module 3 overhead lines version e - transtutors · 2017-12-14 · mechanical design of overhead...

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School of Engineering and Applied Science

Continuing Professional Development

Module 3 – Overhead Lines

Course Co-ordinator Dr. M. Sadeghzadeh

Electricity Distribution Module 3

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INDEX

1. Advantages and Disadvantages of Overhead Line versus Cable

2. Types of Support

3. Earthed and Unearthed Construction

4. Mechanical Design of Overhead Lines

5. Conductor Spacing – Wind Induced Oscillation

6. Conductor Vibration

7. Conductor Characteristics

8. Conductor Current Rating

9. Conductor Sag and Tension

9.1 Overview

9.2 Design Tension Limit

9.3 Conductor Sag

9.4 Equivalent or Ruling Span

9.5 Variation of Sag with Temperature

9.6 Sag/Tension/Temperature Table

9.7 Sag Templates

9.8 Conductor Clearance

9.9 Conductor Uplift

10. Line Survey

11. Route Selection

12. Profiling

13. Insulators

14. Lightning and Overhead Lines

14.1 Use of Arc (Horn) Gaps

14.2 Use of Non-Linear Resistors (Surge Arresters or Surge Diverters)

14.3 Effects of Lightning on Pole Transformers

Appendix A – BS1990 Pole Sizes


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1. Advantages and Disadvantages of Overhead Line versus Cable

Cost - In the UK, Engineering Recommendation P21/5 provides guidance on the comparative costof overhead lines and cables at various system voltages and is reproduced below.

Table 1 U/G Cable versus O/H line

System Voltage Ratio

132kV Double circuit on steel lattice tower 8:1

132kV Single circuit on steel lattice tower 7:1

33kV Double circuit on wood poles 7.5:1

33kV Single circuit on wood poles 7.5:1

11kV Single circuit on wood poles 2.3:1

Low voltage on wood poles 1.7:1

These are generalised values which vary with individual projects, due to the following factors:

1. The underground cable route may be longer than the overhead line route because cables arelaid in publicly owned roads and footpaths which usually do not follow the most direct route,whereas overhead lines follow a more or less direct route, crossing private land to do so.

2. Excavation and surface reinstatement costs for the cable option vary considerably and wherethe route is in roadway, excavation and reinstatement costs will be high. Excavation in rockyground is also costly.

3. For short lengths of cable, for example in the middle of overhead lines, the cost of theterminations predominates, making the per-metre cost of cable high.

4. Where the underground cable option requires the use of ground mounted substations andground mounted switchgear, the cost of these will be greater than their pole mountedequivalents.

Table 1 compares only capital costs. Comparing lifetime costs, the cost of maintenance of theoverhead line will considerably exceed that of the equivalent underground cable. The maintenancecost of an underground cable should be very small or nil (except oil or gas pressure cables).

Returning to the capital costs, the extra cost per additional MVA transmitted is lower with overheadline circuits, particularly as the system voltage increases, because the incremental cost of strongersupports and heavier conductor is small, leading to lower values of cost per MVA transmitted. Incontrast, in underground cable circuits the dielectric loss generates heat which increases as thevoltage squared (Module 2), hence is more important at higher system voltages. In addition, athigher system voltages the insulation must be thicker increasing the thermal blanket effect, so theconductor must be larger to compensate, increasing capital cost.

Overloading is more critical for underground cables because as described in Module 2, thermalexpansion of the conductor can cause permanent deformation and damage, whereas withoverhead lines thermal expansion under overload causes only excessive sag, which should be fullyreversed when the load is reduced.

Repair times for overhead lines are, on average, much shorter than for underground cablesbecause most faults are visible from the ground and no preliminary fault location procedures needbe carried out. Also, repairs are normally straightforward, requiring only replacement of a short


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length of conductor or a few insulators – however this advantage is completely eliminated ifdamage occurs over a large geographical area due to severe weather, when as described inModule 1, multiple failures can overwhelm available repair resources.

Installation – Overhead lines are quick to install and cause less disturbance than the installation ofunderground cables, where disruption to traffic flows and pedestrian access can be severe andextend over long periods of time, especially in congested urban areas.

Security – Attempted theft of overhead line conductor can occur, theft of underground cable isalmost unknown. Theft of bonding conductor from overhead line supports also occurs and this cancause hazardous situations.

Safety - Overhead lines are subject to contact by vehicles and by tools and equipment held bypersons (carbon fibre fishing rods, irrigation pipes, scaffolding poles, etc.). In these circumstancesfault current passes through the person to earth, hence they are frequently fatal or result in severeinjuries. To some extent, the contact hazard may be reduced by use of insulated conductor as laterdiscussed, although this will be at extra cost. There is also the hazard of a broken conductor lyingon the ground, which may remain live if the earth fault current is insufficient to trip the controllingcircuit breaker. Underground cables are subject to damage mainly by mechanical excavators, lessoften by hand held digging tools. In this type of incident injury to persons is caused by heat andblast, injuries which are seldom fatal but can result in severe burns.

Considering the safety of operative personnel, work on underground cables is completely safe ifappropriate procedures are observed and precautions taken – the main hazards are opening thewrong cable in an excavation where more than one cable is exposed and work on live low voltagecables. Work on overhead lines always carries the risk of a fall from height and there are otherrisks notably that of induced voltage when working on one (dead) circuit of a double circuit line,when the other is energised. Helicopter operations, live line and rubber glove working involve theirown special risks. Once again if appropriate precautions are taken, overhead line work should becompletely safe but in practice accidents occur more frequently with work on overhead linescompared to work on cables.

Environmental – The environmental implications of overhead lines are as follows:

1. They are visually intrusive and for this reason consents for construction may prove difficult toobtain. Transmission voltage overhead lines supported on tall steel lattice structures withconductors in vertical formation are more intrusive than primary distribution or secondarytransmission voltage lines on shorter supports with conductors in flat formation.

2. They adversely affect agricultural operations, although compensation is paid. To a majorextent, interference with agricultural operations may be reduced by optimum siting of supports,especially with primary distribution or secondary transmission voltage lines supported on woodpoles.

3. Where an overhead line passes through a wooded area, it is necessary to fell a clear path oneither side to at least the overturning distance of the support, causing environmental damagethat may be seen from a long distance.

4. They generate audible noise in wet weather and can generate electromagnetic interference.

5. They are a hazard to wildlife due to impact of large birds.

6. They inhibit the development of land in terms of requiring safety clearances to structures andbuildings, leading to a loss of land value. Again, compensation may be paid.


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7. The electromagnetic field is held by some to be hazardous to health.

2. Types of Support

Figure 1 shows a few support designs suitable for use at Primary distribution and Secondarytransmission voltage.

In northern Europe creosoted softwood poles still offer an economical means of support, eventhough their life is limited. Starting from the left on the upper row, the ‘Wishbone’ crossbar designseen on the extreme left was in use over the first half of the twentieth century and is stilloccasionally seen, but has been superseded by the horizontal crossarm type on its immediateright. This was introduced in the late 1940’s along with unearthed construction and is the type usedin the most recent UK standard, EATS 43-40. When earthed construction was used, crossarmswere sloped to deter birds from perching, but this is not necessary with unearthed construction asthe pole insulates the crossarm from earth.

The centre ‘Portal’ support is popular for larger cross section conductors and longer spans. It alsoremains in use, although its suspension insulators require longer poles of greater cost and thesuspension (cap and pin) insulators themselves are more expensive than fixed pin or post types.The centre Steel bracing may be omitted in some versions. It is possible to substitute glass fibrereinforced plastic poles for wood in all cases, the material offering an indefinite life and goodresistance to impulse voltages but at much greater cost.

The braced Steel channel tower (PU tower in the UK) and reinforced concrete support seen on theright of the upper row are mainly used in continental Europe and in warmer climates where woodeating insects prevent the use of wood supports. BS 607 (now withdrawn) was the relevant UKstandard for concrete poles. Both these two designs require an over running earth wire – they areearthed construction whereas the left hand trio require no earth wire because they are unearthed.

Fig.1 Overhead line support designs


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On the lower row, the left hand ‘H’ support offers a double circuit capability at a lower costcompared with that of two single circuits and is particularly suitable for the double Secondarytransmission circuits described in Module 1. Some early versions used suspension insulatorsrather than the post insulators shown. Finally, the right hand illustration on the lower row shows‘Trident’ construction, a design dating from the 1980’s allowing a single 132kV circuit on woodpoles.

Tubular steel poles were once widely used for low voltage purposes because their great strength inbending allowed substantial side loads to be imposed without having to fit stay wires. This was amajor advantage in the urban distribution situation where high density of development and closeproximity of traffic combine to make stay positioning extremely difficult. Unfortunately in a similarmanner to wood poles rotting at the ground line, steel poles tend to corrode at the ground line,often on the inside, leading to catastrophic failure. In addition the failure of an insulator may allowthe pole to become live, although with insufficient earth fault current to blow the supply fuse. This isa major danger to the public.

Though not shown in the illustration, all of these supports require protection against unauthorisedclimbing, taking the form of barbed wire either a number of rows simply wrapped around thesupport or fixed to a rectangular Steel frame. In some countries including the UK legal warningnotices informing the public against climbing are also required.

Wood poles - Most softwood poles used in northern Europe are Scots Pine (Pinus Sylvestris)grown in Baltic countries and felled in the winter months, conforming and graded according to BS1990. Typically they are 50 - 80 years old when felled. Other species have been used, notablyEuropean larch (Larix Decidua), Douglas Fir (Pseudotsuga Taxifolia) and Western Red Cedar(Thuja Plicata). In tropical countries Eucalyptus (Eucalyptus Globulus) is used. Appendix A givesdimensions for BS1990 poles.

Table 2 Properties of pole timbers

Species Ultimate FibreStress in

bending N/m²

Ultimate ShearStress N/m²

Modulus ofElasticity N/m²

Pinus Sylvestris 53x106 6.8 x106 10,250 x106

Larix Decidua 68 x106 7.8 x106 10,900 x106

Pseudotsuga Taxifolia 61 x106 7.2 x106 9,500 x106

Thuja Plicata 45 x106 6.1 x106 5,500 x106

Pole preservation - Decay due to biological rot is the major cause of deterioration in wood poles.Creosote, a by product of coal gasification, is used as preservative, through a variety of vacuumand pressure impregnation processes. Providing that the pole has low moisture content and hasbeen allowed to season under cover in the pole yard for some months, impregnation shouldcompletely saturate the outer softwood but Creosote will not penetrate the inner hardwood. Ideally,poles should be scarfed and drilled for equipment bolts, etc. before impregnation.

Other treatments against decay are used notably CCA (Copper Chrome Arsenic) which being awater based impregnant can be used for low voltage supports but not high voltage because thesalts impair the insulation value. Surprisingly, CCA has proven no more prone to leaching thanCreosote, even though the process is water based. CCA poles are clean to work with but recentlycreosoted poles often have black, highly adhesive surface deposits that contaminate clothing andtools; for this reason it is good practice to allow them to weather for a few months in an outdoorstore before transport to site.


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Unfortunately over time the creosote leaches away and the outer softwood cracks, allowing rotspores to reach the inner parts of the pole. This can lead to the dangerous situation where the poleappears visually satisfactory on the outside, but is completely rotten on the inside. Biological rotrequires both air and water, which are available close to the ground line where the decay processtends to be fastest and also where stress is greatest, at least the stress caused by bending loads.In some countries poles are not planted directly in the ground but bolted to metal or concrete stubs,so that the possibility of ground level rot is eliminated. At high level there tends not to be enoughwater to promote rot, except at the very top where rain collects or drains from the crossarm. Lowerthan 300mm below ground level air is not present and the rotting process is extremely slow, for thisreason pole and stay baulks generally last much longer than the poles. Even so, baulks should bepressure creosoted as a precaution.

Poles can be ‘sounded’ for rot; that is, struck with a hammer and the resulting sound used as anaudible assessment of condition. The technique requires a good deal of experience to give reliableresults; it is routinely carried out as a safety precaution by linesman prior to climbing and has theadvantage that it is quick and easy. A more time consuming method of detecting and assessing rotis the Mattson borer; this tool is screwed into the pole to remove small diameter (3-5mm) samplesfor visual examination and at least one sample should be taken from below ground level.Alternatively a Pole Ultrasonic Rot Locator (PURL) may be used, giving a visual display of thesampled section. These tools, particularly the latter, allow a company’s pole population to beinspected periodically and candidates for replacement identified in a planned and financiallymanaged manner.

Rot is impossible to repair and difficult to remedy; tubular splints either of metal or fibreglass canbe used close to the base of the pole and clamped on to reinforce partially rotted areas, but thesetend to trap moisture, accelerating the rotting process. Established rot can be stopped or at leastslowed down by inserting biological poisons. The Cobra process, invented in Germany in the1920’s, involved the pressure injection of poisonous salts (Sodium Flouride, Dinotrophenol andArsenious acid) into the timber above and below the ground line. This preservative method waswidely used in the middle of the twentieth century but later discontinued because of concern thatthe chemicals were toxic to humans as well as wood. Today’s preservative of choice is Boron inrod form, in which the active constituents have been fused together to give the appearance ofglass. This biological poison is slowly dissolved by the moisture content of the pole and effectivelyhalts the decay, but only for a limited time, around 8 years, so the treatment must be renewed.

3. Earthed and Unearthed Construction

Early overhead lines were earthed construction the term means that an earth conductor is providedin addition to the phase conductors. The earth conductor provides a low impedance path for earthfault current, ensuring that the protection system operates quickly and reliably. Where the earthwire is positioned above the phase conductors it also acts as a lightning shield, although theshielding angle that the conductor provides is subject to opinion. Most early designs of overheadline had the earth wire positioned above the phase conductors, but not in every case.

Where supports are made of conducting material for example steel or reinforced concrete aconnecting earth wire must be provided and this includes all modern transmission lines supportedon Steel lattice towers.

