eeRyanC
Member
- Location
- Seattle, WA, USA
Hello,
I am having a difficult time with an example provided with the '08 NEC handbook under article 210.20(A) (I do not have a previous example of the handbook to check how long this example has been around). The example states to determine the min. size overcurrent protective device and min. conductor size for the following ckt.: 25A continuous load, 60 deg. C overcurrent device terminal rating, THWN conductors, and 4 current, carrying cu. conductors in a raceway.
I don't understand the methodology in the solution where the bundled conductors are accounted for. In the example, the calculated load of the circuit is adjusted by the .8 derating factor to get 25A/.8 = a conductor ampacity of 31.25A. My understanding was that the bundling derating had to be applied to the conductor ampacity (not limited by termination rating) rather than the calculated load. Doesn't the derating have to be figured into the conductor ampacity in order to protect the conductors against overcurrent protection per 240-4? In this particular example, the resultant conductor size is still correct because the THWN insulators are rated at 75 deg which yields 50*.8 = 40A which is greater than the 35A breaker. However, if type TW or UF insulators were used, we would get 40*.8 = 32A which would not be adequate. Is this correct?
-Ryan
I am having a difficult time with an example provided with the '08 NEC handbook under article 210.20(A) (I do not have a previous example of the handbook to check how long this example has been around). The example states to determine the min. size overcurrent protective device and min. conductor size for the following ckt.: 25A continuous load, 60 deg. C overcurrent device terminal rating, THWN conductors, and 4 current, carrying cu. conductors in a raceway.
I don't understand the methodology in the solution where the bundled conductors are accounted for. In the example, the calculated load of the circuit is adjusted by the .8 derating factor to get 25A/.8 = a conductor ampacity of 31.25A. My understanding was that the bundling derating had to be applied to the conductor ampacity (not limited by termination rating) rather than the calculated load. Doesn't the derating have to be figured into the conductor ampacity in order to protect the conductors against overcurrent protection per 240-4? In this particular example, the resultant conductor size is still correct because the THWN insulators are rated at 75 deg which yields 50*.8 = 40A which is greater than the 35A breaker. However, if type TW or UF insulators were used, we would get 40*.8 = 32A which would not be adequate. Is this correct?
-Ryan