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Old   September 24, 2008, 05:25
Default Hi I believe that the gradi
Markus Weinmann
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I believe that the gradient of a vector field grad() is not computed consistent with the common definition.
The definition of grad() in OF follows equation 2.3 in the programmers guide. The correct definition of the velocity gradient tensor is V_ij=grad(U)=du_i/dx_j. OF defines the gradient as V_ij=du_j/dx_i, which is the transpose of the correct definition.

The wrong definition of grad(U) result for example in a wrong definition of the skew symmetric part in OF: W_ij=0.5(grad(U) - grad(U).T())=0.5(du_j/dx_i - du_i/dx_j) which should actually be W_ij=0.5(du_i/dx_j - du_j/dx_i).

Is there any reason why OF uses a different definition of grad(). Does this mean that the user alwasy needs to use grad().T() in OF to get the correct definition of the gradient?

I am somewhat confused about this issue.
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Old   September 24, 2008, 05:47
Default The "definition" of grad in Op
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The "definition" of grad in OpenFOAM is not a definition but simply the result of the grad vector operator operating from the left of the field as it is written, i.e. it is entirely consistent with the definition of the vector multiply operator.

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Old   July 16, 2009, 18:54
Default Definition of skew symmetric part
Sven Winkler
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Where in OpenFOAM (e.g. in which file) can I find the above mentioned skew symmetric part W_ij=0.5(grad(U) - grad(U).T())??
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