 2 r = 3121103 2 lb4 sin 30° = 61103 2 lb = 6 kip

Ans.

12 psf FR 5 12(103) lb

30°

(a)

982

M11_HIBB9290_01_SE_C11_ANS.indd 982

Ans: 1FD 2 p = 6 kip

16/03/17 10:38 AM

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*11–48. Air pressure acting on the inclined surfaces is approximated by the linear distributions shown. Determine the resultant horizontal force acting on the surface if it is 3 m wide.

3 kPa 3 kPa 45°

5 kPa

6m

20° 6m

SOLUTION The air is considered to be incompressible. The flow is steady.

(FR)y

The resultant force of the trapezoidal and triangular pressure prism are Ftrap = Ftri =

1 N c (5 + 3) ( 103 ) 2 d 36 m(3 m)4 = 72 ( 103 ) N = 72 kN 2 m

FPD = (FR)x Ftri = 27 kN 45°

1 N c 3 ( 103 ) 2 d 36 m(3 m)4 = 27 ( 103 ) N = 27 kN 2 m

Ftrap = 72 kN 20 ° 3 kPa 5 kPa

Referring to Fig. a, + S

(FR)x = ΣFx ;

FPD = (72 kN) sin 20° + (27 kN) sin 45° = 43.7 kN

Ans.

(a)

Ans: FPD = 43.7 kN 983

M11_HIBB9290_01_SE_C11_ANS.indd 983

16/03/17 10:38 AM

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11–49. The front of the building is subjected to wind that exerts a pressure of p = 10.25y1>2 2 lb>ft2, 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 incompressible. The flow is steady. The force of the wind pressure on a differential area dA = b dy = (80 ft)dy is 1

1

dF = p dA = ( 0.25y2 ) (80 dy) = 20y2dy. Thus, the resultant force on the entire windward surface is FR =

LA

dF =

L0

30 ft

30 ft 3 1 2 20y2dy = 20 a b y2 ` 3 0

Ans.

= 2190.89 lb = 2.19 kip

Ans: FR = 2.19 kip 984

M11_HIBB9290_01_SE_C11_ANS.indd 984

16/03/17 10:38 AM

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11–50. 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 CD = 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>ft3 for air at T = 40°F . FD = CDApr

U2 2

= 1.43 380 ft(30 ft)4( 0.00247 slug>ft3 ) £

( 80 ft>s ) 2 2

§

= 27.13 ( 103 ) lb = 27.1 kip

Ans.

Ans: FD = 27.1 kip 985

M11_HIBB9290_01_SE_C11_ANS.indd 985

16/03/17 10:38 AM

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11–51. Determine the moment developed at the base A of the circular sign due to the 14 m>s wind. The air is at 20°C. Neglect the drag on the pole.

0.75 m

14 m!s

3m

SOLUTION We will consider the flow as steady and air as incompressible. From the table in Appendix A, ra = 1.202 kg>m3 and va = 15.1110-6 2 m2 >s for air at T = 20°C. The characteristic length for the circular sign is its diameter, d = 1.5 m. Thus, the Reynolds number of the flow is Re =

114 m>s211.5 m2 Ud = = 1.39071106 2 v 15.1110-6 2 m2 >s

Since Re 7 104, the value of CD = 1.1 for the circular disk can be used. Here, AP = p10.75 m2 2 = 0.5625p m2. FD = CDAP a

raV 2 b 2

= 1.110.5625p m2 2 c = 228.98 N

FD 5 228.98 N

3.75 m 3

11.202 kg>m 2114 m>s2 2

2

d

Ax

Here, FD acts through the center of the circular signboard, as shown on its FBD, Fig. a. Consider the moment equilibrium about point A. + ΣMA = 0;

A

MA - 1228.98N213.75 m2 = 0

MA = 858.67 N # m = 859 N # m

Ay

MA

(a)

Ans.

Ans: MA = 859 N # m 986

M11_HIBB9290_01_SE_C11_ANS.indd 986

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*11–52. 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 in still air 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>ft3 and ma = 0.374 ( 10-6 ) lb # s>ft2 at T = 60°F. The Reynolds number is raUD Re = = ma

( 0.00237 slug>ft3 )( 30 ft>s )a

0.75 ft b 12

0.374 ( 10-6 ) lb # s>ft2

= 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: FD = 11.3 lb 987

M11_HIBB9290_01_SE_C11_ANS.indd 987

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11–53. 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

6m

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

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

Ans: FD = 20.1 N 988

M11_HIBB9290_01_SE_C11_ANS.indd 988

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11–54. Determine the pressure drag acting on the side of the truck if the air is at 60°F and it is blowing at 30 mi>h. Take CD = 1.3. Note that 1 mi = 5280 ft .

8 ft

40 ft

30 mi!h

SOLUTION We will consider the flow as steady and the air as incompressible. From the table in Appendix A, ra = 0.00237 slug>ft3 and va = 0.158110-3 2 ft2 >s for air at T = 60°F. mi 5280 ft 1h Here, the free-stream velocity is U = a30 ba ba b = 44 ft>s. h 1 mi 3600 s The projected area of the side of container perpendicular to the free stream is AP = (40 ft)(8 ft) = 320. Thus, the drag acting on the container’s side can be determined. FD = CDAp a

raU 2 b 2

= 1.3(320 ft2) c

0.00237 slug>ft3)(44 ft>s)2

= 954.37 lb = 954 lb

2

d

Ans.

Ans: FD = 954 lb 989

M11_HIBB9290_01_SE_C11_ANS.indd 989

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11–55. The antenna on the building is made from two smooth cylindrical sections as shown. Determine the restraining moment at its base to hold it in equilibrium if it is subjected to a wind having an average speed of 25 m>s. The air is at a temperature of 10°C.

200 mm

10 m

400 mm

15 m

SOLUTION We will consider the flow as steady and the air as incompressible. From the table in Appendix A, ra = 1.247 kg>m3 and va = 14.2110-6 2 m2 >s for air at T = 10°C. Thus, the Reynolds numbers for the top and bottom section are (Re)b = (Re)t =

(FD)t 5 779.375 N

(25 m>s)(0.4 m)

UDb = = 7.0423(105) va 14.2(10-6) m2 >s (25 m>s)(0.2 m) UDt = = 3.5211(105) va 14.2(10-6) m2 >s

Since the antenna is of two different-diameter smooth cylinders, the drag coefficient can be determined by entering the respective Re’s into Fig. 11–31, which gives (CD)b _ 0.35 and (CD)t _ 0.8. Also, (Ap)b = (0.4 m)(15 m) = 6 m2 and (Ap)t = (0.2 m)(10 m) = 2 m2. Thus, the drag on the top and bottom sections of the antenna are (FD)t = (CD)t (Ar)t a = (0.8)(2 m2) c

raU 2 b 2

(1.247 kg>m3)(25 m>s)2 2

= 623.6 N

(FD)b = (CD)b (Ar)b a = (0.35)(6 m2) c = 818.34 N

(FD)b 5 818.34 N

20 m 7.5 m

Ax Ay

MA

d

raU 2 b 2

(1.247 kg>m3)(25 m>s)2 2

d

Here, (FD)t and (FD)b act through mid-height of their respective sections as shown in the free-body diagram of the antenna, Fig. a. + ΣMA = 0;

MA - (818.34 N)(7.5 m) - (623.6 N)(20 m) = 0 MA = 18.608(103) N # m = 18.6 kN # m

Ans.

Ans: MA = 18.6 kN # m 990

M11_HIBB9290_01_SE_C11_ANS.indd 990

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*11–56. 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. + ΣMA = 0;

MA - 3397.28 N(1.25 m) = 0

MA = 4246.6 N # m = 4.25 kN # m

Ans.

Ans: MA = 4.25 kN # m 991

M11_HIBB9290_01_SE_C11_ANS.indd 991

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11–57. The truck has a drag coefficient of CD = 1.12 when it is traveling 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 ba ba b = 22.22 m>s . U = a80 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

Ans: # W = 80.5 kW 992

M11_HIBB9290_01_SE_C11_ANS.indd 992

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11–58. The truck has a drag coefficient of CD = 0.86 when it is traveling 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: # W = 26.1 kW 993

M11_HIBB9290_01_SE_C11_ANS.indd 993

16/03/17 10:38 AM

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11–59. The motorcycle and passenger has a projected front area of 6.76 ft2. Determine the power required to travel at a constant velocity of 70 mi>h if the drag coefficient is CD = 0.64 and the air is at 80°F. Note that 1 mi = 5280 ft.

70 mi!h

SOLUTION

W

We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 0.00228 slug>ft3 for air at T = 80°F. The free-stream mi 5280 ft 1h velocity is U = a70 ba ba b = 102.67 ft>s. Here, h 1 mi 3600 s FD = CDAr a

raU 2 b 2 2

= (0.64)(6.76 ft ) c

(0.00228 slug>ft3)(102.67 ft>s)2 2

= 51.986 lb

FD 5 21.22 miylb

F N2

d

N1 (a)

Referring to the free-body diagram of the motorcycle shown in Fig. a, + d

ΣFx = max;

F - 51.986 lb = 0

F = 51.986 lb

Subsequently, the power needed to produce this driving force is # W = F # U = (51.986 lb)(102.67 ft>s2) = (5337.28 lb # ft>s) a = 9.70 hp

1 hp

550 lb # ft>s

b

Ans.

Ans: # W = 9.70 hp 994

M11_HIBB9290_01_SE_C11_ANS.indd 994

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*11–60. 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 = 4p m2. 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)

Ans: U = 9.27 m>s 995

M11_HIBB9290_01_SE_C11_ANS.indd 995

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11–61. 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 = 1.36 a

Ans.

d = 3.71 m

FD

a= 0

90(9.81) N (a)

Ans: d = 3.71 m 996

M11_HIBB9290_01_SE_C11_ANS.indd 996

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11–62. The man and the parachute have a total mass of 90 kg. If the parachute has an open diameter of 6 m and the terminal velocity is 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

CD = 2.08

2

d

Ans.

FD

a= 0

90(9.81) N (a)

Ans: CD = 2.08 997

M11_HIBB9290_01_SE_C11_ANS.indd 997

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11–63. 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.

60 mi!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 ba ba b = 44.44 m>s. U = a160 h 1 km 3600 s

From

FD = CDAP r

air

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, + F = 831.01 N F - 831.01 N = 0 S ΣFx = max; Subsequently, the power that must be supplied by the engine to produce this drive force is # W = 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: # W = 36.9 kW 998

M11_HIBB9290_01_SE_C11_ANS.indd 998

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*11–64. 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 =

1h km 1000 m 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

Ans.

Ans: FD = 17.6 N 999

M11_HIBB9290_01_SE_C11_ANS.indd 999

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11–65. A boat tows a half-submerged 600-lb oil drum having an approximate diameter of 2.5 ft. If the drag coefficient is CD = 0.76, determine the tension in the horizontal tow rope at the instant the boat is traveling at 20 ft>s and has an acceleration of 4 ft>s2. Take rw = 1.94 slug>ft3.

20 ft!s

SOLUTION

a 5 4 ftys2

We consider the relative flow steady, and the water is incompressible, since the drum is half-submerged and the flow is along the length of the drum, p11.25 ft2 2 AP = = 0.78125p ft2. Thus, the drag on the drum is 2 FD = cD AP a

rwU 2 b 2

= (0.76)(0.78125p ft2) c

x

W 5 600 lb T FD 5 723.74 lb

1.94 slug>ft3 2

= 723.74 lb

FB (a)

d

Referring to the free-body diagram of the drum in Fig. a, + S ΣFx = max;

T - 723.74 lb = a

600 lb b14 ft>s2 2 32.2 ft>s2

T = 798.28 lb = 798 lb

Ans.

Ans: T = 798 lb 1000

M11_HIBB9290_01_SE_C11_ANS.indd 1000

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11–66. The uniform crate has a mass of 50 kg and rests on a surface having a coefficient of static friction of ms = 0.5. If the speed of the wind is 10 m>s, determine whether it will cause the crate to either tip over or slide. The air temperature is 20°C. Take CD = 1.06.

2m 2m

2m

SOLUTION

W 5 490.5 N

We will consider the flow as steady and air as incompressible. From the table in Appendix A, ra = 1.202 kg>m3 and va = 15.1110-6 2 m2 >s for the air at T = 20°C. Thus, the Reynolds number is Re = 4

110 m>s212 m2 Ua = = 1.32451106 2 va 15.1110-6 2 m2 >s

FD 5 254.824 N 1m

O

Since Re 7 10 and we consider the crate as a cube, CD = 1.06 Here, AP = (2 m)(2 m) = 4 m2. FD

(1.202 kg>m2)(10 m>s)2 rU 2 = CDAp a b = 1.06(4 m2) c d 2 2

F

x N (a)

= 254.824 N

The free-body diagram of the crate is shown in Fig. a. Here, FD acts at the mid-height of the crate. Considering the equilibrium of the crate, + c ΣFy = 0; + S ΣFx = 0; + ΣM0 = 0;

N - 50(9.81) N = 0

N = 4.90.5 N

254.824 N - F = 0

F = 254.824 N

350(9.81)N4x - 254.824(1 m) = 0

x = 0.520 m

Since F 7 F max = ms N = 0.5(490.5 N) = 245.25 N, and x 6 1 m, the crate will slide before it tips. Ans.

Ans: The crate will slide. 1001

M11_HIBB9290_01_SE_C11_ANS.indd 1001

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11–67. The rocket has a 60° nose cone 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: FD = 2.20 kN 1002

M11_HIBB9290_01_SE_C11_ANS.indd 1002

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*11–68. 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 airflow.

SOLUTION The air is considered to be incompressible. The flow is steady. From Appendix A, r = 0.00237 slug>ft3 and n = 0.158 ( 10-3 ) ft2 >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 ft2. Normal:

( 50 ft>s ) U2 = 1.1 ( 1 ft2 )( 0.00237 slug>ft3 ) 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 ReL =

( 50 ft>s ) (1 ft) UL = 3.165 ( 105 ) = n 0.158 ( 10-3 ) ft2 >s

Since ReL 6 (Rex)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 2ReL

= 0.0140 lb

= 2c

0.664(1 ft) ( 0.00237 slug>ft3 )( 50 ft>s ) 2(1 ft) 23.165 ( 105 )

d

Ans.

Ans: Normal: FD = 3.26 lb Parallel: FD = 0.0140 lb 1003

M11_HIBB9290_01_SE_C11_ANS.indd 1003

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11–69. A ball has a diameter of 40 mm and falls in honey with a terminal velocity of 0.3 m>s. Determine the mass of the ball. For honey, take rh = 1360 kg>m3 and nh = 0.04 m2 >s. Note: The volume of a sphere is V = 43pr 3.

SOLUTION We will consider the relative flow as steady and the honey as incompressible. The Reynolds number is Re =

(0.3 m>s)(0.04 m) UD = = 0.3 vh 0.04 m2 >s

FB 5 0.44708 N

W 5 m(98.1)

Since Re 6 1, the drag coefficient for the sphere can be determined using Stoke’s equation. CD =

24 24 = = 80 Re 0.3 FD 5 6.1525 N

Here, AP = p(0.02 m)2 = 0.4(10-3)p m2. So FD = CDAp a

a50

(a)

(1360 kg>m3)(0.3 m>s)2 r hU 2 b = 8030.4(10-3)p m2 4 c d 2 2 = 6.1525 N

The volume of the ball is V =

4 3 4 pr = p(0.02 m)3 = 10.667(10-6)p m3. 3 3

Thus, the buoyant force is FB = rhgV = (1360 kg>m3)(9.81 m>s2)310.667(10-6)p m3 4 = 0.44708 N

Consider the force equilibrium along the vertical by referring to the free-body diagram of the ball, Fig. a. + c ΣFy = may;

6.1525 N + 0.44708 N - m(9.81) = 0 m = 0.6727 kg = 673 g

Ans.

Note: The density of the ball, 20.08(103) kg>m3, is just a little less than that of osmium, a bluish-white metal that is the densest known naturally occurring element at 22.59(103) kg>m3.

Ans: m = 673 g 1004

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11–70. 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. Assume the body of liquid remains at rest as it is being stirred.

T v

L L

w

SOLUTION The liquid is considered to be incompressible. The relative flow is steady. The drag on the differential area dA = w dr is shown on the free-body diagram of the blade, Fig. a. dFD = CDAP r

U 2

z

2

dr

Here, AP = dA = w dr and U = vr. Thus, dFD = CD(w dr)rc

(vr)2 2

d =

T

1 C rwv2r 2 dr 2 D

dr

dFD

Since the blade rotates with a constant angular velocity, moment equilibrium exists about the z axis. Thus, L

r r

dFD

2 dFD(r) - T = 0 L0 L

L

1 T = 2 dFD(r) = 2 CD rwv2r 3 dr L0 L0 2 = CDrwv2

L0

L

r 3 dr

r4 2 b 4 0

L

= CDrwv2 a =

1 C rwv2L4 4 D

Ans.

Ans: T =

1 C rwv2L4 4 D

1005

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11–71. A solid ball has a diameter of 20 mm and a density of rb = 3.00 Mg>m3. Determine its terminal velocity if it is dropped into a liquid having a density of rl = 2.30 Mg>m3 and a kinematic viscosity of nl = 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

a= 0

4 3 4 D 3 p pr = p a b = D3. Thus, the weight of the 3 3 2 6

p p W = mg = rbVg = rb c D3 d g = rb gD3 6 6

FD = 3

L

vUD

(a)

p p FB = rLVg = rL a D3 bg = rL gD3 6 6

Referring to the free-body diagram in Fig. a, p p + c ΣFy = may; 3prLnUD + rLgD3 - rbgD3 = 0 6 6 U =

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

= 0.4907 ( 10-3 ) 6 1

(O.K.)

Ans: U = 0.00128 m>s 1006

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*11–72. A ball has a diameter of 8 in. When it is kicked, it is given 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>ft3 and n = 0.158 ( 10-3 ) ft2 >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 ) ft2 >s

Entering this Re into the graph for a sphere, CD ≅ 0.5 (approx.). Also, 2 4 AP = p a ft b = 0.1111p ft2. 12

( 18 ft>s ) U2 § = 0.5 ( 0.1111p ft2 )( 0.00237 slug>ft3 ) £ 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: FD = 0.0670 lb 1007

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11–73. 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 42.5110-12 2 g, estimate the time needed for it to settle to the earth. Assume gravity is constant, and for air, r = 1.202 kg>m3 and m = 18.1110-6 2 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 43 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 is 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: t = 114 days 1008

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11–74. A sprayer ejects a coating of particles at 25 m>s. Determine their velocity 8 μs after leaving the nozzle. Assume the average diameter of the particles is 0.6 μm and each has a mass of 0.8110-12 2 g. The air is at 20°C. Neglect the vertical component of the velocity. Note: The volume of a sphere is V = 43pr 3.

25 m!s

SOLUTION

mg

We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 1.202 kg>m3 and ma = 18.1110-6 2 N # s>m2 for air at T = 20°C. Thus, the maximum Reynolds number is 1Re2 max

11.202 kg>m3 2125 m>s2306110-6 2 m4 raUmaxD = = = 0.9961 ma 18.1110-6 2 N # s>m2

a5

dv dt x

FD 5 32.58(10 –12)pV (a)

Since 1Re2 max 6 1, the drag on the particle can be determined using FD = 3pmaVD

= 3p318.1110-6 2 N # s>m2 41V230.6110-6 2m4 = 32.58110-12 2pV

Referring to the free-body diagram of the particle in Fig. a, S

+ ΣFx = max;

-32.58(10-12)pV = c

- 40.725(103)p

L0

8(10 -6) s

dt =

0.8(10-12) 1000 V

L25 m>s V

dV

kg d a

dV b dt

-0.3258p = ln V 0 V 25 m>s -0.3258p = ln

V 25

V 2.5 V = 8.98 m>s

e -0.3258p =

Ans.

Ans: V = 8.98 m>s 1009

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11–75. Impure water at 20°C enters the retention tank and rises to a level of 2 m. Determine the shortest time needed for all spherical 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

FB =

6

Water 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 a= 0

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

The volume of the particles is V =

W = mg = rVg = rb a

p 3 p D bg = rbgD3 6 6

FD = 3

L

vUD

(a)

p p FB = rLVg = rL a D3 bg = rLgD3 6 6

Referring to the free-body diagram in Fig. a, p p + c ΣFy = may ; 3prLnUD + rLgD3 - rbgD3 = 0 6 6 U =

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 = = 2435.42 s a b -3 U 60 s ( ) 0.8212 10 m>s

Ans.

= 40.6 min

Ans: t = 40.6 min 1010

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*11–76. A styrofoam ball having a diameter of 3 in. and a weight of 0.05 lb is dropped from a high building. Determine its terminal velocity. The temperature of the air is 60°F. Note: The volume of a sphere is V = 43pr 3.

SOLUTION

FB 5 0.6243(10–3) lb

We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 0.00237 slug>ft3 and va = 0.158110-3 2 ft2 >s for air at T = 60°F. Thus, the Reynolds number is Re =

3 Va ft b 12

VD = = 1.58231103 2V va 0.158110-3 2 ft2 >s

The projected area perpendicular to the air stream is Ar = pa

FD = CDAP a = CD a

raV 2 b 2

W 5 0.05 lb

y a50

(1) FD 5 58.17(10–6)CD V2 (a)

2 1.5 p 2 ft b = ft 12 64

3 2 p 2 10.00237 slug>ft 2V ft b c d 64 2

= 58.17110-6 2CDV 2

3 4 4 15 p 3 The volume of the ball is V = pr 3 = pa ft b = ft . Thus, the buoyant 3 3 12 384 force is

FB = raVg = (0.00237 slug>ft3) a

p 3 ft b(32.2 ft>s2) = 0.6243(10-3) lb 384

Referring to the free-body diagram of the ball, Fig. a, + c ΣFy = may;

58.17110-6 2CDV 2 + 0.6243110-3 2 lb - 0.05 lb = 0 29.13 V = 2CD

(2)

Trial and error iterative procedure is required. The iterations are tabulated below. Iteration

Assumed CD

V(ft>s); (Eq. 2)

Re; (Eq. 1)

CD from Fig. 11-31

1

0.6

37.61

0.5

2

0.5

41.20

5.951104 2

0.5

6.521104 2

Since the assumed CD is almost the same as that obtained from Fig. 11-31 in the 2nd iteration, the result of V in this iteration is acceptable. Thus, Ans.

V = 41.2 ft>s

Ans: V = 41.2 ft>s 1011

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11–77. A raindrop has a diameter of 1 mm. Determine its approximate terminal velocity as it falls. Assume that for air ra = 1.247 kg>m3 and na = 14.2110-6 2 m2 >s. Neglect buoyancy. Note: The volume of a sphere is V = 43pr 3.

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

FD (a)

0.001 m 2 b = 2.5 ( 10-7 ) p m2. 2 U2 U2 b = CDAP r = CD 3 2.5 ( 10-7 ) p m2 4( 1.247 kg>m3 ) a 2 2

Here, AP = p a FD

= 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 = 3.98 m>s

Ans: U = 3.98 m>s 1012

M11_HIBB9290_01_SE_C11_ANS.indd 1012

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11–78. 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. Since we have “creeping flow,” the path of descent is practically a straight line. 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

(O.K.)

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: d = 8.88 m 1013

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11–79. A 2-mm-diameter sand particle having a density of 2.40 Mg>m3 is released from rest at the surface of oil. 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 = 30.2110-3 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 =

W (a)

30.2 ( 10 - 3 ) N # s>m2

m0 = r oD

( 900 kg>m3 ) (0.002 m)

= 0.016778 m>s = 16.8 mm>s Ans.

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 ) L 0

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 1014

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*11–80. The spherical balloon filled with helium has a weight of 0.02 lb. Determine its terminal velocity of ascent. The air temperature is 60°F. Note: The volume of a sphere is V = 43 pr 3.

6 in.

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 0.00237 slug>ft3 and va = 0.158(10-3) ft2 >s for air at T = 60°F. Thus, the Reynolds number is V(1 ft) VD Re= = = 6.3291(103)V va 0.158(10-3) ft2 >s

(1)

4 The volume of the balloon is V = p(0.5 ft)3 = 0.1667p ft3. Thus, the buoyant 3 force is

FD 5 0.29625(10–3)pCDV2 W 5 0.02 lb

y a50

Fb 5 0.03996 lb (a)

Fb = raVg = (0.00237 slug>ft3)(0.1667p ft3)(32.2 ft>s2) = 0.03996 lb Here, AP = p(0.5 ft)2 = 0.25p ft2. Then the friction drag on the balloon is FD = CDAP a

(0.00237 slug>ft3)V 2 raV 2 b = CD(0.25p ft2) c d = 0.29625(10-3)pCDV 2 2 2

When the balloon is ascending, with the terminal velocity, its acceleration is a = 0. Referring to the free body of the balloon, Fig. a, + c ΣFy = may;

0.03996 lb - 0.02 lb - 0.29625(10-3)pCDV 2 = 0 4.6308 V = 2CD

(2)

Trial and error iterative procedure is required. The iterations are tabulated below. Iteration

Assumed CD

V(ft>s); (Eq. 2)

1

0.6

5.9783

2

0.5

6.5489

Re; (Eq. 1)

CD from Fig. 11-31

4

0.5

4

0.5

3.78(10 ) 4.14(10 )

Since the assumed CD in the 2nd iteration is almost the same as that obtained from Fig. 11-31, the result of V in this iteration is acceptable. Thus, Ans.

V = 6.55 ft>s

Ans: V = 6.55 ft>s 1015

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11–81. The parachutist and chute 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 become 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 Vr dV + VrV dA 0t Lcy Lcs

The control volume considered is the parachute and the parachutist. Since there is no opened control surface,

VrV dA = 0. Also, Vr can be factored out from the Lcs

FD

dV = V since the volume of the Lcy control volume is fixed. Realizing that rV = m, the above equation reduces to

integral since it is independent of V . Also

+ T ΣFy =

d(my) dt

= m

dy dt

Referring to the FBD shown in Fig. a, and realizing that FD = CDAP a mg - CDAP a

raV 2 dy b = m 2 dt

raV 2 b, 2

2mg - CDAP raV 2 dy = m 2 dt

With the initial condition at t = 0, V = Vo, L0

t

V

dt = 2 m

t =

t =

dV 2mg cDAP raV 2 LV0 2m

2( 22mg) 2CDAp ra m 22mgCDAP ra

£ ln °

ln°

22mg + 2CDAP raV

22mg - 2CDAP raV

22mg + 2CDAP raV

22mg - 2CDAP raV

V

¢†

mg (a)

V0

¢ - ln°

22mg + 2CDAP raV0

22mg - 2CDAP raV0

¢§

(1)

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, V0 = 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

1016

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11–81.

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

Ans.

Ans: t = 0.805 s Vt = 12.0 m>s 1017

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11–82. 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 rr = 2400 kg>m3. Note: The volume of a sphere is V = 43 pr 3.

600 mm

SOLUTION

Fb

d ( mV )

+ T ΣFy =

dt

= m

dV V dV = m dt ds

FD

Referring to the FBD shown in Fig. a, and realizing that Fb = rw g V, rwV 2 FD = CDAp a b and W = rsg V, 2 rwV 2 V dV rsg V - rwg V - CDAr a b = rsV 2 ds 2(rs - rw)Vg - CDAprwV 2 = 2rsV

V dV ds

W

(a )

With the initial condition at s = 0, V = 0, L

s

V

ds = 2rsV

0

V dV 2 L 2(rs - rw)Vg - CDAprwV 0

Let a = 2(rs - rw)V g and b = CDAprw. Then L

V

s

ds = 2rsV

0

V dV 2 L a - bV 0

s = s =

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 =

bs rsV

a ( 1 - e -bs>rsV ) Ab

(1)

1018

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11–82. 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,

Ans.

V = 1.34 m>s

Ans: V = 1.34 m>s 1019

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11–83. A smooth ball has a diameter of 43 mm and a mass of 45 g. Determine its initial deceleration when it is thrown vertically upward with a speed of 20 m>s. The temperature is 20°C.

SOLUTION

FD = 0.1746 N

The air is considered to be incompressible. The relative flow is steady.

y

From Appendix A, r = 1.202 kg>m3 and y = 15.1 ( 10-6 ) m2 >s for air at T = 20°C . Thus, the initial Reynolds number is Re =

a

(20 m>s)(0.043 m)

UD = 5.695 ( 104 ) = y 15.1 ( 10-6 ) m2 >s

Entering this Re into the graph for a sphere, CD ≅ 0.5 (aprox.). Here, 0.043 m 2 AP = p a b = 0.46225 ( 10-3 ) p m2. 2 FD = CDApr

0.045(9.81) N (a)

U2 = 0.5 3 0.46225 ( 10-3 ) p m2 4( 1.202 kg>m3 ) J 2

( 20 m>s ) 2 2

R

= 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: a = -13.7 m>s2 1020

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*11–84. The smooth cylinder is suspended from the rail and is partially submerged in 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

0.5 m

0.25 m

SOLUTION The fluids are considered incompressible. The relative fluid is steady. From Appendix A, ra = 1.202 kg>m3 and ya = 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

(1)

V0(0.25 m) UwD = = 2.5 ( 105 ) V0 nw 1.00 ( 10-6 ) m2 >s

N1

a= 0 x

( FD( a

( FD ) a = ( CD ) a ( AP )a ra

2

= ( CD ) a(0.25 m) ( 1.202 kg>m3 ) J

W

(2)

( FD ( k

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. Ua2

N2

( 8 - V0 ) 2 2

(a)

R

= 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 - 16y0 + 64 ) - 415.27 ( CD ) wV 02 = 0 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) 1.4

1.4

0.3742

1.26 ( 105 )

(3)

9.36 ( 104 )

( CD ) a

( CD ) w

1.4

1.4

Since the assumed CD is about 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: V0 = 0.374 m>s 1021

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11–85. A 5-Mg airplane is flying at a speed of 60 m>s. If each wing can be assumed to be rectangular of span 6 m and cord distance of 1.5 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 = 1.21 kg>m3.

6m

6m

SOLUTION We will consider the relative flow as steady and the air as incompressible. For two wings, A = 211.5 m216 m2 = 18 m2 Thus, the lift is FL = CLAa

11.21 kg>m3 2160 m>s2 2 rV 2 b = C1 118 m2 2 c d = 39,204CL 2 2

Equilibrium along the vertical requires + c ΣFy = 0;

FL - W = 0 39,204 CL - 500019.812 N = 0 CL = 1.2511.

Enter this value of CL into Fig. 11–44. a ≈ 15°

Ans.

Ans: a ≈ 15° 1022

M11_HIBB9290_01_SE_C11_ANS.indd 1022

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11–86. The 4-Mg airplane has wings that are each 6 m long and 1.5 m wide. It is flying horizontally at an altitude of 2 km with a speed of 450 km>h. Determine the lift coefficient. 6m

6m

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, ra = 1.007 kg>m3 for air at an attitude of 2 km. Here, the freekm 1000 m 1h stream velocity is V = a450 ba ba b = 125 m>s. For two wings, h 1 km 3600 s A = 2(1.5 m)(6 m) = 18 m2. Thus, the lift is FL = CLAa

11.007 kg>m3 2(125 m>s)2 rV 2 b = CL 118 m2 2 c d = 141.611103 2CL 2 2

Equilibrium along the vertical requires

c ΣFy = 0;

FL - W = 0 141.611103 2CL - 400019.812 N = 0

Ans.

CL = 0.277

Ans: CL = 0.277 1023

M11_HIBB9290_01_SE_C11_ANS.indd 1023

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11–87. An 8000-lb airplane is flying at a speed of 180 mi>h. If each wing can be assumed to be rectangular of span 18 ft and cord distance of 4 ft, determine the section 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 = 0.00190 slug>ft3.

SOLUTION We will consider the relative flow as steady and the air as incompressible. Here, mi 5280 ft 1h the free-stream velocity is V = a180 ba ba b = 264 ft>s. For two h 1 mi 3600 s wings, A = 2(4 ft)(18 ft) = 144 ft2. Thus, the lift is FL = CLAa

(0.00190 slug>ft3)(264 ft>s)2 raV 2 b = CL 1144 ft2 2 c d = 9.53441103 2CL 2 2

The equilibrium along the vertical requires + c ΣFy = 0;

FL - w = 0 3

9.5344110 2CL - 8000 lb = 0

CL = 0.839

Enter this value of CL into Fig. 11–44. The angle of attack is a = 9°1approx.2 Using this result, Fig. 11–39 gives 1CD2 ∞ ≈ 0.045. For each wing, Apl = 14 ft2118 ft2 = 72 ft2 . Since the wing tip effect will be neglected, CD = 1CD2 ∞ = 0.045. Thus, the drag on each wing is FD = CDApl a

10.00190 slug>ft3 21264 ft>s2 2 raV 2 b = 0.045172 ft2 2 c d 2 2 = 214.52 lb

Ans.

= 215 lb

Ans: FD = 215 lb 1024

M11_HIBB9290_01_SE_C11_ANS.indd 1024

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*11–88. The plane can take off at 120 mi>h when it is at an airport located at an elevation of 2500 ft. Determine the takeoff speed from an airport at sea level.

V

SOLUTION We will consider the relative flow as steady and the air as incompressible. From the table in Appendix A, 1ra 2 1 = 0.002218 slug>ft3 and 1ra 2 2 = 0.002375 slug>ft3 for air at an altitude of 2500 ft and sea level, respectively. Here, it is required that 1CL 2 1A1 c

1FL 2 k = 1FL 2 2

1ra 2 1 1V 12 2 2

d = 1CL 2 2A2 c c

V2 = a

1ra 2 2 1V 22 2

2 1CL 2 1A1 1ra 2 1

B 1CL 2 2A2 1ra 2 2

bV1

dd

Since A1 = A2 and the angle of attack is the same for both cases, 1CL 2 1 = 1CL 2 2. Thus, V2 = a

1ra 2 1

B 1ra 2 2

bV1 = a

0.002218 slug>ft3

B 0.002375 slug>ft3

b1120 mi>h2 = 115.97 mi>h = 116 mi>h Ans.

Ans: V2 = 116 mi>h 1025

M11_HIBB9290_01_SE_C11_ANS.indd 1025

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11–89. 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 ft2, 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: u = 20.8° 1026

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11–90. The glider has a weight of 160 kg. If the drag coefficient is CD = 0.316, the lift coefficient is CL = 1.20, and the total area of the wings is A = 6 m2, determine if it can land on a 1.5-km-long landing strip that is located 5 km away from where its altitude is 1.5 km. Assume the density of the air remains constant.

u

SOLUTION

= 14.75′

1.5 Km

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)

d (a )

Dividing Eq. (1) by Eq. (2), FD W sin u = = tan u FL W cos u

(3)

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.

Ans: The glider can land. 1027

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11–91. 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 can be considered negligible to that on the wings, since the glider has a very long wingspan.

8 m!s

u

SOLUTION

y

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

W sin u - FD = 0

raV 2 W sin u - CDAP a b = 0 2 W sin u = CDAP a

+ c ΣFy = 0;

FL - W cos u = 0

CLAP a

raV 2 b 2

raV 2 b - W cos u = 0 2 W cos u = CLAP a

Dividing Eq. (1) by Eq. (2), W sin u = W cos u

tan u =

CDAP a CLAP a

CD CL

u = tan-1 a

FL

FD

(1) (a)

raV 2 b 2

x

(2)

raV 2 b 2

raV 2 b 2

CD 0.04 b = tan-1 a b = 3.27° CL 0.7

Ans.

Ans: u = 3.27° 1028

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*11–92. 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: U2 = 130 mi>h 1029

M11_HIBB9290_01_SE_C11_ANS.indd 1029

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11–93. 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 3.5 ft

SOLUTION Relative to the airplane, the flow is steady. Also, air is assumed to be incompressible. Appendix A gives ra = 2.043 ( 10-3 ) slug>ft3. 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 •

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.) 1030

M11_HIBB9290_01_SE_C11_ANS.indd 1030

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11–94. 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, and 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 stall occurs?

16 ft 3.5 ft

SOLUTION Relative to the airplane, the flow is steady. Also, air is assumed to be incompressible. Appendix A gives ra = 2.043 ( 10-3 ) slug>ft3. Since the air plane is flying at a constant altitude, equilibrium exists along the vertical. Thus, + c ΣFy = 0;

FL - W = 0 2

CLAp a

raV b - W = 0 2

CL[2(16 ft)(3.5 ft)] •

(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)2 >(16 ft)(3.5 ft)

Thus, the drag force on the airplane caused by the wing is FD = CD AP a

raV 2 b 2

= 0.023332(16 ft)(3.5 ft)4 • = 135 lb

3 2.043 ( 10-3 ) slug>ft3 4 ( 225 ft>s ) 2 2

¶

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 AP a

raV 2 b - W = 0 2

1.5[2(16 ft)(3.5 ft)] •

3 2.043 ( 10-3 ) slug>ft3 4 V 2s

Vs = 107.95 ft>s = 108 ft>s

2

¶ - 2000 lb = 0

Ans.

Ans: FD = 135 lb a = 20° Vs = 108 ft>s

1031

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11–95. A 15,000-lb airplane has two wings, each having a span of 18 ft and average cord distance of 4 ft. If the total drag on each wing is 850 lb when it is flying at 300 ft>s, determine the total drag on each wing when the plane is flying at the same angle of attack and the same altitude with a speed of 450 ft>s. Assume an elliptical lift distribution. Take ra = 1.78110-3 2 slug>ft3.

SOLUTION We will consider the flow as steady and the air as incompressible. Since the plane is flying at a constant altitude, it is in vertical equilibrium. This means that the lift is equal to its weight; i.e., FL = 15 000 lb. Here, for two wings, Ar = 214 ft2118 ft2 = 144 ft2. Then, the lift coefficients for V1 = 300 ft>s and V2 = 450 ft>s are given by 1CL 2 1 = 1CL 2 2 =

FL Ap a

raV 12 2 FL

= b

raV 12 Ap a b 2

=

1144 ft2 2 e 1144 ft2 2 e

15 000 lb 31.78110-3 2 slug>ft3 41300 ft>s2 2 2

15 000 lb 31.78110-3 2 slug>ft3 41450 ft>s2 2 2

= 1.3005 f

= 0.5780

f

For V1 = 300 ft>s, the given total drag on each wing with Ap = 14 ft2118 ft2 = 72 ft2 gives 1FD 2 1 = CDAp a

raV 12 b; 2

850 lb = 1CD 2 1 172 ft2 2 e

1CD 2 1 = 0.1474

178110-3 2 slug>ft3 1300 ft>s2 2 2

f

Using this total drag coefficient,

1CD 2 1 = 1CD 2 ∞ + 0.1474 = 1CD 2 ∞ +

1CD 2 ∞ = 0.02776

1CL 2 12

pb2 >Ap

1.30052 p(18 ft)2 >72 ft2

Since the angle of attack is the same for both cases, 1CD 2 ∞ remains constant. Thus, for V2 = 450 ft>s, the total drag coefficient is 1CD 2 2 = 1CD 2 ∞ +

= 0.02776 +

Thus, the total drag is 1FD 2 2 = 1CD 2 2Ap a

1CL 2 22

pb2 >Ap

0.57802 = 0.05139 p118 ft2 2 >72 ft2

31.78110-3 2 slug>ft3 41450 ft>s2 2 raV 22 b = 10.051392172 ft2 2 e f 2 2 Ans.

= 666.82 lb = 667 lb

1032

M11_HIBB9290_01_SE_C11_ANS.indd 1032

Ans: 1FD 2 2 = 667 lb

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*11–96. 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 v = 400 rad>s. Determine its horizontal deviation d from striking a target a distance of 10 m away. Use Fig. 11–50 and neglect the effect on lift caused by the vertical component of velocity. Take ra = 1.20 kg>m3 and na = 15.0110-6 2 m2 >s.

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 ) ya 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 = CLAp a

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: d = 31.8 mm 1033

M11_HIBB9290_01_SE_C11_ANS.indd 1033

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11–97. A smooth ball has a diameter of 80 mm. If it is hit with a speed of 10 m>s and given an angular velocity of 80 rad > s, determine the lift on the ball. Take ra = 1.23 kg>m3 and na = 14.6110-6 2 m2 >s .

80 rad!s 10 m!s

SOLUTION For the given data, 180 rad>s210.08 m2 vD = = 0.32 2V 2110 m>s2 Re =

110 m>s210.08 m2 VD = = 5.481104 2 va 14.6110-6 2 m2 >s

Since Re … 61104 2, Fig. 11–50 can be used to determine the lift coefficient, which is CL ≈ 0.1. Here, AP = p10.04 m2 2 = 0.0016p m2. Thus, the lift can be determined using FL = CLAP a

raV 2 b 2

= 0.110.0016p m2 2 c

= 0.0309 = 0.03 N

11.23 kg>m3 2110 m>s2 2 2

d

Ans.

Ans: FL = 0.03 N 1034

M11_HIBB9290_01_SE_C11_ANS.indd 1034

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12–1. A canal is 2 m deep and 3 m wide, and the water flows with a velocity of 1.5 m>s. If a stone is thrown into it, determine how fast the waves will travel upstream and downstream. What is the Froude number?

1.5 m!s

2m 3m

SOLUTION We will consider the flow is steady and water is incompressible. The speed of the wave in still water is c = 2gy

c = 2 ( 9.81 m>s2 ) (2 m) = 4.429 m>s

Thus, applying the relative velocity equation with vw>r = c the speed of a wave traveling upstream is (Vw)u = vr + vw>r Ans.

+ b(vw)u = -1.5 m>s + 4.429 m>s = 2.93 m>s and the speed of a wave traveling downstream is (Vw)d = vr + vw>r

Ans.

+ Q(vw)d = 1.5 m>s + 4.429 m>s = 5.93 m>s The Froude number of the flow is Fr =

1.5 m>s V V = = = 0.339 4.429 m>s 1gy C

Ans.

Ans: Vup = 2.93 m>s Vdown = 5.93 m>s Fr = 0.339 1035

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12–2. A large tank contains water having a depth of 3 ft. If the tank is on an elevator that is descending, determine the horizontal speed of a wave created on its surface if the rate of descent is (a) constant at 3 ft>s, (b) accelerated at 2 ft>s2, (c) accelerated at 32.2 ft>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 = 32.2 ft>s . Then,

c = 2 ( 32.2 ft>s2 ) (3 ft) = 9.83 ft>s

Ans.

c = 2 ( 30.2 ft>s2 ) (3 ft) = 9.52 ft>s

Ans.

