How to calculate heat in wires.
- Thread starter extar
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Why not just use a wire ampacity chart? The rate that wire heats at will depend on several things including wire alloy and wire insulation used. Wire charts provide maximum ampacity as well as resistance so loss over distance can be calculated.
Ron
Ron
Just as a back of the envelope calculation, let us assume that we want the power dissipated by the wire to be less than 60 watts. You can almost put your hand on a 60 watt incandescent light bulb. You might not want to keep it there however.
80 Amps * 80 Amps * R = 60 Watts
implies R <= .009375 Ohms
There are approximately 305 meters in 1000 feet, so
R/1000' = .009375 * 305 = 2.859 Ω/1000'
As you can see, any AWG guage smaller than 14 guage should meet the requirement.
There is also the change in resistance with an increase in temperature to consider.
This table indicates you might need something a bit larger in diameter (smaller in AWG), maybe #6
http://en.wikipedia.org/wiki/American_wire_gauge
And this table says that #6 in free air can carry 101 amps
http://wiki.xtronics.com/index.php/Wire-Gauge_Ampacity
80 Amps * 80 Amps * R = 60 Watts
implies R <= .009375 Ohms
There are approximately 305 meters in 1000 feet, so
R/1000' = .009375 * 305 = 2.859 Ω/1000'
As you can see, any AWG guage smaller than 14 guage should meet the requirement.
There is also the change in resistance with an increase in temperature to consider.
This table indicates you might need something a bit larger in diameter (smaller in AWG), maybe #6
http://en.wikipedia.org/wiki/American_wire_gauge
And this table says that #6 in free air can carry 101 amps
http://wiki.xtronics.com/index.php/Wire-Gauge_Ampacity
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I don´t think that AWG 14 is a good choice, according to wikipedia it is rated for 15 to 25A. The isolation would burn off at 80A.
You definitley don´t want to be loosing 60W over a meter long piece of wire, unless you want it to glow red.
It seems that AWG 6 to 4 would be about right.
You definitley don´t want to be loosing 60W over a meter long piece of wire, unless you want it to glow red.
It seems that AWG 6 to 4 would be about right.
I agree in that using AWG 14 for anything 80 amps is going to be a fusible link.
There are countless ampacity charts out there and formulas. There are also variables the original poster fails to mention. Single wire or multiple? Enclosed or non enclosed> If enclosed the enclosure material and pipe diameter? Wire insulation type? All of this makes it very difficult to calculate the heat rise as the original poster mentions. That said for a 1 meter or roughly 3 foot run of 80 amp service I would be going with AWG 6 or AWG 4 and likely AWG 4 for several conductors in a pipe. While AWG 6 would be fine I like a margin of at least 20%. To refine this for the original poster since AWG is wonderful where I live I would be using Cu (Copper) conductor wire that is 0.162 inch diameter or 4.12 mm diameter, that would be AWG 6. If the ambient temperatures are warm then derate that. If using Aluminum conductors with a Cu wash or plating I would derate further.
Here in the US the NEC (National Electric Code) calls the shots. That is for residential as well as industrial. I just got done arguing this with my neighbor. He plans to upgrade his residential service from the old, very old, 60 Amp fuse box to a new 200 Amp service entry. I told him AWG 4/0 for the entry and weather head using IMT tube and he wanted plastic PVC. Sorry the weatherhead comes down the driveway side of the house so IMT or it will never pass the inspection and 4/0 Cu in a 2" pipe or it will never pass muster.
Ron
This issue is not as straightforward as it appears on the surface. I said the inital calculation was back of the envelope. I specifically did not account for the fact that the resistance of copper goes up as the temperature increases. This is a bad thing for our purposes. It also does not account for the fact that this meter of wire has three possible means of dissipating the heat namely: radiation, conduction, and convection. What will happen depends on what the temperature of the wire is at thermal equilibrium. I think the the characterization of 14 AWG as a fusible link is hyperbole for effect. I don't disagree that it will get quit warm and that is why I amended my initial analysis after further research.
Turns out I don't think there is a straightforward way to calculate the temperature rise without a great deal more information. Turns out the tables are kinda like the "Pirate's Code"; they're more like guidelines than actual rules.
BTW the melting point of copper is 1,085 °C or 1,358 °K. At that temperature it is going to be radiating heat as a function of the 4th power of the temperature times a constant to account for the fact it is not a Black Body.
Turns out I don't think there is a straightforward way to calculate the temperature rise without a great deal more information. Turns out the tables are kinda like the "Pirate's Code"; they're more like guidelines than actual rules.
BTW the melting point of copper is 1,085 °C or 1,358 °K. At that temperature it is going to be radiating heat as a function of the 4th power of the temperature times a constant to account for the fact it is not a Black Body.
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Hello,
In power heating, a one square inch surface area can dissipate about 1 watt of power continuously while maintaining a 60 degree C temperature rise.
An AWG #10 wire has about 1 square inch of surface area every 3.3 inches, so every 10 inches can dissipate 3 watts, and with insulation we should make that 12 inches. So every foot dissipates 3 watts with a temperature rise of 60 degrees C.
