How To Calculate Max Ideal Ampacity of Single Conductor

Status
Not open for further replies.
C

circuit_girl

Member
Location
GA
Occupation
Circuit Design Engineer
Hi Guys,

I've been beating my head against a wall for 4 days, with cmils, meter Kelvins, Coulombs, and so forth. I've spent so much time on Google, I have a headache. I just feel numb :/ - I am trying to determine the maximum amount of current an AWG sized conductor can handle before it melts. I don't care about derating, or temperature coefficients, or all of that. I just want an ideal starting point and then I can work on derating and all the rest. I apologize that this is a bit all over the place, I'm frazzled :(

I know that NEC ampacity tables are very conservative. Essentially (for copper), they are 1A per 700cmil. From what I can guestimate, it looks like their rule-of-thumb is really about insulation, and anything the conductor is surrounded with or attached to- A #10AWG wire rated at 15A (20C) is possibly more likely able to handle 100A before it melts (ideally)... but how do I calculate what that actually is?

Unfortunately, NEC deals primarily with Copper and Aluminum... I need to be able to calculate this stuff for other metals, like Nichrome, and Tungsten, etc. - I want to do it mathematically, not rely on an Ohmeter or actually frying a wire with a current generator to empirically determine an answer. I know how to calculate Specific Resistance from AWG size if I know the Specific Resistivity (or 'K' value) of the conductor and it's material.... but I haven't figured out how the K value is calculated. I know the tables are based on 20C, but what if it's 25C, or 75C? How is K calculated?

I'm at a point where I think the only way I can figure it out is count electrons in a given material and work that back to Amps, but.... there has to be a smarter way.... :/

Can anyone help this poor girl out? To Sum:

o how to calculate ideal amps to melt a conductor of given AWG and material?
o how to calculate Specific Resistivity (aka 'K' value) of a given material at a given temp (20C, 50C, 75C)?

_Thank you for *any* help_
C
 
Interesting question. I don't have a definite answer for you but I do have just a couple thoughts. We would need to make some assumptions. Presumably we would just assume conductor in free air with no air movement. It's easy to calculate the Watts lost in the conductor for a given current, the harder part is calculating how that heat is dissipated into the surrounding medium (air). Also the insulation will have a little bit of insulative value. Defining "melt" is a bit tricky too. Generally I think a plastic just gets softer and softer as temperature increases......I guess you would want the point at which it starts to permanently deform from gravity? Or maybe you need a lesser temperature if you are assuming there are spots where it is say going around a corner and has a force of a corner or edge on the insulation. you would need to define some sort of compressive yield strength and its associated temperature.

This sort of thing may be easier to just test in the real world instead of trying to find a theoretical value.
 
Wow, you don't start small with your questions :)

I won't touch the question about calculating specific conductivity, my quantum theory isn't close to good enough to understand or explain that.

You also don't ask how to calculate the melting point of the conductive material, which is also relevant to your question, and I don't know the theory used to predict melting points.

So let's assume that you already have conductivity information, what next?

The current carrying capacity is limited by conductor temperature.

By simple ohm's law, you can calculate the heat produced per length of conductor. (Current^2) * (Specific resistivity) * (length) / (cross section) = (heat produced)

The conductor steady state temperature is reached when the heat carried away equals the heat produced. The calculation here depends upon the surrounding ambient temperature and the thermal conductivity of the surrounding materials. If all you know is the specific thermal conductivity ( W/m K ) then you need to calculate the actual thermal conductivity of the surroundings including their geometry. Explaining that is also beyond me; I work with situations where the geometry is solved and I know W/K. If you know W/K then you can simply figure the temperature difference required for a given thermal flow in W.

So if a length of wire is carrying current and dissipating W watts of heat, then the thermal resistance in W/K tells you the temperature difference K between the wire and the surrounding ambient for that heat to dissipate.

Note in your calculation that resistance changes with temperature, so when you calculate the heat produced by the wire you need to know its temperature. Often you need to iterate, assume one temperature, calculate the heat produced, use this to calculate temperature and then use this new temperature to calculate resistance and heat produced.

