Sub panel and 2-2-4-6 Aluminum cable

[david luchini]

In D2b they added the AC units 10080va and used the number in the neutral contribution

I don't see the AC unit load being included in the neutral calculation in Example D2(b).

 

[dm9289]

Also it's not only L-L loads, equal 120 volt loads on each ungrounded conductor will cancel the neutral load. At 120 volts, 75 amps on each ungrounded conductor will yield a 0 amp neutral load

i totally agree with you from a safety standpoint that 120v loads on opposite phase will have cancelling affect, and 240v L-L loads should not be counted but cant find it in code or examples.

 

[dm9289]

I don't see the AC unit load being included in the neutral calculation in Example D2(b).

Upon further review see that. There has to be a code ref I'm missing on L-L loads

 

[dm9289]

If you truly do not know what will be connected to this feeder, then you are correct. Perhaps the home owner will install a set of 120V bitcoin rigs, and then only operate half of them at a time.

Absent any other instructions, (for example the 70% demand reduction for ranges or electric dryers), then you need to do the article 220 calculation for the loads connected to the neutral. 220.61(B)2 gives a demand reduction factor for the neutral when the feeder is larger than 200A, but that demand factor doesn't apply.

The only reduction that you can apply on a 90A feeder to a garage is eliminating any loads not actually connected to the neutral. All you really have is the experience of the members here who say that it is very unlikely that any residential use of this 90A circuit won't have some significant L-L loads.

-Jon

Agreed and i would say as close to impossible as you can get

 

[david luchini]

There has to be a code ref I'm missing on L-L loads

It's 220.61(A).

If you have a L-L load...say a 240V water heater, then there is not unbalanced load for the water heater between the neutral and either of the ungrounded conductors....because there is no connection to the neutral conductor.

 

[david luchini]

240v L-L loads should not be counted but cant find it in code or examples.

D2(b) is an Example of the L-L loads not being counted in the neutral calculation. Both the 240V Air conditioner load and 240V Water Heater (you have to make an assumption that it's 240V) are not included.

 

[david luchini]

Thats where i have been looking seems like only cooking units and dryers. I do know this should not be a problem, long as something is on both phases. I cant find anything on load balancing that would make it OK. The condition where i could see a safety issue is if someone purposely put all the load on one phase.

The 220.61 neutral calculation assumes that the load is balanced on the phases. It's not for a worst-case situation where all of the loads are on one phase.

It takes into account that loads are constant.

 

[dm9289]

Thank you everyone for your help, sometimes I'm looking for too much in writing and loose common sense hopefully this comment will help others like me that were having problems. I know you all tried valiantly to get this thru my thick head it finally did

1.Example my home AC compressor (although not written in code) IT HAS NO NEUTRAL WIRE SO CANT HAVE NEUTRAL CURRENT. LOL code should not have to write this. The inside blower may come into play in another area but its a small 120v motor in my case.
2. Baseboard heating also no neutral wire so no neutral current.
3. Hot tubs I have limited experience but if it has a neutral would probably have to look at manual for guidance to learn max unbalanced

 

[wwhitney]

The 220.61 neutral calculation assumes that the load is balanced on the phases.

I wouldn't say that 220.61 directs you to assume that. I would say it references the actual division of the L-N loads between ungrounded conductors:

"220.61(A) . . . The maximum unbalanced load shall be the maximum net calculated load between the neutral conductor and any one ungrounded conductor."

So if you go ahead and install them balanced, you can calculate it that way. If you install all L-N loads on one ungrounded leg, you are required to calculate it for that worst case.

Cheers, Wayne

 

[infinity]

So if you go ahead and install them balanced, you can calculate it that way. If you install all L-N loads on one ungrounded leg, you are required to calculate it for that worst case.

I agree and that's why I commented earlier that would be up to the next guy to figure out not the installer of the feeder who may have no knowledge of the future loads.

 

[LarryFine]

Thats where i have been looking seems like only cooking units and dryers.

Those are L-N-L loads with known-reduced neutral currents, not L-L-only loads.

Can you conceive having 90a of neutral current, even if you intentionally try to?

The electrician's responsible for balance-loading of L-N loads. Hey, that's you!

 

[winnie]

Those are L-N-L loads with known-reduced neutral currents, not L-L-only loads.

I think that this is a distinction that I missed above. The permissions in 220.61(B) apply to L-N-L loads, allowing one to reduce the size of the neutral relative to the hots even though the N is connected to the load.

The calculation in 220.61(A) allows one to exclude L-L loads, because L-L loads by definition cannot load the neutral at all.

Can you conceive having 90a of neutral current, even if you intentionally try to?

The electrician's responsible for balance-loading of L-N loads. Hey, that's you!

IMHO if one were to only have L-N loads (say 120V receptacle circuits on a subpanel), code does not provide a way to do the calculation to reduce the neutral until the calculated load exceeds 200A. No matter how 'balanced' the installation is.

