40 



SYNCHRONOUS MOTORS 



motor (not including resistance-losses in the armature) will be right 

 lines perpendicular to YA 2 Y r . In the diagram, several of these right 

 lines have been drawn, corresponding to powers increasing each time 

 by 5 k.w. Let us complete the diagram by describing, from A 2 as a 

 center, with the rad'us ZI = 100X18.5 A, the circle of normal current. 

 The point AI, which depends on the load, can be displaced anywhere 

 in the whole space inside the circle of normal current- value ; but it 

 cannot be displaced outside this space, except for a short time, owing 

 to the heating of the machine. For each position of the load-point A i , 

 its distance from the point O, measured by the scale of volts, represents 

 the corresponding E.M.F. of the generator, and the distance AiA 2 , 

 divided by Z, represents the value of the corresponding current. 



If we suppose the generator to be exactly like the motor, i.e., if its 

 E.M.F. is 2000 volts, the diagram shows immediately that the maximum 

 power practicably obtainable cannot exceed 30 k.w. To attain 37.5 

 k.w. without exceeding the normal current, it would be necessary to 

 raise the E.M.F. of the generator, E\, to 2960 volts. 



Diagram of the Second Kind. The Vector of the Generator 



E.M.F., Ei, as a Fixed Axis. 

 Let the vectors corresponding to 

 the generator and motor E.M.F.'s 

 be represented by two right lines, 

 OA i and OA 2 , respectively (both 

 being taken with the signs they 

 present when the closed circuit is 

 followed around). The angle 

 ""-A AiOA 2 (Fig. 25), which we will 

 FlG - 2 5- still call 0, now measures the lag 



of E 2 behind EI. 



The current 7 is equal to the resultant E.M.F. AiA 2 divided by 

 the impedance, 



(12) 



and it lags by the angle f behind AiA 2 . Its vector A 1 C=RI is there- 

 fore the projection of AiA 2 over that angle, and the phase-difference 

 (f> between the current and the active E.M.F. E l is the angle OA^C. 

 In the case shown in the diagram, the current lags behind A^O, i.e., 

 behind the E.M.F. E\ measured at the terminals of the generator. 

 The E.M.F. OA\ can be decomposed into two components, of 



