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Old   June 27, 2003, 08:24
Default Isotropic Turbulence
  #1
gorka
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Hi,

I'm a bit confused with the isotropic assumption in turbulence modelling. I've read in some references that in isotropic turbulence: (Tau = Reynolds stresses)

Tau(1,1) = Tau(2,2) = Tau(3,3) and Tau(i,j) = 0 for any i different from j.

but I can't understand how this result can be achieved with the Boussinesq Hypothesis:

Tau(i,j)= Mu_t*(diff(Ui)/diff(xj) + diff(Uj)/diff(xi)) -2/3*rho*k*D_Kronecker(i,j)

Moreover, Wilcox refers to isotropy in this sense:

Tau(i,j)/S(i,j) = C(mu_t) (S = Mean Strain Tensor)

that has nothing to do with the former definition!!

Can anyone help me, please?

Thank you very much,

Gorka
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Old   June 30, 2003, 15:39
Default Re: Isotropic Turbulence
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hall
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I think you maybe mess up several stress tensor definition.

In isotropy turbulence, stress tensor = <u_i*u_j>, it is a seconde-order correlation. And the tau(1) = tau(2) = tau(3) is true

In Boussinesq Hypothesis, it is a one-equation model of RANS(Raynolds averaged N-S), the stress tensor is the averaged fluctuating velocity (overbar of (u'u') )

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Old   June 30, 2003, 15:41
Default Re: Isotropic Turbulence
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hall
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In isotropy turbulence, stress tensor = average of (u_i*u_j)

huh, the system eat my equation
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