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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Answer Posted / Vijendra Singh
To calculate the ampacity after applying an additional safety factor of .25%, we can multiply the derated ampacity (922amps) by 1 - .25%:nn922 * (1 - .0025) = 922 * .9975 ≈ 918.68 amps.nnTherefore, the final cable ampacity would be approximately 919 amps.
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