For a normal shock wave in air Mach number is 3.

If the atmospheric pressure and air density are $26.5~ KN/m^2 $ and $0.413 ~kg/m^3$ respectively, determine the flow Conditions before and after the shock wave. Take $\gamma = 1.4$

Solution:

Let, subscripts 1 and 2 represent the flow conditions before and after the shock wave.

Mach number, $ \quad M_{1}=2 $

Atmospheric pressure, $ \quad p_{1}=26.5 \mathrm{kN} / \mathrm{m}^{2} $

Air density, $ \rho_{1}=0.413 \mathrm{~kg} / \mathrm{m}^{3} \\ $

Mach number, $M_{2}$ :

$$ \begin{aligned} M_{2}^{2} &=\frac{(\gamma-1) M_{1}^{2}+2}{2 \gamma M_{1}^{2}-(\gamma-1)} \\\\ &=\frac{(1.4-1) \times 3^{2}+2}{2 \times 1.4 \times 3^{2}-(1.4-1)}=\frac{3.6}{11.2-0.4}=0.2016 \\\\ &=M_{2}=0.448....(1) \\ \end{aligned} $$

Pressure, $\mathbf{p}_{2} : \\$ $$ \frac{p_{2}}{p_{1}}=\frac{2 \gamma M_{1}^{2}-(\gamma-1)}{(\gamma+1)} \ $$

$$ =\frac{2 \times 1.4 \times 3^{2}-(1.4-1)}{(1.4+1)}=\frac{11.2-0.4}{2.4}=10.33 $$

$$ p_{2}=26.5 \times 10.33=273.74 \mathrm{kN} / \mathrm{m}^{2}......(2) \\ $$

Density, $\rho_{2}: \\$ $$ \frac{\rho_{2}}{\rho_{1}}=\frac{(\gamma+1) M_{1}^{2}}{(\gamma-1) M_{1}^{2}+2} \ $$

$$ =\frac{(1.4+1) 3^{2}}{(1.4-1) 3^{2}+2}=\frac{21.6}{3.6+2}=3.857.....(3) \\ $$

$$ \therefore \quad \rho_{2}=0.413 \times 3.857=1.592 \mathrm{~kg} / \mathrm{m}^{3} $$

Temperature, $ \mathbf{T}_{1}: \\ $

Since, $ p_{1}=\rho_{1} R T_{1}, \quad \therefore T_{1}=\frac{p_{1}}{\rho_{1} R}=\frac{26.5 \times 10^{3}}{0.413 \times 287}=223.6 \mathrm{~K}$ or $-\mathbf{4 9 . 4 ^ { \circ } \mathbf { C }} \\ $ (Ans.)

Temperature, $ \mathrm{T}_{2}: \\ $

$$ \frac{T_{2}}{T_{1}}=\frac{\left[(\gamma-1) M_{1}^{2}+2\right]\left[2 \gamma M_{1}^{2}-(\gamma-1)\right]} \\{(\gamma+1)^{2} M_{1}^{2}}=\frac{\left[(1.4-1) 3^{2}+2\right]\left[2 \times 1.4 \times 3^{2}-(1.4-1)\right]}{(1.4+1)^{2} 3^{2}} \\ $$

$$ =\frac{(3.6+2)(25.2-0.4)}{51.84}=2.679 \\ $$

$ \therefore \quad T_{2}=223.6 \times 2.679=\mathbf{599.02 ~ K}$ or $\mathbf{1 0 4 . 3}^{\circ} \mathrm{C} \quad \\ $

Velocity, $V_{1}: \\$ : $$ C_{1}=\sqrt{\gamma R T_{1}}=\sqrt{1.4 \times 287 \times 223.6}=299.7 \mathrm{~m} / \mathrm{s} \ $$

Since, $ \quad \frac{V_{1}}{C_{1}}=M_{1}=2 \therefore V_{1}=299.7 \times 3=\mathbf{899. 1} \mathrm{m} / \mathbf{s} \quad \\ $ (Ans.)

Velocity, $V_{2} \\$ : $$ C_{2}=\sqrt{\gamma R T_{2}}=\sqrt{1.4 \times 287 \times 599.02}= 157.11\mathrm{~m} / \mathrm{s} \ $$

Since,

$$ \frac{V_{2}}{C_{2}}=M_{2}=0.577 \quad \therefore V_{2}=389.35 \times 0.487=\mathbf{189.61} \mathbf{m} / \mathbf{s} \quad \text { (Ans.) } \\ $$