# Ueqn.A() extract the diag. term from the Ueqn??

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 July 4, 2013, 05:40 #2 Senior Member   Dongyue Li Join Date: Jun 2012 Location: Torino, Italy Posts: 757 Rep Power: 10 If A()=diag()/mesh.V(), this is my mesh: Code: convertToMeters 0.03; vertices ( (0 0 0) (1 0 0) (1 1 0) (0 1 0) (0 0 0.1) (1 0 0.1) (1 1 0.1) (0 1 0.1) ); blocks ( hex (0 1 2 3 4 5 6 7) (3 3 1) simpleGrading (1 1 1) ); So V is 0.003*0.01*0.01=0.000003. but 0.00012/0.000003 does not equal to 800. utkunun likes this.

 July 24, 2013, 11:08 #3 Member   Join Date: Aug 2010 Posts: 31 Rep Power: 9 Hi, Ueqn.A()=diag()/mesh.V() holds true only for cells, which do not contact the boundary. E.g. in your example: 600=0.00018/(0.01*0.01*0.003) [and be careful mesh.V()=3e-7 (not 3e-6)]. At the moment I'm also trying to clarify details about the "Ueqn.A()". B.r. Martin. sharonyue and utkunun like this.

July 24, 2013, 11:12
#4
Senior Member

Dongyue Li
Join Date: Jun 2012
Location: Torino, Italy
Posts: 757
Rep Power: 10
Quote:
 Originally Posted by Martin80 Hi, Ueqn.A()=diag()/mesh.V() holds true only for cells, which do not contact the boundary. E.g. in your example: 600=0.00018/(0.01*0.01*0.003) [and be careful mesh.V()=3e-7 (not 3e-6)]. At the moment I'm also trying to clarify details about the "Ueqn.A()". B.r. Martin.
So this is only true for inner cells, not boundary cells. Thats clear! Thx very much!

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