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I am taking some introductory fluid dynamic classes, and have become very confused by the Kutta-Joukowski theorem. One of the conclusions that can be derived by applying Kutta-Joukowski is that a spinning cylinder in an inviscid flow will produce lift.

But how can this be the case? Let's consider this diagram of lift on a spinning cylinder in viscous flow:

enter image description here

The mechanism by which this spinning cylinder would produce lift would be the unequal viscous shear on either side (due to different relative velocities to the fluid). The air on the bottom of the cylinder would be accelerated by viscous shear, whereas the air on the top would be retarded. This moves the downstream stagnation point (or equivalently, causes a deflected downstream wake), and the rest is Newton's third law.

But in an inviscid flow, there is no way for the airflow to be entrained by any viscous effects! Even if we accept the no-slip condition at the boundary of the cylinder, how does the air traveling at the boundary of the cylinder accelerate the air near it without relying on viscous shear?

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    $\begingroup$ You must have misunderstood something (or your professor). A spinning cylinder in inviscid flow does not produce lift. $\endgroup$
    – Peter Kämpf
    Commented Mar 24, 2015 at 12:12

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Inviscid flow doesn't exist.

That's so important to understand, I'll say it again:

Inviscid flow doesn't exist!

However, we use it all the time. So what gives? It turns out that many of the effects we are interested in are viscous, but the viscous effects can be modeled various other ways. This is effectively the same type of question as Does a wing in a potential flow have lift?

Just like with a wing, we can ignore the starting problem by saying all of the initial transients and vorticity generated by viscous effects are far from the spinning cylinder. But we know circulation must be conserved, so that starting vortex that was generated by the viscous effects causes us to attach a vortex rotating in the opposite direction to the cylinder (or the wing, airfoil, etc.). This is the bound vortex that we encounter all the time in aerodynamics. It allows us to model the generation of lift around a body in an inviscid flow.

But it's really important to remember that lift does not exist without viscosity and likewise inviscid fluids do not exist! They are a mathematical construction that enables us to model phenomenon we are interested in modelling.

When we model this flow, we super-impose a vortex solution with a uniform flow solution. This gives different velocities on top and bottom and this results in different pressures and thus lift. But it only exists because we impose this vortex solution where the cylinder is. And we can do that because it is maintaining the requirement that there be zero circulation throughout the entire fluid (which includes the initial starting vortex). My very poor drawing of this for an airfoil found on my answer to why a wing has lift illustrates this:

enter image description here

When we solve using the potential flow equations, we essentially cut our domain along that dashed line, removing the starting vortex from the domain.

Okay, superfluids are essentially inviscid. But we're not flying through super cold helium (usually) so we can ignore that exception.

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    $\begingroup$ Interesting. You assert that inviscid flow does not exist, then devote the longest paragraph of your answer to arguing away the complications associated with viscosity, to finally give a counterexample where inviscid flow does exist (and not just essentially, but absolutely so, by the way). Is there also a relevant aspect to your answer that I missed? $\endgroup$
    – user73762
    Commented Mar 23, 2015 at 21:52
  • $\begingroup$ @pyramids I'm not sure I understand your objection/question. We use inviscid assumptions along with other tools (ie. the idea of a bound vortex, or the Kutta condition for airfoils) to mimic the effects that viscosity would have to complete our model of the real flow. $\endgroup$
    – tpg2114
    Commented Mar 23, 2015 at 21:54
  • $\begingroup$ My whole point is you can't just look at an "inviscid" solution and have it make any physical sense without considering how you got there from the real problem, nor without recognizing what things we introduce as a model to make sure we get a good answer. $\endgroup$
    – tpg2114
    Commented Mar 23, 2015 at 21:55
  • $\begingroup$ @tpg2114 I don't understand this. Obviously inviscid fluids don't exist. But as a mathematical construction they are supposed to help us understand the behavior of actual fluids in the limiting case of viscosity approaching zero. Starting in the high Reynolds number regime, if I decreased viscosity, I would expect the lift to decrease, until finally at infinitesimal viscosity I would expect infinitesimal lift, and in the unphysical case of zero viscosity I would expect my mathematical model to yield zero lift. This is not the case. $\endgroup$
    – Asad Saeeduddin
    Commented Mar 23, 2015 at 22:27
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    $\begingroup$ It's the exact same situation I depicted with the airfoil. And it has absolutely nothing to do with assuming $\nu \rightarrow 0$. Instead, you are modeling the flow about a body using the linear potential flow equations which allows you to superimpose elementary solutions to construct a flow that satisfies the boundary conditions you would get in a viscous flow. Here, you add a uniform flow to a vortex which represents the cylinder rotating. $\endgroup$
    – tpg2114
    Commented Mar 23, 2015 at 22:41
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The mechanism is known as Bernoulli's law (see e.g. wikipedia or scienceworld): A moving fluid, regardless of viscosity, has a different apparent pressure along the direction of motion and perpendicular to it. Along perpendicular directions, the pressure is reduced. This is the situation seen by your cylinder, which (when rotating as in your diagram) will see a higher relative flow speed on its upper side than on its lower side. The difference in pressure is lift.

EDIT: A perhaps even better pointer on Bernoulli's law than the links I gave above is hyperphysics. To communicate about the particular problem (lift generation from a rotating cylinder moving through a fluid), it helps to state its historic name, Magnus effect (wikipedia link).

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  • $\begingroup$ "Along perpendicular directions, the pressure is reduced". I'm afraid I don't undestand this. Reduced by what? $\endgroup$
    – Asad Saeeduddin
    Commented Mar 23, 2015 at 22:36
  • $\begingroup$ By the directional effect of superimposing a collective motion. Think of pressure as energy density (same units!). Adding velocity leads to a nonlinear increase of it in the forward direction (individual molecule speed squared enters), which must be compensated by a decrease elsewhere for energy conservation. Bernoulli worked out the details for us. $\endgroup$
    – user73762
    Commented Mar 23, 2015 at 22:48
  • $\begingroup$ But "adding velocity" only suggests that a pressure gradient exists along a streamline, or at least that's my understanding of Bernoulli's law. You can't compare the velocities of fluid at points in different streamlines and claim there exists a pressure gradient between the points because the velocities are different, right? $\endgroup$
    – Asad Saeeduddin
    Commented Mar 23, 2015 at 22:59
  • $\begingroup$ Why not? If you let a parallel fast and a slow stream interact, the slow stream will get sucked into the fast stream. You seem to already agree if I introduce a thin grating to separate them (and perhaps make one of them fast by acceleration in a gravitational field), but the same effect happens if I make the grating arbitrarily sparse, even in the limit of letting it vanish $\endgroup$
    – user73762
    Commented Mar 23, 2015 at 23:10
  • $\begingroup$ "If you let a parallel fast and a slow stream interact, the slow stream will get sucked into the fast stream." I seriously doubt this would happen if you discount the effect of viscous shear. Eg. when you have air flow in a pipe, there is a parabolic radial velocity distribution, but the static pressure is the same throughout the cross section, all the way from the zero velocity edge to the high velocity center. $\endgroup$
    – Asad Saeeduddin
    Commented Mar 24, 2015 at 1:34

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