DE 



TAILED STUDY OF OPERATION WITH NORMAL LOAD 69 



and that, if sufficient compensating effect has been obtained at light 

 loads, the compensating effect will be still better with heavier loads. 

 First Numerical Example. Assume a distribution- system supply- 

 ing 500 k.w. of transformers at 5000 volts. The magnetizing current 

 is a small fraction of the full-load current, varying usually from 2 per 

 cent to 5 per cent, according to whether the transformers are large or 

 small. Let the average be taken at 3.5 per cent. The full- load current 

 being 100 amperes, the total magnetizing current, which is approximately 

 constant at all hours of the day, will be 3.5 amperes. The power- 

 factor with no load being, on the average, 0.70, the active current with 



o f 



no load will also be 3.5, and the resultant current will be = 5 amperes. 



To neutralize the lag by means of a synchronous motor having a rated 

 efficiency of 0.80 when running without load, it would be necessary 

 to produce a reactive current of 3.5 amperes at the expense of 0.7 ampere 

 of active current corresponding to the resistance-losses. The current 

 taken from the distribution-system would then be 3. 5 + 0. 7 4.2 amperes, 

 instead of 5 amperes. This saving is not sufficient to justify the pur- 

 chase and maintenance of a motor intended to serve solely as a com- 

 pensator; but the use of a condenser would not be much more advan- 

 tageous, as the current would always remain slightly above 3.5 in 

 consequence of dielectric hysteresis. 



Second Numerical Example. Assume a distribution-system com- 

 prising, at the end of the line, where the tension is 2000 volts, some 

 translating devices composed of 200 k.w. of induction-motors and 100 

 k.w. of synchronous motors, the latter having a total reactance of 15 

 ohms and a resistance of 0.8 ohm. Assume the synchronous motors 

 to have an efficiency of 80 per cent, the induction-motors to have a 

 magnetizing current equal to one-third of the normal current, an 

 efficiency of 85 per cent, and a power-factor of 0.95, at full load. 



[The magnetizing current generally varies between one-third and 

 one-quarter of the load-current in polyphase motors and between one- 

 half and one-third in single-phase motors.] 



The problem assumed is: to compensate the effects of the lagging 

 current in the induction-motors when running with no load, and also 

 with a quarter-load, it being assumed that all the motors are constantly 

 in operation. 



The total current necessary for the induction-motors at full load is 



200000 



-=124 amperes. 



0.85X0.95X2000 



