THE CONTINUOUS CURRENT MOTOR 157 



There are two things to find the peripheral pull in 

 Ibs. and the peripheral velocity in ft. per min. The prod- 

 uct of these two quantities divided by 33000 (the number 

 of ft.-lb./min. in 1 h.p.) will give the h.p. of the motor. 



As the armature is lap wound and the machine has 12 

 poles the winding must have 12 paths. Therefore the 

 current per path = 480/12 = 40 amperes, and this is the cur- 

 rent in each conductor. The active length of each con- 

 ductor is 10 inches. The total number of conductors 

 = 240X4X2 = 1920. Of these 60% (i.e., 1152), lie in a 

 field of 60,000 lines per sq.in. and 15% (i.e., 288) lie in a 

 field of 35,000 lines per sq.in. 



The force in Ibs, is therefore equal to 



.885XlO~ 7 X40l (1152X10X60000) 



+ (288 X 10 X 35000) } = 2970 



The peripheral speed in feet per minute = 4XzX200 = 2515 



2515X2970 



Ihe horsepower developed therefore = = 226. 



ooUUU 



32. Current-torque Curves. The relation between the 

 torque developed by a motor and the current flowing 

 through its armature winding is shown by a current-torque 

 curve. This curve has different forms in motors with 

 different styles of field windings. 



Shunt Motor. The simplest case is that of the shunt 

 motor. The field current of this machine is independent 

 of the current flowing through the armature because its 

 field windings are connected directly across the supply 

 line, the voltage of which is assumed constant. Now if 

 the field current of a motor is constant, the strength of its 

 magnetic field is practically constant. Hence from equa- 

 tion (25 ) it is seen that the current-torque curve of a shunt 

 motor must be a straight line, the torque being directly 

 proportional to the armature current. This is shown 



