Material Ampacity Comparisons

We frequently use bare conductors in our products. We use steel, copper, nickel-plated steel, stainless steel, and brass connections.

The IACS lists copper wire at 100% conductivity.

NEC Table 310.21 lists 8 gauge bare copper wire for 98 amps based on a 40deg. C ambient in free air.

The circular mil of 8 gauge bare copper wire is 16509.

Compared to copper the IACS lists bare steel at 3% to 15% conductivity.

Does the 3 to 15% conductivity mean that 33 times (for 3%) and 7 times (for15%) as much circular mil of bare steel is needed in order to carry the same ampacity as copper?

Is 3.6 times as much nickel required to carry the same current as copper? (nickel is 22% IACS compared to copper).

Dale Hayes

My percentage was wrong for nickel - it should be 4.5 times as much nickel compared to copper.

Dale Hayes

It's less confusing to express resistivity in ohms/1000'. Resistivity for metals is sometimes expressed as ohm-cm but I've never converted it to ohms/foot.

Not quite.

The ampacity of a conductor is the current that the conductor can carry on a continuous basis without being damaged; usually we mean the 'thermal ampacity' which is the current that the conductor can carry without exceeding its temperature limits.

When current flows through the wire, resistance of the wire causes heat generation. Conductors with greater resistance will produce greater heating for the same current flow; and if you were to scale the conductors as you suggest (increasing area in proportion to resistance), then your various wires would have the same resistance and produce the same total heating.

However heat generation alone does not determine the temperature of the wire. You must also consider heat removal, which depends upon the temperature of the surroundings and the thermal resistance between the conductor and the surroundings. As a rough approximation, the larger conductors will have more heat dissipation capacity than the smaller conductors, and thus could tolerate more heating with the same temperature rise.

Additionally, I presume that different materials have different maximum allowed temperatures. A conductor which can tolerate operating at higher temperature will have a greater ampacity, simply because that higher temperature means that it can tolerate more heat production and thus higher current.

Thus you will find that the conductor cross sectional area for a given ampacity is rather different than a simple inverse proportion to the conductivity.

-Jon

P.S. Larger conductors have greater heat dissipation surface, increasing in proportion to the diameter. But the cross section is increasing as the _square_ of the diameter, meaning that you have more circular mils for a given amount of heat dissipation surface. This means that when looking at a single material and asking what is the ampacity of conductors of different sizes, you will find that larger conductors must operate at lower current densities (fewer amps per circular mil). But when you are looking at _different_ materials and asking 'how much cross section is required to carry a given current', then the cross section will increase more slowly than suggested by the inverse proportion to conductivity.

I stand corrected.

Remember that the heat in a conductor is proportional to the Square of the current.
That is one of the reasons why the current permitted in Table 310.16 is less in proportion to the size of the conductor as the conductor size increases.

Smaller conductors carry more current in proportion to the size of the conductor.
Hence, for larger current, we parallel conductors.

I knew I'd get into this someday, but I kept putting it off.

Let's say some version of the electrical code says that the ampacity of #10 copper wire in an ambient of 30C and insulation rating of 90C is 40 amps, giving a rise above ambient of 90-30=60C.
What is the AWG # for an insulated steel wire of the same ampacity in the same environment?

From a table, I find the resistivity of some type of steel is 120 "units" and the resistivity of some type of copper is 10.4 "units",
so steel has 120/10.4 = 11.5 as much ohms per foot as copper for the same AWG #.

My table also says that some alloy of #10 awg copper wire heat treated in a certain way has a resistance of 1.0 milliohm per foot at 20C. I know the tempco of copper is 0.0039, so at 90C this resistance value would be 0.001(1 + .0039[90-20]) = .00127 ohms.

Heating per foot is proportional to (I^2)R per foot and for copper we have 40 amps giving a rise of 60C. Heating per foot is (40^2)(0.00127) = 2.032 watts/foot. I'd think 2 watts dissipated in this volume of copper would feel slightly warm to the touch, so I think this is a reasonable value, so I might be on the right track so far.

