Answer
a) $x_{6}=0.2899$
b) $q_{l}=116.2\text{ kJ/kg}$
$w_{in}=42.72\text{ kJ/kg}$
c) $\mathrm{COP}_{\mathrm{R}}=2.72$
Work Step by Step
(a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13)$$
\begin{array}{ll}
h_1=236.99 \mathrm{~kJ} / \mathrm{kg}, & h_2=266.29 \mathrm{~kJ} / \mathrm{kg} \\
h_3=259.36 \mathrm{~kJ} / \mathrm{kg}, & \\
h_5=127.25 \mathrm{~kJ} / \mathrm{kg}, & h_6=127.25 \mathrm{~kJ} / \mathrm{kg} \\
h_7=73.32 \mathrm{~kJ} / \mathrm{kg}, & h_8=73.32 \mathrm{~kJ} / \mathrm{kg}
\end{array}
$$ The fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6 ,
$$
x_6=\frac{h_6-h_f}{h_{f g}}=\frac{127.25-73.32}{186.04}=\mathbf{0 . 2 8 9 9}
$$ (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber: $$
\begin{aligned}
& \dot{E}_{\text {in }}-\dot{E}_{\text {out }}=\Delta \dot{E}_{\text {system }}^{90 \text { (steady) }}=0 \rightarrow \dot{E}_{\text {in }}=\dot{E}_{\text {out }} \\
& \begin{aligned}
\sum \dot{m}_e h_e & =\sum \dot{m}_i h_i \\
(1) h_9 & =x_6 h_3+\left(1-x_6\right) h_2 \\
h_9 & =(0.2899)(259.36)+(1-0.2899)(266.29)=264.28 \mathrm{~kJ} / \mathrm{kg}
\end{aligned} \\
& \left.\begin{array}{rl}
P_9= & 0.5 \mathrm{MPa} \\
h_9 & =264.28 \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} s_9=0.94108 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{aligned}
$$ Also, $$
\left.\begin{array}{l}
P_4=1.4 \mathrm{MPa} \\
s_4=s_9=0.94108 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array}\right\} h_4=286.19 \mathrm{~kJ} / \mathrm{kg}
$$ Then the amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are $$
\begin{aligned}
q_L & =\left(1-x_6\right)\left(h_1-h_8\right)=(1-0.2899)(236.99-73.32) \mathrm{kJ} / \mathrm{kg}=116.2 \mathrm{~kJ} / \mathrm{kg} \\
w_{\mathrm{in}} & =w_{\text {complini }}+w_{\text {compll,in }}=\left(1-x_6\right)\left(h_2-h_1\right)+(1)\left(h_4-h_9\right) \\
& =(1-0.2899)(266.29-236.99) \mathrm{kJ} / \mathrm{kg}+(1)(286.19-264.28) \mathrm{kJ} / \mathrm{kg} \\
& =\mathbf{4 2 . 7 2} \mathbf{~ k J} / \mathbf{k g}
\end{aligned}
$$ (c) The coefficient of performance is determined from $$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{116.2 \mathrm{~kJ} / \mathrm{kg}}{42.72 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{2 . 7 2}
$$
