Using the 90C column in the Ampacity Table

[FaradayFF]

Greetings,
Do we use the cable ampacity in the 90C column only when the cable would be terminated at the lugs rated for 90C on both ends?
Thanks,
EE

 

[augie47]

1st , don't let the term "lugs" mislead you. The actual lug many be 90° rated but to terminate at 90° the device with the lug must be 90° rated.. rare..
You can also use the 90° rating for conductors with that insulation for de-rating purposes (fill, ambient, etc) as long as in the end you don't exceed the rated temperature at terminations.

 

[Jraef]

1st , don't let the term "lugs" mislead you. The actual lug many be 90° rated but to terminate at 90° the device with the lug must be 90° rated.. rare..
You can also use the 90° rating for conductors with that insulation for de-rating purposes (fill, ambient, etc) as long as in the end you don't exceed the rated temperature at terminations.

Which is pretty much the only way you can use it in the field, which is what the NEC is all about. 90C ratings are used inside of equipment like switchboards and MCCs where they are making connections directly to bus bars, then the entire ASSEMBLY gets tested and listed again by someone like UL. But it would be wildly cost prohibitive to re-test something in the field just to use one size smaller wire.

 

[Carultch]

Greetings,
Do we use the cable ampacity in the 90C column only when the cable would be terminated at the lugs rated for 90C on both ends?
Thanks,
EE

In practice, equipment with 90C terminations is rare. Most equipment you'll find in reality, has 75C terminations. The entire assembly has to be listed and labeled for 90C terminations, not just the lugs on the device. It is very common that lugs carry a 90C marking, but you don't necessarily get to take credit for it.

You can use the 90C ampacity values for terminations if you terminate on separately-installed connectors. This does require it happens in an otherwise-empty enclosure, with all wiring and connections within it rated for 90C. One reason you might do this, is if you are recovering from a mistake. Another reason is a value-engineering decision, so the majority length can be reduced by a size or two. I would expect that voltage drop would become significant enough to not do this, any time there is an advantage to the extra complexity. Thus, it is more of a value-engineering option that exists in theory, that I expect to be rare in practice.

The most common value to the installer of the wire with a 90C-rating, is that it gives you headroom for your derate calculations. I.e. ampacity adjustments for conductor bundling, and temperature corrections. These calculations don't apply to termination ampacity, but just for the wire ampacity. As an example, consider a 60A circuit with a 0.8 derate for bundling and 0.96 derate for ambient temperature. This means #6 Cu wire has a derated ampacity of 57.6A, and a terminal ampacity of 65A. This can be protected by the 60A circuit per 240.4(B), assuming there is 57.6A or less of full load on it (which there probably is). If you had to start this calculation with the 75C rating, you'd have to upsize it to #4 Cu. The 90C rating lets the #6 Cu wire be OK, because it gives you headroom for this calculation when starting with 75A instead of 65A.

For over 100A, 75C is the default rating. For 100A and less, you have a burden of proof if you want to use the 75C terminations, but most of the time, you will find 75C terminals in practice. The default is 60C for 100A and less. It is usually academic examples where you encounter the need for 60C termination sizing, since equipment listed otherwise for 75C is most common. Certain cable types like NM cable (Romex) require 60C terminations.

 

[tom baker]

Nive explanation.

 

[Carultch]

just one question when you make the calculation the result of your adjustment factor or ampacity correction. that result is gonna be base in your gage of your conductor.
Tutuapp 9apps Showbox

Please rephrase your question. I don't understand what you are trying to ask. I find it may help to give an example problem, and show what part of the solution is in question.

 

[Vines41]

In practice, equipment with 90C terminations is rare. Most equipment you'll find in reality, has 75C terminations. The entire assembly has to be listed and labeled for 90C terminations, not just the lugs on the device. It is very common that lugs carry a 90C marking, but you don't necessarily get to take credit for it.

You can use the 90C ampacity values for terminations if you terminate on separately-installed connectors. This does require it happens in an otherwise-empty enclosure, with all wiring and connections within it rated for 90C. One reason you might do this, is if you are recovering from a mistake. Another reason is a value-engineering decision, so the majority length can be reduced by a size or two. I would expect that voltage drop would become significant enough to not do this, any time there is an advantage to the extra complexity. Thus, it is more of a value-engineering option that exists in theory, that I expect to be rare in practice.

