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September 24, 2008, 05:25 |
Hi
I believe that the gradi
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Member
Markus Weinmann
Join Date: Mar 2009
Location: Stuttgart, Germany
Posts: 77
Rep Power: 17 |
Hi
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. Markus |
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