The relationship between motor power, speed and torque

The concept of power is the work done per unit time. Under the condition of a certain power, the higher the speed, the lower the torque, and vice versa. For example, the same 1.5kw motor, the output torque of the 6th stage is higher than that of the 4th stage. The formula M=9550P/n can also be used for rough calculation.

For AC motors: rated torque = 9550* rated power/rated speed; for DC motors, it is more troublesome because there are too many types. Probably the rotational speed is proportional to the armature voltage and inversely proportional to the excitation voltage. Torque is proportional to field flux and armature current.

• Adjusting the armature voltage in DC speed regulation belongs to constant torque speed regulation (the output torque of the motor is basically unchanged)
• When adjusting the excitation voltage, it belongs to constant power speed regulation (the output power of the motor is basically unchanged)

T = 9.55*P/N, T output torque, P power, N speed, the motor load is divided into constant power and transverse torque, constant torque, T remains unchanged, then P and N are proportional. The load is constant power, then T and N are basically inversely proportional.

Torque=9550*output power/output speed

Power (Watts) = Speed ​​(Rad/sec) x Torque (N.m)

In fact, there is nothing to discuss, there is a formula P=Tn/9.75. The unit of T is kg·cm, and torque=9550*output power/output speed.

The power is certain, the speed is fast, and the torque is small. Generally, when a large torque is required, in addition to a motor with high power, an additional reducer is required. It can be understood in this way that when the power P remains unchanged, the higher the speed, the smaller the output torque.

We can calculate it like this: if you know the torque resistance T2 of the equipment, the rated speed n1 of the motor, the speed n2 of the output shaft, and the drive equipment system f1 (this f1 can be defined according to the actual operation situation on site, most of the domestic ones are above 1.5 ) and the power factor m of the motor (that is, the ratio of active power to total power, which can be understood as the slot full rate in the motor winding, generally at 0.85), we calculate its motor power P1N. P1N>=(T2*n1)*f1/(9550*(n1/n2)*m) to get the power of the motor you want to select at this time.
For example: the torque required by the driven equipment is: 500N.M, the work is 6 hours/day, and the driven equipment coefficient f1=1 can be selected with an even load, the reducer requires flange installation, and the output speed n2=1.9r/min Then the ratio: