# kappatWallFunction

 Register Blogs Members List Search Today's Posts Mark Forums Read

 December 9, 2010, 20:08 kappatWallFunction #1 New Member   Hagen Müller Join Date: Nov 2010 Posts: 27 Rep Power: 6 Hello, has anybody experience with the kappatJayatillekeWallFunction? I think the wall function prescribes the Temperature boundary layer with the P-function: P = 9.24{r3 - r} and T+ = σT(U+ + P) I found the relation here: http://dma.ing.uniroma1.it/Hanjalic/...t-flow-cfd.pdf But it is still not clear how OpenFOAM exactly predicts the Temperature value at the wall adjacent cell. In my opinion further quantities like the temperature gradient or the wall heat flux are needed. Furthermore, I don't know how to set up a case with this wall Function. It does not work like for example nutWallFunction is used (boundary patch type). Thank you for your help

 January 19, 2012, 10:39 #2 Senior Member   Anne Gerdes Join Date: Aug 2010 Location: Hamburg Posts: 152 Rep Power: 6 Hey, in my opinion the boundary condition for kappa_t models the effect of the thermal law of the wall, using the linear difference quotient for the approximation of the heat flux at the wall, that means: q = kappa_t (dT/dx) \approx \kappa_t T_p / y_p From the definition of T+ T+ = [(T_w -T_p) rho c_p u_tau]/ q we obtain an equation for heat flux q which depends on T+: q=[(T_w -T_p) rho c_p u_tau] /T+ The thermal law of the wall T+ = Pr_t ( 1/kappa ln(Ey+) + P) gives an equation for T+ which can be inserted into the equation for the flux. With the definition of the flux and the approximation we obtain q = kappa_t T_p / y_p = [(T_w -T_p) rho c_p u_tau] /T+ and by inserting the logarithmic law of the wall for T+ we obtain an equation for kappa_t which models the effect of the logarithmic law of the wall. In OpenFOAM the implementation in kappatJayatillekeWallFunction is kappa_t = nu * (y+ / (Pr_t * (log(Ey+)/kappa + P) - 1/Pr) When I derive this boundary condition from the equations which I mentioned before I obtain kappa_t = mu *cp (y+ / (Pr_t * (log(Ey+)/kappa + P) - 1/Pr) Has anyone made similar experiences? If you are interested I could also post a more detailed deduction of the boundary condition in order to see where might be the error.

January 19, 2012, 10:41
#3
Senior Member

Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 152
Rep Power: 6
Quote:
 Originally Posted by Hagen Furthermore, I don't know how to set up a case with this wall Function. It does not work like for example nutWallFunction is used (boundary patch type).
I think you just use it as boundary condition for kappa_t at the wall.
The wall function for kappa_t is analogous to the wall function for nu_t.
In my opinion it should be possible to use both, nutWallFunction for nut at the wall and JayatillekeWallFunction for kappa_t at the wall.

 February 7, 2012, 18:44 #4 Member   Jubayer Join Date: Oct 2009 Location: The University of Western Ontario, London, Ontario Posts: 42 Blog Entries: 1 Rep Power: 7 Hi Anne, As you said, "In OpenFOAM the implementation in kappatJayatillekeWallFunction is kappa_t = nu * (y+ / (Pr_t * (log(Ey+)/kappa + P) - 1/Pr) When I derive this boundary condition from the equations which I mentioned before I obtain kappa_t = mu *cp (y+ / (Pr_t * (log(Ey+)/kappa + P) - 1/Pr) Has anyone made similar experiences? If you are interested I could also post a more detailed deduction of the boundary condition in order to see where might be the error."" Actually there is nothing wrong between the above two equations. In everywhere kappat is actually turbulent thermal conductivity with unit J/kg-K, but in openFoam kappat is something with unit m^2/s (you can look into hot room tutorial to see the unit of kappat). So the real_kaapat = openFoam_kappat*Cp*rho. So if you divide your real_kappat equation (2nd one) with Cp and rho you will get openFoam_kappat equation (1st one) Now I have a question. What happens to this wall function if I want to use low-Re modelling for heat transfer (y+<1). Law of the wall says, T+ = y+ for y+<11. What happens to this JayatillekeWallFunction for y+<11? Do I have to modify the wall function? Thanks.

February 9, 2012, 09:01
#5
Senior Member

Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 152
Rep Power: 6
Hey Jubayer,

thank you for the hint with the kappat_t used in OpenFOAM. Now everything makes sense to me.

Quote:
 Now I have a question. What happens to this wall function if I want to use low-Re modelling for heat transfer (y+<1). Law of the wall says, T+ = y+ for y+<11. What happens to this JayatillekeWallFunction for y+<11? Do I have to modify the wall function? Thanks.
For thermal simulations you have to compute thermal y+. If thermal y_T+ < y+ then the wall function approach is used.

When yT+>y+ in OpenFOAM/JayatillekeWallFunction kappa_t is set to zero

Code:
 if (yPlus > yPlusTherm)
{
scalar nu = nuw[faceI];
scalar kt = nu*(yPlus/(Prt_*(log(E_*yPlus)/kappa_ + P)) - 1/Pr);
kappatw[faceI] = max(0.0, kt);
}
else
{
kappatw[faceI] = 0.0;
}
But I do not yet understand how the thermal y+ is computed. You can find the iteration inside the code

HTML Code:
scalar kappatJayatillekeWallFunctionFvPatchScalarField::yPlusTherm
(
const scalar P,
const scalar Prat
) const
{
scalar ypt = 11.0;

for (int i=0; i<maxIters_; i++)
{
scalar f = ypt - (log(E_*ypt)/kappa_ + P)/Prat;
scalar df = 1.0 - 1.0/(ypt*kappa_*Prat);
scalar yptNew = ypt - f/df;

if (yptNew < VSMALL)
{
return 0;
}
else if (mag(yptNew - ypt) < tolerance_)
{
return yptNew;
}
else
{
ypt = yptNew;
}
}

return ypt;
}
Regards
Anne

 May 22, 2012, 04:56 #6 New Member   chkath Join Date: May 2012 Posts: 2 Rep Power: 0 Hi, can anybody tell me how Anne derived the boundary condition: kappa_t = mu *cp (y+ / (Pr_t * (log(Ey+)/kappa + P) - 1/Pr) from the equation: q = kappa_t T_p / y_p = [(T_w -T_p) rho c_p u_tau] /T+ Thank you for your reply!

