Expansion waves

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

$$d\theta =\sqrt{M^2-1}\frac{dV}{V}$$

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

$$\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}d\theta =\underset{M_1}{\overset{M_2}{\int }}\sqrt{M^2-1}\frac{dV}{V}$$

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

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

Differentiate to get

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

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

$$\frac{T_o}{T}=1+\frac{\gamma -1}{2}{M}^2$$

For a calorically perfect gas $a=\sqrt{\gamma RT}$ and $a_o=\sqrt{\gamma R{T}_o}$ and thus

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

$${\left(\frac{a_o}{a}\right)}^2=1+\frac{\gamma -1}{2}{M}^2$$

Solve for $a$ gives

$$a=a_o{(1+\frac{\gamma -1}{2}{M}^2)}^{-1/2}$$

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

$$da=a_o(-\frac{1}{2})\left(\frac{\gamma -1}{2}\right)2M{(1+\frac{\gamma -1}{2}{M}^2)}^{-3/2}dM$$

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

$$\frac{da}{a}=-\left(\frac{\gamma -1}{2}\right){(1+\frac{\gamma -1}{2}{M}^2)}^{-1}MdM$$

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

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

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

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

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

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

$$\underset{{\theta }_1}{\overset{{\theta }_2}{\int }}d\theta =\underset{M_1}{\overset{M_2}{\int }}\sqrt{M^2-1}{(1+\frac{\gamma -1}{2}{M}^2)}^{-1}\frac{dM}{M}$$

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

$$\nu (M)=\sqrt{\frac{\gamma +1}{\gamma -1}}{\tan }^{-1}\sqrt{\frac{\gamma -1}{\gamma +1}(M^2-1)}-{\tan }^{-1}\sqrt{M^2-1}$$

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

$${\theta }_2-{\theta }_1=\nu (M_2)-\nu (M_1)$$

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:

1. Calculate $\nu M_1$ using Eqn. \ref{eq:prandtl:meyer} or tabulated values
2. Calculate $\nu (M_2)$ as $\nu (M_2)={\theta }_2-{\theta }_1+\nu (M_1)$
3. Calculate $M_2$ from the known $\nu M_2$ 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

$$\frac{T_{o_1}}{T_1}=1+\frac{\gamma -1}{2}{M}_1^2$$

$$\frac{T_{o_2}}{T_2}=1+\frac{\gamma -1}{2}{M}_2^2$$

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

$$\frac{T_2}{T_1}=\frac{T_2}{T_{o_2}}\frac{T_{o_1}}{T_1}=\frac{1+\frac{\gamma -1}{2}{M}_1^2}{1+\frac{\gamma -1}{2}{M}_2^2}=\frac{2+(\gamma -1)M_1^2}{2+(\gamma -1)M_2^2}$$

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

$$\frac{T_2}{T_1}=\frac{2+(\gamma -1)M_1^2}{2+(\gamma -1)M_2^2}$$

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

$$\frac{p_2}{p_1}={\left[\frac{2+(\gamma -1)M_1^2}{2+(\gamma -1)M_2^2}\right]}^{\gamma /(\gamma -1)}$$

$$\frac{{\rho }_2}{{\rho }_1}={\left[\frac{2+(\gamma -1)M_1^2}{2+(\gamma -1)M_2^2}\right]}^{1/(\gamma -1)}$$