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Fieldfunction du/dy - du/dx

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Old   October 4, 2011, 08:21
Default Fieldfunction du/dy - du/dx
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Hi guys,

I have got a little problem. I want to set up a field function where velocity gradients are included:

e.g:

du/dx -> CCM Code: grad($$Velocity[0]). I think this is for du/dx right, where u is the first component of the velocity vector.

How does this work for du/dy or du/dz??

Thanks for help
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Old   October 4, 2011, 11:34
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Check out the help file:

Using Tools > Using Field Functions > Field Function Programming Reference > Vector Functions, Miscellaneous Functions
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Old   October 5, 2011, 19:10
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"grad($$Velocity[0])" will get you a vector field function consisting of (du/dx, du/dy, du/dz). If you just want du/dx, then you need to do "grad($$Velocity[0])[0]"

To be clear...

du/dx = grad($$Velocity[0])[0]
du/dy = grad($$Velocity[0])[1]
dv/dx = grad($$Velocity[1])[0]
dv/dy = grad($$Velocity[1])[1]
dw/dx = grad($$Velocity[2])[0]
etc...
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Old   October 6, 2011, 02:25
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Ah Kyle, thank u! :-) This is it what I was looking for....
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