Electrical Power Systems

under the action of T, H and wx.
or equilibrium,
T x = H and T y = wx
Taking moments about P,
I
HG KJ
H.y = wx x
2
\ y = wx
2H
2
Analysis of Sag and Tension 383
...(15.53)
or short span with small sag, T max – T min can be considered as small. Therefore T max »
T min = H. or T = T max = T min = H.
Therefore, eqn. (15.53) can be written as:
2
y = wx
2T
when x = L
, y = d
2
\ d = wL
8T
Since T = H,
Also
2
2
d = wL
8H
rom eqns. (15.13) and (15.56), we get
l = L
HG
2
8d
1 +
3L
2
I
KJ
...(15.54)
...(15.55)
...(15.56)
...(15.57)
Example 15.1: A transmission line conductor has been suspended freely from two towers and
has taken the form of a catenary that has c = 487.68 m. The span between the two towers is
152 m, and the weight of the conductor is 1160 kg/km. Calculate the following:
(a) Length of the conductor (b) Sag
(c) Maximum and minimum value of conductor tension using catenary method.
(d) Approximate value of tension by using parabolic method.
Solution:
(a) rom eqn. (15.12),
l = 2H
w
since c = H
w
I
HG KJ
wL
sinh
2H