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boundary integration
Posted 2009年7月17日 GMT+2 14:551 Reply
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Hi everyone,
I have a quick question about boundary integration. Lets say we have a 2D rectangle in electrostatics, with the top and bottom of the rectangle acting as electrodes and the sides as insulating. Say a voltage of +/-5 is applied to the top/bottom electrodes, and the subdomain is conductive. After solving, you can perform a line integral of the normal current density along the top/bottom electrode, which gives you a value in A/m. Does this value indicate the average current density flowing through that boundary? So if you multiply the A/m value by the length of the boundary, will this give you the average current?
Or, is it the thickness in the undefined dimension (because we are in 2D) that is multiplied by the A/m value to get the current? So if in reality the square is a box 0.5 m thick, the current on the face of the box would be the A/m value multiplied by 0.5m?
Thanks,
Ryan
I have a quick question about boundary integration. Lets say we have a 2D rectangle in electrostatics, with the top and bottom of the rectangle acting as electrodes and the sides as insulating. Say a voltage of +/-5 is applied to the top/bottom electrodes, and the subdomain is conductive. After solving, you can perform a line integral of the normal current density along the top/bottom electrode, which gives you a value in A/m. Does this value indicate the average current density flowing through that boundary? So if you multiply the A/m value by the length of the boundary, will this give you the average current?
Or, is it the thickness in the undefined dimension (because we are in 2D) that is multiplied by the A/m value to get the current? So if in reality the square is a box 0.5 m thick, the current on the face of the box would be the A/m value multiplied by 0.5m?
Thanks,
Ryan
1 Reply
Last Post 2009年7月18日 GMT+2 01:22
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Posted:
2 decades ago
Hi Ryan
what about trying to use a little analytical calcutations on your case ? just to check:
If you take a
L=1 [m] long rectanglular wire of
w=100 [mm] width (your rectangle) and
d=1 [um] "thickness",
then you have a section of A=0.1*1e-6=1e-7[m^2],
using standard copper of conductivity around 6e7 [S/m] (or 1/6e7 [Ohm*m] resisitivity),
you should know that the total resistance is
R [Ohm] = resistivity[Ohm*m]*L[m]/A[m^2] = 1/6e7/1e-7 = 1/6 Ohms.
(the longer the cable, the higher is the resistance, the larger the area, the lower is the resistance, just as for a water pipe)
If you apply 1 V across the total length of your "cable", the Ohms law stating
U=R*I or I=U/R = 1/(1/6) = 6 [A].
This again leads to a current density of:
I/A= 6/1e-7 = 60e6 [A/m^2] = 6 [A/mm^2] or for your width of 6 [A/um] per um thickness
by the way, 6 [A/mm^2] is a safe maximum-rating for an electric cable.
So try it out in COMSOL, you will see that integrating over a length (ds) multiplies the integrant by the length dimension, and that the thickness "d" is required in addition to the integration over ds to multiply by an area.
You can try to integrate :
(int) nJ_emdc*d_emdc (ds)
Pls note that the "int" or "sum over" and "ds" are implict in COMSOL,
and I assume you are using the _emdc physics.
You will see that COMSOL will give you the total current.
Just an advise:
Always cross check your FEM calculation with a simplified Analytical approximation of your model,
and when really you must stand up and trust the FEM, have the model consolidated and validated by representative Measurements.
So long
what about trying to use a little analytical calcutations on your case ? just to check:
If you take a
L=1 [m] long rectanglular wire of
w=100 [mm] width (your rectangle) and
d=1 [um] "thickness",
then you have a section of A=0.1*1e-6=1e-7[m^2],
using standard copper of conductivity around 6e7 [S/m] (or 1/6e7 [Ohm*m] resisitivity),
you should know that the total resistance is
R [Ohm] = resistivity[Ohm*m]*L[m]/A[m^2] = 1/6e7/1e-7 = 1/6 Ohms.
(the longer the cable, the higher is the resistance, the larger the area, the lower is the resistance, just as for a water pipe)
If you apply 1 V across the total length of your "cable", the Ohms law stating
U=R*I or I=U/R = 1/(1/6) = 6 [A].
This again leads to a current density of:
I/A= 6/1e-7 = 60e6 [A/m^2] = 6 [A/mm^2] or for your width of 6 [A/um] per um thickness
by the way, 6 [A/mm^2] is a safe maximum-rating for an electric cable.
So try it out in COMSOL, you will see that integrating over a length (ds) multiplies the integrant by the length dimension, and that the thickness "d" is required in addition to the integration over ds to multiply by an area.
You can try to integrate :
(int) nJ_emdc*d_emdc (ds)
Pls note that the "int" or "sum over" and "ds" are implict in COMSOL,
and I assume you are using the _emdc physics.
You will see that COMSOL will give you the total current.
Just an advise:
Always cross check your FEM calculation with a simplified Analytical approximation of your model,
and when really you must stand up and trust the FEM, have the model consolidated and validated by representative Measurements.
So long