| Author | Message | |||
     sk484@cam.ac.uk (seongmin) Junior Member Username: seongmin Post Number: 3 Registered: 03-2006 |
Hi, Can you tell me how to modify the code for transient viscous flow using 3d cylindrical coordinates? There was some error message when just putting time factor like dt(vr) and dt(vz) ..etc pls sort this out. | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 598 Registered: 06-2003 |
If you want to do 3D, you don't need cylindrical coordinates. The advantage of cylindrical coordinates in 2D is that you can build in the r-dependence of volume and surface area without doing a full 3D model. Cartesian coordinates in 3D fully capture the volume and area factors. If you have an equation expressed in 3D cylindrical coordinates and you don't know how to express the same thing in Cartesian coordinates, you can use the definition facilities of FlexPDE to convert the derivative operators. (See "Samples | Misc | Polar.pde" for an example of how to do this.) There are simple algebraic conversions from cartesian positions and velocities to cylindrical positions and velocities. You can apply these to convert your equations, or you can solve in cartesian geometry and convert to cylindrical for display. Alternatively, you can tell FlexPDE that the coordinate names are "R,Z,Theta" and write the appropriate correction terms in your pde's. Your plots will look funny, because R,Z and Theta will be plotted as cartesian coordinates. But the solution will be correct if you write the equations correctly. | |||
| sk484@cam.ac.uk (seongmin) Member Username: seongmin Post Number: 4 Registered: 03-2006 |
Thanks but I mean this I have a error message in 3d cylindrical coordinates. Time step has fallen below 1e-009 of starting value. you may have a temporal discontinuity in parameters. etc.... HOw can I sort this out? | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 599 Registered: 06-2003 |
See my response to the 4/28/06 posting "reducing errlim below 1e-8 gives error: time step fallen below 1e-9" for a discussion of what this error means. But I don't know what you mean by 3d cylindrical coordinates. FlexPDE doesn't support 3d cylindrical coordinates directly. | |||
| sk484@cam.ac.uk (seongmin) Member Username: seongmin Post Number: 5 Registered: 03-2006 |
Hi, I mean this: (rho, z) coordinate is giving 3d result so that's why I called it 3d cyclindrical. Anyway. I am attaching file, pls check this out. I could not get time plot for history of time. popattach{1623,transient flow file.} | |||
| sk484@cam.ac.uk (seongmin) Member Username: seongmin Post Number: 6 Registered: 03-2006 |
There is something wrong. I am attaching again!!
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 600 Registered: 06-2003 |
FlexPDE will try to solve the problem you pose. If this posed problem has no solution, then FlexPDE will fail to invent one. We cannot try to second-guess you and alter your prescription to try to find something that actually has a solution. This is your job. You should reconsider what you have provided as boundary conditions on the pressure equation. You have said the initial value of pressure is p1, and that the outward normal derivative of pressure on all boundaries is also p1. 1) a PDE with all natural boundary conditions and no internal source or sink dependent on the value is ill-posed. It has an infinite number of solutions differing by an additive constant. 2) The pressure equation is merely a diffusion of Div(V). To insist that the pressure be always rising at the boundaries demands an infinitely rising value in time. Since there is no initial timestep at which these problems can be resolved, FlexPDE cuts the timestep and retries until the cutoff stepsize is reached. If you repose the problem as one which actually has a solution, I am sure you have a more satisfying experience. |
Transient flow in closed tube