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Sag in transmission lines

Overhead power lines (transmission and distribution lines) are suspended on pole/tower supports. Suspended conductors are subjected to mechanical tension which must be under safe value. Excess mechanical tension may break the conductor. Therefore, a conductor between two supports must not be fully stretched and allowed to have a dip or sag.
The difference in level between the points of support and the lowest point on the conductor is called as sag.
Keeping the desired sag in overhead power lines is an important consideration. If the amount of sag is very low, the conductor is exposed to a higher mechanical tension which may break the conductor. Whereas, if the amount of sag is very high, the conductor may swing at higher amplitudes due to the wind and may contact with alongside conductors. Lower sag means tight conductor and higher tension. Higher sag means loose conductor and lower tension. Therefore, a suitable value of sag is calculated so that the conductor remains in safe tension limit keeping the sag minimum.
  • The tension at any point on the conductor acts tangentially. Therefore, the tension at the lowermost point on the conductor is horizontal.
  • The horizontal component of tension at any point on the conductor is constant.
  • The tension at the support points is nearly equal to the horizontal component of tension at any point on the conductor.

Calculation of sag

The tension on a suspended conductor is governed by the conductor weight, effects of wind, ice loading and temperature variations. Generally, conductor tension is kept less than 50% of its ultimate tensile strength. The value of sag is calculated for two different scenarios - (i) supports are at equal levels and (ii) supports are at unequal levels.

When supports are at equal levels

Consider a conductor suspended at supports of equal heights as shown in the figure below. A and B are the support points and O is the lowest point on the conductor.

sag in power lines when supports are at equal levels


Let,
  • l = length of the conductor span
  • w = weight per unit length of the conductor
  • T = tension in the conductor
Consider a point P on the conductor. Considering the lowest point O as the origin, let the coordinates of point P be x and y. Assume the curvature is so small that the curved length is equal to its horizontal projection (i.e. OP = x). The forces acting on the conductor portion OP are -
  1. the weight w.x acting at a distance x/2 from the point O
  2. the tension T acting at the point O
Equating the moments of the two forces about point O, we get,
T.y = w.x * x/2
or, y = w.x2 / 2T
The maximum sag (dip) is represented by the value of y at either of the support points. At support point A,
x = l/2 and y = S (sag)
therefore, sag S = w(l/2)2 / 2T
therefore, sag S = w.l2/8T

When supports are at unequal levels

We generally encounter supports of unequal heights in hilly areas. The following figure shows a conductor suspended on supports at unequal heights.

sag in power lines when supports are at unequal levels


Let,
  • l = length of the conductor span
  • w = weight per unit length of the conductor
  • T = tension in the conductor
  • h = difference in levels between the two supports
  • x1 = distance of support at lower level (A) from the origin O
  • x2 = distance of support at higher level (B) from the origin O
From the above calculation of sag in the previous point, y = w.x2 / 2T
Now, at support A, x = x1 and y = S1
Therefore, S1 = w.x12 / 2T
sag in power lines equation


Values of S1 and S2 can be easily calculated from x1 and x2.

[Also read: Insulators used in overhead power lines]

Effect of wind and ice loading

The above calculations of sag are only true for still air and normal temperature conditions. However, in actual practice conductors are suspended to wind pressures and often loaded with ice coating in cold areas. The force due to wind is assumed to act horizontally to the conductor. And the force applied by ice loading is vertically downwards. Therefore, the total force on the conductor is the vector sum of vertical and horizontal forces.

In this case, weight of the conductor per unit length will be,
wt = sqrt.[(w + wi)2 + (ww)2]
where,
  • w = weight of the conductor per unit length
  • wi = weight of the loaded ice per unit length
  • ww = wind force per unit length
When a conductor has wind as well as ice loading,
  • The conductor sets itself in a plane at an angle θ to the vertical plane.
    Where, tan θ = ww / (w + wi)
  • In this case, sag is given by S = wt.l2 / 2T. But, here, S represents the slant sag, i.e. sag is in the plane where the conductor has set itself. The vertical sag is equal to S.cosθ