Motor mechanical power calculation method - News - Global IC Trade Starts Here Free

High-power led lamp beads 1W white light Sanan 30mil high power LED lamp beads imitation lumens

Shunluo Inductance Power Inductor SWPA Series First Class Agent

Probe current voltage pin 420*4450 head diameter 5.0 over current current and voltage pin

Single chip microcomputer STM32L151CCU6

Mobile phone crystal 3.2*2.5mm 3225 26M (26.000MHZ)

The power of the motor. It should be selected according to the power required by the production machine, and the motor should be operated under the rated load as much as possible. Pay attention to the following two points when choosing:

(1) If the motor power is too small, there will be a phenomenon of “small horse-drawn cart”, causing the motor to be overloaded for a long time. The insulation is damaged by heat. Even the motor is burnt.
(2) If the motor power is selected too large. There will be a "big horse car" phenomenon. The output mechanical power cannot be fully utilized, and the power factor and efficiency are not high (see table), which is not only bad for users and the power grid. It also causes a waste of electricity.
To properly select the power of the motor, the following calculations or comparisons must be made:
(1) For a constant load continuous operation mode, if the power of the load (ie, the power on the production machine shaft) is known (Pl(kw). The power P(kw) of the required motor can be calculated as follows: P = P1/n1n2 where n1 is the efficiency of the production machine; n2 is the efficiency of the motor. That is, transmission efficiency.
The power obtained by the above formula is not necessarily the same as the power of the product. therefore. The rated power of the selected motor should be equal to or slightly greater than the calculated power.

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 this specification. So choose 7.5kw motor.
(2) Short-time work quota motor. Compared with a motor with the same continuous working rating. The maximum torque is large, the weight is small, and the price is low. Therefore, when conditions permit, try to use a motor with a short-time working quota.
(3) For the motor with intermittent working quota, the power selection should be based on the magnitude of the load duration, and the motor specially used for the intermittent operation mode should be selected. The formula for calculating the load continuous string Fs% is
FS%=tg/(tg+to)×100%
Where tg is the working time, t. To stop the time min; tg ten to is the duty cycle time min.

In addition. The analogy can also be used to select the power of the motor. The so-called analogy. It is compared with the power of a motor similar to that used in production machinery. The specific method is to know how much power the similar production machinery uses in this unit or other nearby units, and then use a similar power motor for the test. The purpose of the test run is to verify that the selected motor matches the production machine. The verification method is to make the motor drive the production machinery, measure the working current of the motor with a clamp-type 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 much different from the rated current marked on the spleen. This indicates that the power of the selected motor is appropriate. If the actual operating current of the motor is about 70% lower than the rated current marked on the nameplate. It means that the power of the motor is too large (that is, the "large horse-drawn car" should be replaced with a motor with a smaller power. If the measured motor current is more than 40% larger than the rated current indicated on the nameplate, it indicates the power of the motor. Choosing too small (ie "small horse-drawn cart"), should replace the motor with higher power.

table:
Load condition no load 1/4 load 1/2 load 3/4 load full load power factor 0.2 0.5 0.77 0.85 0.89
Efficiency 0 0.78 0.85 0.88 0.895