Choosing the right motor power is crucial for ensuring efficient and safe operation of machinery. The motor must be selected based on the actual power requirements of the production equipment, and it should ideally operate under its rated load to avoid unnecessary wear and tear.
When selecting a motor, there are two key points to consider:
(1) If the motor's power is too low, it may lead to a situation known as "small horse pulling a cart," where the motor becomes overloaded for extended periods. This can cause overheating, damage to insulation, and even motor failure.
(2) On the other hand, if the motor is oversized—commonly referred to as "big horse pulling a cart"—it may not use its full mechanical output capacity. This results in lower power factor and efficiency, which is inefficient for both the user and the power grid, leading to unnecessary energy waste.
To properly select the motor’s power, you need to perform calculations or comparisons. For constant load and continuous operation, if the load power (Pl, in kW) is known, the required motor power (P, in kW) can be calculated using the formula: P = Pl / (n1 × n2), where n1 is the efficiency of the production machine and n2 is the motor's efficiency.
The calculated value might not match an available motor specification, so it's recommended to choose a motor with a rated power that is equal to or slightly higher than the calculated value.
For example, if a production machine requires 3.95 kW of power, with a mechanical efficiency of 70% and a motor efficiency of 80%, the required motor power would be: P = 3.95 / (0.7 × 0.8) = 7.1 kW. Since 7.1 kW is not a standard size, a 7.5 kW motor would be the best choice.
For short-time duty motors, they typically have higher maximum torque, are lighter, and more cost-effective compared to continuously rated motors. Therefore, when possible, it's better to use a short-time duty motor.
For intermittent duty motors, the selection should be based on the load duration. A specific formula is used to calculate the duty cycle (FS%): FS% = (tg / (tg + to)) × 100%, where tg is the working time and to is the rest time. This helps determine the appropriate motor for the application.
Another practical method is to use analogy—comparing the motor power used in similar machines. By checking how much power similar equipment uses in your facility or nearby, you can select a comparable motor and test it. During testing, measure the motor’s operating current with a clamp meter and compare it to the rated current on the nameplate. If the measured current is close to the rated value, the motor is appropriately sized. If it's significantly lower, the motor may be oversized ("big horse"), and if it's much higher, it may be undersized ("small horse").
Table: Load Conditions vs. Efficiency and Power Factor
No Load: Power Factor 0.2, Efficiency 0
1/4 Load: Power Factor 0.5, Efficiency 0.78
1/2 Load: Power Factor 0.77, Efficiency 0.85
3/4 Load: Power Factor 0.85, Efficiency 0.88
Full Load: Power Factor 0.89, Efficiency 0.895
Parallel Actuator,Mini Electric Actuator,Linear Drive Actuator,Electric Actuator Controller
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