How do we calculate the normal Reynolds stresses? (Linear eddy viscosity models)

userid42 · August 18, 2017, 13:19

I have read in multiple places that the normal Reynolds stresses are equal when the linear eddy viscosity model is used.
If they are equal, per the definition of turbulent kinetic energy:

$k = \frac{1}{2} (\overline{u'u'} + \overline{v'v'} + \overline{w'w'})$

$\overline{u'u'} = \frac{2}{3}k$ .

From what I understand, we can calculate the Reynolds stresses from the Boussinesq approximation. In this approximation, the eddy viscosity is assumed to be isotropic (not turbulence).

The Boussinesq approximation is:

$-\rho\overline{u^{'}_{i}u^{'}_{j}}=\mu_t\left(\frac{\partial\overline{u_i}}{\partial{x_j}}+\frac{\partial\overline{u_j}}{\partial{x_i}}-\frac{2}{3}\frac{\partial\overline{u_k}}{\partial{x_k}}\delta_{ij}\right)-\frac{2}{3}\rho k\delta_{ij}$ .

From this equation, the normal stresses (for incompressible flows) can be calculated by:

$\overline{u^{'}u^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{u}}{\partial{x}}+\frac{\partial\overline{u}}{\partial{x}}\right),$

$\overline{v^{'}v^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{v}}{\partial{y}}+\frac{\partial\overline{v}}{\partial{y}}\right),$

and

$\overline{w^{'}w^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{w}}{\partial{z}}+\frac{\partial\overline{w}}{\partial{z}}\right)$ .

From the expressions above, I see that the normal stresses are not equal because the velocity gradients will be different. Also, this approach introduces an extra term " $-\frac{\mu_t}{\rho}(...)$ ".

Which approach is correct? Am I missing something? Or is this a misconception of what is isotropic?
Feel free to correct me if I stated something wrong.

Thanks!

Hamidzoka · August 20, 2017, 09:54

hi
Boussinesq approximation as you have shown in your post implies a relation between mean parameters of a flow and reynolds stresses and can be applied to both compressible and incompressible flows. turbulence models that use this hypothesis are categorized as eddy viscosity models.
Some eddy viscosity-based turbulence models like standard k-epsilon use further simplifications to Boussinesq approximation to get equal values for normal components of reynolds stress tensor (isotropic turbulence). This provides lower computational costs with less accurate results.
Therefore, according to your post both are true depending on the selected turbulence model.
Note that K-epsilon is not necessarily an isotropic model. one example is non-linear k-epsilon models.

LuckyTran · August 20, 2017, 11:43

Quote:

Originally Posted by userid42

I have read in multiple places that the normal Reynolds stresses are equal when the linear eddy viscosity model is used.

I really must applaud your willingness to learn what is going on. If one simply writes down the eddy viscosity model then it is immediately apparent that something does not make sense! Unfortunately these misconceptions are widespread. Even if you call Fluent and CD-Adapco support you will hear these answers from them. I also would not be surprised to hear it from a Dept Head or journal editor either. The problem is few people have studied turbulence. More people now are interested in applied turbulence, that is more people use turbulence models to get a result than the number of people that have any clue what any of it means. Part of the problem is that we use imprecise language. The same occurs in incompressible flows. When you say incompressible flow, do you mean a constant density, a temperature dependent density, or a divergence free velocity? They're not equivalent. What exactly do you mean by incompressible flow? What exactly is an isotropic turbulence model? Which part of it is isotropic?

An isotropic turbulence model usually means that the eddy viscosity is isotropic, i.e. a scalar constant value (a uniform matrix) instead of a matrix with different values for each element. The confusion begins because nearly all popular turbulence models are eddy viscosity models (linear or non-linear). Even non-linear eddy viscosity models will still use an isotropic eddy viscosity. The eddy viscosity model is too prevalent.

The first approach is just plain wrong as a model. What you have written is of course correct if one assumes that the normal stresses are all equal, but they can't be in general. Near boundaries for example, the wall-normal stresses must vanish due to the kinematic condition. And this consequence has nothing to do with turbulence modelling, it's just pure kinematics. Secondly, note that this assumption leads you only to a relation about the normal stresses and nothing about the remaining stresses (all of them appear in the momentum equation). Even if it was correct, it's an incomplete model.

