Mechanical Design of Transmission LinesFarhan MahmoodEED, UET Lahore.

Main Considerations in the Mechanical DesignThe main considerations in the mechanical design of an overhead transmission line are:

Adequate clearance between conductor and groundHigh mechanical strength of the conductorsTension or working stress of the conductor < ultimate tensile strength

Ultimate tensile strength = F.O.S working stress

Basic Design ConsiderationsWhile erecting an overhead line, it is important that conductors are under safe tension. If the conductors are too much stretched between supports to save the conductor material, the stress in the conductor may reach unsafe value and in certain cases, the conductor may break due to excessive tension.In order to permit safe tension in the conductors, they are not fully stretched but are allowed to have a dip.

Few Important Terms in the Mechanical DesignTENSION, a force tending to stretch or elongate a conductor.

ULTIMATE TENSILE STRENGTH, maximum stress, which a conductor can withstand without failure.

Few Important Terms in the Mechanical DesignSAG, the vertical distance (d) between the mid-point of a conductor to the line joining the two supports level.

Few Important Terms in the Mechanical DesignCATENARYS CURVE, When the conductor is suspended between two supports at the same level, it takes the shape of catenary's curve. However, if the sag is very small as compared with the span, then sag-span curve is parabola.

Few Important Terms in the Mechanical DesignSPAN, the horizontal distance (L) between the two adjacent supports.

Points to RememberThe following points are to be noted,The tension at any point on the conductor is tangent to that pointThe horizontal component of the tension is constant throughout the length of wire. The tension will be maximum at the supports and minimum at the lowest point of the curve.

Factors affecting SagSAG plays a very important role in the mechanical design of an overhead line. It is not a good practice to provide either too high or too low sag.

It is always desired that tension and sag should be as low as possible, which is not possible simultaneously.Low Sag ------> tight wire & high tensionHigh Sag ------> loose wire and low tensionTherefore, a COMPROMISE is made between the two.

Sag (Too Low)Sag (Too High)1. Tension in the conductor is too high1. Tension in the conductor is too low2. Less conductor length is required2. More conductor length is required3. Lower supports are required3. Higher supports are required

Factors affecting SagThe factors affecting the sag of a conductor strung between supports are

Weight of conductor Distance between the supports (span length) Working tensile strengthTemperature

Sag CalculationsA conductor AOB of length l is suspended at two towers A and B and are spaced L unit apart. Let O is the lowest point of the wire. Consider a length OP of the curve length s.w = weight/unit length, H = tension at point O T = tension at point P,

AxH

Sag CalculationsThree forces are acting on it

Horizontal tension H at the lowest point Weight ws of OP acting through its center of gravityTension T at point P along tangent to the curve at P. For equilibrium, horizontal forces in one direction must be balanced by horizontal forces in the other direction. Same is true for vertical forces.

Sag CalculationsT cos T sin wsLet be the angle which the tangent at P makes with the horizontal.

Components of Tension(1).(2)Dividing equation 1 by 2 we get.(3)

Length..(4)

LengthFrom equation 3 we haveSo eq. 4 becomes

LengthIntegrating both sides , we haveWhere A is the integration constant

LengthUsing intial values as x = 0 s = 0 we get A = 0, So we have,

Length.(5)

Calculation of SagAs we know that So,Since

Calculation of SagorIntegrating both sides we have

Sag CalculationsWhere B is the Integration constantUsing initial values x = 0 y = 0 We get B = -H/wSo,

Sag CalculationsThis equation is called the equation of catenaryOn Expanding we get

Calculation of TensionNeglecting high powered terms we getThe tension at point P is,

Calculation of Tension When x=L/2 , y is equal to the sag or deflection d

Calculation of LengthUsing the equations,

After putting the value of H from expression of d into l,

Solve Example The weight of a overhead conductor of a line is 4.0 N/m. The ultimate strength is 8000 N. If safety factor is 4 and span length is 160 m, find (a) sag and (b) total length of the line between spans.

Solution

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Supports at different levels (unsymmetrical span)

ContinuedLet P1 and P2 are two points at heights h1 and h2 from the ground respectively,

where h is difference between the elevations of two supports. If the span length is L, x1 + x2 = L. Therefore,

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Solve ExampleAn overhead transmission line conductor has the following dataWeight = 0.35 kg/m;Maximum allowable strength = 800 kg;Safety factor = 2;Span length = 160 m. Supports are at different levels where one support is at 70 m from the ground. Find the minimum clearance from the ground and the minimum point of the catenary from the supports when the second support is at (a) 40 m and (b) 65 m.

Solution

ContinuedThis shows that the minimum point lies outside of the span is 134.29 m from the lower span. Therefore, the minimum ground clearance is 40 m that is the height of lower support.

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EFFECT OF ICE- AND WIND-LOADING The sag and tension of lines are different in normal weather conditions. Since lines are designed for all the conditions, it is important to calculate the sag and tension during the ice- and wind-loading conditions.Conductor weightIce LoadingWind Loading

Conductor WeightThe weight of the conductor acts vertically downwards and depends upon the type of the conductor used. The weight of the conductor per unit length is available from the table giving the mechanical characteristics of the conductor.

Ice LoadingIn snowy areas, ice is deposited on the conductors and its accumulation on the conductor affects the design of line, By increasing the weight per unit length.By increasing the projected surface area subjected to wind pressureIn calculations, it will be assumed that ice is uniformly on the surface of the conductor.

Ice LoadingCross-section area of conductor = d2 / 4

Cross-section area of ice coated conductor = (d + 2t)2 / 4

Cross-section area of ice

Ice Loading If density of ice = i Weight of ice Wi = Ai i

Total weight of conductor per unit length WT = W + Wi

Combined effect of wind & iceProjected area of the conductor = (D+ 2t) x 1ww = (D + 2t) kg/mWhere is the wind pressure per unit area acting in a direction normal to the direction of span

Effective Loading

Assignment An ACSR conductor has the following data: normal copper area = 120 mm2 size = (30 + 7)16.30 mm; weight = 0.4 kg/rn, tensile strength = 1250 kg, safety factor = 5. If span length is 200 m, find Sag in still air Sag, if the conductor is covered with 0.5-cm thick ice (ice density of 915 kg/rn3) Sag (total and vertical), if the conductor is covered with ice of 0.5-cm thickness and a wind pressure of 10 kg/rn2 is acting on the projected area.

Solution

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