[blyatman]

In the equation for in the following link: https://www.afs.enea.it/project/nept...de61.htm#eq-gk, is equal to the product of the Reynolds stress and the velocity gradient. However, it then says , where is the mean rate of strain tensor, which I'm assuming is just .

Now, the full Reynolds stress term contains a term for the TKE, as shown here:
https://www.cfd-online.com/Wiki/Bous...ity_assumption

In the Fluent theory, clearly does not contain the TKE term. As far as I can tell, it does not seem to absorbed into another term in the TKE equation. Why is the TKE term not present in the definition of ? Is it neglected because it is insignificant?

[LuckyTran]

A proper tensorial closure model should give you 4.4-21.

However, the Boussinesq hypothesis deals only with the traceless part of the Reynolds stress tensor and that's why it is written that way (with the -2/3 k magically showing up out of nowhere).

The (2/3)k is often absorbed into the pressure term explicitly or implicitly since there's no way for a solver to differentiate between pressure and 2/3 k unless you couple k into the momentum equations. It usually is very small. So yes, it magically disappears as well.

k is proportional to UI (mean velocity x turbulence intensity) whereas the dynamic pressure is like U^2/2

To fully track across different sources where is the two=thirds k is a bit tedious because you have to write down the full NS equations and all the constitutive relations on the same page and their definitions without vague references on relying on people to know what is a traceless strain rate tensor (because that could be any of 4 different tensors).

[blyatman]

Thanks, this mostly makes sense. I'm a bit confused about the part where you said another reason it's done is to decouple the momentum equation from the k equation. From the Boussinesq hypothesis, the Reynolds stress term is contains the eddy viscosity, which depends on k. Thus, aren't the momentum and k equations coupled? Intuitively, this is what I would expect since the turbulence affects the mean flow field. However, I understand the part about how k is small, so it is neglected anyway (regardless of whether it's coupled or not).

[LuckyTran]

Originally Posted by blyatman

I'm a bit confused about the part where you said another reason it's done is to decouple the momentum equation from the k equation.

Coupling between independent transport equations is different. Of course everything is coupled otherwise we would not need to solve them at the same time (temperature/energy also affects the flow field even though temperature never appears anywhere in the momentum equations). I'm talking about k appearing explicitly in the momentum equations.


If i have a system that I think is:

ax+by+(p1)=0

but in reality the real system is:

ax+by+(p2+2/3 k)=0

where the a, x, b, y are the same...

If I solve the first system for p1, I will still end up solving for p2+2/3 k whether I like it or not. Even if I use a magic wand to make the two-thirds k disappear analytically, numerically it shows ups because the a,x,b,y are the same. So the pressure that you solve for from your CFD code isn't actually pressure unless you explicitly put the k in it and solve the second system.