Electrical Transmission and Distribution System and Design

Electrical Transmission and Distribution System and Design Power System System Structure Generation, Transmission and Distribution systems are the main components of an electric power system. Generatin Gen eratingg stations and and distribution systems systems are connected through transmission lines. Normally, transmission lines implies the bulk transfer of power by high – voltage links Figure 1 Power 1 Power System Diagram between main load centers. On the other hand, distribution system is mainly responsible for the conveyance of this power to the consumers by means of lower voltage networks. Electric power is generated in the range of 11 kV to 25 kV, kV, which whic h is is increase by stepped up transformers ot the main transmissio transm issionn voltage. At At sub - stations, the c onnect onne ctio ionn between various various compone co mponent ntss are made, for example, examp le, lines lines and transformers and switching of these components is carried out. Transmission level voltages are in the range of 66 kV to 400 kV (or higher). Large amounts of power are transmitted from the generating Figure 2 Power 2 Power System Structure stations to the load centers cen ters at 220 kV or higher. The power supply network can be divided int two parts, transmission and distribution d istribution systems. systems. T he transmissio transmissionn system may be dividided into primary and secondary distribution system. Most of the distribution networks operate radially for less short circuit current and better protective coordinatio coordina tion. n.

Figure 3 Components 3 Components of Power System

Figure 4 Transmission 4 Transmission System

Distribution networks are different than transmission networks in many ways, quite apart from voltage magnitude. The general structure or topology of the distribution is different and the number of branches and sources is m uch higher. A typic typical al distribution system system consists con sists of a step – down transformer at a bu bulk lk supply point feeding a num ber of lines with wi th vary varying ing length from from a few hundred of meters to several kilometers. Several three – phase step – down transformes are spaced along these feeders and from these, three – phase pha se four – wire networks of c onsumers are supplied suppl ied which gives 230 V, V, single singl e phase phase supply to houses and simular loads.

Figure 5 Distribution 5 Distribution System

A com plete circ uit diagram of a power powe r system for all the three – phases phase s is very c ompli om plicated. cated. It is very very muc m uchh practica practicall to represent power system system using simple sim ple symbol symbolss – for each component resulting in what is called a Single Line Diagram.. The single line diagram of a power system Diagram network shows the m ain connections c onnections and arrangement of the system components along with their data (such as output rating, voltage, resistance and reactance, etc.). In case of transmission lines sometimes sometime s the conduc con ductor tor size size and spacings spac ings are given. It is i s not necessary tos tos how all the components of the system on a single line diagram, e.g.

Electrical Enginee ring Department | Engr. Engr. Gerard Francesco Francesco DG. Apoli nario

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Electrical Transmission and Distribution System and Design circui ci rcuitt breakers breakers need not be shown in a load loa d flow study but are must for protection protec tion study. study. In a single si ngle line li ne diagram, the the system system c omponent ompon entss are usually drawn in the form of their symbols. Generators and transformer connections – star, delta and neutral earthing are indicated indi cated by symbol symbolss drawn by the side of the representation of these these elements. eleme nts. Circui Circ uit breakers are represented by rectangular blocks.

(input line voltage voltage must be high so that the losses will will decrease dec rease and the area of conductor cond uctor will also al so decreas decrease)

Figure 6 Single 6 Single Line Diagram of Power System

Power system engineers have devised the per – unit system such that different physical quatities such as current, voltage, voltage, power po wer and impedanc im pedanc e are expressed expressed as a decimal fraction or multiple of base quantities. In this system, system, the different voltage voltage levels l evels disapper and a power po wer network c onsisting synchronous generators, transformers transformers and lines reduc es to a system system of simple impedances. im pedances. Major Maj or Sources of Elec Electrical trical Energy 1. The Sun 2. T he Wind Wind 3. Water 4. Fuels a. Fossil Fuels i. Coal ii. Oil iii. Natural Gas b. Nuclear Nuc lear Energy 5. T hermal and Geothermal Energy Importance of Electrical Electric al Energy 1. Convenient Form – it can be easily converted into other forms of energy such as li ght, heat, mec hanica hanicall energy, etc. (with the help of T URBINES) 2. Easy Control – the electrically operated machines have simple simpl e and convenient c onvenient starting, starting, control con trol and operation. (switches) 3. Greater Flexibility Flexibility – it can ca n be easily easil y transported from from one place to another with the help of conductors. (ASCR – Aluminum Alumin um Conductor Steel Reinforced) 4. Cheapness – it is overall ec onomical onom ical to use use this form form of energy for domestic, commercial and industrial purposes. 5. Cleanliness – electrical electric al energy is not associated associ ated wit with smoke, fumes, and poisonous poisonou s gases. 6. High Hig h Transmissi Tra nsmission on Efficiency – the electric elec trical al energy energy can be transmitted transmi tted conveniently conveniently and efficiently from from the center cen ter of generation to the the c onsumer onsume r with the the help of overhead conductors known as transmission lines.

Figure 7 How 7 How Electricity is transfer?

(1) Elec tricity leaves power plant, (2) its voltage is i s inc reas reasee at a step – up transformer, transformer, (3) the elec tricity trav travels els along along the transmission line to the area where the power is needed, (4) there, in the substation, the voltage is decreased dec reased with the help of a step – down transformer, (5) again, transmission transmissio n line carri ca rries es the elec tricity tricity,, (6) Electric Ele ctricitity reaches reach es the the final consumption c onsumption points Electric Supply System – the conveyance of electric power from from a power station station to to consumer co nsumer ’s premises. Electric Electri c Supply System Components Components 1. Power Station Station (Generators) 2. T ransmission ransmission Lines 3. Distribution System System – connects conne cts other part of the power system system such as transmission line through the help of SUBSTATION. Classification Class ification of Electric Supply System 1. AC AC or DC system system 2. Overhead or Underground Unde rground system system Parts of a typical AC Power Supply Scheme 1. Generating Stations 2. Primary Transmission Transm ission 3. Secondary Secon dary T ransmission 4. Primary Distribution 5. Secondary Secon dary Distribution 6. System System of Supply to to individual consum ers A system network (or grid grid)) is the name given to to that part of power system which consists of the sub – stations and transmission lines line s of various voltage rating. rating . T ransmission Systems Systems It incl i ncludes udes all land, land , conversion structures structures and equipment e quipment at at the primary sources of supply, line switching and conversions stations between a generating or receiving point and the entrance to a distribution c enter or wholesale wholesale point, all l ines and equipm ent whose primary purpose is to augment, augm ent, integrate integrate or tie together sources of power supply. sup ply.


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Electrical Transmission and Distribution System and Design T wo Types of Electric al Transmission Transmission 1. Electrical Elec trical Power Transmission Transmission 2. Electrical Elec trical Com municat munication ion Transmission T ypic pical al Electric Elec tric Power Transmission and Distribution Scheme

Figure 8 Electric 8 Electric Power Transmission and Distribution Scheme

Purposes of T ransmission 1. T o transmit power from a water power site to to a m arket. 2. For bulk supply of power load center from outlying steam stations. These Th ese are likely to be relativ rel atively ely short. sho rt. 3. For interconnection purposes, that is, for transfer of energy from one system to another in case of emergency or in response to diversity in the system peaks. T wo Methods of T ransmi ransmission ssion 1. Overhead System – transmission lines lines - power is conv con veyed by bare conduc con ductors tors of copper co pper or aluminum aluminum which are strung between wooden or steel poles erected at convenient distances along a route. The bare copper coppe r or aluminum alum inum wire is fixed fixed to an insulat i nsulator or which whic h is itself fixed onto a cross – arm on the pole. The number numbe r of cross – arms carried ca rried by a pole depends on on

the number of wires it should carry. Line supports consist con sist of: a. pole structures structures which are made of wood, reinforced c oncrete or steel are used up to 66 kV b. steel towers are towers are used for higher voltages. 2. Underground System – this being especially suited for densely populated areas though it is somewhat costlier costlie r than than the first method. - employs insulated cables which may be single, double or triple – c ore etc. etc. T ransmi ransmission ssion Lines T hese are connecting conn ecting links between the generating stating stating and distribution d istribution system. system. Principle Princi ple Element Eleme ntss of High Voltage AC AC T ransmission Lines Lines 1. Conduc tors (R, L, C) 2. Step – Up and Step – Down Transformers 3. Line Insulators 4. Supports 5. Protective Device (fuses, relays, breakers) 6. Voltage Regulating Dev D evic icee (surge protective devic device) e) CHOICE OF SUPPLY Comparisons Comp arisons of DC and AC T ransmission and Distribution

Figure 9 Transmission 9 Transmission and Distribution by means of AC and DC

DC Transmission A. Advantages 1. It requires only two conductors as compared to three for ac transmission. It is also possible to transmit power through only one conductor by using earth as returning conductor, hence much copper cop per is saved. saved. 2. T here is no induc i nductance, tance, capaci cap acitance tance,, phase displacement and surge problem in dc transmission. 3. Due to absence absenc e of inductance, the voltage drop in a dc transmission transmission line is less than the than the ac for the same load loa d and sending end voltage. 4. T here is no skin effect in effect in a dc system. T herefore, herefore, the entire cross section of the line conductor is utilized.


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Electrical Transmission and Distribution System and Design

√ 

5. For the sam e working voltage, the potential potential stress on the insulation is times less in case of dc system system than that in ac system. system. T herefore, dc line line requires less insulation. 6. A dc line has less corona co rona loss and reduced reduced interference with c ommunication circuits. 7. In dc transmission, transmissio n, there there is no stability problems problems and synchronizing difficulties. 8. T he high voltage dc transm transmission ission is free free from the the dielec tric losses, partic particularly ularlyin case of cables. c ables. 9. Charging current, whic h contrib c ontributes utes to continuous continuous loss even on no load, is elim inated. 10. Underground c ables can be used because of less potential stress stress and negligible negli gible dielectric loss. B. Disadvantages 1. Electric power cannot be generated at high dc dc voltage due to com mutation mutation problems. 2. The dc voltage cannot be stepped up for transformation of power at high voltages. 3. T he dc switches and circuit circ uit breakers breakers have have their own limitation. AC A C T ransmission A. Advantages 1. T he power can be generated at high hig h voltages. voltages. 2. Maintenance of ac substation is easy and cheaper. 3. The ac voltage can be stepped up or stepped down by transformers transformers with ease and efficiency effic iency.. This permits to transmit power at high voltages and distribute it at safe potentials. B. Disadvantages 1. An An AC line requires more copper than than DC. 2. T he construction of an AC line l ine is more more compl co mplicat icated ed than a DC transmission transmi ssion line. 3. Due to skin effect in the ac system, system, the effective resistance of the the line is inc reased. 4. An An AC line has capacitance, cap acitance, thus, there is a continuous loss of power due to the charging current cu rrent even even when the line lin e is open. ope n. 5. In case of overhead lines spacing between the conduc co nductors tors is to be kept more to prov p rovide ide adequat adeq uate insulation and to avoid avoid c orona loss. 6. The alternators are to be synchronized before putting them them in parallel. 7. The variation in speeds of alternators are to be controlled co ntrolled within very very low limits. lim its. AC A C Distribution is undoubtedly superior to that by DC Distribution in terms of voltage control by the means of a transformer.

