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Old   June 28, 2009, 08:55
Default Smagorinsky model details
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Daniel WEI (老魏)
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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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Old   August 19, 2009, 12:42
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Hi Daniel,
Did you find any answer to this question?. in the paper:
"A SUBGRID-SCALE MODEL FOR LARGE-EDDY SIMULATION OF
PLANETARY BOUNDARY-LAYER FLOWS
PETER E SULLIVAN, JAMES C. McWILLIAMS, and CHIN-HOH MOENG" 1994, they defined Cs as:

Cs=(Ck*(Ck/Ce)^0.5)^0.5

So, I think you are right...

Now, I'm confused, why it is defined in Smagorinsky.H like 2*Ck/Ce ??

Is it because of the symm(grad(U)) definition???

It would be great if you could share your opinion...

Gaby
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Old   August 20, 2009, 08:35
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Sorry, you see, no one come and help.

And you have noticed, I have done a detailed deduction in my top post, I still don't know why they use
  1. 2.0*ck_/ce_
  2. magSqr(dev(symm(gradU))) instead of symm(gradU) && symm(gradU)
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Old   August 20, 2009, 19:59
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Quote:
Originally Posted by gaby View Post
Hi Daniel,
Did you find any answer to this question?. in the paper:
"A SUBGRID-SCALE MODEL FOR LARGE-EDDY SIMULATION OF
PLANETARY BOUNDARY-LAYER FLOWS
PETER E SULLIVAN, JAMES C. McWILLIAMS, and CHIN-HOH MOENG" 1994, they defined Cs as:
Gaby
Hi Gaby, can you send a copy of the paper to me? I am finding the expression to estimate the k inlet boundary condition in oneEqEddy model. Thanks.

Sandy
sandy.lee37@gmail.com
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Old   October 6, 2010, 05:00
Default tensor norm definition
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Hallo,
I was also trying to understand the implementation of Smagorinsky model.
As it is said here
definition norm of strain rate
the definition of the norm of a tensor differs from what is computed in OF, so in my opinion the 2 before ck_/ce_ is exaclty the missing sqr(2) in the definition.
so, if
|S| = sqrt(2 S:S)
then
2 magSqr(S) = 2 sqrt(S:S)^2 = sqrt(2*S:S)^2 = |S|^2

Am I right?

PS: default Smagorinsky constant should be Cs =0.1677 (in Pope's book is said to be around 0.17) given ck=0.094, ce= 1.048
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Old   March 23, 2011, 16:31
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Hello

I am also trying to understand how the Smagorinsky model is coded, for the incompressible version and also for the compressible version.

And It seems that for the incompressible Smagorinsky model, the default constant Cs is equal to
Cs=sqrt(ck*sqrt(2*ck/ce))

If I define Cs such that the eddy viscosity is equal to
nuSgs=( Cs *delta)^2 * ||D||

if ck=0.094 and ce=1.048 then Cs=0.1995.. ~ 0.2

Same question as Andrea : am I right?
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Old   April 22, 2011, 08:13
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Hi,

I have same questions too.
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Old   April 22, 2011, 23:22
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Reference: http://pof.aip.org/resource/1/phfle6/v9/i5/p1416_s1
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Old   May 1, 2011, 11:53
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Thanks Alberto for the useful reference.
As MaximeIST and lakeat told if we compare what is stated in your reference and openFoam smagorinsky with Pope's book we reach to Cs=0.2.
As stated in Pope's book this constant can be 0.1~0.2 and using 0.2 can be cause of high diffusivity. Please tell me if this is true.
My cyclone simulation with Smagorinsky has high diffusivity and i want use 0.1 for Cs, How can i do that? Which one of Ck or Ce should be changed? I mean are they be used elsewhere or not?

