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volScalarField::Internal vs. volScalarField

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Old   January 7, 2018, 18:40
Default volScalarField::Internal vs. volScalarField
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Hi!

In kEpsilon.C, G is defiened as:

Code:
    volScalarField::Internal G
    (
        this->GName(),
        nut.v()*(dev(twoSymm(tgradU().v())) && tgradU().v())
    );
Why is it defined as "volScalarField::Internal" instead of just a simple "volScalarFiled"? And have something like this:

volScalarField G1= this->nut_*(dev(twoSymm(tgradU())) && tgradU());


Thanks,

Mary
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Old   December 1, 2020, 08:44
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René Thibault
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I got the same question lately. Did you get an answer for your question?

Regards,
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Old   December 5, 2020, 04:02
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Mark Olesen
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Quote:
Originally Posted by Tibo99 View Post
I got the same question lately. Did you get an answer for your question?
The question to ask yourself: if it were a volScalarField, which is an internal field (with dimensions) and a boundary - what would you like to calculate on the boundaries?
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Old   December 26, 2020, 10:33
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This is what I have understood.
The volScalarField is a typedef of GeometricField<>. So volScalarField contains an internal field a boundaryfield etc. Now, the volScalarField::Internal is the 'type' of the internal field in the volscalarField. It's defined by typedefing of Diemnsionedfield.
For volScalarField the 'Internal' translates to:
Code:
typedef DiamensionedField<Scalar, volMesh>    Internal;
So, the code
Code:
    volScalarField::Internal G
    (
        this->GName(),
        nut.v()*(dev(twoSymm(tgradU().v())) && tgradU().v())
    );
Can be seen as
Code:
    DimensionedField<scalar, volField> G
    (
        this->GName(),
        nut.v()*(dev(twoSymm(tgradU().v())) && tgradU().v())
    );
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