Hi, thanks for your reply.
I copied the files in the install dir (right subdir) and did a new install of fipy (completely new install on other pc), it should be ok but I tried running the code and the result is the same as before and so is wrong. I am doing some error? Thanks > Date: Thu, 13 May 2010 16:51:39 -0400 > From: [email protected] > To: [email protected] > Subject: Re: Problem in solving Poisson equation with 1D Cylindrical mesh > > > With any luck this has now been dealt with. Latest versions of > branches/version-2_1 and trunk/ should now have the fixes for > <http://matforge.org/fipy/ticket/294> > > On Tue, May 11, 2010 at 5:44 PM, Eduard Manley <[email protected]> wrote: > > Hi, > > thanks for your reply. > > > > You can find attached a sample problem. > > > > The equation is a heat equation, variable is temperature,other and transient > > and diffusion term, there is a spatially varying source term (q) computed > > by analytical expression which is and hyperbola. > > > > We should reach about 100°C in 30-35 sec and the shape of the solution > > should be similiar to and hyperbola. > > > > As it is now the file works fine (mesh declared with: mesh = > > CylindricalGrid1D(dr=dr, nr=(len(DR))) + (r_int,) ) but if you define the > > mesh with ( mesh = CylindricalGrid1D(dx=DR) + (r_int,) ) the result you > > obtain are wrong. > > > > The 2 declaration of CylindricalGrid1D should be the same, the face and cell > > centers are the same, but the result are different. > > > > You can observe the same problem also when solving the same equation > > but without transient term. > > > > Thanks > > Eduard > > > > > > > >> Date: Tue, 11 May 2010 10:45:24 -0400 > >> From: [email protected] > >> To: [email protected] > >> Subject: Re: Problem in solving Poisson equation with 1D Cylindrical mesh > >> > >> > >> Can you put together the simplest script possible that demonstrates > >> the problem and I'll try and debug it? Thanks. > >> > >> On Thu, May 6, 2010 at 2:01 PM, Eduard Manley <celez1...@hotmailcom> > >> wrote: > >> > Thanks for your reply. > >> > > >> > I probably found the reason of the problem. > >> > > >> > As said before I'm trying to solve an eq of this type: > >> > > >> > A(d phi/d t) = div (D grad phi) + q > >> > > >> > where A and D are costant coefficient and q is a spatially varying heat > >> > source. > >> > > >> > The problem is in how I create the cylindrical 1D mesh (origin of the > >> > mesh > >> > is not in 0.). > >> > > >> > It doesn't matter if the discretization is logarithmic or uniform but > >> > how I > >> > declare it: > >> > > >> > (using a uniform spacing:) > >> > > >> > ** mesh = CylindricalGrid1D(dx=dr, nx=(len(DR))) + (r_int,) ** > >> > SHOULD BE EQUAL TO: > >> > > >> > ** mesh = CylindricalGrid1D(dx=DR) + (r_int,) ** > >> > > >> > [dr = 5e-04, nr=58, r_int=0.00125, DR is a list which contains the > >> > various > >> > dx(58 elements of value dr for uniform grid)] > >> > > >> > BUT It is NOT. > >> > > >> > The mesh (cell centers, facecenters) is ok but the results are NOT. > >> > The results are right only if I create the mesh using dx=dr and nx=.. . > >> > And this is why before I thought the problem was the logaritmic > >> > discretization (must use DR=[...]). > >> > > >> > As said before with Cylindrical 2D mesh results are instead correct. > >> > Is this a bug? > >> > > >> >> Date: Thu, 6 May 2010 11:11:10 -0400 > >> >> From: [email protected] > >> >> To: [email protected] > >> >> Subject: Re: Problem in solving Poisson equation with 1D Cylindrical > >> >> mesh > >> >> > >> >> > >> >> Edward, This may be to do with having a very small volume (or area or > >> >> line or point) for the inner most element of the domain. It should be > >> >> the same whether one is using a 1D or 2D mesh. Since you are getting > >> >> differences in the 1D and 2D case, it should be relatively easy to > >> >> debug and figure out what's going on It could also be that the > >> >> boundary condition on the inner boundary as zero area and this is > >> >> causing issues. Try shifting the grid by a small value away from the > >> >> zero point and see if things are improved. I have always had this > >> >> issue with cylindrical grids and have never really had a satisfactory > >> >> solution (other than shifting away from the zero point). If you > >> >> discover a better way to handle this, let me know. Cheers. > >> >> > >> >> > >> >> If you can't debug it, then send me the most minimalist scripts that > >> >> show the issue and I'll give it a shot. > >> >> > >> >> On Wed, May 5, 2010 at 9:53 PM, Eduard Manley <[email protected]> > >> >> wrote: > >> >> > Problem partially solved: > >> >> > > >> >> > I'm using a logarithmic discretization (first dr= 5e-04 and next > >> >> > dr increasing as 1.05)(with internal radius= 0.00125, external > >> >> > radius=0.03) > >> >> > and, > >> >> > for some unknown reason this create problems and wrong result with > >> >> > cylindrical 1D mesh. I tried using a uniform discretization (dr=5e-04 > >> >> > nx=58) > >> >> > and now the result is correct. > >> >> > > >> >> > However I need to use the logarithmic discr. so after some hours of > >> >> > sleep > >> >> > I'll think about the reason..... > >> >> > > >> >> > ________________________________ > >> >> > Hotmail: Trusted email with powerful SPAM protection. Sign up now. > >> >> > >> >> > >> >> > >> >> -- > >> >> Daniel Wheeler > >> >> > >> >> > >> > > >> > ________________________________ > >> > Your E-mail and More On-the-Go. Get Windows Live Hotmail Free Sign up > >> > now. > >> > >> > >> > >> -- > >> Daniel Wheeler > >> > >> > > > > ________________________________ > > Hotmail: Trusted email with Microsoft’s powerful SPAM protection. Sign up > > now. > > > > -- > Daniel Wheeler > > _________________________________________________________________ Hotmail: Free, trusted and rich email service. https://signup.live.com/signup.aspx?id=60969
