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Unformatted text preview: Chapter 8 Power and Refrigeration Cycles The Reversed Carnot Cycle 8122C Because the compression process involves the compression of a liquidvapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of highmoisture content refrigerant. 8123 A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis ( a ) Noting that T H = 30 ° C = 303 K and T L = T sat @ 120 kPa = 22.36 ° C = 250.6 K, the COP of this Carnot refrigerator is determined from ( 29 ( 29 4.78 = = = 1 K 6 . 250 / K 303 1 1 / 1 COP C R, L H T T ( b ) From the refrigerant tables (Table A11), h h h h g f 3 30 4 30 263 50 91 49 = = = = ° ° @ @ . . C C kJ / kg kJ / kg Thus, and ( 29 kJ/kg 142.3 = = = → = = = = kJ/kg 172.01 K 303 K 250.6 kJ/kg 01 . 172 49 . 91 50 . 263 4 3 H H L L L H L H H q T T q T T q q h h q ( c ) The net work input is determined from /kg kJ 71 . 29 = = = 3 . 142 01 . 172 net L H q q w 896 s T Q H Q L 120 kPa 1 2 3 4 30 ° C Chapter 8 Power and Refrigeration Cycles 8124E A steadyflow Carnot refrigeration cycle with refrigerant134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heatabsorption process, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis ( a ) Noting that T H = T sat @ 90 psia = 72.83 ° F = 532.8 R and T L = T sat @ 30 psia = 15.38 ° F = 475.4 R. ( 29 ( 29 8.28 = = = 1 R 475.4 / R 532.8 1 1 1 COP C R, L H T / T ( b ) Process 41 is isentropic, and thus ( 29 ( 29 ( 29 0.237 = =  = ⋅ = + = + = = 0364 . 2209 . 0364 . 0801 . R Btu/lbm 0.0801 0729 . 2172 . 05 . 0729 . psia 30 @ 1 1 psia 90 @ 4 4 1 fg f fg f s s s x s x s s s ( c ) Remembering that on a Ts diagram the area enclosed represents the net work, and s 3 = s g @ 90 psia = 0.2172 Btu/lbm·R, ( 29 ( 29 ( 29 [ ]( 29 Btu/lbm 7.88 R Btu/lbm 0801 . 2172 . R 15.38 72.83 4 3 in net, = ⋅ = = s s T T w L H 897 s T Q H Q L 1 2 3 4 Chapter 8 Power and Refrigeration Cycles Ideal and Actual VaporCompression Cycles 8125C Yes; the throttling process is an internally irreversible process. 8126C To make the ideal vaporcompression refrigeration cycle more closely approximate the actual cycle. 8127C No. Assuming the water is maintained at 10 ° C in the evaporator, the evaporator pressure will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design refrigeration or airconditioning devices that involve such extremely low pressures....
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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.
 Spring '08
 Chung

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