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September 7, 2012, 09:42 
Istropic part of Reynolds stress tensor

#1 
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Robert Sawko
Join Date: Mar 2009
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Dear All,
I have a fairly fundamental question about RANS modelling. It occurred to me only recently although I was aware of it for a while. It's just that I never felt bold enough to ask. If this is covered in one of the standard references for turbulence modelling please let me know and I would be happy to do more reading. My main question is: where is the isotropic part of Reynolds stress tensor? Why do momentum equations in majority (all?) of the codes neglect it. Here's the background. Boussinesq hypothesis defines RST in such a way that tr(R)=2k where R is RST and tr is trace. However the codes I work with seem to put into the momentum equations the following expression: div(dev(R)) where div divergence and dev dev(R) = R  1/3tr(R)I Why can we do that? Can we do that? The argument I encountered speaks about similarity to pressure and that the pressure term somehow takes into account the effect of normal fluctuations but I am not sure of this. For instance in fully developed pipe/channel you would expect k to vary strongly in the transverse direction, especially close to the walls. On the other hand pressure gradient is only in the streamwise direction. That's surely not the same behaviour! Is there any good justification for removing this term then? 

September 7, 2012, 12:01 

#2  
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Filippo Maria Denaro
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Quote:
I am not sure to understand your question but if you solve incompressible flows all isotropic terms of the tensor can be added to the "pressure" ... at the end the do not appear in the momentum equation but are englobed in a modified pressure variable. This aspect happens also more in generla, for example when solving the projection method with an implicit time marching scheme you have a modification inglobed in the isotropic tensor.. Conversely, in compressible flow they need of some modeling ... 

September 11, 2012, 06:50 

#3 
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Robert Sawko
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Thanks a lot for the reply. I think this is exactly what I wanted to know, but please let me reiterate just to see if I understand this right.
Let's focus on incompressible flows only and no gravity or other external forces. Just mean and turbulent flow+RANS model. So you're saying that the "pressure" variable is actually a combination of actual pressure and k. In other words \hat p = p + 1/3*k (provided I haven't made a mistake in +/ and coefficients) and than we solve momentum equations with \hat p as an independent variable. Is that right? This should affect my postprocessing somehow. 

September 11, 2012, 07:01 

#4 
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Filippo Maria Denaro
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for incompressible flows, the "pressure" variable is only a lagrangian multiplier... any scalar function that ensure the divergencefree velocity is your "pressure". Thus, since there is no pressure state equation, your postprocessing in terms of the "pressure" is not meaningful...
As example, if you solve the problem (projection method) with an implicit timemarching method (such as CrankNicolson), the "pressure" is further modified 

September 11, 2012, 09:11 

#5 
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Robert Sawko
Join Date: Mar 2009
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Again, thanks for taking your time and writing these clarifications for me. As you see my knowledge of fundamental fluid mechanics is a bit lacking. One of reasons why I am asking these questions is to improve it so if you feel tired of answering please point me out to some sort of book that will tie the mathematical concepts with fluid dynamics.
Right, I admit that I forgot about the lack of physical interpretation for pressure in incompressible flow. I understand that it is there just to secure that the velocity field remains solenoidal. I see it is not tied to any of the thermodynamic quantities etc. The only postprocessing I do on "pressure" is obtaining pressure gradients. So I guess it is reasonable to ask this question for compressible flows only. When I solve for U, p, k, T and the rest how do I include the effect of the isotropic part of RST? This is perhaps a bit off topic but I would also like to know why you can use the word "Lagrangian multiplier". Surely you need some optimization problem to speak about LM and although I see that it kind of make sense to say that I do not see the full picture. When you solve optimization problems with LM you just construct a certain expression and investigate its gradient for sufficient conditions. How do I convert my PDE problem to such form? 

September 11, 2012, 09:37 

#6  
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Filippo Maria Denaro
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Quote:
I suggest to gieve a look to the book of Peric and Ferziger, as well as that of Pope 

September 11, 2012, 09:58 

#7 
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Robert Sawko
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Thanks again. Yes, I have read a substantial part of Pope but I don't know Ferzieger very well. A quick look and I managed to find the answer to the question about LM. If anyone was looking for it's
Section 7.6 page 202. This was exactly what I wanted to know and leaves the matter plain as day. I'll have a more thorough read on RANS section later today. Perhaps I'll find something about isotropic part of RST. 

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