1
$\begingroup$

I am trying to understand what the 3D shock structure would be for a wing in supersonic flow. Suppose we have a wing which is made of double-wedge airfoil cross-sections.

If we view the shock structure over the airfoil (side view), it would look like so:

Supersonic flow over a symmetric double wedge airfoil. Source: Houghton

Now, if we look at the top view, we would see the following structure (top view): Shock structure over top of wing. Source: Houghton

Questions:

  1. Is the shock seen in the side view (airfoil shock structure) related to the one seen in the top view?
  2. How would the shock structure change if the airfoil is kept the same but the wing is swept more? (If you can show some 3D plots for the shock structure, it would really be helpful!)
  3. Can you explain what determines the shock angles in this picture (is it sweep or airfoil shape or both - and which is more dominant)?

enter image description here

| improve this question | |
$\endgroup$
1
$\begingroup$

Is the shock seen in the side view (airfoil shock structure) related to the one seen in the top view?

No, not at all. The first picture (side view airfoil) shows an unswept wing in supersonic flow and is not very precise. The shock angles of the LE shock and the TE shock should be identical.

The second picture (top view wing) shows a swept wing at bit above the speed of sound (almost half the speed in the first picture). Here, the flow pattern around the airfoil is fully subsonic (that's why the wing is swept!) and only its most forward tip will cause a shock which is almost straight because flow speed is only a bit higher than the speed of sound.

How would the shock structure change if the airfoil is kept the same but the wing is swept more? (If you can show some 3D plots for the shock structure, it would really be helpful!)

That depends on the ratio between speed and sweep angle. As long as the sweep angle exceeds the shock angle, flow around the airfoil behaves like in the subsonic case with a stagnation line, flow around the round leading edge and no bow shock or expansion fans. When the shock angle is greater, the airfoil will show a flow pattern like in the top picture. Sorry, no fancy 3D plots!

Can you explain what determines the shock angles in this picture (is it sweep or airfoil shape or both - and which is more dominant)?

The first and foremost factor for the shock angle is speed. Mach 1 will cause a straight shock and higher speeds will bend the shock backwards into an oblique shock with a shock angle $\varphi$ according to the equation $$\varphi = arccos\left(\frac{1}{Ma}\right)$$ This angle will become smaller with increasing bluntness of the body causing it.

shock angle diagram

Shock angle diagram (from Wikipedia, source). The caption is: This chart shows the oblique shock angle, β, as a function of the corner angle, θ, for a few constant M1 lines. The red line separates the strong and weak solutions. The blue line represents the point when the downstream Mach number becomes sonic. The chart […] is valid for an ideal diatomic gas.

Explanation: M1 = Mach number ahead of the shock, M2 = Mach number past the shock. β is the half cone angle such that β + $\varphi$ = 90°.

Consequently, the angle of the shocks emanating from that T-38 show that it flies at around Mach 1.09. In Schlieren images like this darker areas indicate higher density and lighter ones lower density. The two dark lines coming from the fuselage tip and engine exhausts are the two compression shocks which are audible as a double boom when a supersonic aircraft passes overhead.

| improve this answer | |
$\endgroup$
  • $\begingroup$ If we look at the picture of the wing from the top view, we see that the sweep causes the supersonic free stream to turn through a finite angle (specifically, the sweep angle). Since a supersonic flow turned in on itself generates a shock wave, can you explain why the sweep angle would not change the shock which emanates from the leading edge of the wing? Basically, why is the shock angle determined by the formula you gave (for Phi) (which is indep on sweep angle) and not by the graph you showed? $\endgroup$ – Nick Hill Jul 12 at 19:59
  • $\begingroup$ @NickHill "we see that the sweep causes the supersonic free stream to turn through a finite angle" – do we? That drawing isn't very helpful. The Schlieren photo with the T-38 is much better and here you see the shock angle and can use it to directly determine speed. The bow shock has a slightly rounded tip. The local angle close to the nose is what can be read from the diagram, the far-field angle is given in the formula. $\endgroup$ – Peter Kämpf Jul 13 at 1:55
  • $\begingroup$ Suppose we have a flying wing with a certain sweep angle (lambda). Would the shock emanating from the apex depend on lambda at all? I'm just a little confused by why the shock angle would not be changing if the flow is changing direction starting at the apex of the wing and is being turned in on itself. Should this flow turning not create a shock wave that depends on lambda? $\endgroup$ – Nick Hill Jul 14 at 2:18
  • $\begingroup$ @NickHill: How should that work? Pressure changes travel with the speed of sound, so any change to the air cannot happen ahead of the shock front, independent of what comes after the apex as long as Lambda is larger than the Mach cone. There simply is no way the wing can change the shock angle as long as sweep prevents it from doing so. $\endgroup$ – Peter Kämpf Jul 14 at 3:44
  • $\begingroup$ I clarified my question further here: aviation.stackexchange.com/questions/79593/… $\endgroup$ – Nick Hill Jul 17 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.