N (a)

b.) For a = 2 ft>s2, g′ = g - 2 ft>s2 = 30.2 ft>s2. Then,

c.) For a = 32.2 ft>s2, g′ = g - 32.2 ft>s2 = 0. Then,

Ans.

c = 20(3 ft) = 0

Ans: a) 9.83 ft>s b) 9.52 ft>s c) 0 1036

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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 1037

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*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

Ans: Fr = 0.434, subcritical Vc = 2.91 m>s 1038

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12–5. Water flows with a velocity 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 ft3 >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 1039

M12_HIBB9290_01_SE_C12_ANS.indd 1039

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12–6. Water flows in a rectangular channel with a velocity of 2 m>s and depth of 4 m. What other possible depth of flow provides the same specific energy? Plot the specific energy diagram.

SOLUTION We will consider the flow as steady and the water as incompressible. The flow rate is Q = VA = 12 m>s23b14 m24 = 8 b. For the rectangular channel, the specific energy can be determined by E =

Q2

+ y=

2 2

2gb y

18b2 2 2

2

2

219.81 m>s 21b 2y

+ y=

3.2620 + y y2

(1)

For the depth of y = 4 m, Eq. (1) gives E =

3.2620 + 4 m = 4.2039 m 14 m2 2

Substitute this result into Eq. (1),

4.2039 m =

3.2620 + y y2

y3 - 4.2039y2 + 3.2620 = 0 Solving numerically, Ans.

y = 1.01072 m = 1.01 m y= 4m

(The given depth)

y = - 0.8068 m

(not realistic)

In this case, the critical depth can be determined by yc = a

Q2 gb2

b

1>3

= c

18b2 2

19.81 m>s2 2b2

d

1>3

y(m) y5E

= 1.8685 m

For y = 4 m 7 yc, the flow is subcritical.

4

For y = 1.01 m 6 yc, the flow is supercritical. The minimum specific energy can be determined by E min =

yC 5 1.869

3 3 y = 11.8685 m2 = 2.803 m 2 c 2

1.011

The plot of depth yvs specific energy is shown in Fig. (a)

4.204

E(m)

Emin 5 2.803 (a)

Ans: y = 1.01 m 1040

M12_HIBB9290_01_SE_C12_ANS.indd 1040

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12–7. Water flows in a rectangular channel with a velocity of 4 m>s. If the depth is 1.5 m, what other possible velocity of flow provides the same specific energy?

SOLUTION We will consider the flow as steady and the water as incompressible The flow rate is Q = VA = 14 m>s23b11.5 m24 = 6 b. For the rectangular channel, the specific energy can be determined by E =

Q2

2 2

2gb y

+ y=

16b2 2

+ y=

219.81 m>s2 21b2 2y2

1.8349 + y y2

(1)

For the depth of y = 1.5 m, Eq. (1) gives

1.8349 + 1.5 m = 2.3155 11.5 m2 2

E =

Substitute this result into Eq. (1)

2.3155 =

1.8349 + y y2

y3 - 2.3155y2 + 1.8349 = 0 Solving by trial and error, y = 1.5865 m y2 = 1.5 m

(The given depth)

y = -0.7710

(Not realistic)

In this case, the critical depth can be determined. yc = a

Q2 2

gb

b

1>3

= c

16b2 2

2

2

19.81 m>s 2b

For y = 1.5 m 6 yc, the flow is supercritical.

d

1>3

= 1.5425 m

For y = 1.5865 m 7 yc, the flow is subcritical. Using the continuity equation, 0 r dV + r V # dA = 0 0 t Lcv Lcs 0 - V1A1 + V2A2 = 0 - 14 m>s23b11.5 m24 + V2 3b11.5865 m24 = 0 V2 = 3.7819 m>s = 3.78 m>s

Ans.

Ans: V2 = 3.78 m>s 1041

M12_HIBB9290_01_SE_C12_ANS.indd 1041

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*12–8. A rectangular channel having a width of 8 ft is transporting water at a rate of 36(103) ft3 >min. If the depth is 8 ft, determine the specific energy at this depth, and also the specific energy at the critical depth.

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the specific energy can be determined. Here, Q = c 361103 2

ft3 1 min da b = 600 ft3 >s. min 60 s E = = =

For y = 8 ft, E =

Q2

2gb2y2

+ y

1600 ft3 >s2 2

+ y

2132.2 ft>s2 218 ft2 2y2 87.3447 + y y2

(1)

87.3447 + 8 ft = 9.3648 ft = 9.36 ft 18 ft2 2

Ans.

The critical depth for a rectangular channel can be determined by yc = a

Q2 2

gb

b

1>3

= c

1600 ft3 >s2 2 2

132.2 ft>s 218 ft2

2

d

1>3

= 5.5901 ft

Thus, the specific energy when the flow depth is at yc, which happens to be its minimum, can be determined using Eq. (1) by substituting y = yc = 5.5901 ft. Emin =

87.3447 + 5.5901 ft = 8.3852 ft = 8.39 ft 15.5901 ft2 2

Ans.

Alternatively, Emin can be determined by Emin =

3 3 y = 15.5901 ft2 = 8.3852 ft = 8.39 ft 2 c 2

Ans: E = 9.36 ft, E min = 8.39 ft 1042

M12_HIBB9290_01_SE_C12_ANS.indd 1042

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12–9. The channel transports water at 8 m3 >s. If the depth 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 rectangular, 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 Emin =

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, subcritical Emin = 1.52 m At y = 1.5 m, E = 1.73 m 1043

M12_HIBB9290_01_SE_C12_ANS.indd 1043

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12–10. The rectangular channel transports water at 12(103) ft3 >min. Determine the critical depth yc, plot the specific energy diagram for this flow, and calculate E for y = 0.75yc and y = 1.25yc.

y

6 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determine. Here, Q = c 121103 2

ft3 1 min da b = 200 ft3 >s. min 60 s

yc = a

Q2

b gb2

1>3

1200 ft3 >s2 2

= c

2

132.2 ft>s 216 ft2

Thus, the minimum specific energy is Emin =

2

d

1>3

= 3.2556 ft. = 3.26 ft

3 3 yc = 13.2556 ft2 = 4.8834 ft. 2 2

Ans.

y(m) y5E

The specific energy in terms of depth yis E = =

Q2

+ y 2gb2y2 1200 ft3 >s2 2

2132.2 ft>s2 216 ft2 2y2

+ y

4.07 yC 5 3.26 2.44

17.2533 = + y y2 For y =

For y =

3 3 y = 13.2556 ft2 = 2.4417 ft, 4 c 4 17.2533 E = + 2.4417 ft = 5.3356 ft. 12.4417 ft2 2

Emin 5 4.88

5.34 5.11

E(m)

(a)

5 5 y = 13.2556 ft2 = 4.0695 ft, 4 c 4 E =

17.2533 + 4.0695 ft = 5.111 ft. 14.0695 ft2 2

The plot of yvs E is shown in Fig. a

Ans: yc = 3.26 ft For y = 0.75yc, E = 5.34 ft For y = 1.25yc, E = 5.11 ft 1044

M12_HIBB9290_01_SE_C12_ANS.indd 1044

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12–11. The rectangular channel transports water at 48 (103) ft3 >min. Determine the critical depth and the minimum specific energy. If the specific energy is 12 ft, what are the two possible flow depths? y 10 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here, Q = c 481103 2

ft3 1 min da b = 800 ft3 >s. min 60 s

yC = a

Q2

2

gb

b

1>3

= c

(800 ft3 >s)2 2

(32.2 ft>s )(10 ft)

2

d

1>3

= 5.8359 ft = 5.84 ft

Ans.

Then, the minimum specific energy is 3 3 Emin = yc = 15.8359 ft2 = 8.7539 ft = 8.75 ft Ans. 2 2 The specific energy as a function of depth ycan be determined. For E = 12 ft by E = 12 ft = 12 =

Q2

+ y 2gb2y2 1800 ft3 >s2 2

2132.2 ft>s2 2110 ft2 2y2

+ y

99.3789 + y y2

y3 - 12y2 + 99.3789 = 0 Solving by trial and error, y = 11.2090 ft = 11.2 ft 7 yc

(subcritical)

Ans.

y = 3.3992 ft = 3.40 ft 6 yc

(supercritical)

Ans.

y = - 2.6082 ft

(not realistic)

Ans: yc = 5.84 ft, Emin = 8.75 ft y = 3.40 ft or y = 11.2 ft 1045

M12_HIBB9290_01_SE_C12_ANS.indd 1045

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*12–12. The rectangular channel transports water at 8 m3 >s. Plot the specific energy diagram for the flow and indicate the depth yfor E = 3 m.

y

1.5 m

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

2.82

( 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

y c = 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) 1046

M12_HIBB9290_01_SE_C12_ANS.indd 1046

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12–13. The rectangular channel passes through a transition that causes its width to narrow from 8 ft to 6 ft. If the flow is 18(103) ft3 >min and yA = 6 ft, determine the depth at B.

8 ft A 6 ft B

yA

yB

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined using Eq. 12–7. Here, Q = c 181103 2

ft3 1 min da b = 300 ft3 >s. At section A, min 60 s

1yC 2 A = a

Q2

gbA2

and at section B,

1yC 2 B = a

Q2 gbB2

b b

1>3

1>3

= c = c

1300 ft3 >s2 2

2

1300 ft3 >s2 2

2

2

132.2 ft>s 218 ft2 2

132.2 ft>s 216 ft2

d d

1>3

1>3

= 3.5216 ft

= 4.2661 ft

The specific energy of the flow can be determined at section A, E =

Q2 2gbA2yA2

+ yA =

1300 ft3 >s2 2

2132.2 ft>s2 218 ft2 2 16 ft2 2

+ 6 ft = 6.6066 ft

E remains constant throughout the channel. Then, at section B, E = 6.6066 ft =

Q2 2gbB2yB2

+ yB

1300 ft3 >s2 2

2132.2 ft>s2 216 ft2 2yB2

+ yB

yB3 - 6.6066yB2 + 38.8199 = 0 Solving by trial and error; yb = 5.13345 ft 7 1yc 2 B

(subcritical)

yB = -2.11031 ft

(not realistic)

yB = 3.58342 ft 6 1yc 2 B

(supercritical)

Since yA = 6 ft 7 1yc 2 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 must also be subcritical. This requires yB 7 1yc 2 B. Thus

Ans.

yB = 5.13 ft

Ans: yB = 5.13 ft 1047

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12–14. The rectangular channel has a transition that causes its width to narrow from 8 ft to 6 ft. If the flow is 18(103) ft3 >min and yB = 3.5 ft, determine the depth at A.

8 ft A 6 ft B

yA

yB

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here, ft3 1 min Q = c 181103 2 da b = 300 ft3 >s. At section A, min 60 s 1yc 2 A = a

Q2

gbA2

And at section B,

1yc 2 B = a

b

Q2

b 2

1>3

1>3

gbB

= c

= c

1300 ft3 >s2 2 2

132.2 ft>s 218 ft2 1300 ft3 >s2 2

132.2 ft>s2 216 ft2

2

d

d 2

1>3

1>3

= 3.5216 ft

= 4.2661 ft

The specific energy of the flow can be determined at section B, E =

Q2 2gbB2yB2

+ yB =

1300 ft3 >s2 2

+ 3.5 ft = 6.6690 ft

2132.2 ft>s2 216 ft2 2 13.5 ft2 2

E remain constant throughout the channel. Then at section A, E = 6.6690 ft =

Q2 2gbA2yA2

+ yA

1300 ft3 >s2 2

2132.2 ft>s2 218 ft2 2yA2

+ yA

yA3 - 6.6690yA2 + 21.8362 = 0 Solving by trial and error, yA = 6.0778 ft 7 1yc 2 A

(subcritical)

yA = - 1.6228 ft

(not realistic)

yA = 2.2139 ft 6 1yc 2 A

(supercritical)

Since yB = 3.5 ft 6 1yc 2 B, the flow at section B is supercritical. Since the type of flow must remain the same throughout the channel, the flow at section A must also be supercritical. This requires yA 6 1yc 2 A. Thus, Ans.

yA = 2.21 ft

Ans: yA = 2.21 ft 1048

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12–15. The volumetric flow in the channel is measured using the venturi segment. If yA = 8 ft and yB = 7.6 ft, determine the flow through the channel.

8 ft

3 ft A B

yA yB

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the specific energy can be determined. At sections A and B, EA = EB =

Q2 2gbA2yA2 Q

+ yA =

2

2gbB2yB2

+ yB =

Q2 2

2

2132.2 ft>s 218 ft2 18 ft2

2

+ 8 ft =

Q2

9132.2 ft>s2 213 ft2 2 17.6 ft2 2

Q2 + 8 263782.4

+ 7.6 ft =

Q2 + 7.6 33477.696

Since the head losses are neglected, specific energy remains constant throughout the channel. Thus, EA = EB 2

Q2 Q + 8 = + 7.6 263782.4 33477.696 Q = 123.85 ft3 >s = 124 ft3 >s

Ans.

1049

M12_HIBB9290_01_SE_C12_ANS.indd 1049

Ans: Q = 124 ft3 >s

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*12–16. The rectangular channel tapers gradually from 10 ft wide to 6 ft wide, with a transition as shown. If water flows with a velocity of 3 ft>s on the 10-ft-wide side, determine the depth y2 on the 6-ft-wide side. Also, what is the horizontal force directed downstream that the water exerts on each of the flared wing walls?

10 ft

6 ft

y2 4 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Here the flow rate is

F1 2F

3

Q = V1A1 = 13 ft>s23110 ft214 ft24 = 120 ft >s

Since the channel is rectangular, the critical depth can be determined. For 10-ft-wide channel, 1>3 1120 ft3 >s2 2 Q2 1>3 1yc 2 1 = a 2 b = c d = 1.648 ft gb1 132.2 ft>s2 2110 ft2 2

w1

F2 w2

(a)

Since y1 = 4 ft 7 1yc 2 1, the flow is subcritical. For 6-ft-wide channel 1yc 2 2 = a

Q2

gb22

b

1>3

= c

1120 ft3 >s2 2 2

132.2 ft>s 216 ft2

2

d

1>3

= 2.316 ft

Since the channel base remains horizontal, the specific energy for 10-ft-wide and 6-ft-wide sections must be the same. For 10-ft-wide section, E =

Q2 2gb12y12

+ y1 =

For 6-ft-wide section, E =

Q2 2gb22y22

+ y2;

1120 ft3 >s2 2

2132.2 ft>s2 2110 ft2 2 14 ft2 2

4.1398 ft =

1120 ft3 >s2 2

+ 4 ft = 4.1398 ft

2132.2 ft>s2 216 ft2 2 y22

+ y2

y23 - 4.1398y22 + 6.2112 = 0 Solving numerically, y2 = 3.6815 ft 7 1yc 2 2

(subcritical)

y2 = - 1.0898 ft

(not possible)

y2 = 1.5481 ft 6 1yc 2 2

(supercritical)

1050

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12–16. Continued

The subcritical flow must be maintained downstream. Thus Ans.

y2 = 3.6815 ft = 3.68 ft The continuity equation can be simplified as 0 r dV + r V # dA = 0 0 t Lcv Lcs 0 - V1A1 + V2A2 = 0 -120 ft3 >s + V2 316ft213.6815 ft24 = 0

V2 = 5.4326 ft>s

The fixed control volume considered contains the water between the upstream and downstream cross sections of the channel, Fig. a. The intensities of the distributed loads acting on these cross sections are W1 = gwy1b1 = 162.4lb>ft3 214 ft2110 ft2 = 2.4961103 2 lb>ft

W2 = gwy2b2 = 162.4lb>ft3 213.6815 ft216 ft2 = 1.378 1103 2 lb>ft

Thus, the resultants of the distributed loads are

1 1 w y = 32.496 1103 2 lb>ft414 ft2 = 4.992 1103 2 lb 2 1 1 2 1 1 F2 = w2 y2 = 31.378 1103 2lb>ft413.6815 ft2 = 2.5372 1103 2 lb 2 2 F1 =

The FBD of the control volume is shown in Fig. a. Since the flow is steady and incompressible, the linear momentum equation can be simplified as ΣF =

0 Vr dV + Vr V # dA 0 t Lcv Lcs

+ ΣFx = 0 + 1V1 2 xrw 1 -Q2 + 1V2 2 xrwQ S + S ΣFx = rwQ31V2 2 x - 1V1 2 x4

4.9921103 2 lb - 2.53721103 2 lb - 2F = a

62.4 lb>ft3 32.2 ft>s2

b 1120 ft3 >s215.4326 ft>s - 3 ft>s2

F = 944.57 lb = 945 lb

Ans.

Ans: y2 = 3.68 ft F = 945 lb 1051

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12–17. Water flows at 25 ft3 >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 transition. 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 =

( 25ft3 >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 =

( 25ft3 >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 7 ( 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, Emin = E =

3 y 2 c

3 y 2 c yc = 2.0200 ft

3.0300 ft =

1052

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12–17. 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 ft3 >s = ( 8.0649 ft>s ) (bc)(2.0200 ft)

Ans.

bc = 1.53 ft

Ans: y2 = 2.96 ft, b2 = 1.53 ft 1053

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12–18. The rectangular channel has a width of 15 ft and transports water at 36(103) ft3 >min. If the depth at A is 8 ft, determine the minimum height h of the channel bed at B in order to produce critical flow at B.

A B 8 ft

yB h

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here, ft3 1 min Q = c 361103 2 da b = 600 ft3 >s. It is required that min 60 s yB = yC = a

Q2

b gb2

1>3

= c

Thus, the minimum specific energy EB = Emin =

1600 ft3 >s2 2

132.2 ft>s2 2115 ft2

d 2

1>3

= 3.6764 ft

3 3 y = 13.6764 ft2 = 5.5146 ft 2 c 2

The specific energy at section A can be determined by EA =

Q2 2gb2yA2

+ yA =

1600 ft3 >s2 2

2132.2 ft>s2 2115 ft2 2 18 ft2 2

+ 8 ft = 8.3882 ft

Thus, the minimum height of the channel’s bed is hmin = EA - Emin

= 8.3882 ft - 5.5146 ft = 2.8736 ft Ans.

= 2.87 ft If h 7 hmin , the flow is choked and will remain critical after being lifted.

Ans: hmin = 2.87 ft 1054

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12–19. The rectangular channel has a width of 8 ft and transports water at 18(103) ft3 >min. If the elevation of the bed drops by 0.75 ft, determine the depth y2 after the depression.

6 ft

y2

0.75 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here, ft3 1 min Q = c 181103 2 da b = 300 ft3 >s. min 60 s yc = a

Q2

2

gb

b

1>3

= c

1300 ft3 >s2 2 2

132.2 ft>s 218 ft2

2

d

1>3

= 3.5216 ft

The specific energy at sections (1) and (2) can be determined E1 = E2 =

Q2 2gb2y12 Q2 2

2gb

y22

+ y1 = + y2 =

1300 ft3 >s2 2

2132.2 ft>s2 218 ft2 2 16 ft2 2 1300 ft3 >s2 2

2132.2 ft>s2 218 ft2 2y22

+ 6 ft = 6.6066 ft

+ y2 =

21.8362 + y2 y22

Since the elevation of the bed is decreasing, h = - 0.75 ft. Here E1 = E2 + h 6.6066 ft =

21.8362 + y2 + 1 - 0.75 ft2 y22

y23 - 7.3566y22 + 21.8362 = 0 Solving by trial and error, y2 = 6.8976 ft 7 yc

(subcritical)

y2 = 2.0235 ft 6 yc

(supercritical)

y2 = -1.5645 ft

(not realistic).

Since y1 = 6 ft 7 yc, the flow at section (1) is subcritical and this type of flow must remain the same throughout the chain. The flow at section (2) also is subcritical. This requires y2 7 yc. Thus Ans.

y2 = 6.90 ft

Ans: y2 = 6.90 ft 1055

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*12–20. 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: y2 = 1.78 m V2 = 5.07 m>s 1056

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12–21. Water flows at 10.8(103) ft3 >min through a rectangular channel having a width of 6 ft. If yA = 5 ft, show that yB = 4.25 ft is possible. What is the necessary increase in elevation h of the channel bed to create this depth?

A B yA

yB

h

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here, ft3 1 min Q = c 10.81103 2 da b = 180 ft3 >s. min 60 s yc = a

Q2

b gb2

1>3

= c

1180 ft3 >s2 2

132.2 ft>s2 216 ft2

d 2

1>3

= 3.0348 ft

Since yA = 5 ft 7 yc, the flow at section A is subcritical. This requires yB 6 yA and also yB 7 yC to maintain subcritical flow. In this case, yB = 4.25 ft indeed satisfies these two requirements. The specific energies at sections A and B can be determined by EA = EB =

Q2 2gb2yA2 Q2 2

2gb

yB2

+ yA = + yB =

1180 ft3 >s2 2

2132.2 ft>s2 216 ft2 2 15 ft2 2

+ 5 ft = 5.5590 ft

1180 ft3 >s2 2

2132.2 ft>s2 216 ft2 2 14.25 ft2 2

+ 4.25 ft = 5.0237 ft

From the requirement of the rising flow,

EA = EB + h 5.5590 ft = 5.0237 ft + h Ans.

h = 0.5353 ft = 0.535 ft

Ans: h = 0.535 ft 1057

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12–22. Water flows at 10.8(103) ft3 >min through a rectangular channel having a width of 6 ft. If yA = 2 ft, show that yB = 2.75 ft is possible. What is the necessary increase in elevation h of the channel bed to create this depth?

B A

yB

yA

h

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here 1 min Q = 310.81103 2 ft3 >min4 a b = 180 ft3 >s. 60 s y= a

Q2

b gb2

1>3

= c

1180 ft3 >s2 2 2

132.2 ft>s 216 ft2

2

d

1>3

= 3.0348 ft.

Since yA = 2 ft 7 yc, the flow at section A is supercritical. This requires yB 7 yA and also yB 6 yC to maintain supercritical flow. In this case, yB = 2.25 ft indeed satisfies these two requirements. The specific energies at section A and B can be determined by EA = EB =

Q2 2

2gb

yA2

Q2 2gb2yB2

+ yA = + yB =

1180 ft3 >s2 2

2132.2 ft>s2 216 ft2 2 12 ft2 2

+ 2 ft = 5.4938 ft

1180 ft3 >s2 2

2132.2 ft>s2 216 ft2 2 12.75 ft2 2

+ 2.75 ft = 4.5980 ft

From the requirement of the rising flow,

EA = EB + h 5.4938 ft = 4.5980 ft + h Ans.

h = 0.8958 ft = 0.896 ft

Ans: h = 0.896 ft 1058

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12–23. Water flows within the 4-m-wide rectangular channel at 20 m3 >s. Determine the depth 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

1059

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*12–24. Water flows within the 4-m-wide rectangular channel at 20 m3 >s. Determine the depth 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

1060

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12–25. The rectangular channel is 5 ft wide, and the depth of the water is 6 ft as it flows with a velocity of 4 ft>s. Determine if the flow is initially subcritical or supercritical. What is the required height h of the properly designed bump so that the flow can change to another type of flow after it passes over the bump? What is the new depth y2 for this flow?

4 ft!s y15 6 ft

y2 h

SOLUTION We will consider the flow as steady and the water as incompressible. Here, the flow rate is Q = VA = 14 ft>s2315 ft216 ft24 = 120 ft3 >s.The critical depth of the flow can be determined by yc = a

Q2 2

gb

b

1>3

= c

1120 ft3 >s2 2 2

132.2 ft>s 215 ft2

2

d

1>3

= 2.6153 ft

Since y1 7 yc, the initial flow is subcritical. The minimum specific energy is Emin =

3 3 y = 12.6153 ft2 = 3.9230 ft 2 c 2

The specific energy can be determined. In general, E = E =

Q2 2gb2y2

+ y=

8.9441 + y y2

1120 ft3 >s2 2

2132.2 ft>s2 215 ft2 2y2

+ y

At section (1) where y = y1 = 6 ft, E1 =

8.9441 + 6 ft = 6.2484 ft 16 ft2 2

Then, the required height of the bump is

Ans.

h = E1 - Emin = 6.2484 ft - 3.9230 ft = 2.3255 ft = 2.33 ft At section (2), E2 = E1 = 6.2484 ft. Then E2 =

8.9441 + y2 y22

6.2484 ft =

8.9441 + y2 y22

y23 - 6.2484y22 + 8.9441 = 0 Solving by trial and error, y2 = 6 ft 7 yc

(subcritical)

y2 = 1.3515 ft 6 yc

(supercritical)

y2 = - 1.1030 ft

(not realistic)

Here, the bump is properly designed. Hence the flow will change to supercritical after the peak of the bump. Thus Ans.

y2 = 1.35 ft

Ans: h = 2.33 ft y2 = 1.35 ft 1061

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12–26. The rectangular channel is 1.5 m wide, and the water flows with a velocity of 3.5 m>s. If y1 = 0.9 m, show that this flow is supercritical. Determine the required height h of the bump so that the flow can change to subcritical flow after it passes over the properly designed bump. What is the depth y2 for this flow?

3.5 m!s y2

y1 h

SOLUTION We will consider the flow as steady and the water as incompressible. Here the flow rate is Q = VA = 13.5 m>s2311.5 m210.9 m24 = 4.725 m3 >s. The critical depth can be determined yc = a

Q2

2

gb

b

1>3

= c

14.725 m3 >s2 2 2

19.81 m>s 211.5 m2

2

d

1>3

= 1.0038 m

Since y1 6 yc, the flow at section (1) is supercritical. The minimum specific energy is Emin =

3 3 y = 11.0038 m2 = 1.5057 m 2 c 2

The specific energy can be determined. In general, E =

Q2 2 2

2gb y

+ y=

14.725 m3 >s2 2

219.81 m>s2 211.5 m2 2y2

+ y=

0.5057 + y y2

At section (1) where y = y1 = 0.9 m, E1 =

0.5057 + 0.9 m = 1.5244 m 10.9 m2 2

Then the required height of the bump is

h = E1 - Emin = 1.5244 m - 1.5057 m = 0.01865 m = 18.7 mm

Ans.

At section (2), E2 = E1 = 1.5244 m. Then E2 =

0.5057 + y2 y22

1.5244 m =

0.5057 + y2 y22

y23 - 1.5244y22 + 0.5057 = 0 Solving by trial and error, y2 = 0.9 m 6 yc

(supercritical)

y2 = 1.1242 m 7 yc

(subcritical)

y2 = - 0.4998

(not realistic)

Here the bump is properly designed. Hence the flow will change to subcritical after the peak of the bump. Thus, Ans.

y2 = 1.12 m

Ans: h = 18.7 mm y2 = 1.12 m 1062

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12–27. 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 subcritical, and determine the required depth of the water, y′, at the top of the bump so that the downstream flow can be transformed to supercritical flow. What is the downstream depth?

4 ft

y9

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)

y2 = 1.18 ft 6 2.0509 ft

(rapid flow)

y2 = -0.912 ft

(unrealistic)

Ans.

The water surface profiles are

4 ft

4 ft

2.05 ft

4 ft

Tranquil

2.05 ft Rapid

Ans: y′ = 2.05 ft y2 = 1.18 ft 1063

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*12–28. A flow of 60 m3 >s passes under the sluice gate. If y1 = 6 m, determine the depth y2 and the type of flow downstream. Also, what is the maximum volumetric flow that can pass through the gate?

4 ft

y1

y2

SOLUTION We will consider y1 as constant so that the flow is steady and the water is incompressible. The critical depth of the flow is yc =

2 2 y = 16 m2 = 4 m 3 1 3

Here, y1 and y2 can be related using y1 = 6m =

Q2 2gb2y22

+ y2

160 m3 >s2 2

219.81 m>s2 214 m2 2y22

+ y2

y23 - 6y22 + 11.4679 = 0 Solving by trial and error, y2 = 5.6394 m 7 yc

(subcritical)

y2 = 1.6177 m 6 yc

(supercritical)

y2 = - 1.2571 m

(not realistic)

So the depths and type of flow are Supercritical: y2 = 1.62 m Subcritical:

y2 = 5.64 m

The maximum volumetric flow rate for this sluice gate is Q max = =

8 gb2y13 A 27

8 19.81 m>s2 214 m2 2 16 m2 3 A 27

= 100.23 m3 >s = 100 m3 >s

Ans.

Ans: Q max = 100 m3 >s Supercritical: y2 = 1.62 m Subcritical: y2 = 5.64 m 1064

M12_HIBB9290_01_SE_C12_ANS.indd 1064

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12–29. Determine the volumetric flow of water through the gate if y1 = 6 m, y2 = 2.5 m. What type of flow occurs?

4 ft

y1

y2

SOLUTION We will consider y1 as constant so that the flow is steady and the water is incompressible. Using the relation between y1 and y2, y1 = 6m =

Q2 2gb2y22

+ y2 Q2

219.81 m>s2 214 m2 2 12.5 m2 2

+ 2.5 m

Q = 82.87 m3 >s = 82.9 m3 >s

Ans.

The critical depth of the flow downstream from the gate is yc =

2 2 y = 16 m2 = 4 m 3 1 3

Since y2 = 2.5 m 6 yc; the flow is supercritical.

1065

M12_HIBB9290_01_SE_C12_ANS.indd 1065

Ans: Q = 82.9 m3 >s, supercritical

16/03/17 4:48 PM

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12–30. If the depths y0 = 6 ft, where the water is essentially at rest, and y2 = 2 ft, determine the volumetric flow through the sluice gate and the depth y1 just before the gate.

4 ft

y0

y2

SOLUTION We will consider y1 as constant so that the flow is steady and the water is incompressible. Using the relation between y0 and y2, y0 = 6 ft =

Q2 2gb2y22

+ y2 Q2

2132.2 ft>s2 214 ft2 2 12 ft2 2

+ 2 ft

Q = 128.40 ft3 >s = 128 ft3 >s

Ans.

Using this result and the relation between y0 and y, y0 = 6 ft =

Q2

+ y 2gb2y2 1128.40 ft3 >s2 2

2132.2 ft>s2 214 ft2 2y2

+ y

y3 - 6y2 + 16 = 0 Solving by trial and error, y = 2 ft y = 5.4641 ft y = - 1.4641 ft

(not realistic)

The first root is the flow depth downstream from the gate, and the second root is the flow depth just before the gate. Thus, Ans.

y1 = 5.46 ft

Ans: Q = 128 ft3 >s y1 = 5.46 ft 1066

M12_HIBB9290_01_SE_C12_ANS.indd 1066

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12–31. If the flow is 7200 ft3 >min and y0 = 8 ft, where the water is essentially at rest, determine the depth y2 and the depth y1 just before the gate.

4 ft

y0

y2

SOLUTION We will consider y1 as constant so that the flow as steady and the water as ft3 1 min incompressible. Here, Q = a7200 ba b = 120 ft; In general, min 60 s y0 =

8 ft =

Q2

2gb2y2

+ y

1120 ft3 >s2 2

2132.2 ft>s2 214 ft2 2y2

+ y

y3 - 8y2 + 13.9752 = 0 Solving by trial and error, y = 7.7684 ft y = 1.4620 ft y = -1.2305 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, y2 = 1.46 ft

Ans.

y1 = 7.77 ft

Ans.

Ans: y2 = 1.46 ft y1 = 7.77 ft 1067

M12_HIBB9290_01_SE_C12_ANS.indd 1067

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*12–32. If the depth y0 = 12 ft, where the water is essentially at rest, determine the volumetric flow through the channel as a function of depth y2. Plot the results of flow on the vertical axis, versus depth y2 for 0 … y2 … 10 ft . Use increments of ∆y2 = 2 ft .

4 ft

y0

y2

SOLUTION We will consider y1 to be constant so that the flow is steady and the water is incompressible. Using the relation between y0 and y2, Q2

y0 =

2gb2y22

+ y2 Q (ft3ys)

Q2

12 ft =

2132.2 ft>s2 214 ft2 2y22

+ y2

600 500

Q = 21030.4y22 112 - y2 2

400

The maximum flow occurs when y2 =

300

2 2 y = 112 ft2 = 8 ft 3 0 3

200

The values of Q for 0 … y2 … 10 ft with increment of Ay2 = 2 ft are tabulated below and the plot of Q vs y2 is shown in Fig. a. y2 1ft2

Q 1ft3>s2

0

2

4

6

0

203

363

472

8

10

514 (Max.) 454

100 0

2

4

6

8

10

y2 (ft)

(a)

1068

M12_HIBB9290_01_SE_C12_ANS.indd 1068

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12–33. Determine the hydraulic radius for each channel cross section.

30° 30° h

l

b (a)

(b)

30°

SOLUTION l

a.) For the rectangular section, A = bh

p = 2h + b Rh =

30° l b

bh 2h + b

Ans.

(c)

b.) For the triangular section, A =

1 1 1 23 23 2 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 =

bh 2h + b 23 l 8 23l1l + 2b2 412l + b2

1069

M12_HIBB9290_01_SE_C12_ANS.indd 1069

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12–34. The channel has a triangular cross section. Determine the critical depth y = yc in terms of u and the flow Q. y u

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 )

Ans: 2Q2 tan2 u 1>2 a b g 1070

M12_HIBB9290_01_SE_C12_ANS.indd 1070

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12–35. The drainage canal has a downward slope of 0.003. If its bottom and sides have weed growth, determine the volumetric flow of water when the depth is 8 ft. Take n = 0.030.

12 ft

8 ft

SOLUTION We will consider the flow as steady and the water as incompressible. A = 112 ft218 ft2 = 96 ft2

P = 218 ft2 + 12 ft = 28 ft For FPS units, k = 1.486 in the Manning equation Q =

kA5>3S 01>2 n P 2>3

=

1.486196 ft2 2 5>3 10.0031>2 2 10.030 s>ft1>3 2128 ft2 2>3

.

= 592.20 ft3 >s = 592 ft3 >s

1071

M12_HIBB9290_01_SE_C12_ANS.indd 1071

Ans.

Ans: Q = 592 ft3 >s

16/03/17 4:49 PM

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*12–36. The rectangular channel has a finished concrete surface and has a slope of 0.0015. If it is to discharge 8.75 m3 >s, determine the depth for uniform flow. Is the flow subcritical or supercritical? Take n = 0.012.

y

2m

SOLUTION We will consider the flow as steady and water as incompressible. The cross sectional area and the wetted perimeter are A = (2 m)(y) = (2y) m2

P = 2y + 2 m

For SI units, k = 1. Then, applying the Manning equation using the given flow rate, Q =

kA5>3S 01>2 n P 2>3

; 8.75 m3 >s =

112y2 5>3 10.00151>2 2

10.012 s>m1>3 212y + 22 2>3

1.2672y5>2 - 2y - 2 = 0

Solving numerically, Ans.

y = 1.81603 m = 1.82 m The mean velocity of the flow is 8.75 m3 >s Q V = = = 2.4091 m>s A 211.81603 m2

Thus, the Froude number is Fr =

V 2gy

=

2.4091 m>s 219.81 m>s2 211.81603 m2

= 0.5708

Since Fr 6 1, the flow is subcritical

Ans: y = 1.82 m, subcritical 1072

M12_HIBB9290_01_SE_C12_ANS.indd 1072

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12–37. The channel is made of wood and has a downward slope of 0.0015. Determine the volumetric flow of water if y = 2 ft. Take n = 0.012. y

h 3 ft

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From Table 12–2, n = 0.012 s>m 3 for a wood surface. Here, A = (3 ft)(2 ft) = 6 ft2 P = 2(2 ft) + 3 ft = 7 ft For FPS units, 5

Q =

2

nP 3

5

1

1.486A3S 02

=

1

1.486 ( 6 ft2 ) 3 ( 0.0015 2 ) 1

2

( 0.012 s>m 3 ) (7 ft)3

= 26.0 ft3 >s

Ans.

Ans: Q = 26.0 ft3 >s 1073

M12_HIBB9290_01_SE_C12_ANS.indd 1073

17/03/17 12:00 PM

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12–38. 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? Take n = 0.012.

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 Table 12–2, n = 0.012 s>m 3 for a wood surface. For FPS units, 5

Q =

2

nP 3

5

1

1.486A3S 02

=

1

1.486 ( 4.5 ft2 ) 3 ( 0.0015 2 ) 1

2

( 0.012 s>m 3 ) (6 ft)3

= 17.8 ft3 >s

1074

M12_HIBB9290_01_SE_C12_ANS.indd 1074

Ans.

Ans: y = 1.5 ft Q = 17.8 ft3 >s

16/03/17 4:49 PM

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12–39. 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. Take n = 0.014. R

SOLUTION 1

From Table 12–2, n = 0.014 s>ft 3 for an unfinished concrete surface. Here, p A = R2 P = pR 2 For FPS units, 5

Q =

1

1.486A3S 02 2

nP 3 5

60 ft3 >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: R = 2.74 ft 1075

M12_HIBB9290_01_SE_C12_ANS.indd 1075

16/03/17 4:49 PM

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*12–40. A 1.5-m-diameter unfinished-concrete drain pipe is required to transport 4.25 m3 >s of water at the depth shown. Determine the required downward slope of the pipe. Take n = 0.014. 1.125 m

0.75 m

SOLUTION We will consider the flow as steady and the water as incompressible. From the geometry shown in Fig. a, u = cos -1 a

Then

0.375 m b = 60° 0.75 m

a = 360° - 2u = 360° - 2160°2 = 1240°2 a

The cross-sectional area and the wetted perimeter are

p rad 4 b = p rad 180° 3

2

10.75 m2 4 R2 4 1a - sin a2 = c p - sin a pb d = 1.4217 m2 2 2 3 3 4 P = aR = a pb 10.75 m2 = p m 3

0.375 m

A =

0.75 m

For SI units, k = 1. Then Manning equation becomes Q = 4.25 m3 >s =

0.75 m

u

u

a

(a)

A5>3S 01>2 nP 2>3 11.4217 m2 2 5>3S 01>2

10.014 s>m1>3 21p m2 2>3

Ans.

S0 = 0.0050417 = 0.00504

Ans: S0 = 0.00504 1076

M12_HIBB9290_01_SE_C12_ANS.indd 1076

16/03/17 4:49 PM

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12–41. Water flows through the 1-m-diameter culvert at a depth of 0.25 m. If the culvert has an unfinished-concrete surface and has a slope of 0.002, determine the volumetric flow. Take n = 0.014.

0.5 m 0.25 m

SOLUTION We will consider the flow as steady and the water as incompressible. From the geometry shown in Fig. a, u = cos -1 a

Then

0.25 m b = 60° 0.5 m

p rad 2 b = p rad. 180° 3 The cross-sectional area and the wetted perimeter are a = 2u = 2160°2 = 1120°2 a

0.5 m

u u

a

0.25 m 0.25 m

10.5 m2 2 2 R2 2 A = 1a - sin a2 = c p - sin a pb d = 1.1535 m2 2 2 3 3 2 P = aR = a pb 10.5 m2 = 0.3333 p m 3

(a)

For SI units, k = 1. Then the Manning equation gives Q = =

kA5>3S 01>2 nP 2>3 110.1535 m2 2 5>3 10.0021>2 2 0.01410.3333p2 2>3

= 0.1364 m3 >s = 0.136 m3 >s

Ans.

1077

M12_HIBB9290_01_SE_C12_ANS.indd 1077

Ans: Q = 0.136 m3>s

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12–42. A 3-ft-diameter unfinished-concrete drain pipe is required to transport 80 ft3 >s of water at a depth of 2.25 ft. Determine the required downward slope of the pipe. Take n = 0.014.

SOLUTION Water is considered to be incompressible. The flow is steady. From the geometry shown in Fig. a,

Then,

u = cos -1 a

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 R2 4 (a - sin a) = c p rad - sin p d = 5.6867 ft2 2 2 3 3

P = aR =

4 p(1.5 ft) = 2p ft 3 1

From Table 12–2, n = 0.014 s>ft3 for an unfinished concrete surface. For FPS units, 5

Q =

1

1.486A3S 02 2

nP 3 5

3

80 ft >s =

1

1.486 ( 5.6867 ft2 ) 3S 02 1

2

( 0.014 s>ft3 ) (2p ft) 3 Ans.

S0 = 0.0201

0.75 ft

1.5 ft

1.5 ft (a )

Ans: S0 = 0.0201 1078

M12_HIBB9290_01_SE_C12_ANS.indd 1078

16/03/17 4:49 PM

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12–43. The channel is made of unfinished concrete and has the cross section shown. If the downward slope is 0.0006, determine the flow of water through the channel when y = 6 ft. Take n = 0.014.

6 ft

y

2 ft

6 ft

8 ft

6 ft

SOLUTION We will consider the flow as steady and the water as incompressible. A = 120 ft214 ft2 + 18 ft212 ft2 = 96 ft2 P = 214 ft + 6 ft + 2 ft2 + 8 ft = 32 ft For FPS units, k = 1.486. Then the Manning equation becomes. Q = =

1.486 A5>3S 01>2 nP 2>3 1.486196 ft2 2 5>3 10.00061>2 2 10.014 s>ft1>3 2132 ft2 2>3

= 519.18 ft3 >s = 519 ft3 >s

Ans.

1079

M12_HIBB9290_01_SE_C12_ANS.indd 1079

Ans: Q = 519 ft3 >s

16/03/17 4:49 PM

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*12–44. The channel is made of finished concrete and has the cross section shown. If the downward slope is 0.0006, determine the flow of water through the channel when y = 5 ft. Take n = 0.012.

6 ft

y

2 ft

6 ft

8 ft

6 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Here A = 120 ft213 ft2 + 18 ft212 ft2 = 76 ft2

P = 213 ft + 6 ft + 2 ft2 + 8 ft = 30 ft

For FPS units, k = 1.486. Then the Manning equation becomes 1

Q = =

1.486A5>3S 02 nP 2>3 1.486176 ft2 2 5>3 10.00061>2 2 10.012 s>ft1>3 2130 ft2 2>3

= 428.41 ft3 >s = 428 ft3 >s

Ans.

1080

M12_HIBB9290_01_SE_C12_ANS.indd 1080

Ans: Q = 428 ft3 >s

16/03/17 4:49 PM

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12–45. Water flows uniformly through 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. Take n = 0.012.

