Answer
a) $\dot m=0.04200\text{ kg/s}$
$\dot{W}_{\text {in }}= 2.40\text{ kW}$
b) $\dot{Q}_L=6.17\text{ kW}$
c) $\dot{Q}_{\text {gain }}=0.203\text{ kW}$
Work Step by Step
(a) From the refrigerant tables (Tables A-12 and A-13), $$
\begin{aligned}
& \left.\begin{array}{l}
P_1=100\ \mathrm{kPa} \\
T_1=-20^{\circ} \mathrm{C}
\end{array}\right\} \begin{array}{l}
h_1=239.52 \mathrm{~kJ} / \mathrm{kg} \\
s_1=0.97215 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
v_1=0.19841 \mathrm{~m}^3 / \mathrm{kg}
\end{array} \\
& \left.\begin{array}{l}
P_2=0.8\ \mathrm{MPa} \\
s_{2 s}=s_1
\end{array}\right\} h_{2 s}=284.09 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_3=0.75 \mathrm{MPa} \\
T_3=26^{\circ} \mathrm{C}
\end{array}\right\} h_3 \cong h_{f @ 26^{\circ} \mathrm{C}}=87.83 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=87.83 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) } \\
& \left.T_5=-26^{\circ} \mathrm{C}\right\rangle P_5=0.10173\ \mathrm{MPa} \\
& \text { sat. vapor }\} h_5=234.70 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Then the mass flow rate of the refrigerant and the power input becomes $$
\begin{aligned}
\dot{m} & =\frac{\dot{V_1}}{v_1}=\frac{0.5 / 60 \mathrm{~m}^3 / \mathrm{s}}{0.19841 \mathrm{~m}^3 / \mathrm{kg}}=0.04200 \mathrm{~kg} / \mathrm{s} \\
\dot{W}_{\text {in }} & =\dot{m}\left(h_{2 s}-h_1\right) / \eta_C=(0.04200 \mathrm{~kg} / \mathrm{s})((284.09-239.52) \mathrm{kJ} / \mathrm{kg}] /(0.78)=2.40 \mathrm{~kW}
\end{aligned}
$$ (b) The rate of heat removal from the refrigerated space is $$
\dot{Q}_L=\dot{m}\left(h_5-h_4\right)=(0.04200 \mathrm{~kg} / \mathrm{s})(234.70-87.83) \mathrm{kJ} / \mathrm{kg}=6.17 \mathrm{~kW}
$$ (c) The pressure drop and the heat gain in the line between the evaporator and the compressor are $$
\Delta P=P_5-P_1=101.73-100=\mathbf{1 . 7 3}
$$ and $$
\dot{Q}_{\text {gain }}=\dot{m}\left(h_1-h_5\right)=(0.04200 \mathrm{~kg} / \mathrm{s})(239.52-234.70) \mathrm{kJ} / \mathrm{kg}=\mathbf{0 . 2 0 3} \mathbf{~ k W}
$$
