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March 21, 2022, 10:30 |
lagrangian cloud::SU() returns different
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#1 |
Member
Rishikesh
Join Date: Apr 2016
Posts: 63
Rep Power: 10 |
Hi all,I noticed the following when coupling lagrangian cloud with eulerian continuum. In the momentum source calculator function, MomentumCloud::SU() (for OF8+, otherwise it would probably be KinematicCloud::SU()), we see the following code:
Code:
if (solution_.semiImplicit("U")) { const volScalarField::Internal Vdt(this->mesh().V()*this->db().time().deltaT()); return UTrans()/Vdt - fvm::Sp(UCoeff()/Vdt, U) + UCoeff()/Vdt*U; } Where UCoeff and UTrans are the self-explanatory fields, derived by adding up the interaction from all member particles of the cloud. The returned object is an fvVectorMatrix, with the first and last are explicit components, and the middle term, the coefficient term. Now, if there is no "semi-implicit" flag, ie the source is purely an explicit term, the code returns: Code:
else { tmp<fvVectorMatrix> tfvm(new fvVectorMatrix(U, dimForce)); fvVectorMatrix& fvm = tfvm.ref(); fvm.source() = -UTrans()/(this->db().time().deltaT()); return tfvm; } Which is quite analogous to the explicit part of semi-implicit source above, but it does not have a division by cell volume. It appears to me that in general, we supply fvVectorMatrix terms on a per unit volume basis. What is then the reason behind difference in the divisor in the two cases. Is the second formulation correct? If it is, can someone help me understand why it is so? I would expect that the two should not even run with the same solver/case, based on dimensional considerations, but they do. However, I cannot find any other point in lagrangian libraries, where the semi-implicit flag is used to change the way in which UTrans is calculated differently depending of the flag. I would appreciate any help in enlightening me about this. Thanks |
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Tags |
lagrangian openfoam, semi-implicit-source |
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