Answer
$w_{\mathrm{T}, \text { out }}=1176 \text{ kJ/kg}$
$\eta_{\text {th }}=35.8\%$
Work Step by Step
From the steam tables (Tables A-4, A-5, and A-6),
$$
\begin{aligned}
& h_1=h_{f @20 \mathrm{kPa}}=251.42 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 20 \mathrm{kPa}}=0.001017 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{p, \text { in }}=v_1\left(P_2-P_1\right) \\
& =\left(0.001017 \mathrm{~m}^3 / \mathrm{kg}\right)(6000-20 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
& =6.08 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{p, \text { in }}=251.42+6.08=257.50 \mathrm{~kJ} / \mathrm{kg} \\
& P_3=6 \mathrm{MPa} \mid h_3=3178.3 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_3=400^{\circ} \mathrm{C}\right\} s_3=6.5432 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_4=2 \mathrm{MPa} \\
s_4=s_3
\end{array}\right\} h_4=2901.0 \mathrm{~kJ} / \mathrm{kg} \\
& \left.P_5=2 \mathrm{MPa}\right\} h_5=3248.4 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_5=400^{\circ} \mathrm{C}\right\} s_5=7.1292 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
&
\end{aligned}
$$ $$
\left.\begin{array}{l}
P_6=20 \mathrm{kPa} \\
s_6=s_5
\end{array}\right\} \begin{aligned}
& x_6=\frac{s_6-s_f}{s_{f 8}}=\frac{7.1292-0.8320}{7.0752}=0.8900 \\
& h_6=h_f+x_6 h_{f 8}=251.42+(0.8900)(2357.5)=2349.7 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The turbine work output and the thermal efficiency are determined from $$
w_{\mathrm{T}, \text { out }}=\left(h_3-h_4\right)+\left(h_5-h_6\right)=3178.3-2901.0+3248.4-2349.7=1176 \mathrm{~kJ} / \mathrm{kg}
$$ and $$
q_{\text {in }}=\left(h_3-h_2\right)+\left(h_5-h_4\right)=3178.3-257.50+3248.4-2901.0=3268 \mathrm{~kJ} / \mathrm{kg}
$$ $$
w_{\text {net }}=w_{\text {T, out }}-w_{\text {p,in }}=1176-6.08=1170 \mathrm{~kJ} / \mathrm{kg}
$$ Thus, $$
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{1170 \mathrm{~kJ} / \mathrm{kg}}{3268 \mathrm{~kJ} / \mathrm{kg}}=0.358=35.8 \%
$$
