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Nian: Created page with "Category:Compressible flow Category:Two-dimensional flow Category:Inviscid flow TOC \section{Prandtl-Meyer Expansion Waves} \begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter05/pdf/Mach-wave.pdf} \caption{Mach wave flow turning} \label{fig:machwave} \end{center} \end{figure} \noindent A single Mach wave has a insignificant effect on the flow passing it but an expansion region constitutes an infinite number of Mach..."

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Created page with "Category:Compressible flow Category:Two-dimensional flow Category:Inviscid flow __TOC__ \section{Prandtl-Meyer Expansion Waves} \begin{figure}[ht!] \begin{center} \includegraphics[]{figures/standalone-figures/Chapter05/pdf/Mach-wave.pdf} \caption{Mach wave flow turning} \label{fig:machwave} \end{center} \end{figure} \noindent A single Mach wave has a insignificant effect on the flow passing it but an expansion region constitutes an infinite number of Mach..."

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[[Category:Compressible flow]]
[[Category:Two-dimensional flow]]
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__TOC__

\section{Prandtl-Meyer Expansion Waves}

\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter05/pdf/Mach-wave.pdf}
\caption{Mach wave flow turning}
\label{fig:machwave}
\end{center}
\end{figure}

\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\\

\begin{equation}
d\theta=\sqrt{M^2-1}\frac{dV}{V}
\label{eq:mach:turning}
\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.\\

\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

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

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

\[V=Ma \Rightarrow \ln V=\ln M + \ln a\]\\

\noindent Differentiate to get\\

\begin{equation}
\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}
\label{eq:mach:turning:c}
\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\\

\begin{equation}
\frac{T_o}{T}=1+\frac{\gamma-1}{2}M^2
\label{eq:adiatbatic:energy}
\end{equation}\\

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

\[\frac{T_o}{T}=\left(\frac{a_o}{a}\right)^2\Rightarrow\]\\

\begin{equation}
\left(\frac{a_o}{a}\right)^2=1+\frac{\gamma-1}{2}M^2
\label{eq:adiatbatic:energy}
\end{equation}\\

\noindent Solve for $a$ gives\\

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

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

\begin{equation}
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}
\end{equation}\\

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

\begin{equation}
\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}
\end{equation}\\

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

\[\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}\]\\

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

\[\frac{dV}{V}=\frac{dM}{M}-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM\]\\

\[\frac{dV}{V}=dM\left[\frac{\left(1+\dfrac{\gamma-1}{2}M^2\right)-\left(\dfrac{\gamma-1}{2}\right)M^2}{\left(1+\dfrac{\gamma-1}{2}M^2\right)M}\right]=\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M}\]\\

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

\begin{equation}
\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}
\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\\

\begin{equation}
\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}
\end{equation}\\

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

\begin{equation}
\theta_2-\theta_1=\nu(M_2)-\nu(M_1)
\label{eq:prandtl:meyer:c}
\end{equation}\\

\newpage

\subsection{Solving Problems using the Prandtl Meyer Function}

\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}.\\

\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}

\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter05/pdf/expansion-corner.pdf}
\caption{Expansion corner with known net flow turning}
\label{fig:expansion:corner}
\end{center}
\end{figure}

\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\\

\begin{equation}
\frac{T_{o_1}}{T_1}=1+\frac{\gamma-1}{2}M_1^2
\label{eq:toa}
\end{equation}\\

\begin{equation}
\frac{T_{o_2}}{T_2}=1+\frac{\gamma-1}{2}M_2^2
\label{eq:tob}
\end{equation}\\

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

\[\frac{T_2}{T_1}=\frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\]\\

\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\\

\begin{equation}
\frac{T_2}{T_1}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}
\label{eq:tr}
\end{equation}\\

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

\begin{equation}
\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}
\end{equation}\\

\begin{equation}
\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}
\end{equation}

\begin{figure}[ht!]
\begin{center}
\includegraphics[]{figures/standalone-figures/Chapter05/pdf/Prandtl-Meyer-function.pdf}
\caption{Asymptotic behavior of the Prandlt-Meyer function}
\label{fig:Prandtl:Meyer:function}
\end{center}
\end{figure}

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