Answer
$\mathrm{COP}_{\mathrm{R}}=4.60$
Work Step by Step
From the refrigerant tables (Tables A-11, A-12, and A-13),
$$
\begin{aligned}
& \left.\begin{array}{l}
\begin{array}{l}
P_3=900\ \mathrm{kPa} \\
\text { sat. liquid }
\end{array}
\end{array}\right\} h_3=h_{f @ 900 \mathrm{kPa}}=101.62 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{rl}
T_4 & =T_{\text {sat } @ 900 \mathrm{kPa}}-11.3 \\
& =35.51-5.51=30^{\circ} \mathrm{C} \\
P_4 & =900\ \mathrm{kPa}
\end{array}\right\} h_4 \equiv h_{f @ 30^{\circ} \mathrm{C}}=93.58 \mathrm{~kJ} / \mathrm{kg} \\
& h_5 \equiv h_4=93.58 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) } \\
& \left.\begin{array}{l}
T_6=-10.09^{\circ} \mathrm{C} \\
\text { sat. vapor }
\end{array}\right\} \begin{array}{l}
h_6=h_{[email protected]^{\circ} \mathrm{C}}=244.50 \mathrm{~kJ} / \mathrm{kg} \\
P_6=P_{\text {sat }@-10.09 \mathrm{C}}=200\ \mathrm{kPa}
\end{array}
\end{aligned}
$$ An energy balance on the heat exchanger gives $$
\dot{m}\left(h_1-h_6\right)=\dot{m}\left(h_3-h_4\right) \longrightarrow h_1=h_3-h_4+h_6=101.62-93.58+244.50=252.54 \mathrm{~kJ} / \mathrm{kg}
$$ Then, $$
\begin{aligned}
& \left.\begin{array}{l}
P_1=200\ \mathrm{kPa} \\
h_1=252.54\ \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} s_1=0.9679 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_2=900\ \mathrm{kPa} \\
s_2=s_1
\end{array}\right\} h_2=285.37 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The COP of this refrigeration system is determined from its definition,$$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{h_6-h_5}{h_2-h_1}=\frac{244.50-93.58}{285.37-252.54}=\mathbf{4 . 6 0}
$$
