Answer
$P_{x}=4829\text{ kPa}$
$T_{3}=1309\text{ K}$
$q_{\text {out }}=170.2\text{ kJ/kg}$
Work Step by Step
The specific volume of the air at the start of the compression is $$
v_1=\frac{R T_1}{P}-\frac{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K} \times 253 \mathrm{~K}\right)}{80 \mathrm{kPa}}=0.9076 \mathrm{~m}^3 / \mathrm{kg}
$$ and the specific volume at the end of the compression is $$
v_2=\frac{v_1}{r}=\frac{0.9076 \mathrm{~m}^3 / \mathrm{kg}}{14}=0.06483 \mathrm{~m}^3 / \mathrm{kg}
$$ The pressure at the end of the compression is $$
P_2=P\left(\frac{v_1}{v_2}\right)^t=P_1 r^t=(80 \mathrm{NPa})(14)^{1 .+}-3219 \mathrm{kPa}
$$ and the maximum pressure is $$
P_x=P_3=r_P P_2=(1.5)(3219 \mathrm{kPa})=4829\ \mathrm{kPa}
$$ The temperatare at the end of the compression is $$
\begin{aligned}
& T_2=T_1\left(\frac{v_1}{v_2}\right)^{k-1}-T_1 r^{k-1}=(253 \mathrm{~K})(14)^{\gamma^{-4-1}}=727.1 \mathrm{~K} \\
& \text { and } T_x=T_2\left(\frac{P_2}{P_2}\right)=(727 . \mathrm{K} \mathrm{K})\left(\frac{4 \times 29 \mathrm{kPa}}{3219 \mathrm{LPa}}\right)=1091 \mathrm{~K} \end{aligned}
$$ From the definition of cutoff ratio $$
\left.v_3=r_2 v_x=r_2 v_2=(1.2) 0.06483 \mathrm{~m}^3 / \mathrm{kg}\right)=0.07780\ \mathrm{~m}^3 / \mathrm{kg}
$$ The remaining state temperabures are then $$
\begin{aligned}
& T_3=T_4\left(\frac{v_3}{v_x}\right)-(1091 \mathrm{~K})\left(\frac{0.07790}{0.06483}\right)=1309 \mathrm{~K} \\
& T_4=T_3\left(\frac{v_3}{v_4}\right)^{t-1}=(1309 \mathrm{~K})\left(\frac{0.07780}{0.9076}\right)^{1.4-1}=490.0 \mathrm{~K}
\end{aligned}
$$ Applying the first law and work expression to the heat addition peocesses gives $$
\begin{aligned}
\_{\mathrm{m}} & =c_v\left(T_x-T_2\right)+c_p\left(T_3-T_x\right) \\
& =(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(109 \mathrm{l}-727.1) \mathrm{K}+(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1.309-1091) \mathrm{K} \\
& =480.4 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The hear rejected is $$
q_{\text {out }}=c_v\left(T_4-T_i\right)=(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(4900-253) \mathrm{K}=170.2 \mathrm{~kJ} / \mathrm{kg}
$$
