Wire Size

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ramsy said:
Example D3(a) omits Adjustments & Corrections.

1) Adjustment for 8 current-carrying conductors (CCC) @ 70% is omitted from Overcurrent calculation.
2) Continuous Load correction of 125% is omitted from Feeder calculation.
Click to expand...

Example D3(a) includes both a correction factor for temperature and an adjustment factor for number of current carrying conductors, in the proper place.

1) The overcurrent device doesn't care how many current carrying conductors in are the same raceway.
2) The 125% of continuous load is factored into both the OCPD size and the minimum conductor size calculation.
 
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david luchini said:
No, that is not correct.



You didn't mention if this was a feeder or branch circuit, but the rules are similar for both. 215.2 & 215.3 cover feeders and 210.19 & 210.20 cover branch circuits.

Your calculated load is 75A. Per 215.2(A)(1) (assuming feeder,) the conductors shall have an ampacity not less than required to serve the load...that is 75A. So you need a conductor that has an ampacity of at least 75A.

The second thing 215.2(A)(1) requires is that the conductor have a minimum SIZE, before the application of any adjustment or correction factor, that has an allowable ampacity of not less than the noncontinuous load plus 125% of the continuous load. This is where the 125% comes in. 75A*125%=93.75A. Table 310.15(B)(16) tells us that the smallest allowable conductor will be #3 AWG.

Next, you are applying an adjustment factor for 9 CCCs, which is 70%. That means you need a conductor that has an ampacity of at least 75A/0.7=107.1A. Since you are using THHN, the 90deg column can be used. T310.15(B)(16) shows that #3 AWG, THHN will have the proper starting ampacity, 115A. (Or going in the other direction, #3 THHN has an allowable ampacity of 115A at 90deg, applying the 70% adjustment factor makes the ampacity 80.5A, which is larger than the load of 75A from the first line of 215.2(A)(1)).

Next, 215.3 says the conductor must be protected against overcurrent per Part I of Art 240, and also says the rating of the ocpd shall not be less than the noncontinuous load plus 125% of the continuous load, or 93.75A, so a 100A ocpd would be the minimum size allowed, so let's assume the ocpd is 100A. The #3 AWG adjusted for 9 CCCs has an ampacity of 80.5, so it would not be properly protected by a 100A ocpd using the next size up rule.

#2 AWG would have an adjusted ampacity of 91A, so #2 THHN would have an ampacity large enough for the load (first part of 215.2(A)(1)), would be larger than the minimum allowable conductor size (second part of 215.2(A)(1)), and would be properly protected by a 100A ocpd (per 215.3 and 240.)
Click to expand...
Apparently I am doing the math wrong myself... Ok, the 125 % took us to 94 amps... but why did we go back to the 75 amps again to do the derating? Are we allowed to figure the worst case of several and use that, rather than calculating the 125 % then derating at 70%??

So we basically did the first calculation only to figure the OCPD? and the derating for conductors gets us the wire size?

Beginning to wonder if I have been using too much wire a lot of times... because I was looking at a 0 or 1 wire for here...and based on availability was writing a 0 wire down...lol
 
david luchini said:
Example D3(a) includes both a correction factor for temperature and an adjustment factor for number of current carrying conductors, in the proper place..
Click to expand...

Example D3(a) uses remote building Feeders with no continuous loads, the continuous lighting & process dryers are in main building.
 
ramsy said:
Example D3(a) uses remote building Feeders with no continuous loads, the continuous lighting & process dryers are in main building.
Click to expand...

Example D3(a) uses two feeders to remote buildings in a single conduit. The load for each building includes 38,900VA non-continuous and 56,600VA continuous. The example contains a 0.96 temperature correction factor and a 0.70 adjustment factor.
 
Steelhead said:
Again, I'm confused.
Click to expand...

Maybe this will simplify..

Calculate load.

Apply correction/adjustment factors to determine required conductor ampacity.

Find min. OCPD size using 125% of continuous load.

Verify that conductor from step 2 is properly protected by OCPD from step 3. Upsize conductor if required.

Verify that conductor size meets min required conductor size based on 125% of continuous load.
 
david luchini said:
..2) The 125% of continuous load is factored into both the OCPD size and the minimum conductor size calculation.
Click to expand...

Example D3(a) Overcurrent Protection applies (25% of 56,600 VA) (See 215.3 ) = 14,200 VA.
But, the Ungrounded Feeder Conductors subtract this 14,200 VA from the Feeder calculation.

How is +25% "continuous loads" factored into conductor size if only calculated for Feeder OCP (215.3), and Service Entrances (§230), but not calculated for Feeder & Branch conductors?
 
ramsy said:
Example D3(a) Overcurrent Protection applies (25% of 56,600 VA) (See 215.3 ) = 14,200 VA.
But, the Ungrounded Feeder Conductors subtract this 14,200 VA from the Feeder calculation.

