Answer
$\dot{Q}_L=213.7\text{ kJ/s}$
$\dot{W}_{\mathrm{in}} =94.9\text{ kW}$
Work Step by Step
From the refrigerant tables (Tables A-11, A-12, and A-13),
$$
\begin{aligned}
& \left.T_1=8.9^{\circ} \mathrm{C}\right\} \quad h_1=h_{g @ 8.9{ }^{\circ} \mathrm{C}}=255.60 \mathrm{~kJ} / \mathrm{kg} \\
& \text { sat. vapor }\} s_1=s_{g @ 8.9^{\circ} \mathrm{C}}=0.92711 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_2=1400\ \mathrm{kPa} \\
s_2=s_1
\end{array}\right\} h_2=281.56 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
\begin{array}{l}
P_3=1400\ \mathrm{kPa} \\
\text { sat. liquid }
\end{array}
\end{array}\right\} h_3=h_{f @ 1400 \mathrm{kPa}}=127.25 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=127.25 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) } \\
& \left.\begin{array}{l}
T_5=8.9^{\circ} \mathrm{C} \\
\text { sat. liquid }
\end{array}\right\} h_5=h_{f @ 8.9^{\circ} \mathrm{C}}=63.91 \mathrm{~kJ} / \mathrm{kg} \\
& h_6 \cong h_5=63.91 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) } \\
& \left.T_7=-32^{\circ} \mathrm{C}\right\} \quad h_7=h_{g @-32^{\circ} \mathrm{C}}=230.93 \mathrm{~kJ} / \mathrm{kg} \\
& \text { sat. vapor }\} s_7=s_{g @-32^{\circ} \mathrm{C}}=0.95819 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_8=P_{\text {sat }} @ 8.9^{\circ} \mathrm{C}=400 \mathrm{kPa} \\
s_8=s_7
\end{array}\right\} h_8=264.51 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ An energy balance on the separator gives $$
\dot{m}_6\left(h_8-h_5\right)=\dot{m}_2\left(h_1-h_4\right) \longrightarrow \dot{m}_6=\dot{m}_2 \frac{h_1-h_4}{h_8-h_5}=(2 \mathrm{~kg} / \mathrm{s}) \frac{255.60-127.25}{264.51-63.91}=1.280\ \mathrm{~kg} / \mathrm{s}
$$ The rate of cooling produced by this system is then $$
\dot{Q}_L=\dot{m}_6\left(h_7-h_6\right)=(1.280 \mathrm{~kg} / \mathrm{s})(230.93-63.91) \mathrm{kJ} / \mathrm{kg}=\mathbf{2 1 3 . 7} \mathrm{kJ} / \mathrm{s}
$$ The total power input to the compressors is $$
\begin{aligned}
\dot{W}_{\mathrm{in}} & =\dot{m}_6\left(h_8-h_7\right)+\dot{m}_2\left(h_2-h_1\right) \\
& =(1.280 \mathrm{~kg} / \mathrm{s})(264.51-230.93) \mathrm{kJ} / \mathrm{kg}+(2 \mathrm{~kg} / \mathrm{s})(281.56-255.60) \mathrm{kJ} / \mathrm{kg} \\
& =\mathbf{9 4 . 9}\ \mathbf{k W}
\end{aligned}
$$
