# Particle Reynolds number

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?

Homework Helper
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.

min_ht
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.

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 :(

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.

min_ht and BvU
Gold Member
in my case it is liquid droplet, so it has μ, and that's why I don't really understand

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

min_ht and Chestermiller