How to estimate the Kolmogorov length scale

Question

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.

Answer 1

The size of the Kolmogorov 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. Therefore, as a very crude estimate, for a system of length $L$ and velocity $U$ (and dimensional grounds, on this scale viscosity does not play a role!), one could argue that

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

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

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

Answer 2

Here's an empirical way to roughly estimate ε for a natural stream.

You need:

• an audio recorder
• a thermometer
• a buoyant object
• tape measure
• a stopwatch
• freeware audio software (like Audacity)

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1. In the field, use the themometer to measure the temperature of the stream.

Use the buoyant object (i.e., stick), tape measure, and stopwatch to estimate the velocity of the water: throw the object in and measure how long it takes to travel a certain distance. Repeat this several times to get a typical value. Try to toss the float in different parts of the current to capture some of the variability.

You can correct for the velocity at the surface if you want, but to keep things simple I'm not going to do that here.

Make an audio recording a few feet away from the surface of the water. It's better to have a higher sampling rate.

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2. Use the measured temperature of the water to estimate the absolute dynamic viscosity.

$$ \mu(T) = A \cdot 10^{B/T - C} $$

where $T$ is the temperature in Kelvin, $ A = 2.414 \cdot10^{-5} $, $B = 247.8 $ K, and $ C = 140 $ K.

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2. Next, use the measured temperature to calculate the density of the water. $$ \rho(T) = e \cdot[1 - \frac{(T + a)^2 \cdot (T + b)}{c \cdot (T + d)}] $$

where the constant $ a = -4.0 $ C, $ b = 301.8 $ C, $ c = 522528.9 $ C, $ d = 69.3 $ C, $ e = 1000.0 \frac{kg}{m^2} $.

You can correct for atmospheric pressure as well, but again, I'm not going to do that here. You'd need a barometer to supply that information.

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3. The kinematic viscosity of water is given by

$$ \nu = \frac{\mu}{\rho} $$

Half-way there.

4. Now we'll leverage one of Kolmogorov's ideas in a practical sense.

The length microscale, $ η $ represents the typical length at which turbulent energy is dissipated. Acoustic energy is one way in which turbulence is dissipated, so we can estimate this length by:

$$ η \sim \frac{v}{f} $$

Where $ f $ is the dominant frequency in our audio recording in $ \frac{1}{s}$, and $ v $ is the velocity of the stream in $ \frac{m}{s} $.

You can use Audacity to load an audio recording, then use Analyze -> Plot Spectrum to get an idea of where the dominant frequency is.

It's interesting to see for natural streams that there is usually a wide spread of dissipated energy. The lower peak is an artifact of the recorder. The second peak at 567 Hz is the value we want.

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5. Then, finally we can estimate the energy microscale:

$$ ε = \frac{\nu ^3}{η^4} $$

I used this method for a small alpine stream and arrived at a value of $ ε = 6 \cdot 10^{-8} \frac{J}{kg} $

Answer 3

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

Answer 4

The answers given above seem intriguing but not well supported. Here's my computation for $\epsilon$ in a whitewater river rapid.

Consider mass $m$ of water. As it flows down the rapid, it converts its potential energy to turbulence energy, and hence to dissipation. At steady-state, the rate of potential energy loss is equal to the dissipation. From basic physics, the rate of potential energy loss is $$\frac{dE}{dt} = mgv \sin \theta,$$ where $g$ is the acceleration of gravity ($g \approx 10$ m/s$^2$), $v$ is the river velocity, and $\theta$ is the slope of the river surface. Divide both sides by the mass to give $\epsilon$ in units of W/kg, which then leads to $$\epsilon = gv \sin \theta.$$ In a typical rapid (e.g. a long class II gravel bar rapid), $v \approx 3$ m/s, and $\theta \approx 5$ degree. Plugging in these numbers gives $$\epsilon \approx 2.5 \textrm{ m}^2/\textrm{s}^{-3}.$$ Note that $1 \textrm{ m}^2/\textrm{s}^{-3} = 1 \textrm{ W/kg} = 1 \textrm{ J/s/kg}$. Then, use the relation $$\eta=\left( \frac{\nu^3}{\epsilon} \right)^{1/4}$$ along with $\nu\approx 10^{-6} \textrm{ m}^2/\textrm{s}$ (see Wikipedia "Viscosity") to find that $$\eta \approx 25 \textrm{ }\mu\textrm{m}$$ This seems reasonable to me. However, note that this is not a fundamental property of water as one might expect, but that it depends on the system. For example, if the water had been deeper (i.e. the same rapid, but the river is in flood), then $\theta$ would be unchanged but $v$ would be larger. This would increase $\epsilon$ and decrease $\eta$.