Electrical Engineering is a free introductory textbook to the basics of electrical engineering. See the editorial for more information....  # Calculation of Torque in a Motor

Author: E.E. Kimberly

As shown by equation (11-1), the force on a conductor carrying a current is The torque on one conductor at a radius of r centimeters is Then the total torque, in pound-feet, on all active conductors (those under poles and so capable of producing torque) is (H-2)

in which T = torque, in pound-feet;

Z = total number of active conductors; B = air-gap flux density, in lines per square inch; I = armature-conductor current, in amperes; I = active length of one conductor, in inches; r = average lever arm, or radius, in inches.

Example 11-1. - A 4-pole 230-volt shunt motor has a lap-wound armature 8 in. long with 800 conductors. The air-gap density is 50,000 lines per sq in. The mean radius from the armature center to 'the conductors is 6 in. The pole faces cover 65 per cent of the conductors. Calculate the gross torque when the armature draws 100 amp from the line.

Solution. - For purposes of substitution in equation (11-2), Hence, the torque is The method described for calculating the torque of a motor is correct and is one that is generally used, but the theory of the method includes the assumption that the conductors lie in a field of average intensity under the pole faces. Actually, most of the field flux follows the lower-reluctance paths through the armature teeth and so leaves the conductors in a relatively weak field in the higher-reluctance paths through the slots of the armature. However, the ampere-turns of the armature produce poles on the armature midway between the field poles. It is the mutual attraction and repulsion between field-pole fluxes and armature-pole fluxes that produces the torque of the motor. Although this theory of torque production is correct in fact, its application in calculations is very difficult.

Last Update: 2010-10-06