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Why do the shocks on the upper and lower surfaces of a wing stop at the trailing edge? When going Mach 0.82, the shocks on the upper (and lower) surface don't go the whole chord length of the wing. (In this picture, the upper shock stops around half way)

When going around Mach 0.95, the shocks make it all the way to the trailing edge. I was told that if you go faster, the shocks won't move back more past the trailing edge, like they would if you were speeding up from Mach 0.82. Why is that?

enter image description here

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    $\begingroup$ I think that @RobMcDonald has already answered you this question $\endgroup$
    – sophit
    Commented Jan 31 at 18:59
  • $\begingroup$ @sophit Oh, would you mind telling me what part of that answer answers my question? I can't seem to recognize it. $\endgroup$
    – Wyatt
    Commented Jan 31 at 19:07
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    $\begingroup$ "When flow regions 1 and 3 come together, they can't continue in a straight line. They must turn. That turning is accomplished by a shock." and "So, you can see that the shock 'stops' at the trailing edge because it must be there to turn the flow. After the TE, there is no need to turn the flow." are the relevant parts. $\endgroup$ Commented Jan 31 at 19:33

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The shock on the surface of the airfoil is a normal shock. It exists because the flow is supersonic, but is facing a back pressure such that it must shock down to subsonic. The flow does not turn at that shock because the surface is locally smooth.

Zoom in on the TE of the airfoil...

enter image description here

Flow regions 1 and 3 are supersonic. 2 and 4 may or may not be -- depending on the velocities and angles involved.

When flow regions 1 and 3 come together, they can't continue in a straight line. They must turn. That turning is accomplished by a shock.

The dotted line between regions 2 and 4 is called a slip line. Velocity across the slip line must be parallel (but not necessarily equal in magnitude). Likewise, pressure across the slip line must be equal. You can think of the slip line as a thin flexible membrane -- flow must be parallel to the slip line and if pressure was higher on one side, it would push the membrane until pressure was equal again.

So, you can see that the shock 'stops' at the trailing edge because it must be there to turn the flow. After the TE, there is no need to turn the flow.

There are situations where there can be a third shock in regions 2 & 4, but that is actually a small normal shock that serves to slow the flow down to subsonic.

EDIT: Adding some more details inspired by followup questions...

Even a thin boundary layer can complicate matters dramatically.

The shock causes an abrupt rise in pressure. This is an 'adverse' pressure gradient. Boundary layers tend to separate in adverse pressure gradients, so we often get separation at the point of the shock.

This separation bubble 'looks like' a bump in the surface to the flow outside the boundary layer. So you now need some turning where before you didn't. This causes the shock to change shape.

Shock-boundary layer interaction is really complex. You can see evidence of shock-boundary layer interaction in the Mach 0.84 to 0.98 images below.

enter image description here

Here is an extensive paper on the subject. Check out figures 9, 10, 11 in the paper. They focus specifically on the difference between an idealized inviscid transonic flow and a transonic flow with a boundary layer.

This paper calls part of this phenomena a delta shock (referring specifically to the compression waves at the front of the boundary layer interaction. I am less familiar with that name -- I usually see this whole arrangement (slightly bigger scale) referred to as a lambda-shock because it looks like the greek letter lambda $\lambda$.

The cartoon I drew before was an inviscid idealization. You can see how a little bit of boundary layer can significantly complicate matters.

Here is another idealized cartoon, but zoomed way in, 'just before' the shocks reach the trailing edge...

enter image description here

Here the upper and lower surface shocks are normal shocks, they meet the surface at 90 degrees and do not turn the flow. Also, the regions downstream of them (2 and 4) are now subsonic flows.

Now, the only streamline that needs to make a sharp turn at the corner is the streamline that lies exactly on the surface. Everywhere else, in a subsonic flow, the streamlines can make a gradual turn.

Of course, the limiting streamline can't make an instantaneous turn, so in the subsonic inviscid approximation, we actually have the limiting streamlines go to zero velocity at a sharp trailing edge like this -- that forms the aft stagnation point.

If this airfoil had a cusped trailing edge, things would be a little different.

Of course, all real flows have a boundary layer (even if it is thin), and the limiting streamline at the surface obeys a no-slip condition -- so its velocity is zero. The rest of the streamlines are able to make the TE turn without a singularity.

So, in our inviscid idealization, I would expect that the normal shock would stay a normal shock until it reached the trailing edge. At that point, like the limiting streamline in the subsonic case, the on-body streamline needs to instantly turn and it would do this by going to zero velocity at the trailing edge.

As the freestream Mach is increased further, the shock should lay over and become the oblique shock solution that we expect.

I'm not exactly sure what happens during that transition. Since this only applies to our idealized model, I'm not sure that those details are relevant at all. After all:

  1. boundary layers happen. They allow the limiting streamline to turn and solve the stagnation streamline problem.

  2. real shocks are not infinitely thin

  3. real trailing edges are not infinitely sharp, they have some bluntness to them.

  4. Myriad other complicating factors

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  • $\begingroup$ I see, thanks for the quick reply. So let me make sure I have this right : The trailing edge shock is the same as the shock on the top of the wing. It just stops at the trailing edge because of the air coming in contact with the air from the lower surface. $\endgroup$
    – Wyatt
    Commented Jan 31 at 21:56
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    $\begingroup$ No. The shock on the top of the wing does not cause the flow to turn. So it is a normal shock, while the TE shock is an oblique shock. The shocks on the surface move aft and when they get to the TE, they change from normal to oblique. Does that transition happen just before it reaches the TE, or does it happen at the TE as it speeds up, I'm not sure. In reality, the airfoil will have a boundary layer that will make some of these distinctions more messy. $\endgroup$ Commented Jan 31 at 22:10
  • $\begingroup$ Ah I see, that edit cleared some things up. One last little thought I had : If the flow behind the normal shock is subsonic (in sections 2 & 4), how would an oblique shock form where the TE converges to a point? Does the flow have to be supersonic to create an oblique shock? $\endgroup$
    – Wyatt
    Commented Feb 1 at 0:09
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    $\begingroup$ The flow upstream of a shock must be supersonic. The flow downstream of a normal shock is always subsonic. The flow downstream of an oblique shock can be subsonic or supersonic -- depending on the angle of the shock. There are two solutions, the 'strong' and 'weak' shock solution. In the second drawing, there would not be a second set of shocks at the trailing edge. $\endgroup$ Commented Feb 1 at 0:50
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    $\begingroup$ When they meet at that point, their velocities must be parallel. That means that something must be done away from that point to accelerate them (turn them or slow them down). In a supersonic flow, the shock happens to 'instantly' turn them parallel. In an inviscid subsonic flow, they slow down as they approach that point to reach zero velocity and then accelerate back up away from that point. Your wall analogy at best applies to the supersonic case -- something abruptly must happen, so you get a shock to turn the flow. In subsonic flow, abrupt changes are impossible. $\endgroup$ Commented Feb 1 at 6:22

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