Question 8.20: Air approaches a wedge of half wedge angle 15° with a Mach n......

Given: M_1=2.0 ; \delta=15^{\circ}

From the graph showing oblique shock functions (Appendix A5), corresponding to M_1  = 2.0 and δ = 15°, we get,

\beta=80^{\circ} \text { and } 45^{\circ}

Since the shock formed is a strong shock wave, the higher angle is to be taken.

\beta=80^{\circ}

The problem can be solved using normal shock tables. Corresponding to M_1=2,\left(M_1\right)_{\text {normal }} is:

\left(M_1\right)_n=M_1 \sin \beta=2 \times \sin 80=1.97

From normal shock tables (γ = 1.4), for (M_1)_n = 1.97,

\begin{aligned}& \left(M_2\right)_n=0.582 \\& \left(\frac{p_2}{p_1}\right)_n=4.357 \\& \left(\frac{\rho_2}{\rho_1}\right)_n=2.622 \\& \left(\frac{T_2}{T_1}\right)_n=1.659\end{aligned}

Hence, for the oblique shock, the downstream Mach number and other property ratios are:

\begin{aligned}& M_2=\frac{\left(M_2\right)_n}{\sin (\beta-\delta)}=\frac{0.582}{\sin 65}=0.644 \\& \frac{p_2}{p_1}=\left(\frac{p_2}{p_1}\right)_n=4.357 \\& \frac{T_2}{T_1}=\left(\frac{T_2}{T_1}\right)_n=1.659 \\& \frac{\rho_2}{\rho_1}=\left(\frac{\rho_2}{\rho_1}\right)_n=2.622\end{aligned}