Moving expansion waves
\section{Moving Normal Shock Waves}
\noindent The starting point is the governing equations for stationary normal shocks (repeated here for convenience).
\begin{equation} \rho_1 u_1 = \rho_2 u_2 \label{eq:stationary:cont} \end{equation}
\begin{equation} \rho_1 u_1^2+p_1 = \rho_2 u_2^2 + p_2 \label{eq:stationary:mom} \end{equation}
\begin{equation} h_1 + \frac{1}{2}u_1^2 = h_2 + \frac{1}{2}u_2^2 \label{eq:stationary:energy} \end{equation}
\noindent Shock moving to the right with the constant speed $W$ into a gas that is standing still. Moving with the shock, we would see a gas velocity ahead of the shock $u_1=W$, and the gas behind the shock moves to the right with the velocity $u_2=W-u_p$. Now, let's insert $u_1$ and $u_2$ in the stationary shock relations \ref{eq:stationary:cont} - \ref{eq:stationary:energy}.
\begin{equation} \rho_1 W = \rho_2 (W-u_p) \label{eq:unsteady:cont} \end{equation}
\begin{equation} \rho_1 W^2+p_1 = \rho_2 (W-u_p)^2 + p_2 \label{eq:unsteady:mom} \end{equation}
\begin{equation} h_1 + \frac{1}{2}W^2 = h_2 + \frac{1}{2}(W-u_p)^2 \label{eq:unsteady:energy} \end{equation}
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\noindent Rewriting Eqn. \ref{eq:unsteady:cont}
\begin{equation} (W-u_p) = W \frac{\rho_1}{\rho_2} \label{eq:unsteady:cont:mod} \end{equation}\\
\noindent Inserting Eqn. \ref{eq:unsteady:cont:mod} in Eqn. \ref{eq:unsteady:mom} gives\\
\begin{equation*} p_1+\rho_1 W^2 = p_2+\rho_2 W^2\left(\frac{\rho_1}{\rho_2}\right)^2 \Rightarrow p_2-p_1 = \rho_1W^2\left(1-\frac{\rho_1}{\rho_2}\right) %\label{eq:unsteady:mom:mod} \end{equation*}\\
\begin{equation} W^2=\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_2}{\rho_1}\right) \label{eq:unsteady:mom:mod} \end{equation}\\
\noindent From the continuity equation \ref{eq:unsteady:cont}, we get \\
\begin{equation} W = (W-u_p) \left(\frac{\rho_2}{\rho_1}\right) \label{eq:unsteady:cont:modb} \end{equation}\\
\noindent Inserting Eqn. \ref{eq:unsteady:cont:modb} in Eqn. \ref{eq:unsteady:mom:mod} gives\\
\begin{equation} (W-u_p)^2=\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_1}{\rho_2}\right) \label{eq:unsteady:mom:modb} \end{equation}\\
\noindent Now, let's insert Eqns. \ref{eq:unsteady:mom:mod} and \ref{eq:unsteady:mom:modb} in the energy equation (Eqn. \ref{eq:unsteady:energy}).\\
\begin{equation} h_1 + \frac{1}{2}\left[\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_2}{\rho_1}\right)\right] = h_2 + \frac{1}{2}\left[\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_1}{\rho_2}\right)\right] \label{eq:unsteady:energy:mod} \end{equation}
\begin{equation} h=e+\frac{p}{\rho} \label{eq:enthalpy} \end{equation}
\begin{equation} e_1 + \frac{p_1}{\rho_1} + \frac{1}{2}\left[\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_2}{\rho_1}\right)\right] = e_2 + \frac{p_2}{\rho_2} + \frac{1}{2}\left[\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_1}{\rho_2}\right)\right] \label{eq:unsteady:energy:mod:b} \end{equation}\\
\noindent which can be rewritten as\\
\begin{equation} e_2-e_1=\frac{p_1+p_2}{2}\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right) \label{eq:unsteady:hugonoit} \end{equation}\\
\noindent Eqn \ref{eq:unsteady:hugonoit} is the same Hugoniot equation as we get for a stationary normal shock. The Hugoniot equation is a relation of thermodynamic properties over a shock. As the shock in the unsteady case is moving with a constant velocity, the frame of reference moving with the shock is an inertial frame and thus the same physical relations apply in the moving shock case as in the stationary shock case. The fact that the Hugoniot relation does not include any velocities or Mach numbers but only thermodynamic properties, the relation will be unchanged for a moving shock.