How to get the same heating effect with steel? With steel we're stuck with an R/foot at 90C of .00127(11.5) = .0146 ohms per foot. Therefore, I = (2.032/.0146)^(0.5) = 11.8 amps for AWG #10 steel wire.

A pretty lousy ampacity, for all that work. I hope I didn't make any math errors or grossly wrong assumptions.

And it all changes with Frequency, but that is another can of worms.

I don't know enough to know if you've made any grossly wrong assumptions.

You used a simplifying assumption of ignoring 'skin effect'. My guess is that with a _magnetic_ conductor, you'd need to consider skin effect with smaller conductors than normal.

You also assumed that running a conductor at its table 310.16 ampacity would raise its temperature to the limit. This is how the table is supposed to work, but the table itself is based upon very conservative assumptions about thermal resistance; in most real world situations you would end up with much lower temperature rise then expected from the table.

I think that your calculation of the ampacity of the steel conductors given the ampacity of the copper is correct; simplifying your equations, you could just take the ampacity of the copper conductor and divide it by the square root of the resistance ratio: 40 / sqrt(11.5) = 11.8 given your numbers.

-Jon

So I went around the block to get next door. . .break egg, put on face. . .

Yea, but sometimes you need to go around the block to figure out that you could have just gone next door. Your way would be how I figured it out, I just took the next step.

-Jon

-This talk about skin effect made me wonder if frequency is mentioned anywhere in the NEC. Certainly it has an effect on ampacity.
-So, on page 70-125 of the 1999 NEC (section 310-15) I found a formula that is to be applied "under engineering supervision." The formula is relatively simple but finding values for all the factors is probably like doing income tax.
-I don't know if this is in the current NEC.
-It takes into account way more factors than I considered and it's probably backed up with real data from a lab, and appendix B offered usage guidelines.
-So, bottom line, ditch my post (which was also done under engineering supervision [me]).
-As my own supervisor, I am now going to give myself a coffee break. . .

Nickel Plated Steel

Nickel Plated Steel

Using the same comparison of the 10ga. copper and 10ga. steel wire (10,400cm), what happens to the 11.8 amp rating of the steel wire if we were to nickel plate it? Nickel plated steel can operate at 900 Deg. F., but will the nickel plating increase the amp capacity of the steel wire?

Note: I keep referring to steel because we use steel and nickel-plated steel in certain applications. The 10ga. steel wire is a solid wire.

Dale Hayes

Metallurgy is getting a bit away from my experience and training, but as to conductivity, figure out the skin depth. 1-(1/e), 63.2%, of the current is flowing between this depth and the conductor surface. How thick is the plating, and its conductivity? With some trouble you can probably calculate the advantage of plating.

Skin depth = 1/(pi x f x mu x sigma)^0.5, where mu is permeability of nickel and sigma is the conductivity of nickel, f is 60 Hz and pi is 3.14.

Or, if you have samples, measure them both by passing about an amp of 60 Hz through the conductor by putting it in series with the AC line and a 100 Watt bulb. Without touching the wires that bring the current in (a Kelvin measurement) measure the voltage drop on each wire sample. It's the ratio of resistances that you're looking for, so the current doesn't need to be precise, but it does need to be steady. A steadier source can be had by using car battery in series with a headlight (about 3a) but it's DC.

The tradeoff is that more current will cause heating but it will also let you see the voltage drop to more accuracy, and you may need to resolve this voltage down to millivolts or microvolts. Keithley makes instruments that measure microohms, but they probably use DC.

This gets close to solving the conductivity problem, but ampacity also depends on that formula in the NEC.

By now, I'm sure wire makers (is that your company?) have software that calculates this better than any single human can, but there might be a problem with you getting access to this valuable tool. If it is available somewhere, it probably costs kilobucks.

Metallurgy is getting a bit away from my experience and training, but as to conductivity, figure out the skin depth. 1-(1/e), 63.2%, of the current is flowing between this depth and the conductor surface. How thick is the plating, and its conductivity? With some trouble you can probably calculate the advantage of plating.

