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Reynolds averaging and k-epsilon

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Old   August 4, 2003, 03:54
Default Reynolds averaging and k-epsilon
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Jaap Hoffmann
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Is there a cut-off beyond which one should not refine meshes when working with a k-epsilon model? My concern is that the Reynolds averaging process filters out turbulence, which is then addressed via a turbulence model. In itself, this is to me a bit of an abstract concept, since a link between the velocity and time scales are usually not stated in the Reynolds averaging process. If one keeps on refining a mesh, one may reach a point where one is actually (trying to) solving some detail of the (larger) eddies, that is supposed to be the domain of the turbulence model. This in itself poses a problem, since the turbulence is transient, whilst the mean flow might be steady. Or will the contribution of the turbulence diminish with mesh size and automatically approach a LES solution - with a "wrong" turbulence closure model for the smallest eddies?
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Old   August 4, 2003, 04:57
Default Re: Reynolds averaging and k-epsilon
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Tom
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I think you are misinterpreting the basic problem here (you're confusing two different issues).

Firstly the what is meant by Reynolds averaging is that you average the Navier-Stokes equations over a period of time which is long enough to bring the flow into equilibrium. This averaging requires closure conditions (k-e is one such closure) which results in a closed set of partial differential equations for the mean flow properties.

To solve these equations in practice you need to introduce a mesh on which the equations are to be approximated. Convergence of the numerical discretization to the solution of the original partial differential equations requires that the mesh should be sufficiently fine that the solution does not depend upon the mesh spacing; i.e. further refinements to the mesh do not significantly change the answer.

Thus, if you solve your k-e problem on different meshes you should observe that the solution approaches a steady state which is independant of the mesh as the mesh is refined.

Tom.
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