Smagorinsky model details

lakeat · June 28, 2009, 08:55

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

gaby · August 19, 2009, 12:42

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

lakeat · August 20, 2009, 08:35

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

2.0*ck_/ce_
magSqr(dev(symm(gradU))) instead of symm(gradU) && symm(gradU)

sandy · August 20, 2009, 19:59

Quote:

Originally Posted by gaby

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

andrea · October 6, 2010, 05:00

Hallo,
I was also trying to understand the implementation of Smagorinsky model.
As it is said here
http://www.cfd-online.com/Forums/ope...rain-rate.html
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

MaximeIST · March 23, 2011, 15:31

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?

maysmech · April 22, 2011, 08:13

Hi,

I have same questions too.

alberto · April 22, 2011, 23:22

Reference: http://pof.aip.org/resource/1/phfle6/v9/i5/p1416_s1

maysmech · May 1, 2011, 11:53

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,

alberto · May 1, 2011, 18:27

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.

yingkun · May 2, 2011, 21:59

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

maysmech · May 12, 2011, 05:48

So, what is Cs value in OpenFOAM's Smagorinsky? 0.2 or 0.167?

yingkun · May 12, 2011, 06:50

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?

maysmech · May 12, 2011, 08:30

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.

andrea · May 12, 2011, 08:33

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.

MaximeIST · May 12, 2011, 09:12

Quote:

Originally Posted by yingkun

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

yingkun · May 13, 2011, 08:23

Quote:

Originally Posted by MaximeIST

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.

MaximeIST · May 13, 2011, 11:10

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

yingkun · May 14, 2011, 01:01

Quote:

Originally Posted by MaximeIST

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

MaximeIST · May 16, 2011, 06:13

Quote:

Originally Posted by yingkun

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!

June 28, 2009, 08:55	Smagorinsky model details	#1
lakeat Senior Member Daniel WEI (老魏) Join Date: Mar 2009 Location: Beijing, China Posts: 689 Blog Entries: 9 Rep Power: 20	Sorry, I do not understand, I saw in "Smagorinsky.H", Code: tmp<volScalarField> k(const tmp<volTensorField>& gradU) const { return (2.0ck_/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 makaveli_lcf, maysmech, MaximeIST and 1 others like this. __________________ ~ Daniel WEI ------------- Boeing Research & Technology - China Beijing, China Email

August 19, 2009, 12:42		#2
gaby New Member Gabriela Bracho Join Date: Mar 2009 Location: Valencia, Valencia, Spain Posts: 14 Rep Power: 16	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 2Ck/Ce ?? Is it because of the symm(grad(U)) definition??? It would be great if you could share your opinion... Gaby makaveli_lcf likes this.

August 20, 2009, 08:35		#3
lakeat Senior Member Daniel WEI (老魏) Join Date: Mar 2009 Location: Beijing, China Posts: 689 Blog Entries: 9 Rep Power: 20	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 2.0*ck_/ce_ magSqr(dev(symm(gradU))) instead of symm(gradU) && symm(gradU) __________________ ~ Daniel WEI ------------- Boeing Research & Technology - China Beijing, China Email

October 6, 2010, 05:00	tensor norm definition	#5
andrea Member Andrea Petronio Join Date: Mar 2009 Location: Trieste, Italy Posts: 43 Rep Power: 16	Hallo, I was also trying to understand the implementation of Smagorinsky model. As it is said here http://www.cfd-online.com/Forums/ope...rain-rate.html 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 makaveli_lcf and wenxu like this.

April 22, 2011, 23:22		#8
alberto Senior Member Alberto Passalacqua Join Date: Mar 2009 Location: Ames, Iowa, United States Posts: 1,912 Rep Power: 35	Reference: http://pof.aip.org/resource/1/phfle6/v9/i5/p1416_s1 makaveli_lcf likes this. __________________ Alberto Passalacqua GeekoCFD - A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM - An open-source implementation of quadrature-based moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using.

March 23, 2011, 15:31		#6
MaximeIST New Member Join Date: Feb 2010 Posts: 8 Rep Power: 15	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(cksqrt(2ck/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, 08:13		#7
maysmech Senior Member Join Date: Jan 2010 Posts: 347 Blog Entries: 2 Rep Power: 16	Hi, I have same questions too.

May 1, 2011, 11:53		#9
maysmech Senior Member Join Date: Jan 2010 Posts: 347 Blog Entries: 2 Rep Power: 16	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
alberto Senior Member Alberto Passalacqua Join Date: Mar 2009 Location: Ames, Iowa, United States Posts: 1,912 Rep Power: 35	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. __________________ Alberto Passalacqua GeekoCFD - A free distribution based on openSUSE 64 bit with CFD tools, including OpenFOAM. Available as in both physical and virtual formats (current status: http://albertopassalacqua.com/?p=1541) OpenQBMM - An open-source implementation of quadrature-based moment methods. To obtain more accurate answers, please specify the version of OpenFOAM you are using.

May 2, 2011, 21:59		#11
yingkun Member 桂莹 Join Date: Apr 2011 Posts: 36 Rep Power: 14	Hi,Andrea: I think what you have said is right,the reason is just the different expression of Vsgs

May 12, 2011, 05:48		#12
maysmech Senior Member Join Date: Jan 2010 Posts: 347 Blog Entries: 2 Rep Power: 16	So, what is Cs value in OpenFOAM's Smagorinsky? 0.2 or 0.167?

May 12, 2011, 06:50		#13
yingkun Member 桂莹 Join Date: Apr 2011 Posts: 36 Rep Power: 14	hi,Maysam: since Cs=sqrt(cksqrt(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? makaveli_lcf likes this.

May 12, 2011, 08:30		#14
maysmech Senior Member Join Date: Jan 2010 Posts: 347 Blog Entries: 2 Rep Power: 16	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.

May 12, 2011, 08:33		#15
andrea Member Andrea Petronio Join Date: Mar 2009 Location: Trieste, Italy Posts: 43 Rep Power: 16	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.

May 13, 2011, 11:10		#18
MaximeIST New Member Join Date: Feb 2010 Posts: 8 Rep Power: 15	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(cksqrt(2ck/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

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Multiphase flow. Dispersed and free surface model	Luis	CFX	8	May 29, 2007 18:13
Kato-Launder model	sam	Main CFD Forum	13	September 21, 2006 10:15
Kinetic air props model - conductivity HALF?	Chris Bailey	FLUENT	1	March 7, 2006 10:38
Reply to Tim Re Dynamic Smagorinsky model	Ajay S. Parihar	Main CFD Forum	9	June 2, 2002 16:24
Biharmonic Viscosity with Smagorinsky model	Wen Long	Main CFD Forum	4	May 15, 2002 05:54