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258 Chapter 5 ■ Finite Control Volume Analysis<br />

but with a well-rounded entrance (see Fig. P5.102). The room is<br />

held at a constant pressure of 1.5 psi above atmospheric. Both vents<br />

exhaust into the atmosphere. The loss in available energy associated<br />

with flow through the cylindrical vent from the room to the<br />

vent exit is 0.5V 2 2/2, where V 2 is the uniformly distributed exit velocity<br />

of air. The loss in available energy associated with flow through<br />

the rounded entrance vent from the room to the vent exit is<br />

0.05V 2 2/2, where V 2 is the uniformly distributed exit velocity of air.<br />

4 in.<br />

Gas<br />

Oil<br />

Piston<br />

Channel<br />

1-in. diameter<br />

p = 0<br />

4 in.<br />

200 lb<br />

F I G U R E P5.107<br />

F I G U R E P5.102<br />

Section 5.3.2 Application of the Energy Equation–With<br />

Shaft Work<br />

5.108 What is the maximum possible power output of the hydroelectric<br />

turbine shown in Fig.P5.108?<br />

5.103 A gas expands through a nozzle from a pressure of 300 psia<br />

to a pressure of 5 psia. The enthalpy change involved, ȟ 1 ȟ2,<br />

is 150 Btu/lbm. If the expansion is adiabatic but with frictional effects<br />

and the inlet gas speed is negligibly small, determine the exit<br />

gas velocity.<br />

5.104 For the 180° elbow and nozzle flow shown in Fig. P5.104,<br />

determine the loss in available energy from section 112 to section 122.<br />

How much additional available energy is lost from section 122 to<br />

where the water comes to rest?<br />

50 m<br />

1 m<br />

6 in.<br />

6 m/s<br />

p 1 = 15 psi<br />

V 1 = 5 ft/s<br />

Section (2)<br />

12 in.<br />

F I G U R E P5.104<br />

Section (1)<br />

y<br />

x<br />

Turbine<br />

F I G U R E P5.108<br />

5.109 The pumper truck shown in Fig. P5.109 is to deliver<br />

1.5 ft 3 /s to a maximum elevation of 60 ft above the hydrant. The<br />

pressure at the 4-in.-diameter outlet of the hydrant is 10 psi. If head<br />

losses are negligibly small, determine the power that the pump<br />

must add to the water.<br />

5.105 An automobile engine will work best when the back pressure<br />

at the interface of the exhaust manifold and the engine block is minimized.<br />

Show how reduction of losses in the exhaust manifold, piping,<br />

and muffler will also reduce the back pressure. How could<br />

losses in the exhaust system be reduced? What primarily limits the<br />

minimization of exhaust system losses?<br />

†5.106 Explain how, in terms of the loss of available energy involved,<br />

a home sink water faucet valve works to vary the flow<br />

from the shutoff condition to maximum flow. Explain how you<br />

would estimate the size of the overflow drain holes needed in the<br />

sink of Video V5.1 (Video V3.9 may be helpful).<br />

5.107 (See Fluids in the News article titled “Smart shocks,” Section<br />

5.3.3.) A 200-lb force applied to the end of the piston of the shock absorber<br />

shown in Fig. P5.107 causes the two ends of the shock absorber<br />

to move toward each other with a speed of 5 fts. Determine the head<br />

loss associated with the flow of the oil through the channel. Neglect<br />

gravity and any friction force between the piston and cylinder walls.<br />

60 ft<br />

10 psi<br />

4-in.<br />

diameter<br />

Hydrant<br />

F I G U R E P5.109

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