Answer
$16.5\%$
Work Step by Step
In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. $$
\left.\begin{array}{l}
T_1=20^{\circ} \mathrm{F} \\
\text { sat. vapor }
\end{array}\right\} \begin{aligned}
& h_1=h_{\mathrm{g}@ 20^\circ \mathrm{F}}=106.00\ \mathrm{Btu} / \mathrm{bm} \\
& s_1=s_{\mathrm{g}@ 20^\circ \mathrm{F}}=0.22345\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}
\end{aligned}
$$
$\left.\begin{array}{l}P_2=300 \text { psia } \\ s_2=s_1\end{array}\right\} h_2=125.70$ Btu/lbm
$\left.P_3=300 \mathrm{psia}\right\} \quad h_3=h_{f@ 300\text{ psia}}=66.347\ \mathrm{Btu} / \mathrm{lbm}$ $ s_3=s_{f@ 300\text{psia}}=0.12717\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}$ $h_4 \simeq h_3=66.347\ \mathrm{Btu} / \mathrm{lbm}$ (throttling)
$$
\left.\begin{array}{l}
T_4=20^{\circ} \mathrm{F} \\
s_4=s_3
\end{array}\right\} \begin{aligned}
& h_{4 s}=59.81\ \mathrm{Btu} / \mathrm{lbm} \\
& x_{4 s}=0.4724
\end{aligned} \text { (isentropic expansion) }
$$
The COP of the refrigerator for the throuling case is
$$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{h_1-h_4}{h_2-h_1}=\frac{106.00-66.347}{125.70-106.00}=\mathbf{2 . 0 1 3}
$$ The COP of the refrigerator for the isentropic expansion case is $$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{h_2-h_{4 x}}{h_2-h_1}=\frac{106.00-59.81}{125.70-106.00}=\mathbf{2 . 3 4 4}
$$ The increase in the COP by isentropic expansion is Perount Increase in COP $=\frac{2.344-3.013}{2.013}=0.165=16.5 \%$
