Reverse Joule Heating

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Simon Sønderskov

Simon Sønderskov

November 15, 2012 1:48pm UTC

Reverse Joule Heating

Hi.

This is my first time using Comsol so please bare with me :)

I'm doing a model of a HV power cable, that heats up due to the current flowing through it.
I know that the temperature of the isolation material can not exceed 90degC so i want to figure out how much current this allows.

It's a 2D model where i have a circular crosssection of the cable.

I set the temperature at the boundry of the conductor to 90degC. I put a 1.3m thick layer of soil around the cable (also circular) which is 15degC at the outer boundry.

Then I set a point potential in the middle of the conductor to 150kV and put ground at the outer boundry of the isolation layer.

When i run this i get some very low current densities. Maybe my problem is that the current i flowing i random directions and not into or out of the 2D plane as was intended.

And i'm not sure that it even depends on the temperature i specified or just calculates from the voltage and the resistance of the material.

\Simon

PS. i attatched my comcol file

Attachments:   2Dkabel_42_4.mph  

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Eric Lämmel

Eric Lämmel

November 17, 2012 8:43am UTC in response to Simon Sønderskov

Re: Reverse Joule Heating

Hi,

what you did is to give a disc of copper a potential of 15 kV and sourround it with some grounded, insulating material. Electrically-physically thinking, not simulation-minded thinking, what would happen?

Nothing.

There is no way the current could pass. Resulting in no influence to the thermodynamics.

You have a cable, that is something like an ideal geometry for an axisymmetric model. Try it this way. 2D axisymmetric model with a reasonable voltage drop along the middle of the cable, so that the current has a way to follow.

Good luck!

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Simon Sønderskov

Simon Sønderskov

November 17, 2012 5:32pm UTC in response to Eric Lämmel

Re: Reverse Joule Heating

Hi Eric.

Thank you for the answer!
I did actually think of this, but by spicifying a voltage drop wouldn't i just inderectly specify i current since V=I*R or as would do it in comsol V=I/length/conductivity?

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Eric Lämmel

Eric Lämmel

November 19, 2012 8:25am UTC in response to Simon Sønderskov

Re: Reverse Joule Heating

Hello,

yes and no. COMSOL solves for a V- and T-field and the resulting current must fit to both of them.

I would think the first guess of the current is from voltage drop and material, which results in a rise of temperature. Which in turn is constrained by your temperature boundary condition. The change of temperature yields a different resistance and therefore a different current.

But you are right, maybe electric potential and ground aren't the most suitable BCs for your model. This is up to you, you have to evaluate that.

As a matter of fact, you need a source and a sink for your current, just to determine the direction of the current. So one possibility that comes to my mind is to use ground as sink and a boundary current source as source. The amount of current flowing out of that source might be given by some coupling operator that integrates the current flowing through a cross section of your wire.

This doesn't determine the current by voltage and resistance in advance...

Keep us up to date on this! I'm interested in how you solve that problem.

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Simon Sønderskov

Simon Sønderskov

November 19, 2012 2:23pm UTC in response to Eric Lämmel

Re: Reverse Joule Heating

An update on what i've done:

I made a 2D axissymmetric model of 1 meter of the cable.
Then I set a 150 kV potential in one end of the cable and calculated the potential in the other end of the cable. I did this using a current from a matlab simulation of the same problem.
Then i set a heat source to the total power dissapation.

This is a bit different from what i originally wanted to do. But couldn't figure out how to do it.

But there are still some mysteries to me.
First of all the current I is 2363 Amps. But when i use this to calculate the voltage drop, the temp. increase is very small and when i plot the current (J*area) i get something way smaller than what i originally put in.
So what i did was that i multiplied the current with the cross section of the concucor and used this to calculate the voltage drop. And somehow it works! Also when i plot the current now i get exactly the current i put in..

Why is this?

(See the attached model)

\Simon

Attachments:   2Daxissymm_2.mph  

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Ivar Kjelberg

Ivar Kjelberg

November 20, 2012 6:54am UTC in response to Simon Sønderskov

Re: Reverse Joule Heating

Hi

first a trick: in V4 for a rectangle, or any geometrical shape, with "shell items, you should use the "layer" functionality of the geometry node, its far easier and quicker to set up a series of rectangles as you have

You should check your units, COMSOL flags them orange, these could be wrong in the set-up of the lower voltage which could explain some differences.

Then in 2D-axi you should distinguish current density calculations on the exit boundary and the loop length term 2*pi*r that must be added to transform a edge calculation in 2D-axi into a true surface calculation in the 3D equivalent representation. There is a tab under integration advanced to turn it on automatically, or you should add the 2*pi*r manually to your equations (but not both ;)

You have not fully got the subtleties of 2D axi, try it out on a simple case, and calculate volumes and surfaces and check manually until you have the concept in full control.

Finally, as you have a good GND shielding, your E field in DC will never pass, so you do not really need to calculate the Electric currents in the concrete. This might however change slightly in a AC case, but then you need also to define the GND or equivalent impedance on the external concrete side

--
Good luck
Ivar

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Simon Sønderskov

Simon Sønderskov

November 20, 2012 2:37pm UTC in response to Ivar Kjelberg

Re: Reverse Joule Heating

Thank you for the tips! Nice to hear from some people with more experiance than i have.

I found out what i was doing wrong in the 2D axissymmetric model i posted earlier. I used the resistivity (ohm*m) where i should have used the resistance (ohm/m). But then i devided the current by the area which canceled out my mistake and that is why i got the correct results inspite of the wrong approach. (Let me know if anyone wants to see the final model)

The next step for me is to transfer this to my next problem.
What i want to do now is to make a model of the same cabel that has a bend and see what effect that bend has on the temperature compared to the straight one i already did.
I figure that after doing the geometry i'll use the same approach as in the 2D axissymm. model.

Do you have any alternative ideas to solving this new problem?

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Ivar Kjelberg

Ivar Kjelberg

November 20, 2012 5:59pm UTC in response to Simon Sønderskov

Re: Reverse Joule Heating

Hei

bare hyggelig

Bt pls check carefully the 2D-axi symmetric loop length 2*pi*r in your integrations, I'm not sure you have them all right

integration of a surface = domain in 2D-axi is no volume you must integrate 2*pi*r*integrand_over_suface to get a true 3D volumetric value, the same for an boundary=edge in 2D-axi, it becomes a surface via the integration of 2*pi*r*integrand_over an edge in 3D

--
Good luck
Ivar

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