Answer
$q_{\text {in }} =18.07\text{ Btu/lbm}$
$η_{th}=74.9\%$
$\eta_{\mathrm{th}, \mathrm{C}}=77.5\%$
Work Step by Step
(a) Process 1-2: isentropic compression. $$
T_2=T_1\left(\frac{v_1}{v_2}\right)^{k-1}=(540 \mathrm{R})(8)^{0.667}=2161\ \mathrm{R}
$$ Process 2-3: $v=$ constant heat addition. $$
\begin{aligned}
q_{\text {in }} & =u_3-u_2=c_v\left(T_3-T_2\right) \\
& =(0.0756 \mathrm{Btu} / \mathrm{lbm} . \mathrm{R})(2400-2161) \mathrm{R} \\
& =18.07\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ (b) Process 3-4: isentropic expansion. $$
T_4=T_3\left(\frac{\boldsymbol{v}_3}{\boldsymbol{v}_4}\right)^{\mathrm{k}-1}=(2400 \mathrm{R})\left(\frac{1}{8}\right)^{0.667}=600\ \mathrm{R}
$$ Process 4-1: $\boldsymbol{v}=$ constant heat rejection. $$
\begin{aligned}
& q_{\text {out }}=u_4-u_1=c_v\left(T_4-T_1\right)=(0.0756 \mathrm{Btu} / \mathrm{bm} . \mathrm{R})(600-540) \mathrm{R}=4.536\ \mathrm{Btu} / \mathrm{lbm} \\
& \eta_{\mathrm{th}}=1-\frac{q_{\text {out }}}{q_{\mathrm{is}}}=1-\frac{4.536 \mathrm{Btu} / \mathrm{bm}}{18.07 \mathrm{Btu} / \mathrm{lbm}}=74.9 \%
\end{aligned}
$$ (c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is $$
\eta_{\mathrm{th}, \mathrm{C}}=1-\frac{T_L}{T_H}=1-\frac{540 \mathrm{R}}{2400 \mathrm{R}}=\mathbf{7 7 . 5} \%
$$
