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Particle Reynolds number

  1. Oct 14, 2016 #1

    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?

    Thanks in advance.
  2. jcsd
  3. Oct 14, 2016 #2


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    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. :smile:
  4. Oct 14, 2016 #3
    Thank you for your reply
    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 :(
  5. Oct 14, 2016 #4
    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.
  6. Oct 14, 2016 #5


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    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.

  7. Oct 14, 2016 #6


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    If your goal is to study the flow around a particle, why would you use conditions inside the particle in your study?
  8. Oct 15, 2016 #7
    Thank you all
    I think I got your points and they help a lot
    very appreciated!
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