Answer
$\eta_{\mathrm{th}}=39.4\%$
$\dot{m}_{\text {oool }}=641.0\text{ lbm/s}$
Work Step by Step
(a) From the steam tables (Tables A-4E, A-5E, and A-6E),
$$
\begin{aligned}
& h_1=h_{\text {sat }@ 1 \text { psia }}=69.72\ \mathrm{Btu} / \mathrm{lbm} \\
& v_1=v_{\text {sat } @1 \text { psia }}=0.01614\ \mathrm{ft}^3 / \mathrm{lbm} \\
& T_1=T_{\text {sin } 1 \text { psia }}=101.69^{\circ} \mathrm{F} \\
& w_{p, \text { in }}=v_1\left(P_2-P_1\right) \\
& \begin{array}{l}
=\left(0.01614 \mathrm{ft}^3 / 1 \mathrm{bm}\right)(800-1 \mathrm{psia})\left(\frac{1 \mathrm{Btu}}{5.4039 \mathrm{psia} \cdot \mathbf{f}^3}\right) \\
=2.39\ \mathrm{Btu} / \mathrm{bm}
\end{array} \\
& h_2=h_1+w_{\rho \text { in }}=69.72+2.39=72.11\ \mathrm{Btu} / \mathrm{lbm} \\
& P_3=800\ \mathrm{psia} \quad h_3=14560\ \mathrm{Btu} / \mathrm{lbm} \\
& \left.T_3=900^{\circ} \mathrm{F} \quad\right\} s_3=1.6413\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
& \left.\begin{array}{l}
s_4=s_3 \\
\text { (sat.vapor) }
\end{array}\right\} \begin{array}{l}
h_4=h_{g @ s_g=s_4}=11785\ \mathrm{Btu} / \mathrm{lbm} \\
P_4=P_{\text {sat}@ s_g=s_4}=62.23\ \mathrm{psia} \text { (thereheatpressure) }
\end{array} \\
& P_5=62.23\ \mathrm{psia}, h_5=1431.4\ \mathrm{Btu} / \mathrm{lbm} \\
& T_5=800^{\circ} \mathrm{F} \quad s_s=1.8985\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
&
\end{aligned}
$$ $$
\left.\begin{array}{l}
P_6=1 \text { psia } \\
s_6=s_5
\end{array}\right\} \begin{aligned}
& x_6=\frac{s_6-s_f}{s_{f g}}=\frac{1.8985-0.13262}{1.84495}=0.9572 \\
& h_6=h_f+x_6 h_{f g}=69.72+(0.9572)(10357)=10610\ \mathrm{Bt} \mathrm{w} / \mathrm{lbm}
\end{aligned}
$$ (b) $$
\begin{aligned}
& q_{\text {in }}=\left(h_3-h_2\right)+\left(h_5-h_4\right)=1456.0-72.11+1431.4-1178.5=1636.8\ \mathrm{Btu} / \mathrm{lbm} \\
& q_{\text {out }}=h_6-h_1=1061.0-69.72=991.3\ \mathrm{Btu} / \mathrm{bm}
\end{aligned}
$$ Thus, $$
\eta_{\mathrm{th}}=1-\frac{q_{\text {out }}}{q_{\mathrm{in}}}=1-\frac{991.3 \mathrm{Btw} / \mathrm{lbm}}{1636.8 \mathrm{Btu} / \mathrm{lbm}}=39.4 \%
$$ (c) The mass flow rate of the cooling water will be minimum when it is heated to the temperature of the steam in the condenser, which is $101.7^{\circ} \mathrm{F}$, $$
\begin{aligned}
& \dot{Q}_{\text {out }}=\dot{Q}_{\text {in }}-\dot{W}_{\text {net }}=\left(1-\eta_{\text {th }}\right) \dot{Q}_{\text {in }}=(1-0.3943)\left(6 \times 10^4 \mathrm{Btu} / \mathrm{s}\right)=3.634 \times 10^4 \mathrm{Btu} / \mathrm{s} \\
& \dot{m}_{\text {oool }}=\frac{\dot{Q}_{\text {out }}}{c \Delta T}=\frac{3.634 \times 10^4 \mathrm{Btu} / \mathrm{s}}{\left(1.0 \mathrm{Btu} / \mathrm{lbm}\cdot ^\circ \mathrm{F}\right)(101.69-45)^\circ \mathrm{F}}=641.0\ \mathrm{lbm} / \mathbf{s}
\end{aligned}
$$
