FTFS Chap08 P090 - Chapter 8 Power and Refrigeration Cycles...

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Unformatted text preview: Chapter 8 Power and Refrigeration Cycles Carnot Vapor Cycle 8-90C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 8-91C The Carnot cycle is not a realistic model for steam power plants because (1) limiting the heat transfer processes to two-phase systems to maintain isothermal conditions severely limits the maximum temperature that can be used in the cycle, (2) the turbine will have to handle steam with a high moisture content which causes erosion, and (3) it is not practical to design a compressor that will handle two phases. 8-92E A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the quality at the end of the heat rejection process, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis ( a ) We note that and T T T T T T H L L H = = = = = = =- =- = sat @120 psia sat @14.7 psia th, C F R F R R R 341 3 801 3 212 672 1 1 672 801 3 . . . 16.1% ( b ) Noting that s 4 = s 1 = s f @ 120 psia = 0.49201 Btu/lbmR, x s s s f fg 4 4 0 49201 0 31212 1 4446 =- =- = . . . 0.1245 ( c ) The enthalpies before and after the heat addition process are ( 29 ( 29 Btu/lbm 25 1147 5 878 95 67 312 Btu/lbm 67 312 2 2 psia 120 1 . . . . h x h h . h h fg f @ f = + = + = = = Thus, and, ( 29 ( 29 Btu/lbm 134.4 = = = =- =- = Btu/lbm 58 . 834 161 . Btu/lbm 58 . 834 67 . 312 25 . 1147 in th net 1 2 in q w h h q 8-75 14.7 psia 120 psia 3 2 4 1 s T q in Chapter 8 Power and Refrigeration Cycles 8-93 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis ( a ) Noting that T H = 250C = 523 K and T L = T sat @ 20 kPa = 60.06C = 333.1 K, the thermal efficiency becomes 36.3% =- =- = K 523 K 333.1 1 1 C th, H L T T ( b ) The heat supplied during this cycle is simply the enthalpy of vaporization , Thus, ( 29 kJ/kg 1093.1 = = = = = = kJ/kg 2 1716 K 523 K 333.1 kJ/kg 2 1716 in out 250 in . q T T q q . h q H L L C @ fg ( c ) The net work output of this cycle is ( 29 ( 29 kJ/kg 623.0 kJ/kg 2 1716 363 in th net = = = . . q w 8-94 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the amount of heat rejected, and the net work output are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible....
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This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

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FTFS Chap08 P090 - Chapter 8 Power and Refrigeration Cycles...

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