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June 28, 2009, 08:55 
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))); } Question 1: Why using magSqr(dev(symm(gradU))) instead of symm(gradU) && symm(gradU) to get ???? Question 2: If magSqr(dev(symm(gradU))) = symm(gradU) && symm(gradU) = , then But I saw in "Smagorinsky.C" Code:
nuSgs_ = ck_*delta()*sqrt(k(gradU)); Then, replace K with Compare with We'll get But I heard somone said 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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~ Daniel WEI  NatHaz Modeling Laboratory Department of Civil & Environmental Engineering & Earth Sciences University of Notre Dame, USA Email  My Personal CFD Blog 

August 19, 2009, 12:42 

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Gabriela Bracho
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Hi Daniel,
Did you find any answer to this question?. in the paper: "A SUBGRIDSCALE MODEL FOR LARGEEDDY SIMULATION OF PLANETARY BOUNDARYLAYER FLOWS PETER E SULLIVAN, JAMES C. McWILLIAMS, and CHINHOH 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 

August 20, 2009, 08:35 

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Daniel WEI (老魏)
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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
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~ Daniel WEI  NatHaz Modeling Laboratory Department of Civil & Environmental Engineering & Earth Sciences University of Notre Dame, USA Email  My Personal CFD Blog 

August 20, 2009, 19:59 

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Sandy Lee
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Quote:
Sandy sandy.lee37@gmail.com 

October 6, 2010, 05:00 
tensor norm definition

#5 
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Andrea Petronio
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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 

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? 

April 22, 2011, 23:22 

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Alberto Passalacqua
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May 1, 2011, 11:53 

#9 
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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. 

May 1, 2011, 18:27 

#10 
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Alberto Passalacqua
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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.


May 2, 2011, 21:59 

#11 
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桂莹
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Hi,Andrea:
I think what you have said is right,the reason is just the different expression of Vsgs 

May 12, 2011, 06:50 

#13 
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桂莹
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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? 

May 12, 2011, 08:33 

#15 
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Andrea Petronio
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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 rewrote our Smag. with just Cs required.


May 12, 2011, 09:12 

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Quote:
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 rootmean 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 

May 13, 2011, 08:23 

#17  
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桂莹
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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. 

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 

May 14, 2011, 01:01 

#19  
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桂莹
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Quote:
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 

May 16, 2011, 06:13 

#20  
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Quote:
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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