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Modern Engineering Thermodynamics

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188 CHAPTER 6: First Law Open System Applications<br />

1<br />

2<br />

W p<br />

2<br />

(a) Centrifugal pump 1<br />

W M<br />

1<br />

2<br />

(b) Hydraulic pump<br />

W p<br />

(c) Hydraulic motor<br />

2<br />

1<br />

W C<br />

W T<br />

1 2<br />

(d) Axial flow compressor<br />

(e) Turbine<br />

W E<br />

W G<br />

W E<br />

W EM<br />

(f) Electric generator<br />

(g) Electric motor<br />

1<br />

W<br />

2<br />

FIGURE 6.11<br />

Graphical symbols for common shaft work machines.<br />

(h) Piston-type compressor or engine<br />

EXAMPLE 6.5<br />

Nearly every urban home has water supplied from a local water main. This water is used in washing and cooking, but its<br />

pressure could also be used as an energy supply. Suppose you installed a small hydraulic motor or turbine on the inlet<br />

water pipe of your house. Every time water is used in the house, the motor or turbine produces shaft work that could be<br />

used to run a small appliance or drive an electric generator and charge a battery. How much power could you realize in this<br />

way if you use an average of 20.0 gal of water over an eight (8.00) hour period, with an inlet water pressure of 85.0 psig<br />

and an exit water pressure of 10.0 psig?<br />

Solution<br />

First, draw a sketch of the system (Figure 6.12).<br />

The system is open, and the unknown is _W shaft : The material is liquid water, an incompressible fluid, and we ignore any<br />

changes in specific kinetic or potential energy of the flow stream plus any heat transfer that may occur. For steady state,<br />

steady flow, isothermal conditions, our modified energy rate balance (Eq. (6.12)) becomes Eq. (6.32), or<br />

_W shaft = _mvðp out − p in Þ<br />

where _mv = 20.0 gal/8.00 h = 2.50 gal/h (on average).

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