Annex D Example D3(a)

Xptpcrewx

Senior Member
I'm working through Annex D Example D3(a) and need help understanding why the 3,960 VA of receptacle load is assumed to be equally distributed between all three phases when performing the conversion to amperes if we are dealing with 22 receptacles... If we assume near equal balancing, why wouldn't one phase have 180 VA more than the other two phases? and why isn't this reflected in the neutral load even if its fed from a SDS?

Secondly, why are they using 99,000 VA instead of 113,200 VA for determining the ungrounded feeder conductor size?
In the example, 195 amperes X 0.96 X 0.7 = 131 A does not have sufficient ampacity to carry 136A.
Wouldn't this require 3/0 instead of 2/0 since 225 amperes X 0.7 X 0.96 = 151 A and is greater than 136 A?

Thanks in advance.
 

david luchini

Moderator
Staff member
I'm working through Annex D Example D3(a) and need help understanding why the 3,960 VA of receptacle load is assumed to be equally distributed between all three phases when performing the conversion to amperes if we are dealing with 22 receptacles... If we assume near equal balancing, why wouldn't one phase have 180 VA more than the other two phases? and why isn't this reflected in the neutral load even if its fed from a SDS?
The extra receptacle would be less than half an amp imbalance. The receptacles aren't reflected in the neutral calculation because they are not carried by the 480/277 neutral.

Secondly, why are they using 99,000 VA instead of 113,200 VA for determining the ungrounded feeder conductor size?
In the example, 195 amperes X 0.96 X 0.7 = 131 A does not have sufficient ampacity to carry 136A.
Wouldn't this require 3/0 instead of 2/0 since 225 amperes X 0.7 X 0.96 = 151 A and is greater than 136 A?

Thanks in advance.
They use 113,200VA to determine the minimum conductor size before the application of any correction/adjustment factors. 113,200va/480/1.732=136A = 1/0 AWG conductor

They use 99,000VA (119A) to determine the ampacity of the conductor after the correction/adjustment factors. 195Ax0.7x0.96= 131A, which is greater than the 119A load, and is protected by a 150A c/b.

Therefore, a 2/0Awg conductor will handle the load for the conditions specified.
 
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Xptpcrewx

Senior Member
Response

Response

The extra receptacle would be less than half an amp imbalance. The receptacles aren't reflected in the neutral calculation because they are not carried by the 480/277 neutral.
Well 180 VA / 277V = 0.6498 A ≈ 1 A rounded to nearest whole ampere per section 220.5(B). Why would this not be carried by the 480/277Y neutral? It is a 3ph/4w feeder with unbalanced load....

They use 113,200VA to determine the minimum conductor size before the application of any correction/adjustment factors. 113,200va/480/1.732=136A = 1/0 AWG conductor
Actually, the example only uses 113,200 VA / (480V X SQRT(3)) = 136 A for the overcurrent protection calculation.

They use 99,000VA (119A) to determine the ampacity of the conductor after the correction/adjustment factors. 195Ax0.7x0.96= 131A, which is greater than the 119A load, and is protected by a 150A c/b.

Therefore, a 2/0Awg conductor will handle the load for the conditions specified.
Again, this is the part that doesn't make any sense to me. Can you elaborate why they aren't using 136 A to size the feeder conductors? The calculated load is 119 A, but 136 A reflects the 125% required by section 215.2(A)(1) and is what you need to size the feeder conductors. Note: This example isn't using a 100% rated breaker. Thanks in advance.
 

david luchini

Moderator
Staff member
Well 180 VA / 277V = 0.6498 A ≈ 1 A rounded to nearest whole ampere per section 220.5(B). Why would this not be carried by the 480/277Y neutral? It is a 3ph/4w feeder with unbalanced load....
The feeder to the transformer to supply the 208/120V loads is a 3ph/3w feeder. There is no neutral connection.

