Unsteady Rankine-Hugoniot relations. Is it applicable?

For a unsteady normal shock moving in a medium, the Rankine-Hugoniot relations can be derived, by converting the problem into a steady by fixing the frame of reference on the shock. This idea is easy to follow, given the shock is uniform and steady, meaning the shock strength ($\rho_2/\rho_1$) does not change with time.

Recently, I came across a relation for density and velocity as a function of time for a shock whose pressure jump decays exponentially with time. How can one derive this?

\begin{align} \rho(t)&=\rho_0\frac{(\gamma+1)p(t)+2\gamma p_0}{(\gamma-1)p(t)+2\gamma p_0} \\ u(t)&=\sqrt{\frac{2}{\gamma p_0}}\frac{a_0p(t)}{\sqrt{(\gamma+1)p(t)+2\gamma p_0}} \end{align}

• Recently, I came across... Can you provide the reference? – Kyle Kanos Nov 12 '15 at 12:42
• Also, are $a_0$ and $p_0$ the downstream speed of sound and pressure, respectively? – Kyle Kanos Nov 12 '15 at 12:45
• @kyleKanos a0 and p0 are upstream(I guess, otherwise the equation becomes implicit). The expression was given in a thesislink page 21 – prasanna Nov 12 '15 at 19:02
• @kyleKanos also check this paper. They have expressed velocity as a function of time. – prasanna Nov 12 '15 at 19:11

Background

The link to the dissertation you provided gives a derivation in Appendix A through the following steps.

First, start with the general Rankine-Hugoniot relations for a simple normal incidence hydrodynamic shock wave, where: $$\frac{ \rho_{dn} }{ \rho_{up} } = \frac{ u_{up} }{ u_{dn} } = \frac{ \left( \gamma + 1 \right) M_{up}^{2} }{ \left( \gamma + 1 \right) + \left( \gamma - 1 \right) \left( M_{up}^{2} - 1 \right) } \\ \frac{ P_{dn} }{ P_{up} } = \frac{ \left( \gamma + 1 \right) + 2 \gamma \left( M_{up}^{2} - 1 \right) }{ \left( \gamma + 1 \right) }$$ where $\rho$ is the mass density, $u$ is the flow speed along the shock normal vector, $\gamma$ is the ratio of specific heats, $M$ is the Mach number, and $up$($dn$) corresponds to the average value in the upstream(downstream) regions.

Derivation of Density Ratio

If we rewrite the second equation above by solving for $M_{up}$ we get: $$M_{up}^{2} = \frac{ P_{up} \left( \gamma - 1 \right) + P_{dn} \left( \gamma + 1 \right) }{ 2 \ \gamma \ P_{up} }$$

You will notice that this can be rewritten as: $$M_{up}^{2} = \left( \frac{ P_{dn} }{ P_{up} } - 1 \right) \frac{ \gamma + 1 }{ 2 \ \gamma } + 1$$

The physically important leap comes when you recognize that the 1 in the ()'s above represents a system no different than the original, namely undisturbed. That means that all other terms are part of the disturbed media, which the authors call the overpressure part of the expression. Thus, we can define the overpressure as $P_{s} = P_{dn} - P_{up}$ so that we have: $$M_{up}^{2} = \frac{ P_{s} }{ P_{up} } \left( \frac{ \gamma + 1 }{ 2 \ \gamma } \right) + 1$$

So if we substitute this version of $M_{up}$ into our expression for the density ratio above we get: $$\frac{ \rho_{dn} }{ \rho_{up} } = \frac{ \left( \gamma + 1 \right) P_{s} + 2 \ \gamma \ P_{up} }{ \left( \gamma - 1 \right) P_{s} + 2 \ \gamma \ P_{up} }$$ then we can see that we just replace $Q_{s} \rightarrow Q\left( t \right)$ and $Q_{up} \rightarrow Q_{o}$ to get your expression, where $Q$ is any of the changing quantities.

Derivation of Speed Ratio

The first thing to do is to just invert the expression for $\rho_{dn}$ above to get: $$\frac{ u_{dn} }{ u_{up} } = \frac{ \left( \gamma - 1 \right) P_{s} + 2 \ \gamma \ P_{up} }{ \left( \gamma + 1 \right) P_{s} + 2 \ \gamma \ P_{up} }$$ and we then note that $u_{up} = M_{up} a_{up}$, where $a_{j}$ is the sound speed in the $j^{th}$ region. If we substitute the new expression for $u_{up}$ and the overpressure version of $M_{up}$ we get: $$u_{up} = a_{up} \sqrt{ \frac{ P_{s} }{ P_{up} } \left( \frac{ \gamma + 1 }{ 2 \ \gamma } \right) + 1 } \\ = a_{up} \sqrt{ \frac{ P_{s} \left( \gamma + 1 \right) + 2 \ \gamma \ P_{up} }{ 2 \ \gamma \ P_{up} } }$$ which gives us: $$u_{dn} = \frac{ a_{up} }{ \sqrt{ 2 \ \gamma \ P_{up} } } \left[ \frac{ \left( \gamma - 1 \right) P_{s} + 2 \ \gamma \ P_{up} }{ \sqrt{ \left( \gamma + 1 \right) P_{s} + 2 \ \gamma \ P_{up} } } \right]$$

Differences

I am fairly certain they have missed or dropped a term in their expression for $u\left( t \right)$ because even MathematicaTM cannot equate the terms in their equation A.8. I tried deriving their results several different ways, even letting Mathematica do all the work just in case I made a mistake and I continually get my result shown above.

• Thank you so much for sharing the knowledge. I cannot up vote you as I have insufficient rep. – prasanna Jan 25 '16 at 9:38