# Sag in Transmission Line and its Calculations

Sag in a transmission line is the vertical difference in level between the conductor support points and the lowest point of the conductor.

Overhead conductors are used in the transmission of power from the power station to the distribution end as well as from the distribution end to the consumers. Hence, this is one of the critical infrastructures in the power system.

The overhead conductors are supported on two ends by transmission towers. These conductors aren’t stretched between the supports but are hanging between them with a slight curvature around their midpoint. The vertical distance between the support points (which are parallel and equal) and this point of curvature is known as sag.

The picture above shows the sag of an overhead conductor.

The picture also shows something known as span. It is nothing but the distance between the parallel conductor-support structures or the transmission towers. Span is of two types

1. Equal or Level span: When both of the support structures are at the same base level and of the same height, it is said to have an equal span.
2. Unequal span: When both of the support structures aren’t on the same base level (like in the case of a hill slope) or at the same height or both, it is called an unequal span.

## Importance of Sag in a Transmission Line

The picture below denotes sag in an overhead conductor.

Here ‘S’ denotes sag which is the distance between the support points (X and Y) and the curve point ‘O’.

Sag is an important feature of an overhead line. This is because the transmission line can’t be stretched to its limits between the support points as it experiences varying physical conditions like wind, rain, snowfall, heat from the sun, etc. If the overhead line remains stretched and at tension, it may break down and can cause related problems.

That’s why, the overhead lines are provided a certain curve at a point between the two support points which is termed as sag. However, too much sag is also undesirable as it may decrease the distance between the ground and the conductor thus causing a potential threat to the nearby locations.

Thus, it is important to have a nominal value of sag that can be derived mathematically.

## Calculating Sag in a Transmission Line

Before we go for the calculation of sag, certain points need to be noted.

1. When the overhead conductor is supported on two ends at equal spans, a curve is formed at a point almost at the mid of the conductor. This is known as sag. Sag is much smaller than the span.
2. The tension in the conductor at each point acts tangentially.
3. The sag span curve is parabolic.
4. The tension at support is almost equal throughout the conductor.
5. The horizontal component of this tension is almost constant throughout the conductor

There are two different conditions for the calculation of sag

a. Sag for equal span
b. Sag for an unequal span

## Sag for Equal Span conductor in transmission line

Let the span is ‘L’ between the two support points denoted by X and Y respectively. ‘T’ denotes the tension in the conductor acting tangentially to the conductor.

w’ is the weight per unit length of the conductor.

We assume any point on the conductor let’s say ‘P’. The vertical distance of this point from O is denoted by y, where ‘O’ is the lowest point of the conductor.

‘S’ is the sag which is the vertical distance from support points to O.

‘x’ is the horizontal distance between points O and P.

Under balanced conditions, the two moments of force about point O remain equal. Therefore we can equate both as

Putting, and,

We get;

Therefore, The formula of sag for equal-span overhead conductor is;

## Sag for Unequal Spanconductor in transmission line

The above picture represents the unequal support level situation. In this case, there are two types of sags – one for each of the support levels shown by X and Y. Both the support levels are not on the same base or at the same height.

Here,

L is the span of the conductor
S1 and S2 = Two sags each from two support levels X and Y.
AB is the conductor terminal with O being the lowest point between them.
X1 = Horizontal Distance of point O from the lower-level point X.
X2 = Horizontal Distance of point O from the upper-level point Y.
h = Difference in height of both levels.
T is the tension of the conductor acting tangentially and w is the weight of the conductor per unit length.

Therefore, both the sags S1 and S2 can be given as

And,

From the diagram, we can say,

And,

Again from the diagram we can say that

Therefore,

Solving both equations (i) and (ii), we get

Therefore, by knowing the conductor span (L), Tension in the conductor (T), and the difference in the heights of both the levels (h) and weight of the conductor per unit length (w), we can calculate X1 and X2 and thus putting the value of X1 and X2 in equations (a) and (b), we can get the value of both the sags S1 and S2.

So,

Putting the value of X1 in equation(1) we get;

It should be noted that the weight per unit length of the conductor taken here is its own length assuming the normal condition of temperature, wind speed, and other physical conditions. However, the sag in an overhead conductor is also affected due to snow (during snowfall) and wind (during strong winds).

## Effect of Snow and Wind on Sag

1. The effective weight per unit length of the conductor changes due to wind or the accumulation of ice during snowfall.
2. The wind flow affects the conductor weight per unit length in the horizontal direction or along the direction of its flow.
3. The snowfall causes an accumulation of ice on the conductor that tends to increase the conductor weight per unit length in the vertically downward direction or in the direction of the snowfall.
4. When the snowfall and the flow of wind are simultaneous, there is a resultant force that is applied on the overhead conductor at a certain angle whose horizontal component is along the direction of the flow of the wind and vertical component is along the direction of the snowfall.

The picture above shows the combined effect of snowfall and wind on the conductor.

Here

Therefore,

And,

So, we need to calculate ww and wi.

### Calculation of wi

The picture above represents the overhead conductor coated with ice where d is the diameter of the conductor and ‘t’ is the thickness of the ice sheet.

Therefore,

Therefore,

### Calculation of ww

Similarly, the wind force per unit length of the conductor can be calculated as

We assume that the conductor is stretched along its effective diameter (that is with ice sheet included) due to the wind such that it forms a rectangle whose length is equal to the diameter of the conductor and has unit width.

Therefore,

Therefore,

Thus we can calculate the value of both wi and ww and get the value of the resultant force wand the resultant angle θ using the equations that we derived at the beginning.

Therefore, the effective sag in such a conductor is given as