Actually, the example only uses 113,200 VA / (480V X SQRT(3)) = 136 A for the overcurrent protection calculation.
The section under "ungrounded conductors" says the minimum conductor size for the breaker terminations from the 75deg col. Is #1/0. That is based on the 136A calculation.
Again, this is the part that doesn't make any sense to me. Can you elaborate why they aren't using 136 A to size the feeder conductors? The calculated load is 119 A, but 136 A reflects the 125% required by section 215.2(A)(1) and is what you need to size the feeder conductors. Note: This example isn't using a 100% rated breaker. Thanks in advance.
The load is only 119A. The conductor needs to have an ampacity sufficient for the load. #2/0 conductor has the sufficient ampacity after the correction and adjustment factors have been applied. In other words, the #2/0 has sufficient ampacity to carry the 119A load. The conductor also cannot be smaller than the #1/0 conductor calculated from the continuous load factor. #2/0 is not smaller than #1/0, so it meets that requirement.
 

Xptpcrewx

Senior Member
The feeder to the transformer to supply the 208/120V loads is a 3ph/3w feeder. There is no neutral connection.
Ok. Good point about no neutral connection but this doesn't answer the question about the 1 A difference.

The section under "ungrounded conductors" says the minimum conductor size for the breaker terminations from the 75deg col. Is #1/0. That is based on the 136A calculation.
I get that (even though it is very poorly structured in the example); but I am not concerned about the minimum conductor size for the "breaker terminations".

The load is only 119A. The conductor needs to have an ampacity sufficient for the load. #2/0 conductor has the sufficient ampacity after the correction and adjustment factors have been applied. In other words, the #2/0 has sufficient ampacity to carry the 119A load. The conductor also cannot be smaller than the #1/0 conductor calculated from the continuous load factor. #2/0 is not smaller than #1/0, so it meets that requirement.
But what about 125% for continuous loads and Article 215 section 215.2(A)(1)?? You are basically telling me this does not apply...
 

david luchini

Moderator
Staff member
But what about 125% for continuous loads and Article 215 section 215.2(A)(1)?? You are basically telling me this does not apply...
The 125% for continuous loads applies to the minimum conductor size. But the actual load is only 119A, not 136A.
The conductor needs to have an ampacity sufficient to carry the load of 119A...this requirement is the first sentence of 215.2(A)(1)
 

Xptpcrewx

Senior Member
I really appreciate your help here but I am still getting hung up.... Thanks for your patience.

The 125% for continuous loads applies to the minimum conductor size. But the actual load is only 119A, not 136A.
The conductor needs to have an ampacity sufficient to carry the load of 119A...this requirement is the first sentence of 215.2(A)(1)
I agree but why are you neglecting the second sentence of 215.2(A)(1)? The 136 A is the non-continuous load + the continuous load...

Also, 180 VA / 277V = 0.6498 A ≈ 1 A rounded to nearest whole ampere per section 220.5(B) for the receptacle load is still bothering me. Can you please address this?
 

david luchini

Moderator
Staff member
I agree but why are you neglecting the second sentence of 215.2(A)(1)? The 136 A is the non-continuous load + the continuous load...
I'm not neglecting the second sentence of 215.2(A)(1). It tells me a need a minimum conductor SIZE of #1/0. The #2/0 that has the proper ampacity for the actual load is LARGER than the minimum conductor size of #1/0. You have to meet both requirements in 215.2(A)(1). I

Also, 180 VA / 277V = 0.6498 A ≈ 1 A rounded to nearest whole ampere per section 220.5(B) for the receptacle load is still bothering me. Can you please address this?
The receptacle load isn't 180va/277v. The receptacles are supplied by 480V, 3W feeder. The imbalance is miniscule and won't change the feeder size.
 

tortuga

Senior Member
I'm not neglecting the second sentence of 215.2(A)(1). It tells me a need a minimum conductor SIZE of #1/0. The #2/0 that has the proper ampacity for the actual load is LARGER than the minimum conductor size of #1/0. You have to meet both requirements in 215.2(A)(1).
Yeah in other words when your derating for the 8CC's in the same raceway you only need to use the actual load. No need to add the 125%.
 

