Answer Posted / gaju
The Planck current is the unit of electric current, denoted
by Ip, in the system of natural units known as Planck units.
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We have drive feeding an output contactor for a 5000 hp motor @ 5kV application. The motor FLA is 629amps x 1.25% service factor = 786.25 amps The cable that we current have in place between the drive and the output contactor is 350kcmil- MV-105 (parallel cables). Per the NEC table 310.60 (C) (69). This show the cable ampacity is at 615 amps per cable for a rating of 1230 amps. We have derated the cable for the cable tray and the multiple cables application and our calculations show that after the correction factor factor we are at 922amps. 1230 amps x .75% correction factor factor = 922.5 amps. The customer is driving us to then use another safety factor of .25% off of the already derated cable. Can you please provide a longhand calculation for your solution on this application?
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