# Turbulence models

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 May 29, 2001, 13:18 Turbulence models #1 pop Guest   Posts: n/a Hi everyone, I have a question regarding turbulence models, is it possible to solve a laminar problem using any of the known turbulence models?? Or in other words, if I have a big model that contains laminar regions and hi Re regions what I am supposed to do to solve it? Is it laminar or turbulent?? Thanks Pop

 May 29, 2001, 17:24 Re: Turbulence models #2 K. Tsai Guest   Posts: n/a Large eddy simulation with dynamic sub-grid scale models.

 May 30, 2001, 01:10 Re: Turbulence models #3 John C. Chien Guest   Posts: n/a (1). You need to "model it first", meaning that you need to focus on your goal, then eliminate unnecessary features of the problem. (2). Once you have the model created, you can proceed to solve it. (3). For example, if your goal is to find out something in the laminar flow region, then you can assume that it is laminar flow. (4). On the other hand, if the boundary layer is not important, you might as well model it as inviscid flow. (5). It is possible to use turbulence model to cover both the laminar flow and turbulent flow regions. But, you need to validate the model against the test data first.

 May 31, 2001, 00:16 Re: Turbulence models #4 Steve Collie Guest   Posts: n/a If you are using 2-eqn models then it is better to use k-w based models as for k=0 (i.e. laminar flow) the w-equations uncouples from the k-equation and has a laminar solution. If you use k-e then for k=0 the e-equation is not well posed. While the k-w does have a laminar solution transistion is known to occur at a Reynolds number of about an order of magnitude to small (k-e has the same problem). So really you need a transistion model [see D. C. Wilcox 1994]. However if you know where the turbulent regions are then you can just turn the turbuelent source terms off where the flow is laminar and on in the turbulent regions. By doing this you neglect to model the development of turbulence in the transistion region - so expect some error. As was mentioned LES is a more accurate approach, but much more expensive.

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