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LaunderGibson - turbulent viscosity in epsilon equation

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Old   November 14, 2012, 06:21
Default LaunderGibson - turbulent viscosity in epsilon equation
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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
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Old   March 31, 2015, 10:49
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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}

Last edited by rakendd; March 31, 2015 at 19:27.
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