Answer
$\frac{T_2}{T_0}= 0.740$
$T_{2}= 378 K$
$P_{2}= 82.4 K$
Work Step by Step
The schematic of the duct is shown in Fig. 12-25. For isentropic flow through a duct, the area ratio $A / A^*$ (the flow area over the area of the throat where $\mathrm{Ma}=1$ ) is also listed in Table $\mathrm{A}-32$. At the initial Mach number of $\mathrm{Ma}=$ 0.5 , we read $$
\frac{A_1}{A}=1.3398, \quad \frac{T_1}{T_0}=0.9524, \text { and } \quad \frac{P_1}{P_0}=0.8430 $$
With a 20 percent reduction in flow area, $A_2=0.8 A_1$, and $$
\frac{A_2}{A^*}=\frac{A_2}{A_1} \frac{A_1}{A^*}=(0.8)(1.3398)=1.0718
$$ For this value of $A_2 / A^*$ from Table A-32, we read
$$
\frac{T_2}{T_0}=0.9010, \quad \frac{P_2}{P_0}=0.6948, \text { and } \mathrm{Ma}_2=\mathbf{0 . 7 4 0}
$$ 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
$$
\begin{aligned}
& \frac{T_2}{T_1}=\frac{T_2 / T_0}{T_1 / T_0} \quad \rightarrow \quad T_2=T_1\left(\frac{T_2 / T_0}{T_1 / T_0}\right)=(400 \mathrm{~K})\left(\frac{0.9010}{0.9524}\right)=378 \mathrm{~K} \\
& \frac{P_2}{P_1}=\frac{P_2 / P_0}{P_1 / P_0} \quad \rightarrow \quad P_2=P_1\left(\frac{P_2 / P_0}{P_1 / P_0}\right)=(100 \mathrm{kPa})\left(\frac{0.6948}{0.8430}\right)=82.4 \mathrm{~K}
\end{aligned}
$$ which are the temperature and pressure at the desired location.
