Expansion waves: Difference between revisions

Latest revision as of 10:05, 1 April 2026

Prandtl-Meyer Expansion Waves

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

d θ = \sqrt{M^{2} - 1} \frac{d V}{V}

(Eq. 4.1)

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.

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

\int_{θ_{1}}^{θ_{2}} d θ = \int_{M_{1}}^{M_{2}} \sqrt{M^{2} - 1} \frac{d V}{V}

(Eq. 4.2)

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

V = M a \Rightarrow \ln V = \ln M + \ln a

(Eq. 4.3)

Differentiate to get

\frac{d V}{V} = \frac{d M}{M} + \frac{d a}{a}

(Eq. 4.4)

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

\frac{T_{o}}{T} = 1 + \frac{γ - 1}{2} M^{2}

(Eq. 4.5)

For a calorically perfect gas $a = \sqrt{γ R T}$ and $a_{o} = \sqrt{γ R T_{o}}$ and thus

\frac{T_{o}}{T} = {(\frac{a_{o}}{a})}^{2} \Rightarrow

(Eq. 4.6)

{(\frac{a_{o}}{a})}^{2} = 1 + \frac{γ - 1}{2} M^{2}

(Eq. 4.7)

Solve for $a$ gives

a = a_{o} {(1 + \frac{γ - 1}{2} M^{2})}^{- 1 / 2}

(Eq. 4.8)

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

d a = a_{o} (- \frac{1}{2}) (\frac{γ - 1}{2}) 2 M {(1 + \frac{γ - 1}{2} M^{2})}^{- 3 / 2} d M

(Eq. 4.9)

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

\frac{d a}{a} = - (\frac{γ - 1}{2}) {(1 + \frac{γ - 1}{2} M^{2})}^{- 1} M d M

(Eq. 4.10)

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

\frac{d V}{V} = \frac{d M}{M} + \frac{d a}{a}

(Eq. 4.11)

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

\frac{d V}{V} = \frac{d M}{M} - (\frac{γ - 1}{2}) {(1 + \frac{γ - 1}{2} M^{2})}^{- 1} M d M

(Eq. 4.12)

\frac{d V}{V} = d M [\frac{(1 + \frac{γ - 1}{2} M^{2}) - (\frac{γ - 1}{2}) M^{2}}{(1 + \frac{γ - 1}{2} M^{2}) M}] = {(1 + \frac{γ - 1}{2} M^{2})}^{- 1} \frac{d M}{M}

(Eq. 4.13)

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

\int_{θ_{1}}^{θ_{2}} d θ = \int_{M_{1}}^{M_{2}} \sqrt{M^{2} - 1} {(1 + \frac{γ - 1}{2} M^{2})}^{- 1} \frac{d M}{M}

(Eq. 4.14)

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

ν (M) = \sqrt{\frac{γ + 1}{γ - 1}} \tan^{- 1} \sqrt{\frac{γ - 1}{γ + 1} (M^{2} - 1)} - \tan^{- 1} \sqrt{M^{2} - 1}

(Eq. 4.15)

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

θ_{2} - θ_{1} = ν (M_{2}) - ν (M_{1})

(Eq. 4.16)

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}.

A problem of that type can be solved as follows:

Calculate $ν M_{1}$ using Eqn. \ref{eq:prandtl:meyer} or tabulated values
Calculate $ν (M_{2})$ as $ν (M_{2}) = θ_{2} - θ_{1} + ν (M_{1})$
Calculate $M_{2}$ from the known $ν M_{2}$ using Eqn. \ref{eq:prandtl:meyer} or tabulated values

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

\frac{T_{o_{1}}}{T_{1}} = 1 + \frac{γ - 1}{2} M_{1}^{2}

(Eq. 4.17)

\frac{T_{o_{2}}}{T_{2}} = 1 + \frac{γ - 1}{2} M_{2}^{2}

(Eq. 4.18)

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 + \frac{γ - 1}{2} M_{1}^{2}}{1 + \frac{γ - 1}{2} M_{2}^{2}} = \frac{2 + (γ - 1) M_{1}^{2}}{2 + (γ - 1) M_{2}^{2}}

(Eq. 4.19)

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

\frac{T_{2}}{T_{1}} = \frac{2 + (γ - 1) M_{1}^{2}}{2 + (γ - 1) M_{2}^{2}}

(Eq. 4.20)

