# Fieldfunction du/dy - du/dx

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 October 4, 2011, 08:21 Fieldfunction du/dy - du/dx #1 Senior Member   Join Date: Dec 2010 Posts: 135 Rep Power: 7 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

 October 4, 2011, 11:34 #2 Senior Member   Join Date: Apr 2009 Posts: 129 Rep Power: 9 Check out the help file: Using Tools > Using Field Functions > Field Function Programming Reference > Vector Functions, Miscellaneous Functions

 October 5, 2011, 19:10 #3 Senior Member   Join Date: Mar 2009 Location: Austin, TX Posts: 137 Rep Power: 10 "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... eRzBeNgEl likes this.

 October 6, 2011, 02:25 #4 Senior Member   Join Date: Dec 2010 Posts: 135 Rep Power: 7 Ah Kyle, thank u! :-) This is it what I was looking for....

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