Regards,

Last edited by maysmech; May 1, 2011 at 14:03.
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Old   May 1, 2011, 18:27
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I'd just use the dynamic Smagorinsky model, so that you do not have to fiddle with the coefficient, and you do not need to play with dumping functions.
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Old   May 2, 2011, 21:59
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Hi,Andrea:
I think what you have said is right,the reason is just the different expression of Vsgs
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Old   May 12, 2011, 05:48
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So, what is Cs value in OpenFOAM's Smagorinsky? 0.2 or 0.167?
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Old   May 12, 2011, 06:50
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hi,Maysam:
since Cs=sqrt(ck*sqrt(*ck/ce), in smagorinsky.c Ck=0.094 , in GenEddyVisc.c Ce=1.048, then Cs=0.167 as we all know ,am I right?
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Old   May 12, 2011, 08:30
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I am in doubt because of first post of this thread that it becomes 0.2.
I don't know. Any idea will be appreciated.
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Old   May 12, 2011, 08:33
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It's Cs \approx 0.17. You need just to follow the way it's computed. The question is why the simple Smagorinsky model is implemented in a such confusing way? We re-wrote our Smag. with just Cs required.
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Old   May 12, 2011, 09:12
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Quote:
Originally Posted by yingkun View Post
hi,Maysam:
since Cs=sqrt(ck*sqrt(*ck/ce), in smagorinsky.c Ck=0.094 , in GenEddyVisc.c Ce=1.048, then Cs=0.167 as we all know ,am I right?
Hello

I may keep on confusing people, but the way it is coded, if I am not doing mistake is
Cs=sqrt(ck*sqrt(2*ck/ce))
in the incompressible Smagorinsky.H line 114.
There is a factor 2 added in the root-mean squared.

And in the case where Ce=1.048 and Ck=0.094, and with this factor 2, we obtain Cs=0.1995.

May be I miss something?

Maxime
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Old   May 13, 2011, 08:23
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Quote:
Originally Posted by MaximeIST View Post
Hello

I may keep on confusing people, but the way it is coded, if I am not doing mistake is
Cs=sqrt(ck*sqrt(2*ck/ce))
in the incompressible Smagorinsky.H line 114.
There is a factor 2 added in the root-mean squared.

And in the case where Ce=1.048 and Ck=0.094, and with this factor 2, we obtain Cs=0.1995.

May be I miss something?

Maxime
Hi,Maxime,
In the incompressible Smagorinsky.H line 114,it's the expression of K,not Ck,it means k=2Ck/Ce*(delta^2)*(Sof^2),may be my blog can help you: http://blog.sina.com.cn/s/blog_6d9c27ab0100u9ez.html
it's just what I think, I'm not sure.
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Old   May 13, 2011, 11:10
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Hello Yingkun

I agree with you that it is the expression of k which is in line 114 of the Smagorinsky.H.

But the factor 2 is still there and as lakeat shown in the first message of this thread, we still get
Cs=sqrt(ck*sqrt(2*ck/ce))

I have been in your blog, and I think you lost this factor 2 in the passage of equation (7) to the last equation of the page.

Maxime
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Old   May 14, 2011, 01:01
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Quote:
Originally Posted by MaximeIST View Post
Hello Yingkun

I agree with you that it is the expression of k which is in line 114 of the Smagorinsky.H.

But the factor 2 is still there and as lakeat shown in the first message of this thread, we still get
Cs=sqrt(ck*sqrt(2*ck/ce))

I have been in your blog, and I think you lost this factor 2 in the passage of equation (7) to the last equation of the page.

Maxime
hi,
I think you don't understand exactly what I mean,the difference is just caused by the different expressions of S between turbulent therory and Openfoam,there is a factor of sqrt(2) difference,lakeat just regards them the same
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Old   May 16, 2011, 06:13
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Quote:
Originally Posted by yingkun View Post
hi,
I think you don't understand exactly what I mean,the difference is just caused by the different expressions of S between turbulent therory and Openfoam,there is a factor of sqrt(2) difference,lakeat just regards them the same
OK! that's what I missed!
It's true I don't understand what is written in your blog, chinese is not easy for me !

So you said that the sqrt(2) is a consequence of the definition of S in OpenFoam.
In this case, the default constant in the incompressible Smagorinsky is 0.167.
good to know!
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