10 5 pa 997 kg m 3 1 2 parenleftbig 1 m s

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× 10 5 Pa ) 997 kg m 3 + 1 2 parenleftBig 1 m s parenrightBig 2 = 24 . 27 m s . (9.84) The exit velocity is barely changed from our earlier analysis. Now determine the new height. Let us again apply Eq. (9.44): P 1 ρ + 1 2 v 2 1 + gz 1 = P 3 ρ + 1 2 v 2 3 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright =0 + gz 3 , (9.85) P 1 ρ + 1 2 v 2 1 + gz 1 = P atm ρ + gz 3 , (9.86) P 1 P atm ρ + 1 2 v 2 1 = g ( z 3 z 1 ) , (9.87) z 3 z 1 = P gauge ρg + 1 2 g v 2 1 . (9.88) Substituting numbers, we get z 3 z 1 = 2 . 934 × 10 5 Pa parenleftBig 997 kg m 3 parenrightBig ( 9 . 81 m s ) + 1 2 ( 9 . 81 m s 2 ) parenleftBig 1 m s parenrightBig 2 = 30 . 05 m. (9.89) The extra boost in height comes from accounting for the initial kinetic energy of the water. 9.3 Component efficiency Recall for cycles, we define a thermal efficiency as η = W net /Q H = what you want/what you pay for. We can further define efficiencies for components. For a component, we will generally take an efficiency to be η component = what we get the best we could get . (9.90) CC BY-NC-ND. 2011, J. M. Powers.
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9.3. COMPONENT EFFICIENCY 279 The best we could get is generally an isentropic device. So, for example, for a turbine, we say η turbine = actual work work done by an isentropic turbine = w w s . (9.91) Here the subscript “ s ” denotes isentropic. For a nozzle, we would like to maximize the kinetic energy of the working fluid, so we say η nozzle = v 2 v 2 s . (9.92) However for pumps and compressors, the isentropic pump requires the least work input. So we take instead η pump,compressor = w s w . (9.93) Example 9.5 N 2 is adiabatically compressed from T 1 = 300 K , P 1 = 100 kPa to P 2 = 1000 kPa . The compressor efficiency is η c = 0 . 9. Find the final state and the compression work per unit mass. Assume N 2 is a CIIG. The first law for adiabatic compression gives w = h 2 h 1 . (9.94) Table A.8 from BS tells us that at T 1 = 300 K , h 1 = 311 . 67 kJ/kg , s o T 1 = 6 . 8463 kJ/kg/K . But we are not sure what state 2 is, and the first law does not help yet, as we do not know either w or h 2 . Let us calculate state 2 assuming an isentropic process, and then use our knowledge of compressor efficiency to correct for real effects. From Eq. (8.95), we can conclude that for an isentropic process in which s 2 = s 1 that s 2 s 1 = 0 = s o T 2 s o T 1 R ln P 2 P 1 , (9.95) s o T 2 = s o T 1 + R ln P 2 P 1 , (9.96) = parenleftbigg 6 . 8463 kJ kg K parenrightbigg + parenleftbigg 0 . 287 kJ kg K parenrightbigg ln 1000 kPa 100 kPa , (9.97) = 7 . 50714 kJ kg K . (9.98) Knowing s o T 2 , we next interpolate Table A.8 from BS to get T 2 s = 554 . 4 K, h 2 s = 578 . 471 kJ kg . (9.99) (Note the CPIG assumption would have yielded T 2 s = (300 K )(10) 0 . 286 = 538 K ). So the work for the isentropic compressor is w s = h 2 s h 1 = parenleftbigg 578 . 471 kJ kg parenrightbigg parenleftbigg 311 . 67 kJ kg parenrightbigg = 266 . 8 kJ kg . (9.100) CC BY-NC-ND. 2011, J. M. Powers.
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280 CHAPTER 9. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Now consider the compressor efficiency: η c = w s w , (9.101) w = w s η c , (9.102) w = 266 . 8 kJ kg 0 . 9 , (9.103) w = 296 . 4 kJ kg . (9.104) Thus we have the actual work per unit mass. So the actual enthalpy at state 2 can be derived from the first law: h 2 = h 1 + w = parenleftbigg 311 . 67 kJ kg parenrightbigg + parenleftbigg 296 . 4 kJ kg parenrightbigg = 608 . 12 kJ kg . (9.105) Now knowing h 2 , we can again interpolate Table A.8 of BS to find the final temperature T 2 to be T 2 = 582 . 1 K. (9.106)
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