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Changing the way COMSOL calculates the resistance of a conductor 2

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Jensdsds

Jensdsds

Electrical
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Jun 9, 2021
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Hello,

I'am doing a Joule heating study of a simple cable model.
The cable has a terminal of 50A in one end, and a ground in the other end.
A heat flux is added to the surface.

My problem is:
I am simulating a Stranded cable as a solid cable in comsol for time saving and mesh optimisation.
This however makes it so that the solid cable in the simulation has more copper than the real life stranded cable using the same diameter.

I want to change the simulation so that i dictacte the resistivity from the standard 1.72 e-7 to a value of my choice.
Is there a way to do this? Or should i go with another aproach to manually decide the resistance of the cable?

Right now the linearized resistivity value isn't used by COMSOL. Unless i choose linearized resistivity in the constitutive relation.
But this makes the model take 10 years to compute.

Thanks in advance!
Kind regards - Jens
Sk%C3%A6rmbillede_2022-05-04_150421_sg76wa.png
Sk%C3%A6rmbillede_2022-05-04_150118_nnone9.png
 
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Perhaps you could account for too high amount of copper in the cable model by adjusting the electrical conductivity of the material (a bit like adjusting elastic properties for a homogenized porous structure). But it's hard to say what value to use so a simple test run with a comparison of a cabled modeled as stranded and simplified could be useful. Or maybe you will find such data in the literature.
 
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You care about the physical dimensions of the outside of the cable (for heat transfer) and not the thermal conductivity (and/or thermal mass) of the copper, I assume? If so, just adjust the current flow along the cable to get the heat dissipation to match what you want to simulate.

If you do want to look at transients, i.e. the real thermal conductivity and mass of the copper is important, then model the cable as multiple (straight) strands with contact at the tangent points, or perhaps even model the stands as a continuous solid chunk with appropriate voids in it to get the right ratios.
 
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Hello,

Thank you both for the advice, i will try looking into the solutions.
Yes btrueblood you are correct. I intend to keep the same physical parameters for the cable, so that the heat flux matches the real test stand. So my only wish is to correct the resistance of the wire.

My study intends to anticiate the temperature rise of a real life cable during a heat run with a specefic ampere value, using a COMSOL model. But so far my simple model overshoots the temperature a bit.

So far the way to make my Model match reality has been to tune in the heat transfer coeffecient in the heat flux boundary on the cable surface. e.g. raise it from 5W/m^2*K to 9W/m^2*K.
All the other parameters of the model seems to be tuned to match reality.

I have one question for the heat transfer coeffecient HTC:
If the HTC from the cable surface to air is 5W/m^2*K in an enclosed box, is this a semi constant value for htc?
Or does it change with the material type and cable arrangement.

Thank you in advance.
Kind regards - Jens
 
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The resistance problem should be fixed now, calculating a new electrical conductivity value.
Thank you
 
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If the HTC from the cable surface to air is 5W/m^2*K in an enclosed box, is this a semi constant value for htc?
Or does it change with the material type and cable arrangement.
Click to expand...

Probably more the cable arrangement vs. material. The convection coefficient is dependent on the air flow around the cable, so blockages to air flow will make a big difference and can turn convection into conduction only, i.e., if vertical air flow is restricted, for example.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
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"If the HTC from the cable surface to air is 5W/m^2*K in an enclosed box, is this a semi constant value for htc?"

Nothing in heat transfer is constant, LOL. And if this is natural convection (no forced air/fan) it's pretty normal to have to tweak book values to get a good correlation. As IRstuff said, the configuration of the cable, size of the box, etc. will make a difference. So will conductive heat transfer at the boundaries where the real cable is terminated.
 
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Hello,

Alright, thank you guys for the answers.
It's been very helpful!

Kind regards - Jens.
 
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