[FluidKo]

I'm reading paper in below link.
https://www.researchgate.net/publica..._of_Turbulence

I've uploaded picture that is describing my question.
I'm wondering why green lined term is expressed like this picture.
I can understand red line part(Mixing length theory).
But I can't understand for green lined part
Thanks

[LuckyTran]

Because a is the anisotropic part of the Reynolds stress tensor (and here it is actually also normalized by k).

At the same time, the trace of the Reynolds stress tensor (uu+vv+ww) is 2k (twice the turbulent kinetic energy). Hence, subtracting 2/3-k from the Reynolds stress tensor is a convenient way of arriving at the anisotropic part.

The next line is a general (up to quadratic) Stress-Strain model and has all the allowed permutations of the Strain rate tensor (S, S-S, S-Omega, Omega-Omega). The only term missing is the linear Omega term. For example, if I give you x and y and tell you to build a quadratic model based off of it you would return to me: a+bx+cy+dx^2+ex*y+fy2

[FluidKo]

Originally Posted by LuckyTran

Because a is the anisotropic part of the Reynolds stress tensor (and here it is actually also normalized by k).

At the same time, the trace of the Reynolds stress tensor (uu+vv+ww) is 2k (twice the turbulent kinetic energy). Hence, subtracting 2/3-k from the Reynolds stress tensor is a convenient way of arriving at the anisotropic part.

The next line is a general (up to quadratic) Stress-Strain model and has all the allowed permutations of the Strain rate tensor (S, S-S, S-Omega, Omega-Omega). The only term missing is the linear Omega term. For example, if I give you x and y and tell you to build a quadratic model based off of it you would return to me: a+bx+cy+dx^2+ex*y+fy2

I can under stand it is a convenient way of arriving at the anisotropic part.
But there are 2 rested wordering things.

1. '2k/3' part
Reynolds stress is anisotropic so we can't do like below

'2k/3' part should be just '(2/3)*Kronecker delta'.

We should express '2k/3' part by 3 green lines

Is it right that I understand well?

2. What is the reason there is a turbulence viscosity, dissipation, deformation and Vorticity?
Where is these driven from?
Or is it just assumption for this author's dimensional analysis and his own modeling?

Thanks

[LuckyTran]

1. Yes the Reynolds stress tensor is not necessarily isotropic. Hence, we want to separate it into its isotropic part (the dilatation) and the anisotropic part. See, How to get the anisotropic part of a tensor. You subtract the trace. It just so happens that the trace of the Reynolds stresses is something we like to call turbulent kinetic energy. Actually k is half of the trace of the Reynolds stress tensor. Hence, 1/3-rd the trace of the Reynolds stresses is (2/3)k. And no it is not just (2/3)*kronecker delta. kronecker delta has no units so you would just be subtracting 2/3 from a dimensional quantity which makes no sense. It is (2/3)k*kronecker delta. The part with the 1 green line is an identity resulting from the definitions of Reynolds stresses, the anisotropic part of the Reynolds stresses (the a thingy), and k; hence, the triple bar equality. You can always write this down, it has nothing to do with modeling.

Everything in Eq (1), the single red line and three green lines all of them together, is a model for the anisotropic part of the Reynolds stress tensor. The three green lines is NOT a substitution for 2/3-k. I would recommend changing the color of your underline to not red and not green so you don't continue to confuse yourself. That is, your red underlines are not equivalent, and your green underlines are not equivalent in the two equations.

2. Yes dimensional analysis has been invoked and all the details have been skipped. The turbulent viscosity appears because that how we typically model the linear part of the Stress-Strain relation (e.g. the Boussinesq eddy viscosity hypothesis). The k and epsilon look like they appear out of nowhere but if you do dimensional analysis, these are their scalings. Keep in mind the trace of the Reynolds stresses we've already mentioned (many many times now) is k. So the linear terms will be normalized by k. Quadratic terms are quadratic (i.e. SS, and SS is none other than dissipation) so they get normalized by epsilon. You could opt to not normalize a and just write down dimensional coefficients.

Any quadratic relation between Stress and Strain will have an equivalent form. You can write 1+x+y+x^2+x*y+y^2 however you want, they're all the same form. What differs from model to model is the value of the turbulent viscosity, C1, C2, and so on.

[FluidKo]

Originally Posted by LuckyTran

1. Yes the Reynolds stress tensor is not necessarily isotropic. Hence, we want to separate it into its isotropic part (the dilatation) and the anisotropic part. See, How to get the anisotropic part of a tensor. You subtract the trace. It just so happens that the trace of the Reynolds stresses is something we like to call turbulent kinetic energy. Actually k is half of the trace of the Reynolds stress tensor. Hence, 1/3-rd the trace of the Reynolds stresses is (2/3)k. And no it is not just (2/3)*kronecker delta. kronecker delta has no units so you would just be subtracting 2/3 from a dimensional quantity which makes no sense. It is (2/3)k*kronecker delta. The part with the 1 green line is an identity resulting from the definitions of Reynolds stresses, the anisotropic part of the Reynolds stresses (the a thingy), and k; hence, the triple bar equality. You can always write this down, it has nothing to do with modeling.

Everything in Eq (1), the single red line and three green lines all of them together, is a model for the anisotropic part of the Reynolds stress tensor. The three green lines is NOT a substitution for 2/3-k. I would recommend changing the color of your underline to not red and not green so you don't continue to confuse yourself. That is, your red underlines are not equivalent, and your green underlines are not equivalent in the two equations.

2. Yes dimensional analysis has been invoked and all the details have been skipped. The turbulent viscosity appears because that how we typically model the linear part of the Stress-Strain relation (e.g. the Boussinesq eddy viscosity hypothesis). The k and epsilon look like they appear out of nowhere but if you do dimensional analysis, these are their scalings. Keep in mind the trace of the Reynolds stresses we've already mentioned (many many times now) is k. So the linear terms will be normalized by k. Quadratic terms are quadratic (i.e. SS, and SS is none other than dissipation) so they get normalized by epsilon. You could opt to not normalize a and just write down dimensional coefficients.

Any quadratic relation between Stress and Strain will have an equivalent form. You can write 1+x+y+x^2+x*y+y^2 however you want, they're all the same form. What differs from model to model is the value of the turbulent viscosity, C1, C2, and so on.

Thanks for your sincere answer