Answer
$P_{x}=4829\text{ kPa}$
$T_{3}=1516\text{ K}$
$ q_{\text {in }}=556.5\text{ kJ/kg}$
$\eta_{\text {is }}=0.646$
Work Step by Step
The specific volume of the air at the start of the compression is $$
v_1=\frac{R T_1}{R_1}=\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(293 \mathrm{~K})}{80 \mathrm{kPa}}=1.051 \mathrm{~m}^3 / \mathrm{kg}
$$ and the specific volume at the end of the compression is $$
v_2=\frac{v_1}{r}=\frac{1.051 \mathrm{~m}^3 / \mathrm{kg}}{14}=0.07508 \mathrm{~m}^3 / \mathrm{kg}
$$ The pressure at the end of the compression is $$
P_2=P_1\left(\frac{v_1}{v_2}\right)^k=P_1 r^k=(80 \mathrm{kPa})(14)^{1,4}=3219\ \mathrm{kPa}
$$ and the maximum pressure is $$
P_x=P_3=r_p P_2=(1.5)(3219\ \mathrm{kPa})=4829\ \mathrm{kPa}
$$ The temperature at the end of the compression is $$
\begin{aligned}
& T_2=T_1\left(\frac{v_1}{v_2}\right)^{i-1}=T_1 r^{i-1}-(293 \mathrm{~K})(14)^{1.4-1}-842.0 \mathrm{~K} \\
& T_x=T_2\left(\frac{p_3}{P_2}\right)=(842.0 \mathrm{~K})\left(\frac{4829 \mathrm{kPa}}{3219 \mathrm{kPa}}\right)-1263 \mathrm{~K}
\end{aligned}
$$ From the definition of cutoff ratio $$
v_1=r_c v_x=r_c v_2=\left(1.2 \times 0.07508 \mathrm{~m}^3 / \mathrm{kg}\right)=0.09010\ \mathrm{~m}^3 / \mathrm{kg}
$$ The remaining state temperatures are then $$
\begin{aligned}
& \left.T_3=T_x\left(\frac{v_3}{v_x}\right)-01263 \mathrm{~K}\right)\left(\frac{0.09010}{0.07508}\right)=1516 \mathrm{~K} \\
& T_4=T_3\left(\frac{v_3}{v_4}\right)^{1-1}-(1516 \mathrm{~K})\left(\frac{0.09010}{1.051}\right)^{1.4-1}=567.5 \mathrm{~K}
\end{aligned}
$$ Applying the first law and arork expression to the heat addition processes gives $$
\begin{aligned}
& q_{\text {in }}=c_v\left(T_x-T_2\right)+c_p\left(T_3-T_x\right) \\
& =(0.7 \mathrm{Is} \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K})(1263-842.0) \mathrm{K}+(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} . \times 15 \mathrm{l} 6-1263) \mathrm{K} \\
& =556.5 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The heat rejected is $$
q_{\text {cut }}=c_v\left(T_4-T_1\right)=(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(567.5-293) \mathrm{K}=197.1 \mathrm{~kJ} / \mathrm{kg}
$$ Then, $\quad \eta_{\text {is }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{197.1 \mathrm{~kJ} / \mathrm{kg}}{556.5 \mathrm{~kJ} / \mathrm{kg}}=0.646$
