Ground Wire K Factor

[mbrooke]

Where do I find the K factor for various size AWG copper wire? I'm not finding it in the NEC's tables.

 

[winnie]

Isn't the 'K' factor simply the resistivity of the conductor expressed in ohms per K circular mil per 1000 feet?

At 20C, the resistivity of copper is 1.68×10−8 ohm * meter. There are 1.97 * 10^9 circular mil per square meter, and 304.8 meters in 1000 feet, so the resistivity of copper at 20C is 10.1 ohm * kcmil / 1000 feet.

You can adjust for different temperatures using the tempco of the conductor, and adjust for different conductors by looking up their standard resistivity expressed in ohm * meter.

Calculating the K factor for hot dogs is left as an exercise for the student http://tuhsphysics.ttsd.k12.or.us/Research/IB07/JenkLaToLive/index.htm

 

[Mouser]

Based on information found in Table 8 in Chapter 9.

K = cm x Ohms per foot

 

[mbrooke]

Isn't the 'K' factor simply the resistivity of the conductor expressed in ohms per K circular mil per 1000 feet?

At 20C, the resistivity of copper is 1.68×10−8 ohm * meter. There are 1.97 * 10^9 circular mil per square meter, and 304.8 meters in 1000 feet, so the resistivity of copper at 20C is 10.1 ohm * kcmil / 1000 feet.

You can adjust for different temperatures using the tempco of the conductor, and adjust for different conductors by looking up their standard resistivity expressed in ohm * meter.

Calculating the K factor for hot dogs is left as an exercise for the student http://tuhsphysics.ttsd.k12.or.us/Research/IB07/JenkLaToLive/index.htm

Oh! Forgot about that K factor. My bad, I meant this one:

https://imgur.com/qmBqEnD

 

[winnie]

Ahh. I don't know the answer in this case, but I can guess since it is related to something that I have done for electric motors.

The power dissipated in a conductor is R * I^2. The energy dissipated is t * R * I^2. If you consider short duration events, then all of this energy goes to heating up that conductor.

The conductor has a certain heat capacity that tells you how much temperature change you get when you dump a certain amount of energy into a mass of conductor.

Take a specific example: 1000 feet of #10 Cu, with 200A flowing in it for 5 seconds.

The resistance is 1 Ohm, so you are dissipating 40kW in that wire. 5 seconds means 200kJ.

The mass of this wire is 14.2 kg, and the specific heat capacity of the wire is 385 J/kg/K.

So we get 200000 / ( 14.2 *385) = 36.6 K

So when you dump this huge overload current through this #10 conductor, you can expect it to heat up by 37K in 5 seconds.

Now I played fast and loose with the approximations, didn't include the temperature coefficient of resistance or heat capacity, but you get the basic idea: you are dumping a lump of energy into a lump of mass and can calculate how much energy (I^2 * R * t) and how much temperature rise (energy / heat capacity).

My guess is that the 'k' factor generalizes the above. You consider the resistance, mass, and heat capacity of a _unit_ conductor, one of 'unit' cross section.

The length doesn't matter, because as length changes both the total energy dissipated and the total mass to handle that energy change in the same fashion.

So consider a 'unit wire' exactly 1mm^2 in cross section and 1 m long. This has a resistance of 0.0168 Ohm, a mass of 0.00896 kg, and a heat capacity of 3.5J/k. Say for conditions of use we permit the temperature of the conductor to rise by 100K during a fault. Then this unit conductor can be permitted to absorb 350J

K is proportional to heat capacity, proportional to allowed temperature rise, inversely proportional to resistance, and you need to take the square root of the whole mess to match the square root on top.

This makes K = sqrt(3.5 * 100 / 0.0168) = 144

Again just an approximation ignoring the way the various factors change with temperature.

-Jon



Say your conductor has a resistance of 1 ohm

 

[mbrooke]

Ahh. I don't know the answer in this case, but I can guess since it is related to something that I have done for electric motors.

The power dissipated in a conductor is R * I^2. The energy dissipated is t * R * I^2. If you consider short duration events, then all of this energy goes to heating up that conductor.

You've got!

The assumption is the all the heat will be contained to the conductor and none will radiate into the environment. The ending temperature can not go over the rating of the insulation (90*C).

The conductor has a certain heat capacity that tells you how much temperature change you get when you dump a certain amount of energy into a mass of conductor.

Take a specific example: 1000 feet of #10 Cu, with 200A flowing in it for 5 seconds.

The resistance is 1 Ohm, so you are dissipating 40kW in that wire. 5 seconds means 200kJ.

The mass of this wire is 14.2 kg, and the specific heat capacity of the wire is 385 J/kg/K.

So we get 200000 / ( 14.2 *385) = 36.6 K

So when you dump this huge overload current through this #10 conductor, you can expect it to heat up by 37K in 5 seconds.

Now I played fast and loose with the approximations, didn't include the temperature coefficient of resistance or heat capacity, but you get the basic idea: you are dumping a lump of energy into a lump of mass and can calculate how much energy (I^2 * R * t) and how much temperature rise (energy / heat capacity).

My guess is that the 'k' factor generalizes the above. You consider the resistance, mass, and heat capacity of a _unit_ conductor, one of 'unit' cross section.

The length doesn't matter, because as length changes both the total energy dissipated and the total mass to handle that energy change in the same fashion.

Ah, but length does matter. The longer the run the longer it takes to trip the breaker. The longer is takes to trip the breaker the more heat is dissipated into the wire. Beyond a certain length the insulation will in theory melt off.

So consider a 'unit wire' exactly 1mm^2 in cross section and 1 m long. This has a resistance of 0.0168 Ohm, a mass of 0.00896 kg, and a heat capacity of 3.5J/k. Say for conditions of use we permit the temperature of the conductor to rise by 100K during a fault. Then this unit conductor can be permitted to absorb 350J

K is proportional to heat capacity, proportional to allowed temperature rise, inversely proportional to resistance, and you need to take the square root of the whole mess to match the square root on top.

This makes K = sqrt(3.5 * 100 / 0.0168) = 144

Again just an approximation ignoring the way the various factors change with temperature.

-Jon

Hmmmm, got me thinking... R goes up as the wire gets hotter. Sigma calculus needed on this one?

 

[winnie]

Ah, but length does matter. The longer the run the longer it takes to trip the breaker. The longer is takes to trip the breaker the more heat is dissipated into the wire. Beyond a certain length the insulation will in theory melt off.

Hmmmm, got me thinking... R goes up as the wire gets hotter. Sigma calculus needed on this one?

I agree, length does matter in a real circuit because it changes fault current and clearing time.

IMHO it doesn't matter for this equation because current and time are givens.

I could imagine needing to use an iterative method where you pick a wire size, calculate the I and t, use this to calculate a new wire size, rinse and repeat. Or perhaps current and clearance are dominated by other circuit elements.

Jon

 

[mbrooke]

I agree, length does matter in a real circuit because it changes fault current and clearing time.

IMHO it doesn't matter for this equation because current and time are givens.

I could imagine needing to use an iterative method where you pick a wire size, calculate the I and t, use this to calculate a new wire size, rinse and repeat. Or perhaps current and clearance are dominated by other circuit elements.

Jon

I'll look at Annex A a bit more and give it a few more runs.

With that said I'm trying to figure out how table 250.122 was derived.