Answer
$0.7457\text{ KW}$
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: $$
\begin{aligned}
& P_1=50\ \mathrm{psia} \quad h_1=h_{g @ 50 \mathrm{psia}}=108.83\ \mathrm{Btu} / \mathrm{lbm} \\
& \text { sat. vapor }\} s_1=s_{g @ 50 p s i a}=0.22192\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
& \left.\begin{array}{l}
P_2=120\ \mathrm{psia} \\
s_2=s_1
\end{array}\right\} h_2=116.64\ \mathrm{Btu} / \mathrm{lbm} \\
& \left.\begin{array}{l}
P_3=120 \text { psia } \\
\text { sat. liquid }
\end{array}\right\} h_3=h_{f @ 120 \text { psia }}=41.79\ \mathrm{Btu} / \mathrm{lbm} \\
& h_4 \cong h_3=41.79\ \mathrm{Btu} / \mathrm{lbm} \text { }
\end{aligned}
$$ The mass flow rate of the refrigerant and the power input to the compressor are determined from
$$
\dot{m}=\frac{\dot{Q}_H}{q_H}=\frac{\dot{Q}_H}{h_2-h_3}=\frac{60,000 / 3600 \mathrm{Btu} / \mathrm{s}}{(116.64-41.79) \mathrm{Btu} / \mathrm{lbm}}=0.2227\ \mathrm{lbm} / \mathrm{s}
$$ $$
\begin{aligned}
\dot{W}_{\text {in }} & =\dot{m}\left(h_2-h_1\right)=(0.2227 \mathrm{~kg} / \mathrm{s})(116.64-108.83) \mathrm{Btu} / \mathrm{lbm} \\
& =1.739\ \mathrm{Btu} / \mathrm{s}=\mathbf{2 . 4 6} \mathrm{hp} \text { since } 1 \mathrm{hp}=0.7068\ \mathrm{Btu} / \mathrm{s}
\end{aligned}
$$ The electrical power required without the heat pump is $$
\dot{W}_e=\dot{Q}_H=(60,000 / 3600 \mathrm{Btu} / \mathrm{s})\left(\frac{1 \mathrm{hp}}{0.7068 \mathrm{Btu} / \mathrm{s}}\right)=23.58\ \mathrm{hp}
$$ $$
\begin{aligned}
\dot{W}_{\text {saved }} & =\dot{W}_e-\dot{W}_{\text {in }}=23.58-2.46 \\
& =\mathbf{2 1 . 1} \mathbf{h p}=15.75 \mathrm{~kW} \text { since } 1 \mathrm{hp}=0.7457 \mathrm{~kW}
\end{aligned}
$$
