K FORMULAS 37 



Conducted, 266,800 c.m. aluminum cable. 



Load at receiver end of line, 9000 K.V.A., 80% P.P. (lagging), 



100,000 volts, three phase, 60 cycles. 



Load taken by a substation at the middle of the line, 150 miles 

 from either end, 2000 K.V.A. at the line voltage and at 70% 

 P.P. (lagging). 

 Solution of first section of line. 



r = 0.3410 x = 0.791 / = 150, 

 P = 72 Q = 54. 



From the K formulas, 



A = 100,000 4860 + 3560 + 6360 



= 105,060 volts. 

 B = 2100 -+ 8470 2670 



= 7900 volts. 

 C = 72 - 3.50 + 1.13 - 0.57 



= 69.06 amperes. 

 D = 1.51 - 54 + 2.63 + 80.59 

 = 30.73 amperes. 



B 2 



A -\ = EI = 105,360 volts. 



2 A 



In-phase current = 



Reactive current = 



71.15 amps 

 EC- AD 



= 25.46 amps 

 Solution of second section of line. 

 Conditions at middle of line : 



EI = 105,360. 

 In-phase current of substation load 



loop X 2000 X 0.70 



= = 13.29 amps. 



105,360 



Reactive current of substation load 



1000 X 2000 X 0.7141 



105,360 -='3-56 amps. 



