CONVERGENT SERIES 49 



Load taken by a substation at the middle of the line, 1 50 miles 

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

 P.P. (lagging). (Prob. B, Chap. V.) 

 Solution of first section of line: 



r = 0.3410, x = 0.791, b = 5.44 X io~ 6 , / = 150. 

 Z = 51. 15 + j 118.65. 

 Y = +j 0.000816. 

 FZ = 0.09682 +j 0.04174. 



(FZ F 2 Z 2 \ _ 



2 2-3-4 7 



I 1 h etc. ) = 0.01607 +j 0.00689. 



\2-3 2-3-4-5 / 



E = + 100,000. 



(FZ \ 

 h etc. ) = 4810 +j 2050. 



(P-jQ)Z= 10,090 +75780. 



200 j 20. 



A+jB= 105,080 +^"7810. 



B 2 



A H -- - = EI = 105,370 volts. 

 2 A 



In a similar manner, it is found that 



C + jD = 69.08 +j 30.36 amps. 



In-phase current, - = 71.15 amps. 



Reactive current, - - = 25.15 amps. 



A+ < 



2 A 



Solution of second section of line: 

 Conditions at middle of line, 



E! = 105,370 volts. 

 In-phase current of substation load 



1000 X 2000 X 0.70 

 = - - *- = 13.29 amps. 



105,370 



Reactive current of substation load 



looo X 2000 X 0.7141 



105,370 -='3.55 amps. 



