The power of the motor should be selected according to the power required by the production machinery, so as to make the motor run under the rated load as far as possible. The following two points should be noted in the selection:
(1) if the power of the motor is too small, the phenomenon of “small horse pulling big cart” will appear, which will cause long-term overload of the motor and damage its insulation due to heat. Even the motor will be burned.
(2) If the motor power is too large, the phenomenon of “big horse pulling small car” will appear. Its output mechanical power can not be fully utilized, and the power factor and efficiency are not high (see table), which is not only unfavorable to users and power grid. And it’s a waste of electricity.
In order to select the power of the motor correctly, the following calculations or comparisons must be made:
(1) For the constant load continuous operation mode, if the power of the load (i.e. the power on the shaft of the production machinery) pl (kw) is known, the power of the required motor P (kw) can be calculated according to the following formula: P = P1 / N1N2, where N1 is the efficiency of the production machinery; N2 is the efficiency of the motor. That is, transmission efficiency.
The power calculated according to the above formula is not necessarily the same as that of the product. Therefore, the rated power of the selected motor should be equal to or slightly greater than the calculated power.
For example: the power of a production machine is 3.95kw. The mechanical efficiency is 70%. If a motor with an efficiency of 0.8 is selected, how much kW should the power of the motor be?
Solution: P = P1 / N1N2 = 3.95 / 0.7 * 0.8 = 7.1kw. Since there is no 7.1kw specification, the 7.5kW motor is selected.
(2) Compared with the motor with the same power and continuous rating, the motor with short-time rating has larger maximum torque, smaller weight and lower price. Therefore, the motor with short-term working quota should be selected as far as possible when conditions permit.
(3) For the motor with intermittent work quota, the power should be selected according to the load duration rate, and the motor specially used for intermittent operation mode should be selected. The calculation formula of FS% is
FS％＝tg/（tg+to） × 100％
Where TG is working time, t. The stop time is min; TG 10 to is the working cycle time min.
In addition, the power of motor can be selected by analogy. The so-called analogy. It is to compare with the power of motors used in similar production machinery. The specific method is: to know how high-power motors are used in similar production machinery of the unit or other nearby units, and then select motors with similar power for test run. The purpose of test run is to verify whether the selected motor matches the production machinery. The verification method is: make the motor drive the production machinery to run, measure the working current of the motor with a clamp ammeter, and compare the measured current with the rated current marked on the motor nameplate. If the actual working current of the electric machine is not different from the rated current marked on the motor, it indicates that the power of the selected motor is appropriate. If the actual working current of the motor is about 70% lower than the rated current marked on the rating plate, it indicates that the power of the motor is too large (that is, the motor with smaller power should be replaced by the motor with bigger power). If the measured working current of the motor is more than 40% higher than the rated current marked on the nameplate, it indicates that the power of the motor is too small (i.e. “small”
The motor with higher power should be replaced when the horse pulls the cart.
Editor in charge: LQ