Oblique shocks and expansion waves

Oblique shocks

The Shock Polar

The shock polar is a graphical representation of all possible flow deflection angles for a given Mach number. The shock polar is generated by plotting the normalized axial velocity component downstream of an oblique shock versus the normalized vertical velocity component. In essence the ratio of vertical and axial velocity components gives the deflection angle. Figure~\ref{fig:shock:polar:a} shows a set of shock polars generated for different upstream Mach numbers (indicated in the figure). The vertical and axial velocity components are normalized by the characteristic speed of sound (the speed of sound at sonic conditions). Each of the shock polars have two solutions where the vertical velocity component is zero, i.e. zero-deflection solutions. The zero-deflection solution furthest to the right represents the Mach wave solution and the solution to the left represents a normal shock. This is easy to realize as these are the only possible solutions that would result in zero flow deflection and the Mach wave solution is located where $V_x/a^{\ast }$ is greater than one, which means that the flow is supersonic on the downstream side, whereas the normal shock results in a subsonic flow on the downstream side ($V_x/a^{\ast }<1.0$). Figures~\ref{fig:shock:polar:b}-{fig:shock:polar:f} shows only the shock polar corresponding to an upstream mach number of 2.5. In Figure~\ref{fig:shock:polar:b}, a unit half-circle is added to indicate which parts of the shock polar that represents supersonic solutions and which parts that represent supersonic solutions. Everything that falls inside of the unit circle represents subsonic solutions. i.e. the downstream Mach number is subsonic and all solutions that are outside of the circle represents solutions for which the flow is supersonic. Figure~\ref{fig:shock:polar:c} shows how the maximum possible flow deflection relates to the shock polar for a specific upstream Mach number. The line in the figure represents the flow deflection and when the line is tangent to the shock polar, the maximum flow deflection is reached. Further increase of the flow deflection angle leads to that the flow deflection line is outside of the shock polar and thus there are no possible solutions for angles greater than ${\theta }_{max}$. For the maximum deflection angle there is only one possible solution since the flow deflection line is a tangent to the shock polar. For all angles smaller than ${\theta }_{max}$, there are, however, two possible solutions (see Figure~\ref{fig:shock:polar:d}). In most cases there is one supersonic solution (the weak solution) and one subsonic solution (the strong solution) as indicated in Figure~\ref{fig:shock:polar:d}). However, for angles close to ${\theta }_{max}$ both solutions may fall inside of the unit circle and thus both solutions will be subsonic. This is in line with what we saw for the $\theta$-$\beta$-Mach relation earlier (see section~\ref{sec:theta:beta:Mach}). Finally, we will have a look at how the shock angle ($\beta$) is related to the shock polar. Figure~\ref{fig:shock:polar:e} shows how the shock angle ($\beta$) is found for the weak solution of a given upstream mach number and a given flow deflection ($\theta$). Draw a line starting at the Mach wave solution going through the the downstream solution (in this case the weak solution). Now, make another line starting at the origin that is perpendicular to the first line. The angle of the second line is the shock angle ($\beta$). In analogy, figure~\ref{fig:shock:polar:f} shows how to obtain the shock angle for the strong solution.

Shock Reflection at a Solid Boundary

What happens when an oblique shock reaches a solid wall? To sort this out we will analyze the schematic flow situation illustrated in Figure~\ref{fig:regular:reflection}. The figure shows a supersonic flow through a channel where there is a sudden bend of the lower wall leading to the generation of an oblique shock in order to deflect the flow such that it follows the wall downstream of the corner. The shock angle ${\beta }_a$ is a function of the upstream Mach number $M_1$ and the flow deflection $\theta$. A bit further downstream the shock will reach the upper wall of the channel. The question now is what will happen at the point reaches the upper wall. As indicated in the figure the shock will deflect but in what way will it deflect. Will it be a specular reflection, i.e. will the angle of the shock be the same but in the other direction? To answer this question let's sort out why the shock will reflect. After the first shock, the flow will be deflected the angle $\theta$, which means that, at the upper wall, the flow must be deflected again such that it follows the direction of the upper wall. Hence, the deflection angle will again be $\theta$ but in the opposite direction. So, the deflection angle is the same, does that mean that the shock angle will be the same? The answer is no, but why? Passing the first shock, the Mach number is reduced and thus the $\theta$-$\beta$-Mach relation will give us another shock angle ${\beta }_2$ for the second shock. Hence, the shock is not reflected specularly since ${\beta }_1\ne {\beta }_2$.

The situation discussed in the previous paragraph assumed that the flow deflection at the upper wall was less than the maximum flow deflection possible for the Mach number ahead of the reflection (downstream of the first shock). If, however, the flow deflection that must take place exceeds the maximum possible flow deflection angle, it will not be possible to generate an oblique shock that fulfills the requirements. Instead, a normal shock will be generated at the upper wall (a so-called Mach reflection) that will be gradually converted into an oblique shock. This situation is depicted in Figure~\ref{fig:Mach:reflection}. The slip line indicated in the figure is a consequence of the fact that the flow the goes through the shock system experiences different entropy increases depending on if the flow passes a single stronger shock that generate higher losses or a set of weaker oblique shocks that will generate less losses.

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

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

(Eq. 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

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

(Eq. 2)

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

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

(Eq. 3)

Differentiate to get

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

(Eq. 4)

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

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

(Eq. 5)

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

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

(Eq. 6)

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

(Eq. 7)

Solve for $a$ gives

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

(Eq. 8)

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

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

(Eq. 9)

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

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

(Eq. 10)

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

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

(Eq. 11)

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

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

(Eq. 12)

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

(Eq. 13)

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

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

(Eq. 14)

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

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

(Eq. 15)

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

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

(Eq. 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:

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

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

(Eq. 17)

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

(Eq. 18)

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

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

(Eq. 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

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

(Eq. 20)

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

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

(Eq. 21)

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

(Eq. 22)