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Would an airplane flying through superfluid helium experience lift and drag? The airplane is presumed cold enough to not heat up the helium.

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  • $\begingroup$ shouldn't it be a boat? fluid ? $\endgroup$ – anna v May 7 '14 at 18:30
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    $\begingroup$ @annav In general "boats," even submarines, have nonnegative buoyancy, while aircraft, even in liquid (remember, gas is a fluid too!) depend on dynamic lift. $\endgroup$ – Carl Witthoft May 7 '14 at 19:38
  • $\begingroup$ Your vehicle could not have a power source. It couldn't have a combustion engine (or the helium would evaporate). Chemical reactions (for a battery) would be very sluggish (probably not at all). All that is left is the momentum it had when it entered the helium. $\endgroup$ – LDC3 Jul 12 '14 at 22:47
  • $\begingroup$ @LDC3: I think the question could be better formulated as placing a fuselage with wings in a fluid tunnel, in stead of a wind tunnel, where there is relative motion between the fluid and the fuselage and then measuring the lift and drag of the fuselage sustained from the fluid. That should resolve the problem of having to have an engine on the plane. $\endgroup$ – Hans Jul 12 '14 at 23:03
  • $\begingroup$ @LDC3 You could have a propeller blade rotated by a spring hidden inside the fuselage. $\endgroup$ – mpv Jul 13 '14 at 12:25
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Viscosity is necessary in order for the wing to generate lift. Without the change in circulation caused by flow separation from the trailing edge, there will be no lift. In an inviscid fluid there will be no separation, and hence no lift. A similar flow pattern can be observed in viscous fluids when the Reynolds number is extremely low (Re<<1), and you can see for yourself that there is a conspicuous absence of downwash behind the wing in this case.

enter image description here

$(Re<<1$, No Flow separation, $Lift=0)$

enter image description here

$(Re>>1$, Flow Separation, $Lift\not=0)$

So, based on my understanding of superfluids I think the flow pattern will be similar. Also, I believe an experiment was done with a micro-turbine in supercooled Helium in which the turbine produced near zero torque (implying that no lift was generated). So the answer to your question is, in principle, the plane should experience neither lift nor drag. However, I can imagine scenarios in which this would not be strictly true due to the non-ideal nature of real flows.

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  • $\begingroup$ I'd love to see a citation for the superfluid turbine experiment you mention. $\endgroup$ – rob Jul 13 '14 at 3:34
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    $\begingroup$ The included paper is the Ph.D thesis for Paul P. Craig of Caltech circa 1960, and (as a note of trivia) it turns out that one of the members of his thesis committee was none other than one Richard Feynman. Who knew!!! $\endgroup$ – Bryson S. Jul 13 '14 at 3:52
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Superfluids have zero or near-zero viscosity so the drag is likely :-) to be near zero. However, $lift$ is primarily due to conservation of momentum. As the aircraft wing presumably has a nonzero angle of attack, it will be forcing helium atoms downwards, thus forcing the wing upwards.

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    $\begingroup$ Wrong. The flow will not be directed downwards in a flow with zero viscosity (see Kutta condition). A similar thing happens with creeping flow (Re<<1). $\endgroup$ – Bryson S. Jul 13 '14 at 2:49
  • $\begingroup$ Ok, in a zero viscosity fluid, when I push a flat plate @ 45 degrees, where will the molecules go? For that matter, there is rather a lot of disagreement about Kutta. E.g. secretofflight.wordpress.com/why-kz-is-incorrect $\endgroup$ – Carl Witthoft Jul 13 '14 at 20:32
  • $\begingroup$ I know it seems counter-intuitive because we rarely see this type of thing happening at everyday Reynolds numbers, but it's true. The shape of a 747 will not make lift if it were moving through honey, it just won't. It's the same reason why sharks and dolphins have fins to propel themselves but bacteria must use a flagellum or corkscrew type of method. No separation = no lift. $\endgroup$ – Bryson S. Jul 13 '14 at 20:35
  • $\begingroup$ @CarlWitthoft: In a superfluid the molecules will first go down and then up producing exactly zero net lift. (see the image in Bryson's answer). But it turns out that real superfluids aren't theoretical superfluids. They always have a real part so they can impart lift - just very, very weak. $\endgroup$ – slebetman Jul 14 '14 at 4:02
  • $\begingroup$ @slebetman Thanks for the detailed info. I always like to learn new stuff. $\endgroup$ – Carl Witthoft Jul 14 '14 at 11:13
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Presumably superfluid is still largely incompressible under small pressure, so it obeys Bernoulli's principle. With the normal shape of the wing, it will still generate lifts due to the pressure difference between the top and bottom wing surfaces.


Edit: I was wrong above. If the superfluid is inviscid and irrotational, the D'Alembert's paradox states that the net force is zero. However, the plane is in three dimension, there will be vortices spilling from the tips of the wing. The flow is thus rotational, violating the premise of D'Alembert's paradox. There will be induced lift and drag. A phenomenological theory called Prandtl's lift-line theory can be used to compute the net force induced by the vortices.

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    $\begingroup$ Is this the infamous simple and wrong airplane explanation? $\endgroup$ – André Chalella Jul 13 '14 at 3:13
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    $\begingroup$ There is nothing wrong with using Bernoulli's principle to relate pressure to velocity. The reason this principle is commonly abused is because a spurious explanation is given for why the velocity is higher in one area versus another... $\endgroup$ – Bryson S. Jul 13 '14 at 12:19
  • $\begingroup$ @BrysonS.: I have corrected my answer. Please review. $\endgroup$ – Hans Apr 21 at 2:48

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