For a normal shock wave in air Mach number is 2. If the atmospheric pressure and air density are 26.5 kN/m2 and 0.413 kg/m3 respectively, determine the flow conditions before and after the shock wave. Take γ = 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, $ p_{1}=26.5 \mathrm{kN} / \mathrm{m}^{2} \\ $

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

Mach number, $M_{2}$ :

$$ \begin{aligned} M_{2}^{2} &=\frac{(\gamma-1) M_{1}^{2}+2}{2 \gamma M_{1}^{2}-(\gamma-1)} \\\\ &=\frac{(1.4-1) \times 2^{2}+2}{2 \times 1.4 \times 2^{2}-(1.4-1)}=\frac{3.6}{11.2-0.4}=0.333 \end{aligned} \\ $$

$$ M_{2}=\mathbf{0 . 5 7 7}....(1) \\ $$

Pressure, $\mathbf{p}_{2}$ :

$$ \frac{p_{2}}{p_{1}}=\frac{2 \gamma M_{1}^{2}-(\gamma-1)}{(\gamma+1)}....(2) \\ $$

$$ =\frac{2 \times 1.4 \times 2^{2}-(1.4-1)}{(1.4+1)}=\frac{11.2-0.4}{2.4}=4.5 \\ $$

$$ p_{2}=26.5 \times 4.5=119.25 \mathrm{kN} / \mathrm{m}^{2} \text { (Ans.) } $$

Density, $\rho_{2}:$

$$ \begin{aligned} \frac{\rho_{2}}{\rho_{1}} &=\frac{(\gamma+1) M_{1}^{2}}{(\gamma-1) M_{1}^{2}+2} \\\\ &=\frac{(1.4+1) 2^{2}}{(1.4-1) 2^{2}+2}=\frac{9.6}{1.6+2}=2.667...(3) \end{aligned} \\ $$

$$ \rho_{2}=0.413 \times 2.667=1.101 \mathrm{~kg} / \mathrm{m}^{3} \text { (Ans.) } \\ $$

Temperature, $T_{1}$ :

Since ,

$$ p_{1}=\rho_{1} R T_{1} \\ $$

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

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

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

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

$$ =\frac{(1.6+2)(11.2-0.4)}{23.04}=1.6875 $$

$$ \therefore \quad T_{2}=223.6 \times 1.6875=377.3 \mathrm{~K}\ or \ \mathbf{1 0 4 . 3}^{\circ} \mathrm{C} \\ $$

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,

$$ \quad \frac{V_{1}}{C_{1}}=M_{1}=2 \therefore V_{1}=299.7 \times 2=599.4 \mathrm{~m} / \mathbf{s} \quad \\ $$

Velocity, $V_{2}$ :

$$ C_{2}=\sqrt{\gamma R T_{2}}=\sqrt{1.4 \times 287 \times 377.3}=389.35 \mathrm{~m} / \mathrm{s} \\ $$

Since,

$$ \frac{V_{2}}{C_{2}}=M_{2}=0.577 \therefore V_{2}=389.35 \times 0.577=224.6 \mathrm{~m} / \mathbf{s} \quad \text { (Ans.) } \\ $$