A refrigerator operating on a vapor-compression cycle is considered. The rate of refrigeration, the power input, the compressor efficiency, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis 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 liquid and enters the compressor as superheated vapor. The enthalpies of the refrigerant at various states are determined from the refrigerant tables to be
\left.\begin{array}{l}P_{1}=0.14 \mathrm{MPa} \\T_{1}=-10^{\circ} \mathrm{C}\end{array}\right\} h_{1}=246.37 \mathrm{~kJ} / \mathrm{kg}
\left.\begin{array}{l}P_{2}=0.8 \mathrm{MPa} \\T_{2}=50^{\circ} \mathrm{C}\end{array}\right\} h_{2}=286.71 \mathrm{~kJ} / \mathrm{kg}
\left.\begin{array}{l}P_{3}=0.72 \mathrm{MPa} \\T_{3}=26{ }^{\circ} \mathrm{C}\end{array}\right\} h_{3} \cong h_{ f @ 26^{\circ} \mathrm{C}}=87.83 \mathrm{~kJ} / \mathrm{kg}
h_{4} \cong h_{3}(\text { throttling }) \rightarrow \quad h_{4}=87.83 \mathrm{~kJ} / \mathrm{kg}
(a) The rate of heat removal from the refrigerated space and the power input to the compressor are determined from their definitions:
\dot{Q}_{L}=\dot{m}\left(h_{1}-h_{4}\right)=(0.05 \mathrm{~kg} / \mathrm{s})[(246.37-87.83) \mathrm{kJ} / \mathrm{kg}]=7.93 \mathrm{~kW}
and
\dot{W}_{\text {in }}=\dot{m}\left(h_{2}-h_{1}\right)=(0.05 \mathrm{~kg} / \mathrm{s})[(286.71-246.37) \mathrm{kJ} / \mathrm{kg}]=2.02 \mathrm{k} \mathrm{W}
(b) The isentropic efficiency of the compressor is determined from
\eta_{C} \cong \frac{h_{2 s}-h_{1}}{h_{2}-h_{1}}
where the enthalpy at state 2 s\left(P_{2 s}=0.8 \mathrm{MPa}\right. and \left.s_{2 s}=s_{1}=0.9724 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right) is 284.20 \mathrm{~kJ} / \mathrm{kg}. Thus,
\eta_{C}=\frac{284.20-246.37}{286.71-246.37}=0.938 \text { or } 93.8 \%
(c) The coefficient of performance of the refrigerator is
\mathrm{COP}_{\mathrm{R}}=\frac{\dot{Q}_{L}}{\dot{W}_{\mathrm{in}}}=\frac{7.93 \mathrm{~kW}}{2.02 \mathrm{~kW}}=3.93
Discussion This problem is identical to the one worked out in Example 11–1, except that the refrigerant is slightly superheated at the compressor inlet and subcooled at the condenser exit. Also, the compressor is not isentropic. As a result, the heat removal rate from the refrigerated space increases (by 10.3 percent), but the power input to the compressor increases even more (by 11.6 percent). Consequently, the COP of the refrigerator decreases from 3.97 to 3.93.
