Kolmogorov scale - dissipation?

The rate at which energy moves down a turbulent cascade is given by: $$\newcommand{\p}{\frac{\partial #1}{\partial #2}} \newcommand{\f}{\frac{ #1}{ #2}} \newcommand{\l}{\left(} \newcommand{\r}{\right)} \newcommand{\mean}{\langle #1 \rangle}\newcommand{\e}{\varepsilon} \e\sim \f{v_l^3}{l}$$ As far as I know $\e$ is a constant, but I am confused by the nature of this expression. Does it hold for any scale $l$ and corresponding velocity scale $v_l$ or only for a particular one? If the former please explain and if the latter what length and velocity scales are chosen and why does it not hold for any?

• Have you looked at introductory turbulence textbooks that derive this expression and where it comes from? It's usually pretty clearly laid out what these variables mean. Dec 13 '16 at 19:40
• @tpg2114 I've looked in Landau and Lifshitz and several online sources. From what I have found they are all a bit ambiguous about it. e.g. they say $l$ is a 'characteristic vortex length' which could be interpreted as meaning the vortex length I am looking at or a specific one. Hence the confusion. Dec 13 '16 at 19:43
• You may want to clarify then because "any length scale" is very different from "any of the big scales" or "any of the small scales." A vortex scale is radically different than a Kolmogorov scale and the derivations (and resulting definitions) are different. Dec 13 '16 at 19:45

According Kolmogorov's theory, that expression holds at any scale in the inertial range, where dissipation rate $\epsilon$ and length scale $l$ are the only parameters that determine what happens at that scale. I think it was G. I. Taylor who later extended above expression to large energy containing scales (integral scale, to be specific), and when done so it indeed becomes a very convenient way of calculating $\epsilon$. But at Kolmogorov scales that expression cannot hold because fluid viscosity becomes important.