Electrical Power Systems

where
Dl = l0. DT
MA
DT = T1 – T0 T 0 = initial tension of conductor (kg)
DT = change in conductor tension (kg)
M = modulus of elasticiy of conductor (kg-m)
A = actual metal cross section of conductor (m 2 ).
15.3 CALCULATIONS O LINE SAG AND TENSION
Analysis of Sag and Tension 375
...(15.4)
igure 15.1 shows a conductor suspended freely from two supports, which are at the same lavel
and spaced L meter, takes the form of a catenary curve providing the conductor is perfectly
flexible and conductor weight is uniformly distributed along its length. When sag (d) is very
small in comparison to span L (i.e., the conductor is tightly stretched), the resultant curve can
be considered as parabola. If d < 0.06 L, the error, the error in sag computed by the parabolic
equations is less than 0.5 per cent. If 0.06 L < d < 0.1 L, the error in sag computed by the
parabolic equations is about 2 per cent.
15.3.1 Catenary Method: Supports at Same Level
igure 15.1 shows a span of conductor with two supports at the same level and separated by a
horizontal distance L. Let O be the lowest point on the catenary curve l be the length of the
conductor between two supports. Let W is the weight of the conductor per unit length (kg/m),
T is the tension of the conductor (kg) at any point P in the direction of the curve, and H is the
tension (kg) at origin O.
urther, s be the length of the curve between points O and P, thus the weight of the portion
s is ws.
ig. 15.1: Conductor suspended between supports at same level.
Tension T can be resolved into two components, T x , the horizontal component and T y , the
vertical component. Then, for equilibrium,