Moving shock waves: Difference between revisions
Created page with "Category:Compressible flow Category:Unsteady waves Category:Inviscid flow __TOC__ \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 =..." |
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\lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)\simeq 1.89 | \lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)\simeq 1.89 | ||
\end{equation*}\\ | \end{equation*}\\ | ||
\section{Shock Wave Reflection} | |||
\noindent When the incident shock wave reaches the wall, a shock propagating in the opposite direction is generated with a shock strength such that the velocity of the induced flow behind the incident shock is reduced to zero. The flow can not go through the wall and thus the velocity must be zero in the vicinity of the wall. The properties of the incident shock wave are directly related to the pressure ratio over the shock wave. Therefore, it would be convenient to have a relation between the reflected shock wave and incident shock wave. | |||
\vspace{1cm} | |||
\begin{figure}[ht!] | |||
\begin{center} | |||
\includegraphics[]{figures/standalone-figures/Chapter07/pdf/shock-reflection.pdf} | |||
\caption{shock reflection at solid wall (located at $x$=4.0)} | |||
\label{fig:reflection} | |||
\end{center} | |||
\end{figure} | |||
\subsection{The Incident Shock Wave} | |||
\noindent The pressure ratio over the incident shock in Fig.~\ref{fig:reflection} can be obtained as \\ | |||
\begin{equation} | |||
\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}\left(M_s^2-1\right) | |||
\label{eq:incident:pr} | |||
\end{equation}\\ | |||
\noindent where $M_s$ is the wave Mach number, which is calculated as \\ | |||
\begin{equation} | |||
M_s=\frac{W}{a_1} | |||
\label{eq:incident:Mach:def} | |||
\end{equation}\\ | |||
\noindent In Eqn.~\ref{eq:incident:Mach:def}, $W$ is the speed with which the incident shock wave travels into region 1 and $a_1$ is the speed of sound in region 1 (see Fig.~\ref{fig:reflection}). | |||
\noindent Solving Eqn.~\ref{eq:incident:pr} for $M_s$, we get \\ | |||
\begin{equation} | |||
M_s=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_2}{p_1}-1\right)+1} | |||
\label{eq:incident:Mach} | |||
\end{equation}\\ | |||
\noindent Anderson derives the relations for calculation of the ratio $T_2/T_1$ \\ | |||
\begin{equation} | |||
\frac{T_2}{T_1}=\frac{p_2}{p_1}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}\right) | |||
\label{eq:incident:tr} | |||
\end{equation}\\ | |||
\noindent From Eqn.~\ref{eq:incident:tr} it is easy to get the corresponding relation for $\rho_2/\rho_1$\\ | |||
\begin{equation} | |||
\frac{\rho_2}{\rho_1}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}} | |||
\label{eq:incident:rr} | |||
\end{equation}\\ | |||
\noindent Anderson also shows how to obtain the induced velocity, $u_p$, behind the incident shock wave, {\emph{i.e.}} the velocity in region 2 (see Fig.~\ref{fig:reflection}).\\ | |||
\begin{equation} | |||
u_p=W\left(1-\frac{\rho_1}{\rho_2}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right) | |||
\label{eq:incident:up} | |||
\end{equation}\\ | |||
\subsection{The Reflected Shock Wave} | |||
\noindent The pressure ratio over the reflected shock can be obtained from Eqn.~\ref{eq:incident:pr} by analogy\\ | |||
\begin{equation} | |||
\frac{p_5}{p_2}=1+\frac{2\gamma}{\gamma+1}\left(M_r^2-1\right) | |||
\label{eq:reflected:pr} | |||
\end{equation}\\ | |||
\noindent where $M_r$ is the Mach number of the reflected shock wave defined as\\ | |||
\begin{equation} | |||
M_r=\frac{W_r+u_p}{a_2} | |||
\label{eq:reflected:Mach:def} | |||
\end{equation}\\ | |||
\noindent where $W_r$ is the speed of the reflected shock wave and $a_2$ is the speed of sound in region 2 (see Fig.~\ref{fig:reflection}).\\ | |||
\noindent Solving Eqn.~\ref{eq:reflected:pr} for $M_r$ gives\\ | |||
\begin{equation} | |||
M_r=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_5}{p_2}-1\right)+1} | |||
\label{eq:reflected:Mach} | |||
\end{equation}\\ | |||
\noindent The ratios $T_5/T_2$ and $\rho_5/\rho_2$ can be obtained from Eqns.~\ref{eq:incident:tr} and \ref{eq:incident:rr} by analogy\\ | |||
\begin{equation} | |||
\frac{T_5}{T_2}=\frac{p_5}{p_2}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}\right) | |||
\label{eq:reflected:tr} | |||
\end{equation}\\ | |||
\begin{equation} | |||
\frac{\rho_5}{\rho_2}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}} | |||
\label{eq:reflected:rr} | |||
\end{equation}\\ | |||
\noindent The velocity in region 2 which is the same as the induced flow velocity behind the incident shock wave can be obtained as\\ | |||
\begin{equation} | |||
u_p=W_r\left(\frac{\rho_5}{\rho_2}-1\right)=M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right) | |||
\label{eq:reflected:up} | |||
\end{equation}\\ | |||
