RSM in 2D Flow
 

Hi,

I was wondering why specifically we require 4 additional transport equations for the Reynolds stresses in 2D flow??

(For the 3D case, it makes perfect sense to take all 6 Reynolds stress components). I understand that turbulence is a 3D phenomenon. Hence, for 2D simulations, we can allow a zero mean W (out of the page) with a fluctuating velocity w'.

But then it raises the question:
Why do we solve for (w'w')?? but not (u'w') or (v'w') then?

Thank you very much everyone!

This is a duplicate of https://www.cfd-online.com/Forums/ma...e-2d-flow.html, I hope an administrator is able to merge the threads and put them all into the FLUENT forum.

hope this stays on the main forum as its a general theory question and doesn't only apply to fluent. thank you

So I answered this question also in the other thread. But I must say this is a very good question!

In 2D you have a zero mean w but non-zero w'. That is correct.

So for 2D mean-flow, you don't ever need to solve for u'w' or v'w', you can skip them. They can be anything and it would not affect your solution.

So the question is why w'w'?

1. If you solve for the Reynolds stresses via transport equations. However, you still have a closure problem for the Reynolds stress equations. Eventually you bring in a transport equation for k and a scalar dissipation to arrive at closure. To get k though, you need w'w'.

2. In most formulations w'w' is hiding in the x-momentum and y-momentum equation in the turbulent pressure because there is an isotropic part of the Reynolds stresses hiding in there. Fluent uses this formulation. However, fundamentally, there is a w'w' even in the x and y momentum equations.

Hi, thanks so much for your great answer! I'm learning a lot from this

Correct me if I'm wrong:
- d(u'w')/dz and d(v'w')/dz do actually appear in the 2D momentum equations (see attached)
- However, we can skip these because we impose that these quantities have no variation in the z direction when solving a 2D flow. Hence, it doesn't matter what u'w' and v'w' are.

Now, just one final thing for me to figure out..

Main question: Where is w'w' fundamentally in the x and y momentum equations? (see attached image)

Especially for RSM where we don't use the Boussinesq hypothesis, w'w' would only show up in the transport equation for k as you mentioned earlier, which I completely agree with. But that's not really fundamental then if I understand correctly?

Thank you again :)

That is correct. d(u'w')/dz and d(v'w')/dz are automatically zero in 2D, otherwise you are talking 3D.

Regarding k, I will double-check. You may be right and it only appears when you apply the Boussinesq hypothesis. My initial thoughts however is that you have shown the incompressible RANS. There is only 1 set of terms involving the fluctuating velocities (when there can be up to 3 coming from the advection part) along with the fluctuating viscous stresses. When you try to solve RANS analytically you often do order of magnitude analysis and kill a lot of these terms, but when you want to derive the correct RANS equations, you should keep them.

Ah-hah. I found my mistake. You are right, the w'w' does not appear in the momentum equations until you start modelling.

Where they do appear is in the energy equation. See Eqns. (14-16)
https://www.cfd-online.com/Wiki/Favr...okes_equations

Some concluding remarks:

Depending on the school of thought, the "Navier-Stokes equations" may refer to all three equations (continuity, momentum, & energy) together. Here it is not incorrect to say that w'w' appears somewhere in the 2D Reynolds-Averaged Navier-Stokes.

But to be even more fundamental, there is only 1 Reynolds Stress tensor. In cartesian this tensor has 9 terms, of which 6 are independent. Really there's only 1 tensor and you need to solve a transport equation for the Reynolds stress. You shouldn't need to worry about which of those 9 or which of those 6 need to be solved because always you solve a transport equation for the entire transported quantity: mass, momentum, energy, turbulent kinetic energy, Reynolds stress, etc.

For example, there's only 1 velocity vector. It's not important that the velocity can be referred to by u,v,w in x,y,z coordinates. Decomposing these vectors and tensors into their components is just from human convenience of looking at the terms. The numerical solver can directly solve for these vectors and tensors quite easily (users of OpenFOAM are aware of how easy it is to modify the governing equations this way). One can makeup a coordinate system with 4 non-orthogonal dimensions and end up with 4 velocity components and many more parts of Reynolds stresses. The wrong & hasty interpretation of this 4-D Navier-Stokes would be that we suddenly have more unknowns than equations. However, that is not the case as long as one does the proper algebra.