Isotropic EV

mcclud · July 22, 2013, 12:59

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 $\frac{2}{3}k$ with turbulence models relying on the Boussinesq approximation for closure. But when I finally looked at the approximation myself (incompressible):
$\overline{u_i u_j} = \frac{2}{3}k \delta_{ij} - \nu_t(\frac{\partial U_i}{\partial x_j} +\frac{\partial U_j}{\partial x_i})$
It gives
$\overline{u^{2}} = \frac{2}{3}k - 2\nu_t\frac{\partial U}{\partial x}$
$\overline{v^{2}} = \frac{2}{3}k - 2\nu_t\frac{\partial V}{\partial y}$
$\overline{w^{2}} = \frac{2}{3}k - 2\nu_t\frac{\partial W}{\partial z}$

What assumption or piece of logic allows each of the normal Re stresses to equal $\frac{2}{3}k$ ?

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.

Thanks,
Darryl

FMDenaro · July 22, 2013, 13:49

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 ...

July 22, 2013, 12:59	Isotropic EV	#1
mcclud New Member Darryl McClure Join Date: Dec 2011 Posts: 10 Rep Power: 14	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 $\frac{2}{3}k$ with turbulence models relying on the Boussinesq approximation for closure. But when I finally looked at the approximation myself (incompressible): $\overline{u_i u_j} = \frac{2}{3}k \delta_{ij} - \nu_t(\frac{\partial U_i}{\partial x_j} +\frac{\partial U_j}{\partial x_i})$ It gives $\overline{u^{2}} = \frac{2}{3}k - 2\nu_t\frac{\partial U}{\partial x}$ $\overline{v^{2}} = \frac{2}{3}k - 2\nu_t\frac{\partial V}{\partial y}$ $\overline{w^{2}} = \frac{2}{3}k - 2\nu_t\frac{\partial W}{\partial z}$ What assumption or piece of logic allows each of the normal Re stresses to equal $\frac{2}{3}k$ ? 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. Thanks, Darryl

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July 22, 2013, 13:49		#2
FMDenaro Senior Member Filippo Maria Denaro Join Date: Jul 2010 Posts: 6,768 Rep Power: 71	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 ...