[pawarit28]

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!

[Светлана]

Hello, in 3d it is

u'u' u'v' u'w'

v'u' v'v' v'w'

w'u' w'v' w'w'

but w'u' = u'w' and so on so the matrix is symmetric. We are left with 6 unique items which I highlighted in bold. In 2d it is four stresses in total of which only three are unique. The 2d version is underlined.

[pawarit28]

Hi, thanks for answering!
The Reynolds Stress Models (RSM) in 2D CFD simulations (such as in Fluent) will solve the transport equations for the three unique stresses: u'u', u'v', v'v'
as well as an additional transport equation for: w'w'

This final term in bold is what confuses me. I don't understand why CFD packages solve for w'w' in 2D exactly..Do you have any more insight on this?

(PS. and since w' is indeed non-zero..then why can we ignore u'w' and v'w' in 2D simulations but not w'w'?)

[Светлана]

Interesting. I apologize that I did not catch this from your original question. Where is this a quote from?

[pawarit28]

Apologies if I was unclear at first.

Here's a brief overview:
https://www.sharcnet.ca/Software/Ans...ss_vs_rst.html
"The alternative approach, embodied in the RSM, is to solve transport equations for each of the terms in the Reynolds stress tensor. An additional scale-determining equation (normally for epsilon or omega) is also required. This means that five additional transport equations are required in 2D flows"
(i.e. 4 equations for reynolds stresses + 1 scale-dissipation)

(if you google "Reynolds stress additional transport equations", all sources/literature agree that four unique stresses are solved for 2D flow. We also agree that u'v' = v'u'. Therefore, we've only counted the three that you first mentioned)

Then when running any 2D Reynolds Stress Model (RSM) simulations in Fluent, I always observed that w'w' was always being solved for. Leading me to believe that this is the fourth and final stress solved for in 2D

Thanks again!

[Светлана]

Perhaps due to lack of focus (although I tried to focus for about 2 minutes), I do not see your quote in the page linked. Please highlight it and take a screenshot, or change the link. Thank you.

[pawarit28]

"The alternative approach, embodied in the RSM, is to solve transport equations for each of the terms in the Reynolds stress tensor. An additional scale-determining equation (normally for epsilon or omega) is also required. This means that five additional transport equations are required in 2D flows"

these five equations consist of: 4 reynolds stresses + 1 scale-dissipation

[FMDenaro]

I have not experience in such a type of modelling but I can suppose that the 2D case is actually to be intepreted as a statistically averaged of a 3D turbulent case. The third component of the fluctuations has therefore an impact on the 2D averaged flow.
Someone more expert in this model can highlight the issue

[Hamidzoka]

Hi;
have you checked the calculated w'w' value or compared its magnitude with v'v' and u'u'?
To my understanding if you consider w'w' term in a 2D simulation it means that a fraction of turbulent kinetic energy (K=(u'u'^2+v'v'^2+w'w'^2)^0.5) is distributed in w direction and therefore the portions of u'u' and v'v' goes down. in other words turbulent kinetic energy leaks from the 2D plane of the domain and the true kinetic energy level is underestimated. since in a pure 2D turbulent flow no normal to plane velocity component exists.
This does not make sense in general unless the magnitude of w'w' term is quite small in comparison to v'v' and u'u'.
one more point to add is that second moment closure models are numerically unstable and solving the w'w' term may be an strategy to add more stability to the solution.

Regards

[LuckyTran]

For 2D, you need to also solve for w'w' because it is part of k since k = 1/2 (u'u' + v'v' + w'w'). You use k as part of your closure model for the Reynolds stresses so that you actually do need (u'u', v'v', and u'v').

Btw, k appears in the energy equation.

Originally Posted by Hamidzoka

In other words turbulent kinetic energy leaks from the 2D plane of the domain and the true kinetic energy level is underestimated. since in a pure 2D turbulent flow no normal to plane velocity component exists.
This does not make sense in general....

This "leakage" property is from where we say that turbulence is 3D. Any turbulence that lacks this characteristic we call non-physical turbulence. It's very physical, with kinematic origins. If you squeeze a tube of toothpaste, the toothpaste comes out the end. You apply an in-plane force but you get an out-of-plane motion!

[FMDenaro]

I agree...as I wrote above, 2D turbulence actually does not exist but in same cases it is possible to solve it in a statistical meaning. But you cannot disregard the contribution of the part of the third component on the motion on the plane.