Answer
a) $\dot{m}=0.6749\text{ kg/s}$
b) $\dot{W}_{\text {in }}=32.9\text{ kW}$
c) $\dot{X}_{\text {destroyed }}=7.80\text{ kW}$
Work Step by Step
(a) 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),$$
\begin{aligned}
& P_1=120 \mathrm{kPa}, h_1=h_{g @ 120 \mathrm{kPa}}=236.99 \mathrm{~kJ} / \mathrm{kg} \\
& \text { sat. vapor }\} s_1=s_{g @ 120 \mathrm{kPa}}=0.94789 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_2=0.7 \mathrm{MPa} \\
s_2=s_1
\end{array}\right\} h_{2 s}=273.52 \mathrm{~kJ} / \mathrm{kg} \quad\left(T_2=34.95^{\circ} \mathrm{C}\right) \\
& \left.\begin{array}{l}
\begin{array}{l}
P_3=0.7 \mathrm{MPa} \\
\text { sat. liquid }
\end{array}
\end{array}\right\} h_3=h_{f @ 0.7 \mathrm{MPa}}=88.82 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=88.82 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) }
\end{aligned}
$$ The mass flow rate of the refrigerant is determined from $$
\dot{m}=\frac{\dot{Q}_L}{h_1-h_4}=\frac{100 \mathrm{~kJ} / \mathrm{s}}{(236.99-88.82) \mathrm{kJ} / \mathrm{kg}}=0.6749 \mathrm{~kg} / \mathrm{s}
$$ (b) The actual enthalpy at the compressor exit is $$
\begin{aligned}
\eta_C=\frac{h_{2 s}-h_1}{h_2-h_1} \longrightarrow h_2 & =h_1+\left(h_{2 s}-h_1\right) / \eta_C\\&=236.99+(273.52-236.99) /(0.75) \\
& =285.70 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Thus, $$
\dot{W}_{\text {in }}=\dot{m}\left(h_2-h_1\right)=(0.6749 \mathrm{~kg} / \mathrm{s})(285.70-236.99) \mathrm{kJ} / \mathrm{kg}=\mathbf{3 2 . 9} \mathbf{k W}
$$ (c) The mass flow rate of the cooling water is determined from $$
\dot{Q}_H=\dot{m}\left(h_2-h_3\right)=(0.6749 \mathrm{~kg} / \mathrm{s})(285.70-88.82) \mathrm{kJ} / \mathrm{kg}=132.9 \mathrm{~kW}
$$ $$
\dot{m}_{\text {cooling }}=\frac{\dot{Q}_H}{\left(c_p \Delta T\right)_{\text {water }}}=\frac{132.9 \mathrm{~kJ} / \mathrm{s}}{\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(8^{\circ} \mathrm{C}\right)}=3.97 \mathrm{~kg} / \mathrm{s}
$$ The exergy destruction associated with this adiabatic compression process is determined from $$
\dot{X}_{\text {destroyed }}=T_0 \dot{S}_{\text {gen }}=T_0 \dot{m}\left(s_2-s_1\right)
$$ where $$
\left.\begin{array}{l}
P_2=0.7 \mathrm{MPa} \\
h_2=285.70 \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} s_2=0.98665 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
$$ Thus, $$
\dot{X}_{\text {destroyed }}=(298 \mathrm{~K})(0.6749 \mathrm{~kg} / \mathrm{s})(0.98665-0.94789) \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}=\mathbf{7 . 8 0}\ \mathrm{kW}
$$
