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\section{Prandtl-Meyer Expansion Waves}
=== Prandtl-Meyer Expansion Waves ===


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\noindent A single Mach wave has a insignificant effect on the flow passing it but an expansion region constitutes an infinite number of Mach waves and the integrated effect is significant. The net turning of the flow by a single Mach wave is depicted schematically in Fig. \ref{fig:machwave}. It can be shown geometrically that\\
A single Mach wave has a insignificant effect on the flow passing it but an expansion region constitutes an infinite number of Mach waves and the integrated effect is significant. The net turning of the flow by a single Mach wave is depicted schematically in Fig. \ref{fig:machwave}. It can be shown geometrically that


\begin{equation}
<math display="block">
d\theta=\sqrt{M^2-1}\frac{dV}{V}
d\theta=\sqrt{M^2-1}\frac{dV}{V}
\label{eq:mach:turning}
</math>
\end{equation}\\


\noindent Since Eqn. \ref{eq:mach:turning} is derived from the flow turning geometry with the assumption that the net flow tuning is small, it is valid for all gas models.\\
Since Eqn. \ref{eq:mach:turning} is derived from the flow turning geometry with the assumption that the net flow tuning is small, it is valid for all gas models.


\noindent To get the integrated effect of all Mach waves in the expansion region, we integrate Eqn. \ref{eq:mach:turning} over the expansion region
To get the integrated effect of all Mach waves in the expansion region, we integrate Eqn. \ref{eq:mach:turning} over the expansion region


\begin{equation}
<math display="block">
\int_{\theta_1}^{\theta_2}d\theta=\int_{M_1}^{M_2}\sqrt{M^2-1}\frac{dV}{V}
\int_{\theta_1}^{\theta_2}d\theta=\int_{M_1}^{M_2}\sqrt{M^2-1}\frac{dV}{V}
\label{eq:mach:turning:b}
</math>
\end{equation}\\


\noindent To be able to do the integration, we need to rewrite it\\
To be able to do the integration, we need to rewrite it


\[V=Ma \Rightarrow \ln V=\ln M + \ln a\]\\
<math display="block">
V=Ma \Rightarrow \ln V=\ln M + \ln a
</math>


\noindent Differentiate to get\\
Differentiate to get


\begin{equation}
<math display="block">
\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}
\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}
\label{eq:mach:turning:c}
</math>
\end{equation}\\


\noindent Each Mach wave is isentropic and thus the expansion is an isentropic process, which means that we can use the adiabatic energy equation\\
Each Mach wave is isentropic and thus the expansion is an isentropic process, which means that we can use the adiabatic energy equation


\begin{equation}
<math display="block">
\frac{T_o}{T}=1+\frac{\gamma-1}{2}M^2
\frac{T_o}{T}=1+\frac{\gamma-1}{2}M^2
\label{eq:adiatbatic:energy}
</math>
\end{equation}\\


\noindent For a calorically perfect gas $a=\sqrt{\gamma RT}$ and $a_o=\sqrt{\gamma RT_o}$ and thus\\
For a calorically perfect gas <math>a=\sqrt{\gamma RT}</math> and <math>a_o=\sqrt{\gamma RT_o}</math> and thus


\[\frac{T_o}{T}=\left(\frac{a_o}{a}\right)^2\Rightarrow\]\\
<math display="block">
\frac{T_o}{T}=\left(\frac{a_o}{a}\right)^2\Rightarrow
</math>


\begin{equation}
<math display="block">
\left(\frac{a_o}{a}\right)^2=1+\frac{\gamma-1}{2}M^2
\left(\frac{a_o}{a}\right)^2=1+\frac{\gamma-1}{2}M^2
\label{eq:adiatbatic:energy}
</math>
\end{equation}\\


\noindent Solve for $a$ gives\\
Solve for <math>a</math> gives


\begin{equation}
<math display="block">
a=a_o \left(1+\frac{\gamma-1}{2}M^2\right)^{-1/2}
a=a_o \left(1+\frac{\gamma-1}{2}M^2\right)^{-1/2}
\label{eq:adiatbatic:energy:b}
</math>
\end{equation}\\




