How to Calculate 1-Phase and 3-phase Motor Full Load Current (FLC)

Custom Search

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:

In Europe, shaft power is usually measured in kilowatts. In the USA, however, shaft power is measured in terms of horsepower (HP).

Shaft Power delivered by a single-phase motor is given by:

Where:

U = Line voltage

I = Line current or Full Load Current

CosØ = Power Factor

ɳ = Motor efficiency


Shaft Power delivered by a 3-phase motor is given by:

3-phase Motor Shaft Power in kW   =

U * I *CosØ*√3*ɳ/1000


Where:

U = Line voltage

I = Line current or Full Load Current

CosØ = Power Factor

ɳ = Motor efficiency


How to calculate 1-phase & 3-phase Motor Load current 
To calculate Motor Load current applying the above formulas, here we present some sample calculations:
Suppose we have a 1.5HP, 1-phase AC motor powered by 240V source with a design power factor of 0.8 and efficiency of 85%, the full load current will be arrived at by manipulating the power formula for 1-phase motors to give the motor full load current as :

Full Load Current, I   =

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


Suppose we have a 3-phase, 10KW, AC Motor with 415V, line voltage 3-phase power source and power factor 0.8 and efficiency of 88%, the full load current will be arrived at by manipulating the power formula for 3-phase motors to give the motor full load current as:

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


Important notes on Motor power calculations
The electrical power consumed by a single-phase motor in KW = U * I *CosØ/1000

Where the current here, I , is the full load or line current which MUST be given to calculate the power consumed by the motor. Unless otherwise clearly stated, the power rating of a motor in KW or Horsepower is always the shaft power in which case, the efficiency of the motor will be required to calculate the input power and thereafter the actual current the motor takes can then be calculated

Similarly, the electrical power consumed by a 3-phase motor in KW is given by:
U * I *CosØ*√3/1000

Where the current here, I , is the full load or line current which MUST be given to calculate the power consumed by the motor. Unless otherwise clearly stated, the power rating of a motor in KW or Horsepower is always the shaft power in which case, the efficiency of the motor will be required to calculate the input power and thereafter the actual current the motor takes can then be calculated

You May Also Like: