Answer
$\operatorname{COP}_{\mathrm{R}} =1.56$
$\dot{Q}_{\text {Refrig }}=0.124\text{ kg/s}$
Work Step by Step
From the isentropic relations, $$
\begin{aligned}
& T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(249 \mathrm{~K})(3)^{0.4 / 1.4}=340.8 \mathrm{~K} \\
& T_4=T_3\left(\frac{P_4}{P_3}\right)^{(k-1) / k}=(278 \mathrm{~K})(3)^{0.4 / 1.4}=380.5 \mathrm{~K} \\
& T_6=T_5\left(\frac{P_6}{P_5}\right)^{(k-1) / k}=(278 \mathrm{~K})\left(\frac{1}{9}\right)^{0.4 / 1.4}=148.4 \mathrm{~K}
\end{aligned}
$$ The COP of this ideal gas refrigeration cycle is determined from
$$
\begin{aligned}
\operatorname{COP}_{\mathrm{R}} & =\frac{q_L}{w_{\text {net,in }}}=\frac{q_L}{w_{\text {comp in }}-w_{\text {turb,out }}} \\
& =\frac{h_1-h_6}{\left(h_2-h_1\right)+\left(h_4-h_3\right)-\left(h_5-h_6\right)} \\
& =\frac{T_1-T_6}{\left(T_2-T_1\right)+\left(T_4-T_3\right)-\left(T_5-T_6\right)} \\
& =\frac{249-148.4}{(340.8-249)+(380.5-278)-(278-148.4)} \\
& =\mathbf{1 . 5 6}
\end{aligned}
$$ The mass flow rate of the air is determined from $$
\dot{Q}_{\text {Refrig }}=\dot{m} c_p\left(T_1-T_6\right) \longrightarrow \dot{m}=\frac{\dot{Q}_{\text {Refrig }}}{c_p\left(T_1-T_6\right)}=\frac{(45,000 / 3600) \mathrm{kJ} / \mathrm{s}}{(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(249-148.4) \mathrm{K}}=\mathbf{0 . 1 2 4 \mathrm { kg } / \mathrm { s }}
$$
