88 SYNCHRONOUS MOTORS 



two scales being therefore in the ratio of 15 to i). We will now study 

 this example in detail. 



Numerical Example. Suppose a case where energy is to be dis- 

 tributed for mechanical purposes by means of a line supplying a great 

 number of induction-motors. During the day, when a portion of these 

 motors are running at light load, their mean power-factor is about 

 0.70, i.e., the active current J w is approximately equal to the reactive 

 current. Let us assume that it has been decided that the total power 

 thus distributed will be about 100 kw. 



The distribution-voltage is 2000 volts; the line-current has been 

 predetermined on the assumption of a power-factor of 0.70; and, there- 

 fore, assuming an average efficiency of 0.75, the current would be 



i ooooo 



/= =95 amperes. 



2000X0.75X0.70 



The line, having one ohm of resistance (its efficiency being about 

 95 per cent), and being supposed to have negligible inductance, 

 gives a voltage-drop of 100 volts. An alternator has therefore been 

 provided, of sufficient capacity to produce 95X2100=200 k.w., and 

 it is believed that ample allowance has been made. This alternator 

 will have, for example, Xi=i8; Ri=i.6; whence Z\\T ) ohms, 

 and tan 7-1 = 8.3. The machine furnished by the manufacturer is 

 regulated for a dead-resistance load and, consequently, it is capable 

 of giving, with its fields almost saturated, the E.M.F. which seems 

 necessary for the full load, by making OA=U=2ooo volts, and EQ= 

 O A =2 580 volts. 



On putting the machine in operation an unexpected drop of voltage 

 is noted, and it is impossible to obtain the 95 amperes calculated. 

 In fact, if we draw the segment A A \ at an angle of 45, starting from 

 OA i, it is not 2580 volts which would be necessary, but practically, 



EI = OA i = 3 230 volts. 



Since the machine can only produce 2580 volts, the load-point 

 will have to be moved to the point of intersection, a\, of the angle- 

 line 7-, with the circle A C. The machine cannot, therefore, deliver 

 more than 45 amperes at the voltage of 2100 volts; and, since the 

 corresponding active current is only 32 amperes, it will be seen that 

 the alternator, although bought for 200 k.w. and expected to be good 

 for 100 k.w., can give, in actual service, only 21 + 2100=46 k.w.; 

 and yet the manufacturer is not to blame. 



