Supersonic Speeds, Shock Waves — Question 10

Question 10

Shock-frame conservation separates front speed from material speed

A planar shock travels right at $D=400\,\mathrm{m/s}$ into stationary fluid of density $\rho_1=1.20\,\mathrm{kg/m^3}$. Behind it, $\rho_2=1.25\rho_1$ and material speed is $u_2$ in the laboratory. In the shock frame, use mass balance $\rho_1D=\rho_2(D-u_2)$ and momentum balance $p_1+\rho_1D^2=p_2+\rho_2(D-u_2)^2$.

1. Find the downstream material speed and its speed relative to the front.

2. Find the pressure jump and show that it equals $\rho_1Du_2$. Explain which additional information would be needed for a full thermodynamic shock state.

Question 10 - Solution

Shock-frame conservation separates front speed from material speed

(a) Method and calculation

From mass conservation,$$D-u_2=\frac{\rho_1}{\rho_2}D=\frac{400}{1.25}=320\,\mathrm{m/s},$$$$u_2=D(1-1/1.25)=80.0\,\mathrm{m/s}.$$The front outruns the compressed material; the two velocities are physically distinct.

(b) Interpretation and further calculation

Using the common mass flux $j=\rho_1D=\rho_2(D-u_2)$,$$p_2-p_1=j[D-(D-u_2)]=\rho_1Du_2$$$$=(1.20)(400)(80)=38400\,\mathrm{Pa}.$$Direct substitution gives the same result: $1.20(400)^2-1.50(320)^2=38400\,\mathrm{Pa}$. Energy conservation and an equation of state are also needed to determine temperature and thermodynamic admissibility; mass and momentum alone are not a complete shock model.

Physical check

Compression gives $u_2>0$ and $p_2>p_1$. As $\rho_2/\rho_1\to1$, these jumps vanish in the supplied balances.