Jeffcot r — Electrical Design of A.C. High Tension Transmission Lines. 83 

 1 



722 



•978. 



39,300 

 P^ = 3^8:00()=^^^^- 

 Thus these results are in close agreement with those of paragraph 13. 



15. As an example of long distance transmission, we will take the same data 

 as in paragraph 13, with the exception of the length of line, which we will now 

 suppose to be 200 miles. 



The calculation of the several quantities proceeds exactly as before till we come 

 to ml. Then 



(3) ml = 200 X (-000137 + i x -00177) = -0874 + i x -354. 

 cosh ml = cosh -0874 cos •35l + i sinh -0874 sin -354. 



cosh -0874 = 1 + -003819 + -0000006 = l-0038-<!. 

 sinh -0874 = -0874 (1 + -001273) = -08751. 

 cos -354 = -9380. 

 sin -354 = -3467. 

 .-. coshm^ = 1-00382 x -938 + i x -08751 x -3467 = -9416 + i x -03034. 

 sinh ml = sinh -0874 cos '354 + i x cosh -0874 x sin -354 



= -08751 X -938 + i x 1-00382 x -3467 = -0821 + i x -348. 

 n sinh vil = (388 - i x 96-2) (-0821 + i x -348) = 65-32 + i x 127-1 

 sinhm^ -0821 +tx -348 , ,n->o • „^ . ..x -, ^ . 

 -^ = 388 - z x 96-2 = (- l^-l^ + ' ^ S94-3) x 10- 



(4) Calculation of electrical quantities. 



Fj, Po. C's are the same as in paragraph 13. 



Fa cosh ml = (32,300 + i x 20,000) (-9416 + i x -03034) = 29,810 + i x 19,810. 



Cg?!, sinh 7)1/ = 85 x (65-32 + i x 127-1) = 5,550 + i x 10,800. 



.-. Fj = 35,360 + i X 30,610 = 46,750 volts to neutral at sending end. 



F, 

 C, = C, cosh ml + — ^ sinh ml. 



^ - n 



C'a cosh ml = 85 (-9416 + i x -03034) = 8003 + i x 2-579. 



SI w n 7/? / 

 Fa = (32,300 + i x 20,000) ( - 10-18 + i x 894-3) x lO"' 



= - 18-22 + i X 28-68. 

 .-. C'j = 61-81 + I X 31-26 = 69-3 amps, per phase at sending end. 



Pj = 35,360 X 61-81 + 30,610 x 31-26 = 2,183,000 + 957,000 watts 



=■ 3,140 kw. 

 2,750 



" = 3:no = ^^^• 



, 35,360 +ix 30,610 

 ^^ = ^94i6"^^T^03034 = ^^'^^'^ +^x 31,250. 



1 38,470- + 31, 

 "' V32,300^ + 20, 



250"- 



000-' = i-^°s- 