20°

20° y

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From Table 12–2, 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

A3S 02 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

1081

M12_HIBB9290_01_SE_C12_ANS.indd 1081

Ans.

Ans: Q = 2.51 m3 >s

16/03/17 4:49 PM

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12–46. The channel has a triangular cross section and is flowing full. Determine the volumetric flow of water if the sides are made of wood and the downward slope is 0.002. Take n = 0.012. 3 ft 50°

50°

SOLUTION Water is considered to be incompressible. The flow is steady. 1

From Table 12–2, n = 0.012 ft>m 3 for a wood surface. Also, with a = 50° and y = 3 ft for a triangular channel, A =

(3 ft)2 y2 = = 7.5519 ft2 tan a tan 50°

P =

2(3 ft) 2y = = 7.8324 ft sin a sin 50°

For FPS units, 5

Q =

2

nP 3

5

1

1.486 A3S 02

=

1

1.486 ( 7.5519 ft2 ) 3 ( 0.002 2 ) 1

2

( 0.012 ft>m 3 ) (7.8324 ft)3

= 40.8 ft3 >s

1082

M12_HIBB9290_01_SE_C12_ANS.indd 1082

Ans.

Ans: Q = 40.8 ft3 >s

16/03/17 4:49 PM

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12–47. The drainage pipe is made of finished concrete and is sloped downward at 0.002. Determine the volumetric flow from the pipe if the center depth is y = 1 m. Take n = 0.012. 0.6 m

y

SOLUTION We will consider the flow as steady and the water as incompressible. From the geometry shown in Fig. a, u = cos -1 a

Then

0.4 m b = 0.8411 rad 0.6 m

0.4 m

0.6 m

u

u

a = 2p - 210.84112 = 4.6010 rad The cross-sectional area and the wetted perimeter are A =

0.6 m

10.6 m2 2 R2 1a - sin a2 = 34.6010 - sin14.601024 = 1.0071 m2 2 2

a

(a)

P = aR = 14.6010210.6 m2 = 2.7606 m

For SI units, k = 1. Then Manning equation becomes 1

Q = =

A5>3S 02 nP 2>3 11.0071 m2 2 5>3 10.0021>2 2

10.012 s>m1>3 212.7606 m2 2>3

= 1.9162 m3 >s = 1.92 m3 >s

Ans.

Ans: Q = 1.92 m3 >s 1083

M12_HIBB9290_01_SE_C12_ANS.indd 1083

16/03/17 4:49 PM

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*12–48. The drainage pipe is made of finished concrete and is sloped downward at 0.002. Determine the volumetric flow from the pipe if the center depth is y = 0.3 m. Take n = 0.012. 0.6 m

y

SOLUTION We will consider the flow as steady and the water as incompressible. From the geometry shown in Fig. a, u = cos -1 a

Then

0.3 m b = 60° 0.6 m

0.3 m

p rad 2 a = 2u = 2160°2 = 1120°2 a b = p rad 180° 3 The cross-sectional area and the wetted perimeter are

0.6 m

u u

a

0.3 m

(a)

10.6 m2 2 2 R2 2 A = 1a - sin a2 = c p - sin a pb d = 0.2211 m2 2 2 3 3 2 P = aR = a pb 10.6 m2 = 0.4p m 3

For SI units, k = 1. Then Manning equation becomes Q = =

A5>3S 01>2 nP 2>3 10.2211 m2 2 5>3 10.0021>2 2

10.012 s>m1>3 210.4p m2 2>3

= 0.2587 m3 >s = 0.259 m3 >s

Ans.

1084

M12_HIBB9290_01_SE_C12_ANS.indd 1084

Ans: Q = 0.259 m3 >s

16/03/17 4:49 PM

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12–49. The culvert is at a downward slope S0. Determine the depth y of water that will produce the maximum volumetric flow. R u

SOLUTION

y

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 =

1 2 1 R (2u) - (2R sin u)(R cos u) 2 2

A = R2u A = R2u -

1 3 R2 (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>3S 01>2 nP 2>3

= a

A5 1>3 kS 01>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

R

Here,

y

A5 dA dP d a 2b P 2 1 5A4 2 - A5(2P) du du P = = 0 du 2 P4 dA dP 5P 2A4 - 2PA5 = 0 du du PA4 a5P

Since PA4 ≠ 0; 5P

(a)

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° 1085

M12_HIBB9290_01_SE_C12_ANS.indd 1085

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12–49. Continued

Finally, y = R - R cos u = R - R cos 151.21° = 1.8764 R Ans.

= 1.88 R

Ans: y = 1.88R 1086

M12_HIBB9290_01_SE_C12_ANS.indd 1086

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12–50. The culvert is at a downward slope S0. Determine the depth y of water 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 (a)

Referring to the Manning equation, V =

kRh 2>3S0 1>2 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° = 1.6256R Ans.

= 1.63R

Ans: y = 1.63R

1087

M12_HIBB9290_01_SE_C12_ANS.indd 1087

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12–51. The channel is made of unfinished concrete and has a downward slope of 0.005. Determine the volumetric flow if the depth is y = 6 ft. Take n = 0.014.

y

5

3

5

4

3 4

5 ft

SOLUTION We will consider the flow as steady and the water as incompressible. 3 3 Here, tan a = , sin a = , y = 6 ft, and b = 5 ft. Then 4 5 y 6 ft + bb = 16 ft2 c + 5 ft d = 78 ft2 tan a 3>4 216 ft2 2y P = + b = + 5 ft = 25 ft sin a 3>5

A = ya

For FPS units, k = 1.486. Then Manning equation becomes Q = =

1.486A5>3S 01>2 nP 2>3 1.486178 ft2 2 5>3 10.0051>2 2 10.014 s>ft1>3 2125 ft2 2>3

= 1249.99 ft3 >s = 1250 ft3 >s

Ans.

1088

M12_HIBB9290_01_SE_C12_ANS.indd 1088

Ans: Q = 1250 ft3>s

16/03/17 4:49 PM

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*12–52. The channel is made of unfinished concrete. Determine the critical depth if the volumetric flow is 1200 ft3 >s. Also, what is the corresponding critical slope? Take n = 0.014.

y

5

3

5

4

3 4

5 ft

SOLUTION 3 We will consider the flow as steady and the water as incompressible. Here, tan a = , 4 3 sin a = , b = 5 ft and y = yc. 5 btop = 2a

yc yc 8 b + b = 2a b + 5 ft = yc + 5 tan a 3>4 3

Ac = yc a Pc =

yc yc 4 + bb = yc a + 5b = yc a yc + 5b tan a 3>4 3

2yc 2yc 10 + b = + 5 = y + 5 sin a 3>5 3 c

(1) (2) (3)

For the critical flow to occur, gAc3 Q2btop

= 1

3 4 132.2 ft>s2 2 c yc a yc + 5b d 3 = 1 8 11200 ft3 >s2 2 a yc + 5b 3

3 4 c yc a yc + 5b d 3 = 44720.50 8 yc + 5 3

Solving numerically, Ans.

yc = 7.0790 ft = 7.08 ft Substitute this result into Eqs. (1), (2), and (3), btop = 23.8773 ft

A c = 102.2109 ft2

P c = 28.5966 ft

Then 1Rh 2 c =

Ac 102.2109 ft2 = = 3.5742 ft Pc 28.5966 ft

For FPS units, k = 1.486. The critical slope is Sc =

n2gAc 1.4862btop 1Rh 2 4>3 c

=

10.014 s>ft1>3 2 2 132.2 ft>s2 21102.2109 ft2 2 1.4862 123.8773 ft213.57422 4>3

= 0.002239 = 0.00224

Ans.

Ans: yc = 7.08 ft Sc = 0.00224 1089

M12_HIBB9290_01_SE_C12_ANS.indd 1089

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12–53. The channel is made of unfinished concrete. Determine the required slope if the flow is to be 500 ft3 >s when the depth is y = 6 ft. Take n = 0.014.

1

1

y

1

1

8 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Also, with a = 45°, y = 6 ft and b = 8 ft. A = ya P =

y 6 ft + bb = 16 ft2 c + 8 ft d = 84 ft2 tan a tan 45°

216 ft2 2y + b = + 8 ft = 24.9706 ft sin a sin 45°

For FPS units, k = 1.486. Then the Manning equation becomes Q = 500 ft3 >s =

1.486A5>3S 01>2 np2>3 1.486184 ft2 2 5>3S 01>2

10.014 s>ft1>3 2124.9706 ft2 2>3

Ans.

S0 = 0.000624

Ans: S0 = 0.000624 1090

M12_HIBB9290_01_SE_C12_ANS.indd 1090

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12–54. The channel is made of unfinished concrete. If the slope is 0.003, determine the volumetric flow when the depth y = 8 ft. Is the flow subcritical or supercritical? Take n = 0.014.

1

1

y

1

1

8 ft

SOLUTION We will consider the flow is steady and the water is incompressible. Also, with a = 45°, y = 8 ft and b = 8 ft, then A = ya P =

y 8 ft + bb = 18 ft2 c + 8 ft d = 128 ft2 tan a tan 45°

218 ft2 2y + b = + 8 ft = 30.6274 ft sin a sin 45°

For FPS unit, k = 1.486. The Manning equation becomes Q =

1.486A5>3S 01>2 nP 2>3

Thus, the Froude number is Fr =

1.4861128 ft2 2 5>3 10.0031>2 2

10.014 s>ft1>3 2130.6274 ft2 2>3

= 1930.76 ft3 >s = 1931 ft3 >s

The average velocity of the flow is V =

=

Ans.

1930.76 ft3 >s Q = = 15.084 ft>s A 128 ft2

V 2gy

=

15.084 ft>s 2132.2 ft>s2 218 ft2

= 0.9398

Since Fr 6 1, the flow is subcritical.

1091

M12_HIBB9290_01_SE_C12_ANS.indd 1091

Ans: Q = 1931 ft3 >s, subcritical

16/03/17 4:49 PM

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12–55. The channel is made of unfinished concrete. If it is required to transport water at 12 m3 >s, determine the critical depth y = yc and the critical slope. Take n = 0.014.

60°

60°

3m

SOLUTION We will consider the flow is steady and the water is incompressible. Here a = 60°, b = 3 m and y = yc. Then btop = 2a

yc yc 2yc b + b = 2a b + 3m = + 3 tan a tan 60° 23

yc yc yc + bb = yc a + 3 mb = yc a + 3b tan a tan 60° 23 2yc 2yc 4yc Pc = + b = + 3m = + 3 sin a sin 60° 23

Ac = yc a

(1) (2) (3)

For critical flow to occur,

gAc3

= 1

Q2btop 19.81 m>s2 2 c yc a 3

2

112 m >s2 a yc3 a

yc

23 2yc

23

Solving numerically,

yc

23 2yc

23

+ 3b

+ 3b d

3

= 1

+ 3b

3

= 14.6789 + 3

Ans.

yc = 1.0933 m = 1.09 m Substitute this result into Eqs. (1), (2), and (3) btop = 4.2624 m

A c = 3.9699 m2

Pc = 5.5248 m

Then 1Rh 2 c =

Ac 3.9699 m2 = = 0.7186 m Pc 5.5248 m

For SI unit system, k = 1. The critical slope is Sc =

n2gAc btop 1Rh 2 c4>3

=

10.014 s>m1>3 2 2 19.81 m>s2 213.9699 m2 2 14.2624 m210.7186 m2 4>3

= 0.0027825 = 0.00278

Ans.

Ans: yc = 1.09 m Sc = 0.00278 1092

M12_HIBB9290_01_SE_C12_ANS.indd 1092

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*12–56. The channel is made of unfinished concrete. If it is required to transport water at 12 m3 >s, determine its downward slope when the depth of the water is 2 m. Also, what is the critical slope for this depth, and what is the corresponding critical flow? Take n = 0.014.

60°

60°

3m

SOLUTION We will consider the flow is steady and the water is incompressible. Here, a = 60°, b = 3 m, and y = 2 m. btop = 2a A = ya P =

y 2m b + b = 2a b + 3 m = 5.3094 m tan a tan 60°

y 2m + bb = 12 m2 a + 3 mb = 8.3094 m2 tan a tan 60°

212 m2 2y + b = + 3 m = 7.6188 m sin a sin 60°

For SI units, k = 1. Then the Manning equation becomes Q = 12 m3 >s =

A5>3S 01>2 nP 2>3 18.3094 m2 2 5>3S 01>2

10.014 s>m1>3 217.6188 m2 2>3

S0 = 0.00036411 = 0.364110-3 2

Ans.

For the critical depth yc = 2 m, the corresponding critical flow rate can be determined, gAc3 Qc2btop

= 1

19.81 m>s2 218.3094 m2 2 3 Qc2 15.3094 m2

= 1

Qc = 32.56 m3 >s = 32.6 m3 >s

Ans. Ac = Then, the corresponding critical slope can be determined with 1Rh 2 c = Pc 8.3094 m2 = 1.0906 m. 7.6188 m Sc =

n2gAc btop 1Rh 2 4>3 c

=

10.014 s>m1>3 2 2 19.81 m>s2 218.3094 m2 2 15.3094 m211.0906 m2 4>3

= 0.0026804 m>m = 0.00268

Ans.

Ans: S0 = 0.364110-3 2 Qc = 32.6 m3 >s Sc = 0.00268 1093

M12_HIBB9290_01_SE_C12_ANS.indd 1093

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12–57. Determine the volumetric flow of water through the channel if the depth is y = 1.25 m and the downward slope of the channel is 0.005. The sides of the channel are finished concrete. Take n = 0.012.

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 =

1.25 m = 0.625 m 2

2

1.25 m

1

Thus, the area and the wetted perimeter are A =

1 3 2 m + 2 m + 2(0.625 m) 4 (1.25 m) = 3.28125 m2 2

a

2m

a

(a)

P = 2 m + 22(1.25 m)2 + (0.625 m)2 = 4.7951 m 1

From Table 12–2, n = 0.012 s>m 3 for finished concrete. Applying Manning’s equation, 5

Q =

1

A3S 02 2

nP 3

( 3.28125 m2 ) 3 1 0.0052 2 5

=

1

1

2

( 0.012 s>m 3 ) (4.7951 m)3

= 15.0 m3 >s

Ans.

1094

M12_HIBB9290_01_SE_C12_ANS.indd 1094

Ans: Q = 15.0 m3 >s

16/03/17 4:49 PM

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12–58. 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. Take n = 0.012.

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

2 1

1 a = y 2 a

Thus, the area and the wetted perimeter are A =

y

2m

a

(a )

1 1 1 c 2 m + 2 m + 2 a yb d y = a y2 + 2yb m2 2 2 2

P = 2m + 2

1 2 y2 + a yb = ( 2 + 15y ) m 2 B 1

From Table 12–2, 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 =

3 1 1 a y2 + 2yb 1 0.0052 2 2 1

5

3 1 a y2 + 2yb 2 2

Solving by trial and error,

2

( 0.012 s>m 3 )( 2 + 15y ) 3

( 2 + 15y ) 3

= 2.5456

Ans.

y = 1.249 m = 1.25 m

Ans: y = 1.25 m 1095

M12_HIBB9290_01_SE_C12_ANS.indd 1095

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12–59. The channel is made of unfinished concrete and has a slope of 0.004. Determine the volumetric flow and the corresponding critical slope. Specify whether the flow is subcritical or supercritical. Take n = 0.012.

2m

0.75

0.75 1

1

3m

SOLUTION 3 We will consider the flow as steady and the water as incompressible. Here tan a = , 4 3 sin a = , b = 3 m and y = 2 m. Then 5 A = ya P =

y 2m + bb = 12 m2 a + 3 mb = 11.3333 m2 tan a 3>4

212 m2 2y + b = + 3 m = 9.6667 m sin a 3>5

For FPS units, k = 1.486. Then the Manning equation becomes Q = =

1.486A5>3S 01>2 nP 2>3 1.486111.3333 m2 2 5>3 10.0042 1>2 10.012 s>m1>3 219.6667 m2 2>3

= 98.69 m3 >s = 98.7 m3 >s

Ans.

When the critical flow occurs, y = yc. Then btop = 2a

yc yc b + b = 2a b + 3 m = 12.6667yc + 32 m tan a 3>4

Ac = yc a Pc =

yc yc + bb = yc a + 3 mb = 11.3333yc2 + 3yc 2 m2 tan a 3>4

2yc 2yc + b = + 3 m = 13.3333yc + 32 m sin a 3>5

(1) (2) (3)

For critical flow to occur,

gAc3 Q2btop

= 1

19.81 m>s2 211.3333yc2 + 3yc 2 3 198.69 m3 >s2 2 12.6667yc + 32

11.3333yc2 + 3yc 2 3 2.6667yc + 3

= 1

= 992.86

1096

M12_HIBB9290_01_SE_C12_ANS.indd 1096

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12–59. Continued

Solving numerically, yc = 3.1263 m Substitute this result into Eqs. (1), (2), and (3) btop = 11.3369 m Then

The critical slope is Sc =

1Rh 2 c = n2gAc

btop 1Rh 2 c4>3

=

A c = 22.4110 m2

Pc = 13.4211 m

Ac 22.4110 m2 = = 1.6698 m Pc 13.4211 m 10.012 s>m1>3 2 2 19.81 m>s2 2122.4110 m2 2 111.3369 m211.66982 4>3

= 0.0014096 = 0.00141

Ans.

Since S0 = 0.004 7 Sc, the flow is supercritical.

Ans: Q = 98.7 m3>s, Sc = 0.00141, supercritical 1097

M12_HIBB9290_01_SE_C12_ANS.indd 1097

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*12–60. The finished concrete channel has a flow of 20 m3 >s. Determine the critical slope. Take n = 0.012.

yc

2

2 1

1

3m

SOLUTION We will consider the flow as steady and the water as incompressible. Here, tan a = 2, 2 sin a = , b = 3 m, and y = yc. 25 yc yc btop = 2a b + b = 2a b + 3 m = 1yc + 32m (1) tan a 2 Ac = yc a

yc yc yc + bb = yc a + 3 mb = 1yc + 62m2 tan a 2 2

2yc 2yc + b = + 3 m = 25yc + 3 sin a 2> 25 For the critical flow to occur, Pc =

gAc3 Q2btop 19.81 m>s2 2 c

120 m3 >s2 2 1yc + 32 3yc 1yc + 624 3

Solving numerically,

(3)

= 1

3 yc 1yc + 62 d 2

yc + 3

(2)

= 1

= 326.20

Ans.

yc = 1.51405 m Substitute this result into Eqs. (1), (2), and (3) btop = 4.51405 m

A c = 5.6883 m2

P c = 6.3855 m

Then 1Rh 2 c =

Ac 5.6883 m2 = = 0.89082 m Pc 6.3855 m

For SI units, k = 1. The critical slope is Sc =

n2gAc k 2btop 1Rh 2 4>3 c

=

10.012 s>m1>3 2 2 19.81 m>s2 215.6883 m2 2 112 214.51405 m210.89082 m2 4>3

= 0.002077 = 0.00208

Ans.

Ans: Sc = 0.00208 1098

M12_HIBB9290_01_SE_C12_ANS.indd 1098

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12–61. 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. Take n = 0.014.

4m 60°

60°

b

SOLUTION 1

From Table 12–2, 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 =

(9.2376 + 4b)3 1 0.002 2 2 5

1

A3S 02 2

nP 3

=

1

1

5

(9.2376 + 4b)3 2

(9.2376 + b)3

2

( 0.014 s>m 3 ) (9.2376 + b)3

= 100 m3 >s

= 31.305

Solving by trial and error, Ans.

b = 3.08 m

Ans: b = 3.08 m 1099

M12_HIBB9290_01_SE_C12_ANS.indd 1099

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12–62. Determine the length of the sides a of the channel in terms of its base b, so that for 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

SOLUTION

b

The cross-sectional area and wetted perimeter of the trapezoidal channel written in terms of flow depth ywith a = 60° are y y y2 + bb = ya + bb = + by tan a tan 60° 23

A = ya

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 equations, A = Py - 23y2

(1)

Taking Q as fixed, we rewrite Q =

5 3

1 2

A S0 2

nP 3

as 2

A = kP 5 Where K is constant. Substituting from Eq. (1), 2

Py - 23y2 = kP 5

dP Taking the derivative of both sides with respect to y and setting = 0 to obtain dy the minimum P, 3 dP dP 2 P + y - 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

Ans: a = b 1100

M12_HIBB9290_01_SE_C12_ANS.indd 1100

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12–63. 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 ywith a = u are A = ya P =

y y y2 + bb = ya + bb = + by tan a tan u tan u

2y 2y + b = + b sin a sin u

Expressing A in terms of P by eliminating b from the above two equations, y2 2y2 + Py tan u sin u

A =

Taking Q as fixed, we rewrite 1 2

5

A3S 0

Q =

2

nP 3

as

2

A = kP 5 and so y2 2y2 2 + Py = kP 5 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 y2 sin2 u

cos u =

+ +

2y2cos u 2

sin u 2y2 cos u sin2 u

0P = 0 to 0u

3 0P 0P 2 + y = kP - 5 0u 5 0u

= 0

1 2 Ans.

u = 60° Substituting this result into Eq. (1), 2

Py - 23y2 = kP 5

(2)

1101

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12–63. Continued

0P Taking the partial derivative of both sides with respect to y and setting = 0 to 0y obtain the minimum P, 0P 2 -3 0P P + y - 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 1102

M12_HIBB9290_01_SE_C12_ANS.indd 1102

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*12–64. Show that the width b = 2h(csc u - cot u) in order to minimize the wetted 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 l

h = tan u a = h cot u a Thus, the area and the wetted perimeter are 1 (b + b + 2h cot u)h = bh + h2 cot u 2 P = b + 2h csc u

(1)

A =

(2)

a

h

l

b

a

(a )

Using Eq. (1) to express b in term of h and A A - h2 cot u h Substituting this result into Eq. (2), b =

A - h2 cot u + 2h csc u h A P = + h(2 csc u - cot u) h P =

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 Substituting 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) du

h ( csc 2u - 2 csc u cot u ) = 0

( -csc 2u ) 4 = 0

1103

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12–64. 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 lengths of the sides are

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.

Ans: u = 60° 1104

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12–65. 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 =

R

R

R2 R2 (a - sin a) = (2u - sin 2u) 2 2

y

P = aR = (2u)R = 2Ru

R

(a )

Expressing A in terms of P by eliminating R from the above two equations, A = Then,

P2 2 sin 2u a b 8 u u2 2

A = kP 5 2 P2 2 sin 2u a b = kP 5 8 u u2 -8 2 sin 2u = 8kP 5 2 u u

dP = 0 to obtain Taking the derivative of both sides with respect to u and setting du the minimum P, -

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

1105

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12–66. A rectangular channel is made of unfinished concrete, and it has a width of 6 ft and an upward slope of 0.008. Determine the type of surface profile for a flow of 200 ft3 >s if the depth of the water at a specific location is y = 2.5 ft. Sketch this profile. Take n = 0.014.

SOLUTION

Type A3 profile

We will consider the flow as steady and the water as incompressible. For the rectangular channel, the critical depth can be determined by yC 5 3.2556 ft yc = a

Q2 2

gb

b

1>3

= c

1200 ft3 >s2 2 2

132.2 ft>s 216 ft2

2

d

1>3

= 3.2556 ft

Since y = 2.5 ft 6 yc, the flow is supercritical. Also, the channel has an adverse slope. Referring to Table 12–2 with these two conditions, the surface profile is of type A3. The plot of this profile is shown in Fig. a.

y 5 2.5 ft

8 1000 Adverse slope (a)

Ans: A3 1106

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12–67. A rectangular channel is made of unfinished concrete, and it has a width of 1.5 m and a downward slope of 0.06. Determine the type of surface profile for a flow of 8.5 m3 >s if the depth of the water at a specific location is y = 2 m. Sketch this profile. Take n = 0.012.

SOLUTION

Type S1 profile

We will consider the flow as steady and the water as incompressible. For the rectangular channel, the critical depth can be determined. yc = a

Q2 gb2

b

1>3

= c

18.5 m3 >s2 2

19.81 m>s2 211.5 m2 2

d

1>3

y52m

= 1.4848 m

Since y = 2 m 7 yc, the flow is subcritical. The depth yn of the uniform flow can be determined using the Manning equation. Here A = byn = 11.5 m2yn = 1.5yn

yC 5 1.485 m

yn 5 0.5835 m

6

100

(a)

P = 2yn + 1.5 m

For SI units, k = 1.

Q = 8.5 m3 >s =

Solving numerically,

A5>3S 01>2 nP 2>3 11.5yn 2 5>3 10.061>2 2

10.012 s>m1>3 212yn + 1.52 2>3 yn5>3

12yn + 1.52 2>3

= 0.21186

yn = 0.5834 m The critical slope can now determined with A c = 11.5 m211.4848 m2 = 2.22717 m2, A c 2.22717 m2 btop = 1.5 m, Pc = 211.4848 m2 + 1.5 m = 4.46956 m and 1Rh2 c = = = Pc 4.46956 m 0.49830 m Sc =

n2gAc btop 1Rh 2 4>3 c

=

10.0125 s>m1>3 2 2 19.81 m>s2 212.22717 m2 2

= 0.005309

11.5 m210.49830 m2 4>3

Here, S0 = 0.06 7 Sc and y = 2 m 7 yc 7 yn. Referring to Table 12–2 with these two conditions, the surface profile is of type S1. The plot of this profile is shown in Fig. a.

Ans: S1 1107

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*12–68. A rectangular channel is made of finished concrete, and it has a width of 10 ft and a downward slope of 0.0002. Determine the type of surface profile for a flow of 150 ft3 >s and the depth of the water at a specific location is 4 ft. Sketch this profile. Take n = 0.012.

SOLUTION

Type M2 profile

We will consider the flow as steady and the water as incompressible. For the yn 5 4.736 ft rectangular channel, the critical depth can be determined by yc = a

Q2 gb2

b

1>3

= c

1150 ft3 >s2 2

y 5 4 ft

132.2ft>s2 2110 ft2 2

d

1>3

= 1.9118 ft

Since y = 4 ft 7 yc, the flow is subcritical. The depth yn of the uniform flow can be determined using the Manning equation. Here A = byn = 110 ft2yn = 10yn

yC 5 1.912 ft

2

10,000

(a)

p = 2yn + 10 ft.

For FPS units, k = 1.486. Then

Q = 150 ft3 >s =

yn5>3

12yn + 102 2>3

1.486A5>3S 01>2 nP 2>3 1.486110yn 2 5>3 10.00021>2 2

10.012 s>ft1>3 212yn + 102 2>3

= 1.8453

Solving numerically,

yn = 4.7359 ft The critical slope can now be determined using A c = 110 ft211.9118 ft2 = Ac 19.1180 ft2, btop = 10 ft, Pc = 211.9118 ft2 + 10 ft = 13.8236 and 1Rh 2 c = = Pc 2 19.1180 ft = 1.3830 ft. 13.8236 ft Sc =

n2gAc btop 1Rh 2c4>3

=

10.012 s>ft1>3 2 2 132.2 ft>s2 2119.1180 ft2 2

= 0.005753

110 ft211.3830 ft2 4>3

Here, S0 = 0.0002 6 Sc and yc 6 y = 4 ft 6 yn. Referring to Table 12–2 with these two conditions, the surface profile is of type M2. The plot of this profile is shown in Fig. a.

Ans: M2 1108

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12–69. 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. Take n = 0.012.

SOLUTION 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

Type S1

1>3

= 0.3469 m

y = 0.6 m 1

yn = 0.245 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) m2

100

yc = 0.347 m Steep slope (a)

P = ( 2yn + 1.25 ) m From the Table 12–2, n = 0.012 for a finished concrete surface. For SI units to determine the normal depth, Q =

A5>3S 01>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 Table 12–2 with these two conditions, the surface profile is of type S1. The plot of this profile is shown in Fig. a.

Ans: S1 1109

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12–70. 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. Take n = 0.012.

SOLUTION Water is considered to be incompressible. The flow is steady. The critical depth is yc = °

type S2

yc = 0.347 m

1 3

y = 0.3 m

1 3

( 0.8 m3 >s ) 2 ¢ = £ § = 0.3469 m gb2 9.81 m>s2 ( 1.25 m ) 2 Q2

1

100

yn = 0.245 m

Steep slope (a)

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 = (1.25 m)yn = (1.25yn) m2 P = ( 2yn + 1.25 ) m From the Table 12–2, n = 0.012, for a finished concrete surface. For SI units to determine the normal depth, 5

Q =

1

A3S 02 2

nP 3 5

0.8 m3 >s =

1

( 1.25yn ) 3 ( 0.01 ) 2 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 4

btop ( Rh ) c3

=

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 Table 12–2 with these two conditions, the surface profile is of type S2. The plot of this profile is shown in Fig. a.

Ans: S2

1110

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12–71. 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 a 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. Take n = 0.014.

SOLUTION

yB = 0.896 m A

Water is considered to be incompressible. The flow is steady. The critical depth is yc = °

1 3

( 4 m3 >s ) 2 ¢ = £ § = 0.7415 m gb2 ( 9.81 m>s2 )( 2 m ) 2 Q2

1

yA = 0.9 m

1 3

x (m) 0

Since yA = 0.9 m 7 yc the flow is subcritical. Also, the channel bed is horizontal. Referring to Table 12–2 with these two conditions, the surface profile is of type H2 as shown in Fig. a. Thus, the flow depth decreases as x increases. At the control section A, yA = 0.9 m. Then, VA =

yC = 0.892 m yD = 0.888 m B y = 0.884 m C D E E 2 3 4

0.663

1.306

1.930

2.536

(a)

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 =

( Rh ) B =

4 m3 >s Q = = 2.2321 m>s AB (2 m)(0.896 m)

(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

( Rh )m3

=

=

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

1111

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12–71. 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: y = 0.888 m 1112

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*12–72. Water flows at 300 ft3 >s along a horizontal rectangular channel made of finished concrete. If the channel has a width of 10 ft and the water depth at the control section A is 2 ft, determine the approximate distance x from A to where the depth becomes 2.5 ft. Use ∆y = 0.1 ft and plot the surface profile. Take n = 0.012.

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined by Eq. 12–7. yc = a

Q2 2

gb

b

1>3

= c

1300 ft3 >s2 2 2

132.2 ft>s 2110 ft2

2

d

1>3

= 3.0348 ft

Since yA = 2 ft 6 yc, the flow is supercritical. Also, the channel bed is horizontal. Referring to Table 12–2 with these two conditions, the surface profile is of type H3 as shown in Fig. a where the flow depth increases with x. At the control station A, yA = 2 ft . Then VA = 1Rh 2 A =

300 ft3 >s Q = = 15 ft>s AA 110 ft212 ft2

110 ft212 ft2 AA = = 1.4286 ft PA 212 ft2 + 10 ft

Here, Ay = 0.1 ft is chosen. Then at station B, yB = 2 ft + 0.1 ft = 2.1 ft. Thus VB = 1Rh 2 B =

300 ft3 >s Q = = 14.2857 ft>s AB 110 ft212.1 ft2 110 ft212.1 ft2 AB = = 1.4789 ft PB 212.1 ft2 + 10 ft

Thus the mean values for segment (1) between station A and B are VA + VB = 14.6429 ft>s 2 1Rh 2 A + 1Rh 2 B 1.4286 ft + 1.4789 ft = = = 1.4537 ft 2 2

Vm = 1Rh 2 m

Sfm =

∆x =

n2Vm2 1.4862 1Rh 2 m4>3

=

10.012 s>ft1>3 2 2 114.6429 ft>s2 2 1.4862 11.4537 ft2 4>3

1yB - yA 2 + 1VB2 - VA2 2 >2g S0 - Sfm

=

= 0.0084905

0.1 ft + 3114.2857 ft>s2 2 - 115 ft>s2 2 4 > 32132.2 ft24 0 - 0.0084905

= 26.48 ft

1113

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12–72. Continued

Using the same procedure, the computation for the stations that follow are tabulated below. Station

Segment

V(ft , s)

y(ft) 2

A

15

Vm(ft , s)

1

Rh(ft)

2.1

14.2857

2.2

13.6364

2.3

13.0435

13.9610

12.5 12.25

2.5

26.48 0.0073806

24.60 51.08

0.0064606

22.54 73.62

1.5985

0.0056913

20.31

1.6441

0.0050429

17.89

1.6216

5 12.0

x(ft) 0

1.5516

12.7717

F

26.48

1.5753

4 2.4

0.0084905

1.5033

13.3399

E

1.4537

1.5278

3 D

∆x(ft)

1.4789

2 C

Sfm

1.4286

14.6429

B

(Rh)m (ft)

93.92

1.6667

111.81

Thus, the location of station having a depth of y = 2.5 ft is Ans.

x = 112 ft The plot of the surface profile is shown in Fig. a yD 5 2.30 ft yE 5 2.40 ft F yC 5 2.20 ft E D yB 5 2.10 ft C B A yA 5 2 ft

yF 5 2.50 ft

1 0

2 26.48

3

4

5

51.08 73.62 93.92 111.81 (a)

x (ft)

Ans: x = 112 1114

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12–73. Water flows at 150 ft3 >s down a rectangular channel made of finished concrete. The channel has a width of 10 ft and a downward slope of 0.0002, and the water depth is 4 ft at the control section A. Determine the distance xfrom A to where the depth becomes 3.5 ft. Use ∆y = 0.1 ft and plot the surface profile. Take n = 0.012.

A B

4 ft

x

SOLUTION We will consider the flow as steady and the water as incompressible. For the rectangular channel, the critical depth can be determined using Eq. 12–7 yc = a

Q2 2

gb

b

1>3

= c

1150 ft3 >s2 2 2

132.2 ft>s 2110 ft2

2

d

1>3

= 1.9118 ft

Since yA = 5 ft 7 yc, the flow is subcritical. The depth yn of the uniform flow can be determined using Manning equation. Here, A = byn = 110 ft2yn = 10yn

P = 2yn + 10 ft

For FPS units, k = 1.486. Then

Q = 150 ft3 >s =

1.486A5>3S 01>2 nP 2>3

1.486110yn 2 5>3 10.00021>2 2

yn5>3

yF 5 3.50 ft

yA 5 4 ft

10.012 s>ft1>3 212yn + 102 2>3

12yn + 102 2>3

Solving numerically,

yB 5 3.90 ft yC 5 3.80 ft yD 5 3.70 ft yE 5 3.60 ft A B C D E F

= 1.8453

1

0

yn = 4.7359 ft

2 696

3

4

5

1275 1763 2176 2528 (a)

x (ft)

The critical slope can now be determined with A c = 110 ft211.9118 ft2 = Ac 19.1180 ft2, btop = 10 ft, Pc = 211.9118 ft2 + 10 ft = 13.8236 ft and 1Rh 2 c = = Pc 2 19.1180 ft = 1.3830 ft 13.8236 ft Sc =

n2gAc k 2btop 1Rh 2 c4>3

=

10.012 s>ft1>3 2132.2 ft>s2 2119.1180 ft2 2 11.4862 2 110 ft211.3830 ft2 4>3

= 0.002605

Here, SB = 0.0002 6 Sc and yc 6 yA = 4 ft 6 yn. Referring to Table 12–2 with these two conditions, the surface profile is of type M2 as shown in Fig. a where the flow depth decreases as xincreases. At the control station A, yA = 4 ft. Then VA = 1Rh 2 A =

150 ft3 >s Q = = 3.75 ft>s AA 110 ft214 ft2 110 ft214 ft2

214 ft2 + 10 ft

= 2.2222 ft

1115

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12–73. Continued

Here, ∆y = 0.1 ft is chosen. Then at station B, yB = 4 ft - 0.1 ft = 3.90 ft. Thus, 150 ft3 >s Q = = 3.8462 ft>s AB 110 ft213.90 ft2

VB =

110 ft213.90 ft2

1Rh 2 B =

213.90 ft2 + 10 ft

= 2.1910 ft

Thus the mean values for segment 1 between stations A and B Vm = 1Rh 2 m = Sfm = Ax =

3.75 ft>s + 3.8462 ft>s VA + VB = = 3.7981 ft>s 2 2 1Rh 2 A + 1Rh 2 B 2

n2Vm2 2

1.486

1Rh 2 4>3 m

=

=

2.2222 ft + 2.1910 ft = 2.2066 ft 2

10.012 s>ft1>3 2 2 13.7981 ft>s2 2 1.4862 12.2066 ft2 4>3

1yB - yA 2 + 1VB2 - VA2 2 >2g S0 - Sfm

=

= 0.00032745

0.1 ft + 313.8462 ft>s2 2 - 13.75 ft>s2 2 4 > 32132.2 ft>s2 24 0.0002 - 0.00032745

= 695.61 ft Using the same procedure, the computation for the stations that follow are tabulated below. Station

Segment

y(ft) 4

A

V(ft , s) 3.75

1 3.8462

2 3.9474

3 D

4.0541

3.60

4.1667

3.50

4.2857

4 E

∆x(ft)

0.00032745

695.61 2.1751

0.00035138

579.69

2.1428

0.00037783

487.79

1275.30

2.1264

1763.09 2.1097

0.00040718

413.30

2.0930 4.2262

x(ft)

695.61

2.1591

4.1104

5 F

Sfm

0 2.2066

4.0007 3.70

(Rh)m (ft)

2.1910 3.8968

3.80

C

Rh(ft) 2.2222

3.7981 3.90

B

Vm(ft , s)

2176.38 2.0759

2.0588

0.00043982

351.83 2528.21

Thus, the location of station having a depth of y = 3.50 is Ans.

x = 2528 ft The plot of the surface profile is shown in Fig. a

Ans: x = 2528 ft 1116

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12–74. The unfinished concrete channel has a width of 3 m and slopes upward at 0.008. At control section A, the water depth is 2 m and the velocity is 6 m>s. Determine the depth of the water yat the section 5.5 m downstream from A. Use ∆y = 0.03 m and plot the surface profile. Take n = 0.014.

6 m!s

B

A

y

2m 5.5 m

SOLUTION We will consider the flow as steady and the water as incompressible. Since the channel is rectangular, the critical depth can be determined. Here Q = VAAA = 16 m>s2313 m212 m24 = 36 m3 >s. yc = a

Q2

2

gb

b

1>3

= c

136 m3 >s2 2 2

19.81 m>s 213 m2

2

d

1>3

= 2.4485 m

Since yA = 2 m 6 yc, the flow is rapid. Also, the channel has an adverse slope. Referring to Table 12–2 with these two conditions, the surface profile is of type A3 as shown in Fig. a where the flow depth increases with x. At the control station A, yA = 2 m. Then VA = 6 m>s 1Rh 2 A =

13 m212 m2 AA = = 0.8571 m PA 212 m2 + 3 m

Here, Ay = 0.03 m is chosen. Then, at station B, yB = 2 m + 0.03 m = 2.03 m. Thus, VB = 1Rh 2 B =

36 m3 >s Q = = 5.9113 m>s AB 13 m212.03 m2

13 m212.03 m2 AB = = 0.8626 m PB 212.03 m2 + 3 m

The mean values for segment 1 between station A and B are Vm = 1Rh 2 m = Sfm = ∆x =

6 m>s + 5.9113 m>s VA + VB = = 5.9557 m>s 2 2 1Rh 2 A + 1Rh 2 B 2

n2Vm2

1Rh 2m4>3

=

=

0.8571 m + 0.8626 m = 0.8599 m 2

10.014 s>m1>3 2 2 15.9557 m>s2 2 10.8599 m2 4>3

1yB - yA 2 + 1VB2 - V 2A 2 >2g S0 - Sfm

=

= 0.0085023

0.03 m + 3159113 m>s2 2 - 16 m>s2 2 4 > 3219.81 m>s2 24 - 0.008 - 0.0085023

= 1.4441 m

1117

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12–74. Continued

Station

Segment

V(ft , s)

y(ft) 2

A

6

Vm(ft , s)

1

Rh(ft)

(Rh)m (ft)

5.9557 2.03

B

5.9113

2 2.06

5.8252

3 2.09

5.7416

2.12

5.6604

2.15

5.5814

4 5

0.0085023

1.4441

0.8653

0.0081858

1.3281

0.8706

0.0078860

1.2146

1.4441 2.7723 3.9869 0.8758

0.0076018

1.1036

0.8785 5.6209

F

0.8599

0.8733 5.7010

E

x(ft) 0

0.8680 5.7834

D

∆x(ft)

0.8626 5.8683

C

Sfm

0.0571

5.0904 0.8810

0.0073321

0.8836

0.9949 6.0854

Since x = 5.5 m is between station E and F, its flow depth can be determined by interpolation of the flow depth between these two stations. y = 2.12 m + a

2.15 m - 2.12 m b 15.5 m - 5.0904 m2 6.0854 m - 5.0904 m

= 2.1323m = 2.13 m

Ans.

The plot of the surface profile is shown in Fig. a yD 5 2.09 m yC 5 2.06 m yE 5 2.12 m yB 5 2.03 m F E D C B A yA 5 2 m yF 5 2.15 m

1 0

2 1.444

3

4

5

2.772 3.987 5.090 6.085 (a)

x (ft)

Ans: y = 2.13 m 1118

M12_HIBB9290_01_SE_C12_ANS.indd 1118

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12–75. 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: yes, yc = 4.74 m 1119

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*12–76. The sill at A causes a hydraulic jump to form in the channel. If the channel width is 1.5 m, determine the upstream and downstream velocities 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 2 y2 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 1120

M12_HIBB9290_01_SE_C12_ANS.indd 1120

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12–77. Water flows at 650 ft3 >s over the 12@ft-wide spillway of the dam. If the depth of the water at the bottom apron is 2 ft, determine the depth y2 of the water after the hydraulic jump.

y2 2 ft

SOLUTION Water is considered to be incompressible. The flow is steady. The velocity of the flow at section (1) of depth y1 = 2 ft is V1 =

650 ft3 >s Q = = 27.083 ft>s A1 (12 ft)(2 ft)

Then, the Froude number of the flow at section (1) is

( Fr ) 1 =

V1 2gy1

=

27.083 ft>s 2 ( 32.2 ft>s2 ) (2 ft)

= 3.3749

Since ( Fr ) 1 7 1, the flow at section (1) is supercritical, which means that a hydraulic jump can occur. y2 1 = c 21 + 8Fr12 - 1 d y1 2

y2 1 = c 21 + 8 ( 3.3749 ) 2 - 1 d 2 ft 2

Ans.

y2 = 8.60 ft

Ans: y2 = 8.60 ft 1121

M12_HIBB9290_01_SE_C12_ANS.indd 1121

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12–78. 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 amount of energy head is lost within 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 V1 =

8 m3 >s Q = = 16 m>s A1 (2 m)(0.25 m)

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 + 8 Fr12 - 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 1122

M12_HIBB9290_01_SE_C12_ANS.indd 1122

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12–79. The hydraulic jump has a depth of 10 ft at the downstream end, and the velocity is 4 ft>s. If the channel is 6 ft wide, determine the depth y1 of the water before the jump and the head loss within the jump.

