ThermoSolutions-CHAPTER11

ThermoSolutions-CHAPTER11 - 11-1 Chapter 11 REFRIGERATION...

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11-1 Chapter 11 REFRIGERATION CYCLES The Reversed Carnot Cycle 11-1C Because the compression process involves the compression of a liquid-vapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of high-moisture content refrigerant. 11-2 A steady-flow Carnot refrigeration cycle with refrigerant-134a 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 q C = 303 K and T L = T sat @ 160 kPa = -15.60 q C = 257.4 K, the COP of this Carnot refrigerator is determined from +, + , 5.64 0 0 1 K 4 . 257 / K 303 1 1 / 1 COP C R, L H T T ( b ) From the refrigerant tables (Table A-11), kJ/kg 58 . 93 kJ/kg 66 . 266 C 30 @ 4 C 30 @ 3 q q f g h h h h Thus, and kJ/kg 147.03 ¸ ¸ ¹ · ¨ ¨ © § ¡o ¡ 0 0 kJ/kg 173.08 K 303 K 257.4 kJ/kg 08 . 173 58 . 93 66 . 266 4 3 H H L L L H L H H q T T q T T q q h h q 30 q C 43 2 1 160 kPa Q L Q H T s ( c ) The net work input is determined from kJ/kg 26.05 0 0 03 . 147 08 . 173 net L H q q w

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11-2 11-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heat-absorption 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 = 72.78 q F = 532.8 R and T L = T sat @ 30 psia = 15.37 q F = 475.4 R. +, 8.28 0 0 1 R 475.4 / R 532.8 1 1 / 1 COP C R, L H T T ( b ) Process 4-1 is isentropic, and thus + , + , 0.2374 0 ¸ ¸ ¹ · ¨ ¨ © § 0 ¢ . . 18589 . 0 03793 . 0 08207 . 0 R Btu/lbm 0.08207 14525 . 0 05 . 0 07481 . 0 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 T-s diagram the area enclosed represents the net work, and s 3 = s g @ 90 psia = 0.22006 Btu/lbm·R, + , + , Btu/lbm 7.92 R Btu/lbm 08207 . 0 22006 . 0 ) 37 . 15 78 . 72 ( 4 3 in net, ¢ 0 0 0 0 s s T T w L H Ideal and Actual Vapor-Compression Cycles 11-4C Yes; the throttling process is an internally irreversible process. 11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle. 11-6C No. Assuming the water is maintained at 10 q 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 air-conditioning devices that involve such extremely low pressures. 11-7C Allowing a temperature difference of 10 q C for effective heat transfer, the condensation temperature of the refrigerant should be 25 q C. The saturation pressure corresponding to 25 q C is 0.67 MPa. Therefore, the recommended pressure would be 0.7 MPa. 11-8C The area enclosed by the cyclic curve on a T - s diagram represents the net work input for the reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known.
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ThermoSolutions-CHAPTER11 - 11-1 Chapter 11 REFRIGERATION...

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