Answer
$q_{\text {in }}=241.42\text{ Btu/lbm}$
$η_{th}=53.0\%$
$\eta_{\mathrm{th}, \mathrm{C}}=77.5\%$
Work Step by Step
(a) Process 1-2: isentropic compression.
$$
\begin{aligned}
& T_1=540 \mathrm{R} \longrightarrow \begin{array}{l}
u_1=92.04\ \mathrm{Btu} / \mathrm{lbm} \\
v_{r_1}=144.32
\end{array} \\
& v_{r_2}=\frac{v_2}{v_1} v_{r_2}=\frac{1}{r} v_{r_2}=\frac{1}{8}(144.32)=18.04 \longrightarrow u_2=211.28\ \mathrm{Btw} / \mathrm{bm}
\end{aligned}
$$ Process 2-3: $\boldsymbol{v}=$ constant heat addition. $$
\begin{aligned}
& T_3=2400\ \mathrm{R} \longrightarrow \begin{array}{l}
u_3=452.70\ \mathrm{Btu} / \mathrm{lbm} \\
v_{r_3}=2.419
\end{array} \\
& q_{\text {in }}=u_3-u_2=452.70-211.28=241.42\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ (b) Process 3-4: isentropic expansion. $$
\boldsymbol{v}_{r_4}=\frac{\boldsymbol{v}_4}{\boldsymbol{v}_3} \boldsymbol{v}_{r_3}=r v_{r_3}=(8)(2.419)=19.35 \longrightarrow u_4=205.54\ \mathrm{Btu} / 1 \mathrm{bm}
$$ Process 4-1: $v=$ constant heat rejection. $$
\begin{aligned}
& q_{\text {out }}=u_4-u_1=205.54-92.04=113.50\ \mathrm{Btu} / \mathrm{lbm} \\
& \eta_{\mathrm{th}}=1-\frac{q_{\text {out }}}{q_{\text {is }}}=1-\frac{113.50 \mathrm{Btu} / \mathrm{lbm}}{241.42 \mathrm{Btw} / \mathrm{lbm}}=53.0 \%
\end{aligned}
$$ (c) The thermal efficiency of a Carnot cycle operating between the same temperature limits is $$
\eta_{\mathrm{th}, \mathrm{C}}=1-\frac{T_L}{T_H}=1-\frac{540 \mathrm{R}}{2400 \mathrm{R}}=\mathbf{7 7 . 5} \%
$$