4 ft!s

y1

10 ft

SOLUTION We will consider the flow as steady and the water as incompressible. Applying the continuity equation 0 r dV + rV # dA = 0 0t Lcv Lcs 0 - V1A1 + V2A2 = 0 - V1 316 ft2y1 4 + 14 ft>s2316 ft2110 ft24 = 0 V1 = a

Then

40 b ft>s y1

y2 2 y2 2V12 = a b + gy1 y1 y1

2a

40 2 b y1 2

132.2 ft>s 2y1

= a

10 ft 2 10 ft b + y1 y1

y12 + 10y1 - 9.9379 = 0

Solving for the positive root, Ans.

y1 = 0.9108 ft = 0.911 ft The head loss during the jump is hL =

1y2 - y1 2 3 4y1y2

=

110 ft - 0.9108 ft2 3 410.9108 ft2110 ft2

Ans.

= 20.61 ft = 20.6 ft

Ans: y1 = 0.911 ft hL = 20.6 ft 1123

M12_HIBB9290_01_SE_C12_ANS.indd 1123

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*12–80. 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 ft3 >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 1124

M12_HIBB9290_01_SE_C12_ANS.indd 1124

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12–81. 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 cs L 0 - V1A1 + V2A2 = 0 - V1 3 2 m ( y1 ) 4 + ( 1.25 m>s ) 3 (2 m)(5 m) 4 = 0

V1 =

6.25 y1

Thus, y2 2 y2 2V12 = a b + gy1 y1 y1 2a

6.25 2 b y1

( 9.81 m>s ) y1 2

= 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 1125

M12_HIBB9290_01_SE_C12_ANS.indd 1125

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12–82. Water flows down the 10-ft-wide spillway, and at the bottom it has a depth of 2 ft and attains a speed of 30 ft>s. Determine if a hydraulic jump will form, and if so, determine the velocity of the flow and its depth after the jump.

2 ft

y2

SOLUTION We will consider the flow as steady and the water as incompressible. The Froude number of the flow at section (1) where y1 = 2 ft is 1Fr2 1 =

V1 2gy1

=

30 ft>s 2132.2 ft>s2 212 ft2

= 3.7383

Since 1Fr2 1 7 1, the flow at section (1) is supercritical, which means that a hydraulic jump can occur. y2 1 = 3 21 + 81Fr2 12 - 14 y1 2 y2 1 = 3 21 + 813.73832 2 - 14 2 ft 2

Ans.

y2 = 9.6208 ft = 9.62 ft

Applying the continuity equation, 0 r dV + rV # dA = 0 0 t Lcv Lcs 0 - V1A1 + V2A2 = 0 - 130 ft>s23110 ft212 ft24 + V2 3110 ft219.6208 ft24 = 0 V2 = 6.236 ft>s = 6.24 ft>s

Ans.

Ans: y2 = 9.62 ft V2 = 6.24 ft>s 1126

M12_HIBB9290_01_SE_C12_ANS.indd 1126

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12–83. The rectangular channel has a width of 3 m and the depth is 1.5 m. Determine the volumetric flow of water over the rectangular sharp-crested weir if the depth of the water over the top of the weir plate is H = 1 m. Take Cd = 0.83.

3m

1.5 m

SOLUTION Water is considered to be incompressible. The flow is steady. 3 2 C 22g bH 2 3 d 3 2 = (0.83) 22 ( 9.81 m>s2 ) (3 m)(1 m)2 3

Qactual =

= 7.35 m3 >s

Ans.

1127

M12_HIBB9290_01_SE_C12_ANS.indd 1127

Ans: Qactual = 7.35 m3 >s

16/03/17 4:51 PM

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*12–84. The rectangular channel is fitted with a 90° triangular weir plate. If the upstream depth of the water is 6 ft, and the bottom of the weir plate is 4.8 ft from the bottom of the channel, determine the volumetric flow of water passing over the weir. Take Cd = 0.72.

90° 6 ft 4.8 ft

SOLUTION We will consider the flow as steady and the water as incompressible. For a triangular weir, the volumetric flow rate can be determined. Using u = 90°, H = 6 ft - 4.8 ft = 1.2 ft and Cd = 0.72. Qactual = =

8 C 22gH 5>2 tan u>2 15 d

8 90° 10.722 a 22132.2 ft>s2 211.2 ft2 5>2 tan a b 15 2

= 4.8610 ft3 >s = 4.86 ft3 >s

1128

M12_HIBB9290_01_SE_C12_ANS.indd 1128

Ans.

Ans: Qactual = 4.86 ft3 >s

16/03/17 4:51 PM

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12–85. Determine the volumetric flow of water passing over the broad-crested weir if it is in a channel having a width of 3 m. Take Cw = 0.87.

1.5 m 0.75 m

SOLUTION We will consider the flow as steady and the water as incompressible. For the broad-crested weir, the volumetric flow rate can be determined. Using, H = 1.5 m - 0.75 m = 0.75 m, b = 3 m and Cw = 0.87. 3>2 2 Qactual = Cwb2ga Hb 3

3>2 2 = 10.87213 m2 1 29.81 m>s2 2 c 10.75 m2 d 3

= 2.890 m3 >s = 2.89 m3 >s

Ans.

Ans: Qactual = 2.89 m3 >s 1129

M12_HIBB9290_01_SE_C12_ANS.indd 1129

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12–86. 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 (3 m) 2 ( 9.81 m>s2 ) c ( y - 1) d 3

Ans.

y = 3.3775 m = 3.38 m

Ans: y = 3.38 m 1130

M12_HIBB9290_01_SE_C12_ANS.indd 1130

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13–1. Oxygen is contained in a vessel under an absolute pressure of 120 psi. If the temperature increases from 40°F to 120°F, determine the changes in pressure and entropy.

SOLUTION Oxygen is considered to be compressible. From the table in Appendix A, R = 1554 ft # lb>slug # °R and k = 1.40 for oxygen since the oxygen is contained in a rigid vessel, the mass and volume remain constant throughout the process. Therefore the density r is a constant. Here, T1 = 40°F + 460 = 500°R and T2 = 120°F + 460 = 580°R. Applying the ideal gas law, p1 rRT1 = ; p2 rRT2

120 psi p2

=

rR1500°R2 rR1580°R2

p2 = 139.2 psi Thus, the change in pressure is ∆p = p2 - p1 = 139.2 psi - 120 psi = 19.2 psi

Ans.

For the change in entropy, cp has to be computed first cp = Then,

1.411554 ft # lb>slug # °R2 kR = = 5439 ft # lb>slug # °R k - 1 1.4 - 1 s2 - s1 = cp ln

∆s = 15439 ft # lb>slug # °R2 c ln a

p2 T2 - R ln T1 p1

139.2 psi 580°R b d - 11554 ft # lb>slug # °R2 c ln a bd 500°R 120 psi

= 576.61 ft # lb>slug # °R = 577 ft # lb>slug # °R

Ans.

The positive result indicates that entropy increases during the process meaning, that heat is being added.

Ans: p2 - p1 = 19.2 psi ∆s = 577 ft # lb>slug # °R 1131

M13_HIBB9290_01_SE_C13_ANS.indd 1131

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13–2. As nitrogen flows from a point A to a point B its temperature increases from 40°F to 70°F and its absolute pressure decreases from 30 psi to 25 psi. Determine the change per unit mass in internal energy, enthalpy, and entropy between the two sections.

SOLUTION The flow is considered steady and the nitrogen is considered compressible. From the table in Appendix A, R = 1775 ft # lb>slug # °R and k = 1.40 for nitrogen. Then, cv = cp =

1775 ft # lb>slug # °R R = = 4437.5 ft # lb>slug # °R k - 1 1.40 - 1

1.4011775 ft # lb>slug # °R2 kR = = 6212.5 ft # lb>slug # °R k - 1 1.40 - 1

Here, TA = 40°F + 460 = 500°R, TB = 70°F + 460 = 530°R, pA = 30 psi and pB = 25 psi. Then, the changes in internal energy, enthalpy, and entropy are ∆u = cv ∆T = 14437.5 ft # lb>slug # °R21530°R - 500°R2

= 133.125 1103 2 ft # lb>slug = 1331103 2 ft # lb>slug

∆h = cp ∆T = 16212.5 ft # lb>slug # °R21530°R - 500°R2

= 186.375 1103 2 ft # lb>slug = 1861103 2 ft # lb>slug sB - sA = cp ln

∆s = 16212.5 ft # lb>slug # °R2 c ln a

Ans.

Ans.

pB TB - R ln TA pA

25 psi 530°R b d - 11775 ft # lb>slug # °R2 c ln a bd 500°R 30 psi

= 685.62 ft # lb>slug # °R = 686 ft # lb>slug # °R

Ans.

Ans: ∆u = 133(103) ft # lb>slug ∆h = 186(103) ft # lb>slug ∆s = 686 ft # lb>slug # °R 1132

M13_HIBB9290_01_SE_C13_ANS.indd 1132

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13–3. As oxygen flows from a point A to a point B its temperature increases from 50°C to 65°C and its absolute pressure decreases from 300 kPa to 240 kPa. Determine the change per unit mass in internal energy, enthalpy, and entropy between the two sections.

SOLUTION The flow is considered steady and the oxygen is considered compressible. From the table in Appendix A, R = 259.8 J>kg # K and k = 1.40 for oxygen. Then, cv = cp =

259.8 J>kg # K R = = 649.5 J>kg # K k - 1 1.40 - 1

1.401259.8 J>kg # K2 kR = = 909.3 J>kg # K k - 1 1.40 - 1

Here, TA = 50°C + 273 = 323 K, TB = 65°C + 273 = 338 K, pA = 300 kPa and pB = 240 kPa. Then, the changes in internal energy, enthalpy and entropy are ∆u = cv ∆T = 1649.5 J>kg # K21338 K - 323 K2 = 9742.5 J>kg = 9.74 kJ>kg

Ans.

= 13.63951103 2 J>kg = 13.6 kJ>kg

Ans.

∆h = cp ∆T = 1909.3 J>kg # K21338 K - 323 K2 sB - sA = cp ln ∆s = 1909.3 J>kg # K2 c ln a

pB TB - R ln TA pA

338 K 240 kPa b d - 1259.8 J>kg # K2 c ln a bd 323 K 300 kPa

= 99.249 J>kg # K = 99.2 J>kg # K

Ans.

Ans: ∆u = 9.74 kJ>kg, ∆h = 13.6 kJ>kg ∆s = 99.2 J>kg # K 1133

M13_HIBB9290_01_SE_C13_ANS.indd 1133

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*13–4. Air has a temperature of 15°C and an absolute pressure of 250 kPa at A. As it passes through the transition, its temperature becomes 40°C and the absolute pressure becomes 125 kPa at B. Determine the change in the density and the enthalpy of the air.

B A

SOLUTION The flow is considered steady and the air is considered compressible. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. Here TA = 15°C + 273 = 288 K and TB = 40°C + 273 = 313 K. Applying the ideal gas law pA = rARTA;

250 1103 2 N>m2 = rA 1286.9 J>kg # K21288 K2 rA = 3.0256 kg>m3

pB = rBRTB;

125 1103 2 N>m2 = rB 1286.9 J>kg # K21313 K2 rB = 1.3920 kg>m3

Thus, the change in density is ∆r = rB - rA = 1.3920 kg>m3 - 3.0256 kg>m3 = - 1.6337 kg>m3 = -1.63 kg>m3 Also, cp =

Ans.

1.401286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Then the change in enthalpy is ∆h = cp ∆T = 11004.15 J>kg # K21313 K - 288 K2 = 25.10 1103 2J>kg = 25.1 kJ>kg

Ans.

Ans: ∆r = - 1.63 kg>m3 ∆h = 25.1 kJ>kg 1134

M13_HIBB9290_01_SE_C13_ANS.indd 1134

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13–5. Air has a temperature of 70°F and an absolute pressure of 80 psi at A. If the temperature at B is 55°F and the absolute pressure is 65 psi, determine the changes in the enthalpy and the entropy of the air.

B A

SOLUTION The flow is considered steady and the air is considered compressible. From the table in Appendix A, R = 1716 ft # lb>slug # °R and k = 1.40 for air. Then, cp =

1.4011716 ft # lb>slug # °R2 kR = = 6006 ft # lb>slug # °R k - 1 1.40 - 1

Here, TA = 70°F + 460 = 530°R, TB = 55°F + 460 = 515°R, pA = 80 psi and pB = 65 psi. Then, the changes in enthalpy and entropy are ∆h = cp ∆T = 16006 ft # lb>slug # °R21515°R - 530°R2

= -90.09 1103 2 ft # lb>slug = - 90.11103 2 ft # lb>slug sB - sA = cp ln

∆s = 16006 ft # lb>slug # °R2 c ln a

Ans.

pB TB - R ln TA pA

65 psi 515°R b d - 11716 ft # lb>slug # °R2 c ln a bd 530°R 80 psi

= 183.88 ft # lb>slug # °R = 184 ft # lb>slug # °R

Ans.

Ans: ∆h = -90.1(103) ft # lb>slug ∆s = 184 ft # lb>slug # °R 1135

M13_HIBB9290_01_SE_C13_ANS.indd 1135

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13–6. 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 1136

M13_HIBB9290_01_SE_C13_ANS.indd 1136

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13–7. A 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 change per unit mass of 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, p1 rRT1 = 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, cy =

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 = cy ∆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 1137

M13_HIBB9290_01_SE_C13_ANS.indd 1137

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*13–8. A closed tank contains oxygen at 400°F and under an absolute pressure of 30 lb>in2. 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, p1 rRT1 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,

Ans.

cy =

1554 ft # lb>slug # R R = = 3885 ft # lb>slug # R k - 1 1.40 - 1

cp =

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 = cy ∆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

1138

M13_HIBB9290_01_SE_C13_ANS.indd 1138

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

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13–9. Nitrogen is compressed from an absolute pressure of 200 kPa to 500 kPa, with no change in temperature. Determine the changes in entropy and enthalpy.

SOLUTION Nitrogen is considered to be compressible. From the table in Appendix A, R = 296.8 J>kg # K and k = 1.40 for nitrogen. Since there is no change in temperature, ∆T = 0 or T1 = T2. The change in enthalpy is ∆h = cp AT = 0

Ans.

The change in entropy is s2 - s1 = cp ln

p2 T2 - R ln T1 p1

∆s = cp ln 1 - 1296.8 J>kg # K2 ln a = -271.96 J>kg # K = -272 J>kg # K

500 kPa b 200 kPa Ans.

The negative sign indicates that entropy is decreasing during the process meaning that heat is being released.

Ans: ∆h = 0 ∆s = -272 J>kg # K 1139

M13_HIBB9290_01_SE_C13_ANS.indd 1139

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13–10. 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 +

V out 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 in2 - V out2 2 2

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: Tout = 5.01°C 1140

M13_HIBB9290_01_SE_C13_ANS.indd 1140

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13–11. The Mach cone formed on the top of a rocket has a half angle of 15°. If the air temperature is -20°C, determine the speed of the rocket.

SOLUTION The air is considered to be compressible. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. Here, T = -20°C + 273 = 253 K. The velocity of sound is c = 2kRT = 21.401286.9 J>kg # K21253 K2 = 318.78 m>s

Using the half angle of the Mach cone, 318.78 m>s c sin a = ; sin 15° = V V V = a1231.67

m b s

= 1.231103 2 m>s

1141

M13_HIBB9290_01_SE_C13_ANS.indd 1141

Ans: V = 1.231103 2 m>s

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*13–12. Determine the half angle a of the Mach cone and the speed of a jet plane that flies at M = 2.2 at an altitude of 5 km.

SOLUTION The air is considered to be compressible. From the table in Appendix A, T = - 17.47°C for air at an altitude of 5 km. In kelvins, T = - 17.47°C + 273 = 255.53 K. Also, R = 286.9 J>kg # K and k = 1.40. Then, the half angle of the Mach cone can be determined. sin a =

1 1 = ; M 2.2

Ans.

a = 27.04° = 27.0°

The speed of jet plane is given by V = M 2kRT = 2.221.401286.9 J>kg # K21255.53 K2 = 704.81 m>s = 705 m>s

Ans.

Ans: a = 27.04° = 27.0° V = 705 m>s 1142

M13_HIBB9290_01_SE_C13_ANS.indd 1142

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13–13. Determine the Mach number for a race car that travels at 230 mi>h in 70°F weather. Note: 1 mi = 5280 ft.

SOLUTION The air is considered to be compressible. From the table in Appendix A, R = 1716 ft # lb>slug # °R and k = 1.40 for air. Here, T = 70°F + 460 = 530°R. The speed of the car is V = a230

The Mach number is M =

V 2kRT

mi 5280 ft 1h ba ba b = 337.33 ft>s. h 1 mi 3600s =

337.33 ft>s

21.4011716 ft # lb>slug # °R21530°R2

= 0.299

Ans.

Ans: M = 0.299 1143

M13_HIBB9290_01_SE_C13_ANS.indd 1143

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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 = = 1484.5 m>s = 1.48 km>s B r C 998.3 kg>m3

Ans.

Ans: cair = 343 m>s cw = 1.48 km>s 1144

M13_HIBB9290_01_SE_C13_ANS.indd 1144

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13–15. Determine the speed of sound in water and in air, both at a temperature of 60°F. Take EV = 3111103 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>ft3 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>ft3

cw = 4805.88 ft>s = 4.81 1 103 2 ft>s

Ans.

1145

M13_HIBB9290_01_SE_C13_ANS.indd 1145

Ans: cair = 1.12 1 103 2 ft>s cw = 4.81 1 103 2 ft>s

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*13–16. How fast in miles per hour must a jet plane fly at an altitude of 30 000 ft in order to have a Mach number of M = 1.5? Note: 1 mi = 5280 ft.

SOLUTION The air is considered to be compressible. From the table in Appendix A, T = -47.98°F or T = - 47.98°F + 460 = 412.02°R for air at an altitude of 30 000 ft. Also, R = 1716 ft # lb>slug # R and k = 1.40. The velocity of the sound is c = 2kRT = 21.4011716 ft # lb>slug # K21412.02°R2 = 994.91 ft>s

It is required that M = 1.5.

M =

V ; c

V = a1492.36

1.5 =

V 994.91

ft 1 mile 3600 s ba ba b s 5280 ft 1h

Ans.

= 1017.52 mi>h = 1018 mi>h

Ans: V = 1018 mi>h 1146

M13_HIBB9290_01_SE_C13_ANS.indd 1146

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13–17. A jet plane has a speed of M = 2 at an altitude of 10 km. How long does it take for the sound to travel to an observer on the ground after the plane passes directly overhead?

M52 a a

SOLUTION The air is considered to be compressible. From the table in Appendix A, T = -49.90°C or T = -49.90°C + 273 = 223.1 K for air at an altitude of 10 km Also, R = 286.9 J>kg # K and k = 1.40. The speed of the plane is V = M 2kRT

= 221.41286.9 J>kg # K21223.1 K2

a 5 30° 10 km

= 598.70 m>s

The half angle of the Mach cone is given by

x

1 1 sin a = = ; a = 30° M 2

(a)

The sound of the jet plane can be heard by an observer on the ground if he or she is in the Mach cone extended to the ground as shown in Fig. a, which is at a distance x from the observer when the plane is directly overhead. From the geometry, 10 km = tan 30°; x

x = 17.32 km

Thus, the time take for the sound to be heard is given by x = Vt;

17.321103 2 = 1598.70 m>s2t

Ans.

t = 28.935 s = 28.9 s

Ans: t = 28.9 s 1147

M13_HIBB9290_01_SE_C13_ANS.indd 1147

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13–18. 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>ft3 and EV = 3111103 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>ft3 )

= 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: s = 1.35 mi 1148

M13_HIBB9290_01_SE_C13_ANS.indd 1148

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13–19. A jet plane has a speed of 600 mi>h when flying at an altitude of 10 000 ft. Determine the Mach number. Note: 1 mi = 5280 ft.

SOLUTION The air is considered to be compressible. The speed of the jet is V = a600

mi 1h 5280 ft 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: M = 0.817 1149

M13_HIBB9290_01_SE_C13_ANS.indd 1149

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*13–20. A ship is located where the depth of the ocean is 3 km. Determine the time needed for a sonar signal to reflect off the bottom and return to the ship. Assume the average water temperature is 10°C. Take r = 1030 kg>m3 and EV = 2.111109 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.

Ans: t = 4.19 s 1150

M13_HIBB9290_01_SE_C13_ANS.indd 1150

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13–21. 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

Equating Eqs. (1) and (2),

S = Vt = V(6)

337.15 5000 = V 2(6V)2 + 50002

a 5000 m

Ans.

V = 368.67 m>s = 369 m>s

(6V)2 + 50002 (a)

Ans: V = 369 m>s 1151

M13_HIBB9290_01_SE_C13_ANS.indd 1151

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13–22. The Mach number of the air flow out the nozzle at B is M = 0.4. If the air has a temperature of 8°C and absolute pressure of 20 kPa, determine the absolute pressure and temperature of the air within the large reservoir at A.

B A

SOLUTION We will consider the flow to be steady, subsonic, and compressible, since 0.3 6 M 6 1. The air in reservoir A is at rest. Thus, the temperature and pressure there are T0 (stagnation temperature) and p0 (stagnation pressure), respectively. T0 = T a1 +

k - 1 2 M b; 2

T0 = 318°C + 2732 K4 c 1 + a = 289.99 K = 290 K

k - 1 2 k-1 M b ; 2

A B

Ans.

1.40

k

p0 = p a1 +

1.40 - 1 b 10.42 2 d 2

p0 = 120 kPa2 c 1 + a

1.40 - 1 1.40 - 1 b 10.42 2 d 2

= 22.33 kPa = 22.3 kPa

Ans.

A B

Ans: T0 = 290 K p0 = 22.3 kPa 1152

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13–23. A 300-mm-diameter pipe transports nitrogen at M = 0.85. If the absolute pressure is 280 kPa and the temperature is 10°C, determine the mass flow through the pipe and the stagnation density.

SOLUTION We will consider the flow to be steady, subsonic, and compressible, since 0.3 6 M 6 1. From the table in Appendix A, R = 296.8 J>kg # K and k = 1.40 for Nitrogen. With the absolute temperature of T = 10°C + 273 = 283 K, the velocity of the flow is V = M 2kRT = 0.8521.401296.8 J>kg # K21283 K2 = 291.48 m>s

Applying the ideal gas law, p = rRT;

2801103 2 N>m2 = r1296.8 J>kg # K21283 K2 r = 3.3336 kg>m3

Thus, the mass flow can be determined from # m = rVA = 13.3336 kg>m3 21291.48 m>s23p10.15 m2 2 4 = 68.68 kg>s = 68.7 kg>s Ans. The stagnation density is given by r0 = ra1 +

k - 1 2 1>1k - 12 M b 2

1

1.40 - 1 1.40 - 1 = 13.3336 kg>m 2 c 1 + a b 10.852 2 d 2

3

= 4.6715 kg>m3 = 4.67 kg>m3

Ans.

Ans: . m = 68.7 kg>s r0 = 4.67 kg>m3 1153

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*13–24. A 300-mm-diameter pipe transports air at M = 1.6. If the stagnation temperature is 25°C, and the mass flow through the pipe is 60 kg>s, determine the density of the air and the stagnation pressure.

SOLUTION We will consider the flow is of steady supersonic compressible since M 7 1. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. Here, T0 = 125°C + 2732 k = 298 K. T0 = T a1 + p0 = pa1 +

k - 1 2 M b; 2

k

k - 1 2 k-1 M b ; 2

The velocity of the flow is

298 K = T c 1 + a p0 = pc 1 + a

1.40 - 1 b 11.62 2 d 2

T = 197.09 K

1.40

1.40 - 1 1.40 - 1 b 11.62 2 d = 4.2504 p (1) 2

V = M 2kRT = 11.602 21.401286.9 J>kg # K21197.09 K2 = 450.18 m>s

Using the mass flow, the density of the air is given by # m = rVA; 60 kg>s = r1450.18 m>s2 3 p10.15 m2 2 4 r = 1.8855 kg>m3 = 1.89 kg>m3

Ans.

Applying the ideal gas law

p = rRT = 11.8855 kg>m3 21286.9 J>kg # K21197.09 K2 = 106.621103 2 Pa = 106.62 kPa

Substitute this result into Eq. (1),

p0 = 4.25041106.62 kPa2 = 453.17 kPa = 453 kPa

Ans.

Ans: r = 1.89 kg>m3 p0 = 453 kPa 1154

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13–25. Air at a temperature of 30°C and an absolute pressure of 680 kPa flows through a 200-mm-diameter duct. If the Mach number is M = 0.42, determine the mass flow.

SOLUTION We will consider the flow to be steady, subsonic and compressible since 0.3 6 M 6 1. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. Here, T = 130°C + 2732 k = 303 K. Applying the ideal gas law, p = rRT ;

6801103 2 N>m2 = r1286.9 J>kg # K21303 K2 r = 7.8223 kg>m3

The velocity of the flow can be determined by applying Eq. 13.2 V = M 2kRT = 0.4221.401286.9 J>kg # K21303 K2 = 146.52 m>s

Thus, the mass flow is given by # m = rVA = 17.8223 kg>m3 21146.52 m>s2 3 p10.1 m2 2 4 = 36.01 kg>s = 36.0 kg>s

Ans.

Ans: . m = 36.0 kg>s 1155

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13–26. 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 bM2 d 2

Using this result, T0 = T a1 + a

k - 1 bM2 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

1156

M13_HIBB9290_01_SE_C13_ANS.indd 1156

Ans.

Ans: V = 1.001103 2 ft>s

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13–27. Determine the greatest possible mass flow through the nozzle if the throat has a diameter of 50 mm. The air in the reservoir has an absolute pressure of 400 kPa and a temperature of 30°C.

50 mm

SOLUTION The flow is considered to be of incompressible and isentropic. 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 = 400 kPa and T0 = 30°C + 273 = 303 K. For air k = 1.40 (table in Appendix A). T0 = T * a1 + p0 = p* a1 +

k - 1 2 M b; 2

k

k - 1 2 k-1 M b ; 2

303 K = T * c 1 + a

1.4 - 1 b 112 2 d 2

400 kPa = p* c 1 + a

T * = 252.5 K 1.40

1.40 - 1 1.4 - 1 b 112 2 d 2

p* = 211.31 kPa

From the table in Appendix A, R = 286.9 J>kg # K. Applying the ideal gas law, the density of the air at the throat is given by p* = r*RT *;

211.31 1103 2 N>m2 = r* 1286.9 J>kg # K21252.5 K2 r* = 2.9170 kg>m3

The velocity of the air at the throat is V * = M 2kRT * = 112 21.401286.9 J>kg # K21252.5 K2 = 318.46 m>s

Thus, the greatest mass flow rate through the nozzle is # m = r*V *A = 1 2.9170 kg>m3 2 1318.46 m>s2 3 p10.025 m2 2 4 = 1.8240 kg>s = 1.82 kg>s

Ans.

Ans: . m = 1.82 kg>s 1157

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*13–28. If the Laval nozzle is open to the atmosphere where the absolute pressure is 102 kPa, determine the required absolute pressure of the air in the tank so that isentropic supersonic flow occurs through the divergent section of the nozzle. The nozzle has an outer diameter of 30 mm and the throat has a diameter of 20 mm.

20 mm

30 mm

SOLUTION The flow is considered to be compressible and isentropic. The ratio of the crosssectional area of the exit plane to that of the throat is p10.015 m2 2 A = = 2.25 A* p10.01 m2 2 For isentropic supersonic flow to occur through the divergent portion of the nozzle, M 7 1 at the exit plane and backpressure is p4 = patm = 102 kPa. Here, k = 1.40 for air. k+1

A 1 = ≥ * M A

1 +

21k - 12 1 1k - 12 M2 2 ¥ 1 1k + 12 2

1.40 + 1

2.25 =

2.25 =

1 ≥ M

1 +

211.40 - 12 1 11.40 - 12 M2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 M2 3 a b M 1.2

Solving and choose M 7 1 (supersonic), or with the aid of compressible flow calculator, M = 2.3282. Then, k

p0 = p4 a1 +

k - 1 2 k-1 M b 2

1.40

1.40 - 1 1.40 - 1 = 1102 kPa2 c 1 + a b 12.32822 2 d 2

= 1332.90 kPa = 13.31103 2 kPa

1158

M13_HIBB9290_01_SE_C13_ANS.indd 1158

Ans.

Ans: P0 = 13.31103 2

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13–29. If the Laval nozzle is open to the atmosphere where the absolute pressure is 102 kPa, determine the required pressure of the air in the tank so that isentropic subsonic flow occurs through the divergent section of the nozzle. The nozzle has an outer diameter of 30 mm and the throat has a diameter of 20 mm.

20 mm

30 mm

SOLUTION The flow is considered to be compressible and isentropic. The ratio of the crosssectional area of exit plane to that of throat is p10.015 m2 2 A = = 2.25 A* p10.01 m2 2 For isentropic subsonic flow to occur through the divergent portion of the nozzle, M 6 1 at the exit plane and backpressure is p3 = patm = 102 kPa. Here, k = 1.40 for air. k+1

A 1 = ≥ M A*

2.25 =

2.25 =

1 +

1 ≥ M

1 +

21k - 12 1 1k - 12 M2 2 ¥ 1 1k + 12 2

1.40 + 1

211.40 - 12 1 11.40 - 12 M2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 M2 3 a b M 1.2

Solving and choose M 6 1 (subsonic), or with the aid of compressible flow calculator, M = 0.2685. Then, k

p0 = p3 a1 +

k - 1 2 k-1 M b 2

1.40

1.40 - 1 1.40 - 1 = 1102 kPa2 c 1 + a b 10.26852 2 d 2

= 107.24 kPa = 107 kPa

Ans.

Ans: p0 = 107 kPa 1159

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13–30. 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 bM2 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>s

Ans.

Ans: # m = 0.0547 kg>s

1160

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13–31. The large tank contains air at an absolute pressure of 150 kPa and temperature of 20°C. Air flows out of the tank through the 5-mm-diameter nozzle at A. 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 bM2 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

Ans.

1161

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13–31. (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

( S+ ) F = 0 + ( 253.71 m>s )( 1.3357 kg>m3 )( 253.71 m>s ) 3 p(0.0025 m)2 4

Ans.

= 1.69 N

F

(a )

Ans: . m = 0.00665 kg>s, F = 1.69 N 1162

M13_HIBB9290_01_SE_C13_ANS.indd 1162

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*13–32. The large tank contains air at an absolute pressure of 60 psi and a temperature of 70°F. The nozzle has a throat diameter of 1 in. and an exit diameter of 1.75 in. Determine the absolute temperature and pressure within the connected pipe so that the nozzle chokes, but also maintains isentropic subsonic flow within the divergent section of the nozzle. Also, what is the mass flow into the tank if the absolute pressure within the pipe is 30 psi?

SOLUTION The flow is considered to be compressible and isentropic. The stagnation temperature and pressure are of those in the tank, i.e., T0 = 70°F + 460 = 530°R and p0 = 60 psi. Since the nozzle is required to choke, M = 1 at the throat. Here, p10.875 in.2 2 A = = 3.0625 A* p10.5 in.2 2 From the table in Appendix A, k = 1.40 and R = 1716 ft # lb>slug # °R. Then k+1

1 A = ≥ M A*

1 +

21k - 12 1 1k - 12M2 2 ¥ 1 1k + 12 2

1.40 + 1

3.0625 =

3.0625 =

1 ≥ M

1 +

211.40 - 12 1 11.40 - 12M2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 M2 3 a b M 1.2

Solving and choose M 6 1 since the isentropic subsonic flow is required to occur through the divergent portion of the nozzle. With the aid of compressible flow calculator, M = 0.1932 Then T0 = T a1 +

k - 1 2 M b; 2

530°R = T c 1 + a

T = 526.07°R = 526°R

k - 1 2 k-1 M b ; 2

Ans.

1.40

k

p0 = pa1 +

1.40 - 1 b 10.19322 2 d 2

60 psi = pc 1 + a

1.40 - 1 1.40 - 1 b 10.19322 2 d 2

Ans.

P = 58.46 psi = 58.5 psi

1163

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13–32. (Continued)

Applying the ideal gas law, at the exit plane, p = rRT ;

a58.46

lb 12 in. 2 b a b = r11716 ft # lb>slug # °R21526.07°R2 1 ft in2 r = 0.009325 slug>ft3

The velocity of the flow at the exit plane is V = M 2kRT = 10.19322 21.4011716 ft # lb>slug # °R21526.07°R2 = 217.23 ft>s

Finally, the mass flow is

2

0.875 # m = rVA = 10.009325 slug>ft3 21217.23 ft>s2 c pa ft b d 12 = 0.03383 slug>s = 0.0338 slug>s

Ans.

For the backpressure pb = 30 psi 6 58.46 psi, the nozzle remains choked. Thus, the same mass flow is maintained. # m = 0.0338 slug>s Ans.

Ans: T = 526°R P = 58.5 psi . m = 0.0338 slug>s 1164

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13–33. The large tank contains air at an absolute pressure of 60 psi and temperature of 70°F. The nozzle has a throat diameter of 1 in. and an exit diameter of 1.75 in. Determine the absolute temperature and 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 section of the nozzle.

SOLUTION The flow is considered to be compressible and isentropic. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 70°F + 460 = 530°R and p0 = 60 psi. Since the nozzle is required to choke, M = 1 at the throat. Here, p10.875 in.2 2 A = = 3.0625 * A p10.5 in.2 2 From the table in Appendix A, k = 1.40 and R = 1716 ft # lb>slug # °R. Then k+1

A 1 = ≥ M A*

1 +

21k - 12 1 1k - 12M2 2 ¥ 1 1k + 12 2

1.40 + 1

3.0625 =

3.0625 =

1 ≥ M

1 +

211.40 - 12 1 11.40 - 12M2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 M2 3 a b M 1.2

Solving and choose M 7 1 since isentropic supersonic flow is required to occur through the divergent portion of the nozzle. With the aid of compressible flow calculator, M = 2.6592 Then T0 = T a1 +

k - 1 2 M b; 2

530°R = T c 1 + a

1.40 - 1 b 12.65922 2 d 2

Ans.

T = 219.52°R = 220°R k

k - 1 2 k-1 p0 = pa1 + M b ; 2

1.40

1.40 - 1 1.40 - 1 b 12.65922 2 d 60 psi = pc 1 + a 2

Ans.

P = 2.744 psi = 2.74 psi

1165

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13–33. (Continued)

Applying the ideal gas law, at the exit plane p = rRT;

a2.744

lb 12 in. 2 b a b = r11716 ft # lb>slug # °R21219.52°R2 1 ft in2 r = 0.001049 slug>ft3

The velocity of the flow at the exit plane is V = M 2kRT = 2.659221.4011716 ft # lb>slug # °R21219.52 °R2 = 1931.17 ft>s

Finally, the mass flow is

2

0.875 # m = rVA = 10.001049 slug>ft3 211931.17 ft>s2 c pa ft b d 12 = 0.03383 slug>s = 0.0338 slug>s

Ans.

Ans: T = 220°R p = 2.74 psi . m = 0.0338 slug>s 1166

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13–34. A converging nozzle having an exit diameter of 30 mm is connected to the large tank. If the temperature of the air in the tank is 20°C and the absolute pressure is 580 kPa, determine the mass flow from the tank. The absolute pressure outside the tank is 102 kPa.

SOLUTION The flow is considered to be compressible and isentropic. The stagnation temperature and pressure are those in the tank; i.e., T0 = 20°C + 273 = 293 K and p0 = 580 kPa. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K for air. Then k

p0 = P a1 +

k - 1 2 k-1 M b 2

580 kPa = 1102 kPa2 c 1 + a M = 1.7932

1.40

1.40 - 1 1.40 - 1 b M2 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 there after. Thus, at the exit plane, M = 1. T0 = T * a1 + 293 K = T * c 1 +

k - 1 2 M b 2

1.40 - 1 2 11 2 d 2

T * = 244.17 K

k

P0 = p* a1 +

k - 1 2 k-1 M b 2

580 kPa = P * c 1 + a

1.40

1.40 - 1 1.40 - 1 b 112 2 d 2

P * = 306.40 kPa

Applying the ideal gas law, the density of air at the exit plane is given by p* = r*RT *;

306.40 1103 2 N>m2 = r* 1286.9 J>kg # K21244.17 K2 r* = 4.3740 kg>m3

The velocity of the flow at the exit plane is V * = M* 2kRT * = 112 21.401286.9 J>kg # K21244.17 K2 = 313.16 m>s

Finally, the mass flow is # m = r*V *A* = 14.3740 kg>m3 21313.16 m>s23p10.015 m2 2 4 = 0.9682 kg>s = 0.968 kg>s

Ans.

Ans: . m = 0.968 kg>s 1167

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13–35. A converging nozzle having an exit diameter of 30 mm is connected to the large tank. If the temperature of the air in the tank is 60°C and at the exit it is 10°C, determine the absolute pressure within the tank and the mass flow of the air. The absolute pressure outside the tank is 102 kPa.

SOLUTION The flow is considered to be compressible and isentropic. The stagnation temperature and pressure are of those in the tank. Thus, T0 = 60°C + 273 = 333 K. The temperature at the exit plane is T = 10°C + 273 = 288 K. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K for air. T0 = T a1 +

k - 1 2 M b 2

333 K = 1283 K2 c 1 + a

1.40 - 1 bM2 d 2

M = 0.9399

Since M 6 1, the nozzle does not choke. Then k

p0 = pa1 +

k - 1 2 k-1 M b 2

= 1102 kPa2 c 1 + a

1.40

1.40 - 1 1.40 - 1 b 10.93992 2 d 2

= 180.26 kPa = 180 kPa

Ans.

Applying the ideal gas law, the density of air at the exit plane is given by p = rRT ;

102 1103 2 N>m2 = r1286.9 J>kg # K21283 K2 r = 1.2563 kg>m3

The velocity of the flow at the exit plane is V = M 2kRT = 10.93992 21.401286.9 J>kg # K21283 K2 = 316.88 m>s

Finally, the mass flow is # m = rVA = 11.2563 kg>m3 21316.88 m>s23p10.015 m2 2 4 = 0.2814 kg>s = 0.281 kg>s

Ans.

Ans: p0 = 180 kPa . m = 0.281 kg>s 1168

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*13–36. 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

P0 = pa1 +

k - 1 2 (k - 1) M b 2

700 kPa = 150 kPac 1 + a M = 1.663

1.4

1.4 - 1 1.4 - 1 bM2 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

Ans.

Ans: # m = 1.78 kg>s

1169

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13–37. 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

P0 = pa1 +

k - 1 2 (k - 1) M b 2

700 kPa = 400 kPac 1 + a M = 0.9311

1.4

1.4 - 1 1.4 - 1 bM2 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

Ans.

Ans: # m = 1.77 kg>s 1170

M13_HIBB9290_01_SE_C13_ANS.indd 1170

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13–38. The large tank contains air at an absolute pressure of 400 kPa, and the temperature is 20°C. If the pressure at the entrance A of the nozzle is 300 kPa, determine the mass flow out of the tank.

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 bM2 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: # m = 1.05 kg>s 1171

M13_HIBB9290_01_SE_C13_ANS.indd 1171

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13–39. 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

40 mm

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

k - 1 2 (k - 1) M ≤ 2

p0 = p¢1 +

1.4

103 kPa = 30 kPac 1 + a M = 1.4535

1.4 - 1 1.4 - 1 bM2 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. k - 1 2 M ≤ 2

T0 = T ¢1 +

293 K = T * c 1 + a T * = 244.17 K

1.4 - 1 b(1)2 d 2 k

k - 1 2 (k - 1) M b 2

p0 = p a1 +

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

Ans.

Ans: # m = 0.306 kg>s

1172

M13_HIBB9290_01_SE_C13_ANS.indd 1172

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*13–40. 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 and R = 286.9 J>kg # K 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

p0 = p¢1 +

k - 1 2 (k - 1) M ≤ 2 1.4

1.30 ( 106 ) Pa = p* c 1 + a p* = 686.8 kPa

1.4 - 1 1.4 - 1 b ( 12 ) d 2

Since the back pressure p = 101.3 kPa 6 p*, the nozzle will choke. Therefore 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: M = 2.44 1173

M13_HIBB9290_01_SE_C13_ANS.indd 1173

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13–41. The large cylindrical tank contains air at an absolute pressure of 200 psi and a temperature of 90°F. The throat of the nozzle has a diameter of 0.25 in., and the exit diameter is 0.75 in. Determine the absolute pressure in the pipe required to produce isentropic supersonic flow through the pipe. What is the Mach number of this flow?

SOLUTION The flow is considered to be of compressible and isentropic. The stagnation temperature and pressure are of those in the tank, i.e., T0 = 90°F + 460 = 550°R and p0 = 200 psi. the ratio of the cross-sectional area of the exit plane to that of the throat is p 10.75 in.2 2 A 4 = = 9 p A* 10.25 in.2 2 4

For isentropic supersonic flow to occur through the divergent portion of the nozzle, M 7 1 at the exit plane and the backpressure is at p4. From the table in Appendix A, k = 1.40 for air. Then k+1

A 1 = ≥ * M A

1 +

21k - 12 1 1k - 12M2 2 ¥ 1 1k + 12 2

1.40 + 1

9 =

9 =

1 ≥ M

1 +

211.40 - 12 1 11.40 - 12M2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 M2 3 a b M 1.2

Solving and choose M 7 1 or with the aid of compressible flow calculator, then Ans.

M = 3.8060 = 3.81 And

1.40

k

p0 = pa1 +

k - 1 2 k-1 M b ; 2

200 psi = p4 c 1 + a

p4 = 1.7116 psi = 1.71 psi

1.40 - 1 1.40 - 1 b 13.80602 2 d 2

Ans.

