My understanding of Kolmogorov scales doesn't really go beyond this poem:

Big whirls have little whirls that feed on their velocity,
and little whirls have lesser whirls and so on to viscosity.

The smallest scale according to wikipedia* would be $\eta = (\frac{\nu^3}{\epsilon})^\frac{1}{4}$

But can I assume the same shear across all scales, and hence (for a shear thinning liquid) the same apparent viscosity?
Are there practical observations about this?

Update: Maybe I need to clarify my question. I'm not so much interested in the theory as in one real physical phenomenon this theory describes: That there is a lower limit to the size of a vortex for a given flow, and this size can at least be estimated using above equation. Now, a lot of real fluids are non-Newtonian in one way or the other, I'm asking about shear because the apparant viscosity is (also) shear dependent.
While the theory of Kolmogorv may be hard to translate for non-Newton flow, the actual physical phenomenon of an observable (or evenmeasureable) lower limit for vortex size should still hold - are there any measurements or observations?

  • $\begingroup$ Is there any way to make this question more clear? $\endgroup$
    – mart
    Jul 16 '12 at 8:04
  • $\begingroup$ As I said. The Kolmogorov theory and therefore the minimum size of a vortex does NOT apply to non Newtonian fluids. It is not hard to use it, it is impossible. You can't use a theory in a domain where it is known to be invalid. $\endgroup$
    – Stan Won
    Jun 19 '13 at 9:00
  • $\begingroup$ So, when the fluid is non-Newtonian, you can have eddies of indefinitly small size? Is that what ou are trying to say? $\endgroup$
    – mart
    Jun 19 '13 at 9:43
  • $\begingroup$ No. I am saying that the statistical eddy theory which posits a cascade of eddies from L to µ (aka Kolmogorov) doesn't work for non Newtonian fluids. There is no general Kolmogorov like theory for non Newtonian fluids so nothing general is known about scaling and eddy size properties. As viscosity is variable, the behaviour must be studied case by case with Navier Stokes + rheology considerations. For many plastic like non Newtonian fluids there is no eddy cascade at all so the question about their "size" is not even well defined. $\endgroup$
    – Stan Won
    Jun 24 '13 at 9:03

Yes it is not clear why you are mentionning shear. Hence it is not clear whether your are interested by very exotic cases or on the contrary by the classical Kolmogorov turbulence theory.

I will give an answer assuming the latter. The Kolmogorov theory starts by analogy with statistical mechanics by assuming an isotropic and homogeneous distribution of vortices in a Newtonian fluid. Viscosity is constant for a Newtonian fluid. Then assuming self similarity, Kolmogorov established the energy spectrum in relation to the wave number. Because the wave number is the scale, he established how the energy is distributed on the different scales.

The originality of this theory is that there is a minimum scale, called Kolmogorov length where the dissipation of energy by viscosity happens. Of course this length is not constant but depends on the flow and the largest scales (inertial scale) L are related to the Kolmogorov scale l by L/l = Re^(3/4) where Re is the Reynolds number. The scale invariance of viscosity is simply given by the fact that we deal with a Newtonian fluid.

Now to shear thinning. Shear thinning (or thickening) fluids are non Newtonian. Viscosity depends on stress and even on time. The conditions of isotropy, homogeneity and self similarity are not given so the Kolmogorov theory and its lengths have nothing to say about these exotical substances. I add that you won't observe turbulent vortex cascades in them either.

These substances behave at the same time like solids (plastics) and fluids so that Navier Stokes equations alone are not the best way to study them. The previous sentence is an understatement.

  • $\begingroup$ Thanks for the answer that's actually helpful. I updated the question somewhat - If no other answer shows up, I'll accept yours. $\endgroup$
    – mart
    Jun 11 '13 at 13:54
  • $\begingroup$ This is not the answer I wanted, but it seems to fit the question ... I'll see if I can hammer out a etter question for my needs. $\endgroup$
    – mart
    Jun 24 '13 at 9:07

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