Question 8.16: A compression shock occurs in a divergent air flow passage. ......

Given: p_1=0.2  MPa ; T_1=308  K ; V_1=400  m / s

(i) The Mach number upstream of the shock is calculated from the given data:

M_1=\frac{V_1}{a_1}=\frac{V_1}{\sqrt{\gamma R T_1}}=\frac{400}{\sqrt{1.4 \times 287 \times 308}}=1.14

The normal shock tables (Appendix A4) can be used for finding the downstream parameters. From normal shock tables (γ = 1.4) corresponding to M = 1.14

\begin{aligned}M_2 &=0.882 \\\frac{p_2}{p_1}&=1.3495 \\P_2 &=0.2 \times 1.3495 \\&=0.2699  MPa \\\frac{T_2}{T_1}&=1.0903 \\T_2 &=308 \times 1.0903 \\&=335.8  K\end{aligned}

The velocity of air after the shock wave

\begin{aligned}V_2 &=M_2 a_2=M_2 \sqrt{\gamma R T_2}=0.882 \times \sqrt{1.4 \times 287 \times 335.8}\\&=323.98  m / s\end{aligned}

(ii) From shock tables

\frac{P_{02}}{P_{01}}=0.997

Change in entropy across the shock is:

\begin{aligned}s_2-s_1 &=R \ln \frac{P_{01}}{P_{02}}=287 \ln \frac{1}{0.997}\\&=0.862   J / kg K\end{aligned}