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Consider a cylinder filled partially with a liquid (e.g. water). The cylinder is sealed, and is at held at room temperature (e.g 298K). At equilibrium (or when no external disturbance is imparted to the system), the liquid in the cylinder exists in equilibrium with its vapor at the vapor pressure of the liquid, applicable at room temperature.

Suppose that the bottom of the cylinder is fitted with a piston. Note that the system is still sealed. Now imagine that the piston moves upwards. The liquid column moves upwards, the volume the vapor occupies decreases, but the vapor still exists at the vapor pressure of the liquid (assuming that thermodynamic equilibrium can be achieved fast enough).

What's interesting is when the piston move downwards. If the piston moves at a slow speed, the liquid column should stay "on top of" the piston (without any "gap" between the column and the piston). Now, if the piston moves downwards fast enough (or when the downward acceleration is high enough), the liquid column shall "leave" the piston, and a "packet" of vapor shall be formed between the piston and the liquid column. (Please refute the former claim if you think it's wrong.) Why? if the "friction" between the cylinder walls and the liquid column is negligible, and the liquid column is already in a state of free-falling, there is no mechanism to "pull" it downwards further anymore.

Now, imagine that the piston moves upwards again. Then the liquid column shall "collide" with the piston. Will it "stick" to the piston, as in an inelastic collision, or recoil, as in an elastic collision?

P.S. After thinking for a while, I think it is not at all an easy question to answer. Now whether the liquid will "recoil" (not "slosh", which implies that the liquid changes shape) depends on how momentum is transferred from the piston to the liquid column. Please refer to the youtube video for the "beer bottle trick". If the cavity formed at the bottom is a "vacuum", which is probably the case, then when the liquid "smash back", the atmospheric pressure shall press on the liquid column, and most probably it will not "recoil". On the other hand, if the cavity formed is gas-filled, then the kinetic energy from the liquid (which is trying to "smash back" on the glass bottle) may be dissipated to the gas (in the original gas-filled cavity), through the formation of many tiny cavitations (claim: the latter claim is not sound; I am just guessing). For as long as the impact is "dampened", a recoil will not happen.
Indeed there are many ways in which a liquid may dissipate energy, because it's formless/shapeless, and there's viscosity in the picture. The more "flexible" it is to energy dissipation, the less likely it is to recoil.

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  • $\begingroup$ Check out YouTube for videos of people shattering beer bottles in a similar manner. Fill the bottle half-way, cover the top, and shake vertically very quickly. The liquid rises, pulls a vacuum underneath, and "snaps back" hard enough to break off the bottom of the bottle. $\endgroup$ – Carl Witthoft Aug 22 '14 at 12:17
  • $\begingroup$ Thanks! Where's the video? Indeed I know my claim can be checked with some very simple apparatus... "pulls a vacuum underneath"...how do you "observe" that? Of course when you shake beer hard enough, carbon dioxide will come out, creating a very high pressure pressing down on the beer, which then, together with its weight and kinetic energy, "cracks" through the bottom of the beer can. But again, where's the video? $\endgroup$ – Jamie Aug 22 '14 at 13:27
  • $\begingroup$ Thank you very much! The link is here: youtube.com/watch?v=lj3x2U4CaEs. Check out the slow motion session. It is clear that a "vacuum", or rather vapor bubbles, is/are left behind as the bottle is "accelerated". The liquid then catch up and snap back onto the bottom. The "bubbles" are not waves; if they are they should not appear just at the bottom, but one should see them either travel upwards or downwards. Indeed this is related to the so-called "water hammer" effect. But my original question is still not answered. $\endgroup$ – Jamie Aug 22 '14 at 13:40
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Your question is somewhat more "general" than what is implied. You are essentially talking about cavitation (or getting very close to talking about it). Cavitation does not require slugs of liquid traveling downward. In fact, there are internet pictures showing ship propeller designs where the engineers didn't account for the pressure drop on the downstream side of the propeller. If the pressure gets below the vapor pressure of the liquid, the liquid "boils" over a very short time interval, and when the pressure rises substantially, the associated small vapor bubbles instantly collapse. When the collapse happens, the vapor bubble temperature rises dramatically and a very small shock wave is formed. If this collapse happens to occur on the metal surface, the shock wave WILL remove small bits of metal. If this condition persists, the propeller will be worn away.

If you are wondering how high the bubble temperature can get, there are experiments whereby ultrasound is used to deliberately create cavitation in bulk liquids. When the bubbles collapse, the temperature gets high enough to emit a very small flash of light, and the process is called sonoluminescence. Measurements indicate that the temperatures that produce this small flash of light are higher than the temperature at the surface of the sun, so while the process within one of these bubbles is extremely small, it is also extremely violent.

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Because of gravitational forces you liquid is at the bottom of both phases.

A pressure gradient, due to those forces exist in both phases; the pressure increases linearly through bot fluid, with a discontinuous ramp at the interface liquid/gaz.

So if you push down the piston, by increasing the volume you will modify the equilibrium point and so decrease the equilibrium pressure. The only way you can create an extra gaz volume is by cavitation of the fluid, which can only occur at the minimal pressure point of the liquid i.e the fluid/gaz interface. So I think you can't actually create some kind of "gap" between the piston and the liquid, as far as your piston speed is lower than the sound speed (the pressure information will be faster than the piston and so create evaporation at the interface to balance the volume modification).

If you piston is faster than sound velocity over liquid water (really?) I think you can actually create a local dilatation and so bubbles at the bottom of the fluid. This non stationary evolution is certainly really hard to model.

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  • $\begingroup$ Thanks for answering. "... the pressure increases linearly through bot fluid...". This is true, if the liquid is not accelerating. The critical point is, the liquid accelerates to such a point, such that it is already free-falling. In this case, if a small "bubble" already exists between piston and the liquid column, (ignoring surface tension) can it not grow, by your arguments? $\endgroup$ – Jamie Aug 22 '14 at 8:36
  • $\begingroup$ through "both" fluids sorry. Due to the gravitational forces. In fluid static model you have dp/dz=-rho*g with an ascendant z axis. $\endgroup$ – Nicolas Berteloot Aug 22 '14 at 8:40
  • $\begingroup$ I think that in any case, you create a local pressure/decompression of the fluid which leads to some extra cohesion forces which prevent your fluid from getting away from the piston. It's the same phenomenon that you have when you fill in a pipe, then cover one edge with your hand and put it in a vertical position: the fluid doesn't fall because of this extra force; the motion of the fluid would create a gaz gap. (I have to say I don't know what are the exact cause of this...). So you probably create a large amount of bubbles that go through the fluid up to the interface, but not a real pocket $\endgroup$ – Nicolas Berteloot Aug 22 '14 at 9:11
  • $\begingroup$ Please see the following link: nasa.gov/mission_pages/station/research/news/… Note that the "geometry" of the liquid within a container, even under micro-gravity, is very hard to predict. Unless there is some force which "sticks" the liquid to the container walls, I don't think that the situation is at all obvious, as your reasoning suggests. $\endgroup$ – Jamie Aug 22 '14 at 11:31
  • $\begingroup$ @NicolasBerteloot This question reminded me of the xkcd/what if question and this youtube video linked to in the answer. If I understand the "liquid column/piston"-scenario correctly it is quite simlar to the "hitting the top of the bottle"-scenario in the video. You can clearly see the cavity formed in the video. So according to the reasoning in this answer, that would mean the bottle was moving faster than the speed of sound in water, right? That doesn't seem very likely. $\endgroup$ – jkej Aug 22 '14 at 17:53

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