Electrical Power Systems

Total weight of the conductor
= wl = 0.533 × 300.0355 kg.
= 159.918 kg.
Analysis of Sag and Tension 401
Example 15.9: Calculate the sag and vertical sag of a transmission line having conductor is
diameter 0.93 cm. Weight of conductor is 0.6 kg/m and breaking strength 2000 kg. Assume
factor of safety 2 and span length 200 m and supports at the same level. The line is subjected to
wind pressure of 40 kg/m 2 of the projected area. The radial thickness of the ice is 1.25 cm and
weight of the ice is 912 kg/m 3 .
Solution:
L = 200 m, w = 0.60 kg/m
Weight of the ice per meter length can be obtained using eqn. (15.73),
wi = wcpti(dc + ti) ×10 –4 kg/m
ti = 1.25 cm, dc = 0.93 cm, wc = 912 kg/m 3
\ wi = 912 × p × 1.25 (0.93 + 1.25) ×10 –4 kg/m
\ wi = 0.7807 kg/m.
\ wT = w + wi = 0.60 + 0.7807
\ wT = 1.3807 kg/m
Horizontal force due to wind pressure can be obtained using eqn. (15.82), i.e.,
bdc + 2tig
= p kg/m
100
p = 40 kg/m 2
b093 . + 2´ 125 . g
\ = × 40 kg/m
100
\ = 1.372 kg/m
Effective load acting on the conductor can be obtained using eqn. (15.83)
2
b g
we = + w+ wi
\ we = 2 2
w
\ w e = 1.946 kg/m.
actor of safety = 2.0
T = 2000
2
e 2
2
b g b g
2 2
+ = T 1372 . + 13807 .
= 1000 kg.
b g
\ Sag d = wL
8T
= 1946 200 . ´
1 ´ 1000
\ d = 9.73 m
HG
Vertical sag = d cosq = d × w
w
= 6.9035 m.
T
e
I
2
I
HG KJ
= 9.73 × KJ 13807 .
1946 .