Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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. - Lewis Fry Richardson

Th smallest whirl according to Wikipedia would be that big:

$\eta = (\frac{\nu^3}{\varepsilon})^\frac{1}{4}$

... with $\nu$ beeing kinematic viscosity and $\epsilon$ the rate of energy disspiation.

Since I find no straightforward way to calculate $\epsilon$, I'm completely at loss at what orders of magnitude to expect. Since I imagine this to be an important factor in some technical or biological processes, I assume that someone measured or calculated these microscales for real life flow regimes. Can anyone point me to these numbers?

I'm mostly interested in non-compressible fluids, but will take anything I get. Processes where I believe the microscales to be relevant are communities of synthropic bacteria (different species needing each others metabolism and thus close neighborhood) or dispersing something in a mixture.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

The size of the Kolmogorv scale is not universal, it is dependent on the flow phenomena you are looking at. I don't know the details for compressible flows, so I will give you some hints on incompressible flows.

From the quotes poem, you can anticipate that everything that is dissipated at the smallest scales, has to be present at larger scale first. Therefor, as a very crude estimate, for a system of length $L$ and size $U$ (and dimensional grounds, on this scale viscosity does not play a role!), one could argue that

$$\varepsilon=\frac{U^3}{L}$$

For the crude estimate, one could use this $\varepsilon$ to estimate the Kolmogorov length scale.

To put in numbers, suppose you ($L=1m$) are running ($U=3m/s$) (in air $\nu=1.5\times10^{-5} m^2/s$), then, $\eta=100\mu m$. Which sounds at least reasonable.

share|improve this answer
    
isn't the dimension of $\epsilon$ J/s? because $\frac{U^3}{L} misses a kg to become Power. –  mart Feb 13 '13 at 13:20
    
no, wait, $\epsilon$ has to be J/s kg. Allright! –  mart Feb 13 '13 at 13:21
    
I only remember $m^2/s^3$ :) –  Bernhard Feb 13 '13 at 13:31

Epsilon values range from $10^{-10}$ to $10^{-4} \frac{m^2}{s^3}$ in typical oceanic, lakes and rivers situations, use those values to have an idea of Kolmogorof scale

share|improve this answer
1  
Seems better suited for a comment. –  Brandon Enright Jun 3 at 2:46
    
1) I changed the syntax so the numbers are more readable, please check if still correct, 2) as @BrandonEnright said, this does not answer the question as to how to arrive there. –  mart Jun 3 at 7:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.