Answer
a) $\delta_{\text {regen }}=0.434$
b) $\dot{Q}_L=21.36\text{ kW}$
c) $COP_{}=0.478$
Work Step by Step
(a) From the isentropic relations, $$
\begin{aligned}
T_{2 s} & =T_1\left(\frac{P_2}{P_1}\right)^{(\mathrm{k}-1) / \mathrm{k}}=(273.2 \mathrm{~K})(5)^{0.4 / 1.4}=432.4 \mathrm{~K} \\
\eta_C & =\frac{h_{2 s}-h_1}{h_2-h_1}=\frac{T_{2 s}-T_1}{T_2-T_1} \\
0.80 & =\frac{432.4-273.2}{T_2-273.2} \longrightarrow T_2=472.5 \mathrm{~K}
\end{aligned}
$$ The temperature at state 4 can be determined by solving the following two equations simultaneously: $$
\begin{aligned}
& T_{5 s}=T_4\left(\frac{P_5}{P_4}\right)^{(\mathrm{k}-1) / \mathrm{k}}=T_4\left(\frac{1}{5}\right)^{0.4 / 1.4} \\
& \eta_T=\frac{h_4-h_5}{h_4-h_{5 s}} \rightarrow 0.85=\frac{T_4-193.2}{T_4-T_{5 s}}
\end{aligned}
$$ Now we obtain $T_4=281.3 \mathrm{~K}$.
An energy balance on the regenerator may be written as $$
\dot{m} c_p\left(T_3-T_4\right)=\dot{m} c_p\left(T_1-T_6\right) \longrightarrow T_3-T_4=T_1-T_6
$$ or, $$
T_6=T_1-T_3+T_4=273.2-308.2+281.3=246.3 \mathrm{~K}
$$ The effectiveness of the regenerator is $$
\delta_{\text {regen }}=\frac{h_3-h_4}{h_3-h_6}=\frac{T_3-T_4}{T_3-T_6}=\frac{308.2-281.3}{308.2-246.3}=\mathbf{0 . 4 3 4}
$$ (b) The refrigeration load is $$
\dot{Q}_L=\dot{m} c_p\left(T_6-T_5\right)=(0.4 \mathrm{~kg} / \mathrm{s})(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(246.3-193.2) \mathrm{K}=\mathbf{2 1 . 3 6 k W}
$$ (c) The turbine and compressor powers and the COP of the cycle are $$
\begin{aligned}
& \dot{W}_{\mathrm{C}, \text { in }}=\dot{m} c_p\left(T_2-T_1\right)=(0.4 \mathrm{~kg} / \mathrm{s})(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(472.5-273.2) \mathrm{K}=80.13 \mathrm{~kW} \\
& \dot{W}_{\mathrm{T}, \text { out }}=\dot{m} c_p\left(T_4-T_5\right)=(0.4 \mathrm{~kg} / \mathrm{s})(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(281.3-193.2) \mathrm{kJ} / \mathrm{kg}=35.43 \mathrm{~kW} \\
& \mathrm{COP}=\frac{\dot{Q}_L}{\dot{W}_{\text {net,in }}}=\frac{\dot{Q}_L}{\dot{W}_{\mathrm{C}, \text { in }}-\dot{W}_{\mathrm{T}, \text { out }}}=\frac{21.36}{80.13-35.43}=\mathbf{0 . 4 7 8}
\end{aligned}
$$
