Answer
$n_{T}=0.788$
$\dot{W}_{\text {net }}=20,165\text{ kW}$
$\eta_{\text {th }}=12.4\%$
Work Step by Step
(a) We need properties of isobutane, which are not available in the book. However, we can obtain the properties from EES.
Turbine:
$$
\begin{aligned}
& P_3=3250\ \mathrm{kPa} \mid h_3=761.54 \mathrm{~kJ} / \mathrm{kg} \\
& \left.T_3=147^{\circ} \mathrm{C} \quad\right\} s_3=2.5457 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_4=410\ \mathrm{kPa} \\
s_4=s_3
\end{array}\right\} h_{4 s}=670.40 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_4=410 \mathrm{kPa} \\
T_4=179.5^{\circ} \mathrm{C}
\end{array}\right\} h_4=689.74 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_T=\frac{h_3-h_4}{h_3-h_{4 s}}=\frac{761.54-689.74}{761.54-670.40}=\mathbf{0 . 7 8 8}
\end{aligned}
$$ (b) Pump: $$
\begin{aligned}
& h_1=h_{f$410 \mathrm{kPa}}=273.01 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f@410 \mathrm{kPa}}=0.001842 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{p, \text { in }}=v_1\left(P_2-P_1\right) / \eta_P \\
& =\left(0.001842 \mathrm{~m}^3 / \mathrm{kg}\right)(3250-410) \mathrm{kPa}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) / 0.90 \\
& =5.81 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{\rho \text { in }}=273.01+5.81=278.82 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ $$
\begin{aligned}
& \dot{W}_{\mathrm{T}, \text { out }}=\dot{m}\left(h_3-h_4\right)=(305.6 \mathrm{~kJ} / \mathrm{kg})(761.54-689.74) \mathrm{kJ} / \mathrm{kg}=21,941 \mathrm{~kW} \\
& \dot{W}_{\mathrm{P}, \mathrm{in}}=\dot{m}\left(h_2-h_1\right)=\dot{m} w_{\mathrm{p} \text { in }}=(305.6 \mathrm{~kJ} / \mathrm{kg})(5.81 \mathrm{~kJ} / \mathrm{kg})=1777 \mathrm{~kW} \\
& \dot{W}_{\text {net }}=\dot{W}_{\mathrm{T}, \text { out }}-\dot{W}_{\mathrm{P}, \text { in }}=21,941-1777=\mathbf{2 0 , 1 6 5} \mathbf{k W}
\end{aligned}
$$ Heat Exchanger: $$
\dot{Q}_{\text {in }}=\dot{m}_{\text {geo }} c_{\text {geo }}\left(T_{\text {in }}-T_{\text {eut }}\right)=(555.9 \mathrm{~kJ} / \mathrm{kg})\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(160-90)^{\circ} \mathrm{C}=162,656 \mathrm{~kW}
$$ (c) $$
\eta_{\text {th }}=\frac{\dot{W}_{\text {net }}}{\dot{Q}_{\text {in }}}=\frac{20,165}{162,656}=\mathbf{0 . 1 2 4}=12.4 \%
$$
