Answer
$\dot{W}_{\text {in }}=6.176\text{ kW}$
$\mathrm{COP}_{\mathrm{R}}=1.85$
Work Step by Step
From the refrigerant tables (Tables A-11E, A-12E, and A-13E),
$$
\begin{aligned}
& \left.\begin{array}{l}
P_3=180 \mathrm{psia} \\
\text { sat. liquid }
\end{array}\right\} h_3=h_{f@180 \mathrm{psia}}=51.51\ \mathrm{Btu} / \mathrm{lbm} \\
& h_4=h_6 \cong h_3=51.51\ \mathrm{Btu} / \mathrm{lbm} \text { (throttling) } \\
& \left.\begin{array}{l}
P_5=60 \mathrm{psia} \\
\text { sat. vapor }
\end{array}\right\} h_5=h_{g \text { @ } 60 \mathrm{psia}}=110.13\ \mathrm{Btu} / \mathrm{lbm} \\
& \left.\begin{array}{l}
\begin{array}{l}
P_7=10 \mathrm{psia} \\
\text { sat. vapor }
\end{array}
\end{array}\right\} h_7=h_{g \text { @ } 10 \mathrm{psia}}=98.69\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ The mass flow rates through the high-temperature and low-temperature evaporators are found by
$$
\begin{aligned}
& \dot{Q}_{L, 1}=\dot{m}_1\left(h_5-h_4\right) \longrightarrow \dot{m}_1=\frac{\dot{Q}_{L, 1}}{h_5-h_4}=\frac{15,000 \mathrm{Btu} / \mathrm{h}}{(110.13-51.51) \mathrm{Btu} / \mathrm{lbm}}=255.9\ \mathrm{lbm} / \mathrm{h} \\
& \dot{Q}_{L, 2}=\dot{m}_2\left(h_7-h_6\right) \longrightarrow \dot{m}_2=\frac{\dot{Q}_{L, 2}}{h_7-h_6}=\frac{24,000 \mathrm{Btu} / \mathrm{h}}{(98.69-51.51) \mathrm{Btu} / \mathrm{lbm}}=508.7\ \mathrm{lbm} / \mathrm{h}
\end{aligned}
$$ Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
$$
\dot{m}_1 h_5+\dot{m}_2 h_7=\left(\dot{m}_1+\dot{m}_2\right) h_1 \longrightarrow h_1=\frac{\dot{m}_1 h_5+\dot{m}_2 h_7}{\dot{m}_1+\dot{m}_2}=\frac{(255.9)(110.13)+(508.7)(98.69)}{255.9+508.7}=102.52\ \mathrm{Btu} / \mathrm{lbm}
$$ Then, $$
\begin{aligned}
& \left.\begin{array}{l}
P_1=10 \mathrm{psia} \\
h_1=102.52\ \mathrm{Btu} / \mathrm{lbm}
\end{array}\right\} s_1=0.2382\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
& \left.\begin{array}{l}
P_2=180\ \mathrm{psia} \\
s_2=s_1
\end{array}\right\} h_2=130.08\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ The power input to the compressor is
$$ \dot{W}_{\text {in }}=\left(\dot{m}_1+\dot{m}_2\right)\left(h_2-h_1\right)=(255.9+508.7) \mathrm{lbm} / \mathrm{h}(130.08-102.52) \mathrm{Btu} / \mathrm{lbm}\left(\frac{1 \mathrm{~kW}}{3412.14 \mathrm{Btu} / \mathrm{h}}\right)=6.176 \mathrm{~kW}
$$ The COP of this refrigeration system is determined from its definition, $$
\mathrm{COP}_{\mathrm{R}}=\frac{\dot{Q}_L}{\dot{W}_{\text {in }}}=\frac{(24,000+15,000) \mathrm{Btu} / \mathrm{h}}{6.176 \mathrm{~kW}}\left(\frac{1 \mathrm{~kW}}{3412.14 \mathrm{Btu} / \mathrm{h}}\right)=\mathbf{1 . 8 5}
$$
