Answer
$\eta_{\text {th }}=40.5\%$
$\dot{m}=48.5\text{ kg/s}$
Work Step by Step
(a) From the steam tables (Tables A-4, A-5, and A-6),
$$
\begin{aligned}
& h_1=h_{f @ 30 \mathrm{kPa}}=289.18 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 30 \mathrm{kPa}}=0.001022 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{\mathrm{p}, \text { in }}=v_1\left(P_2-P_1\right) \\
& \begin{array}{l}
=v_1\left(P_2-P_1\right) \\
=\left(0.001022 \mathrm{~m}^3 / \mathrm{kg}\right)(10,000-30 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)
\end{array} \\
& =10.19 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{\mathrm{p}, \text { in }}=289.18+10.19=299.37 \mathrm{~kJ} / \mathrm{kg} \\
& \left.P_3=10 \mathrm{MPa}\right\} h_3=3500.9 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_3=550^{\circ} \mathrm{C}\right\} s_3=6.7561 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_4=4 \mathrm{MPa} \\
s_4=s_3
\end{array}\right\} h_4=3204.9 \mathrm{~kJ} / \mathrm{kg} \\
& P_5=4 \mathrm{MPa}, h_s=3559.7 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_s=550^{\circ} \mathrm{C}\right\} s_s=7.2335 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_6=2 \mathrm{MPa} \\
s_6=s_5
\end{array}\right\} h_6=3321.1 \mathrm{~kJ} / \mathrm{kg} \\
& \left.P_7=2 \mathrm{MPa}\right\} h_7=3578.4 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_7=550^{\circ} \mathrm{C}\right\} s_7=7.5706 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
&
\end{aligned}
$$ $$
\left.\begin{array}{l}
\begin{array}{l}
P_8=30 \mathrm{kPa} \\
s_8=s_7
\end{array}
\end{array}\right\} \begin{aligned}
& x_8=\frac{s_8-s_f}{s_{f g}}=\frac{7.5706-0.9441}{6.8234}=0.9711 \\
& h_8=h_f+x_8 h_{f g}=289.27+(0.9711)(2335.3)=2557.1 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Then, $$
\begin{aligned}
q_{\text {in }} & =\left(h_3-h_2\right)+\left(h_5-h_4\right)+\left(h_7-h_6\right) \\
& =3500.9-299.37+3559.7-3204.9+3578.4-3321.1=3813.7 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {out }} & =h_8-h_1=2557.1-289.18=2267.9 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {net }} & =q_{\text {in }}-q_{\text {out }}=3813.7-2267.9=1545.8 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Thus, $$
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{1545.8 \mathrm{~kJ} / \mathrm{kg}}{3813.7 \mathrm{~kJ} / \mathrm{kg}}=0.4053=\mathbf{4 0 . 5} \%
$$ (b) The mass flow rate of the steam is then $$
\dot{m}=\frac{\dot{W}_{\text {net }}}{w_{\text {net }}}=\frac{75,000 \mathrm{~kJ} / \mathrm{s}}{1545.8 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{4 8 . 5} \mathrm{kg} / \mathrm{s}
$$
