Double de-rating?

[Smart $]

I'm afraid I have to disagree with Charlie here (sorry Charlie - Go Irish.) Per 210.19(A)(1), your branch circuit conductor should have an ampacity not less than the maximum load to be served. You need a conductor with an ampacity of not less than 10. However, the minimum branch-circuit conductor size shall not be less than 125% of the continuous load (the load in this case is completely continuous.) So the minimum branch-circuit conductor size would need an ampacity of 12.5 - or #14.

If we had five circuits (10 CCCs) with a load of 10A each, and wanted to use a 75deg rated #14 awg conductor (20A ampacity) and applied the 50% adjustment factor from T310.15(B)(2)(a), then the #14 would have an ampacity of 10. Since the adjusted ampacity of the #14 is not less than the maximum load to be served (10amps), and the #14 is not smaller than the minimum required branch-conductor size, the ten (10) #14 CCCs would be acceptable. Example D3(a) in the Annex has a good example of this.

I agree with your assessment.

A bit off topic, because OP is regarding lighting circuits, but I'm currently a bit fuzzy regarding motor circuits. Do we compound the factoring, i.e. 125% for single or largest motor ? 125% for continuous... or not...???

 

[david luchini]

I agree with your assessment.

A bit off topic, because OP is regarding lighting circuits, but I'm currently a bit fuzzy regarding motor circuits. Do we compound the factoring, i.e. 125% for single or largest motor ? 125% for continuous... or not...???

210.1 says that Art. 210 doesn't apply to motor only circuits, so no need to apply 125% twice for motor loads.

Combination motor and other loads is a little more tricky, as both 210 and 430 will apply. My reading of 430.24 would be 125% of the largest motor, 100% of other motors and 100% of other loads for the minimum conductor ampacity.

 

[kwired]

Regarding the "?2ckts" ,,,,,,, the efficiency comparison is made using two 3? circuits because it has the same number of wires as three 1? circuits. And the line current of two balanced 3? circuits powering the same loads is half that of a one balanced 3? circuit. :slaphead:

How do you supply same load with two circuits, and comply with NEC?

I understand where you are going with a three phase circuit vs multiples of three single phase circuits. Any remaining load that is not a multiple of three will be an imbalanced three phase circuit.

I still don't see where the "?2ckts" is coming from. If we had 6 loads and balanced them we could take the total VA and divide by 2 to get VA for two three phase circuits. But we have 4 loads. Three of them balance and the fourth has to go between two phases.

I also don't care if individual luminaires are 10 amp or whatever, all we were given to work with by OP was 10 amp continuous load and 8 current carrying conductors. Maybe there is only 10 amps total @ 208 making it 2080 total VA. That would help his voltage drop situation even more. But he further indicated that it is likely 4 different circuits with a net load of 10 amps each. So that is all there is to go with for now.

 

[kwired]

You guys got me rethinking how to properly calculate the minimum needed ampacity. Don't have time to research/verify things at the moment but I will get back with my understanding of how this should be done. I think a few of us have butted heads on some of the details in the past but I think it is good to do that. Sometimes you see something you never thought of before or maybe you did know but maybe forgot. There can be a lot involved with properly determining what minimum is needed sometimes.

 

[Smart $]

210.1 says that Art. 210 doesn't apply to motor only circuits, so no need to apply 125% twice for motor loads.

Combination motor and other loads is a little more tricky, as both 210 and 430 will apply. My reading of 430.24 would be 125% of the largest motor, 100% of other motors and 100% of other loads for the minimum conductor ampacity.

Okay.

So for the latter, 125% for largest motor, operated as a continuous load, would actually have the 125% portion thereof compounded.

 

[Smart $]

How do you supply same load with two circuits, and comply with NEC?

A "load" isn't necessarily an individual load... it can be multiple individual loads combined together to be a "load".

I still don't see where the "?2ckts" is coming from. If we had 6 loads and balanced them we could take the total VA and divide by 2 to get VA for two three phase circuits. But we have 4 loads. Three of them balance and the fourth has to go between two phases.

