Answer
$\eta_{\text {th }}=41.3\%$
$\dot{m}=50.0\text{ kg/s}$
Work Step by Step
(a) From the steam tables (Tables A-4, A-5, and A-6),
$$
\begin{aligned}
& h_1=h_{f @ 10 \mathrm{kPa}}=191.81 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 10 \mathrm{kPa}}=0.00101 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{p, \text { in }}=v_1\left(P_2-P_1\right) \\
& =\left(0.00101 \mathrm{~m}^3 / \mathrm{kg}\right)(10,000-10 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
& =10.09 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{p, \text { in }}=191.81+10.09=201.90 \mathrm{~kJ} / \mathrm{kg} \\
& P_3=10 \mathrm{MPa} \quad h_3=3375.1 \mathrm{~kJ} / \mathrm{kg} \\
& T_3=500^{\circ} \mathrm{C} \quad s_3=6.5995 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_4=1 \mathrm{MPa} \\
s_4=s_3
\end{array}\right\} h_4=2783.8 \mathrm{~kJ} / \mathrm{kg} \\
& P_5=1 \mathrm{MPa} \quad h_{\mathrm{S}}=3479.1 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_5=500^{\circ} \mathrm{C}\right\} s_5=7.7642 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
&
\end{aligned}
$$ $$
\left.\begin{array}{l}
P_6=10 \mathrm{kPa} \\
s_6=s_5
\end{array}\right\} \begin{aligned}
& x_6=\frac{s_6-s_f}{s_{f g}}=\frac{7.7642-0.6492}{7.4996}=\mathbf{0 . 9 4 8 7} \text { (at turbine exit) } \\
& h_6=h_f+x_6 h_{f g}=191.81+(0.9487)(2392.1)=2461.2 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ (b) $$
\begin{aligned}
w_{\mathrm{T}, \text { out }} & =\left(h_3-h_4\right)+\left(h_{\mathrm{5}}-h_6\right)=3375.1-2783.7+3479.1-2461.2=1609.3 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {in }} & =\left(h_3-h_2\right)+\left(h_5-h_4\right)=3375.1-201.90+3479.1-2783.7=3868.5 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {net }} & =w_{\mathrm{T}, \text { out }}-w_{p, \text { in }}=1609.4-10.09=1599.3 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Thus the thermal efficiency is $$
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{1599.3 \mathrm{~kJ} / \mathrm{kg}}{3868.5 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{4 1 . 3} \%
$$ (c) The mass flow rate of the steam is $$
\dot{m}=\frac{\dot{W}_{\text {net }}}{w_{net}}=\frac{80,000 \mathrm{~kJ} / \mathrm{s}}{15993 \mathrm{~kJ} / \mathrm{kg}}=50.0 \mathrm{~kg} / \mathbf{s}
$$
