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March 16, 2011, 01:57 |
nutWallFunction
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#1 |
Member
Kurne
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Dear All
I am using k − ω-SST in which i am unable to understand the wall functions i.e. nutWallFunction,nutRoughWallFunction etc.Will anybody tell me what it is,How to calculate and for what it is. Thanks In Advance.
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Simulation Is Determination Of Imagination Towards Approximation ® Best Regards Mubeen K Kurne |
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March 17, 2011, 03:56 |
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#2 |
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Kurne
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Dear All
Got the answer thanks to all.
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Simulation Is Determination Of Imagination Towards Approximation ® Best Regards Mubeen K Kurne |
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March 18, 2011, 07:52 |
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#3 |
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Claus Meister
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Hey kurne,
wall functions are used to model the velocity profile at the first node to the wall. Model based on magnitudes such as k, epsilon, nut, wall shear stress, friction velocity and so on. Therefore, the implemented function, e.g. epsilonWallFunction, provide the computation of the law of the wall. Instead using them, you need to resolve the boundary layer such that the transition from laminar to turbulent can be captured. Cheers |
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March 18, 2011, 08:05 |
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#4 |
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Kurne
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Dear Claus Meister
Thank you very much for the answer, in the real sense i understood the meaning of wall function but i stuck up with the nut wall function but after going through the forum and upon your reply i got the correct concept. Well, want to ask you one silly question do you know how to calculate the nutWallFunction at the inlet if we are having the data at inlet say velocity,turbulence intensity,turbulence kinetic energy and turbulence energy dissipation rate.If you know then please tell me.But i got the information from the forum that initial value of 0 is generally used. Thanks Once Again.
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Simulation Is Determination Of Imagination Towards Approximation ® Best Regards Mubeen K Kurne |
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March 18, 2011, 09:06 |
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#5 |
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Claus Meister
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The wall function are only used at the wall and not at a inlet or outlet patch. Since the nut value is composed of k and epsilon, you should set a "calculated" BC at the inlet and the outlet, respectively. A "calculated" BC means that the values at the respective patch are computed by other magnitudes.
Cheers |
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March 19, 2011, 00:05 |
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#6 |
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Kurne
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Dear Claus Yes thats right but when i go through the tutorials of OpenFOAM 1.7.1 of mixer in simpleSRFFoam solver. I found the nut directory in 0 file.When i remove the nut file from the 0 directory the solver wont run.
It means that the value of nut is required at 0 directory with the value at inlet,outlet and other patches too,Correct me if i am wrong.So i asked that how to calculate the value of nut at inlet and outlet. Thanks In Advance.
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Simulation Is Determination Of Imagination Towards Approximation ® Best Regards Mubeen K Kurne Last edited by kurne; March 21, 2011 at 01:34. |
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March 20, 2011, 19:15 |
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#7 |
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Claus Meister
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You dont have to calculate the nut values for inlet and outlet. You just have to set:
inlet { type calculated; value uniform 0; } wall { type nutWallFunction; value uniform 0; } etc. hope that helps |
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March 21, 2011, 01:23 |
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#8 |
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Kurne
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Dear Claus
Thank you very much for answering to my query.Well, is it possible to calculate the value of nutWallFunction at the patches other than inlet and outlet.If yes will you please tell me how. I know that my queries are silly hence sorry for it. Thanks In Advance.
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Simulation Is Determination Of Imagination Towards Approximation ® Best Regards Mubeen K Kurne |
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October 28, 2011, 07:01 |
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#9 |
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Anne Gerdes
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Hey Kurne, I am investigating about the nutWallFunction and I do not really understand the formula used in OpenFOAM in nutWallFunction which reads nutw[faceI] = nuw[faceI]*(yPlus*kappa_/log(E_*yPlus) - 1.0); Can you please explain your findings? I found this question in some threads so I think me and other users will be very grateful. Thank you in advance. |
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October 28, 2011, 08:06 |
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#10 |
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Claus Meister
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Loop up in Versteeg pp. 275ff; there is a description about the entire procedure. I used by my self yesterday.
http://books.google.de/books?id=RvBZ...page&q&f=false Cheers |
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November 2, 2011, 04:00 |
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#11 |
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Anne Gerdes
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Thank you very much. I borrowed the book and will post my findings here as soon as possible.
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November 25, 2011, 23:26 |
nutkwallfunction
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#12 |
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pooyan
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Hello everyone
I have recently started working with OpenFoam. I am now solving cavity problem in 3D. for turbulence model, I selected RANS and k-epsilon. for wall function, I selected nutkwallfunction. However, I could not find type of wall functions (how the wall function is defined) anywhere except from the c++ documentary which was not clear to me. I would be grateful if you could tell me what specific wall functions are used in nutkwallfunction. Thanks, |
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November 28, 2011, 04:18 |
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#13 |
Senior Member
Anne Gerdes
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You can have a look at
~/OpenFOAM/OpenFOAM-1.7.0/src/turbulenceModels/incompressible/RAS/derivedFvPatchFields/wallFunctions> There you will find all kind of wall functions. Especially for your wall function it is written in the .H file Class Foam::incompressible::RASModels::nutkWallFunctionF vPatchScalarField Description Boundary condition for turbulent (kinematic) viscosity when using wall functions, based on turbulence kinetic energy. - replicates OpenFOAM v1.5 (and earlier) behaviour At first glance I could not see any differences between nutWallFunction and nutkWallFunction.... |
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November 28, 2011, 04:20 |
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#14 | |
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Roman Thiele
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Quote:
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November 28, 2011, 04:23 |
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#15 |
Senior Member
Anne Gerdes
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What version do you use?
I cannot find nutUWallFunction at OF-1.7. When looking at the source code of nutWallFunction I see the same computation of yPlus like in nutkWallFunction, based on turbulent kinetic energy.... |
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November 28, 2011, 04:25 |
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#16 |
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Roman Thiele
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In 2.0 more nutWallFunction boundary conditions were introduced, one of them nutUWallFunction, where basically the way of how yPlus is calculated is changed.
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November 28, 2011, 04:28 |
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#17 |
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Anne Gerdes
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Okay, thank you for this information!
Can you also explain the computation of nut at the wall function, especially this formula nutw[faceI] = nuw[faceI]*(yPlus*kappa_/log(E_*yPlus) - 1.0); Is it derived from the logLaw? How? |
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November 28, 2011, 05:36 |
log law in compact version
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#18 |
Senior Member
Roman Thiele
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Hej,
I just found how this is derived and calculated it myself, http://www.cfd-online.com/Wiki/Two_e...wall_functions if you see how the rho is eliminated, maybe you can post it here. solution to eliminate rho: Code:
(mut+mu)Up/yP = (rho y uTau kappa )/(log(yPlus E)) =1/rho (mut+mu)Up/yP = ( y uTau kappa )/(log(yPlus E)) = (nut+nu)Up/yP = ( y uTau kappa )/(log(yPlus E))
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~roman Last edited by romant; November 28, 2011 at 05:50. Reason: found solution to eliminate rho |
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November 28, 2011, 06:40 |
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#19 |
Senior Member
Anne Gerdes
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Thank you so much!
Now I understand the derivation. There is one typo on the site, the missing density at the definition of tau_w, But there is no problem as the whole equation rho (nut + nu) * up/yp = (rho *u_tau *kappa *u) / ln (E y+) can simply be divided by rho. |
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November 28, 2011, 06:50 |
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#20 | |
Senior Member
Roman Thiele
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