[arash509]

I have a very simple model. my model is a channel with air as working fluid which there is a constant heat flux of 800 W/m2 in the lower surface of channel,and there is a convection boundary condition for the upper surface of channel.



I wanted to compare the laminar and turbulence models so I have selected the Re less than 2000 so flow in channel should be laminar. but the problem is that when I change model to turbulence the results is totally different.


For laminar outflow temperature is = 460 K


For K-Omega model outflow temperature is = 440 K


For K-epsilon model the outflow temperature is = 420 K



Can you please tell me why there is difference between models?

[LuckyTran]

Which turbulence models exactly? And what wall treatment options did you use? Standard k-omega, standard k-epsilon, k-omega SST, realizable k-epsilon, other k-epsilon? Even amongst these you can expect different results. Furthermore, we have no dimensionality to the info you provided. There's no way for any of us to know how significant the differences are.

First, you should expect a huge difference between laminar and turbulence models. The heat transfer rate going from laminar to turbulent is about an order of magnitude.

So now you should be wondering why k-omega models and k-epsilon models predict different results regardless of what the laminar model is doing. You can't even begin to address this unless you can ensure that your mesh is suitable for all of the models attempted. So get started... Number of cells across the boundary layer? Wall y+? Etc.

[arash509]

Dear LuckyTran


Thanks for response,



1. models are (a) laminar model (b) Standard k-omega (c) standard k-epsilon


2. model is very simple, only a rectangular duct 2m length, with cross section of 1cm*1m. The bottom surface has constant heat flux 800 W/m2, the upper surface has convection heat transfer to environment with heat transfer coefficient of 9.5 W/k.m2


3. my question is why should I expect a huge difference between laminar and turbulence models, because the flow is laminar so even when turbulence model choose it should not have much difference with laminar results because I tried it for low Re numbers around 500-1000. for all of them I can see huge difference


4. I also used boundary layer mesh but still there is difference.

[LoGaL]

The RANS turbulence models alwais assume the flow to be turbulent, even if in reality is laminar, so if in a laminar flow you turn on a turbulence model, it will see some turbulence that in reality does not exist, hence why the results are quite different

[arash509]

Dear Logal

Thanks a lot for your response, however, I don’t think results should have much difference. Because if flow is laminar and we choose turbulence models those terms related to turbulency should be zero then the results shouldn’t have much difference.

[AlexanderZ]

Originally Posted by arash509

Dear Logal

Thanks a lot for your response, however, I don’t think results should have much difference. Because if flow is laminar and we choose turbulence models those terms related to turbulency should be zero then the results shouldn’t have much difference.

why do you post questions if you are an expert?

[LoGaL]

Originally Posted by arash509

Dear Logal

Thanks a lot for your response, however, I don’t think results should have much difference. Because if flow is laminar and we choose turbulence models those terms related to turbulency should be zero then the results shouldn’t have much difference.

Should, should, should, think... Did you check? Open some turbulence model equations, find a way to take the limit Re --> 0 ( not Velocity, Re) and see if all the terms disappear as you think they should.

[LuckyTran]

There are major phenomenological differences between a laminar and turblent flow which we can shorten the discussion by just looking at the effective viscosity. If you use a laminar model, there is no eddy viscosity and the effective viscosity is just the molecular viscosity which is purely dependent on temperature and pressure. When you turn on a turbulence model, you are saying that the eddy viscosity is proportional to the strain rate and effective viscosity is (roughly) the sum of molecular and eddy viscosity. Laminar flows have strain but they should not produce eddy viscosity! So if there's any strain in your flow that should be laminar, it will still produce an eddy viscosity and I would never expect the results to be the same. Now if you have just a uniform undisturbed flow where there isn't any velocity gradients whatsoever, then you could argue the results will match.

Originally Posted by arash509

Dear Logal

Thanks a lot for your response, however, I don’t think results should have much difference. Because if flow is laminar and we choose turbulence models those terms related to turbulency should be zero then the results shouldn’t have much difference.

Velocity gradients don't go to zero even for laminar flows, so the turbulency terms never disappear.

It seems like you don't care why the k-omega and k-epsilon models produce different results, only that the laminar and turbulent ones are different. So I guess we don't need to discuss it any further.

Anyway, using a turbulence model on a low Reynolds number won't produce a laminar solution.