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kEpsilon vs. kOmegaSST

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Old   February 8, 2020, 13:22
Default kEpsilon vs. kOmegaSST
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Hi all,


just a simple question. As we all know, the kOmegaSST model is the extension of the kOmega model. In some posts I read the following which should be correct (as far as I know):


  • kEpsilon predicts well for far-field analysis without high gradients and no flow separation; e.g. freestreams such as Sandia flames etc.
  • kOmega predicts well close to the walls (compared to kEpsilon) and should not be used for far-field calculations
  • kOmegaSST combines the advantages of both above mentioned models using a blending function.


So far so good. I do have a case here, which is a freestream (such as the sandia flame). I calculated the 4 vol.-% species contour for the kEpsilon and kOmegaSST model which is different (that is obvious).









What I figured out

  • The turbulent kinetic energy is much higher close to the pipe outlet for the kOmegaSST model which lead to a higher turbulent viscosity there. Thus, a higher species diffusion compared to the kEpsilon.


So the question in this particular regards is:


  • kEpsilon or kOmegaSST


In general, I would prefer the kEpsilon for the freestream case but for some reason, I could imagine that the kOmegaSST model predicts the turbulent quantities more accurate around the pipe which lead to the higher diffusion...

I know, turbulence modeling is an own big topic and each model predicts its own values. So any hint or knowledge is recommend. Sure, I know that kOmegaSST is used widely in industries but for this problem, I am not sure if the kEpsilon fits better.
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Old   February 9, 2020, 12:24
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Please note that kOmegaSST comprises of the standard kEpsilon in theory. Under certain conditions, it should recover to kEpsilon, therefore. So, I would stick with kOmegaSST unless otherwise suggested.

Having said that the standard kEpsilon is mostly lighter than kOmegaSST to run. Yet since floating-point computation is virtually free in 2020, unlike memory constraints, nobody should give a f.ck. However, numerical stability is another aspect kEpsilon might be less favourable, and this may tradeoff the gains you would get from kEpsilon's lightness.

On the other hand, both models will provide `wrong` results. So, I would go for LES or DES or very large LES, or even LES with RANS grid. In the long term, you end up with immense amount of loss in terms of time when you stick with RANS.

I might also be wrong with the above remarks. So, for them, I only cite my arse, for those who would assume their complete validity.

Hope you find your answer.
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Old   February 9, 2020, 13:13
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Thanks for your thoughts.


Well, modeling in RANS seems to be old-school nowadays as the computers get more powerful and memory as well as cpu power is available and thus, LES gets more interesting. However, the point is always: wasting computational power for a more accurate results or be fine with averaged 'wrong' results.

Thanks. I stay with the SST.
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Old   February 14, 2020, 01:23
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Hi Tobi. For interest sake, have you perhaps considered LRR and compare against k-e and SST?
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Old   February 14, 2020, 16:04
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Could you please elaborate why you specifically consider LRR?
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