Answer
$w_{net}=124.2\text{ Btu/lbm}$
$q_{in}=193.8\text{ Btu/lbm}$
$η_{th}=0.641$
Work Step by Step
Working around the cycle, the germane properties at the various
$$
\begin{aligned}
& T_2=T_1\left(\frac{v_1}{v_2}\right)^{k-1}=T_1 r^{k-1}=(535 \mathrm{R})(15)^{L-4-1}=1580\ \mathrm{R} \\
& P_2=P_1\left(\frac{v_1}{v_2}\right)^k=P_1 r^k=(14.2 \mathrm{psia})(15)^{L-4}=629.2\ \mathrm{psia} \\
& P_x=P_3=r_p P_2=(1.1)(629.2\ \mathrm{psia})=692.1\ \mathrm{psia} \\
& T_x=T_2\left(\frac{P_x}{P_2}\right)=(1580 \mathrm{R})\left(\frac{692.1 \mathrm{psia}}{629.2 \mathrm{psia}}\right)=1738\ \mathrm{R} \\
& T_3=T_x\left(\frac{v_3}{v_x}\right)^{=}=T_x r_c=(1738\ \mathrm{R})(1.4)=2433\ \mathrm{R} \\
& T_4=T_3\left(\frac{v_3}{v_4}\right)^{k-1}=T_3\left(\frac{r_c}{r}\right)^{k-1}=(2433\ \mathrm{R})\left(\frac{1.4}{15}\right)^{1.4-1}\\&=942.2 \mathrm{R}
\end{aligned}
$$ Applying the first law to each of the processes gives $$
\begin{gathered}
w_{1-2}=c_v\left(T_2-T_1\right)=(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(1580-535) \mathrm{R}=178.7\ \mathrm{Btu} / \mathrm{lbm} \\
q_{2-x}=c_v\left(T_x-T_2\right)=(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(1738-1580) \mathrm{R}=27.02\ \mathrm{Btu} / \mathrm{lbm} \\
q_{x-3}=c_p\left(T_3-T_x\right)=(0.240 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(2433-1738) \mathrm{R}=166.8\ \mathrm{Btu} / \mathrm{lbm} \\
w_{x-3}=q_{x-3}-c_v\left(T_3-T_x\right)=166.8 \mathrm{Btu} / \mathrm{lbm}-(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(2433-1738) \mathrm{R}=47.96\ \mathrm{Btu} / \mathrm{lbm} \\
w_{3-4}=c_v\left(T_3-T_4\right)=(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(2433-942.2) \mathrm{R}=254.9\ \mathrm{Btu} / \mathrm{lbm}
\end{gathered}
$$ The net work of the cycle is $$
w_{\text {net }}=w_{3-4}+w_{x-3}-w_{1-2}=254.9+47.96-178.7=124.2\ \mathrm{Btu} / \mathrm{lbm}
$$ and the net heat addition is $$
q_{\text {in }}=q_{2-x}+q_{x-3}=27.02+166.8=193.8\ \mathrm{Btu} / \mathrm{lbm}
$$ Hence, the thermal efficiency is $$
\eta_{\text {th }}=\frac{w_{\text {net }}}{q_{\text {in }}}=\frac{124.2 \mathrm{Btu} / \mathrm{lbm}}{193.8\ \mathrm{Btu} / \mathrm{lbm}}=\mathbf{0 . 6 4 1}
$$
