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 )
W_{C}
= 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.
Force
on the conductor
- The weight W_{C} 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 = W_{C} x ( x / 2 )
y
= W_{C} x^{2} / 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 = W_{C} ( L / 2 )^{2} / 2T
Sag
s = W_{C} L^{2} / 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 W_{C}
- 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 / meter^{2} and D is diameter of conductor.
Area projected per meter length of conductor = D × 1 meter^{2}
Total
wind pressure on the conductor = pD kg
Effect
force on the conductor due to weight of conductor and wind pressure
W_{e}
= √ ( W_{W}^{2} + W_{C}^{2} )
Sag
s = W_{e}L^{2} / 8T
Where
W_{e} = Effective force on conductor
The
effective force on the conductor has two components
(
1 ) W_{e} Cos θ – acts on the vertical downward direction and
(
2 ) W_{e} Sin θ – acts on the positive x axis direction.
Vertical
sag = s Cos θ
= s ( W_{C} / W_{e}
)
( 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.
W_{i}
= Weight ( Mass ) of ice per meter length
= Volume of ice per meter length × density
of ice
= π
/ 4 { ( D + 2t )^{2} – D^{2} }× density
= π / 4 { ( D^{2} + 4Dt + 4t^{2}
– D^{2} }× density of ice
= π
/ 4 { ( 4Dt + 4t^{2} ) }× density of ice
=
π { ( Dt + t^{2} ) }× 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 = ( W_{C} + W_{i} ) L^{2} / 8T
Where
W_{i }= 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 )
W_{i
}= πt { ( D + t ) }× density of ice
Pressure
on conductor due to wind W_{w} = p × ( D + 2t ) × 1
= p × ( D + 2t )... ( 3 )
Effective
force on conductor
W_{e}
= √ ( W_{C} + W_{i} )^{2} + W_{w}^{2}
Sag
s = W_{e}L^{2} / 8T
Where W_{e} = √ ( W_{C} + W_{i} )^{2} + W_{w}^{2}
Vertical
Sag = s Cos θ
= s {( W_{C} + W_{i}
) / W_{e }}
Sag |
Sag Formula |
Conductor
weight only |
Sag
s = W_{C} L^{2} / 8T |
Effect
of wind on sag |
Sag
s = W_{e}L^{2} / 8T Where
W_{e} = √ ( W_{W}^{2} + W_{C}^{2} ) Vertical
sag = s Cos θ = s ( W_{C} / W_{e}
) |
Effect
of ice coating on sag |
Sag
s = ( W_{C} + W_{i} ) L^{2} / 8T Where
W_{i }= πt { ( D + t ) }× density of ice |
Combined
effect of wind and ice on sag |
Sag
s = W_{e}L^{2} / 8T Where W_{e} = √ ( W_{C} + W_{i} )^{2} + W_{w}^{2} Vertical
Sag = s Cos θ = s {( W_{C} + W_{i}
) / W_{e }} |
What
do you mean by voltage harmonics and current harmonics?
Effect
of harmonics on power factor
No comments:
Post a comment