When there is an expansion fan aft of an oblique shock such that its bounding mach waves intersect with the oblique shock, the oblique shock is curved in that region. What is the physical phenomenon which causes this curvature?

Is it due to the static pressure difference between the air on either side of the shock where the expansion fan accelerates the air after the shock and therefore decreases the static pressure?

Or is it due to the fact that for the mach numbers before and after the shock (with the latter being accelerated compared to the case without the expansion fan) to be consistent with the shock angel it has to be more oblique?

  • $\begingroup$ Do you have any diagram of the situation you are imagining? I can't find any diagram with a shock and an expansion fan that would show the shock to curve. $\endgroup$
    – Jan Hudec
    Apr 25, 2023 at 21:19
  • $\begingroup$ Hello Theo, welcome to aviation.stackexchange.com. Very nice first question, we are looking forward to see more of these! $\endgroup$
    – DeltaLima
    Apr 26, 2023 at 8:45

1 Answer 1


enter image description here

The oblique shock exists to turn the flow. Its angle (and strength) determines the turning angle. Along the body (say a 10 deg diamond airfoil), the boundary condition dictates that the shock must turn the flow 10 deg.

The fan exists to turn the flow the other way. At the crest of our 10 deg diamond airfoil, the fan turns the flow 20 deg away. (There will be a final 10 deg TE shock to turn the flow back to zero degrees).

Next, imagine a streamline somewhat off the body -- say an inch or a centimeter if you prefer. That streamline must shock and turn because the 'lower' streamline turned -- it created a new boundary condition. The streamlines can not cross, so if the 'lower' streamlines turn, so must we.

However, when we get to the turn the other way, only the on-body streamline must turn instantly. Higher streamlines can turn more gradually -- this is why an expansion fan has finite width -- it is a fan. The streamlines moving through the fan gradually turn, until at the end of the fan, they have turned the required 20 degrees.

The Mach angle of the leading edge of the fan is such that it will eventually hit the shock. This more complex area of the flowfield is usually not pictured in undergraduate texts as it is more complex. Good for you for having this question.

enter image description here

The streamline just above the blue streamline does not need to follow the body -- it only needs to follow the blue streamline.

When the fan reaches the shock, it relieves the need for the shock to turn the flow as much. Instead of turning 10 degrees, a streamline may only need to turn 9.5 degrees because the fan has turned the next lower streamline. Then, the next streamline only needs to turn 9 degrees, etc.

A shock that turns the flow less (at the same freestream Mach number) will be a weaker shock, and will have a lower shock angle. This results in the curved shock.

This kind of 2D compressible flow is a lot of fun. It relates to not only airfoils, but inlets and nozzles for engines and shock diamond flow in a jet. Everything that happens is a reaction to the boundary conditions. Think of what the boundary forces the flow to do and you'll be able to think through very complex flows.

  • $\begingroup$ Very nice answer 👍 $\endgroup$
    – sophit
    Apr 26, 2023 at 5:23
  • $\begingroup$ Thanks for your clear answer answer Rob. Just a follow up question: what's the mach nb after the curved shock? From my understanding there are 2 ways to define that mach nb: 1) The mach number after the shock with the slightly reduced flow turn angle as you discussed found using the shock equations. 2) The mach nb after right after the leading mach wave of the expansion. I'm asking this question because I'm trying to write some code to compute and plot the curved shocks off of a diamond airfoil and calculating the mach numbers after the shocks using these 2 methods gives different results. $\endgroup$
    – Theo
    Apr 26, 2023 at 23:08
  • 1
    $\begingroup$ I don't think you can use the pure shock/fan equations / tables for this kind of calculation. I've usually seen it completed with the method of characteristics (or supersonic linearized theory). On the one hand, the shock solution to the flow downstream of the slightly weaker shock must be true -- and the inflow condition to the fan will depend on that weakened shock (and so the fan solution must somehow be iterative). So I think I understand why your answers are different and which is 'right'. Unfortunately, that won't help you progress through the rest of the flow as the MOC is needed. $\endgroup$ Apr 27, 2023 at 1:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .