Answer
b) $
\operatorname{COP}_{\mathrm{R}}=0.525 $
c) $
\dot{m}=0.0791\text{ kg/s}$
Work Step by Step
(a) From the isentropic relations,
$$
\begin{aligned}
& T_{2 s}=T_1\left(\frac{P_2}{P_1}\right)^{(\mathrm{k}-1) / \mathrm{k}}=(267 \mathrm{~K})(4)^{0.667 / 1.667}=465.0 \mathrm{~K} \\
& T_{4 s}=T_3\left(\frac{P_4}{P_3}\right)^{(\mathrm{k}-1) / \mathrm{k}}=(323 \mathrm{~K})\left(\frac{1}{4}\right)^{0.667 / 1.667}=185.5 \mathrm{~K}
\end{aligned}
$$ and $$
\begin{aligned}
& \begin{aligned}
\eta_T=\frac{h_3-h_4}{h_3-h_{4 s}}=\frac{T_3-T_4}{T_3-T_{4 s}} \longrightarrow T_4 & =\\T_3-\eta_T\left(T_3-T_{4 s}\right)=323-(0.85)(323-185.5)
& =\mathbf{2 0 6 . 1 ~ K}=T_{\min }
\end{aligned} \\
& \begin{aligned}
\eta_C=\frac{h_{2 s}-h_1}{h_2-h_1}=\frac{T_{2 s}-T_1}{T_2-T_1} \longrightarrow T_2 & \\=T_1+\left(T_{2 s}-T_1\right) / \eta_C=267+(465.0-267) /(0.85)
& =499.9 \mathrm{~K}
\end{aligned}
\end{aligned}
$$ (b) The COP of this gas refrigeration cycle is determined from
$$
\begin{aligned}
\operatorname{COP}_{\mathrm{R}} & =\frac{q_L}{w_{\text {net,in }}}=\frac{q_L}{w_{\text {comp, in }}-w_{\text {turbout }}} \\
& =\frac{h_1-h_4}{\left(h_2-h_1\right)-\left(h_3-h_4\right)} \\
& =\frac{T_1-T_4}{\left(T_2-T_1\right)-\left(T_3-T_4\right)} \\
& =\frac{267-206.1}{(499.9-267)-(323-206.1)} \\
& =\mathbf{0 . 5 2 5}
\end{aligned}
$$ (c) The mass flow rate of helium is determined from $$
\dot{m}=\frac{\dot{Q}_{\text {refig }}}{q_L}=\frac{\dot{Q}_{\text {refig }}}{h_1-h_4}=\frac{\dot{Q}_{\text {refig }}}{c_p\left(T_1-T_4\right)}=\frac{25 \mathrm{~kJ} / \mathrm{s}}{(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(267-206.1) \mathrm{K}}=0.0791 \mathrm{~kg} / \mathrm{s}
$$
