Answer
a) $T_2=54.5^{\circ}C$
b) $\dot{Q}_{\text {people }}=0.665\text{ kW}$
c) $COP_{}=5.87$
d) $ \dot{V}_{1, \min }=15.7\text{ L/min}$
Work Step by Step
(a) From the refrigerant-134a tables (Tables A-11 through A-13)
$$
\begin{aligned}
& \left.\begin{array}{l}
P_1=500\ \mathrm{kPa} \\
x_1=1
\end{array}\right\} \begin{array}{l}
h_1=259.36 \mathrm{~kJ} / \mathrm{kg} \\
v_1=0.04117 \mathrm{~kJ} / \mathrm{kg} \\
s_1=0.9242 \mathrm{~kJ} / \mathrm{kg}
\end{array} \\
& \left.\begin{array}{l}
P_2=1200\ \mathrm{kPa} \\
s_2=s_1
\end{array}\right\} h_{2 s}=277.45 \\
& h_3=h_{f @ 1200 \mathrm{kPa}}=117.79 \mathrm{~kJ} / \mathrm{kg} \\
& h_4=h_3=117.79 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_C=\frac{h_{2 s}-h_1}{h_2-h_1} \\
& 0.75=\frac{277.45-259.36}{h_2-259.36} \longrightarrow h_2=283.48 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_2=1200\ \mathrm{kPa} \\
h_2=283.48 \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} T_2=\mathbf{5 4 . 5}^{\circ} \mathrm{C}
\end{aligned}
$$ (b) The mass flow rate of the refrigerant is $$
\dot{m}=\frac{\dot{V}_1}{v_1}=\frac{(100 \mathrm{~L} / \mathrm{min})\left(\frac{1 \mathrm{~m}^3}{1000 \mathrm{~L}}\right)\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)}{0.04117 \mathrm{~m}^3 / \mathrm{kg}}=0.04048 \mathrm{~kg} / \mathrm{s}
$$ The refrigeration load is $$
\dot{Q}_L=\dot{m}\left(h_1-h_4\right)=(0.04048 \mathrm{~kg} / \mathrm{s})(259.36-117.79) \mathrm{kJ} / \mathrm{kg}=5.731 \mathrm{~kW}
$$ which is the total heat removed from the room. Then, the rate of heat generated by the people in the room is determined from $$
\dot{Q}_{\text {people }}=\dot{Q}_L-\dot{Q}_{\text {heat }}-\dot{Q}_{\text {equip }}=(5.731-250 / 60-0.9) \mathrm{kW}=\mathbf{0 . 6 6 5 k W}
$$ (c) The power input and the COP are $$
\begin{aligned}
& \dot{W}_{\text {in }}=\dot{m}\left(h_2-h_1\right)=(0.04048 \mathrm{~kg} / \mathrm{s})(283.48-259.36) \mathrm{kJ} / \mathrm{kg}=0.9764 \mathrm{~kW} \\
& \mathrm{COP}=\frac{\dot{Q}_{\mathrm{L}}}{\dot{W}_{\text {in }}}=\frac{5.731}{0.9764}=\mathbf{5 . 8 7}
\end{aligned}
$$ (d) The reversible COP of the cycle is $$
\mathrm{COP}_{\mathrm{nev}}=\frac{1}{T_H / T_L-1}=\frac{1}{(34+273) /(26+273)-1}=37.38
$$ The corresponding minimum power input is $$
\dot{W}_{\mathrm{in}, \min }=\frac{\dot{Q}_{\mathrm{L}}}{\operatorname{COP}_{\mathrm{rev}}}=\frac{5.731 \mathrm{~kW}}{37.38}=0.1533 \mathrm{~kW}
$$ The minimum mass and volume flow rates are $$
\begin{aligned}
& \dot{m}_{\min }=\frac{\dot{W}_{\text {in,min }}}{h_2-h_1}=\frac{0.1533 \mathrm{~kW}}{(283.48-259.36) \mathrm{kJ} / \mathrm{kg}}=0.006358 \mathrm{~kg} / \mathrm{s} \\
& \dot{V}_{1, \min }=\dot{m}_{\min } \nu_1=(0.006358 \mathrm{~kg} / \mathrm{s})\left(0.04117 \mathrm{~m}^3 / \mathrm{kg}\right)=\left(0.0002618 \mathrm{~m}^3 / \mathrm{s}\right)=15.7 \mathrm{L/min}
\end{aligned}
$$
