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?
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$\begingroup$ Hint: $l/u$ is a time scale. What other time scale has been mentioned in the text preceding the formula? $\endgroup$– typesanitizerCommented May 1, 2016 at 4:20
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$\begingroup$ Another hint -- production = dissipation. $\endgroup$– tpg2114Commented May 1, 2016 at 4:21
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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. $\endgroup$– dmckee --- ex-moderator kittenCommented May 1, 2016 at 4:33
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$\begingroup$ @dmckee Will keep that in mind and will definitely do so in the future. $\endgroup$– saakCommented May 1, 2016 at 4:36
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1 Answer
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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.
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$\begingroup$ Energy production we are considering here is Kinetic Energy, hence the u^2. correct? $\endgroup$– saakCommented May 1, 2016 at 4:33