\subsection{Reflected Shock Relation} | |||
\noindent With the relations for the incident shock wave and reflected shock wave defined, we now have the tools to derive a relation between the incident and reflected shock waves. The induced flow velocity $u_p$ calculated using the relation obtained for the incident shock wave must of course be the same as when calculated using reflected wave properties, {\emph{i.e.}} the result of Eqn.~\ref{eq:incident:up} is identical to that of Eqn.~\ref{eq:reflected:up}\\ | |||
\begin{equation} | |||
M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right) | |||
\label{eq:relation:a} | |||
\end{equation}\\ | |||
\noindent rewriting gives \\ | |||
\begin{equation} | |||
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \frac{a_1}{a_2} | |||
\label{eq:relation:b} | |||
\end{equation}\\ | |||
\noindent Assuming calorically perfect gas gives $a=\sqrt{\gamma RT}$ and thus\\ | |||
\begin{equation} | |||
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \sqrt{\frac{T_1}{T_2}} | |||
\label{eq:relation:c} | |||
\end{equation}\\ | |||
\noindent Let's first look at the term on the left hand side of Eqn.~\ref{eq:relation:c}\\ | |||
\begin{equation*} | |||
M_r \left(1-\frac{\rho_2}{\rho_5}\right) | |||
\end{equation*}\\ | |||
\noindent Using the $\rho_5/\rho_2$ and $p_2/p_5$ from Eqns.~\ref{eq:reflected:rr} and~\ref{eq:reflected:pr} and simplifying gives\\ | |||
\begin{equation} | |||
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_r^2-1}{M_r}\right) | |||
\label{eq:relation:d} | |||
\end{equation}\\ | |||
\noindent Using the same approach on the corresponding term for the incident shock wave on the right hand side of Eqn.~\ref{eq:relation:c} gives\\ | |||
\begin{equation} | |||
M_s \left(1-\frac{\rho_1}{\rho_2}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_s^2-1}{M_s}\right) | |||
\label{eq:relation:e} | |||
\end{equation}\\ | |||
\noindent Now, inserting~\ref{eq:relation:d} and~\ref{eq:relation:e} in Eqn.~\ref{eq:relation:c} gives\\ | |||
\begin{equation} | |||
\left(\frac{2}{\gamma+1}\right)\left(\frac{M_r^2-1}{M_r}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_s^2-1}{M_s}\right)\sqrt{\frac{T_1}{T_2}} | |||
\label{eq:relation:f} | |||
\end{equation}\\ | |||
\noindent Simplifying and inverting gives\\ | |||
\begin{equation} | |||
\left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{\frac{T_2}{T_1}} | |||
\label{eq:relation:g} | |||
\end{equation}\\ | |||
\noindent The rightmost term in Eqn.~\ref{eq:relation:g} ($\sqrt{T_2/T_1}$) needs to be rewritten. Inserting~\ref{eq:incident:pr} in~\ref{eq:incident:tr} and expanding all terms gives\\ | |||
\begin{align*} | |||
\frac{T_2}{T_1} & =\frac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)+2\gamma(\gamma-1)M_s^2(M_s^2-1)}{(\gamma+1)^2M_s^2} = \\ | |||
& \\ | |||
& =\frac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)}{(\gamma+1)^2M_s^2}+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\gamma = \\ | |||
& \\ | |||
& =\dfrac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)-(2(\gamma-1)(M_s^2-1))}{(\gamma+1)^2M_s^2}+\nonumber\\ | |||
& \dfrac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\dfrac{1}{M_s^2}\right) | |||
\end{align*}\\ | |||
\noindent Finally we end up with the following relation\\ | |||
\begin{equation} | |||
\frac{T_2}{T_1}=1+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\frac{1}{M_s^2}\right) | |||
\label{eq:relation:tr} | |||
\end{equation}\\ | |||
\noindent The temperature ratio over the incident shock wave is now totally defined by the incident Mach number $M_s$ and the ratio of specific heats $\gamma$. With~\ref{eq:relation:tr} in~\ref{eq:relation:g} we get the sought relation between the reflected and incident Mach numbers.\\ | |||
\begin{equation} | |||
\left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{1+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\frac{1}{M_s^2}\right)} | |||
\label{eq:relation:final} | |||
\end{equation}\\ | |||
\noindent It should be noted that Eqn.~\ref{eq:relation:final} is valid for calorically perfect gases only. | |||
Revision as of 15:34, 16 March 2026
\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}
%\newpage
\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*}\\
\section{Shock Wave Reflection}
\noindent When the incident shock wave reaches the wall, a shock propagating in the opposite direction is generated with a shock strength such that the velocity of the induced flow behind the incident shock is reduced to zero. The flow can not go through the wall and thus the velocity must be zero in the vicinity of the wall. The properties of the incident shock wave are directly related to the pressure ratio over the shock wave. Therefore, it would be convenient to have a relation between the reflected shock wave and incident shock wave.
