Answer
$\eta_{\text {th }}=48.6\%$
$\dot{W}_{\mathrm{net}}=19,428\text{ kJ/s}$
$\dot{m}_{\text {cool }} =194.6\text{ kg/s}$
Work Step by Step
(a) From the steam tables (Tables A-4, A-5, and A-6),
$$
\begin{aligned}
& h_1=h_{f @ 7.5 \mathrm{kPa}}=168.75 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 7.5 \mathrm{kPa}}=0.001008 \mathrm{~m}^3 / \mathrm{kg} \\
& T_1=T_{\text {sat } @ 7.5 \mathrm{kPa}}=40.29^{\circ} \mathrm{C} \\
& w_{\mathrm{p}, \text { in }}=v_1\left(P_2-P_1\right) \\
& =\left(0.001008 \mathrm{~m}^3 / \mathrm{kg}\right)(6000-7.5 \mathrm{kPa})\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
& =6.04 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{\mathrm{p}, \mathrm{in}}=168.75+6.04=174.79 \mathrm{~kJ} / \mathrm{kg} \\
& h_4=h_{g @ 7.5 \mathrm{kPa}}=2574.0 \mathrm{~kJ} / \mathrm{kg} \\
& s_4=s_{g @ 7.5 \mathrm{kPa}}=8.2501 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_3=6 \mathrm{MPa} \\
s_3=s_4
\end{array}\right\} \begin{array}{l}
h_3=4852.2 \mathrm{~kJ} / \mathrm{kg} \\
T_3=1089.2^{\circ} \mathrm{C}
\end{array} \\
&
\end{aligned}
$$ (b) $$
\begin{aligned}
q_{\text {in }} & =h_3-h_2=4852.2-174.79=4677.4 \mathrm{~kJ} / \mathrm{kg} \\
q_{\text {out }} & =h_4-h_1=2574.0-168.75=2405.3 \mathrm{~kJ} / \mathrm{kg} \\
w_{\text {net }} & =q_{\text {in }}-q_{\text {out }}=4677.4-2405.3=2272.1 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ and $$
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{2272.1 \mathrm{~kJ} / \mathrm{kg}}{4677.4 \mathrm{~kJ} / \mathrm{kg}}=48.6 \%
$$ Thus, $$
\dot{W}_{\mathrm{net}}=\eta_{\mathrm{th}} \dot{Q}_{\text {in }}=(0.4857)(40,000 \mathrm{~kJ} / \mathrm{s})=19,428 \mathrm{~kJ} / \mathrm{s}
$$ (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 $40.29^{\circ} \mathrm{C}$, $$
\begin{aligned}
\dot{Q}_{\text {out }} & =\dot{Q}_{\text {in }}-\dot{W}_{\text {net }}=40,000-19,428=20,572 \mathrm{~kJ} / \mathrm{s} \\
\dot{m}_{\text {cool }} & =\frac{\dot{Q}_{\text {out }}}{c \Delta T}=\frac{20,572 \mathrm{~kJ} / \mathrm{s}}{\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(40.29-15^{\circ} \mathrm{C}\right)}=194.6 \mathrm{~kg} / \mathrm{s}
\end{aligned}
$$