Quote:

Originally Posted by Hamidzoka

From the expressions above, I see that the normal stresses are not equal because the velocity gradients will be different. Also, this approach introduces an extra term " $-\frac{\mu_t}{\rho}(...)$ ".

It's not an extra term, that term should always be there. That is what makes it a linear eddy viscosity model. It is the most important term. Notice that without that term there is no eddy viscosity and nothing that needs to be modeled. If you keep simplifying, you will eventually arrive at the definition of the turbulent kinetic energy.

FMDenaro · August 20, 2017, 12:06

I agree that some confusions in the nomenclature appear...

Isotropy in the eddy viscosity model means that you have the identity matrix multiplied by the eddy viscosity function (Ini_turb). As you see, that means that the eddy viscosity is not linear, depending ni_turb on the averaged velocity. A non-isotropic model introduces a full matrix for the eddy viscosity Ni_turb.

Furthermore, speaking about the normal stress, for incompressible flow the isotropic part of the tensor is not modelled. It simply enters into the pressure gradient modifying the pressure. A model is strictly necessary for a compressible flow.

A book like that of Wilcox can give many details

userid42 · August 21, 2017, 13:52

Thank you all for your detailed explanations!
I will only calculate the normal Reynolds stresses from the eddy viscosity expression then

!

If anyone has a different perspective, please feel free to comment.

sbaffini · August 25, 2017, 14:00

Quote:

Originally Posted by userid42

I have read in multiple places that the normal Reynolds stresses are equal when the linear eddy viscosity model is used.

Well, can you give any example of relevant source stating this? I guess the main confusion is on this against the concept of isotropic turbulence.

Isotropic turbulence (roughly, equal normal stresses) is sometimes used as hypothesis when deriving boundary conditions for turbulence models (i.e. to directly link turbulence intensity I to k).

Linear eddy viscosity models don't imply equal normal stresses, as you realized. So, the resulting turbulence effect is not isotropic.

It is the model that is isotropic or, better, the model dependency on Sij. So, in a certain sense, isotropy of the turbulence model is intended as isotropy for materials.

However, I agree that this terminology is very confusing and should be removed from most sources not going into the details of it.

userid42 · August 27, 2017, 12:08

Quote:

Originally Posted by sbaffini

Well, can you give any example of relevant source stating this? I guess the main confusion is on this against the concept of isotropic turbulence.

Hello sbaffini,
I agree with your statements. Here is once source stating that the normal stresses are equal (Pg. 28, or see the attachment):
http://www.bakker.org/dartmouth06/engs150/10-rans.pdf.

Statements like this make me think that I am misunderstanding something. It is also possible that I am misinterpreting this person's slide. Feel free to provide any clarifications.

Quote:

Originally Posted by sbaffini

Isotropic turbulence (roughly, equal normal stresses) is sometimes used as hypothesis when deriving boundary conditions for turbulence models (i.e. to directly link turbulence intensity I to k).

Could you give me some more details about this statement (or provide an article)? I am not informed about this.

Thank you for your response!

sbaffini · August 28, 2017, 05:58

Well, that's not exactly what I consider a relevant source

What they mention is not even misleading, just plain wrong.

For what concerns the bc, 2eq models will require k at inlets. If, however, you only have the turbulence intensity, you need a mean to go from one to the other.

Look here for the details (under Estimating Turbulent Kinetic Energy from Turbulence Intensity):
http://www.afs.enea.it/project/neptu....htm#ke-params

basically, you have I that only refers to the streamwise fluctuations and convert it to k using the isotropic turbulence assumption, as that is typically still your best option if you lack any other information.

userid42 · August 28, 2017, 14:11

Quote:

Originally Posted by sbaffini

Well, that's not exactly what I consider a relevant source

The website appears to be put together by someone with an @ANSYS.com e-mail address. For that reason, I thought it could be a good source

.

Thank you for your kind explanations!