Ans: M = 3.81 p4 = 1.71 psi 1174

M13_HIBB9290_01_SE_C13_ANS.indd 1174

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13–42. The large cylindrical tank contains air at an absolute pressure of 200 psi and a temperature of 90°F. The throat of the nozzle has a diameter of 0.25 in., and the exit diameter is 0.75 in. Determine the absolute pressure in the pipe required to choke the nozzle, and also maintain isentropic subsonic flow through the pipe. What is the velocity of the flow through the pipe for this condition?

SOLUTION The flow is considered to be compressible and isentropic. The stagnation temperature and pressure are of those in the tank, i.e., T0 = 90°F + 460 = 55°R and p0 = 200 psi. The ratio of the cross-sectional area of the exit plane to that of the throat is p 10.75 in.2 2 A 4 = = 9 p A* 10.25 in.2 2 4

For isentropic subsonic flow to occur through the divergent portion of the nozzle, M 6 1 at the exit plane. Also, since the nozzle is required to choke, the backpressure pb = p3. From the table in Appendix A, k = 1.40 and R = 1716 ft # lb>slug # °R for air. Then k+1

A 1 = ≥ M A*

1 +

21k - 12 1 1k - 12M2 2 ¥ 1 1k + 12 2

1.40 + 1

9 =

9 =

1 ≥ M

1 +

211.40 - 12 1 11.40 - 12M2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 M2 3 a b M 1.2

Solving and choose M 6 1 or with the aid of compressible flow calculator, then M = 0.06446 Then 1.40

k

p0 = p3 a1 +

k - 1 2 k-1 M b ; 2

T0 = T a1 +

200 psi = p3 c 1 + a

1.40 - 1 1.40 - 1 b 10.64462 2 d 2

Ans.

p3 = 119.41 psi = 199 psi

k - 1 2 M b; 2

550°R = T c 1 + a

T = 549.54°R

1.40 - 1 b 10.64462 2 d 2

Thus, the velocity of the flow at the exit plane is given by V = M 2kRT = 10.064462 21.4011716 ft # lb>slug # °R21549.54°R2 = 74.07 ft>s = 74.1 ft>s

Ans. Ans: p3 = 199 psi V = 74.1 ft>s

1175

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13–43. 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 bM2 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: # m = 0.714 kg>s 1176

M13_HIBB9290_01_SE_C13_ANS.indd 1176

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*13–44. 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 = p 0 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

Ans.

Ans: M = 0.446 # m = 17.2 kg>s 1177

M13_HIBB9290_01_SE_C13_ANS.indd 1177

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13–45. 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

1.4

1.4 - 1 1.4 - 1 b(0.4990)2 d 280 kPa = pc 1 + a 2

Ans.

p = 236.21 kPa = 236 kPa

Ans: p = 236 kPa 1178

M13_HIBB9290_01_SE_C13_ANS.indd 1178

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13–46. 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 bM2 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: # m = 0.0519 kg>s 1179

M13_HIBB9290_01_SE_C13_ANS.indd 1179

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13–47. 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 bM2 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

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

1180

M13_HIBB9290_01_SE_C13_ANS.indd 1180

Ans.

Ans: # m = 4.15110-3 2 kg>s

16/03/17 2:13 PM

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*13–48. 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 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

dt = 0.1966 m = 197 mm

p 2 d b 4 t

Ans.

Also, the pressure at the exit plane must be equal to the backpressure, i.e., pe = 25 kPa. k

(k - 1) k - 1 p0 = pe a1 + Me 2 b 2

400 kPa = (25 kPa)c 1 + a Me = 2.4578

1.4

1.4 - 1 1.4 - 1 bMe2 d 2

1181

M13_HIBB9290_01_SE_C13_ANS.indd 1181

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13–48. (Continued)

Since Me 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 Me 2 2 ¥ k + 1 2 1.4 + 1

=

1 ≥ 2.4578

de = 0.3130 m = 313 mm

1 + a

2(1.4 - 1) 1.4 - 1 b(2.4578)2 2 ¥ 1.4 + 1 b a 2

Ans.

Ans: dt = 197 mm de = 313 mm 1182

M13_HIBB9290_01_SE_C13_ANS.indd 1182

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13–49. 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 is 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 MA = 0.5 and pA = 400 kPa at the entrance plane A, pA = 0.8430; p0

p0 =

AA

AC

A*

= 1.3398;

A*

400 kPa = 474.49 kPa = 474 kPa 0.8430

Ans.

p (0.11 m)2 AC 4 = a b = 1.3398≥ ¥ = 1.1258 p A* AA 2 (0.12 m) 4 AA

AC From Appendix B with * = 1.1258, choose MC 6 1 (subsonic). The interpolation A of the values gives Ans.

MC = 0.6608 = 0.661 Also, AB A*

p (0.2 m)2 AB 4 = * a b = 1.3398≥ ¥ = 3.7218 p A AA (0.12 m)2 4 AA

Since the nozzle will not choke, MB 6 1 (subsonic). From Appendix B and interpolating between the values in the table gives Ans.

MB = 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

Ans: p0 = 474 kPa, MC = 0.661 MB = 0.158, pB = 466 kPa 1183

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13–50. 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 is the stagnation pressure, the pressure at the throat, and the subsonic and supersonic Mach numbers for isentropic 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 MA = 0.1 and pA = 400 kPa, k

k-1 k - 1 p0 = pa1 + MA2 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

2(k - 1) k - 1 1 + a bMA2 AA 2 1 ¥ = ≥ MA k + 1 A* 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.

dt = 0.04949 m = 49.5 mm Using the result of p0 at the throat where Mt = 1, k

p0 = pt a1 +

k - 1 2 k-1 Mt b 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 dt, k+1

AB *

A

=

1 ≥ MB

1 +

p (0.2 m)2 4 p (0.04949 m)2 2

2(k + 1) k - 1 MB2 2 ¥ k + 1 2 1.31 + 1

=

1 ≥ MB

2(1.31 - 1) 1.31 - 1 bMB2 2 ¥ 1.31 + 1 a b 2

1 + a

2.4443MB0.2684 - 0.155MB2 - 1 Solving by trial and error, MB = 0.0358 6 1 (subsonic)

Ans.

MB = 4.07 7 1 (supersonic)

Ans.

Ans: p0 = 403 kPa, dt = 49.5 mm, pt = 219 kPa, MB = 0.0358, MB = 4.07

1184

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13–51. 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 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: # m = 0.964 kg>s 1185

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*13–52. 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 = 4. Then, using the equations or Appendix B nozzle chokes, * = A p(0.01 m)2 gives M = 0.1465 (subsonic) pB = 0.9851; pB = 0.9851(150 kPa) = 147.77 kPa r0 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

Ans.

Ans: p = 90 6 pB, the nozzle will choke. # m = 0.120 kg>s 1186

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13–53. Air flows through the nozzle at 2 kg>s, and the absolute pressure and temperature of the air at A are 650 kPa and 350 K, respectively. Determine the pressure at B to maintain isentropic flow in the divergent section of the nozzle for both subsonic and supersonic flow at B. Also determine the Mach number at the throat C and at B.

80 mm

50 mm

20 mm

A

C B

SOLUTION The flow is considered to be steady, compressible and isentropic. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. Applying the ideal gas law at A, 650 1103 2 N>m2 = rA 1286.9 J>kg # K21350 K2

pA = rARTA;

rA = 6.4731 kg>m3

From the mass flow, # m = rAVAAA;

2 kg>s = 16.4731 kg>m3 21VA 23p10.025 m2 2 4 VA = 157.36 m>s

The Mach number of the flow at A is given by 157.36 m>s = MA 21.401286.9 J>kg # K21350 K2

VA = MA 2kRTA;

MA = 0.4197

Using this result, the stagnation pressure is k

k-1 k - 1 p0 = pA a1 + MA2 b 2

1.40

1.40 - 1 1.40 - 1 b 10.41972 2 d = 1650 kPa2 c 1 + a 2

= 733.73 kPa Also,

k+1

A 1 = ≥ MA A*

= a

1 +

21k - 12 1 1k - 12MA2 2 ¥ 1 1k + 12 2

1 b≥ 0.4197

= 1.5298

1 +

1.40 + 1

211.40 - 12 1 11.40 - 1210.41972 2 2 ¥ 1 11.40 + 12 2

Using this result, AC A*

=

p10.01 M2 2 AA AC a b = 11.52982 c d = 0.2448 6 1 A* AA p10.025 M2 2

(nozzle chokes) Ans.

MC = 1

1187

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13–53. (Continued)

Also, AB A*

=

p10.04 M2 2 AA AB a b = 11.52982 c d = 3.9163 A* AA p10.025 M2 2

Since the nozzle chokes, the flow at B could be subsonic 1MB 6 12 or supersonic 1MB 7 12 depending on the backpressure pB. with this result, k+1

AB *

A

=

1 ≥ MB

1 +

21k - 12 1 1k - 12MB2 2 ¥ 1 1k + 12 2

1.40 + 1

3.9163 =

3.9163 =

1 ≥ MB

1 +

211.40 - 12 1 11.40 - 12MB2 2 ¥ 1 11.40 + 12 2

1 1 + 0.2 MB2 3 a b MB 1.2

Solving or with the aid of compressible flow calculator, MB = 0.1498 6 1

(subsonic)

MB = 2.9180 7 1

(supersonic)

Since the flow is required to be isentropic along the divergent portion of the nozzle, pB = p3 for subsonic flow and pB = p4 for supersonic flow. For subsonic flow 1MB = 0.14982, 1.40

k

p0 = pB a1 +

k-1 1.40 - 1 k - 1 1.40 - 1 MB2 b ; 733.73 kPa = pB c 1 + a b 10.14982 2 d 2 2

Ans.

pB = 722.32 kPa = 722 kPa

For supersonic flow, 1.40

k

p0 = pB a1 +

k-1 1.40 - 1 1.40 - 1 k - 1 b 12.91802 2 d MB2 b ; 733.73 kPa = pB c 1 + a 2 2

Ans.

pB = 22.60 kPa = 22.6 kPa

Ans: MC = 1 Subsonic: MB = 0.150, pB = 722 kPa Supersonic: MB = 2.92, pB = 22.6 kPa 1188

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13–54. Air flows through the nozzle at 2 kg>s, and the absolute pressure and temperature of the air at A are 650 kPa and 350 K, respectively. Determine the temperature at B to maintain isentropic flow in the divergent section of the nozzle for both subsonic and supersonic flow at B.

80 mm

50 mm

20 mm

A

C B

SOLUTION The flow is considered to be steady, compressible and isentropic. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40. Applying the ideal gas law at A, 650 1103 2 N2 >m2 = rA 1286.9 J>kg # K21350 K2

pA = rARTA;

rA = 6.4731 kg>m3

From the mass flow, # m = rAVAAA;

2 kg>s = 16.4731 kg>m3 21VA 23p10.025 m2 2 4 VA = 157.36 m>s

The Mach number of the flow at A is given by 157.36 m>s = MA 21.401286.9 J>kg # K21350 K2

VA = MA 2kRTA;

MA = 0.4197

Using this result, the stagnation temperature is T0 = TA a1 +

k - 1 MA2 b 2

= 1350 K2 c 1 + a

= 362.33 K

1.40 - 1 b 10.41972 2 d 2

Also, applying k+1

A 1 = ≥ MA A*

= a

1 +

21k - 12 1 1k - 12MA2 2 ¥ 1 1k + 12 2

1 b≥ 0.4197

= 1.5298

1 +

1.40 + 1

211.40 - 12 1 11.40 - 1210.41972 2 2 ¥ 1 11.40 + 12 2

Using this result, AC *

A

=

p10.01 M2 2 AA AC a b = 11.52982 c d = 0.2448 6 1 * A A p10.025 M2 2 A

(nozzle chokes)

MC = 1

Ans.

1189

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13–54. (Continued)

Also, AB A*

=

p10.04 M2 2 AA AB a b = 11.52982 c d = 3.9163 A* AA p10.025 M2 2

Since the nozzle chokes, the flow at B, could be subsonic 1MB 6 12 or supersonic 1MB 7 12 depending on the backpressure pB. With this result, k+1

21k - 12 1 1 + 1k - 12MB2 AB 1 2 = ≥ ¥ MB 1 A* 1k + 12 2

1.40 + 1

3.9163 =

1 ≥ MB

1 +

211.40 - 12 1 11.40 - 12MB2 2 ¥ 1 11.40 + 12 2

Solving or with the aid of compressible flow calculator, MB = 0.1498 6 1

(subsonic)

MB = 2.9180 7 1

For subsonic flow 1MB = 0.14982, gives T0 = TB a1 +

k - 1 MB2 b ; 2

362.33 K = TB c 1 + a

k - 1 MB2 b ; 2

1.40 - 1 b 10.14982 2 d 2

Ans.

TB = 360.71 K = 361 K

For supersonic flow 1MB = 2.91802, T0 = TB a1 +

(supersonic)

362.33 K = TB c 1 + a

1.40 - 1 b 12.91802 2 d 2

Ans.

TB = 134.05 K = 134 K

Ans: MC = 1 Subsonic: MB = 0.150, TB = 361 K Supersonic: MB = 2.92, TB = 134 K 1190

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13–55. Air flows into the nozzle at MA = 0.2 and isentropically 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

k - 1 2 2(k - 1) MA 1 + AA 1 2 = = ≥ ¥ MA k + 1 A* 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.

dt = 0.01743 m = 17.4 mm Using this result, k+1

AB A*

=

1 = ≥ MB

1 +

k - 1 2 2(k - 1) MB 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.

dB = 0.02264 m = 22.6 mm At plane A, k

p0 = pA a1 +

k-1 k - 1 MA2 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 = pB a1 +

k-1 k - 1 MB2 b 2

1.4

1.4 - 1 1.4 - 1 308.48 kPa = pB c 1 + a b ( 22 ) d 2

Ans.

pB = 39.43 kPa = 39.4 kPa

Ans: dt = 17.4 mm, dB = 22.6 mm p0 = 308 kPa, pB = 39.4 kPa

1191

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*13–56. Air flows into the nozzle at MA = 0.2 and isentropically 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.

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

k - 1 2 2(k - 1) 1 + MA AA 1 2 = ≥ ¥ MA k + 1 A* 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 d 4 t 2

Ans.

dt = 0.01743 m = 17.4 mm Using this result, k+1

AB *

A

=

1 +

1 = ≥ MB

k - 1 2 2(k - 1) MB 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.

dB = 0.02264 m = 22.6 mm At plane A, T0 = TA a1 +

k - 1 MA2 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 MB2 b 2

302.4 = TB c 1 + a TB = 168 K

1.4 - 1 b ( 2 )2 d 2

Ans.

Ans: dt = 17.4 mm dB = 22.6 mm T0 = 302 K TB = 168 K

1192

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13–57. Nitrogen flows through the pipe. If the stagnation temperature is 40°F, the stagnation pressure is 235 psi, and the absolute pressure is 200 psi, determine the mass flow.

6 in.

SOLUTION The flow is considered to be steady and compressible. The absolute stagnation temperature and pressure are T0 = 40°F + 460 = 500°R and p0 = 235 psi. From the table in Appendix A, R = 1775 ft # lb>slug # °R and k = 1.40 for nitrogen. 1.40

k

p0 = pa1 +

k - 1 2 k-1 M b ; 2

235 psi = 1200 psi2 c 1 + a

Then T0 = T a1 +

k - 1 2 M b; 2

M = 0.4856

500°R = T c 1 + a

T = 477.48 °R

1.40 - 1 1.40 - 1 b M2 d 2

1.40 - 1 b(0.48562) d 2

Applying the ideal gas law, p = rRT;

a200

lb 12 in 2 ba b = r11775 ft # lb>slug # °R21477.48°R2 2 1 ft in r = 0.03398 slug>ft3

The velocity of the flow is V = M 2kRT = 10.48562 21.4011715 ft # lb>slug # °R21477.48°R2 = 528.92 ft>s

Finally, the mass flow is 2 3 # m = rVA = 10.03398 slug>ft3 21528.92 ft>s2 c pa ft b d 12

= 3.5290 slug>s = 3.53 slug>s

Ans.

Ans: . m = 3.53 slug>s 1193

M13_HIBB9290_01_SE_C13_ANS.indd 1193

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13–58. Air at a temperature of 20°C and atmospheric pressure of 102 kPa flows through the nozzle into the pipe where the absolute internal pressure is 45 kPa. Determine the throat diameter of the nozzle so that the mass flow into the pipe is 20 g>s.

d

SOLUTION The flow is considered to be steady, compressible and isentropic. The stagnation temperature and pressure are of those in the atmosphere i.e., T0 = 20°C + 273 = 293 K and p0 = 102 kPa. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. At the exit plane, 1.40

k

p0 = pa1 +

k - 1 2 k-1 M b ; 2

102 kPa = 145 kPa2 c 1 + a M = 1.1476

1.40 - 1 1.40 - 1 b M2 d 2

The flow with M 7 1 (supersonic) at the exit plane is not possible since the nozzle chokes at the exit plane causes an expansion shock wave to form. Thus, at the exit plane, M = 1. Then the absolute temperature and pressure at the exit plane are given by T0 = T * a1 + p0 = p* a1 +

k - 1 2 M b; 2

k

k - 1 2 k-1 M b ; 2

293 k = T * c 1 + a

1.40 - 1 b 112 2 d T * = 244.17 K 2

102 kPa = p* c 1 + a

1.40

1.40 - 1 1.40 - 1 b 112 2 d 2

p* = 53.8847 kPa

Applying ideal gas law using these results, p* = r*RT * ;

53.8847 1103 2 N>m2 = r* 1286.9 J>kg # K21244.17 K2 r* = 0.7692 kg>m3

The velocity of the flow is V * = M 2kRT * = 112 21.401286.9 J>kg # K21244.17 K2 = 313.16 m>s

Finally, from the mass flow # m = r*V *A*;

d 2 0.02 kg>s = 10.7692 kg>m3 21313.16 m>s2 c pa b d 2 d = 0.01028 m = 10.3 mm

Ans.

Ans: d = 10.3 mm 1194

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13–59. 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 d = 10 mm.

d

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

p0 = pa1 +

k - 1 2 k-1 M b 2

101.3 kPa = (80 kPa)c 1 + a M = 0.5906

1.4

1.4 - 1 1.4 - 1 bM2 d 2

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: . m = 0.0155 kg>s 1195

M13_HIBB9290_01_SE_C13_ANS.indd 1195

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*13–60. Air at a temperature of 20°C and atmospheric pressure of 102 kPa flows through the nozzle into the pipe where the absolute internal pressure is 60 kPa. Determine the mass flow into the pipe. The nozzle has a throat diameter of d = 15 mm.

d

SOLUTION The flow is considered to be steady, compressible and isentropic. The stagnation temperature and pressure are of those in the atmosphere i.e., T0 = 20°C + 273 = 293 K and P0 = 102 kPa. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. At the exit plane, 1.40

k

p0 = pa1 +

k - 1 2 k-1 M b ; 2

102 kPa = 160 kPa2 c 1 + a M = 0.9047

1.40 - 1 1.40 - 1 bM2 d 2

The flow with M 6 1 (subsonic) at exit plane is possible (nozzle does not choke). Then T0 = T a1 +

k - 1 2 M b; 2

293 K = T c 1 + a T = 251.78 K

1.40 - 1 b 10.90472 2 d 2

Applying the ideal gas law using this result, p = rRT;

60 1103 2 N>m2 = r1286.9 J>kg # K21251.78 K2 r = 0.8306 kg>m3

The velocity of the flow is V = M 2kRT = 10.90472 21.401286.9 J>kg # K21251.78 K2 = 287.71 m>s

Finally, the mass flow is # m = rVA = 10.8306 kg>m3 21287.71 m>s23p10.0075 m2 2 4 = 0.04223 kg>s = 42.2 g>s

Ans.

Ans: # m = 42.2 g>s 1196

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13–61. The large tank contains air at an absolute pressure of 680 kPa and a temperature of 85°C. If the diameter at the end of the converging nozzle is 15 mm, determine the mass flow out of the tank where the standard atmospheric pressure is 101.3 kPa.

15 mm

SOLUTION The flow is considered to be steady, compressible and isentropic. The absolute stagnation temperature and pressure are of those in the tank; i.e., T0 = 85°C + 273 = 358 K and p0 = 680 kPa. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. At the exit plane, 1.40

k

p0 = pa1 +

k - 1 2 k-1 M b ; 2

680 kPa = 1101.3 kPa2 c 1 + a M = 1.9012

1.40 - 1 1.40 - 1 b M2 d 2

The flow with M 7 1 (supersonic) is not possible at the exit plane, since the nozzle chokes at the exit plane, causes an expansion show wave formed thereafter. Thus, at exit plane, M = 1. Thus, the absolute temperature and pressure at the exit plane are given by T0 = T * a1 +

k - 1 2 1.40 - 1 M b ; 358 K = T * c 1 + a b 112 2 d 2 2

p0 = p* a1 +

k - 1 2 k-1 M b ; 2

T * = 298.33 K 1.40

k

680 kPa = p* c 1 + a

1.40 - 1 1.40 - 1 b 112 2 d 2

p* = 359.23 kPa

Applying the ideal gas law using these results, p* = r*RT *;

359.23 1103 2 N>m2 = r* 1286.9 J>kg # K21298.33 K2 r* = 4.1970 kg>m3

The velocity of the flow is given by V * = M 2kRT * = 112 21.401286.9 J>kg # K21298.33 K2 = 346.16 m>s

Finally, the mass flow is # m = r*V *A* = 14.1970 kg>m3 21346.16 m>s23p10.0075 m2 2 4 = 0.2567 kg>s = 0.257 kg>s

Ans.

Ans: . m = 0.257 kg>s 1197

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13–62. The duct has a diameter of 150 mm. If the average friction factor is f = 0.004, and air is drawn into the duct with an inlet velocity of 125 m>s, a temperature of 285 K, and an absolute pressure of 165 kPa, determine these properties at the exit.

150 mm 1

2

90 m

SOLUTION The flow through the duct is considered as Fanno flow since friction is involved. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. The Mach number at the inlet is given by Mi =

Vi 2kRTi

125 m>s

=

21.401286.9 J>kg # K21285 K2

= 0.3695

The compressible flow calculation gives

f L = 2.9450 D max a

0.004 bL max = 2.9450 0.15 m

(O.K)

L max = 110.44 m 7 L = 90 m Also computed are Ti T* Then T* = a

V* = a

p* = a

= 1.1681

Vi V*

= 0.3993

pi p*

= 2.9254

T* 1 b 1Ti 2 = a b 1285 K2 = 243.98 K Ti 1.1681

V* 1 b 1Vi 2 = a b 1125 m>s2 = 313.05 m>s Vi 0.3993 p* 1 b 1pi 2 = a b 1165 kPa2 = 56.40 kPa pi 2.9254

For the flow properties at the exit with respect to the critical location, f f f 0.004 L′ = L L = 2.9450 - a b 190 m2 = 0.5450 D D max D i-e 0.15 m

Again, using this value, the compressible flow calculator gives Me = 0.5869

Te T*

= 1.1227

Ve V*

= 0.6219

pe p*

= 1.8053

Then Te = 1.12271243.98 K2 = 273.91 K = 274 K

Ans.

Ve = 0.62191313.05 m>s2 = 194.67 m>s = 195 m>s

Ans.

pe = 1.8053156.40 kPa2 = 101.82 kPa = 102 kPa

Ans.

Ans: Te = 274 K, Ve = 195 m>s, pe = 102 kPa 1198

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17/03/17 9:47 AM

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13–63. The duct has a diameter of 150 mm. If the average friction factor is f = 0.004, and air is drawn into the duct with an inlet velocity of 125 m>s, a temperature of 285 K, and an absolute pressure of 165 kPa, determine the mass flow through the duct and the resultant friction force acting on the 90-m length of duct.

150 mm 1

2

90 m

SOLUTION The flow through the duct is considered as Fanno flow since friction is involved. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. Applying the ideal gas law, pi = riRTi;

165 1103 2 N>m2 = ri 1286.9 J>kg # K21285 K2 ri = 2.0179 kg>m3

Thus, the mass flow is # m = riVi A = 12.0179 kg>m3 21125 m>s23p10.075 m2 2 4 = 4.4575 kg>s = 4.46 kg>s Ans.

The Mach number at the inlet is given by Mi =

Vi 2kRTi

=

125 m>s

21.401286.9 J>kg # K21285 K2

= 0.3695

Applying Eq. 13-50 with this Mach number, the compressible flow calculation gives f L = 2.9450 D max a

0.004 bL = 2.9450 0.15 m max

(O.K)

Lmax = 110.44 m 7 L = 90 m Also computed are Ti T*

= 1.1681

Vi V*

= 0.3993

pi p*

= 2.9254

Then T* = a

V* = a

P* = a

T* 1 b 1Ti 2 = a b 1285 K2 = 243.98 K Ti 1.1681

V* 1 b 1Vi 2 = a b 1125 m>s2 = 313.05 m>s Vi 0.3993 P* 1 b 1pi 2 = a b 1165 kPa2 = 56.40 kPa Pi 2.9254

1199

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13–63. (Continued)

For the flow properties at the exit with respect to the critical location, f 1 f f 0.004 L = L L = 2.9450 - a b 190 m2 = 0.5450 D D max D i-e 0.15 m

Again, using this value, the compressible flow calculator gives Me = 0.5869

Te T*

Ve

= 1.1227

V*

= 0.6219

pe p*

= 1.8053

Then Te = 1.12271243.98 K2 = 273.91 K

Ans.

Ve = 0.62191313.05 m>s2 = 194.67 m>s

Ans.

pe = 1.8053156.40 kPa2 = 101.82 kPa

Ans.

Again, applying the ideal gas law 101.82 1103 2 N>m2 = re 1286.9 J>kg # K21273.91 K2

pe = re RTe;

re = 1.2957 kg>m3

Applying the linear momentum equation by referring to the FRD of the control volume shown in Fig. a 0 ΣF = VrdV + VrV # dA 0t Lcv Lcs

1 S+ 2 3

2

pi A - pe A - Ff = 0 + Vi ri 1 -Vi A2 + Ve re 1Ve A2

3165110 2 N>m 43p10.075 m2 24 - 3101.821103 2 N>m2 43p10.075 m2 2 4 - Ff = 1125 m>s212.0179 kg>m3 21 -125 m>s23p10.075 m2 2 4

+ 1194.67 m>s211.2957 kg>m3 21194.67 m>s23p10.075 m2 2 4 Ff = 805.84 N = 806 N

Pi A

Ff

Ans.

Pe A

(a)

Ans: . m = 4.46 kg>s, Ff = 806 N 1200

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*13–64. A large reservoir contains air at a temperature of T = 20°C and absolute pressure of p = 300 kPa. The air flows through the 1.5-m-long, 50-mm-diameter pipe having an average friction factor of 0.03. Determine the mass flow 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 b(1.5 m) = 0.9 Lmax = a D 0.05 m

Using this result and interpolating the values tabulated in the Fanno flow table in Appendix B, select M1 6 1, M1 = 0.5226

T1 T

*

= 1.1378

V1 V

*

= 0.55742

p1 p*

= 2.0414

For the isentropic flow at inlet 1 with M1 = 0.5226, 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 = 286.9 J>kg # K,

249.05 ( 103 ) N>m2 p1 = = 3.1246 kg>m3 = 3.12 kg>m3 RT1 ( 286.9 J>kg # K ) (277.82 K)

r1 =

Ans.

Applying, V1 = M1 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.

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

1201

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13–65. A large reservoir contains air at a temperature of T = 20°C and absolute pressure of p = 300 kPa. The air flows through the 1.5-m-long, 50-mm-diameter pipe having an average friction factor of 0.03. Determine the stagnation temperature and stagnation 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

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 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 M1 6 1, M1 = 0.5226

T1 T

= 1.1378

*

p1 *

p

= 2.0414

( p0 ) 1 p0*

= 1.2990

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 M1 = 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

1202

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13–65. (Continued)

Applying, cy =

286.9 J>kg # K R = = 717.25 J>kg # K k - 1 1.4 - 1

The change in entropy is ∆s = s* - s1 = cy 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 ∆s = 75.1 J>kg # K 1203

M13_HIBB9290_01_SE_C13_ANS.indd 1203

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13–66. 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 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. Here p1 = 90 kPa, p0 = 101 kPa and T0 = (273 + 24) K = 297 K. p1 90 kPa With = = 0.89109, p0 101 kPa M1 = 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*

= 2.6334;

p* =

p* 90 kPa (p1) = = 34.18 kPa p1 2.6334

= 1.1611;

T* =

T* 287.37 K (T ) = = 247.49 K T1 1 1.1611

At section 2, f f f 0.002 L = L L = 2.1483 - a b(150 m) = 0.6483 D D max D 1-2 0.2 m M2 = 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, M2 = 0.565 p2 = 64.3 kPa, T2 = 279 K 1204

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13–67. 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 such that the absolute pressure at section 1 is 90 kPa, determine the mass flow through the duct and calculate 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

P1 A

Since friction is involved, the flow can be considered as Fanno flow. The air is drawn in isentropically. Here, p1 = 90 kPa, p0 = 101 kPa and T0 = (273 + 24) K = 297 K. p1 90 kPa With = = 0.89109, p0 101 kPa T1 = 0.96759; T0

M1 = 0.4092

2

Ff

P *A

(a )

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 = M1 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 = RT1

p1 =

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 M1 = 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 =

1 S+ 2

0 VrdV + VrV # dA 0t Lcv Lcs

p1 A - p*A - Ff = 0 + V1r1( -V1A) + V *r* (V *A)

1205

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13–67. (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 m2 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 1206

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*13–68. The 2-in.-diameter pipe has an average friction factor of f = 0.005. A nozzle on the large tank A delivers air to the pipe at section 1 with M1 = 1.8, a temperature of 95°F, and an absolute pressure of 150 psi. Determine the mass flow. Show that a normal shock forms within the pipe if L0 = 12 ft.

A

2 in.

1

B 2

L L0

SOLUTION The flow through the pipe is considered as Fanno flow since friction is involved. From the table in Appendix A, R = 1716 ft # lb>slug # °R and k = 1.40 for air. Here, lb 12 in. 2 T1 = 95°F + 460 = 555°R and p1 = a150 2 b a b = 21600 lb>ft2. Applying 1 ft in the ideal gas law, p1 = r1RT1;

21600 lb>ft2 = r1 11716 ft # lb>slug # °R21555°R2 r1 = 0.02268 slug>ft3

The velocity of the flow at the entrance is given by V1 = M1 2kRT1 = 1.821.4011716 ft # lb>slug # °R21555°R2 = 2078.46 ft>s

Then, the mass flow is

2 1 # m = r1V1A = 10.02268 slug>ft3 212078.46 ft>s2 c pa ft b d 12

= 1.0284 slug>s = 1.03 slug>s

Ans.

With M1 = 1.8, the compressible flow calculator gives f L = 0.2419 D max £

0.005 Lmax = 0.2419 § 2 a ft b 12

Lmax = 8.0629 ft

Since Lmax 6 L0 = 12 ft and M1 7 1, a normal shock forms within the pipe. Thus, the flow at the exit plane will be choked.

Ans: . m = 1.03 slug>s 1207

M13_HIBB9290_01_SE_C13_ANS.indd 1207

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13–69. The 2-in.-diameter pipe has an average friction factor of f = 0.005. The nozzle on the large tank A delivers air to the pipe at section 1 with M1 = 1.8 and a temperature of 95°F. Determine the velocity and temperature of the air at L = 5 ft if L0 = 6.5 ft.

A

2 in.

1

B 2

L L0

SOLUTION The flow through the pipe is considered as Fanno flow since friction is involved. From the table in Appendix A, R = 1716 ft # lb>slug # °R and k = 1.40 for air. Here, T1 = 95°F + 460 = 555°R. The velocity of the flow at the entrance is given by V1 = M1 2kRT1 = 1.821.4011716 ft # lb>slug # °R21555°R2 = 2078.46 ft>s

With M1 = 1.8, the compressible flow calculator gives f L = 0.2419 D max £

0.005 Lmax = 0.2419 § 2 a ft b 12

Lmax = 8.0629 ft

Since Lmax 7 L0 = 6.50 ft, the pipe will not be choked at the exit Also, T1 T* V1 V

*

= 0.7281;

T* =

= 1.5360;

V* =

555°R = 762.20°R 0.7281 2078.46 ft>s 1.5360

= 1353.18 ft>s

For the flow properties at L = 5 ft with respect to the critical location, f f f 0.005 L′ = Lmax L = 0.2419 15 ft2 = 0.09187 D D D £ 2 § a ft b 12

Again using this value, the compressible flow calculator gives M = 1.3780 T T* V V*

Ans.

= 0.8697;

T = 0.86971762.20°R2 = 662.88°R = 663°R

= 1.2851;

V = 1.2851 11353.18 ft>s2 = 1739.00 ft>s = 1739 ft>s Ans.

Ans: T = 663°R, V = 1739 ft>s 1208

M13_HIBB9290_01_SE_C13_ANS.indd 1208

16/03/17 3:20 PM

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13–70. 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 M1 7 1, and the flow is choked at the exit, section 2, when L = 5 m, determine the mass flow through the pipe. Assume an average 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, M2 = 1 at exit. As a result, Lmax = 5 m. Then f 0.0085 L = a b(5 m) = 0.425 D max 0.1 m

With M1 7 1.

M1 = 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. For M1 = 2.4677 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 = M1 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: # (m)max = 3.16 kg>s 1209

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13–71. 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 M1 6 1, and the flow is choked at the exit, section 2, when L = 5 m, determine the mass flow through the pipe. Assume an average friction factor of 0.0085 throughout the pipe.

100 mm 1

2 L

SOLUTION Here M2 = 1 at the exit. As a result Lmax = 5 m. Then, f 0.0085 L = a b(5 m) = 0.425 D max 0.1 m For M1 6 1, we get M1 = 0.61774 With T0 = (273 + 40) K = 313 K, and p0 = 450 kPa, with M1 = 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 = M1 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)

¢ ( 211.13 m>s ) p(0.05 m)2 Ans.

Ans: # m = 6.91 kg>m3 1210

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*13–72. Outside air at a temperature of 25°C is drawn into the duct and then heated along the duct at 130 kJ>kg. At section l the temperature is T = 15°C and the absolute pressure is 98 kPa. Determine the Mach number, temperature, and pressure at section 2. Neglect friction.

50 mm 1

2 2m

SOLUTION Air is considered to be inviscid and the flow is considered as steady compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. Since the air is drawn in isentropically the isentropic flow formulas can be applied at the entrance. Here, 1T0 2 1 = 25°C + 273 = 298 K , T1 = 15 °C + 273 = 288 K and p1 = 98 kPa. From the table in Appendix A, R = 286.9 J/kg # K and k = 1.40. 1T0 2 1 = T1 a1 +

k - 1 2 M1 b ; 2

298 K = 1288 K2 c 1 + a M1 = 0.4167

1.40 - 1 b 1M12 2 d 2

Using the result of M1 the compressible Rayleigh flow calculator gives p1

= 1.9307

*

p Then

p* = T* = T 0* = Also, cp =

T1 T

*

= 0.6972

1T0 2 1 T *0

= 0.5580

p* 1 b 198 kPa2 = 50.76 kPa 1p 2 = a p1 1 1.9307

T* 1 1T 2 = a b 1288 K2 = 445.01 K T1 1 0.6472

T 0* 1 1T 2 = 1298 K2 = 534.02 K 1T0 2 1 0 1 0.5580

1.401286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Then, ∆Q = cp 31T0 2 2 - 1T0 2 1 4; 1301103 2 J>kg = 11004.15 J>kg # K231T0 2 2 - 298 K4 ∆m 1T0 2 2 = 427.46 K

With

1T0 2 2 T 0*

=

427.46 K = 0.8005, the aerodynamic Rayleigh calculator gives 534.02 K

Ans.

M2 = 0.58346 = 0.583 T2 T* p2 p*

= 0.8993 ; = 1.6254 ;

T2 = 0.8993 1445.01 K2 = 400.21 K = 400 K

P2 = 1.6254 150.76 kPa2 = 82.50 kPa = 82.5 kPa

Ans. Ans.

Since M2 6 1, the flow is not choked.

Ans: M2 = 0.583 T2 = 400 K P2 = 82.5 kPa 1211

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13–73. Outside air at a temperature of 25°C is drawn into the duct and then heated along the duct at 130 kJ>kg. At section 1 the temperature is T = 15°C and the absolute pressure is 98 kPa. Determine the mass flow and the change in internal energy between sections 1 and 2.

50 mm 1

2 2m

SOLUTION Air is considered to be inviscid and the flow is considered as steady compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. Since the air is drawn in isentropically, the isentropic flow formulas can be applied at the entrance. Here, 1T0 2 1 = 25°C + 273 = 298 K, T1 = 15°C + 273 = 288 K, and p1 = 98 kPa. From the table in Appendix A, R = 286.9 J>kg # K and K = 1.40. k - 1 2 1.40 - 1 M1 b ; 298 K = 1288 K2 c 1 + a b 1M12 2 d 2 2

1T0 2 1 = T1 a1 +

M1 = 0.4167

Using the result of M1 the compressible Rayleigh flow calculator gives p1 *

p

= 1.9307

Then p* = T* = T *0 = Also, cp =

T1 T

*

= 0.6472

1T0 2 1 T *0

= 0.5580

p* 1 1p 2 = a b 198 kPa2 = 50.76 kPa p1 1 1.9307

T* 1 1T 2 = a b 1288 K2 = 445.01 K T1 1 0.6472

T *0 1 1T0 2 1 = a b 1298 K2 = 534.02 K 1T0 2 1 0.5580

1.401286.9 J>kg # k2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Then ∆Q = cp 31T0 2 2 - 1T0 2 1 4; ∆m

1301103 2 J>kg = 11004.15 J>kg # K231T0 2 2 - 298 K4 1T0 2 2 = 427.46 K

1212

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13–73. (Continued)

1T0 2 2

with

T *0

=

427.46 K = 0.8005, 534.02 K

The compressible flow calculator gives. M2 = 0.58346 T2 T*

T2 = 0.8993 1445.01 K2 = 400.21 K

= 0.8993;

Since M2 6 1, the flow is not choked. The velocity of the flow at section 1 is

V1 = M1 2kRT1 = 10.41672 21.401286.9 J>kg # K21288 K2 = 141.71 m>s

Applying the ideal gas law at section 1, p1 = r1RT1;

981103 2 N>m3 = r1 1286.9 J>kg # K21288 K2 r1 = 1.1861 kg/m3

Thus, the mass flow is # m = r1V1A = 11.1861 kg>m3 21141.71 m>s23p10.025 m2 2 4 = 0.3300 kg>s = 0.330 kg>s

Ans.

The change in internal energy from section 1 to 2 can be determined using cr =

286.9 J>kg # K R = = 717.25 J>kg # K k - 1 1.40 - 1

Then ∆u = cv ∆T = 1717.25 J>kg # K21400.21 K - 288 K2 = 80.49 1103 2 J>kg = 80.5 kJ>kg

Ans.

Ans: . m = 0.330 kg>s, ∆u = 80.5 kJ>kg 1213

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13–74. Air is drawn into the pipe at M1 = 1.85, T1 = 130°F, and an absolute pressure of p1 = 85 psi. If it exits the pipe at M2 = 1.15, determine the amount of heat per unit mass that is absorbed or released by the air.

1

2

SOLUTION Air is considered to be inviscid and the flow is considered as steady and compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. Since air is drawn in isentropically, the isentropic flow formulas can be applied at the entrance. Here, T1 = 130°F + 460 = 590°R and p1 = 85 psi. From the table in Appendix A, R = 1716 ft # lb>slug # °R and k = 1.40. M1 = 1.85, 1T0 2 1 = T1 a1 +

k - 1 2 M1 b 2

= 1590°R2 c 1 + a = 993.855°R

1.40 - 1 b 11.852 2 d 2

Also, with M1 = 1.85, the compressible Rayleigh flow calculator gives 1T0 2 1 T *0

= 0.8250;

T 0* =

T 0* 1 1T0 2 1 = a b 1993.855°R2 = 1204.62°R 1T0 2 1 0.8250

Since M2 = 1.15 7 1, the flow will not be choked at the exit. The compressible flow calculator gives 1T0 2 2 T *0

Then, cp = Finally,

= 0.9872 ;

1T0 2 2 = 0.987211204.62°R2 = 1189.21°R

1.4011716 ft # lb>slug # °R2 kR = = 6006 ft # lb>slug # °R k - 1 1.40 - 1

∆Q = cp 31T0 2 2 - 1T0 2 1 4 ∆m

= 16006 ft # lb>slug # °R231189.21°R - 993.855°R4 = 1.11331106 2 ft # lb>slug = 1.171106 2 ft # lb>slug

Ans.

The positive result indicates that the heat energy is absorbed by the air.

Ans: ∆Q = 1.171106 2 ft # lb>slug ∆m 1214

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13–75. Air is drawn into the 6-in.-diameter pipe at M1 = 1.85, T1 = 130°F, and an absolute pressure of p1 = 85 psi. If it exits the pipe at M2 = 1.15, determine the stagnation temperatures at sections 1 and 2, and the change in entropy per unit mass between these sections due to uniform heating along the pipe.

1

2

SOLUTION Air is considered to be inviscid and the flow is considered as Rayleigh flow. Since air is drawn isentropically, the isentropic flow formulas can be applied at the entrance. Here, T1 = 130°F + 460 = 590°R and p1 = 85 psi. From the table in Appendix A, R = 1716 ft # lb>slug # °R and k = 1.40. with M1 = 1.85, 1T0 2 1 = T1 a1 +

k - 1 2 M1 b 2

= 1590°R2 c 1 + a

1.40 - 1 b 11.852 2 d 2

Ans.

= 993.855°R = 994°R

Also, with M1 = 1.85, the compressible flow calculator gives p1 p* T1 T

*

1T0 2 1 T 0*

= 0.4144;

p* = a

p* 1 b p1 = a b 185 psi2 = 205.12 psi p1 0.4144

T 0* = c

T 0* 1 d 1T0 2 1 = a b 1993.855°R2 = 1204.62°R 1T0 2 1 0.8250

T* = a

= 0.5877; = 0.8250;

T* 1 bT = a b 1590°R2 = 1003.85°R T1 1 0.5877

Since M2 = 1.15 7 1, the flow will not be choked at the exit. p2 p* T2 T* 1T0 2 2 T 0*

= 0.8417;

p2 = 0.84171205.12 psi2 = 172.64 psi

= 0.9369;

T2 = 0.936911003.85°R2 = 940.46°R

= 0.9872;

1T0 2 2 = 0.987211204.62°R2 = 1189.21°R = 1189°R

Also, cp =

Ans.

