| Author | Message | ||
     Andrey S. Kozhukhov (andrey) New member Username: andrey Post Number: 1 Registered: 04-2008 |
Hello. While trying to solve Richards equation for groundwater, I came across a little problem. While water flows into the domain area, it also should leave it according to the fact that outside my domain there is also ground. And so the flux should match saturation outside. Can you please give me a piece of advice on how to solve it, or I should just expand the domain? The script is in the file | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 1094 Registered: 06-2003 |
In your system, the Natural BC specifies the value of -K*grad(h), the outward water flux at the surface (See "Natural" in the Help Index). Natural(h)=0 therefore means a closed system, with no flow out the boundaries. A nonzero specification for Natural(h) will allow flow across the boundary. Since FlexPDE cannot compute conditions outside the computation domain, you must use some ad hoc approximation to determine the flux. A couple of possibilities come to mind: 1) Natural(h)= P*(h-h0) This formulation will drag h toward h0 with a strength "P". If you think of h0 as being positioned at a distance D from the boundary, then a difference approximation -K*grad(h)~K*(h-h0)/d implies P=K/d. 2) Natural(h)=-Normal(K*grad(h)) This formulation is not really a boundary condition at all, in that it supplies no new information to make the solution unique. It is "tautological", because it merely supplies the value of the surface integrals generated by the Divergence Theorem, and completes the equation system for whatever gradient appears at the boundary. Both of these formulations will work best if the boundary is far away and circular instead of square. There is other trouble here, however. Both formulations run into trouble at late time. You should examine the expressions for K and C to see if there is potential for cancellation errors leading to oscillation. |