[Gerrit]

Hi everyone,

I have a question concerning the difference of the turbulence models k-epsilon and realizable k-epsilon.
I know that different coefficients are used in the equations, but when it is about stability, which one would be the way to go? Especially if the y+ values are because of the complexity not everywhere big enough?

Thanks!

Best
Gerrit

[DarylMusashi]

The realizable k-ɛ model differs from the standard k−ɛ model in two ways. Firstly it contains a new formulation for the turbulent viscosity: Cμ is not a constant like in the standard model but a variable (see https://en.wikipedia.org/wiki/Realiz...lent_Viscosity).
The second difference is a new transport equation for the dissipation rate, ɛ, that is derived from an exact equation for the transport of the mean-square vorticity fluctuation.

As a result it mainly gives improved predictions for the spreading rate of jets, a superior ability to capture the mean flow of complex structures and for flows involving rotation, boundary layers under strong adverse pressure gradients, separation and recirculation. If you predict such flow in your case you should really think about to use this variation of the k-ɛ model.

But both model variants do use wall functions. So no matter if you use the standard or realizable k−ɛ model you must make sure the y+ -values in your first cell near the wall must not be below 30 and should not excessively exceed values of 100. If your mesh does not fulfill these requirements you need to remesh your geometry to get reasonable results.

You cannot overcome a poor quality mesh with changing to another turbulence model, especially not if the other equally uses wall functions.

As a guess I would suggest the standard k-ɛ model runs more stable, as the turbulent viscosity is calculated in a less complex way. But if your mesh fulfills my mentioned requirements stability should not be a problem in both cases.

[Gerrit]

I thanks for the answer! Sounds as if I had to re mesh a bit

Anyway, the realizable k epsilon sound interesting:
http://aerojet.engr.ucdavis.edu/flue.../html/ug/node4


Best

Gerrit

[khaledhmz]

Originally Posted by DarylMusashi

The realizable k-ɛ model differs from the standard k−ɛ model in two ways. Firstly it contains a new formulation for the turbulent viscosity: Cμ is not a constant like in the standard model but a variable (see https://en.wikipedia.org/wiki/Realiz...lent_Viscosity).
The second difference is a new transport equation for the dissipation rate, ɛ, that is derived from an exact equation for the transport of the mean-square vorticity fluctuation.

As a result it mainly gives improved predictions for the spreading rate of jets, a superior ability to capture the mean flow of complex structures and for flows involving rotation, boundary layers under strong adverse pressure gradients, separation and recirculation. If you predict such flow in your case you should really think about to use this variation of the k-ɛ model.

But both model variants do use wall functions. So no matter if you use the standard or realizable k−ɛ model you must make sure the y+ -values in your first cell near the wall must not be below 30 and should not excessively exceed values of 100. If your mesh does not fulfill these requirements you need to remesh your geometry to get reasonable results.

You cannot overcome a poor quality mesh with changing to another turbulence model, especially not if the other equally uses wall functions.

As a guess I would suggest the standard k-ɛ model runs more stable, as the turbulent viscosity is calculated in a less complex way. But if your mesh fulfills my mentioned requirements stability should not be a problem in both cases.

===============
hi,sir I used this two turbulence model in combustion at the same boundary conditions but I get a large difference in the results,for example in the contours temperature I find two different contours...
what's the problem her?