The power of the electric motor should be selected based on the power required by the production machinery, ensuring that the motor operates under its rated load as much as possible. The following two points should be considered when selecting the motor:
(1) If the motor power is selected too small, a "small horse pulling a big cart" phenomenon will occur , causing the motor to be overloaded for a long time, which will damage the insulation due to heat, or even burn out the motor.
( 2 ) If the power of the motor is too large, the phenomenon of "a large horse pulling a small cart" will occur. Its output mechanical power cannot be fully utilized, and the power factor and efficiency are not high (see table). This is not only detrimental to users and the power grid, but also causes energy waste.
To correctly select the power of an electric motor, the following calculations or comparisons must be performed:
( 1 ) For a constant load continuous working mode, if the power of the load (i.e. the power on the production machine shaft) Pl ( kw ) is known, the required power P ( kw ) of the motor can be calculated by the following formula: P=P1/n1n2 where n1 is the efficiency of the production machine; n2 is the efficiency of the motor, i.e., the transmission efficiency.
The power calculated using the above formula may not be the same as the product's power. Therefore, the rated power of the selected motor should be equal to or slightly greater than the calculated power.
Example: A certain production machine has a power of 3.95 kW and a mechanical efficiency of 70 %. If an electric motor with an efficiency of 0.8 is selected , what should the power of the electric motor be in kW ?
Solution: P = P1 / n1n2 = 3.95 / 0.7 * 0.8 = 7.1kw. Since there is no 7.1kw specification , a 7.5kw motor is selected .
( 2 ) Motors with short-time duty rating. Compared with motors with the same power and continuous duty rating, motors with short-time duty rating have a larger maximum torque, lighter weight, and lower price. Therefore, when conditions permit, motors with short-time duty rating should be selected as much as possible.
( 3 ) For motors with intermittent duty ratings, the power selection should be based on the load duty cycle, and a motor specifically designed for intermittent operation should be selected. The formula for calculating the load duty cycle Fs % is as follows:
FS %＝ tg/ （ tg+to ）× 100 %
In the formula, tg is the working time, t0 is the stopping time (min ), and tg + to is the working cycle time (min) .
In addition, the analogy method can also be used to select the power of the motor. The analogy method involves comparing the motor's power with that of similar production machinery. Specifically, this involves finding out the power rating of motors used in similar production machinery in your own unit or nearby units, and then selecting a motor with a similar power rating for a test run. The purpose of the test run is to verify whether the selected motor matches the production machinery. The verification method is as follows: run the production machinery with the motor, measure the motor's operating current with a clamp meter, and compare the measured current with the rated current marked on the motor's nameplate. If the actual operating current of the motor is not significantly different from the rated current marked on the nameplate, it indicates that the selected motor power is appropriate. If the actual operating current of the motor is about 70 % lower than the rated current marked on the nameplate, it indicates that the motor power is too large (i.e., "overpowered" and a smaller motor should be selected). If the measured operating current of the motor is more than 40 % larger than the rated current marked on the nameplate, it indicates that the motor power is too small (i.e., "underpowered" and a larger motor should be selected).
