Does quasi-static motion imply zero energy dissipation?

When a droplet is deposited on a surface with some surface roughness and subsequently tilted it can stick due to pinning (think of droplets on a window after rain).

What I am interested in is how/whether I can show that all the potential energy from tilting the droplet is converted in surface energy needed for the deformation of the droplet. In other words, is there any energy dissipated in tilting the droplet, when I do it slow 'enough'? (ignoring the motor used for tilting)

Intuitively I would expect zero dissipation, because if I tilt the droplet a bit and then back to the initial position it will have exactly the same shape tilting forward as tilting back to horizontal. Thus if I would draw a force-displacement (of the center of mass) curve I would have a single line indicating, in analogy with elastic behaviour, no hysteresis, thus no energy dissipation. Is this reasoning correct? If so, how can I rigorously make this claim?

• I'm not sure how to show that there's no dissipation in this case (I suspect there isn't any), but the argument you give is not valid by itself. After all, if I pour water from a cup into an identical cup below it, it will have exactly its original shape, but I've clearly dissipated some gravitational potential. Commented Dec 10, 2013 at 10:00
• Your thought experiment does not explicitly involve quasi-static motion. Are you maybe thinking of a reversible process? Commented Dec 10, 2013 at 12:45
• @DoruConstantin Not necessarily reversible but as far as I can tell dissipation is would be velocity dependent thus going to zero for quasi-static motion. Commented Dec 10, 2013 at 16:00
• @Nathaniel - I think your counter example is not really the same, because in my case the droplet would go back to the original position (it doesn't slide when tilted) whereas in your case the liquid now has a lower center of mass. The equivalent to the droplet would be to pour water in an identical cup at the same height as the cup originally was. Assuming it is laminar (which is probably a bad assumption, but let's go with it anyway), you would again have the viscous losses of which the energy comes from you lifting the original cup. Commented Dec 10, 2013 at 16:06
• @Michiel sure. Maybe an even better analogy would be tilting the droplet non-quasistatically, so that it wobbles to a stop - in that case it's clear that kinetic energy gets dissipated, even though it ends up back in its initial shape. Commented Dec 11, 2013 at 1:10

Any reshaping of the droplet will require flow of water inside the droplet and there will be viscous losses. Presumably the energy would come from an increased torque on whatever motor was moving the droplet and substrate.

• Ok, that sounds logical. Is there any way I could quantify this and connect it to (lack of) hysteresis? Commented Dec 10, 2013 at 10:19
• There might be simplified models for deformation of a spherical membrane that you could adapt, though this isn't my field so I can't comment. I suspect that doing anything beyond the most basic approximation would require a finite element calculation. Commented Dec 10, 2013 at 10:21
• An order of magnitude estimate would be sufficient, but if I would use the case of a spherical membrane I am already assuming that bulk dissipation is negligible right? Commented Dec 10, 2013 at 16:02
• Because I would like to think of the droplet as a perfectly elastic membrane which stores all potential energy in surface energy, but than I would have to make a case that viscous losses are negligible small. Commented Dec 10, 2013 at 16:10
• If everything is perfectly elastic, the energy needed to deform the drop will be returned to the motor via torque in the opposite direction when you tilt it back again... Commented Dec 11, 2013 at 1:11

If you only consider viscous dissipation within the droplet, this should indeed go to zero in the vanishing velocity limit: the (local) dissipation rate is quadratic in the velocity, so that decreasing the velocity by a factor of $\lambda$ reduces the (local and global) dissipation rate by $\lambda^2$.

Of course, the process takes $\lambda$ times longer, but the overall dissipation still decreases by a factor of $\lambda$.

See Happel & Brenner, "Low Reynolds Number Hydrodynamics" for more details.