|July 22, 2013, 12:59||
Join Date: Dec 2011
Posts: 10Rep Power: 6
I am studying Prandtl's secondary flow of the second kind and have stumbled upon something that I can't make sense of.
Many authors state that the normal Re stresses are all equal to with turbulence models relying on the Boussinesq approximation for closure. But when I finally looked at the approximation myself (incompressible):
What assumption or piece of logic allows each of the normal Re stresses to equal ?
I have also seen a statement in a thread that the shear Re stresses are all zero when using Boussinesq for closure. Any explanation on this point as well? Because the Boussinesq approximation equation seems to refute both of these claims.
Any corrections or comments on my understanding are welcome.
|July 22, 2013, 13:49||
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 2,411Rep Power: 31
I am not sure to have fully understood your question ...Are you considering U as Reynolds averaged? Then
(uu)bar= UU + (u'u')bar
you will model (u'u')bar as supplementary eddy viscosity tensor Td but it must be zero traced. Therefore, you decompose in isotropic and deviatoric tensors
Td=(1/3)I *Trace(Td) + Td0
For incompressible flows, the isotropic part is simply summed to the pressure and you solve for a global scalar field whose gradients ensure the divergence-free velocity.
Sorry if I did not get the right meaning of your question ...
|Thread||Thread Starter||Forum||Replies||Last Post|
|difference between isotropic and homogeneous in turbulence||dut_thinker||Main CFD Forum||7||November 8, 2015 00:44|
|help on isotropic turbulence initial condition||Hall||Main CFD Forum||8||April 28, 2011 16:55|
|Help needed on homogeneous isotropic turbulence an||Guoping Xia||Main CFD Forum||0||March 12, 2006 22:54|
|Isotropic Turbulence||gorka||Main CFD Forum||2||June 30, 2003 15:41|
|isotropic homogeneous turbulence||tingguang||Main CFD Forum||0||August 1, 2002 17:08|