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-   -   Favre average and mean(rho/mean(rho))=1 (http://www.cfd-online.com/Forums/main/96366-favre-average-mean-rho-mean-rho-1-a.html)

 nabla2 January 20, 2012 09:32

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

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 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 . While this looks convincing, I have no idea how to show it. It looks similar to one of the Reynolds assumption but it's not the same as far as I can see.

Any idea?

Cheers

 Sixkillers January 20, 2012 10:43

By using property
then

This is how I understand it, but maybe someone will come up with better solution ;)

 nabla2 January 20, 2012 18:12

Quote:
 Originally Posted by Sixkillers (Post 340314) By using property then This is how I understand it, but maybe someone will come up with better solution ;)
Why is (I ask because of the division)? Is this connected to some Reynolds assumptions? I only know . 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 .

Thanks a lot!

 Sixkillers January 22, 2012 16:27

From my point of view property is like However I don't have any deeper explanation of it. So sorry I didn't satisfied your question :D

 nabla2 January 23, 2012 06:11

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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