284 INDUCTION MOTORS. [Exp. 



equal to RJ^) and can be calculated. (For a 3-phase motor, 

 this loss is 3-R 1 / ( 2 2) .) 



1 6. Separation of Primary and Secondary Losses. If the 



primary copper loss is calculated and laid off as NS, we have SK 

 equal to the secondary loss. This needs to be done for one point 

 only, the line ASG being then drawn as a straight line. The point 

 selected is usually the point G, corresponding to the locked posi- 

 tion, and this can be located in several ways. The various meth- 

 ods for doing this are only approximate* and give slightly vary- 

 ing results, sufficiently accurate, however, for practical results. 



17. First Method. One procedure is to calculate the added 

 primary loss for the locked position and lay this off as HG, thus 

 locating the point G. The line AG is then drawn. For example, 

 in the present test, / (2) on short circuit is ^^" = 50.5 amperes; 

 R = 0.255 ohms. The point G is located so that HG = $ 

 X 0.255 X 5&5 2=I ,9$ watts. This procedure was used in con- 

 structing Fig. i. 



1 8. Second Method. With the motor at standstill, JH has 

 no particular significance; neither has 7 (see 32). Without 

 involving either of these, G is readily, and perhaps more accu- 

 rately, located from the short circuit current /s, by laying off 

 JG = RJs 2 , multiplying by 3 for a 3-phase motor. In the present 

 example, JG = $ X 0.255 X 56^' = 2,451 watts. 



19. The same location for G as found in the preceding para- 

 graph can be obtained by dividing JP" in the ratio R i : R 2 , where 

 R 2 is calculated as in 33. Then JG : GP"= R^ : R 2 =o.2$$ : 0.453. 

 This does not involve any special scale for JG and GP", and it is 



that the last term is neglected as small in the working range of the motor 

 (being zero when 7o and 7<2> are at right angles to one another), or (2) 

 that the loss represented by the last term is included in MN and com- 

 pensates for the decrease in friction and windage as the motor slows down 

 with load. 



* There is no method for determining secondary loss which is exact for 

 all loads and all types of motors ; compare 32. The use of three straight 

 lines AH, AG and AP" radiating from A for defining the various losses is 

 convenient but not exact. 



