# What is Sag & Tension in transmission lines & Formula Calculation

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## Sag & Tension  in transmission line

#### Today we are going to learn a topic called sag and tension in the transmission lines.During the erection of overhead transmission line, the conductors are connected between two tower structures. The conductors connected between the tower structures must not be connected too tight or too loose. If they are connected very tightly, the tension on the conductor will be very high and at some point, it may break. If they are connected very loose, the charging current increases the length of wire to be used and the height of the tower structure increases. Hence, the conductors must be connected in such a way that the tension is minimum and at the same time there should be good clearance between the ground and the conductor. ### Factors Affecting Sag

#### (ii) Location of conductor

Sag also depends on the location of conductors.If the conductors are present in the area where ice formation takes place, then due to the accumulation of ice on the conductor its overall weight increases.This increases the weight of the conductors which in turn increases the value of sag.

(iii) Length of span

Sag is proportional to the square of the length of the span.Hence, longer the span greater will be the sag provided the tension and weight of the conductor are constant.

(Vi) Tension

### Calculation of Sag in Overhead transmission lines:

(i) When supports are at equal levels

#### Let us consider a line conductor between two equal height line supports.Line supports are A and B with O as the lowest point as shown in the figure. Point O will be the lowest point as two levels are equal lowest point will be at the mid-span. #### l = Length of span w = Weight per unit length of conductor T = Tension in the conductor.

Now consider any point on the conductor. Let’s say point ‘P’.By considering lowest point O as the origin, let the coordinates of point P be x and y. Assuming that the curvature is so small that curved length is equal to its horizontal projection (i.e., OP = x), the two forces acting on the portion OP of the conductor are :
(i) The weight wx of conductor acting at a distance x/2 from O.
(ii) The tension T acting at O.

#### y=Tx²/2T

The maximum dip (sag) is represented by the value of y at either of the supports A and B. At support A, x = l/2 and y = S

### (ii) When supports are at unequal levels

The difference in level between points of supports and the lowest point on the conductor is called “sag”.When transmission lines run on steep inclines as in the case of hilly areas, we generally come across conductors suspended between supports at unequal levels.The shape of the conductor between the supports may be assumed to be a part of the parabola. In this case, the lowest point of the conductor will not lie in the middle of the span.

#### Consider a conductor suspended between two supports A and B which are at different levels as shown in the following figure. #### If w is the weight per unit length of the conductor, then, #### The total weight of the conductor i.e., both the conductor weight and the weight of the ice acts vertically downwards whereas the wind force acts horizontally on the conductor. Therefore, the vector sum of horizontal and vertical forces acting on the conductor gives the total force shown in the figure.  #### and Sag S2 = wx2²/2T

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