The most common value to the installer of the wire with a 90C-rating, is that it gives you headroom for your derate calculations. I.e. ampacity adjustments for conductor bundling, and temperature corrections. These calculations don't apply to termination ampacity, but just for the wire ampacity. As an example, consider a 60A circuit with a 0.8 derate for bundling and 0.96 derate for ambient temperature. This means #6 Cu wire has a derated ampacity of 57.6A, and a terminal ampacity of 65A. This can be protected by the 60A circuit per 240.4(B), assuming there is 57.6A or less of full load on it (which there probably is). If you had to start this calculation with the 75C rating, you'd have to upsize it to #4 Cu. The 90C rating lets the #6 Cu wire be OK, because it gives you headroom for this calculation when starting with 75A instead of 65A.

For over 100A, 75C is the default rating. For 100A and less, you have a burden of proof if you want to use the 75C terminations, but most of the time, you will find 75C terminals in practice. The default is 60C for 100A and less. It is usually academic examples where you encounter the need for 60C termination sizing, since equipment listed otherwise for 75C is most common. Certain cable types like NM cable (Romex) require 60C terminations.

This is the same explanation I heard recently and came here to check my understanding.

In this situation we are discussing (3) 240V (unbalanced) circuits with a continuous load of 20.8A (Powerwall ESS Systems) in a conduit that terminate at 30A breakers in a 100A subpanel. Wire ampacity needs to be 26A minimum with the 1.25 factor. The Powerwalls serve unbalanced 120V loads, as well as 240V loads in any mixture

I was using (9) #8 THWN-2 for this circuit, since 50A x 0.7 Fill =35A so it was OK in my mind.

Another pointed out that (9) #10 THWN-2 in conduit would be good to use instead.
Since 40A x 0.7 = 28A it passes the ampacity check inside the conduit, and at the 75C terminals the ampacity is 35A is unfactored so it passes there.

I have had code officials point out that 110.14.c was the reason behind this "Terminal Temperature ampacity limitation"

I sort of feel silly for wasting so much copper but am glad to know something new.

 

[Carultch]

This is the same explanation I heard recently and came here to check my understanding.

In this situation we are discussing (3) 240V (unbalanced) circuits with a continuous load of 20.8A (Powerwall ESS Systems) in a conduit that terminate at 30A breakers in a 100A subpanel. Wire ampacity needs to be 26A minimum with the 1.25 factor. The Powerwalls serve unbalanced 120V loads, as well as 240V loads in any mixture

I was using (9) #8 THWN-2 for this circuit, since 50A x 0.7 Fill =35A so it was OK in my mind.

Another pointed out that (9) #10 THWN-2 in conduit would be good to use instead.
Since 40A x 0.7 = 28A it passes the ampacity check inside the conduit, and at the 75C terminals the ampacity is 35A is unfactored so it passes there.

It's iffy whether the neutral needs to count or not for ampacity adjustment purposes. If it would not need to count, you'd only need the 0.8 derate factor for 6 conductors counting as CCC's, instead of 0.7 for 9 conductors that count. Perhaps others could advise.

A neutral that only serves to carry the imbalance doesn't need to count, because the total current among all lines and neutral, never adds up to what it would be if it were balanced among the lines. The only factor in this situation that would make it need to count, is if it serves harmonic-intensive non-linear loads. Unfortunately, I don't understand exactly how get a quantifiable/objective answer for whether the load situation counts or not. I know non-linear loads are everywhere with our modern technology (computers, non-incandescent lighting, anything with a DC power supply, etc), but I would expect a lot of them are close enough to linear to count as linear. Please, correct me if there is something I am missing.

Even if it did, the person who pointed out that #10 Cu would work in this application, is correct. Provided that the operating current of each circuit never exceeds 28A of total load, whether continuous or non-continuous, the #10 Cu is sufficient. It needs to have an ampacity for the load, have an ampacity for being protected by the OCPD. Both terminal ampacity and wire ampacity. If it intermittently serves 29A or 30A loads, it would not be sufficient, and #8 Cu would be required, assuming neutral counts.

