Sizing 3 phase service conductors

[david luchini]

Really? Is it wrong?

Yes

 

[david luchini]

259 is required for the load after 80% was applied to the conductor original ampacity

So a conductor of 324i is needed but 300 kcmil only shows ampacity of 300i

The load is 234A

 

[Tulsa Electrician]

Yes
1.25 and .80 reciprocate
1/.80= 1.25

So what I did was take a known to find the minimum. For me it is easier math and work for math check.
Since 259 is the total calculated load and four current carrying conductors. I take 259 times 1.25 for minimum ampacity derating for four conductors.

It's easier then doing each at .80
259 * 1.25 is the same as 259/.80.
Both are 323.75
Then scan down each 90 c column and choose the first one over 323.75. then as a math check look left to see if it's 75c rating carries the load.

If the question states using a given temperature rated conductor than I would start there.

It did not. It does throw us a bone and say service conductor.

So we're dealing with calculated load and conductors location.

No mention of wet damp dry etc.
So assume wet for service and use a 90 c rated wet location conductor
THWN- 2/ THHN. There are others

 

[Jpflex]

Yes
1.25 and .80 reciprocate
1/.80= 1.25

So what I did was take a known to find the minimum. For me it is easier math and work for math check.
Since 259 is the total calculated load and four current carrying conductors. I take 259 times 1.25 for minimum ampacity derating for four conductors.

It's easier then doing each at .80
259 * 1.25 is the same as 259/.80.
Both are 323.75
Then scan down each 90 c column and choose the first one over 323.75. then as a math check look left to see if it's 75c rating carries the load.

If the question states using a given temperature rated conductor than I would start there.

It did not. It does throw us a bone and say service conductor.

So we're dealing with calculated load and conductors location.

No mention of wet damp dry etc.
So assume wet for service and use a 90 c rated wet location conductor
THWN- 2/ THHN. There are others

Yes I see what you were doing. When I saw you multiply the total demand by 1.25 I thought you were doing this redundantly for continuous loads which also uses a 1.25 multiplier

1.25 and / 0.80 being reciprocals understood

 

[wwhitney]

So with 75C terminations the wire selected needs to have a 75C unadjusted ampacity (from the table) of at least 259A, and our OCPD also needs to be at least 259A (which means 300A).
. . .
So for 90C insulation and a 0.8 factor for 4 CCCs, we need a wire with a 90C unadjusted ampacity (from the table) of at least 251/0.8 = 314A.

Now that I have a chance to look at Table 310.16, I see that 90C rated 300 kcmil Cu will work, with a 75C base ampacity of 285A > 259A, and a 90C base ampacity of 320A > 314A. The ampacity of 90C rated 300 kcmil Cu in the given conditions would be 0.8*320 = 256A.

Cheers, Wayne

 

[wwhitney]

300 mcm Cu is to small.
320*.80= 256, must meet 259 which is the minimum for derate.

No, you do not ever need to simultaneously use the 125% factor for continuous loads and the ampacity correction and adjustment for temperature and number of CCCs. There are two separate checks--one uses the termination temperature limit, and the 125% continuous factor, but not correction and adjustment. The other uses correction and adjustment, but can use the full insulation temperature, and doesn't use a 125% continuous factor.

300 mcm Cu will work for the given problem.

Cheers, Wayne

 

[Tulsa Electrician]

No, you do not ever need to simultaneously use the 125% factor for continuous loads and the ampacity correction and adjustment for temperature and number of CCCs. There are two separate checks--one uses the termination temperature limit, and the 125% continuous factor, but not correction and adjustment. The other uses correction and adjustment, but can use the full insulation temperature, and doesn't use a 125% continuous factor.

300 mcm Cu will work for the given problem.

Cheers, Wayne

So if I understand what you saying.
The calculated load is 259 amps.
The neutral counts as a current carrying conductor in this case according the the problem in the book.

So we do have to derate for four ccc and not have to meet the calculated load in ampacity. I will go back and read 230.42 (A) (2)

Thank you

 

[david luchini]

So if I understand what you saying.
The calculated load is 259 amps.
The neutral counts as a current carrying conductor in this case according the the problem in the book.

