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

nabla2 · January 20, 2012, 08:32

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

Sixkillers · January 20, 2012, 09:43

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

nabla2 · January 20, 2012, 17:12

Quote:

Originally Posted by Sixkillers

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!

Sixkillers · January 22, 2012, 15:27

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

nabla2 · January 23, 2012, 05:11

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

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

January 20, 2012, 09:43		#2
Sixkillers Member Join Date: Nov 2011 Location: Czech Republic Posts: 97 Rep Power: 14	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

January 22, 2012, 15:27		#4
Sixkillers Member Join Date: Nov 2011 Location: Czech Republic Posts: 97 Rep Power: 14	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

January 23, 2012, 05:11		#5
nabla2 New Member Join Date: Jan 2012 Posts: 3 Rep Power: 14	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