Chapter 10 - Vapor and Combined Power Cycles - Problems - Page 593: 10-34

$\eta_{\text {th }}=35.0\%$

From the steam tables (Tables A-4, A-5, and A-6), $$ \begin{aligned} h_1 & =h_{f @ 20 \mathrm{kPa}}=251.42 \mathrm{~kJ} / \mathrm{kg} \\ v_1 & =v_{f @ 20 \mathrm{kPa}}=0.0010172 \mathrm{~m}^3 / \mathrm{kg} \\ w_{\text {p,in }} & =v_1\left(P_2-P_1\right) \\ & =\left(0.0010172 \mathrm{~m}^3 / \mathrm{kg}\right)(5000-20) \mathrm{kPa}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\ & =5.065 \mathrm{~kJ} / \mathrm{kg} \\ h_2 & =h_1+w_{\mathrm{p}, \text { in }}=251.42+5.065=256.49 \mathrm{~kJ} / \mathrm{kg} \end{aligned} $$ $$ \begin{aligned} & \left.P_4=1200 \mathrm{kPa}\right\} \quad h_4=h_f+x_4 h_{f g}=798.33+(0.96)(1985.4)=2704.3 \mathrm{~kJ} / \mathrm{kg} \\ & \left.x_4=0.96\right\} s_4=s_f+x_4 s_{f g}=2.2159+(0.96)(4.3058)=6.3495 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\ & \left.\begin{array}{l} P_3=5000 \mathrm{kPa} \\ s_3=s_4 \end{array}\right\} \begin{array}{l} h_3=3006.9 \mathrm{~kJ} / \mathrm{kg} \\ T_3=327.2^{\circ} \mathrm{C} \end{array} \\ & \left.P_6=20 \mathrm{kPa}\right\} \quad h_6=h_f+x_6 h_{f g}=251.42+(0.96)(2357.5)=2514.6 \mathrm{~kJ} / \mathrm{kg} \\ & \left.x_6=0.96\right\} s_6=s_f+x_6 s_{f g}=0.8320+(0.96)(7.0752)=7.6242 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\ & \left.\begin{array}{l} P_5=1200 \mathrm{kPa} \\ s_5=s_6 \end{array}\right\} \begin{array}{l} h_5=3436.0 \mathrm{~kJ} / \mathrm{kg} \\ T_5=\mathbf{4 8 1 . 1}{ }^{\circ} \mathrm{C} \end{array} \\ & \end{aligned} $$ Thus, $$ \begin{aligned} q_{\text {in }} & =\left(h_3-h_2\right)+\left(h_5-h_4\right)=3006.9-256.49+3436.0-2704.3=3482.0 \mathrm{~kJ} / \mathrm{kg} \\ q_{\text {out }} & =h_6-h_1=2514.6-151.42=2263.2 \mathrm{~kJ} / \mathrm{kg} \end{aligned} $$ and $$ \eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{2263.2}{3482.0}=0.3500=\mathbf{3 5 . 0} \% $$