392 ALTERNATING CURRENTS 



the total capacitance C to neutral be divided, one-half being 

 concentrated at the sending end and one-half at the receiving 

 end, in parallel with the load, Fig. 355. This assumption intro- 

 duces little or no error in the results, even for the longest existing 

 60-cycle lines. The condenser at the sending end has no effect 

 on the regulation, but its charging current 7 c /2 must be added 

 vectorially to the line current / in order to obtain the total 

 current supplied by the generator. The current 7 c /2 taken by 

 the condenser at the load must be added vectorially to the load 

 current I R in order to obtain the total line current 7. The prob- 

 lem is then treated by the methods already outlined. 



Example. It is required to deliver 30,000 kw. at 0.80 power-factor at a 

 distance of 100 miles, with a line loss not exceeding 10 per cent, of the power 

 delivered. The voltage at the load is 120,000 volts, 60 cycles, and the lines 

 are arranged at the apexes of an equilateral triangle, 12 ft. on a side. Deter- 

 mine: (a) the line regulation; (6) the total power supplied by the generat- 

 ing station. 



The power per phase, 



30000 



The volts to neutral at the load, 



E R = 120,000/ \/3 = 69,300 volts. 

 The current per conductor at the load, 



- 18 ' 5 



The power loss per conductor = 10,000 X 0.10 = 1,000 kw. = 1,000,000 



watts. 



1 000 000 



The conductor resistance R' = / 1cn ' r \' = 30.7 ohms. 



(180. o) 



30 7 

 Res. per mile = ' = 0.307 ojim. 



From Appendix H, page 464, the wire having the next lowest resistance 

 per mile is 0000, the resistance per mile of which is 0.264 ohm. 



The conductor resistance R = 100 X 0.264 = 26.40 ohms. 



From Appendix I, page 465, the reactance per conductor per mile 

 for 0000 wire and 144-in. spacing is 0.810 ohm. 



Total reactance = 100X0.810 = 81.0 ohms. 



The charging current for 0000 wire with 144-in. spacing and 100,000 volts 

 to neutral is from Appendix J, page 466, 0.523 amp. per mile. 



The total charging current for the above line is 



fiQ *}00 



L = 0.523 X - X 1QO = 36.2 amp. 



