340 



ELECTRICAL ENGINEERING 



that is, the rotor efficiency is equal to the rotor speed in per cent 

 of synchronous speed and, therefore, the efficiency of an induction 

 motor is always less than the speed in per cent of synchronous 

 speed. 



215. Modification of Diagram. When an induction motor is 

 running without load, a current J flows in each phase of the 

 stator which has two components, Fig. 320, I M the magnetizing 

 current 90 degrees behind the impressed e.m.f. E\, and IP the 

 power component in phase with EI. 



FIG. 320. 



Oar 

 FIG. 321. Circle diagram of an induction motor. 



The product nEJp (where n is the number of phases) is the 

 power required to supply the no-load losses. These are the iron 

 loss and a small copper loss in the stator and the friction and 

 windage losses of the rotor. The iron loss in the rotor may be 

 neglected since the rotor frequency is low. 



The current required to supply the stator losses has no cor- 

 responding component in the rotor, but the power to overcome 

 the friction and windage losses must be transferred from the 

 stator to the rotor and therefore requires a current in the rotor. 



As the motor is loaded and slows down the stator iron loss re- 

 mains nearly constant, the friction and windage losses decrease 

 and the rotor iron loss increases. At standstill the friction and 

 windage losses are absent but the rotor iron loss is large since the 

 rotor frequency is the same as the stator frequency. The iron 

 friction and windage losses are therefore considered to remain 

 constant and the small component of rotor current required to 

 supply the friction and windage losses is neglected. 



The diagram, Fig. 317, must therefore be changed to Fig. 321 

 by the addition of IP the power component of the stator current 

 per phase at no load. 