Nec tables are not based on a constant circular mils per amp. Small conductors have more surface area per circular mil and dissipate heat better, and thus operate at higher current density.

I hope this hand waving helps you find better direction in your study.

Jon
 
There’s a misconception that the “empirical” method is old and unscientific, and the new “Scientific” method is new and based on math and lab analysis.

Scientific knowledge is ALL empirical knowledge.

That is the definition of science. Observation and Deduction. From that we can extrapolate repeatable events and assign numerical values if desired.

But it all starts with empirical data.

This concept has far reaching implications that I will not go into right now, but my point is that in order to find the melting point from AC current for a titanium rod, it may be necessary to melt a titanium rod with AC current, and measure the value. 👍
 
circuit_girl said:
Hi Guys,

I've been beating my head against a wall for 4 days, with cmils, meter Kelvins, Coulombs, and so forth. I've spent so much time on Google, I have a headache. I just feel numb :/ - I am trying to determine the maximum amount of current an AWG sized conductor can handle before it melts. I don't care about derating, or temperature coefficients, or all of that. I just want an ideal starting point and then I can work on derating and all the rest. I apologize that this is a bit all over the place, I'm frazzled :(

I know that NEC ampacity tables are very conservative. Essentially (for copper), they are 1A per 700cmil. From what I can guestimate, it looks like their rule-of-thumb is really about insulation, and anything the conductor is surrounded with or attached to- A #10AWG wire rated at 15A (20C) is possibly more likely able to handle 100A before it melts (ideally)... but how do I calculate what that actually is?

Unfortunately, NEC deals primarily with Copper and Aluminum... I need to be able to calculate this stuff for other metals, like Nichrome, and Tungsten, etc. - I want to do it mathematically, not rely on an Ohmeter or actually frying a wire with a current generator to empirically determine an answer. I know how to calculate Specific Resistance from AWG size if I know the Specific Resistivity (or 'K' value) of the conductor and it's material.... but I haven't figured out how the K value is calculated. I know the tables are based on 20C, but what if it's 25C, or 75C? How is K calculated?

I'm at a point where I think the only way I can figure it out is count electrons in a given material and work that back to Amps, but.... there has to be a smarter way.... :/

Can anyone help this poor girl out? To Sum:

o how to calculate ideal amps to melt a conductor of given AWG and material?
o how to calculate Specific Resistivity (aka 'K' value) of a given material at a given temp (20C, 50C, 75C)?

_Thank you for *any* help_
C
Click to expand...
FWIW, 1mm^sq would overheat/melt at about 20 A.
 
210410-2222 EDT

My copy of "Reference Data for Radio Engineers", fourth Edition, Sixth Printing 1960, on page 55 has a table of fusing current for various materials based on I = K*d^(3/2). For copper #12 the value is 235 A. This is from K = 10,224, and d = 0.0808".

#14 copper wire is 166 A.

#12 would be roughly equivalent to a square shape of 0.072" by 0.072". One mm is about 0.039". This 1 mm would be about 30 A for fusing from my table.

.
 
Last edited:
I have an old excel workbook based on The Standard Handbook for Electrical Engineers ed.10 but I have problem to upload the file[xls]
 
electrofelon said:
Interesting question. I don't have a definite answer for you but I do have just a couple thoughts. We would need to make some assumptions. Presumably we would just assume conductor in free air with no air movement. It's easy to calculate the Watts lost in the conductor for a given current, the harder part is calculating how that heat is dissipated into the surrounding medium (air). Also the insulation will have a little bit of insulative value. Defining "melt" is a bit tricky too. Generally I think a plastic just gets softer and softer as temperature increases......I guess you would want the point at which it starts to permanently deform from gravity? Or maybe you need a lesser temperature if you are assuming there are spots where it is say going around a corner and has a force of a corner or edge on the insulation. you would need to define some sort of compressive yield strength and its associated temperature.

