Thermodynamics

Chapter 11 | 613EXAMPLE 11–1The Ideal Vapor-Compression RefrigerationCycleA refrigerator uses refrigerant-134a as the working fluid and operates on anideal vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. If themass flow rate of the refrigerant is 0.05 kg/s, determine (a) the rate of heatremoval from the refrigerated space and the power input to the compressor,(b) the rate of heat rejection to the environment, and (c) the COP of therefrigerator.Solution A refrigerator operates on an ideal vapor-compression refrigerationcycle between two specified pressure limits. The rate of refrigeration, thepower input, the rate of heat rejection, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–6.We note that this is an ideal vapor-compression refrigeration cycle, and thusthe compressor is isentropic and the refrigerant leaves the condenser as asaturated liquid and enters the compressor as saturated vapor. From therefrigerant-134a tables, the enthalpies of the refrigerant at all four states aredetermined as follows:P 1 0.14 MPa ¡ h 1 h g @ 0.14 MPa 239.16 kJ>kgs 1 s g @ 0.14 MPa 0.94456 kJ>kg # KTP 2 0.8 MPafhs 2 s 2 275.39 kJ>kg1P 3 0.8 MPa ¡ h 3 h f @ 0.8 MPa 95.47 kJ>kgh 4 h 3 1throttling2 ¡ h 4 95.47 kJ>kg(a) The rate of heat removal from the refrigerated space and the power inputto the compressor are determined from their definitions:Q H0.8 MPa32W inQ # L m # 1h 1 h 4 2 10.05 kg>s231239.16 95.472 kJ>kg4 7.18 kWandW # in m # 1h 2 h 1 2 10.05 kg>s231275.39 239.162 kJ>kg4 1.81 kW0.14 MPa4s 4Q L1(b) The rate of heat rejection from the refrigerant to the environment isQ # H m # 1h 2 h 3 2 10.05 kg>s231275.39 95.472 kJ>kg4 9.0 kWIt could also be determined fromQ # H Q # L W # in 7.18 1.81 8.99 kW(c) The coefficient of performance of the refrigerator isCOP R Q# LW # 7.18 kWin1.81 kW 3.97That is, this refrigerator removes about 4 units of thermal energy from therefrigerated space for each unit of electric energy it consumes.Discussion It would be interesting to see what happens if the throttling valvewere replaced by an isentropic turbine. The enthalpy at state 4s (the turbineexit with P 4s 0.14 MPa, and s 4s s 3 0.35404 kJ/kg · K) is 88.94 kJ/kg,FIGURE 11–6T-s diagram of the idealvapor-compression refrigeration cycledescribed in Example 11–1.s