Unformatted text preview: Square Coax FlexPDE Solution There are a few different ways to solve this problem using FlexPDE. In all cases there is one region that defines the outer conductor boundary. The methods differ in how we describe the inner conductor. The first method is the easiest: just define a constant voltage path inside the first region (called “inner” in the following script). The charge is then evaluated by integrating the normal field component over this inner path: { SquareCoax.PDE } { This problem computes the field and capacitance in a square coaxial line. } Title 'Square Coaxial Line' Variables V ! The electrostatic potential Definitions a=0.03 ! Outer conductor width b=0.03 ! Outer conductor height w=0.01 ! inner conductor width t=0.01 ! inner conductor height il=(a
w)/2; ih=(b
t)/2; V0=1 ! center conductor voltage eps0=8.854e
12 ! Free
space permittivity er=1 ! Dielectric constant of filled region eps=er*eps0 ! permittivity variable used in calculations { Here we find the charge and capacitance } Qt=er*eps0*LINE_INTEGRAL(Normal(grad(V)),'inner') Cap =Qt/V0 Equations DIV(
eps*GRAD(V)) = 0 ! Laplace's equation Boundaries REGION 1 eps=er*eps0 Start(0,0) Value(V)=0 Line to (a,0) to (a,b) to (0,b) to close! The grounded sides Start "inner" (il,ih) Value(V)=V0 Line to (il+w,ih) to (il+w,ih+t) to (il,ih+t) to close ! The inner conductor Plots Contour(V) Report(Cap) End The second method, perhaps the best, is to define the inner conductor using an EXCUDE region. This essentially specifies that it is only the space between the conductors that matters for the computation: Boundaries REGION 1 'box' eps=eps0 Start(0,b) Value(V)=0 Line to (0,0) to (a,0) to (a,b) close ! The grounded sides EXCLUDE 'inner box' Start "inner" (il,ih) Value(V)=V0 Line to (il+w,ih) to (il+w,ih+t) to (il,ih+t) close ! Inner box held at potential V0 Both of the previous scripts give the correct answer: Square Coaxial Line
e3 30. 16:26:33 10/24/10 FlexPDE 5.1.4
V max u: t: s: r: q: p: o: n: m: l: k: j: i: h: g: f: e: d: c: b: a: min 1.00 1.00 0.95 0.90 0.85 0.80 0.75 0.70 0.65 0.60 0.55 0.50 0.45 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 0.00 25. m p 20. g k x o j n s Y 15. t f 10. r q l 5. i e 0. 0. 5. b 10. c o 15. 20. 25. 30. e3 d h X SquareCoaxTwoRegionExclude: Grid#2 P2 Nodes=692 Cells=316 RMS Err= 0.002 Cap= 5.145760e11 Integral= 2.582217e4 The capacitance is reported as 5.14e
11, which is 51.4 pF/m. As a sanity check to see whether this answer makes sense, lets compare with the cylindrical coaxial system which has a known analytic result (take the outer conductor radius as b/2 and inner conductor radius as w/2): C≈ 2πε 0 = 50.6 pF/m ln(b / w ) This gives us some confidence in our answer. The third possible method uses a second Region statement to define the inner conductor. This is not an advisable approach but it can be made to work if the region and paths are defined carefully. Here is the method that works: REGION 2 'inner box' Start "inner" (il,ih) Value(V)=V0 Line to (il,ih+t) to (il+w,ih+t) to (il+w,ih) close ! Inner box held at potential V0 And here is the one that doesn’t: REGION 2 'inner box' Start "inner" (il,ih) Value(V)=V0 Line to (il+w,ih) to (il+w,ih+t) to (il,ih+t) close ! Inner box held at potential V0 If you try to run these scripts you will see that FlexPDE correctly calculates the potential everywhere, but it reports a charge density that is many orders of magnitude lower in the second case relative to the first. Can you spot the difference in the two scripts? It is subtle: in the second script the inner boundary is defined counter
clockwise, and in the first it is clockwise. In vector calculus the normal direction for a close path is defined to the left as one moves along the path. However, here is a case where it looks like we defined everything in the right direction but it still gives the wrong answer. The only difference here is that we define the path separately REGION 2 'inner box' Start (il,ih) Value(V)=V0 Line to (il,ih+t) to (il+w,ih+t) to (il+w,ih) close ! Inner box held at potential V0 Start "inner" (il,ih) Line to (il,ih+t) to (il+w,ih+t) to (il+w,ih) close ! The inner conductor This script gives an answer that is off by a factor of 2. It is not exactly clear why. A “REGION” is an area of the problem where we want the program to solve for the fields. So when we define a path at the interface between two previously
defined regions, it is possible that the program interpolates the field as the average of the field on either side of the boundary. ...
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This note was uploaded on 12/04/2010 for the course ECE 134 taught by Professor York during the Fall '08 term at UCSB.
 Fall '08
 York

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