neutral calc question

[jcgjupiter]

Please give this some thought, its really bugging me. In the Stallcup book, published by the NFPA. In all the examples and questions for Commercial load calculations, the Neutral is only calculated at 100% not 125%.

The load is figured at 125% on the ungrounded conducters for each continuos loaded calc, but the neutral only at 100%.

Could these expensive books published by the same people (NFPA) that put out the NEC have made such a grevious error.

I know what the code says ( maximum unbalance) and all that, but is there something somewhere in my feeble understanding as i am learning that I am missing. I know everyone else does it differently. (Tom henry, Mike Holt,) but this Henry Stallcup didn't start yesterday either.
Go to a Technical Bookstore look at Stallcup's 2002 Master Electrician book (very thick) look at commercial calculations. On page after page
you will see the for continuos loading the load is figured at 125%, the neutral at 100%.

Can anyone share some insight on this. I know in the real world it probably wouldn't make a difference in the size of the neutral, but on a tough test it would. Thanks for hearing me out. Johnny

 

[bphgravity]

Re: neutral calc question

The latest edition (spring 2004) of the "necdigest" has a section called "code issues" by James Stallcup SR. This article is titled "Systematic and Orderly Calculations - Part II." On page 37, he provides a summary of all calculations, and shows all neutral loads at 100% as you are also stating.

Article 220.22 seems pretty clear on the issue. The maximum unbalanced load determined by Article 220 includes 25% increase for continuous loads.

Lets try to contact Mr. Stallcup for an explanation!

 

[bennie]

Re: neutral calc question

When the 125% is applied to the load current it may or may not require the next size conductor.

The phase conductors may require the larger conductor, but the amount in the neutral will be within its normal limits. Therefore no need to upsize.

 

[charlie b]

Re: neutral calc question

Originally posted by bphgravity:Article 220.22 seems pretty clear on the issue. The maximum unbalanced load determined by Article 220 includes 25% increase for continuous loads.

I agree. In fact, what it says is that ?the maximum unbalanced load shall be the maximum net COMPUTED load between the neutral and any one ungrounded conductor. . . ? (my emphasis). The 25% is part of the ?computed load,? so it should be part of the neutral calculation as well.

 

[bphgravity]

Re: neutral calc question

I contacted Mr. Stallcup Sr. via Email to get an explanation of his calculations shown in the necdigest as stated in my above post. Here is the response I have received from him today:

"Bryan,
Section 220.22 of the NEC does not require a 125% to be applied when sizing the load to select
the neutral conductor. Remember, the 125% rule is applied to derate the OCPD 80% so it will not trip open due to continuous loading (three hours or more). The neutral conductor continues to the busbar in the panelboard which under continuous loading does not need derating for it is not capable of tripping open due to a short-circuit condition. I hope this helps.
James Sr"

 

[bphgravity]

Re: neutral calc question

Now Im getting confused.

Lets perform a thought experiment and pretend we have an electrical service with 140-ampere continuous load. Say you don't apply the 25% increase. We install a 150-ampere OCPD and 1/0 conductors. What happens?

If we follow the rule and apply the 25%, we get a 175-ampere service and 2/0 conductors. Is this going to provide less heat overall on the system, which protects insulation and terminations, or is this for OCPD nuisance trip prevention as Mr. Stallcup is suggesting?

Any thoughts out there?

 

[bob]

Re: neutral calc question

BP you wrote
"If we follow the rule and apply the 25%, we get a 175-ampere service and 2/0 conductors. Is this going to provide less heat overall on the system, which protects insulation and terminations, or is this for OCPD nuisance trip prevention as Mr. Stallcup is suggesting?"
If you have continuous load you must rate the OC device at 80% of its listed value. Assume you have
a 100 amp load cont. 100amps/0.80 = 125 amp breaker. 125 amps x 0.80 = 100 amps cont.
You get the same result if you multiply the
100 amps x 1.25 = 125 amp breaker. 1/0.80 = 1.25.
You get another benifit. The larger wire will act as a heat sink and remove heat from the breaker.

 

[bphgravity]

Re: neutral calc question

Okay, so with that said, what about the grounded conductor? No connection to a breaker. Does the 25% increase need to be applied?

 

[charlie b]

Re: neutral calc question

Nothing I have read so far changes my view. Please look at my previous post, and in particular, at the one word that I put in bold and all caps text.

 

[bob]

Re: neutral calc question

Charlie's Post
"I agree. In fact, what it says is that ?the maximum unbalanced load shall be the maximum net COMPUTED load between the neutral and any one ungrounded conductor. . . ? (my emphasis"

If you are saying that you take the max unbalanced
load, which includes the 1.25 factor, and determine the neutral from this figure, I agree. I do not think you get the max unbalanced load and add 25% to obtain the correct neutral conductor size.