T he best method m ethod is to em ployAC system system for generation and and distribution and DC system system for transmission. T ransmission of electric elec tric power by DC system system became became possible because of introduction of mercury arc rectifiers and thyratrons thyratrons which whic h can convert convert AC AC into DC and vice versa efficiently and at reasonabl e cost.

Figure 10 HVDC 10 HVDC Transmission System

The electric power is generated as AC; its voltage is stepped up to high voltage by step – up transformer and and conv con verted into DC by some suitable sui table rectify rec tifying ing device. The transmission of electric power is carried at high DC voltage voltage (500 kV). At rec eiving end, DC power is cconverted onverted back bac k int into three phase AC using suitable c onverters onverters and then stepped stepped down to low voltage for distribution by suitable step – down transformers. The favorable conditions resulting from the use of converters in the above system gives the following additional advantages: advantages: 1. Independence of frequencies at sending end and receiv recei ving end networks makes possible possib le generation ge neration of of power at frequency frequenc y different different from that of load system. system. 2. With the use of suitable converters, power can be reversed, reversed, increased inc reased or decreased in the system system at will and with rapidity i.e., i.e., the c ontrol is easy and rapid. rapid . 3. Because Becau se of the absenc e of cha rging and stability stabili ty problems, there is no limit to distance over which DC power can be transmitted. 4. Keeping in view the c ircuit ircu it breaker capaci c apacities, ties, the shor short circuit currents are not transmitted through lines and hence a tremendous amount of saving is affected by installing smaller sma ller circuit ci rcuit breakers. 5. T he smaller smal ler amount of power can be transmitted transmitted much much more ec onomicall onom icallyy. Comparisons Comp arisons of Overhead Overhead and Underground Systems Systems Transmission and distribution of electric power can be carried carrie d out by overhead overhead as well w ell as underground undergrou nd systems. systems. Comparison Comp arison between the two two is giv gi ven: a. Public Safety: Safety: Underground system is safer than overhead system. b. Initial Cost: Undergroun Cost: Undergroundd system system is more expensive. c. Flexibility: Overhead system is more flexible than underground system. In overhead system new conduc con ductors tors can be laid along the existing ones for load load


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Electrical Transmission and Distribution System and Design expansion. In case of underground system new conductor conduc torss are to to be laid in new channels. d. Working Workin g Voltag Voltage: e: The The underground system c annot annot be operated above 66 kV because bec ause of insulation difficult di fficultie iess but overhead system system can ca n be designed for operation operation up to 400 kV or higher. e. Maintenance Cost: Maintenance Cost: Maintenance c ost of undergrou underg round nd system system is i s very low in c omparison ompari son with that of overhead overhead system. f. Frequency of Faults or Failures: As Failures: As the cables are laid underground, undergroun d, so these are not n ot easily acc essible essible.. The insulation is also better, so there are very few chances of power failures or fault as compared to overhead system. g. Frequency of Accid ents: ents: T T he chances c hances of acci dent dents in underground unde rground system system are very low as c ompared to to overhead system. h. Voltage Drop: In underground underg round system system bec be c ause of less less spacing between the conductors, inductance is very low as a s compa c ompared red to overhead overhead lines, l ines, therefore, therefore, voltage voltage drop is low in underground system. system. i. Appearance: Underground system of distribution or Appearance: Underground system transmission is aesthetic bec ause no wiring is visible. visible. j. Fault Location and Repairs: Though there are very rare chanc ch ances es of occ urring urring fault in underground un derground syste system, m, but if it oc curs cu rs it is very very difficult diffic ult to loc ate the fault and and its repair is difficult difficul t and expensive. k. Charging Charg ing Current: Current: Because Becau se less spacing spac ing between the the conductors the cables have much capacitance, drawing higher charging chargi ng current. current. l. Jointing: Jointing Jointing: Jointing of underground c ables is difficult so so tapping for loads and a nd service mains m ains is not no t conveniently possible in underground und erground system. system. m. Damage Due to Lightning and Thunderstorm: Underground system system is free from interruption i nterruption of service service because thunderstorm, lightning and objects falling across the wires. n. Surge Effect: In Effect: In underground system surge effect is smoothened down dow n as surge energy is absorbed a bsorbed by the sheath. o. Interference to Communication Circuits: In underground system, system, there is i s no interference to com munication circuit ci rcuits. s. High voltage transmission transm ission is carried carrie d out by overhead overhead system system due du e to low cost. However, However, distribution in congest con gested ed areas and in modern mode rn cities are carried c arried out by underground underground cables. The overhead line as a mean of transmitting electrical power over long distances is cheap and efficient. It is reliable – there are only one or two breakdowns due to lightning, and one or two due to other reasons, such as frost

and fog, per 160 km of a line a year. Most of the breakdowns whic h do occur oc cur are transient and none out of of them can be rendered innocuous, as far as system is concerne con cerned, d, by using high speed automatic automa tic reclosing rec losing circu circuitit breakers. Repairs, where necessary nece ssary,, can ca n easily easil y and quickly quickly be carried on site. The only real disadvantage of the overhead overhead line l ine is i s that it needs large la rge electrical clearances clearances and and it is sometimes not possible to get several circuits into congested con gested areas. EHV power po wer cables cab les are used in congested or built – up areas, near airfields, line terminations and transformer banking c onnections where the use of overhead overhead lines lin es is not permitted because of space limitations, safety requirements, amenities, etc. The major factor against a greater use of insulated cables is the high price of the equipment. Influence of Working Voltage on Size of Feeders and Distributors in DC Sy System stem 1. It reduces the size (area of cross section of the core carrying the current) cu rrent) of the feeders and distributors. 2. It inc reases the the effic efficienc iencyy of transmission Examples: 1. What is the percentage perc entage savings savings in feeder copper co pper if the line voltage voltage in i n a 2 wire wi re DC sy system is raised from 220 V to 400 V for the same power po wer transmitted over the the same same distance and having the same power loss? 2. What is the percentage perce ntage saving saving in copper c opper feeder if the the line voltage voltage in i n a 2 wire wi re DC system system is raised from 220 V to 500 V for the same power po wer transmitted over over the same same distance and having the the sam e power loss? Various Systems Systems of Power T ransmission In prac tice, three phase three wire AC system system is universally used for transmission transmi ssion and three phase four wire AC system system is used for distribution of electric power but for special purposes other systems may also be used. The various systems systems of power transmission are: 1. DC System a. DC 2 wire b. DC 2 wire with midpoint mi dpoint earthed c. DC 3 wire 2. Single Sing le Phase AC AC System System a. Single phase, 2 wire b. Single phase, 2 wire with mid – point earthed c. Single phase, 3 wire 3. T wo Phase AC AC System System a. T wo phase, 4 wire b. T wo phase, 3 wire 4. T hree Phase AC System System a. T hree phase, 3 wire b. T hree phase, 4 wire


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Electrical Transmission and Distribution System and Design T he basis for compari c ompariso sonn between the various systems systems of power transmission is usually economy. Since in a transmission system, the cost of the conductor material acc ounts for a major majo r part of the total c ost, the the best system system for transmission of electrical power is that for which the volume olum e of c onductor onduc tor material required is minimum. Thus Thus, the requirement requirem ent of volum volumee of c onductor onduc tor material forms the the basis of comparison co mparison between various various systems. systems. In m aking comparison com parison of the the volume of o f c onductor onduc tor materia materiall required for various various transmission systems, systems, the basis ba sis will be the equal maxim um stress on the dielectric. diel ectric. This is becaus because the voltage is only limited lim ited by the problem proble m of insulating the the conduc con ductors tors against again st disruptive discharge. disc harge. For com paring the the amount amoun t of conduc con ductor tor material required for different system system two case c ase arises: 1. When overhead system is employed for transmission of power. In overhead system, the conductors are insulated from the cross arms and supporting towers and as the towers and cross arms are earthed so the maximum voltage between each conduc con ductor tor and earth forms forms the basis of comparison co mparison of of volume olum e of conductor c onductor material required. required. 2. When underground cables are employed for transmission of power. In underground cables the maximum disruptive stress is between the two conduc con ductors tors of the cable; c able; the maximum m aximum voltage between the c onductors onduc tors forms the basis of c omparis omparison on of volum volumee of conduc c onductor tor material required. required. Comparisons of Various Systems Power Transmission (Overhead) Assumptions: A ssumptions: 1. Same power (P watts) transmitted by each eac h system. system. 2. T he distance distance (l meters) over over which power is transmitted remains the same. 3. T he line losses (WL watts) watts) are the same in each case. 4. The max voltage between any conductor and earth (Vm) is the same in each case. Let: = power factor; VC = volume of conduc co nductor tor material l = length; WL = line li ne losses; Vm = maximum voltage P = power po wer to to be transmitted; ρ = resistivity 1. Two wire dc d c system with one condu ctor ctor earth

cos∅

I =     WA == 2IR =      V = 2Al =   K =   Load Current

Line losses

Area Area of cross section of conduc con ductor tor Volume of conductor material

required Let

Note: T his system system will be the basis basi s for compariso co mparisonn wit with other systems. 2. 2 wire with midpoint midp oint earthed, earthed, DC system

I = W = 2IR =  A =   Load Current

Line losses

Area of

cross section conductor

of

Figure 12 2 12 2 Wire with Midpoint Earthed, DC System

V = 2Al = 

Volume of conductor material

required 3. 3 wire, DC system

I =    WA  = = 2IR =    V = 2.5Al =  K

Figure 13 3 13 3 Wire DC System

Load Current

Line losses

Area Area of cross section of conduc con ductor tor

Assuming Assuming area of cross section of neutral neutral wire as half of that any of the outers, Volume of conductor material required Figure 11 Two 11 Two Wire DC System with One Conductor Earth


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Electrical Transmission and Distribution System and Design 4. Single Phase AC system, 2 wire, with one conductor earth

7. Two phase, 4 wire, AC system

I = √  ∅ = √ ∅

I = √ ∅

Load current

Figure 14 Singl 14 Singlee Phase AC System, ystem, 2 Wire w ith Conductor Earth Earth

WA  == 2IR= 2√ ∅      ∅ V = 2Al = ∅

Figure 17 Two 17 Two Phase, 4 Wire AC Sys tem

Line losses

Area Area of cross section sec tion of conductor cond uctor Volume of conductor material

required 5. Single Phase AC system, 2 wire wire system with with mid – point earthed

