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

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 January 20, 2012, 09:32 Favre average and mean(rho/mean(rho))=1 #1 New Member   Join Date: Jan 2012 Posts: 3 Rep Power: 5 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

 January 20, 2012, 10:43 #2 Member   Join Date: Nov 2011 Location: Czech Republic Posts: 95 Rep Power: 5 By using property then This is how I understand it, but maybe someone will come up with better solution

January 20, 2012, 18:12
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 Originally Posted by Sixkillers 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!

 January 22, 2012, 16:27 #4 Member   Join Date: Nov 2011 Location: Czech Republic Posts: 95 Rep Power: 5 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

 January 23, 2012, 06:11 #5 New Member   Join Date: Jan 2012 Posts: 3 Rep Power: 5 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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