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Old   March 5, 2005, 23:06
Default About the rotation and strain tensor
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maximus
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Hi,I have a question about the rotation and strain tensors. I am trying to use the S-A turbulence model. When I discretzie the source term of S-A model. there is a term about rotation and strain tensors. I have only a little knowledge on tensor analysis, so I am confused how to spread the tensors to discrete terms. The term is S=sqrt(Oij*Oij),where Oij=1./2.*[par(Ui)/par(xj).-(par(Uj)/par(xi))]. If I spread Oij, I find Oij=0 as following: Oij=par(u)/par(x)-par(u)/par(x)+par(u)/par(y)-par(v)/par(x)+par(v)/par(x)-par(u)/par(y)+par(v)/par(y)-par(v)/par(y)=0. So I think I must make a mistake. And I try to spread the source term as following: Oij*Oij=[par(u)/par(x)-par(u)/par(x)]**2+[(par(u)/par(y)-par(v)/par(x))]**2+[(par(v)/par(x)-par(u)/par(y)]**2+[(par(v)/par(y)-par(v)/par(y))]**2.=2*[(par(u)/par(y)-par(v)/par(x))]**2;then S=sqrt(Oij*Oij). The same way is used to the strain term. Am I right. Thanks a lot!!!!
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Old   March 6, 2005, 04:08
Default Re: About the rotation and strain tensor
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Rami
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The rotation is a TENSOR, not a scalar. Using your notation, assuming you use a Cartesian system and 2D flow:

2Oxx = u,x -u,x = 0

2Oxy = u,y - v,x

2Oyx = v,x - u,y (=-2Oxy)

2Oyy =v,y - v,y = 0

Therefore, in general, the diagoal entries are 0, but not the off-daigonal ones. You may also note that you get a skew-symmetric tensor.

From these entries you may easily find the magnitude (S in your notation). In the case of the above assumptions you'll get

2S^2 = (u,y - v,x)^2
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Old   March 6, 2005, 05:18
Default Re: About the rotation and strain tensor
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maximus
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Hi,Rami, thank you very much! I get itŁĄ
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