Where wood (or glass reinforced plastic) poles are used, advantage can be taken of wood’sexcellent insulating properties and omit the earth conductor. However, poles supporting certainequipment such as transformers and switch disconnectors must continue to have their steelworkearthed and this is achieved by local earth systems at the foot of the pole. The flashover value of apin insulator is 115kV, whereas a wood pole has a flashover value of approximately 4kV percentimetre; 27cm of wood pole is equivalent in flashover value to a pin insulator. This allowsunearthed lines to have a much improved lightning performance, because although lightning


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induced voltage surges can still travel along the line, the possibility of flashover to earth and asubsequent power arc is much reduced. Where unearthed construction is adopted, the crossarmsmay become live if an insulator fails leading to a voltage gradient down the pole and a consequentleakage current. In theory this could cause the pole to catch fire but in practice pole fires are veryrare.

The advantages of unearthed construction may be summed up as follows:

(a) First cost and continuing maintenance cost of the earth wire are eliminated.

(b) There is no earth conductor to break and fall into the phase conductors.

(c) Crossarms may be horizontal because perching birds can no longer bridgebetween phase and earth; earthed construction crossarms were mostly sloping toprevent this effect.

(d) Faults due to wind and ice shedding are eliminated.

(e) Poles can be shorter.

(f) In earthed construction, stays can become live and remain live because the earthimpedance is too high and the fault current too low to trip the protection. This is amajor hazard.

A disadvantage of unearthed construction is that stays must be fitted with insulators but the cost ofthis is low. Since the introduction of BS 1320 in 1946 almost all wood pole lines in the UK havebeen unearthed construction and most earlier earthed construction lines have had the earthconductor removed.

4. Mechanical Design of Overhead Lines

Ground conditions - All of the mechanical force exerted on an overhead line by conductortension, weight, ice and wind loading is transmitted to the ground and so soil conditions must beconsidered. In the case of primary transmission voltage linessupported on steel towers it is customary for a geotechnicalsurvey to be carried out at each tower position, but this is noteconomic for lower voltage lines on wood pole or similarsupports, where empirical calculations are used to assess thesituation shown in Figure 2, where the force shown acting on thesupport at the top will tend to pivot the support around a pointbelow ground level that is taken to be the depth of planting

divided by 2 .

Here the maximum stress before movement occurs is given bythe formula:

3

Nm10

g

kDhM

Where:

Mg= Moment of resistance of the ground in Nmk = Rupture force on the ground in N/m²/m depthh = Depth of pole in the ground in metresD = Diameter of pole below ground

Fig. 2 Pole subjected tobending stress


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The value of k varies and can be determined from trial holes but for northern Europe a typical valueis 300,000 N/m²/m and this can be used as a basis for calculation.

Example 1

Pole depth of planting 2mPole diameter 0.3mTotal pole length 11mConductor tension 8000N

Moment of ground resistance Mg =3300,000 0.3 2

10

Nm

= 72,000 Nm

If the pole top load is applied at a distance above the fulcrum point of:

29 10.41

2 m

And the applied load at pole top is 10.41x8000 = 83,280Nm

In this situation, the applied load is greater than the ground resistance and this does not take intoaccount wind load on the support itself, which could in severe weather add another approximately5,000Nm. So some additional method of increasing the ground resistance is required, which couldbe by increasing the depth of pole in the ground or by fitting a large piece of timber to the pole. Theleft hand side of Figure 3 shows an additional piece of timber (a baulk) fitted to a pole. Formaximum effect this should be sited with its upper surface level with the ground, but in practice itmust be fitted at least 0.5m below ground level so that ploughing operations are not impeded.Unfortunately this depth is approximately the pole pivot point and so the improvement in moment ofresistance offered by fitting an additional timber baulk is reduced. A second baulk may be fitted atthe base of the support (shown on the right).

The value of k quoted previously as 300,000 N/m²/m varies with location and equally importantlywith the degree of ground compaction attained around the pole. Historically pole holes wereexcavated by machine or hand in the shape shown in Figure. 4, where a sloping side allows thepole to be slid into position and then raised by pressure against the vertical face at the back.

Fig.3 Poles with below ground timber baulks


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The backfilled ground must be compacted extremely well and even then does not attain themoment of resistance of undisturbed ground.

Comparative tests and operational experience have shown that augured pole holes are muchsuperior to excavated pole holes in that the undisturbed ground provides a moment of resistance atleast twice that of restored ground, no matter how well it has been compacted. Augured holes alsoallow additional pole planting depth to obtain extra moment of resistance at very little extraexcavation cost, although the extra pole length will be more costly.

Support load, bending – Supports are loaded in two ways. In the case of supports in the straightsections of the overhead line, the applied stress is mainly in bending due to wind load and partly incompression due to conductor, insulator and cross arm weight. For supports at the line ends and atthe corners, the effect of fitting a stay or stays is to stress the support wholly in compression, thatis, as a strut.

Considering firstly bending stress, then on one side the wood fibres are in tension and on the otherside they are in compression, creating forces that resist the applied bending moment.

The Second Moment of Area or the Moment of Inertia of a circular section support is:

4

64

DI

Whilst the Modulus of Section is:

3

32

DZ

D is taken as the diameter of the support 1.5m from the butt end. The ultimate fibre stress F forsoftwood poles is 53,800kN/m². Now the imposed Fibre Stress is:

F =Bending Moment

Section Modulus

Fig. 4 Pole excavation in cross section


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Maximum Resisting Moment Section Modulus F

3353,800

5300 kNm32

DD

………….(1)

Example 2

Pole diameter 0.25m

Maximum Resisting Moment 35300 0.25 = 82.8 kNm

The implication of equation (1) is that the ability of wood pole supports to resist bending loads isproportional to the cube of the diameter. The value obtained from equation (1) must be moderatedby a Factor of Safety (FoS) to obtain a design figure; this will be at the designer’s choice buttypically between 2.5 and 4. Note that as mentioned in Module 1 in extreme weather conditions it isdesirable that the more easily replaced conductor fails before the less easily replaced supports. Toachieve this, the designer must take into account that, over time, the strength of wood poles willreduce at a faster rate due to rot than the failure strength of conductor will reduce due to corrosion.This argues for a Factor of Safety that is greater for the supports than for the conductor.

A well known characteristic of wood is that under a continuously applied bending stress it will takeup a permanent deformation. For wind loads this is not a matter of importance because the windwill blow from different directions at different times; however a continuous bending load created bya small angle in the line of conductor may cause permanent deformation; for this reason conductorangles on non-stayed poles should be limited to very small values, typically less than 3°.

Where steel, glass reinforced plastic (GRP) or reinforced concrete structures are used as supports,manufacturers must be consulted as to the maximum acceptable bending loads.

Example 3

A three phase overhead line uses 150mm² Cricket All Aluminium conductor to BS 215 having adiameter of 16mm. The radial ice loading is 19mm and the half spans on either side are 45m and50m. Wind pressure at right angles to the line is 380N/m², pole length 12m with conductors at poletop, pole planting depth 2m (10m of pole above ground level).

Projected area of conductor + ice loading = 16 + (2x19) = 54mm

The total conductor + ice area exposed to wind is 95x0.054x3 = 15.39m²

Wind load on the conductors = 15.39x380 = 5848N

Wind load on the pole itself, assuming 250mm diameter = 10x0.25x380 = 950 N (this is assumedto act at half pole height).

Maximum bending moment at ground level = (5848x10) + (950x5) = 63,230 Nm

Allow a Factor of safety of 2.5, so the maximum allowable fibre stress is53,800

21,5202.5

kN/m².

Moment of Resistance = Bending Moment = F x Section Modulus


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3363,230 21,520 10

32

D From which D = 0.31m

Support load, compression – Supports located at the line angles and at the terminals are fittedwith a stay or stays, so that the conductor tension and wind loads imposed at the pole top areconverted into a compressive load in the pole and tension load in the stay(s) as shown in Figure. 5.

Figure. 6 shows how these forces are resolved.

Under the conditions shown, the pull on the pole is given by

Pole pull = 2 sin cos2 2

A AP WL Newtons

Where A = Angle of line deviationP = Tension of all three conductorsW = Wind load in N/m2

L = Span length in metres (= sum of adjacent spans/2)

This formula is usually simplified to:

Pole pull = 2 sin2

AP WL Newtons

Fig. 5 Line angle loads

Fig. 6 Tension forces


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The stay angle should be 45° to the pole vertical or asnear to that value as site conditions allow. Traditionallythe stay ended below ground at a large timber block, butmodern practice is to use a screw anchor. The tension inthe stay is:

Pull on PoleStay Tension =

Sin

The pole compressive load due solely to conductortension and wind pressure is:

Pole Compressive Load = Stay Tension x cos

The total compressive load on the support must also include the so called ‘weight span’, which isthe downward force (in Newtons) exerted by the (near) half spans on either side (three conductorsfor an unearthed 3 phase line) – the other (distant) half spans exert compressive load on thesupports on either side.

Example 4

Determine the load in the stay for a Dingo 150mm2 conductor having a maximum working tensionof 10kN, an average span on either side of 100m, conductor diameter of 16.75mm, angle of linedeviation 45°, no ice load and wind pressure 380N/m2

P = 3 x 10,000 = 30,000N

Wind load W per metre =16.75

1 380 6.3651000

N

Pole pull = 2 sin2

AP WL Newtons =

452 30,000sin 6.365 300 22,961 1909 24,870N

2

Tension in stay =24,870

35,176N45

Sin

Support compressive load due to conductor tension and wind load = 35,167N x Cos45 = 24,870N

It will be noted that stay tension is relatively high and will be higher still if site conditions result in apole to stay angle less than 45°. Where the ability of the ground to resist stay tension or the safeworking load of the stay wire after the Factor of Safety has been allowed are a problem it may beovercome by providing two or more stays, each sufficiently separated that the ground pressurezone of one does not overlap the ground pressure zone of another (Figure. 8).

Fig. 8 Two and three stay arrangements

Fig. 7 Stay tension & pole load


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Stay wires are galvanised steel, which in the UK are available in two sizes, 7/3.25mm with a MeanFailure load (MFL) of 66.8kN, mainly used for lightly stressed low voltage lines and 7/4.0mm with aMFL of 101kN. Where the overhead line is unearthed construction a stay insulator must be used(Picture 9).

Figure 10 shows the traditional method of installing a stay baulk, using a tee shaped excavation.The forward face of the trench, against which the baulk bears, should be undercut at an angle – atask that can properly be done only by hand, not by machine. However this is difficult to achievebecause, as the excavation is normally 2m deep to reach firm ground, health and safetyconsiderations will require it to be shuttered. Threading a large and heavy timber baulk throughtrench shuttering is also a complex task.

These problems have tended to favour a change to screw anchors (Figure 11), which mayconveniently be installed using the same hydraulic motor used to auger pole holes, as describedearlier. Although the bearing surface is substantially less than the timber baulk, the disc bears onrelatively undisturbed ground and in addition, it may be inserted to a much greater depth, in firmerholding sub strata, at little additional cost. The cost per stay is much less than the timber baulkdesigns hence double or triple stays may be installed relatively cheaply.

Fig. 10 Stay excavation

Pic. 9 Two forms of stay insulators


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Screw anchors also have the advantage that monitoring the torque (fluid pressure) of the hydraulicmotor during installation provides a good indication of ground conditions hence the quality ofanchoring; if the ground is too soft, the motor may be reversed, the anchor extracted and anotherlocation tried.

The total vertical load on the pole is not only that imposed by conductor tension acting through thestay but also that due to the weight of conductors, crossarm(s), insulators and the pole itself. Theweight of conductors is half that of the two spans combined. If the vertical load exceeds thestrength of the pole the failure mode is by buckling and the formula used to determine this criticalload P is Euler’s:

2

2

ELP

I

Where E= Modulus of elasticity, taken as 910.35 10 for wood, I is the Moment of Inertia of a

circular section (4

64

DI

) and L is the pole length. Strictly, this formula is applicable to a beam

pinned at both ends, of uniform cross sectional area and of uniform strength, which a pole is not;nevertheless this formula is the one that is normally applied, taking D as an average value. Wherethe vertical load exceeds the safe limit for a single pole, after taking into account the design Factorof Safety, one of the designs shown in Fig. 12 may be selected.

Fig. 11 Screw anchor

Fig. 12 Double pole supports


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The pole on the left is a ‘Rutter’, that in the centre is an ‘A’ pole whilst that on the right is a braced‘H’ pole. The first two designs are now considered obsolescent but the ‘H’ pole remains inwidespread use, although the modern tendency is to omit the steel bracing and rely only on thebracing provided by the ground in which the poles are erected or a wooden cross brace belowground level. Euler’s formula for these designs is:

2

2

4 ELP

I

Hence compared with the single support there is a fourfold increase in buckling strength and inpractice the buckling mode of failure is eliminated.

Crossarms – Figure 13 shows a typical intermediate support crossarm with post insulators,suitable for use at Secondary transmission voltage. Force P the horizontal component arises fromthe wind pressure on the conductor + radial ice, whose moment is the length of the post insulator.Force W is the conductor + ice load weight, whose moment is the distance from crossarm centre tothe outer insulator position; in certain circumstances there may be an additional force due toconductor down pull as described later in this Module. Both these forces are tending to bend thecrossarm downward from its centre.

If P = Wind pressure on conductor + radial iceW = weight of conductor + ice loadL = crossarm width from centre to insulator

location

Then the total bending moment = WL + PX in Nm

Fig. 13 Forces on crossarm


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Maximum working stress F =Ultimate strength of Steel

Factor of Safety

And the maximum working stress =Bending moment (BM)

Section Modulus(Z)

ThereforeBM

Z =F

Example 5A three phase overhead line uses 150mm² Cricket All Aluminium conductor to BS 215 having adiameter of 16mm and weight 0.432kg/m. The radial ice loading is 19mm and the half spans oneither side are 45m and 50m. Wind pressure at right angles to the line is 380N/m². The conductorspacing is 1.1m and insulator height is 180mm.