I have had code officials point out that 110.14.c was the reason behind this "Terminal Temperature ampacity limitation"

Derate factors do not apply to terminal temperature limitations. Imagine drawing a virtual box very locally at the terminal of the wire. Notice you don't see it bundled with other wires? Bundling only needs to apply once you leave the terminal enclosure, and travel, usually in a raceway. Less intuitive is that the ambient temperature derate also doesn't apply to terminations. Terminations are the non-derated (usually) 75C ratings from Table 310.16.

#10 and smaller is a special case, where you are generally limited to certain amperes of protection, even if you have terminations in excess of 60C. There is the "small conductor rule" for sizes #14, #12, and #10, as discussed in 240.4(D).

 

[wwhitney]

A few comments:

The Powerwall ESS data sheet lists: Real Power, max continuous 5 kW, peak 7 kW; Apparent Power, max continuous 5.8 kVA, peak 7.2 kVA. So the continuous load is 5800/240 = 24A, and the peak load is 7200/240 = 30A.

Seems clear to me that the neutral is not counted as a current carrying conductor in this case, so 3 circuits would be only 6 CCCs.

Cheers, Wayne

 

[Vines41]

Thanks for the expanded explanation Carultch and wwhitney who initiated this discussion over in the Tesla Motor Club Thread. This has ben very informative.

Wayne, I don't fully understand why the neutral is not counted. Theoretically the ESS could supply all 120V loads, which only connect from L1 to neutral.

Also since you brought up the real vs apparent power, I should understand this difference. Do you have any further explanation as to what that means in this context? Does that mean the power factor is greater than 1?

 

[wwhitney]

Wayne, I don't fully understand why the neutral is not counted. Theoretically the ESS could supply all 120V loads, which only connect from L1 to neutral.

Say you have a 30A double pole breaker and a two wire circuit. Worst case for heating is 30A on each wire.

Now you have the same breaker and a 3 wire circuit with neutral. Barring harmonics, the currents N, L1, and L2 will satisfy N = L1 - L2. The worst case is still two wires carrying 30A, and 0A on the third. It could be N and L1 at 30A, or L1 and L2, or N and L2. A typical case of current on all three lines would be L1 = 30A, N = L2 = 15A. That's less resistive heating in the wires, as the heating is I^2 * R, and 30^2 + 30^2 > 30^2 + 15^2 + 15^2.

So when the neutral carries just the unbalanced load between the two ungrounded condutors, you don't have to count it as a current carrying conductor, as the current it carries doesn't contribute to additional heating.

Cheers,
Wayne

 

[wwhitney]

Also since you brought up the real vs apparent power, I should understand this difference. Do you have any further explanation as to what that means in this context? Does that mean the power factor is greater than 1?

The full topic is a little too lengthy to explain here, perhaps someone can suggest a good reference.

Power factor is always <=1, and it's the ratio of (real power)/(apparent power). For the Powerwall, consider the two continuous ratings, 5 kW and 5.8 kVA. That would correspond to a power factor of 0.86. So for a continuous load on one Powerwall, if it has a power factor of greater than 0.86, the load will be limited by the real power rating of 5 kW. If it has a power factor of less than 0.86, the load will be limited by the apparent power rating of 5.8 kVA, and you won't be able to get the full 5 kW of real power out of the Powerwall.

If you like, kVA represents the actual current in the wire, and kW represents the extent to which that current is in phase with the voltage and available to carry energy to the load, versus current just cycling around the circuit (reactive current).

Cheers, Wayne

 

[Carultch]

Also since you brought up the real vs apparent power, I should understand this difference. Do you have any further explanation as to what that means in this context? Does that mean the power factor is greater than 1?

For a simple resistive load, current and voltage are synchronized, and both properties come as a package deal, proportional to each other. Because resistive loads mean V=I*R. You can multiply instantaneous voltage and instantaneous current, and take a time average. Similarly, you can multiply RMS voltage and RMS current, and get the same result. The former calculation gives real power (P in units of kW), the lateral calculation gives apparent power (S in units of kVA). When both kW and kVA are the same, we say the power factor is unity. Power factor is the ratio of kW to kVA.

