THE INDUCTION MOTOR. 



Armature Winding. 240 slots, 2 conductors in each slot; wound 



in three phases in star connection; resistance of each phase, 0.016 

 ohm. 



112. E. M. F. According to (18) we have: 



(18) 



b = 17 cm. 

 / = 29.5 cm. 

 (B = 5600. 



The diminution of the air-gap through the open slots is taken into 

 account by reckoning only 85 per cent of the surface of the air-gap. 

 Thus, we find : 



4> = 1.58 . io* c. g. s. 



1 13. In the three-phase current motor we had : 

 #s = i . 42 . io* c. g. s. 



We insert N = 1.58 . io 8 in formula (19) and get 

 100 . e = 1.85 . ~ . z . $ . io" 6 = 2250 volts. 



Therefore, the terminal voltage should be equal to 2250 volts. 



114. Hysteresis. The loss through hysteresis and eddy currents 

 amounts to 1150 watts, of which 570 is dissipated in the teeth, and 

 580 in the iron ring. 



Magnetizing Current. From (17) follows 



(B . 1.6 A 



tm = 



H V 2 



We insert the following values : 



(B = 5600 c. g. s. 

 A = 0.15 cm. 

 n = 80. 

 and find 



i m = 1 2 amperes. 



64 