Projected area of conductor + ice loading = 16 + (2x19) = 54mmThe total conductor + ice area exposed to wind is 95 x 0.054 = 5.13m²

Horizontal wind load on the conductor acting through the insulator = 5.13 x 380 = 1949N

Bending load on crossarm due to wind = 1949 x 0.18=351Nm

Weight of conductor = 95 x 0.432 = 41kg or 402N.

Now the area of ice is a cylinder with inner radius 8mm and outer radius 27mm = 2.089 x 10-3m²

Volume of ice = 2.089 x 10-3 x 95 = 0.198m3 and the weight of ice = 0.198 x 1000 = 198kg or1942N

Total weight of conductor + ice = 402 + 1942 = 2344N and the bending load on the crossarm dueto weight = 2344 x 1.1 = 2578Nm

Therefore the total bending load = 2578 + 351 = 2929Nm

A typical value for the maximum working stress of steel is 43,200N/m² so allowing a Factor ofSafety of 2.5 this reduces to 43,200/2.5 = 17,200Nm.

BMZ =

FHence Z = 32929

0.17m17200

or a Section Modulus of 17cm3

A suitable Steel crossarm section may now be selected from standard tables of Steel angle.

Intermediate crossarms - In the calculation of crossarm loads on intermediate supports and theselection of a suitable Steel section, it is not customary to design for the bending stress on acrossarm due to the broken conductor condition shown in Figure 14. In the case of Pin or Postinsulators, the bindings at each insulator restrain the conductor very well laterally but poorlylongitudinally, therefore if the conductor breaks, the tension is greatly reduced as the conductorslides back through the bindings.

In the case of suspension (cap and pin) insulators, the set is free to pivot in the vertical plane,though to a limited extent. Note that in the case of Steel transmission towers, the arms should bestrong enough to withstand a broken conductor condition.


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Figure 15 shows how additional downwards bending stress is imposed on a crossarm due tovertical angles in the route of the line. This illustration exaggerates the vertical profile but showshow the conductor tension acting at a downward angle creates a vertical component of load on thecrossarm. The downward pull is proportional to the Sine of the angle between conductors andvertical.

At section points in the line, either straight or angle, crossarms must allow for the broken conductorcondition (Figure 16) which imposes a considerable bending moment on the crossarm as well as atorsion stress on the pole, which is transmitted to the wood through the crossarm and strut bolts.To prevent the wood splitting at the pole top, two additional bolts may be fitted at right angles to thecrossarm and strut bolts and parallel to the crossarm. These Anti-Split bolts are locatedimmediately above and below the crossarm bolt.

Fig. 14 Broken conductor condition

Fig. 15 Conductor down pull

Fig. 16 Conductor broken at an in line section support


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At the angle supports of lines with conductors in horizontal formation, the crossarm length must beincreased to maintain the conductor separation distance. The increase is in proportion to thetangent of the angle of deviation and becomes very large at angles above 45°, as shown in thetable which shows the increase required to maintain a nominal conductor separation distance of 1metre.

Angle of line deviation degrees Separation between insulators tomaintain a 1 metre conductor spacing

45 1.00m

50 1.192m

55 1.43m

60 1.73m

65 2.14m

The rapid increase in crossarm length for line deviation angles greater than 45° sets a practicallimit of deviation angle of approximately 55° - there is in addition the increasing tension in the staysand increasing compressive stress in the support to consider.

On a single pole support a longer crossarm also has the effect of increasing the bending momenton the crossarm in the event of conductor breakage. There are two methods of overcoming thisproblem, the first is to add a second crossarm behind and in parallel with the first and the second isto use the ‘H’ pole design described earlier, which reduces the moment arm considerably. ‘H’ polesupports will also normally include a second crossarm. Where a support is sufficiently robust toresist the effect of conductor breakage, it may be referred to as a Failure Containment support.Picture 17 shows a failure containment support with tension insulators inserted into a long run ofintermediate supports – ideally this should also be fitted with fore and aft stays.

.

The section support in the photograph prevents conductor failure on one section of the lineaffecting the adjacent section. Figure 18 shows an ‘H’ pole twin crossarm support at a line angleposition, fitted with a single stay.

Pic.17 Failure containment section support


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The arrangement shown in Figure 18 is also generally applicable to line terminations, where bothpoles are fitted with stays at the rear of the supports and the jumper connections are taken down toa cable box or cable sealing ends. For line terminations, the crossarms must withstand theconductor tensions plus the combined conductor & ice weight, the former acting horizontally and thelatter vertically.

5. Conductor Spacing – Wind Induced Oscillation

In addition to the wind induced vibration described in the next section, conductors are subject tolarge amplitude oscillation, referred to as Galloping. The natural frequency mode tends to bearound 1 Hz, leading to this slow periodic motion also being known as Dancing. Whereas vibrationamplitudes are measured in millimetres, oscillation amplitudes are much larger, on transmissionlines up to a metre. The conductor moves most often in the vertical plane, although horizontal androtational motion is also possible, all modes giving rise to the possibility of the conductor separationbecoming less than the flashover distance. The large amplitude motion also adds greatly to thestress on insulators and supports, raising the risk of mechanical failure.

The mechanisms that initiate gallop are not always clear, though it may be caused on someoccasions by asymmetric conductor aerodynamics due to ice build up on one side. The profile of iceaccumulated on the downwind side approximates to an aerofoil, altering the normally round profileof the wire to an elliptical shape capable of generating aerodynamic lift. However galloping can alsooccur on ice free conductors.

Figure 19 shows a situation where three horizontally spaced conductors are galloping in a rotationalmode, where the possibility of flashover is inhibited only by the conductor and hence the crossarmspacing. Exactly the same situation can and does occur on vertically separated conductors.

Fig. 18 Section ‘H’ pole angle with twin crossarms


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Determination of an appropriate conductor separation clearly depends upon the amount ofconductor sag and a widely used formula is embodied in the German DIN VDE 0210 specification,as follows:

Mid Span Spacing ( ) AMk Sag I S

Where:

I is the length of the suspension insulator (0 if pin or post insulators are used)Sag is the still air sag at maximum temperature for the longest spanSAM is a voltage related factor (0.1m for 11kV)k is a conductor orientation and type factor (for horizontal conductors up to 150mm2, this is 0.7)

6. Conductor Vibration

Whereas conductor oscillation is a large amplitude effect, conductor vibration is resonant, smallamplitude and high frequency, creating multiple loops and nodes along the length of the span.Typically the frequency is 5 to 100Hz with an amplitude of 10-15mm. Because it occurs at relativelylow wind speeds, generally under 15km/hr, if vibration occurs it will be present for most of the time.It is more prevalent with Steel Cored Aluminium, All Aluminium and Aluminium Alloy conductors ascompared to Copper and Cadmium Copper conductors, possibly due to the lower mass per unitlength of Aluminium. Solid conductors are well known to be subject to vibration.

Vibration problems can lead to conductor failure as fatigue cracks develop, following the naturalcrystalline boundaries within the metal – the eventual sharp breaks are easily distinguished fromtensile failure, because the latter produces a characteristic reduction in diameter at the point offailure (necking) whilst the former does not. This mode of failure generally occurs at the span endsand affects Aluminium to a greater extent compared to Copper, due to the softer metal’s lowerfatigue strength. Vibration problems are greater with long spans, especially long spans supportedby suspension insulators hence they affect some 33kV lines and most 66kV, 132kV and highervoltage lines. On primary distribution voltage lines the rigidity of pin and post insulators tends tosuppress vibration, compared to the flexibility of suspension insulators and this factor reduces theproblem to a point where special precautions against it are not required.

Fig. 19 Conductor gyration


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Control of conductor tension, especially avoiding high conductor tensions can help reduce vibrationand this is discussed later in the Module. The other options to reduce the possibility of vibrationinduced failure are to use well designed conductor clamps and to fit Stockbridge dampers. Figure20 shows a suspension conductor clamp intended to accommodate conductor vibration. The clampmouth is shaped to avoid a sharp edge where the conductor enters and the consequent hammereffect the edge would have on the conductor strands – the left hand side clamp is additionally fittedwith protecting armour rods intended to cushion and dampen any vibration. These rods helpdampen vibration and also protect against lightning flashover.

The Stockbridge damper dates from 1924 and is named after its inventor, George Stockbridge. Thisis the most favoured method of damping vibration, comprising two hollow weights joined to a centralbody by flexible steel cables. In action the weights and central steel cables oscillate around theclamped central body; repeated flexing of these cables generates inter-strand friction and therebyenergy loss. Weight/cable assemblies are tuned to maximise the damping effect at a chosenparticular frequency.

Fig. 20 Suspension clamp with (left) and without armour rods (right)

Fig. 21 Stockbridge vibration damper


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Dampers should be positioned at a point of maximum vibration (an anti-node); for any given modeof vibration there will be one less anti-node than the number of nodes, because the span ends arealways nodes. An empirical expression for the optimum positioning of dampers is:

20.0152 TdD

V w

Where D = distance from suspension clamp in metres, T is the tension in kg, V = critical windvelocity in metres per second, d = conductor diameter and w = conductor mass per in kg per metre.For example consider Lion 225mm2 conductor, 22.26mm diameter, 20kN (2038kg) working tension,1.214 kg/m mass, wind speed 4m/s (14.4km/h). Substituting values,

20.0152 2038 22.26

4 1.214

xD from which D = 3.5m

It is possible to fit more than one damper, the second tuned to a different frequency. Hence thepurpose of the second damper is not to increase the damping effect at a single wind speed, but toflatten the tuned response to obtain damping over a greater range of wind speeds. It is alsopossible to fit dampers at right angles to each other, thereby providing damping in two planes.

7. Conductor Characteristics

Conductors are the largest cost element on Primary distribution and Secondary transmission linesand the overhead line engineer has a wide choice of types. All of them are based upon a centralcore wire surrounded by a number of concentric layers of helically applied wires, each layerreversed in twist with the outer layer being a right hand or ‘Z’ twist.

Stranding improves the conductor flexibility and allows easier storage on, and removal from, adrum. The reverse twist in each layer is intended to inhibit bird caging, that is, the layers unwindingwhen the conductor is cut.

Where all the wires are the same diameter, as is normal, the number increases by 6 in each layer,for example 6 (first layer), 12 (second layer), 18 (third layer), 24 (fourth layer) etc. giving strandtotals of 7, 19, 37, 61 and onwards. Conductors up to 37 strands are used at Primary distributionand Secondary transmission voltages. Strands are normally round, but as for the power cableconductors described in Module 2, they may also be compacted to reduce the overall diameter andhence the wind load.

Five metals are or have historically been used for conductor, as follows:

Copper (Cu) – Copper is used as the comparison standard for all other conductors, which are ratedaccording to their equivalent Copper area. It is used in the hard drawn condition, which increases itstensile strength over the soft form for a 3% loss of conductivity. Very heavy, for the same currentcarrying capacity it is twice the weight of Aluminium. On the other hand, the high conductivity givessmaller diameters and therefore low wind loads. Widely used in the first half of the twentiethcentury, it is seldom used today because of its cost compared with the alternatives. Its high scrapvalue also makes it attractive to thieves. However, Copper conductor does have indefinite life.Some early Copper conductors were single strand and Steel reinforced Copper conductor has alsobeen produced. Solid Copper conductors were well known to be vulnerable to vibration fatiguefailure.


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Cadmium Copper alloy (Cad Cu) – Offers an approximate 50% improvement over Copper intensile strength, but with a loss of conductivity of 18%. Cadmium is highly toxic, an important factoragainst its use. It is also considered too expensive compared with the alternatives, but in commonwith Copper it has an indefinite life. Both Copper and Cadmium Copper were more used in the earlypart of the 20th century than they are today, although Copper can still find application in coastalareas where salt induced corrosion is a problem for other materials.

Aluminium – AAC (All Aluminium Conductor) is used in the hard drawn form when it is referred toas HDA. It offers 61% of the conductivity of Copper at half its tensile strength. Because it ishomogenous, galvanic corrosion is less of a problem however crevice corrosion can occur. AAC isrelatively weak and so span lengths must be restricted especially where a heavy ice load isexpected.

Aluminium Alloy – AAAC (All Aluminium Alloy Conductor). Magnesium and Silicon are added toimprove the tensile strength by 50% with an 11% loss of conductivity compared to commercialpurity Aluminium. In common with Aluminium, it is subject to crevice corrosion. Generally preferredto Aluminium and because it does not have the corrosion problems of Steel is widely used forPrimary distribution and Secondary transmission lines. Silmalec is a proprietary alloy containing0.5% Silicon and 0.5% Magnesium.

Galvanised Steel – This is used in combination with Aluminium (historically also with Copper) toimprove the overall tensile strength, referred to as SCA in the UK (Steel Cored Aluminium) or ACSRin the USA (Aluminium Conductor Steel Reinforced). The Steel part may be single conductor ormulti-strand. Most transmission lines and some Primary distribution voltage lines erected today usethis combination. The life of SCA is limited by corrosion of the Steel to rust and corrosion of theAluminium to Hydroxide, which is a white powder. The Zinc coating goes first as it is sacrificed bygalvanic corrosion to the Aluminium, finally the Aluminium is sacrificed to the Steel. Corroded SCAconductors may be recognised from ground level by characteristic bulges. Larger sizes of SCA, inparticular those used for transmission purposes require connectors that compress separately ontothe Steel and Aluminium sectors. Smaller sizes of connector are Compressed Overall.

