[Manu4CFD]

While performing a steady state RANS CFD simulation with RNG k-ε turbulence model in Ansys Fluent, if I plot the velocity magnitude (V = SQRT(Ux^2 + Uy^2 + Uz^2) ) at a random point within the computational domain, does this velocity magnitude represents the mean and fluctuating velocity components or is it just representing the mean velocity at that particular point?

[LuckyTran]

only the mean

[Manu4CFD]

Originally Posted by LuckyTran

only the mean

So, it means that as the instantaneous velocity "u" is split into mean velocity "U" and fluctuating velocity u' i.e., (u = U + u'), while time averaging over a sufficiently long time, say "T", the fluctuating component u' tends to zero and the time averaged mean velocity U is treated as a constant. So in short, while performing a steady state RANS simulation, the velocity magnitude plotted for a point or even on a contour tends to provide us only the mean velocity "U".
Is my interpretation of your statement correct?

[LuckyTran]

Originally Posted by Manu4CFD

So in short, while performing a steady state RANS simulation, the velocity magnitude plotted for a point or even on a contour tends to provide us only the mean velocity "U".
Is my interpretation of your statement correct?

Yes. The "velocity" that you get when you do RANS is only the mean velocity.


RANS means Reynolds-averaged Navier Stokes. When you solve RANS, you are not solving the time-accurate Navier Stokes equations but the Reynolds-averaged ones, written in terms of the reynolds-averaged variables (which you are calling the time average velocity).


If you do URANS (unsteady RANS) you still get unsteady reynolds-averaged, which is averaged over a finite T instead of infinite T. Here the velocity you get is still the reynolds-averaged velocity (containing no fluctuating part), although it does vary in time slowly.

[Manu4CFD]

Originally Posted by LuckyTran

Yes. The "velocity" that you get when you do RANS is only the mean velocity.


RANS means Reynolds-averaged Navier Stokes. When you solve RANS, you are not solving the time-accurate Navier Stokes equations but the Reynolds-averaged ones, written in terms of the reynolds-averaged variables (which you are calling the time average velocity).


If you do URANS (unsteady RANS) you still get unsteady reynolds-averaged, which is averaged over a finite T instead of infinite T. Here the velocity you get is still the reynolds-averaged velocity (containing no fluctuating part), although it does vary in time slowly.

Thanks a lot for your valuable feedback.