| Author | Message | |||
     Eduards Rens (eduards) Member Username: eduards Post Number: 5 Registered: 12-2005 |
Hello, I was analyzing time dependent 2D diffusion problem for some time now. I decided to solve it in 1D space for more precise comparison with theoretical values using data from elevation plots export data. 1. How can I specify JUMP BCs in 1D space? I cant transfer my 2D problem to much simpler case. I am calculating effective diffusion parameter D from formula F = C0 * erfc(x/2/sqrt(D*t), where I get F and x values from elevation plot export file. When I was in 2D space, I divided all area in small segments and took midpoint as x, and got U as averaging integral integral(U,'region')/integral(1,'region'). 2. It might be pure physics question about this aproach, is this right or wrong? Can I use just elevation line from midpoint?
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| Eduards Rens (eduards) Member Username: eduards Post Number: 6 Registered: 12-2005 |
3. About JUMP BC in general. Documentation says contact(u)=1/Rc * jump(u), where Rc- resistance. I dont understand this concept very well. Lets say, for example, I would like only one half of flux to go thru border, or one tenth, or whatever.. A. Flux in both directions is equal. That means u'(x) at side 1 = u'(x) at side 2. B. There is jump. u(x) at side 1 = phi * u(x) at side 2. How Rc in documentation relate to my constant phi? | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 805 Registered: 06-2003 |
See "Contact Resistance" in the Help Index for a discussion of the theory of the JUMP boundary. | |||
| Eduards Rens (eduards) Member Username: eduards Post Number: 7 Registered: 12-2005 |
Thank you for your answer. Unfortunately, I have read that chapter in documentation numerous times, but cant solve my 3rd question. I came here to forum as last resort. I understand the idea of jump boundaries, but cant figure out how that resistance term Rc relate to how many times bigger is value difference on both sides. | |||
| karunamurthy (karunamurthy_k) New member Username: karunamurthy_k Post Number: 2 Registered: 04-2007 |
hello, For 2d heat conduction in cylidrical coordinates(along r, phi) how to specify boundary conditions along coordinate phi? because phi varies from 0 to 360 and phi=0 and phi=360 are same. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 806 Registered: 06-2003 |
Rens - The rules regarding contact resistance boundaries are the same as for Natural BCs at the same boundary. That is, the surface terms arising from integration by parts (divergence theorem) are continuous across interior boundaries. For the equation div(K*grad(u))+S = 0, k*grad(u) is continuous across interior boundaries. At a Jump boundary, the contact condition defines the boundary flux, and this value is equated to K*grad(u) on both sides of the boundary. If the value is K1 on one side and K2 on the other, then K1*grad(U)1 = K2*grad(U)2 = (U2-U1)/R. [This is merely a restatement of the documentation, but I don't know how else to explain it.] If you want some non-conserving condition, you will have to choose for your variable a quantity for which flux IS conserved and compute your non-conserved value from it using different material parameters. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 807 Registered: 06-2003 |
Karunamurthy - There is no boundary condition on phi. The principle of cylindrical coordinates is that you are computing an (R,Z) cross-section and there is no variation in phi. There are two ways to do (R,PHI) computations. 1) merely rename the coordinates R and PHI and write the correct form of your PDEs. Plots will show R and PHI as cartesian coordinates, but the solutions will be according to the PDE. 2) Use definitions to transform your differential operators to cartesian forms. The computation will be in (X,Y) cartesian geometry and plots will show the correct geometry. See the example "Samples | Misc | Polar.pde". In either case, the boundary conditions are presented as described in the documentation. Just be sure you do the math right. |
Inhomogenoeus diffusion