Electrical Power Systems

396 Electrical Power Systems
Horizontal distance of point, B from O (OB¢ )
= L – x 1 = 250 – (–123.27)
= 373.27 m
Therefore, height of mid point P above O,
2 I
HG KJ 116 2 48 27
=
d1 =
w L - x1
2
2T
. ´ . .
2 ´ 1800
Similarly, height of point B abve O,
2 b g =
= 19.86 m
2 b g = 44.9 m
wL- x1
116 . ´ 37327 .
d2 =
2T
2 ´ 1800
Hence mid point P is (d2 – d1 ) = (44.9 – 19.86)
= 25.04 m below point B.
Height of the mid point P with respect to A
= (19.86 – 4.9) m = 14.96 m
Therefore, clearance between the conductor and the water level mid–way between the
towers will be
s = (40 + 14.96) m = 54.96 m
or
s = (80 – 25.04) m = 54.96 m
Example 15.5: An overhead transmission line at a river crossing is supported from two towers
at heights of 30 m and 70 m above the water level. The horizontal distance between the towers
is 250 m. If the required clearance between the conductors and the water midway between the
towers is 45 m and if both the towers are on the same side of the point of maximum sag, find the
tension in the conductor. The weight of the conductor is 0.80 kg/m.
Solution:
ig. 15.15