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Average shear rate of of particle sureface

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  1. Apr 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Diffusion coefficient (D) of 100 nm particles ( density = 1050 kg/m3) in a viscous solvent at 25 C is 2.2 x 10-13 m3/s

    What is the average shear rate at the particle surface?

    2. Relevant equations

    D = kT/ 6πηa (1)
    where k is Boltzman constant, T = temperature, ηsusp = viscosity and a = radius

    Pe = a2γ /D (2)

    where γ = shear rate

    3. The attempt at a solution
    Assumed particles are in water

    To find shear rate of particle surface I am not too sure that eq2 is the right equation to use or not. Can we assume that Pe = 1 ??

    Pe = 1 means the solvent in the viscous state ???

    Thank you

    Zen
     
  2. jcsd
  3. Apr 24, 2015 #2
    This can't be the complete problem statement. Please provide the complete problem statement, with its exact wording.

    Chet
     
  4. Apr 24, 2015 #3
    Hi Chet, This is the full question
    upload_2015-4-24_22-24-50.png
    upload_2015-4-24_22-22-53.png

    Thanks
     

    Attached Files:

  5. Apr 24, 2015 #4
    If I understand correctly, you have completed parts a, b, and c, and now you want want to determine the answer to part d, correct? If so, what was your answer to part c.

    Chet
     
  6. Apr 24, 2015 #5
    Hi Chet !

    Yes you are right I have done part a-c. My answer on C was 0.05 m/s

    Thanks

    Zen
     
  7. Apr 24, 2015 #6
    Are you sure about that average velocity of the particles? It seems awfully large.

    Chet
     
  8. Apr 25, 2015 #7
    Thanks Chet

    I have changed the radius of particle from 100µm to 50µm but the velocity was larger than 50 µm. it doesn't make sense. :(

    Vav = ((3kT)/m)1/2

    Volume of particle = 4/3 πr3 = 4/3 π (50x10-9)3 = 5.236 x10-22 m3
    Mass of particle = ρV = 1050 x 5.236x 10-22 = 5.498 x 10-19 kg

    Vav = ((3x1.38 x 10-23 x 298 )/ 5.498 x 10-19 )1/2 = 0.149 m/s
     
  9. Apr 25, 2015 #8
    I think this one is more reasonable ?

    v = (2r2 Δρg) / 9η
    = ( 2x( 50x10-9 )2 x (1050-1000) x 9.8) / (9 x 8.9x10-4)
    = 3.059 x 10-10 m/s

    THanks
     
  10. Apr 25, 2015 #9
    This is much better, if analyzed correctly. I can't vouch for that part of the calculation (because my experience is not in that area), but I can help you get the average shear rate at the particle surface once the particle velocity is established. Sometimes the shear rate is taken as the radial velocity gradient at the particle surface, and some times it is taken as the square root of the second invariant of the rate of deformation tensor. How is it defined in your situation?

    Do you know the equations for the components in spherical coordinates of the velocity vector in Stokes flow past a moving sphere? If so, please write them down.

    Chet
     
  11. Apr 25, 2015 #10
    Hi Chet.

    I am not really sure about but I can see as the radial velocity gradient at the particle surface (dv/dx)

    Fg-Fb-Fd = 0 at Terminal Velocity
    where Fg = gravity force, Fb = buoyancy force and Fd = drag force.

    Fg = (Vρpg)
    Fb = (Vρfg)
    Fd = 9πηrv

    v = (2r2 Δρg) / 9η

    I am not really sure this is what you asked for.

    Thanks
     
  12. Apr 25, 2015 #11
    I'm looking the equations for the components as a function of r and θ.

    Chet
     
  13. Apr 25, 2015 #12
  14. Apr 25, 2015 #13
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