13
$\begingroup$

While pouring a liquid into a glass from a bottle, some streams have a wavy shape, like the one in the following photo:

example

What causes the stream to be of such a shape?

$\endgroup$
1

2 Answers 2

12
$\begingroup$

On the one hand, as the liquid is not being poured very slowly, there are different velocities in the incoming liquid: it's faster at the top, and, it seems, at the back. On the other hand, the cohesive forces (resulting in particular in surface tension at the surface of the flow) are sufficient here to prevent separation of some liquid from the rest — splattering. Then it is as if you had two masses going at different speeds (the liquid at the back and at the front) linked by some piece of rubber (cohesive forces) : they turn around one another.

The smooth appearance due to surface tension gives the final touch for a beautiful flow. You'll probably enjoy taking a tour in John Bush's gallery if you like this, http://math.mit.edu/~bush/?page_id=252. A very relevant example is the case of colliding viscous jets, pdf and below, which exemplifies beautifully the present discussion at low flow rate (left) but gives much more impressive results at higher flow rates! (right)

http://i.imgur.com/HCHZWo2.jpg

$\endgroup$
1
$\begingroup$

You have made a very interesting observation. In short, this is due to the shape oscillations of the flowing stream of water.

I did the experiment and observed the pattern too. Here you can see the same image from two different angles.

Front view

Front


Side view

side view


When you look carefully at the two images, you would see the flowing water is flattening out in mutually perpendicular directions as it falls down the cup. This works due to a principle similar to the one behind shape oscillations of a freely suspended water droplet. Only difference is that here, you don't have an oscillation in the direction of the flow. So this is actually a 2D case of water drop shape oscillation.

expl

Consider water flowing out from the cup at a uniform rate. We can imagine that the flowing stream of water consists of several cross-sectional films falling down one after another (marked in the picture as green rings). For simplicity, we neglect any interactions between the adjacent films. Let's track the shape of one of the films. When no other force is at play, the film tends to remain in a circular shape.

At the point where water loses contact with the cup (which I shall call 'Base'), the film is stretched out horizontally due to the normal reaction of the cup. When the film leaves the base, surface tension would provide a restoring force for the film. The film would get fully stretched in the perpendicular direction at position A.

base

A

This sets up a 2-D oscillation. The film would oscillate between these two states as it falls down.

oscillation


The oscillations may not be exactly as above but is somewhat similar.

Now imagine all of the films falling one after other down the cup. Note that all the films start out with a horizontal stretch. This would mean that each cross-sectional film would be in the same state of oscillation after falling the same distance from the cup. In other words, initial phase of all the films would be identical.

So in the big picture, as you go down the flow, you are actually observing the different states of the above oscillations with respect to time.


When you plot the different time states of the above depicted 2-D oscillation in along a Z-axis, you get a pattern similar to the one we observe in reality (but less curvy).

gif1


Here we have not included any damping forces between adjacent cross-sectional films. If we had, we would have obtained a much smoother shape.

P.S. This was actually posted as an answer to another question which turned out to be a duplicate. Putting it here as it would be more appropriate for future readers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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