-Jon

 

[LarryFine]

Neutral sizing is based on presuming only circuits on the more-heavily-loaded line are energized; a worst-case situation.

 

[wwhitney]

IMHO if one were to only have L-N loads (say 120V receptacle circuits on a subpanel), code does not provide a way to do the calculation to reduce the neutral until the calculated load exceeds 200A. No matter how 'balanced' the installation is.

If you are saying that you have to assume that all the L-N loads in the entire panel are on one bus, leaving half of the space unused, I disagree. 220.61(A) means that you get to look at the exact division of of the L-N loads among the two or three busses, and just size the neutral for the most heavily loaded bus.

Cheers, Wayne

 

[LarryFine]

If you are saying that you have to assume that all the L-N loads in the entire panel are on one bus, leaving half of the space unused, I disagree. 220.61(A) means that you get to look at the exact division of of the L-N loads among the two or three busses, and just size the neutral for the most heavily loaded bus.

Exactly, which should (theoretically) not be much more than half (third for 3ph) of the feeder's calculated load.

Thus a 90-amp feeder's neutral should (again, theoretically) only need to be sized for, say, 50 to 60 amps.

 

[wwhitney]

Exactly, which should (theoretically) not be much more than half (third for 3ph) of the feeder's calculated load.

Yes, but at half of the voltage, so the same current.

I.e. a 100A 120/240V panel can supply up to 24,000 VA of loads. If those are all 120V loads, split evenly between the two ungrounded conductors, the neutral needs to be 100A.

Cheers, Wayne

 

[winnie]

If you are saying that you have to assume that all the L-N loads in the entire panel are on one bus, leaving half of the space unused, I disagree. 220.61(A) means that you get to look at the exact division of of the L-N loads among the two or three busses, and just size the neutral for the most heavily loaded bus.

I agree with this.

What I was trying to say is that for purposes of calculating the required neutral ampacity you cannot take any credit for the balancing effects that we know will actually occur in normal use. If you have a 100A 120/240V panel, with 40A of calculated L-L load and 60A of balanced L-N load (so 60A on each side), then you need a neutral ampacity of 60A.

220.61 makes this explicitly clear: "The feeder or service neutral load shall be the maximum unbalance of the load determined by this article. The maximum unbalanced load shall be the maximum net calculated load between the neutral conductor and any one ungrounded conductor." You do the article 220 calculation for each phase to neutral, and take the maximum as the neutral load.

Note that the system could in theory load the neutral more than this without tripping any breakers, but really, this is an article 220 calculation and probably pretty conservative.

Again with the 100A panel. You might have a 50A 2 pole breaker for a range, two 30A breakers for dryer and water heater, and ten 20A single pole breakers for receptacles and lighting. But by your article 220 calc you have less than 100A load and less than 60A single phase load. Clearly if you loaded up the 20A circuits on only 1 leg of the service you might get 100A flowing on the neutral. But this is _hugely_ unlikely, and we are not required to provide more than a 60A neutral, nor protect the neutral with a 60A OCPD.

-Jon

So if you have 60A of L-N loads on leg A and 70A

 

[david luchini]

So if you go ahead and install them balanced, you can calculate it that way. If you install all L-N loads on one ungrounded leg, you are required to calculate it for that worst case.

I would think you need to do the load calculation before the installation, so you know what size conductors to install.

Installing first and doing a load calculation afterwards seems counterproductive.

In any event, there are several Annex D examples. They assume the load is balanced.

 

[wwhitney]

I would think you need to do the load calculation before the installation, so you know what size conductors to install.

The two are intertwined.

For a not-too-realistic example, if the load calculation for a 120/240 service ends up with (5) 120V circuits, each with a 20A load, and everything else is a 240V load, then you know that as long as you split the 120V circuits 3/2 between legs, your neutral load is 60A. If you then install them with a 4/1 split between legs, but only a 60A neutral conductor, you've created a code violation.

So as part of the load calculation for the neutral, you need to split the L-N loads into circuits, and distribute those circuits among the ungrounded conductors, and the find the ungrounded conductor most heavily loaded with L-N loads. At least, that's my reading of 220.61(A).

I'll read through Annex D and respond further as required.

Cheers, Wayne

 

[wwhitney]

In any event, there are several Annex D examples. They assume the load is balanced.

Yes, I see that example D(1)(a) assumes the load is perfectly balanced--the neutral load is taken as all of the 120V load (after the 100%/35% computation), plus 70% of the 120/240V load (range/dryer), and then divided by 240V. Which is the same as taking half (perfectly balanced) and then dividing by 120V.

What is the justification for assuming perfectly balanced (both practically and code-wise)? That best case is obviously never achieved in practice.

Of course, if the rest of the computation is sufficiently conservative, this non-conservative assumption won't cause any problems. Just not seeing the code language that allows it, compared to a more detailed analysis of how each leg is loaded. [On a more detailed analysis, it's not clear how handle the 100%/35% computation, though.]

Cheers, Wayne