Answer
$\dot{Q}_{\text {in }}=45,039\text{ kW}$
$\eta_{\text {th }}=42.6\%$
Work Step by Step
(a) From the steam tables (Tables A-4, A-5, and A-6),
$$
\begin{aligned}
& h_1=h_{\text {sat } @ 10 \mathrm{kPa}}=191.81 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{\text {sat } @ 10 \mathrm{kPa}}=0.00101 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{p, \text { in }}=v_1\left(P_2-P_1\right) \\
&=\left(0.00101 \mathrm{~m}^3 / \mathrm{kg}\right)(15,000-10 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
&=15.14 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{p, \text { in }}=191.81+15.14=206.95 \mathrm{~kJ} / \mathrm{kg} \\
& P_3=15 \mathrm{MPa}\} h_3=3310.8 \mathrm{~kJ} / \mathrm{kg} \\
&\left.T_3=500^{\circ} \mathrm{C}\right\} s_3=6.3480 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{aligned}
$$ $$
\begin{aligned}
& \left.\begin{array}{l}
P_6=10 \mathrm{kPa} \\
s_6=s_5
\end{array}\right\} \begin{array}{l}
h_6=h_f+x_6 h_{f g}=191.81+(0.90)(2392.1)=2344.7 \mathrm{~kJ} / \mathrm{kg} \\
s_6=s_f+x_6 s_{f g}=0.6492+(0.90)(7.4996)=7.3988 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array} \\
& \left.\begin{array}{l}
T_5=500^{\circ} \mathrm{C} \\
s_5=s_6
\end{array}\right\} \begin{array}{l}
P_5=\mathbf{2 1 5 0} \mathbf{k P a}(\text { the reheat pressure }) \\
h_5=3466.61 \mathrm{~kJ} / \mathrm{kg}
\end{array} \\
& \left.\begin{array}{l}
P_4=2.15 \mathrm{MPa} \\
s_4=s_3
\end{array}\right\} h_4=2817.2 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ (b) The rate of heat supply is $$
\begin{aligned}
\dot{Q}_{\text {in }} & =\dot{m}\left[\left(h_3-h_2\right)+\left(h_5-h_4\right)\right] \\
& =(12 \mathrm{~kg} / \mathrm{s})(3310.8-206.95+3466.61-2817.2)) \mathrm{kJ} / \mathrm{kg} \\
& =\mathbf{4 5 , 0 3 9} \mathbf{~ k W}
\end{aligned}
$$ (c) The thermal efficiency is determined from $$
\dot{Q}_{\text {out }}=\dot{m}\left(h_6-h_1\right)=(12 \mathrm{~kJ} / \mathrm{s})(2344.7-191.81) \mathrm{kJ} / \mathrm{kg}=25,835 \mathrm{~kJ} / \mathrm{s}
$$ Thus, $$
\eta_{\text {th }}=1-\frac{\dot{Q}_{\text {out }}}{\dot{Q}_{\text {in }}}=1-\frac{25,834 \mathrm{~kJ} / \mathrm{s}}{45,039 \mathrm{~kJ} / \mathrm{s}}=\mathbf{4 2 . 6 \%}
$$
