Jkffcott — Electrical Design of A. C. High Tension Transmission Lines. 81 



(2) Determination of most ceonomic size of eonduetor. 



C = 31 amps. 



A' = -0043 X -61 X / \, = -122 sq. inch. 



\lo X •08 '- 



Choose conductor 19/13, having A = -129 sq. inch; .-. a = 023 inch. 

 The current transmitted at maximum load is 85 amperes, and so the 

 current density in the conductor is allowable. 



(3) Calculation of line constants. 



R = — j- = '34 ohms per mile. 



h = 108 inches, a = -23 inch. 



^ = 470 ; log,o -^ = 2-67. 

 a a 



L =(8 + 74-1 X 2-67) x 10"^ = 2-06 x 10"^ henrys per mile. 



K = b X 10~° mhos per mile. 



„ 3-88 X 10-« , ,n ^aBf A 

 = — 2f7 " X 10 *" farads per mile. 



P = 2tt X 50 = 314-16. 

 pL = -648. 

 pS = 4-56 X 10-'. 



'« =J(-34 + i X -648) (5 x lO"*" + i x 4-56 x 10"''). 



Multiplying and evaluating these complex quantities by the rules given in 

 paragraph 6, we find 



m 



= J - 2-958 + i X 1-558 x 10"^ = -000437 + i x -00177. 



4 



•Si + ix -648 



r— TTTl . ,^^ -.n e = 100 /i4-24 - i x 7-47 



5 X lO"" + I X 4-56 X 10"° '^ 



= 388 - i X 96-2. 

 ml = 25 X (-000437 + i x -00177) = -01092 + i x -0442. 

 cosh ml = cosh -01092 cos -0442 + i sinh -01092 sin -0442. 

 cosh -01092 = 1 + -00006 = 1-00006. 

 sinh -01092 = -01092 (1 + -00002) = -01092. 

 cos -0442 = -999. 

 sin -0442 = -0442. 



Hence 



cosh ml = 1-00006 x -999 + i x -01092 x -0442 = -999 + i x -000483. 

 sinh ml = sinh -01092 cos -0442 + i x cosh -01092 sin -0442 



= -01092 X -999 + i x 1-00006 x -0442 



= -01091 +i X -0442. 

 n sinh ml = (388 - i x 96-2) (-01091 + i x -0442) = 8-49 + i x 16-1. 



smW ^ •01091 + ^ X -0442 ^ .^^^5 + i x 113-9) x 10- 

 n 388 - t X 9o-2 



