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September 1, 2014, 18:12 |
initial value for k and ε
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#1 |
New Member
Anastasios Stampoultzoglou
Join Date: May 2014
Posts: 21
Rep Power: 12 |
Hi all,
I have some questions about the start values of k and ε for internal and boundary field. a)Why do we need start-values for k and ε in a k-ε model? For example if i put 0 in the start values i have an error, why is that happening? b) How can i find the start values for k and ε for my simulation? Is there any equations for that reason or do i have to took the decision by comparing the results with the experiment data? For example i made some tests for k and ε values (0.000001 , 0.0002 and 0.2), and the best results were for the 0.2 value. I wanna know if the correct way to take the start values for k-ε is the way that i took or there is another way? To be more specific, while i was doing my tests i used the same start values both for the internal field and boundary field. P.S. i used interFoam solver and i have a dam break problem, so i don't have start value for velocity. Thank you very much i would appreciate any help. Regards, Tasos. |
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September 2, 2014, 11:31 |
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#2 |
Senior Member
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Have you checked this site: http://www.cfd-online.com/Wiki/Turbu...ary_conditions ?
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September 2, 2014, 11:48 |
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#3 |
New Member
Anastasios Stampoultzoglou
Join Date: May 2014
Posts: 21
Rep Power: 12 |
First of all, thank you for your reply. Yes i saw this yesterday, but the thing is that i have a dam break problem. So i don't have a start value for velocity (inlet).
The boundary conditions that i have are : leftWall, downWall, atmosphere and at the right side outlet. Any ideas? |
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September 2, 2014, 12:48 |
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#4 |
Senior Member
Pablo Higuera
Join Date: Jan 2011
Location: Auckland
Posts: 627
Rep Power: 19 |
Hi Anastasios,
since the case starts from rest I would initialize k and epsilon to very small values (but not zero to avoid zero division error). No velocity means 'zero' turbulent kinetic energy and dissipation rate, and will yield 'zero' turbulent viscosity. If your turbulence is not producing the results you expect there are a number of elements to take a look at: interface compression, mesh, turbulence model constants... Best, Pablo |
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September 2, 2014, 13:17 |
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#5 |
New Member
Anastasios Stampoultzoglou
Join Date: May 2014
Posts: 21
Rep Power: 12 |
Hello Pablo,
Thank you for your answer. As i said before, i had put 3 start values for both k and ε. (k and ε 0.2, 0.0002, 0.000001). One would expect that closer to the experimental results would be the simulation with the start value 0.000001 but this didn't happen. The best results was for the start value 0.2. Thats why i am concerned. What do you think? Best regards Tasos |
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September 6, 2014, 15:20 |
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#6 |
Retired Super Moderator
Bruno Santos
Join Date: Mar 2009
Location: Lisbon, Portugal
Posts: 10,982
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Greetings to all!
@Tasos: I got the PM you sent me and I've given a quick read to this thread. From my experience, the solver usually tries to tell you during the simulation what each turbulence-related field should have for a specific flow situation. It will either crash or have really weird values (e.g. 1e15) if the values are very bad. In addition, the turbulence-related fields are mostly theoretical/empirical models, whose values don't necessarily equate to something physical, since they simply are sort-of of a modelling approximation to how turbulence behaves. In other words, your mileage may vary, depending on your own case. I believe this is explained on the wiki page that jhoepken indicated. And Phicau also gave a good brief answer My take on the initial questions:
But AFAIK, technically only with experimental results can you try and find which values better approximate the simulation to the experimental results. Of course you will also to need to take into account the fact that you need to be careful regarding which turbulence model you're using and if you're using a steady-state solver or transient solver and so on... I guess the easiest way to explain this is: first solve a known and simple experimental case, such as a 2D or 3D cylinder inside a flow, to see the vortices developing behind the cylinder; for example:
Best regards, Bruno
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September 6, 2014, 17:01 |
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#7 |
New Member
Anastasios Stampoultzoglou
Join Date: May 2014
Posts: 21
Rep Power: 12 |
Thank you very much Mr Bruno and all of you, i appreciate your answers. Helped me a lot
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