Motor Feeder Short-Circuit and Ground-Fautl Protection

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mondodoug

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NEC newb here. I'm a litte confused with article 430.62
It says that the rating of a motor feeder protective device is determined by adding the rating of the largest branch-circuit protective device for any motor supplied by the feeder to the sum of the full-load currents of all of the other motors supplied by that feeder.
What I have: 480VAC into a panel with two 25HP soft starts. According to my calculation, each motor has a FLC of 34 amps and should be fused for 51 amps. I'm trying to size the feeder circuit breaker. Do I need an 85 amp breaker or do I use the sum of 85 to size the breaker?
Thanks!
 
I believe you need a 110-ampere OCD.

Following the requirement:

34 FLC x 250% = 85A + 34A = 119A (Can't go up - next size down is 110A)
 
I agree with Bryan, based on the information given. Make sure that you use the amperage give in the 430 tables for the motors and NOT nameplate.
 
Thanks for the help! I use an ElectriCalc Pro for doing the calculations. I think I must have messed up the settings. I reset it and I've got better numbers.
I'm still confused, though. For a 25HP motor with a FLC of 34 amps it calculates a DE Fuse of 59.5 amps or an 85 amp Inverse Time breaker.
Does that sound right? I was planning to use fuses for the branches and a single breaker for the feeder. Would that make my feeder breaker 93.5 amps?
 
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