| Author | Message | |||
     Ranjan Radhamohan (mrranjan2) Member Username: mrranjan2 Post Number: 4 Registered: 07-2004 |
Hi, I have a set of experimental data and another FEA software package, Microwave Studio, that confirms it. I can't seem to get the same results in FlexPDE, and I was wondering if you could see anything clearly wrong in my code. I think my setup is similar to the 3D oil drum example, so I've tried to use it as a guide. My model is inside-out, like the oil drum (the hollow vacuum cavity shows up as solid on a grid plot). I would like the surrounding area to be a conductor, like an oil drum's metal frame. It may be redundant, but I'm also trying to place an electric wall at the top and bottom surfaces (electric field is perpendicular to surface). One of my modes matches the experimental data - the 75 GHz one. But the first mode should be ~61 GHz. Thank you very much, Ranjan
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 205 Registered: 06-2003 |
If you look at the first plot (contour(Ez) on x=0 for mode #1), you will see that the top and bottom faces of the leftmost structure are not zero. If these are supposed to be conductive surfaces, they will need a Value(E)=0 boundary condition. You have imposed Value(E)=0 on the extreme top and bottom surfaces of the figure, but these faces are absent in that leftmost stack, so the BC is not applied. Use SURFACE 2 Value(E)=0 and SURFACE <?> Value(E)=0 in the void region to impose these conditions. | |||
| Ranjan Radhamohan (mrranjan2) Member Username: mrranjan2 Post Number: 5 Registered: 07-2004 |
I tried imposing Value(E) = 0 on all horizontal conductive surfaces, but unfortunately the results are farther off. I'm not sure I understand how FlexPDE deals with vectors. Is 'E' (or 'u' in the oil drum example) a 3D vector? When I have the program tell me XCOMP, YCOMP, and ZCOMP of E, only XCOMP is nonzero (this is the case for all modes). In a 3D problem, especially a asymmetric one, shouldn't I be seeing nonzero components of E for Y and Z, too (not necessarily for all modes, but some)? I've tried using three equations: Ex: del2(Ex) + lambda*Ex = 0 Ey: del2(Ey) + lambda*Ey = 0 Ez: del2(Ez) + lambda*Ez = 0 But I'm not sure how to tell FlexPDE that Ez is a vector only the z-direction, etc. The 'VECTOR' command only seems to work in definitions. And also, with three equations (or three components), I can't see how to set BCs for the walls of a conducting cylinder, since simply setting Value(Ez) = 0 would allow many non-perpendicular vectors with components of Ex and Ey at the conducting surface. Thanks again, Ranjan | |||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 206 Registered: 06-2003 |
1. Examine your output plots to determine if you have in fact imposed the boundary conditions you think you have. 2. All variables in FlexPDE are scalars (See "Problem Descriptor Reference | The Sections of a Descriptor | Variables"). If you have a vector field, you must write a scalar equation for each vector component, as you have shown in your posting. A vector is defined as, for example, as Ex*1x+Ey*1y+Ez*1z, where 1x, 1y and 1z are unit vectors in the coordinate directions. In this definition, Ex, Ey and Ez are scalars. A vector with only a single component, such as Ez, means you have one variable, Ez, and one scalar equation, del2(Ez)+lambda*Ez=0. This is where you started out. The interpretation of Ez as the z component of a vector is manifested in the way you deal with it in creating your component equations and boundary conditions, in summarizing data, etc. FlexPDE sees it as a scalar field. (There is an unfortunate confusion introduced in FlexPDE by the use of the name "SCALAR VARIABLE" to describe single valued,or "Global" variables. Ordinary variables are field quantities represented internally as node-dimensional vectors. Ignore this use of the word SCALAR for now.) |
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