# Calculating neutral size



## shelly41 (Nov 22, 2020)

Hi all, I have a question about calculating the neutral size for service conductors.

My first assumption is that 240V loads like the dryer, stove, hot water heater etc. don't carry current on the neutral (other than a small amount for digital displays/control panels etc.). However, I am a little confused on this point, because this article says in paragraph 5 that "The minimum neutral load on the service for a 5,500-watt clothes dryer is 3,850 volt-amperes..." Can anyone clarify?

Assuming I'm not counting any 240V loads, let me continue the calculation. I followed the NEC standard load calculation and I came up with ~12,200 VA in 120V loads (6675VA general lighting, 1850VA fixed appliances, 3600VA for two space heaters). I used the standard reduction for general lighting, hopefully that was the correct thing to do (100% up to 3000VA, 35% demand beyond 3000VA). So that's a total of a max of 102A @120V. Ideally, that 102A would be split 51A for each leg. But let's say in the real world it's actually 70A between one leg and the neutral (maybe I'm running both space heaters on the same leg?). And then let's say for some reason all the loads on the other leg are off. Is the max unbalanced load 70A? And if so, then 70A would be the current used to size the neutral?

Thanks!


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## HertzHound (Jan 22, 2019)

1850VA for fixed appliances? But you didn’t add in two 1500VA for the small appliance load or the 1500VA for the laundry load. 

The dryer neutral part is easy. 70% reduction on the neutral. 5,500 x .7 = 3,850 neutral load. You take the same reduction on the range, so 70% of the range load would be your neutral current. 

Or you can take 70% of the unbalanced load in excess of 200A. I’m not sure if you can take both 70% reductions. So range/dryer @ 70% or unbalanced in excess of 200A @ 70%. Here’s the code wording. I’d have to look for an example. I separated the part where it says OR. 

220.61 Feeder or Service Neutral Load.
(A)Basic Calculation.
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.
Exception: For 3-wire, 2-phase or 5-wire, 2-phase systems, the maximum unbalanced load shall be the maximum net calculated load between the neutral conductor and any one ungrounded conductor multiplied by 140 percent.

(B)Permitted Reductions.
A service or feeder supplying the following loads shall be permitted to have an additional demand factor of 70 percent applied to the amount in 220.61(B)(1) 

or 

portion of the amount in 220.61(B)(2) determined by the following basic calculations:
(1)A feeder or service supplying household electric ranges, wall-mounted ovens, counter-mounted cooking units, and electric dryers, where the maximum unbalanced load has been determined in accordance with Table 220.55 for ranges and Table 220.54 for dryers
(2)That portion of the unbalanced load in excess of 200 amperes where the feeder or service is supplied from a 3-wire dc or single-phase ac system; or a 4-wire, 3-phase system; or a 3-wire, 2-phase system; or a 5-wire, 2-phase system
Informational Note: See Examples D1(a), D1(b), D2(b), D4(a), and D5(a) in Informative Annex D.


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## shelly41 (Nov 22, 2020)

HertzHound said:


> 1850VA for fixed appliances? But you didn’t add in two 1500VA for the small appliance load or the 1500VA for the laundry load.
> 
> The dryer neutral part is easy. 70% reduction on the neutral. 5,500 x .7 = 3,850 neutral load. You take the same reduction on the range, so 70% of the range load would be your neutral current.
> 
> ...


Hi thanks for the reply. Sorry for the confusion, I had bundled the small appliance and laundry circuits in with the General Lighting load. The fixed appliance amount is for the range hood, bathroom fan, and dishwasher. I think I may be calculating this incorrectly if this page shows the correct way to do the calculation: see here. 

If I follow this calculation and I have 3 general lighting circuits, 3 small appliance, and 1 laundry circuit, plus 2 separate circuits for space heaters (should I separate that from small appliance??) that's a total of nine 15-amp circuits. So if I split these between Phase A and Phase B, that would be four 15-amp circuits for one phase and five 14-amp circuits for the other. Adding in the fixed appliances and distributing, I'm getting 65 amps for the phase with higher current. Here's my question though: when I calculate this for the neutral, do I not follow 220.42 and 220.52(A) and multiply by 35% for loads above 3000VA? I have to use the full load of the receptacles and general lighting?

For the dryer and range, it seems you are right I can't fully discount the neutral current and have to multiply by 70%. So if I have a range at 8kVA and dryer at 5kVA, both x 70% = 38 amps?


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## emtnut (Mar 1, 2015)

Do you not like the answers you got on a few of your questions at Diychatroom ?

If so, call an electrician Shelly


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## shelly41 (Nov 22, 2020)

emtnut said:


> Do you not like the answers you got on a few of your questions at Diychatroom ?
> 
> If so, call an electrician Shelly


I did ask this question on DIY Chatroom, but no one responded to my question until a little bit ago. Just looking for multiple sources of input. Not really necessary for you to be rude.


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## zac (May 11, 2009)

Because your a freeloader.


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## zac (May 11, 2009)

shelly41 said:


> I did ask this question on DIY Chatroom, but no one responded to my question until a little bit ago. Just looking for multiple sources of input. Not really necessary for you to be rude.


And emtnut is not rude! He's just crazy. 

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## shelly41 (Nov 22, 2020)

Alright, well in the spirit of not being a total freeloader: in case anyone else has a similar question in the future, the examples in Annex D D1(a) and D1(b) pretty much answered my questions.

But, even if I am totally freeloading, you can take solace in the fact that there's probably some useful info that others browsing the forum can get from my discussions.


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## zac (May 11, 2009)

I take my freeloader comment back. 

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## shelly41 (Nov 22, 2020)

zac said:


> I take my freeloader comment back.
> 
> Sent from my SM-G970U using Tapatalk


Haha thanks


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## emtnut (Mar 1, 2015)

shelly41 said:


> I did ask this question on DIY Chatroom, but no one responded to my question until a little bit ago. Just looking for multiple sources of input. Not really necessary for you to be rude.


I find DIY'ers coming into an electrician chat room rude as well.

Bite me 😜


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## emtnut (Mar 1, 2015)

shelly41 said:


> Alright, well in the spirit of not being a total freeloader: in case anyone else has a similar question in the future, the examples in Annex D D1(a) and D1(b) pretty much answered my questions.


Post that in DIYchat. There are no electricians that don't know about reduced neutrals.

Freeloader 😗


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## shelly41 (Nov 22, 2020)

emtnut said:


> Post that in DIYchat. There are no electricians that don't know about reduced neutrals.
> 
> Freeloader 😗


Fine, I will leave your elite electrician forum. I do have an electrical engineering degree, but have never really dabbled in electrician work. Thought I could sneak in here 

I did post what I found in DIYChat, so maybe it will help some other noob


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## zac (May 11, 2009)

Sorry shelly, a degree in engineering isn't the same as an electrician. This is what we do. Would you like it if we bluffed with you? 

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## zac (May 11, 2009)

I would like to reinstate my freeloader post.

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