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I'm trying to size wire for UL 508a panels. I have UL's wire gauge requirements, but those requirements are for continuous use. The device I'm designing will only run for two seconds, with minutes or hours between runs. Since currents of interest are 25, 50, 100, and 200 amps, there's a lot to be saved by not using wire rated for continuous use!

Is there a proper way to size wire for pulse applications like this? If the continuous ampacity of (for example) 75C copper stranded 4 AWG is 85 amps, how much can I run for two seconds? Is there some rule of thumb? Some equation? A table? Appropriate application of calculus?

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  • \$\begingroup\$ One question and one comment. First, why don't you just ask UL what their requirements are in this case? Or ask your wire supplier what they recommend? Second, I think what will be the limiting factor here is current density. With such a short duty cycle you might be tempted to go as small as possible with wire gauge but you can get failures if your current density is too high even if your duty cycle is really low. Your wire vendor may have info on max current density. \$\endgroup\$
    – Eric
    Commented Jan 20, 2014 at 15:48
  • \$\begingroup\$ Many wire tables list maximum rated currents from various sources and under various applications (free air, ducted, in equipment, new-moon ...). The highest of these is liable to be safe in your case PROVIDED THAT you can limit duration to a known maximum under fault conditions and that you know the true maximum amplitude. I like Chris Johnson's thermal input in isolation approach as if used with NO cooling (closed system) it gives you a "laws of Physics" absolute maximum tolerable rating. Actual will be some fraction of this. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jan 21, 2014 at 23:28
  • \$\begingroup\$ Interest-only data point: Mains cables for consumer use which are supplied on plastic wind up spools with a handle to allow users to wind them up again after use, melt if used wound up at rated load :-)] ie the long term energy input exceeds the available cooling capacity. Of course, the insulation melts and not the wire. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jan 21, 2014 at 23:30

3 Answers 3

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If this question were in a physics exam, I would answer it as follows; whether this is a sensible idea in practice is entirely another matter. One would have to be pretty certain that no fault state could leave current flowing for more than two seconds.

We know from the specification of the wire the resistance per metre R and mass of copper per metre M. Given the current, I, we know that the power dissipated in the wire is I^2 R per metre. The total heat energy dissipated per metre of wire is therefore E=I^2 R t, where t=2 seconds is the time that the current is active for. We (conservatively) estimate that negligible heat leaves the copper wire during this 2 seconds, and therefore the rise in temperature T is given by

T=E/(MC) = I^2 R t / (MC)

where C is the specific heat capacity of copper. A wire needs to be chosen with R and M such that this temperature rise T is acceptable.

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  • \$\begingroup\$ I took Rth into account, and I've just run a spreadsheet comparing your approach with mine. Rth is, in fact, negligible on these time scales. Five seconds it might make 1% worth of difference, and 30 seconds it makes 5%. \$\endgroup\$
    – Stephen Collings
    Commented Jan 20, 2014 at 16:03
  • \$\begingroup\$ +1 This is half of the analysis needed. However, this is only adequate in the case where there will only ever be a single pulse (i.e. effectively infinite time between pulses). To do the other half of the analysis, you need to determine the amount the temperature declines between pulses. The amount the temperature declines between pulses must be greater than the rise as a result of the pulse. Otherwise each pulse will contribute to increase the temperature to higher and higher values. The cooling almost certainly follow's Newton's law of cooling, so the analysis should be straightforward. \$\endgroup\$
    – alx9r
    Commented Jan 20, 2014 at 18:40
  • \$\begingroup\$ Indeed, this is not the full story. It might not be straightforward to work out appropriate coefficient for Newton's law of cooling, which is going to be a function of the thickness and thermal properties of the wire insulation, perhaps also of the enclosure. \$\endgroup\$
    – Chris Johnson
    Commented Jan 20, 2014 at 20:45
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The resistance of the wire has two primary effects. The first is that it causes a voltage drop at the load, and this is independent of the duty cycle. The second is that it causes the wire to heat up, which can cause it to fail.

In general, wiring should be conservatively rated in all applications, because you really don't want the wire itself to be the point of failure, even under fault conditions, such as excessive duty cycle or overcurrent. The wire has to withstand the fault until the protection gear has time to operate.

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    \$\begingroup\$ The resistance will change with temperature, so it can be indirectly related to duty cycle. \$\endgroup\$
    – Diego C Nascimento
    Commented Jan 20, 2014 at 15:23
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There is an ampacity table (Table 36.1) in the standard that refers to power resistors (as in motor braking resistors). The shortest "on time" (and the lowest duty cycle) shown is 5 seconds on/75 seconds off (6.25% duty cycle). Under those conditions, they allow conductor ampacity of 35% of the motor FLA. There is a bit more information in the lead-in regarding different on/off times, but the meaning seems somewhat muddled.

Now, whether that's applicable or not to your situation, I would not want to speculate. It at least gives you some idea of what UL consider to be safe, and that is certainly necessary, but may not be sufficient.

As others have said, you would have to have some kind of circuit protection appropriate for the wire size that you are actually using, not for the surge currents.

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