The resistance of #10 wire is close to 0.001 ohms per foot, which means for one foot of wire at 80 amps it has to dissipate 6.4 watts, which is about double what the #10 wire can do so the temperature would rise by more than 120 degrees C which is probably unacceptable.
Going down 3 wire gauge sizes results in half the resistance and sqrt(2) times the surface area, so going down 3 wire sizes to #7 wire would result in about 3.2 watts per foot, and because the surface area increases by a factor equal to sqrt(2) we'd only see about a 42 degree C rise in temperature.
So #7 should be able to handle it but yes #6 would be even better.
The short run means some heat will conduct out of the ends of the wire if there is enough heat conduction at those ends.
Very often the real issue is the voltage drop. The voltage drop for a #10 gauge wire (if it could handle it that is) at 80 amps would be about 80 mv per foot, and for #7 wire about 40 mv per foot, so for three feet that's about 240 mv for the #10 wire and 120 mv for the #7 wire. If 120 mv is not acceptable then you still have to go to a lower gauge wire. #6 gauge wire will drop about 95 mv at 80 amps.
In power heating, a one square inch surface area can dissipate about 1 watt of power continuously while maintaining a 60 degree C temperature rise.
An AWG #10 wire has about 1 square inch of surface area every 3.3 inches, so every 10 inches can dissipate 3 watts, and with insulation we should make that 12 inches. So every foot dissipates 3 watts with a temperature rise of 60 degrees C.
The resistance of #10 wire is close to 0.001 ohms per foot, which means for one foot of wire at 80 amps it has to dissipate 6.4 watts, which is about double what the #10 wire can do so the temperature would rise by more than 120 degrees C which is probably unacceptable.
Going down 3 wire gauge sizes results in half the resistance and sqrt(2) times the surface area, so going down 3 wire sizes to #7 wire would result in about 3.2 watts per foot, and because the surface area increases by a factor equal to sqrt(2) we'd only see about a 42 degree C rise in temperature.
So #7 should be able to handle it but yes #6 would be even better.
The short run means some heat will conduct out of the ends of the wire if there is enough heat conduction at those ends.
Very often the real issue is the voltage drop. The voltage drop for a #10 gauge wire (if it could handle it that is) at 80 amps would be about 80 mv per foot, and for #7 wire about 40 mv per foot, so for three feet that's about 240 mv for the #10 wire and 120 mv for the #7 wire. If 120 mv is not acceptable then you still have to go to a lower gauge wire. #6 gauge wire will drop about 95 mv at 80 amps.
Good information. Thanks.
Turns out that bare copper with a heavy oxide layer is a terrific Black Body radiator with an emissivity of about 0.78
0.78(5.67e-8)(1358)^4 = 150410 Watts/M^2
I estimate the meter of #14 AWG would have an area of 8 square inches
so our wire would be radiating 776 watts over 8 sq. in.
My conclusion is that under such conditions the 60 watts would not be able to melt the #14 guage wire.
Turns out that bare copper with a heavy oxide layer is a terrific Black Body radiator with an emissivity of about 0.78
0.78(5.67e-8)(1358)^4 = 150410 Watts/M^2
I estimate the meter of #14 AWG would have an area of 8 square inches
so our wire would be radiating 776 watts over 8 sq. in.
My conclusion is that under such conditions the 60 watts would not be able to melt the #14 guage wire.
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It isn't really clear what environmental conditions were applied to the meter of wire. Certainly if the wire has insulation on it that will effect the range of acceptable outcomes. 45° C certainly appeals to my sense of what might be safe and unsafe in a lab environment for a random 1 Meter of wire. BTW an 80 A supply @ +5V was not an uncommon thing the early days of TTL where a processor might have involved 100 or more IC equivalents over a 1.5 sq. ft. area.
Lundwall_Paul
- Joined Oct 18, 2011
- 236
What is the load? Is it 80 amps pulsed or constant current? What voltage is the supply? Once you select a gauge of wire you should measure to see what the voltage drop is across the 1 meter of wire.
Hello again,
In post #8 i am not sure where the 776 watts came from.
Number 14 (AWG) wire has resistance of about 0.0025 ohms per foot, and the temperature should rise by about 60 degrees with an actual power dissipation of about 0.2 watts per inch, or about 2.4 watts per foot. With 20 amps the power dissipation is about 1 watt per foot so the temperature rise would be about half that, or 30 degrees. That's probably as high as we want to go.
Initial quick calculation:
With 80 amps, we have 16 watts per foot, which is about 6.7 times higher than the reference of 2.4 watts per foot, meaning the temperature would rise to something like 400 degrees C, which is a typical higher than medium soldering iron temperature. Unless there was special insulation the insulation would melt and burn, but in free air the copper, having a much higher melting temperature, would not melt.
What the above quick calculation does not consider is the rise in resistance with temperature. With a 400 degree wire, the resistance will more than double which would mean the power dissipation would go up with a constant 80 amp current. What this means is that the temperature of the wire starts to get close to the melting point of copper which is about 1083 degrees C, which would mean the resistance would increase more and thus dissipate even more power (assuming constant current again) and so i doubt the wire can take that for very long.
