FluidMechanics1-Quiz3Soln

FluidMechanics1-Quiz3Soln - 172 Chapter 3 Integral...

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E3.19 EXAMPLE 3.19 A hydroelectric power plant (Fig. E3.19) takes in 30 m 3 /s of water through its turbine and dis- charges it to the atmosphere at V 2 5 2 m/s. The head loss in the turbine and penstock system is h f 5 20 m. Assuming turbulent flow, a < 1.06, estimate the power in MW extracted by the tur- bine. 172 Chapter 3 Integral Relations for a Control Volume Solution We neglect viscous work and heat transfer and take section 1 at the reservoir surface (Fig. E3.19), where V 1 < 0, p 1 5 p atm , and z 1 5 100 m. Section 2 is at the turbine outlet. The steady-flow en- ergy equation (3.71) becomes, in head form, } p g 1 } 1 } 2 1 V g 1 2 } 1 z 1 5 } p 2 } 1 } 2 2 V g 2 2 } 1 z 2 1 h t 1 h f } p a } 1 } 1 2 . ( 0 9 6 . ( 8 0 1 ) ) 2 } 1 100 m 5 } p a } 1 } 1 2 .0 (9 6 . ( 8 2 1 .0 m m /s /s 2 ) ) 2 } 1 0 m 1 h t 1 20 m The pressure terms cancel, and we may solve for the turbine head (which is positive): h t 5 100 2 20 2 0.2 < 79.8 m The turbine extracts about 79.8 percent of the 100-m head available from the dam. The total power extracted may be evaluated from the water mass flow: P 5 m ˙ w s 5 ( r Q )( gh t ) 5 (998 kg/m 3 )(30 m 3 /s)(9.81 m/s 2 )(79.8 m) 5 23.4 E6 kg ? m 2 /s 3 5 23.4 E6 N ? m/s 5 23.4 MW Ans. 7 The turbine drives an electric generator which probably has losses of about 15 percent, so the net power generated by this hydroelectric plant is about 20 MW.
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This note was uploaded on 10/01/2011 for the course ME 509 taught by Professor Wereley during the Spring '11 term at Purdue University-West Lafayette.

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FluidMechanics1-Quiz3Soln - 172 Chapter 3 Integral...

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