I’m building a CNC router table that has a three-phase 220 V spindle motor, and 120 V computer (and accessories). The spindle’s nameplate:
As you can see it tops out at 20.3 A @ 220 V.
I have three-phase 208 V power. I would like to have a single plug and power cord for the spindle motor and related 120 V accessories (PC, monitor, drive electronics, solenoid valves, etc). The total of these accessory loads is probably less than 500 W.
Can I use 10/5 SJOOW power cord (not sure I can get that, but I know I can get SOOW 10/5 cord), and a 5-conductor locking plug to provide the three phases, neutral, and ground?
The NFPA 2017 NEC Handbook tables 400.5(A)(1) and 400.5(A)(3) seem so suggest I’m limited to 20 A with 10/5 cable.
400.4 also says:
A neutral conductor that carries only the unbalanced current from other conductors of the same circuit shall not be required to meet the requirements of a current-carrying conductor.
That could allow me to consider my cord as having three conductors, and thus not needing derating to 20 A. But it also refers me to b.310.15(b)(2)(11) for derating for load diversity.
What’s my best bet here?