
[Sponsors] 
October 29, 2013, 18:46 
Average and instantaneous variables in turbulent modeling

#1 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Dear friends,
I would appreciate if any CFX expert helps me out. How I can see the instantaneous variables like velocity and vorticity in CFX post? It seems that the variables in CFX post is average value. Thank you, Mehdi 

October 29, 2013, 21:57 

#2 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
By default the instantaneous values are shown. If you are seeing averaged values then you must have activated the transient averaging options.
Or are you talking about Reynolds/Fauve Averaging? That's a totally different question 

October 30, 2013, 02:08 

#3 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Thanks for your reply, lets tell you my perception and please let me know if I am wrong,
U = U(ave) + U(fluctuating) My question is, if in post processing the U(ave) is shown or U by default? There is an option on the CFX pre on the Output Control about the transient statistics, there are some options like arithmetic averaging, minimum …, full. What option should I use in order to have both U(ave) and U(fluctuating) available in CFX post or generally what I should do to have them all in post processing ? And if I forget to set anything in CFX pre, should I start the solution again with new setups ? Thank you Mehdi 

October 30, 2013, 05:58 

#4 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
I presume you asking about Reynolds/Fauve averaging.
The whole point of turbulence models is that U(fluctuating) is not simulatied directly, but its effects on U(ave) are modelled through the Reynolds Stresses or additional dissipation. So a RANS turbulence model will not give you U(fluctuating), instead it will give you k (turbulent kinetic energy) and epsilon or omega; or if you are using a Reynolds Stress model you will get the Reynolds Stresses directly. Alternately, if you have done a LES or DNS model  then you have modelled the turbulent fluctuations and it is up to you to average it to get the average and fluctuating components. But you need to really know what you are doing to do LES and DNS and you would already know all about this basic stuff  so I assume you are not talking about LES or DNS. 

October 30, 2013, 15:43 

#5 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Thanks Glenn,
By your explanation and considering that I am using two eq SST , as it is RANS model then it can not give me the fluctuating parts of variables instead it solves the Reynolds tensor bu using some equations to close the NS equations. Two questions ; Then by considering the RANS equation , the result is the average velocity and pressure ? What is the transient statistics setups(average , max ,... ,full) at CFX pre? Thanks Mehdi 

October 30, 2013, 17:55 

#6 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
I strongly recommend reading a CFD textbook and getting a background on RANS. It is a fundamental technique used in the development of the mathematical model and it has important implications  you really need to understand this to be proficient at CFD.
In CFX for a RANS simulation, the variables are all Reynolds averaged (or Fauve averaged if compressible). The transient statistics then average these Reynolds averaged values. Obviously you have to be careful about your understanding of this double averaging process. 

October 30, 2013, 20:11 

#7 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Dear Glen,
Thank you for your reply. Please let me know if I am wrong in any parts. I have studied some textbook about the RANS models but my knowledge about the LES and DNS is limited. On the RANS , when we substitute the U=Uav+Ufluc in the NS eq. then we have 6 unknown terms known as Reynolds stress tensor . That we use some turbulence modeling to close the NS eq. We can use eddy viscosity models like ke or kw by assuming that the Reynolds tensor members are homogenous or Reynolds stress models that assume heterogeneous members. Then the solution for RANS models should give us the Reynolds average values and we do not have the fluctuating parts. Despite the turbulence modeling that gives us the Reynolds average values, we can have transient simulation and there is an option on the transient statistics for averaging, I guess that on this situation we have the spontaneous variables and average with respect to time in the post processing? Another question is that, when we solve the transient equations we have the results which stand for the spontaneous results, then what does averaging in time mean? Because we progress in time and in every time steps we have new results. Thank you, Mehdi 

October 31, 2013, 05:40 

#8 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
Your description is correct, you understand what is going on with RANS  excellent.
The simplest approach for RANS is to apply it to a steady state simulation. Then you obviously have a Uav which does not change with time and Ufluc which contains all the fluctuations with time. This is easy to comprehend. But for a transient simulation it is a bit tricker when you use RANS. RANS assumes that there is a distinction between the time scale of Uav and Ufluc so it can be separated. So when you do an unsteady RANS simulation you are directly modelling Uav  which implies that Uav is the time scales resolved by your model, and Ufluc is not simulated and it is assumed that this is frequencies faster than your simulation can resolve. This gets tricky when you consider things like mesh resolution and time step resolution, as the time scales you are resolving change, which means the simulation might not converge against mesh or time step size. This is one of the reasons why grid convergence can be difficult in transient flows. It also means that you can have too fine a mesh, and too fine a time step. If you start resolving what physically are turbulent eddiesd that the RANS assumption breaks down. 

November 1, 2013, 22:25 

#9 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Thanks Glen
I have learnt a lot form you and other experts here. Mehdi 

November 6, 2013, 23:25 

#10 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
I can not get the same results and velocity pattern that I have seen on the papers.
On the right at the picture(a) is the instantaneous velocity contour I have found in a paper, (b) is the velocity contour and (c) is the Trnave velocity contour, I have two questions ; 1 is (b) the instantaneous velocity 2 Why cant I derive the same velocity contour pattern like (a) and how I can have the velocity contour like that the physic of the problem and the BC are the same , Thanks Mehdi 

November 6, 2013, 23:46 

#11 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
You are going to have to read the paper to find out exactly what they are showing, I am not sure why they show both velocity and transient averaged velocity.
The random fluctuations which are the turbulent eddies are random. You will not get the same one twice, even when you start with apparently the same conditions. So you will never get a picture which looks the same as a. 

November 7, 2013, 02:48 

#12 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Sorry Glen ,
I forgot to tell you that b and c are my results , but a is from a paper . I have seen in another paper such flow pattern in velocity contour and vorticity , but I could not have the same demonstration and I do not know how to do it. It is a transient solution and I am using SST turbulent modeling and I saw another paper using RANS but I saw different flow pattern on the vorticity like a . But mine are not dispersed. Thanks 

November 7, 2013, 05:15 

#13 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
i would expect a RANS simulation to give you a field like b. To get a field like a you would need LES or one of the related turbulence models. You can also fool RANS into giving a field like a by using a fine mesh and accurate differencing with small time steps.


November 7, 2013, 16:12 

#14 
Member
Join Date: Jul 2013
Posts: 66
Rep Power: 5 
Thanks Glen ,
you are right , RANS compromises some details of the flow pattern . I have another question, Suppose that I am modeling laminar transient flow . When CFX solves the equations , in every time steps I have the instantanous results , then what does the trnave mean here . Thanks Mehdi 

November 7, 2013, 17:24 

#15  
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 12,836
Rep Power: 100 
Quote:
If you are modelling using laminar flow then you are not performing a RANS simulation but are directly modelling all flow features. In this case you can use the trnave functions to give you the average flow field. 

Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Laminar vs Turbulent NavierStokes  truman  Main CFD Forum  6  March 7, 2011 14:44 