I am working on 1-D mass transient diffusion in a radial domain (spherical object) using finite volume method. My equation reads $$\frac{\partial C}{\partial t} = D\left[\frac{\partial^2 C}{\partial r^2} + \frac{2}{r}\frac{\partial C}{\partial r}\right]\, .$$

In a sense of conservative mass within the the radial domain, plus a closed BC ( constant concentration of A species on the surface), does that mean I should end up with concentration gradient among the discretized nodes? In other words, at first time step (forward in time) for internal nodes, Should I get distributed concentrations that the sum of whole internal nodes will give me the concentration on the surface?

  • $\begingroup$ I have added your equation with MathJax, please check that it is OK and keep using MathJax for future equations. $\endgroup$ – nicoguaro Apr 6 at 15:39
  • 2
    $\begingroup$ I am really having hard time understanding the question being asked here, it's hard to parse it... $\endgroup$ – Maxim Umansky Apr 6 at 16:14
  • $\begingroup$ Thank you @nicoguaro for the edit. $\endgroup$ – Matt Apr 6 at 17:47
  • $\begingroup$ With closed boundary conditions if you solve this to steady state the concentration (mass per unit volume) will be constant for all finite volume cells. I think the what you are hinting at is that the volume of cells increases with radius so does that imply a concentration gradient? No. It just means that the mass I’m each cell is proportional to the volume of each cell so that the concentration is constant. There is a bit of guess work in my interpretation here. $\endgroup$ – boyfarrell Apr 9 at 9:55
  • $\begingroup$ Thanks @boyfarrell. I agree with you for solving the generic equation at steady state. Well, I though I would have different concentrations at each cell that the sum of all cells will give me the concentration at the surface knowing that we have limited source over there. $\endgroup$ – Matt Apr 9 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.