Electrical Power Systems

Analysis of Sag and Tension 399
Distance of lowest point of conductor, O (see ig. 15.5) from the support at lower level A can
be obtained using (15.65),
x 1 = L
2
– hT
wL
e
Difference in level of supports h = 10 m,
\ x 1 = 300
HG
2
10 ´ 5764. 88
-
255 . ´ 300
\ x 1 = 74.65 m
or unequal support, sag at lower level can be calculated
Using eqn. (15.58), i.e.
\ d 1 =
d1 = wx e 1 2
I KJ
m
2T
we = 2.55 kg/m, x1 = 74.65 m,
T = 5764.88 kg
2 b g m
255 . ´ 7465 .
2 ´ 5764. 88
\ d 1 = 1.232 m.
I
HG 255 . KJ m
Vertical sag = d1 cosq = 1.232 × 23 .
= 1.111 m Ans.
Example 15.7: An overhead transmission line conductor having weight 1.16 kg/m, diameter
1.7 cm and an ultimate strength 32 × 10 6 kg/m 2 . When erected between supports 300 m apart
and having 12 m difference in height, determine the sag with respect to the taller of the two
supports. Conductor was loaded due to 1 kg of ice per meter and factor of safety is 2.0
Solution:
Span length L = 300 m, w = 1.16 kg/m,
wi = 1 kg/m
wT = w + wi = 1.16 + 1 = 2.16 kg/m.
Difference in level of two supports h = 12 m.
Diameter of the conductor dc = 1.7 cm
cross section area A c = p
4 d c 2
\ Ac = p
4 (1.7)2 ×10 –4 m 2 = 2.27 × 10 –4 m 2
actor of safety = 2
Allowable tension