[zkzkzk010537]

Dear friends,

Here i have a general judgement, would you please give opinions on it?

' In RANs model, as long as your mesh is highly refined (very large mesh size), you will get same result over your case whatever the turbulence model you use for your prediction.'

And if so, any benifits from changing RANs to LES, and sorry i don't have the background of LES, but i did hear that LES and RANs have eventually share the same equations.

Thanks,
Kai.

[LuckyTran]

No and No.

Regardless of how fine the mesh is, the governing equations change depending on which models you use (the extra equations needed to bring closure to the Navier-Stokes). You can get the same result only if the turbulence models predict the same behavior (which is almost impossible). As an example, the popular one and two equations model assume the the turbulence is isotropic whereas the reynolds-stress-models permit anisotropic Reynolds stresses. The models already qualitatively and quantitatively disagree on what the turbulence looks like, and hence in general you should never expect the results to be the same.

RANS & LES starts from the Navier-Stokes equations as a starting point. So sure eventually they share the same equations, that is not incorrect but a naive statement. RANS (Reynolds-Averaged or time-averaged Navier-Stokes) is very different from LES (which is spatially filtered Navier-Stokes).

[zkzkzk010537]

Thanks for your reply, i just heard of the statement from others, maybe i misunderstand that.

Regarding to LES, what's the benefit of it if i use transient RANs and refine my mesh to a scale that is same as lowest scale used in LES except for less computational power needed.
Thanks.

[LuckyTran]

As an analogy. RANS is like classical thermodynamics and LES is like statistical thermodynamics.

steady RANS would be similar to using steady state but classical thermodynamics, unsteady RANS would be similar to doing transient but classical thermodynamics. LES is like doing statistical thermodynamics.

steady RANS only gives you time-averaged behavior. Unsteady RANS gives you some temporal information, but it resolves only the slow effects and does not capture the fast effects (the fast turbulence). An implicit assumption is that the time-scale of the turbulence is so fast that the turbulence instantaneously adjusts to the local-time changes (because the same RANS models are used, e.g. equilibrium thermodynamics applied to transient phenomena).

LES can actually give time-resolve information of both slow and fast effects, e.g. it can solve for time-accurate instantaneous velocities, aka turbulence.

[zkzkzk010537]

Thank you so much for your reply. It helps a lot. It seems to make sense, while, as i have no idea about LES, i will go and read some papers first and come back to you with a few questions probably.