CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > General Forums > Main CFD Forum

Istropic part of Reynolds stress tensor

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   September 7, 2012, 09:42
Default Istropic part of Reynolds stress tensor
  #1
Senior Member
 
Robert Sawko
Join Date: Mar 2009
Posts: 117
Rep Power: 22
AlmostSurelyRob will become famous soon enough
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?
AlmostSurelyRob is offline   Reply With Quote

Old   September 7, 2012, 12:01
Default
  #2
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,764
Rep Power: 71
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by AlmostSurelyRob View Post
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?

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 ...
FMDenaro is offline   Reply With Quote

Old   September 11, 2012, 06:50
Default
  #3
Senior Member
 
Robert Sawko
Join Date: Mar 2009
Posts: 117
Rep Power: 22
AlmostSurelyRob will become famous soon enough
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.
AlmostSurelyRob is offline   Reply With Quote

Old   September 11, 2012, 07:01
Default
  #4
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,764
Rep Power: 71
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
for incompressible flows, the "pressure" variable is only a lagrangian multiplier... any scalar function that ensure the divergence-free velocity is your "pressure". Thus, since there is no pressure state equation, your post-processing in terms of the "pressure" is not meaningful...

As example, if you solve the problem (projection method) with an implicit time-marching method (such as Crank-Nicolson), the "pressure" is further modified
FMDenaro is offline   Reply With Quote

Old   September 11, 2012, 09:11
Default
  #5
Senior Member
 
Robert Sawko
Join Date: Mar 2009
Posts: 117
Rep Power: 22
AlmostSurelyRob will become famous soon enough
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 post-processing 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?
AlmostSurelyRob is offline   Reply With Quote

Old   September 11, 2012, 09:37
Default
  #6
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,764
Rep Power: 71
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by AlmostSurelyRob View Post
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 post-processing 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?

I suggest to gieve a look to the book of Peric and Ferziger, as well as that of Pope
FMDenaro is offline   Reply With Quote

Old   September 11, 2012, 09:58
Default
  #7
Senior Member
 
Robert Sawko
Join Date: Mar 2009
Posts: 117
Rep Power: 22
AlmostSurelyRob will become famous soon enough
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.
AlmostSurelyRob is offline   Reply With Quote

Reply

Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
post-processing utility R, calculating Reynolds stress tensor volker OpenFOAM Post-Processing 17 May 21, 2020 17:36
Reynolds Stress Tensor Sergio13 OpenFOAM Post-Processing 12 May 11, 2020 05:40
Convergence of Reynolds Stress Tensor / Fluctuations in OpenFOAM Anne Lincke OpenFOAM Running, Solving & CFD 0 July 17, 2012 10:50
Reynolds Stress Models Jade M Main CFD Forum 0 April 21, 2010 16:38
Deviatoric Part of the reynolds tress tensor make OpenFOAM Running, Solving & CFD 2 September 22, 2008 14:02


All times are GMT -4. The time now is 16:13.