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Old   August 22, 2007, 20:26
Default K-Epsilon for Vortex Shedding
  #1
Sham
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Hi,

I was told that RKE is a better model to predict vortex shedding than SKE. I did try with RKE and I did get vortex shedding however my Cd value is not right. When I run my model with SKE, my Cd is right but I have not seen vortex shedding. My Re=10000 and I should be getting Cd=1.2 whereas I got 0.55 with RKE. I use EWT and y+~2. I hope someone can give me suggestion how to tackle the problem. I am more incline to use RKE since it gave me good vortex shedding however my Cd is wrong.

Thanks.
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Old   August 23, 2007, 01:22
Default Re: K-Epsilon for Vortex Shedding
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Razvan
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I see that my help on your previous post was not noted/appreciated. And you continue on struggling with the k-epsilon models and the damn EWT. Let me explain in detail why you should not use them:

- all GUI-acessible k-epsilon models are HIGH-Re models, which means that they are not valid all the way to the wall, that's way they need wall-functions to work;

- near-wall accuracy of these models is completely dependant on the wall-function (in your case, because y+ is around 1, only EWT is applicable), which by the way, is not at all recommendable for this particular flow;

- the vortex shedding phenomenon is totally dictated by the shape of the bluff-body and by the boundary layer on it; that means you have to use a B-L accurate turbulence model (all high-Re models are NOT applicable);

- your Re=10,000 is a very difficult one, because at this value, you will have a late transition of the B-L from laminar to turbulent, so you will need a turbulence model that can approximate the transition with some accuracy (all k-epsilon models like ske, RNG-ke or Rke are simply useless, no transition, fully turbulent models).

You see that none of the GUI-accessible k-epsilon models is fit for the job. So you must take into consideration other models like:

- skw or SST-kw ("transitional flow" option activated) which are naturally low-Re models (valid all the way to the wall); SST-kw is much more recommendable then skw;

- low-Re omega-based RSM model, which should give you, at least in theory, better results then SST-kw, due to high flow streamline curvature and strong adverse pressure gradient present;

- LES, which is the best; actually, ALL serious vortex shedding calculations are done using LES (just have a search quick on Google...); LES is the only turbulence model that fulfills all the conditions I have presented.

I hope that this time you will take my comments into consideration, but after all, who should you believe: your professor or a noboby from the CFD-Online forum?

It's your choise,

Razvan
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Old   August 23, 2007, 02:33
Default Re: K-Epsilon for Vortex Shedding
  #3
Sham
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Razvan,

I did not ignore your comments and I am tring what you suggested in the previous post and still waiting for the outcome.

I am just a bit confuse because Fluent User Portal says that RKE is a good tool in predicting vortex shedding and they claim they have validated the results but I found it quite hard to get a a good Cd value for RKE.

So you are saying that kw and RSM are better in modelling turbulence. And you mentioned that k-e is only for HIGH Re and no good to predict vortex shedding. Could you explain on that a bit more. Isn't my Re=10000 is high Re for CFD. I am planning to investigate a certain range of Re.

Thanks and I REALLY appreciate your opinion.

Sham.
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Old   August 23, 2007, 02:45
Default Re: K-Epsilon for Vortex Shedding
  #4
Razvan
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Hi Sham,

high-Re model = not valid near wall (needs wall-functions)

low-Re model = valid near wall (no need for wall function)

Low-Re turbulence models are more accurate for B-L dominated flows (like flow over cylinders, airfoils, etc.).

A Re=10,000 is not high, actually it is quite low, for external flows Re>20,000 is considered turbulent!

The correct prediction of the B-L flow is the key, because Cd&Cl are highly influenced by the B-L separation points on cylinder. So you have to use a model that can calculate separation correctly.

Razvan
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Old   August 23, 2007, 03:02
Default Re: K-Epsilon for Vortex Shedding
  #5
Sham
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Thanks Razvan, I appreciate your help. I shall give it a try. If you are in Australia, probably we should go out for a drink
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Old   August 23, 2007, 03:33
Default Re: K-Epsilon for Vortex Shedding
  #6
Razvan
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Thank you for the invitation, I would love to! Unfortunately I'm too far away wright now (about 10,000 km!!!).

