Answer
$\mathrm{COP}_{\mathrm{R}}=2.07$
Work Step by Step
From the refrigerant tables (Tables A-11E, A$12 \mathrm{E}$, and A-13E),
$$
\begin{aligned}
& \left.\begin{array}{l}
\begin{array}{l}
P_3=160\ \mathrm{psia} \\
\text { sat. liquid }
\end{array}
\end{array}\right\} h_3=h_{f@ 160 \mathrm{psia}}=48.527\ \mathrm{Btu} / \mathrm{lbm} \\
& h_4=h_6 \cong h_3=48.527\ \mathrm{Btu} / \mathrm{lbm} \text { (throttling) } \\
& \left.\begin{array}{l}
T_5=30^{\circ} \mathrm{F} \\
\text { sat. vapor }
\end{array}\right\} h_5=h_{g @ 30^{\circ} \mathrm{F}}=107.42\ \mathrm{Btu} / \mathrm{lbm} \\
& \left.\begin{array}{l}
\begin{array}{l}
T_7=-29.5^{\circ} \mathrm{F} \\
\text { sat. vapor }
\end{array}
\end{array}\right\} h_7=h_{g @-29.5^{\circ} \mathrm{F}}=98.69\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ For a unit mass flowing through the compressor, the fraction of mass flowing through Evaporator II is denoted by $x$ and that through Evaporator I is $y(y=1-x)$. From the cooling loads specification, $$
\begin{gathered}
\dot{Q}_{L, \text { e vapl }}=2 \dot{Q}_{L, \text { e vap } 2} \\
x\left(h_5-h_4\right)=2 y\left(h_7-h_6\right)
\end{gathered}
$$ where $\quad x=1-y$
Combining these results and solving for $y$ gives $$
y=\frac{h_5-h_4}{2\left(h_7-h_6\right)+\left(h_5-h_4\right)}=\frac{107.42-48.527}{2(98.69-48.527)+(107.42-48.527)}=0.3699
$$ Then, $\quad x=1-y=1-0.3699=0.6301$
Applying an energy balance to the point in the system where the two evaporator streams are recombined gives
$$
x h_5+y h_7=h_1 \longrightarrow h_1=\frac{x h_5+y h_7}{1}=\frac{(0.6301)(107.42)+(0.3699)(98.69)}{1}=104.19\ \mathrm{Btu} / \mathrm{bm}
$$ Then, $$
\begin{aligned}
& \left.\begin{array}{l}
P_1=P_{\text {sat } @-29.9 \mathrm{~F}} \cong 10 \mathrm{psia} \\
h_1=104.19\ \mathrm{Btu} / \mathrm{lbm}
\end{array}\right\} s_1=0.2418\ \mathrm{Btu} / \mathrm{bm} \cdot \mathrm{R} \\
& \left.\begin{array}{l}
P_2=160\ \mathrm{psia} \\
s_2=s_1
\end{array}\right\} h_2=131.15\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ The cooling load of both evaporators per unit mass through the compressor is
$$
\begin{aligned}
& q_L=x\left(h_5-h_4\right)+y\left(h_7-h_6\right) \\
& =(0.6301)(107.42-48.527) \mathrm{Btu} / \mathrm{lbm}+(0.3699)(98.69-48.527) \mathrm{Btu} / \mathrm{lbm} \\
& =55.66\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ The work input to the compressor is $$
w_{\text {in }}=h_2-h_1=(131.15-104.19) \mathrm{Btu} / \mathrm{lbm}=26.96\ \mathrm{Btu} / \mathrm{lbm}
$$ The COP of this refrigeration system is determined from its definition, $$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{55.66 \mathrm{Btu} / \mathrm{lbm}}{26.96 \mathrm{Btu} / \mathrm{lbm}}=\mathbf{2 . 0 7}
$$
