I have a 10 hp power water pump which on 420 volts 3 phase, my calcalation would be 12.8 A line current, which mean each phase draw 12.8 A is that correct ? Can any person help me in this question. Thanks Regards
420 volts phase to phase is about 242 volts to neutral. 10 HP is 7460 watts, so total current is 30.76 amps, or 10.25 amps per phase. But there could be power factor and efficiency issues that would cause the current draw to be 12.8 amps with output power of 10 HP.
Hi Paul. Thanks for your reponse. My water pump connection on 3 phase do not have neutral, if I include the 0.8 on power factor and efficiency it will have 16 A line current right ? Regards
Hi Paul. Thanks for your reponse. My water pump connection on 3 phase do not have neutral, if I include the 0.8 on power factor and efficiency it will have 16 A line current right ?
The US National Electrical Code gives 14A, 460V as typical for a 10HP motor. Converting that would be 15.3A at 420V as typical.
Actual values depend on power factor and efficiency as mowhoong posted. They also depend on whether the pump is developing 10HP. The power depends on the head pressure.
Hi Paul. Thanks for your reponse. My water pump connection on 3 phase do not have neutral, if I include the 0.8 on power factor and efficiency it will have 16 A line current right ? Regards
The effective phase-to-neutral voltage (which is the vector component in phase with current at zero PF), is V(phase-phase) / sqrt(3). 10 HP = 7460 watts / 242 V = 30.8 A /3 = 10.2 A per phase. You should never see 16 amps line current except on start-up. But a 0.8 factor on the 10.2 A gives 12.8 amps.
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