Answer
$\eta_{\text {th }}=40.2\%$
$\dot{m}=164.7\text{ kg/s}$
Work Step by Step
(a) From the steam tables (Tables A-4, A-5, and A-6), $$
\begin{aligned}
& h_{\mathrm{l}}=h_{f @ 10 \mathrm{kPa}}=191.81 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 10 \mathrm{kPa}}=0.00101 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{p, \text { in }}=v_1\left(P_2-P_1\right) \\
& =\left(0.00101 \mathrm{~m}^3 / \mathrm{kg}\right)(10,000-10 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
& =10.09 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{p, \text { in }}=191.81+10.09=201.90 \mathrm{~kJ} / \mathrm{kg} \\
& P_3=10 \mathrm{MPa} \quad h_3=3375.1 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_3=500^{\circ} \mathrm{C}\right\} s_3=6.5995 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_4=10 \mathrm{kPa} \\
s_4=s_3
\end{array}\right\} x_4=\frac{s_4-s_f}{s_{f g}}=\frac{6.5995-0.6492}{7.4996}=\mathbf{0 . 7 9 3 4} \\
& h_4=h_f+x_4 h_{f g}=191.81+(0.7934)(2392.1)=2089.7 \mathrm{~kJ} / \mathrm{kg} \\
&
\end{aligned}
$$ (b) $$
\begin{aligned}
q_{\text {in }} & =h_3-h_2=3375.1-201.90=3173.2 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {out }} & =h_4-h_1=2089.7-191.81=1897.9 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {net }} & =q_{\text {in }}-q_{\text {out }}=3173.2-1897.9=1275.4 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ and $$
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{1275.4 \mathrm{~kJ} / \mathrm{kg}}{3173.2 \mathrm{~kJ} / \mathrm{kg}}=40.2 \%
$$ (c) $$
\dot{m}=\frac{\dot{W}_{\text {net }}}{w_{\text {net }}}=\frac{210,000 \mathrm{~kJ} / \mathrm{s}}{1275.4 \mathrm{~kJ} / \mathrm{kg}}=164.7 \mathrm{~kg} / \mathrm{s}
$$
