Answer
$\eta_{\mathrm{th}}=29.3\%$
$\eta_{\mathrm{th}}=34.0\%$
Work Step by Step
(a) Rankine cycle analysis: From the steam tables (Tables A-4, A-5, and A-6), $$
\begin{aligned}
& h_1=h_{f@ 50 \mathrm{~kPa}}=340.54 \mathrm{~kJ} / \mathrm{kg} \\
& v_{\mathrm{1}}=v_{f@ 50 \mathrm{~kPa}}=0.001030 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{\text {p,in }}=v_1\left(P_2-P_1\right) \\
& -\left(0.001030 \mathrm{~m}^3 / \mathrm{kg}\right)(5000-50) \mathrm{kPa}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
& -5.10 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{p \mathrm{~in}}=340.54+5.10=345.64 \mathrm{~kJ} / \mathrm{kg} \\
& P_3=5 \mathrm{MPa} \mid h_3=2794.2 \mathrm{~kJ} / \mathrm{kg} \\
& \left.x_3=1 \quad\right\} x_3=5.9737 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_4=50 \mathrm{kPa} \\
s_4=s_3
\end{array}\right\} x_4=\frac{s_4-s_f}{s_s}=\frac{5.9737-1.09120}{6.5019}=0.7509 \\
& h_4=h_f+x_4 h_8=340.54+(0.7509)(2304.7) \\
& -2071.2 \mathrm{~kJ} / \mathrm{kg} \\
& q_{\text {in }}=h_3-h_2=2794.2-345.64=2448.6 \mathrm{~kJ} / \mathrm{kg} \\
& q_{\text {out }}=h_4-h_1=2071.2-340.54=1730.7 \mathrm{~kJ} / \mathrm{kg} \\
& w_{\text {net }}=q_{\text {ia }}-q_{\text {oat }}-2448.6-1730.7=717.9 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_{\mathrm{a}}=1-\frac{q_{\mathrm{es}}}{q_{\text {is }}}-1-\frac{1730.7}{2448.6}-0.2932-29.3 \% \\
&
\end{aligned}
$$ (b) Canot Cycle analysis: $$
\begin{aligned}
& \left.\begin{array}{l}
T_2=T_3=263.9^{\circ} \mathrm{C} \\
x_2=0
\end{array}\right\}\left\{\begin{array}{l}
h_2=1154.5 \mathrm{~kJ} / \mathrm{kg} \\
s_2=2.9207 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array}\right. \\
& \left.\begin{array}{l}
P_1=50 \mathrm{kPa} \\
s_1=s_2
\end{array}\right\} \begin{array}{l}
x_1=\frac{s_1-s_f}{s_{fg}}=\frac{2.9207-1.0912}{6.5019}=0.2814 \\
h_1=h_f+x_1 h_{fg}
\end{array} \\
& -340.54+(0.2814)(2304.7)=989.05 \mathrm{~kJ} / \mathrm{kg} \\
& q_{\text {ia }}=h_3-h_2=2794.2-1154.5-1639.7 \mathrm{~kJ} / \mathrm{kg} \\
& q_{\text {out }}=h_4-h_1-2071.2-340.54=1082.2 \mathrm{~kJ} / \mathrm{kg} \\
& w_{\text {net }}=q_{\text {in }}-q_{\text {out }}-1639.7-1082.2=557.5 \mathrm{kl} / \mathrm{kg} \\
& \eta_{\mathrm{b}}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{1082.2}{1639.7}=0.3400=34.0 \% \\
&
\end{aligned}
$$
