NEC Annex D, Example D5(b)

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Jacobo

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I was doing the Example D5(b) and I found that for the Main Feeder neutral load, the calculation in D5(b) didn?t take into account 220.55. I checked my calculation (applying the standard method for the neutral) and every time I came to the same conclusion.
I?ve been struggling with this for two days and have no answer. I don?t know if I?m wrong, but, shouldn?t 220.55 be applied for this case?.

Hint: :According to D5(b), the Main Feeder Neutral load = 107,650 VA. This figure comes from D4(b), but D4(b) is served at 120/240 V (where 220.55 doesn?t apply), while D5(b) is served at 208Y/120 V. (where, I suppose, 220.55 does apply).
 

Smart $

Esteemed Member
Location
Ohio
Example D5(b) is an Optional Calculation. 220.55 does not apply. See 220.84 (under Part IV).
 

Jacobo

Member
I was doing the Example D5(b) and I found that for the Main Feeder neutral load, the calculation in D5(b) didn?t take into account 220.55. I checked my calculation (applying the standard method for the neutral) and every time I came to the same conclusion.
I?ve been struggling with this for two days and have no answer. I don?t know if I?m wrong, but, shouldn?t 220.55 be applied for this case?.

Hint: :According to D5(b), the Main Feeder Neutral load = 107,650 VA. This figure comes from D4(b), but D4(b) is served at 120/240 V (where 220.55 doesn?t apply), while D5(b) is served at 208Y/120 V. (where, I suppose, 220.55 does apply).

It's correct that D5(b) is an Optional Calculation, however, the neutral load calculation shall be based on the Standard method [See 220.84(A)(3)].
 

Smart $

Esteemed Member
Location
Ohio
It's correct that D5(b) is an Optional Calculation, however, the neutral load calculation shall be based on the Standard method [See 220.84(A)(3)].
Look at the last section of the example titled "Further Demand Factor". Note it factors the amount over 200A at 70%... then look at 220.61(B) and note it says 70% of either (1) or (2).
 

Jacobo

Member
I was doing the Example D5(b) and I found that for the Main Feeder neutral load, the calculation in D5(b) didn?t take into account 220.55. I checked my calculation (applying the standard method for the neutral) and every time I came to the same conclusion.
I?ve been struggling with this for two days and have no answer. I don?t know if I?m wrong, but, shouldn?t 220.55 be applied for this case?.

Hint: :According to D5(b), the Main Feeder Neutral load = 107,650 VA. This figure comes from D4(b), but D4(b) is served at 120/240 V (where 220.55 doesn?t apply), while D5(b) is served at 208Y/120 V. (where, I suppose, 220.55 does apply).

The application of 220.61(B) is not in discussiont. To understand what I'm questioning is summarized in the Hint of my original question. (Please read it)..
Note: To better understand the core of my question, following is an excerpt from 220.55:
"Where two or more single-phase ranges are supplied by a 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected between any two phases".
 

Smart $

Esteemed Member
Location
Ohio
The application of 220.61(B) is not in discussiont. To understand what I'm questioning is summarized in the Hint of my original question. (Please read it)..
Note: To better understand the core of my question, following is an excerpt from 220.55:
"Where two or more single-phase ranges are supplied by a 3-phase, 4-wire feeder or service, the total load shall be calculated on the basis of twice the maximum number connected between any two phases".
I understand what you are asking. Under 220.61(B), you can reduce the calculated neutral maximum unbalanced current by a demand factor of 70% of either (1) or (2). The example uses "option" (2). Therefore option (1) does not apply... and that is where 220.54/55 comes into the picture.

Now if you want to choose option (1) for your practice calculation, that's your choice... but it cannot be compounded with option (2).

The example has 40 ranges. The maximum number between any two phases will be at a minimum 14. Twice that evaluates as 15kVA + 1kVA ? 28 = 43kVA

The allowed reduction is 30% (the complement of 70%)...

43kVA ? 30% = 12.9kVA

12,900VA ? (208V ? sqrt(3)) = 35A

The example using option (2) has a reduction of 30A. Using option (1) would give you 5A more reduction on the neutral.
 

Jacobo

Member
I was doing the Example D5(b) and I found that for the Main Feeder neutral load, the calculation in D5(b) didn?t take into account 220.55. I checked my calculation (applying the standard method for the neutral) and every time I came to the same conclusion.
I?ve been struggling with this for two days and have no answer. I don?t know if I?m wrong, but, shouldn?t 220.55 be applied for this case?.

Hint: :According to D5(b), the Main Feeder Neutral load = 107,650 VA. This figure comes from D4(b), but D4(b) is served at 120/240 V (where 220.55 doesn?t apply), while D5(b) is served at 208Y/120 V. (where, I suppose, 220.55 does apply).

Smart $
I asked a very simple question in my post:: "shouldn?t 220.55 be applied for this case?.
The case that I'm exposing is not a perticular one. It is in the NEC book and it's a referrence for all us in the industry. Its procedure should be as precise as possible so it's important to have a documented answer to my question. This is what I did in my post and in my opinion, the procedure applied in the Example is incorrect, and if I'm right, it would be important to hear the opinion of the CMP No. 12.
Note: 220.61is not an issue here. Again, the core of the problem is explained in the Hint.

Thank you for your participation.
I
 

Smart $

Esteemed Member
Location
Ohio
Smart $
I asked a very simple question in my post:: "shouldn?t 220.55 be applied for this case?.
The case that I'm exposing is not a perticular one. It is in the NEC book and it's a referrence for all us in the industry. Its procedure should be as precise as possible so it's important to have a documented answer to my question. This is what I did in my post and in my opinion, the procedure applied in the Example is incorrect, and if I'm right, it would be important to hear the opinion of the CMP No. 12.
Note: 220.61is not an issue here. Again, the core of the problem is explained in the Hint.

Thank you for your participation.
I
You keep saying 220.61 is not at issue, but when it comes to the Minimum Neutral Load it is at issue, as that is the only way you are going to get a reduction of the neutral load per 220.61(B)(1). This is the only way 220.55 applies under the Optional Calculation Method of 220.84... at the feeder level. 220.55 has already been applied at the branch-circuit level. We have already discussed 220.61(B)(2) being used in Example D5(b), so 220.55 does not apply at all as calculated. (As a side note, it could be applied to the subpanel feeder neutral load. :happyyes:)


If you want to show the entire calculation from scratch, and note where 220.55 should apply, I'll be glad to comment... :happyyes:


Now with that said, the Annex D has this note at the very beginning:
This informative annex is not a part of the requirements of this NFPA
document but is included for informational purposes only.
It has been not been modified for I don't know how many editions now. It's been that way as long as I can remember. Yes there are few errors... some have been introduced by changes to the requirements in the actual Code... some may have been there from the get go... IDK.

It's quite likely the only way you are going to get the opinion of a CMP is if you make a proposal to change it... and I suggest you be 100% certain before making such a proposal. ;)
 
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