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taalf May 5, 2011 06:36

How to calculate circulation ?
Hello, everybody,

I would like to ask some of you a question which is mine for a long time.

It is about circulation \Gamma.

One can find very easily the definition of circulation : a closed line integral of the velocity dot the elementary length of the line :

\Gamma = \oint \vec{V} \bullet \vec{dl}

And so, one can speak of the circulation around a wing. And then calculate, for example, the lift force thanks to the Kutta-Jukowski theorem.

But I don't understand where should be this line when we are dealing with such a circulation around a wing.

Indeed, if we put the line directly on the solid boundaries of the wing, the velocity is zero, so the circulation is zero too.

In the other hand, if we extend the line far from the wing, the line see a flow which isn't perturbed anymore by the wing, and the result is that the circulation is also zero.

Close to the wing, the circulation can be non-zero, but where exactly should we put this line to calculate the so called "circulation around the wing" ?

Actually, I very wondered in what forum I could speak about that, and it seemed to me that the CFD Online's forum was the best one.

Can somebody help ?


truffaldino May 5, 2011 10:49


Originally Posted by taalf (Post 306341)
Hello, everybody,

In the other hand, if we extend the line far from the wing, the line see a flow which isn't perturbed anymore by the wing, and the result is that the circulation is also zero.


Magnitude of the perturbation velocity far from the airfoil (I assume you are talking about 2D case) is inversely proportional to the distance, so the integral does not vanish there as integration path length is proporional to the distance. So, take any contour far from the airfoil, where flow is almost inviscid.

taalf May 5, 2011 12:25

Hello, @truffaldino,

Thank you for your response.

So, mathematically, the circulation is constant whatever the line if this line enclose the wing, even if I take a circle of radius equal to hundred times the chord of the wing.

I recognize that I am not surprised by your response, because mathematically, I agree with you. But, in a CFD case, when we use a farfield boundary condition, on which we impose the velocity, it is clear that the circulation through this boundary condition is zero, because the velocity vector is constant along this boundary, and so the integral is zero.

So, we could say that we impose a zero circulation around our wing !?

Am I stupid or there is something which I don't well understand ?

truffaldino May 5, 2011 16:45

William: It is good that you are paying attention to fluid dynamics fundamentals.

The difference between "mathematical" case an one with imposed boundary conditions is that in the former the external forces act from the airfoil surface only, while in the latter the external force is imposed from the both boundary and the airfoil.

If your boundary is far enough from the airfoil, the "defect" in circulation close to the airfoil (but far enough from the boundary layer) will be small (proportional to the ratio between the airfoil chord to the distance to the boundary) . Roughly speaking you have to satisfy both Joukowsky-Kutta condition and BC. When you are close to the airfoil Joukowsky-Kutta condition makes much more bigger contribution to the circulation than the BC at far boundaries.

So, choose the contour that is far enough from the external boundary and also where effect of boundary layer/wake is negligible (the flow is practically inviscid at the distance of a few boundary layer sizes from the airfoil surface). This can be contours with distance from the airfoil surface ranging from airfoil chord to several bl sizes.


taalf May 6, 2011 10:13

Thank you very much, @truffaldino,

Yes, it is the never-ending problem of CFD that it is only an approximation of the analytical theory of fluid mechanics. We have to keep this in mind...

Thank you very much for your response which confirms what I thought but what I wasn't able to accept ! Putting words on it sure helped me.

I will make some exercises and tests ot 'see' that you are right. :) Even if I know that you are.

Thank you for your help !


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