IW = =√ 2IR=∅2    √ ∅  A =  ∅

  WA = = 44 √ ∅    ∅  V = 4Al = ∅

Line Losses

Area of cross sec tion of conduct co nductor or Volume olum e of c onductor onduc tor materia materiall

required 8. Two phase, 3 wire, AC system

Load current

Area Area of cross section of conductor Figure 15 Single 15 Single Phase AC Sy stem, 2 Wire with Midpoint Midpoint Earthed

I = √ ∅ I = I  I = √2I

Load current

Neutral current:

Line losses

V = 2Al =  ∅

Load current

Figure 18 Two 18 Two Phase, 3 Wire AC System

  P ρ l P W = 22+√2Vcos ∅ A  √2√2Vcos ∅ Aρl A = √

Line losses:

   ∅ . V = 2Al  √ 2A2Al = ∅

Area of cross sec tion of conduct co nductor or


required 6. Single phase AC system, 3 wire system

I = √ / ∅ 

Volume olum e of conduct co nductor or

material required 9. Three phase, 3 wire, AC system

Load current

W = 22  √∅  

I = √ ∅

Line losses

Load

current per phase

Figure 16 Single 16 Single Phase AC Sys tem, 3 Wire Wire

A =  ∅ V = 2.5Al = ∅

Area Area of cross section sec tion of conductor cond uctor

Assuming Assuming cross section of neutral wire half of that of any of the outers, Volume of conductor material required

Figure 19 Three 19 Three Phase, 3 Wire AC System

    √ WA  = = 33 ∅     ∅


Line losses

Area of cross sec tion of conduct co nductor or

7


V = 3Al = .∅


required 10. Three phase, 4 wire, AC system Assuming Assuming balanced balanced load, there will be no current in neutral wire and copper c opper losses will will be same as in three phase, three wire system

AV = =2A l =      K =  

Area Area of cross section of conduc con ductor tor Volume of conductor material

required Let

Note: T his system system will be the basis basi s for compariso co mparisonn wit with other systems. 2. 2 wire with midpoint midp oint earthed, earthed, DC system This system is the same as a 2 – wire DC system, so volume of conductor material required for this system system is the same as that in a 2 wire wi re DC system. system.

Figure 10 Three 10 Three Phase, 4 Wire AC System

A = ∅ V = 3.5Al =  ∅

Area of cross section of

conductor Taking cross section of neutral wire as half of either outer, Volume of conductor

material required Comparisons Comp arisons of con conduct ductor or m at aterial erial in Underground Syste System m Conditions of Comparisons 1. Same power (P watts) transmitted transmi tted by each system. system. 2. T he distance distance (l meters) over over which power is transmitted remains the same. 3. T he line losses (WL watts) watts) are the same in each case. 4. The max voltage between any conductor and earth (Vm) is the same in each case. Comparisons of Various Systems Power Transmission (Underground) Let: = power factor VC = volum volumee of conduc c onductor tor material l = length WL = line losses Vm = maximum voltage P = power po wer to to be transmitted

cos∅

ρ = resistivity

Figure 22 DC 22 DC 2 Wire Midpoint Earthed (Underground)

V = K

Volume olum e of conduc con ductor tor material required required 3. 3 wire, DC system

I = 

Load Current

    W = 2I R = 

Line losses



Figure 23 3 23 3 Wire DC System (Underground)

A =  V = 2.5Al = 

Area Area of cross section of conduc con ductor tor

Assuming Assuming area of cross section of neutral neutral wire as half of that any of the outers, Volume olum e of conductor cond uctor material material required 4. Single Phase AC system, 2 wire

1. Two wire wire dc system

I =        W = 2IR =  

I = √  ∅ = √ ∅

Load Current

Load current

Line

losses

Figure 21 Two 21 Two Wire DC System (Underground)

Figure 24 Single 24 Single Phase, 2 Wire AC Sy stem (Underground) (Underground)

W = 2IR = 22√ ∅  


Line losses 8


AV = =2A l=∅  ∅

Area Area of cross section sec tion of conductor cond uctor

8. Two phase, 3 wire, AC system


required 5. Single Sing le Phase AC system, 2 wire system sy stem with with mid – point earthed T his system system is i s the same sam e as a 2 – wire single phase AC system, so volume of conductor con ductor material of this case is also the same. Figure 25 Single 25 Single Phase, 2 Wire Wire w ith Midpoint Earthed AC Sy stem (Underground)

V = ∅

Load current phase

per

Current in middle mid dle wire wire

Figure 28 2 28 2 Phase, 3 Wire AC Sy stem stem (Underground)

  W = 22 2 V2 Pcos ∅ Aρl  √2√22V2 Pcos ∅ Aρlρl    +     √    ∅ .    ∅

AV = = 2A l  √ 2A2A l =

Line losses

Volum Volumee of conduc con ductor tor material required required

6. Single phase AC system, 3 wire system Assuming Assuming balanc ed load, the system reduces to a single phase, 2 wire AC system system except excep t that that a neutral neutral conductor of half the cross section is provided in addition. Single Phase, 3 Wire AC

Area Area of cross section sec tion of conduc con ductor tor

Volume olum e of conductor con ductor mate m aterrial ial

required 7. Two phase, 4 wire, AC system

Volume Volum e of conduct co nductor or

material required 9. Three phase, 3 wire, AC system

I = √  ∅

Load current

Figur e 26 Sy stem (Underground) (Underground)

V = 2.5Al = .∅

I =  ∅ ∅ √ 2I2I

Figure 29 Three 29 Three Phase, 3 Wire AC Sy stem (Underground) (Underground)

  W = 3 √  ∅  AV ==3A l=∅ .  ∅

Line Losses

Area Area of cross section sec tion of conduc con ductor tor Volume olum e of conductor con ductor material material

Figure 27 Two 27 Two Phase, 4 Wire AC System (Underground)

In this system, each phase shares the half of the total load. This T his system system is equiv equi valent to two wire wi re AC system. system. In this c ase, cross sec tion area of eac h conductor is taken taken half of that of single sin gle phase two wire AC AC system system but b ut four four wires are required in place of two wires, so the same volume olum e of conductor co nductor material is required. Volume olum e of conduc c onduc tor material required

V = ∅

required 10. Three phase, 4 wire, AC system Assuming Assuming balanc ed load, this system system is i s reduc ed to to a 3 phase AC system except that an additional wire, called the neutral wire, is provided provided with half the cros c rosss section of the outer thus, Figure 30 Three 30 Three Phase, 4 Wire AC

V = 3.5Al = .∅ Sy stem (Underground) (Underground)

Volume

of

conductor

material required T he following points may be noted: 1. T here is a great saving saving in conductor cond uctor material if DC system system is adopted for transmission of electric power po wer..


9

Electrical Transmission and Distribution System and Design However, due to technic al difficulties, DC system system is i s not used for for transmission. transm ission. 2. Two phase, three wire system is obviously quite unsuitable for long lon g distance transmission and need no no further consideration. 3. Considering the AC system, system, the 3 phase AC system system is is most suitable for transmission due to two reasons. Firstly, there is considerable saving in conductor material. Secondly, this system is convenient and efficient. Considering other factors suc h as efficiency effici ency of operation o peration and conv con venience, three phase three wire system system is usually adopted. Examples: 1. A 50 – km long transmission line supplies a load of 5 MVA at 0.8 pf lagging laggin g at 33 kV. T he efficiency efficien cy of transmission is 90%. Calculate the volume of aluminum alum inum c onductor onductor required for the the line li ne when: a. single phase, 2 wire system system is used b. 3 phase, 3 wire system system is used. u sed. T he specific speci fic resistance resistance of aluminum is 2.85 × 10 −8 Ω – m. 2. A 3 – phase, 4 – wire system is used for lighting. Compare Compa re the amount amoun t of copper cop per required with that needed for a 2 – wire D.C. system with same line voltage. Assume the same losses l osses and balanc ba lanced ed load load.. T he neutral is one half the cross – section of one of the respective outers. 3. An An existing single phase AC system system com prising of two two overhead overhead c onductors onduc tors is to be conv c onverted erted into a 3 phase, ph ase, 3 wire system system by providing an additional add itional conduct con ductor or of same size. Calc C alculate ulate the percentage of additional load load that c an be transmitted transmitted by the three – phase system system if the operating voltage and percentage line losses remain the same in both systems. systems. 4. An An existing DC three wire system system is to be c onverted onverted into a three phase 4 wire system by adding a fourth wire equal in cross section to each outer of the dc system. system. Assuming the same supply and load voltages voltages to neutral ne utral and balanc bal anced ed conditions conditions find the extra power power at unity power factor that can be supplied by the AC system. 5. A 3 wire DC system system is i s convert c onverted ed to a 4 wire, 3 phase phase AC AC system system by the addition of another wire equal in section to one of the outers. For the same effective

voltage between be tween outers and neutral at a t the consume con sumerr’s

terminals and the sam e percentage loss, find the percentage additional load that can be supplied. Assume Assume balanc ed load and in the the AC AC system system a power factor of 0.9. 6. A DC 2 wire system system is to be conv con verted into AC 2 phase, 3 wire system system by the addition of a 3 rd conduc con ductor tor of the

same cross section as the two existing conductors. Calculate the perc entage additional additional load whic h can can now supplied if the voltage between wires and the percentage perce ntage power loss in the line remains rema ins unchange unch anged. d. 7. A DC 2 – wire distribution system system is conv con verted into AC 3 phase, 3 wire system system by adding a third conductor con ductor of the same siz si ze as the two existing c onductors. If voltage between conductors and percentage power loss remain the same, calcul cal culate ate the percentage addit ad ditio iona nall balanced load which can now be carried by the conduc con ductors tors at 0.95 pf. 8. A DC 2 wire system system is i s to to be converted converted into 3 phase, 3 wire AC system by adding a third conductor of the same size as the two existing conductors. Calculate the percentage percen tage additional balanced load that can c an now now be carried by the conductors at 0.96 pf lagging. Assume Assume the the same voltage voltage between the conductor cond uctorss and and the same percentage perc entage power loss. 9. A 2 phase, 3 – wire AC system has h as a middl mi ddlee condu con duct ctor or of same cross – sectional area as the outer and supplies a load of 20 MW. T he system system is c onverted onverted into into 3 – phase, 4 – wire w ire system system by running a neutral wire wi re.. Calculate the new power which can be supplied if voltage across c onsumer terminal and percentage lin linee losses remai remainn the the same. same . Assume balanc bal anced ed load. 10. A single – phase AC AC system system supplies a load of 200 kW and if this system system is c onverted onverted to 3 phase, pha se, 3 wire AC system system by running runn ing a third similar simi lar conduct co nductor, or, calculat calculate the 3 – phase load that can now be supplied if the voltage between the conduc tors is the same. sam e. A Assume ssume the power po wer factor factor and transmission effic efficiency iency to to be the same in the two cases. c ases. 11. A single – phase load of o f 5 MW is transmitted transm itted by a pair of overhead conductors. If a third conductor of the same cross – section and material be added and 3 phase supply be thus substituted for the original origi nal single single phase, calculate the 3 – phase load which c an now now be be transmitted if the voltage between the c onductors onduc tors and and the percentage loss in the l ines remains unchanged. 12. A substation substation supplies power at 11 kV, kV, 0.8 pf lagging to a consumer c onsumer through a single – phase transmissi transm ission on line ine

having total resistance resistanc e (both go and return) of 0.15 Ω.

