Menu

Discussion Forum

New Discussion
Filtern

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.

anodic and cathodic currents do not match!!

Posted 25.11.2010, 12:10 GMT-5 12 Replies

Antonio Diez
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Hi there,

Im trying to simulate the behaviour of an electrolyzer. This device is comprised of a cylindrical copper electrode (anode) immersed in a cylindrical recipient filled with sea water. several conductive rings are placed inside the boundaries of the recipient. My goal is to determine the impedances between the anode and the diffferent cathodes.

In order to do that I set the following conditions:
boundaries:
electrode: electric potential 1v
cathode used: ground
non used cathodes: electric insulation

subdomains:
conductive->isotrocip 5.99e7S/m
sea water-> 5 S/m

Im trying to compute z as Z=E/I (i know E is 1V which is the potential applied to the anode, I is to be determined)
to calculate I, I perform a boundary integration in the postprocessing of the results.
To my surprise, when I do that selecting the anode, I get a current bigger than that of the cathode.
Given that I perform the boundary integration over the entire and only boundary connected to ground, I cannot understand where the current is going and why anodic and cathodic currents do not match (aprox 1 A vs 1.2 A)
I tried solving my probling in 2D axial symetry and 3D. Also, i tried to use extra fine mesh but the results where still the same.

some help would be appreciated.

thanks a lot and best regards,

Antonio
12 Replies Last Post 27.11.2010, 04:39 GMT-5
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 04:31 GMT-5
Hi

how much is "bigger" ? 1% or 50% ...

are you sure you do not have any fringe effects in the corners ?

--
Good luck
Ivar
Hi how much is "bigger" ? 1% or 50% ... are you sure you do not have any fringe effects in the corners ? -- Good luck Ivar
Antonio Diez
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 04:54 GMT-5
Bigger is 1.28 versus 0.988, so about 30%. Im not sure about fringe effects in the corners. However I tried with the extremely fine mesh and that didn't change the values (wouldn't that reduce the effect if it exists?).
I'm trying 2D axyal simmetry and 3D, hoping to find a solution but no luck so far.

Regards,

Antonio
Bigger is 1.28 versus 0.988, so about 30%. Im not sure about fringe effects in the corners. However I tried with the extremely fine mesh and that didn't change the values (wouldn't that reduce the effect if it exists?). I'm trying 2D axyal simmetry and 3D, hoping to find a solution but no luck so far. Regards, Antonio
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 07:54 GMT-5
Hi

no, in fact EM artefact by sharp edge singilarities do NOT necessarily improve with the mesh (only on a global way, sometime), this applies also for point loads in ST too by the way

Furthermore are you using physiscs BC's or lumped port/terminal BC's, probably better for your case


And in 2D axi, do not forget the 2*pi*r (or is it only *r now in V4) pls check on simple example by integrating an edge to get corresponding "loop" area

--
Good luck
Ivar
Hi no, in fact EM artefact by sharp edge singilarities do NOT necessarily improve with the mesh (only on a global way, sometime), this applies also for point loads in ST too by the way Furthermore are you using physiscs BC's or lumped port/terminal BC's, probably better for your case And in 2D axi, do not forget the 2*pi*r (or is it only *r now in V4) pls check on simple example by integrating an edge to get corresponding "loop" area -- Good luck Ivar
R. F. Hamou
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 07:56 GMT-5
Hi,

I wonder if the anode and the cathode have the same surface area!! Check your "current densities" in both electrodes!!

Hope it helps

Good luck
Hi, I wonder if the anode and the cathode have the same surface area!! Check your "current densities" in both electrodes!! Hope it helps Good luck
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 08:25 GMT-5
Hi

Indeed, that is why one should integrate out the full current in Amps and not A/m^2 (good for element physics but not for the end user) as this differs on how to calculate it, and in non obvious ways in 3D, 2D, 2D-axi all depending on the geoemtry and the location of the boundaries

--
Good luck
Ivar
Hi Indeed, that is why one should integrate out the full current in Amps and not A/m^2 (good for element physics but not for the end user) as this differs on how to calculate it, and in non obvious ways in 3D, 2D, 2D-axi all depending on the geoemtry and the location of the boundaries -- Good luck Ivar
R. F. Hamou
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 09:13 GMT-5
Hi again,

I have uploaded a model that I have just set using Comsol 3.5a (conductive media DC)
It has two geometry 1 (different electrode surface area) and geometry 2 (same electrode area)

Try to compare the current and the current densities in both cases you will notice that is the same for geom2 (same electrode area) and slightly different for geom1 (different area)

I hope it helps,

Good luck
Hi again, I have uploaded a model that I have just set using Comsol 3.5a (conductive media DC) It has two geometry 1 (different electrode surface area) and geometry 2 (same electrode area) Try to compare the current and the current densities in both cases you will notice that is the same for geom2 (same electrode area) and slightly different for geom1 (different area) I hope it helps, Good luck
Antonio Diez
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 10:44 GMT-5

Hi

no, in fact EM artefact by sharp edge singilarities do NOT necessarily improve with the mesh (only on a global way, sometime), this applies also for point loads in ST too by the way

Furthermore are you using physiscs BC's or lumped port/terminal BC's, probably better for your case


And in 2D axi, do not forget the 2*pi*r (or is it only *r now in V4) pls check on simple example by integrating an edge to get corresponding "loop" area

--
Good luck
Ivar



Im using physics BC's (ground, electric potential...) I wasn't aware of the existance of lumped port/terminal BC's. Where can I find those?

