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Turbulence Dissipation *Rate*

  1. Dec 10, 2014 #1


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    I've been reading about dissipation of energy in turbulent flows. I've read in various places about Richardson's idea of Energy cascade. But one of the biggest problems I have is understanding how the books (or texts online) all refer to ϵ, the rate of dissipation of kinetic energy, being independent of viscosity, that is:

    ϵ ~ u³/l ...............................................EQ1

    where u and l denote the characteristic length and velocity scales of the large eddies.

    My problem is as follows. The texts ALL say that the energy dissipation RATE is independent of viscosity. Okay. But then the majority of books then say, at the small scales:

    ϵ = 2*v*Sij*Sij ...................................................................EQ2

    So how can ϵ, the rate of dissipation of kinetic energy, be independent of viscosity, if at the small scales it can be calculated from viscosity?

    The book I'm reading also says sometimes that EQ 2 is simply "dissipation" rather than "dissipation rate" or that other times ϵ represents "dissipation of turbulent energy". It seems to regularly use different terminology for the same equation and same symbol. So I need some clarity here.

    Any help?


    Some text alluded to the fact that the dissipation stage is last process in the entire energy cascade of processes and therefore the dissipation rate must be calculated from the first process in the cascade processes, hence ϵ ~ u³/l but I didn't understand this because, although I agree that the rate of dissipation may depend on how quickly energy cascades to the smaller scale, it must surely also depend on how quickly the energy is transformed to heat at the smallest Kolmogorov scale? I understand that the inertial forces dominate over viscous forces for large eddies, but I don't see how viscosity can be neglected from a rate calculation once you get down to the small scales.
  2. jcsd
  3. Dec 11, 2014 #2


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    Compare the magnitudes of the "dissipation" rates.
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