Answer
$η_{th}=81.67\%$
$\dot{Q}_{\text {regen }} =42,875\text{ Btu/s} $
$\dot{W}_{\text {net}, \text { out }}=35,384\text{ Btu/s}$
Work Step by Step
(a) The thermal efficiency of this totally reversible cycle is $$
\eta_{\mathrm{th}}=1-\frac{T_L}{T_H}=1-\frac{550 \mathrm{R}}{3000 \mathrm{R}}=\mathbf{8 1 . 6 7 \%}
$$ (b) The amount of heat transferred in the regenerator is $$
\begin{aligned}
\dot{Q}_{\text {regen }} & =\dot{Q}_{41 \text { in }}=\dot{m}\left(h_1-h_4\right)=\dot{m} c_p\left(T_1-T_4\right) \\
& =(14 \mathrm{lbm} / \mathrm{s})(1.25 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(3000-550) \mathrm{R} \\
& =\mathbf{4 2 , 8 7 5 ~B t u} / \mathrm{s}
\end{aligned}
$$ (c) The net power output is determined from $$
\begin{aligned}
& s_2-s_1=c_p \ln \frac{T_2}{T_1} \quad-R \ln \frac{P_2}{P_1}=-(0.4961 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}) \ln \frac{25 \mathrm{psia}}{200 \mathrm{psia}}=1.0316\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \\
& \dot{Q}_{\text {in }}=\dot{m} T_H\left(s_2-s_1\right)=(14 \mathrm{lbm} / \mathrm{s})(3000 \mathrm{R})(1.0316 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})=43,328\ \mathrm{Btu} / \mathrm{s} \\
& \dot{W}_{\text {net }, \text { out }}=\eta_{\text {th }} \dot{Q}_{\text {in }}=(0.8167)(43,328)=\mathbf{3 5}, 384\ \mathrm{Btu} / \mathbf{s}
\end{aligned}
$$
