Question 8.19: Air having an initial Mach number M1 = 2.0 is deflected thro......

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

In order to solve the problem, the shock wave angle β has to be initially found out. β can be obtained in two ways, by solving Eq. 8.97 or from the curves or tables showing δ as a function of M_1 and β . From the graph showing oblique shock functions (Appendix A5), corresponding to δ = 15° and from the line of M_1 = 2.0, we get,

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

The lower value of β corresponds to weak shock wave.

\beta=45^{\circ}

The remaining part of the problem can be solved using normal shock tables (Appendix A4). For solving oblique shock problems using normal shock tables, the corresponding (M_1)_{normal} of the moving co-ordinate system has to be calculated.

M_{1 n}=M_1 \sin \beta=2 \times \sin 45=1.414

From normal shock tables (γ = 1.4) corresponding to M_{1 n}=1.414,

\begin{gathered}M_{2 n}=0.733 \\\left(\frac{p_2}{p_1}\right)_n=2.1685 \\\left(\frac{\rho_2}{\rho_1}\right)_n=1.714 \\\left(\frac{T_2}{T_1}\right)_n=1.262\end{gathered}

Since the static property ratios does not change because of the moving co-ordinate system, across the oblique shock we get,

\begin{aligned}& \frac{p_2}{P_1}=2.1685 \\& \frac{T_2}{T_1}=1.262 \\& \frac{\rho_2}{\rho_1}=1.714\end{aligned}

The Mach number downstream of the oblique shock is:

\begin{aligned}M_2 &=\frac{\left(M_2\right)_n}{\sin (\beta-\delta)}=\frac{0.733}{\sin 30}\\&=1.466\end{aligned}