All the best,

Razvan
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Old   August 23, 2007, 21:32
Default Re: K-Epsilon for Vortex Shedding
  #7
Sham
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Razvan,

I gave it a try for the models that you suggested and here are the results:

SST-KW : with transitional flow activated, the results dis not converge and the residuals values fluctuate about the same point and id did not decrease.

SKW : The Cl value oscillate about 0 and drag oscillate about 1. However, residuals are flat and not decreasing.

Low Re RSM: Not converged at all.

Low Re KE : Converged but Cd value is 0.6.

The Cd value should be about 1.2 according to exp. value and my Y+ value is about 2.5 which is way below 30.

It looks that the models other than ke seem to be quite hard to achieve convergence. Only SKE gives me 1.2 drag value. I use PISO and 2nd order discretisation and standard pressure.

I hope you can help me here as I am still investigating as to the appropriate model flow over cylinder to analyse Cd, Cl and vortex shedding.

How can I incrtease Y+ to 30-40. I can make the mesh very coarse but just afraid on not resolving the vorticities correctly. I use structured grid. My domain is O-type grid with half of circle is vel inlet and teh other half is outflow with 25xOD of domain size.

Thanks for your advise in advance.

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Old   August 24, 2007, 01:33
Default Re: K-Epsilon for Vortex Shedding
  #8
Razvan
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Now I see that you have skipped one of the most important remarks I made: the unsteady solver.

This type of problem is one of the best examples of unsteady flow. If you try to simulate vortex shedding using the steady solver, you will certainly obtain the wrong results! The steady solver is just trying to "calm down" the flow, to force the unsteadiness to become steady. By doing so, it heavily alters the solution.

BUT, if the turbulence model used is not sufficiently diffusive, the steady solver will not be able to converge to a steady solution, not in a milion iterations! You correctly observed that only ske allowed a converged steady solution. That is because ske is the most diffusive turbulence model in FLUENT! SST-kw and low-Re omega-based RSM are much more accurate and much less difusive then ske (they "capture" more physics from the flow), so the steady solver is unable to damp the unsteady behaviour that they reveal in the flow field. This is actually one way to verify how diffusive a turbulence model is!

Coarsening the mesh will certainly damp some of the unsteady behaviour of these models, because less cells mean less physics (only large vorticies will be captured). Also, the grid has a major importance in the accuracy and economy of your simulation, so I would suggest a different type of grid for this problem: a combined O-H fully structured grid (with less cells in front of the cylinder and more cells behind it).

For pressure discretisation use PRESTO, QUICK for momentum, Second Order for turbulence.

So, switch to the unsteady solver. You will find detailed instructions in the FLUENT's doc.

All the best,

Razvan
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Old   August 24, 2007, 02:06
Default Re: K-Epsilon for Vortex Shedding
  #9
Sham
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Razvan,

I am not sure if I am botheirng you now with all these questions. However, I still hope you can answer very quick questions below?

1. What do you actually mean O-H fully structured grid? Will thsi help in my y+ values.

2. How to get the right time step size and max iterations per time step? Fluent says that the solution should converge in 5-10 time steps and I dont actually undersatnd what it means.

3. How to correctly monitor the residuals plot? How should it behave i.e. the corect manner?

I really appreciate you help and it makes more sense now after reading some of your posts in this forum for other people.

Sham.
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Old   August 24, 2007, 02:34
Default Re: K-Epsilon for Vortex Shedding
  #10
Razvan
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Hi Sham,

Here are your answers:

1. O-H grid = "O" (circular grid) around the cylinder, up to a certain distance, let's say one diameter, combined with a H (normal parallel strucured grid) in the rest of the domain, extended let's say 10 diam in front, up and down from the cylinder, and 20-25 diam behind the cylinder. If you want, I can send you such a grid to better understand what I'm talking about.

2. The right time step size should be the one that ensures a CFL=1 at most; the easiest way to accurately determine it, is by trial-and-error: just give a certain start value for the time step, then iterate for 2-3 time steps (with maybe 5 iterations/time-step), Display/Contours/Velocity/Cell Courant number, verify that the max. value is around 1, if not decrease the time step proportionally to obtain a CFL value near 1. For the time step value that gives a CFL near 1, you should not make more then 5 iterations/time-step. "Fluent says that the solution should converge in 5-10 time steps", it's not 5-10 time steps, but 5-10 iterations per time step!