1.4011716 ft # lb>slug # °R2 kR = = 6006 ft # lb>slug # °R k - 1 1.40 - 1

The change in entropy from section 1 to 2 is ∆s = s2 - s1 = cp ln

p2 T2 - R ln T1 p1

= 16006 ft # lb>slug # °R2 ln a

940.46°R b 590°R

- 11716ft # lb>slug # °R2 ln a

172.64 psi 85 psi

= 1584.40 ft # lb>slug # °R = 1584 ft # lb>slug # °R

b Ans. Ans: 1T0 2 1 = 994°R, 1T0 2 2 = 1189°R ∆s = 1584 ft # lb>slug # °R

1215

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*13–76. Nitrogen having a temperature of T1 = 270 K and absolute pressure of p1 = 330 kPa flows into the pipe at M1 = 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. M1 = 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 (p ) = = 154.83 kPa p1 1 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 = M1 2kRT1 = 0.3c 21.4 ( 296.8 J>kg # K)(270 K) d = 100.48 m>s V* =

100.48 m>s V* ( V1 ) = = 523.80 m>s V1 0.19183

Since nitrogen is drawn in to section 1 isentropically, M1 = 0.3 T1

( T0 ) 1

= 0.98232; ( T0 ) 1 =

( T0 ) 1 T1

( T1 ) =

270 K = 274.86 K 0.98232

Then T0* = cp =

T0*

( 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

( T0 ) 2 T *0

=

659.92 K = 0.83279 792.42 K

M2 = 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.

Ans: p2 = 242 kPa V2 = 310 m>s

1216

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13–77. Nitrogen having a temperature of T1 = 270 K and absolute pressure of p1 = 330 kPa flows into the pipe at M1 = 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

2

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. M1 = 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, M1 = 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

=

Ans.

659.92 K = 0.83279 792.42 K

M2 = 0.6131 6 1 (pipe is not choked) p2 p* T2 T*

= 1.5724; p2 = 1.5724(154.83 kPa) = 243.46 kPa = 0.92946; T2 = 0.92946(660.36 K) = 613.78 K

1217

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13–77. (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.

Ans: 1T0 2 1 = 275 K, 1T0 2 2 = 660 K ∆s = 943 J>kg # K 1218

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13–78. Air from a large reservoir is at a temperature of 300 K and an absolute pressure of 200 kPa. If 65 kJ>kg of heat is added to the air as it flows from 1 to 2, determine the density and pressure at section 1 if the duct is choked at section 2. The backpressure at 2 causes M1 6 1.

1

2

SOLUTION Air is considered to be inviscid and the flow is considered as steady compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. cp =

1.401286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Also, 1T0 2 1 = 300 k and 1p0 2 1 = 200 kPa. Then ∆Q = cp 31T0 2 2 - 1T0 2 1 4; ∆m

651103 2 J>kg = 11004.15 J>kg # K231T0 2 2 - 300 K4

1T0 2 2 = 364.73 K

It is required that the duct be choked at section 2, M2 = 1. The compressible flow calculator gives 1T0 2 2 T 0* = 1T0 2 2 = 364.73 K = 1; T 0* 1T0 2 1

300 K = = 0.8225 and choose M1 6 1 (requirement), the compressible 364.73 K T 0* flow calculator gives M1 = 0.6033

With

Since the air is drawn isentropically into section 1, the isentropic flow formulas can be applied at section 1. 1T0 2 1 = T1 a1 +

k - 1 2 M1 b ; 2

300 K = T1 c 1 + a

T1 = 279.64 K

1.40

k

1p0 2 1 = p1 a1 +

k - 1 2 k-1 M1 b ; 2

1.40 - 1 b 10.60332 2 d 2

200 kPa = p1 c 1 + a

1.40 - 1 1.40 - 1 b 10.60332 2 d 2

p1 = 156.39 kPa = 156 kPa

Ans.

Thus, the density of the air can be determined by applying the ideal gas law at section 1. p1 = r1RT1;

156.391103 2 N>m2 = r1 1286.9 J>kg # K21279.64 K2 r1 = 1.949 kg>m3 = 1.95 kg>m3

Ans.

Ans: p1 = 156 kPa, r1 = 1.95 kg>m3 1219

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13–79. Air from a large reservoir is at a temperature of 300 K and an absolute pressure of 200 kPa. If 65 kJ>kg of heat is added to the air as it flows from 1 to 2, determine the temperature and pressure at 2 if it is choked there. The backpressure at 2 causes M1 6 1.

1

2

SOLUTION Air is considered to be inviscid and the flow is considered as steady compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. cp =

1.401286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Here, 1T0 2 1 = 300 K and 1p0 2 1 = 200 kPa. Then, ∆Q = cp 31T0 2 2 - 1T0 2 1 4; ∆m

651103 2 J>kg = 11004.15 J>kg # K231T0 2 2 - 300 K4 1T0 2 2 = 364.73 K

It is required that the duct choked at section 2, M2 = 1. The compressible flow calculator gives. 1T0 2 2 T 0*

With

1T0 2 1 T 0*

=

T 0* = 1T0 2 2 = 364.73 K

= 1;

300 K = 0.8225 and choosing M1 6 1 (requirement), the 364.73 K

compressible flow calculator gives M1 = 0.6033

T1 T

*

= 0.9200

p1 p*

= 1.5898

Since the air is drawn isentropically into section 1, the isentropic flow formulas can be applied at section 1. 1T0 2 1 = T1 a1 +

k - 1 2 M1 b ; 2

300 K = T1 c 1 + a

T1 = 279.64 K

1.40 - 1 b 10.60332 2 d 2 1.40

k

1p0 2 1 = p1 a1 +

1.40 - 1 k - 1 2 k-1 1.40 - 1 M1 b ; 200 kPa = p1 c 1 + a b 10.60332 2 d 2 2

p1 = 156.39 kPa

Then using M1 = 0.6033, the compressible flow calculator gives T1 T

*

p1 *

p

= 0.9200;

T* = a

= 1.5898; p* = a

T* 1 b 1T1 2 = a b 1279.64 K2 = 303.94 K = 304 K T1 0.9200

Ans.

*

p 1 b 1p1 2 = a b 1156.39 kPa2 = 98.37 kPa = 98.4 kPa p1 1.5898

Ans.

Ans: T * = 304 K, p* = 98.4 kPa 1220

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*13–80. Air having a temperature of 280 K and pressure of 320 kPa flows from a large reservoir into the duct. As it flows, 122 kJ>kg of heat is added. Determine the greatest velocity it can have at section 1. The backpressure at 2 causes M1 6 1.

200 mm 1

2

SOLUTION Air is considered to be inviscid and the flow is considered as steady compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. cp =

1.401286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Here, 1T0 2 1 = 280 K and 1p0 2 1 = 320 kPa. Then,

∆Q = cp 31T0 2 2 - 1T0 2 1 4; 1221103 2 J>kg = 11004.15 J>kg # K231T0 2 2 - 280 K4 ∆m 1T0 2 2 = 401.50 K

Since it is required to have the greatest velocity at the entrance (section 1) the duct has to be choked at exit (section 2) which means M2 = 1. The compressible flow calculator gives. 1T0 2 2 T 0* = 1T0 2 2 = 401.50 K = 1; T 0* With

1T0 2 1 T 0*

=

280 K = 0.6974 and choose M1 6 1 (requirement), the compressible 401.50 K

flow calculator gives M1 = 0.5041 Since the air is drawn isentropically into section 1, the isentropic flow formulas can be applied at section 1. 1T0 2 1 = T1 a1 +

k - 1 2 M1 b ; 2

280 K = T1 c 1 + a

T1 = 266.46 K

1.40 - 1 b 10.50412 2 d 2

Thus, the velocity of the flow at section 1 can be determined from V1 = M1 2kRT1 = 10.50412 21.401286.9 J>kg # K21266.46 K2 = 164.93 m>s = 165 m>s

Ans.

Ans: V1 = 165 m>s 1221

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13–81. Air having an absolute temperature of 450 K and pressure of 600 kPa flows from a large reservoir into a duct. As it flows, 150 kJ>kg of heat is added. Determine the mass flow if the backpressure causes M1 7 1 and the flow chokes at section 2.

200 mm 1

2

SOLUTION Air is considered to be inviscid and the flow is considered as steady compressible. Since heat energy is being added, the flow can be considered as Rayleigh flow. From the table in Appendix A, R = 286.9 J>kg # K and k = 1.40 for air. cp =

1.401286.9 J>kg # K2 kR = = 1004.15 J>kg # K k - 1 1.40 - 1

Here, 1T0 2 1 = 450 K and 1p0 2 1 = 600 kPa. Then,

∆Q = cp 31T0 2 2 - 1T0 2 1 4; 1501103 2 J>kg = 11004.15 J>kg # K231T0 2 2 - 450 K4 ∆m 1T0 2 2 = 599.38 K

Since it is required that the duct choked at section 2, M2 = 1. The compressible flow calculator gives 1T0 2 2 T 0*

With

1T0 2 1 T 0*

=

= 1;

T 0* = 1T0 2 2 = 599.38 K

450 K = 0.7508 and choose M1 7 1 (requirement), the 599.380 K

compressible flow calculator gives M1 = 2.2315 Since the air is drawn isentropically into section 1, the isentropic flow formulas can be applied at section 1. 1T0 2 1 = T1 a1 +

k - 1 2 M1 b ; 2

450 K = T1 c 1 + a T1 = 225.47 K

1.40

k

1p0 2 1 = p1 a1 +

k - 1 2 k-1 M1 b ; 2

1.40 - 1 b 12.23152 2 d 2

600 kPa = p1 c 1 + a p1 = 53.42 kPa

1.40 - 1 1.40 - 1 b 12.23152 2 d 2

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13–81. (Continued)

Then, the velocity of the flow at section 1 is given by V1 = M1 2kRT1 = 12.23152 21.401286.9 J>kg # K21225.47 K2 = 671.52 m>s

Applying the ideal gas law, p1 = r1RT1;

53.421103 2N>m2 = r1 1286.9 J>kg # K21225.47 K2 r1 = 0.8258 kg>m3

Finally, the mass flow is # m = r1V1A = 10.8258 kg>m3 21671.52 m>s23p10.1 m2 2 4 = 17.421 kg>s = 17.4 kg>s

Ans.

Ans: . m = 17.4 kg>s 1223

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13–82. The bottle tank contains 0.13 m3 of oxygen at an absolute pressure of 900 kPa and temperature of 20°C. If the exit nozzle has a diameter of 15 mm, and when it is opened, determine the time needed to drop the absolute pressure in the tank to 300 kPa. 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

( p0)2 = pa1 +

k - 1 2 k-1 M b 2

1.4

1.4 - 1 1.4 - 1 300 kPa = (101.3 kPa)c 1 + a bM2 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

k - 1 2 k-1 p0 = p a1 + 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

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13–82. (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: t = 4.28 s 1225

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13–83. 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

4 MPa = 0.1 MPac 1 + a

M = 3.0448

1.38

1.38 - 1 1.38 - 1 bM2 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. 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 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 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

Ans. Ans: # m = 183 kg>s

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*13–84. 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 backpressure is a vacuum, p = 0. k

k - 1 2 k-1 p0 = p a1 + M b 2 4 MPa = 0 c 1 + a M = ∞

1.38

1.38 - 1 1.38 b M2 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. 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

k - 1 2 (k - 1) p0 = p a1 + 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: # m = 183 kg>s 1227

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13–85. The jet engine has a converging–diverging exhaust nozzle. The absolute pressure at the inlet to the nozzle is 900 kPa, and the temperature of the fuel–air mixture is 1850 K. If the mixture flows at 125 m>s into the nozzle, determine the diameters of the throat and exit plane for isentropic flow. The inlet has a diameter of 600 mm. The outside absolute pressure is 102 kPa. The fuel mixture has k = 1.4 and R = 265 J>kg # K.

SOLUTION The fuel mixture is considered compressible and the flow is steady relative to the rocket. The Mach number of the inlet flow is given by Min =

Vin

=

2kRTin

125 m>s

21.401265 J>kg # K211850 K2

= 0.1509

Since the process is isentropic, the isentropic flow formulas can be applied with Min = 0.1509 at the inlet, k

k-1 k - 1 Min2 b p0 = pin a1 + 2

1.40

1.40 - 1 1.40 - 1 = 1900 kPa2 c 1 + a b 10.15092 2 d 2

= 914.42 kPa

k+1

Ain *

A

1 ± Min

=

1 +

21k - 12 k - 1 Min2 2 ≤ k + 1 2 1.40 + 1

p10.3 m2 2 1 = ≥ p * 2 0.1509 1d 2 4

1 + a

211.40 - 12 1.40 - 1 b 10.15092 2 2 ¥ 1.40 + 1 2

d * = 0.3043 m

At the exit, k

p0 = pout a1 +

k-1 k - 1 Mout2 b 2

914.42 kPa = 1102 kPa2 c 1 + a

1.40

1.40 - 1 1.40 - 1 b Mout2 d 2

Mout = 2.0873 7 1(supersonic)

k+1

Aout A*

=

1 ± Mout

1 +

21k - 12 k - 1 Mout2 2 ≤ k + 1 2 1.40 + 1

p 2 d out 4 p 10.3043 m2 2 4

=

1 D 2.0873

1 + a

211.40 - 12 1.40 - 1 b 12.08732 2 2 T 1.40 + 1 2

Ans.

dout = 0.4102 m = 410 mm

Since subsonic flow in convergent portion transits to supersonic in divergent portion, the flow is choked at the throat. Thus dt = d * = 0.3043 m = 304 mm

Ans.

Ans: dout = 410 mm, dt = 304 mm

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13–86. The jet engine has a converging–diverging exhaust nozzle. The absolute pressure at the inlet to the nozzle is 900 kPa, and the temperature of the fuel mixture is 1850 K. If the mixture flows at 125 m>s into the nozzle and exits with isentropic flow, determine the diameters of the throat and exit plane, and the mass flow through the nozzle. The inlet has a diameter of 600 mm. The outside absolute pressure is 50 kPa. The fuel mixture has k = 1.4 and R = 265 J>kg # K.

SOLUTION The fuel mixture is considered compressible and the flow is steady relative to the rocket. The Mach number of the inlet flow is given by Min =

Vin 2kRTin

=

125 m>s

21.401265 J>kg # K211850 K2

= 0.1509

Since the process are isentropic, the isentropic flow formulas can be applied. With Min = 0.1509 at the inlet k

p0 = pin a1 +

k-1 k - 1 Min2 b 2

= 1900 kPa2 c 1 + a

= 914.42 kPa

1.40

1.40 - 1 1.40 - 1 b 10.15092 2 d 2

k+1

21k - 12 k - 1 1 + Min2 Ain 1 2 = ± ≤ Min k + 1 A* 2 1.40 + 1

p10.3 m2 2 1 = ≥ p * 2 0.1509 1d 2 4

1 + a

211.40 - 12 1.40 - 1 b 10.15092 2 2 ¥ 1.40 + 1 2

d * = 0.3043 m

At the exit, k

p0 = pout a1 +

k - 1 2 k-1 Mout b 2

1.40

1.40 - 1 1.40 - 1 914.42 kPa = 150 kPa2 c 1 + a b M2out d 2

Mout = 2.5438 7 1 (supersonic)

k+1

Aout A*

=

1 ± Mout

1 +

21k - 12 k - 1 2 Mout 2 ≤ k + 1 2 1.40 + 1

p 2 d out 4 p 10.3043 m2 2 4

=

1 ≥ 2.5438

1 + a

211.40 - 12 1.40 - 1 b 12.54382 2 2 ¥ 1.40 + 1 2

Ans.

dout = 0.5043 m = 504 mm

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13–86. (Continued)

Since the subsonic flow in convergent portion transits to supersonic in divergent portion, the flow choked at the throat. Thus, dt = d * = 0.3043 m = 304 mm

Ans.

Applying ideal gas law at the inlet plane, pin = rinRTin;

9001103 2 N>m2 = rin 1265 J>kg # K211850 K2 rin = 1.8358 kg>m3

Then, the mass flow is given by # m = rin Vin Ain = 11.8358 kg>m3 21125 m>s23p10.3 m2 2 4 = 64.88 kg>s = 64.9 kg>s

Ans.

Ans: dout = 504 mm, dt = 304 mm, . m = 64.9 kg>s 1230

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13–87. 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 isentropically with supersonic flow, determine the velocity of the exhaust. Take k = 1.4 and R = 249 J>kg # K. Assume isentropic flow.

300 mm dt

de

SOLUTION The mixture is considered to be compressible. The flow is steady. The Mach number of the inlet flow is Min =

Vin 2kRTin

At the inlet plane,

=

250 m>s

21.40 ( 249 J>kg # K ) (800 K)

= 0.4734

k - 1 Min2 b 2

T0 = Tin a1 +

T0 = (800 K) c 1 + a

1.4 - 1 b(0.4734)2 d = 835.86 K 2 k

k-1 k - 1 p0 = pin a1 + Min2 b 2

1.4

1.4 - 1 1.4 - 1 p0 = (300 kPa) c 1 + a b(0.4734)2 d 2

= 349.76 kPa

At the exit plane, k

p0 = pe a1 +

k - 1 2 k-1 Me b 2

349.76 kPa = (101.3 kPa)c 1 + a Me = 1.4574

T0 = Te a1 +

k - 1 2 Me b 2

835.86 K = Te c 1 + a Te = 586.64 K

1.4

1.4 - 1 1.4 - 1 bMe2 d 2

1.4 - 1 b(1.4574)2 d 2

The velocity of the flow at the exit plane is Ve = Me 2kRTe = (1.4574) 21.40 ( 249 J>kg # K ) (586.64 K) = 659.08 m>s = 659 m>s

Ans.

Ans: Ve = 659 m>s 1231

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*13–88. 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 Min =

Vin

=

2kRTin

p0 = pin a1 +

250 m>s

21.40 ( 249 J>kg # K ) (800 K)

= 0.4734

k

k-1 k - 1 Min2 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 ± Min

1 +

2(k - 1) k - 1 Min2 2 ≤ k + 1 2

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 dt 4 2

Ans.

dt = 0.2541 m = 254 mm At the exit plane, k

p0 = pe a1 +

k - 1 2 k-1 Me b 2

1.4

349.76 kPa = (101.3 kPa)c 1 + a Me = 1.4574

1.4 - 1 1.4 - 1 bMe2 d 2

k+1

Ae A*

=

1 ± Me

1 +

p 2 d 4 e p (0.2541 m)2 4

k - 1 2 2(k - 1) Me 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.

de = 0.2723 m = 272 mm

Ans: dt = 254 mm de = 272 mm

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13–89. The air in the large tank to the left of the nozzle is at a temperature of 80°C and an absolute pressure of 630 kPa. Determine the mass flow from the nozzle if the backpressure is 350 kPa.

60 mm 25 mm

SOLUTION The flow can be considered as steady and compressible. The stagnation temperature and pressure are T0 = 80°C + 273 = 353 K and p0 = 630 kPa. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K. Assuming that the nozzle chokes at the throat. Then, with

Ae *

=

p10.03 m2 2

A p10.0125 m2 2 the compressible flow calculator gives

= 5.76 and choose M 6 1 (subsonic),

Me = 0.1011 Also, calculated, assuming isentropic flow, is p3 = 0.9929; p3 = 0.9929 1630 kPa2 = 625.51 kPa p0

Since the backpressure pb = 350 kPa is less than p3, the nozzle indeed chokes. Thus, Mt = 1 at the throat. T0 = T * a1 +

k - 1 2 1.40 - 1 Mt b; 353 K = T * c 1 + a b 112 2 d 2 2

T * = 294.17 K

k

1.40

1.40 - 1 k - 1 2 k-1 1.40 - 1 p0 = p a1 + Mt b ; 630 kPa = p* c 1 + a b 112 2 d 2 2

*

Applying the ideal gas law, p* = r*RT *;

p* = 332.82 kPa

332.821103 2 N>m2 = r* 1286.9 J>kg # K2 1294.17 K2 r* = 3.9435 kg>m3

The velocity of the flow at the throat is V * = Mt 2kRT * = 112 21.401286.9 J>kg # K21294.17 K2 = 343.74 m>s

Finally, the mass flow is # m = r*V *A* = 13.9435 kg>m3 21343.74 m>s23p10.125 m2 2 4 = 0.6654 kg>s = 0.665 kg>s

Ans.

Ans: . m = 0.665 kg>s 1233

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13–90. The air in the large tank to the left of the nozzle is at a temperature of 80°C and an absolute pressure of 630 kPa. Determine the two values of the backpressure that will choke the nozzle yet produce isentropic flow. Also, what is the maximum exit velocity of the isentropic flow?

60 mm 25 mm

SOLUTION The flow can be considered as steady and compressible. The stagnation temperature and pressure are T0 = 80°C + 273 = 353 K and p0 = 630 kPa. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K for air. Here, it is required that the nozzle chokes at the throat. Then with

Ae A*

=

p10.03 m2 2 p10.0125 m2 2

= 5.76, for isentropic

flow the compressible flow calculator gives two values of Me, which are 1Me 2 sub = 0.1011 6 1 (subsonic)

1Me 2 sup = 3.3245 7 1 (supersonic)

Also calculated for subsonic flow, 1pe 2 sub p0

= 0.9929;

1pe 2 sub = 1630 kPa210.99292 = 625.51 kPa = 626 kPa Ans.

And for supersonic flow, 1pe 2 sup p0

= 0.01687; 1Te 2 sup T0

1pe 2 sup = 1630 kPa210.016872 = 10.63 kPa = 10.6 kPa Ans.

= 0.3115;

1Te 2 sup = 1353 k210.31152 = 109.95 K

The maximum velocity of the isentropic flow (supersonic) is Vmax = 1Me 2 sup 2kR1Te 2 sup

= 13.32452 21.401286.9 J/kg # K21109.95 K2

= 698.65 m>s = 699 m>s

Ans.

Ans: 1pe 2 sub = 626 kPa, 1pe 2 sup = 10.6 kPa Vmax = 699 m>s 1234

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13–91. A jet plane creates a shock wave that forms in front of the plane in air having a temperature of 10°C and an absolute pressure of 60 kPa. If the plane travels at M = 2.3, determine the pressure and temperature just behind the shock wave.

SOLUTION The air is considered compressible and the flow is steady relative to the plane. The static temperature and pressure of the air just before the shock are T1 = 10°C + 273 = 283 K and p1 = 60 kPa. From the table in Appendix A, k = 1.40 for air. The static pressure after the shock can be determined with M1 = 2.3. With the aid of compressible flow calculator, p2 = 6.005; p1

p2 = 6.005 160 kPa2 = 360.3 kPa = 360 kPa

Ans.

For the Mach number just after the shock wave. The aerodynamic calculator gives M2 = 0.5344 Using this result, with the aid of compressible flow calculator T2 = 1.9468; T1

T2 = 1.9468 1283 K2 = 550.94 K = 551 K

Ans.

Ans: p2 = 360 kPa, T2 = 551 K 1235

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*13–92. The jet plane travels at M = 2.5 in still air at an altitude of 25 000 ft. If a shock forms at the air inlet of the engine, determine the stagnation pressure within the engine just before the shock and the stagnation pressure a short distance within the chamber.

SOLUTION The air is considered compressible and the flow is steady relative to the plane. In the reference frame of the plane, the air is moving, so that the stagnation pressure will be higher than the static pressure. The static pressure just before the shock is the air pressure at an altitude of 25 000 ft. From the table in Appendix A, this value is p1 = 786.3 lb>ft2 For air, k = 1.40 (table in Appendix A). Here, M1 = 2.5. The stagnation pressure is found using Table B–1 or a compressible flow calculator: p1 = 0.05853 1p0 2 1

1p0 2 1 =

1 1786.3 lb>ft2 2 = 13434.7 lb>ft2 0.05853

= 13.4 1103 2 lb>ft2

Ans.

With the aid of compressible flow calculator the stagnation pressure just after the shock is given by 1p0 2 2 1p0 2 1

= 0.4990;

1p0 2 2 = 0.4990 113434.7 lb>ft2 2

= 6704.1 lb>ft2 = 6.701103 2 lb>ft2

Ans.

Since no shock wave forms in the engine chamber, the stagnation pressure is constant throughout the chamber in front of the shock.

1236

M13_HIBB9290_01_SE_C13_ANS.indd 1236

Ans: (P0)2 = 6.70 1103 2 lb>ft2 (P0)1 = 13.4 1103 2 lb>ft2

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13–93. A normal shock is produced at the nose of a jet plane flying with M = 2.2. If the air is at a temperature of 20°F and an absolute pressure of 6 lb >in2, determine the pressure and the stagnation pressure just in front of the shock.

SOLUTION The air is considered compressible and the flow is steady relative to the plane. Then, the static temperature and pressure just before the shock relative to the plane are T1 = 20°F + 460 = 480°R and p1 = 6 psi. From the table in Appendix A, k = 1.40 for air. The stagnation pressure before the shock can be determined since the flow here can be considered isentropic. k

1p0 2 1 = p1 a1 +

k-1 k - 1 M1 2 b 2

= 16 psi2 c 1 + a = 64.16 psi

1.40

1.40 - 1 1.40 - 1 b 12.22 2 d 2

The Mach number just in front of the shock wave can be determined M1 = 2.2. The compressible flow calculator gives, M2 = 0.5471 Also, p2 = 5.4800; p1 1p0 2 2 1p0 2 1

= 0.6281;

p2 = 5.4800 16 psi2 = 32.88 psi = 32.9 psi

1p0 2 2 = 0.6281164.16 psi2 = 40.30 psi = 40.3 psi

1237

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Ans. Ans.

Ans: p2 = 32.9 psi, 1p0 2 2 = 40.3 psi

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13–94. A normal shock is produced at the nose of a jet plane flying with M = 2.2. If the air is at a temperature of 20°F and an absolute pressure of 6 lb>in2, determine the velocity of the air relative to the plane and its temperature just in the front the shock.

SOLUTION The air is considered compressible and the flow is steady relative to the plane. Then the static temperature just before the shock relative to the plane is T1 = 20°F + 460 = 480°R. From the table in Appendix A, k = 1.40 and R = 1716 ft # lb>slug # °R. The Mach number just in front of the shock wave can be determined with M1 = 2.2. The aerodynamic calculator gives M2 = 0.5471 Using this result, with the aid of compressible flow calculator, T2 = 1.8569; T1

T2 = 1.8569 1480°R2 = 891.29°R = 891°R

Ans.

The velocity of the flow in front of the shock can be determined using V2 = M2 2kRT2 = 10.54712 21.4011716 ft # lb>slug # °R21891.29°R2 = 800.51 ft>s = 801 ft>s

Ans.

Ans: T2 = 891°R, V2 = 801 ft>s 1238

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13–95. A standing shock occurs in the pipe. At section 1 upstream the air has an absolute pressure of p1 = 80 kPa, temperature T1 = 75°C, and velocity V1 = 700 m>s. Determine the pressure, temperature, and velocity of the air at section 2. Also, what is the Mach number at section 1 and 2?

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.

M2 = 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; T2 = 1.5864(273 + 75°C) = 552.09 K = 552 K Ans. T1 p2 = 3.9232; p1

p2 = 3.9232(80 kPa) = 313.85 kPa = 0.314 MPa

V2 = 0.4044; V1

V2 = 0.4044 ( 700 m>s ) = 283.07 m>s = 283 m>s Ans.

Ans.

Ans: M1 = 1.87, M2 = 0.601 T2 = 552 K, p2 = 314 kPa, V2 = 283 m>s 1239

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*13–96. A Laval nozzle is connected to the large tank. If the temperature of the air in the tank is 375 K and the absolute pressure is 480 kPa, determine the range of backpressures that will cause expansion shock waves to form at the exit of the nozzle.

40 mm

75 mm

SOLUTION The flow can be considered as steady and compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 375 K and p0 = 480 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. From table in Appendix A, k = 1.40 for air. p 175 mm2 2 4 = = 3.515625 p A* 140 mm2 2 4

Ae

With the aid of compressible flow calculator and choose M 7 1 (supersonic) Me = 2.8046 Using this result, p4 = 0.03659; p0

p4 = 0.03659 1480 kPa2 = 17.56 kPa = 17.6 kPa

Thus, for the expansion shock waves to form,

Ans.

pb 6 17.6 kPa

Ans: pb 6 17.6 kPa 1240

M13_HIBB9290_01_SE_C13_ANS.indd 1240

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13–97. A Laval nozzle is connected to the large tank. If the temperature of the air in the tank is 375 K and the absolute pressure is 480 kPa, determine the range of backpressures that will cause oblique compression shock waves to form at the exit of the nozzle.

40 mm

75 mm

SOLUTION The flow can be considered as steady and compressible. The stagnation temperature and pressure are of those in the tank; ie., T0 = 375 K and p0 = 480 kPa. For an oblique compression shock wave to form at the exit plane, the backpressure must be in between p4 and p6, which are the backpressure at which isentropic supersonic flow occurs and the backpressure at which the standing normal shock wave form at the exit plane. From the table in Appendix A, K = 1.40 for air. For isentropic flow, p 175 mm2 2 4 = = 3.515625 p A* 140 mm2 2 4

Ae

With the aid of compressible flow calculator and choose M 7 1 (supersonic) Me = 2.8046 Using this result p4 = 0.03659; p0

p4 = 0.03659 1480 kPa2 = 17.56 kPa = 17.6 kPa

When the standing normal shock wave formed at the exit plane, the pressure behind and in front of the normal shock are p4 and p6, respectively and the corresponding Mach number is 1Me 2 1 = 2.8046. The compressible flow calculator gives p6 = 9.0104; p4

p6 = 9.0104 117.56 kPa2 = 158.24 kPa = 158 kPa

Thus, an oblique compression shock formed if

Ans.

17.6 kPa 6 pb 6 158 kPa

Ans: 17.6 kPa 6 pb 6 158 kPa 1241

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13–98. A 200-mm-diameter pipe contains air at a temperature of 10°C and an absolute pressure of 100 kPa. If a shock forms 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.

M 5 1.3

0.4 m 0.6 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 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. M2 = 0.4772 Using the results of M1 and M2 to apply the equation or using the table, V2 = 0.2614; V1

V2 = 0.2614 ( 1000 m>s ) = 261.39 m>s = 261 m>s

Ans.

Ans: V2 = 261 m>s 1242

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13–99. The jet is flying at M = 1.3, where the absolute air pressure is 50 kPa. If a shock forms at the inlet of the engine, determine the Mach number of the air just within the engine where the diameter is 0.6 m. Also, what is the pressure and the stagnation pressure in this region? Assume isentropic flow within the engine.

M 5 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 M1 = 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

( p0 ) 1

( p0 ) 1 =

= 0.3609;

50 kPa = 138.54 kPa 0.3609

The Mach number after the shock can be determined by applying the equation or using the table. M2 = 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 M2 = 0.7860, the equation or table gives A2 A*

= 1.0443

Here, the area of the exit plane is A3. Then A3 A*

= a

p(0.3 m)2 A3 b a b = 1.0443 J R A* A2 p(0.2 m)2 A2

= 2.3497

Applying the equation or using the table and choosing M3 6 1 (the flow must be subsonic since the nozzle does not choke), Ans.

M3 = 0.2561 Using this result to apply the equation at the exit plane, p3

( p0 ) 2

= 0.9554;

p3 = 0.9554(135.68 kPa) = 129.63 kPa = 130 kPa

Ans.

Ans: p0 = 136 kPa M = 0.256 p = 130 kPa 1243

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*13–100. A large tank supplies air at a temperature of 275 K and an absolute pressure of 560 kPa to the nozzle. Determine the backpressure that will cause the nozzle to choke. What range of backpressures will cause expansion waves to form at the exit?

25 mm

55 mm

SOLUTION The flow can be considered as steady and compressible. The stagnation temperature and pressure are of those in the tank; i.e., To = 275 K and p0 = 560 kPa. For the nozzle to choke, pb … p3 and for the expansion wave to form at the exit plane pb 6 p4 where p3 and p4 are the backpressures that cause isentropic subsonic and supersonic flow respectively along the divergent portion of the nozzle. From the table in Appendix A, k = 1.40 for air. For isentropic flow, p 155 mm2 2 4 = = 4.84 p A* 125 mm2 2 4

Ae

with the aid of compressible flow calculator,

Me = 0.1206 6 1 (subsonic) Me = 3.1405 7 1 (supersonic) For the subsonic flow, p3 = 0.9899; p0 For the supersonic flow, p4 = 0.02208; p0 Thus, to choke the nozzle,

p3 = 0.9899 1560 kPa2 = 554.33 kPa p4 = 0.02208 1560 kPa2 = 12.37 kPa Ans.

pb … 554 kPa To form expansion shock waves just after the exit plane,

Ans.

pb … 12.4 kPa

Ans: to choke the nozzle, pb … 554 kPa To form expansion shock waves pb … 12.4 kPa 1244

M13_HIBB9290_01_SE_C13_ANS.indd 1244

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13–101. A large tank supplies air at a temperature of 275 K and an absolute pressure of 560 kPa to the nozzle. Determine the range of backpressures that will cause a standing shock to form within the nozzle.

25 mm

55 mm

SOLUTION The flow can be considered as steady and compressible. The stagnation temperature and pressure are of those in the tank; i.e., T0 = 275 K and p0 = 560 kPa. For the standing normal shock wave to form within the divergent portion, the backpressure must be in between p6 and p3 where p6 is the backpressure at which the standing normal shock formed at exit plane and p3 is the backpressure that causes isentropic subsonic flow along the divergent portion of the nozzle. From the table in Appendix A, k = 1.40 for air. For isentropic flow, p 155 mm2 2 4 = = 4.84 p A* 125 mm2 2 4

Ae

With the aid of compressible flow calculator,

Me = 0.1206 6 1 (subsonic) Me = 3.1405 7 1 (supersonic) For subsonic flow, p3 = 0.9899; p0 For the supersonic flow,

p3 = 0.9899 1560 kPa2 = 554.33 kPa = 554 kPa

p4 = 0.02208; p0

p4 = 0.02208 1560 kPa2 = 12.37 kPa

When a standing normal shock wave formed at the exit plane, the pressure behind and in front of the shock are p4 and p6, respectively and the corresponding Mach number is 1Me 2 1 = 3.1405. The compressible flow calculator gives p6 = 11.3399; p4

p6 = 11.3399 112.37 kPa2 = 140.23 kPa

Thus, for the standing normal shock wave to from within the divergent portion of the nozzle Ans.

140 kPa 6 pb 6 554 kPa

Ans: 140 kPa 6 pb 6 554 kPa 1245

M13_HIBB9290_01_SE_C13_ANS.indd 1245

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13–102. 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 MA = 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 MA = 3.0 into the isentropic flow tables, pA

( p0 ) A

( p0 ) A =

= 0.02722;

( p0 ) A pA

( pA ) =

15 kPa = 551.07 kPa 0.02722

Next, we will consider the flow across the shock. Enter M1 = 3.0 into the normal shock tables, M2 = 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 M2 = 0.47519. Enter this value into the isentropic flow tables. Interpolation gives AC

( A* ) ′

= 1.3905

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 MB 6 1 and performing the interpolation, MB = 0.1890

pB

( p0 ) B

= 0.97539

Then pB = 0.97539 (180.94 kPa) = 176.49 kPa = 176 kPa

Ans.

Ans: pB = 176 kPa 1246

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13–103. 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, p(0.025 m)2 A1 = = 6.25. Enter this value into isentropic flow tables, and after A* p(0.01 m)2 interpolation, select M1 7 1, M1 = 3.4114 Next consider the flow across the normal shock. Enter M1 = 3.4109 into normal shock tables. Interpolation gives M2 = 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 M2 = 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 Me 6 1. Performing the interpolation, Me = 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: pe = 40.7 kPa 1247

M13_HIBB9290_01_SE_C13_ANS.indd 1247

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*13–104. If the absolute pressure of air within the large tank A is 80 psi, determine the range of backpressures at B that will cause an oblique compression shock wave to form at the exit plane.

4 in. 2.5 in.

A

B

SOLUTION The flow can be considered as steady compressible. The stagnation pressure is of that in the tank; i.e., p0 = 80 psi. For an oblique compression shock wave to form at the exit plane, the backpressure must be in between p4 and p6, which are the backpressure at which the isentropic supersonic flow occurs and the backpressure at which the standing normal shock wave formed at the exit plane. From the table in Appendix A, k = 1.40 for air. For isentropic flow, p 14 in2 2 4 = = 2.56 p A* 12.5 in2 2 4

Ae

With the aid of compressible flow calculator and choose M 7 1 (supersonic) Me = 2.4683 Using this result, p4 = 0.06149; p0

p4 = 0.06149 180 psi2 = 4.9190 psi

When the standing normal shock wave formed at the exit plane the pressure behind and in front of the normal shock are p4 and p6, respectively and the corresponding Mach number is 1Me 2 1 = 2.4683. The compressible flow calculator, p6 = 6.9413; p4

p6 = 6.941314.9190 psi2 = 34.14 psi

Thus, an oblique compression shock is formed if Ans.

4.92 psi 6 pb 6 34.1 psi

Ans: 4.92 psi 6 pb 6 34.1 psi 1248

M13_HIBB9290_01_SE_C13_ANS.indd 1248

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13–105. If the absolute pressure of air within the large tank A is 80 psi, determine the range of backpressures at B that will cause a standing normal shock wave to form within the divergent section of the nozzle.

4 in. 2.5 in.

A

B

SOLUTION The flow can be considered as steady and compressible. The stagnation pressure is of that in the tank; i.e., p0 = 80 psi. For the standing normal shock wave to form within the divergent portion, the back pressure must be in between p6 and p3 where p6 is the backpressure at which the standing normal shock formed at the exit plane and p3 is the backpressure that cause isentropic subsonic flow along the divergent portion of the nozzle. From the table in Appendix A, k = 1.40 for air. For isentropic flow, with p 14 in2 2 4 = = 2.56 p A* 12.5 in2 2 4

Ae

With the aid of compressible flow calculator

Me = 0.2335 6 1 (subsonic) Me = 2.4683 7 1 (supersonic) For subsonic flow, p3 = 0.9627; p0

p3 = 0.9627180 psi2 = 77.02 psi

For supersonic flow, p4 = 0.06149; p0

p4 = 0.06149180 psi2 = 4.9190 psi

When a standing normal shock wave formed at the exit plane, the pressure behind and in front of the shock are p4 and p6 respectively and the corresponding Mach number is 1Me 2 1 = 2.4683. The compressible flow calculator gives p6 = 6.9413; p4

p6 = 6.941314.9190 psi2 = 34.14 psi

Thus, for the standing normal shock wave to form within the divergent portion of the nozzle, Ans.

34.1 psi 6 pb 6 77.0 psi

Ans: 34.1 psi 6 pb 6 77.0 psi 1249

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13–106. If the absolute pressure within the large tank A is 80 psi, determine the range of backpressures at B that will cause expansion shock waves to form at the exit plane.

4 in. 2.5 in.

A

B

SOLUTION The flow can be considered as steady and compressible. The stagnation pressure is of that in the tank; i.e., p0 = 80 psi. For an expansion shock wave to form at the exit plane, the backpressure pb 6 p4 where p4 is the backpressure at which isentropic supersonic flow occurs. From the table in Appendix A, k = 1.40 for air. p 14 in2 2 4 = = 2.56 p A* 12.5 in2 2 4

Ae

With the aid of compressible flow calculator and choose M 7 1 (supersonic), Me = 2.4683 Using this result, p4 = 0.06149; p0

p4 = 0.061491802 = 4.9190 psi

Thus, for expansion shock waves to form, Ans.

pb 6 4.92 psi

Ans: pb 6 4.92 psi 1250

M13_HIBB9290_01_SE_C13_ANS.indd 1250

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13–107. 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 ft

0.150 ft

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 backpressure pb must be between p6 and p3, which are the backpressures 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 ft)2 4 A = = 1.7778 p A* (0.15 ft)2 4 We obtain, M1 = 0.3500 (subsonic) M2 = 2.0618 (supersonic) Using the result of M1 = 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 M1 = 2.0618, p4 = 0.1161; p0

p4 = 0.1161(70 psi) = 8.1248 psi

Using the result of M1 = 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 1251

M13_HIBB9290_01_SE_C13_ANS.indd 1251

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*13–108. 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 ft

0.150 ft

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 backpressure must be between p6 and p3, which are the backpressures 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 ft)2 4 A = = 1.7778 p A* (0.15 ft)2 4 Choose M 7 1 (supersonic), M1 = 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 M1 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 1252

M13_HIBB9290_01_SE_C13_ANS.indd 1252

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13–109. 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 ft

0.150 ft

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 backpressure p6 6 p4, where p4 is the backpressure at which isentropic supersonic flow occurs. For air k = 1.40. Enter the table or apply the equation with p (0.2 ft)2 4 A = = 1.7778 p A* (0.15 ft)2 4 Choose M 7 1 (supersonic), M1 = 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

Ans: pb 6 8.12 psi 1253

M13_HIBB9290_01_SE_C13_ANS.indd 1253

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13–110. 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

( d+ ) - Vp = -VS + (Va>cv)2

(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 M1 - 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) M12 p2 = p1 (k - 1)M12 + 2

(2)

Equating Eqs. (1) and (2), (k + 1) M12 M1 = M1 - Vp >c1 (k - 1)M12 + 2

(k - 1)M12 + 2 = (k + 1)M12 - (k + 1)(Vp >c1) M1 M12 - a

k + 1 Vp b a b M1 - 1 = 0 2 c1

1254

M13_HIBB9290_01_SE_C13_ANS.indd 1254

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13–110. (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. equation becomes M12 - a

Thus,

the

above

150 m>s 1.4 + 1 b a b M1 - 1 = 0 2 343.05 m>s

M12 - 0.5247 M1 - 1 = 0 Solving for the positive root, M1 = 1.2962 Thus,

Vs = M1c1 = 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 = M12 ; = c d ( 1.2962 2 ) p1 k + 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 1255

M13_HIBB9290_01_SE_C13_ANS.indd 1255

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13–111. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel is represented by the 8° wedge. Determine the pressure created on its top surface if the angle of attack is set at a = 1.5°.