How is +25% "continuous loads" factored into conductor size if only calculated for Feeder OCP (215.3), and Service Entrances (§230), but not calculated for Feeder & Branch conductors?
Click to expand...

Example D3(a)

Ungrounded Feeder Conductors

Minimum size conductors (per 215.2(A)(1)) at overcurrent device termination... #1/0 AWG



The calculated load is 95,500VA, or 115A: That would be a #2 AWG

Noncontinuous load plus 125% of continuous load is 109,700V, or 132A: That would be #1/0 AWG

They have applied the 125% of continuous load to determine the minimum conductor size.
 
david luchini said:
No, that is not correct.



You didn't mention if this was a feeder or branch circuit, but the rules are similar for both. 215.2 & 215.3 cover feeders and 210.19 & 210.20 cover branch circuits.

Your calculated load is 75A. Per 215.2(A)(1) (assuming feeder,) the conductors shall have an ampacity not less than required to serve the load...that is 75A. So you need a conductor that has an ampacity of at least 75A.

The second thing 215.2(A)(1) requires is that the conductor have a minimum SIZE, before the application of any adjustment or correction factor, that has an allowable ampacity of not less than the noncontinuous load plus 125% of the continuous load. This is where the 125% comes in. 75A*125%=93.75A. Table 310.15(B)(16) tells us that the smallest allowable conductor will be #3 AWG.

Next, you are applying an adjustment factor for 9 CCCs, which is 70%. That means you need a conductor that has an ampacity of at least 75A/0.7=107.1A. Since you are using THHN, the 90deg column can be used. T310.15(B)(16) shows that #3 AWG, THHN will have the proper starting ampacity, 115A. (Or going in the other direction, #3 THHN has an allowable ampacity of 115A at 90deg, applying the 70% adjustment factor makes the ampacity 80.5A, which is larger than the load of 75A from the first line of 215.2(A)(1)).

Next, 215.3 says the conductor must be protected against overcurrent per Part I of Art 240, and also says the rating of the ocpd shall not be less than the noncontinuous load plus 125% of the continuous load, or 93.75A, so a 100A ocpd would be the minimum size allowed, so let's assume the ocpd is 100A. The #3 AWG adjusted for 9 CCCs has an ampacity of 80.5, so it would not be properly protected by a 100A ocpd using the next size up rule.

#2 AWG would have an adjusted ampacity of 91A, so #2 THHN would have an ampacity large enough for the load (first part of 215.2(A)(1)), would be larger than the minimum allowable conductor size (second part of 215.2(A)(1)), and would be properly protected by a 100A ocpd (per 215.3 and 240.)
Click to expand...

So basically, when I have a continuous load and derate for number of conductors (or other conditions of use), I only need to end up with greater than the original load value, not greater than after the continuous adder correct?
 
electrofelon said:
So basically, when I have a continuous load and derate for number of conductors (or other conditions of use), I only need to end up with greater than the original load value, not greater than after the continuous adder correct?
Click to expand...

Yes, that is correct.

(The conductor still must be as large as the minimum conductor size required with the continuous load adder, and must still be properly protected by the OCPD.)

That is why a #2/0 AWG XHHW-2 is acceptable in Example D3(a), and why a #2 AWG THHN would be acceptable in the OP's question in this post.
 
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ramsy said:
We were missing (a) or (b): "worse case senario" described in D3(a)
215.2(A)(1)(a) or (b), 210.19(A)(1)(a) or (b)
Click to expand...
\

No we weren't. We got #1/0 for the first and #2/0 for the second. Therefore, #2/0 XHHW-2 is the correct conductor size for the load in the conditions described in Example D3(a).
 
Thanks for your persistence David.
I believe you are correct, and properly follow Ex. D(3)(a).

Some of us were missing the important (a) or (b) choice:
Choosing "worst case" between options:
215.2(A)(1)(a) or (b) for Feeders
210.19(A)(1)(a) or (b) for Branches.

(a) Results of "Continuous Load" factor from 75°c column
or
(b) Results of "Derating Adjustments" from 90°c column.

Some of us mistakenly assume we must factor everything with results checked from 90°c column:
(Incorrect): Continuous Load x 1.25 / CCC-factor (0.7) / Ambient-Correction (0.96) + Motors x 1.25, etc. = Conductor size. (Incorrect)

Confusing the code's use of Booleen logic "or" with "and" calculates more conservative sized conductors than required by NFPA-70.

Thats why you make the big bucks, and win more bids, because you pass inspections with less material costs.
 
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