\noindent Differentiate Eqn \ref{eq:adiatbatic:energy:b} to get\\
Differentiate Eqn \ref{eq:adiatbatic:energy:b} to get


\begin{equation}
<math display="block">
da=a_o\left(-\frac{1}{2}\right)\left(\frac{\gamma-1}{2}\right)2M\left(1+\frac{\gamma-1}{2}M^2\right)^{-3/2}dM
da=a_o\left(-\frac{1}{2}\right)\left(\frac{\gamma-1}{2}\right)2M\left(1+\frac{\gamma-1}{2}M^2\right)^{-3/2}dM
\label{eq:adiatbatic:energy:c}
</math>
\end{equation}\\


\noindent Eqn. \ref{eq:adiatbatic:energy:b} in Eqn. \ref{eq:adiatbatic:energy:c} gives\\
Eqn. \ref{eq:adiatbatic:energy:b} in Eqn. \ref{eq:adiatbatic:energy:c} gives


\begin{equation}
<math display="block">
\frac{da}{a}=-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM
\frac{da}{a}=-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM
\label{eq:adiatbatic:energy:d}
</math>
\end{equation}\\


\noindent From Eqn. \ref{eq:mach:turning:c}, we have\\
From Eqn. \ref{eq:mach:turning:c}, we have


\[\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}\]\\
<math display="block">
\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}
</math>


\noindent With $da/a$ from Eqn. \ref{eq:adiatbatic:energy:d}, we get\\
With <math>da/a</math> from Eqn. \ref{eq:adiatbatic:energy:d}, we get


\[\frac{dV}{V}=\frac{dM}{M}-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM\]\\
<math display="block">
\frac{dV}{V}=\frac{dM}{M}-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM
</math>




\[\frac{dV}{V}=dM\left[\frac{\left(1+\dfrac{\gamma-1}{2}M^2\right)-\left(\dfrac{\gamma-1}{2}\right)M^2}{\left(1+\dfrac{\gamma-1}{2}M^2\right)M}\right]=\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M}\]\\
<math display="block">
\frac{dV}{V}=dM\left[\frac{\left(1+\dfrac{\gamma-1}{2}M^2\right)-\left(\dfrac{\gamma-1}{2}\right)M^2}{\left(1+\dfrac{\gamma-1}{2}M^2\right)M}\right]=\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M}
</math>


\noindent Now, insert $dV/V$ in Eqn. \ref{eq:mach:turning:b} to get\\
Now, insert <math>dV/V</math> in Eqn. \ref{eq:mach:turning:b} to get


\begin{equation}
<math display="block">
\int_{\theta_1}^{\theta_2}d\theta=\int_{M_1}^{M_2}\sqrt{M^2-1}\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M}
\int_{\theta_1}^{\theta_2}d\theta=\int_{M_1}^{M_2}\sqrt{M^2-1}\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M}
\label{eq:mach:turning:c}
</math>
\end{equation}\\


\noindent The integral on the right hand side of Eqn. \ref{eq:mach:turning:c} is the Prandtl-Meyer function, which is usually denoted $\nu$. The Prandtl-Meyer function evaluated for Mach number $M$ becomes\\
The integral on the right hand side of Eqn. \ref{eq:mach:turning:c} is the Prandtl-Meyer function, which is usually denoted <math>\nu</math>. The Prandtl-Meyer function evaluated for Mach number <math>M</math> becomes


\begin{equation}
<math display="block">
\nu(M)=\sqrt{\frac{\gamma+1}{\gamma-1}}\tan^{-1}\sqrt{\frac{\gamma-1}{\gamma+1}(M^2-1)}-\tan^{-1}\sqrt{M^2-1}
\nu(M)=\sqrt{\frac{\gamma+1}{\gamma-1}}\tan^{-1}\sqrt{\frac{\gamma-1}{\gamma+1}(M^2-1)}-\tan^{-1}\sqrt{M^2-1}
\label{eq:prandtl:meyer}
</math>
\end{equation}\\


\noindent and thus the net turning of the flow can be calculated as\\
and thus the net turning of the flow can be calculated as


\begin{equation}
<math display="block">
\theta_2-\theta_1=\nu(M_2)-\nu(M_1)
\theta_2-\theta_1=\nu(M_2)-\nu(M_1)
\label{eq:prandtl:meyer:c}
</math>
\end{equation}\\


\newpage
==== Solving Problems using the Prandtl Meyer Function ====


\subsection{Solving Problems using the Prandtl Meyer Function}
A typical problem is one where we know the net flow turning and the upstream flow conditions and want to calculate the flow conditions downstream of the expansion region. An example of such a problem is given in Fig. \ref{fig:expansion:corner}.