3. The "correct" behaviour" of the residual plots is dropping for at least 3 orders of magnitude every time step. At the beggining of the next time step, the residuals will jump approx. to the same value from the beggining of the previous time step, and drop again, and so on.

Razvan
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Old   August 24, 2007, 02:58
Default Re: K-Epsilon for Vortex Shedding
  #11
Sham
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Razvan,

It is most appreciated if you can send me the grid to my e mail celakadotcom@yahoo.com.

I am still waiting your explanataion on my y+ values being way below 30. How can I deal with that?

Sham.
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Old   August 24, 2007, 04:04
Default Re: K-Epsilon for Vortex Shedding
  #12
Razvan
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There is nothing to deal with. The standard and non-equilibrium wall-functions are appropriate if y+ is between 30 and 300, and the EWT is valid if y+ is around 1. All you have to do is to try to avoid the values in between, especially values around 11. Your y+ is 2.5, that is still good enough for EWT, if you use one of the high-Re k-epsilon models. If you use a low-Re model, you do not need any wall functions at all.

I sent you an e-mail with a sample case of a 2d cylinder using a O-H grid, in laminar flow regime. You will be able to see the settings used and also the results, the vortex shedding especially.

Razvan
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Old   August 25, 2007, 17:13
Default K-Epsilon for Nozzle flow saparation
  #13
Eilon
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Hi Razvan, From your posts I see that you are very familiar with turbulance modeling in Fluent, I can asure you that none of my profesors even come close. I want to hear your recomandation for a good model for predicting nozzle flow saparation at Mach numbers above 3. The length reynolds number is high (10^6), the flow is with shock waves and the BL is important in determening the saparation location. ske gave saparation results but I am not sure they are correct. Rke gave no saperation at all (steady simulation). Any thoughts?

Eilon
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Old   August 26, 2007, 00:17
Default Re: K-Epsilon for Vortex Shedding
  #14
Sham
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Razvan,

From FLUENT Manual:

"For time-periodic calculations, you should choose the time step based on the time scale of the periodicity. For a rotor/stator model, for example, you might want 20 time steps between each blade passing. For vortex shedding, you might want 20 steps per period.

To determine a proper choice of $\Delta t$, you can plot contours of the Courant number within the domain. To do so, select Velocity... and Cell Courant Number from the Contours of drop-down lists in the Contours panel. For a stable, efficient calculation, the Courant number should not exceed a value of 20-40 in most sensitive transient regions of the domain."

Could you explain in relation to this when you say that CFL has to be about 1 to get the right time step. It is mentioned here the CFL between 20-40 and for vortex shedding we need 20 steps per period?

I just need to get my understanding right.

Thanks once again for your kind help.

Sham.
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Old   August 26, 2007, 03:57
Default Re: K-Epsilon for Vortex Shedding
  #15
Razvan
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Hi Sham,

Let's look at it this way:

- in an system in pefect equilibrium nothing happens, there is no change in time (the numerical solution of such a system is steady and "fully converged", if I am allowed to say so...);

- in a system that is not in equilibrium, solution changes with time, the physical perturbations travel inside from one boundary to another with a certain speed. Numerically, we can control the speed of the solution advancement in time, using the Courant number. Thus, a CFL=1 means that the perturbations are allowed to travel one cell lenght during a time step.

Stability analysis theory determines that the stability condition for a numerical system is CFL<1. That is true especially for explicit methods. For implicit methods, stability analysis indicates that there is no stability limit. In practice, there is a stability limit for implicit methods, due to the mathematical nonlinearities (usually no more than CFL=20-100). Also, the explicit solvers of FLUENT can run at a CFL>1 (up to 2.5), but that only because of numerical "tricks" implemented.

When using the steady solvers, we are interested in a very short solution time, so forcing convergence using a CFL>1 is perfectly justified, as we are not interested in the intermediar solutions, but only in the final solution (which theoretically is not affected by the forced convergence process).

