A Quick Guide on motor full load current calculation
One of the crucial devices used in various industries today is the Electric motor. It is used to convert electrical energy into a mechanical form of energy. These motors work on the principles of electromagnetism.
There are various factors affecting the functioning and durability of electric motors, out of which full load current is one of the most significant parameters. The motor full load current calculation must be considered for determining the suitable electric motor for a specific application.
Each electric motor is intended to operate in a definite range of conditions. Also, every motor has maximum operating conditions, which the power supply should match and not exceed.
If you want to know more about motor full load current calculation, let us explore further!
What is the Full Load Current in an Electric Motor?
Before jumping onto the calculation aspect, let us first quickly recall what is a full load current in the context of an electric motor.
The Full Load Current (FLC) of a motor is the maximum operation rating or the amount of current it can draw under specific conditions. It is also referred to as Full Load Amperage (FLA) or Full Load Amps.
If the motor supplies the rated current equivalent to the maximum current at the rated voltage, the load connected is known as a full load. Since different motors have different power ratings, the full load for all motors is not fixed.
Hence, it is significant to do a motor full load current calculation to determine its suitability per requirement. Notably, the motor full load current is an important factor for selecting apt fuses, cables, overload relays, circuit breakers, and other equipment required to prevent the motor from damage.
It is not considered favourable to operate a motor above the full or rated load. It can affect the durability and damage the motor. To ascertain a full load of a single-phase or three-phase AC motor, you need to do the motor full load current calculation.
Some Important Concepts For Motor Full Load Current Calculation
The motor full load calculation for an AC motor is not very complicated. For this, you first need to understand the concept of input power and shaft power of the motor.
When an electric motor is connected to a single-phase or three-phase voltage source, the power it draws is its input power. Then, the motor boosts its speed, generates torque, and yields shaft power.
The shaft power is essentially the output of mechanical power by the motor. Usually, the shaft power is measured in KW (kilowatts) in Europe, whereas it is measured in terms of HP (horsepower) in the US.
The relationship between Input power (electrical) and shaft power (mechanical) is presented as:
Output Shaft Power of a Motor (KW) = Input Electrical Power (KW) x Motor Efficiency (%)
Hence, shaft power for single and three-phase can be calculated as —
|Single-phase Motor Shaft Power (KW) = (V * A * PF * ɳ) / 1000||Three-phase Motor Shaft Power (KW)= (V * A * PF * ɳ * √3) / 1000|
- – V = Voltage (or line voltage)
- – A = Full Load Current (line current or rating)
- – PF = Power Factor Rating (technically denoted as CosØ)
- – ɳ = Motor efficiency
Understanding Motor Full Load Current Calculation
From the above formula of output shaft power of a motor, we can easily derive the formula for calculating the motor full load current for single-phase & three-phase motors.
- – Full Load Current for Single-phase motor:
A = Single-phase power in KW *1000 / (V * PF * ɳ)
- – Full Load Current for Single-phase motor:
A = Three-phase power in KW *1000 / (V * PF * ɳ * √3)
Let us take a quick example of a 10KW 3-phase AC motor, with the following ratings:
- Line voltage: 415V with a 3-phase power source
- Motor efficiency: 88%.
- Power factor: 0.8
The motor full load current calculation is as follows:
Full Load Current, A = Three-phase motor power in KW multiplied with 1000 / (V * PF * ɳ * √3)
Here, 3-phase motor power= 10KW; V= 415; PF= 0.8; ɳ= 0.88
A = 10 * 1000 / (415 * 0.8 * 0.88 * √3)
A = 10000 / (506.035)
A = 19.76 Amps
*(value of √3= 1.732)
#1. These PF, Efficiency, and power ratings are present on the nameplate of the motor.
#2. If the motor power is in HP instead of KW, you can convert it to KW in this way:
1HP = 746Watts and 1KW = 1000W
Power in KW = Power in HP *746/1000
Hence, the full load current for this 3-phase motor is 19.76 Amps.
Motor Full Load Current Calculation Via App or Websites
Other than manually doing the motor full load current calculation, you can also take the help of certain smartphone applications or websites available nowadays.
Various FLC calculators can help you in motor full load current calculation. You just have to open the app and enter the following figures in their respective fields.
- Select Motor Type – Single-phase or three-phase.
- Motor-rated power in kW or HP
- Voltage Supply
- Motor Efficiency (as per the nameplate)
- Rated power factor (if known)
After entering this data, tap on the Calculate button to find out the full load current of a motor.
The motor full load current is the maximum current the motor windings are designed to draw under specific conditions. It is simply the rated current of the motor at rated load and voltage conditions.
With most electric motors, the power rating in KW (or HP) is commonly the output shaft power of the motor. Hence, you need to do a motor full load current calculation to determine the actual current the motor takes.
The above article thoroughly explains this calculation, so you can clearly understand the concept.
Q. What are the essential factors for motor full load current calculation?
The crucial factors are:
- – The phase-to-phase or line voltage (V)
- – The power rating of the motor in kW
- – Rated power factor of the motor (CosØ)
- – Motor Efficiency (ɳ)
Q. From where can we get the ratings required for motor full load current calculation?
You can get the information required for the motor full load current calculation on the motor nameplate.
Q. Why does an electric motor not be operated with a load exceeding the full load?
You must not apply a current load exceeding the motor full load for a considerable duration, as the motor will start overheating, which can damage the windings and insulation of the electric motor.