Answer
a) $V_{1}=7.87\text{ L/s}$
b) $ \dot{W}_{\text {in }}=4.75\text{ kW}$
$COP_{y}=2.79$
c) $\dot{E} x_{\dot{Q}_L}=1.81\text{ kW}$
$\dot{E} x_{\text {dest, }, \text { tatal }}=2.94\text{ kW}$
Work Step by Step
(a) The properties of ammonia are given in problem statement. An energy balance on the condenser gives $$
\begin{aligned}
& q_H=h_1-h_4=1439.3-437.4=1361 \mathrm{~kJ} / \mathrm{kg} \\
& \dot{m}=\frac{\dot{Q}_H}{q_H}=\frac{18 \mathrm{~kW}}{1361 \mathrm{~kJ} / \mathrm{kg}}=0.01323 \mathrm{~kg} / \mathrm{s}
\end{aligned}
$$ The volume flow rate is determined from $$
\begin{aligned}
\dot{V}_1 & =m v_1=(0.01323 \mathrm{~kg} / \mathrm{s})\left(0.5946 \mathrm{~m}^3 / \mathrm{kg}\right) \\
& =0.007865 \mathrm{~m}^3 / \mathrm{s}=7.87 \mathrm{~L} / \mathrm{s}
\end{aligned}
$$ (b) The power input and the $\mathrm{COP}$ are $$
\begin{aligned}
& \dot{W}_{\text {in }}=\dot{m}\left(h_2-h_1\right)=(0.01323 \mathrm{~kg} / \mathrm{s})(1798.3-1439.3) \mathrm{kJ} / \mathrm{kg}=\mathbf{4 . 7 5} \mathrm{kW} \\
& \dot{Q}_L=\dot{m}\left(h_1-h_4\right)=(0.01323 \mathrm{~kg} / \mathrm{s})(1439.3-437.4) \mathrm{kJ} / \mathrm{kg}=13.25 \mathrm{~kW} \\
& \mathrm{COP}=\frac{\dot{Q}_L}{\dot{W}_{\text {in }}}=\frac{13.25 \mathrm{~kW}}{4.75 \mathrm{~kW}}=\mathbf{2 . 7 9}
\end{aligned}
$$ (c) The exergy of the heat transferred from the low-temperature medium is
$$
\dot{E} x_{\dot{Q}_L}=-\dot{Q}_L\left(1-\frac{T_0}{T_L}\right)=-(13.25 \mathrm{~kW})\left(1-\frac{300}{264}\right)=1.81 \mathrm{~kW}
$$ The second-law efficiency of the cycle is $$
\eta_{\mathrm{I}}=\frac{\dot{E} x_{\dot{Q}_{\perp}}}{\dot{W}_{\mathrm{in}}}=\frac{1.81}{4.75}=0.381=\mathbf{3 8 . 1} \%
$$ The total exergy destruction in the cycle is the difference between the exergy supplied (power input) and the exergy recovered (the exergy of the heat transferred from the low-temperature medium): $$
\dot{E} x_{\text {dest }, \text { total }}=\dot{W}_{\text {in }}-\dot{E} x_{\dot{Q}_L}=4.75-1.81=\mathbf{2 . 9 4}\ \mathbf{kW}
$$