Table 3 Conductor Characteristics

Property Units CopperHard

Drawn

CadmiumCopper

HardDrawn

Aluminium

AluminiumAlloy

GalvanisedSteel

Conductivity %Copper

97 79.2 61 53 9

Resistance Ω m2/km 17.71 21.769 28.264 32.5 192

Temperaturecoefficient

Per °C 0.00381 0.0031 0.00403 0.0036 0.0054

Linearexpansion

°C x 10-6 17 17 23 23 11.5

Linear Mass Kg/mm2

km8.89 8.945 2.703 2.7 7.8

UTS N/mm2 414 621 160-200 295 1320-1700


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The elasticity, tensile strength and weight of SCA vary considerably according to the relativeproportions of Steel and Aluminium and values will need to be obtained from manufacturers. Forexample there are three forms of 125 mm² conductor, as shown in Table 4.

The designer must select the type most suited to the anticipated span lengths and the expectedenvironmental conditions, with a greater proportion of Steel allowing longer spans and greater windand ice loadings.

As an alternative to galvanising, the Steel central conductors may be Aluminium clad, which offersgreater conductivity though at a greater cost – this conductor is referred to as ACSAR (AluminiumConductor Steel/Aluminium Reinforced)

Table 4 Versions of 125 mm² SCA conductor

Name AluminiumStranding

SteelStranding

Weight

Kg/km

Diameter mm

Cougar 18/3.05 1/3.05 419 15.25

Leopard 6/5.82 7/1.75 493 15.81

Tiger 30/2.36 7/2.36 603 16.52

Creep is a well known characteristic of Aluminium conductor and is the term used to describe itspermanent extension over time. Partly this is due to mechanical effects and partly due to crystallinemovement under the influence of tension and temperature. Mechanical creep arises from onestrand impacting slightly on the strand above or below, with which it is in line contact, so slightlyreducing the helical length – this mostly takes place early in the life of a conductor. The effect maybe reduced by over tensioning the conductor for a few hours after initial erection - typically 10%over-tension for 24 hours.

The amount of creep to be expected over long periods of time is a complex matter; predictorequations are available in specialist literature. Approximate 10 year extension values in mm/km areAAC - 800, AAAC – 500, SCA – 500.

Greasing – To guard against corrosion and prolong the life of Aluminium conductors, conductorgrease may be applied, so that water does not penetrate and there is therefore no electrolyte. Twoforms of greasing are used:

Hot applied – Here a heated mixture of wax and oil, formulated to remain solid to a temperatureabove the expected maximum conductor operating temperature, is applied to each layer ofconductor.

Cold applied – Here a high viscosity, thixotropic mixture of oil and thickener is injected into theconductor under pressure. Proprietary greases are available that include corrosion inhibitors.Different levels of cold greasing are available, for example Steel core only, core and first layer, coreand second layer, all layers greased.

Greasing levels

In addition to the selection between hot and cold applied, the user may also specify different levelsof greasing – in the UK these are described in Engineering Recommendation L38/1. The choicedepends upon the required level of protection against corrosion and to some extent the tolerance oflinesmen to grease contamination of hands, clothes, tools, etc.


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Steel core onlygreased

Inner layers greasedexcept the outer

surface of penultimatelayer

All inner layersgreased

Fully greased exceptthe outer surface of the

outer layers

Conductor greasing forms an integral part of conductor manufacture, which is carried out on largehorizontal machines of the type shown in Figure 22. The ’54 bobbin’ equipment shown is capable ofproducing distribution and secondary transmission conductors having up to 61 strands, startingfrom an initial 7 strand input conductor. There are three stations; the first with 12 spools, the secondwith 18 spools and the third with 24 spools. Each station rotates in the opposite direction to itsneighbour and the outer layer of conductor is always ‘Z’ laid – that is, the strands appear to slope tothe right. After each layer is applied the conductor is drawn through a die which both compressesthe strands and in the example illustrated also injects the grease – however injection through thedie is not always implemented, sometimes grease is flood applied immediately before the die.Control over the stranding process is obtained through the machine’s gearing, that is, the speedand direction of rotation of the three spool stations in relationship to the longitudinal speeddetermined by the drive capstan. Post forming rollers are positioned after the final die and beforethe take up drum. Their function is to mechanically stress the finished conductor in such a way as toremove any tendency for the strands or layers to spring apart when the conductor is cut – this isimportant for ease of work on site. The relevant standards for conductor are:

IEC1089 Round wire concentric lay overhead electrical conductors and

BS EN 501182 Aluminium Conductors and Aluminium Conductors Steel Reinforced

Covered conductor – The use of Aerial Bundled Conductor (ABC) has been described in Module1. Over the past 20 years there has also been a growing interest in the use of covered conductor at

Fig. 22 54 bobbin conductor laying up machine


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Primary distribution voltage, mainly because of its enhanced safety. PVC can be used as theinsulator, but it becomes brittle at low temperatures and softens excessively at high temperatures.XLPE is a better covering material; 2.3mm is the normal thickness for use at 11kV. In Scandinaviawhere covered conductor lines were developed, they are known as BLX. In the UK, EATSstandards 43-119, 43-120 and 43-121 describe covered conductors, fittings and designrespectively.

8. Conductor Current Rating

The current rating of overhead line conductors is controlled by the following factors, whichdetermine the heat balance:

1. The design operating temperature of the line Tm in °C, which is closely related to thepermissible sag. Normally this is 50°C.

2. The ambient temperature Ta in °C.

3. The solar heat gain S in W/m².

4. Wind speed V, which controls the wind cooling effect. This is normally taken as 0.447m/s(1 mile per hour)

The following empirical formulae were first published in 1949:

Convection loss is given by:

0.448

Convection loss = 387 W/km m aT T V D

Where D = Conductor diameter in mm.

Radiation loss is given by:

4 47Radiation loss = 1.791 10 273 273 W/km

m aE D T T

Where E is a value for the surface emissivity. For weathered conductors after a few months servicethis has a value of 0.9.

The Solar Gain is given by:

Solar Gain Y S D W/m

Where Y is the solar absorption coefficient; for weathered conductors this is taken to have the samevalue as E, that is, 0.9.

The value of S for northern Europe is taken as (in summer) 580w/m². This value must be adjustedfor tropical and semi tropical countries. In winter, solar gain may be ignored as a factor in northernEurope.

The Heat Gain from the passage of current is:

220Electrical energy gain 1 20mI R T


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Where R20 is the conductor resistance at 20°C

So the basic heat balance equation is:

Electrical energy gain + Solar Gain = Radiation loss + Convection loss

220

0.448 4 47

1 20

387 1.791 10 273 273 W/km

m

m a m a

I R T Y S D

T T V D E D T T

From which:

0.448 7 4 4

20

387( )( ) 1.791 10 ( 273) ( 273)

1 20

m a m a

m

T T V D E D T T Y S DI

R T

Example 6

Determine the current rating of 75mm² ‘Willow’ 7/4.04 AAAC having a resistance at 20°C of0.3669Ω/km and a diameter of 12mm. The ambient temperature is 25°C, wind speed 0.447m/s andthe emissivity/absorption coefficient is 0.9. The operating temperature is 50°C, solar radiation is580W/m² and the coefficient of electrical resistance with temperature is 0.0036 per °C.

First determine the Convection loss from:

0.448

Convection loss = 387 W/km m aT T V D

Substituting values:

Convection loss = 387[50 – 25][0.447 x 12]0.448

= 20,533W/km

Next determine the Radiation loss from:

4 47Radiation loss = 1.791 10 273 273 W/km

m aE D T T


4 47Radiation loss = 1.791 10 0.9 12 50 273 25 273 W/km

= 5799W/km

The Solar gain is:

Solar Gain Y S D


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Solar gain = 0.9 x 580 x 12 = 6264 W/km (taking into account that the solar radiation is specified inwatts per square metre and the diameter D is specified in mm)

The change in resistivity is calculated from:

020Resistance at 50 C = 1 20 mR T

In this example the resistance at 500C is:

0.3669[1 + 0.0036(50-20)] = 0.4065 Ohms

Finally:

20,533 5,799 6,264222Amps

0.4065

I

The above is an important method of calculation because, unlike cable makers, conductor makersdo not publish current ratings for conductors. It is the task of the utility engineer to calculate them,taking into account local factors. It is best to calculate the values of convection loss, radiation lossand solar gain separately to ensure that the values determined are of the right magnitude, eventhough this may increase the small rounding errors.

Short circuit current rating – The duration of short circuit current, normally not more than 3seconds, is insufficient for radiation or convection loss to have any effect; rather the entire electricalheat input raises the conductor temperature in accordance with its specific heat. There is a strictlimit to the allowable temperature rise, above which the conductors will permanently anneal(soften). This is 200°C for Aluminium and its alloys and 210°C for Copper. Hence for lines alreadyoperating at the maximum permissible 50°C limit the additional temperature rise under short circuitconditions is 150°C for Aluminium and 160°C for Copper. The line sag will also increase temporarilyalthough this effect is not normally allowed for Primary distribution or Secondary transmissionvoltage lines, which is a consideration in not designing to the minimum allowed ground clearance.

The equation describing the rise in temperature due to electrical energy input is:

2 1( )2 6 30 1 010 (1 ) 10T Tt I R dt WS d

Where I = Short circuit current t = duration of short circuit current

R1 = Conductor resistance at T1 in Ω/km T1 = Conductor Initial temperature

= Temperature coefficient of resistance T2 = Conductor final temperature

= Conductor temperature rise above T1 W = Conductor mass

S = Specific heat of conductor in joules/°C.gram


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The initial equation may be rearranged as follows:

2 1

2 3( )1

0 0

10

(1 )T TtI R d

dtWS

Therefore 2 3

2 11log 110 e T TI R t

WS

And 3

22 1

1

10log 1 ( )e

WSI t T T

R

The expression 2I t is the normal method of expressing short circuit capability because from it theallowable short circuit current for any duration of fault may easily be determined. For mixed metalconductors, the constants must be aggregated or averaged as follows:

1 2 3 4W W W W W 1 1 2 2 3 3 n nW S W S W S W SS

W

1 1 2 2 3 3 n nA A A A

A

1 2 3

1

1 1 1 1

n

R

R R R R

Table 5 specifies the necessary constants:

Metal Temperature coefficient ofResistance per °C

Specific Heat S(J/°C.g)

Aluminium 0.00403 0.932

Aluminium Alloy 0.0036 0.932

Steel 0.00537 0.5215

Copper 0.00393 0.385

Example 7

Determine the I2t current rating of 60mm² ‘Pine’ conductor 7/3.61 AAAC having a resistance at 20°Cof 0.4594 Ω/km and a weight of 196 kg/km. The initial temperature is 50°C, the allowable maximumtemperature is 200°C, the specific heat is 0.932 J/°C.g and the coefficient of electrical resistancewith temperature is 0.0036 per °C.

The conductor resistance R1 at 50°C is

1 20 1 (50 20) 0.4594(1 (0.0036 30) R R = 0.5090Ω/km

Hence


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3

2 196 0.932 10log 1 0.0036(200 50)

0.5090 0.0036eI t

= 43.04kA2s

From which the short circuit current for 3 seconds43.04

3.78kA3

I or approximately 72MVA

at 11kV three phase.

9. Conductor Sag and Tension

9.1 Overview

The tension of overhead line conductors is normally between 15 and 20% of the conductor UltimateTensile Strength (UTS) at the minimum operating temperature, depending upon the conductormaterial and composition.

The design procedure is to determine an appropriate tension at the minimum operatingtemperature, from which the tensions and sags can be calculated for other temperatures. From thecalculated sag value, templates can be produced to help determine the support heights, supportspacing, etc. as later described.

In the UK, the traditional minimum operating temperature has been -5°C (22°F) and the maximum50°C (122°F). In recent years some countries have raised the upper temperature limit for Aluminiumto 90°C, however this temperature should not exceeded, because above 90°C and over a longperiod of time the metal starts to anneal (soften). Whether the higher operating temperature iseconomic depends upon the relative cost of energy and conductor (see Kelvin’s Law, Module 1).

In deciding on a low temperature tension limit, two factors must first be considered, wind and iceload and Aeolian vibration.

Wind and Ice load - For any chosen conductor and span, the combined wind and ice load must becompatible with the strength of the conductor and span length. These loads comprise theconductor weight plus the weight of the design ice load (if ice is expected in the environment wherethe line is to be constructed) acting vertically, together with the wind load, which acts horizontally.

They combine vectorially to give the Maximum Conductor Resultant (MCR) as follows and asshown in Figure 23:

2 2

MCR Conductor + Ice load + Maximum wind load

In the UK, the design maximum ice load for wood pole lines is normally taken as 9.5mm radialthickness and for steel tower lines as 12.5mm radial thickness at a density of 0.913g/cm3, which isthe value for pure ice (water = 1.0gm/cm3).


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Again in the UK, the wind pressure on bare conductor for lines operating below 45kV is taken as380N/m2 (velocity = 24m/s) for ice loaded conductor and 760N/m2 (velocity = 34m/s) for bareconductor, as it is considered that ice accretion is less likely in high wind speeds. Wind pressurevaries as the square of wind speed and may be calculated from:

P = 0.634 V2 in N/mm2

Where V is in m/s.

Under the specified conditions the total loaded weight for any ice thickness (in mm) is given by:

3Ice loaded Weight 2 (2 2 )kg/m

10b

KW R R D

Where Wb = Conductor weight in kg/m, R is the radial thickness of ice in mm and D is theconductor diameter in mm. Factor K relates to the ice density with:

K = 0.717 for pure ice at 0.913g/cm2 and

K = 0.668 for wet snow at 0.850g/cm2.

Where ice is not experienced in the local environment, the conductor weight only need beconsidered.

The wind load is given by:

2Wind Load kg/m

1000

P D R

Where P is the wind pressure in kg that is, 77.5kg/m2 for an ice loaded conductor with wind velocityof 24m/s or 38.75kg/m² for a bare conductor with wind velocity of 34m/s.

Example

Determine the MCR for ‘Wasp’ 100mm2 All Aluminium conductor with diameter 13.17mm andweight 0.29kg/m. The pure ice radial thickness is 9.5mm and the wind speed is 24m/s.

