100 CONSTANT-VOLTAGE TRANSMISSION 



Draw a straight line at an angle below the base 

 line, where cos 6 is the P. F., lagging, of the load. By 

 means of a pair of dividers, add the ordinates of the 

 straight line to those of the circle, thus plotting the 

 ellipse, giving the Kva. of synchronous motors required. 



(2) Theoretical Limit of Load, in Kilowatts. 

 The maximum load is c f + a' 



E ( E s E' R' + E" X' 



>r 1000 \ V#'2 + X' 2 R' 2 + X'* 



This is numerically less than c' since a 1 is negative. It 



may be read from the circle diagram as it is the farthest 



distance to the right reached by the circle or the ellipse. 



(3) Reactive Kva. in the Line at Receiver End. 



For a more precise value than that obtained from 

 the circle diagram, solve the quadratic equation: 



02 Q 2 2 b z Q + c z = o 

 where a* = R' 2 + X'* 



b 2 = E' X' - E" R' 

 and c 2 = P 2 (R' 2 + X'*) + 2 P (' R' + E" X'} + E' 2 



Then Q = 



Use only the negative value of the radical. Q is the 



E O 



reactive current, and -- , the reactive Kva., in the line 



p n 



at the receiver, for a given Kw. load -- . When Q, as 



found above, is positive, it represents leading reactive 

 Kva. in the line at the receiver, and lagging, when "it is 

 negative. 



