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November 27, 2012, 12:43 
Continuity equation in implicit form

#1 
Senior Member
HECKMANN Frédéric
Join Date: Jul 2010
Posts: 236
Rep Power: 8 
Dear everyone,
I'm looking for a way of converting my continuity equation in an implicit form. This equation correspond to a particle tracking using an Eulerian approach. "rho" is not a gas, it is my particle concentration. Right now I have : Continuity equation : Code:
Info<< "Solve Continuity" << endl; tmp<fvScalarMatrix> CEqn ( fvm::ddt(rho) + fvc::div(phi) ); CEqn().solve(); I'm a little bit confused by the OpenFoam documentation : http://www.foamcfd.org/Nabla/guides/...sGuidese9.html It says that the convection term can be Imp/Exp with the following form: div(psi,scheme)* where psi is a surfaceScalarField (just like my "phi"). However, if I try the following code, I get a mistake while compiling: Code:
word scheme("div(phi)"); Info<< "Solve Continuity" << endl; tmp<fvScalarMatrix> CEqn ( fvm::ddt(rho) + fvm::div(phi, scheme) ); CEqn().solve(); Thanks in advance. 

November 27, 2012, 13:21 

#2 
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Meindert de Groot
Join Date: Jun 2012
Location: Netherlands
Posts: 34
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Did you already try fvm::div(phi,rho)? I think that should do the trick for you.


November 27, 2012, 20:54 

#3 
Senior Member
HECKMANN Frédéric
Join Date: Jul 2010
Posts: 236
Rep Power: 8 
Well, if I use fvm::div(phi,rho) , it would imply that phi is an interpolation of U only. However, in my solver, phi is an interpolation of U and rho.
I'm not sure, can I use 2 different phi ?  phiU = (interpolation of U) for the continuity equation  phiRhoU = (phiU * interpolation of Rho) for the momentum equation Actually, your advice rise more questions than answers ^^ For more details, my model is defined as followed: Nomenclature :  volume fraction of the particles  velocity of the particles  function of the relative Reynolds number  time respond of the particle  velocity of the fluid (constant) Continuity equation : Momentum equation 

November 28, 2012, 06:02 

#4 
Member
Meindert de Groot
Join Date: Jun 2012
Location: Netherlands
Posts: 34
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Hi Frédéric,
Defining two different phi's would indeed be the way to go. Otherwise, you would end up with two unknowns in the continuity equation ( and ). You would not be able to solve for both. I would be careful with the definition of phiRhoU. It might be better to do an interpolation of U * Rho, but this is something you could test of course. How do you solve the momentum equation? Is given from solving the continuity equation? I have one small remark. You seem to be using both rho and alpha for the volume fraction of the particles. Correct me if I am wrong. I think it would be better to be consistent to avoid confusion. Good luck. Last edited by meindert; November 28, 2012 at 06:23. 

November 28, 2012, 06:48 

#5 
Senior Member
HECKMANN Frédéric
Join Date: Jul 2010
Posts: 236
Rep Power: 8 
I have applied your method and it seems to work perfectly. I made a short run and the results look similar to my explicit solver.
To gives some details, here is roughly what happen in my code Code:
//// phi = fvc::interpolate(U) & mesh.Sf(); //// Info<< "Solve Continuity" << endl; tmp<fvScalarMatrix> CEqn ( fvm::ddt(rho) + fvm::div(phi,rho) ); CEqn().solve(); //// phiR = fvc::interpolate(rho)*(fvc::interpolate(U) & mesh.Sf()); //// Info<< "Solve Momentum" << endl; tmp<fvVectorMatrix> UEqn ( fvm::ddt(rho, U) + fvm::div(phiR, U) == fRer/tau*rho*Uf  fvm::Sp(fRer/tau*rho, U) ); UEqn().solve(); Thanks again 

November 28, 2012, 06:53 

#6 
Member
Meindert de Groot
Join Date: Jun 2012
Location: Netherlands
Posts: 34
Rep Power: 5 
What happens if you try
Code:
phiR = fvc::interpolate(rho*U) & mesh.Sf() 

November 28, 2012, 07:18 

#7  
Senior Member
HECKMANN Frédéric
Join Date: Jul 2010
Posts: 236
Rep Power: 8 
It also works and it simplify my code
I have another question related to the field management. Right now I have a RelativeReynolds number stored in a volScalarField and a drag value in another volScalarField. Until now, I use the following code : Code:
// Compute the Relative Reynolds number field Rer = mag( Uf  U ) // Compute the drag value field fRer = 1.0 + 0.15 * pow(Rer, 0.687) + ... Quote:


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