But, when we are using the unsteady solvers, we might be interested in the time-accuracy of the solution (although we can use the unsteady solvers as an alternative when the steady solvers are not numerically stable). When using the implicit unsteady solvers, we can force a CFL>1 and the calculation could be stable, but the time-accuracy of the solution will be compromised, and large numerical errors will be propagated in the flow field every time step (proportional to the CFL increase). This happens because a perturbation wave travelling more than one cell lenght within one time step is not physical.

The deterioration rate of the solution accuracy is not linearily proportional to the CFL increase (because usually CFL is not constant in all flow domain, due to different cell sizes and velocities values), so if using a max. CFL=5, in your case there should be no too large error in the solution.

But you have to consider this too: a time step that ensures a CFL<=1 will need 5 or less iterations for sufficient convergence, but a larger time step will always need more iterations to reach the same "level of convergence". For example, at a CFL=10 you could need 20 iterations or more. But you must understand that no matter how many iterations you make for such a time step, you will NEVER reach the same solution accuracy as for a CFL=1 time step.

In conclusion, the time step size must be carefully chosen based on the problem to be solved. If there is only a small portion of the flow domain where high velocities are occuring in small sized cells, you could increase the max. CFL number upto let's say 5-10, without experiencing unacceptable overall solution deterioration. But, if the problem to be solved is simpler, with a more uniform flow field and constant cell size, a CFL>1 will heavily affect solution accuracy.

All the best,

Razvan
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Old   August 26, 2007, 04:27
Default Re: K-Epsilon for Nozzle flow saparation
  #16
Razvan
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Hi Elion,

One short comment on the professors: no matter where I went in this world, I've always seen the same problem, the majority of the professors are good theoreticians but extremely lousy CFD users. There are good users among the professors too, I have seen that also. But the rest of them are simply too proud to admit that many times a poor student can use CFD better. I do not want to say that you don't need theoretical knowledge to make CFD, on the contrary, but sometimes a very good instinctive understanding of the involved phenomena and a good "hand&eye" combination, are more important than a handful of equations on a piece of paper. No offence intended to anyone at all!

From what you have written, I understand that you want to simulate an over-expanded supersonic jet (the outlet static pressure of the nozzle is smaller than the external pressure, which pushes the shockwave system inside the nozzle). Is that correct?

This type of problem isn't completely dependant on the turbulence model, but there are other factors to consider before that:

- the flow domain considered and the mesh (extremely important);

- boundary conditions (usually a pressure-inlet + pressure-outlet combination is the most stable)

- numerical solver (usually a density-based coupled solver must be used);

- finally, the turbulence model coupled with the wall-treatment - which links back to the mesh step (Realizable k-e with non-equilibrium wall functions for y+>30, or SST-kw for y+=1, should do a good job).

If you could give us more details, you should receive a better answer.

All the best,

Razvan
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Old   August 26, 2007, 05:42
Default Re: K-Epsilon for Vortex Shedding
  #17
Sham
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Hi Razvan,

I am impressed by how prompt your response is and I appreciate it a lot. Right now I am trying with SST k-w and it looks like it is giving me some promising results even though the Cd is a bit high, probably due to the 2D effect. I will also try with Low Re RSM and Low Re K-Epsilon model to see the differences between these models.

I just need a quick clarification from you here;

1. From FLUENT Manual:

"Both k-omega models available in FLUENT are available as low-Reynolds-number models as well as high-Reynolds-number models. If the Transitional Flows option is enabled in the Viscous Model panel, low-Reynolds-number variants will be used, and, in that case, mesh guidelines should be the same as for the enhanced wall treatment. However, if this option is not active, then the mesh guidelines should be the same as for the wall functions."

If this is the case, when using SST-kw with transitional flow option and my Y+ value is 2.5, that means I am getting the right results? What about Low Re RSM and Low Re K-Epsilon? Are the Y+ approach the same?

2. What is your opinion on V2F model? Is it suitable for predicting vortex shedding for this kind of problem?

3. For the RSM model, is it appropriate for 2D because I read somewhere that it predicts 3D behavior, even though it is available in 2D solver?

Sorry Razvan if I start to ask so many questions. It seems that you know what you are talking about and can make things clear for us all here.

