View source for Moving expansion waves

[[Category:Compressible flow]]
[[Category:Unsteady waves]]
[[Category:Inviscid flow]]
<noinclude>
[[Category:Compressible flow:Topic]]
</noinclude>

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==== Moving Expansion Waves ====

The expansion wave propagation into the driver section in a shock tube can be described using characteristic lines.

<!--
\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter07/pdf/expansion-region.pdf}
\caption{Expansion fan centered at $(x,t)=(0.0,0.0)$}
\label{fig:characteristics}
\end{center}
\end{figure}
-->

The expansion is propagating into stagnant fluid in region four (the driver section), which means that the flow properties ahead of the expansion wave are constant.

<math display="block">
J^+_a=J^+_b
</math>

<math>J^+</math> invariants constant along <math>C^+</math> characteristics

<math display="block">
J^+_a=J^+_c=J^+_e
</math>

<math display="block">
J^+_b=J^+_d=J^+_f
</math>

Since <math>J^+_a=J^+_b</math> this also implies <math>J^+_e=J^+_f</math>. In fact, since the flow properties ahead of the expansion are constant, all <math>C^+</math> lines will have the same <math>J^+</math> value.

<math>J^-</math> invariants constant along <math>C^-</math> characteristics

<math display="block">
J^-_c=J^-_d
</math>

<math display="block">
J^-_e=J^-_f
</math>

<math display="block">
\left.
\begin{aligned}
&u_e=\frac{1}{2}(J^+_e+J^-_e)\\
&u_f=\frac{1}{2}(J^+_f+J^-_f)\\
&J^-_e=J^-_f\\
&J^+_e=J^+_f
\end{aligned}
\right\}\Rightarrow u_e=u_f \Rightarrow a_e=a_f
</math>

Due to the fact the <math>J^+</math> is constant in the entire expansion region, <math>u</math> and <math>a</math> will be constant along each <math>C^-</math> line.

The constant <math>J^+</math> value can be used to obtain relations for the variation of flow properties through the expansion region. Evaluation of the <math>J^+</math> invariant at any position within the expansion region should give the same value as in region 4.

<math display="block">
u+\frac{2a}{\gamma-1}=u_4+\frac{2a_4}{\gamma-1}=0+\frac{2a_4}{\gamma-1}
</math>

and thus

<math display="block">
\frac{a}{a_4}=1-\frac{\gamma-1}{2}\left(\frac{u}{a_4}\right)
</math>

Eqn. \ref{eq:expansion:a} and <math>a=\sqrt{\gamma RT}</math> gives

<math display="block">
\frac{T}{T_4}=\left[1-\frac{\gamma-1}{2}\left(\frac{u}{a_4}\right)\right]^2
</math>

Using isentropic relations, we can get pressure ratio and density ratio

<math display="block">
\frac{p}{p_4}=\left[1-\frac{\gamma-1}{2}\left(\frac{u}{a_4}\right)\right]^{2\gamma/(\gamma-1)}
</math>

<math display="block">
\frac{\rho}{\rho_4}=\left[1-\frac{\gamma-1}{2}\left(\frac{u}{a_4}\right)\right]^{2/(\gamma-1)}
</math>