Xptpcrewx

Senior Member
I'm not neglecting the second sentence of 215.2(A)(1). It tells me a need a minimum conductor SIZE of #1/0. The #2/0 that has the proper ampacity for the actual load is LARGER than the minimum conductor size of #1/0. You have to meet both requirements in 215.2(A)(1).
Not sure what differences in interpretation are but the misunderstanding has to be there.... Lets try this another way. Here's my understanding:

Section 215.2 Minimum Rating and Size

The minimum feeder-circuit conductor size (which we are trying to find), before the application of any adjustment or correction factors (in this case 0.7 X 0.96 = 0.672), shall have an allowable ampacity not less than the non-continuous load (42,400 VA) plus 125 percent of the continuous load (1.25 X 56,600 VA).

In other words, not less than 42,400 VA + 1.25 X 56,600 VA = 113,150 VA.
Which converts to: 113,150/3/277 = 136 A

So, the minimum feeder-circuit conductor size, before the application of any adjustment or correction factors (none applied yet), shall have an allowable ampacity not less than 136 A. This correlates to a 1/0 AWG conductor based on table 310.15(B)(16) 75 degree C column.

Feeder conductors shall have an ampacity not less than required to supply the load calculated in Parts III, IV, and V of article 220.
(No problem here since 42,400 VA + 56,600 VA = 99,000 VA and this converts to 119 A which can be carried by 1/0 AWG)


Now by applying adjustment and correction factors to 2/0 AWG (table 310.15(B)(16) 90 degree C column)... 0.672 X 195 A = 131 A

The ampacity of 131 A is NOT sufficient to carry 136 A.

Now by applying adjustment and correction factors to 3/0 AWG (table 310.15(B)(16) 90 degree C column)... 0.672 X 225 A = 151 A

The ampacity of 151 A is sufficient to carry 136 A.
 

david luchini

Moderator
Staff member
(No problem here since 42,400 VA + 56,600 VA = 99,000 VA and this converts to 119 A which can be carried by 1/0 AWG)


Now by applying adjustment and correction factors to 2/0 AWG (table 310.15(B)(16) 90 degree C column)... 0.672 X 195 A = 131 A

The ampacity of 131 A is NOT sufficient to carry 136 A.
You were good up until this point. The ampacity of the #2/0, 131A, is sufficient to carry the load of the feeder, which is 119A.

The maximum load that the feeder will see is 119A, not 136A.
 

Xptpcrewx

Senior Member
You were good up until this point. The ampacity of the #2/0, 131A, is sufficient to carry the load of the feeder, which is 119A.

The maximum load that the feeder will see is 119A, not 136A.

You say I was good up to this point, but that seems like a contradiction because that includes all this AND the statement in bold:

"The minimum feeder-circuit conductor size (which we are trying to find), before the application of any adjustment or correction factors (in this case 0.7 X 0.96 = 0.672), shall have an allowable ampacity not less than the non-continuous load (42,400 VA) plus 125 percent of the continuous load (1.25 X 56,600 VA)."

"In other words, not less than 42,400 VA + 1.25 X 56,600 VA = 113,150 VA."
"Which converts to: 113,150/3/277 = 136 A"

"So, the minimum feeder-circuit conductor size, before the application of any adjustment or correction factors (none applied yet), shall have an allowable ampacity not less than 136 A."
 

david luchini

Moderator
Staff member
You say I was good up to this point, but that seems like a contradiction because that includes all this AND the statement in bold:

"The minimum feeder-circuit conductor size (which we are trying to find), before the application of any adjustment or correction factors (in this case 0.7 X 0.96 = 0.672), shall have an allowable ampacity not less than the non-continuous load (42,400 VA) plus 125 percent of the continuous load (1.25 X 56,600 VA)."

