INDUCTION MOTOR 341 



The diameter of the circle is raised through the distance of = Ip 

 and of now represents not the magnetizing current I M but the no- 

 load current 7 = V/ M 2 -+- I P 2 . 



If os represents the stator current per phase at standstill and 

 sw is its power component, then, since there is no output, the 

 power input is consumed by the losses. Therefore, input = losses 

 = nEiSW] constant losses = nEiIp = nEiVW and copper losses = 

 nEiSV = n (7 /2 ri + 1^. 



The stator copper loss is taken as nl'Vi because the stator copper 

 loss at no load is included in the constant losses. It is therefore 

 assumed that 7i 2 ri = Jo 2 /! + 7'Vi which is approximately correct 

 up to full load. 



Divide sv at t so that st : tv = 7 2 2 r 2 : 7 /2 ri, then, stator load 

 copper loss = nEitv and rotor copper loss = nEist. 



Join ft and from d any point on the circle to the left of s 

 draw dmpqr perpendicular to the diameter fk. It is to be shown 

 that mp is the stator current required to supply the rotor copper 

 loss for a rotor current represented by fd and that pq is the cur- 

 rent to supply the corresponding stator load copper loss. 



The rotor copper loss is 



ni 2 7*2 = n I 



= K X fq (fq + qK) = K X fq X D = fq X a constant, 



since K and D are both constants. 



Therefore the rotor copper loss is proportional to fq; but 



mp _ st mp st 



~W~(/s) : 



/-.> or V>j\2 ffo\2' 



and since st represents the rotor copper loss for a current fs, mp 

 represents the rotor copper loss for a current fd. 



Similarly pq represents the stator copper loss for stator cur- 

 rent fd. 



216. Interpretation of Diagram. At any value of stator cur- 

 rent od = 1 1, Fig. 321, 

 nEi dr = stator input in watts, 



nEiqr = constant losses, 



nE-^pq = stator copper loss, 

 nEimp = rotor copper loss, 

 nEi dm = rotor output in watts = mechanical load, 