So we do have to derate for four ccc and not have to meet the calculated load in ampacity. I will go back and read 230.42 (A) (2)

Thank you

The calculated load is 234A.

84,320VA @ 208V, 3ph = 234A.

 

[augie47]

Now that I have a chance to look at Table 310.16, I see that 90C rated 300 kcmil Cu will work, with a 75C base ampacity of 285A > 259A, and a 90C base ampacity of 320A > 314A. The ampacity of 90C rated 300 kcmil Cu in the given conditions would be 0.8*320 = 256A.

Cheers, Wayne

Correct me if I'm wrong...
230.42 states:
(1) Where the service-entrance conductors supply continuous loads or any combination of noncontinuous and continuous loads, the minimum service-entrance conductor size shall have an allowable ampacity not less than the sum of the noncontinuous loads plus 125 percent of continuous loads.
From that I think we agree the calculated load is 259 amps.
300 kcmil 90° is 320 amps.... now that we know the load is non-linear we have to apply a 80 factor or 320 x .8 = 256... shy of the required 259.
300 kcmil 75° is 285 but, again, the .8 factor would give us an ampacity of 228, shy of the required 259.

 

[david luchini]

Correct me if I'm wrong...
230.42 states:
(1) Where the service-entrance conductors supply continuous loads or any combination of noncontinuous and continuous loads, the minimum service-entrance conductor size shall have an allowable ampacity not less than the sum of the noncontinuous loads plus 125 percent of continuous loads.
From that I think we agree the calculated load is 259 amps.
300 kcmil 90° is 320 amps.... now that we know the load is non-linear we have to apply a 80 factor or 320 x .8 = 256... shy of the required 259.
300 kcmil 75° is 285 but, again, the .8 factor would give us an ampacity of 228, shy of the required 259.

The calculated load is 234A. After any adjustment/ correction factors, the conductor must have an ampacity not less than 234A...NEC 230.42(2).

Then select the larger conductor from 230.42(1) and (2). (The conductor also must be protected by the ocpd)

 

[wwhitney]

230.42 states:
(1) Where the service-entrance conductors supply continuous loads or any combination of noncontinuous and continuous loads, the minimum service-entrance conductor size shall have an allowable ampacity not less than the sum of the noncontinuous loads plus 125 percent of continuous loads.
From that I think we agree the calculated load is 259 amps.

The load for 230.42(A)(1) is 259A, the "calculated load" is 234A (no 125% factor).

230.42(A) and many of the similar sections are poorly written; see 690.8(B) for a well written version of the ideas here. I have various PIs in for the 2026 NEC to correct this issue. You have to read the above section 230.42(A)(1) in context with 240.42(A)(2), which says:

(2) The minimum service-entrance conductor size shall have an ampacity not less than the maximum load to be served after the application of any adjustment or correction factors.

So what's the difference between (1) and (2)? One obvious difference is that (2) doesn't have the 125% factor. Another difference is that (2) explicitly says "after the application of any adjustment or correction factors."

In context, then that means that (1) is "before the application of any adjustment or correction factors." That phrase deserves to be in part (1), particularly given the definition of ampacity, hence my PIs. But it only makes sense to read (1) that way, as otherwise (2) becomes redundant and never controls.

The result is that in (1), we use the 125% continuous use factor, but don't use the 0.8 CCCs factor, nor would we use any temperature correction factor. We also are limited to the termination temperature ampacity by 110.14(C).

While in (2), we don't use the 125% continuous use factor, but do use the ampacity adjustment and correction factors. We are also allowed to use the insulation temperature rating per 110.14(C)'s sentence "Conductors with temperature ratings higher than specified for terminations shall be permitted to be used for ampacity adjustment, correction, or both."

So there you have, two different checks, with two different sets of modifiers in each check. Plus we also have to keep 240.4 in mind. For that section the word "ampacity" is used without modification (even implicit), so it uses the value calculated in 230.42(A)(2) and the like.

