Answer
$r_{\mathrm{bw}}=48.5\%$
$ \eta_{\text {th }}=46.5\%$
Work Step by Step
(a) Noting that process 1-2 is isentropic, $$
\begin{aligned}
& T_1=520\ \mathrm{R} \longrightarrow \begin{array}{l}
h_1=124.27\ \mathrm{Btu} / \mathrm{lbm} \\
P_{r_1}=1.2147
\end{array} \\
& P_{r_2}=\frac{P_2}{P_1} P_{r_1}=(10)(1.2147)=12.147 \longrightarrow \begin{array}{l}
T_2=996.5\ \mathbf{R} \\
h_2=240.11\ \mathrm{Btu} / \mathrm{lbm}
\end{array}
\end{aligned}
$$ (b) Process 3-4 is isentropic, and thus $$
\begin{aligned}
T_3 & =2000\ \mathrm{R} \longrightarrow \begin{array}{l}
h_3=504.71\ \mathrm{Btu} / \mathrm{lbm} \\
P_{r_3}=174.0
\end{array} \\
P_{r_4} & =\frac{P_4}{P_3} P_{r_3}=\left(\frac{1}{10}\right)(174.0)=17.4 \longrightarrow h_4=265.83\ \mathrm{Btu} / \mathrm{lbm} \\
w_{\mathrm{C}, \text { in }} & =h_2-h_1=240.11-124.27=115.84\ \mathrm{Btu} / \mathrm{lbm} \\
w_{\mathrm{T}, \text { out }} & =h_3-h_4=504.71-265.83=238.88\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ Then the back-work ratio becomes $$
r_{\mathrm{bw}}=\frac{w_{\mathrm{C}, \text { in }}}{w_{\mathrm{T}, \text { out }}}=\frac{115.84 \mathrm{Btu} / \mathrm{lbm}}{238.88 \mathrm{Btu} / \mathrm{lbm}}=\mathbf{4 8 . 5} \%
$$ (c) $$
\begin{aligned}
& q_{\text {in }}=h_3-h_2=504.71-240.11=264.60\ \mathrm{Btu} / \mathrm{lbm} \\
& w_{\text {net,out }}=w_{\mathrm{T}, \text { out }}-w_{\mathrm{C}, \text { in }}=238.88-115.84=123.04\ \mathrm{Btu} / \mathrm{lbm} \\
& \eta_{\text {th }}=\frac{w_{\text {net,out }}}{q_{\text {in }}}=\frac{123.04 \mathrm{Btu} / \mathrm{lbm}}{264.60 \mathrm{Btu} / \mathrm{lbm}}=46.5 \%
\end{aligned}
$$
