[Khaza]

Hi. I have tried running the transient analysis of my tower model yesterday. I have used the following parameters.

Model : 3D

1.Diameter of tower = 75m
2.height of tower = 350m
3.domain size : ( xmin: -1750m,Xmax: 5250m , ymin:0m ,ymax:2100m,zmin:-1750,zmax: 1750
4.mesh : surface mesh: 0.5, body mesh : 0.85

As far as I know the quality of mesh is very good with 2 million elements.

I have run the steady analysis and got the convergence. Now the problem is I am not getting symmetric shape for lift coefficient in transient analysis. I have patched the region for lift and used the time step size as 0.1

Parameters I have used for Transient analysis:
model : transient
Viscous model : k-epsilon model ( Realizable,scalable wall functions and curvature correction)
Medium : Fluid ( air) , viscosity : 1.6 * 10^-5 approx
RE : 3 *10 ^8 ( I am skeptical about the viscous model ( should I use SST or SAS or LES model since the reynolds number is high and my model is fully turbulent )
boundary conditions: inlet (used wind profile here : peak velocity - 64.47m/s)
solution methods ( simple and second order implicit)
solution controls ( courant number did not change -200 ( I think courant number play a role in transient analysis)
time step : 0.085

I am looking for this kind of solutions for my model as shown in the below link for unsteady cylinder model .

Can anybody help me how to achieve the symmetric shape when the region is patched ?

https://confluence.cornell.edu/pages...geId=144972457

Thanks so much

[FMDenaro]

your question confused me... what do you mean for symmetric lift coefficient??

if your solution is unsteady and you have turbulent flow separation around the bluff body (generating a wide range of scales) it is quite normal you get oscillating lift. Thus, for symmetric you mean that the lift oscillates around a constant average value?

PS: LES is not a model but a different formulation...

[adrin]

Khaza, how did you get a CFL (Courant #) of 200 ?!! This is huge; even if the solution is stable (assuming you're using a fully implicit method; i.e., including convection) your time accuracy would be way off. That is, your time resolution may be too low to get any time-accurate results.

As I had recommended earlier, you should discuss (and think about) these topics in non-dimensional units. Physical units, especially when mixed with non-dimensional parameters (such as Reynolds number, CFL, etc) get to be _very_ confusing. I can tell from the physical units that your Reynolds number is based on the max free stream velocity and the tower diameter, but I don't have the time to guess how you got CFL.

Finally, why do you expect your lift coefficient to be symmetric?

adrin

[adrin]

"LES is not a model but a different formulation"

Unless one's doing DNS, everything is a model. Well, actually, to be accurate, even DNS is a model because the Navier Stokes equations are themselves a model (with a bunch of assumptions that may or may not be valid for all flow regimes)

adrin

[Khaza]

Originally Posted by adrin

"LES is not a model but a different formulation"

Unless one's doing DNS, everything is a model. Well, actually, to be accurate, even DNS is a model because the Navier Stokes equations are themselves a model (with a bunch of assumptions that may or may not be valid for all flow regimes)

adrin

Hi adrin

I didn't calculate courant number . It was default. If I have to calculate it ,it should be (D*V/mesh element size ) I think.

What I am trying to achieve is lift coefficient and drag coefficient and I would like to know how to get the right answers for this coeffecients.

I would like to know 3 things.

1. I have calculated the Reynolds number Re = 3 * 10^8 , from parameters viscosity = 1.6*10^-5, diameter of my tower =15m ,velocity peak = 64.47m/s) Can u tell me whether this calculation is correct???

2. I have used k-epsilon model . Is it right model for the Reynolds of 3*10^8 ? Should I change the model to SAS ? ( I would appreciate if u can give more info or any material which clearly explains this except fluent guide.)

3. Do I need to calculate the courant number to my model ? or just take the default value?

4.Finally, how do I know my unsteady solution is correct? or how do I know whether my lift and drag coefficeints I achieved are correct?

Thanks so much for your input

[FMDenaro]

Originally Posted by adrin

"LES is not a model but a different formulation"

Unless one's doing DNS, everything is a model. Well, actually, to be accurate, even DNS is a model because the Navier Stokes equations are themselves a model (with a bunch of assumptions that may or may not be valid for all flow regimes)

adrin

well, let me explain better my idea....

model is a "physical" model, assuming you are doing some assumption on the physics. In this general sense, NS equations are a model for the continuoum fluid-mechanics, you're right.