\subsection{Moving Shock Relations}
\noindent For a calorically perfect gas we have $e=C_v T$. Inserted in the Hugoniot relation above this gives\\
\begin{equation} C_v(T_2-T_1)=\frac{p_1+p_2}{2}\left(\nu_1-\nu_2\right) \label{eq:unsteady:hugonoit:b} \end{equation}\\
where $\nu=1/\rho$\\
\noindent Now, using the ideal gas law $T=p\nu/R$ and $C_v/R=1/(\gamma-1)$ gives\\
\begin{equation*} \left(\frac{1}{\gamma-1}\right)(p_2\nu_2-p_1\nu_1)=\frac{p_1+p_2}{2}\left(\nu_1-\nu_2\right) \end{equation*}
\begin{equation*} \Leftrightarrow \end{equation*}
\begin{equation*} p_2\left(\frac{\nu_2}{\gamma-1}-\frac{\nu_1-\nu_2}{2}\right)=p_1\left(\frac{\nu_1}{\gamma-1}+\frac{\nu_1-\nu_2}{2}\right) \end{equation*}\\
\noindent From this result, we can derive a relation for the pressure ratio over the shock as a function of density ratio\\
\begin{equation} \frac{p_2}{p_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{\nu_1}{\nu_2}\right)-1}{\left(\dfrac{\gamma+1}{\gamma-1}\right)-\left(\dfrac{\nu_1}{\nu_2}\right)} \label{eq:unsteady:hugonoit:c} \end{equation}\\
\noindent $\nu=RT/p$ and thus
\begin{equation} \frac{\nu_1}{\nu_2}=\frac{T_1}{T_2}\frac{p_2}{p_1} \label{eq:unsteady:density:ratio} \end{equation}\\
\noindent Eqn. \ref{eq:unsteady:density:ratio} in Eqn. \ref{eq:unsteady:hugonoit:c} gives\\
\begin{equation} \frac{p_2}{p_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{T_1}{T_2}\dfrac{p_2}{p_1}\right)-1}{\left(\dfrac{\gamma+1}{\gamma-1}\right)-\left(\dfrac{T_1}{T_2}\dfrac{p_2}{p_1}\right)} \label{eq:unsteady:hugonoit:c} \end{equation}\\
\noindent Now, we can get a relation for calculation of the temperature ratio over the moving shock as function of the shock pressure ratio\\
\begin{equation} \frac{T_2}{T_1}=\frac{p_2}{p_1}\left[\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right] \label{eq:unsteady:temperature:ratio} \end{equation}\\
\noindent Once again using the ideal gas law\\
\begin{equation} \frac{\rho_2}{\rho_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)} \label{eq:unsteady:density:ratio} \end{equation}\\
\noindent Going back to the momentum equation\\
\begin{equation*} p_2-p_1 = \rho_1W^2\left(1-\frac{\rho_1}{\rho_2}\right)=\left\{W=M_s a_1\right\}=\rho_1M_s^2a_1^2\left(1-\frac{\rho_1}{\rho_2}\right) \end{equation*}\\
\noindent with $a_1^2=\gamma p_1/\rho_1$, we get\\
\begin{equation} \frac{p_2}{p_1} = \gamma M_s^2\left(1-\frac{\rho_1}{\rho_2}\right)+1 \label{eq:unsteady:Mach:a} \end{equation}\\
\noindent From the normal shock relations, we have\\
\begin{equation} \frac{\rho_1}{\rho_2} = \frac{2+(\gamma-1)M_s^2}{(\gamma+1)M_s^2} \label{eq:unsteady:Mach:b} \end{equation}\\
\noindent Eqn. \ref{eq:unsteady:Mach:b} in \ref{eq:unsteady:Mach:a} gives\\
\begin{equation} \frac{p_2}{p_1} = 1 + \left(\frac{2\gamma}{\gamma+1}\right)(M_s^2-1) \label{eq:unsteady:Mach:c} \end{equation}\\
or\\
\begin{equation} M_s=\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1} \label{eq:unsteady:Mach} \end{equation}\\
\noindent Eqn. \ref{eq:unsteady:Mach} with $M_s=W/a_1$\\
