Answer
$\frac{A_2}{A^*}= 1.6281$
$T_{2}=399K$
$P_{2}=95.7K$
Work Step by Step
$$
\frac{A_1}{A^*}=2.0351, \quad \frac{T_1}{T_0}=0.9823, \text { and } \quad \frac{P_1}{P_0}=0.9395
$$ With a 20 percent reduction in flow area, $A_2=0.8 A_1$, and $$
\frac{A_2}{A^*}=\frac{A_2}{A_1} \frac{A_1}{A^*}=(0.8)(2.0351)=1.6281
$$ For this value of $A_2 / A^*$ from Table $\mathrm{A}-32$, we read
$$
\frac{T_2}{T_0}=0.9791, \quad \frac{P_2}{P_0}=0.8993 \text {, and } \mathrm{Ma}_2=0.391
$$ Here we chose the subsonic Mach number for the calculated $A_2 / A^*$ instead of the supersonic one because the duct is converging in the flow direction and the initial flow is subsonic. Since the stagnation properties are constant for isentropic flow, we can write $$
\begin{aligned}
& \frac{T_2}{T_1}=\frac{T_2 / T_0}{T_1 / T_0} \quad \rightarrow \quad T_2=T_1\left(\frac{T_2 / T_0}{T_1 / T_0}\right)=(400 \mathrm{~K})\left(\frac{0.9791}{0.9823}\right)=399 \mathrm{~K} \\
& \frac{P_2}{P_1}=\frac{P_2 / P_0}{P_1 / P_0} \quad \rightarrow \quad P_2=P_1\left(\frac{P_2 / P_0}{P_1 / P_0}\right)=(100 \mathrm{kPa})\left(\frac{0.8993}{0.9395}\right)=95.7 \mathrm{~K}
\end{aligned}
$$ which are the temperature and pressure at the desired location.