Talking about "model in turbulence" as in this post, we restricted ourselves to some functional model that mimics the behaviour of the unresolved scales. In this sense LES is not a model like the RANS/URANS is not a model. They are formulations because they "transform" mathematically the NS equations in different equations, each one using a mathematical operator (filtering for LES, statistical average for RANS).
Therefore, each formulation has its specific model for the unresolved terms.

Often I read people writing about LES, RANS, URANS indistinctly as "turbulence models". That is simply wrong and generates many confusion in interpreting the results for each of these approaches.

[adrin]

Khaza,

I don't use commercial codes to know the details of your work. If the CFL is set to 200 by default you need to find out how that impacts your solution. Is this just an output from your run (which implies your tilmestep size is too large) or is it used to control your tilmestep size based on the local values of velocity and mesh size? I would start by changing it to something like O(1); for example, start with 0.5.

1) I think you mean 75m not 15. Your Re is correct.

2) k-e is generally useless, especially if you have recirculation zones. It's "good enough" for fully turbulent homogenous flow with not much of flow curvature. I doubt you want to do LES because you'll need far more mesh points than you're already using. Just use k-e with whatever correction (fudge factors) the code offers you, as a first step. You can then go to more complex models (i.e., more CPU time) after you figure out everything else.

3) Please see my explanation above about Courant. You need to check the manual and find out how it's used (is it just an output of the run or is it an input to it?)

4) Your last question is very difficult to answer. You build on experience (with the particular problem) one step at a time. Understanding the physics of the flow is often the key to success. Unfortunately, many people these days think of CFD as a toy that you can just push buttons here and there and get results - I think without physical understanding of the problem one's CFD is doomed to failure. Having said that, you need to make sure you have small enough of a timestep to capture the frequency of oscillation, and small enough of a mesh to capture the flow resolution (do a parametric investigation and see how your results change). Also, check the literature for experiments that are sufficiently similar, which you can use for "sanity check"; that is, check that you've set your parameters properly and accurately enough.

adrin

[adrin]

Filippo,

Though this might be outside the interest of this thread I still don't understand how you propose to not think of LES as a model any more than (U)RANS! Of course the starting points for the two routes are different, but in the end there is _physical_ modeling involved in both cases. In the case of statistical models we have tons of global "constants", which are the artifact of how the models are formulated. In the case of LES we often have (a) local constant, which is generally obtained based on some physical modeling assumptions (for example, local homogeneity, etc). Even the dynamic models still have to rely on physical modeling assumptions. I agree the level of rigor in LES is far far greater than (U)RANS, but that doesn't obviate the need for SGS _physical modeling_!

adrin

[FMDenaro]

Originally Posted by adrin

Filippo,

Though this might be outside the interest of this thread I still don't understand how you propose to not think of LES as a model any more than (U)RANS! Of course the starting points for the two routes are different, but in the end there is _physical_ modeling involved in both cases. In the case of statistical models we have tons of global "constants", which are the artifact of how the models are formulated. In the case of LES we often have (a) local constant, which is generally obtained based on some physical modeling assumptions (for example, local homogeneity, etc). Even the dynamic models still have to rely on physical modeling assumptions. I agree the level of rigor in LES is far far greater than (U)RANS, but that doesn't obviate the need for SGS _physical modeling_!

adrin

well, I don't want to confute the requirement of a model to close the equations...
I am just stating that if you take the original NS equations you can chose different mathematical formulations to work with. If you write the filtered NS equations they are exact, if you write the statistically averaged NS equations they are exact. But, this way, you have described different formulations as the solution fields belong to a different class of trasformation. And, depending on the formulation, the physical meaning of each solution is very different, irrespective of the chosen turbulence model.
Is the fact that we have not a closed equation the key which forces us to use " a model" (Actually, even the simple disregarding of any unresolved terms is a model...)
In conclusion, the word "turbulence model" relies on the closure problem, the word "formulation" relies on the kind of field we want to solve.

For a practical aspect, if you run a LES code you should check the solution very differently from the URANS one, owing to the different formulation not for the different model.

[adrin]

>>In conclusion, the word "turbulence model" relies on the closure problem, the word "formulation" relies on the kind of field we want to solve.

I see now what you mean(t), and agree

adrin

[Khaza]

Originally Posted by FMDenaro

your question confused me... what do you mean for symmetric lift coefficient??

if your solution is unsteady and you have turbulent flow separation around the bluff body (generating a wide range of scales) it is quite normal you get oscillating lift. Thus, for symmetric you mean that the lift oscillates around a constant average value?

PS: LES is not a model but a different formulation...

Yes I for symmetric I meant the lift oscillates around a constant average value.