But the OP doesn't say these are 4 individual loads. Could be 36 individual loads wired as four loads.

I also don't care if individual luminaires are 10 amp or whatever, all we were given to work with by OP was 10 amp continuous load and 8 current carrying conductors. Maybe there is only 10 amps total @ 208 making it 2080 total VA. That would help his voltage drop situation even more. But he further indicated that it is likely 4 different circuits with a net load of 10 amps each. So that is all there is to go with for now.

I opened the 3? circuit discussion by questioning applicability. OP'er has not responded, so it is you that have expanded the discussion into this area.

Additionally, OP'ers don't always cover all aspects or possibilities in their query. Usually that's why they are asking to begin with... and quite often and inadvertently narrow the scope of their own dilemma.

 

[david luchini]

How do you supply same load with two circuits, and comply with NEC?

You don't. You supply separate loads with two circuits. Just like the OP is supplying separate loads on four circuits.

I understand where you are going with a three phase circuit vs multiples of three single phase circuits. Any remaining load that is not a multiple of three will be an imbalanced three phase circuit.

I still don't see where the "?2ckts" is coming from. If we had 6 loads and balanced them we could take the total VA and divide by 2 to get VA for two three phase circuits. But we have 4 loads. Three of them balance and the fourth has to go between two phases.

We were given four circuits each with a total load current of 10A. Given that this is parking lot lighting and I'm not aware of a single fixture that has a 2080 watt input, it is likely that each circuit has multiple lighting fixtures on it. For example, a 200W pulse start metal halide fixture with a ballast with a 232 watt input at 208V would give a load current of 1.11 amps per fixture. Nine fixture on each circuit of the four circuits in the OP would give you 10 Amps per circuit. The four circuits together would give you 36 fixtures total. Recircuiting these 36 fixtures into two (2) three phase circuits of 18 fixtures each with 6 fixtures balanced between each phase would give you a line current on each circuit of 11.55 Amps.

I also don't care if individual luminaires are 10 amp or whatever, all we were given to work with by OP was 10 amp continuous load and 8 current carrying conductors. Maybe there is only 10 amps total @ 208 making it 2080 total VA. That would help his voltage drop situation even more. But he further indicated that it is likely 4 different circuits with a net load of 10 amps each. So that is all there is to go with for now.

That is why Smart$ asked the question:

Is it possible to put the lighting on a 3? circuit?

I think he is suggesting that supplying the parking lot lighting with 3 phase circuits would reduce the number of current carrying conductors. Though to be honest, with having to increase the conductor size for the voltage drop at 550ft, the adjustment factor for # of CCCs really doesn't come into play.

With the reduced number of conductors and with the long length, there could be a material cost savings in going to two 3 phase circuits. There may be a material savings using one 3 phase 30A circuit to feed all the lights.

 

[david luchini]

So for the latter, 125% for largest motor, operated as a continuous load, would actually have the 125% portion thereof compounded.

No, I read the section for motor plus other loads as sizing the branch conductors at 125% of the largest motor plus 100% of the other loads.

I read "ampacity required for the other loads" in 430.24 to mean "shall have an ampacity not less than the maximum load to be served," in 210.19(A)(1).

In other words, the combination of sections in 430 and 210 only require an additional 25% for the largest motor, and no additional 25% for continuous load.

 

[david]

I'm afraid I have to disagree with Charlie here (sorry Charlie - Go Irish.) Per 210.19(A)(1), your branch circuit conductor should have an ampacity not less than the maximum load to be served. You need a conductor with an ampacity of not less than 10. However, the minimum branch-circuit conductor size shall not be less than 125% of the continuous load (the load in this case is completely continuous.) So the minimum branch-circuit conductor size would need an ampacity of 12.5 - or #14.

If we had five circuits (10 CCCs) with a load of 10A each, and wanted to use a 75deg rated #14 awg conductor (20A ampacity) and applied the 50% adjustment factor from T310.15(B)(2)(a), then the #14 would have an ampacity of 10. Since the adjusted ampacity of the #14 is not less than the maximum load to be served (10amps), and the #14 is not smaller than the minimum required branch-conductor size, the ten (10) #14 CCCs would be acceptable. Example D3(a) in the Annex has a good example of this.

