How To Calculate Max Ideal Ampacity of Single Conductor

Status
Not open for further replies.
circuit_girl said:
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?
Click to expand...
Just a thought........
Can you try it in SI units? It is simpler than Imperial.
 
K8MHZ said:
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.
Click to expand...
@K8MHZ No, no... not trying to come up with a fuse. What I'm wanting is to determine a value for any size AWG conductor for any material, that ballparks the highest current it can withstand before it fails. I'm trying to determine my ideal 'outside' value for such. So that in other calculations I'm doing for a larger project, I know if I've got a flyer, or if my equations are generating reasonable results that I can then go test and/or work backwards from in terms of derating etc.

I'm looking for an ideal value because at a minimum it simplifies (or should simplify) initial calculations. If the max ideal amperage a 14AWG Copper wire can take is actually 166A, I want to know how that was calculated so I can repeat it. And I want to know how to do that with other metals.

As an example- what is max the current a 42AWG 1-foot long piece of Nichrome60 wire can handle? Nobody knows. is it 10mA? 500mA? Millions of feet of this wire are made every year... and nobody can tell me how much current it can handle before it fries ideally or with everything else calculated in.

I'm not really concerned larger conductors. It's the small ones, beyond 40AWG that are giving me the biggest problem because there is very little information in any table for wiring of this size.

-C
 
winnie said:
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.



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
Click to expand...
@winnie I just saw this- let me digest this... _thanks!_
 
circuit_girl said:
@K8MHZ No, no... not trying to come up with a fuse. What I'm wanting is to determine a value for any size AWG conductor for any material, that ballparks the highest current it can withstand before it fails. I'm trying to determine my ideal 'outside' value for such. So that in other calculations I'm doing for a larger project, I know if I've got a flyer, or if my equations are generating reasonable results that I can then go test and/or work backwards from in terms of derating etc.

I'm looking for an ideal value because at a minimum it simplifies (or should simplify) initial calculations. If the max ideal amperage a 14AWG Copper wire can take is actually 166A, I want to know how that was calculated so I can repeat it. And I want to know how to do that with other metals.

As an example- what is max the current a 42AWG 1-foot long piece of Nichrome60 wire can handle? Nobody knows. is it 10mA? 500mA? Millions of feet of this wire are made every year... and nobody can tell me how much current it can handle before it fries ideally or with everything else calculated in.

I'm not really concerned larger conductors. It's the small ones, beyond 40AWG that are giving me the biggest problem because there is very little information in any table for wiring of this size.

-C
Click to expand...

Understood.

One thing I noticed is no attention paid to the duration of the current. Are you just assuming it to be the NEC definition of 'constant'? Which would be in line with an 'outside' value.

Some other variables to be considered....solid vs. stranded. AC vs. DC. 120hz vs. other frequencies. 12 volt vs. higher.....free air or enclosed?

Repeating the experiment to run 166 amps through a 14AWG copper wire for 3+ hours would be fun. Then you would have to do 167 amps to make sure the 166 was accurate.
 
K8MHZ said:
Understood.

One thing I noticed is no attention paid to the duration of the current. Are you just assuming it to be the NEC definition of 'constant'? Which would be in line with an 'outside' value.

Some other variables to be considered....solid vs. stranded. AC vs. DC. 120hz vs. other frequencies. 12 volt vs. higher.....free air or enclosed?

Repeating the experiment to run 166 amps through a 14AWG copper wire for 3+ hours would be fun. Then you would have to do 167 amps to make sure the 166 was accurate.
Click to expand...
I'll say continuous on duration- either it will handle it or it will melt. If it's at the melt temp, then I know to stay below that. Open air. Solid core. DC only. This is purely about direct maximum energy the conductor can handle in open air to be at the fry point, so I know to stay below it by a good margin. Right now the ONLY definitive thing I have is the melt-point of copper (or any other metal/alloy I look at).
 
circuit_girl said:
@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:

Click to expand...

I see that 700 circular mil per amp, but the only place where I've seen something similar is for designing windings in electric motors, where you have lots of wire packed together and so the surface area of an individual wire is quite meaningless.

The above table simply does not match the electric code table, do a search for NEC 310.16 and you will see what I mean.

circuit_girl said:
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?
Click to expand...

I was using the K factor for the electricial resistivity of copper, expressed in ohms for 1 foot length and 1 circular mil cross section. I derived it from the specific resistivity of copper given in ohm meters. That value can be calculated using quantum mechanics, but I don't understand that aspect well enough to explain it.

I also used K to mean degrees kelvin, and tried to make it clear the difference.

-Jon
 
circuit_girl said:
I've tried both and neither comes up with the 10,224 value for K used by @gar, I believe.
Click to expand...

I believe I can tell you where that number comes from.

My guess is that it is ohms cmil / kilofoot at 20C.

R = 1.68*10^-8 (ohm meter) * 1000 feet * (0.3048 meter/foot) / ( 1 cmil * 5.06707*10^-10 (m^2/cmil)) = 10106.

Okay, I missed by a percent, but you get the idea.

-Jon
 
winnie said:
I believe I can tell you where that number comes from.

