Answer
a) $x_{4}=0.323$
b) $COP_{R}=3.373$
c) $\dot{X}_{\text {destroyed }}=0.210\text{ kW}$
Work Step by Step
(a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),
$$
\begin{aligned}
& P_1=140 \mathrm{kPa} \mid h_1=h_{g @ 140\ \mathrm{kPa}}=239.19 \mathrm{~kJ} / \mathrm{kg} \\
& \text { sat . vapor }\} s_1=s_{g@ 140\ \mathrm{kPa}}=0.94467 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_2=0.8 \mathrm{MPa} \\
s_{2 \pi}=s_1
\end{array}\right\} h_{2 x}=275.40 \mathrm{~kJ} / \mathrm{kg} \\
& \begin{aligned}
\eta_C=\frac{h_{2 x}-h_1}{h_2-h_1} \longrightarrow h_2 & =h_1+\left(h_{2 x}-h_1\right) / \eta_C \\
& =239.19+(275.40-
\end{aligned} \\
& =239.19+(275.40-239.19) /(0.85) \\
& =281.79 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_3=0.8\ \mathrm{MPa} \\
\text { sat. liquid }
\end{array}\right\} h_3=h_{f @0.8\ \mathrm{MPa}}=95.48 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \approx h_3=95.48 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) }
\end{aligned}
$$ The quality of the refrigerant at the end of the throttling process is
$$
x_4=\left(\frac{h_4-h_f}{h_{f g}}\right)_{@ 140\ \mathrm{kPa}}=\frac{95.48-27.06}{212.13}=\mathbf{0 . 3 2 3}
$$ (b) The COP of the refrigerator is determined from its definition, $$
\operatorname{COP}_{\mathrm{R}}=\frac{q_L}{w_{\mathrm{in}}}=\frac{h_1-h_4}{h_2-h_1}=\frac{239.19-95.48}{281.79-239.19}=\mathbf{3 . 3 7 3}
$$ (c) The power input to the compressor is determined from $$
\dot{W}_{\text {in }}=\frac{\dot{Q}_L}{\mathrm{COP}_R}=\frac{5 \mathrm{~kW}}{3.373}=1.48 \mathrm{~kW}
$$ The exergy destruction associated with the compression process is determined from
$$
\dot{X}_{\text {destroyed }}=T_0 \dot{S}_{g e \mathrm{n}}=T_0 \dot{m}\left(s_2-s_1+{\frac{q_{\text {surr }}}{T_0}}^{+0}\right)=T_0 \dot{m}\left(s_2-s_1\right)
$$ where $$
\begin{aligned}
& \dot{m}=\frac{\dot{Q}_L}{q_L}=\frac{\dot{Q}_L}{h_1-h_4}=\frac{5 \mathrm{~kJ} / \mathrm{s}}{(239.19-95.48) \mathrm{kJ} / \mathrm{kg}}=0.03479 \mathrm{~kg} / \mathrm{s} \\
& \left.\begin{array}{l}
P_2=0.8\ \mathrm{MPa} \\
h_2=281.79 \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} s_2=0.96494 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{aligned}
$$ Thus, $$
\dot{X}_{\text {destroyed }}=(298 \mathrm{~K})(0.03479 \mathrm{~kg} / \mathrm{s})(0.96494-0.94467) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}=\mathbf{0 . 2 1 0}\ \mathrm{kW}
$$