How are you going samller than a min cercuit size of 14 AWG with an ampacity less than 20 amps at 75 Deg. C.
Table 310.16 ampacity shows a 75 Deg. C. conductors has to have a min ampacity of 20 amps to be used on a 15 amp lighting branch cercuit
210.23 or am i misunderstanding what your saying.

It sounds like your saying you can use a 14 AWG with a 12.5 ampacity on a lighting branch cercuit?

 

[Smart $]

No, I read the section for motor plus other loads as sizing the branch conductors at 125% of the largest motor plus 100% of the other loads.

I read "ampacity required for the other loads" in 430.24 to mean "shall have an ampacity not less than the maximum load to be served," in 210.19(A)(1).

In other words, the combination of sections in 430 and 210 only require an additional 25% for the largest motor, and no additional 25% for continuous load.

Works for me.

 

[david]

How are you going smaller than a min circuit size of 14 AWG with an ampacity less than 20 amps at 75 Deg. C.
Table 310.16 ampacity shows a 75 Deg. C. conductors has to have a min ampacity of 20 amps to be used on a 15 amp lighting branch circuit
210.23 or am i misunderstanding what you’re saying.

It sounds like your saying you can use a 14 AWG with a 12.5 ampacity on a lighting branch circuit?

Let me rephrase this it sounds like you're saying you can take a 14 AWG 75 Deg. C. rated conductor cut the ampacity of this wire in half and still use this 10 amp 14 Awg conductor on a 15 amp rated lighting branch circuit.
It looks to me that you would have to provide 10 amp OCP for this 14AWG conductor and the min lighting branch circuit would have to be a 15 amp circuit 210.23

 

[Smart $]

...

I think he is suggesting that supplying the parking lot lighting with 3 phase circuits would reduce the number of current carrying conductors. Though to be honest, with having to increase the conductor size for the voltage drop at 550ft, the adjustment factor for # of CCCs really doesn't come into play.

With the reduced number of conductors and with the long length, there could be a material cost savings in going to two 3 phase circuits. There may be a material savings using one 3 phase 30A circuit to feed all the lights.

That's what it amounts to... reducing four 1? to two 3? circuits. The voltage drop difference would have been more noticeable had it been three 1? to two 3?.

 

[kwired]

But the OP doesn't say these are 4 individual loads. Could be 36 individual loads wired as four loads.

That's what it amounts to... reducing four 1? to two 3? circuits. The voltage drop difference would have been more noticeable had it been three 1? to two 3?.

Well neither one of us can divide this into balanced three phase loads until we know what there is to divide up.

Like you said it could be 4 individual loads or 36 individual loads divided across 4 circuits.

4 individual loads will not evenly balance across three phases.

36 can if it is not already split into 4 circuits, otherwise it is either hook it up as 4 circuits or rework something if you want it evenly balanced.

Obviously I did not interpret what he has the same way you did. But he was not specific enough for either interpretation. All we really know is 8CCC and 10 amps of load on something, as it is now.

 

[david luchini]

How are you going samller than a min cercuit size of 14 AWG with an ampacity less than 20 amps at 75 Deg. C.
Table 310.16 ampacity shows a 75 Deg. C. conductors has to have a min ampacity of 20 amps to be used on a 15 amp lighting branch cercuit
210.23 or am i misunderstanding what your saying.

It sounds like your saying you can use a 14 AWG with a 12.5 ampacity on a lighting branch cercuit?

Let me rephrase this it sounds like you're saying you can take a 14 AWG 75 Deg. C. rated conductor cut the ampacity of this wire in half and still use this 10 amp 14 Awg conductor on a 15 amp rated lighting branch circuit.
It looks to me that you would have to provide 10 amp OCP for this 14AWG conductor and the min lighting branch circuit would have to be a 15 amp circuit 210.23

I'm afraid after reading both posts, I don't understand what you are asking, especially regarding 210.23. Maybe you can clarify.