My guess is that it is ohms cmil / kilofoot at 20C.

R = 1.68*10^-8 (ohm meter) * 1000 feet * (0.3048 meter/foot) / ( 1 cmil * 5.06707*10^-10 (m^2/cmil)) = 10106.

Okay, I missed by a percent, but you get the idea.

-Jon
Click to expand...
Okay, let me play with that and see what I can come up with
-C
 
winnie said:
I see that 700 circular mil per amp, but the only place where I've seen something similar is for designing windings in electric motors, where you have lots of wire packed together and so the surface area of an individual wire is quite meaningless.

The above table simply does not match the electric code table, do a search for NEC 310.16 and you will see what I mean.



I was using the K factor for the electricial resistivity of copper, expressed in ohms for 1 foot length and 1 circular mil cross section. I derived it from the specific resistivity of copper given in ohm meters. That value can be calculated using quantum mechanics, but I don't understand that aspect well enough to explain it.

I also used K to mean degrees kelvin, and tried to make it clear the difference.

-Jon
Click to expand...
@winnie My problem with NEC 310.16 (and much of NEC's other documentation on the matter) is that is not inclusive. It covers a few AWG sizes, none that I'm really interested in, and for others they pick arbitrary values based on fuzzy logic only they know of- not published formulas. My frustration stems from the fact that once people start using tables, the original knowledge, logic, and JUSTIFICATION for the values in the tables is lost- which should *NEVER* happen from an engineering standpoint. Tables are meaningless without published, reviewable, equational and empirical data and reasoning to back the table values up.

I'm still sort of where I started at. The only actual way, based on pure physics, to find maximum amperage a bare conductor of any material of a given AWG size can handle before it melts is to determine maximum free electrons in 1cmil of conductor and do a coulomb to amp conversion and scale it. Derating is irrelevant. Dissipation is irrelevant. All that is tertiary to simple physics. This will provide the absolute maximum amperage value that is possible in an ideal condition for 1 second or more through a conductor. I can derate from there.

----

I want to thank all you really cool guys for helping this little geeky girl out.

xoxoxo -C
 
circuit_girl said:
@winnie My problem with NEC 310.16 (and much of NEC's other documentation on the matter) is that is not inclusive. It covers a few AWG sizes, none that I'm really interested in, and for others they pick arbitrary values based on fuzzy logic only they know of- not published formulas. My frustration stems from the fact that once people start using tables, the original knowledge, logic, and JUSTIFICATION for the values in the tables is lost- which should *NEVER* happen from an engineering standpoint. Tables are meaningless without published, reviewable, equational and empirical data and reasoning to back the table values up.
Click to expand...

The entire code itself does not contain justification for most of its requirements. But it should be noted that the code is _not_ for engineers. It is intended to provide a set of practical methods that a tradesperson can used to put together a safe system. I believe that NEC's tables do have some basis in the Neher-McGrath equations, and in fact 'under engineering supervision' this equation may be used to calculate conductor ampacity.

The history of the justification of the requirements does exist in the records of the code making panels. The NEC does not exist in a vacuum, and you can look up the reasoning behind every code requirement. I do however agree that basis for code requirements must be available and not simply 'that's the way its always been done'.

circuit_girl said:
I'm still sort of where I started at. The only actual way, based on pure physics, to find maximum amperage a bare conductor of any material of a given AWG size can handle before it melts is to determine maximum free electrons in 1cmil of conductor and do a coulomb to amp conversion and scale it. Derating is irrelevant. Dissipation is irrelevant. All that is tertiary to simple physics. This will provide the absolute maximum amperage value that is possible in an ideal condition for 1 second or more through a conductor. I can derate from there.
Click to expand...

I am going to suggest that dissipation is the most important factor, not 'irrelevant'. If you assume that all heat produced by current flowing in the wire stays in the wire, then any amount of current will eventually cause the wire to overheat and melt.

IMHO you need to break the problem down into reasonable parts.

'determine maximum free electrons in 1cmil of conductor and do a coulomb to amp conversion and scale it' sounds like a an approach to calculating the resistance of the material from first principals. Sounds like fun, but keep in mind that resistivity of various materials is tabulated.

Similarly you should in theory be able to calculate the melting point of various materials, but that is also tabulated.

From the melting point and geometry you can calculate the heat dissipation at the melting point of the material. Dissipation will include conduction, convection, and radiation.

From the geometry and the resistivity of the material you can calculate the current flow which will result in the allowed heat dissipation.

Solving (or looking up or measuring) each of these aspects is IMHO necessary for determining current handling capacity.

circuit_girl said:
I want to thank all you really cool guys for helping this little geeky girl out.

xoxoxo -C
Click to expand...

Best of luck. It sounds like you are working on a cool problem. Please let us know how it turns out, we are all here to learn.

-Jon
 
@winnie @SG-1 @Julius Right - Thank you guys. Much appreciated; I will digest this info as well, and yes I will report back with what I eventually find. I am think that ultimately I'm going to have to determine at least 1 factor empiraclly- like resistance of 1 foot of a given wire in Ohms, and then calculate Resistivity from that.

-C
 
Status
Not open for further replies.
Top