August 18, 2017, 13:19	How do we calculate the normal Reynolds stresses? (Linear eddy viscosity models)	#1
userid42 New Member Daniel Join Date: Aug 2017 Posts: 7 Rep Power: 9	I have read in multiple places that the normal Reynolds stresses are equal when the linear eddy viscosity model is used. If they are equal, per the definition of turbulent kinetic energy: $k = \frac{1}{2} (\overline{u'u'} + \overline{v'v'} + \overline{w'w'})$ $\overline{u'u'} = \frac{2}{3}k$ . From what I understand, we can calculate the Reynolds stresses from the Boussinesq approximation. In this approximation, the eddy viscosity is assumed to be isotropic (not turbulence). The Boussinesq approximation is: $-\rho\overline{u^{'}_{i}u^{'}_{j}}=\mu_t\left(\frac{\partial\overline{u_i}}{\partial{x_j}}+\frac{\partial\overline{u_j}}{\partial{x_i}}-\frac{2}{3}\frac{\partial\overline{u_k}}{\partial{x_k}}\delta_{ij}\right)-\frac{2}{3}\rho k\delta_{ij}$ . From this equation, the normal stresses (for incompressible flows) can be calculated by: $\overline{u^{'}u^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{u}}{\partial{x}}+\frac{\partial\overline{u}}{\partial{x}}\right),$ $\overline{v^{'}v^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{v}}{\partial{y}}+\frac{\partial\overline{v}}{\partial{y}}\right),$ and $\overline{w^{'}w^{'}}=\frac{2}{3}k - \frac{\mu_t}{\rho}\left(\frac{\partial\overline{w}}{\partial{z}}+\frac{\partial\overline{w}}{\partial{z}}\right)$ . From the expressions above, I see that the normal stresses are not equal because the velocity gradients will be different. Also, this approach introduces an extra term " $-\frac{\mu_t}{\rho}(...)$ ". Which approach is correct? Am I missing something? Or is this a misconception of what is isotropic? Feel free to correct me if I stated something wrong. Thanks! Ashkan Kashani likes this. Last edited by userid42; August 18, 2017 at 22:07.

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August 20, 2017, 09:54		#2
Hamidzoka Senior Member Hamid Zoka Join Date: Nov 2009 Posts: 293 Rep Power: 19	hi Boussinesq approximation as you have shown in your post implies a relation between mean parameters of a flow and reynolds stresses and can be applied to both compressible and incompressible flows. turbulence models that use this hypothesis are categorized as eddy viscosity models. Some eddy viscosity-based turbulence models like standard k-epsilon use further simplifications to Boussinesq approximation to get equal values for normal components of reynolds stress tensor (isotropic turbulence). This provides lower computational costs with less accurate results. Therefore, according to your post both are true depending on the selected turbulence model. Note that K-epsilon is not necessarily an isotropic model. one example is non-linear k-epsilon models.

August 20, 2017, 12:06		#4
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,897 Rep Power: 73	I agree that some confusions in the nomenclature appear... Isotropy in the eddy viscosity model means that you have the identity matrix multiplied by the eddy viscosity function (Ini_turb). As you see, that means that the eddy viscosity is not linear, depending ni_turb on the averaged velocity. A non-isotropic model introduces a full matrix for the eddy viscosity Ni_turb. Furthermore, speaking about the normal stress, for incompressible flow the isotropic part of the tensor is not modelled. It simply enters into the pressure gradient modifying the pressure. A model is strictly necessary for a compressible flow. A book like that of Wilcox can give many details

August 21, 2017, 13:52		#5
userid42 New Member Daniel Join Date: Aug 2017 Posts: 7 Rep Power: 9	Thank you all for your detailed explanations! I will only calculate the normal Reynolds stresses from the eddy viscosity expression then ! If anyone has a different perspective, please feel free to comment.

August 28, 2017, 05:58		#8
sbaffini Senior Member Paolo Lampitella Join Date: Mar 2009 Location: Italy Posts: 2,195 Blog Entries: 29 Rep Power: 39	Well, that's not exactly what I consider a relevant source What they mention is not even misleading, just plain wrong. For what concerns the bc, 2eq models will require k at inlets. If, however, you only have the turbulence intensity, you need a mean to go from one to the other. Look here for the details (under Estimating Turbulent Kinetic Energy from Turbulence Intensity): http://www.afs.enea.it/project/neptu....htm#ke-params basically, you have I that only refers to the streamwise fluctuations and convert it to k using the isotropic turbulence assumption, as that is typically still your best option if you lack any other information.