Answer
$408.6\text{ kJ/kg}$
$47.9\%$
Work Step by Step
The properties of air at various states are $$
\begin{aligned}
& T_1=295 \mathrm{~K} \longrightarrow \begin{array}{l}
h_1=295.17 \mathrm{~kJ} / \mathrm{kg} \\
P_{r_1}=1.3068
\end{array} \\
& P_{r_2}=\frac{P_2}{P_1} P_{r_1}=\frac{600 \mathrm{kPa}}{100 \mathrm{kPa}}(1.3068)=7.841 \longrightarrow \begin{array}{l}
u_2=352.29 \mathrm{~kJ} / \mathrm{kg} \\
T_2=490.3 \mathrm{~K}
\end{array} \\
& T_3=1500 \mathrm{~K} \longrightarrow \begin{array}{l}
u_3=1205.41 \mathrm{~kJ} / \mathrm{kg} \\
P_{r_3}=601.9
\end{array} \\
& \frac{P_3 v_3}{T_3}=\frac{P_2 v_2}{T_2} \longrightarrow P_3=\frac{T_3}{T_2} P_2=\frac{1500 \mathrm{~K}}{490.3 \mathrm{~K}}(600 \mathrm{kPa})=1835.6 \mathrm{kPa} \\
& P_{r_4}=\frac{P_4}{P_3} P_{r_3}=\frac{100 \mathrm{kPa}}{1835.6 \mathrm{kPa}}(601.9)=32.79 \longrightarrow h_4=739.71 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ From energy balances, $$
\begin{aligned}
& q_{\text {in }}=u_3-u_2=1205.41-352.29=853.1 \mathrm{~kJ} / \mathrm{kg} \\
& q_{\text {out }}=h_4-h_1=739.71-295.17=444.5 \mathrm{~kJ} / \mathrm{kg} \\
& w_{\text {net,out }}=q_{\text {in }}-q_{\text {out }}=853.1-444.5=\mathbf{4 0 8 . 6 k J} / \mathbf{k g}
\end{aligned}
$$ Now the thermal efficiency becomes $$
\eta_{\text {th }}=\frac{w_{\text {net }, \text { out }}}{q_{\text {in }}}=\frac{408.6 \mathrm{~kJ} / \mathrm{kg}}{853.1 \mathrm{~kJ} / \mathrm{kg}}=0.479=47.9 \%
$$