T he voltage voltage drop in the line is 15%. If the same power is to be supplied to the same c onsumer by two wire DC system system by a new line having a total resistance of 0.05 0.05 Ω and if the allowable voltage drop is 25%, calculate the DC supply voltage. Choic e of Working Voltage Voltage for T ransm ransmission ission Advantages Advan tages of Hig h Transmission Voltag Voltagee 1. Reduces the volume of conductor material and reduces reduc es the the cost of the supporting supp orting structure structure materials.


10

Electrical Transmission and Distribution System and Design 2. Increase transmi ssion efficiency since line l ine losses and and line current c urrentss are reduced. 3. Decreases percentage line drop and leads to better voltage regulation. regu lation. Limi tations of High Hig h Transmission Transmission Voltag Voltagee 1. T he increase inc rease cost of insulating the conductors conduc tors and the he earthed tower inc reases. This increas i ncreases es the cost c ost of line ne supports. 2. More clearance is required between conductors and ground; hence, higher towers are required. 3. More distance is required between the conductors; therefore, longer cross arms are required. 4. The increased cost of transformers, switchgear and other terminal apparatus. Economics of Power T ransmi ransmission ssion 1. Economic Econo mic Choice of Transmission Voltage Voltage 2. Economic Econo mic Choice of Conductor Size Size Economic Econo mic Choice of Transmission Voltage Voltage T he method m ethod of finding the ec onomical onom ical transmission transmission voltag voltagee is as follows. Power to be transmitted, generation voltage voltage and length of transmission line li ne are assumed to be known. known. We c hoose some som e standard transmission vvoltage oltage and a nd work work out the following costs: 1. T ransformers, at the generating and receiv recei ving ends of transmission line. For a given power, this cost increases slowly with the increase in transmission voltage. 2. Switchgear. Switchgea r. This c ost also increases with the inc reas reasee in transmission voltage. voltage. 3. Lightning arrestor. T his cost increases inc reases rapidly with the increase inc rease in transmission voltage. voltage. 4. Insulation and supports. This cost increases sharply with the increase i ncrease in transmission transmissio n voltage. voltage. 5. Conductor. T his cost decreases dec reases with the increase in transmission voltage. The sum of all above costs gives the total cost of transmission for the voltage considered. Similar calcul cal culat ations ions are made m ade for other other transmission voltages. T hen, a curve is drawn for total cost of transmission against voltage as shown. T he lowest point (P) on the curve gives the economical transmission Figure 31 Capital 31 Capital Cost vs voltage. Thus, in the present Transmis sion Voltage Voltage c ase, OA is the optimum optimum transmission voltage. This method of finding the economic eco nomical al transmission transmission voltage is rarely used in pract prac tice as different di fferent costs co sts cannot be determined determi ned with a fair degree degree of accuracy.

The present – day trend is to follow certain empirical formula for finding the ec onomical onom ical transmission voltage voltage

KV = 5.5 L   KV = 5.5 .    or

where: KW – power to be transmitted transmission line in miles L – length of transmission With the increase in distance of transmission, the cost of terminal apparatus ap paratus is reduced resulting in higher hi gher econo ec onomi micc transmission voltage. Similarly, if the power to be transmitted becom es large, the the cost per kW of the termina termi nall station station equipm ent is reduced. As As a rough guide the the voltage voltage for transmission is chosen cho sen as 625 volts per km though in practice the voltage per km varies from about 400 to 900 volts for longer to shorter distances. distances. T he choice c hoice is usually limited in prac tice by the the requirement requirem ent of standardization standardization and for satisfactory regulation without excessive equipm ent cost. T he voltages voltages normally normal ly adopted for transmission are giv gi ven: Distance in km Transmission voltage in kV

15 – 30

30 – 60

60 – 100

100 – 200

200 – 300

300 – 400

11

33

66

132

220

400

Table 1 Ty pical Voltage Voltage Rating for given distances distances

T he most com mon mon transmission voltages voltages are 33 kV, kV, 66 kV, kV, 132 kV, kV, 220 kV, 400 kV and a nd 765 kV. kV. Examples: 1. Estimate the weight of c opper required requi red to supply a load load of 100 MW at unity pf by a 3 – phase, 380 kV system over 100 km. The neutral point is earthed. The resistance resistance of the conductor conduc tor is 0.045 0.045 ohm/cm 2/km. The weight of copper is 0.01 kg/cm 3. The efficiency of transmission transmission can be assumed to be 90 percent. 2. 30,000 kW at power factor 0.8 lagging is being transmitted over a 220 kV, three phase transmission line. The T he length of the line is 275 km and the efficiency effic iency of transmission is 90%. C Calc alculate ulate the weight of coppe copperr required. Also, calcu cal culate late the weight of copper c opper had the the power been transmitted over a single – phase transmission line for the same li ne voltage voltage and losses l osses.. Assume Assume that that the the resistance of 1 km long conducto con ductorr and and 2 1 cm is 0.173 Ω and specific speci fic gravity gravity of copper c opper is 8.9. 8.9. 3. Electric Elec tric power of 50 M W is to be transmitted ov over er a 132 KV, KV, 3 phase, 3 wire transmission line. The length of the the line is 300 km and the efficiency of transmission is 85%. Aluminum is used for conductor material which has resistivity of 3 × 10 −9 Ω – m. Calculate Calc ulate the volume of conduc c onductor tor material required required for a power factor of 0.8 lagging. Economic Econo mic Choice Choi ce of Conduc tor Size Size The cost of conductor material is generally a very considerable part of the total cost of a transmission line.


11

Electrical Transmission and Distribution System and Design T herefore, the determination of proper size size of conduc con ductor tor for for the line li ne is of vital importance. imp ortance. T he mos m ostt economical area area of of conductor is that for which the total annual cost of transmission transmission line is minimum. minim um. T his is known as Kelvin’s Law aft Law after er Lord Kelvin who first stated it in 1881. T he total total annual c ost of transmission line lin e can be divided broadly into into two parts, annual annua l charge on capital outlay and annual cost of energy wasted in the conduc co nductor. tor. Annual Charge on Capital Outlay Outlay bec ause interest and depreciation on the c apital - T his is because cost of complete installation of transmission line. In case of overhead overhead system, system, it will be the annual interest and depreciation on the capital cost of conductors, supports and insulators and the cost co st of their erection e rection.. Now, for an overhead line, insulator cost is constant, the c onductor cost co st is proportional to the the area of cross sec tion and the the c ost of supports and their erection erec tion is is – section partly constant c onstant and partly proportional proportion al to area of c ross ross sec tion of the conduc con ductor. tor. T herefore, herefore, annual char charge – section on an overhead overhead transmission line can ca n be expressed as: as:

Annual Annual charge = P  Pa

where P1 and P2 are constants and a is the area of cross – section of the conductor relationship exists for underground system. system. In - A similar relationship this system, system, the annual ann ual c harge is because bec ause interest interest and and depreciation deprec iation on the cost co st of conductors, insulation insulation and and the cost of laying the cables. ca bles. Now, the cost co st of insulatio insulationn is c onstant and the cost of c onduc tor is proportional to area of cross – section sec tion of conductor. Annual Cost of Energy Wasted Wasted - This is because energy lost mainly in the conductor due to I2R losses. l osses. Assumin Assumingg a constant con stant current in the the conductor throughout through out the year, year, the energy lost in the conductor is proportional to resistance. As resistance is inversely proportional to the area of cross – Figure 32 Illustration 32 Illustration of Kelvin's Law section of the conduc con ductor, tor, therefore, the energy lost in the conduct con ductor or is inversely proportional to area of cross – section. T hus, the annual c ost of energy wasted in an ov o verhead transmission line can c an be expressed as:

Annual Annual coscost of energy wasted = Pa

where P3 is constant and a is the area of cross – section of the the c onductor

Limi tations of Kelvin’s Law

1. T he assumption that annual cost co st because interest and and depreciation on the capital outlay is in the form P 1 + P2a is i s strictly stric tly speaking not true. true. For Fo r instance, in cabl c ables es neither the c ost of cable c able dielectric diel ectric and sheath nor the the cost of laying vary vary in i n this this manner. ma nner. 2. T his law does not consider c onsider several several physical factors like safe current c urrent density, density, mec hanical strength, corona co rona loss loss etc. 3. The conductor size determined by this law may not always be practicabl prac ticablee one because it may be too smal sm all for the safe carry ca rrying ing of necessary nec essary current. cu rrent. 4. Interest and depreci de preciation ation on the the capital c apital outlay cannot cannot be determined de termined ac curately curately.. 5. It is not easy to estimate the energy loss in the line without actual load curves, which are not available at the time of estimation. 6. In the case of cables ca bles there are sheath losses and with high voltages dielectric losses also. Dielectric loss occ urs continuously therefore load factor of dielec tric loss is 100%. Hence the cost per unit of energy lost as dielec tric loss is less than cost per unit supplying line loss. Examples: 1. A 2 – conductor cond uctor cable 1 km long is required to to supply a c onstant current of 200 A throughout througho ut the the year. year. The cost of cable cab le inc luding installation installation is Php (20a + 20) per per meter where ‘a’ is the area of cross – section of the conductor conduc tor in cm 2. The cost of energy is 5P per kWh and interest and depreci de preciation ation charges amount to to 10%. Calculate the most economical conductor size. Assume Assume resistivity resistivity of conduc c onductor tor material material to be 1.73 μΩ – cm. 2. Determine the best current density in A/mm A/mm 2 for a 3 – phase overhead overhead line l ine if the line is in use for 2000 hours per year and if the conductor costing Php 3.0 per kg has a specific resistance of 1.73 Ω – m and weighs 6200 kg/m 3. Cost of energy is 10 P/unit. Interest and depreciation is 12% of c onductor cost. 3. A 3 – core, c ore, 11 kV cable cab le supplies a load of 1500 kW at 0.8 pf lagging l agging for 300 days days in a year at an average of 8 hours per day. The capital cost per km of cable is Php 8000 +20000a. T he resistance per km of a cable cable 2 of cross – sectional area 1 cm is 0.173 0.17 3 Ω. If the energy loss cost is 2 centav cen tavos os per unit, and the rate of interest and depreciation is 12%, calculate the most economical current density and diameter of the conductor. 4. T he cost per km for each of the copper conductor of a section a cm 2 for a transmission transm ission line is Php (2800a + 1300). The Th e load factor of the the load current c urrent is 80% and and


12

Electrical Transmission and Distribution System and Design the load loa d factor of the the losses is 65%. 65 %. The rate of o f interest interest and depreciation is 10% and the cost of energy is 50 centav cen tavos os per kWh. Find the most economi eco nomical cal curren currentt

1.78 ×10− Ω m

density for transmission line using Kelvin’s law.

