Electrical Power Systems

400 Electrical Power Systems
T =
6 -4
32 ´ 10 ´ 2. 27 ´ 10
2
\ T = 3632 kg.
Distance of the lowest point of the conductor from the taller support can be obtained using
eqn. (15.66), i.e.,
x 2 = L
2
\ x 2 = 300
HG
2
+ hT
w L
T
12 ´ 3632
+
2. 16 ´ 300
Vertical sag can be obtained using eqn. (15.59) i.e.,
d 2 =
w x
2T
T 2 2
= 216 217 26
I KJ
. ´ .
2 ´ 3632
kg
= 217.26 m
2 b g = 14.03 m. Ans.
Example 15.8: An overhead transmission line has a span of 300 m. Ultimate strength is
6000 Kg and factor of safety is 2.0. If the sag is 2m, determine (a) weight of the conductor
(b) length of the line.
Solution:
(a) Span length L = 300 m
Allowable tension T =
Ultimate strength
actor of safety
\ T = 6000
= 3000 kg.
2
Using eqn. (15.55), sag expression is written as:
2
d = wL
8T
d = 2 m
2 b g = 2
w ´ 300
\
8 ´ 3000
\ w = 0.533 kg/m.
\ Weight of the conductor = 0.533 kg/m.
(b) Length of the line can be obtained using eqn. (15.57), i.e.,
l = L
HG
8d
1 +
3L
2
2
I
KJ
\ l = 300.0355 m
HG
2 bg
b g
= 300 1
8´ 2
+
3 ´ 300
2
I
KJ