In a normal shock wave occurring in a helium ( k = 1.66) the density downstream of the shock is three times that on the upstream. Calculate the corresponding pressure ratio and velocity ratio. What are the Mach numbers upstream and downstream of the shock?

To find:

Pressure Ratio $(P_2/P_1)$

Mach number at upstream $(M_1)$and downstream $(M_2)$of the shock

We know that,

$(\frac{P_2}{P_1})=(\frac{ρ_2}{ρ_1})^γ$

$∴(\frac{P_2}{P_1})=(3)^1.66$

$∴(\frac{P_2}{P_1}=6.1947$

For Normal Shock,

$(\frac{P_2}{P_1})=1+\frac{2γ}{γ+1}(M_1^2-1)$

$∴(M_1^2-1)=\frac{(γ+1)}{2γ} \{(\frac{P_2}{P_1})-1\}$

$∴(M_1^2-1)=\frac{(1.66+1)}{2(1.66}) {(6.1947)-1}$

$∴(M_1^2-1)$=4.162

$∴M_1^2=5.162$

$∴M_1=2.272$

Now, By Continuity equation

$ρ_1 U_1=ρ_2 U_2$

$ρ_1 U_1=ρ_2 U_2$

$∴ρ_2 M_2 C_2=ρ_1 M_1 C_1$

Since, $M=U/C$

$∴M_2=\frac{ρ_1 M_1 C_1}{ρ_2 C_2}$

But, C=$\sqrt{γRT}$

$∴M_2=\frac{ρ_1}{ρ_2} (\frac{T_1}{T_2})^(1/2) M_1$

But,

$(\frac{T_1}{T_2})=(\frac{ρ_1}{ρ_2})^(γ-1)$

$∴M_2=\frac{ρ_1}{ρ_2} (\frac{ρ_1}{ρ_2})^(\frac{γ-1)}{2} M_1$

$∴M_2=(\frac{ρ_1}{ρ_2})^(\frac{γ+1)}{2} M_1$

$∴M_2=(\frac{1}{3})^(\frac{1.66+1)}{2}$ 2.272

$∴M_2=0.527$

…which are the required Mach numbers