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While learning about shock waves in an introductory Gas Dynamics course, it was revealed that normal shocks are unstable if formed in a converging channel. Even if the local conditions ostensibly require the presence of a shock in the converging section, the flow instead chooses to reinvent itself, moving the shock wave to a diverging section while simultaneously altering the upstream conditions. I can verify that this is a genuine phenomenon, but is there any formal explanation in terms of the underlying flow physics?

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    $\begingroup$ I found an example (while with a diverging channel...) in this paper. If one look at the equations $(45),(46)$, and the little discussion on the same page, one sees that the sign of the parameter $\tau$ gives the stability or the unstability. Of course, the value of $\tau$ depends on your particular model, so one has to read the paper from the beginning, to understand from where this value of $\tau$ is coming. I suppose an analogeous model exists for the converging channel, which provides a negative value for $\tau$ $\endgroup$
    – Trimok
    Sep 10 '14 at 9:33
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    $\begingroup$ This paper by Kantrowitz is referenced in the paper you provided and is most directly purposed towards answering the question, though I am still perusing it. naca.central.cranfield.ac.uk/reports/1947/naca-tn-1225.pdf $\endgroup$
    – Bryson S.
    Sep 10 '14 at 14:18
  • $\begingroup$ In the Kantrowitz paper, you might see directly formula $(31)$ page $25$, with $3$ different cases (formulae $(32)(33)(34$) page $26$, and the discussion at the bottom of page $26$), going with Fig .$8$ at the end of the document. I am not a specialist, so I suggest you to ask a new question by describing precisely what is not clear or what you don't understand in this paper. $\endgroup$
    – Trimok
    Sep 16 '14 at 9:07
  • $\begingroup$ I think the rate of convergence (defined by the shock speed and angle relative to shock normal) is important here too. If the channel converges faster than the shock can propagate, there will be significant effects from reflected waves interfering with the incident shock. By "reinvent" do you mean wave breaking or gradient catastrophe? If so, then that just means the pressure pulse producing the shock is steepening faster than the shock can dissipate energy. $\endgroup$ Sep 25 '14 at 13:48
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The answer is due to the area-Mach number relation for hydrodynamic shocks. G.B. Whitham has a great book (check out Chapter 8) on all sorts of various waves and has a good discussion of this topic.

The idea is that one can define the Mach number as a function of the cross-sectional area of a ray tube. The simple form is: $$ \frac{ 1 }{ A } \frac{ d A }{ d M } = -g\left( M \right) \tag{1} $$ where g(M) is given by: $$ \begin{align} g\left( M \right) & = \frac{ M }{ M^{2} - 1 } \left( 1 + \frac{ 2 }{ \gamma + 1 } \frac{ 1 - \mu^{2} }{ \mu } \right) \left( 1 + 2\mu + \frac{ 1 }{ M^{2} } \right) \tag{2a} \\ \mu^{2} & = \frac{ \left( \gamma - 1 \right) M^{2} + 2 }{ 2\gamma M^{2} - \left( \gamma - 1 \right) } \tag{2b} \end{align} $$ where $\gamma$ = ratio of specific heats, $M$ = Mach number. The idea is to find equations for flow velocity, pressure, and density as functions of initial conditions. We can rewrite the first equation using the chain rule to find: $$ g\left( M \right) \frac{ d M }{ d x } + \frac{ 1 }{ A } \frac{ d A }{ d x } = 0 \tag{3} $$ Then the idea is to parameterize the Mach number as a function of area. In the case you describe, the Mach number is inversely proportional to the scale length used to define the area (e.g., radius if spherical channel). So you can see that this would cause $M$ to diverge as A $\rightarrow$ 0.

Since the bulk flow is not going to accelerate to keep $M$ = constant, it is unlikely that a shock will maintain itself.

Unfortunately, this is does not discuss everything. As I mentioned in my comment, you need to worry about reflection off the channel walls, which can cause shock-shock interactions. I did find a few papers that argued stability could be found (e.g., pdf found here), but I doubt that is a general solution.

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