\begin{equation} W=a_1\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1} \label{eq:unsteady:W} \end{equation}\\
\subsection{Induced Flow Behind Moving Shock}
\noindent Let's try to find a relation for calculation of the induced velocity behind the moving shock. Once again, the starting point is the continuity equation for moving shocks (Eqn. \ref{eq:unsteady:cont}) repeated here for convenience\\
\begin{equation*} \rho_1 W = \rho_2 (W-u_p) \end{equation*}\\
\noindent The induced velocity appears on the right side of the continuity equation\\
\begin{equation*} W (\rho_1-\rho_2) = -\rho_2 u_p \end{equation*}\\
\begin{equation} u_p = W \left(1-\frac{\rho_1}{\rho_2}\right) \label{eq:unsteady:up:a} \end{equation}\\
\noindent From before we have a relation for $W$ as a function of pressure ratio and one for $\rho_1/\rho_2$, also as a function of pressure ratio.\\
Eqn. \ref{eq:unsteady:up:a} togheter with Eqns. \ref{eq:unsteady:W} and \ref{eq:unsteady:density:ratio} gives\\
\begin{equation} u_p=a_1\underbrace{\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}}_{I}\underbrace{\left[1-\dfrac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]}_{II} \label{eq:unsteady:up:b} \end{equation}\\
\noindent The equation subsets I and II can be rewritten as:\\
Term I:
\begin{equation*} \sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}=\sqrt{\frac{\gamma+1}{2\gamma}\left[\left(\frac{p_2}{p_1}\right)+\left(\frac{\gamma-1}{\gamma+1}\right)\right]} \end{equation*}\\
Term II:
\begin{equation*} \left[1-\dfrac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]=\frac{1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)} \end{equation*}\\
\noindent With the rewritten terms I and II implemented, Eqn. \ref{eq:unsteady:up:b} becomes\\
\begin{equation} u_p=\frac{a_1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\sqrt{\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}} \label{eq:unsteady:up} \end{equation}\\
\noindent Since the region behind the moving shock is region 2, the induced flow Mach number is obtained as\\
\begin{equation*} M_p=\frac{u_p}{a_2}=\frac{u_p}{a_1}\frac{a_1}{a_2}=\frac{u_p}{a_1}\sqrt{\frac{\gamma R T_1}{\gamma R T_2}}=\frac{u_p}{a_1}\sqrt{\frac{T_1}{T_2}} \end{equation*}\\
\noindent With $up/a_1$ from Eqn. \ref{eq:unsteady:up} and $T_1/T_2$ from Eqn. \ref{eq:unsteady:temperature:ratio}\\
\begin{equation} M_p=\frac{1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\left(\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}\right)^{1/2} \left(\frac{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)+\left(\dfrac{p_2}{p_1}\right)^2}\right)^{1/2} \label{eq:unsteady:Mp} \end{equation}\\
\noindent There is a theoretical upper limit for the induced Mach number $M_p$\\
\begin{equation*} \lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)=\sqrt{\frac{2}{\gamma(\gamma-1)}} \end{equation*}\\
\noindent As can be seen, at the upper limit the induced Mach number is a function of $\gamma$ and for air ($\gamma=1.4$) we get\\
\begin{equation*} \lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)\simeq 1.89 \end{equation*}\\