\vspace{1cm}
\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter07/pdf/shock-reflection.pdf} \caption{shock reflection at solid wall (located at $x$=4.0)} \label{fig:reflection} \end{center} \end{figure}
\subsection{The Incident Shock Wave}
\noindent The pressure ratio over the incident shock in Fig.~\ref{fig:reflection} can be obtained as \\
\begin{equation} \frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}\left(M_s^2-1\right) \label{eq:incident:pr} \end{equation}\\
\noindent where $M_s$ is the wave Mach number, which is calculated as \\
\begin{equation} M_s=\frac{W}{a_1} \label{eq:incident:Mach:def} \end{equation}\\
\noindent In Eqn.~\ref{eq:incident:Mach:def}, $W$ is the speed with which the incident shock wave travels into region 1 and $a_1$ is the speed of sound in region 1 (see Fig.~\ref{fig:reflection}).
\noindent Solving Eqn.~\ref{eq:incident:pr} for $M_s$, we get \\
\begin{equation} M_s=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_2}{p_1}-1\right)+1} \label{eq:incident:Mach} \end{equation}\\
\noindent Anderson derives the relations for calculation of the ratio $T_2/T_1$ \\
\begin{equation} \frac{T_2}{T_1}=\frac{p_2}{p_1}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}\right) \label{eq:incident:tr} \end{equation}\\
\noindent From Eqn.~\ref{eq:incident:tr} it is easy to get the corresponding relation for $\rho_2/\rho_1$\\
\begin{equation} \frac{\rho_2}{\rho_1}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}} \label{eq:incident:rr} \end{equation}\\
\noindent Anderson also shows how to obtain the induced velocity, $u_p$, behind the incident shock wave, {\emph{i.e.}} the velocity in region 2 (see Fig.~\ref{fig:reflection}).\\
\begin{equation} u_p=W\left(1-\frac{\rho_1}{\rho_2}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right) \label{eq:incident:up} \end{equation}\\
\subsection{The Reflected Shock Wave}
\noindent The pressure ratio over the reflected shock can be obtained from Eqn.~\ref{eq:incident:pr} by analogy\\
\begin{equation} \frac{p_5}{p_2}=1+\frac{2\gamma}{\gamma+1}\left(M_r^2-1\right) \label{eq:reflected:pr} \end{equation}\\
\noindent where $M_r$ is the Mach number of the reflected shock wave defined as\\
\begin{equation} M_r=\frac{W_r+u_p}{a_2} \label{eq:reflected:Mach:def} \end{equation}\\
\noindent where $W_r$ is the speed of the reflected shock wave and $a_2$ is the speed of sound in region 2 (see Fig.~\ref{fig:reflection}).\\
\noindent Solving Eqn.~\ref{eq:reflected:pr} for $M_r$ gives\\
\begin{equation} M_r=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_5}{p_2}-1\right)+1} \label{eq:reflected:Mach} \end{equation}\\
\noindent The ratios $T_5/T_2$ and $\rho_5/\rho_2$ can be obtained from Eqns.~\ref{eq:incident:tr} and \ref{eq:incident:rr} by analogy\\
\begin{equation} \frac{T_5}{T_2}=\frac{p_5}{p_2}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}\right) \label{eq:reflected:tr} \end{equation}\\
\begin{equation} \frac{\rho_5}{\rho_2}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}} \label{eq:reflected:rr} \end{equation}\\
\noindent The velocity in region 2 which is the same as the induced flow velocity behind the incident shock wave can be obtained as\\
\begin{equation} u_p=W_r\left(\frac{\rho_5}{\rho_2}-1\right)=M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right) \label{eq:reflected:up} \end{equation}\\
\subsection{Reflected Shock Relation}
