338 ELECTRICAL ENGINEERING 



therefore, 



p r = E'l' cos 6 2 = -E 2 -I 2 cos 2 = #2/2 cos 2 . . (307) 

 n 2 H! 



Thus the power input to the rotor per phase is the product of the 

 e.m.f. which would be generated in the rotor by the flux $ g at 

 standstill and the power component of the rotor current. 

 The total power input to the n phases of the rotor is 



P r = nEzIz cos 2 ....... (308) 



212. Rotor Copper Loss and Slip. The power consumed by 

 the rotor copper loss per phase is 



7 2 V 2 = s# 2 / 2 cos 2 , ....... (309) 



and for the n phases it is 



L r = n/a 2 ^ = nsE 2 I 2 cos 2 . ... (310) 



01 . , . rotor copper loss L r snE 2 I 2 cos 2 



Slip = the ratio- ~ = s. (311) 



rotor input P r nE 2 I 2 cos 2 



213. Rotor Output and Torque. The rotor output per phase 

 is 



p p r / 2 V 2 = #2/2 cos 2 s# 2 / 2 cos 02 



= (1 - ) # 2 7 2 cos 2 ..... ; . . (312) 

 and the total rotor output is 



P = np = n (i _ 8 ) E 2 I 2 cos 2 watts .... (313) 



n (1 s) #2/2 cos 2 , 



' * -horsepower. . . . (314) 



From equations 308 and 313 the rotor output is 



and it is equal to the rotor input multiplied by the rotor speed in 

 per cent of synchronous speed. 



If T is the torque in foot pounds the output may be expressed as 



Thus the torque is 



T _ n(l - s) #2/2 cos 2 33,000 _ 7 n , n# 2 / 2 cos 2 

 746 ' 2irS S 



1 -s 



= 7.04 ^ = 7.04 rot rinpu ^ ft. Ibs (315) 



JV sync, speed 