This sort of thing may be easier to just test in the real world instead of trying to find a theoretical value.
Click to expand...
@electrfelon To clarify- I just mean 'open air'. no insulation. Copper begins to deform- I view melting as the point at which one is trying to move more electrons than are available to be moved, so the constituent ability of the material to hold together basically fails at the atomic level.

To further clarify- here is the chemical definition of melting:

As a solid is heated, its particles vibrate more rapidly as the solid absorbs kinetic energy. At some point the amplitude of vibration becomes so large that the atoms start to invade the space of their nearest neighbors and disturb them and the melting process initiates. The melting point is the temperature at which the disruptive vibrations of the particles of the solid overcome the attractive forces operating within the solid.
 
Last edited:
Seven-Delta-FortyOne said:
There’s a misconception that the “empirical” method is old and unscientific, and the new “Scientific” method is new and based on math and lab analysis.

Scientific knowledge is ALL empirical knowledge.

That is the definition of science. Observation and Deduction. From that we can extrapolate repeatable events and assign numerical values if desired.

But it all starts with empirical data.

This concept has far reaching implications that I will not go into right now, but my point is that in order to find the melting point from AC current for a titanium rod, it may be necessary to melt a titanium rod with AC current, and measure the value. 👍
Click to expand...
I'm not against empircal. The thing is, with everything known in chemistry and physics, we should be able to calculate this from an ideal standpoint.
 
circuit_girl said:
@electrfelon To clarify- I just mean 'open air'. no insulation. Copper begins to deform- I view melting as the point at which one is trying to move more electrons than are available to be moved, so the constituent ability of the material to hold together basically fails at the atomic level.
Click to expand...
Ah ok, so no insulation. I still think that is a fuzzy target. As temperature rises the conductor will get softer and the tensile strength will go down. You would still have to choose a temperature at which you are declaring the material "melted".

Edit: for example consider an incandescent light bulb filament. It is white hot, very delicate, requires many supports along its length to support it, and a simple knock will break it. With less supports and or a longer filament running at the same temperature, it would have already melted/broken
 
winnie said:
Wow, you don't start small with your questions :)

I won't touch the question about calculating specific conductivity, my quantum theory isn't close to good enough to understand or explain that.

You also don't ask how to calculate the melting point of the conductive material, which is also relevant to your question, and I don't know the theory used to predict melting points.

So let's assume that you already have conductivity information, what next?

The current carrying capacity is limited by conductor temperature.

By simple ohm's law, you can calculate the heat produced per length of conductor. (Current^2) * (Specific resistivity) * (length) / (cross section) = (heat produced)

The conductor steady state temperature is reached when the heat carried away equals the heat produced. The calculation here depends upon the surrounding ambient temperature and the thermal conductivity of the surrounding materials. If all you know is the specific thermal conductivity ( W/m K ) then you need to calculate the actual thermal conductivity of the surroundings including their geometry. Explaining that is also beyond me; I work with situations where the geometry is solved and I know W/K. If you know W/K then you can simply figure the temperature difference required for a given thermal flow in W.

So if a length of wire is carrying current and dissipating W watts of heat, then the thermal resistance in W/K tells you the temperature difference K between the wire and the surrounding ambient for that heat to dissipate.

Note in your calculation that resistance changes with temperature, so when you calculate the heat produced by the wire you need to know its temperature. Often you need to iterate, assume one temperature, calculate the heat produced, use this to calculate temperature and then use this new temperature to calculate resistance and heat produced.

Nec tables are not based on a constant circular mils per amp. Small conductors have more surface area per circular mil and dissipate heat better, and thus operate at higher current density.

I hope this hand waving helps you find better direction in your study.

Jon
Click to expand...
@winnie Jon, thanks- this is really useful. Can you elaborate, please on your temperature equation?

(Current^2) * (Specific resistivity) * (length) / (cross section) = (heat produced)

What units are we talking about? I have not been able to find any real information on how to determine specific resistivity ('K)' of ambient air surrounding a conductor, in part because of the complexity of air layering around a conductor. But I'm not interested (for the moment) in a perfect answer. I simply want an _ideal_ answer, and then I can work backwards from that.

