Note: This discussion is about an older version of the COMSOL Multiphysics® software. The information provided may be out of date.

Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.

Comsol 4.0 : electric currents - el. potential V continuity between two adjacent domains?

Please login with a confirmed email address before reporting spam

Hello all,

I have a fairly simple yet bizarre problem with my Comsol model.

I am using Comsol 4.0, 2D, Electric Currents.

I have an annulus (a ring) that's split in two domains, each with a different conductivity and permittivity (but constant).
On one side of the annulus for the boundaries there I have an Electric Potential set, and on the other side of the annulus for the boundaries there I have a Ground set.

That's pretty much my model. When I run it in stationary mode, I have a bizzare solution, which can only makes sense if the potential isn't continuous between the internal boundaries inside the ring, between the two domains. Even more so, if I do a plot of Electric Field I notice the huge discrepancies at the borders of the domains, like E(domain1) ~= E(domain2) which should be false!

Analitically, I should have the potential as:
For the innermost part of the ring, the one that has a boundary with electric potential, the potential inside this domain is defined as:
V1(r) = V_boundary + (V_boundary - V_ground) * log(r / r_min) / log (r_min/ r_max )
For the outermost part of the ring, the one that has a boundary with ground set, the potential inside this domain is defined as:
V2(r) = V_ground + (V_boundary - V_ground) * log( r / r_max ) / log ( r_min / r_max )

Where V_boundary is the potential, V_ground is the ground, r is the radius of the point where I want to calculate the potential, r_min is the innermost part of the ring and r_max is the outermost part of the ring.

Plotting this for
1) V_boundary of 40.000 V,
2) r_min = 0.02 m,
3) r_max = 0.10 m
4) each domain having 0.04 m
(so 0.02 -> 0.06 m is domain 1, and 0.06 -> 0.10 m is domain 2)
5) sigma_domain1 = 1e-15 S/m
6) sigma_domain2 = 1e-16 S/m
7) epsilon_r_domain1 = 2.2 -
8) epsilon_r_domain2 = 3.5 -

I **should** have something like this:
-- Attached file : Image_analytical


Using this model, or using its counterpoint axi-symmetric 2D model, I keep getting this result:
-- Attached file : Image_comsol

I've tried using Union mode and Assembly mode for creating the geometry, but alas, it's still the same. I've attached all 3 models which basically all give the same result, different than the one I should be getting analytically.

Am I doing something wrong?

Please advise.
Thank you!


5 Replies Last Post 2014年6月13日 GMT-4 10:36
Edgar J. Kaiser Certified Consultant

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2014年6月12日 GMT-4 12:27

Hi,

actually I would expect the COMSOL result and not your analytical calculation. The potential is obviously continuous at the internal boundary, but with a kink,which is to be expected, because the conductivity changes. And thus the field is discontinuous.

Cheers
Edgar

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Hi, actually I would expect the COMSOL result and not your analytical calculation. The potential is obviously continuous at the internal boundary, but with a kink,which is to be expected, because the conductivity changes. And thus the field is discontinuous. Cheers Edgar -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2014年6月13日 GMT-4 05:20
Dear Edgar,

Thank you for your reply.

However, I can not agree with you on this matter, considering the following:
- the conductivity, as its constant in regards to space and time (does not depend on "r" or on "t"), does not interfere in the expression of potential V, nor in the expression of field intensity E.

I may be wrong, but can you please correct me?
The regime I am using (electric currents, stationary) isn't governed by these expressions? :
1. E = - grad V
2. div J = 0
3. J = sigma * E

And that my 2D geometry has symmetry and can be considered as a 1D geometry with variations of just "r", my expression: [ div ( sigma * ( - grad V ) ) = 0 ] turns into ( in cylindrical coordinates, with variations just on the "r" axis ):

( 1 / r ) * [ (d / dr) * r * sigma * [ ( d/dr) * - V ] ] = 0.


if I am considering that expression and the expressions I've mentioned above in the post regarding properties of material and values of expressions at geometric borders, more specifically:
i) V_domain1 ( r = r_min ) = V_potential
ii) V_domain2 ( r = r_max ) = V_ground
iii) V_domain1 (r = r_med ) = V_domain2 ( r = r_med)
iv) E_domain1 (r = r_med ) = E_domain2 ( r = r_med)
with r_med being the radii of the border separating the two domains within the ring
( from 0.02 m to 0.06m the first domain, and from 0.06 to 0.10 m the 2nd domain )

if I integrate considering all of the above I will have:

( 1 / r ) * [ (d / dr) * r * sigma * [ ( d/dr) * - V ] ] = 0
=> [ d/dr * r * sigma * [ (d/dr) - V(r) ] ] = 0 ( simplifications by 1/r as the expression in the right equals to 0 )
=> [ d/dr * r * sigma * [ (d/dr) V(r) ] ] = 0 ( simplifications by a minus sign)
=> r * sigma * [ d/dr V(r) ] = - ConstantA ( integrating the previous expression, on the left removing the d/dr and on the right integrating leaves a constant )
=> [ d/dr V(r) ] = ConstantA / ( r*sigma ) ( defining d/dr V(r) as whatever else remains)
And considering that E(r) equals to minus d/dr V(r) then we'll have E(r) equals to minus the expression on the right.
Then, if we integrate another time, we'll have V(r) as
=> V(r) = ConstantA * log(r) /sigma + ConstantB

Thus, I'll be left with:
E_domain1(r) = - Constant1 / (sigma_domain1 * r)
V_domain1(r) = Constant1 * log(r) / sigma_domain1 + Constant2

E_domain2(r) = - Constant3 / (sigma_domain2 * r)
V_domain2(r) = Constant3 * log(r) / sigma_domain2 + Constant4

With the expressions of the constants as following: (deduced from the system of equations i)- iv) )
Constant1 = (V_potential - V_ground) * sigma_domain1 / log(r_min / r_max)
Constant2 = V_potential - (V_potential - V_ground) * log(r_min) / log (r_min / r_max)
Constant3 = (V_potential - V_ground) * sigma_domain2 / log (r_min / r_max )
Constant4 = V_ground - (V_potential - V_ground) * log(r_max) / log(r_min / r_max)

And introducing the constants in the expressions of E and V I will be left with:

V_domain1(r) = V_potential + (V_potential - V_ground) * log(r / r_min) / log(r_min / r_max)
V_domain2(r) = V_ground + (V_potential - V_ground) * log(r / r_max) / log(r_min / r_max)

E_domain1(r) = E_domain2(r) = - (V_potential - V_ground) / ( r * log (r_min / r_max) )


Thus, as the material properties aren't varying with coordinates they will not interfere in this regime and I should have a smooth potential curve from one side to the other of the ring, governed only by the potentials of the two extremities, the fraction between the r_min and r_max and by the "r" of the point where the potential is to be found.

Are you saying that Comsol is using other expressions, and not these that I have mentioned above?
Or are you saying that I may be mistaking the regime that I am using for another one, and that my analytically expressions above aren't the correct expressions for this regime?