4° 700 m!s

a 4°

SOLUTION The flow is considered as steady and compressible. The forming of oblique shock is an adiabatic nonisentropic process. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K. Here, T1 = 273 + 15°C = 288 K and p1 = 70 kPa. Then, the Mach number behind the shock can be determined from 700 m>s = M1 21.401286.9 J>kg # K21288 K2

V1 = M1 2kRT1;

M1 = 2.0581

From the geometry shown in Fig. a, ut = 4° - 1.5° = 2.5° tan ut = tan 2.5° =

2 cot b t 1M12 sin2 b t - 12 M12 1k + cos 2 bt 2 + 2

2 cot bt 12.05812 sin2 bt - 12

2.05812 11.40 + cos 2bt 2 + 2

With the aid of oblique shock calculator and choose the weak shock angle, bt = 31.1046°

ut

4°

ub

4°

Then, the normal component of M1 is

Then

(a)

31M1 2 n 4 t = M1 sin bt = 2.0581 sin 31.1046° = 1.0632 1p2 2 t p1

1p2 2 t

70 kPa

=

2k k - 1 31M1 2 n 4 t2 k + 1 k + 1

= c

211.402

1.40 + 1

d 11.06322 2 -

1.40 - 1 1.40 + 1 Ans.

1p2 2 t = 80.65 kPa = 80.7 kPa

1256

M13_HIBB9290_01_SE_C13_ANS.indd 1256

a 5 1.5°

Ans: 1p2 2 t = 80.7 kPa

16/03/17 2:16 PM

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*13–112. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel is represented by the 8° wedge. Determine the pressure created on its bottom surface if the angle of attack is set at a = 1.5°.

4° 700 m!s

a 4°

SOLUTION The flow is considered as steady and compressible. The forming of oblique shock is an adiabatic nonisentropic process. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K for air. Here, T1 = 273 + 15°C = 288 K and p1 = 70 kPa. Then, the Mach number behind the shock can be determined from 700 m>s = M1 21.401286.9 J>kg # K21288 K2

V1 = M1 2kRT1;

M1 = 2.0581

From the geometry shown in Fig. a, ub = 4° + 1.5° = 5.5°. tan ub = tan 5.5° =

2 cot bb 1M12 sin2 bb - 12 M12 1k + cos 2 bb 2 + 2

2 cot bb 12.05812 sin2 bb - 12

2.05812 11.40 + cos 2bb 2 + 2

With the aid of oblique shock calculator and choose the weak shock angle, bb = 33.7469°

ut

4°

ub

4°

Then, the normal component of M1 is (a)

31M1 2 n 4 b = M1 sin bb = 2.0581 sin 33.7469° = 1.1433 1p2 2 b p1

1p2 2 b

70 kPa

=

2k k - 1 31M1 2 n 4 b2 k + 1 k + 1

= c

211.402

1.40 + 1

d 11.14332 2 -

1.40 - 1 1.40 + 1 Ans.

1p2 2 b = 95.09 kPa = 95.1 kPa

1257

M13_HIBB9290_01_SE_C13_ANS.indd 1257

a 5 1.5°

Ans: 1p2 2 b = 95.1 kPa

16/03/17 2:16 PM

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13–113. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel can be represented by the 8° wedge. Determine the pressure created on its top surface if the angle of attack is set at a = 6.5°.

4° 700 m!s

a 4°

SOLUTION The flow is considered as steady and compressible. The expansion waves will be formed at top surface and the process is isentropic. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K. Here, T1 = 273 + 15° = 288 K and 1p1 2 t = 70 kPa. Then, the Mach number behind the expansion wave can be determined from V1 = M1 2kRT1;

ut 4° ub

700 m>s = M1 21.401286.9 J>kg # K21288 K2

a 5 6.5°

4°

(a)

M1 = 2.0581

For M1 = 2.0581, its corresponding deflection angle with respect to the reference position can be determined using the Prandtl–Meyer expansion function, v = v1M1 2 =

k + 1 k - 1 tan-1 c 1M2 - 12 d - tan-1 1 2M2 - 12 Ak - 1 Ak + 1

1.40 + 1 1.40 - 1 tan-1 c 12.05812 - 12 d - tan-1 1 22.05812 - 12 = 27.9690° A 1.40 - 1 A 1.40 + 1

From the geometry shown in Fig. a, the deflection angle of the streamline is equal to the angle of deflection of the surface then is ut = 6.5° - 4° = 2.5°. Thus, with respect to the reference position v1M2 2 = v1M1 2 + ut = 27.9690° + 2.5° = 30.4690°

Use this value to solve for M2. The expansion waves calculator gives M2 = 2.1517

Since the expansion is an isentropic process, the isentropic flow formulas can be applied. For M1 = 2.0581, 1.40

k

1p0 2 t = 1p1 2 t a1 +

k-1 k - 1 M1 2 b ; 2

1p0 2 t = 1p1 2 t c 1 + a

1p1 2 t 1p0 2 t

For M2 = 2.1517,

1.40 - 1 1.40 - 1 b 12.05812 2 d 2

= 0.1167

1.40

k

1p0 2 t = 1p2 2 t a1 +

k-1 k - 1 M2 2 b ; 2

1p0 2 t = 1p2 2 t c 1 + a

1p2 2 t 1p0 2 t

Then 1p2 2 t 1p1 2 t

= c

1p2 2 t 1p0 2 t

d c

1p0 2 t 1p1 2 t

d;

1.40 - 1 1.40 - 1 b 12.15172 2 d 2

= 0.1009

1p2 2 t

70 kPa

= 10.10092 a

1p2 2 t = 60.48 kPa = 60.5 kPa

1 b 0.1167

1258

M13_HIBB9290_01_SE_C13_ANS.indd 1258

Ans. Ans: 1p2 2 t = 60.5 kPa

16/03/17 2:16 PM

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13–114. Air flows at 700 m>s through a long duct in a wind tunnel, where the temperature is 15°C and the absolute pressure is 70 kPa. The leading edge of the wing in the tunnel can be represented by the 8° wedge. Determine the pressure created on its bottom surface if the angle of attack is set at a = 6.5°.

4° 700 m!s

a 4°

SOLUTION The flow is considered as steady and compressible. The oblique shock will be formed at the bottom surface and the process is adiabatic and isentropic. From the table in Appendix A, k = 1.40 and R = 286.9 J>kg # K for air. Here, T1 = 273 + 15°C = 288 K and 1p1 2 b = 70 kPa. Then, the Mach number behind the shock can be determined from 700 m>s = M1 21.401286.9 J>kg # K21288 K2

V1 = M1 2kRT1;

ut 4° ub

a 5 6.5°

4°

(a)

M1 = 2.0581

From the geometry shown in Fig. a, ub = 4° + 6.5° = 10.5° tan ub = tan 10.5° =

2 cot bb 1M12 sin2 bb - 12 M12 1k + cos 2 bb 2 + 2

2 cot bb 12.05812 sin2 bb - 12

2.05812 11.40 + cos 2bb 2 + 2

With the aid of oblique shock calculator and choose the weak shock angle, bb = 38.7241° Then, the normal component of M1 is

Then

31M1 2 n 4 b = M1 sin bb = 2.0581 sin 38.7241° = 1.2875 1p2 2 b 1p1 2 b

1p2 2 b

70 kPa

=

2k k - 1 31M1 2 n 4 b2 k + 1 k + 1

= c

211.402

1.40 + 1

d 11.28752 2 -

1.40 - 1 1.40 + 1 Ans.

1p2 2 b = 123.71 kPa = 124 kPa

1259

M13_HIBB9290_01_SE_C13_ANS.indd 1259

Ans: 1p2 2 b = 124 kPa

16/03/17 2:16 PM

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13–115. 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 weak 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 5 2.4 d

SOLUTION Here u =

d 16° = = 8°. 2 2

tan u = tan 8° =

2 cot b ( M12 sin2 b - 1 ) M12 (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 M1 is (M1)n = M1 sin b = 2.4 sin 31.1489° = 1.2414 Thus, (M2)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.

(M2)n = M2 sin (b - u);

0.81751 = M2 sin (31.1489° - 8°) M2 = 2.0795

The velocity of the air in front of the shock is V2 = M2 2kRT2 = 2.0795 c 21.4 ( 286.9 J>kg # K)(317.34 K) d = 742.42 m>s = 742 m>s

Ans.

1260

M13_HIBB9290_01_SE_C13_ANS.indd 1260

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13–115. (Continued)

The separation of shock wave occurs when d = 2umax. Here, umax can be determined by plotting u vs b. For M1 = 2.4, this yields tan u = b 24.62°

2.42(1.4 + cos 2b) + 2

30°

40°

50°

60°

70°

80°

90°

6.72° 16.56° 23.78° 28.09° 27.76 19.17°

0°

u

2 cot b (2.42 sin2 b - 1)

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° 1261

M13_HIBB9290_01_SE_C13_ANS.indd 1261

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*13–116. The leading edge on the wing of an aircraft can be approximated by the wedge 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 a weak 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°

900 m!s

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 ( M12 sin2 b - 1 ) M12 ( 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 M1 is (M1)n = M1 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.

Ans: b = 25.6° p2 = 84.5 kPa T2 = 307 K 1262

M13_HIBB9290_01_SE_C13_ANS.indd 1262

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13–117. A 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 the wedge shape shown, determine the pressure and temperature of the air just in front or to the right of the expansion waves.

4° 4°

15°

700 m!s

SOLUTION 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

= 11° a = 15 ° 4°

= 2.0836

4°

For this Mach number, its corresponding deflection angle with respect to the reference state can be determined using the Prandtl-Meyer expansion function.

(a )

k + 1 k - 1 ( M2 - 1) d - tan - 1 ( 2M2 - 1 ) v = tan - 1 c Ak - 1 Ak + 1 v(M1) =

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(M2) - v(M1);

11° = v(M2) - 28.6569° v(M2) = 39.6569°

Thus, M2 = 2.5230 The expansion is an isentropic process. For M1 = 2.0836, T1 = 0.53525 T0

p1 = 0.11219 p0

For M2 = 2.5230, T2 = 0.43993 T0

p2 = 0.056476 p0

Using these ratios, T0 T2 T2 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 1263

M13_HIBB9290_01_SE_C13_ANS.indd 1263

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13–118. 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 weak oblique shock wave that forms at the leading edge.

A

3°

B

3°

a 5 2°

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 ( M1 2 sin2 bA - 1 )

3°

M12 (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 M1 is (M1)n = M1 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 1264

M13_HIBB9290_01_SE_C13_ANS.indd 1264

16/03/17 2:16 PM

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13–119. 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 weak oblique shock wave that forms at the leading edge.

A

3°

B

3°

a 5 2°

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 3°

For the lower surface b, uB = 5° (geometry shown in Fig. a). tan uB = tan 5° =

2 cot bB ( M12 sin2 bB - 1 )

a = 2°

M12 ( k + cos 2bB ) + 2

3°

2 cot bB (2.38132 sin2 bB - 1)

B = 3° + 2° = 5°

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 M1 is (M1)n = M1 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 p1

Ans.

T2 = 1.09361; TB = T2 = 1.09361 (281 K) = 307.30 K = 307 K T1

Ans.

Ans: pB = 123 kPa, TB = 307 K 1265

M13_HIBB9290_01_SE_C13_ANS.indd 1265

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*13–120. 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 weak oblique shock at A and in front or to the right of the expansion waves at B.

B A

3° 3°

900 m!s

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

=

Here, u = 3°. tan u = tan 3° =

900 m>s

21.4 ( 286.9 J>kg # K )( 273 + 8 ) K

= 2.6789

2 cot b (M12 sin2 b - 1) M12(k + cos 2b) + 2 2 cot b (2.67892 sin2 b - 1) 2.67892 (1.4 + cos 2b) + 2

Solving numerically to find the weak shock angle, b = 24.1083° The normal component of M1 is (M1)n = M1 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.

(M2)n = 0.91634 (M2)n = M2 sin (b - u);

0.91634 = M2 sin (24.1083° - 3°) M2 = 2.5445

The expansion waves will occur at the corner of surface A and B. For M2 = 2.5445, its corresponding deflection angle with respect to the reference state can be determined. v(M2) = 40.1534° The deflection angle of the surface is u = 3° + 3° = 6°. Then u = v ( M3 ) - v ( M2 ) ;

6° = v(M3) - 40.1534° v(M3) = 46.1534°

Thus, M3 = 2.8196 Since the expansion is an isentropic process, for M2 = 2.5445, p2 = 0.054618 p0

1266

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13–120. (Continued)

For M3 = 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 1267

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13–121. Oxygen at a temperature of 25°C and an absolute pressure of 200 kPa flows through the rectangular duct at 900 m>s. When it comes to the transition, it is redirected as shown. Determine the angle b of the weak oblique shock that forms at A, and the temperature and pressure of the oxygen just in front or to the right of the wave.

15°

B

b 15° A

SOLUTION The flow is considered as steady and compressible. The oblique shock wave will be formed at the bottom inner wall and the process is adiabatic. From the table in Appendix A, k = 1.40 and R = 259.8 J>kg # K for oxygen. Here, T1 = 25°C + 273 = 298 K and p1 = 200 kPa. Then, the Mach number behind the shock can be determined from V1 = M1 2kRT1;

900 m>s = M1 21.401259.8 J>kg # K21298 K2 M1 = 2.7337

Here, angle of deflection of the streamline is ub = 15°. tan ub = tan 15° =

2 cot bb 1M12 sin2 bb - 12 M12 1k + cos 2 bb 2 + 2

2 cot bb 12.73372 sin2 bb - 12

2.73372 11.40 + cos 2 bb 2 + 2

With the aid of oblique shock calculator and choose the weak shock angle, Ans.

bb = 34.4632° = 34.5° Also, 1p2 2 A

= 2.6252;

1T2 2 A

= 1.3517;

p1

T1

1p2 2 A = 2.62521200 kPa2 = 525.04 kPa = 525 kPa

Ans.

1T2 2 A = 1.35171298 K2 = 402.81 K = 403 K

Ans.

31M2 2 n 4 A = 0.6851

1268

M13_HIBB9290_01_SE_C13_ANS.indd 1268

Ans: bb = 34.5° 1p2)A = 525 kPa 1T2 2 A = 403 K

16/03/17 2:16 PM

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13–122. Oxygen at a temperature of 25°C and an absolute pressure of 200 kPa flows in the rectangular duct at 900 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 in the duct at B.

15°

B

b 15° A

SOLUTION The flow is considered as steady and compressible. The expansion wave will be formed at top inner wall and the process is isentropic. From the table in Appendix A, k = 1.40 and R = 259.8 J>kg # K for oxygen. Here, T1 = 25°C + 273 = 298 K and p1 = 200 kPa. Then, the Mach number behind the expansion wave can be determined from V1 = M1 2kRT1;

900 m>s = M1 21.401259.8 J>kg # K21298 K2 M1 = 2.7337

For this Mach number its corresponding deflection angle with respect to the reference position can be determined using the Prandtl–Meyer expansion function, v = v1M1 2 =

k + 1 k - 1 tan-1 c 1M2 - 12 d - tan-1 1 2M2 - 12 Ak - 1 Ak + 1

1.40 + 1 1.40 - 1 tan-1 c 12.73372 - 12 d - tan-1 1 22.73372 - 12 = 44.346° A 1.40 - 1 A 1.40 + 1

The deflection angle of the streamline is equal to the angle of deflection of the wall, that is u = 15°. Thus, with respect to the reference position v1M2 2 = v1M1 2 + u = 44.3465° + 15° = 59.3465°

Now solve for M2. The expansion wave calculator gives M2 = 3.5518

Since the expansion is an isentropic process, the isentropic flow formulas can be applied. For M1 = 2.7337, T0 = T1 a1 +

k - 1 M1 2 b ; 2

k

k-1 k - 1 p0 = p1 a1 + M1 2 b ; 2

T0 = T1 c 1 + a

T1 = 0.4009 T0

1.40 - 1 b 12.73372 2 d 2

1.40

1.40 - 1 1.40 - 1 p0 = p1 c 1 + a b 12.73372 2 d 2

p1 = 0.04078 p0

1269

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13–122. (Continued)

SOLUTION For M2 = 3.5518, T0 = 1T2 2 B a1 +

k - 1 M2 2 b ; 2

T0 = 1T2 2 B c 1 + a

1T2 2 B T0

= 0.2838

1.40

k

p0 = 1p2 2 B a1 +

k-1 k - 1 M2 2 b ; 2

p0 = 1p2 2 B c 1 + a

1p2 2 B

= 0.01218

T0 b; T1

1T2 2 B

p0

Then,

1T2 2 B T1

1p2 2 B p1

= c

= c

1T2 2 B T0

d a

1.40 - 1 b 13.55182 2 d 2

298 K

1.40 - 1 1.40 - 1 b 13.55182 2 d 2

= 10.28382 a

1T2 2 B = 211.01 K = 211 K

1p2 2 B p0

d a

p0 b; p1

1p2 2 B

200 kPa

1 b 0.4009

= 10.012182 a

1p2 2 B = 59.747 kPa = 59.7 kPa

1 b 0.04078

1270

M13_HIBB9290_01_SE_C13_ANS.indd 1270

Ans.

Ans.

Ans: 1T2 2 B = 211 K, 1p2 2 B = 59.7 kPa

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14–1. The axial flow pump produces a discharge of 0.095 m3 >s . If the impeller is rotating at 80 rad>s, and it has a mean radius of 75 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.

200 mm

b1

45°

75 mm

80 rad!s

SOLUTION

(Vrel)1

We will consider the relative flow as steady and the water as incompressible. The speed at the midpoint on the impeller is U = vrm = 180 rad>s210.075 m2 = 6 m>s

The axial velocity of the water through the impeller is given by Va =

a1 5 90°

0.095 m3 >s Q = = 4.0319 m>s A p310.1 m2 2 - 10.05 m2 2 4

4.0319 m>s 6 m>s

V 5 4.0319 mys

b1

Since it is required that a1 = 90°, 1Vt 2 1 = 0. Hence, V = Va = 4.0319 m>s. Using these results, the kinematic diagram of the water at the lead of the blade shown in Fig. a can be drawn. From the geometry, tan b1 =

b1

b1 = 33.90° = 33.9°

Ans.

1Vrel 2 1 = 7.229 m>s = 7.23 m>s

Ans.

U 5 6 mys (a)

Using this result, cos 33.90° =

6 m>s 1Vrel 2 1

1271

M14_HIBB9290_01_SE_C14_ANS.indd 1271

Ans: b1 = 33.9° 1Vrel 2 1 = 7.23 m>s

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14–2. The axial flow pump produces a discharge of 0.095 m3 >s . If the impeller is rotating at 80 rad>s, and it has a mean radius of 75 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.

200 mm

b1

45°

75 mm

80 rad!s

SOLUTION

(Vrel)2

b2 5 45°

We will consider the relative flow as steady and the water as incompressible. The speed at the midpoint on the impeller blade is Va 5 4.0319 mys

U = vrm = 180 rad>s210.075 m2 = 6 m>s

(Vt )2

a2

The axial velocity of the water through the blade is given by Va =

0.095 m3 >s Q = = 4.0319 m>s A p310.1 m2 2 - 10.05 m2 2 4

V2

45°

Here, b2 = 45°. Using these results, the kinematic diagram of the water at the tail of the blade shown in Fig. a can be drawn. From the geometry,

U 5 6 mys (a)

1Vt 2 2 = U - Va cot b2 = 6 m>s - 14.0319 m>s2 cot 45° = 1.9681 m>s

Using this result,

V2 = 2Va 2 + 1Vt 2 22 = 214.0319 m>s2 2 + 11.9681 m>s2 2 = 4.487 m>s = 4.49 m>s

Ans.

Also, Va = sin b2; 1Vrel 2 2

4.0319 m>s 1Vrel 2 2

= sin 45°

1Vrel 2 2 = 5.702 m>s = 5.70 m>s

1272

M14_HIBB9290_01_SE_C14_ANS.indd 1272

Ans.

Ans: V2 = 4.49 m>s 1Vrel 2 2 = 5.70 m>s

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14–3. The axial flow pump produces a discharge of 275 ft3 >min. If the impeller has an angular velocity of v = 150 rad>s, determine the velocity of the water when it is delivered to the stator blades. What is the relative velocity of the water as it flows off the impeller? The impeller blades have a mean radius of 3 in. and the angles shown.

8 in.

Impeller

Stator 3 in.

45° 30°

SOLUTION

v

U 5 37.5 ftys

We will consider the relative flow as steady and the water as incompressible. The speed at the midpoint on the impeller blade is U = vrm = 1150 rad>s2a

3 ft b = 37.5 ft>s 12

4.5833 ft3 >s

= 17.5070 ft>s 2 2 4 2 ft b - a ft b d 12 12 Here, b2 = 90° - 30° = 60°. Using these results, the kinematic diagram of the water at the tail of the blade shown in Fig. a can be drawn. From the geometry, pc a

Va = sin b2; 1Vrel 2 2

17.5070 ft>s 1Vrel 2 2

V2

a2

ft3 1 min Here, the flow rate is Q = a275 ba b = 4.5833 ft3 >s. Then, the axial min 60 s velocity of the water through the blade is given by Q = Va = A

60° (Vt )2

30° b2 5 60°

Va 5 17.5070 ftys (Vrel)2 (a)

= sin 60° Ans.

1Vrel 2 2 = 20.22 ft>s = 20.2 ft>s 1Vt 2 2 = U - Va cot b2

= 37.5 ft>s - 117.5070 ft>s2 cot 60° = 27.3923 ft>s

Thus, V2 = 2Va 2 + 1Vt 2 22 = 2117.5070 ft>s2 2 + 127.3923 ft>s2 2 = 32.5090 ft>s = 32.5 ft>s Ans.

Ans: 1Vrel 2 2 = 20.2 ft>s V2 = 32.5 ft>s 1273

M14_HIBB9290_01_SE_C14_ANS.indd 1273

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*14–4. The axial flow pump produces a discharge of 275 ft3 >min. Determine the power supplied to the water by the pump when v = 150 rad>s . The impeller blades have a mean radius of 3 in. and the angles shown.

8 in.

Impeller

Stator 3 in.

45° 30°

v

SOLUTION We will consider the relative flow as steady and the water as incompressible. The speed at the midpoint on the impeller blade is U = vrm = 1150 rad>s2a

3 ft b = 37.5 ft>s 12

ft3 1 min ba b = 4.5833 ft3 >s. Then, the axial min 60 s velocity of the water through the blade is given by Here, the flow rate is Q = a275 Va =

Q = A

pc a

4.5833 ft3 >s

2 2 4 2 ft b - a ft b d 12 12

= 17.5070 ft>s

Here, b2 = 90° - 30° = 60°. Using these results, the kinematic diagram of the water at the tail of the blade shown in Fig. a can be drawn. From the geometry, 1Vt 2 2 = U - Va cot b2 = 37.5 ft>s - 117.5070 ft>s2 cot 60° = 27.3923 ft>s

Also, b1 = 90° - 45° = 45°. Then, the kinematic diagram of the water at the leading edge of the blade shown in Fig. b can be drawn. From the geometry, 1Vt 2 1 = U - Va cot b1 = 37.5 ft>s - 117.5070 ft>s2 cot 45° = 19.9930 ft>s

The power supplied to the water can be determined. # Wpump = rQU31Vt 2 2 - 1Vt 2 1 4 = a

62.4 lb> ft3 32.2 ft>s2

b14.5833 ft3 >s2137.5 ft>s2127.3923 ft>s - 19.9930 ft>s2

= 12464.53 ft # lb>s2a

1 hp

550 ft # lb>s

= 4.48 hp

b

Ans. U 5 37.5 ftys

U 5 37.5 ftys

45°

60° V2

(Vt )2

V1

(Vt )1

a1 a2 30° b2 5 60°

Va 5 17.5070 ftys (Vrel)2 (a)

45° b1 5 45°

Va 5 17.5070 ftys

(Vrel)1 (b)

Ans: # Wpump = 4.48 hp 1274

M14_HIBB9290_01_SE_C14_ANS.indd 1274

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14–5. 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. 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

b2 = 35°

Thus, the speed at the midpoint on the impeller is

U = 1.6875 m/s

U = vrm = ( 30 rad>s )( 0.05625 m ) = 1.6875 m>s

Va = 0.3018 m/s

V2

(Vrel(2

a2

(Vt( 2

b2 = 35°

(a )

Here, the cross-sectional area within the open region of the impeller is A = p 3 (0.075 m)2 - (0.0375 m)2 4 = 4.21875 ( 10 - 3 ) p m2

Thus, the axial flow velocity can be determined from

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: V2 = 1.29 m>s ( Vt ) 2 = 1.26 m>s 1275

M14_HIBB9290_01_SE_C14_ANS.indd 1275

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14–6. If the axial-flow pump produces a discharge of 350 ft3 >min, determine the required blade angle b1 so a1 = 90° . Also what is the power supplied to the water by the pump? The impeller blades have a mean radius of 2.5 in. and v = 200 rad>s.

v 60° b1 2.5 in. 8 in.

SOLUTION V1 5 Va 5 17.8254 ftys

We will consider the flow as steady and the water as incompressible. The speed at the midpoint on the impeller blade is U = vrm = 1200 rad>s2a

2.5 ft b = 41.6667 ft>s. 12

b1

ft3 1 min ba b = 5.8333 ft3 >s. Then, the axial Here, the flow rate is Q = a350 min 60 s velocity of water through the blade is given by Q Va = = A

5.8333 ft3 >s

2 2 4 1 pc a ft b - a ft b d 12 12

17.8254 ft>s Va = U 41.6667 ft>s

b1

a1

U 5 41.6667 ftys (a) V2

Va 5 17.5070 ftys (Vrel)2

= 17.8254 ft>s 60°

Here, it is required that a1 = 90°. Using this result, the kinematic diagram of the water at the lead of the blade shown in Fig. a can be drawn. From the geometry, tan b1 =

b1 = 23.16° = 23.2°

(Vrel)1

a2

b2 5 60°

U 5 41.6667 ftys (V2 )t (b)

Ans.

1V1 2 t = 0

Also, b2 = 60°. The kinematic diagram of the water at the lead of the blade shown in Fig. b can be drawn. From the geometry, 1V2 2 t = U - Va cot b2;

1V2 2 t = 41.6667 ft>s - 117.8254 ft>s2 cot 60° = 31.3752 ft>s

The power supply to the water can be determined. # Wpump = rQU31Vt 2 2 - 1Vt 2 1 4 = a

62.4 lb>ft3 32.2 ft>s2

b15.8333 ft3 >s2141.6667 ft>s2131.3752 ft>s - 02

= 114778.17 ft # lb>s2a

1 hp

550 ft # lb>s

Ans.

b = 26.9 hp

Ans: b#1 = 23.2° Wpump = 26.9 hp 1276

M14_HIBB9290_01_SE_C14_ANS.indd 1276

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14–7. An axial-flow pump has an impeller with a mean radius of 100 mm. It rotates at 1200 rev>min. At the exit the stator blade angle is 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

1 min 2p rad rev 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 V2 = 8 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,

( Vt ) 2 = V2 cos a2 = ( 8 m>s ) cos 70° = 2.7362 m>s = 2.74 m>s Ans. Va = ( Vt ) 2 tan a2 = (2.7362 m>s ) tan 70° = 7.5175 m>s tan b2 = =

b2

a2 = 70 °

U = 12.566 m/s

Va U - ( Vt ) 2

Va

(Vrel(2

b2

(Vt (2 (a )

7.5175 m>s 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: (Vt)2 = 2.74 m>s Vrel = 12.4 m>s 1277

M14_HIBB9290_01_SE_C14_ANS.indd 1277

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*14–8. The radial fan is used to force air into the ducts of a building. If the air is at a temperature of 30°C, and the impeller is rotating at 80 rad>s, determine the power output of the motor. The blades have a width of 20 mm. Air enters the blades in the radial direction and is discharged with a velocity of 30 m>s at the angle shown.

30 m!s 40°

80 rad!s

200 mm

SOLUTION We will consider the relative flow as steady and the air as incompressible. The kinematic diagram of the water on the blade tail is shown in Fig. a. From the geometry, 1Vt 2 2 = 130 m>s2 sin 40° = 19.2836 m>s

1Vr 2 2 = 130 m>s2 cos 40° = 22.9813 m>s

V2 5 30 mys

(Vr)2

The flow rate in the pump is

Q = 1Vr 2 2 A2 = 122.9813 m>s232p10.2 m210.02 m24 = 0.57758 m3 >s

From the table in Appendix A, ra = 1.164 kg>m3 for air at 30°C. Since the air enters the impeller blade radially, 1V1 2 t = 0. Thus, the torque supplied the motor can be determined

40°

(Vt )1

(a)

T = raQ3r2 1Vt 2 2 - r1 1Vt 2 1 4

= 11.164 kg>m3 210.57758 m3 >s2310.2 m2119.2836 m>s2 - 04 = 2.5929 N # m

Thus, the power onput of the motor is # Wpump = Tv = 12.5929 N # m2180 rad>s2 = 207.43 W = 207 W

Ans.

Ans: # Wpump = 207 W 1278

M14_HIBB9290_01_SE_C14_ANS.indd 1278

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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.

V2 = 20 m/s

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)]

a1

(Vt (2

( 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,

( Vt ) 2 = 2V 22 - ( Vr ) 22 = 2 ( 20 m>s ) 2 - (8.4883 m>s ) 2 = 18.1094 m>s

(Vr(2 = 8.4883 m/s (a)

Since the flow enters the impeller blade radially, ( Vt ) 1 = 0. The torque applied to the pump shaft is

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

Ans.

Ans: T = 1.81 kN # m 1279

M14_HIBB9290_01_SE_C14_ANS.indd 1279

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© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. b2 5 40°

14–10. Air at a temperature of 80°F enters the 4-in.-wide blades of the blower in the radial direction and is discharged from the blades at an angle of b2 = 40°. Determine the power required to turn the blades at 120 rad>s, producing a discharge of 150 ft3 >s.

120 rad!s

9 in. 15 in.

SOLUTION We will consider the relative flow is steady and the air is incompressible. The speed of a point on the blade tail is U2 = vr2 = 1120 rad>s2a

15 ft b = 150 ft>s 12

Since the air enters the blades radially where a1 = 90° and 1Vt 2 1 = 0, the pump head is hpump = =

U2Q cot b2 U 22 g 2pr2bg 1150 ft>s2 2 32.2 ft>s

2

-

1150 ft>s21150 ft3 >s2 cot 40°

2pa

= 380.67 ft

15 4 ft ba ft b132.2 ft>s2 2 12 12

From the table in Appendix A, ra = 0.00228 slug> ft3 for air at 80°F. Then, the required power is

#

Wpump = gQhpump = 10.00228 slug>ft3 2132.2 ft>s2 21150 ft3 >s21380.67 ft2 = 14192.11 ft # lb>s2a

= 7.622 hp = 7.62 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Wpump = 7.62 hp 1280

M14_HIBB9290_01_SE_C14_ANS.indd 1280

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14–11. The radial-flow pump impeller rotates at 900 rev >min. If the width of the blades is 3 in., and the blade head and tail angles are as shown, determine the power developed by the pump. Water is initially guided radially onto the blades.

b2 5 308

b1 5 458 3 in. 900 rev!min 1.5 in.

SOLUTION We will consider the relative flow as steady and the water as incompressible. The angular velocity of the impeller is rev 1 min 2p rad v = a900 ba ba b = 30p rad>s min 60 s 1 rev

V1

b1 5 45° a1

Thus, the speed of points on the lead and tail of the blades are

1.5 U1 = vr1 = 130p rad>s2a ft b = 3.75p ft>s 12 U2 = vr2 = 130p rad>s2a

(Vrel)1

45°

U1 5 37.5p ftys (a)

3 ft b = 7.5p ft>s 12

Since the water is initially guided radially onto the blades V1 = 1Vr 2 1 and 1Vt 2 1 = 0. The kinematic diagram of the water on the lead of the blade is shown in Fig. a. From the geometry, V1 = U1 tan b1 = 13.75p ft>s2 tan 45° = 3.75p ft>s.

Thus, the flow rate is

Q = V1A1 = 13.75p ft>s2 c 2pa

1.5 3 ft b a ft b d = 2.3132 ft3 >s 12 12

Since the water enters the blades radially where a1 = 90° and 1Vt 2 1 = 0, the pump head is hpump = =

U2Q cot b2 U 22 g 2pr2bg 17.5p ft>s2 2 32.2 ft>s2

= 9.7755 ft

-

17.5p ft>s212.3132 ft3 >s2 cot 30° 2pa

3 3 ft ba ft b132.2 ft>s2 2 12 12

Thus, the power generated by the pump is # W = gQhpump = 162.4 lb>ft3 212.3132 ft3 >s219.7755 ft2 = 11411.03 ft # lb>s2a

= 2.566 hp = 2.57 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # W = 2.57 hp 1281

M14_HIBB9290_01_SE_C14_ANS.indd 1281

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*14–12. The radial-flow pump impeller rotates at 900 rev>min. If the width of the blades is 3 in., and the blade head and tail angles are as shown, determine the relative velocity of the water as it flows onto and off the blades. Water is initially guided radially onto the blades.

b2 5 308

b1 5 458 3 in. 900 rev!min 1.5 in.

SOLUTION We will consider the relative flow as steady and the water as incompressible. The angular velocity of the impeller is

(Vrel)1

rev 1 min 2p rad v = a900 ba ba b = 30p rad>s min 60 s 1 rev

b1 5 45° a1

Thus, the speed of points on the lead and tail of the blades are U1 = vr1 = 130p rad>s2a U2 = vr2 = 130p rad>s2a

V1

1.5 ft b = 3.75p ft>s 12

45°

U1 5 37.5p ftys (a)

3 ft b = 7.5p ft>s 12

Since the water is initially guided radially onto the blades, V1 = 1Vr 2 1, 1Vt 2 1 = 0 and a1 = 90°. The kinematic diagram of the lead of the blade is shown in Fig. a. From the geometry, V1 = U1 tan b1 = 13.75p ft>s2 tan 45° = 3.75p ft>s

1Vrel 2 1 =

3.75p ft>s U1 = = 16.66 ft>s = 16.7 ft>s cos b1 cos 45°

(Vrel)2

(Vr)2 5 1.875 p ftys V2

Ans. b2 5 30°

Using the continuity condition,

30°

a2 (Vt )2

U2 5 7.5 p ftys

(b)

1Vr 2 1r1 = 1Vr 2 2r2

13.75p ft>s211.5 in.2 = 1Vr 2 2 13 in.2 1Vr 2 2 = 1875p ft>s

Together with b2 = 30° and U2 = 7.5p ft>s, the kinematic diagram of the water at the tail of the blade shown in Fig. b can be drawn. From the geometry, 1Vrel 2 2 =

1Vr 2 2

sin b 2

=

1.875p ft>s sin 30°

= 11.78 ft>s = 11.8 ft>s

1282

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Ans.

Ans: 1Vrel 2 1 = 16.7 ft>s 1Vrel 2 2 = 11.8 ft>s

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14–13. The blades on the radial-flow pump rotate at 180 rad>s. If the blades are 2 in. wide, determine the discharge if the water enters each blade in the radial direction.

30° 60° 2 in.

SOLUTION We will consider the relative flow as steady and the water as incompressible. The speed of a point on the lead of the blade is

180 ra d!s 8 in.

2 U1 = vr1 = 1180 rad>s2a ft b = 30 ft>s 12

Here, the water is required to enter the impeller blades radially, then V1 = 1Vr 2 1, 1Vt 2 1 = 0 and a1 = 90°. Also, b1 = 60°. The kinematic diagram of the water on the lead is shown in Fig. a. From the geometry, V1 = U1 tan b1 = 130 ft>s2 tan 60° = 3023 ft>s

Thus, the discharge in the pump is

Q = V1A1 = 13023 ft>s2 c 2pa

2 2 inb a ft b d = 9.069 ft3 >s = 9.07 ft3 >s Ans. 12 12

1283

M14_HIBB9290_01_SE_C14_ANS.indd 1283

(Vrel)1 V1 b1 5 60° a1 60° U1 5 30 ftys (a)

Ans: Q = 9.07 ft3 >s

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14–14. The blades on the radial-flow pump rotate at 180 rad>s. If the discharge is 240 ft3 >min, determine the ideal pump head. The blades are 2 in. wide.

30° 60° 2 in.

SOLUTION

180 ra d!s

We will consider the relative flow as steady and the water as incompressible. The speed of the points on the lead and tail of the blade are

8 in.

2 U1 = vr1 = 1180 rad>s2a ft b = 30 ft>s 12 U2 = vr2 = 1180 rad>s2a 3

8 ft b = 120 ft>s 12

(Vrel)1 (Vr)1 5 22.9183 ftys

1 min ft ba b = 4 ft3 >s. From the flow rate the min 60 s radial components of water velocity on the lead and tail of the blade are

Here, the flow rate is Q = a240

2 2 4 ft >s = 1Vr 2 1 c 2pa ft b a ft b d 12 12 8 2 3 4 ft >s = 1Vr 2 2 c 2pa ft b a ft b d 12 12 3

Q = 1Vr 2 1A1;

Q = 1Vr 2 2A2;

V1

b1 5 60° a1

1Vr 2 1 = 22.9183 ft>s

60°

1Vr 2 2 = 5.7296 ft>s

(Vt )1

U1 5 30 ftys (a)

Here, b1 = 60° and b2 = 30°. Thus, the kinematic diagram of water on the lead and tail of the blade are shown in Figs. a and b, respectively. From the geometry of Fig. a, 1Vr 2 1 = 3U1 - 1Vt 2 1 4 tan b1; From the geometry of Fig b, 1Vr 2 2 = 3U2 - 1Vt 2 2 4 tan b2; The ideal pump head is hpump =

U2 1Vt 2 2 - U1 1Vt 2 1 g

=

22.9183 ft>s = 330 ft>s - 1Vt 2 1 4 tan 60° 1Vt 2 1 = 16.7681 ft>s

(Vrel)2 (Vr)2 55.7296 ftys

5.7296 ft>s = 3120 ft>s - 1Vt 2 2 4 tan 30° 1Vt 2 2 = 110.0761 ft>s

32.2 ft>s

V2 30°

1120 ft>s21110.0761 ft>s2 - 130 ft>s2116.7681 ft>s2 2

30°

U2 5 120 ftys

a2 (Vt )2 (b)

Ans.

= 394.60 ft = 395 ft

Ans: hpump = 395 ft 1284

M14_HIBB9290_01_SE_C14_ANS.indd 1284

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14–15. The blades on the radial-flow pump rotate at 180 rad>s. If the discharge is 240 ft3 >min, determine the velocity of the water as it flows onto and off the blades. The blades are 2 in. wide.

30° 60° 2 in.

SOLUTION

180 ra d!s

We will consider the relative flow as steady and the water as incompressible. The speed of the points on the lead and tail of the blade are 2 U1 = vr1 = 1180 rad>s2 a ft b = 30 ft>s 12 8 U2 = vr2 = 1180 rad>s2 a ft b = 120 ft>s 12

8 in.

(Vrel)1 (Vr)1 5 22.9183 ftys

1 min ft3 ba b = 4 ft3 >s. From the flow rate the min 60 s radial components of water velocity on the lead and tail of the blade are

V1

Here, the flow rate is Q = a240 Q = 1Vr 2 1A1;

Q = 1Vr 2 2A2;

4 ft3 >s = 1Vr 2 1 c 2pa

4 ft3 >s = 1Vr 2 2 c 2pa

2 2 ft ba ft b d 12 12

8 2 ft ba ft b d 12 12

b1 5 60° a1 60°

1Vr 2 1 = 22.9183 ft>s

(Vt )1

U1 5 30 ftys

1Vr 2 2 = 5.7296 ft>s

(a)

Here, b1 = 60° and b2 = 30°. Thus, the kinematic diagram of water on the lead and tail are shown in Figs. a and b, respectively. From the geometry of Fig. a, 1Vr 2 1 = 3U1 - 1Vt 2 1 4 tan b1;

22.9183 ft>s = 330 ft>s - 1Vt 2 1 4 tan 60°

(Vrel)2

1Vt 2 1 = 16.7681 ft>s

Then

(Vr)2 55.7296 ftys

V1 = 21Vt 2 12 + 1Vr 2 12 = 2116.7681 ft>s2 2 + 122.9183 ft>s2 2 = 28.3975 ft>s = 28.4 ft>s Ans. 22.9183 ft>s 1Vr 2 1 a1 = tan -1 c d = tan -1 a b = 53.8090° = 53.8° Ans. 1Vt 2 1 16.7681 ft>s 1Vr 2 2 = 3U2 - 1Vt 2 2 4 tan b2;

Then

V2 30°

U2 5 120 ftys

From the geometry of Fig. b,

30°

a2 (Vt )2 (b)

5.7296 ft>s = 3120 ft>s - 1Vt 2 2 4 tan 30° 1Vt 2 2 = 110.0761 ft>s

V2 = 21Vt 2 22 + 1Vr 2 22 = 21110.0761 ft>s2 2 + 15.7296 ft>s2 2 = 110.2251 ft>s = 110 ft>s Ans. a2 = tan -1 c

1Vr 2 2 1Vt 2 2

d = tan -1 a

5.7296 ft>s

110.0761 ft>s

b = 2.9796° = 2.98°

Ans.

Ans: V1 = 28.4 ft>s, a1 = 53.8° V2 = 110 ft>s, a2 = 2.98° 1285

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*14–16. Water flows through the 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 blade is 20 mm, 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

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)]

(Vt (1

( Vr ) 2 = 1.8189 m>s

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,

(a)

(Vr(2 = 1.8189 m/s

( Vt ) 1 = 2V12 - ( Vr )12 = 2 ( 6 m>s ) 2 - ( 3.9789 m>s ) 2 = 4.4909 m>s

V2 = 10 m/s

From the geometry of Fig. b,

a2

( Vt ) 2 = 2V22 - ( Vr ) 22 = 2 ( 10 m>s ) 2 - ( 1.8189 m>s ) 2 = 9.8332 m>s

(Vt (2

(b)

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)( 9.8332 m>s ) - (0.08 m)( 4.4909 m>s ) 4

= 54.46 N # m = 54.5 N # m

Ans.

