When we stir a glass or water, the dregs move into the middle of glass. Why?
Hmmmmm.... I can think of 2 possible reasons:
1) When you stir water in a glass, the water is accelerating towards the centre. This centripetal acceleration causes a pressure gradient and therefore the water surface is no longer flat. It is deeper at the outside and shallower at the centre. (Think of a tub of water which you cause to accelerate - the water's inertia 'means that it doesn't 'want' to move causing a deep and shallow end. With this pressure gradient (due to different heights of water, it is higher pressure at the outside and lower pressure at the centre. Therefore there is a net movement of water molecules towards the centre. Any dregs in the water will experience the same effect.
2) Boundary layers: at the edge of the glass the water will be stationary. The water at the centre will therefore be at a lower pressure than the water at the edge, as it is moving (relatively fast). This pressure gradient is in the same direction as, and would have the same effects as the above. I'm guessing though that the boundary layer will be only a few millimetres thick in this glass though so I think it's probably #1 mainly.
Consider a cylindrical tub with water in it. Place a cork in the water. Now rotate the tub forever. Where will the cork go?
If the higher pressure near the outside actually created a net movement of water towards the centre then the centre wouldn't be shallow for very long, would it?
Who said so? Do you think the phenomenon will disappear if I rotate the container instead of stirring it?
The real reason:
The circular shape and rotation of the water are irrelevant to the phenomena. The same effect can be witnessed in a straight river that has varying depths.
The effect is due to the speed of the water. Faster water will carry more and larger sedimentary particles than slower water. As the particles are tossed about in the water, they will remain suspended until they encounter a volume of slower water, where they will settle. Fast movement, and they will become suspended again.
The slower water in the centre of the glass (or in the deep parts of a river) creates an area where the particles will have a chance to settle.
BTW, without this phenomenon, panning for gold would be fruitless. The gold is carried down fast-moving rivers and is deposited where the rivers slow briefly and can no longer suspend the heavier gold grains. This is a natural filtering-by-weight effect.
(Hah! I knew it would have a name!)
Stokes Law: http://en.wikipedia.org/wiki/Sediment#Sediment_transport": "If a fluid, such as water, is flowing, it can carry suspended particles. The settling velocity is the minimum velocity a flow must have in order to transport, rather than deposit, sediments..."
Eventually the water will form a depression, and the system will be stable again. So the cork will probably find a resting place where the centrifugal force equals the pull of gravity. So that doesn't answer the question.
I need to replace the cork with an object that doesn't float....
Hmmm... my thoughts are this:
The swirling water in a confined containment increases peripheral pressure which forces suspended particulates to seek an area of least pressure, the center.
Jeez. Look up 'Stokes Law' awready.
I've been thinking about this problem some more. My original model of the rotating tub of water was missing something. I believe the phenomenon we are trying to explain will occur only after the tub STOPS rotating.
Then the outer layers of water will slow down while the center will continue spinning around. This gives us the "whirlpool" effect we need.
Here is as far as I can get by myself. Maybe somebody else can contribute.
We start with a rotating cylindrical tub of water, and assume that the situation is static (all the water is rotating in sync). Then we stop the tub.
Viscosity will eventually leave us with the water in the center of the tub rotating more quickly, with the rotation decreasing as we move away from the center.
Will the velocity/pressure differences be enough to suck any particulate into the center? Or is pressure not the mechanism?
Look, my pool is 13ft. across and I can't start it rotating or stop it rotating. All I can do is start and stop the circulation of the water with the pump. And the exact same thing happens - all the dirt collects in a small mound at the centre. There's no vortex and there's no "slope" to the water's surface (it's quite a slow rotation).
If I disturb this central mound of debris, it will eventually collect again, as long as the water is still moving. Actually, the dirt will collect ANYWHERE in the pool where the speed of the moving water is disrupted, such as near the ladder or behind wrinkles in the fabric.
Debris settles where the water velocity slows.
Holy Jeez, why are you trying to make this more complicated than it is?
They're just pulling your chain, Dave. Which, in Britain, would of course cause exactly the type of vortex that they're talking about. :tongue:
Okay Dave, chill out... surely it doesn't matter if everyone's line of thought coincides with yours or not, right or wrong regardless?
When I posted I was taking Stoke's law (I didn't know it was called that though) for granted - I think it's pretty intuitive that larger particles can be suspended in fast flowing water than in slow flowing water. How though does this on its own explain the movement of sedimentary particles towards the middle? For something to move, there has to be a force acting on it. In this pool / glass / or whatever you want to call it, this will be a pressure field. As far as I can see, Stoke's law doesn't predict a pressure field. That's what my initial posts attempted to explain.
When you said "If the higher pressure near the outside actually created a net movement of water towards the centre then the centre wouldn't be shallow for very long, would it?", you missed my point: I clarify it below. But whilst we're on the topic, if something is rotating then it has to move towards the centre of rotation. That doesn't necessarily mean that it will move radially relative to the rotating object, hence why the depth isn't constantly changing in different parts of the glass / pool.
I had also already discounted the effect of the boundary layer. What I was saying though was that relative to the container, fluid velocity is always zero at the wall.
