# Airfoil lift in superfluid flow

A superfluid has zero viscosity.

Lift of an airfoil (and any shape for that matter) is zero without fluid viscosity. So, superfluids shouldn't generate lift when flowing around an airfoil.

Is that true? I think it is impossible for a superfluid not to generate lift on an airfoil, because, as the theory goes, the fluid particles would have to have infinite velocity at the sharp trailing edge.

What do you think?

## Answers and Replies

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K^2
Superfluids have superfluid component and normal component. While superfluid component generates no lift, normal component does.

I saw a video on Youtube that shows propeller turning in superfluid helium. I'll try to find it.

DrDu
Yes, I think that similar things happen to that in the rotating bucket experiments. I.e. the formation of strings of normal fluid which carry rotation and compensate for the rotation around the wing which is responsible for the lift.

Any fluid flow will create lift if the aerofoil is at a positive angle of attack.

Gold Member
xxChrisxx said:
Any fluid flow will create lift if the aerofoil is at a positive angle of attack.
That is not true. With no viscosity in a steady flow, lift will not be generated no matter what. You need viscosity. d'Alembert's paradox is a good thing to look at to illustrate this.

inviscid-lift said:
Lift of an airfoil (and any shape for that matter) is zero without fluid viscosity. So, superfluids shouldn't generate lift when flowing around an airfoil.

Is that true? I think it is impossible for a superfluid not to generate lift on an airfoil, because, as the theory goes, the fluid particles would have to have infinite velocity at the sharp trailing edge.
Theoretically, this is true. As mentioned before, most superfluids have a superfluid part and an ordinary part, so you will have some viscosity that way. As I stated in the PM I sent you, in theory you would have an infinite velocity around that trailing edge, but even if you had a perfect superfluid, you can't get a trailing edge that is any sharper than the radius of an atom (any you really can't even get that small). That leads to a huge but finite velocity.

DrDu
This is a good set of experiments on liquid helium. Video quality is not very good, but it covers all the important points without getting too scientific.

Oh, what a wonderful video from old times when physicists still wore shirts and ties instead of safety gogles and spilt t-shirts!

K^2
That is not true. With no viscosity in a steady flow, lift will not be generated no matter what. You need viscosity. d'Alembert's paradox is a good thing to look at to illustrate this.
Just to add to this, the separation layer formation, necessary for lift, depends on viscosity. In an ideal superfluid, layers would not separate, meaning that circulation would always be precisely zero, and no lift would be produced.

I can't comment directly about super fluids, however the theory of lift for nonviscous fluids requires a bound vortex around the wing. Lift develops in a uniform flow field with a superimposed vortex flow. A foil initially immersed in a nonviscous, uniform flow field will not develop a bound vortex. There is no torsional force on the fluid to generate one. In real life, the fluid viscosity serves this role.

However, there is a caveat. The flow in the above, simple, superposition model is two dimensional, so the vortex field doesn't need to terminate. In three dimensions, I'm unsure whether a vortex that terminates upon itself (a 'smoke ring') can be produced. The idea, here, being that a closed loop vortex filament could have zero total angular momentum.

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rcgldr
Homework Helper
This is a good set of experiments on liquid helium. Video quality is not very good, but it covers all the important points without getting too scientific.

I watch the video segments and one thing I don't quite understand is why the experiment with the pourous material made up of very tiny pathways is considered to exhibit zero viscosity? Isn't it possible that an interaction between the helium and the other material or something like a surface tension effect is what is different between stage I and II of liquid helium as opposed to the viscosity within the helium itself?

Here's an image--crude but informative, of combining uniform translational flow together with rotational flow about the foil, in a nonviscous fluid to generate life.

[PLAIN]http://www.centennialofflight.gov/essay/Theories_of_Flight/Airfoils/TH13G4.jpg [Broken]

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K^2
^ That's a very good illustration.
I watch the video segments and one thing I don't quite understand is why the experiment with the pourous material made up of very tiny pathways is considered to exhibit zero viscosity? Isn't it possible that an interaction between the helium and the other material or something like a surface tension effect is what is different between stage I and II of liquid helium as opposed to the viscosity within the helium itself?
Because these aren't just straight channels. These are pores. Even if you take away the condition that boundary layer has zero velocity, which comes from particle interactions, and leave only that the velocity in normal direction is zero, just due to the geometry of pores, this is sufficient to say that a flow with non-zero viscosity cannot flow at any reasonable speed.

DrDu