Answer
$\dot{W}_{\text {net }}=3422\text{ kW}$
$\dot{Q}_{\text {process }} =32,365\text{ Btu/s}$
Work Step by Step
(a) From the steam tables (Tables A-4E, A-5E, and A-6E),
$$
\begin{aligned}
& h_1 \equiv h_{f@ 240^\circ \mathrm{F}}=208.49\ \mathrm{Btu} / \mathrm{lbm} \\
& h_2 \equiv h_1 \\
& \left.P_3=600\ \mathrm{psia}\right\} h_3=1321.3\ \mathrm{Btu} / \mathrm{lbm} \\
& T_3=650^{\circ} \mathrm{F} \quad s_3=s_5=s_7=1.5614\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
& h_3=h_4=h_5=h_6 \\
& \left.\begin{array}{l}
P_7=120 \text { psia } \\
s_7=s_3
\end{array}\right\} h_7=1169.2 \mathrm{Btu} / \mathrm{lbm} \\
& \dot{W}_{\text {net }}=\dot{m}_5\left(h_5-h_7\right) \\
& =(2 / 3 \times 32 \mathrm{lbm} / \mathrm{s})(1321.3-1169.2) \mathrm{Btu} / \mathrm{lbm} \\
& =3244\ \mathrm{Btu} / \mathrm{s}=3422 \mathrm{~kW}
\end{aligned}
$$ (b) $$
\begin{aligned}
\dot{Q}_{\text {process }} & =\sum \dot{m}_i h_i-\sum \dot{m}_e h_e \\
& =\dot{m}_6 h_6+\dot{m}_7 h_7-\dot{m}_1 h_1 \\
& =(1 / 3 \times 32)(1321.3)+(2 / 3 \times 32)(1169.2)-(32)(208.49) \\
& =32,365\ \mathbf{B t u} / \mathbf{s}
\end{aligned}
$$ (c) $\quad \varepsilon_u=1$ since all the energy is utilized.
