Electrical Power Systems

386 Electrical Power Systems
\ d1 = wx1
2
Similarly
Therefore,
2T
d2 = wx2
2
2T
...(15.58)
...(15.59)
h = d2 – d1 Using eqns. (15.60), (15.59) and (15.58), we get,
...(15.60)
h = w
2T
x2x 2
1 2 e - j ...(15.61)
Also
rom eqn. (15.61),
x 2 – x 1 =
L = x 1 + x 2
2TH
wx + x
rom eqns. (15.63) and (15.62), we get,
x 2 – x 1 = 2TH
wL
Solving eqns. (15.62) and (15.64), we get
In eqn. (15.65),
if L
2
if L
2
if L
2
> hT
wL , then x 1 is positive
= hT
wL , then x 1 is zero
< hT
wL , then x 1
x 1 = L
2
x 2 = L
2
is negative
...(15.62)
...(15.63)
b 1 2g
– hT
wL
+ hT
wL
...(15.64)
...(15.65)
...(15.66)
If x 1 is negative, the lowest point (point 0) of the imaginary curve lies outside the actual
span as shown in ig. 15.6.