I'm very new to turbulence.I'm reading this book on "Turbulence: An introduction for scientist and Engineers" by Davidson. I came across this formula for energy dissipation. This is in chapter one. I cannot seem to make sense of how they obtained this formula. Can anyone explain this formula to me?Screenshot of Davidson's book Page 19

  • $\begingroup$ Hint: $l/u$ is a time scale. What other time scale has been mentioned in the text preceding the formula? $\endgroup$ May 1, 2016 at 4:20
  • $\begingroup$ Another hint -- production = dissipation. $\endgroup$
    – tpg2114
    May 1, 2016 at 4:21
  • 1
    $\begingroup$ Please copy the text (and mark it up as a quote, of course) rather than posting images of it. This makes a difference in several ways: (a) it makes the text searchable, (b) it makes the text available to users who use screen readers, and (c) it makes a fine-grained edit history available (though that is unlikely to be a major concern in this case). You'll want to use MathJax for the math, of course. $\endgroup$ May 1, 2016 at 4:33
  • $\begingroup$ @dmckee Will keep that in mind and will definitely do so in the future. $\endgroup$
    – saak
    May 1, 2016 at 4:36

1 Answer 1


The idea is all based on scaling laws, which is a very common theme in turbulence. As I mentioned in a comment, this assumes that the energy produced is equal to the energy dissipated. The energy produced is proportional to $u^2$.

The turbulence is assumed to dissipate according to some time scale. A reasonable time scale is the eddy-turnover time, or $u/l$.

The dissipation rate is then the amount of energy dissipated per unit time. So let's take the energy produced and the timescale of the eddy to get:

$$\epsilon \propto \frac{u^2}{u/l} = \frac{u^3}{l}$$

and you arrive at your estimate.

  • $\begingroup$ Energy production we are considering here is Kinetic Energy, hence the u^2. correct? $\endgroup$
    – saak
    May 1, 2016 at 4:33

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