NEC Cable Ampacity Calculation 1

[NickParker]

a) Required Cable Ampacity = FLC X 1.25
(OR)
b) Required Cable Ampacity = FLC / DF (Ambient temp) x DF (Cables in raceway)
(OR)
C) Required Cable Ampacity = FLC x 1.25 / DF (Ambient temp) x DF (Cables in raceway)
(OR)
d) Required Cable Ampacity = Largest of a), b) & C).
(OR)
e) Required Cable Ampacity = Largest of a) & b)

Which one of the above is the correct method to calculate the required ampacity?

 

[waross]

1.25 times ampacity.

More than 3 wires in a raceway or cable the ampacity is derated.
So; derated ampacity times 1.25.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[xnuke]

For general loads, it is the largest of one of three determined sizes: a) or b) or the size required by 110.14C (unless modified by a relevant NEC article, like 430). The statement by waross is not correct.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[wroggent]

For a feeder, the ampacity shall be the larger of 215.2(A)(1)(a) and 215.2(A)(1)(b), but also it is required that 110.14(C) is satisfied.

For 215.2(A)(1)(a):

Load = 100% * noncontinuous + 125% * continuous

Conductor ampacity (at any rated temperature) must be >= load


For 215.2(A)(1)(b)

Load = 100% * (continuous + noncontinuous)

Cable_rating_new = Cable_rating_old (at any rated temperature) * DF(amb temp) * DF (cables in raceway)

Cable_rating_new must be >= load


For 110.14(C)

The cable ampacity (without application of derating factors) at the termination's temperature rating (usually 60C or 75C but not always) must be >= the equipment termination's rating. For a OCPD on a feeder, see 215.3 for calculating the rating: it is 100% * noncontinous + 125% * continuous.



So I would say e) is the closest.

 

[waross]

a) Required Cable Ampacity = FLC X 1.25

Pretty clear.
Under the codes that i am subject to:
Cable ampacity = listed ampacity x derating for more than four cables in a raceway or cable and any further derating for temperature.
This is the new ampacity and this is the ampacity that that factor of 1.25 will be applied to.
Possibly I did not explain myself well.



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[xnuke]

Bill, he asked about the NEC, not the CEC. Could that be why our answers differ?

In 2014, the NEC clarified that it wasn't 1.25 AND application of derating factors, but rather 1.25 OR derating factors for continuous loads, like wroggent and I stated. A lot of people were using the first method for sizing. I don't know if the CEC followed suit with that clarification.

We also both mentioned that per the NEC, terminal temperature ratings must also be accounted for when sizing (which was added to the CEC in 2012, I believe).

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[davidbeach]

Maybe I’m repeating what’s been said or just saying it differently, but the way it works is you apply the derating factors to the ampacity table. Then you take the load times 1.25 and find a derated conductor large enough. If you’re using a 90C conductor/insulation in a condition that allows 90C (dry for instance) in an application that only allows 60C or 75C (lug ratings for instance) you can derate the 90C ampacity but you can then compare that to the 60C or 75C ratings. Then you have to derate the grounding conductor similarly.