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How do we calculate the normal Reynolds stresses? (Linear eddy viscosity models) |
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#1 |
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Daniel
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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: ![]() ![]() 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: ![]() From this equation, the normal stresses (for incompressible flows) can be calculated by: ![]() ![]() and ![]() 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 " ![]() 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! Last edited by userid42; August 18, 2017 at 21:07. |
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#2 |
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Hamid Zoka
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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. |
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#3 | |
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Lucky
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Quote:
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. 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. |
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#4 |
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Filippo Maria Denaro
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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 |
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#5 |
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Daniel
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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. |
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#6 | |
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Quote:
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. |
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#7 | ||
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Daniel
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Quote:
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:
Thank you for your response! |
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#8 |
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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. |
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#9 |
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Daniel
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