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     Mahir Celik (mahir) Member Username: mahir Post Number: 6 Registered: 06-2003 |
Hello Mr. Nelson, I want to simulate for a simple case the potential distrubition in an inhomogenous object considering displacement current. I assumed, that my current source is Sin and harmonic . As a result the potential-distrubition will complex. So I have seperated my Equation in real an imagine part for potential.But the results are strange.I suppose the error may be by definition of natural Bc. I hope you can help me.Below is the pde file and i also attached a word document including the physical equation. Thanks for your Help Mahir Celik TITLE 'Considering displacement current ' SELECT errlim 0.1 COORDINATES CARTESIAN3 VARIABLES Vr Vi DEFINITIONS Ke=56e4 { Conductivity of Cu S/cm } K1=100e-8 { Conductivity of material 1 S/cm } K2=50e-8 { Conductivity of material 2 S/cm } K Eps1=5 { Diel. material 1 F/cm} Eps0=8.85e-14 { Diel. Vakum F/cm } Eps2=10 { Diel. material 2 F/cm} Eps L=5 { Length of metal plate } d=0.1 { Thickness of metal Plate } H=10 { Thicknes of Material between Plates } I=10e-3 { Electrical Current } En=I/(Ke*L^2) { Normal component of E-field } omega=2*Pi*50e3 { Angular Frequency } Er=-grad(Vr) { Real part E-field strength } Erm=magnitude(Er) Ei=-grad(Vi) { Imagine part E-field strength } Eim=magnitude(Ei) E=vector(Erm,Eim) { Total E-field strength } Em=magnitude(E) Jr=K*E { Real Part Current Densinity } Jrm=magnitude(Jr) Ji=omega*Eps*E { Imagine Part Current Densinity } Jim=magnitude(Ji) Ir=SINTEGRAL(Jrm,"BOTTOM") { Calculated realt Part of el. Current } Iim=SINTEGRAL(Jim,"BOTTOM") { Imagine Part of el. Current } Iges=sqrt(Ir^2+Iim^2) { Total current, should be equal to I=10e-3} U1=vector(VAL(Vr,0,0,20),Val(Vi,0,0,20)) { Total Potential at the first metall plate } U2=vector(VAL(Vr,0,0,10),VAL(Vi,0,010)) { Total Potential at the borderline between the two diffrent materials} U3=Vector(VAL(Vr,0,0,0),VAL(Vi,0,0,0)) { Total Potential at the first metall plate } Zimp=(U1-U3)/Iges { Impedance of the system } PhaseI=ATAN2(Jim,Jrm)*180/PI { Phase of the current } PhaseV=ATAN2(Vi,Vr)*180/PI { Phase of Potential } PhaseZ=PhaseV-PhaseI { Pahse of Impedance } Pthermu=integral(K*E^2,1,"MATERIAL 1" ) { Heating Power in material 1 } Rwu=Magnitude((U2-U3))^2/Pthermu { resistance of material 1 } Pthermo=integral(K*E^2,1,"MATERIAL 2" ) { Heating Power in material 1 } Rwo=(Magnitude(U1-U2))^2/Pthermo { resistance of material 1 } Wel1=integral(Eps*E^2,1,"MATERIAL 1" ) { stored energy in material 1} C1=2*Wel1/(Magnitude(U2-U3))^2 { Capacitance of material 1 } Wel2=integral(Eps*E^2,1, "MATERIAL 2" ) { stored energy in material 2} C2=2*Wel2/(Magnitude(U1-U2))^2 { Capacitance of material 2 } EQUATIONS div(-K*grad(Vr))+div(Eps*omega*grad(Vi))=0 div(-K*grad(Vi))-div(Eps*omega*grad(Vr))=0 EXTRUSION SURFACE "BOTTOM" Z=0 LAYER "PLATE 1" SURFACE "METAL" Z=d LAYER "MATERIAL 1" SURFACE "PLATE2 BOTTOM" Z=d+H LAYER "MATERIAL 2" SURFACE "METALL 2" Z=d+H*2 LAYER "PLATE 2" SURFACE "TOP" Z=2*(d+H) BOUNDARIES SURFACE "TOP" NATURAL(Vr)=I/25 NATURAL(Vi)=0 SURFACE "BOTTOM" NATURAL(Vr)=-I/25 NATURAL(Vi)=0 REGION 1{ Describes the Glas} LAYER "PLATE 1" Eps=0 K=Ke LAYER "MATERIAL 1" Eps=Eps1*Eps0 K=K1 LAYER "MATERIAL 2" Eps=Eps2*Eps0 K=K2 LAYER "PLATE 2" Eps=0 K=Ke start(0,0) Line TO (L,0) LINE TO(L,L) LINE TO(0,L) Line TO FINISH PLOTS GRID(X,Z) ON SURFACE Y=0 CONTOUR(Vr) ON SURFACE Y=0 AS "POTENTIAL Vr" CONTOUR(Vi) ON SURFACE Y=0 AS "POTENTIAL Vi" CONTOUR(Magnitude(Zimp)) ON SURFACE Y=0 AS "IMPEDANCE" CONTOUR(Eps) ON SURFACE Y=0 PAINTED AS "MATERIAL" CONTOUR(Iges) ON SURFACE Y=0 CONTOUR(Ir) ON SURFACE Y=0 VECTOR(Jr) norm ON SURFACE Y=0 VECTOR(Ji) norm ON SURFACE Y=0 CONTOUR(C1) ON SURFACE Y=0 CONTOUR(C2) ON SURFACE Y=0 END
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| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 32 Registered: 06-2003 |
1. Applying the divergence theorem to your equations produces the meaning of the Natural boundary conditions for your equations: Natural(Vr) means the outward normal component of -K*grad(Vr)+Eps*omega*grad(Vi). Natural(Vi) means the outward normal component of -K*grad(Vi)-Eps*omega*grad(Vr). This same process determines the interior interface conditions, which is by default that the sum of the Naturals=0 (remember the direction of the normal is reversed as seen from the two sides of a boundary). In other words, the interior continuity is: Normal(K*grad(Vr)+Eps*omega*grad(Vi)) continuous and Normal(-K*grad(Vi)-Eps*omega*grad(Vr)) continuous. If this is not correct for your problem, then you must rearrange your equations, or insert interior and exterior BCs to produce the correct values of these quantities. 2. You have specified only natural BC's, which means that strictly speaking the system is ill-posed. When you specify only derivative boundary conditions, there are an infinity of solutions, differing by an additive constant. Sometimes FlexPDE can succeed in spite of this fact, because it uses a perturbation method to find the changes necessary to bring the initial guess into agreement with the PDE. But there are no guarantees in this case. If you really intend your problem to be totally floating, you should put in an integral constraint or two to make the solution unique. |
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