I think I'm making some progress by refining the mesh.
I set up a simpler model of my problem (so that it doenst take so much memory) and tried different meshes. Both in 3D and ax sym 2D.
The best result i got in 3D was: anodic current 0.808 A, cathodic current: 0.695A. As you can see they differ 0.1 A, but it tells me that the current must be between those two values (0.7,08)
Now, I do the 2D ax sym model, and I refine the mesh as much as I can, ending up with 0.792 A for the cathodic current. (may be right)

However, in 2D I don't know how to calculate the anodic current since it is the integral of a line parallel to the r axis.
In this case I cannot multiply by 2*pi*r (see attached pic). Any ideas?

[QUOTE] Hi no, in fact EM artefact by sharp edge singilarities do NOT necessarily improve with the mesh (only on a global way, sometime), this applies also for point loads in ST too by the way Furthermore are you using physiscs BC's or lumped port/terminal BC's, probably better for your case And in 2D axi, do not forget the 2*pi*r (or is it only *r now in V4) pls check on simple example by integrating an edge to get corresponding "loop" area -- Good luck Ivar [/QUOTE] Im using physics BC's (ground, electric potential...) I wasn't aware of the existance of lumped port/terminal BC's. Where can I find those? I think I'm making some progress by refining the mesh. I set up a simpler model of my problem (so that it doenst take so much memory) and tried different meshes. Both in 3D and ax sym 2D. The best result i got in 3D was: anodic current 0.808 A, cathodic current: 0.695A. As you can see they differ 0.1 A, but it tells me that the current must be between those two values (0.7,08) Now, I do the 2D ax sym model, and I refine the mesh as much as I can, ending up with 0.792 A for the cathodic current. (may be right) However, in 2D I don't know how to calculate the anodic current since it is the integral of a line parallel to the r axis. In this case I cannot multiply by 2*pi*r (see attached pic). Any ideas?
Antonio Diez
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 10:47 GMT-5
heres the pic.

regards,

Antonio
heres the pic. regards, Antonio

Attachments:

Antonio Diez
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 10:56 GMT-5

Hi,

I wonder if the anode and the cathode have the same surface area!! Check your "current densities" in both electrodes!!

Hope it helps

Good luck



Anode and cathode have different surface area. See pic attached in my last post (it is a 2d Axial sym). Current densities vary along the surface of both cathode and anode since distance between them varies too. NEVERTHELESS, total current should match (as in any series circuit) (integral of the current densities of the anode for all its surface, should equal the intergral of the current densities of the cathode for all its surface).

thanks,

Atonio
[QUOTE] Hi, I wonder if the anode and the cathode have the same surface area!! Check your "current densities" in both electrodes!! Hope it helps Good luck [/QUOTE] Anode and cathode have different surface area. See pic attached in my last post (it is a 2d Axial sym). Current densities vary along the surface of both cathode and anode since distance between them varies too. NEVERTHELESS, total current should match (as in any series circuit) (integral of the current densities of the anode for all its surface, should equal the intergral of the current densities of the cathode for all its surface). thanks, Atonio
Antonio Diez
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 11:18 GMT-5
sorry, i cant open the model (using 3.2 here).
sorry, i cant open the model (using 3.2 here).
R. F. Hamou
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 26.11.2010, 12:17 GMT-5

NEVERTHELESS, total current should match (as in any series circuit) (integral of the current densities of the anode for all its surface, should equal the intergral of the current densities of the cathode for all its surface).


##Dear Antonio, I suggested to you to check the current densities just to see how smooth is your J along the boundaries!!


However, in 2D I don't know how to calculate the anodic current since it is the integral of a line parallel to the r axis.
In this case I cannot multiply by 2*pi*r (see attached pic). Any ideas?


Of course you can!! Why not?? The revolution of (2*pi*R) of this line which is parallel to the axis would give you a cylinder surface!! NO??

I have built the same model including your geometry (I am sorry but the 3.2 is in the Comsol museum ;-) )

And the currents are
1.542469 A for the top electrode.
1.542634 A for the bottom electrode.

Good luck man,
[QUOTE] NEVERTHELESS, total current should match (as in any series circuit) (integral of the current densities of the anode for all its surface, should equal the intergral of the current densities of the cathode for all its surface). [/QUOTE] ##Dear Antonio, I suggested to you to check the current densities just to see how smooth is your J along the boundaries!! [QUOTE] However, in 2D I don't know how to calculate the anodic current since it is the integral of a line parallel to the r axis. In this case I cannot multiply by 2*pi*r (see attached pic). Any ideas? [/QUOTE] Of course you can!! Why not?? The revolution of (2*pi*R) of this line which is parallel to the axis would give you a cylinder surface!! NO?? I have built the same model including your geometry (I am sorry but the 3.2 is in the Comsol museum ;-) ) And the currents are 1.542469 A for the top electrode. 1.542634 A for the bottom electrode. Good luck man,

Attachments:

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 27.11.2010, 04:39 GMT-5
Hi

pls be aware then when I load this model into v4.1 I get some funny results, probably due to some translations issues.
The workaround is to remove all "point" electric potential BC's that COMSOL adds in v4.1, or better to reconstruct it fully its quickly done.

calculating on the imported modl gives wrong results

--
Good luck
Ivar
Hi pls be aware then when I load this model into v4.1 I get some funny results, probably due to some translations issues. The workaround is to remove all "point" electric potential BC's that COMSOL adds in v4.1, or better to reconstruct it fully its quickly done. calculating on the imported modl gives wrong results -- Good luck Ivar

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.

Suggested Content