AHJ and derated conductor sizing

[beechigby]

I have had this discussion here before, trying to understand the correct way to computre wire size.
My AHJ does it differently. Is he correct?
I have 16 microinvertes each putting out .9 amps at 240Vac = 14.4A
x 125% for continous duty= = 18 A
/ .65 (factor derived from temp addr of 33deg C to Ambinet temp of 30deg C= =27.69A
my AHJ says i need to use #10 THWN-2 wire cause #12 wire in the 75deg C col is rated for only 25A
Is that correct
I thought you derate the wire to find out allowable ampacity.
please help

 

[bob]

I have 16 microinvertes each putting out .9 amps at 240Vac = 14.4A
x 125% for continous duty= = 18 A / .65 (factor derived from temp addr of 33deg C to Ambinet temp of 30deg C= =27.69A
my AHJ says i need to use #10 THWN-2 wire cause #12 wire in the 75deg C col is rated for only 25A
Is that correct

90 C Ampacity rating of #12 THWN-2 = 30 amps at 30C. Correction factor 30C to 33C = 0.96
Adjusted Ampacity rating = 30 amps x 0.96 = 28.8 amps. Max 75C amperage rating #12 = 25 amps.
The adjusted amperage is greater than the listed 75C rating, therefore 75C rating is allowed.
If the adjusted amperage had been lower than the 75C rating then you would have to use the lower rating.
However the max breaker size for #12 CU is 20 amps.

Not sure where the 0.65 factor came from.

 

[Smart $]

...
my AHJ says i need to use #10 THWN-2 wire cause #12 wire in the 75deg C col is rated for only 25A
Is that correct
...

Highlighted appears to be regarding terminal temperature limitations under 110.14(C). For this determination you use the actual load current, or in this case, the actual source current of 14.4A. Your wire size must equal or exceed 14.4A in the 75?C column of Table 310.15(B)(16) [formerly Table 310.16]. That means #14 Cu or larger meets the 110.14(C) requirement.

 

[Smart $]

What year Code edition?

 

[beechigby]

We are now using the 2011 code.
In my calculation, the .65 factor is derived from the temp addr for conduit on the rooftop
30degC ambient temp +33 deg addr = 63degC to enter table 310.15 B 2 A yielding .65 in 90deg col.
It is my understanding that you derate the wire for temp to discover the max real ampacity that is allowed.
so I think 14.4 x 125% = 18 amps, (max inv. amps). and then, 30 amps x .65 = 19.5 amps= max wire is allowed to carry therefore, by inspection of 70deg col of T. 310.15 B 16 I see that the limitation on #12 wire is 25A so my 19.5, my 18, and my 14.4amps are all ok. do I have this right. if so, how to i convince the AHJ?
Max amps possible is 14.4 (inverter output, ) but that is calculated to be 125% more due to 3hr + operation, ie 18 amps
the AHJ is using the .65 inversely and applying it to the 18amps, to discover what the wire must carry, and looking comparitivelt to the 70deg col of table 310.15 B 16. to come up with #10 wire
Is he correct ?
really appreciate the input

 

[david luchini]

Max amps possible is 14.4 (inverter output, ) but that is calculated to be 125% more due to 3hr + operation, ie 18 amps
the AHJ is using the .65 inversely and applying it to the 18amps, to discover what the wire must carry, and looking comparitivelt to the 70deg col of table 310.15 B 16. to come up with #10 wire
Is he correct ?

No, he is not correct.

215.2(A)(1) tell you that the minimum feeder conductor size, before any adjustment or correction factor, shall have an allowable ampacity of not less than the non continuous load plus 125% of the continuous load. In your case: 14.4 * 1.25 = 18A. Per Table 310.15(B)(16), #14 Awg (75degC) would be the minimum conductor size.

215.2(A)(1) also says that the conductors shall have an ampacity not less than required to supply the calculated load. The calculated load is 14.4A. #12 Awg THWN-2 has an (unadjusted) ampacity of 30. Applying your 0.65 correction factor for the ambient temperature gives: 30 * 0.65 = 19.5A. 19.5 is larger than 14.4, so the ampacity of #12 THWN-2 corrected for 63degC ambient is not less than required to supply the calculated load. In addition, it is not smaller than the minimum allowable conductor size (#14). And finally, the #12 THWN-2 was a corrected ampacity of 19.5A will be properly protected by a 20A OCPD per 240.4(B).

#12 THWN-2 would be acceptable in your situation.

 

[Smart $]

No, he is not correct.

215.2(A)(1) tell you that the minimum feeder conductor size, before any adjustment or correction factor, shall have an allowable ampacity of not less than the non continuous load plus 125% of the continuous load. In your case: 14.4 * 1.25 = 18A. Per Table 310.15(B)(16), #14 Awg (75degC) would be the minimum conductor size.

215.2(A)(1) also says that the conductors shall have an ampacity not less than required to supply the calculated load. The calculated load is 14.4A. #12 Awg THWN-2 has an (unadjusted) ampacity of 30. Applying your 0.65 correction factor for the ambient temperature gives: 30 * 0.65 = 19.5A. 19.5 is larger than 14.4, so the ampacity of #12 THWN-2 corrected for 63degC ambient is not less than required to supply the calculated load. In addition, it is not smaller than the minimum allowable conductor size (#14). And finally, the #12 THWN-2 was a corrected ampacity of 19.5A will be properly protected by a 20A OCPD per 240.4(B).

#12 THWN-2 would be acceptable in your situation.

Under 2011 NEC the determination would be per 690.8(B)(2)... but the requirements are essentially identical.

(2) Conductor Ampacity. Circuit conductors shall be sized
to carry not less than the larger of 690.8(B)(2)(a) or (2)(b).

(a) One hundred and twenty-five percent of the maximum
currents calculated in 690.8(A) without any additional
correction factors for conditions of use.

(b) The maximum currents calculated in 690.8(A) after
conditions of use have been applied.

(c) The conductor selected, after application of conditions
of use, shall be protected by the overcurrent protective
device, where required.

 

[Smart $]

...how to i convince the AHJ?
...

Well to get his brain questioning how he is doing it, point him to Example D3(a) in Annex D, and therein to the following:

Total VA 113,200 VA
Conversion to amperes using three significant figures:

113,200 VA / (480V ? √3) = 136 A​

Minimum size overcurrent protective device: 136 A
Minimum standard size overcurrent protective device (see 240.6): 150
amperes

Ungrounded Feeder Conductors
The conductors must independently meet requirements for (1) terminations,
and (2) conditions of use throughout the raceway run.

Minimum size conductor at the overcurrrent device termination
[see 110.14(C) and 215.2(A)(1), using 75?C ampacity column in Table
310.15(B)(16)]: 1/0 AWG.

Minimum size conductors in the raceway based on actual load [see
Article 100, Ampacity, and 310.15(B)(3)(a) and correction factors to Table
310.15(B)(16)]:
99,000 VA / 0.7 / 0.96 = 147,000 VA
[70% = 310.15(B)(3)(a)] & [0.96 = Correction factors to Table
310.15(B)(16)]
Conversion to amperes:
147,000 VA / (480V ? √3) = 177 A

The example doesn't actually match your situation, but note the conditions of use have no bearing on the "terminations" determination.

 

[david luchini]

Under 2011 NEC the determination would be per 690.8(B)(2)... but the requirements are essentially identical.

Thanks, gotta remember to read the full context.:slaphead: