272 ALTERNATING CURRENTS 



V 



PK = - , where V is the phase voltage and x\ and x 2 are 

 x\ -\- x z 



the respective stator and rotor reactances per phase, referred to 

 the stator. 



A perpendicular HJ is then dropped from H to OL. The line 

 HF is divided by G into two segments such that HG:GF = 7 2 2 # 2 : 

 Ii 2 Ri, that is, in proportion to the secondary and primary resist- 

 ances, respectively, as a one-to-one ratio of rotor to stator turns 

 is assumed. Line PG is then drawn. 



At any load current 7, 7 2 (= PE) is the secondary current, 

 being equal to 7 7 vectorially. EA is the energy component 

 of the current 7, and therefore the total power input per phase, 



Pa = EA X V 

 The core and friction losses 



P c = BA X V per phase 



The primary copper loss Ii z Ri = BC X V per phase 

 The secondary copper loss 7 2 2 / 2 = CD X V per phase 

 The output P = DE X V per phase 



DE 



The efficiency = -7- 



The torque T = CE (to scale) 



CD 



The slip, s = 



EA 



The power-factor = cos 8 = -y 



Draw P'G' parallel to PG and tangent to the circle at E'. 



Break-down torque T B = C'E' (to scale). 



The above diagram is drawn for but one phase of the motor. 

 The values of power, losses, and torque must be multiplied by n 

 if the motor has n phases. 



The torque scale may be found as follows: 



The torque is equal to a constant times the power, divided by 

 the speed, the value of the constant depending on the units 

 adopted. The power output per phase is P = V X DE. The 

 rotor speed Nz = N (1 s) where N is the synchronous speed 

 in r.p.m. 



/ CD\ N(CE -CD) _NXDE 



Nz N \ I ~CE) ~CE~ CE 



