Answer
$q_{out}=114.2\text{ Btu/lbm}$
$\eta_{\text {th }}=64.7\%$
Work Step by Step
(a) Process 1-2: isentropic compression.
$$ T_2=T_1\left(\frac{\boldsymbol{v}_1}{\boldsymbol{v}_2}\right)^{k-1}=(580 \mathrm{R})(18.2)^{0.4}=1851\ \mathrm{R}
$$ Process 2-3: $P=$ constant heat addition.$$
\frac{P_3 \boldsymbol{v}_3}{T_3}=\frac{P_2 \boldsymbol{v}_2}{T_2} \longrightarrow \frac{\boldsymbol{v}_3}{\boldsymbol{v}_2}=\frac{T_3}{T_2}=\frac{3200 \mathrm{R}}{1851 \mathrm{R}}=\mathbf{1 . 7 2 9}
$$ (b) $q_{\text {in }}=h_3-h_2=c_p\left(T_3-T_2\right)=(0.240 \mathrm{Btu} / \mathrm{lbm} . \mathrm{R})(3200-1851) \mathrm{R}=323.7\text{ Btu/lbm}$.
Process 3-4: isentropic expansion.$$
T_4=T_3\left(\frac{\boldsymbol{v}_3}{\boldsymbol{v}_4}\right)^{k-1}=T_3\left(\frac{1.729 \boldsymbol{v}_2}{\boldsymbol{v}_4}\right)^{k-1}=(3200 \mathrm{R})\left(\frac{1.729}{18.2}\right)^{0.4}=1248\ \mathrm{R}
$$ Process 4-1: $v=$ constant heat rejection.$$
\begin{aligned}
& q_{\text {out }}=u_4-u_1=c_v\left(T_4-T_1\right) \\
& =(0.171 \mathrm{Btu} / \mathrm{lbm} . \mathrm{R})(1248-580) \mathrm{R} \\
& =114.2\ \mathrm{Btu} / \mathrm{lbm} \\
&
\end{aligned}
$$ (c) $$
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{114.2\ \mathrm{Btu} / \mathrm{lbm}}{323.7\ \mathrm{Btu} / \mathrm{lbm}}=0.6472=64.7 \%
$$
