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Dean Schrage September 5, 2001 07:14

is it laminar or is it not (links included)...
I have implemented a solution to the K-E transport equations to solve for turbulent viscosities. My question involves activation. When do I turn on this model? Or do I leave it on all the time? That is, the K-E solutions yields a turbulent viscosity coefficient. It doesn't really know the state of the flow. It's just a dumb advective diffusive solver with sources and boundary conditions. Well not really dumb but sort of.

For lamina flow cases, I can still generate turbulent viscosity coefficients near to the molecular values and generally a fraction smaller. If I am to turn off the turbulence model, I would need to decide if turbulence is present prior to simulating these quantities for more complex flow systems. This seems to be defeating the purpose. If the K-E simulation is correct then I suppose it should yield eddy diffusivities which apply properly. On the other hand if I leave the model running, am I generating correct values. The following links show viscosity ratios for laminar flow cases, backstep and natural convection cavity

And perhaps that forms the question which is more academic than applied. For example, in supposed laminar flows (2D duct, Re <2300) is there a degree of turbulence which is not registered in engineering calculations but can still exists at a smaller intensity and can be determined with these 2 equation models.




John C. Chien September 5, 2001 12:05

Re: is it laminar or is it not (links included)...
(1). CFD is Numerical Analysis and Mathematical Modeling in Fluid Mechanics. (2). Turbulence model is the result of turbulence modeling. And whether to "turn it on" or to "keep it running" all the time, is the decision of your numerical analysis. (3). You will have to try different approaches like the inventor Edison, or the British carpenter who invented very accurate clocks. (4). CFD is not the way to discover new physics, it is your activities to model the your fluid mechanics problem. (5). If you are traveling by plane, and the turbine blade failed, then you don't care whether it is academic or applied, whether the model is turned on or off.

sylvain September 6, 2001 04:19

Re: is it laminar or is it not (links included)...
k-eps model(s) do(es) NOT handle laminar to turbulent transition. If you turn on the turbulence model, then the result is turbulent even if the Reynolds Number is lower than the Critic Reynolds Number, which could be "sometimes" possible but most of the time, in this case, the results will be far from reality.



Peter liang September 6, 2001 04:26

Re: is it laminar or is it not (links included)...
So, how about first order upwind scheme? Does it handle both turbulent n' laminar?

Dean Schrage September 6, 2001 06:05

Re: is it laminar or is it not (links included)...
I don't understand your question about upwinding

sylvain September 6, 2001 07:50

Re: is it laminar or is it not (links included)...
I was just talking about k-eps turbulence models used to close Reynolds Average Navier Stokes equations. This doesn't depend on the numerical scheme used, does it ?

Jonas Larsson September 6, 2001 10:20

Re: is it laminar or is it not (links included)...
You're asking a difficult question - transition modeling is very much still an area of active research. Simulating transitional and intermittent flows is very difficult, and most people don't even try this in engineering calculcations.

A few words of advice - if you know the flow is laminar keep the model turned off and if you know the flow is turbulent then turn it on, simple eh? ;-)

k-epsilon models are not capable of predicting transtition in general. How they behave in this type of flows is also often very dependent on the numerics in your solver. Most often you get turbulent flow immediatly if you turn on k-epsilon, with eddy-viscosities much higher than your molecular viscosity. If your eddy-viscosity stays below the molecular viscosity it should be okay to keep it on also for laminar flows though.

Sometimes you see people trying to predict by-pass transition with two-equation models, sometimes even with some success. By-pass transition is transition in the boundary-layer caused by a high level of free-stream turbulence, essentially diffusion of turbulent energy into the boundary layer - this is very much different from a "natural transition", caused by growing instabilities. Don't try to predict natural transition with k-epsilon. You have to use some kind of specialized transition model for this.

You asked about if there is some degree of turbulence in a laminar flow - the answer is no, by definition, and using a k-epsilon model in a laminar flow-case does not predict anything additionaly useful than running it without the turbulence model.

Transition modeling has been discussued on these forums many times - click the "Search" button in the top right corner and search for "transition" - gives more than 240 hits.

Btw, I read your post above about the low quality of the responses you got. I'm sorry that you feel that way. You are asking a very difficult question though, and a question to which it is difficult to give a clear a concise answer - that might explain the quality of the responses. Better luck next time!

John C. Chien September 7, 2001 19:57

3 cells in inlet; 6x9 cavity flow
(1)I was finally able to take a look at the webpage, by moving the address one subdirectory up. (2). There are results of calculations one for the backstep flow, with 3 cells at the inlet, and the other for the cavity flow, with 6x9 cells. (3). In my opinion, the inlet location must be at least several step height away from the step corner, and the inlet mesh size must be or order 30 cells. Using only 3 cells at the inlet is just a JOKE. (4). For the cavity flow (I think it is a cavity flow. I could be wrong), more than twenty years ago, I was using 125x125. And most people are using 51x51 nowadays. Using 6x9 size is just CRAZY.

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