To calculate the full load current of a single phase or 3-phase AC motor is pretty straight forward. However, the terms input power and shaft power of the motor must be fully understood otherwise errors will be made in this supposed simple calculation.
The input power of an AC motor is the power it draws when it connected to a 1-phase or 3-phase voltage source. The motor then speeds up, develops a torque and outputs a shaft power.
The shaft power however is the mechanical power output from the motor after the motor losses in the stator, rotor, windings and other losses have been accounted for. The relationship between Input electrical power and shaft power which is mechanical is given by:
Output Shaft Power of a Motor in KW = Input Electrical Power in KW x Motor Efficiency
It is therefore possible to calculate the input electrical power when we know the details about the motor’s power supply, i.e., voltage, power factor, absorbed current and efficiency.
However as is common with most electric motors, the power rating in either KW or HP is usually the shaft power the motor can deliver to a load. This shaft power is dependent on the line voltage, power factor, full load current and the motor efficiency as illustrated below:
U * I *CosØ*ɳ/1000 |
Where:
U = Line voltage I = Line current or Full Load Current CosØ = Power Factor ɳ = Motor efficiency |
3-phase
Motor Shaft Power in kW = |
U * I *CosØ*√3*ɳ/1000 |
U = Line voltage I = Line current or Full
Load Current CosØ = Power Factor ɳ = Motor efficiency |
Single Power in KW *1000/(U* CosØ* ɳ) Here: U = 240 CosØ = 0.8 ɳ = 0.85 Power in KW = 1.5 *746/1000 = 1.119 Note 1 HP = 746W |
|
Therefore, I = |
(1.119 * 1000)/(240 * 0.8*0.85) = 6.86
Amps |
Full Load Current, I = |
3-phase Motor Power
in KW *1000/( U * I *CosØ*√3*ɳ) Here: U = 415 CosØ = 0.8 ɳ = 0.88 |
Therefore, I = |
(10 * 1000)/(415
* 0.8*0.88*√3) = 19.76 Amps |