What is Sag and tension ?In a overhead transmission line conductor are supported at the towers or at the pole for low voltage transmission lines. The conductor are pulled and string effected. When the conductor are supported in this manner ,it will sag or dip under its own weight and it takes the shape of centenary.
Definition of sag and span
“The difference in level between the poins of sipport and the tower point is known as” Sag”
“The distance between the adjacent supporting tower is called the” span”.
Factor effect Sag
The following factor effect the sag in overhead transmission line
- Weight of conductor .The sag is directly affected by this factor ;Heavier the conductor ,greater will be the sag.
- Span length.Sag is directly proportional to the square of the span length other conditions remaining unchanged.
- Working Tensile strength. Other condition remaining the same ,the sag is inversely proportional to the working tensile strength.
Under varying the weather condition of ambient temperature ,the conductor tension which is maximum near the tower ends, should not exceed the permissible limit (=breaking strength of conductor factor of safety of 2 to 2.5 ).
- Temerature.sag increase with increase in temperature.
In order to reduce the conductor material required and to avoid extra pole height for sufficient clearance above the ground level,the sag should kept to a minimum.
The following point are worth noting:
- The conductor’s mechanical loading is due to its own weight(Vertical) ,weight of ice (vertical load ) and wind load (Horizontal).
- The knowledge of maximum sag is essential in determining the ground clearance of conductor i.e least height of the conductor above the ground which is 7 m for voltage more than 166KV for a span of 300m.
- The span is decide by the economic factor with due consideration to the line and the conductor size.
- The number of tower for a given line length are determined by the span which also indirectly governs the sag height of the tower
Calculation of Sag and Tension
To calculated the sag and tension following two point should be noted
- When supports are at equal levels
- When supports are at unequal levels
When supports are at equal levels:
When support are at equal level between the two point A and B of conductor and the lowest point is O as shown figure.
Let, l=span length
w=weght per unit lenth of conductor
T=Tension in the conductor
Consider a Point P on the conductor. Its coordinates are x and y ,taking O as the origin.If the curvature is considered so small that curved length is equal to its horizontal projection.i-e OP=x then force acting on the portion OP are :
Weight w.x of the conductor acting at a distance x/2 from O.
The tension T acting at O.
Taking moment of these forces about point P ,we get
T.x.y=w.x.x.x/2 or y=wx2 /2T
The maximum sag (dip ) is represented by the value of y at either of the supports A and B.At supports A (or B), x=l/2 and y=S
Sag, S =w (l/2)2/2T =wl2/8T
When Supports are at unequal levels;
Consider an overhead line AOB supported over the supports A and B at unequal levels as shown in fig.O being the lowest point.
Let, l=span length ,
h=Difference in levels between two supports,
x1=Distance of support A (Lower level from O)
x2=Distance of support B (Lower level from O)
w=weight per unit length of conductor and ,
T=Tension in the conductor.
Now Sag S1=wx12/2T and S2=wx22/2T ,
Also x1 + x =l ………………(i)
S2 –S1 = wx22/2T –wx12/2T =w/2T (x22-x12)-w/2T(x2+x1)(x2-x1)
0r, S2 –S1=wl/2T(x2-x1) : (x1+x2=l)
or , (x2-x1)=2Th/wt ………….(ii)
Adding (i) and (ii) we get
2x2=l+2Th/wt or x2=l/2+ Th/wt
And , x1 =l – (1/2+Th/wt)=1/2-Th/wt After finding x1 and x2 the value of S1 and S2 can be calculated.
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