0
$\begingroup$

If a wing is stalled, will the wing not exhibit any circulation as predicted by the Kutta-Joukowski theorem? If so how would this affect the pressure distribution between the top and bottom portions of the wings? Would the top portion of the wing (low pressure) carry most of the lift?

$\endgroup$
1
$\begingroup$

If there is no circulation, there would be no lift. In reality, circulation stops to increase linearly when the angle of attack nears the stall range and drops beyond stall. However, there is plenty of lift left beyond stall.

180° polar for several NACA 4-digit airfoils

180° polar for several NACA 4-digit airfoils, from Hoerners Fluid Dynamic Lift.

how would this affect the pressure distribution […]?

Think of the stalled airfoil as of an entity which comprises the airfoil itself, the boundary layer around it and the separated flow in its wake. This will still have nearly-inviscid flow around it, and the wake will make it "look" to the outer flow like an airfoil with a lower angle of attack and a really drawn-out pressure recovery. There is still a suction peak at the nose, albeit much weaker than for the clean airfoil in attached flow, and constant suction past the separation point. On the bottom, however, flow is almost unchanged except for a region near the trailing edge where the low pressure in the wake speeds up the local flow.

Would the top portion of the wing (low pressure) carry most of the lift?

The answer to the question on which side more lift is created depends a lot on how lift is interpreted. In the end, lift is the difference between the pressures on the bottom and the top sides, and cannot meaningfully be split into a top or bottom contribution. What can be done is to compare the pressure on one side with static pressure, which will yield a side-specific lift fraction. But then you can arbitrarily compare surface pressures with any pressure level found inside the wing box - it will cancel out once you calculate total lift.

$\endgroup$

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.