26/02/2021

Sag in Transmission Line

 

The calculation of sag for the transmission line is done by considering

( 1 ) Supports are at the same level

( 2 ) Supports are at the unequal level

We will calculate sag for supports are at the same level.

Assumptions

  • When the conductor is erected between two transmission line supports, the curve formed by it is in the shape of catenary. 
  • Actually the sag is less as compared to the span therefore it can be assumed as parabola. 
  • The tension on the conductor at any point is considered is in the horizontal direction. 
  • The tension near the supports is equal and it is in the horizontal direction.

Calculation of sag

Let

L = length of span ( meter )

WC = Weight of conductor per meter ( kg / meter )

T = Tension on the conductor ( kg )

  • The shape of the sag is parabola therefore the calculation of sag is done by considering only small portion of conductor. 
  • The sag is maximum at lowest point ( midway ) of the conductor. 
  • Let us consider that the lowest point on conductor is O. 
  • Take point P on the conductor whose co-ordinates are ( x , y ). 
  • As the curvature is small, the curved length OP is considered as horizontal distance x.

sag-in-the-transmission-line.png


Force on the conductor

  • The weight WC acts on the vertical downward direction. It is acted at a distance x / 2 from point O or point P. 
  • Horizontal tension T at any point on the conductor ( It is shown in the figure at point O. )

Taking moment about point P

Horizontal force × Vertical Distance = Vertical force × Horizontal Distance

T × y = WC x ( x / 2 )

y = WC x2 / 2T  ….. ( 1 )

The value of sag becomes maximum at centre point of conductor

Putting x = L / 2 and y = s ( sag ) in the equation ( 1 )

Sag s = WC ( L / 2 )2 / 2T

Sag s = WC L2 / 8T

Following points to be noted

  • The sag is directly square of proportional to square of the length of conductor. If the length of conductor is doubled, the sag is increases four times.

       S α ( L )2

  • The sag is directly proportional to weight of conductor per meter.

        S α WC

  • The sag is inversely proportional to inversely proportional to the tension on the conductor.

        S α ( 1 / T )

( 1 ) Effect of wind on the sag

  • The wind pressure acts in the horizontal direction of the conductor. 
  • The effective force on the conductor due to wind pressure and weight of conductor is in the transverse direction as shown in the figure. 
  • Let the intensity of wind pressure is p kg / meter2 and D is diameter of conductor.

Area projected per meter length of conductor = D × 1 meter2

Total wind pressure on the conductor = pD kg

Effect force on the conductor due to weight of conductor and wind pressure

We = √ ( WW2 + WC2 )

Sag s = WeL2 / 8T

Where We = Effective force on conductor

The effective force on the conductor has two components

( 1 ) We Cos θ – acts on the vertical downward direction and

( 2 ) We Sin θ – acts on the positive x axis direction.

Vertical sag = s Cos θ

                    = s ( WC / We )


combined-effect-of-wind-and-ice-on-sag.png


( 2 ) Effect of ice coating on sag

Let us consider

T = Thickness of ice coating over the conductor

D = Diameter of conductor

The weight of ice act on the vertically downward direction.

Wi = Weight ( Mass ) of ice per meter length

     = Volume of ice per meter length × density of ice

     = π / 4 { ( D + 2t )2 – D2 }× density

     = π / 4 { ( D2 + 4Dt + 4t2 – D2 }× density of ice

     = π / 4 { ( 4Dt + 4t2 ) }× density of ice

     = π  { ( Dt + t2 ) }× density of ice

     = πt { ( D + t ) }× density of ice .....( 2 )

The density of ice is taken in kg per meter cube if the conductor length is given in the meter.

Sag s = ( WC + Wi ) L2 / 8T

Where Wi = πt { ( D + t ) }× density of ice

( 3 ) Combined effect of wind and ice on sag

The effect force on the conductor due to combined effect of wind and ice is shown in the figure.

Weight of ice per meter length from equation ( 2 )

Wi = πt { ( D + t ) }× density of ice

Pressure on conductor due to wind Ww = p × ( D + 2t ) × 1

                                                               = p × ( D + 2t )... ( 3 )

Effective force on conductor

We = √ ( WC + Wi )2 + Ww2

Sag s = WeL2 / 8T

Where We = √ ( WC + Wi )2 + Ww2

Vertical Sag = s Cos θ

                     = s {( WC + Wi ) / We }

Sag

Sag Formula

Conductor weight only

Sag s = WC L2 / 8T

 

Effect of wind on sag

Sag s = WeL2 / 8T

Where We = √ ( WW2 + WC2 )

Vertical sag = s Cos θ

                    = s ( WC / We )

Effect of ice coating on sag

 

Sag s = ( WC + Wi ) L2 / 8T

Where Wi = πt { ( D + t ) }× density of ice

 

Combined effect of wind and ice on sag

 

Sag s = WeL2 / 8T

Where We = √ ( WC + Wi )2 + Ww2

Vertical Sag = s Cos θ

                     = s {( WC + Wi ) / We }

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