# Isotropy/Anisotropy in Eddy-Viscosity models

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 June 12, 2015, 11:25 Isotropy/Anisotropy in Eddy-Viscosity models #1 New Member   Carla Join Date: Jun 2015 Posts: 16 Rep Power: 11 Hi to everyone, I am new in CFD-online. I hope this is the right place to post a theoretical question. Here it goes. I've heard many times about an "intrinsic" assumption of isotropic turbulence in linear eddy-viscosity-based turbulence models. However, I do not quite understand this assertion. From what I understand, mathematically, isotropic turbulence means that the Reynolds stress is a diagonal matrix where every element is equal. In the eddy viscosity hypothesis for incompressible flows: The first term in the RHS is isotropic, but the second term it appears to me that is modelling anisotropy, as it has non-zero terms outside the diagonal. Then, what do people mean by "assumption of isotropic turbulence"? Do they refer to the transport equation for turbulent kinetic energy? Here is stated that turbulent diffusion is what is assumed to be isotropic. (I wouldn't understand anisotropic diffusion, though. What would make diffusion to have a preferential direction?) Could someone help me with that? Thank you for your time. Carla

June 12, 2015, 11:44
#2
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Filippo Maria Denaro
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Quote:
 Originally Posted by cvillescas Hi to everyone, I am new in CFD-online. I hope this is the right place to post a theoretical question. Here it goes. I've heard many times about an "intrinsic" assumption of isotropic turbulence in linear eddy-viscosity-based turbulence models. However, I do not quite understand this assertion. From what I understand, mathematically, isotropic turbulence means that the Reynolds stress is a diagonal matrix where every element is equal. In the eddy viscosity hypothesis for incompressible flows: The first term in the RHS is isotropic, but the second term it appears to me that is modelling anisotropy, as it has non-zero terms outside the diagonal. Then, what do people mean by "assumption of isotropic turbulence"? Do they refer to the transport equation for turbulent kinetic energy? Here is stated that turbulent diffusion is what is assumed to be isotropic. (I wouldn't understand anisotropic diffusion, though. What would make diffusion to have a preferential direction?) Could someone help me with that? Thank you for your time. Carla

Think as example about the assumption of Newtonian fluid, it links the stress tensor to the velocity gradient by means of a linear functional relation in which you have the isotropic physical viscosity.

Similarily, you can build the SGS tensor and decide to use in the relation with the averaged velocity gradient some functional relation in which the eddy viscosity can be expressed in an isotropic or anysotropic assumption.
In the most general case you could suppose to build a 3x3 matrix of SGS viscosities entry.

You can read much more details in many books (Wilcox, Sagaut, Pope, ecc.)

 June 12, 2015, 14:02 #3 Member   robo Join Date: May 2013 Posts: 47 Rep Power: 13 The portion of that expression that is anisotropic is only a measure of the anisotropy of the mean flow and not of the anisotropy of the turbulence itself. If you expand the expression for the turbulent diffusion it will be written in terms of k, which only has information about the isotropic part of the turbulence. So nowhere in that expression is there actually any information about the anisotropic turbulent motion, merely an assumed relation based on the isotropic turbulent motion and the anisotropic mean motion. cvillescas likes this.

 June 13, 2015, 03:43 #4 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,811 Rep Power: 72 Just a furhter comment... the formula you wrote highlights the isotropic (note that for incompressible flows, any isotropic tensor simply add to the pressure term Ip) and deviatoric (with zero trace) of the modelled tensor. The possible implementation of the anysotropic model acts on the last term by considering a matrix of the eddy viscosity. This can be justified by the fact that the real effect of the unresolved fluctuations (LHS in the above expression) has no theoretical reasons to act only in isotropic way. You can also search for some old paper of Cercignani et al. cvillescas likes this.

July 8, 2015, 07:19
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Carla
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Thank you for your answers and sorry for the late reply. I have been going through Pope and there, a non-linear eddy viscosity model is suggested (the one in the image). aij is the deviatoric Reynolds stress tensor.

If I understand correctly what you are saying, this suggested turbulence model allows for misalignment between aij and S but this does not mean that it takes anisotropy into account because:
• From what FMDenaro said, the eddy viscosity is not a matrix
• And from what robo said, S and Omega only reflect the already existing anisotropy in the mean flow. And do not take into account the anisotropy of the turbulence itself.
Is that correct? Or the fact that you have omega in there also allows for anisotropy of turbulence itself?

Carla
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 July 8, 2015, 07:24 #6 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,811 Rep Power: 72 have a look here http://citeseerx.ist.psu.edu/viewdoc...=rep1&type=pdf other infos are illustrated in the book of Sagaut cvillescas likes this.

 July 8, 2015, 07:48 #7 New Member   Carla Join Date: Jun 2015 Posts: 16 Rep Power: 11 Thank you!

 June 21, 2019, 05:30 #8 New Member   Join Date: Jun 2019 Posts: 1 Rep Power: 0 Hi Carla and everyone, who google led here for answers, I had the exact same question as Carla today and didn't really understand it from the previous answers, so I looked a bit further into it. I think the key point is to look into the derivation of viscous stresses for Newtonian fluids (https://ceprofs.civil.tamu.edu/ssoco...onianFluid.PDF), which was already mentioned by FMDenaro. The viscous stresses are assumed to be a linear function of the strain rate. But since the viscous stresses are a second order tensor and the strain rate is a second order tensor, there are 81 unknown coefficients. Only the assumption of an isotropic fluid (and a symmetric strain rate tensor) reduces the number of coefficients to 2. Afterwards, the hypothesis of Stokes is applied to yield the dynamic viscosity. Thus, the dynamic viscosity is an isotropic scalar quantity. Overall, this means that the fluid property (dynamic viscosity) is isotropic, but not the resulting viscous stresses. Now, the Boussinesq approximation is closely related to the definition of Newtonian fluids. Thus, the conclusion should be that the eddy viscosity and thus the turbulence properties are isotropic, but not the Reynolds stresses due to turbulence. (Please, feel free to correct me in case of any mistakes.) Cheers, Fabian

 June 21, 2019, 06:37 #9 Senior Member   Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,811 Rep Power: 72 We can look to this matter in a matrix notation: T = matrix for the stress tensor K = matrix for the viscosities S = matrix for the velocity gradient The Newtown linear relation assumes T = K.S (1) As you wrote, we need to introduce some assumptions. The isotropy assumption on the viscosity matrix says T = (mu I).S (2) being I the identity matrix. From these expressions, we can assume - anisotropy as the (1) instead of (2) - non-linearity as the (1) wherein K=K(S)

 Tags anisotropy, eddy viscosity, turbulence

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