5.

6.

7.

8.

9.

ρ=

. T he cost of a 3 – phase overhead overhead transmission transm ission line is Php (25000a + 2500) per km where ‘a’ is the area of cross – section sec tion of of each conductor c onductor in cm 2. The Th e line is is supplying a load of 5 MW at 33kV and 0.8 pf laggin lagg ingg assumed to be constant throughout the year. Energy costs 4 centavos per kWh and interest and depreciation total 10% per annum. Find the most economical eco nomical size size of the conductor. con ductor. Given that that specif spec ific ic 6 − resistance of conductor c onductor material is 10 Ω – cm. A 2 – wire feeder carries a constant current of 250 A throughout the year. year. T he portion of capital c apital cost c ost whi which ch is proportional prop ortional to area of cross – section is Php 5 per kg of copper copp er conduc tor. T he interest and deprec iation ion total 10% per annum and the cost of energy is 5 centav cen tavos os per kWh. Find the most m ost economical ec onomical area of of cross – section of the conductor. cond uctor. Given that the density of copper c opper is 8.93 g/cm 3 and its specific resistance is 1.73×10−8 Ω – m. Determine the most economical econ omical cross – section for a 3 phase transmission line, 1 km long to supply at a constant con stant voltage of 110 kV for the followi ng daily load cycle: 6 hours – 20 MW at pf 0.8 lagging 12 hours – 5 MW at pf 0.8 lagging laggin g 6 hours – 6 M W at pf 0.8 lagging T he line is i s used for 365 days days yearly. yearly. T he cost c ost per per km km of line including inc luding erection is Php (9000 + 6000a) where where sec tion of conductor in cm2. ‘a’ is the area of cross – section The annual rate of interest and depreciation is 10% and the energy costs 6 centavos per kWh. The resistance per km of each ea ch conductor c onductor is 0.176/a. Determine the most economical size size of a 3 – phase line which whic h supplies the following load s at 10 kV: 100 kW at 0.8 pf (lag) for 10 hours 500 kW at 0.9 pf (lag) for 8 hours 100 kW at unity pf for 6 hours. T he above above gives the daily load l oad cycle. The c ost per km of the completely erected line is Php (8000a + 1500) where a is the area of c ross-section of eac h conduct co nductor. or. T he combined interest interest and depreciation is 10% per per annum of capital cost. Cost of energy losses is 5 centav cen tavos os per kWh. Resistiv Resi stivity ity of conduc co nductor tor material al = 6 − 1.72× 10 Ω – cm. If the cost c ost of an overhead line is Php 2000 A (where A is the cross – section in cm 2) and if the interest and depreciation deprec iation charges c harges of the the line are 8%, estimate estim ate the the

most economical current density to use for a transmission requiring requiri ng full load current c urrent for 60% of the year. The cost of generating electrical energy is 5 centav cen tavos/kW os/kWh. h. T he resistance resistanc e of the conductor one one km 2 long and 1 cm cross – section is 0.18 Ω.

10.

Describe Descri be the main mai n limitations of Kelvin’s law. T he cost cost of a 3 – phase overhead overhead transmission line having c ross oss 2 – sectional area A cm is Php (500+2000A) per km.

Calculate Calc ulate the most mo st economical curre cu rrent nt density for the conduc con ductor tor if the rate of interest and deprec iation is is 12% 12% per annum. annum . The cost of energywaste is Php 0.05/kWh. 0.05/kWh. The resistance of e ach conductor is 0.17/A Ω/km. T ake the the load factor factor for losses = 12%. 1 2%. Requirem ents of Satisfactory Electric Supply T he power station delivers delivers power to consumers con sumers through through its its transmission and distribution systems. systems. T he power pow er deliv d eliver ered ed must be characterized by constant or nearly constant voltage, dependability of service, balanced voltage, and efficiency to give minimum annual cost, sinusoidal waveform and freedom from inductive interference with telephone lines. Voltage regulation. regulation . A voltage variation has a large effect upon the operation ope ration of both both power p ower machi m achinery nery and lights. A motor is designed to have its best characteristics at the rated voltage voltage and consequently c onsequently a voltage voltage that is too high hi gh or too low will result in a decrease in efficiency. If the fluctuations in i n the voltage are sudden, these may c ause the the tripping of circui ci rcuitt breakers and consequent interruptions to to service. Usually the voltage at the generator terminals, where this is done, in some cases c ases the the voltage variations variations at the load m ay be made sufficiently sufficientl y small by keeping the resistance and reactance reac tance of the the lines and feeders low. imp ortant requirement of electr el ectric ic Dependability . One important supply is to furnish uninterrupted uni nterrupted servic service. e. The T he losses which which an industrial c onsumer sustains due to the failure of elec tric power powe r supply suppl y are usually usual ly vastly vastly greater than the the ac tual tual value of the power p ower that he would woul d use during this period. ItIt is because bec ause the expense of idle workmen w orkmen and machines machines and and other overhead charges. Interruptions to service cause irritation and are sometimes som etimes p ositively ositivelydangerous to life and and property. For example, failure of power in hospitals, in crowded crowd ed theatres and stores m ay lead to very grave grave consequences. con sequences. Therefore, it is the duty of elec tric supply company to keep the power system going and to furnish uninterrupted service. Balanced volta vol tage ge.. It is very important that the polyphaser voltage should be balanced. If an unbalanced polyphase voltage is i s supplied to a consumer co nsumer operating synchrono synchronous us or induction motors, it will result in a decrease in the efficienc y of his m achinery ach ineryand a decrease dec rease in its maximum maximum power output. Motors cal led upon to deliver full load when when


13

Electrical Transmission and Distribution System and Design their terminal voltages are unbalanced are liable to considerable con siderable damage due to overheating. overheating. One method of maintaining mai ntaining balance bal ance of voltage voltage is by having balanced balanc ed loads loads connected connec ted to to the circuit. circ uit. Efficiency. Efficiency. The T he effici efficiency ency of a transmission system system is i s not of m uch impor imp ortance. tance. The important economic feature feature of the the design being the layout of the system to perform the requisite function of generating and delivering del ivering power with a minimum overall annual cost. The annual cost can be minimi min imizzed to a considerable con siderable extent extent by taking taking care c are of power factor of the system system.. It is bec ause losses in the lines line s and and machinery are largely determined by power factor. Therefore, it is important that consumers having loads of low power factor fac tor should be penali zed by being charged c harged at at a higher hi gher rate per kWh kWh than those those who wh o take power at high power factors. Loads of low power factor also require greater generator capac ity than those of high p ower factor factor (for the sam e amount amoun t of power) and produce larger voltage voltage drops in the lines and transformers. Frequency. Frequency. The frequency of the supply system must be maintained mai ntained constan con stant.t. It is because bec ause a change cha nge in frequency frequency would change the motor speed, thus interfering with the manufacturing manu facturing operations. operations. Sinusoidal waveform. waveform . The T he alternating voltage voltage suppli ed to the consumers c onsumers should have a sine waveform. It is i s bec aus ause any harmonics which might be present would have detrimental effect upon the effici ency and maximu m aximum m power power output of the connected machinery. Harmonics may be avoided by using generators of good design and by avoidanc avoidanc e of high flux densities de nsities in transformers. transformers. Freedom from inductive interference . Power lines running parallel to telephone telep hone lines produce electrostat electrostatic ic and and electromagnetic field disturbances. These fields tend to cause cau se objectionable objectio nable noises and hums in the apparatus apparatus connec con nected ted to communication communication circuits. circuits. Induc tive tive interferen interference ce with telephone lines lin es may be avoided avoided by limiting lim iting as much as possible the amount of zero – sequence and harmonic current and by the proper transposition of both power lines and telephone lines. Mec hanical Design of Overhead Lines T ransmission and distribution lines are vital links between between generating stations and consumers as power from generating stations is transmitted at high voltage (such as 132, 220, 400 or 765 7 65 kV) kV) over over long distanc es to to major m ajor load load centers and then the power is distributed to various substations located loca ted at various various places plac es and local lo calitities ies through through distribution li nes. Because Becau se of tremendous tremendo us industrial growth, growth, requirement of power has increased manifold. Hence it becomes imperative that transmission and distribution of power from the generating stations to the various consumers is carried out with minimum possible loss and

disturbance. This objective can be achieved only if the transmission and distribution system is so designed and constructed that it is an efficient, technically sound and reliable system. The line should have sufficient current carrying capacity to transmit the required power over a given distance without an excessive voltage drop and overheating. overheating. T he line losses l osses should be small sm all and insulat i nsulatio ionn of the line should be adequate to cope with the system voltage. T he line should have have sufficient mechan mech anica icall strength to c ope with wi th the worst probable weather condit c onditio ions ns and provide satisfactory serv service ic e over a long period without the necessity nece ssity of too much muc h maintenance. Electric Elec tric power c an be transmitted transmitted by Underg Und erground round Cable Cabless or by Overhead Lines. Lines. Two main mai n reasons re asons why undergrou u nderground nd cables are rarel rarelyy used: 1. The insulation cost for underground transmission is very high. 2. It is very difficult difficul t to prov p rovide ide proper prope r insulation to to the cable cab le to withstand high hi gh voltage required for ec onomic onomic transmission. Main Mai n Components of Overhead Lines 1. Supports – this this may m ay be poles or towers and keep the the conductor at the suitable level above the ground. It depends on the working voltage and region where these are used. 2. Cross arms and Clamps – this provides support to the insulator and made of either wood or steel angle section and are used on pole structures. 3. Insulators – which are a re attached to support, take strain strain and insulate the conductors cond uctors from the ground. It c an be be pin, strain or suspension suspensio n type. type. 4. Conductors – which carry electric power from the sending end station to the rec eiving end station. station. It can be composed of copper, aluminum, ACSR or of any other composition depending upon the current to be carried carrie d and the span of the line. l ine. 5. Guys and Stays – braces or cables are fastened to the pole at the termination or angle poles to resist lateral forces. 6. Lightning Arrestors – to discharge excessive voltages built buil t upon the the line, lin e, to earth, ea rth, due to lightning. l ightning. 7. Fuses and Isolating I solating Switches Switches – to isolate different parts of the overhead system. 8. Continuous Earth Wire – is run on the top of the towers to protect the line against again st lightning disc harges. 9. Vee Guards Gu ards – are often provided below be low bare ba re overhea overheadd lines running runni ng along alon g or across public streets to to make ma ke the the line safe if it should break. 10. Guard Wires – are provided above or below power lines while whi le crossing c rossing telephone or telegraph lines li nes.. The


