Shock-tube relations

\subsection{The Shock Tube Relations}

\begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter07/pdf/shock-tube.pdf} \caption{traveling waves in a shock tube} \label{fig:shocktube} \end{center} \end{figure}

\noindent From the analysis of the incident shock, we have a relation for the induced flow behind the shock\\

\begin{equation} u_2=u_p=\frac{a_1}{\gamma_1}\left(\frac{p_2}{p_1}-1\right)\left(\frac{\left(\dfrac{2\gamma_1}{\gamma_1+1}\right)}{\left(\dfrac{\gamma_1-1}{\gamma_1+1}\right)+\left(\dfrac{p_2}{p_1}\right)}\right)^{1/2} \label{eq:shocktube:up:a} \end{equation}\\

\noindent The velocity in region 3 can be obtained from the expansion relations\\

\begin{equation} \frac{p_3}{p_4}=\left[1-\frac{\gamma_4-1}{2}\left(\frac{u_3}{a_4}\right)\right]^{2\gamma_4/(\gamma_4-1)} \label{eq:shocktube:up:b} \end{equation}\\

\noindent Solving for $u_3$ gives\\

\begin{equation} u_3=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_3}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right] \label{eq:shocktube:up:c} \end{equation}\\

\noindent There is no change in pressure or velocity over the contact surface, which means $u_2=u_3$ and $p_2=p_3$.\\

\begin{equation} u_2=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_2}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right] \label{eq:shocktube:up:d} \end{equation}\\

\noindent Now, we have two ways of calculating $u_2$. Setting Eqn. \ref{eq:shocktube:up:a} equal to Eqn. \ref{eq:shocktube:up:d} leads to the shock tube relation\\

\begin{equation} \frac{p_4}{p_1}=\frac{p_2}{p_1}\left\{ 1 -\frac{(\gamma_4-1)(a_1/a_4)(p_2/p_1-1)}{\sqrt{2\gamma_1\left[2\gamma_1+(\gamma_1+1)(p_2/p_1-1)\right]}}\right\}^{-2\gamma_4/(\gamma_4-1)} \label{eq:shocktube:relation} \end{equation}