\noindent With the relations for the incident shock wave and reflected shock wave defined, we now have the tools to derive a relation between the incident and reflected shock waves. The induced flow velocity $u_p$ calculated using the relation obtained for the incident shock wave must of course be the same as when calculated using reflected wave properties, {\emph{i.e.}} the result of Eqn.~\ref{eq:incident:up} is identical to that of Eqn.~\ref{eq:reflected:up}\\
\begin{equation} M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right) \label{eq:relation:a} \end{equation}\\
\noindent rewriting gives \\
\begin{equation} M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \frac{a_1}{a_2} \label{eq:relation:b} \end{equation}\\
\noindent Assuming calorically perfect gas gives $a=\sqrt{\gamma RT}$ and thus\\
\begin{equation} M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \sqrt{\frac{T_1}{T_2}} \label{eq:relation:c} \end{equation}\\
\noindent Let's first look at the term on the left hand side of Eqn.~\ref{eq:relation:c}\\
\begin{equation*} M_r \left(1-\frac{\rho_2}{\rho_5}\right) \end{equation*}\\
\noindent Using the $\rho_5/\rho_2$ and $p_2/p_5$ from Eqns.~\ref{eq:reflected:rr} and~\ref{eq:reflected:pr} and simplifying gives\\
\begin{equation} M_r \left(1-\frac{\rho_2}{\rho_5}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_r^2-1}{M_r}\right) \label{eq:relation:d} \end{equation}\\
\noindent Using the same approach on the corresponding term for the incident shock wave on the right hand side of Eqn.~\ref{eq:relation:c} gives\\
\begin{equation} M_s \left(1-\frac{\rho_1}{\rho_2}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_s^2-1}{M_s}\right) \label{eq:relation:e} \end{equation}\\
\noindent Now, inserting~\ref{eq:relation:d} and~\ref{eq:relation:e} in Eqn.~\ref{eq:relation:c} gives\\
\begin{equation} \left(\frac{2}{\gamma+1}\right)\left(\frac{M_r^2-1}{M_r}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_s^2-1}{M_s}\right)\sqrt{\frac{T_1}{T_2}} \label{eq:relation:f} \end{equation}\\
\noindent Simplifying and inverting gives\\
\begin{equation} \left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{\frac{T_2}{T_1}} \label{eq:relation:g} \end{equation}\\
\noindent The rightmost term in Eqn.~\ref{eq:relation:g} ($\sqrt{T_2/T_1}$) needs to be rewritten. Inserting~\ref{eq:incident:pr} in~\ref{eq:incident:tr} and expanding all terms gives\\
\begin{align*} \frac{T_2}{T_1} & =\frac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)+2\gamma(\gamma-1)M_s^2(M_s^2-1)}{(\gamma+1)^2M_s^2} = \\
& \\
& =\frac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)}{(\gamma+1)^2M_s^2}+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\gamma = \\
& \\
& =\dfrac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)-(2(\gamma-1)(M_s^2-1))}{(\gamma+1)^2M_s^2}+\nonumber\\
& \dfrac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\dfrac{1}{M_s^2}\right)
\end{align*}\\
\noindent Finally we end up with the following relation\\
\begin{equation} \frac{T_2}{T_1}=1+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\frac{1}{M_s^2}\right) \label{eq:relation:tr} \end{equation}\\
\noindent The temperature ratio over the incident shock wave is now totally defined by the incident Mach number $M_s$ and the ratio of specific heats $\gamma$. With~\ref{eq:relation:tr} in~\ref{eq:relation:g} we get the sought relation between the reflected and incident Mach numbers.\\
\begin{equation} \left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{1+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\frac{1}{M_s^2}\right)} \label{eq:relation:final} \end{equation}\\
\noindent It should be noted that Eqn.~\ref{eq:relation:final} is valid for calorically perfect gases only.