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I’m trying to size some welding conductors per the following information. The welder will be used once a day for 2 seconds only. 6000A of current will be drawn through the conductors during this time. The power supply provides 10v. The conductors shall be 8 feet in length.

It seems like the largest welding cables are 500mcm. I would like to do a calculation that shows the temperature rise in the cable would not exceed the maximum rating of the cable, even though the cable is not rated for 6000A. Can anyone help me with this?

The cable I’m looking at is general cable carolprene 500mcm (500kcmil) stranded copper welding cable.

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    \$\begingroup\$ what is 500mcm ? \$\endgroup\$ – Solar Mike Apr 9 at 14:23
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    \$\begingroup\$ Circular Mil as a way of measuring large diameter wires. Link \$\endgroup\$ – Tyler Apr 9 at 14:35
  • \$\begingroup\$ It’s the same thing as kcmil, a measurement of wire size. \$\endgroup\$ – amantonas Apr 9 at 14:35
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    \$\begingroup\$ What’s that in mm^2? Either way, 2 seconds should be below the thermal time constant of the cable. \$\endgroup\$ – winny Apr 9 at 15:23
  • \$\begingroup\$ So should be MCM and used in USA and Canada... \$\endgroup\$ – Solar Mike Apr 9 at 15:30
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When a cable dissipates electrical energy, two things limit the temperature rise.

a) energy going into heating the copper
b) energy being lost to the surroundings

Although both matter to some extent, most interesting cases involve very short times, where (a) is the only thing that matters, this is called adiabatic heating. For continuous operation, only (b) matters, this is called isothermal heating.

For a 2 second pulse, you are probably in the adiabatic regime. In this regime, we rate conductors by their \$I^2t\$. If they are safe at some current for a given pulse length, then if we double the current, we double the voltage drop, four times the power means we can only take one quarter the pulse length, for a constant \$I^2t\$.

An insulated cable will be limited by the thermal rating of the insulation. It might also be limited by electromagnetic forces.

To work out the adiabatic rating of the cable, take a unit length of cable, say 1m, and then

a) compute its thermal capacity
b) compute its resistance
c) determine the \$I^2t\$ needed to raise its temperature from ambient to what the insulation will tolerate.

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To assist with the other answers, the rise in temperature, ΔT, will be given by

$$ \Delta T = \frac {P \cdot t}{m \cdot SHC} $$

where P is power dissipated in the cable - not in the load - (watts), t is the time (s), m is the mass (kg) and SHC is the specific heat capacity (J/kg.K).

This ignores heat loss to ambient via radiation, conduction and convection so it should be worst case. Note also that cable resistance will rise with temperature so you may need to factory that in.

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    \$\begingroup\$ Can you add that this is the core of any transient calculation but can’t be used for static operation? It’s clear to people who know it but perhaps not for the occasional visitor. \$\endgroup\$ – winny Apr 9 at 16:56
  • \$\begingroup\$ Please edit it in, Winny, as I'm not sure where this leads. What's the issue with static operation? Unending temperature rise? \$\endgroup\$ – Transistor Apr 9 at 17:08
  • \$\begingroup\$ Exactly. OP didn’t seem to fully gasp this and I’m afarair someone else will do the same and take your education at face value without constraints. Heck, if I had the time I would make an answer with your equation derived for time (English? Derivative?) as the slope at the beginning of the time-temperature curve and the black body+convection equations at t=infinity and have my “students” sketch the complete curve from these two points/asymptotes. \$\endgroup\$ – winny Apr 9 at 17:27
  • \$\begingroup\$ Errrr. That’s afraid and equation of course. Damn you autocorrect! \$\endgroup\$ – winny Apr 9 at 17:38
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    \$\begingroup\$ I think you might misunderstand our relationship. I give my time freely to answer questions of interest but have no obligations regarding how much information I supply. The fundemental law of physics in this case is that when energy is put into matter then temperature rises. Energy = power by time so that's the top of the equation. Temperature rise will be inversely proportional to the mass and the specific heat capacity so that explains the bottom of the equation. \$\endgroup\$ – Transistor Apr 9 at 18:24
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You know the CSA and the length, so you can calculate the resistance and hence the power dissipation given the bulk conductivity of copper and $$P = I^2R$$. Note that with 6kA and only 10V pushing it, you will need to ensure total cable and connection resistance is well below 1.6 milli ohms!

You know the power, and can calculate the mass of copper in play, so by looking up the specific heat capacity of copper you can calculate the temperature rise during a weld. Then you decide if the temperature rise is acceptable.

One thing to watch is the mechanical fixing of the cables there will be large Lorenz forces trying to push those wires apart during a pulse, so you will need things to be mechanically secure, and will probably want to minimise loop area to keep inductance down (It don't take much with a maximum loop impedance of 1.6 milli ohms).

I would be surprised if you actually hit 6kA with only 10V behind it, possible, but I doubt it.

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