If we consider the rpm to be 1450, then the full load current of 3 phase, 415V motor is approx 190 amps.
Now, the starting current of a induction motor is 5-7 times the full load current.
Hence the starting current will be approx 950 - 1330 amps.
Similar calculations can be made for below mentioned ratings also:
110 KW, 220 V ------F.L. current: 350 amps
110KW, 240 V-------F.L. current: 325 amps
the starting amp load is about 5 to 6 times the motor amp rating
starting current of 3 phase 75 KW induction motor
Assuming the Motor is IE1 Efficiency and 4 Poles, the in rush current will be 1183 A
The starting current is high because the motor's rotor winding has very low resistance. It's similar to a transformer with a shorted secondary windings. As the motor accelerates,the back emf increases which resist the flow of current in the rotor winding. Hence,the current drop to the rated full-load value.
Maximum current is defined by the rated KW of the device, say Motor. The motor may not run at its maximum rated capacity all the time. Nominal current is drawn when motor runs at nominal load.
FLA: Full Load Ampere. Its the current drawn by the motor when the motor is running at full load FLA = kW / (1.732*V*pf*efficiency*100)
starting current of 3 phase 75 KW induction motor
Assuming the Motor is IE1 Efficiency and 4 Poles, the in rush current will be 1183 A
depending on kva class of the motor between 4-8 times the running current
110 volts divided by 1,300 watts(1.3 kw) = .09 kw or 900 watts.
The cable size depends on a few factors, such as the supply voltage, the method of starting the motor, (direct-on-line, or star delta) and the length of cable required to compensate for voltage drop. Also, if the cable is copper or aluminum.110 kW motor has a rated current of 200 amps at 415 volts, so the correct copper cable size would have a cross-sectional area of 75 square mm., if the motor is started direct-on-line.At 660 volts, the rated current would be 118 volts and the cable size would be 35 square mm.
As the motor is drawing 9.7×110 = 1,067 watts (or 1.067 kW), and delivering 1.25×746 watts (or .9325 kW) of mechanical energy, it should release 1,067-932.5 = 134.5 watts (or .1345 kW) of heat.
The current depends on the supply voltage.
The starting current is high because the motor's rotor winding has very low resistance. It's similar to a transformer with a shorted secondary windings. As the motor accelerates,the back emf increases which resist the flow of current in the rotor winding. Hence,the current drop to the rated full-load value.
3phese 460 volt 75 kw a/c drivi full load current
KW = (sqrt 3) X Line voltage X Line current X Power factor Line current is proportional to Line voltage for same KW load so if 220 motor is connected to 110 v the Line current will increase. see Example: A 5 KW motor is connected to 220 V & then 110 v ( assume power factor is 0.850) Current ( I with 220 v) = (5 X 1000) / (1.73 X 220 X 0.850) = 15.5 Amp Current ( I with 110 v) = (5 X 1000) / (1.73 X 110 X 0.850) = 30.9 Amp
you need the current of motor or the KW/HP rating
It depends on what you mean by 'converter'; I'm not aware of such a machine. <<>> No, a 4 kW rotophase will not handle the start up current of a 4 kW 3 phase motor.