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Laplacian of a TensorField ?

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Old   September 23, 2010, 08:11
Post Laplacian of a TensorField ?
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Hi,

How to calculate laplacian of A*T
if A is a scalar function of another scalar Main field c and , T is a 2nd order tensor

example

when i put

"
solve
(
fvm::ddt(c) + fvm::div(phi,c) == fvm::laplacian(A,T)
);

A=2*pow(c,2)+5;
"

it doesnot work!!!!?

Help lease??
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Old   September 23, 2010, 08:18
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Quote:
Originally Posted by T.D. View Post
solve
(
fvm::ddt(c) + fvm::div(phi,c) == fvm::laplacian(A,T)
);
so what you are saying is
Code:
scalar + scalar == tensor
I hope you see the mistake here
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Old   September 23, 2010, 09:02
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Execuse me,
In another words
the last term to the write is

∇·(A*(∇.T)) which must be a scalar ok to go on with the c equation.


where A is a scalar and T is a second order tensor


so ∇·T is div(T) and it is a vector ok (since divergence of a tensor) ??

and V=A*∇·T is a vector then

so ∇·(A*(∇.T)) must be written something like div(A,V) ?? knowing that A is always a function of c.


helpme
Thanks
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Old   September 23, 2010, 09:14
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Quote:
Originally Posted by T.D. View Post
∇·(A*(∇.T))
ah ok, so its 2 div operations...
∇·(A*(∇.T)) = fvc::div(A*fvc::div(T))

but this will make this term fully explicit.
I think it will be hard to treat it fully implicit.
If you run into stability problems you should try to split it up into implicit/explicit parts.
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Old   September 23, 2010, 10:43
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Hi Niklas thanks,


BUT A is a function of the scalar field c.

so it stays like that ∇·(A*(∇.T)) = fvc::div(A*fvc::div(T)) ??


another thing please: do you know how to turn off all dimensions?

thank you so much
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Old   September 23, 2010, 10:54
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yes of course it does.

and why do you want to turn off dimensions,
it has saved me on numerous occasions.

but ok, I can see that the exp-function can give you problems.

if you are using 1.7.1 it is in
OpenFoam-1.7.1/etc/controlDict

lookup the line
dimensionSet 1;

and change the 1 to 0.
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Old   September 23, 2010, 13:40
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Thanks a lot Niklas
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