 

[charlie b]

Re: neutral calc question

Bob: I agree, for I think you and I are saying the same thing. That is, when you calculate the maximum imbalance, part of that comes from continuous loads (at 125%) and part from non-continuous loads (at 100%). You would certainly not tack on an additional 25% on top of that result.

 

[bphgravity]

Re: neutral calc question

But, what is the 25% for? Is for the OCPD? If so, the neutral is continuous. Obviously, if the ungrounded conductor is recieving heavier loading, than the grounded conductor will as well and should require the same adjustments.

As my previous posts have shown, I tend to agree that the continuous load computation needs to be carried over to the neutral, so where did someone with James Stallcup SR. reputaion and experience get his philosphy?

 

[charlie]

Re: neutral calc question

Please keep in mind that the people on the Code Making Panels do not agree with each other all the time. IMO, the neutral may be sized to carry the current as calculated without the additional 25%.

Ampacity. The current, in amperes, that a conductor can carry continuously under the conditions of use without exceeding its temperature rating.

I feel like the conductor will not be damaged even for continuous loads. The 125% is for the overcurrent protection, not for the conductors. For instance, 210.20(A) ". . . rating of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load." 215.3 says the same thing. The theme is overcurrent protection, not conductor protection.

Don't let my last statement lead you to believe the conductors do not need protection, they do. The rest of the Code provisions must also be followed.

 

[bob]

Re: neutral calc question

BP Posted
"But, what is the 25% for? Is for the OCPD? If so, the neutral is continuous. Obviously, if the ungrounded conductor is recieving heavier loading, than the grounded conductor will as well and should require the same adjustments."
The extra 25% is for the breaker. If the load is continuous, then the breaker rating is reduced to 80% of its rating. Suppose the continuous load is equal to 100 amps. We need to determine the breaker size for this load.
breaker size x 0.80 = 100 amps
breaker size = (100 amps)/(0.80)
breaker size = 125 amps.
You get the same answer if you take the 100 amps
continuous and multiply by 1.25. #3 thhn 75C can
carry this load forever and not have the insulation damaged.
If you have a breaker rated for 100% load then obviously you would not use the 1.25 factor.
Since the neutral is not connected to the breaker,
then the 80% factor is not included.

 

[bphgravity]

Re: neutral calc question

Bob, I understand the math, I just don't understand the intent.

 

[bphgravity]

Re: neutral calc question

Im asumming that a 100-ampere continuous load operates with the same effects to conductors and terminals as does a 125-ampere non-continuous load. So basically, the temperature rise and insulation stress on a 125-ampere conductor and OCPD will be the same be it a 125-ampere non-continuous load or 100-ampere continuous load.

So back to the original question. If the continuous loads happen to be line-to-neutral loads, then the neutral conductor would also experience the increase effects of the continuous loading and therefore should also be increased to accommodate.

 

[roger]

Re: neutral calc question

Let's simplify. How can a "Neutral" see the same current as any ungrounded conductor, so why would it be calculated the same?

Roger

 

[bob]

Re: neutral calc question

Im asumming that a 100-ampere continuous load operates with the same effects to conductors and terminals as does a 125-ampere non-continuous load.

That would not be true. This is a heat transfer problem. I used to know how to caculate this but have long since forgot. Using #3 thhn 75C, a continuous load of 100 amps would cause the conductor to reach 75C in 3 hours. A
non-continuous load of 125 amps(need #1 thhn)
would cause the conductor to reach a temp of some where between the ambient 30C and 75C. Don't know where because we do not know how long the load was on.

So basically, the temperature rise and insulation stress on a 125-ampere conductor and OCPD will be the same be it a 125-ampere non-continuous load or 100-ampere continuous load.

That would also not be true.

[ April 28, 2004, 08:15 PM: Message edited by: bob ]

 

[bphgravity]

Re: neutral calc question

I have done some research on this topic and found some interesting information. The first time the 125% wording showed up in the code was way back in 1965. Both conductors and OCPD were required to be increased under continuous loading.

In 1993, the wording for conductors dropped and only OCPD were required to be increased. It changed again for the 1996 code. It had been learned that even though the code had changed allowing 100% rated conductors but still requiring 125% rating of the OCPD, UL/NEMA were still testing OCPD with the increased conductors at 125%. This meant that the code allowed smaller conductor sizes on OCPD that were tested with the larger conductor sizes.

A proposal in the 1995 ROP showed that if the insulation on 90 degree conductors were being stressed, what was happening to the 75 degree rated breakers. The point was well taken and now the rule is for both conductor and breaker to be increased.

Also, it appears the exception to 210.20(A) is a waste of code due to the fact that UL does not list 100% rated OCPD under 250-ampere and 250-volts. That pretty much eliminates any typical branch circuit.

Thank to Brian McPartland for his perspective.

 

[eonebrandon]

Re: neutral calc question

..on the subject of neutrals, I have been taught and have therefore assumed that in the case of a 120v load, the neutral was concidered a "current carring conductor when appling table 310.15 (B)(2)(A). Is this not the case?