# Particle Reynolds number

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1. Oct 14, 2016

### min_ht

Hello,

The particle Reynolds number makes me confused and I hope someone can help me on this please!

Normally (as I read in every books and papers) that when a bubble or drop rises in a fluid, the bubble/drop Reynolds number is calculated by:

Re = ρUD/μ

where U is particle velocity, D can be particle diameter, and ρ and μ are density and viscosity of continuous fluid

my question is why don't use ρ and μ of bubble/drop? why use them of surrounding fluid?
what is the physical meaning of this Re?

2. Oct 14, 2016

### BvU

The particle is supposed to be solid, so it has no $\mu$: the fluid has to move around the particle and not the other way around.

3. Oct 14, 2016

### min_ht

Actually the term "particle" refers to all solid particle, liquid drop or gas bubble (book: Bubbles, Drops, and Particles of Clift et al. 1978)
just normally people use particle for solid body :)
and that equation for Re applies for all of them
in my case it is liquid droplet, so it has μ, and that's why I don't really understand :(

4. Oct 14, 2016

### Staff: Mentor

The resistance to the particle movement is caused by the fluid deforming. As reckoned from the frame of reference of particle, the fluid is flowing past. So it is the fluid deformation and flow around the particle that determines the drag on the particle. That's why the focus is on the fluid.

5. Oct 14, 2016

### Mech_Engineer

In the case of something like a liquid droplet falling through air, you would make the assumption that the droplet is a spherical particle with no deformation. The Reynolds number still applies to the surrounding fluid flowing around the particle, not the other way around.

https://en.wikipedia.org/wiki/Sediment_transport#Particle_Reynolds_Number

6. Oct 14, 2016

If your goal is to study the flow around a particle, why would you use conditions inside the particle in your study?

7. Oct 15, 2016

### min_ht

Thank you all
I think I got your points and they help a lot
very appreciated!