14

Electrical Transmission and Distribution System and Design guard wires and steel structures are solidly connect c onnected ed to earth. 11. Phase Plates – to distinguish the various various phases. p hases. 12. Bird Guards – a stick of ebonite with rounded top is fixed near the insulator on the cross arm to prevent flashover due to birds pecking on the conductors (on lines with pin insulators). insul ators). 13. Danger Plate – it is provided on each pole, as a warning measure me asure indicat indic ating ing the working voltage voltage of the a t a height of line and the word “danger”. ItIt is i s provided at 2.5 m from the ground. 14. Barbed Wire – is wrapped on a pole at a height of about 2.5 m from the ground g round for for at least 1 m eter. T his prevents prevents climb cl imbing ing byunauthorized persons. 15. Miscellaneous Items – such as vibration dampers, top hampers, ham pers, beads for jumpers jum pers etc. T ransmission Line Supports - These are the supporting structures of overhead line conduc co nductors tors on various types types of poles and towers. - It can carry the load due to insulators and conductors including inc luding the ic e and wind loads on the conductors along along with the wi nd load on the support suppo rt itself. lie in - A distinction is drawn between straight poles which lie direct direc t line of transmission and norm ally only support the conduc co nductors tors and the special spec ial poles which may m ay carry some load due du e to conductor co nductor tension. These T hese latter supports may may be angle towers, terminal towers, towers at tee – off parts, anchor ancho r towers, or towers for some special spec ial such as for c rossing the rivers. rivers. In case of telegraph or o r railway line crossing spec ial requirements are to be met with. - T he choice choi ce of line supports for a situation depends depen ds upon upon the line span, cross – sectional area, line voltage, voltage, cos co st and local c onditions. onditions. - T he design of an overhead line supports depends upon upon the fact whether the support is rigid or has a certain amount amou nt of flexibility in the direction direc tion of the line. lin e. Wooden Wooden poles and some special types of steel structures are of latter type and only the transverse wind pressure occurring upon the conductors and upon the support itself is usually considered in their design. The longitudinal longitudi nal pull of the conductors co nductors is normally normal ly balanced balanced on either side si de of the support but in the event event of breakin b reakingg of one or more conductors c onductors on one side, there there will be an an unbalanced load which may be far in excess than transverse wind pressure. With flexible supports this unbalanced load is quickly absorbed by an increase in sag in undamaged span because of bending of the supports on each side if the wrecked wrec ked span towards the the adjoining adjoin ing spans. After After three or four spans the longitudin longi tudinal al pull becomes negligible. A certain general rigidity in the longitudinal longitudi nal direction is provided for in prac tice by using using

rigid anchoring towers at an interval of 1.5 km or so. These anchoring towers are designed to withstand the breaking of one wire in three on one side as well a s the transverse transverse load. l oad. - In case of rigid supports such as the lattice lattic e – steel broad – base structures equal strength is usually provided in both the longitudinal l ongitudinal and transverse transverse directions direc tions and every tower is designed to withstand the unbalanced load because of breaking of one wire in three on the same side. In addition, anchor towers are often provided, in which the support can withstand the failure of two conductors out of three, or even all conductors on one side. Characteristics Chara cteristics of Transmission Line Supports 1. High mechanical strength to withstand the weight of conductor conduc tors, s, wind loads etc. 2. Light in weight without loss of mec hanical strength. strength. 3. Cheap in cost c ost and econom ical to maintain. mai ntain. 4. Longer life. 5. Good looking 6. Easy acc essibility of conductors cond uctors for maintenance. mai ntenance. Types Type s of Transmissi Transmi ssion on Line Supports 1. Wooden Poles - these are made of seasoned wood and are suitable for lines of moderate cross – sectional area and of relatively short span up to 60 m. - Such support is cheap, easily available, provide insulating properties and therefore widely used for distribution purposes in rural areas. - In districts having a Figure 33 Single 33 Single Pole Construction Show ing Installation Installation of Three – plentiful supply of Phase, 4 Wire Cable in Vertical timber and where the Formation. A 2 Wire Serv ice Tapping Supplies Consum ers Homes cost of transporting

steel towers is high h igh single and ‘H’ ‘H ’ poles have been been

used for overhead lines l ines operating at voltages up to 130 kV and av a verage span lengths of 150 m eters. Sal Sal or chir wooden poles up to 11 meter length with minimu min imum m circumference circumference of 38 cm at the top top and 66 cm at the bottom bo ttom are used. wooden - are very eleastic and l ines employing wooden supports are often designed throughout for the transverse transverse load. Longitudinal Longi tudinal strength strength at terminals


15

Electrical Transmission and Distribution System and Design and for anc hor support is provided provided by b y means of guys. guys. Double pole structures of A and H types are often employ empl oyed ed by obtaining a higher h igher transverse transverse strength strength than that could cou ld be econom ically provided provided by mean m eanss of single poles. - The height of a wooden pole depends upon clearance cl earance above the the ground g round surface and secondly sec ondly,, the number numbe r of cross – arms arm s and other equipment equip ment to to be attached. attached . Normally, Normal ly, the height of wooden wood en pole is 10 to 12 m. Types of Wooden Wooden Poles 1. Single member poles are ordinary poles and are used in all positions posi tions where there is no undue stress or tension and where no transformer or switchgear switchge ar are to be mounted mounted on them. 2. A ‘A’ poles are used mainly where bends in lines cause strain and single poles are not suitable. “A” poles consist of two m ember embe r poles spaced spaced at Figure 34 "A" 34 "A" Type Pole the base and joined at the Utilized for 3 Phase 3 top, held together by cross – Wire HT Trans mission w here route route of cables bars in the form of letter A. A. turns at an angle 3. ‘H’ poles comprising two single poles pol es strapped together together by steel or wooden wooden cross – pieces are used mainly where transformers and switchgear are to be mounted on them. 4. Four member poles comprise com prise of two two ‘H’ units in in the form of a square joined by cross – bars. They are used where extra heavy transformers and switchgear are required, usually at the junction of number of circuits. ci rcuits.

e. Required periodic inspection

Figure 37 4 37 4 - Member Pole Construction with Transformer, Switchgear and Fuses . LT Four Wire, Wire, 3 Phase Distributor Distributor is Tapped Off Off Transformer Secondar Secondaryy to Supply a Village

2. Steel Poles - are often used to substitute for wooden poles. - It possesses greater mechanical me chanical strength, longer life and permit permi t longer span to to be b e used (50 – 80m). 8 0m). Such Such poles are general ly used for distribution in ci ties. - It needs to be galvanized or painted or prolong its life. - T he average life of steel steel poles pol es is more than 40 years.

Figure 35 "H" 35 "H" Pole Supporting Transformer

DISADVANTAGES: a. Tendency to rot below ground level b. Comparatively smaller/short small er/shorter er life c. Less mec hanical strength d. Cannot be used for higher voltage above 22 kV

Figure 36 Wood 36 Wood Pole w ith Protection Protection Cap


Figure 38 Steel 38 Steel Poles

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Electrical Transmission and Distribution System and Design Types of Steel Poles 1. Tubular poles are of round cross section. The advantages of tubular poles are that these are lighter in weight and easy to install though initial cost co st is little more as c ompared to wooden poles. po les. ItIt does not no t require special speci al equipment for its erection. erection. T ubular poles in height of 9 to 11 m are generally used for distribution purposes in citie ci tiess to give good good appearance. 2. Rail poles are of the shape of the track used for railways. Steel Steel rail poles in height of 11 m and 13 m are used for transmission purposes at 11 kV and 33 kV respecti respe ctivvely. 3. Rolled steel joints are of I cross c ross section. 3. RCC Poles rei nforced c oncrete poles es - the reinforced become very popular as line supports in i n recent rec ent years. years. - are extensively used for low voltage and high voltage distribution lines up to 33 kV. kV. - It has a greater mechanical strength, longer life and permit permi t longer span (80 – 200 m) than steel poles. gi ves good outlook, require require - It giv little maintenance and have good insulating properties. Figure 39 Concrete 39 Concrete Pole Because of its heavy weight, such poles are often manufactured at the site to avoid avoid cost transportation. - T heir consruction should conform to the standard standard specification for RCC work, but in no case the dimensions shall be less 25 cm × 25 cm at the bottom bottom and 13 cm × 13 cm at the top. - Prestressed concrete poles, called PCC poles, are less bulky and lighter than RCC poles. PCC poles are extensively used on 11 kV and its lines. li nes. Types of RCC Poles 1. Square Cross Section 2. Rectangular bottom and square top with rectangular holes in it to facilitate insulating properties and resistance against chemical chemic al action. action. 4. Lattice Steel Tower - Wooden poles are generally used for distribution purposes in rural areas, the steel tubular poles and and concrete poles are usually used for distribution in urban area to give good appearance and steel rails ls or narrow – base, lattice – steel towers are used for transmission at 11 kV and 33 kV and broad – base

lattice – steel towers are used for transmission purposes at 66 kV and above. - The height of the tower depends dep ends on the the line voltage and length of span. The legs of the towers are set in special concrete foundations. The forces to be considered in the design of a tower are 40 500 kV DC Lattice Lattice T ower vertical loads of Figure 40 500 conductors, insulators, fittings and tower itself, wi nd pressure on conductors and wind pressure on tower itself. For protection against corrosion the steel towers are periodically painted or galvanized. The life of steel towers can be made almost indefinitely large by a reasonable amount of attention to their Figure 41 Single 41 Single Circ uit Tower maintenance. Characteristics of Broad – Base, Lattice Steel Towers 1. Mec hanically stronger and got longer life. 2. Long spans (300 m and above) can be used and are much useful for crossing fields, valleys, railway lines, rive etc. 3. Even though these are two to four time costlier costlie r than wooden wooden poles, yet for tall supports and longer longer spans these prove Figure 42 Double 42 Double Circuit Tower more ec onomical. 4. Reliability Reliab ility is of a high hi gh degree. 5. Capable of withstanding the most severe climatic conditions, and immune from destruction by forest fires. 6. T he risk of service interruptions, due to broken broken or punctured pun ctured insulators, is considerab con siderably ly reduced reduced owing to use of large spans.


