Switching on LES in 2D mode of fluent
 

Can anybody tell me, how to switch on (activate, as desribed in tutorials) the LES in 2D mode of Fluent 6.0? In the tutorials for the new accoustic module it just sais: 1. Activate the LES turbulence model for 2D calculations (rpsetvar 'les-2d? #t)

2. Select the Large Eddy Simulation turbulence model.

Define -> Models -> Viscous...........

Thank you very much

Re: Switching on LES in 2D mode of fluent
 

Just write (rpsetvar 'les-2d? #t), as it say, in your text interface(Plus "return").

Regards

Re: Switching on LES in 2D mode of fluent
 

I am sorry, I am just a new user of Fluent - I have already tried to write this command in the command line in Fluent before I posted my question to this forum, but I always get message saying "invalid command [(rpsetvar]". Maybe I am doing something wrong - I am just getting used to using Fluent now.

Re: Switching on LES in 2D mode of fluent
 

Perfect, I have already managed to go through this. I could not believe it.If you copy this command from the help and paste it in the command line, it does not work, but if you type it - EXACTLY THE SAME - and press enter again, it works. I do not know, what difference it makes, whether you copy the text or type it yourself, but the copy method simply does not work

Thanks :)

hello
I think you have just to type the following message (with parenthesis) in the console window of fluent and it will work:
>(rpsetvar `les-2d? #t)
there is just a trick in the apostrophe : type AltGr + 7 and then to see the apostrophe type the next letter "l".good luck

2d-LES starting
 

I am using the command (rpsetvar 'les-2d? #t)
the following message comes in dialog box
Error: rp-var-value-set!: undefined variable
Error Object: 2d-les?

can anybody help. does my system should have visual studio then only it will work

Regards

I wrote this command (rpsetvar 'les-2d? #t)
and LES is activated. But only problem is, if you run LES, you will never get unsteady statistics (mean and rms) in 2D case. So, it will not helpful and it's better to use LES in 3D.

Les
 

Thanks Gautam. Does using axis symmetric condition will be same as 3d

I used 2D axi-symmetric case. I found that unsteady RANS models perform better than LES. LES tooks long time to converge and I will suggest you to use 3D geometry when using LES. It will give you better accuracy.

LES in 2D is highly non-physical. Same goes for 3D LES with any type of symmetry conditions. Using LES in either of these circumstances should be avoided. If you choose to do it anyway, then be ready for wildly different results.

On the other hand, 2D / 3D RANS should yield same results.

As LuckyTran stated before, because turbulence is a 3D physical phenomenon (e.g. vortex stretching is impossible in two dimensions), you are not going to get sensible results from a 2D (or axisymmetric) LES calculation

Hello
I m trying to start LES in 2D Fluent-15(Double precision)using above command.But it is giving error.please help me.

Hi. I think you have to retype: define/models/viscous> (rpsetvar `les-2d? #t) rather than copying it.
NB: note in (rpsetvar `les-2d? #t) (`) is not an apostrophe, it is an inverted apostrophe (AltGr 7).good luck.

2D-LES in Solid Rocket Motor
 

Hello Everyone

i am new user of Fluent-LES. I am trying to simulate vortex shedding in the Solid Rocket Motor using 2D-LES.Can anybody tell me,how to select time step and is it necessary to do steady state simulation before using LES?Also there is drop in pressure across the chamber while using LES and not getting any vortex shedding over there.can any body tell me the reason behind it...

Thank you in advance............

Also please suggest me,For how many iterations do i need to run the above LES simulation as the residuals are going to get steady and oscillate about some average value in LES and how can i do Frequency analysis of vortex shedding using Fluent or is there any other way.Please help...

Hi, its just a suggestion:
according to fluent user's guide, LES is sensitive to initial conditions so fluent propose to start a k-Epsln or k-Omga turbulence model as a steady-state until convergence then switch to LES model (which is an unsteady model).

about time step the Courant-Friedrichs-Lewy (CFL) does not exceed 1 (u can make a study of CFL effect ex: CFL=1,2,3,... until no change )CFL=(v*dt)/dx
then : dt=(dx*CFL)/v
where:
dx:is the smallest distance in the studied domain;
v: reference velocity
u can start with CFL=1 or less and then if there is no difference increase the value of CFL.
good luck

farouksdz

Thank You Very much...

Hello All...I am using LES in 2d fluent to simulate turbulence inside Solid Rocket Motor. My pressure is varying as i am solving it for particular pressure boundary condition and it should not vary that much..so please tell me, am i going through right direction? or should i rethink about all the procedure...please tell me as i am new to LES.

Thank you...

Hello
Can anyone tell me that how to find exact mesh size for LES and how to do mesh independence study for it? As I am working on Solid Rocket Motor flow dynamics using LES....

Thank You...

You need at least 2 points across the structure you are trying to resolve (preferably 4 or more). Mesh independence is same as for any numerical computation. Look-up grid convergence index by Roache if you are totally lost.

There is no such a thing as mesh independence for a LES performed with an implicit filter (which is the most common approach when it comes to LES), since the size of the filter is directly related to the mesh size the more you refine your grid the more you tend towards DNS (for a well posed LES), so in this case is more correct to talk about grid convergence rather than grid independence.

It is possible to obtain a mesh independent LES by switching to explicit filtering, and thus eliminating the direct relationship between filtering and discretization, as explained here: https://web.stanford.edu/group/ctr/S...0/3_04_you.pdf

hi, according to fluent a mesh corresponds to y+=1 (or less) is sufficient for LES turbulence model. to see the independance mesh simulation you could make several meshes correspond to different y+ values (ex: y+=1; 0.8; 0.6; 0.4; 0.2) and you could find the optimal mesh where there is no difference between two successive y+ values. good luck

Thank You Very Much all of you for reply.....

To complete my previous postI would suggest you to have a look at Celik et al, Index of resolution quality for Large Eddy Simulations, Journal of Fluids Engineering 127 (2005) 949-958

For wall bounded flows (and wall-resolved LES) the y+ requirement near the walls is the most limiting factor and satisfying the y+ and some maximum cell aspect ratio (15,20,25 etc) usually dictates the largest cell size in the domain.

But if you wanted to, you could estimate the "turbulent length scale" of interest and use that as your base cell size, whether the length scale of interest is the geometric scale, integral scale, Taylor scale, or Kolmogorov scale is up to you.

I also want to correct myself.

Yes, implicitly filtered LES is by definition grid dependent. Grid convergence is much better way of talking about mesh dependency rather than grid independence. Technically your solution field is always different with any different grid.

But depending on what variables it is you are trying to resolve, those can become mesh independent or mesh insensitive (but I still prefer to refer to this as grid convergence rather than grid independent). While detailed local variables such as (velocity, temperature) may be grid sensitive, you can still achieve grid insensitive results for bulk quantities like overall pressure drop, pumping power, overall heat flux, etc.

Thank You very much.
one more query is there. How to determine time step size for LES(is it using CFL criteria) and is this CFL criteria is important for implicit LES formulation?

I think you meant implicit time-stepping and not implicit LES (implicit LES means LES with an implicit filter).

The CFL criteria is for determining stability which you generally do not have trouble with in implicit time-stepping schemes. But the Courant number does tell you a lot about temporal resolution. Stability also does not guarantee accuracy and you still need small enough time-steps to ensure accuracy. Hence, you should still target sufficiently small time-step (Courant number approximately 0-1) to ensure your results are accurate and ensure you are not temporally clipping the results.