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Problem calculating the current with a 3 phase motor?
<sigh> i know its trivial but i'm no more used calculating it, but now i need to and i'm unsure how it works.
i know P=U*I
but i know there was a factor for 3 phase engines like SQR(3)
now i'm confused how to proper calculate the current for a motor.
The plate on it reads 1.7 kw / 5.9 kw as i assume this is the wattage in star/delta config.
how much current do i have to expect in Delta or Star configuration at a voltage of 400 Volts phase to phase 230 Volts phase to neutral. how much current do i have to expect for startup ?
and how do i get there (thats even more important to know)
can someone help me ?
2 Answers
- Steve WLv 51 decade agoFavorite Answer
For three-phase motors, the calculations are:
P = sqrt(3) * U * I * cosphi * efficiency, where
P is rated power in Watts
U is line-line rms voltage, in Volts
I is line current in Amps,
cosphi is the power factor (i.e. cos(phi)) at rated power, and
efficiency is at rated power.
This formula is the same for delta and star-connected machines.
Typical starting currents are approximately 6 times the rated full load current, but the nameplate should provide some indication, either as a multiplier, an sKVA (starting kVA), or as a "code letter" (for ANSI/NEMA rated machines).
The 1.7 / 5.9 kW is a bit difficult to know for certain what the machine produce. If there are also comparable rated voltages, this could be understandable, since the motor rating is most likely based on a rated winding current.
At 5.9 kW, on a 400 V system, with an assumed cos(phi) = 0.85, and efficiency = 0.9, the motor full load current would be 11.1 Amps. Starting current would be approximately 66 Amps.
- GailLv 45 years ago
volts times amps = watts. A 300 kW (300,000 watts) at 240 volts can use up to 300,000/240 (roughly 1000) amps. that is, to put out the rated power requires than much current (or a ramping up of voltage inside the system by a transformer). frequency is a measure of the rate of back and forth motion in the AC circuit. Components tend to be "sized" to work at a certain frequency. Mismatch of frequency requirements can cause faulty operation or even failure of the equipment. As a very simple example, a conversion of AC to DC may use a diode and capacitor system to filter and flatten the output signal. The components are designed for a certain rate of signal and will not do as designed (and could fry) if input frequency is not as expected.
