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nutWallFunction boundary condition

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Old   October 26, 2009, 11:19
Lightbulb nutWallFunction boundary condition
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Hi all,
I use a turbulence model (RANS) and I need you for one thing:
I know that "nut" is calculated by the formula nut=C*k^2/epsilon where C=0.09.
I've seen over some tutorials:
" type nutWallFunction;
value uniform 0;
"
and my problem is here, why does the value set to 0 ? Indeed, nut is not equal to 0 near the wall if we consider the formula above.
Could somebody explain that to me ?
thank you in advance
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Old   October 27, 2009, 16:37
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Paulo Alexandre Costa Rocha
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Dear AirS,

I think nut goes to zero when the turbulence kinetic energy (k) goes, based on the zero velocity on the wall.

Best Regards,

Paulo
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Old   November 4, 2009, 07:10
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Thanks for answering but I'm sorry to not understand:

the turbulent kinetic energy near the wall is modelled by the kqRWallFunction and its value is set to uniform but it's not equal to zero, it's even equal to the intial condition that I could compute before (cf. tutorials)
So, is that correct if I say that NEAR the wall the turbulent kinetic is not equal to zero on account of the model but that ON the wall it's fixed to zero by the condition "value uniform 0" ?? it doesn't make sense but I don't know what it could be otherwise.

Hope you can help me guys to sort this out.
Thanks!
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Old   January 22, 2010, 10:49
Question nutWallFunction or calculated nut
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Hi FOAMers,
this seems the good thread for posting a question that have been turning round into my mind for a couple of days now.
I am wandering what is the difference in using nutWallFunction or a
{
type calculated;
value uniform 0;
}

condition for nut if k and epsilon are defined as kQRWallFunction and epsilonWallFunction in the same boundaries. The only explanation I see is that:
  • if nutWallFunction is applied, than a specific wall function is used, depending from the y+ value. y+ is taken as an output parameter and can be obtained using yPlusRAS.
  • if a uniform 0 condition is used, then a standard wall function is used, y+ is internally calculated and cannot obtained as output.
Am I right? And what is the phisical meaning of that?

Thanks a lot to those that will shed some light.

Maddalena
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Old   March 2, 2010, 08:56
Thumbs down ??
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Then, what is the conclusion? ...

Who can give light?
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Old   March 3, 2010, 07:28
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The

value uniform 0;

entry is simply a placeholder for that gets filled with the correct value of nut by the code once you run it. To be clear, the initial value of nutw = 0 has no effect on the solution, it is where the result of calculating nutw will get written in the result fields. It is not an input field in the normal sense of the word, but rather an output field that just needs to be there in the beginning because of the way the code is written.
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Old   March 3, 2010, 08:31
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Quote:
Originally Posted by sandy View Post
Then, what is the conclusion? ...

Who can give light?

Hello Sandy,
besides Eugene's explanations, you can find some more information about the question I posted above on this thread: http://www.cfd-online.com/Forums/ope...tml#post243652.
Hope this help.

maddalena
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