Electrical Power Systems

Analysis of Sag and Tension 397
Assuming parabolic configuration as shown in ig. 15.15.
L = 250 m, w = 0.80 kg/m.
Difference in level between the two supports
h = 70 – 30 = 40 m.
Note that both the towers are on the same side of the point of maximum sag. Hence x 1 is
negative.
Using eqn. (15.61),
As x 1 is negative, x 2 = L – x 1
h = w
2T
\ h = w
2T
\ h = w
2T
or points A and B, h = 40 m
0. 8 ´ 250
\ b250 - 2x1g = 40
2T
\
250 - 2x1 T
or points A and P, h = 45 – 30 = 15 m,
Horizontal distance between A and P
Using eqn. (i)
\
x2x 2
1 2 e - j
b g
bL-2x1g ...(i)
2 2 { L - x1 - x1}
= 0.40 ...(ii)
= 250
2
0. 8 ´ 125
(125 – 2x1 ) = 15
2T
125 - 2x1 T
Dividing eqn. (ii) by eqn. (iii), we get
250 - 2x
125 - 2x
1
1
= 125 m.
= 0.3 ...(iii)
04 .
=
03 .
= 4
3
\ x 1 = – 125 m
substituting x 1 = – 125 in eqn. (ii), we get
b g
T
250 -2-125 = 0.40
\ T = 1250 kg. Ans.