Fig. 23 MCR


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First determine the conductor weight plus the ice loaded weight per

metre3

0.7170.29 13.17 (13.17 19)kg/m

10

= 0.29 + 0.303 = 0.593kg/m

In this example the wind load is

2Wind Load kg/m

1000

P D R

77.5 13.17 19Wind Load kg/m

1000

= 2.49kg/m

Hence adding the wind load and the weight load vectorially the MCR = 2 20.593 2.49 = 2.56kg/m

For a 120m span and a conversion of 9.81 Newtons/kg, the conductor loading due to conductorweight, ice weight and wind pressure is 2.56 x 9.81 x 120 = 3013 Newtons. Note that the MCRvaries as follows:

1. It increases in direct proportion to the span

2. It increases in direct proportion to ice accretion

3. It increases as the square of the wind speed

Aeolian vibration – This should not be confused with conductor oscillation described earlier in thismodule. As wind blows across a conductor, eddies are created first in one direction then in theother, above and below the conductor, forming what is known as a Karman vortex street. Thesealternating eddies set up corresponding vibration in the conductor which if it matches one of theconductor resonant frequencies induces large standing waves in the conductor at 50-100Hz. Thevibration formula below gives the frequency in Hertz:

1

2

Tf

L M

Where L is the node length in metres (span, half span, third or quarter span, etc.), T is the tensionin Newtons and M is the mass in kg/metre.

The mode of vibration is in the vertical plane and may be two wavelengths per span, three, four ormore – steady winds up to 35km/h cause the most vibration, turbulent winds experienced in hillyterrain are less of a problem. Aeolian vibration can quickly result in broken strands and conductorfailure especially if the fixed end of the conductor coincides with a vibration antinode. It isdependent upon conductor resonant frequency which in turn is related to conductor tension. Inpractice it has been found that Aeolian vibration will not occur if the conductor working tension isreduced below:

20% of conductor Ultimate Tensile strength (UTS) for Aluminium based conductors33% of conductor Ultimate Tensile strength (UTS) for Copper based conductors

The above are the values generally accepted in the UK.


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Covered (insulated) conductors are more prone to Aeolian vibration than stranded designs becausethere are no surface imperfections, due to the strands, to dampen the wind eddying effect. For thisreason the working tension limit is further reduced to 10-12% of UTS. Further mitigating measuresare to use dampers clamped to the conductor at the span ends or to use an SCA conductor with ahigh Steel content, as Steel has energy absorbing characteristics compared to Aluminium.

9.2 Design Tension Limit

An overhead line can be constructed to only one specified tension limit. In the UK, two alternativetension limits are generally recognised, as follows:

1. That a factor of safety of 2.5 (40% of UTS) is maintained under conditions of 9.5mmradial thickness of ice and a wind pressure of 380N/m2 at a temperature of -50C.Stated another way, this means that the applied tension at -5°C cannot exceed 40% ofUTS and this tension includes the value of the MCR.

2. That in still air conditions at a temperature of 50C the conductor will be below thetension limit to avoid Aeolian vibration. This 50C temperature is sometimes called theEvery Day Temperature (EDT) because it is the temperature that most overhead linesexperience most of the time, at least in temperate countries.

These limits are normally mutually exclusive and the engineer must decide which is to apply (‘rule’or ‘govern’ are the terms often used) to any particular overhead line. Different tension limits can intheory be applied to different sections of the same line, but this is an undesirable conditionbecause it places avoidable and permanent bending loads on the section angle supports.

In other countries, other tension limits can be applied according to the regulatory requirements andcodes. For example in the USA the NESC recommends limits on the tension of bare overheadconductors as a percentage of the conductor’s UTS. These USA tension limits which may also‘govern’ or ‘rule’ are:

1. 60% of UTS under maximum ice and wind load,

2. 33.3% of UTS initial unloaded (when installed) at 600F,

3. 25% of final UTS unloaded (after maximum loading has occurred) at 600F.

9.3 Conductor Sag

Conductor sag depends upon the tension, the conductor weight and the span length. Only themathematics for supports at the same height is described in this Module, which is satisfactory formost projects, even where there is a modest height difference. Consider a conductor stretchedbetween two horizontal supports as shown in Figure 24. The variables are:

S = True conductor length, m

D = Conductor sag, m

W = Weight per unit length, kg/m

L = Span length, m

Th = Horizontal conductor tension, kg

c= Distance between origin and point C


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The shape of the curve is a catenary and the Sag D for the catenary equation of the half span C toB is given by:

cosh 12

h

h

T WSD

W T

The above equation may be simplified by using MacLaurin’s infinite series for hyperbolic functions.In this form, each added term in the series increases the accuracy. The first three terms of theseries are:

2 2 32 2 24

8 6 8 10 6 8

h h h h h

WS W WS W WSD

T T T T T

As the examples below demonstrate, use of only the first term is considered sufficiently accurate forDistribution and Secondary transmission voltage lines up to 300m span – it is the equation of aparabola rather than a catenary and is sometimes referred to as the ‘parabolic simplification’. Use ofthe second term is customary for long span, large sag transmission lines whilst use of all threeterms is necessary only for the most exacting problems and is usually performed by a computerprogram rather than manually.

Example of Moderate Span

Consider a 120m span of Cricket 150mm2 AAC having a weight of 0.432kg/m – typical of PrimaryDistribution and Secondary transmission lines. The designer has selected a tension of 438kg or3985N, which is 18% of UTS (2432kg).

The first term2

8

h

WSD

Tgives a result of 1.77m

The second term

22

6 8

h h

W WSD

T Tadds an extra 0.51mm

Fig. 24


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The third term

2 324

10 6 8

h h

W WSD


Example of Longer Span

Now consider a 400m span of Zebra 400mm2 ACSR having a weight of 1.622kg/m - typical ofPrimary transmission lines. The designer has selected a tension of 2421kg or 23,750N, which is18% of UTS (13450kg).

The first term2

8

h

WSD

Tgives a result of 13.34m

The second term

22

6 8

h h

W WSD

T Tadds an extra 21mm

The third term

2 324

10 6 8

h h

W WSD


It will be seen that for Primary Distribution and Secondary transmission lines the first term of theexpansion (which is the expression for a parabola) is sufficiently accurate:

2

8

h

WSD

T

Conductor weight W is known and for short spans the distance between supports L ≈ S. Hence forany value of Th, Sag D can be determined or alternatively if D is assumed Th can be determined.

Non uniform tension - An implication of the diagram shown in Fig 24 is that tension is not uniformalong the conductor. It is at a minimum at mid span, where it is loaded to the imposed tension onlyand at a maximum adjacent to the support, where the applied tension adds vectorially to the weightof half the span. This is the reason why, when conductors fail due to excess mechanical load, theytend to do so close to the support. This is also the location where the conductor is most stressed bywind induced oscillation and the consequent metal fatigue.

9.4 Equivalent or Ruling Span - Overhead lines are seldom single span as shown in Figure xx,rather they consist of multiple spans all of them somewhat different, as the supports are located intheir best or most economical positions on the ground. This leads to the concept of Equivalent Spansometimes called Ruling Span. During conductor installation, several spans of conductor areinstalled at the same time between the fixed points at each end (the Dead Ends). By using pulleyblocks at the intermediate supports the conductor is free to move and tension in the conductor isthe same at all locations. Where the spans are unequal length and the supports are at varyingelevations, the mathematics becomes too complex for easy calculation and so some simplificationsand assumptions are made, leading to the Ruling Span Theory and the following equation:

3 3 3 31 2 3

1 2 3

......

......n

r

n

L L L LL

L L L L


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Where Lr is the Ruling Span. Clearly the sag will not be the same in all spans, but it can be

calculated for any selected span using the basic formula2

8

h

WLD

T. This allows the erection team to

obtain the correct sag and tension for the whole section between dead ends by measuring the sagin any selected span.

Example - In Figure 25 a straight line section is made up of four intermediate spans, so that theRuling Span will be:

3 3 3 3110 100 120 108112m

110 100 120 108rL

If the section is to use ‘Oak’ AAAC conductor with a bare conductor (no ice) weight of 0.325kg/mthen at a tension of 650kg (20% of UTS) the sag will be:

Sag D =2 20.325 112

0.78m8 8 650

h

WL

T

It is possible to have different equivalent or ruling spans in different sections of a single overheadline, but this is bad practice because it causes differing tensions and therefore permanent bendingloads at the points where the conductor is fixed that is, the section angles.

The following expression allows the sag Da in any actual span to be calculated from the sag D forthe ruling span:

2

a

r

SD D

S

Where Sr is the ruling span in metres and S is the actual span in metres.

For the purpose of creating sag templates, the following expression gives the sag Dx at anydistance (x) from the mid-span position:

22

1x

xD D

S

…………………………….(c)

9.5 Variation of Sag with Temperature

Conductors expand and contract with change of temperature, leading to changes of sag. Inaddition, the consequent changes in tension cause changes in mechanical stress in the conductor,which due to the metal elasticity also lead to changes in sag. These two effects must bedetermined for each combination of temperature, conductor, span and design tension using the‘Change of State Equation’. At this point, the mathematics becomes complex because cubic

Fig. 25


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expressions must be solved – hence most work in this area is carried out with the aid of computerprograms. However for completeness, a manually solved example will be given.

The change of state from change of temperature and change of tension is:

Start conductor length ± Change in length = Final conductor length

From the parabolic expression:

2 3 2 31 2 1 2 1 2

2 21 2

( ) ( )

( )24 24

W L L t t T T W LL L

aE LT T

………………..(a)

This can be rearranged to:

2 2 2 23 2 1 2

2 2 2 1 121

( ) 02424

W L aEW LT T aE t t T

T

……………..(b)

Where:

E is the modulus of elasticity of the conductor in kg/mm2

a is the conductor total cross sectional area in mm2

∝ is the coefficient of linear expansion (metre extension per metre of span per °C)

t1 and t2 are the initial and final temperatures in °C

W1 is the initial unit weight (which is the MCR if the conductor is ice and wind loaded) see section 9

in kg/m

W2 is the final weight (the no-ice weight in still air) in kg/m

L is the span length in metres

T1 and T2 are the start tension and final tensions at t1 and t2 respectively, both in kg

W1 and W2 may be obtained from maker’s literature - they will be the same in the absence of iceand wind loading. If ice and wind loading is to be taken into account, the value of W1 will includethe MCR (section 9.1). In order to find final tension T2 from a starting tension of T1, the ‘Change ofState Equation’ needs to be solved which will then allow the new sag to be determined.

The coefficient of thermal expansion of conductor metals per °C is as follows:

Aluminium Hard Drawn 23 x 10-6

Aluminium Alloy 23 x 10-6

Galvanised Steel 11.5 x 10-6

Strictly, these values are correct only at 20°C because there is a small variation of the coefficientwith temperature, which is normally neglected in calculation. It will be noted that Steel has a highercoefficient than Aluminium and this implies that with ACSR (SCA) conductors as the temperaturerises, tension is gradually transferred from the Steel to the Aluminium strands, which will eventuallycarry the entire mechanical load. Load transfer occurs at approximately 38°C for 6/1 and 26/7conductors and at approximately 75°C for 18/1 conductors.

The value of Young’s Modulus E is as follows:


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Aluminium Hard Drawn 70 x 103

Aluminium Alloy 70 x 103

Galvanised Steel 200 x 103

For ACSR (SCA) conductors, E varies with the Steel to Aluminium ratio and the manufacturershould be consulted. Because all the other values are known, it is possible to determine finaltension T2 in equation (a) by substituting values for T2 by trial and error until the equation balances– though this is a rather tedious process.

Example

Find the tension at 50°C for ‘Bee’ 125mm2 All Aluminium Conductor (AAC) erected in still air at 5°Con a 120m span under a tension of 366kg (18% of its 2033kg UTS). The modulus of elasticity E ofthe conductor is 7 x 103 kg/mm2, the coefficient of thermal expansion of Aluminium is 23x10-6 per°C and both W1 and W2 are 0.361kg/m.

Equation (b) is reproduced below

2 2 2 23 2 1 2

2 2 2 1 121

( ) 02424

W L aEW LT T aE t t T

T

Substituting values, this becomes:

2 2 3 2 23 2 3 6

2 2 2

0.361 120 125 7 10 0.361 120125 7 10 23 10 (50 5) 366 0

2424 366

x x x x xT T x x x x

x

Which simplifies to:

3 22 21050 68,418,525T T

By trial and error substitution of a value into the above equation T2 = 231kg. From this value, a newvalue of sag may be calculated.

Hence in the above example, the rise in temperature from 5°C to 50°C has caused the conductortension to fall from 366kg to 231kg. Note that the calculation adds in the temperature rise term – itwould be a subtraction if the temperature change was a fall.

The calculation must be performed for each proposed tension limit, to determine which one ‘rules’or ‘governs’ and it must also be performed many times to produce the temperature versus tensionand sag table which the construction team will need, because the actual ambient temperature atwhich the conductor will be erected cannot be known in advance. Hence the popularity of computerprograms for this task (for example SAG10 and PLS-CAD).

Newton – Raphson Iteration

The Newton – Raphson iteration offers a more methodical procedure for solving the cubic equationthan simple trial and error. It is performed as follows:

1. Set the simplified expression to zero2. Differentiate the result3. Divide the equation by its differential to obtain a value V


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4. Subtract V from the initial assumed value of T2 and repeat the process repeated until Vtends to zero

The procedure for the previous equation is as follows.

Step 1 – the equation 3 22 2 1050 68,418,525T T is reset to 3 2

2 2 1050 68,418,525 0T T

Step 2 – the differential is 22 23T 2 x 1050 x T

Step 3 – the division is3 2

2 22

2 2

1050 68,418,525

3 (2 1050 )

T TV

T x xT

Step 4 – Assume an initial value for T2 of (say) 320 and calculate the first value of V. The result isV = 73. Subtract 73 from 320 = 247.

