The Simple Ideal Gas Refrigeration Cycle
An ideal gas refrigeration cycle using air as the working medium is to maintain a refrigerated space at 0°F while rejecting heat to the surrounding medium at 80°F. The pressure ratio of the compressor is 4. Determine (a) the maximum and minimum temperatures in the cycle, (b) the coefficient of performance, and (c) the rate of refrigeration for a mass flow rate of 0.1 lbm/s.
An ideal gas refrigeration cycle using air as the working fluid is considered. The maximum and minimum temperatures, the COP, and the rate of refrigeration are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.
Analysis The T-s diagram of the gas refrigeration cycle is shown in Fig. 11–22. We note that this is an ideal gas-compression refrigeration cycle, and thus, both the compressor and the turbine are isentropic, and the air is cooled to the environment temperature before it enters the turbine.
(a) The maximum and minimum temperatures in the cycle are determined from the isentropic relations of ideal gases for the compression and expansion processes. From Table A–17E,
T_{1}=460 \mathrm{R} \longrightarrow \quad h_{1}=109.90 \mathrm{Btu} / \mathrm{lbm} \quad \text { and } \quad P_{r 1}=0.7913
P_{r 2}=\frac{P_{2}}{P_{1}} P_{r 1}=(4)(0.7913)=3.165 \longrightarrow \begin{array}{l}h_{2}=163.5 \mathrm{Btu} / \mathrm{lbm} \\T_{2}=\mathbf{6 8 3} \mathrm{R}\left(\text { or } 223^{\circ} \mathrm{F}\right)\end{array}
T_{3}=540 \mathrm{R} \longrightarrow h_{3}=129.06 \mathrm{Btu} / \mathrm{lbm} \quad \text { and } \quad P_{r 3}=1.3860
P_{r 4}=\frac{P_{4}}{P_{3}} P_{r 3}=(0.25)(1.386)=0.3456 \longrightarrow \begin{array}{l}h_{4}=86.7 \mathrm{Btu} / \mathrm{lbm} \\T_{4}=363 \mathrm{R}\left(\text { or }-97^{\circ} \mathrm{F}\right)\end{array}
Therefore, the highest and the lowest temperatures in the cycle are 223 and -97^{\circ} \mathrm{F}, respectively.
(b) The COP of this ideal gas refrigeration cycle is
\mathrm{COP}_{\mathrm{R}}=\frac{q_{L}}{w_{\text {net,in }}}=\frac{q_{L}}{w_{\text {comp,in }}-w_{\text {turb }, \mathrm{out}}}
where
\begin{array}{c}q_{L}=h_{1}-h_{4}=109.9-86.7=23.2 \mathrm{Btu} / \mathrm{lbm} \\\\w_{\text {turb,out }}=h_{3}-h_{4}=129.06-86.7=42.36 \mathrm{Btu} / \mathrm{lbm} \\\\w_{\text {comp,in }}=h_{2}-h_{1}=163.5-109.9=53.6 \mathrm{Btu} / \mathrm{lbm}\end{array}
Thus,
\mathrm{COP}_{\mathrm{R}}=\frac{23.2}{53.6-42.36}=2.06
(c) The rate of refrigeration is
\dot{Q}_{\text {refrig }}=\dot{m} q_{L}=(0.1 \mathrm{lbm} / \mathrm{s})(23.2 \mathrm{Btu} / \mathrm{lbm})=2.32 \mathrm{Btu} / \mathrm{s}
Discussion It is worth noting that an ideal vapor-compression cycle working under similar conditions would have a COP greater than 3 .
