Jkffcott — Electrical Design of A . C. High Tension Transmission Lines. 79 



^ 3-88 X 10-^ „ , ., 



5 = — -— farads per mile. 



log.o - 



P = 27r /. 



m = y/{R + ipL) (K+ ipS). 



-^i 



R + ipL 



\K+ ipS 



cosh vil = cosh (x + iy) = cosh x eos y + i sinh x sin y = ^ + i^, 

 sinh ml = sinh {x + iy) = sinli x cos y + -i cosh x sin ?/. 



a;- X* x" 

 cosh a; =l+2~i'^Fi"''ft"t''''''" 



x^ x^ X' 

 sinh a;=a!+^+^ + s-j+.... , 



n sinh ?;!Z. 

 sinh tnl 

 n 



(4) Calculation of electrical quantities. 



Eefer all vector quantities to the phase of the eui-rent C'a as standard. 



The phase angles of the various vector quantities may readily be obtained 



if desired, since we know the components of the complex quantities. 

 For a single-phase line : — 



F2 = ^ X voltage between wires at receiving end = A + i/.i, where /n = A tan ^. 

 Pa = -^ X total power delivered. 



Go = ^^ 



V2 cos (p 



In a symmetrical three-phase line : — 

 V2 = voltage to neutral at receiving end = — - x voltage between wires at 



receiving end = \ + i/x, where ^ = X tan (p. 

 Pi = power transmitted per phase = | x total power delivered. 



p 

 C, = current per phase delivered = -= — - — . 



K2 cos <|) 



Then 



Fj = Facosh ml + nC^ sinhm/ = a -v i\i. 

 Voltage to neutral at sending end = ^/a^ + |3^ 



v. 



C, = d cosh ml + — -" sinh ml = v I- iiL. 

 Current per phase at sending end = ^-^ + ,^2. 



^1 = «X + i^- 



p 

 jj = efficiency = ^. 



-^ 1 