Ans: T = 54.5 N # m 1286

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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 torque the pump shaft 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

(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°

U2 = vr2 = ( 60 rad>s ) (0.25 m) = 15 m>s

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 the pump is

V2

(Vt (2

b2 = 40° U2 = 15 m/s

(b)

Q = ( Vr ) 1A1 = ( 1.6729 m>s ) [2p(0.1 m)(0.04 m)] = 0.04204 m3 >s

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 =

1 Vt 2 2

3 15 m>s

-

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.

Ans: T = 130 N # m 1287

M14_HIBB9290_01_SE_C14_ANS.indd 1287

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14–18. 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

(Vt (1

U2 = vr2 = ( 60 rad>s ) (0.25 m) = 15 m>s

110° b1 = 50°

a1 = 20°

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°

(a)

U1 = 6 m/s

V1 = 4.8912 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

Continuity: 0 r dV + 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, ( 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: hpump =

=

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: hpump = 18.9 m 1288

M14_HIBB9290_01_SE_C14_ANS.indd 1288

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14–19. Water flows through the 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

0.04 m3 >s = ( Vr ) 1[2p(0.08 m)(0.02 m)]

V1

( Vr ) 1 = 3.9789 m>s

Q = ( Vr ) 2A2

(Vr(1 = 3.9789 m/s

a1 = 45 °

0.04 m3 >s = ( Vr ) 2[2p(0.175 m)(0.02 m)]

( Vr ) 2 = 1.8189 m>s

(Vt (1 (a)

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,

( Vt ) 1 = ( Vr ) 1 cot a1 = ( 3.9789 m>s )(cot 45°) = 3.9789 m>s

(Vr(2 = 1.8189 m/s V2 a2 = 10 °

From the geometry of Fig. b,

(Vt (2

(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: T = 59.5 N # m 1289

M14_HIBB9290_01_SE_C14_ANS.indd 1289

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*14–20. Water flows radially onto the blades of the impeller and exits with a velocity of 40 ft>s . If the impeller is turning at 1200 rev>min and the discharge is 630 ft3 >min, determine the tail blade angle b2 and the power supplied to the pump if it has an efficiency of h = 0.65. The blades are 2 in. wide.

2 in.

1.5 in.

6 in. 1200 rev!min.

1.5 in.

6 in.

SOLUTION We will consider the relative flow as steady and the water as incompressible. Here, the angular speed of the impeller is

(Vr)2 5 20.0535 ftys

(Vrel)2

rev 1 min 2p rad v = a1200 ba ba b = 40p rad>s min 60 s 1 rev

Then, the speed of a point on the tail of the blade is

b2

b2

a2 (Vt )2

6 U2 = vr2 = 140p rad>s2a ft b = 20p ft>s. 12

V2 5 40 ftys

U2 5 20p ftys

(a)

ft3 1 min ba b = 10.5 ft3 >s. From the discharge min 60 s the radial component of the water velocity at the tail of the blade is

Here, the discharge is Q = a630

10.5 ft3 >s = 1Vr 2 2 c 2pa

Q = 1Vr 2 2A2;

6 2 ft b a ft b d 12 12

1Vr 2 2 = 20.0535 ft>s

It is required that V2 = 40 ft>s. Using these results, the kinematic diagram of the water on the tail shown in Fig. a can be drawn. From the geometry, a2 = sin -1 c

1Vr 2 2 V2

d = sin -1 a

20.0535 ft>s 40 ft>s

b = 30.0886°

1Vt 2 2 = V2 cos a2 = 140 ft>s2 cos 30.0886° = 34.6101 ft>s b2 = tan -1 c

1Vr 2 2

U2 - 1Vt 2 2

d = tan -1 a

20.0535 ft>s

20p ft>s - 34.6101 ft>s

b = 35.3964° = 35.4° Ans.

Since the water enters the blade radially, a1 = 90°, V1 = 1Vr 2 1 and 1Vt 2 1 = 0, the power onput of the pump is # Wout = rQ3U2 1Vt 2 2 - U1 1Vt 2 1 4 = a

62.4 lb>ft3 32.2 ft>s2

b110.5 ft3 >s23120p ft>s2134.6101 ft>s2 - 04

= 44,248.67 ft # lb>s

Thus, the power input to the pump is # 44,248.67 ft # lb>s # Wout Win = = h 0.65 = 168,074.88 ft # lb>s2a = 123.77 hp = 124 hp

1 hp

550 ft # lb>s

b

Ans. Ans: # Win = 124 hp 1290

M14_HIBB9290_01_SE_C14_ANS.indd 1290

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14–21. Show that the ideal head for a radial-flow pump can be determined from hpump = (U2V2 cos a2)>g, where V2 is the velocity of the water leaving the impeller blades. Water enters the blades in the radial direction.

SOLUTION

(Vrel(2

(Vr(2

V2

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

b2

a2

(Vt (2

b2 U2

(a )

Thus, ∆H = =

3 U2 ( Vt ) 2

- U1 ( Vt ) 1 4 g

=

U2 ( V2 cos a2 ) - 0 g

U2V2 cos a2 g

(Q.E.D)

1291

M14_HIBB9290_01_SE_C14_ANS.indd 1291

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14–22. The impeller 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 blades, 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 kinematic diagram at the lead, Fig. a,

(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,

(Vrel(1

60°

(a) Lead

( Vt ) 2 = U2 - ( Vr ) 2 cot b2 = 6p m>s - ( 1.592 m>s ) cot 30° = 16.093 m>s

(Vr(2 =

4.775 m/s

a1

b1 = 60°

b2 = 30°

U1 = 2 m/s

(V (1 = 4.775 m/s

Thus, the speed of the water leaving the blade isr

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. V1

The ideal head supplied by the pump is hpump = =

U2 ( Vt ) 2 - U1 ( Vt ) 1

a1

gb1 = 60°

(Vt ()1 - ( 2p m>s )( 3.527 m>s ) ( 6p m>s )( 16.093 m>s 60°

= 28.66 m = 28.7 m

9.81 m>s2 (a) Lead

(Vr(2 = 1.592 m/s

4.775 m/s

U1 = 2 m/s

b2 = 30°

Ans.

(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

1292

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14–23. 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 = rQVf>cs(1 + cos u)r

#

= ( 1000 kg>m3 )( 0.2513 m3 >s )( 7 m>s ) (1 + cos 30°)(2.5 m) = 8207.2 N # m = 8.21 kN # m

Ans.

Wturb = rQVf>cs U(1 + cos u) = ( 1000 kg>m3 )( 0.2513 m3 >s )( 7 m>s )( 25 m>s )(1 + cos 30°) = 82,072 W = 82.1 kW

Ans.

Ans: T# = 8.21 kN # m Wturbine = 82.1 kW 1293

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*14–24. The buckets of the Pelton wheel deflect the 4-in.diameter water jet 120° as shown. If the velocity of the water from the nozzle is 72 ft>s, determine the power delivered to the shaft of the wheel if the wheel is rotating at a constant angular velocity of 30 rev>min.

v 8 ft

72 ft!s 120°

SOLUTION We will consider the relative flow as steady and the water as incompressible. The discharge from the nozzle is Q = VA = 172 ft>s2 c pa

The angular velocity of the wheel is v = a30

2 2 ft b d = 2p ft3 >s 12

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

Thus, the speed of the bucket is

U = vr = 1p rad>s218 ft2 = 8p ft>s

Thus, the relative velocity is

Vj>cs = V - U = 72 ft>s - 8p ft>s = 46.8673 ft>s With u = 180° - 120° = 60°, the power delivered to the shaft of the wheel is # Wturb = r QV j>csU11 + cos u2 = a

62.4 lb>ft3 32.2 ft>s2

b12p ft3 >s2146.8673 ft>s218p ft>s211 + cos 60°2

= 121 513.41 ft # lb>s2a = 39.12 hp = 39.1 hp

1 hp

550 ft # lb>s

b

Ans.

Ans: # Wturb = 39.1 hp 1294

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14–25. The buckets of the Pelton wheel deflect the 4-in.diameter water jet 120° as shown. If the velocity of the water from the nozzle is 72 ft>s, determine the angular velocity of the wheel to produce a torque of T = 6.851103 2 lb # ft on the shaft. How fast must the wheel be turning to maximize the power developed by the wheel?

v 8 ft

72 ft!s 120°

SOLUTION We will consider the relative flow as steady and the water as incompressible. The discharge from the nozzle is Q = VA = 172 ft>s2 c pa

The speed of the bucket is

U = vr;

2 2 ft b d = 2p ft3 >s 12

U = v18 ft2 = 8v

Thus, the relative velocity is Vj>cs = V - U = 72 ft>s - 8v = 72 - 8v

With u = 180° - 120° = 60° and the required torque T = 6.851103 2 lb # ft, for T = 6.851103 2 lb # ft,

T = r Q Vj>cs 11 + cos u2r

6.851103 2 lb # ft = a

62.4 lb>ft3 32.2 ft>s2

b12p ft3 >s2172 - 8v211 + cos 60°218 ft2

rad 1 rev 60 s ba ba b s 2p rad 1 min rev = 29.98 min = 30.0 rev> min

v = a3.1398

Ans.

From the text, the maximum power generated by the wheel occurs when the bucket has a speed of U =

72 ft>s V = = 36 ft>s 2 2

# # Thus, the corresponding wheel’s angular velocity is, for W = Wmax, v =

36 ft>s U rad 1 rev 60 s = = a4.50 ba ba b r 8 ft s 2p rad 1 min rev = 42.97 min rev = 4.30 min

Ans.

Ans: For T = 6.851103 2 lb # ft, v = 30.0 rev>min # # For W = Wmax, v = 43.0 rev>min 1295

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14–26. 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 + 15V 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

Vp = 0.04Vj

(2)

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 # Wturbine = 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 Wturbine = 10.3 kW

1296

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14–27. The blades of a gas turbine are rotating at 80 rad>s. If the mean radius of the blades is 1.25 ft, and the horizontal velocity of the flow entering the blades is Va = 20 ft>s, determine the relative velocities of the gas as it flows smoothly onto and off the blades. Also, what is the required angle b1 at the entrance of the blades, and exit angle a2?

20°

15° 20 ft!s

SOLUTION We will consider the relative flow as steady and the gas as incompressible. The speed of the midpoint on the turbine’s blade is

Stator

Turbine

U = vrm = 180 rad>s211.25 ft2 = 100 ft>s

With this result, Va = 20 ft>s, a1 = 20° and b2 = 15°, the kinematic diagram of the gas at the lead and tail of the blade are shown in Figs. a, and b, respectively. From the geometry of Fig. a, Va = 1Vt 2 1 tan a1;

b1 = tan -1 c

20 ft>s = 1Vt 2 1 tan 20°

20 ft>s Va d = tan -1 a b = 23.9387° = 23.9° U - 1Vt 2 1 100 ft>s - 54.9495 ft>s

Ans.

Va sin b1 = ; 1Vrel 2 1

sin 23.9387° =

From the geometry of Fig. b, sin b2 = tan b2 = Then

1Vt 2 1 = 54.9495 ft>s

Va ; 1Vrel 2 2

sin 15° =

Va ; U - 1Vt 2 2

a2 = tan -1 c

20 ft>s 1Vrel 2 1

20 ft>s 1Vrel 2 2

tan 15° =

1Vrel 2 1 = 49.2904 ft>s = 49.3 ft>s

1Vrel 2 2 = 77.2741 ft>s = 77.3 ft>s 20 ft>s

100 ft>s - 1Vt 2 2

1Vt 2 2 = 25.3590 ft>s

20 ft>s Va d = tan -1 a b = 38.2620° = 38.3° 1Vt 2 2 25.3590 ft>s

U 5 100 ftys

V1

a1 5 20°

Ans.

U 5 100 ftys

b1

(Vt )1

Ans.

b2 5 15° a2 (Vt )2

V2 Va 5 20 ftys

Va 5 20 ftys

b1

(Vrel)1 (a)

b2 5 15°

(Vrel)2 (b)

Ans: b1 = 23.9° 1Vrel 2 1 = 49.3 ft>s 1Vrel 2 2 = 77.3 ft>s a2 = 38.3° 1297

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*14–28. Water flowing at Va = 20 ft>s is directed from the stator onto the blades of an axial flow turbine at an angle of a1 = 30° . If the turbine blades are rotating at 60 rad>s, determine the required angle b1 of the blades so that they properly accept the flow. What should be the angle a2 of the stator so that water is delivered to it from the turbine at an angle of b2 = 15°? The turbine has a mean radius of 1.5 ft.

b2 b1

a2

a1

SOLUTION We will consider the relative flow as steady and the water as incompressible. The speed of the midpoint on the turbine’s blade is

Stator blades

U = vrm = 160 rad>s211.5 ft2 = 90 ft>s

Turbine blades

Stator blades

With this result, Va = 20 ft>s, a1 = 30° and b2 = 15°, the kinematic diagram of the water at the lead and tail of the blade shown in Figs. a and b, respectively, can be drawn. From the geometry of Fig. a,

Then

Va = 1Vt 2 1 tan a1;

tan b1 =

20 ft>s = 1Vt 2 1 tan 30°

20 ft>s Va = U - 1Vt 2 1 90 ft>s - 34.6410 ft>s

1Vt 2 1 = 34.6410 ft>s b1 = 19.8637° = 19.9°

Ans.

From the geometry of Fig. b, tan b2 = Then

Va ; U - 1Vt 2 2

a2 = tan -1 c

tan 15° =

20 ft>s 90 ft>s - 1Vt 2 2

1Vt 2 2 = 15.3590 ft>s

20 ft>s Va d = tan -1 a b = 52.4776° = 52.5° 1Vt 2 2 15.3590 ft>s

b1

a1 5 30° (Vt )1

(Vrel)1

Va 5 20 ftys V1

b1

U 5 90 ftys (a)

Ans.

(Vrel)2

b2 5 15°

Va 5 20 ftys

(Vt )2 a2

V2 b2 5 15°

U 5 90 ftys (b)

Ans: b1 = 19.9° a2 = 52.5° 1298

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14–29. Water flowing at Va = 20 ft>s is directed from the stator onto the blades of the axial-flow turbine at an angle of a1 = 30°, where the mean radius of the blades is 1.5 ft. If b2 = 15°, the turbine blades are rotating at 60 rad>s, and the flow is 350 ft3 >s, determine the torque produced by the water.

b2 b1

a2

a1

SOLUTION We will consider the relative flow as steady and the water as incompressible. The speed of the midpoint on the turbine’s blade is

Stator blades

U = vrm = 160 rad>s211.5 ft2 = 90 ft>s

Turbine blades

Stator blades

With this result, Va = 20 ft>s, a1 = 30° and b2 = 15°, the kinematic diagram of the water at the lead and tail of the blade shown in Figs. a and b, respectively, can be drawn. From the geometry of Fig. a, Va = 1Vt 2 1 tan a1;

From the geometry of Fig. b, Va tan b2 = ; U - 1Vt 2 2

20 ft>s = 1Vt 2 1 tan 30° tan 15° =

1Vt 2 1 = 34.6410 ft>s

20 ft>s 90 ft>s - 1Vt 2 2

Va = 4 m/s

1Vt 2 2 = 15.3590 ft>s

The torque produced by water and exerted on the turbine’s shaft is

a1 = 28 °

62.4 lb>ft3 32.2 ft>s2

a2 = 43 °

(Vt (2

T = - rQrm 31Vt 2 2 - 1Vt 2 1 4 = -a

Va = 4 m/s

V2 (b)

(Vt (1

b1350 ft3 >s211.5 ft23115.3590 ft>s2 - 134.6410 ft>s24

= 19.6171103 2 lb # ft = 19.61103 2 lb # ft

(a)

V1

Ans.

Ans: T = 19.6(103) lb # ft 1299

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14–30. 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

SOLUTION 0.6 m

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 = V1 cos a1 = ( 20 m>s ) cos 35° = 16.3830 m>s Then, the discharge is

(Vt (1

Q = (Vr)1 A1 = ( 11.4715 m>s ) 3 2p(0.6 m)(0.05 m) 4 = 2.1623 m >s 3

a1 = 35 °

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>m

3

(Vr(1 V1 = 20 m/s (a )

)( 2.1623 m >s )3 0 - (0.6 m) ( 16.3830 m>s ) 4 3

= - 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: # Ws = 167 kW 1300

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14–31. 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

SOLUTION 0.6 m

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 = V1 cos a1 = ( 20 m>s ) cos 35° = 16.3830 m>s h(m)

Then, the discharge is

800

Q = ( Vr ) 1 A1 = ( 11.4715 m>s ) 3 2p(0.6 m)(0.05 m) 4 = 2.1623 m3 >s

700

Since the water exits the blade radially, (Vt)2 = 0. The torque the water exerts on the turbine’s shaft is 480

T = rQ 3 r2(Vt)2 - r1(Vt)1 4

Pump performance curve

500 400 300

= ( 1000 kg>m3 )( 2.1623 m3 >s ) 3 0 - (0.6 m) ( 16.3830 m>s )4

200

= - 21.26 1 103 2 N # m

100 0

Q(m3/s( 05 0.0 10 0.0 15 0.0 20 0.0 25 0.0 30 0.0 35 0.0 40

rev 1 min 2p rad ba ba b2p rad>s min 60 s 1 rev

0.0

Here, v = a75

hreg

600

Then, the power the water supplies to the turbine is # Ws = Tv = 3 -21.26 1 103 2 N # m 4 ( 2p rad>s )

(a)

0.03375

= - 166.94 1 103 2 W

The ideal head the turbine draws from the water is # 166.94 ( 103 ) W Ws hturbine = = rQg ( 1000 kg>s )( 2.1623 m3 >s )( 9.81 m>s2 ) = 7.869 m = 7.87 m

Ans.

Ans: hturbine = 7.87 m 1301

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*14–32. The blades of a Kaplan turbine have a mean radius of 0.6 m and are rotating at 50 rad>s, which provides a flow of 40 m3 >s. If the angles for each blade are a1 = 35°, b1 = 70°, and b2 = 40°, determine the ideal power supplied to the turbine.

v

SOLUTION We will consider the relative flow as steady and the water as incompressible. The speed of the midpoint on the turbine’s blade is

a1 5 35°

U = vrm = 150 rad>s210.6 m2 = 30 m>s

75° (Vrel)1

Va

30 m>s V1 = sin 70° sin 75° b2 5 40°

Va = V1 sin a1 = 129.1852 m>s2 sin 35° = 16.740 m>s

16.7400 m>s

The ideal power supplied by the turbin’s shaft is # Wturb = rQU31Vt 2 2 - 1Vt 2 1 4

b2 5 40°

Va 5 16.740 mys

Using the result of Va, U and b2 = 40°, the kinematic diagram of the water at the tail of the blade shown in Fig. b can be drawn. From the geometry of this figure, 30 m>s - 1Vt 2 2

U 5 30 mys

a2

(Vrel)2

1Vt 2 1 = V1 cos a1 = 129.1852 m>s2 cos 35° = 23.9072 m>s tan 40° =

V1

(Vt )2

Then

Va ; U - 1Vt 2 2

b1 5 70°

(a)

V1 = 29.1852 m>s

tan b2 =

U 5 30 mys

b1 5 70°

Using this result, a1 = 35°, and b1 = 70°, the kinematic diagram of the water at the lead of the blade shown in Fig. a can be drawn. From the geometry of this figure, V1 U = ; sin p1 sin 1180° - a1 - b1 2

(Vt )1

V2

(a)

1Vt 2 2 = 10.0501 m>s

= 11000 kg>m3 2140 m3 >s2130 m>s2110.0501 m>s - 23.9072 m>s2 = - 16.6281106 2 W = - 16.6 MW

Ans.

The negative sign indicates that the power is withdrawn from the water.

Ans: # Wturb = -16.6 MW 1302

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14–33. The velocities on and off the 90-mm-wide blades of the Francis 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 30° b1

b2 V2 80 rad!s

150 mm

275 mm

SOLUTION U1 = 22 m/s

Water is considered to be incompressible. The relative flow is steady.

(Vt (1

a1 = 30 °

The speed of points on the head and tail of the impeller blade are

b1

U1 = vr1 = ( 80 rad>s )(0.275 m) = 22 m>s

V1 = 18 m/s

U2 = vr2 = ( 80 rad>s )(0.15 m) = 12 m>s

U2 = 12 m/s

( Vt ) 1 = V1 cos a1 = ( 18 m>s ) cos 30° = 15.5885 m>s

(Vr(1

(Vrel(1

(a)

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

b1

b2 b2

(Vrel(2

Then, tan b1 =

( Vr ) 1 9 m>s = 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 + U22 = 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 1303

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14–34. The velocities on and off the 90-mm-wide blades of a Francis 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 30° b1

b2 V2 80 rad!s

150 mm

275 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

( Vr ) 1 = 9.0027 m>s

a1 = 30°

(Vr(1 = 9.0027 m/s

V1

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;

(Vt (1

(a )

9.0027 m>s = ( Vt ) 1 tan 30°

( 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 =

3 - 6.0034 ( 103 ) N # m 4( 80 rad>s )

= - 480.27 ( 103 ) W = - 480 kW

Ans.

Ans: # Ws = 480 kW 1304

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14–35. 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 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

V1

90°

(Vt (1

a2

105 ft>s V1 = sin 60° sin 90°

Va = V1 sin a1 = ( 90.9327 ft>s ) sin 30° = 45.4663 ft>s

45.4663 ft>s =

3 105 ft>s

( Vt ) 2 = 26.25 ft>s

U = 105 ft/s

b1 = 60°

(Vt (1

- ( Vt ) 2 4 tan 30°

Thus, the magnitude of V2 is V2 = 2 (

) + Va = 2 ( 26.25 ft>s ) + ( 45.4663 ft>s )

Vt 22

2

2

b2 = 30°

(Vrel(2

2

= 52.5 ft>s

(b)

Ans.

90°

U = 105 ft/s V1

Using this result, U, and b2 = 30°, the velocity diagram of the water at the blade’s a1 = 30° tail is shown in Fig. b. From the geometry of this figure, - ( Vt ) 2 4 tan b2;

(Vrel(1

Va = 45.4663

(a)

Then,

3U

(Vt (2

Va b1 = 60°

V1 = 90.9327 ft>s = 90.9 ft>s

Va =

b2 = 30°

a1 = 30°

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)

U = 105 ft/s

Va b1 = 60°

(Vrel(1

b2 = 30° a2

(Vt (2

Va = 45.4663

b2 = 30°

(Vrel(2

(a) (b)

Ans.

Ans: V1 = 90.9 ft>s V2 = 52.5 ft>s 1305

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*14–36. Water enters the blades of the Francis turbine with a velocity of 16 m>s as shown. If the blades are rotating at 120 rev>min, and the flow off the blades is radial, determine the power the water supplies to the turbine. The blades are 100 mm wide.

16 m!s 40°

120 rev !min

0.8 m

SOLUTION We will consider the relative flow as steady and the water as incompressible. Using V1 = 16 m>s and a1 = 40°, the kinematic diagram of the water at the lead of the blade shown in Fig. a can be drawn. From the geometry of this figure, 1Vr 2 1 = V1 sin a1 = 116 m>s21sin 40°2 = 10.2846 m>s

1Vt 2 1 = V1 cos a1 = 116 m>s21cos 40°2 = 12.2567 m>s

(Vt)1

Then the discharge is

a1 5 40°

Q = 1Vr 2 1A1 = 110.2846 m>s232p10.8 m210.1 m24 = 5.1696 m3 >s

Since the water exits the blade radially, 1Vt 2 2 = 0. The torque the water exerts on the turbine’s shaft is

V1 5 16 mys

(Vr)1

T = -rQ3r2 1Vt 2 2 - r1 1Vt 2 1 4

= - 11000 kg>m3 215.1696 m3 >s230 - 10.8 m2112.2567 m>s24 = 50.68991103 2 N # m

rev 1 min 2p rad ba ba b = 4p rad>s. Then the power the water min 60 s 1 rev supplies to the turbine is Here, v = a120

# Wturb = Tv350.68991103 2 N # m414p rad>s2 = 636.991103 2 W = 637 kW

Ans.

Ans: # Wturb = 637 kW 1306

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14–37. Water enters the blades of the Francis turbine with a velocity of 16 m>s as shown. If the blades are rotating at 120 rev>min, and the flow off the blades is radial, determine the ideal head the turbine draws from the water. The blades are 100 mm wide.

16 m!s 40°

120 rev !min

0.8 m

SOLUTION We will consider the relative flow as steady and the water as incompressible. Using V1 = 16 m>s and a1 = 40°, the kinematic diagram of the water at the lead of the blade shown in Fig. a can be drawn. From the geometry of this figure, 1Vr 2 1 = V1 sin a1 = 116 m>s21sin 40°2 = 10.2846 m>s

1Vt 2 1 = V1 cos a1 = 116 m>s21cos 40°2 = 12.2567 m>s

(Vt)1

Then, the discharge is

a1 5 40° 3

Q = 1Vr 2 1A1 = 110.2846 m>s232p10.8 m210.1 m24 = 5.1696 m >s

Since the water exit the blade radially, 1Vt 2 2 = 0. The torque the water exerts on the turbine’s shaft is

V1 5 16 mys

(Vr)1

T = -rQ3r2 1Vt 2 2 - r1 1Vt 2 1 4

= - 11000 kg>m3 215.1696 m3 >s230 - 0.8 m112.2567 m>s24 = 50.68991103 2 N # m

rev 1 min 2p rad ba ba b = 4p rad>s. Then the power the water min 60 s 1 rev supplies to the turbine is

Here v = a120

# Wturb = Tv350.68991103 2 N # m414p rad>s2 = 636.991103 2W

Ans.

= 12.5604 m = 12.6 m

Ans.

The ideal head the turbine draws from the water is # 636.991103 2W Wturb hturbine = = rgQ 11000 kg>m3 219.81 m>s2 215.1696 m3 >s2

Ans: hturbine = 12.6 m 1307

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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, 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

a1 = 30°

Ans.

V1

(Vt (1

(a)

Then, the power the water supplies to the turbine is

#

Ws = Tv =

( -229.72 N # m )( 40 rad>s )

= - 9.1888 ( 103 ) W = -9.19 kW

Ans.

Ans: T# = 230 N # m Ws = 9.19 kW 1308

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14–39. 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 0.5 m3 >s = ( Vr ) 1 3 2p(1.2 m)(0.3 m) 4

Q = ( Vr ) 1A1;

( 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 =

∆H 1.8734 m = = 0.6244 = 62.4% ∆Ht 3m

Ans.

Ans: h = 62.4% 1309

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*14–40. 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;

(Vt (1

8 m3 >s = ( Va ) 1 3 p 3 (0.6 m)2 - (0.2 m)2 4 4

a1 = 50 °

( Va ) 1 = 7.9577 m>s

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;

Va = 7.9577 m/s

V1 (a )

7.9577 m>s = ( Vt ) 1 tan 50°

( Vt ) 1 = 6.6773 m>s The mean radius of the impeller blade is rm =

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

Ans.

Ans: # W = 598 kW 1310

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14–41. The pipe system consists of a 3-in.-diameter, 100-ftlong galvanized iron pipe, a fully opened gate valve, four elbows, a flush entrance, and a pump with the pump head curve shown. If the friction factor is f = 0.022, estimate the flow and corresponding pump head generated by the pump.

B

18 ft

A

SOLUTION We will consider the flow as steady and the water as incompressible. The loss coefficient for the flush entrance, fully opened gate valve and elbow are 0.5, 0.19, and 0.9, respectively. Thus, the total head loss is hL = af

VB2

VB2

100 ft L d + ΣkLb = 0.022 + 0.5 + 0.19 + 410.92 c D 2g £ ° 3 ¢ § 2132.2 ft>s2 2 ft 12 = 0.20326VB2

(1)

hpump (ft) 160 140 120 100 80 60

Using the discharge, Q = VBAB;

40 20

2

1.5 ft b d 12 VB = 20.3718Q Q = VB c pa

(2)

Q gal!min 50 100 150 200 250 300 350 400 450 500

Substitute Eq. (2) into (1). hL = 84.3556Q2 Write the energy equation between A and B. Realizing that pA = pB = patm = 0 (A and B opened to atmosphere), zA = 0 and zB = 18 ft (datum through A), VA _ 0 (large reservoir) and hturb = 0, pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 0 + 0 + 0 + hpump = 0 +

120.3718Q2 2

2132.2 ft>s2 2

+ 18 ft + 0 + 84.3556Q2

hpump = 190.7999Q2 + 182 ft

Q (gal/min)

0

50

100

150

200

250

300

350

400

450

500

Q (ft3/s)

0

0.111

0.223

0.334

0.446

0.557

0.668

0.780

0.891

1.003

1.114

hpump (ft)

18

19.13

22.51

28.14

36.03

46.17

58.57

73.22

90.13

109.29 130.70

a

gal min

ba

1 ft3 1 min ba b = ft3 >s 7.48 gal 60 s

hpump (ft)

The plot of these data on the pump performance curve is shown in Fig. a. By locating the operating point, Q = 420 gal>min and hpump = 97.5 ft

Ans.

160 140 120

97.5

Operating point

100 80 60 40 20 0

420 50

100 150 200 250 300 350 400 450 500 (a)

Q (galymin)

Ans: Q = 420 gal>min hpump = 97.5 ft

1311

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14–42. Water at T = 25°C is drawn from an underground detention tank and discharged into a drain using a 6-m-long pipe and pump. The pipe has a diameter of 100 mm and a friction factor of f = 0.02. Determine if cavitation occurs when the velocity through the pipe is 4 m>s and h = 3 m. Use the pump performance curves in Fig. 14–16. The atmospheric pressure is 101.3 kPa. Neglect minor losses.

B

h A

SOLUTION We will consider the flow as steady and the water as incompressible. Write the energy equation between A and B, realizing that pA = patm = 101.3 kPa (A is opened to atmosphere and the absolute pressure will be used), zA = 0 and zB = 3 m (datum through A), VA ≃ 0 (large reservoir) and hpump = hturb = 0. First, it is necessary to calculate the head loss. Also, from the table in Appendix A, rw = 997.1 kg>m3 and pr = 3.17 kPa for water at T = 25°C. L V2 + ΣkL b D 2g 14 m>s2 2 6m d = 0.9786 m = c 10.022a b + 0d c 0.1 m 219.81 m>s2 2

hL = af Then,

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 101.31103 2 N>m2 3

2

1997.1 kg>m 219.81 m>s 2

+ 0 + 0 + 0 =

pB 3

2

1997.1 kg>m 219.81 m>s 2

pB = 54.4061103 2 Pa

+

14 m>s2 2

219.81 m>s2 2

+ 3 m + 0 + 0.9786 m

Therefore, the available suction head at the pump inlet is

Then

14 m>s2 2 54.4061103 2 Pa pB VB2 + = + = 6.3776 m gw 2g 219.81 m>s2 2 1997.1 kg>m3 219.81 m>s2 2 1NSPH2 Avail = 6.3776 m -

Here, the discharge is

3.171103 2 N>m2

1997.1 kg>m3 219.81 m>s2 2

= 6.0536 m

Q = VA = 14 m>s23p10.05 m2 2 4 = a0.01p

3 7.48 gal 1 ft m3 60 s ba b a b = 497.92 gal>min ba s 0.3048 m 1 min 1 ft3

Enter this value of Q into Fig. 14–16.

Since 1NSPH2 Avail

0.3048 m b = 5.79 m ft 7 1NSPH2 req‘d, the cavitation will not occur.

1NSPH2 req‘d = 119 ft2 a

Ans.

Ans: Cavitation will not occur. 1312

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14–43. Water at T = 25°C is drawn from an underground detention tank and discharged into a drain using a 6-m-long pipe and pump. The pipe has a diameter of 100 mm and a friction factor of f = 0.02. Determine if cavitation occurs when the velocity through the pipe is 4 m>s and h = 4 m. Use the pump performance curves in Fig. 14–16. The atmospheric pressure is 101.3 kPa. Neglect minor losses.

B

h A

SOLUTION We will consider the flow as steady and the water as incompressible. Write the energy equation between A and B, realizing that pA = patm = 101.3 kPa (A is opened to atmosphere and the absolute pressure will be used), zA = 0 and zB = 4 m (datum through A), VA ≃ 0 (large reservoir) and hpump = hturb = 0. First, it is necessary to calculate the head loss. Also, from the table in Appendix A, rw = 997.1 kg/m3 and pv = 3.17 kPa for water at T = 25°C. hL = af

L V2 + ΣkL b D 2g

= c 0.02a

Then

14 m>s2 2 6m d = 0.9786 m b + 0d c 0.1 m 219.81 m>s2 2

pA pB VA2 VB2 + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g 101.31103 2 N>m2 3

2

1997.1 kg>m 219.81 m>s 2

+ 0 + 0 + 0 =

pB 3

2

1997.1 kg>m 219.81 m>s 2

pB = 44.6241103 2 Pa

+

14 m>s2 2

219.81 m>s2 2

+ 4 m + 0 + 0.9786 m

Therefore, the available suction head at the pump inlet is

Then

44.6241103 2 N>m2 14 m>s2 2 pB VB2 + = + = 5.3776 m gw 2g 219.81 m>s2 2 1997.1 kg>m3 219.81 m>s2 2 1NSPH2 Avail = 5.3776 m -

Here, the discharge is

3.171103 2 N>m2

1997.1 kg>m3 219.81 m>s2 2

= 5.0536 m

Q = VA = 14 m>s23p10.05 m2 2 4 = a0.01p

3 7.48 gal 1 ft m3 60 s ba b a b = 497.92 gal>min ba s 0.3048 m 1 min 1 ft3

Enter this value of Q into Fig. 14–16.

1NSPH2 req‘d = 119 ft2a

0.3048 m b = 5.79 m ft

Since 1NSPH2 Avail 6 1NSPH2 req‘d, the cavitation will occur.

Ans.

Ans: Cavitation will occur. 1313

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*14–44. 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 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 yw = 1.00 ( 10-6 ) m2 >s at 20°C. The average velocity is

A

Q Q 1600 = = Q p A p(0.025 m)2

V = Thus, the Reynolds number is

1600 Qb(0.05 m) p

=

1.00 ( 10-6 ) m2 >s

80 ( 106 ) Q p

(1) h(m)

The loss coefficient for an elbow is 0.9. Thus, the total head loss is

800

L V2 V2 hL = f + ΣkL D 2g 2g

700

L V2 = af + ΣkL b D 2g

500 480 400 2 1600 Qb p

300

50 m = cf a b + 5(0.9) d 0.05 m 2 ( 9.81 m>s2 )

200 100 01 5 0. 02 0 0. 02 0. 5 03 0 0. 03 5 0. 04 0

Q (m3/s( 5

0.

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

)

01

= (1000f + 4.5) ( 13 220.30 Q

2

Pump performance curve

00

a

VA2

hreg

600

0.

a

0.

VD Re = = yw

(a)

0.03375

VB2

pA pB + + zA + hpump = + + zB + hturb + hL gw gw 2g 2g

0 + 0 + 0 + hreq = 0 +

a

2 1600 Qb p

2 ( 9.81 m>s2 )

+ 8 m + 0 + (1000f + 4.5)( 13 220.30Q2 )

hreq = (1000f + 5.5) ( 13 220.30Q2 ) + 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.

1314

M14_HIBB9290_01_SE_C14_ANS.indd 1314

Ans: Q = 0.03375 m3 >s

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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, p in 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 1315

M14_HIBB9290_01_SE_C14_ANS.indd 1315

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14–46. 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 pipe.

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 + 5 ft + hpump = 0 + 0 + 115 ft + 0 + 0.02£ 3 ≥ + 0 ft 2(32.2 ft>s) 12 hpump = 110 + 0.1242 V 2 Q = VA = Vp a

Thus,

2 1.5 ft b = 0.04909V 12

hpump = 110 + 51.554Q2 This equation must be solved along with the curve in Fig. 14–16 for the 6-in. impeller. 1 min 1 ft3 Assume Q = 300 gal>mina ba b = 0.6684 ft3 >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: Q = 400 gal>min 1316

M14_HIBB9290_01_SE_C14_ANS.indd 1316

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14–47. 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 level is at h = 80 ft. Neglect friction losses in the 3-in.-diameter hose, 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 + 0.05434V 2 Q = VA = Vpa Thus,

2 1.5 ft b = 0.04909V 12

hpump = 75 + 22.55Q2 This equation is solved along with the curve in Fig. 14–16 for the 5.5-in. impeller. Assume Q = 500 gal>mina

1 min 1 ft3 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: Q = 500 gal>min 1317

M14_HIBB9290_01_SE_C14_ANS.indd 1317

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*14–48. A pump delivers 200 gal>min. when operating at 1770 rev>min. to supply a total head of 80 ft. If the flow is held constant but the head is increased to 120 ft, determine the speed at which a substitute pump should operate and yet provide the same efficiency?

SOLUTION From the head coefficient, h1 v12D12

h2 v22D22

D1 h2 v2 = v1 D2 A h1

From the flow coefficient, Q1 v1D13

Substitute Eq. (2) into (1).

=

=

(1)

Q2 v2D23

D1 v2 1>3 Q1 1>3 = a b a b v1 D2 Q2

(2)

v2 1>3 Q1 1>3 h2 v2 = a b a b v1 v1 Q2 B h1 v2 =

Q1 h2 3>4 a b v1 A Q2 h1

Here, Q1 = Q2, v1 = 1770 rev>min, h1 = 80 ft and h2 = 120 ft. Then v2 = 112a

120 ft 3>4 b 11770 rev>min2 = 2399.06 rev>min = 2399 rev>min 80 ft

Ans.

Ans: v2 = 2399 rev>min 1318

M14_HIBB9290_01_SE_C14_ANS.indd 1318

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14–49. A radial flow pump whose 150-mm-diameter impeller rotates at 200 rad>s produces a change in ideal head of 300 mm. Determine the change in head for a geometrically similar pump that has an impeller diameter of 300 mm and operates at 90 rad>s.

SOLUTION We will consider the relative flow as steady and the water as incompressible. Dynamic similitude of the head coefficient gives h1 v12D12

=

h2 v22D22

h2 0.3 m = 1200 rad>s2 2 10.15 m2 2 190 rad>s2 2 10.3 m2 2 h2 = 0.243 m

Ans.

Ans: h2 = 0.243 m 1319

M14_HIBB9290_01_SE_C14_ANS.indd 1319

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14–50. A radial flow pump whose 100-mm-diameter impeller rotates at 120 rad>s discharges water at 0.25 m3 >s. Determine the discharge of a geometrically similar pump having a 150-mm-diameter impeller that rotates at 90 rad>s.

SOLUTION We will consider the relative flow as steady and the water as incompressible. Dynamic similitude of the flow coefficient gives Q1 v1D13

=

3

0.25 m >s

1120 rad>s210.1 m2

3

=

Q2 v2D23 Q2 190 rad>s210.15 m2 3

Q2 = 0.633 m3 >s

1320

M14_HIBB9290_01_SE_C14_ANS.indd 1320

Ans.

Ans: Q2 = 0.633 m3 >s

20/03/17 5:45 PM

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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 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 b a b v1 Q1 D2 650 gal>min 900 gal>min

= 1264 rpm

b ( 13 ) (1750 rpm)

Ans.

Ans: v2 = 1264 rpm 1321

M14_HIBB9290_01_SE_C14_ANS.indd 1321

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*14–52. 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

D2 = a = a

1

Q2 3 v1 3 b a b D1 v2 Q1

650 gal>min 900 gal>min

= 5.38 in.

1 3

1

b (1)3 (6 in.)

Ans.

Ans: D2 = 5.38 in. 1322

M14_HIBB9290_01_SE_C14_ANS.indd 1322

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14–53. A pump provides a discharge of 300 gal>min when operating at a head of 25 ft with its impeller rotating at 900 rev>min. Determine the discharge and head if the same pump is used with its impeller rotating at 1200 rev>min.

SOLUTION We will consider the relative flow as steady and the water as incompressible. Since the same pump is used for both cases, D1 = D2 = D. Dynamic similitude of the flow coefficient gives Q1 v1D13 300 gal>min 1900 rev>min2D3

= =

Q2 v2D2 3 Q2 11200 rev>min2D3

Ans.

Q2 = 400 gal>min

Dynamic similitude of the head coefficient gives h1 v12D12

=

h2 v22D22

h2 25 ft = 2 2 1900 rev>min2 D 11200 rev>min2 2D2 h2 = 44.4 ft

Ans.

Ans: Q2 = 400 gal>min h2 = 44.4 ft 1323

M14_HIBB9290_01_SE_C14_ANS.indd 1323

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14–54. A pump has a 6-in. diameter impeller that discharges 150 gal>min. If the power required is 3 hp, determine the discharge of a similar pump having an impeller diameter of 8 in. that requires a power of 5 hp.

SOLUTION We will consider the relative flow as steady and the water as incompressible. Dynamic similitude of the power coefficient gives # # W2 W1 = 3 5 v13D15 v 2 D2 # W2 v2 3 D2 5 (1) # = a b a b v1 D1 W1 Dynamic similitude of the flow coefficient gives Q1 v1D13

=

Q2

v2D2 3 v2 D1 3 Q2 = a b a b v1 D2 Q1

(2)

Substituting Eq. (2) into (1), # W2 D1 4 Q2 3 # = a b a b D2 Q1 W1 # 1>3 W2 D2 4>3 Q2 = a # b a b Q1 D1 W1 = a

5 hp

3 hp

1>3

b

a

8 in. 4>3 b 1150 gal>min2 6 in.

Ans.

= 260.99 gal/min = 261 gal/min

Ans: Q2 = 261 gal>min 1324

M14_HIBB9290_01_SE_C14_ANS.indd 1324

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14–55. 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: # W2 = 1.31 hp 1325

M14_HIBB9290_01_SE_C14_ANS.indd 1325

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*14–56. 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

2

Dm

b = c

(∆H)p (∆H)m

da

vm 2 b vp

(1)

Dynamic similitude of the flow coefficient gives Qp vp Dp3

=

Qm vmDm3

Dp 3 vm Qm = a ba b vp Qp Dm

(2)

Substituting Eq. (2) into Eq. (1), and rearranging, a

Dp Dm

4

b = c

Dp = c

(∆H)m (∆H)p (∆H)m

= a

(∆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: Dp = 7.00 in. 1326

M14_HIBB9290_01_SE_C14_ANS.indd 1326

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