Answer
$\eta_{\text {th }}=58.6\%$
Work Step by Step
(a) Process 1-2: isentropic compression.$$
\begin{aligned}
& T_1=580] \mathrm{R} \longrightarrow \begin{array}{l}
u_1=98.90\ \mathrm{Btu} / \mathrm{lbm} \\
v_{r_1}=120.70
\end{array} \\
& \boldsymbol{v}_{r_2}=\frac{\boldsymbol{v}_2}{v_1} v_{r_1}=\frac{1}{r} v_{r_1}=\frac{1}{18.2}(120.70)=6.632 \longrightarrow \begin{array}{l}
T_2=1725.8\ \mathrm{R} \\
h_2=429.56\ \mathrm{Btu} / \mathrm{lbm}
\end{array}
\end{aligned}
$$ Process 2-3: $P=$ constant heat addition. $$
\begin{aligned}
& \frac{P_3 \boldsymbol{v}_3}{T_3}=\frac{P_2 v_2}{T_2} \longrightarrow \frac{v_3}{v_2}=\frac{T_3}{T_2}=\frac{3200\ \mathrm{R}}{1725.8 \mathrm{R}}=1.854 \\
& T_3=3200\ \mathrm{R} \longrightarrow \begin{array}{l}
h_3=849.48\ \mathrm{Btu} / \mathrm{lbm} \\
v_{r_3}=0.955
\end{array} \\
& q_{\text {in }}=h_3-h_2=849.48-429.56=419.9\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ (b) Process 3-4: isentropic expansion. $$
\boldsymbol{v}_{r_4}=\frac{\boldsymbol{v}_4}{\boldsymbol{v}_3} \boldsymbol{v}_{r_5}=\frac{\boldsymbol{v}_4}{1.854 v_2} \boldsymbol{v}_{r_3}=\frac{r}{1.854} \boldsymbol{v}_{r_3}=\frac{18.2}{1.854}(0.955)=9.375\ \longrightarrow u_4=
$$ Process 4-1: $v=$ constant heat rejection: $$
q_{\text {out }}=u_4-u_1=272.58-98.90=173.7\ \mathbf{B t u} / \mathbf{l b m}
$$ (c) $$
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{173.7\
\mathrm{Btu} / \mathrm{lbm}}{419.9\ \mathrm{Btu} / \mathrm{lbm}}=0.586=\mathbf{5 8 . 6 \%}
$$