I * K * <meter?> / <cmil> = Watt?

The reason I'm looking for an ideal answer, is because it gives me the upper end of current possibility so if I calculate anything above that I know it's obviously wrong.

_Thank you_
C
 
electrofelon said:
Ah ok, so no insulation. I still think that is a fuzzy target. As temperature rises the conductor will get softer and the tensile strength will go down. You would still have to choose a temperature at which you are declaring the material "melted".
Click to expand...
Well, I think the known 'melting point' of copper is a good starting point. I'm sure I can find a table for that- but how did _they_ calculate that number?

I just discovered this (which at least establishes an upper limit from temperature standpoint):

www.periodic-table.org

Copper - Melting Point

Copper - Melting Point. Melting point of Copper is 1084.62°C. The melting point of a substance is the temperature at which this phase change occurs.
www.periodic-table.org
 
gar said:
210410-2222 EDT

My copy of "Reference Data for Radio Engineers", fourth Edition, Sixth Printing 1960, on page 55 has a table of fusing current for various materials based on I = K*d^(3/2). For copper #12 the value is 235 A. This is from K = 10,224, and d = 0.0808".

#14 copper wire is 166 A.

#12 would be roughly equivalent to a square shape of 0.072" by 0.072". One mm is about 0.039". This 1 mm would be about 30 A for fusing from my table.

.
Click to expand...
@gar Thank you very much . So in comparison to your table, if I do the calculations based on AWG and Specific Resistivity for 20C (and using NEC's published rule of thumb of 1 Amp per 700cmils for copper) -these calculated values match the published AWG charts:

12AWG:
AWG:
12​
Specific Resistivity*:
10.73​
Wire Diameter (mm):
2.052525388​
Wire Diameter (Inches):
0.080808086​
cmil(s):
6529.946789​
Specific Resistance 'R' (Ohms/Ft):
0.001643199​
NEC Ampacity Rule of Thumb Value for Cu:
700​
NEC Ampacity Rating (Amps):
9.328495413​

So Fuse Amperage for 12AWG:

I = K*d^1.5
I = 6529.946789 * 0.080808086^1.5
I = 150A

and for

14AWG:
AWG:
14​
Specific Resistivity*:
10.73​
Wire Diameter (mm):
1.627726634​
Wire Diameter (Inches):
0.064083726​
cmil(s):
4106.723905​
Specific Resistance 'R' (Ohms/Ft):
0.002612788​
NEC Ampacity Rule of Thumb Value for Cu:
700​
NEC Ampacity Rating (Amps):
5.866748435​

So Fuse Amperage for 14AWG:

I = K*d^1.5
I = 4106.723905 * 0.064083726^1.5
I = 66.6A

I wasn't sure about your K value of 10,224- so I used what calculates to match current AWG tables for copper.

Does this look OK?
-C
 
circuit_girl said:
@winnie Jon, thanks- this is really useful. Can you elaborate, please on your temperature equation?

(Current^2) * (Specific resistivity) * (length) / (cross section) = (heat produced)
Click to expand...

My apologies for not being clear. The above is not a temperature equation, it is simply Ohm's law re-written for power dissipation:

E = I*R (Voltage = current times resistance)
P = E * I (power = current * voltage)
P = I^2 * R (power = I square R)
R = (specific resistivity) * (length) / (cross section) ( resistance is proportional to length and inversely proportional to cross section)

The SI unit for specific resistivity is the ohm meter. At 20C copper has a specific resistivity of 1.68*10^-8 ohm meter
(also 2.043*10^-8 ohm meter at 75C)
Consider a wire 1 foot long with a cross section of 1 circular mil
R = 2.043*10^-8 (ohm*meter) * 1 foot * 0.3048 (meter / foot) / ( 1 cmil * 5.06707*10^-10 (m^2/cmil)) = 12.3 ohm
This is where the 'K' factor of 12.3 comes from in many voltage drop equations. It is simply the resistivity of copper expressed in ohm cmil per foot