17

Electrical Transmission and Distribution System and Design 7. Lightning troubles are also minimized as each tower is a lightning conductor, whereas on wooden pole pol e lines shattered poles and wreck w recked ed line sections sec tions are not infrequent. 8. Fabricated from painted or galvanized angle section which can be transported separately and the erection erec tion done on site. 9. At At a m oderate cost these these c an be designed desig ned for for double circ c ircuit uit giving giving a further further insurance insuranc e agains against discontinuity disc ontinuity of supply. In case of breakdown brea kdown to to one circui c ircuitt it is possible to carry out repairs while le maintaining mai ntaining the c ontinuityof supply on the other other circuit. The advantages and disadvantages of single circuit and double circui ci rcuitt designs are given given in the table: Sing le Circuit Design Its structure is lighter in weight and required less strong foundation because it is subjected to low wind pressure on conductors and structure itself. It need much lower support for equal conductor clearance to earth but it requires more way leave for same number of circuits. Two earth wires are required for single circuit as these cannot be disposed at the top. Danger of flashover is most unlikely and repairs can be carried out without danger to workmen from other circuits. Reliability regarding continuity continuity of supply supply is less. It is more expensive for two circuits than the double circuit design. Greater spacing of conductors is required resulting resulting in greater greater inductiv inductiv e reactance. The phase performance along the line is unbalanced as the central conductor passes at the top of the support, which is an obvious drawback.

Double Circuit Design Its structure structure is heav ier in weight and of more height. It requires relatively stronger foundation. It is subjected to more wind pressure. It needs taller structure but less way leave for equal number of circuits.

Only one earth wire is required requ ired for for two two circuits and a nd more protection agains t lightning is had due to its disposition on at the top. op. There is always danger from the other live circuit.

Reliability regarding continuity continuity of supply is more. more. It is most economical econo mical and cheaper. Lesser spacing of conductor conductor s is required; hence the inductive reactance is less. It gives give s better better approach appr oach to the triangular arrangement; hence the phase performance performance will be more balanced.

Table 2 Comparis 2 Comparis on of Single Circuit to to Double Circuit Design

Classifi Clas sificat cation ion of Steel Towers 1. T angent towers can be used for straight runs of the line and up to 2° - line deviation from the straight run. The line is straight or along the tangent to the line route. In such towers the stress is because of the weight of the conductors, ice and wind loads. Figure 43 Steel 43 Steel Tower In addition, extra forces due to break brea k in the line lin e on one side of the tower is also to be considered in the design of towers. The base of such a steel tower may be square or rectangular. rectangul ar. Insulators used wit wi th such such towers are suspension types. types. 2. Deviation tower are special angle towers with line deviation exceeding 2°. They are used where the transmission transm ission line lin e changes cha nges direction. direction. Such towers have broader base and stronger mem bers as they are to withstand the resultant force force due to change in direction in addition to to the forces to which the tangent towers are subjected. subjec ted. Insulators Insulators used with such towers are of strain type. The cost of deviation tower is larger than that of a tangent tower becau be cause se it is designed to withstand heavy lo ading as compared to standard or tangent tower. Deviation towers are further further classified c lassified as: a. Small angle towers (2° to 15° change in direction) b. Medium Medi um angle towers (15° to 30° c hange hange in in direction) c. Large angle towers (30° to 60° change in direction and dead end) Conductor Mater M aterials ials The conductor is one of the important items in a transmission and a nd distribution system of elec tric power, po wer, the cost of the conductor cond uctor material accounts for a m ajor part of the total c ost. So, proper choic ch oicee of conductor cond uctor material and and size size of the conductor condu ctor is of utmost importance. imp ortance. Characteristics Chara cteristics of Conductor 1. High electric elec tric conductivity conductivity or low specific resistanc e 2. High tensile strength to withstand the mechanical stresses 3. Low specific gravity gravity to give low weight per unit volum volumee 4. Low cost to be used over long distances usually involved involved in transmission transm ission lines


18

Electrical Transmission and Distribution System and Design 5. Easy availability 6. Should not be brittle T he m ost commonly used conduct con ductor or materi m aterials als for overhea overheadd lines are copper, aluminum, steel – cored aluminum, galvanized steel and cadmium copper. Aluminum has replaced replac ed copper c opper as the the most m ost commo com monn conduct c onductor or metal for for overhead transmission. Although larger aluminum cross sectional area is required to obtain the same loss as in a copper cop per conductor, aluminum alum inum has a lower cost, co st, light weigh, and the supply of alum inum is abundant. Common Conductor Co nductor Types 1. Stranded Hard Drawn Copper is the best conductor owing to its high electrical conductivity and great tensile strength for al l types types of transmission. transmissio n. Thr Th rough ough hard drawing (cold rolling and drawing) reduces the conductivity slightly but increase the tensile strength considerably. con siderably. However, However, medium hard – drawn dra wn c oppe opperr is suitable for distribution lines li nes and soft – drawn c opper opper wires are suitable for secondary distribution circuits, and for service connections to buildings. Though copper is ideally suitable for transmission and distribution, but due to scarc ity of materials, m aterials, the trend trend nowadays is to use aluminum alum inum in place pl ace of coppe cop perr. The use of cupper cupp er is being restricted for manufacturin man ufacturingg of of the mac hines only. only. Characteristics: a. It does doe s not c orrode in normal atmosphere and is is not subjected subjec ted electroly elec trolytic tic troubles. b. It has higher current density so lesser cross – sectional area of conductor is required and so lesser area is subjec ted to to wind loads. c. It is quite hom ogenous, durable and of high hig h scrap scrap value. d. It has long life and ease of jointing. 2. Alumi Aluminum num is cheaper c heaper in cost and lighter in weight but but is poor in conductivity and tensile strength as com pared to copper. T hese days days there is great trend towards the aluminum alum inum as conductor co nductor material, becaus becausee of its greater availabi availability lity and cheapness c heapness in c omparis omparison on to copper. Characteristics: a. Its conductivity is 60% of that of copper and density is 0.303 time timess that of copper. c opper. b. Has a diameter about 1.26 times that of copper conductor of equal resistance but due to its low density only half weight we ight of alum inum is required to to that of copper. c. The tensile strength is much lower than that of copper (45% to that of copper), but the larger cross – sectional area if metal neutralizes the difference to some extent, and an aluminum

conduc con ductor tor has about 75% of the ultimate strength strength of the equivalent copper conductor. d. For the same conduc tivity tivity aluminum alumi num conduc con ducttor having 1.66 times the cross section of copper is required thus causes a greater surface for wind pressure and supporting structures struc tures are required requi redto be designed for greater transverse transverse strength. e. High towers must be employed with aluminum conductors than would be required with copper conduc con ductors tors having the the same length of span. f. The sag is greater in aluminum wires due to reduced reduce d working stress permissible and be beca caus usee the linear l inear c oefficient of expansion of aluminu alum inum m is 1.4 times of the copper. g. Alum Aluminum inum conductors con ductors being liable liabl e to swing, requires larger cross arms/ h. Low melting point thus it cannot withstand short circuits. i. Jointing Jointing is also difficult difficult compared c ompared to copper. AA AAC (All (All Alum Aluminum inum Conductor) weighs only half as the equivalent copper conductor and cost per unit length of prevailing prevailing market m arket rates, is c onsiderably less. T hey are m ainly used for for low voltage distribution overhead overhead lines having having short spans of up to 65 m . 3. AA AAAC (All (All Aluminum lumi num Alloy Alloy Conduc tor) 4. ACA ACAR R (Alum (Aluminum inum Conductor Alloy Reinforced) 5. Steel – Cored Alumi Aluminum num or ACSR (Aluminum (Alumin um Copper Copper Steel Reinforced) or Alumoweld (Aluminum – Clad Steel Conductor) consists of layers of aluminum strands surrounding a central core of steel strands. The Steel strands are galvanized to prevent rusting and elec trolytic trolytic corrosion; the cross section sec tion are in the the ratio 1:6 but in c ase of high strength c onductors thei their ratio may be 1:4. Characteristics: a. Steel core takes a greater percentage of mechanical stresses while the aluminum carries the bulk of current. cu rrent. b. Has the largest diameter than any other type of conductor of same resistance, so corona losses are reduced, but stronger supports are required for a given span. c. Has high tensile strength and lighter in weight produces small sag and therefore longer spans can be used thus number of line supports m ay be be reduced by 25% and frequency of occurrence of fault is reduced. d. Skin effect is i s very predomi pred ominating nating hence henc e the resistance of the composite conductor is taken equal to that of alumi num cov co vering alone.


19

Electrical Transmission and Distribution System and Design e. Reactance Reac tance is taken to be equal to to that of o f a non – magnetic magn etic c onductor onductor having having diam eter equal to that outside of the conductor. f. Critical Critica l voltage limit lim it of the conductor co nductor can be raised raised by 30 to 50% 50 % as c ompare ompa redd to to copper conductors. cond uctors. g. Gets deteriorated in service to the atmospheric corrosion due to bim etallic action of zinc and the aluminum, alum inum, electrochemical in nature. nature. 6. Galvanized Galvanized Steel – have been used to advantage a dvantage for extremely long spans, or for short line sections exposed to normally high stresses due to climatic conditions. These are found most suitable for lines supplying rural areas and operating at voltages of about 11 kV, where cheapness is the main consideration. con sideration. They T hey are not suitable for EHT lines for transmitting large amounts of power over a long distance due to poor conductivity (13 % of copper), high internal reactance and due to eddy current and hysteresis. Nowadays, their use is limited to telecommunication teleco mmunication lines, stay wires, earth wi res and guard wires. 7. Cadmium Copper – addition of 1 or 2% cadmium in copper cop per increases inc reases the tensile strength by about 40% 40% and reduc es the c onductivit onductivityy by 17% below that of pure pure copper. cop per. Use of Cadm ium Copper will be economical economical for for a line with wi th long spans and smal l cross section, sec tion, wher where the c ost of conductor c onductor material is comparatively comparatively small in in comparison to that of supports, etc. They are also employed for telephone and telegraph lines where currents involved are quite small. However, owing to scarc ity of copper, cop per, cadmium cadm ium copper conductors con ductors on com munication lines are being replaced replac ed by ACSR ACSR conductors. 8. Copper – Clad Steel – is i s obtained by welding we lding a c opper opper coating coa ting on a steel wire core. co re. Line c onductors onduc tors made of of copper – clad c lad steel are preferably stranded, and have a considerably large tensile strength than the equivalent all – c opper conductors. c onductors. The proportion proportion of copper cop per and steel is c hosen that the the c onductiv onduc tivity ity of the com posite wire is 30 to 40% of that of c opper c onduct onductor or of equal diameter. Such material appears to be very suitable for river river – crossings c rossings or other places where an an extremely long span is involved. involved. 9. Phosphor Bronze – use if harmful gases such as amm onia are present in atmosphere and the spans are are extremely long. In this conductor, some strands of phosphor bronze bronze are added to the c admium admi um copper. 10. Expanded ACSR – has filler such as fiber or paper between alum inum and steel strands. Filler Fill er are used to i ncrease the conduc tor diameter diam eter that will reduce reduc e magnetic effects and reduce c orona power loss.

