Answer
$\dot{W}_{\text {net,out }} =87.6\text{ kW}$
Work Step by Step
Process 1-2: isentropic compression.$$
T_2=T_1\left(\frac{\boldsymbol{V}_1}{\boldsymbol{V}_2}\right)^{k-1}=(343 \mathrm{~K})(22)^{0.4}=1181 \mathrm{~K}
$$ Process 2-3:
$P= $constant heat addition.$$
\frac{P_3 \boldsymbol{v}_3}{T_3}=\frac{P_2 \boldsymbol{v}_2}{T_2} \longrightarrow T_3=\frac{\boldsymbol{v}_3}{\boldsymbol{V}_2} T_2=1.8 T_2=(1.8)(1181 \mathrm{~K})=2126 \mathrm{~K}
$$ Process 3-4: isentropic expansion.$$
T_4=T_3\left(\frac{\boldsymbol{V}_3}{\boldsymbol{V}_4}\right)^{k-1}=T_3\left(\frac{2.2 \boldsymbol{V}_2}{\boldsymbol{V}_4}\right)^{k-1}=T_3\left(\frac{2.2}{r}\right)^{k-1}=(2216 \mathrm{~K})\left(\frac{1.8}{22}\right)^{0.4}=781 \mathrm{~K}
$$For the cycle:$$ \begin{aligned}
m & =\frac{P_1 V_1}{R T_1}=\frac{(97 \mathrm{kPa})\left(0.0024 \mathrm{~m}^3\right)}{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(343 \mathrm{~K})}=0.002365 \mathrm{~kg} \\
Q_{\text {in }} & =m\left(h_3-h_2\right)=m c_p\left(T_3-T_2\right) \\
& =(0.002365 \mathrm{~kg})(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(2216-1181) \mathrm{K} \\
& =2.246 \mathrm{~kJ} \\
Q_{\text {out }} & =m\left(u_4-u_1\right)=m c_v\left(T_4-T_1\right) \\
& =(0.002365 \mathrm{~kg})(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(781-343) \mathrm{K} \\
& =0.7438 \mathrm{~kJ} \\
W_{\text {net,out }} & =Q_{\text {in }}-Q_{\text {out }}=2.246-0.7438=1.502 \mathrm{~kJ} / \mathrm{rev} \\
\dot{W}_{\text {net,out }} & =\dot{n} W_{\text {net,out }}=(3500 / 60 \mathrm{rev} / \mathrm{s})(1.502 \mathrm{~kJ} / \mathrm{rev})=87.6 \mathrm{~kW}
\end{aligned}
$$
