How quickly should a fluid come to hydrostatic equilibrium?

Question

Let's say I'm holding a one-liter water bottle, full of water, which I then drop.

Before dropping the water bottle, the equilibrium is for there to be a pressure gradient in the water canceling the gravitational force on the water. While the bottle is in free fall, the new equilibrium is constant pressure everywhere. Should I expect the water to come to this new equilibrium in the few tenths of a second it takes the water bottle to fall?

I expect the answer is basically yes, because density changes (and therefore pressure changes) should propagate at around the speed of sound, and p-waves might bounce around a few times while exponentially dying away (depending on boundary conditions created by the material of the bottle?), at the end of which we have equilibrium. So for a 30-cm bottle with sound speed 1500 m/s, I might guess the time is a few times .02s, which is longer than the ~.5s it takes for the bottle to fall from my hand to the ground.

Does this sort of reasoning make any sense? How can I justify it in a less handwavy manner?

Answer 1

My problem with the assumption is that sound is quite poorly absorbed in water. The $$30 cm = 1/4 \lambda$$ size means you'd look at waves of about 120 cm = 12 Hz. Absorption at those frequencies is measured in deciBels per kilometer. If we'd model the bottle as a cylinder, we might get a standing wave pattern that could persist for several kilometers (i.e. seconds).

Of course, there's likely going to be some non-axis aligned component, the bottle won't be a cylinder, so there's energy spilling over to other wave components, you get turbulence, and that does dissipate energy quickly. But quantifying that turbulence and its energy loss is a pain in the backside.

Also this is entirely ignoring cavitation.

Answer 2

You could start with a pack of cards and ask how long does it takes for the whole pack to free fall after

• the bottom card supporting the rest of the pack is released
• all cards are individually held from the sides, and then released at the same time.

From this, I think the answer to your original question is that it depends upon

• the shape of the bottle which influences how it initially supports the liquid
• the speed of the shock wave through the sides of the bottle when it's released from rest
• the speed of sound in the liquid

I'd expect a 'v' shaped bottle to reach equilibrium in a "significantly" different way to one with straight sides.

Answer 3

The Question is basically "how Quickly"? It's definetly not "instantly", and there is some delay, which causes movement inside the fluid and it must take some time before viscous forces kills these movements. And as they are, by nature exponential, means the lower the velocity, the smaller the losses and thus the time can be considerably different according to what exactly is meant by "hydrostatic equilibrium?"

Wikipedia says that this is the state "when it is at rest". I propose you take a beer bottle, and hit it hard on a table. In 0.2 seconds the fluid is back in it's "final postion" and in few seconds you don't see any waves. But it takes minutes before you can open it without explosive spilling. And it might take hour or two to be able to open it without any spilling. And it might take over night to have it as "rest" as it was be fore the hit.

I have looked these Old MIT video series completely, and though they work with water, it's often stated that the tank was left over night or over 24 h to bring it definitely in rest. In this video, MIT Cavitation they even use this difference of fresly filled fluid as reference. Cause cavitation is really sensitive to nucleoiding points.

So the my answer is, you need hours to reach hydrostatic equilibrium. In any case few seconds are not enough, if you are about to make some serious science around this theme.

Answer 4

How quickly should a fluid come to hydrostatic equilibrium?

All depends on how fast the mechanical waves attenuate in that fluid.

If a spring is compressed by a mass M, like in the attached picture, the system being in static equilibrium, and suddenly the gravity field disappears (the table remaining in its initial position, it is not movable) then the system reaches again a static equilibrium configuration when the oscillations stop. In an ideal case the mass will oscillate indefinitely. In practice the settling time is a function of how fast the spring dissipates its initial energy, $ky^2/2$.