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A moment ago, I was emptying bottles filled with water (2 liters) that are on the terrace of my house. As I did so I remembered something I saw on TV a some time ago (I don't remember when or where or exactly what) but I made a move on the horizontal plane as if the bottle drew a circle several times.

To my surprise, after several laps, a vortex was formed and the bottle did empty in half the time than those that are simply put upside down.

I repeated the experiment 20 times to be sure, and always got the same results.

  • Bottle upside down: 20 seconds. (approx)
  • Bottle upside down and rotation to form a vortex: 10 seconds (approx)

Clearly the vortex allows air to enter the bottle faster than in the other case.

I would like to understand what happens in physical terms.

EDIT:

Another interesting fact is that in the first 6/7 seconds just the first liter is gone. And in just two seconds (after the vortex is formed) the remaining water goes fast, very fast.

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  • $\begingroup$ The fasted way it to blow in a straw you put through he opening. $\endgroup$
    – ratchet freak
    Commented Dec 5, 2014 at 10:12
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    $\begingroup$ Is it still faster than holding the bottle at an angle to allow air in as you pour? $\endgroup$
    – Superbest
    Commented Dec 5, 2014 at 12:51
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    $\begingroup$ @Superbest in a full filled bottle, although it is difficult to maintain backward optimal angle to avoid bubbles occur, it is even slower. Right now I did some experiments and takes between 20 and 25 seconds. $\endgroup$
    – rnrneverdies
    Commented Dec 5, 2014 at 13:03
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    $\begingroup$ @ratchetfreak: depending on straw length you don't need to blow, just let air go in the old fashioned way. See Strawpedoing as a good example - youtu.be/dSr3AbY4HtM?t=22s $\endgroup$
    – Chris
    Commented Dec 5, 2014 at 15:09

3 Answers 3

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Assuming you start with a full bottle of water, when you tip the bottle upside down, a 'partial vacuum' (ie below atmospheric pressure) is created at top of the bottle as the water pours out the bottom. Atmospheric air then 'bubbles through' the mouth of the bottle to compensate. This slows down the flow of water through the mouth of the bottle. Each time air is 'bubbling' its way into the mouth, it impedes the flow of water out the bottle.

Once some air gets into the bottle, this air can 'expand' to let some water out of the mouth, until the air pressure inside drops sufficiently low that atmospheric air can 'bubble through' the mouth again. The more air that is inside the bottle, the more this air can 'expand' before bubbling through. That's why the water pours out faster as the bottle gets emptier.

When you swirl the water, the vortex forms a 'gap' in the centre of the mouth of the bottle through which air can flow freely from atmosphere into the bottle. As long as this continuous 'gap' of air is maintained from the mouth of the bottle to the 'top' of the water inside the bottle, there is no 'partial vacuum' above the water, so it doesn't 'bubble' and the water can pour out more evenly & hence faster.

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    $\begingroup$ I think angular momentum also plays a role in maintaining the vortex, since normally if you hold a bottle up right and induce a vortex, it will not last very long (at least for the column of air) due to friction, while with the upside-down bottle while the water is poring out its radius is reduced which would mean its angular velocity would have to increase, counteracting the decrease of friction. $\endgroup$
    – fibonatic
    Commented Dec 4, 2014 at 23:03
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    $\begingroup$ The use of the term 'vacuum' here worries me. Some will get the wrong idea. $\endgroup$
    – Almo
    Commented Dec 5, 2014 at 8:44
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When water leaves the bottle, the pressure above it drops. This reduces the net force pushing the water out of the opening, until it stops and a bubble can rise up. When the bubble has left the mouth of the bottle, the water can start flowing again. The stop-start of the water, and the reduced pressure inside the bottle, contribute to the lower flow rate in the bubbling case. We can actually estimate the difference in efficiency.

Case 1: vortex.

Simplifying assumptions: - the water comes down half the aperture and the air comes up the other half. - Air pressure is maintained inside the bottle at atmospheric pressure. - The tangential water flow velocity was generated by swirling the bottle initially and we only concern ourselves with vertical velocity.

For water height in the bottle of $h$, the vertical velocity is given (from conservation of energy) by

$$v = \sqrt{2gh}$$

And the mass flow rate $M = \rho v \frac{A}{2}$ where $\rho$ is the density and $A$ the full area of the bottle opening.

Case 2: bubbles.

When the bottle is bubbling the water will keep stopping and starting - but when it flows it has the whole aperture available. But since the water needs to accelerate, then decelerate (as the pressure above the water drops, the velocity goes back down to zero) we can see that the mean velocity will be quite a bit lower. In fact, if the time between bubbles is $T$ and we assume that the pressure above the water is lowered by the weight of the column of water below it when the next bubble is formed, we can write an approximate expression for the net pressure:

$$P_{net} = \rho g h (1 - \frac{t}{T})$$

On average, this is half the pressure experienced by the free flowing water. And using this mean pressure, the water needs to first accelerate, then decelerate. This means that the average velocity of the water will be half what it would otherwise be (if it didn't keep stopping and starting), and half again since the mean pressure difference is halved. This gets a mean flow velocity that is $\frac14$ of the free flowing value - but we have twice the aperture. The net result is that the flow with bubbles is about half as fast as with the vortex. Which, coincidentally, is exactly what you observed.

Note that the above uses many simplifying assumptions - but the basic mechanics I described is plausible. If anyone has a more complete mathematical description I invite them to offer it up.

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  • $\begingroup$ I think this also answers the question I had while thinking about it of what about pouring a bottle normally but making sure that there is always an air path. It should probably use similar calculations to the first but with a lower h due to the more horizontal angle of the bottle. Does that sound about right? $\endgroup$
    – Chris
    Commented Dec 5, 2014 at 14:45
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    $\begingroup$ @Chris yes - when the air flows continuously you maintain the constant flow but if you don't have the $|Delta h$ the liquid will move slowly. The swirl gets you the best of both worlds. $\endgroup$
    – Floris
    Commented Dec 5, 2014 at 15:01
  • $\begingroup$ About your third sentence: I don't think anything particularly interesting happens when the bubble reaches the water level at the top. It is on entering the bottle that the air (not yet a bubble) counters the low pressure that was keeping the water from coming out. And exactly because the water then starts pouring again the air flow is interrupted and an air bubble formed, which can only expand a bit, so the pressure starts dropping again. If you had a very tall bottle (so bubbles take long to get to the surface) things would not evolve mush slower, as your answer suggests. $\endgroup$
    – Marc van Leeuwen
    Commented Dec 6, 2014 at 12:52
  • $\begingroup$ Relevant video for the vortex case: twitter.com/AMAZlNGSCIENCE/status/1303013180692553728?s=20 $\endgroup$
    – Wolpertinger
    Commented Sep 8, 2020 at 9:02
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I think you've understood it all, air gets into the bottle faster. Without the vortex, the air is able to pull on the liquid, preventing it from escaping.

This is why you can pour orange juice faster if the opening is at the top, rather than the bottom. It also stops it splashing.

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