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Smagorinsky model details

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Old   June 28, 2009, 08:55
Default Smagorinsky model details
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Sorry, I do not understand, I saw in "Smagorinsky.H",
Code:
tmp<volScalarField> k(const tmp<volTensorField>& gradU) const
{
    return (2.0*ck_/ce_)*sqr(delta())*magSqr(dev(symm(gradU)));
}
As I remember:

\begin{array}{l}
 {\nu _{SGS}} = {\left( {{C_S}\Delta } \right)^2}\left| {{\bf{\bar S}}} \right| \\ 
 K = {\left( {{C_I}\Delta } \right)^2}{\left| {{\bf{\bar S}}} \right|^2} \\ 
 \left| {{\bf{\bar S}}} \right| = {\left( {{\bf{\bar S}}{\rm{:}}{\bf{\bar S}}} \right)^{{1 \mathord{\left/
 {\vphantom {1 2}} \right.
 \kern-\nulldelimiterspace} 2}}} \\ 
 \end{array}


Question 1: Why using magSqr(dev(symm(gradU))) instead of symm(gradU) && symm(gradU) to get {{\bf{\bar S}}{\rm{:}}{\bf{\bar S}}} ????

Question 2: If magSqr(dev(symm(gradU))) = symm(gradU) && symm(gradU) = {{\bf{\bar S}}{\rm{:}}{\bf{\bar S}}}, then

K = \frac{{2{C_K}}}{{{C_\varepsilon }}}{\Delta ^2}{\left| {{\bf{\bar S}}} \right|^2}

But I saw in "Smagorinsky.C"
Code:
nuSgs_ = ck_*delta()*sqrt(k(gradU));
Which means

{\nu _{SGS}} = {C_K}\Delta \sqrt K

Then, replace K with K = \frac{{2{C_K}}}{{{C_\varepsilon }}}{\Delta ^2}{\left| {{\bf{\bar S}}} \right|^2}

{\nu _{SGS}} = {C_K}\Delta \sqrt K  = {C_K}\Delta \sqrt {\frac{{2{C_K}}}{{{C_\varepsilon }}}{\Delta ^2}{{\left| {{\bf{\bar S}}} \right|}^2}}  = {C_K}\sqrt {\frac{{2{C_K}}}{{{C_\varepsilon }}}} {\Delta ^2}\left| {{\bf{\bar S}}} \right|


Compare with {\nu _{SGS}} = {\left( {{C_S}\Delta } \right)^2}\left| {{\bf{\bar S}}} \right|

We'll get

{\left( {{C_S}} \right)^2} = {C_K}\sqrt {\frac{{2{C_K}}}{{{C_\varepsilon }}}}

But I heard somone said {\left( {{C_S}} \right)^2} = {C_K}\sqrt {\frac{{{C_K}}}{{{C_\varepsilon }}}}

So, I'm puzzled, I wonder if it was a mistake, that k should be written as
Code:
tmp<volScalarField> k(const tmp<volTensorField>& gradU) const
{
    return (ck_/ce_)*sqr(delta())*magSqr(dev(symm(gradU)));
}

Thank you
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