416 THE INDUCTION MOTOR. 



Several ways of calibrating the diagram will suggest 

 themselves for instance, we can proceed thus : The 

 diagram having been constructed in the manner indicated 

 above, take any point B which is evidently (from the 

 appearance of the diagram) in the neighbourhood of 

 full load we could, for instance, draw a line from the 

 point O which would-be a tangent to the semi-circle 

 A B F, thus giving the ^value of stator current at which 

 the power factor has its maximum value. In the present 

 case draw a vector of stator current, B (Fig. 198) 

 6-4 cms. in length, thus representing 128 amperes. 

 The watt component of this current is the vector B G 

 which is found to be 5-6 cms. in length, and con- 

 sequently 112 amperes. At this particular load on the 

 motor, therefore, the watt input is 



(112 x 202 x 3) - 68,000 watts (approximate). 

 Now at this load the total motor losses are as follows : 



(1) The light-load losses are 1,680 watts per phase 

 (from the measurement already made), or 5,040 watts 

 total. 



(2) The stator copper loss for the current OjB =128 

 amperes is given by the resistance per phase and by 

 the component F B of the stator current, for J the C~ R 

 loss caused by the* component F has already been 

 reckoned in with the no-load losses above, seeing that 

 O F is the (no-load) magnetising current of the motor. 

 From the diagram we scale off F B as 5-7 cms. thus 

 114 amperes. The stator copper loss is therefore 



(3 x 114 x 114 x 0.032) = 1,250 watts. 

 In the diagram, the line B H is proportional to this 

 loss and scales 0.6cm. Thus we have the scale from 

 which we can determine the losses at any other load. 



(3) The rotor copper loss is proportional to the line 

 H N in the diagram, which scales 0-9 cm. Consequently 

 we get the rotor loss from the known stator loss by 

 proportion or 



rotor copper loss = 1,250 x $ = 1,875 watts. 

 The total losses in the motor at the load corresponding 

 to the point B are therefore 5,040 + 1,250 + 1,875 =8,165 

 watts. The output is thus 68,000 -- 8,165 watts = 

 59,835 watts == 80 b.h.p. (approx.) 



