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Wall effect on falling sphere of small Reynolds number

  1. Oct 2, 2015 #1
    If I were to drop a sphere down a tube of fluid. What are the effects of the wall of the tube on the sphere?
    Assuming the sphere has an extremely low Reynolds number of less than 1.

    Is it right to say that the ratio of the radius of the tube and the radius of the sphere has an effect on the terminal velocity of the falling sphere? At what ratio of the the two radius will the effect be negligible and at what ratio will the effect be apparent?
     
  2. jcsd
  3. Oct 2, 2015 #2
    I did a school project on this once. The terminal velocity as a function of r/R has a maximum somewhere in the 0.25-0.5 area as I recall. When r/R is very small the TV can be calculated by consideration of the viscous forces. When r/R is close to 1 you can write a mass conservation equation (assuming incompressible fluid) which explains the attenuation of the TV as the effect of the wall becomes important.
     
  4. Oct 3, 2015 #3
    Thank you davidmoore63@y. Can you elaborate more on how do I calculate the change in terminal velocity taking the r/R into account?

    I have 3 sphere where I measured their terminal velocity and calculated the viscosity of the fluid using Stokes' Law.

    1st sphere have a r/R ratio of around 0.13 and the calculated viscosity is around the same as the given value with less than 1% error.
    2nd sphere have a r/R ratio of around 0.18 and this time the calculated viscosity have a percentage error of 18%.
    The percentage error increases when my 3rd sphere, which have a r/R ratio of 0.3 have an error of 42%.

    My calculated viscosity is higher than the correct value and using Stokes' Law, I assume its because terminal velocity drops due to some unseen factors which I guess has to relates to their radius ratio which in turn causes some sort of turbulent flow?
     
  5. Oct 3, 2015 #4
    Let me understand. As you increase the radius ratio, the apparent viscosity that you measure seems to increase. This of course is the result of the additional drag on the sphere because of the closer proximity of the wall. It has nothing to do with turbulent flow (unless the calculated Reynolds number is high, which it is not). I can't believe that there are not fluid mechanics solutions of this problem in the open literature. Falling ball viscometry is a very widely used technique, especially for very viscous fluids. Have you checked the open literature? Have you checked the literature provided by manufacturers of falling ball viscometers?

    Chet
     
  6. Oct 3, 2015 #5
    Thank you Chestermiller. I tried searching for various literature and papers about this but I couldn't find any equation relating the ratio and the drop in terminal velocity. Many literature make reference to Faxen's drag correction which I tried searching about it but still couldn't get a straight forward equation relating both radius with either terminal velocity or drag force.
     
  7. Oct 3, 2015 #6

    Nidum

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