Oblique Shockwaves

In this post I go over the theory of oblique shocks by building on our understand of normal shockwaves! I will give a short overview of the theory and strategy for analysis.

Here is a summary of the contents of this post:

1. Introduction to oblique shocks
2. Analysis of the oblique shock
3. Working relations for an oblique shock
4. Important properties of an oblique shock
5. Conclusions and Useful References

Introduction to Oblique Shocks

What is an oblique shock wave? It is a shock wave that is oblique …. (i.e. on an angle to the incoming flow). Because of this angle, an oblique shock has slightly different properties than a normal shock, and analysis results in a different set of equations (still derived from the continuity, momentum and energy equations though). An oblique shock can be straight or curved.

When do oblique shocks occur? An oblique shock typically occurs when supersonic flow encounters an instantaneous flow deflection (δ). Because the flow cannot make this angle change gradually, it forms a shock wave. Oblique shocks occur in a wide variety of situations and are extremely important to understanding supersonic flow. A nice example for visualizing an oblique shock wave is on a supersonic vehicle:

Analysis of Oblique Shocks

The analysis of an oblique shock is very similar to that of a normal shock. However, now the problem is two-dimensional, and there is an extra momentum equation to be solved.

The figure above shows the geometric layout and control volume used to analyze an oblique shockwave. The relationships between the velocities shown are:

$V_{t1}=V_1 cos \theta$

$V_{n1}=V_1 sin \theta$

$V_{t2}=V_2 cos (\theta-\delta)$

$V_{n2}=V_2 cos (\theta-\delta)$

For the control volume shown in the figure, we can start by analyzing the continuity equation.

$\oint \limits_{cs} \rho V \cdot dA = 0$        (1a)

This gives us:

$\rho_1 V_{n1} = \rho_2 V_{n2}$         (1b)

If we analyze the momentum equation in the normal direction next:

$\Sigma F =\oint \limits_{cs} V_n(\rho V \cdot dA) = 0$        (2a)

This gives us:

$p_1-p_2=\rho_2V_{n2}^2-\rho_1V_{n1}^2$        (2b)

Next we analyze the momentum equation in the tangential direction:

$\Sigma F =\oint \limits_{cs} V_t(\rho V \cdot dA) = 0$        (3a)

This yields the very important feature of oblique shockwaves:

$V_{t1} =V_{t2}$         (3b)

The energy equation yeilds:

$h_1+\frac{V_1^2}{2} =h_2+\frac{V_2^2}{2}$      (4a)

Which is the same as:

$h_1+\frac{V_{n1}^2+V_{t1}^2}{2} =h_2+\frac{V_{n2}^2+V_{t1}^2}{2}$       (4b)

and by applying 3b we get:

$h_1+\frac{V_{n1}^2}{2} =h_2+\frac{V_{n2}^2}{2}$       (4c)

We now have a set of equations (1b, 2b, 3b, 4c, plus the perfect gas equation of state) that can be solved.

Working Relations for Oblique Shocks

Pressure Ratio

$\frac{p_2}{p_1}=\frac{2\gamma M_1^2 \sin^2 \theta}{\gamma+1}-\frac{\gamma-1}{\gamma+1}$

Density and Normal Velocity Ratio

$\frac{\rho_2}{\rho_1}=\frac{V_{n1}}{V_{n2}}=\frac{(\gamma+1)M_1^2\sin^2\theta}{(\gamma-1)M_1^2\sin^2\theta+2}$

Temperature Ratio

$\frac{T_2}{T_1}=\frac{(1+\frac{\gamma-1}{2}M_1^2\sin^2\theta)(\frac{2\gamma}{\gamma-1}M_1^2\sin^2\theta-1)}{\left[\frac{(\gamma+1)^2}{2(\gamma-1)}\right]M_1^2\sin^2\theta}$

Stagnation Pressure Ratio

$\frac{p_{o2}}{p_{o1}}=\left[\frac{\frac{\gamma+1}{2}M_1^2 \sin^2\theta}{1+\frac{\gamma-1}{2}M_1^2 \sin^2 \theta}\right]^{\frac{\gamma}{\gamma-1}}\left[\frac{2\gamma}{\gamma+1}M_1^2\sin^2\theta-\frac{\gamma-1}{\gamma+1}\right]^{\frac{-1}{\gamma-1}}$

The working relations above are all a function of $M_1$ and $\theta$. Unfortunately, there are no algebraic solutions for these functions as a function of the deflection angle $\delta$. As a result, in the case where the wave angle is not known a priori (usually the case) a numerical method must be used to solve for the wave angle. This is easily done using the Newton-Raphson method. See the example in my relevant blog post:

Important Properties of Oblique Shocks

The general properties of an oblique shockwave are the same as for normal shock-waves. Feel free to check out properties of shockwaves section of my relevant blog post: Stationary Normal Shockwaves.

• The tangential velocities across an oblique shock are equal. This was the result shown in Equation 3b.
• Oblique shock waves can result in subsonic or supersonic post-shock Mach numbers. Oblique shock-waves that result in a subsonic post-shock velocity are termed strong oblique shock waves. Oblique shock-waves that result in supersonic post-shock velocity are termed weak oblique shock waves. These shock-waves are illustrated in the figure below:
• When the deflection angle of an obstacle exceeds the $\delta_{max}$ for a given Mach number, this flow will result in a detached shock-wave. A good example of this is shown in the simulation in the relevant blog post: Mach 1.5 flow over 23 degree wedge – rhoCentralFoam

Conclusion and Useful References

Hopefully somebody learned something! In this post I attempted to explain in my own words the difference between normal and oblique shocks. I showed the governing equations from which the working relations for an oblique shock are derived. I then summarized the working relations and summarized some important properties of oblique shocks.

Let me know if I screwed up anywhere please!

Here are some excellent textbooks that have chapters on the subject:

[1] Anderson, J. D. (1990). Modern compressible flow: with historical perspective (Vol. 12). McGraw-Hill.

[2] John, J. E. A., & Keith, T, G. (2006) Gas Dynamics, 3rd Edition, Pearson Prentice Hall

[4] Zucrow, M. J., & Hoffman, J. D. (1976). Gas dynamics. New York: Wiley, 1976, 1.

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Basic Gas Dynamics

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