Cheers, Wayne

 

[augie47]

The calculated load is 234A. After any adjustment/ correction factors, the conductor must have an ampacity not less than 234A...NEC 230.42(2).

Then select the larger conductor from 230.42(1) and (2). (The conductor also must be protected by the ocpd)

I guess that falls back to the original post. Is the 36,,200 the load with the 125% factor already included or do we need to add 125% for a continuous load of 36,200 ??

 

[wwhitney]

Correct me if I'm wrong...
. . .
300 kcmil 90° is 320 amps.... now that we know the load is non-linear we have to apply a 80 factor or 320 x .8 = 256... shy of the required 259.

The upshot is that for this check, you don't need the 125% factor, so the check is 256A > 234A, OK.

300 kcmil 75° is 285 but, again, the .8 factor would give us an ampacity of 228, shy of the required 259.

And the upshot for this check is that we don't use the 0.8 factor, so the check is 285A > 259A, OK.

Cheers, Wayne

 

[augie47]

For the moment hopefully we can agree to disagree... To me (1) says the load must include the 125% and (2) discusses factors such as fill and ampacity.

 

[david luchini]

230.42(A) and many of the similar sections are poorly written;

These sections used to be simpler, then some genius came up with an idea to change the language make it "more clear"....

And it's been confusing people ever since.

 

[david luchini]

For the moment hopefully we can agree to disagree... To me (1) says the load must include the 125% and (2) discusses factors such as fill and ampacity.

230.42 says the load shall be as calculated in Art 220. That's 84,320VA....234A.

 

[david luchini]

For 324 amperes 350 OCPD (Article 240(B)). Conductor size for this service is 400kcmil cu. or 600kcmil alm. Should be sizing conductors at service equipment terminal rating which should be 75 degree.(Article 110.14(C)(1). 90 degree column is used for ambient temperature correction factors or separately installed connectors.
Sound right?

The OCPD is 300A.

 

[wwhitney]

For the moment hopefully we can agree to disagree... To me (1) says the load must include the 125% and (2) discusses factors such as fill and ampacity.

The way you are reading (1), while admittedly understandable, renders (2) completely meaningless. If we apply the same factors in (2) as in (1), except for the 125% continuous use factor, then (1) will always control. This is a hint that your reading of (1) is not correct.

Here's 690.8(B), which actually uses clear language:

(B) Conductor Ampacity
Circuit conductors shall be sized to carry not less than the larger ampacity calculated in accordance with 690.8(B)(1) or (B)(2).

(1) Without Adjustment and Correction Factors
The maximum currents calculated in 690.8(A) multiplied by 125 percent without adjustment or correction factors.
Exception: Circuits containing an assembly, together with its overcurrent device(s), that is listed for continuous operation at 100 percent of its rating shall be permitted to be used at 100 percent of its rating.

(2) With Adjustment and Correction Factors
The maximum currents calculated in 690.8(A) with adjustment and correction factors.

This is the same idea that is behind 230.42(A), 215.2(A)(1), and 210.19(A)(1), just more clearly. In all cases the section that has the 125% factor for continuous currents (which in the context of 690.8 is all the current, so they left out the "plus 100% of the non-continuous" part) is "without adjustment and correction factors," while the section that is "with adjustment and correction factors" does not have any 125% factor for continuous currents.

Cheers, Wayne

 

[wwhitney]

For the moment hopefully we can agree to disagree... To me (1) says the load must include the 125%

Yes, but in context, since it doesn't not include the language in (2) calling for the application of adjustment and correction factors, it does not require applying those factors.

Terribly written, but that's the intention.

Cheers, Wayne

 

[augie47]

Look at example D3 in the Annex...
Particularity :"Minimum size service conductor"

Minimum Size Feeder (or Service) Overcurrent Protection (see 215.3 or 230.90) Subtotal noncontinuous loads 12,200 VA Subtotal continuous load at 125% (16,200 VA × 1.25) 20,250 VA Total 32,450 VA

It seems to support the fact that the load is 234 but the conductors need to address the 125% continuous as shown in 230.42)A)(1)