In my equation above if you have 30A of current flowing in a wire 1 meter long with a cross section of 6 mm^2:
30A^2 * 2.043*10^-8 (ohm meter) * 1 meter / ( 6*10^-6 meter^2) = 3.065 A^2 ohm = 3.065W

6mm^2 wire carrying 30A has to dissipate about 3 W / meter. Since resistance changes with temperature, the exact dissipation to carry this current will change with temperature.

circuit_girl said:
What units are we talking about? I have not been able to find any real information on how to determine specific resistivity ('K)' of ambient air surrounding a conductor, in part because of the complexity of air layering around a conductor. But I'm not interested (for the moment) in a perfect answer. I simply want an _ideal_ answer, and then I can work backwards from that.

I * K * <meter?> / <cmil> = Watt?
Click to expand...

A bit of confusion above: The 'K' that I was talking about is the electrical resistivity of the copper, not the thermal resistivity of the air. Since the general concept and the math is the same for both, and this discussion involves both, I should have more clearly distinguished the two. And to add even more confusion the unit for thermal conductivity is watts per meter per kelvin, or W/ m K...so that letter K shows up in three different usages (electrical resisitivity constant, thermal conductivity constant, and degrees Kelvin)

When current flows through an electrical conductor, there is some resistance and overcoming that resistance requires voltage. Voltage * current = power, and the resulting thermal power needs to be dissipated or the conductor will heat up. Electrical resistance determines the amount of voltage and thus the power dissipated to permit a given current to flow.

When heat flows through a thermal conductor, there is some thermal resistance and overcoming that resistance requires a temperature difference. Thermal resistance determines the temperature difference needed to permit a given amount of heat to flow.

The thermal conductivity of air is 0.024 W / m K (note: this is a conductivity, not a resistance, so watch out for the trap of confusing the two) This same value could be expressed as 41.7 "thermal ohm" meter where watts per kelvin show up in the same way that amps/volt show up for electrical conductivity

To figure this from basic equations requires solving Fourier's law of heat conduction, but for your purposes you are going to run into a significant complication: the thermal resistance of still air doesn't help you when the air can move, and because air is subject to convection from heat it will move. I can only point you in the direction of the issue, I don't know how to solve for heat conduction with moving air.

-Jon
 
circuit_girl said:
@gar Thank you very much . So in comparison to your table, if I do the calculations based on AWG and Specific Resistivity for 20C (and using NEC's published rule of thumb of 1 Amp per 700cmils for copper) -these calculated values match the published AWG charts:
Click to expand...

I am confused by the 1 amp per 700 cmils. I do not believe that is an NEC rule anywhere for conductors. It certainly doesn't match the values for the ampacity tables.

-Jon
 
In essence, are you trying to design a fuse?

It's been my experience that the results from a 10 minute test will trump the results from 3+ hours of math, every time. :)

I don't think it's possible to come up with a simplistic answer like such and such a metal of a certain diameter will melt at exactly a calculated current. There are other factors do be concerned with. What if you somehow come up with a value for a certain diameter, how will that transfer to a different shape?

I spent a couple years in R&D as a metallurgical technician. One of the projects designing superalloys for jet engines. EVERYTHING was tested. The math just got us in the ballpark. You will need to do a gradient test and identify the incipient melt point to come up with an exact phase change temp and that will only work for that specific shape / size, etc.

Also, your tests must be repeatable. Once you are comfortable with the results, then do the math. That is now that you have the answer, create the math that explains it. That will get you closer for the next test. Sorry, but that is just how it works. Math alone won't do it.
 
winnie said:
I am confused by the 1 amp per 700 cmils. I do not believe that is an NEC rule anywhere for conductors. It certainly doesn't match the values for the ampacity tables.

-Jon
Click to expand...
@winnie I beg your pardon- I've seen this as a foot not in several AWG tables (like this) regarding power transmission as opposed to chassis value for ampacity:


What exactly is your 'K' value (I've seen at as 'A' elsewhere, referred to as a constant). If this is NOT a cmil value, then what is it exactly, and how is it derived?
 
Status
Not open for further replies.
Top