Step 5 – Repeat Step 4 using 247 as the new value of T2 and recalculate. The result is V = 15.Subtract 15 from 247 = 232.

Step 6 Repeat Step 4 using 232 as the new value of T2 and recalculate. The result is V = 0.9.Subtract 0.9 from 232 = 231kg.

Iteration is completed when V tends to zero. In this example a result was achieved in two iterationsbut more may be needed, depending on the accuracy of the initial assumed value of T2. Theprocess could be specified in a spreadsheet.

9.6 Sag/Tension/Temperature Table

The example erection sag/tension/temperature table below, intended for use by a constructioncrew, is for a 120m basic span using ‘Dog’ 100mm2 ACSR (SCA) conductor having a UTS of32,700N (3593kg). The engineer has set the anti vibration tension limit to 20% of UTS at 5°C, thedesign ice to 20mm diameter, the design wind pressure to 380N/m2 and the freezing point limitingtension to 7188N (196kg). This table allows the appropriate conductor tension to be set for anyambient temperature, either by dynamometer (tension) or by sag (using a sag board).

Temp°C

TensionkN

Sag m for Span Length m

60 70 80 90 100 110 120

0 7.88 0.23 0.31 0.40 0.51 0.63 0.75 0.90

2 7.60 0.23 0.32 0.41 0.53 0.65 0.78 0.93

4 7.32 0.24 0.33 0.43 0.54 0.67 0.81 0.97

6 7.06 0.25 0.34 0.45 0.57 0.70 0.84 1.01

8 6.79 0.26 0.36 0.46 0.58 0.73 0.88 1.04

10 6.54 0.27 0.37 0.48 0.61 0.75 0.91 1.08

12 6.30 0.28 0.38 0.50 0.63 0.78 0.95 1.13

14 6.06 0.29 0.40 0.52 0.66 0.81 0.98 1.17

16 5.61 0.30 0.41 0.54 0.68 0.84 1.02 1.22

18 5.83 0.32 0.43 0.56 0.71 0.88 1.06 1.28

20 5.40 0.33 0.45 0.58 0.74 0.91 1.10 1.31

22 5.21 0.34 0.46 0.61 0.77 0.95 1.15 1.36

24 5.02 0.35 0.48 0.63 0.80 0.98 1.19 1.41

26 4.84 0.37 0.50 0.65 0.83 1.02 1.23 1.47


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The sag and tension equations described refer to supports on level ground although the errorintroduced by supports at different levels, up to a difference equal to the pole height is acceptable.For lines in mountainous terrain involving supports at widely different levels, the mathematicsbecomes extremely complex; for further information the student is referred to ‘Sag and Tension inMountainous Terrain, Bradbury, Kuska and Tarr, IEE Proceedings Volume 129, Part C, No.5September 1982’

9.7 Sag Templates

Once the task of determining the minimum and maximum sags for the chosen conductor at theRuling Span has been completed, sag templates are prepared from transparent plastic sheet, if thedesign process is to continue by the manual method. The necessary ground clearance ofconductors determines the height of supports and in most countries legal requirements will need tobe fulfilled in terms of clearance over open ground, clearance over roads, etc. For clearance overrailways and navigable rivers, agreement will need to be reached with the owners or the controllingauthority. Figure 26 shows a completed template, the material used is 3mm thick clear acrylic.Three curves are shown,

1. The maximum operating temperature (50°C) curve.

2. The associated minimum ground clearance curve – in the example this is 6m.

3. The minimum operating temperature (-5°C) curve.

The template is cut to curve 1, whilst curves 2 and 3 are scribed on, along with the grid lines, whichallow the template to be correctly oriented on the profile. Some engineers prefer separate hot andcold templates. The cold curve is used to check for uplift as later described.

It is customary to extend both curves outwards for some distance at both ends beyond the nominalspan for which the template is created, to allow for its use when the actual span is greater than thedesigned template value. Clearly the vertical and horizontal scales must be the same as the scalesfor the plot of the ground profile (including the normal 10 to 1 vertical exaggeration). To plot thecurve(s) expression (c) defined earlier is used.

The required support height is:

The indicated support height on the profile at minimum ground clearance distance(according to national law) + the designer’s chosen margin of safety + (in the case ofwood poles, the length in the ground) ± a small distance above or below the top of thesupport depending upon the cross arm position and the insulator height (length in thecase of suspension insulators).

For example if the indicated support height is 7.5m (on a required minimum ground clearancedistance of 5.8m), the designer’s margin of safety is 0.8m and the pole is planted 2m into theground, the support length will be 7.5 + 0.8m + 2.0m = 10.3m or erring on the safe side, a standard10.5m pole. Where extra clearance is required for example over roads, railways or navigable riversthis will have to be taken into account manually. The design process for using templates on groundprofiles is described in greater detail later in this Module. There will also be minor adjustments toallow for example for the height of insulators (only for pin or post, not tension) and the smalldistance that the crossarm is located below the pole top – these tend to cancel out. For twin threephase circuits arranged in vertical formation on either side of single supports the extra phase tophase (crossarm) separation distances will also need to be taken into account.


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Fig 26 Example of a Sag Template


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9.8 Conductor Clearance

Conductor clearance to ground and nearby objects will be affected by long term conductorextension due to creep. The following factors will also need to be allowed for and will be describedlater in this module:

Crossings above or below other power lines

Sideways movement of conductors under wind pressure (Blowout)

Crossing over or under an existing power line needs to be carefully considered. The best practice isto cross a distribution line under a transmission line close to a support position, but not so close thatthe pole is inside its overturning distance. Close to the support the conductor to ground clearancewill be close to its maximum and relatively less affected by sag changes with load and ambienttemperature (Picture 27).

9.9 Conductor Uplift

Uplift is an undesirable condition in which at low temperatures, the conductor attempts to lift fromthe insulators rather than be supported by them. Consider Figure 28.

Pic. 27 Wood pole line crossing a transmission line


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The red line indicates the hot sag and the blue line the cold sag. Support B will certainly requiretension insulators because the cold sag curve between support A and support C shows that it willotherwise be subject to the possibility that the conductors will be pulled upwards off the insulators incold conditions. At support C, the conductors exert a considerable downward pull hence lifting andbinding them onto pin or post insulators will be difficult and from ease of construction and allowedinsulator vertical load standpoints another set of tension insulators may be appropriate.

9.10 Initial Basic Span for Design Purposes

In the process of designing a particular overhead line, the proposed route over ground terrain is firstsurveyed then modelled, either on graph paper or in a computer. The positions and heights ofsupports must then be decided, a process requiring use of a sag template as earlier described inthe manual process or a modelled sag in the computer. In either method, the actual, measuredRuling Span of a section cannot be calculated until the support positions are finalised. Therefore anarbitrary value of ruling span, based upon experience, must be chosen and the support positionsdesigned to that value – this is the design Basic Span. At the end of the design process, the actualRuling Span for each section should be checked with the Basic Span used for design to ensure thatthey are not too widely separated.

A further consequence of Ruling Span is that if it varies greatly between sections of the same line,then for the same tension the sags will be different in adjacent sections of the line. It also meansthat with conductors fixed on either side, section supports are subjected to a bending load actingtowards the side with the longer Ruling Span. This argues for spans along the whole length of theline that do not differ too greatly from each other or from the design Basic Span.

Where an exceptionally long single span is required, for example across a river or a wide road theends should be sectioned off and fore and aft stays fitted, stabilising the supports so that the highertensions in the long span do not affect the tensions in adjacent spans or impose excessive bendingload on the support (Picture 29).

Fig. 28 Conductors routed up a steep slope


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The greater the difference between design Basic Span and actual as constructed Ruling Span, thegreater will be the difference between actual and predicted sag values for the same tension. If theBasic Span is less than the Ruling Span, actual sags will be less than predicted. If the Basic Spanis more than the Ruling Span, actual sags will be more than predicted, resulting in reduced groundclearance.

Table 6 Suggested values of span for Secondary transmission and Primary distributionvoltage lines supported on wood poles.

Conductor Minimum span m Basic span m Maximum spanm

50mm2 SCA Rabbit 60 100 120

100mm2 SCA Dog 60 100 120

150mm2 SCA Dingo 70 130 150

300mm2 AAC Butterfly 70 100 150

200mm2 SCA Jaguar 80 110 150

Dimensioning the Supports

The final stage in the design process is to determine the required pole diameters, based on theloading criteria for intermediate supports and for terminal and section angle supports describedearlier in the module.

10. Line Survey

Survey is carried out straight line section by straight line section, starting at the first terminalposition and working through to the final terminal position. Where survey occurs before formal legal

Pic.29 In line section with stays at a river crossing


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consents for the new line are obtained, the preliminary consent of the landowner(s) to carry out thesurvey will be required and appropriate care taken with loose livestock. Survey over railway land willrequire the paid attendance of a safety lookout from the railway company. In difficult country, morethan one possible route may be surveyed before a final choice is made.

Wooden pegs driven into the ground are used to define the terminal and line angle positions and itis best if these are sited away from agricultural operations – if this is not possible, accurate andrepeatable tying in measurements from fixed points will be required so that the pegs can beprecisely positioned on a temporary basis as and when required. In order to keep the survey on thecorrect line, a number of temporary ‘ranging rods’ are set up – in a long section, several will berequired. These rods may be positioned using a theodolite set up directly over an angle or terminalpeg, with its sights aligned on a rod held over the next angle or terminal peg. Using portable radios,an assistant then places as many intermediate rods as may be required in the correct straight lineposition(s) as directed by the theodolite operator.

The theodolite method works well in open country but less well where tall vegetation such ashedges, trees and shrubs obscure the view. Hilly terrain can also present problems. In these casesthe method of lining in using ranging rods as shown in Figure 30 may be used. The procedure is forthe two ranging rod holders to move side to side until each has his own ranging rod, the ranging rodof his colleague and the ranging rods over the terminal or angle pegs in line. Ranging rods capableof being fitted with multiple extensions are particularly useful although care will need to be takennear existing power lines. Fibreglass rods that are inherently insulating are preferable to metal androds of this type may also be used to safely measure the conductor height of any existing powerlines along the proposed route – this information will certainly be required.

Once the temporary alignment rods are in position, the survey proper of the section may begin. Thesurvey procedure described here is the so called ‘offset’ method. This is a three man task andrequires a measuring chain, a measuring staff and an optical survey instrument. Two portableradios are helpful. The survey instrument may vary from a complex theodolite with a data loggingfacility down to a simple dumpy level – which is entirely satisfactory for most projects. Theprocedure for an offset survey using a dumpy level is described as follows:

Fig. 30 Lining in a straight line section using ranging rods


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Step 1 – The level is set up at a position some distance to the side of the route from which theoperator can see the peg marking one end of the section and some distance along the route. Thestart level is given an arbitrary value (say 100m).

Step 2 – The measuring staff is placed by the first peg and the first reading taken.

Step 3 – The measuring chain is laid out along the route with one end at the first peg.

Step 4 – The measuring staff is moved along the chain and the second and subsequent readingstaken which should be as frequently as required, normally at least every 50m and certainly at everycrossed feature such as field boundaries, ditches, road edges and centres. As the surveyprogresses, the measuring chain is moved along the route using the temporary ranging rods as aguide. Where the route passes close to fixed objects such as buildings, street lamps, etc, offsetdistances and heights should be measured so that safety clearance can be reliably determinedwhen setting out the route profile. Similarly, trees that are likely to need felling should also be noted.Care will need to be exercised near existing overhead power lines, whose positions should also benoted.

Step 5 – As the measuring staff moves further from its start position, a point will be reached whereeither 1. The measurement is moving too close to the upper end of the measuring staff (fallingground) or 2. The measurement is moving too close to the lower end of the measuring staff (risingground) or 3. The distance from staff to instrument is becoming excessive. When this occurs achange point must be implemented; to do this, the staff is held in position whilst the level is moved,set up and a new reading taken. The difference in reading must be taken into account in the surveybook.

As the survey proceeds, the survey book is filled out with the distances and instrument readingsfrom which the route profile will eventually be constructed – the table below represents a sample,the red values are the calculated ground level, which is best worked out back in the office.

Fig. 31 Survey using a measuring chain, measuring staff and dumpy level


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Survey Book for Route_________________ Section_______________________________

Position m Staff Readingm

Calculatedlevel m

Comment

Start terminal peg 1.3m fromwest field boundary 2.2mfrom south field boundary incorner by S/S

1.2 100 Line terminal, 100m initial levelis nominal. First section,proceeding north east

50 0.9 100.3 In pasture land



170 0.4 100.8 In hedge line

210 0.3/1.4 101.1 Change level position. In arableland

250 1.1 101.4 In arable land


335 0.6 101.9 Low voltage line to cross





513 1.0 101.5 Large Ash tree 3m right of routeto be felled


591 1.4 101.2 Minor road boundary hedge

597 1.4 101.2 Edge of road

604 1.2 101.4 Road centre

609 1.4 101.2 Edge of road

612 1.5 101.1 Minor road boundary hedge

650 1.7 100.9 Fence line post and wire

700 1.7/0.3 100.9 Change level position. Roughpasture land

750 0.7 100.5 1m drainage ditch water level

800 1.1 100.1 Rough pasture land


900 1.8/0.3 99.4 Change level position. In Roughpasture land



1046 line section peg 1.7 98.6 Section end turn right 34°


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11. Route Selection

For Primary distribution overhead lines, route selection is rarely an issue because the route isdetermined by the location of the rural loads. Similarly, obtaining consents for the proposed route isnormally free of problems as the owner of the land is, in most cases, also the customer for the load.

Only for Primary and Secondary transmission lines does route selection require seriousconsideration and planning. The following guidelines should be followed:

(1) Areas of high amenity value should be avoided completely, even if this increases the routelength. Sites of Special Scientific Interest (SSSIs) and public parks should definitely be avoidedas consent will be almost impossible to obtain.