I have a long way to go in this simulation. I am investigating VIV problem. I will need to model LES in 3D and I do not know how hard its gonna be. Right now, I start with RANS to get the feeling of turbulence modeling. If you have experience in modelling VIV and using LES, I hope you do not mind guiding me and you will be greatly acknowledged.

Thanks once again.

Sham.
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Old   August 26, 2007, 08:39
Default Re: K-Epsilon for Vortex Shedding
  #18
Razvan
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Hi Sham,

1. skw or SST-kw (no "transitional flow" option activated) = y+>30

skw or SST-kw + "transitional flow" = y+ near 1 (2.5 is still OK)

Low-Re omega-based RSM is similar.

All low-Re k-epsilon models are designed for y+ near 1.

2. v2f is an excellent turbulence model, good for vortical flows also, but it needs y+=1 meshes. I also needs separate licensing, which could be a problem...

3. Practically all engineering flows are 3D, turbulence itself is a 3D phenomenon. The 2D assumptions work only for RANS models and only for a very small number of problems. Vortex shedding is NOT a 2D phenomenon, so the 2D calculations cannot predict it very accurately.

RSM is essentially a 3D turbulence model. It does not work very well in 2D, and if you remember, I recommended it to you for 3D calculations!

What you are doing is correct, starting from RANS to "get the feeling" about turbulence, and then going to LES. But only from a certain POV. There are several BIG differences between RANS and LES:

- LES meshing is extremely difficult, it requires a LOT of skill and time (this does not mean that RANS meshing is easy, but there are several restrictive conditions to obey with LES);

- LES is always 3D and unsteady, wich means very high computational resources are needed (grids are always bigger, required time steps are very small);

- instantaneous LES results are useless most of the times, so you need to perform time-averaging, which also translates into very long computational time.

Yes, it is going to be really hard. But do not give up, it's very rewarding and full of satisfaction once mastered! And remember: LES is the future!

Razvan
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Old   August 26, 2007, 15:45
Default Re: K-Epsilon for Nozzle flow saparation
  #19
Eilon
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Hi Razvan, I agree with your impressions though I will admit that my advisor can look at a contour plot for one minute and spot some flow feature that I haven't seen after looking at it for a week. My computational domain consists of a contoured nozzle (2D) with a rectangular area spanning 7.5*2.5 nozzle-exit heights at the outlet of the nozzle.

I use quad grid with about 30 cells across the nozzle throat and exit, in all about 30,000 cells. As you guest the BC are pressure inlet and pressure outlet. I then do adaptive mesh refinement based on density gradients limiting the cells to 100,000.

I haven't paid much attention to the meshing at the wall of the nozzle. Along most of the wall it's been refined two times. At the lip of the nozzle where the shock-wave starts its been refined 5 times. The wall Y+ is well above 30. I used density-based coupled solver with Realizable k-e and standard wall functions. How much should I refine the mesh at the wall? What do Non equilibrium wall functions mean physically?

Eilon
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Old   August 28, 2007, 08:57
Default Re: K-Epsilon for Nozzle flow saparation
  #20
Razvan
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Hi Eilon,

My comment on professors was quite mean, I admit. I had some very painful experiences with professors suffering from what I like to call "the superiority sindrom". But I am glad that you are dealing with someone that helps you. I only hope he also teaches you something, not just points a flow feature in a contour plot within a minute... That is just a matter of exercise, and will come for you too, in time.

Your flow domain is a bit small I think. Even if you are calculating a separated, shock-dominated flow which means that the resulting jet velocities are not extremely high, the influence of the outlet boundary might be too big. So I would recommend you to construct a bigger external flow domain (something like 30(L)x15(H) nozzle diam.). Also, you should pay attention to the mesh especially in the divergent part of the nozzle.

Here is an example:

http://img213.imageshack.us/img213/2...dnozzleui9.png

As you can see, skewness is very low and all cells have near-square shapes.

The most recommendable turbulence model is SST-kw (without "transitional flow" option for y+>30). Even if the y+>>30, do not adapt the near-wall mesh using the y+ adaption criterion. Calculate a fully converged solution and then adapt using normalized pressure gradient and reconverge.

All the best,

Razvan

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