Answer
a) $\dot{m}=0.6749\text{ kg/s}$
b) $\dot{W}_{\text {in }}=24.7\text{ kW}$
c) $\dot{m}_{\text {cooling }}=3.73\text{ kg/s}$
Work Step by Step
(a) 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),$$
\begin{aligned}
& h_4 \cong h_3=88.82 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) }
\end{aligned}
$$ 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 power input to the compressor is $$
\dot{W}_{\text {in }}=\dot{m}\left(h_2-h_1\right)=(0.6749 \mathrm{~kg} / \mathrm{s})(273.52-236.99) \mathrm{kJ} / \mathrm{kg}=\mathbf{2 4 . 7} \mathbf{~ k W}
$$ (c) The mass flow rate of the cooling water is determined from $$
\begin{aligned}
\dot{Q}_H & =\dot{m}\left(h_2-h_3\right)=(0.6749 \mathrm{~kg} / \mathrm{s})(273.52-88.82) \mathrm{kJ} / \mathrm{kg}\\&=124.7 \mathrm{~kW} \\
\dot{m}_{\text {cooling }} & =\frac{\dot{Q}_H}{\left(c_p \Delta T\right)_{\text {water }}}=\frac{124.7 \mathrm{~kJ} / \mathrm{s}}{\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(8^{\circ} \mathrm{C}\right)}=3.73 \mathrm{~kg} / \mathrm{s}
\end{aligned}
$$
