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Favre average and mean(rho/mean(rho))=1

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Old   January 20, 2012, 09:32
Default Favre average and mean(rho/mean(rho))=1
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Hi,

I'm rather new to the CFD field so I have a rather basic and probably really simple question:
How can I show that the Favre average of the Favre fluctuation is zero? While I have found some help, all seem to skip one issue I have. People show \overline{\rho \psi''} = 0 where the line is Reynolds average and the '' is the Favre fluctuation. But how is this done? While I can do that in principle, one little detail is always skipped. Everything reduces to \overline{\rho/\overline{\rho}}=1. While this looks convincing, I have no idea how to show it. It looks similar to one of the Reynolds assumption \overline{\overline{\phi}\psi}=\overline{\phi}\overline{\psi} but it's not the same as far as I can see.

Any idea?

Cheers
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Old   January 20, 2012, 10:43
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By using property \overline{\rho}=\overline{\overline{\rho}}
then

\overline{\rho/\overline{\rho}}=\overline{\rho}/\overline{\overline{\rho}}=\overline{\rho}/\overline{\rho}=1

This is how I understand it, but maybe someone will come up with better solution
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Old   January 20, 2012, 18:12
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Quote:
Originally Posted by Sixkillers View Post
By using property \overline{\rho}=\overline{\overline{\rho}}
then

\overline{\rho/\overline{\rho}}=\overline{\rho}/\overline{\overline{\rho}}=\overline{\rho}/\overline{\rho}=1

This is how I understand it, but maybe someone will come up with better solution
Why is \overline{\rho/\overline{\rho}}=\overline{\rho}/\overline{\overline{\rho}} (I ask because of the division)? Is this connected to some Reynolds assumptions? I only know \overline{\overline{\phi}\psi}=\overline{\phi}\overline{\psi}. Maybe this can easily be modified to a division but I don't see it (doing actual simulations (and then really coarse mesoscale) for too long... ). As far as I can see \overline{1/\rho} \neq 1/\overline{\rho}.

Thanks a lot!
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Old   January 22, 2012, 16:27
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From my point of view property \overline{\overline{\phi}\psi}=\overline{\phi}\overline{\psi} is like \overline{\overline{\phi}/\psi}=\overline{\phi}/\overline{\psi} However I don't have any deeper explanation of it. So sorry I didn't satisfied your question
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Old   January 23, 2012, 06:11
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Ok, thanks, I guess I'll just accept it that way, as you said it's somehow similar! If someone asks, I'll just refer to Sixkillers
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