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nuclear power plant
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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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sir i ve applied for hindustan petroleum and power grid india ltd..and the exams on feb 22.can you pls send model papers and tell how to prepare for those.i need to get into any of these for sure.....
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