| Author | Message | |||
     Matthew Graham (grahamm2) New member Username: grahamm2 Post Number: 2 Registered: 08-2004 |
-------------------------------------------------------------------------------- Hello! I am attempting to solve a two variable system of PDE (Te & Tl) that is dependent on x & t. I am imposing value boundary condition at each spatial end: Value (dx(Te)) = 0, x=0 Value (dx(Te)) = 0, x=20 Value (dx(Tl)) = 0, x=0 Value (dx(Tl)) =0, x=20 I don't understand how to use 'line to' here in one dimenstion. (N.B. solving a two step lattice heat conduction eqution : need solution for Te(x,t) & Tl(x,t), Cc*dt(Te)=dx((Te/Tl)*K*dx(Te))-G(Te-Tl)+S dt(Tl)=G*(Te-Tl)/Cl I have seen the one_dim.pde file, but do not know what to set the y boundries to (I don't care about them, but physical dy(Te)=0 is incorrect- not insulating in y direction) THANKS | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 202 Registered: 06-2003 |
FlexPDE does not have a one-dimensional mode, so we customarily use a thin strip in 2D to solve one-dimensional systems. This means that the y-dimension of the domain is something small compared to the x-dimension, and you lay it out as a long, thin rectangle. Yheight=Xlength/100 is a reasonable size. "START(0,-Yh/2) LINE TO (Xl,-Yh/2) TO (Xl,Yh/2) TO (0,Yh/2) TO FINISH" works well, allowing Y=0 to be the center line. Your remark about Value(dx(Te)) is confusing. In FlexPDE parlance, a VALUE boundary condition is a Dirichlet condition, or a forced value of the variable. You cannot apply a VALUE condition to a derivative. Derivative boundary conditions are controlled by the NATURAL boundary condition statement (See NATURAL BOUNDARY CONDITIONS in the Help index). In the notes at the top of the ONE_DIM.PDE script, we point out that there is an added dyy() term to generate the reflective natural boundary condition in the y direction. Since the default boundary condition is NATURAL()=0, reflection in y is automatic with the inclusion of the dyy() term. Your equation dt(Tl)=G*(Te-Tl)/Cl is a point equation, and as such can be subject to oscillation. This is because the finite element discretization of the system uses integrals over the mesh cells. In the absence of derivative terms, the integral equations can accept a large number of oscillatory solutions. It is therefore advisable to add a small div(grad(Tl)) term to damp the oscillation. This also provides meaning to a NATURAL BC for Tl, allowing reflective boundaries top and bottom. | |||
| Matthew Graham (grahamm2) Junior Member Username: grahamm2 Post Number: 3 Registered: 08-2004 |
Thank you for your advice, things are starting to make more sense. I am still having t-dep issues. A t=1 there is an expected temperature jump, as my gaussian source term (S) dumps heat in. FlexPDE just registers a delta-like jump at that time, there is no useful information before or after t=1. I solved a uncoupled PDE version of the above in Maple numerically & analytically, so I know what to expect/have constants correct. Do you think that FlexPDE is having trouble solving about t=1. THANKS
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| Matthew Graham (grahamm2) Member Username: grahamm2 Post Number: 4 Registered: 08-2004 |
I have not been able to resolve my problem, it isn't respecting my initial conditions. The general shape is correct, but it is jumping to Te=1e7 range, when I set Te=293 as the initial value. Any thoughts?
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 209 Registered: 06-2003 |
If you add a history plot on Te, as for example HISTORY(Te) AT (10,0) You will see that Te stays at your initial 293 until time=0.88, at which point it begins to climb like a rocket. Check the arithmetic and units of your source term. Also, you have the variable assignments of the equations backward. The equation with dt(Tl) and div(grad(Tl)) should be assigned to Tl. |