"In other words, not less than 42,400 VA + 1.25 X 56,600 VA = 113,150 VA."
"Which converts to: 113,150/3/277 = 136 A"

"So, the minimum feeder-circuit conductor size, before the application of any adjustment or correction factors (none applied yet), shall have an allowable ampacity not less than 136 A."
I'm not sure how else to say the same thing.

The minimum feeder conductor size shall have an allowable ampacity before the application of any adjustment or correction factors of 136A. In the example, they have determined that that is #1/0.

Then the feeder conductor shall have sufficient ampacity to supply the calculated load ( after the application of any adjustment or correction factors.). The calculated load is 119A. In the example, a #2/0 conductor has sufficient ampacity for the load after the application of the correction and adjustment factors.

And the #2/0 conductor is larger than the minimum conductor size from the first step.

The last step would be to make sure that the conductor is properly protected by the OCPD. In the example, the #2/0 conductor has an ampacity if 131, so it is protected by the 150A c/b using the next size up rule.
 

david luchini

Moderator
Staff member
Maybe this approach will work...

The calculated load is 119A.

The minimum conductor size before the application of adjustment/ correction factors must have allowable ampacity if 136A, factoring 125% of continuous load. That is #1/0.

Now figure the ampacity of the 1/0 after the application of adjustment and correction factors. 170A×0.96×0.7= 114A. 114A is too low for the 119A load. So try the next larger conductor, 2/0. The ampacity is 195A×0.96×0.7= 131A. That has sufficient ampacity for the load.
 

tortuga

Senior Member
Xptpcrewx what code cycle are you on?
Not sure what differences in interpretation are but the misunderstanding has to be there.... Lets try this another way. Here's my understanding:

Section 215.2 Minimum Rating and Size

The minimum feeder-circuit conductor size (which we are trying to find), before the application of any adjustment or correction factors
215.2 has not contained the wording
before the application of any adjustment or correction factors
since the 2011 NEC.
In 2014 they clanged the before to after in 215.2(1)(b)
Cheers
 

Xptpcrewx

Senior Member
David, thanks again for your continued response.

Maybe this approach will work...

The calculated load is 119A.

The minimum conductor size before the application of adjustment/ correction factors must have allowable ampacity if 136A, factoring 125% of continuous load. That is #1/0.

Now figure the ampacity of the 1/0 after the application of adjustment and correction factors. 170A×0.96×0.7= 114A.
Yes. This makes sense I agree 100% with this line of reasoning.

114A is too low for the 119A load. So try the next larger conductor, 2/0. The ampacity is 195A×0.96×0.7= 131A. That has sufficient ampacity for the load.
I think I now understand what the issue is.... 215.2(A)(1) doesn't make mention about the minimum conductor size AFTER the application of adjustment/correction factors; and this is what I keep trying to rectify. This may be an entirely different issue now but what is the point of 125%? Is it not to account for the amount of heat generated for overloads and additional ampacity required to prevent thermal damage to the insulation?

If yes, then why in the world would we neglect this in the end result for conditions of use when adjusting/correcting? The 90 degree C wire would need to be rated for 136A after adjustments and correction to prevent thermal damage...
 

david luchini

Moderator
Staff member
If yes, then why in the world would we neglect this in the end result for conditions of use when adjusting/correcting? The 90 degree C wire would need to be rated for 136A after adjustments and correction to prevent thermal damage...

The 90degC wire has an ampacity of 131A and is carrying a load of 119A. How would it face thermal damage? It's operating temperature will be lower than it's insulation rating.
 

Xptpcrewx

Senior Member
The 90degC wire has an ampacity of 131A and is carrying a load of 119A. How would it face thermal damage? It's operating temperature will be lower than it's insulation rating.
The load could potentially be 136A for overload. 2/0 AWG... 195 A X 0.96 X 0.7 = 131 A. So thermal capacity is insufficient when current is at overload of 136A.
 
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