Answer
a) $\dot{m}_A=0.1689 \mathrm{~kg} / \mathrm{s}$
b) $\dot{Q}_L=18.46\text{ kW}$
c) $COP_{y}=2.12$
Work Step by Step
(a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):
\begin{aligned}
& h_1=h_{g @ 160 \mathrm{kPa}}=241.14 \mathrm{~kJ} / \mathrm{kg} \\
& s_1=s_{g @ 160 \mathrm{kPa}}=0.9420 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_2=500 \mathrm{kPa} \\
s_2=s_1
\end{array}\right\} h_{2 s}=264.55 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_C=\frac{h_{2 s}-h_1}{h_2-h_1} \\
& 0.80=\frac{264.55-241.14}{h_2-241.14} \longrightarrow h_2=270.41 \mathrm{~kJ} / \mathrm{kg} \\
& h_3=h_{f @ 500 \mathrm{kPa}}=73.32 \mathrm{~kJ} / \mathrm{kg} \\
& h_4=h_3=73.32 \mathrm{~kJ} / \mathrm{kg} \\
& h_5=h_{g @ 400 \mathrm{kPa}}=255.61 \mathrm{~kJ} / \mathrm{kg} \\
& s_5=s_{g @ 400 \mathrm{kPa}}=0.9271 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
\begin{array}{l}
P_6=1400 \mathrm{kPa} \\
s_6
\end{array}=s_5
\end{array}\right\} h_{6 s}=281.56 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_C=\frac{h_{6 s}-h_5}{h_6-h_5} \\
& 0.80=\frac{281.56-255.61}{h_6-255.61} \longrightarrow h_6=288.04 \mathrm{~kJ} / \mathrm{kg} \\
& h_7=h_{f @ 1400 \mathrm{kPa}}=127.25 \mathrm{~kJ} / \mathrm{kg} \\
& h_8=h_7=127.25 \mathrm{~kJ} / \mathrm{kg} \\
&
\end{aligned} The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on the heat exchanger $$
\begin{aligned}
\dot{m}_A\left(h_5-h_8\right) & =\dot{m}_B\left(h_2-h_3\right) \\
\dot{m}_A(255.61-127.25) \mathrm{kJ} / \mathrm{kg} & =(0.11 \mathrm{~kg} / \mathrm{s})(270.41-73.32) \mathrm{kJ} / \mathrm{kg} \longrightarrow \dot{m}_A\\=\mathbf{0 . 1 6 8 9 k g} / \mathbf{s}
\end{aligned}
$$ (b) The rate of heat removal from the refrigerated space is
$$
\dot{Q}_L=\dot{m}_B\left(h_1-h_4\right)=(0.11 \mathrm{~kg} / \mathrm{s})(241.14-73.32) \mathrm{kJ} / \mathrm{kg}=18.46 \mathrm{~kW}
$$ (c) The power input and the COP are $$
\begin{aligned}
\dot{W}_{\text {in }} & =\dot{m}_A\left(h_6-h_5\right)+\dot{m}_B\left(h_2-h_1\right) \\
& =(0.11 \mathrm{~kg} / \mathrm{s})(288.04-255.61) \mathrm{kJ} / \mathrm{kg}+(0.1689 \mathrm{~kg} / \mathrm{s})(270.41-241.14) \mathrm{kJ} / \mathrm{kg} \\
& =8.698 \mathrm{~kW} \\
\mathrm{COP} & =\frac{\dot{Q}_{\mathrm{L}}}{\dot{W}_{\text {in }}}=\frac{18.46}{8.698}=\mathbf{2 . 1 2}
\end{aligned}
$$
