Favre average and mean(rho/mean(rho))=1 -- CFD Online Discussion Forums

January 20, 2012, 08:32	Favre average and mean(rho/mean(rho))=1	#1
nabla2 New Member Join Date: Jan 2012 Posts: 3 Rep Power: 14	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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