Shock-tube relations - Revision history

Nian at 13:37, 1 April 2026

2026-04-01T13:37:14Z

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Nian at 13:33, 1 April 2026

2026-04-01T13:33:47Z

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	[[Category:Compressible flow]]		<!--
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	\begin{figure}[ht!]		\begin{figure}[ht!]
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	From the analysis of the incident shock, we have a relation for the induced flow behind the shock		From the analysis of the incident shock, we have a relation for the induced flow behind the shock

	<math ~~display="block"~~>		{{NumEqn\|<math>
	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}		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}
	</math>		</math>}}

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

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\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)}		\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)}
	</math>		</math>}}

	Solving for <math>u_3</math> gives		Solving for <math>u_3</math> gives

	<math ~~display="block"~~>		{{NumEqn\|<math>
	u_3=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_3}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right]		u_3=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_3}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right]
	</math>		</math>}}

	There is no change in pressure or velocity over the contact surface, which means <math>u_2=u_3</math> and <math>p_2=p_3</math>.		There is no change in pressure or velocity over the contact surface, which means <math>u_2=u_3</math> and <math>p_2=p_3</math>.

	<math ~~display="block"~~>		{{NumEqn\|<math>
	u_2=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_2}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right]		u_2=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_2}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right]
	</math>		</math>}}

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

	<math ~~display="block"~~>		{{NumEqn\|<math>
	\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)}		\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)}
	</math>		</math>}}

Nian at 19:57, 20 March 2026

2026-03-20T19:57:57Z

← Older revision		Revision as of 19:57, 20 March 2026
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	[[Category:Unsteady waves]]		[[Category:Unsteady waves]]
	[[Category:Inviscid flow]]		[[Category:Inviscid flow]]
			<noinclude>
			[[Category:Compressible flow:Topic]]
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Nian at 16:38, 16 March 2026

2026-03-16T16:38:43Z

← Older revision		Revision as of 16:38, 16 March 2026
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	~~\subsection{The Shock Tube Relations}~~		<!--

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

	~~\begin{equation}~~		<math display="block">
	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}		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}~~		</math>
	~~\end{equation}\\~~

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

	~~\begin{equation}~~		<math display="block">
	\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)}		\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}~~		</math>
	~~\end{equation}\\~~

	~~\noindent~~ Solving for $u_3$ gives\\		Solving for <math>u_3</math> gives

	~~\begin{equation}~~		<math display="block">
	u_3=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_3}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right]		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}~~		</math>
	~~\end{equation}\\~~

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

	~~\begin{equation}~~		<math display="block">
	u_2=\frac{2a_4}{\gamma_4-1}\left[1-\left(\frac{p_2}{p_4}\right)^{(\gamma_4-1)/(2\gamma_4)}\right]		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}~~		</math>
	~~\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\\		Now, we have two ways of calculating <math>u_2</math>. Setting Eqn. \ref{eq:shocktube:up:a} equal to Eqn. \ref{eq:shocktube:up:d} leads to the shock tube relation

	~~\begin{equation}~~		<math display="block">
	\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)}		\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}~~		</math>
	~~\end{equation}~~

Nian: Created page with "Category:Compressible flow Category:Unsteady waves Category:Inviscid flow TOC \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=\f..."

2026-03-16T15:35:14Z

Created page with "Category:Compressible flow Category:Unsteady waves Category:Inviscid flow __TOC__ \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=\f..."

New page

[[Category:Compressible flow]]
[[Category:Unsteady waves]]
[[Category:Inviscid flow]]

__TOC__

\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}