Answer
c) $24.2\%$
Work Step by Step
b) The properties of air at various states are $$
\begin{gathered}
T_1=540 \mathrm{R} \longrightarrow u_1=92.04 \mathrm{Btu} / \mathrm{lbm}, \quad h_1=129.06 \mathrm{Btu} / \mathrm{lbm} \\
q_{\mathrm{in}, 12}=u_2-u_1 \longrightarrow \begin{array}{l}
u_2=u_1+q_{\mathrm{in}, 12}=92.04+300=392.04 \mathrm{Btu} / \mathrm{lbm} \\
T_2=2116 \mathrm{R}, h_2=537.1\ \mathrm{Btu} / \mathrm{lbm}
\end{array} \\
\frac{P_2 v_2}{T_2}=\frac{P_1 v_1}{T_1} \longrightarrow P_2=\frac{T_2}{T_1} P_1=\frac{2116 \mathrm{R}}{540 \mathrm{R}}(14.7 \mathrm{psia})=57.6 \mathrm{psia} \\
T_3=3200 \mathrm{R} \longrightarrow \begin{array}{l}
h_3=849.48\ \mathrm{Btu} / \mathrm{lbm} \\
P_{r_3}=1242
\end{array} \\
P_{r_4}=\frac{P_4}{P_3} P_{r_3}=\frac{14.7 \mathrm{psia}}{57.6 \mathrm{psia}}(1242)=317.0 \longrightarrow h_4=593.22\ \mathrm{Btu} / \mathrm{lbm}
\end{gathered}
$$ From energy balance, $$
\begin{aligned}
q_{23 \text { in }} & =h_3-h_2=849.48-537.1=312.38\ \mathrm{Btu} / \mathrm{lbm} \\
q_{\text {in }} & =q_{12 \text { in }}+q_{23 \text { in }}=300+312.38=612.38\ \mathrm{Btu} / \mathrm{lbm} \\
q_{\text {out }} & =h_4-h_1=593.22-129.06=464.16\ \mathrm{Btu} / 1 \mathrm{bm}
\end{aligned}
$$ c) Now the thermal efficiency becomes $$
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{464.16 \mathrm{Btu} / \mathrm{lbm}}{612.38 \mathrm{Btu} / \mathrm{lbm}}=\mathbf{2 4 . 2} \%
$$