 May 22, 2012, 06:32 #7 Senior Member   Anne Gerdes Join Date: Aug 2010 Location: Hamburg Posts: 152 Rep Power: 6 Hey, actually there was a little mistake, i.e. I derived the equation from q = kappa_eff T_p / y_p = [(T_w -T_p) rho c_p u_tau] /T+ which is equal to q= (kappa + kappa_t) T_p / y_p = [(T_w -T_p) rho c_p u_tau] /T+ (I will correct this in my preceding post) Starting with (kappa + kappa_t) T_p / y_p = [(T_w -T_p) rho c_p u_tau] /T+ you shift the term kappa T_p/y_p to the right-hand side. Then you obtain kappa_t T_p / y_p = [(T_w -T_p) rho c_p u_tau] /(T+) - kappa*T_p/y_p For u_tau you use the equation u_tau = (y+/y_p) *nu kappa_t T_p / y_p = [(T_w -T_p) rho c_p y+*nu] /(T+ y_p) - kappa*T_p/y_p T_w=0 at the wall, setting mu= rho*nu and multiplying with y_p/T_p leads to kappa_t = (mu c_p y+) /(T+) - kappa For the Prandtl-number holds Pr= mu*cp/kappa -> kappa= (mu*cp)/Pr kappa_t = (mu c_p y+) /(T+) - (mu*cp)/Pr So kappa_t= mu*cp (y+/T+ -1/Pr) Finally inserting T+ according to the thermal law of the wall T+= Pr_t * (log(Ey+) /kappa + P) leads to the final equation for kappa_t kappa_t= mu *cp (y+ / (Pr_t * (log(Ey+)/kappa + P) - 1/Pr) If you note more mistakes do not hesitate to inform me

 May 22, 2012, 07:48 #8 New Member   chkath Join Date: May 2012 Posts: 2 Rep Power: 0 Thank you for your reply! It looks much more clear for me, but I still have questions: Why is T_w = 0 at the wall? And you wrote: "multiplying with y_p/T_p leads to: kappa_t = (mu c_p y+) /(T+) - kappa" In the equation before you have a minus in front of T_p. Why does this minus vanishes?

May 24, 2012, 07:22
low-Re Buoyant Boussinesq Heat Transfer
#9
New Member

Will Logie
Join Date: Sep 2010
Location: Canberra, Australia
Posts: 19
Rep Power: 6
I'd like to join the conversation around about here if I may:

Quote:
 Originally Posted by cm_jubayer Now I have a question. What happens to this wall function if I want to use low-Re modelling for heat transfer (y+<1). Law of the wall says, T+ = y+ for y+<11. What happens to this JayatillekeWallFunction for y+<11? Do I have to modify the wall function?
In my experience (very low-Re Boussinesq), the incompressible turbulent thermal diffusivity approaches zero with decreasing y+, thus removing a need for the currently implemented kappatJayatillekeWallFunction. This does not mean however that some thermal law of wall is not necessary, especially for average to high Prandtl numbers (decreasing ratio of yPlusTherm to yPlus). The wallHeatFlux:
Code:
volScalarField kappaEff("kappaEff", RASModel->nu()/Pr + kappat);

surfaceScalarField heatFlux
(
Cp * rho * fvc::interpolate(kappaEff) * fvc::snGrad(T)
);
may still tend to be underestimated (unless yPlus << 1); leading to higher laminar convective heat transfer coefficients (and Nusselt numbers) than one might expect (e.g. the attachment shows free convection from horizontal cylinders in water where yPlus<<1). Alas, this results in long simulation times...

I don't know how this low-Re thermal wall function might look (i.e. which modifications one could make to JayatillekeWallFunction) nor if there are other workarounds (e.g. NuFOAM).

Any thoughts on the matter?

Regards,
Will.
Attached Images
 ISES2011_NuVSRa.png (55.6 KB, 34 views)

May 30, 2012, 06:35
#10
Senior Member

Anne Gerdes
Join Date: Aug 2010
Location: Hamburg
Posts: 152
Rep Power: 6
Hey chkath,

Quote:
 Originally Posted by chkath Why is T_w = 0 at the wall?
This condition is valid when using a homogeneous Dirichlet boundary condition for the temperature at the wall, i.e. the temperature is set to zero, so Tw=0
Quote:
 Originally Posted by chkath And you wrote: "multiplying with y_p/T_p leads to: kappa_t = (mu c_p y+) /(T+) - kappa" In the equation before you have a minus in front of T_p. Why does this minus vanishes?
You are right, this is a mistake (again, I am sorry for this).
The heat flux is defined, according to Fourier's law as

q=- kappa_eff T_p / y_p

Setting this equal to the highRe equation leads to

- kappa_eff T_p / y_p = [(T_w -T_p) rho c_p u_tau] /T+

and this is equal to

kappa_eff T_p / y_p = [(T_p - T_w) rho c_p u_tau] /T+

So the final result remains the same.

Kind Regards
Anne

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are On Forum Rules

All times are GMT -4. The time now is 15:42.