156 ELECTRICAL MACHINERY 



those lying m the pole fringe where the field is weaker and 

 nonuniform. The same division of conductors is advisable 

 when calculating the torque of a motor. 



Example of Torque Calculation. Suppose we wish to 

 calculate the peripheral force on the armature shown in 

 Fig. 96. The length of the pole face we take as 15 cm; 

 the field we consider uniform and of a density equal to 

 8000 lines per sq.cm. There are 200 conductors on the 

 armature of which 60% lie under a pole face and the current 

 in each conductor is 20 amperes. The force (in dynes) 

 per conductor is given by 



f=Hl 7/10 = 8000X15X20X20/10 

 = 240000 dynes 

 = .245 kg. 



There are 200X60% = 120 active conductors. 



The peripheral force on the armature is therefore 

 .245X120 = 29.4 kg. 



Calculation of H.P. of a Motor. If the peripheral speed 

 of the armature is known, the output of the motor in ft.- 

 Ibs. per minute or horsepower is readily determined. If a 

 body is exerting a force /, and moving in the direction in 

 which this force is acting with a velocity v, then the rate 

 of doing work is equal to fv. Hence, if we multiply the 

 peripheral pull on the armature by the velocity of the 

 armature periphery, the product obtained is equal to the 

 amount of power which the motor is giving. 



Let us consider a lap wound armature 4 ft. in diameter 

 having 12 poles. The winding consists of 240 coils of 4 

 turns each and the length of the pole face is 10 inches. 

 60% of the conductors lie under the pole face where the 

 flux density is 60000 lines per sq.in., and 15% lie in the pole 

 fringe where the average density is 35000 lines per sq.in. 

 What h.p. is the motor developing if the current flowing 

 into the armature is 480 amperes and the machine is rotating 

 200 r.p.m ? 



