This is based on a problem that came up today. During the course of this problem I realized that I wasn't so sure I understood the relationship between wattage and heat produced.

In the past we did a test in the lab using .305 Ω /ft wire. The jacket is rated for 150C. We were able to get about 6.5A (at 3.66V) out of it at 24C ambient, with out exceeding the jacket rating. I want to estimate what the ampacity of 0.027 Ω/ft wire is. So I am wondering if I did it correctly, because this amperage seems a little high for the wire to handle, then again most copper wire is only rated at 90C.

So the math I did on it was this

So you do .305 * 2 = .61 Ω /ft 6A^2 * .61 Ω = 21.62W (I^2*r = W) 21.96W * / 2ft = 10.98 W/ ft

So would it be safe to assume that if I did the same with a 0.027w/ft wire with the same jacked I would arrive at this amperage?

If you start with 11W/ft * 2ft = 22W 0.027 Ω * 2ft = .054 Ω Sqrt(22W/(0.054 Ω)) = 20.18A

ETA: we are planning on testing this tomorrow when we get some wire in. So we shall find out.


I don't quite follow your calculations. But if you are doing I^2 *R as the power, and assuming the same max power for each wire.. then that's what I would have done.

However I went here and it looks like I*R is about constant. (?)
(I had to plot it.) Still looks linear.

Maybe someone can tell us both why.

Edit: I*R dependence. (Thanks Spehro, it was suddenly obvious on the drive home.) No matter the thermal loss mechanism (convection, radiation..) It will go as the area of the wire. 2 * pi *r * l (r - radius and l - length), so bigger wire will need more heat to get to a given temperature. (more later)

enter image description here

  • \$\begingroup\$ Well, if you assume the limiting factor is watts/in^2 of conductor surface area, then I think you'll find I*R is constant for a given \$\rho\$ (resistivity). \$\endgroup\$ – Spehro Pefhany Oct 29 '14 at 20:31
  • \$\begingroup\$ Yes I was using I^2*R = P \$\endgroup\$ – TheColonel26 Oct 30 '14 at 1:09
  • \$\begingroup\$ @SpehroPefhany, Well, I spent several hours last night trying to make it work and always found that Imax goes as r^3/2 (where r is the wire radius) or as R^-3/4 (where R is resistance.) or Imax. * R^3/4 is constant. (color me confused.) \$\endgroup\$ – George Herold Oct 30 '14 at 17:38

Your calculation appears to be correct, assuming the wires were the same dimensions. In practice, the lower resistance one would tend to be larger, so it would run cooler.

However, I doubt your measurement, so I would call the calculation based on it into question.

The figures I use show a maximum current of about 11A for AWG 24 copper wire (actually a bit higher resistance than yours) and allowing a temperature rise of 125°C .. for a single wire in open air at sea level (best possible case). In practice because you usually have to allow for higher ambients, more wires in close proximity and possibly higher altitude and (often the dominant concern) less voltage drop you can't get anywhere near that current. AWG 24 is also too frail to leave flapping around in applications involving vibration.

  • \$\begingroup\$ Well the official rating from the Manufacture on the .305 wire is 3.2A. (just found that out) That rating obviously has a large safety factor built into. As heater wire usually operates at -40C to 0C ambient. We were running ours at 24C I believe it was stable for 6 hours before we shut if off. \$\endgroup\$ – TheColonel26 Oct 30 '14 at 0:58
  • \$\begingroup\$ Oh, this is a heat tracing application? Probably a reasonable safety factor, yes. \$\endgroup\$ – Spehro Pefhany Oct 30 '14 at 4:55
  • \$\begingroup\$ @SpehroPefhany, the area is the outside area of the wire (2 * pi * r * l) not the cross sectional area (A = pi*r^2) Search for Preece's law. As far as I can tell the wire tables are not "right". Maybe they are good in the middle range, but you should be able to push more current through thin wires. (And if anyone has 0000 gauge and does 300A, I'm guessing it will get fairly hot.) If this was at all important to me I'd be tempted to try some measurements. \$\endgroup\$ – George Herold Oct 30 '14 at 18:15
  • \$\begingroup\$ You're right, of course. I'll put a comment in your question. \$\endgroup\$ – Spehro Pefhany Oct 30 '14 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.