What does 'energy dissipation' in the description of the Kolmogorov Microscales in fluid dynamics mean?

I've asked about microscales before, and thinking more about the problem turned up more basic questions. Consider the length scale of a microscale: $$\eta = \left(\frac{\nu^3}{\epsilon}\right)^\frac{1}{4}$$ With $\nu$ being the viscosity, and $\epsilon$ the energy dissipation. I understand this to mean the energy transfer from turbulent kinetic flow to thermal energy (at this scale). The unit of $\epsilon$ is Watts. What's the reference volume?

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1 Answer

The unit of $\epsilon$ is not Watts, it's Watts per kilogram. That is it is a dissipation in the unit of mass, to get dissipation density (per volume), do $\rho \epsilon$.

For who might be interested [W / kg] is actually [m^2 / s^3], and the units of the dynamic viscosity $\nu$ are squared meters per second [m^2 / s]

Indeed it is quite standard to use units per mass in hydrodynamics, not per volume, because for mass we have mass conservation law, which come in hand when rearranging the equations.

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thx. After posting this question I did some dimensional lanalysis and saw that it actually had to be W/kg –  mart Aug 24 '12 at 9:46