Calculating Voltage Drop for 200HP Motor at 480V

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SUMMARY

This discussion focuses on calculating voltage drop for a 200HP motor operating at 480V, located 300 feet from the starter. The formula used for calculating percentage voltage drop (%VD) in a three-phase system is provided: %VD(3phase) = [12(1.73)(300ft.)(240A)]/[(Circular mil) * (480V)] x100. The user specifies using 12 for copper and notes that the full load current is 240A. Guidance is given to refer to the American Wire Gauge (AWG) chart for determining the circular mil value.

PREREQUISITES
  • Understanding of three-phase electrical systems
  • Familiarity with voltage drop calculations
  • Knowledge of American Wire Gauge (AWG) standards
  • Basic electrical engineering principles
NEXT STEPS
  • Research the American Wire Gauge (AWG) chart for circular mil values
  • Learn about voltage drop calculations for three-phase systems
  • Explore NEC (National Electrical Code) guidelines related to voltage drop
  • Investigate the impact of wire size on voltage drop in electrical installations
USEFUL FOR

Electrical engineers, electricians, and technicians involved in motor installation and maintenance, particularly those focused on optimizing voltage drop in three-phase systems.

vptran84
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Hi,

I was wondering if anyone can help me figure out how much voltage drop is happening in this particular scenario.

Here is my work:

Given 200HP motor at 480V that is 300 ft. away from the starter.

%VD(3phase) = [12(1.73)(300ft.)(240A)]/[(Circular mil) * (480V)] x100

I use 12 for copper, and with 200HP the Full Load Amp is 240A.

I have no idea where i can get the circular mil. Can someone please help me. I know that it's somewhere in the NEC.

thanks.
 
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