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TOPIC: wall mesh size in k-epsilon

wall mesh size in k-epsilon 7 years 7 months ago #26112

  • sous
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Hello,

As far as I know, when using K-epsilon (or any turbulence) models near a wall, the standard practice is to fix the first point of the mesh at y+=30 (base of the log layer). This allows to use equilibrium expression of K and epsilon. If the mesh point in closer to the wall, the usual assumptions of the law of the wall can not be used and one should instead use, e.g., the Taylor expansions of fluctuating velocities, etc. Or another solution is to use a very fine mesh about y+<=1 such as to be able to resolve the full boundary layer.

In the JMH's book about the TELEMAC k-epsilon model, it is written that the delta distance in eq. 2.283 (boundary equation for epsilon) is chosen at 1/10 of the local mesh size. To my understanding, this implies that:

- either the first point of the mesh is not at y+=30, but much higher in the boundary layer (indeed at least at y+=300, i.e. nearly outside the log layer) such as delta remains in the turbulent log layer and the hypotheses remain valid,

- or the first point of the mesh is near y+=30, but delta is thus within the viscous layer and the equation for epsilon (eq. 2.283) does not apply. Do this mean that Taylor expansions are used (never mentioned) or that the first point should be at y+=1 in order to fully resolve the boundary layer ?

That is quite unclear for me... Am I missing something ? How do you choose your mesh size at the wall if not using the standard recommendations for turbulent boundary layer ?

Thanks a lot for your help !

Damien
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wall mesh size in k-epsilon 7 years 7 months ago #26116

  • jmhervouet
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Hello,

We assume that the first layer is always in the logarithmic profile, it is used e.g. for the Nikuradse friction law. As for the value 1/10, I have a personal theory that it should be 1/(e**2), where e=2.718281828... Assuming that the first layer is in the logarithmic profile and that the bottom layer is in fact at 1/(e**2) of the first layer height we can then show that with linear functions in finite elements, the integral of the flow in the first layer is correct (equal to the integral of the flow obtained with a logarithmic profile), and that the stress computed with a centered derivative in the first layer is also correct, which is an interesting coincidence. 1/10 was an empirical estimation done without underlying theory and you see that it is not very far from e**2.

With best regards,

Jean-Michel Hervouet
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wall mesh size in k-epsilon 7 years 7 months ago #26121

  • sous
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Hello Jean-Michel,

Thanks a lot for your answer. I think I understand now the basics but, so far, I am not familiar with these numerical treatments. Do you think I can find it (the flow integration and stress computation in FE) written done somewhere in a proceeding or a user manual ?

Thanks again for all your work

Damien
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wall mesh size in k-epsilon 7 years 6 months ago #26246

  • jmhervouet
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Hello,

I'm afraid it has never been published. It was just an exercice for students when I did lectures. I will try to write it in English and post it soon.

Regards,

JMH
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wall mesh size in k-epsilon 7 years 6 months ago #26257

  • sous
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Hello

That would be very useful ! Thanks in advance

Damien
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wall mesh size in k-epsilon 7 years 6 months ago #26270

  • jmhervouet
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Hello,

You can look at the latest post in the "documentation" section of this forum.

With best regards,

Jean-Michel Hervouet
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wall mesh size in k-epsilon 7 years 5 months ago #26680

  • mike96
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Very informative thread.

Thanks
Mike Craig
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