Is there a physical argument for the Kutta-Joukowski theorem?

Question

I am doing the following exercise:

Consider inviscid, incompressible, steady flow. The Kutta-Joukowski theorem $$L' = \rho_\infty V_\infty\Gamma$$ where

• $L'$ is the lift per unit span
• $\rho_\infty$ is the freestream pressure
• $V_\infty$ is the velocity
• $\Gamma$ is the circulation taken around the body

was derived exactly for the case of the lifting cylinder. Equation $L' = \rho_\infty V_\infty\Gamma$ also applies in general to a $2$-dimensional body of arbitrary shape. Although this general result can be proven mathematically, it also can be accepted by making a physical argument as well. Make this physical argument by drawing a closed curve around the body where the closed curve is very far away from the body, so far away that in perspective the body becomes a very small speck in the middle of the domain enclosed by the closed curve.

And here is the solution from the author:

The flow over the airfoil can be syntheized by a proper distribution of singularities, i.e, point sources or vortices. The strength of vortices, added together, give the total circulation, $\Gamma$, around the airfoil. This value of $\Gamma$ is the same along all the closed curves around the airfoil. In this case, the airfoil become a speck on the page, and the distributed point vortices appears as one stronger point vortex with strength $\Gamma$. This is exactly equivalent to the single point vortex for the circular-cylinder case, and the lift on the airfoil where the circulation is taken as the total $\Gamma$ is the same as for a circular cylinder, namely, equation $L' = \rho_\infty V_\infty\Gamma$.

Author's figure:

I understand the author's solution until the part that said:

the lift on the airfoil where the circulation is taken as the total $\Gamma$ is the same as for a circular cylinder.

The equation $L' = \rho_\infty V_\infty\Gamma$ is derived from the circular-cylinder case which we have to do the integral of the pressure distribution over the cylinder surface. But I have not understood why the lift on the airfoil is the same for a circular cylinder like author said.

Answer 1

The question as asked in the title is one of the great debates of the discipline of aerodynamics (and you can see by the number of times I've edited this answer that it's still bouncing around in my own head). If you've gotten this far in Anderson and are making this kind of inquiry, you must read McClean. The first 300-or-so pages are devoted basically to answering this very question. Basically: Yes, there is a physical argument (see Peter Kämpf's answer and Anderson 4.5), but saying that lift is caused by circulation is too much of a stretch (which Anderson notes in section 3.16).

Let me turn to your specific example. Unfortunately, I think Anderson is trying to force the mathematics into the physics rather than start with the physics and describe why the mathematics is a good representation (as McClean sets out to do from the beginning). "Make this physical argument by drawing a closed curve around the body..." basically tells us to make a physical argument by first using a mathematical construction, which to me doesn't actually shed any light on the physics. Further, then we're supposed to be thinking about infinite distances (not physically useful), a "speck on a page" (which has nothing to do with airflow), and really just make use of $\Gamma$ (which is defined purely mathematically) rather than describe a physical flowfield. In short, the closest thing to a "physical" argument here is saying more or less that any two things look the same when viewed from far enough away. This explanation does not seem convincing to me. If you put lots of little vortices around the profile of a cow then reduce it to a "speck on the page," will the cow generate the same lift as an airfoil? To get at the physical generation of lift, we need to remain close to the lifting bodies and determine why their specific geometries create flowfields that produce lift. (Consider, for example, why we need a Kutta condition on an airfoil but not on a cylinder.)

Actually, the mathematical proof (based on conformal mapping) uses the same basic "physical" argument that Anderson makes. Conformal mapping tends to be more of a graduate-level topic, though, so beware of feeling like you need to understand all the underlying math at this point.

In summary, Anderson does not do a good job here of explaining physically why the airfoil has the same lift as the cylinder. If you truly want to understand the physics, read McClean. Anderson's "physical" argument is really just a boiled-down mathematical one.

Answer 2

In a talk I attended the author made the convincing argument that only when the Kutta theorem is fulfilled will flow leave the airfoil parallel to the direction of the trailing edge. Every other value of $\Gamma$ would result in flows that try to flow around the trailing edge, and doing so would be physically impossible without separation. If one tried to calculate flow which follows the contour of a regular trailing edge, suction of a magnitude would be required which could not be created even by a vacuum. Only by fulfilling the Kutta theorem could physically observable flow be mathematically described.

If you replace the airfoil with a cylinder and assume the same inflow and outflow angles as on the airfoil, it will have the same lift. In reality, only a rotating cylinder will show something resembling this kind of flow and now the speeds induced by the central vortex are the same that the surroundings of the rotating cylinder show in a crossflow.