48 TRANSMISSION LINE FORMULAS 



Now E = 66,000. 



(YZ \ 



+ etc.J = -0.08543 +70.05772. 



E ( + etc.) = -5640+73810. 



P -JQ = 54-55-7 40.91- 

 Z = 708.24 +7 156.8. 



+ 5900-74430. 

 + 6420+78550. 



(P -70 Z = + 12320+74120. 



/ YZ \ 



'- + etc. 1 = - 0.02856 +7 0.01947. 

 \2-3 ' 



80+7*240. 



350 7 * 20. 



etc. = - 40+* 120. 



.) = - 



E = 66,000. 



+ etc> ) = ~ 5<54 +j 38ia 



-JQ)Z= I2320+J4I20. 



(P - JQ) z + etc. = - 430 +j 120. 



A+jB = 78,320 - 6070 +.; 8050 

 = 72,250+^8050. 



A H = 72,700 volts. 



2 /I 



Line drop = 6700 volts. 



PROBLEM B. 



Find, by the convergent series, the voltage at the supply end of 

 the following line: 



Total length of line 300 miles. 



Spacing 12 feet. 



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



Load at receiver end of line, 9000 K.V.A., 80% P.F. (lag- 

 ging), 100,000 volts, three phase, 60 cycles. 



