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meinenn

November 14, 2012 05:21

LaunderGibson - turbulent viscosity in epsilon equation

Hej
I'm inserting the LaunderGibson tubulence model to a two phase solver. Now I remarked that in LaunderGibson, in the transport equation for epsilon, a term DepsilonEff is used which contains the turbulent viscosity. Can someone explain why we have this viscosity? In my opinion, the great advantage of reynold stress models is the fact that we do not need the eddy viscosity assumption. So I'm a bit confused why there still is this turbulent viscosity...
Thanks for help in advance
Nik

rakendd

March 31, 2015 10:49

The Dissipation tensor in RST Model is assumed to be modeled by isotropy of small dissipative eddies.

$\varepsilon_{ij} =\dfrac{2}{3}\varepsilon\delta_{ij}$
where $\varepsilon$ is the dissipation rate of turbulent kinetic energy.
This is found from $\varepsilon$ equation from $k-\varepsilon$ model, where we model the transport of $\varepsilon$ using turbulent viscosity $\nu_{T}$

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