# Vibration and separation

• July 17th, 2017, 09:27 AM
Zwolver
Vibration and separation
Imagine you have a beaker of glass beads. You have two sizes, one size slightly bigger than the other. Not a big enough difference for the beads to fall in between each other.

Which beads would end up on top if you put in a vibrating machine that makes them act as a liquid. Would the larger ones, or the smaller ones end on top? Or they would randomly disperse?

Density and shape would be identical of course.
• July 17th, 2017, 02:35 PM
pyoko
How "slightly" do you mean? Also are you ignoring other numbers for a perfect simulation? These things matter.
• July 18th, 2017, 03:02 AM
Zwolver
Quote:

Originally Posted by pyoko
How "slightly" do you mean? Also are you ignoring other numbers for a perfect simulation? These things matter.

Slightly, say a 1% difference between diameter.

Then i would first assume it is a vacuum, and that temperature is stable, gravity is 1g, or 9.81m/s2.

Then i would do the same, but with a normal earth atmosphere. What kind of effect would the atmosphere have?

( my hypothesis is that the smaller ones will go down, and the larger ones will go up.. But vibration has a bigger and more energetic effect on the smaller ones, so maybe exactly the opposite will happen )
• August 26th, 2017, 05:25 PM
Rarry
The beads on the bottom of the beaker would be on a hard surface and so as they hit each other the smaller ones would have a net downward force on them and the larger ones a similar slight net upward force. I'd guess that this difference would permeate up through the beaker to make the larger ones rise. That's dependant on the hard surface at the bottom of course. I wonder if it would make a difference if the vibration was vertical or horizontal? Perhaps vertical vibration would tend to drive the smaller ones up? Without that hard base surface, sand on the beach or a desert usually has the lighter grains ones on top, as does the moon, though probably for different reasons.
• August 28th, 2017, 07:03 AM
Zwolver
Maybe i should just test this in the lab...
• August 28th, 2017, 08:16 AM
exchemist
Quote:

Originally Posted by Rarry
The beads on the bottom of the beaker would be on a hard surface and so as they hit each other the smaller ones would have a net downward force on them and the larger ones a similar slight net upward force. I'd guess that this difference would permeate up through the beaker to make the larger ones rise. That's dependant on the hard surface at the bottom of course. I wonder if it would make a difference if the vibration was vertical or horizontal? Perhaps vertical vibration would tend to drive the smaller ones up? Without that hard base surface, sand on the beach or a desert usually has the lighter grains ones on top, as does the moon, though probably for different reasons.

I should have thought that the amount of empty space per unit volume between small particles would be less than for large particles, which would allow an ensemble of small particles to assume a higher density. This would eventually lead, I would imagine, to the smaller particles congregating below the large ones due to this density difference. To look at it another way, a close packed arrangement with the smaller particles at the bottom will have a lower centre of gravity, and thus less GPE, than the other way round. Won't it?
• August 28th, 2017, 10:51 AM
Rarry
Quote:

Originally Posted by exchemist
I should have thought that the amount of empty space per unit volume between small particles would be less than for large particles

With the same particle shape the percentage of the volume taken would be the same. So the percentage volume of randomly arranged spheres will be the same (about 65%) regardless of the size of the spheres, so the density would be the same. It's like the sugar lump problem where the question is about the volume taken by sugar cubes or cube shaped granules. They would occupy the same space. The only difference would occur at the edges but that's only significant if the space occupied is small compared to the size of the objects.

As an example, suppose you want to know the percentage of a square that is occupied by a exact fitting circle, the answer (78.5%) doesn't depend on the size.

But anyway, Zwolver is right, this is the sort of question that would be best to test out rather than guess wrong.
• August 28th, 2017, 04:06 PM
exchemist
Quote:

Originally Posted by Rarry
Quote:

Originally Posted by exchemist
I should have thought that the amount of empty space per unit volume between small particles would be less than for large particles

With the same particle shape the percentage of the volume taken would be the same. So the percentage volume of randomly arranged spheres will be the same (about 65%) regardless of the size of the spheres, so the density would be the same. It's like the sugar lump problem where the question is about the volume taken by sugar cubes or cube shaped granules. They would occupy the same space. The only difference would occur at the edges but that's only significant if the space occupied is small compared to the size of the objects.

As an example, suppose you want to know the percentage of a square that is occupied by a exact fitting circle, the answer (78.5%) doesn't depend on the size.

But anyway, Zwolver is right, this is the sort of question that would be best to test out rather than guess wrong.

Yes on reflection I suppose that must be right.