Thermodynamics

616 | ThermodynamicsAnalysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–8.We note that the refrigerant leaves the condenser as a compressed liquidand enters the compressor as superheated vapor. The enthalpies of therefrigerant at various states are determined from the refrigerant tables to beP 1 0.14 MPaT 1 10°CfP 2 0.8 MPafT 2 50°Ch 1 246.36 kJ/kgh 2 286.69 kJ/kgP 3 0.72 MPaf h 3 h f @ 26°C 87.83 kJ/kgT 3 26°Ch 4 h 3 (throttling) ⎯→ h 4 87.83 kJ/kgT0.72 MPa26°C340.15 MPaQ LQ H2sFIGURE 11–8T-s diagram for Example 11–2.12 0.8 MPa50°CW in0.14 MPa–10°Cs(a) The rate of heat removal from the refrigerated space and the power inputto the compressor are determined from their definitions:andQ # L m # 1h 1 h 4 2 10.05 kg>s231246.36 87.832 kJ>kg4 7.93 kWW # in m # 1h 2 h 1 2 10.05 kg>s231286.69 246.362 kJ>kg4 2.02 kW(b) The isentropic efficiency of the compressor is determined fromwhere the enthalpy at state 2s (P 2s 0.8 MPa and s 2s s 1 0.9724kJ/kg · K) is 284.21 kJ/kg. Thus,h C h C h 2s h 1h 2 h 1284.21 246.36 0.939 or 93.9%286.69 246.36(c) The coefficient of performance of the refrigerator isCOP R Q# LW # 7.93 kWin2.02 kW 3.93Discussion This problem is identical to the one worked out in Example11–1, except that the refrigerant is slightly superheated at the compressorinlet and subcooled at the condenser exit. Also, the compressor is not isentropic.As a result, the heat removal rate from the refrigerated spaceincreases (by 10.4 percent), but the power input to the compressor increaseseven more (by 11.6 percent). Consequently, the COP of the refrigeratordecreases from 3.97 to 3.93.11–5 ■ SELECTING THE RIGHT REFRIGERANTWhen designing a refrigeration system, there are several refrigerants fromwhich to choose, such as chlorofluorocarbons (CFCs), ammonia, hydrocarbons(propane, ethane, ethylene, etc.), carbon dioxide, air (in the air-conditioning ofaircraft), and even water (in applications above the freezing point). The right