| written 7.9 years ago by | modified 5.5 years ago by |
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What are the Mach numbers upstream and downstream of the shock?
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| written 7.0 years ago by |
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 …
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