Answer
a) $COP_{R,C}=5.060$
b) $P_{evap}=132.82\text{ kPa}; P_{cond}=770.64\text{ kPa}$
c) $w_{\text {net}, \text { in }}=27.36\text{ kJ/kg}$
Work Step by Step
(a) The COP of this refrigeration cycle is determined from $$
\mathrm{COP}_{\mathrm{R}, \mathrm{C}}=\frac{1}{\left(T_H / T_L\right)-1}=\frac{1}{(303 \mathrm{~K})(253 \mathrm{~K})-1}=\mathbf{5 . 0 6 0}
$$ (b) The condenser and evaporative pressures are (Table A-11) $$
\begin{aligned}
& P_{\text {evap }}=P_{\text {sat } @-20^{\circ} \mathrm{C}}=132.82\ \mathrm{kPa} \\
& P_{\text {cond }}=P_{\text {sat } @ 30^{\circ} \mathrm{C}}=770.64\ \mathrm{kPa}
\end{aligned}
$$ (c) The net work input is determined from $$
\begin{aligned}
& h_1=\left(h_f+x_1 h_{f g}\right)_{@-20^{\circ} C}=25.47+(0.15)(212.96)=57.42 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=\left(h_f+x_2 h_{f g}\right)_{@-20^{\circ} \mathrm{C}}=25.47+(0.80)(212.96)=195.84 \mathrm{~kJ} / \mathrm{kg} \\
& q_L=h_2-h_1=195.84-57.42=138.4 \mathrm{~kJ} / \mathrm{kg} \\
& w_{\text {net}, \text { in }}=\frac{q_L}{\mathrm{COP}_{\mathrm{R}}}=\frac{138.4 \mathrm{~kJ} / \mathrm{kg}}{5.060}=\mathbf{2 7 . 3 6}\ \mathbf{k J} / \mathbf{k g}
\end{aligned}
$$