\noindent A typical problem is one where we know the net flow turning and the upstream flow conditions and want to calculate the flow conditions downstream of the expansion region. An example of such a problem is given in Fig. \ref{fig:expansion:corner}.\\
A problem of that type can be solved as follows:
 
\noindent A problem of that type can be solved as follows:\\
 
\begin{enumerate}
\item Calculate $\nu{M_1}$ using Eqn. \ref{eq:prandtl:meyer} or tabulated values
\item Calculate $\nu(M_2)$ as $\nu(M_2)=\theta_2-\theta_1+\nu(M_1)$
\item Calculate $M_2$ from the known $\nu{M_2}$ using Eqn. \ref{eq:prandtl:meyer} or tabulated values
\end{enumerate}
 
\vspace*{1cm}


# Calculate <math>\nu{M_1}</math> using Eqn. \ref{eq:prandtl:meyer} or tabulated values
# Calculate <math>\nu(M_2)</math> as <math>\nu(M_2)=\theta_2-\theta_1+\nu(M_1)</math>
# Calculate <math>M_2</math> from the known <math>\nu{M_2}</math> using Eqn. \ref{eq:prandtl:meyer} or tabulated values


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The aim is to derive relations of temperature, pressure and density over an expansion wave. Total temperature upstream and downstream of the expansion wave is calculated as


\noindent The aim is to derive relations of temperature, pressure and density over an expansion wave. Total temperature upstream and downstream of the expansion wave is calculated as\\
<math display="block">
 
\begin{equation}
\frac{T_{o_1}}{T_1}=1+\frac{\gamma-1}{2}M_1^2
\frac{T_{o_1}}{T_1}=1+\frac{\gamma-1}{2}M_1^2
\label{eq:toa}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\frac{T_{o_2}}{T_2}=1+\frac{\gamma-1}{2}M_2^2
\frac{T_{o_2}}{T_2}=1+\frac{\gamma-1}{2}M_2^2
\label{eq:tob}
</math>
\end{equation}\\


\noindent The temperature ratio over the expansion wave may now be calculated as\\
The temperature ratio over the expansion wave may now be calculated as


\[\frac{T_2}{T_1}=\frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\]\\
<math display="block">
\frac{T_2}{T_1}=\frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}
</math>


\noindent The expansion is isentropic and thus total temperature and both total pressure are unaffected by the expansion. Therefore, $T_{o_1}=T_{o_2}$ and thus\\
The expansion is isentropic and thus total temperature and both total pressure are unaffected by the expansion. Therefore, <math>T_{o_1}=T_{o_2}</math> and thus


\begin{equation}
<math display="block">
\frac{T_2}{T_1}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}
\frac{T_2}{T_1}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}
\label{eq:tr}
</math>
\end{equation}\\


\noindent The pressure and density ratios can be obtained from Eqn. \ref{eq:tr} using the isentropic relations\\
The pressure and density ratios can be obtained from Eqn. \ref{eq:tr} using the isentropic relations


\begin{equation}
<math display="block">
\frac{p_2}{p_1}=\left[\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\right]^{\gamma/(\gamma-1)}
\frac{p_2}{p_1}=\left[\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\right]^{\gamma/(\gamma-1)}
\label{eq:pr}
</math>
\end{equation}\\


\begin{equation}
<math display="block">
\frac{\rho_2}{\rho_1}=\left[\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\right]^{1/(\gamma-1)}
\frac{\rho_2}{\rho_1}=\left[\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\right]^{1/(\gamma-1)}
\label{eq:pr}
</math>
\end{equation}


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