50 TRANSMISSION LINE FORMULAS 



Current of substation load = 13.29 j 13.55. 

 Current of first section = 71.15 +j 25.15. 



p i - JQi = 84-44 +j 1 1. 60. 

 E l = + 105,370. 



(YZ \ 



+ etc.) 5070+^2,160. 



(Pi - jQi) Z = 2,940 +j 10,610. 



(Pi -JQi) Z (^ + etc.) = - 120 -j 150. 



= 103,120 -\-j 12,620 volts. 



AI -> ~ = 103,900 volts 



= voltage at the supply end of the line. 



PROBLEMS, CHAP. VI. 



(CONVERGENT SERIES.) 



1. Find, by the convergent series, the voltage drop of the follow- 

 ing line: 



Length of line 80 miles. 



Spacing 10 feet. 



Conductor No. oo aluminum cable. 



Load (at receiver end), 15,000 K.V.A., 100,000 volts, 95% 

 P.F., two phase, 25 cycles. (Prob. C, Chap. III.) 



Ans. 8810 volts. 



2. Find, by the convergent series, the per cent line drop and the 

 per cent regulation of the following line: 



Length of line 100 miles. 



Spacing 8 feet. 



Conductor No. 3 copper cable. 



Load (at receiver end), 3000 K.V.A., 66,000 volts, 90% P.F., 

 three phase, 60 cycles. (See Prob. A., Chap. Ill, and Prob. A, 

 Chap. V.) 



Ans. 7-08% drop, 9.40% reg'n. 



3. Find, by the convergent series, the K.V.A. and voltage at the 

 supply end, and the efficiency of the following line: 