For a more complicated load that contains elements of energy storage (capacitance and inductance), there is a time delay or phase shift between voltage and current. The apparent power you get from multiplying nominal volts and amps will be greater than the real power you deliver to the load.

There is also distortion power factor associated with nonlinear loads. That is, loads where the relation between voltage and current is more than just a proportionality and phase offset. Typically loads involving diodes, transistors, and other semiconductor technology, rather than only using basic Ohm's law circuit elements. These cause the current waveform to be a different shape than the voltage waveform, which also causes kVA to not equal kW. Even if the peaks and zeros of both waveforms are synchronized in phase, the change in shape is what causes a power factor less than unity.

 

[Vines41]

Say you have a 30A double pole breaker and a two wire circuit. Worst case for heating is 30A on each wire.

Now you have the same breaker and a 3 wire circuit with neutral. Barring harmonics, the currents N, L1, and L2 will satisfy N = L1 - L2. The worst case is still two wires carrying 30A, and 0A on the third. It could be N and L1 at 30A, or L1 and L2, or N and L2. A typical case of current on all three lines would be L1 = 30A, N = L2 = 15A. That's less resistive heating in the wires, as the heating is I^2 * R, and 30^2 + 30^2 > 30^2 + 15^2 + 15^2.

So when the neutral carries just the unbalanced load between the two ungrounded condutors, you don't have to count it as a current carrying conductor, as the current it carries doesn't contribute to additional heating.

Cheers,
Wayne

Lots of work this weekend, just now have some free time for myself to dig deeper into this. Thanks to both of you for taking the time to educate me.

So basically it sounds like this current carrying conductor question all comes back to the same reason why the neutral in a multiwire branch circuit is not oversized? The currents from L2 and L1 are phase opposite, so cancel each other out in the neutral? The example of 30A on L1 and 15A on each of the others is a good one.

Now to the second part of the question, how to size this breaker in a Powerwall system, and how does real vs apparent power matter to the circuit sizing?
Wayne pointed out on using an 80A breaker for the subfeed for 3 Powerwalls was an incorrect choice, and that we should be using a 90A breaker instead.

If the Powerwall real power is 5 kW and the Apparent Power is 5.8 kVA, can you further describe why the circuit needs to be sized for 30A?

The math is
5800kVA / 240V = 24.2A x 1.25=30.2A
Instead of
5000W / 240V = 20.8A x 1.25 = 26A

Bring it back to the real world, 3 Powerwalls are pumping out the absolute maximum 5000w each for a bunch of loads, but the 90A breaker trips, what real world condition would have pushed this over the edge of what an 80A breaker can handle?

Does this have to do with the ability of the Powerwall to surge for 10 seconds?
Does this represent a time delay of the voltage and current relationship, and this could be as bad as 800w of heat in its worst case?

 

[Carultch]

So basically it sounds like this current carrying conductor question all comes back to the same reason why the neutral in a multiwire branch circuit is not oversized? The currents from L2 and L1 are phase opposite, so cancel each other out in the neutral? The example of 30A on L1 and 15A on each of the others is a good one.

That's correct. When the L1 and L2 currents are equal, they are also opposite. All the current that flows outbound on the black wire returns on the red wire. Current among the three wires that could carry current, will add up to zero. Since it already adds up to zero on L1 and L2 alone, no current ends up on the neutral. With the exception of harmonics (the components of a waveform that reshape it from a simple sine wave), neutral only carries current when there is an imbalance.

In the case of the three wire split phase system, it is even harmonics (120 Hz, 240 Hz, 360 Hz, etc) that add up rather than cancel, on the neutral. In the case of a three phase/four-wire system, it is triplen harmonics (180 Hz, 360 Hz, 540 Hz, etc) that add up rather than cancel on the neutral.

Now to the second part of the question, how to size this breaker in a Powerwall system, and how does real vs apparent power matter to the circuit sizing?

