maximum current density

I need information about the "maximum current density" of the conductors.

(not superconductors).

I am not sure how this limit is defined.

Data, but also concept, definitions, standards, etc.

Appreciate for any info, data, tips, web-link, .comparison of conductors..

Aran

Reply to
aran
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In the US and Canada the standard is the American Wire Gauge (AWG) standard. Here is a link to get you started (scroll down to table and calculator).

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Dwayne

Reply to
Dwayne

| I need information about the "maximum current density" of the conductors. | | (not superconductors). | | | | I am not sure how this limit is defined. | | Data, but also concept, definitions, standards, etc. | | | | Appreciate for any info, data, tips, web-link, .comparison of conductors.. | | Aran

Make sure you consider thermal issues. You can't just take 100 times the amps and put it through 100 times the cross section of wire, unless you space the wire out for heat dissipation. Ratings take into consideration the typical heat dissipation. De-rating is required when multiple wires are bundled together because the cross section increases faster than the surface area to dissipate from. So if you have 100 times the amps you'll need more than 100 times the cross section. If you're dealing with high frequencies, you have to also consider skin effect (making much of the inner cross section useless).

Reply to
phil-news-nospam

There are cases where it makes sense to use the concept of current density, but it is sometimes misunderstood. You can take the maximum current for a conductor under conditions you choose and divide it by the cross sectional area of the conductor. If the maximum continuous current for a 1.0 mm2 PVC insulated cable is 10A, the maximum current density is 10A/mm2. The problems start if you try to use this number to find the ampacity for other sizes. The power lost in the form of heat in a conductor is proportional to I^2, not I. Therefore, a 50 mm2 conductor will not allow a continuous current of 50 x 10 A = 500A as you would expect.

/Clas-Henrik

Reply to
C-H Gustafsson

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