(2) A balance should be sought between straight line sections comprising intermediate supportswhich are less obtrusive and line angles with stays, which though more visible are the strongestpoints in a route and will better resist severe weather conditions. Section angles should notexceed 45° unless there are compelling reasons.

(3) Choose tree or hill backgrounds rather than sky backgrounds. When the line must cross a ridge,secure the opaque background as long as possible and cross at an oblique angle when a dip inthe ridge provides an opportunity.

(4) Where possible, route along open valleys with trees where the apparent height of supports willbe reduced.

(5) Routing through woodland will require cutting a wide clear corridor which will be visually veryintrusive.

(6) In flat country try to avoid other overhead lines so as to avoid a concentrated ‘wirescape’.

(7) Approach urban areas through industrial zones where possible and avoid residential areas –residents may raise concerns over electromagnetic fields which are best avoided.

(8) Be aware that where overhead lines run parallel to telecommunication circuits or railways, thetelecommunication circuits (including railway telecommunications) and even the railsthemselves may be adversely affected by electromagnetic induction under short circuitconditions, especially earth fault conditions. This can require costly remedial action.

(9) Cross under any existing tower line close to a support position as described earlier in thisModule.

(10) Where crossing motorways, major railways and navigable rivers consider sectioning off theline on either side. A greater conductor Factors of Safety may be appropriate in thesecircumstances.

(11) The chosen route should give due regard to ease of access for initial construction and forsubsequent maintenance.

12. Profiling

After the route of the overhead line has been decided, permissions to erect the line across privateland must be negotiated and agreed, which will involve payment, either in the form of an annualrental or as a lump sum for permanent permission. In the UK, Primary distribution and Secondarytransmission lines are normally agreed on the basis of an annual rental (a Wayleave) whereasPrimary transmission lines are agreed on a lump sum payment for the right to use the route in


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perpetuity (a Permanent Easement or Right of Way). The total cost whether annual payment orlump sum will be calculated from:

1. The number and position of supports (costs may be greater if the support is located in anagricultural field rather than at the boundary).

2. The number and position of stays – with then same cost considerations as 1.

3. Sometimes payment may also be made for buried cables in private land and even whereconductors only cross private land (wires only).

In many countries including the UK, the formal consent of a government agency may also berequired.

The first task in profiling the proposed line is to position the intermediate supports in each section,starting with minimum height supports. Bear in mind that span lengths between intermediatesupports should not deviate too greatly from the Ruling Span of the section (Table 6 suggests arange between +20% and -40%).

It is also good practice, if possible, to position supports and any necessary stays at land boundariesso reducing disturbance to agricultural operations. Once the support positions have been decided, asag template for the maximum sag, maximum temperature condition is applied to determine thesupport heights (Figure 32). Here the vertical distances L1, L2, etc. can be scaled off the graph ormeasured on screen and the required support heights obtained by adding in the necessary groundclearance. It is unwise to design to minimum ground clearance; an additional margin of 8-10%should always be allowed. Wood pole overhead line construction is not high precision civilengineering and the pole planting depth achieved in the field may be plus or minus 0.2m of thatspecified on the construction schedule.

Once the support positions and heights have been determined, the proposed profile must bechecked for any possible Uplift condition. Uplift was briefly mentioned earlier and occurs whenconductors attempt to lift off an intermediate support at minimum temperature and minimum sag, asshown in Figure 33.

Fig. 32 Using a sag template


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In Figure 33, the conductor is above the height of the centre support at the minimum temperature,minimum sag condition, hence it will be attempting to pull vertically upward from the insulators. Toprevent the conductor detaching from the insulators at low temperature, either the outer supportsmust be repositioned, or the centre support increased in height, or an appropriate combination ofboth. Alternatively and more expensively, the centre support may be fitted with tension insulators(sectioned off). Every set of supports along a straight section of line will need to be checked foruplift, other than where the topography is flat or nearly so. Every movement or change in height of asupport affects the adjacent spans, so that the optimum positioning of supports to obtain thenecessary ground clearance, consistent with minimum support lengths and therefore cost togetherwith the elimination of uplift is a lengthy process. Over the past 20 years computer programs haveautomated the design process notably those produced by the Optimal Company.

Table 7 UK overhead line clearances

Description of clearance Minimum clearance metres

LowVoltage

11kV 33kV 132kV

At any point not over a road 5.2 5.8 5.8 6.7

To a road surface 5.8 5.8 5.8 6.7

Line conductor to road surface ofdesignated 6.1m high load route

6.9 6.9 7.1 7.8

Over motorway 8.2 8.2 8.2 8.8

The values given in Table 7 are provided only as a basic guide, safety clearances vary according toeach country’s national recommendations or legislation, which must be consulted and compliedwith. Many other safety clearances will also need to be observed including clearances to buildings,structures (such as bridges) and not least, other power lines as previously mentioned.

Erecting an overhead line effectively sterilises a wide strip of land on either side of the route inwhich future development is either prohibited or at least, severely restricted. This represents a costto the landowner for which the electricity distribution company may have to pay compensation.

Figure 30 shows conductor blowout on a Primary distribution (wood pole) and Primary transmission(steel tower) lines.

Fig. 33 Showing the effect of conductor uplift


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Transmission lines with long suspension insulators and much greater conductor sag present thegreater blowout problem but all types of power line need to be checked for this condition at thedesign stage and also when new development is proposed in the vicinity of existing lines. It will beappreciated that erecting overhead lines especially large tower lines prevents or at least severelyinhibits future development in the strip of land below the conductors – the land owner may well haveto be compensated for this loss of development potential.

Where an overhead line passes though forested land, major tree clearance operations will berequired on either side of the route, over a width equivalent to at least the overturning distance ofthe supports – this cleared route will need to be maintained free of tree re-growth over the lifetimeof the line. The initial and lifetime costs are large and this suggests selecting routes that avoidforested land if at all possible.

Fig. 34 Conductor blowout


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13. Insulators

Where the insulator is in tension, the cap and pin type is used. The design dates from the early partof the twentieth century and is a means of overcoming the poor tensile strength of the insulatingmaterial by placing it in compression, where the strength is improved by a factor of 10.

The underside of the insulator has deep concentric rings to increase the creepage distance andthereby increase the resistance to flashover under polluting atmospheric conditions. Cap and pininsulators may be porcelain or glass and may be used in strings; the higher the voltage the longerthe string, each unit having an operating voltage of approximately 10kV.

If porcelain, the applied glaze may be any desired colour but is usually red-brown or occasionallygrey. Glass insulators have the advantage that they resist the melting effect of power arcs ratherbetter than porcelain; the glass is toughened by blowing cold air over it as it solidifies, forming atension layer in the surface. This improves the impact strength but damaged glass insulators tend toshatter completely whereas porcelain insulators will often merely chip – an advantage whereinsulator shooting is a local sport. Even where a disc is completely broken away, a cap and pininsulator will normally resist flashover in dry conditions but will fail at the first shower of rain.

There are limits to the permissible applied voltage and tension and manufacturer’s advice should besought and a suitable Factor of Safety applied. Over long strings used in transmission linesuniformity of voltage gradient can be a problem, but not at distribution voltages.

Where a rigid support insulator is required, the Pin or Post (sometimes referred to as Lapp) typesshown in Figure 32 may be used. These are generally brown glazed porcelain, although othercolours are available and polymeric versions of the post insulator are marketed, having beenoriginally developed for railway applications.

The pin type is available made either as a single moulding or multi-part; it is used up to 11kV and isdesigned so that the spark over voltage, across the rain sheds, is lower than the puncture voltage.The post type is used up to 33kV; specials are available for Trident 132kV lines.

Fig. 35 Section through cap and pininsulator


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Conductors may be bound into the transverse groove at the top or into the circular groove below.Pin and post insulators have a limited shear stress capability, allowing sideways loads of only afew hundred Newtons and therefore very limited conductor deviation angles; manufacturer’sliterature should be consulted and a suitable Factor of Safety applied.

14. Lightning and Overhead Lines

As previously described, overhead lines are subject to many adverse environmental factors ofwhich lightning is one of the most important. Lightning particularly affects overhead lines supportedon wood poles because the shielding effect of an over-running earth conductor, present in alloverhead lines with conducting supports, is absent.

Lightning strikes are not all the same, being divided into two main categories, positive and negative– the latter being more common. Most strikes have currents in the range 10 to 30kA althoughindividual strikes can have currents up to 80kA. In energy terms, the positive strike has a slowerrise time and a longer decay time so that the total energy delivered, measured in Coulombs, ismuch greater, perhaps as much as 20 times greater. Strikes do not need to connect directly with aconductor to create a system over voltage indeed most over voltages are induced, arising fromstrikes to nearby objects or to adjacent ground up to a distance of 500m. Direct strikes deliver thegreatest energy but induced over voltages from nearby strikes are much more frequent, due to thelarge collection area. The voltage generated by the current in the lightning strike depends upon theSurge Impedance sometimes called the Characteristic Impedance of the line. This can bedetermined as follows:

Consider a voltage impulse with an infinitely steep front and an infinitely long tail. This wavefrontwill travel along the line at a uniform velocity of x km/s and (ignoring the effects of conductorresistance) as it does so the line will be charged to E Volts at the same rate. In any second thelength of line charged is x kilometres. If the line capacitance is C Farads/km then the chargerequired is ECx Coulombs.

Fig. 36 Pin (left) and Post (right) insulators


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Rate of chargeCoulombs

Surge Current Amps............(1)Second

ECx I

Let the line have an inductance of L Henrys/km, then a length x km has inductance Lx Henry. Bydefinition, the current in this length is changed from zero to I Amperes in 1 second and the voltagerequired to overcome the back EMF generated by the inductance is Inductance x Rate of Changeof Current = x VoltsLx I . If conductor resistance is ignored

= Volts.......................(2)LxI E

Substituting equation (2) into (1) gives the speed of charging and hence the speed of thewavefront:

1km/sec

( )x

LC ……………(3)

Dividing equation (2) by equation (1) gives:

E LxI

I ECx and therefore

E L

I C …….(4)

The quantityE

Iis the surge impedance Zo of the overhead line and substituting typical values of

inductance and capacitance for wood pole overhead lines yields a value between 300 and 500Ω,

depending upon the conductor spacing. For a concentric cable with an insulation r of 4, equation

(4) gives a surge impedance of approximately 50Ω. Substituting typical values into equation (3)always gives the speed of light, 3 x 108 metres/sec.

If wood pole lines have a surge impedance in the range 300 to 500Ω, then taking an average valueof 400Ω a direct 20kA strike will generate an over voltage of 8 million volts, far above the basicinsulation level (BIL) of the line insulators, etc. which for an 11kV line is normally 95kV. The overvoltage must be dissipated somewhere and that somewhere is typically:

1. Over the insulators2. Over the insulator arc gaps3. In the non linear resistors (surge arresters) fitted for this purpose4. In the resistance of the conductor

Direct strikes to conductor resulting in megavolt surges usually result in equipment damage unlessprotective equipment (surge arresters) is fitted nearby. Whether it is economic to do so will dependupon many factors, including the economic importance of supply reliability, which will vary fromcircuit to circuit. For example a set of surge arresters fitted at every support would reduce lightningdamage appreciably, but this policy is completely uneconomic. Note that lightning strikes may be‘attracted’ to the metallic conductors of an overhead line, which might otherwise have struck theground or a nearby object – as a rule of thumb, strikes will become direct if they would have struckthe ground within a horizontal distance equal to twice the line height. On this basis, the number ofstrikes per year N as

0.4 gN N h


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Where gN the strike density per km2 per year for the area (figures are published for countries and

regions) and h is the line height. In practice and as would be expected, strikes are more frequentfor lines placed at the tops of hills or along ridges than they are for lines at the valley bottom, asmuch as three times as many.

Better measures of the attraction effect and the level of induced voltage from nearby strikes aredescribed by Ruscks equations; they generally predict induced voltages less than 300kV which aremore easily dealt with.

As well as the Basic Insulation Level (BIL) of a line, the Critical Flashover voltage (CFO) is alsoimportant, being the crest voltage value that causes flashover on a specified percentage ofoccasions. Actual CFO depends upon many factors including support geometry but for wood polelines an approximate value is 250kV per metre of pole plus the value of the insulator CFO. Thisvalue will be reduced if the pole is rotten or wet or if earthed equipment (stays, transformers, etc.)is at high level. It will also be greatly reduced if the crossarm is earthed. Insulator flashover maynot always occur at the support closest to the strike; it may well be at the next earthed crossarmwhere the BIL and CFO are lower. If flashover occurs a power arc can follow which may extend toinclude not only the insulator but possibly a pole mounted transformer as well.

In practice, lightning surges are dissipated fairly rapidly with distance however damage is likely ifthe direct or indirect strike is within 75m of vulnerable equipment.

14.1 Use of Arc (Horn) Gaps

Arc gaps are the traditional method of protecting equipment from voltage surge because they arereasonably effective and low cost. The gap is wide enough to prevent flashover at the systemoperating voltage but narrow enough to spark-over on the arrival of a voltage surge. For optimumperformance, the gap must be positioned as close as possible to the equipment it protects. AtPrimary Distribution voltage, duplex gaps may be fitted because otherwise the gap can be smallenough to be bridged by a bird. Often duplex gaps will be arranged as two sections on oppositesides of the insulator.

Pic. 37Arc gaps on an open terminal transformer


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In arc gaps, flash over on the arrival of the voltage surge is not instantaneous but requires adefinite time period, depending upon several factors:

1. The molecules of solid insulation material need time to rearrange themselves readyfor conduction; in air the molecules need time to ionise.

2. The shape and spacing of arc gap conductors also has an effect. For examplesphere gaps where the gap is small relative to the diameter break down almostinstantly but needle gaps take some time because spherical corona dischargemust be first established.

3. The amplitude of the over-voltage, large values flashing over more rapidly.

The above factors imply that different parts of the network have different breakdown voltages. Theratio

Breakdown voltage at surge frequency

Breakdown voltage at power frequency=

p

w

U

Uis known as the Impulse Ratio.

