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how symmetry bc is applied for a tensor ?

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Old   January 12, 2012, 22:55
Default how symmetry bc is applied for a tensor ?
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Hello,
I want to understand how the symmetry boundary condition is applied for a tensors of rank 1 and above in OF.
Say at a wall with normal n and a velocity vector V at adjacent cell, I think it would be like
Code:
   Vsymm =  V - dot(V,n);
   Vsymm = mag(Vsymm) / mag(V)  * V;
How do you transform a 2 rank and above tensor (f.i 3x3 stress tensor ) at a symmetric wall boundary in a similar way?

thanks

Last edited by dshawul; January 12, 2012 at 23:56.
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Old   January 19, 2012, 17:15
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Ok I found how it is applied in OF
Code:
tmp<vectorField> nHat = this->patch().nf();

    Field<Type>::operator=
    (
        (
            this->patchInternalField()
          + transform(I - 2.0*sqr(nHat), this->patchInternalField())
        )/2.0
    );

    transformFvPatchField<Type>::evaluate();
This averages the reflection of the tensor and itself to get a vector parallel to the wall.
But then it doesn't adjust its magnitude to equal the original vector, like I tried to do in my post above. Is this correct ?
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