Answer
a) $\dot{W}_{\text {in }}=3.55\text{ kW}$
b) $\dot{Q}_L=13.12\text{ kW}$
c) $\dot{W}_{\text {increase }}=13.12\text{ kW}$
Work Step by Step
(a) From the refrigerant tables (Tables A-12 and A-13), $$
\begin{aligned}
& \left.\begin{array}{l}
P_1=280\ \mathrm{kPa} \\
T_1=0^{\circ} \mathrm{C}
\end{array}\right\} h_1=250.85 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_2=1.0\ \mathrm{MPa} \\
T_2=60^{\circ} \mathrm{C}
\end{array}\right\} h_2=293.40 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_3=1.0\ \mathrm{MPa} \\
T_3=30^{\circ} \mathrm{C}
\end{array}\right\} h_3 \cong h_{f @ 30^{\circ} \mathrm{C}}=93.58 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=93.58 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) }
\end{aligned}
$$ The mass flow rate of the refrigerant is $$
\dot{m}_R=\frac{\dot{Q}_H}{q_H}=\frac{\dot{Q}_H}{h_2-h_3}=\frac{60,000 / 3,600 \mathrm{~kJ} / \mathrm{s}}{(293.40-93.58) \mathrm{kJ} / \mathrm{kg}}=0.08341 \mathrm{~kg} / \mathrm{s}
$$ Then the power input to the compressor becomes $$
\dot{W}_{\text {in }}=\dot{m}\left(h_2-h_1\right)=(0.08341 \mathrm{~kg} / \mathrm{s})(293.40-250.85) \mathrm{kJ} / \mathrm{kg}=\mathbf{3 . 5 5}\ \mathbf{ kW} $$ (b) The rate of hat absorption from the water is
$$
\dot{Q}_L=\dot{m}\left(h_1-h_4\right)=(0.08341 \mathrm{~kg} / \mathrm{s})(250.85-93.58) \mathrm{kJ} / \mathrm{kg}=13.12\ \mathbf{k W}
$$ (c) The electrical power required without the heat pump is $$
\dot{W}_e=\dot{Q}_H=60,000 / 3600 \mathrm{~kJ} / \mathrm{s}=16.67 \mathrm{~kW}
$$ Thus, $$
\dot{W}_{\text {increase }}=\dot{W}_e-\dot{W}_{\text {in }}=16.67-3.55=13.12\ \mathbf{k W}
$$
