Full text: Power distribution for electric railroads

  
  
   
  
  
  
  
  
  
  
  
  
  
   
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
   
    
TRANSMISSION OF POWER FOR SUBSTATIONS. 161 
ance in circuit and leave the line current nearly or quite 
in phase with its E. M. F. 
As the motor field is gradually strengthened, the cur- 
rent lags less and less, the apparent energy comes to ap- 
proximate more closely the real energy, and the current 
on the line consequently grows smaller. When the lag 
disappears the apparent energy (i. e. volts multiplied by 
amperes) coincides with the real energy, and the line cur- 
rent is a minimum. As the motor field is strengthened 
still more, the current begins to lead the E. M. F., the ap- 
parent energy increases, and the line current also in- 
creases. ¢ 
It is desirable to keep the power factor of the circuit 
(i. e., the ratio between real and apparent energy) as near 
unity as possible, since when this condition is fulfilled the 
energy delivered per ampere of line current is a maximum, 
and all the apparatus gives its best performance. Hence, 
the field of the motor or converter should be kept at such a 
noint that the line current for normal output, as shown by 
the ammeter, shall be a minimum. 
As the output varies, the current will lag or lead 
somewhat, but if the output for which the lag vanishes is 
properly chosen, the power factor at all working loads will 
still be high, say, within ten per cent of unity. 
The net result of the adjustment of the motor field 
with reference to the inductance in circuit is practically the 
maintenance of a power factor very near unity under all 
normal conditions, so that the circuit behaves almost as if 
it were carrying continuous current. Except for a small 
allowance for changes in the power factor the line may be 
computed much like a continuous current line. In fact 
the formulze of Chap. I may be used unchanged for fig- 
uring single phase and two phase transmission circuits, 
assuming that C in these formulae equals the watts deliv- 
ered divided by the voltage of delivery, as with continuous 
current, 
If, for example, we wish to deliver 450 k.w. 100,000 
ft. from our station, using 10,000 volts on the line with ten
	        
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