THE BULK OF WHAT I WANT TO SAY (!):
Correct me if I'm wrong but I see a fundamental difference between a river and a glass. Sure, they both have fast moving water in the centre which causes sedimentary particles to be suspended, but in a river at the edge, particles are deposited and there they stay, which is why it's usually deeper in the middle of a river. However, in a glass with circulating water, the particles which find themselves near the edge of the glass will also move inwards, where they will become suspended again. The original question by lanovia was why? One thing maybe misleading about my first post was when I said that 'particles move inwards'. They can do. But then they can also move outwards. They just find their equilibrium point where the centrifugal force of rotation equals the pressure forces in the fluid. It's just because pressure in a rotating fluid increases according to the square of the radius, so pressure increases very quickly with radius, which causes the equilibirum point for even quite sizable sediments to be quite near the centre. Hence I just said that they move towards the centre.
Please help me clarify something in my mind. In a rotating fluid that has
reached equilibrium, there is only one component to the pressure - centrifugal force. But in the non-equilibrium case, there are two components to the pressure - centrifugal force AND bernoulli's principle.
Is my mental picture missing something?
Not at the edge, I'm talking about a river with shallow and deep sections along its length. Debris will be lifted and suspended in areas of high river velocity and sedimented in areas of low river velocity.
I put forth that Stokes Law describes the effect so well, that you would be hard-pressed to eliminate it and still have the effect.
So I challenge you create a thought experiment that eliminates Stokes Law and yet still produces the phenomenon. (Well, I guess it'd have to be a real experiment, since otherwise how would we observer the results?)
Hey I actually haven't tried this experiment so I don't know for sure what really happens... maybe I'm getting this wrong? I was under the impression that if you stir a glass of water with sediments in it, they become suspended in the middle of the glass. Any sediments near the edge move into a position of dynamic equilibirum near the centre, rather than settling. If this is the case then something besides stoke's law is required to explain the movement of sediments, i.e. a pressure gradient. From what you're saying though Dave, it sounds like the only phenomemon is suspended particles in the centre, and deposited particles near the edge of the glass, in which case only stoke's law is required to explain this.
Either way, I'm not disputing stoke's law or the important part it has to play here. If though what is seen when performing the experiment is as I initially assumed, then something else is required to complete the explanation.
The challenge is to NOT do the experiment, except in your head. Doing the experiment is like looking in the back of the book.
WHAT??!! :surprised :surprised
By doing the experiment, we see exactly what happens, and then we KNOW what happens, and needn't argue. Arguing about what SHOULD happen makes us no better scientists than the philosophers who argued about how many teeth horses ought to have, and it makes our conclusions no more valid than theirs.
Yep AI. Teeg's right. The Greek philosophers thought the secrets of the universe could be reveaad through thought experiment alone.
It was "common knowledge" that heavier objects fell faster than lighter objects. At least ... until Galileo climbed the Leaning Tower of Pisa.
I would bet my paycheque the following happens:
Upon stirring, all particles are randomly distributed throughout the fluid. There will be no particular concentration anywhere (that's kind of the whole point of stirring :rofl: If stirring didn't evenly distribute stuff, we wouldn't bother stirring.)
As the stirring stops, and the water begins to slow, the suspended particles will settle down inversely proportional to the speed of the water they're moving in. Much of the particles will settle evenly on the bottom, but the particles near the outside - where the water is still moving - will bounce along the bottom. Note that Stokes Law covers the whole range of water velocity vs. sedimentation with three distinct states: suspended, sedimented and an in-between state of bouncing/rolling.
Here's the self-organizing part:
Particles that land in the middle will not be disturbed by any moving water, and thus will generally remain where they land: in the middle.
Particles that land nearer the edge of the container will still be in moving water, and thus will continue to be buffeted. If they're buffeted outwards, they end up in even faster water, the buffeting gets worse, making this a very untenable place to be for any length of time. If they're buffeted inwards, well, see above.
You see, any place except the centre of the container is unstable. The centre is stable. The particle distribution is self-organizing.
I guess this is a form of chaotic attractor!
Everything you are saying sounds reasonable, but I am slow to pick up on one point...wouldn't the water by the side of the glass be the FIRST to slow down because it is rubbing against the glass?
Well, any doubt can be removed by experiment, but...
For the weater to slow down in one part without slowing down elsewhere will create a lot of friction within the water itself.
Note that, except for the friction you mention, the entire mass of water can circulate while maintaining with theoretically ZERO friction. (i.e. if the entire mass of water were frozen solid where internal frictyion were maximized, you could still rotate the whole mass within the glass).
Even frozen solid (maximum internal friction), the edges of the water mass are moving quickly while the centre is moving slowly.
In fact, as far as the suspended particles go, the water is perfectly still and it is the GLASS that is rotating around it. (we're not looking at centrifugal effects here, that effect is too small).
If the water is perfectly still and the glass is rotating, we can now watch all the suspended particles settle straight to the bottom (because they're all in "still" water). Yet when they touch bottom, the particles near the edge will STILL be jostled about by contact with the glass. They'll still be unstable until they settle on a part of the glass where the movement is minimal (i.e. the centre).
What would happen, would you think, if a vibrational field was applied to the water - without stirring?
In other words, place a beaker on a vibrational shaker, place floating items on top of it, set very small shaking frequency & observe.
You would get sedimentation in the nodes.
Aren't there always shear forces (if that's the correct term) involved in stirring a liquid, since the parts farther from the stirrer lag behind the closer bits? (And I don't want to hear anything about your pool and skinny-dipping. :tongue: )
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