ThermoSolutions-CHAPTER10

# ThermoSolutions-CHAPTER10 - 10-1 Chapter 10 VAPOR AND...

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10-1 Chapter 10 VAPOR AND COMBINED POWER CYCLES Carnot Vapor Cycle 10-1C Because excessive moisture in steam causes erosion on the turbine blades. The highest moisture content allowed is about 10%. 10-2C 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. 10-3E 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 19.3% 0 0 K q q R 1 . 833 R 0 . 672 1 1 R 0 . 672 F 0 . 212 R 1 . 833 F 1 . 373 C th, psia 14.7 @ sat psia 180 @ sat H L L H T T T T T T T 14.7 psia 180 psia q in 1 4 2 3 ( b ) Noting that s 4 = s 1 = s f @ 180 psia = 0.53274 Btu/lbm·R, 0.153 0 0 44441 . 1 31215 . 0 53274 . 0 4 4 fg f s s s x s ( c ) The enthalpies before and after the heat addition process are +, + , Btu/lbm 2 . 1112 16 . 851 90 . 0 14 . 346 Btu/lbm 14 . 346 2 2 psia 180 @ 1 . . fg f f h x h h h h Thus, and, + , Btu/lbm 148.1 0 0 Btu/lbm 0 . 766 1934 . 0 Btu/lbm 0 . 766 14 . 346 2 . 1112 in th net 1 2 in q w h h q K

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10-2 10-4 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 = 250°C = 523 K and T L = T sat @ 20 kPa = 60.06°C = 333.1 K, the thermal efficiency becomes 36.3% 0 0 3632 . 0 K 523 K 333.1 1 1 C th, H L T T K T q out q in 1 4 2 3 20 kPa ( b ) The heat supplied during this cycle is simply the enthalpy of vaporization , Thus, +, kJ/kg 1092.3 ¸ ¸ ¹ · ¨ ¨ © § kJ/kg 3 . 1715 K 523 K 333.1 kJ/kg 3 . 1715 in out 250 @ in q T T q q h q H L L C fg \$ 250 q C s ( c ) The net work output of this cycle is + , kJ/kg 623.0 kJ/kg 3 . 1715 3632 . 0 in th net q w 10-5 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 = 250°C = 523 K and T L = T sat @ 10 kPa = 45.81°C = 318.8 K, the thermal efficiency becomes 39.04% 0 0 K 523 K 318.8 1 1 C th, H L T T T q out q in 1 4 2 3 10 kPa ( b ) The heat supplied during this cycle is simply the enthalpy of vaporization , Thus, kJ/kg 1045.6 ¸ ¸ ¹ · ¨ ¨ © § q kJ/kg 3 . 1715 K 523 K 318.8 kJ/kg 3 . 1715 in out C 250 @ in q T T q q h q H L L fg 250 q C s ( c ) The net work output of this cycle is + , kJ/kg 669.7 kJ/kg 3 . 1715 3904 . 0
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## This note was uploaded on 09/21/2009 for the course ME 311 taught by Professor Ferrenberg during the Fall '09 term at Nevada.

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ThermoSolutions-CHAPTER10 - 10-1 Chapter 10 VAPOR AND...

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