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For Newtonian fluids, the sinking (or rising) speed of a particle is (reasonably) well known. What is a relationship for shear thinning fluids?

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  • $\begingroup$ What do you mean by "particle"? Are you looking for something truly as small as a dust mote or just anything that's small in relation to the fluid volume and relatively spherical? $\endgroup$ Feb 10, 2015 at 16:11
  • $\begingroup$ Small particles (bacteria flocs) and not quite so small particles (grit, <100µm) $\endgroup$
    – mart
    Feb 11, 2015 at 6:40
  • $\begingroup$ On one hand, I don't want to join the edit war. On the other hand, this quesiton is relevant for process engineering wich is part of mechanical engineering (and environmental engineering, and chemical engineering sometimes) $\endgroup$
    – mart
    Feb 16, 2015 at 17:02
  • $\begingroup$ While it's relevant to those fields, the question itself is only concerned with fluid dynamics. The tags should describe the content of the question, not the context. $\endgroup$ Feb 16, 2015 at 17:05

1 Answer 1

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Doing a brief review of the literature, for spherical particles at high Reynolds numbers, it appears that you can use standard drag curves to get the drag coefficient as long as you use the following Reynolds number:

$$\mathrm{Re}_\text{PL} \equiv \frac{\rho V^{2-n} d^n}{m}$$

This Reynolds number assumes the fluid follows a power-law. $n$ is the flow behavior index, which is less than 1 for a shear thinning fluid. $m$ in the above equation is $K$ on Wikipedia.

So, if you just want the terminal velocity, you can plug this in the standard equations for terminal velocity, as all they assume is that you know the drag coefficient at the terminal velocity state. (You can derive them from a steady force balance or from the unsteady case. I recommend trying both just to get a feel for these sorts of problems.)

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    $\begingroup$ So m is the viscosity (not quite), d the diameter, V speed and rho density? (Just to be sure) $\endgroup$
    – mart
    Feb 11, 2015 at 6:43
  • $\begingroup$ Yes, that is correct. $\endgroup$ Feb 11, 2015 at 16:07

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