The maximum load on the 5 circuits is 10Amps (continuous). From the first sentence of 210.19(A)(1), the conductors must have an ampacity not less than the max load. From T310.16, the ampacity of #14 awg (75 deg) is 20, and I have to apply an adjustment of 50% for 10 current carrying conductors which leaves an ampacity of 10. This conductor meets the first part of 210.19(A)(1).

The second sentence of 210.19(A)(1) says that the ampacity before any adjustments shall be not less than 125% of the continuous load, so 10*1.25=12.5 Amps. The ampacity of the #14 before any adjustment factor is 20, so the conductor meets the second part of 210.19(A)(1).

210.20(A) says the OCPD for the circuit must not be less than 125% of the continuous load (12.5 Amps) so I need a c/b of at least 15A. 210.20(B) says the conductors must be protected in accordance with 240.4.

240.4(B) allows me to protect the conductors with the next higher up standard size OCPD. For my #14 awg with an ampacity of 10, the next standard size c/b is 15, so a 15A c/b would be the max OCPD size.

240.4(D) also requires that #14 awg be protected at not more than 15A, so a 15A c/b would be the max OCPD size.

So for my 5 circuits (10 CCCs) each with a continuous load of 10A all in the same raceway, I could use 75deg rated #14awg on a 15A c/b's.

 

[david]

I'm afraid after reading both posts, I don't understand what you are asking, especially regarding 210.23. Maybe you can clarify.

240.4(B) allows me to protect the conductors with the next higher up standard size OCPD. For my #14 awg with an ampacity of 10, the next standard size c/b is 15, so a 15A c/b would be the max OCPD size.

lets start here
Additional standard ampere ratings for fuses shall be 1, 3, 6, 10, and 601.

would you agree that 10 amps is a standard OC protection device

 

[david]

I'm afraid after reading both posts, I don't understand what you are asking, especially regarding 210.23. Maybe you can clarify.

210.23 (A) 15- and 20-Ampere Branch Circuits. A 15- or 20-ampere branch circuit shall be permitted to supply lighting units or other utilization equipment, or a combination of both, and shall comply with 210.23(A)(1) and (A)(2)

It doesn’t say a 15 amp rated branch circuit.
It does say 15 and 20 amp branch circuits

I see a 15 amp branch circuits as one having 15 amp over- current protection and conductors with an ampacity in accordance with table 310.16..

The end result would mean a 15 amp branch circuit would need a conductor with adjusted 20 amp capacity table 310.16

 

[david]

From T310.16, the ampacity of #14 awg (75 deg) is 20, and I have to apply an adjustment of 50% for 10 current carrying conductors which leaves an ampacity of 10. This conductor meets the first part of 210.19(A)(1)..

Unless it under supervised engineering
I do not see where the table ampacities are allowed to be reduced.

I do see where the ampacities in these tables have to be maintained by adjusting the conductor sizes.

 

[kwired]

lets start here
Additional standard ampere ratings for fuses shall be 1, 3, 6, 10, and 601.

would you agree that 10 amps is a standard OC protection device

Note the word 'fuses' in that statement.

If they intended to cover both fuses and circuit breakers they would have included these sizes in the first sentence of that section.

 

[david]

Note the word 'fuses' in that statement.

If they intended to cover both fuses and circuit breakers they would have included these sizes in the first sentence of that section.

The type of over current protection that you provide a design choice.

240.4 Protection of Conductors.
(2) The ampacity of the conductors does not correspond with the standard ampere rating of a fuse or a circuit breaker without overload trip adjustments above its rating (but that shall be permitted to have other trip or rating adjustments).

But you already new it said that it includes both fuses or circuit breakers.
Further when they wanted to be specific about fuses verses circuit breaker for 16AWG and 18AWG conductor sizes they spelled it out.

The language says if there is a standard over current protection device that matches the ampacity tables you do not exceed it.
Other wise you can round up to the next higher up to 800 amps

 

[kwired]

The type of over current protection that you provide a design choice.

That is correct. If your design choice is circuit breakers then 1, 3, 6, 10, and 601 are not standard sizes.