Answer
$\dot{W}_{\text {net }} =62.1\text{ hp}$
Work Step by Step
The specific volume of the air at the start of the compression is $$
v_1=\frac{R T_1}{P_1}=\frac{\left(0.3704 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{lbm} \cdot \mathrm{R}\right)(510 \mathrm{R})}{14.4 \mathrm{psia}}=13.12\ \mathrm{ft}^3 / \mathrm{lbm}
$$ The total air mass taken by all 8 cylinders when they are charged is $$
m=N_{\text {cyl }} \frac{\Delta V}{v_1}=N_{\text {cyl }} \frac{\pi B^2 S / 4}{v_1}=(8) \frac{\pi(4 / 12 \mathrm{ft})^2(4 / 12 \mathrm{ff}) 4}{13.12 \mathrm{ft}^3 / \mathrm{lbm}}=0.01774\ \mathrm{lbm}
$$ The rate at which air is processed by the engine is determined from $$
\dot{m}=\frac{m \hat{n}}{N_{\text {rev }}}=\frac{(0.01774\ \mathrm{lbm} / \mathrm{cycle})(1800 / 60 \mathrm{rev} / \mathrm{s})}{2 \mathrm{rev} / \mathrm{cycle}}=0.2661 \mathrm{lbm} / \mathrm{s}=958.0\ \mathrm{lbm} / \mathrm{h}
$$ since there are two revolutions per cycle in a four-stroke engine. The compression ratio is $$
r=\frac{1}{0.045}=22.22
$$ At the end of the compression, the air temperature is $$
T_2=T_1 r^{k-1}=(510 \mathrm{R})(22.22)^{1.4-1}=1763\ \mathrm{R}
$$ Application of the first law and work integral to the constant pressure heat addition gives $$
q_{\text {is }}=c_p\left(T_3-T_2\right)=(0.240 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(2760-1763) \mathrm{R}=239.3\ \mathrm{Btu} / \mathrm{lbm}
$$ while the thermal efficiency is $$
\eta_{\mathrm{h}}=1-\frac{1}{r^{k-1}} \frac{r_c^k-1}{k\left(r_c-1\right)}=1-\frac{1}{22.22^{1.4-1}} \frac{1.4^{1.4}-1}{1.4(1.4-1)}=0.6892
$$ The power produced by this engine is then $$
\begin{aligned}
\dot{W}_{\text {net }} & =m w_{\text {net }}=m \eta_{\mathrm{th}} q_{\mathrm{in}} \\
& =(958.0\ \mathrm{lbm} / \mathrm{h})(0.6892)\left(239.3 \mathrm{Btu} / \mathrm{lbm}\left(\frac{1 \mathrm{hp}}{2544.5 \mathrm{Btu} / \mathrm{h}}\right)\right. \\
& =62.1\ \mathrm{hp}
\end{aligned}
$$
