Answer
$\eta_{\mathrm{th}}=61.4\%$
$MEP_{}=660.4\text{ kPa}$
Work Step by Step
(a) Process 1-2: isentropic compression.$$
T_2=T_1\left(\frac{v_1}{v_2}\right)^{k-1}=(300 \mathrm{~K})(16)^{0.4}=909.4 \mathrm{~K}
$$ Process 2-3: $P=$ constant heat addition. $$
\frac{P_3 v_3}{T_3}=\frac{P_2 v_2}{T_2} \longrightarrow T_3=\frac{v_3}{v_2} T_2=2 T_2=(2)(909.4 \mathrm{~K})=1818.8 \mathrm{~K}
$$ (b) $\quad q_{\text {is }}=h_3-h_2=c_p\left(T_3-T_2\right)=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1818.8-909.4) \mathrm{K}=913.9 \mathrm{~kJ} / \mathrm{kg}$
Process 3-4: isentropic expansion.
$$
T_4=T_3\left(\frac{v_3}{v_4}\right)^{k-1}=T_3\left(\frac{2 v_2}{v_4}\right)^{k-1}=(1818.8 \mathrm{~K})\left(\frac{2}{16}\right)^{0.4}=791.7 \mathrm{~K}
$$ Process 4-1: $\boldsymbol{v}=$ constant heat rejection. $$
\begin{aligned}
& q_{\text {out }}=u_4-u_1=c_v\left(T_4-T_1\right)=(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(791.7-300) \mathrm{K}=353 \mathrm{~kJ} / \mathrm{kg} \\
& \eta_{\mathrm{th}}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{353 \mathrm{~kJ} / \mathrm{kg}}{913.9 \mathrm{~kJ} / \mathrm{kg}}=61.4 \%
\end{aligned}
$$ (c) $$
\begin{aligned}
& w_{\text {net. out }}=q_{\text {in }}-q_{\text {out }}=913.9-353=560.9 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=\frac{R T_1}{P_1}=\frac{\left(0.287 \mathrm{kPa}-\mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(300 \mathrm{~K})}{95 \mathrm{kPa}}=0.906 \mathrm{~m}^3 / \mathrm{kg}=v_{\max } \\
& v_{\text {min }}=v_2=\frac{v_{\max }}{r} \\
& \mathrm{MEP}=\frac{w_{\text {net, out }}}{v_1-v_2}=\frac{w_{\text {net,out }}}{v_1(1-1 / r)}=\frac{560.9 \mathrm{~kJ} / \mathrm{kg}}{\left(0.906 \mathrm{~m}^3 / \mathrm{kg}\right)(1-1 / 16)}\left(\frac{\mathrm{kPa}-\mathrm{m}^3}{\mathrm{~kJ}}\right)=660.4\ \mathbf{k P a}
\end{aligned}
$$