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

\frac{p_{2}}{p_{1}} = {[\frac{2 + (γ - 1) M_{1}^{2}}{2 + (γ - 1) M_{2}^{2}}]}^{γ / (γ - 1)}

(Eq. 4.21)

\frac{ρ_{2}}{ρ_{1}} = {[\frac{2 + (γ - 1) M_{1}^{2}}{2 + (γ - 1) M_{2}^{2}}]}^{1 / (γ - 1)}

(Eq. 4.22)

Expansion waves: Difference between revisions

Latest revision as of 10:05, 1 April 2026

Contents

Prandtl-Meyer Expansion Waves

Solving Problems using the Prandtl Meyer Function

Navigation menu

@@ Line 1: / Line 1: @@
-[[Category:Compressible flow]]
+<!--
-[[Category:Two-dimensional flow]]
+-->[[Category:Compressible flow]]<!--
-[[Category:Inviscid flow]]
+-->[[Category:Two-dimensional flow]]<!--
+-->[[Category:Inviscid flow]]<!--
+--><noinclude><!--
+-->[[Category:Compressible flow:Topic]]<!--
+--></noinclude><!--
-__TOC__
+--><nomobile><!--
+-->__TOC__<!--
+--></nomobile><!--
-\section{Prandtl-Meyer Expansion Waves}
+--><noinclude><!--
+-->{{#vardefine:secno|4}}<!--
+-->{{#vardefine:eqno|0}}<!--
+--></noinclude><!--
+-->
+=== Prandtl-Meyer Expansion Waves ===
+<!--
 \begin{figure}[ht!]
 \begin{center}
@@ Line 14: / Line 26: @@
 \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\\
+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}
+{{NumEqn|<math>
 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}
+{{NumEqn|<math>
 \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
-\[V=Ma \Rightarrow \ln V=\ln M + \ln a\]\\
+{{NumEqn|<math>
+V=Ma \Rightarrow \ln V=\ln M + \ln a
+</math>}}
-\noindent Differentiate to get\\
+Differentiate to get
-\begin{equation}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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
-\[\frac{T_o}{T}=\left(\frac{a_o}{a}\right)^2\Rightarrow\]\\
+{{NumEqn|<math>
+\frac{T_o}{T}=\left(\frac{a_o}{a}\right)^2\Rightarrow
+</math>}}
-\begin{equation}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 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}
+{{NumEqn|<math>
 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}
+{{NumEqn|<math>
 \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
-\[\frac{dV}{V}=\frac{dM}{M}+\frac{da}{a}\]\\
+{{NumEqn|<math>
+\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
-\[\frac{dV}{V}=\frac{dM}{M}-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM\]\\
+{{NumEqn|<math>
+\frac{dV}{V}=\frac{dM}{M}-\left(\frac{\gamma-1}{2}\right)\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}MdM
+</math>}}
-\[\frac{dV}{V}=dM\left[\frac{\left(1+\dfrac{\gamma-1}{2}M^2\right)-\left(\dfrac{\gamma-1}{2}\right)M^2}{\left(1+\dfrac{\gamma-1}{2}M^2\right)M}\right]=\left(1+\dfrac{\gamma-1}{2}M^2\right)^{-1}\frac{dM}{M}\]\\
+{{NumEqn|<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>}}
-\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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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
+<!--
 \begin{figure}[ht!]
 \begin{center}
@@ Line 136: / Line 142: @@
 \end{center}
 \end{figure}
+-->
+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\\
+{{NumEqn|<math>
-\begin{equation}
 \frac{T_{o_1}}{T_1}=1+\frac{\gamma-1}{2}M_1^2
-\label{eq:toa}
+</math>}}
-\end{equation}\\
-\begin{equation}
+{{NumEqn|<math>
 \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
-\[\frac{T_2}{T_1}=\frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}=\frac{1+\dfrac{\gamma-1}{2}M_1^2}{1+\dfrac{\gamma-1}{2}M_2^2}=\frac{2+(\gamma-1)M_1^2}{2+(\gamma-1)M_2^2}\]\\
+{{NumEqn|<math>
+\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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+{{NumEqn|<math>
 \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}
+<!--
 \begin{figure}[ht!]
 \begin{center}
@@ Line 180: / Line 184: @@
 \end{center}
 \end{figure}
+-->

Expansion waves: Difference between revisions

Latest revision as of 10:05, 1 April 2026

Prandtl-Meyer Expansion Waves

Solving Problems using the Prandtl Meyer Function

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