Thank you for your answer again, and hopefully you (or someone else) can provide an additional answer for what I wrote in this message.
Dear Edgar, Thank you for your reply. However, I can not agree with you on this matter, considering the following: - the conductivity, as its constant in regards to space and time (does not depend on "r" or on "t"), does not interfere in the expression of potential V, nor in the expression of field intensity E. I may be wrong, but can you please correct me? The regime I am using (electric currents, stationary) isn't governed by these expressions? : 1. E = - grad V 2. div J = 0 3. J = sigma * E And that my 2D geometry has symmetry and can be considered as a 1D geometry with variations of just "r", my expression: [ div ( sigma * ( - grad V ) ) = 0 ] turns into ( in cylindrical coordinates, with variations just on the "r" axis ): ( 1 / r ) * [ (d / dr) * r * sigma * [ ( d/dr) * - V ] ] = 0. if I am considering that expression and the expressions I've mentioned above in the post regarding properties of material and values of expressions at geometric borders, more specifically: i) V_domain1 ( r = r_min ) = V_potential ii) V_domain2 ( r = r_max ) = V_ground iii) V_domain1 (r = r_med ) = V_domain2 ( r = r_med) iv) E_domain1 (r = r_med ) = E_domain2 ( r = r_med) with r_med being the radii of the border separating the two domains within the ring ( from 0.02 m to 0.06m the first domain, and from 0.06 to 0.10 m the 2nd domain ) if I integrate considering all of the above I will have: ( 1 / r ) * [ (d / dr) * r * sigma * [ ( d/dr) * - V ] ] = 0 => [ d/dr * r * sigma * [ (d/dr) - V(r) ] ] = 0 ( simplifications by 1/r as the expression in the right equals to 0 ) => [ d/dr * r * sigma * [ (d/dr) V(r) ] ] = 0 ( simplifications by a minus sign) => r * sigma * [ d/dr V(r) ] = - ConstantA ( integrating the previous expression, on the left removing the d/dr and on the right integrating leaves a constant ) => [ d/dr V(r) ] = ConstantA / ( r*sigma ) ( defining d/dr V(r) as whatever else remains) And considering that E(r) equals to minus d/dr V(r) then we'll have E(r) equals to minus the expression on the right. Then, if we integrate another time, we'll have V(r) as => V(r) = ConstantA * log(r) /sigma + ConstantB Thus, I'll be left with: E_domain1(r) = - Constant1 / (sigma_domain1 * r) V_domain1(r) = Constant1 * log(r) / sigma_domain1 + Constant2 E_domain2(r) = - Constant3 / (sigma_domain2 * r) V_domain2(r) = Constant3 * log(r) / sigma_domain2 + Constant4 With the expressions of the constants as following: (deduced from the system of equations i)- iv) ) Constant1 = (V_potential - V_ground) * sigma_domain1 / log(r_min / r_max) Constant2 = V_potential - (V_potential - V_ground) * log(r_min) / log (r_min / r_max) Constant3 = (V_potential - V_ground) * sigma_domain2 / log (r_min / r_max ) Constant4 = V_ground - (V_potential - V_ground) * log(r_max) / log(r_min / r_max) And introducing the constants in the expressions of E and V I will be left with: V_domain1(r) = V_potential + (V_potential - V_ground) * log(r / r_min) / log(r_min / r_max) V_domain2(r) = V_ground + (V_potential - V_ground) * log(r / r_max) / log(r_min / r_max) E_domain1(r) = E_domain2(r) = - (V_potential - V_ground) / ( r * log (r_min / r_max) ) Thus, as the material properties aren't varying with coordinates they will not interfere in this regime and I should have a smooth potential curve from one side to the other of the ring, governed only by the potentials of the two extremities, the fraction between the r_min and r_max and by the "r" of the point where the potential is to be found. Are you saying that Comsol is using other expressions, and not these that I have mentioned above? Or are you saying that I may be mistaking the regime that I am using for another one, and that my analytically expressions above aren't the correct expressions for this regime? Thank you for your answer again, and hopefully you (or someone else) can provide an additional answer for what I wrote in this message.

Edgar J. Kaiser Certified Consultant

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2014年6月13日 GMT-4 05:31
But the conductivity is different in the two domains, so it is dependent on a spatial coordinate. A simple analogy would be a serial circuit of 2 resistors. Say you have 1 Ohm and 9 Ohm in series and 1V attached to the terminals. This will result in 0.1 V and 0.9 V voltage drop.

Or maybe I misunderstand you setup completely,

Cheers
Edgar

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
But the conductivity is different in the two domains, so it is dependent on a spatial coordinate. A simple analogy would be a serial circuit of 2 resistors. Say you have 1 Ohm and 9 Ohm in series and 1V attached to the terminals. This will result in 0.1 V and 0.9 V voltage drop. Or maybe I misunderstand you setup completely, Cheers Edgar -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com

Jeff Hiller COMSOL Employee

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2014年6月13日 GMT-4 08:31
Your equation iv should read
iv) J_domain1 (r = r_med ) = J_domain2 ( r = r_med)
not
iv) E_domain1 (r = r_med ) = E_domain2 ( r = r_med)
Jeff
Your equation iv should read iv) J_domain1 (r = r_med ) = J_domain2 ( r = r_med) not iv) E_domain1 (r = r_med ) = E_domain2 ( r = r_med) Jeff

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 2014年6月13日 GMT-4 10:36
Ahh, that explains some things.

Thank you very much for the answers!
Ahh, that explains some things. Thank you very much for the answers!

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.