Answer
b) $\mathrm{COP}_{\mathrm{R}}=2.74$
c) $\dot{Q}_{\text {refrig }}=2.49\text{ kJ/s}$
Work Step by Step
(a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17)$$
\begin{aligned}
& T_1=250 \mathrm{~K} \longrightarrow h_1=250.05 \mathrm{~kJ} / \mathrm{kg} \\
& P_{r_1}=0.7329 \\
& T_1=300 \mathrm{~K} \quad \longrightarrow \quad h_3=300.19 \mathrm{~kJ} / \mathrm{kg} \\
& P_{r_3}=1.386
\end{aligned}
$$ Thus, $$
\begin{aligned}
P_{r_2}=\frac{P_2}{P_1} P_{r_1}=(3)(0.7329)=2.1987 \longrightarrow T_2 & =T_{\max }=342.2 \mathrm{~K} \\
h_2 & =342.60 \mathrm{~kJ} / \mathrm{kg} \\
P_{r_4}=\frac{P_4}{P_3} P_{r_3}=\left(\frac{1}{3}\right)(1.386)=0.462 \longrightarrow T_4 & =T_{\min }=219.0 \mathrm{~K} \\
h_4 & =218.97 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ (b) The COP of this ideal gas refrigeration cycle is determined from $$
\mathrm{COP}_{\mathrm{R}}-\frac{q_L}{w_{\text {net, in }}}-\frac{q_L}{w_{\text {comp, in }}-w_{\text {turb, out }}}
$$ where $$
\begin{aligned}
q_L & =h_1-h_4=250.05-218.97=31.08 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {comp, in }} & =h_2-h_1=342.60-250.05=92.55 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {turb, out }} & =h_3-h_4=300.19-218.97=81.22 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Thus, $$
\mathrm{COP}_{\mathrm{R}}=\frac{31.08}{92.55-81.22}=\mathbf{2 . 7 4}
$$ (c) The rate of refrigeration is determined to be $$
\dot{Q}_{\text {refrig }}=\dot{m}\left(q_L\right)=(0.08 \mathrm{~kg} / \mathrm{s})(31.08 \mathrm{~kJ} / \mathrm{kg})=\mathbf{2 . 4 9} \mathbf{k J} / \mathbf{s}
$$
