TftK INDUCTION MOTOR. 



armature copper losses) was - 1 * x 10 watts. 



_L,oZO 



The no-load driving current at this speed is (see Fig. 187) 

 3'4 amps. Thus the power developed by the motor was 



13 ' 4 Watts = 2 ' 48 



This power is equivalent to a torque of 151b.-ft. 

 This may be arrived at as follows : 



watts 2.480 

 H.P. developed = 



Also HP - 

 so, L.J-. - 



where T = torque in Ib.-ft. 



and n = revolutions per minute. 



H.P. x 33,000 3-32 x 33,000 

 Hence torque = T = 3 TI, /, 



2 TT n 2 IT x 1,1(50 



= 15 approximate. 



It will now be convenient to give a short discussion 

 of each of the quantities shown in the curves of Figs. 

 185 and 186, taking the stator current, power absorbed, 

 efficiency, power factor, and slip separately, and pointing 

 out some of the chief conclusions and reasons for their 

 variation as shown on the curves. 



Primary Current The current supplied to the stator is 

 determined by two factors as in the case of a loaded trans- 

 former. It has, firstly, to supply the varying magnetising 

 current in each phase required to maintain the rotating 

 field, corresponding to the no-load current of a transformer 

 which we may call the no-load component. Secondly, 

 the current has a component which produces a field equal 

 and opposite to that formed in the stator by the rotor 

 currents, corresponding to the component of the primary 

 transformer current which overcomes the demagnetising 

 action of the secondary circuit. This we may distinguish 

 as the load component. 



As already explained, the component of the rotating field 

 due to the rotor currents differs in direction from the 

 main primary field by 90; that is, it will act upon the 

 conductors one-quarter period later than the stator com- 

 ponent of the rotating field. Consequently, the load 

 component of the primary current will be one-quarter 

 period in phase behind the no-load component. 