The impulse ratio for suspension insulators is 1.2 to 1.6 and that for pin insulators 1.5 to 2.3. Theratio for an arc gap should be as close to 1 as possible; note that the shape of an arc gap tends toextinguish the arc by thermal and magnetic effects. Paschen’s Law describes the breakdownvoltage of an arc gap in a gas and is expressed as

( )

ln( )

a pdV

pd b

where p id the pressure in atmospheres, d = gap distance in metres and a and b are constants a =43.66 and b = 12.8 for air at standard temperature and pressure. Table 8 below specifies the twingap distances used a by utility and includes an extra percentage of the calculated distance as asafety measure.

Nominal systemvoltage kV

Basic Insulation Levelof system (BIL)

Arc gap distance mm

11 95 2x32

33 200 2x95

Arc gaps are slow to operate and because the ionised gas takes time to disperse, they can continueto arc under normal supply voltage, even after the voltage surge has passed, often resulting inburning or even melting of the insulator surface. To reduce this type of damage the arc gap shouldbe separated from the surface of the insulator by at least one third of the arc gap distance (21mm@ 11kV, 63mm @ 33kV in the table examples). The delay in operation also means that theequipment they are intended to protect may be damaged.

The problem of continued arcing at power frequency causes a high rate of erosion in operation. For11kV networks the rate can be as high as 35mm a second, so that on the next impulse, the gap istoo great and the protection provided is reduced. This is an inherent problem with arc gapprotection. To some extent, the erosion problem can be reduced by the use of crossed rod gaps –Figure 38 shows normal twin gap and twin crossed rod gaps with distances suitable for 11kV use.


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14.2 Use of Non-Linear Resistors (Surge Arresters or Surge Diverters)

Surge arresters offer effective protection against both switching surges and lightning surgeshowever switching surges are important only on transmission circuits over 132kV. Several typesand classes of surge arrester are in use, as follows:

Distribution class (normal and heavy duty) Intermediate class Station class

The initial selection of a surge arrester is made on the basis of cost versus performance, both ofwhich increase progressively down the list. For overhead line purposes, distribution andintermediate class arresters are considered appropriate – station class devices being generallyused to protect high value substation equipment such as circuit breakers and transformers. Table 9below shows lower voltage part of a range of arresters available from a single manufacturer.

Distribution Class kV Intermediate Class kV Station Class kV

1 3 3

3 6 6

6 9 9

9 Not made Not made

12 12 12

15 15 15

18 Not made Not made

21 21 21

25 24 24

27 Not made Not made

30 30 30

Not made 36 36

Fig. 38 Twin gap arc gaps, normal; and crossed rod types


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Distribution class arresters, normal and heavy duty, are the lowest cost, have the highestdischarge voltage and lack any pressure relief feature.

Intermediate class arresters are mid range in cost, have improved protection characteristics, aremore durable and include a pressure relief feature.

Station class arresters are highest in cost, have the best protection characteristics and durability,the greatest energy dissipating capability and include a pressure relief feature.

Within these voltage ratings the user is also able to choose between current ratings of 5, 10 and20kA; for primary distribution circuits 5kA may be used although 10kA tends to be the more popularchoice and this rating may also be selected for secondary transmission voltage. 20kA rated unitsmay be reserved for duty within substations. Due to the unpredictable nature of lightning strikes,especially their current magnitude, random arrester failures may be expected – some designsincorporate a lower earth lead that detaches on failure, allowing identification of units requiringreplacement from ground level.

In the table, the voltage rating is referred to as Uc, the Maximum Continuous Operating Voltage(MCOV). This is not the nominal system voltage but the phase to neutral (earth) voltage and up to5% over voltage is normally allowed. This is the main selection criterion. For example at 11kV thephase to neutral voltage is 6.35kV and a 9kV rated arrester should be selected (a 6kV ratedarrester at 6 x 1.05 = 6.3kV is inadequately rated). Surge arresters that have a higher thanappropriate rating should not be selected; this is because under lightning surge conditions they holdthe voltage surge to a level of approximately 3 times the nominal rating – this is referred to as Ures .Hence the higher the MCOV, the closer Ures is to the system Basic Insulation Level (BIL) and theworse the level of protection. There is another caution here - the selection criterion described issatisfactory for systems that are earthed through a neutral earthing resistor or directly (solidly)earthed but for systems where the neutral is not earthed, the phase to neutral (earth) voltage canrise to the full phase to phase value.

Two forms of arrester are in use, the older Silicon Carbide and the modern Zinc Oxide which is nowthe preferred type. The former comprises a stack of discs within a ribbed porcelain body, completewith a small spark gap. The gap prevents the small leakage current that would otherwise flowduring normal service. On the arrival of a voltage surge the gap flashes over and because the stackresistance is low, it allows heavy currents to be passed. After the surge has passed, resistancereturns to its normal value. The small gap in this type of arrester has the same delaying effect onoperation as does the gap in the arc (horn) gap. Figure 39 compares the characteristics of the twotypes of arrester.

Fig. 39Characteristics of surge arresters


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Zinc Oxide surge arresters, specified by IEC 600099-4, also comprise a stack of discs inside ahousing, which may be porcelain or polymeric. This type of arrester has a different VI characteristicallowing it to operate without a series gap, with improved speed of operation, however thereremains a steady if small leakage current.

Whether Silicon Carbide or Zinc Oxide, porcelain housings are not nowadays favoured, because ifthey shatter on failure the small pieces can fall to the ground and be ingested by farm animals.

In surge arrester standards the rating is proven by the duty cycle test which for Silicon Carbidetypes comprises the maximum voltage which can be applied to the arrester and still discharge itsrated current. In the test the arrester is energised at the specified voltage and a current of definedmagnitude is applied once a minute for 22 or 24 minutes depending on arrester type (SiliconCarbide or Zinc Oxide). No arrester should be applied to a system where the applied voltage couldexceed its rated voltage – in the Silicon Carbide type this would prevent the gaps re-sealing and theunit failing. In Zinc Oxide arresters there are no gaps so the factor determining rating is thermalheating, which is defined by the temporary over voltage curve (Figure 40).

The curve defines the maximum time an over voltage condition can exist whilst the arrester stillcontinues to perform satisfactorily. In the event of an arrester failure, a pressure relief vent shouldoperate, preventing catastrophic failure of the arrester body and the consequent scattering of smallpieces over agricultural land.

Like arc gaps, surge arresters should be sited as close as possible to the equipment they protect(Picture 41). It is important to keep the lead length to a minimum – for larger GIS installations thiscan include siting within the SF6 filled enclosure.

Fig. 40


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Arresters are more effective but more expensive than arc gaps, hence their use tends to berestricted to the protection of costlier equipment such as switchgear, transformers and cabledsections inset into overhead lines. Figure 42 compares the operating time of gaps and arresterswith the rise time of a typical lightning surge.

Pic.41 Surge arresters on a 33kV line termination – the post insulators at thefront are to mechanically stabilise the heat shrink cable terminations

Fig.42 Comparison of arc gap andarrester characteristics


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Protection of cables by surge arresters – Whatever the length of cable, due to their low BIL theonly certain means of protection against lightning surges is to fit arresters at both ends (in the caseof cables inset into overhead lines). For cables leading from an overhead line termination to asubstation sited disconnector, circuit breaker or transformer, arresters should be fitted close to theline and cable transition point. In the case of very short cables inset into overhead lines it is possibleto achieve a reasonable degree of protection using arresters at only one end. An empiricalexpression from a paper published by CIRED for assessing the risk level Cf, the number offlashovers per year, is:

50.3 10L p

f c d

c p

U UC x xL xN x

U U

Where UL is the BIL (Basic Insulation Level) of the overhead line, Uc is the BIL of the cable, UP isthe voltage protection level of the arrester, Lc is the cable length and Nd is the number of lightingstrikes per 100km of line per year.

14.3 Effects of Lightning on Pole Transformers

Lightning caused pole transformer failures are, or at least were, a major problem in utilities and itwas not unusual in past times for a dozen or more transformers to fail in the course of a singleheavy storm, exhausting transformer stocks and overwhelming the repair teams.

Figure 43 shows the earthing arrangements for a pole transformer; unlike ground mounteddistribution substations the HV earth (the steelwork earth) and the neutral earth are separated toavoid the possibility of the ground rise of potential during an HV side earth fault being transferred tothe customer’s neutral conductor.

Several mechanisms of failure due to lightning strikes are possible. The first possibility in the caseof a direct lightning strike is that the over voltage surge arrives on one phase only. If only an arc gapis fitted to the bushing, its tardy operation will allow at least some of the surge into the HV sidewindings. Because the impedance of the windings is high at the (effectively) high frequency of thesurge, it is attenuated over only a small part of the winding, leading to excessive inter-turn voltagesand a consequent inter-turn insulation failure. Even if this failure is avoided, there is still the

Fig. 43 Separated earths at a pole transformer showing ground rise ofpotential during HV side earth fault


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possibility of mechanical damage to the core and windings from the very strong electromagneticforces generated by the current surge. This can result in the physical displacement of the core orsevering of the connections.

If the incoming surge affects all three phases (which is likely if it is an induced surge) then thevoltage has a slower rise time giving a better opportunity for the arc gaps to operate correctly.However the surge passes through the arc gaps to the transformer tank and onwards to thesteelwork earth, which may have a typical value of 10Ω. If the surge current is 2kA, the tank will riseto 200kV above true earth potential, which voltage will also appear across the tank to LV neutralconnection, because the LV neutral is always separately earthed to protect customers from HV sidevoltage surges (unless the value of the combined steelwork and neutral earth connection is verylow, less than 1Ω, which is seldom the case). This voltage can be transformed up to cause similardamage as would occur for the fault scenario described earlier.

Considering the above, it will be seen that more effective protection can be obtained using surgearresters rather than arc gaps, three (for 3 phase units) connected across the bushings tosupplement the arc gaps and a fourth arrester positioned between the LV neutral and the tank.Figure 44 shows this arrangement; there are of course four arresters, one for each bushing and onefor the neutral to tank position.

This policy has been proven to considerably reduce transformer faults due to lightning.

Fig.44 Surge arrester protecting a poletransformer (neutral to tank arrester not

shown)


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Further Reading

Students wishing to undertake further study of overhead lines may find the following books useful:

1. Overhead Line Practice by John McCombe and F. R. Haigh, published by MacDonald.This book is out of print but copies may sometimes be found on the used market. Wellworth purchasing if a copy can be located.

2. Wood Pole Overhead Lines ISBN 0 86341 356 0 by Dr. Brian Wareing, published bythe IEE.

3. Overhead Power Lines: Planning, Design, Construction (Power Systems) by FriedrichKiessling Peter Nefzger, Joao F. Nolasco and Ulf Kaintzyk. Published by Springer andfocussed mainly on transmission voltage lines.


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Appendix A – BS1990 Pole Sizes

Length m Grade Minimum Diameter1.5m from Butt End

mm

Minimum Diameterat Top mm

8 Light 170 125

8 Medium 215 150

8 Stout 265 190

9 Light 180 125

9 Medium 242 150

9 Stout 275 190

10 Light 185 125

10 Medium 230 150

10 Stout 285 190

11 Light 195 125

11 Medium 240 150

11 Stout 295 190


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Notes on Student’s answers to questions

Marks will be awarded for presentation and answers should be typed in WORD. Students mayconvert their WORD .doc into .pdf to reduce the file size if preferred. In the mathematical questions,students should describe and comment upon the practical implications of their answers.

Mathematical equations should be created using the WORD equation editor. If this is not availableon your PC, go to TOOLS - CUSTOMISE then select INSERT. Scroll down the right hand selectionuntil you see EQUATION EDITOR then drag and drop the symbol onto any place in the tool bar.You will then have the necessary mathematical commands available for use.

Questions on Module 3

Question 1 The map on the following page (you may request a copy of this illustration in jpgformat, please contact the course tutor) shows a geographical area crossed by a Primarytransmission line (shown in black), and a Primary distribution line (shown in red). Rivers are shownin blue, woodland in green, paved roads in dark grey and land contours at 25m intervals inmagenta. Select a route for a Secondary transmission line from Substation A in the North West toSubstation B in the South East. Give reasons for your choice of route, which should, as far aspossible, place supports in the land boundaries. The design (basic) span is to be 110m. Calculatethe ruling span for each section of your line and compare it with the basic span.

Calculate the cost of your design from the following:

Terminal support complete £890 ($1250)Single support section angle up to 30° complete £514 ($720)‘H’ support section angle 30° to 55° complete £742 ($1040)Intermediate support complete £370 ($520)Conductor 3 phase per route metre (3 conductors complete) £8 ($11)

Question 2 Determine the I2t current rating of 80mm² ‘Grasshopper’ conductor 7/3.9mm AAAChaving a resistance at 20°C of 0.3406Ω/km and a weight of 230 kg/km. The initial temperature is15°C, the allowable maximum temperature is 200°C and the coefficient of electrical resistance withtemperature is 0.0036 per °C.

Question 3 Determine the current rating of 200mm² ‘Chafer’ 19/3.78 AAAC having a resistance at20°C of 0.1349Ω/km and a diameter of 18.9mm. The ambient temperature is 35°C, wind speed0.447m/s and the emissivity/absorption coefficient is 0.9. The operating temperature is 75°C, solarradiation is 1200W/m² and the coefficient of electrical resistance with temperature is 0.0036 per °C.

Question 4 Determine the stay tension and pole compression load for a 32° angle deviation in athree phase, three wire overhead line whose conductor is ‘Tiger’ 125mm² SCA, weight 603kg/km,16.5mm diameter tensioned at 9.5kN. The adjacent spans are 110m and 95m and the wind load is380N/m². The stay angle is 450


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module 3 overhead lines version e - transtutors · 2017-12-14 · mechanical design of overhead...

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