Size it according to the amperes on its nameplate. This ultimately will relate to the apparent power it can deliver, rather than the real power. Current is what heats up a wire thus governing its sizing, and it heats up the wire just as much as if it is synchronized with voltage, as it does when it has a phase offset.

The Powerwall is not necessarily producing a reactive power by its own design. It is the when the loads on the system have reactive power, that it drives the powerwall to produce reactive power to offset the capicitative or inductive nature of the loads.

 

[wwhitney]

Does this represent a time delay of the voltage and current relationship, and this could be as bad as 800w of heat in its worst case?

Yes, but it's not 800w of heat. It's 800 VA/120 V = 6.6 A of current that's shuttled around the circuit each cycle without doing any useful work. That's the nature of less than 1 power factor. It's still current that travels through the circuit and causes heating of the wires and in particular the thermal trip mechanism of the circuit breaker.

Cheers, Wayne

 

[Vines41]

With the exception of harmonics (the components of a waveform that reshape it from a simple sine wave), neutral only carries current when there is an imbalance.

In the case of the three wire split phase system, it is even harmonics (120 Hz, 240 Hz, 360 Hz, etc) that add up rather than cancel, on the neutral. In the case of a three phase/four-wire system, it is triplen harmonics (180 Hz, 360 Hz, 540 Hz, etc) that add up rather than cancel on the neutral.

Size it according to the amperes on its nameplate. This ultimately will relate to the apparent power it can deliver, rather than the real power. Current is what heats up a wire thus governing its sizing, and it heats up the wire just as much as if it is synchronized with voltage, as it does when it has a phase offset.

The Powerwall is not necessarily producing a reactive power by its own design. It is the when the loads on the system have reactive power, that it drives the powerwall to produce reactive power to offset the capicitative or inductive nature of the loads.

Thanks for those explanations, I wasn't looking to continue to undersize the breaker, but to evaluate any danger we might have in the field.

I have had one plan checker ever ask about harmonics, and would like to have some reference that describes how an average 2-400A residential service is, or is not subject to them and under what assumptions that conclusion arrived at. I will do some research and see if I come up with anything.

Wayne, I am not sure I understand the difference there, unless you are just being specific about the location of the heat.
In effect I understand at the worst possible circumstances (max load and worst power factor) the powerwall can deliver 5000W of real power to a device but up to 800w shuttled around could be producing heat spread around the circuit, assuming you weren't just driving a constant load to a resistive heater with a power factor of 1. Did I get that right?

 

[wwhitney]

It's not 800W of heat, heat would be a real power, not reactive power. It's an extra 6.6A of current due to the power factor (compared to a different load of the same real power with power factor 1). If the circuit were a superconductor (no resistance in the wires), there'd be no heat, just the extra current.

If the circuit has a resistance of 0.01 ohm, then the current is 24A at 5800 kVA instead of 17.4A at 5000 kVA if the power factor were 1. The heating from the circuit resistance is 5.8 W instead of 3.0W (that's from P = I^2 R).

So that's my point, it does have a side effective of some heating, but it's not 800W of heat, it's a couple of orders of magnitude less. You can't turn VA into W, they are different.

Cheers, Wayne

 

[Vines41]

Got it thanks. This seems like a deep enough explanation. Getting further into this power factor seems like would be fun but not critical to understanding the question, or the OP actual question

I'll size my circuits for a 30A load going forward, and then focus on lower hanging fruit.

 

[Vines41]

I agree its rare for a residence, but a commercial system I am working on has (4) 100 kW inverters landing in a 600A subpanel , and then the 600A run to a 600A H366NR fused disconnect in the electrical room then to a load side tap at the switchboard.

The inverters terminals are rated 105C, the breakers in the subpanel 75C and the Main lugs on the subpanel as well as the 600A AC disconnect and the lugs for the load side tap are all rated 90C.

I think I can use 90 C ampacity for all wiring except the inverters to subpanel connection, as they land in a 150A 3p breaker rated at 75 C. What do others think?

I know there was some discussion about lug ratings vs enclosure ratings, and how both need to be 90C to use that rating. I cannot find those enclosure ratings easily though. I have confirmed the lugs are rated 90C.