It would start to depend a lot on other things like heat conduction at the ends, but it looks bad. The voltage drop would be terrible anyway.
Maybe if you are designing a space heater it's ok
Note that the 80 amp fusing time is about 1 minute, so the wire will take a little while to heat up. There is a chance that because it takes so long that the conduction at the ends will prevent it from actually melting, but it would be so hot it would be dangerous anyway.
In post #8 i am not sure where the 776 watts came from.
Number 14 (AWG) wire has resistance of about 0.0025 ohms per foot, and the temperature should rise by about 60 degrees with an actual power dissipation of about 0.2 watts per inch, or about 2.4 watts per foot. With 20 amps the power dissipation is about 1 watt per foot so the temperature rise would be about half that, or 30 degrees. That's probably as high as we want to go.
Initial quick calculation:
With 80 amps, we have 16 watts per foot, which is about 6.7 times higher than the reference of 2.4 watts per foot, meaning the temperature would rise to something like 400 degrees C, which is a typical higher than medium soldering iron temperature. Unless there was special insulation the insulation would melt and burn, but in free air the copper, having a much higher melting temperature, would not melt.
What the above quick calculation does not consider is the rise in resistance with temperature. With a 400 degree wire, the resistance will more than double which would mean the power dissipation would go up with a constant 80 amp current. What this means is that the temperature of the wire starts to get close to the melting point of copper which is about 1083 degrees C, which would mean the resistance would increase more and thus dissipate even more power (assuming constant current again) and so i doubt the wire can take that for very long.
It would start to depend a lot on other things like heat conduction at the ends, but it looks bad. The voltage drop would be terrible anyway.
Maybe if you are designing a space heater it's ok
Note that the 80 amp fusing time is about 1 minute, so the wire will take a little while to heat up. There is a chance that because it takes so long that the conduction at the ends will prevent it from actually melting, but it would be so hot it would be dangerous anyway.
Hi,
Ok, thanks for clearing that up.
A #14 wire has only about 2.4 square inches so the estimate of the wattage that way would be around in the neighborhood of 200 watts. I am not sure i can trust that estimate though, and the wire has no coating. That estimate tells us that three feet of #14 wire can take over 160 amps without melting. That's got to easily be welding current.
I think the 10 second fusing current would be 160 amps, so the wire would melt in roughly about 10 seconds.
Hi,
Yes i am sorry, i calculated the surface area for one foot not for one meter or even three feet as i had done in the past.
Anyway, i used a more accurate calculation to calculate the final temperature of the wire, taking into account that the resistance goes up as the temperature rises. There are various factors that can change this calculation but the basic calculation turns up some interesting things.
Here is the graph of the temperature of wires in AWG gauges from 2 through 14 and their max estimated temperature, assuming the liquid phase resistance is the same as the solid state phase. It wont really be that way but that only affects wire gauges at the top end where the max temperature goes over about 1083 degrees C approximately.
We knew that the temperature rises and that causes an increase in resistance, and that increase causes more power dissipation because of the constant current, and that causes a higher temperature, etc., and so this causes an exponential effect and for certain wire sizes they just cant take temperatures over a certain limit because they will most likely enter the liquid phase. For a wire hanging from the ends in a horizontal position, this would be even more interesting and even funny because the wire would at first droop taking on the form of a catenary and that would lead to even higher resistance distributed in an unusual way across the now drooping and stretching wire length because of the unequal weight distribution, which would cause yet another increase in temperature, and the more the wire droops and stretches the higher the resistance increases yet still, until it breaks apart under it's own weight.
Normally when the resistance rises the current decreases due to more drop in voltage, but with constant current this indirect 'cooling' effect does not happen.
As you can see from the plot number 14 gauge wire doesnt look like it has much chance to survive.
Yes i am sorry, i calculated the surface area for one foot not for one meter or even three feet as i had done in the past.
Anyway, i used a more accurate calculation to calculate the final temperature of the wire, taking into account that the resistance goes up as the temperature rises. There are various factors that can change this calculation but the basic calculation turns up some interesting things.
Here is the graph of the temperature of wires in AWG gauges from 2 through 14 and their max estimated temperature, assuming the liquid phase resistance is the same as the solid state phase. It wont really be that way but that only affects wire gauges at the top end where the max temperature goes over about 1083 degrees C approximately.
We knew that the temperature rises and that causes an increase in resistance, and that increase causes more power dissipation because of the constant current, and that causes a higher temperature, etc., and so this causes an exponential effect and for certain wire sizes they just cant take temperatures over a certain limit because they will most likely enter the liquid phase. For a wire hanging from the ends in a horizontal position, this would be even more interesting and even funny because the wire would at first droop taking on the form of a catenary and that would lead to even higher resistance distributed in an unusual way across the now drooping and stretching wire length because of the unequal weight distribution, which would cause yet another increase in temperature, and the more the wire droops and stretches the higher the resistance increases yet still, until it breaks apart under it's own weight.
Normally when the resistance rises the current decreases due to more drop in voltage, but with constant current this indirect 'cooling' effect does not happen.
As you can see from the plot number 14 gauge wire doesnt look like it has much chance to survive.
