Wire size for 80V, 800 Amperes DC

Good day, i've been trying to find good calculators to calculate this for me. The only one for DC i've found is

DC Cable Sizing Tool - Wire Size Calculator - MM2 & AWG - solar-wind.co.uk

which states that for a 1 meter cable, 3% loss, 80V DC and 800 Amperes a 5 gauge wire will do (16mm^2).

Any tips on this or where to find the proper equations. All feedback appreciated.

800 amps in 5awg? That's completely absurd. It would melt.

At 800A, there's no "wire" that will work. You're talking thick copper busbar or very heavy duty cable (like 0000Awg or something).

Also your link doesn't work.

link worked for me. I am assuming the 800A is continuous?

The continuous current is 2x240A = 480 . But with peak currents of 2x350A = 700A

I'm powering two DC motors. The connections on the motor controllers sure don't imply a wire anywhere near 0000AWG. They are rated for 350A max each.

Please can you demonstrate by equations or calculators that it will indeed melt, or at least point me in the right direction ?

the link path is. The original link i posted works for me. I left the h out at the beginning of the path to prevent the autolink highlight feature.
"ttp://www.solar-wind.co.uk/cable-sizing-DC-cables.html".

---------- Post added at 20:05 ---------- Previous post was at 19:48 ----------

Judging from the wikipedia article on AWG it seems i need 0000Awg for 350A. Then that calculator i found is really flawed. Which means 2x0000Awg to each motor may be suffecient.

steinar96 said:

The continuous current is 2x240A = 480 . But with peak currents of 2x350A = 700A

I'm powering two DC motors. The connections on the motor controllers sure don't imply a wire anywhere near 0000AWG. They are rated for 350A max each.

Please can you demonstrate by equations or calculators that it will indeed melt, or at least point me in the right direction ?

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Well perhaps if they have no insulation on them and are blown with a big fan, they won't melt...

As for calculations, the two things that matter are power dissipation (dependent on the copper) and thermal conductivity (dependent on the thickness and material of the insulation). Unless you know that info you can't calculate anything.

[/COLOR]Judging from the wikipedia article on AWG it seems i need 0000Awg for 350A. Then that calculator i found is really flawed. Which means 2x0000Awg to each motor may be suffecient.

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The link is working now. The calculator isn't really flawed, and that 3% loss number is believable. But that calculator won't tell you anything about temperature rise, which is what really matters. 3% of that power is about 2KW, which is a huge amount of power for such a small piece of copper.

Probably the only way you'll get that kind of current through standard cable is to not use insulation on it, and probably cool it with air. Cable ratings generally allow something like a 35C temperature rise, and are pretty conservative.

Haha The answer is given by you only. As per the site for 1Mtr length it is 50 Square MM which 1AWG. So for the safer side the idea would be to use paired cables means 3 or 4 (twisted) of 4AWG so that it hold more AMP and it will not heat. The principle is same as the fuse wire.

Thankyou

If you really want to use 3 or 4 cables then don't twist them at all, the current are all flowing in the same direction anyway so there is no magnetic field cancellation. Leave them open so the air can circulate to cool them. Could be a good argument to water cool the cables.
Frank

There is something very wrong here.

steinar96 said:

. . . for a 1 meter cable, 3% loss, 80V DC and 800 Amperes . . . .

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What are you doing with your power that starts with a 3% loss over just one meter?
I have to tell you that I find that irresponsible.

mtwieg said:

Well perhaps if they have no insulation on them and are blown with a big fan, they won't melt...

. . .

3% of that power is about 2KW, which is a huge amount of power for such a small piece of copper.

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Exactly!

If this is anything other than a very short term fix for some local issue, then I strongly oppose this approach. Cable diameters are chosen using a number of criteria: heat and voltage drop being top of the list. Anyway, when you get round to designing the protection for your 1 meter circuit, if you calculate it properly, you'll find that it trips out nicely before the cable gets too hot, so you won't need the fan coolng which mtwieg suggested.

Use a steel bar 25mm x 25mm.

There are standards for electrical installation cables, and they say for single conductor in air you'll need 400 mm².

DXNewcastle said:

Use a steel bar 25mm x 25mm.

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Well, I don't know about steel... low conductivity and all that. A bar of aluminum would work great and shouldn't be hard to find.

mtwieg said:

Well, I don't know about steel... low conductivity and all that. A bar of aluminum would work great and shouldn't be hard to find.

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You are, of course, quite correct.
My "use a steel bar" was meant more as a throw-away remark (but I should avoid attempts at humour on the internet!).

But this all assumes that the incoming supply is also suitable rated. I'd probably be as concerned about overheating the supply cables concealed in trunking or wherever, as I would be about this short length over the last 1 meter.

Usualy i use upto 6Amp's per square mm of cable (upto 36 Amps for a 6mm2 cable,upto 60 for 10mm2 cable etc).
I know this is a conservative number but it survives prolonged use(which i assume its meanth to be)
You can ofcourse use copper or aluminium bars or anything like that but thats only if superman is available to bend it into a usable form while using it.
The voltage is your lesser concern ,what matters is the amount of amps You need, thats what will cause the voltage drop over the cable and what will heat your cable.

Usualy i use upto 6Amp's per square mm of cable

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The calculation doesn't work for higher currents, because the heat is emitted according to the cable surface area, not the cross-section. So you get 400 mm² for 840 A or 185 mm² for 515 A in the standards. Bus bars can have a slightly lower cross-section.