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Monday, August 8, 2016
Question:
Question Re: CQD answer published March 10, 2011 - Feeder Conductors to panel with multiple VFDs I'm confused by the answer to Mark Chatterton's CQD on 3/10/11. Mark's question was asking specifically about sizing the feeder not the branch circuit. However, the last sentence of the answer uses the term "branch circuit conductors". Can you review Mark's 3/10/11 CQD again and clarify how the feeder should be sized for a panel that is powering multiple VFDs? If you are not able to post the clarification as part of the CQD forum, can you provide me a response directly to my email address? Any help/clarification you can provide regarding this topic would be greatly appreciated. Thanks, Chris DeWeese
Note: Here is the original question again:
Would you help clarify how 430.122 (branch/feeder conductors) for adjustable-speed drive systems) affects the minimum feeder conductor size for multi-motor control panel installations where each motor is controlled by a variable frequency drive (VFD)?
• NEC sections 215.2(A)(1), 220.40, 220.50, 220.12(C), 430.24, and 409.20 seem to say we should use 125% of the full-load current rating of the largest motor plus the sum of the full-load current ratings of all the other motors.
• 430.122 seems to say we should use 125% of the rated inputs to the VFDs. These give significantly different values for minimum ampacity of feeder conductors because the VFDs tend to be oversized due to the desire to minimize the number of spare VFD sizes stocked in our plant storerooms. So, what’s the correct method: size the feeder based on the motor full-load current values or based on the VFD input current ratings?
Thanks, and best regards! Mark Chatterton
A
Answer:
Hey Chris thanks for your question, that was quite a while ago but you make a good point. The wording in 430.122(A) uses "Branch/Feeder Circuit Conductors" and "circuit conductors" so the 125% factor applies to both branch circuits and feeders. This is considerably different to the rules in 430.22 and 430.24 and requires adding up all of the rated input currents and multiplying them by 125% to determine the minimum ampacity.