Answer
a) $\dot{Q}_H=38.8\text{ kW}$
b) $\dot{V}_1=0.0159\text{ m}^{3}\text{/s}$
c) $\mathrm{COP}_{\mathrm{R}}=5.07$
Work Step by Step
In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, 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), $$
\begin{aligned}
& \left.\begin{array}{l}
P_1=320\ \mathrm{kPa} \\
\text { sat. vapor }
\end{array}\right\} \begin{array}{l}
h_1=h_{g @ 320 \mathrm{kPa}}=251.93 \mathrm{~kJ} / \mathrm{kg} \\
s_1=s_{g @ 320 \mathrm{kPa}}=0.93026 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
\boldsymbol{v}_1=\boldsymbol{v}_g @ 320 \mathrm{kPa}=0.063681 \mathrm{~m}^3 / \mathrm{kg}
\end{array} \\
& \left.\begin{array}{l}
P_2=1.4 \mathrm{MPa} \\
s_2=s_1
\end{array}\right\} h_2=282.60 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
\begin{array}{l}
P_3=1.4 \mathrm{kPa} \\
\text { sat. liquid }
\end{array}
\end{array}\right\} h_3=h_{f @ 1.4 \mathrm{kPa}}=127.25 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=127.25 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) }
\end{aligned}
$$ The rate of heat supply to the house is determined from
$$
\dot{Q}_H=\dot{m}\left(h_2-h_3\right)=(0.25 \mathrm{~kg} / \mathrm{s})(282.60-127.25) \mathrm{kJ} / \mathrm{kg}=38.8 \mathrm{~kW}
$$ (b) The volume flow rate of the refrigerant at the compressor inlet is $$
\dot{V}_1=\dot{m} v_1=(0.25 \mathrm{~kg} / \mathrm{s})\left(0.063681 \mathrm{~m}^3 / \mathrm{kg}\right)=\mathbf{0 . 0 1 5 9} \mathbf{m}^3 / \mathbf{s}
$$ (c) The COP of t his heat pump is determined from $$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{h_2-h_3}{h_2-h_1}=\frac{282.60-127.25}{282.60-251.93}=\mathbf{5 . 0 7}
$$
