What is Sag and Tension in Transmission Lines | Formula Calculation

What is sag and tension in transmission lines? Have you ever wondered what it is? If you are an engineering student, then you must have such questions in mind. Let’s learn in brief today.

Jabed and Rahim decided to visit Bali, Indonesia. So, they left for Bali. When they entered Bali, Jabed noticed that the 132 kV transmission lines were standing nicely on the hill. He noticed that the two towers were not the same height. In fact, the wire sag or, the hanging was much higher. What is sag and tension in transmission lines? This question arose in his mind. In this case, will the calculation of the sag be the general formula? He asked Rahim about the matter. Because Rahim is a top student in the class. Then Rahim explained the matter to Jabed. I will share Rahim’s advice with you today.

Tower Wire Sag Diagnosis Technique in Hilly Areas

What is Sag and Tension in Transmission Lines- Transmission Tower
Transmission Tower

The supports usually lie on an uneven surface while drawing overhead lines in mountainous areas. Suppose the two towers A and B lie on an uneven surface. The minimum point of the conductor is O. Let us suppose,

L = span or spread

h = height difference between two parts

x1 = from the bottom to the lowest point of the conductor distance

x2 = Distance up to 0 points from the higher surface

T = conductor tension

w = weight of an unit length conductor

S1 = w×1^2 / 2T

Moreover, S2 = W×2^2 / 2T

x1 + x2 = L ——–(1)

Since, S2 – S1 = W / 2T (x2 ^ 2 – x1 ^ 2) = W / 2T (x2 – x1)(x2- x1)

S2 – S1 = W / 2T (x2 ^ 2 – x1 ^ 2) = WL / 2T (x2 – x1)(x2- x1) because, x1 + x2 = L

But, S2 – S1 = h

Thus, h = WL / 2T (x2 – x1)

or, (x2 – x1) = 2Th/WL ———–(2)

From (1) and (2) we get,

x1 = L / 2 – Th / WL

x2 = L / 2 + Th / WL

The Calculation of the Sag when only the Effect of Wind Speed or Storm on the Conductor works

As we know, the wire has its own weight vertically downwards and the air pressure acts horizontally on the wire.

Suppose, wire diameter = Dm

Cable has its own weight of W Kg per meter

Air pressure P = Ww kg / m^2 on the projected area of ​​wire per meter

Total weight of wire per meter = w1 kg

Ww = P × Dm × L

Again, w1 = sqrt (W ^ 2 + Ww^2) and if this pressure is perpendicular to the angle and if the span length is L meters, then the maximum sag = w (L / 2)^2 / 2T = wL ^ 2 / 8T meter

Calculation of Sag when Ice and Storm act simultaneously on the Conductor

Ice Covered Area
Ice Covered Area

Not only the weight of the conductor works in the transmission line but also the effect of ice and wind works on it. This effect is more noticeable in winter prone countries like Canada, USA and UK. The weight of the conductor increases due to ice. Again, the sag also increases. On the other hand the vertical position of the conductor shifts due to the wind. This means that ice and wind have a combined effect on the conductor. So, the combined pressure of ice and wind needs to be brought in to calculate the sag.

Analysis of sag when ice and storm act simultaneously on the conductor:

Suppose, 

d = which is expressed as the density of ice

P = air pressure

w = weight of conductor per unit meter (kg)

wi = weight per unit length (kg)

wi = weight of ice per unit meter of conductor (kg)

Ww = air pressure on the conductor

wt = sqrt[(w + wi)^2 + Ww^2]

tan = Ww / w + wt or, cos = w + wi / wt

Sag = Wt × L ^ 2 / 2T

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Courtesy: Iqbal Mahmood