12
$\begingroup$

Liquid helium (and other similar fluids) can "climb up" the walls of their containers. Who does the work in this case, and what kind of energy does it use? I'm sure we can't make a perpetuum mobile out of this, so I guess some kind of energy must somehow be expended to make the fluid "climb up" the wall.

$\endgroup$
8
$\begingroup$

Courtesy of the book Carl found we have an answer!

Helium

Consider the element of the liquid helium at a height $h$ above the fluid surface and distance $y$ from the wall. To raise that element above the fluid surface costs an energy $mgh$, but because there is a Van der Waals attraction between the helium atoms and the wall you get back an energy $E_{VdW}$. Dzyaloshinskii et al give the energy change per unit mass as:

$$ \Delta E = gh - \frac{\alpha}{y^n} $$

where $\alpha$ is constant giving the strength of the Van der Waals attraction and $n$ is in the range 3 - 4 depending on the film thickness. So it is energetically favourable to lift the fluid up the wall if the Van der Waals attraction outweighs the gravitational potential energy making $\Delta E$ negative. Since $y$ can be taken arbitrarily small (well, at least down to a few times the He atom size) $\Delta E$ will be negative for all heights $h$ and the film covers the whole wall.

The resulting equation for the film thickness $d$ as a function of height is given (without derivation) as:

$$ d \approx \left( \frac{\alpha}{gh} \right)^{1/n} $$

Since the liquid film will have a non-zero thickness at the top of the container wall it can flow over the wall and then down the outside. Even though the film thicknesses work out to be only a few tens of nanometres the zero viscosity of the superfluid helium allows an appreciable flow rate. Indeed, later in the book flow velocities of 30 cm/s are mentioned.

In principle this would apply to all fluids, however for normal fluids the flow rate in a film a few tens of nanometres thick would be infinitesimally small so the climbing is never observed.

A few comments of my own: I note that this derivation ignores the interfacial tensions of the helium/air, helium/wall and air/wall interfaces. I have no figures for what these would be for superfluid helium and possibly they are negligable. The predictions of the Dzyaloshinskii theory are claimed to agree well with experiment. Also you should note that one of the references provided by Carl challenges the above explanation, though without coming to any firm conclusions.

$\endgroup$
  • $\begingroup$ So, if I understood it correctly, from the point of view of entropy, it's basically climbing the wall because having a layer all over the wall is its most stable state? (just as water in zero gravity forms a sphere) $\endgroup$ – vsz Dec 17 '13 at 19:13
  • $\begingroup$ @vsz: If you replace entropy by energy then yes your summary is correct. Though water in zero g forms a sphere due to surface tension, and surfacer tension is ignored in the calculation of the He film. $\endgroup$ – John Rennie Dec 17 '13 at 19:15
2
$\begingroup$

Possibly helpful:

http://arxiv.org/ftp/arxiv/papers/1103/1103.0517.pdf

www.paper.edu.cn/download/downPaper/200812-856‎

The bizarre behaviour of superfluids! Climbing up walls and geting out of glass beakers

EDIT: A googlebooks excerpt seems more useful:

$\endgroup$
  • $\begingroup$ I found the paper you cite when searching for a copy of the Dzyaloshinskii paper, but since its going in point is "the accepted theories are wrong" I found myself a bit suspicious if it. $\endgroup$ – John Rennie Dec 17 '13 at 16:28
  • $\begingroup$ @JohnRennie fair enough. If you want to edit &delete the reference, go ahead. $\endgroup$ – Carl Witthoft Dec 17 '13 at 16:29
  • 2
    $\begingroup$ No, I'd leave the reference in and let readers decide. I do think your answer should really be a comment though as it doesn't answer the question. $\endgroup$ – John Rennie Dec 17 '13 at 16:33
  • $\begingroup$ Aha, that book does the trick. I've simplified its explanation and posted it as an answer. $\endgroup$ – John Rennie Dec 17 '13 at 17:11
0
$\begingroup$

the same kind of phenomenon appears for normal liquid but the explanation is different. for normal liquid, this phenomenon is called capillarity. In short, capillarity can be explained because the liquid like more to be in contact with the wall of the container than with the air.

for suprafluidity, heat creates the phenomenon. As a superfluid can't transmit entropy, if you heat it, then it will dissipate the energy by "climbing up" the walls. a similar experiment is called the fountain effect. you can easily google it...

$\endgroup$
  • $\begingroup$ I'm unconvinced that your first password is correct. The capillary rise doesn't depend on the fluid viscosity - well, not if you wait long enough. From what I can clean by Googling Dzyaloshinskii's paper then phenomenon is more complex than just wetting. $\endgroup$ – John Rennie Dec 17 '13 at 12:04
  • $\begingroup$ @JohnRennie i agree it doesn't depend on the fluid viscosity. i never said it does. i said that you can observe the same behavior, but with different explanation... $\endgroup$ – PinkFloyd Dec 17 '13 at 14:04
  • $\begingroup$ Do you mean that it climbs up the walls only if heated, and behaves as a normal fluid when kept at constant temperature? $\endgroup$ – vsz Dec 17 '13 at 15:01
  • $\begingroup$ @vsz: you need heat to get the fountain, but not for the fluid to climb out of a container. $\endgroup$ – John Rennie Dec 17 '13 at 16:29
0
$\begingroup$

The critical point for that phenomenon is the superfluid's viscosity being zero. For that, the deposit formed on the wall doesn't prevent the fluid from flowing out as it does for the normal fluid case.

$\endgroup$
  • $\begingroup$ This doesn't answer the question. You don't say how the increase in gravitational potential energy is balanced out. $\endgroup$ – John Rennie Dec 17 '13 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.