Skin depth = 1/(pi x f x mu x sigma)^0.5, where mu is permeability of nickel and sigma is the conductivity of nickel, f is 60 Hz and pi is 3.14.

Or, if you have samples, measure them both by passing about an amp of 60 Hz through the conductor by putting it in series with the AC line and a 100 Watt bulb. Without touching the wires that bring the current in (a Kelvin measurement) measure the voltage drop on each wire sample. It's the ratio of resistances that you're looking for, so the current doesn't need to be precise, but it does need to be steady. A steadier source can be had by using car battery in series with a headlight (about 3a) but it's DC.

The tradeoff is that more current will cause heating but it will also let you see the voltage drop to more accuracy, and you may need to resolve this voltage down to millivolts or microvolts. Keithley makes instruments that measure microohms, but they probably use DC.

This gets close to solving the conductivity problem, but ampacity also depends on that formula in the NEC.

By now, I'm sure wire makers (is that your company?) have software that calculates this better than any single human can, but there might be a problem with you getting access to this valuable tool. If it is available somewhere, it's probably costs kilobucks.

You guys are leaving something very important out of this conversation: the insulation. Most overcurrent damage causes insulation degradation, which leads to fire. (Thus the F in NFPA.)

Dale Hayes said:

Nickel plated steel can operate at 900 Deg. F., but will the nickel plating increase the amp capacity of the steel wire?

Click to expand...

Sure, if your nickel-plated wire has insulation that is also rated for 900F.

Dale Hayes said:

Using the same comparison of the 10ga. copper and 10ga. steel wire (10,400cm), what happens to the 11.8 amp rating of the steel wire if we were to nickel plate it? Nickel plated steel can operate at 900 Deg. F., but will the nickel plating increase the amp capacity of the steel wire?

Note: I keep referring to steel because we use steel and nickel-plated steel in certain applications. The 10ga. steel wire is a solid wire.

Dale Hayes

Click to expand...

What is the engineering profession coming to?

The inquiry is apparently related to a business that requires running currents through various materials in systems that are being operated near the limits of capability of the system. A competent engineer would do analyses and tests to determine the performance, safety, reliability, and life of the system. The system would be designed considering the mounting, detailed material properties, environment, and other factors affecting performance of the system.

Nevertheless, the question is posed here seeking answers from people who have no knowledge of the application or the environment, and probably have little or no professional knowledge of the science of heat transfer or properties of the particular materials.

Presumably, the decisions based on the inquiry will lead to delivery of a product to a customer.

In the event of failure of the product, perhaps resulting serious property damage or serious injury or death of someone, the engineer responsible for designing the product will be able to cite the information obtained from this forum.

I could go on, but I think most people reading this will understand the implications.

Pierre C Belarge said:

Remember that the heat in a conductor is proportional to the Square of the current.
That is one of the reasons why the current permitted in Table 310.16 is less in proportion to the size of the conductor as the conductor size increases.

Smaller conductors carry more current in proportion to the size of the conductor.
Hence, for larger current, we parallel conductors.

Click to expand...

Makes sense but I have heard conductors are run in parallel due to the skin effect.

tom baker said:

Makes sense but I have heard conductors are run in parallel due to the skin effect.

Click to expand...

I've heard they're paralleled for several reasons:

Yes, skin effect reduces ampacity as size increases.

Easier to install (weight, flexibility) and terminate.

More surface area per cross-section area for heat.

Lack of availability of materials in extreme sizes.


Besides, we wouldn't be able to create those works of art known as parallel conduits.

Bare Metal - No Insulation

Bare Metal - No Insulation

The "conductors, connectors, jumpers, bussing, etc." that I am refering to is bare metal - no insulation - and will never be insulated.

High temperature is sometimes a factor for us and when it is we use pure nickel connections.

Listed below is info from a Table of operating temperatures of various metals:

Copper: up to 222F

Brass: up to 225F

Tin plated brass: up to 300F

Tin-plated copper: up to 350F

Nickel-plated copper: 650F

Nickel (pure): up to 1000F

Nickel-plated steel: up to 900F


And Gentlemen,

Thank you for the formulas - they have been quite helpful.