Calculate the sag for a span of 200 m if the ultimate tensile strength of the conductor is 6000 Kg. Allow a factor of safety of 2. The weight of the conductor is 900 kg/km.

Question

Question

This question was previously asked in

UPPSC AE Electrical 2019 Official Paper I (Held on 13 Dec 2020)

Answer 1

Concept:

Sag (s):

The distance between the highest point of electric poles or towers and the lowest point of a conductor connected between two poles or towers.

Span length:

It is the shortest distance between two towers or poles.

Sag

\(s = \frac{{W{l^2}}}{{8T}}\)

Where,

S is the sag of the conductor

W is the weight of the conductor

l is the span length of the conductor

T is the working tension on the conductor

SSC JE Electrical 23 20Q Power Systems 2 Hindi - Final images Q8

Calculation:

Span length = 200 m

Breaking strength (Ultimate strength) = 6000 kg

Weight of conductor W = 900 kg/km

= 900 kg/1000m (1 km = 1000 m)

W = 0.9 kg/m

Working tension, T = ultimate strength/safety factor

\( = \frac{{6000}}{2}\)

T = 3000 kg

Sag \(s = \frac{{W{l^2}}}{{8T}}\)

= \(\frac{{\left( {0.9 \times {{200}^2}} \right)}}{{8 \times 3000}}\)

= 1.5 m