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I Higbie's penetration theory in a bubble

  1. Nov 23, 2017 #1
    Hi pf,

    I was wondering about a bubble moving with constant velocity in a liquid, and how the motion affects the mass transfer. Since the viscosity of the gas is significantly smaller than the viscosity of the liquid, the condition ## \tau_{gas} = \tau_{liquid} ## tells me that
    $$ \mu_{gas} ( \frac {dv_{gas}} {dy} )_{boundary} = \mu_{gas} ( \frac {dv_{liquid}} {dy} )_{boundary} → (\frac {dv_{gas}} {dy} )_{boundary} >> ( \frac {dv_{liquid}} {dy} )_{boundary} $$
    So we can assume the system is static (because there is no significant variation from the velocity of the bubble and the velocity of the fluid in the penetration layer). Now my problem arise:
    my professor and all the books I read, introduce the concept of "residence time" ##tc = x/v##, but what exactly is ##x##? I figured out it has to be the diameter of the bubble, but why?? Can someone please explain me this, I spent 2 hours thinking on it and still I don't get it.... I feel like I'm close to the solution, but I'm not 100% sure
  2. jcsd
  3. Nov 27, 2017 #2
    From the rest frame of reference of the bubble, the fluid velocity at the surface of the bubble is on the order of the upward bubble velocity (the latter reckoned from the laboratory frame of reference). This fluid moves like a slab over the perimeter of the bubble (neglecting the radius of curvature, which is large compared to the concentration boundary layer thickness). So the mass transfer to the fluid is like transient mass transfer to a semi-infinite slab. The contact time is on the order of the bubble diameter divided by the upward bubble velocity.
  4. Nov 27, 2017 #3
    HI, thank you for your replay. Sorry if I bother you with one more question bout it, but I really need to know an other thing about this. When I solve the indefinite equation of mass-transfer (I apologize if the name is wrong, but I don't know how to say it in english, anyway I refero to ## \frac {D\rho_i} {Dt} = -λ_i∇^2\rho_i+ r_i##
    i find that the mass transfer coefficient is $$K_c(t) = \sqrt \frac {λ_i} {πt} $$ (##λ_i## = diffusion coefficient). Then I find the average coefficient: $$ K_{c, av} = \frac 1 t ∫K_c(t)dt = 1.12\sqrt \frac {λ_i} {t}$$.
    Now my question is: in the "macroscopic" equation:
    $$ \frac {dm_i} {dt} = -Kc*S*Δ\rho_i $$
    should I use ##K_c(t)## or ##K_{c,av}## ?

    Again sorry for my english
  5. Nov 27, 2017 #4
    You would use the average value. The final mass transfer coefficient only applies toward the trailing edge of the bubble, and the mass transfer coefficient becomes infinite near the leading edge. So the correct value to use is the average value over the contact length.
  6. Nov 28, 2017 #5
    HI, sorry for the late replay I had hard times trying to imagine this phenomenon. I think I figured it out now... probably. Thank you :)
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