| Author | Message | ||
     Philip (riche) Member Username: riche Post Number: 8 Registered: 06-2006 |
Hello Dr Nelson, In cylindrical coordonnee (r,z). When I have a equation: -dz(1/a*dz(Et))-dr(1/(a*r)*(Et + r*dr(Et)))-b*Et=0 Could you tell me the Natural Condition is: Natural=-nz*(1/a*dz(Et))- nr*(1/(a*r)*(Et + r*dr(Et))) or simply: Natural=-nz*(1/a*dz(Et))- nr*(1/(a*r)* r*dr(Et)) (not componant Et inside because her first derivate?) Thank you? Mai | ||
| Philip (riche) Member Username: riche Post Number: 9 Registered: 06-2006 |
I have a more question, please: In cylindrical coordonnee (r,z), Is 2*pi*r included automatically in the expression of Natural ? Best regard | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 674 Registered: 06-2003 |
The term inside the first dr() is of second order because of the dr(Et). Therefore, the term that is integrated by parts is dr((Et+r*dr(Et))/(a*r)) and the natural has both parts, ie, Et+r*dr(Et). Another way to look at this is that your equation is -Div[dz(Et)/a, (Et+r*dr(Et))/(a*r)] - b*Et = 0 Applying the Divergence Theorem gives the natural BC as the outward normal component of the argument of the divergence. Notice that this equation does not conserve Et, since it is not of the form Div(grad(Et)/a) in cylindrical geometry. The argument of the divergence is the conserved flux. The volume element implied in cylindrical integrations is 2*pi*r*dr*dz. The value of a Natural is a flux per unit area. It is not integrated over area, and you should not put any 2*pi*r's in it. |