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how to calculate current density correctly?

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Hello!

I have two metallic electrodes immersed into salt. I want to solve the problem of electrical current passing through the salt. The electrodes are half-cylindrical in shape (D-shape in cross section). How can I calculate current density correctly escaping from curved side of the D?

Boundary surface average as sqrt(ec.Jx^2+ec.Jy^2+ec.Jz^2) or ec.normJ gives exactly the same result: it is way too high and it is very wrong one.

Boundary surface integration as sqrt(ec.Jx^2+ec.Jy^2+ec.Jz^2) or ec.normJ gives also too high (two orders of magnitude) value than comparing with input current.

Am I missing something basic?


6 Replies Last Post Jun 9, 2016, 2:50 a.m. EDT

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Posted: 8 years ago Jun 8, 2016, 2:06 a.m. EDT
Hi

Your current is too high compared to which value? Experimental one? What physic did you use? Electrostatics and you just gave the resistivity of the salt?

I do not know if this is relevant to you, but at the salt-electrode boundary an electrochemical reaction takes place which can be approximated with a resistor, obviously giving lower currents. I would take the Electrochemistry module (if you have access to it) and use Primary Current Distribution physics.

BR
Lasse
Hi Your current is too high compared to which value? Experimental one? What physic did you use? Electrostatics and you just gave the resistivity of the salt? I do not know if this is relevant to you, but at the salt-electrode boundary an electrochemical reaction takes place which can be approximated with a resistor, obviously giving lower currents. I would take the Electrochemistry module (if you have access to it) and use Primary Current Distribution physics. BR Lasse

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Posted: 8 years ago Jun 8, 2016, 3:08 a.m. EDT
Thank you for your reply! Answering your questions:

1) The total current (integrated over a boundary) is much higher (2 orders of magnitude) than the input current on the other side of the electrode. This is the biggest issue, I do not need to compare anything with experiment because current conservation law does not work here.
2) Physics I use: Electric currents (ec); study is: frequency domain
3) Yes, salt properties I put there.

As you suggested I will try electrochemistry module and use Primary Current Distribution physics.
Thank you for your reply! Answering your questions: 1) The total current (integrated over a boundary) is much higher (2 orders of magnitude) than the input current on the other side of the electrode. This is the biggest issue, I do not need to compare anything with experiment because current conservation law does not work here. 2) Physics I use: Electric currents (ec); study is: frequency domain 3) Yes, salt properties I put there. As you suggested I will try electrochemistry module and use Primary Current Distribution physics.

Jeff Hiller COMSOL Employee

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Posted: 8 years ago Jun 8, 2016, 1:18 p.m. EDT
It is the integral of the normal component of the current density that is conserved, not the integral of the norm of the current density.
Best,
Jeff
It is the integral of the normal component of the current density that is conserved, not the integral of the norm of the current density. Best, Jeff

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Posted: 8 years ago Jun 8, 2016, 4:13 p.m. EDT

It is the integral of the normal component of the current density that is conserved, not the integral of the norm of the current density.
Best,
Jeff


Do you mean one should integrate ec.nJ instead of ec.normJ?

Well, in this case I get NaN.

See attached file. When you run it the "Table of values" will have the total current values obtained by integration along the boundary of each electrode. One is integrated over ec.normJ and gives 249A, the other is integrated over ec.normJ and gives NaN. The input current is only 10A.

I am definitely missing something.

[QUOTE] It is the integral of the normal component of the current density that is conserved, not the integral of the norm of the current density. Best, Jeff [/QUOTE] Do you mean one should integrate ec.nJ instead of ec.normJ? Well, in this case I get NaN. See attached file. When you run it the "Table of values" will have the total current values obtained by integration along the boundary of each electrode. One is integrated over ec.normJ and gives 249A, the other is integrated over ec.normJ and gives NaN. The input current is only 10A. I am definitely missing something.


Jeff Hiller COMSOL Employee

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Posted: 8 years ago Jun 8, 2016, 5:21 p.m. EDT
If I may, you're trying to go about it a weird way!
See my modification of your file. I get within 2% of 10 amps on both boundaries, and you could refine your mesh to improve accuracy further if needed.
Note that you can't use ec.nJ on internal boundaries, hence my use of ec.Jx*nx+ec.Jy*ny+ec.Jz*nz . 3D Plot Group 6 shows you the direction of the normal vector on the boundaries in question (On external boundaries, n points outwards, but on internal boundaries you need to check visually).
See also www.comsol.com/support/knowledgebase/973/ .
Best,
Jeff
If I may, you're trying to go about it a weird way! See my modification of your file. I get within 2% of 10 amps on both boundaries, and you could refine your mesh to improve accuracy further if needed. Note that you can't use ec.nJ on internal boundaries, hence my use of ec.Jx*nx+ec.Jy*ny+ec.Jz*nz . 3D Plot Group 6 shows you the direction of the normal vector on the boundaries in question (On external boundaries, n points outwards, but on internal boundaries you need to check visually). See also https://www.comsol.com/support/knowledgebase/973/ . Best, Jeff


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Posted: 8 years ago Jun 9, 2016, 2:50 a.m. EDT
Thanks for help!
Yeah, I should have read knowledgebase before jumping into this business...
Thanks for help! Yeah, I should have read knowledgebase before jumping into this business...

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