Answer
$MEP_{}=49.0\text{ psia}$
Work Step by Step
We begin by determining the temperatures of the cycle states using the process equations and component efficiencies. The ideal temperature at the end of the compression is then $$
T_{2 \mathrm{~s}}=T_1\left(\frac{v_1}{v_2}\right)^{k-1}=T_1 r^{k-1}=(520 \mathrm{R})(8)^{1.4-1}=1195\ \mathrm{R}
$$ With the isentropic compression efficiency, the actual temperature at the end of the compression is $$
\eta=\frac{T_{2 \mathrm{~s}}-T_1}{T_2-T_1} \longrightarrow T_2=T_1+\frac{T_{2 s}-T_1}{\eta}=(520 \mathrm{R})+\frac{(1195-520) \mathrm{R}}{0.85}=1314 \mathrm{R}
$$ Similarly for the expansion, $$
\begin{aligned}
& T_{4 s}=T_3\left(\frac{v_3}{v_4}\right)^{k-1}=T_3\left(\frac{1}{r}\right)^{k-1}=(2300+460 \mathrm{R})\left(\frac{1}{8}\right)^{1.4-1}=1201\ \mathrm{R} \\
& \eta=\frac{T_3-T_4}{T_3-T_{4 s}} \longrightarrow T_4=T_3-\eta\left(T_3-T_{4 s}\right)=(2760 \mathrm{R})-(0.95)(2760-1201) \mathrm{R}=1279\ \mathrm{R}
\end{aligned}
$$ The specific heat addition is that of process 2-3, $$
q_{\text {if }}=c_v\left(T_3-T_2\right)=(0.171 \mathrm{Btw} / \mathrm{bm} \cdot \mathrm{R})(2760-1314) \mathrm{R}=247.3\ \mathrm{Btu} / \mathrm{lbm}
$$ The net work production is the difference between the work produced by the expansion and that used by the compression, $$
\begin{aligned}
w_{\text {net }} & =c_v\left(T_3-T_4\right)-c_v\left(T_2-T_1\right) \\
& =(0.171 \mathrm{Btu} / \mathrm{lbm}-\mathrm{R})(2760-1279) \mathrm{R}-(0.171 \mathrm{Btw} / \mathrm{lbm} \cdot \mathrm{R})(1314-520) \mathrm{R} \\
& =117.5\ \mathrm{Btw} / \mathrm{lbm}
\end{aligned}
$$ The thermal efficiency of this cycle is then $$
\eta_{\text {th }}=\frac{w_{\text {nat }}}{q_{\text {it }}}=\frac{117.5 \mathrm{Btu} / \mathrm{lbm}}{247.3 \mathrm{Btu} / \mathrm{bmm}}=\mathbf{0 . 4 7 5}
$$ At the beginning of compression, the maximum specific volume of this cycle is $$
v_1=\frac{R T_1}{P_1}=\frac{\left(0.3704 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{bm} \cdot \mathrm{R}\right)(520 \mathrm{R})}{13 \mathrm{psia}}=14.82\ \mathrm{ft}^3 / \mathrm{lbm}
$$ while the minimum specific volume of the cycle occurs at the end of the compression $$
v_2=\frac{v_1}{r}=\frac{14.82 \mathrm{f}^3 / 1 \mathrm{bm}}{8}=1.852\ \mathrm{f}^3 / 1 \mathrm{bm}
$$ The engine's mean effective pressure is then $$
\mathrm{MEP}=\frac{w_{\text {et }}}{v_1-v_2}=\frac{117.5 \mathrm{Btu} / \mathrm{bm}}{(14.82-1.852) \mathrm{ft}^3 / \mathrm{bmm}}\left(\frac{5.404 \mathrm{psia} \cdot \mathrm{f}^3}{1 \mathrm{Btu}}\right)=\mathbf{4 9 . 0}\ \mathrm{psia}
$$