All All conduc con ductors tors used for overhead overhead lines are preferably stranded to increase flexibility. Solid wires, except of smaller small er sizes, sizes, are difficult diffic ult to handle and when w hen employed em ployed for for long spans tend to crystallize at the points of support because of swinging in winds. Stranded conductors usually have a central wire around whic h these these are succ essive layers layers of 6, 12, 18, 24 wires wi res.. For n layers, the total total number n umber of indiv indi vidual wire wi re is 3n(n+1) 3n(n+1) +1. If the diameter di ameter of each eac h strand strand is d, then diame d iameter ter of the the stranded conductor will be (2n+1) d. In the process of manufacture adjacent layers are spiraled in opposite directions so that the layers are bound together. The

method of construction c onstruction is called call ed as ‘concrete lay’. lay’.

With c onductors onduc tors of large cross section, sec tion, however, however, another method known as ‘rope lay’ is som etimes employ emp loyed ed to give more flexibility flexibi lity.. Mechanical Properties

Copper

Aluminum

Steel Steel

Specific Weight (kg/m3 ) Young Modulus (kg/mm 2) Ultimate Tensile Strength (kg/mm 2) Specific resistance (Ω m/mm2) Resistance Temperature Coefficient

8900

2700

7860

Aluminum Aluminum and and Steel 1:3 1:4 3450 3700

13000

5600

20700

7500

8300

40

18

40 to 320

120

120

0.01786

0.0287

0.178

0.0038

0.004

0.00496

Table 3 Properties 3 Properties of Conductor Materials of Various Types

Examples: 1. A given amount a mount of power is i s to be transmitted transmi tted by an overhead line. Compare the diameter and weight of aluminum alum inum conduct co nductor or with those of copper coppe r for the same power loss in the line. The following data may be assumed: Specific resistance of aluminum = 2.85 µΩ - cm Specific resistance of copper = 1.70 1.70 µΩ - cm Specific Speci fic gravity gravity of alum inum = 2.71 Specific Speci fic gravity gravity of copper c opper = 8.89 8.89 2. Electric Elec tric power of 30 MW at a pf of 0.8 lagging laggin g is to be transmitted by a 132 kV, 3 phase, ph ase, 3 – wire transmissi transm ission on line over a distance of 120 km. The efficiency of transmission is 90 %. Calc ulate: a. Weight of material required in case of copper having resistivity of 1.78 x 10 -8 Ω - m and specific specific gravity of 8.9. b. Weight of m aterial required in case of aluminu al uminum m -8 having resistivity of 2.6 x 10 Ω - m and a nd specific specific gravity of 2. Cross Arms T he function of a line lin e support (pole) is to support supp ort the line conductors at a safe distance from ground whereas the


20

Electrical Transmission and Distribution System and Design function of cross c ross arms it to keep the conductors cond uctors at a safe distance from eac h other and from from the pole. pol e. Cross arm is a cross – piece fit piece fitted ted to the pole top end portion by means mean s of of brackets, known as pole as pole brackets brackets,, for supporting suppo rting insula in sulators. tors.

with disc insulators. insula tors. The cross arm is fixed fixed to the pole in such a manner that the load of the conductors is taken by the c ross arm and not the clamp cl amp or bolt that that fixes the cross arm to the pole. pol e. Shape of Cross Arms 1. U – Shaped 2. V – Shaped 3. Zig – Zag Shaped T o prevent prevent arcing, arci ng, the c onstruction of the cross arms should be such that under the worst c onditions, the spacing spacing between conductors, c onductors, when swinging, would woul d never be less less than that given given in T able 4. Working Voltage Spacing 6.6 kV 76 mm 11 kV 101 mm 33 kV 190 mm and so on Table 4 Spacing 4 Spacing for Cross Arms dependent dependent on Working Voltage

Pole Brackets Brackets and Clam ps Pole brackets of different types types are shown in the Figure Fi gure 45.

Figure 44 Cross 44 Cross Arms

Types of Cross Arms 1. Wooden Cross Arms are c omm only employ empl oyed ed on 11 kV and 33 kV lines. These are made of sal wood, seasoned sheesam wood or creosoted fire wood. It should be lengthwise and completely free from knots and are preferred owing to their insulating property which provides safety to line staff and minimizes flashover flashover due to birdage. T he usual lengths and cross sections in use are: are: 1.5 1.5 mm x 125 mm x 125 mm for 11 kV lines li nes and 2.1 m m x 125 mm x 125 mm m m for 33 kV lines. Wooden Cross Arms Arms need replac ement owing to to decay every 5 – 7 years depending upon weather conditions. 2. Steel Cross Arms are stronger and an d are generally general ly used on steel poles. For LV distribution, the angle iron or channel cha nnel iron c ross arms shall have have a size size not less than than 50 mm x 50 m m x 6.4 mm and 76 mm x 38 m m. Its Its length shall be suitable for the spacing of the conduc con ductors tors and strong enough to withstand the the resultant forces caused by insulators, their pins and dead weight of insulator attachments etc. To avoid birdage on HT lines, l ines, V – shaped cross arms are used used

Figure 45 Various 45 Various Types of Pole Brackets

Clamps are made of flat iron and are used for fixing or holding holdin g serv service ic e line stay wires, wi res, earth wire, shac kle insulators, cross arms etc. In c ase of servic servicee lines, one end

of the clamp is made longer and provided with an ‘eye’

section, as shown in Figure Fi gure 46.

Figure 46 Various 46 Various Types of Clamps


21


Figure 47 Method 47 Method of Fitting Cross Arm to Pole

Guys and Stays It bec omes essential essential to to stay ov o verhead line supports at angle and termi nal positions as the poles takes the the pull due to the conduc con ductors. tors. T he theoretical theoretical angle between the pole pole and stay should shoul d be 45°, but in general gen eral practice practic e it is not always possible to obtain this, and so stay design is generally based on a m inimum inim um angle of 30° between be tween stay stay and pole.

enough above the ground level so that the guy wire w ire does not meet the soil of the ground. An An egg type type strain insulator insul ator is inserted in the the guy wire for safety. It isolates stay wire electrically from metal support. T he two ends of guy are threaded through the insul ator in in such a way that the porcelain of the insulator is under compression, making it possible to withstand a large pull. In case of breakage of porcelain, the guy will still be effective due to linking of two ends. Stay wires are galvanized steel wires having a tensile strength of 7 tons/cm 2 and are usually of stranded section. T he stay stayss are a re provided provided to angle angl e poles, termina terminall poles, H – pole structures, where the span on the two sides of the pole are very much mu ch different and to supports subjec ted to uprooting pressure. Stay Stay wires wi res are placed place d on the poles and stay rods fixed before stretchi stretching ng of line conduc con ductors tors on the he poles. This T his is very important, impo rtant, otherwise, in c ase the line conduc con ductors tors are placed plac ed first, the poles pol es would be pulled pulle d out of the position. p osition. The T he stay wire is held he ld up on the pole either by m eans of guy hook or through the the bolt or may m ay be fixed to the clamp. The stay wire is drawn up until the pole is pulled over slightly towards the stay to keep the proper tension. Different stay arrangements arrangemen ts are shown shown in Figure Figu re 49.

Figure 48 Guys 48 Guys and Stays

Stay set c onsists of MS M S rod of 19 m m diameter, stay bow, checknut, che cknut, timbles, timble s, stay wire, stay clam c lampp and CI ancho anc horr plate 450 x 450 mm having 4.8 mm hole in the center. Except the plate all al l other parts are galvaniz g alvanized. ed. Stay rod is embedded embe dded in c ement ement concrete 1:3:6 1:3:6 to a depth of not les l esss than 1.67 m keeping length of 46 cm of the rod projected above the ground level. One end of stay wire is fixed to stay rod at the bottom and to the stay clam c lampp to the pole by means mean s of well splic ed joints with a strain insulator i nsulator and turn buckle buc kle inserted in the middl e and near the top respec tively. tively. T he stay or guy is tightened by means of stay stay bow and and anchor anc hor rod to the required tension. Mild M ild steel thim bles bles are are used at both ends of the stay wi re, otherwise strands of stay wire m ay get damaged. T he anchor rod rod is projected projec ted far far Figure 49 Stay 49 Stay Arrangements


22

Electrical Transmission and Distribution System and Design Conduc tors Configuration, Spaci Spacing ng and Clearances Conductor Cond uctor Configuration Configuration Several conductor configurations are possible, but three most common com mon conf c onfigurations igurations are: a. Horizontal Configuration (horizontal disposition of conductors) – all the conductors c onductors are mount m ounted ed over one one cross arm and needs supports supp orts of smaller height heig ht but but a wider right of way. M ost Economical Econom ical for Single Circuit Circuit Lines. b. Vertical Configuration – use in congested c ongested areas where where horizontal horizontal arrangem a rrangement ent is not possible. The drawbac drawbackks are taller towers and more lightning hazards. Most Economical Econo mical for Double Circuit Lines. c. Symm Symmetrical etrical Delta or T riangular Configuration Configuration

Figure 50 Triangular 50 Triangular and Horizontal Formation

In unsymm unsymmetrical etrical arrangem arrangement ent of conduc con ductors, tors, the conductors are usually transposed at regular intervals to balanc e the electrical characteristics of various phases, and and prevent inductive interference with neighboring com munication circuit ci rcuits. s. Conductor Spacings Spacings T he spacing spac ing of conductors is determi ned by considerat conside ratio ions ns partly electrical and partly mechanical. Larger spacing causes increase in inductance of the line and hence the the voltage drop, so that to keep kee p the latter within a reasonable reasonable value the conductors should be as close together as is consistent con sistent with the prevention prevention of c orona. The basic Consideration regarding Minimum Spacing between conductors are electrical clearances between conductors under the worst condition (maximum temperature and wind pressure) p ressure) shall not be less than the the limits lim its of safety safety,, particularly particul arly at the m id spans. Owing to the action of the gusts of wind, conductor c onductor has tendency to move move about in an a n elliptic ell iptical al path, therefore, in case of suspension suspension insulators, the mechanical clearance to supporting structures should be calculated with a 45° swing of the suspension string toward the the structure.

Spacing = √SV/150

T able 5 shows some typical typical spacing values. values. Line Voltage in kV Spacing in meters

0.4

11

33

66

132

220

400

765

0.2

1.2

2.0

2.5

3.5

6.0

11.5

14

Table 5 Conductor Spacings

Conductor Cond uctor Clearances T he minimum mini mum vertical vertical c learances between between the ground and the conduc tor are are shown in Table 6. Line Voltage in kV Clearance to Ground in meters Across Stree Streett Along Along Str S tree eett Other Areas

0.4

11

33

66

132

220

400

5.8 5.5 4.6

5.8 5.5 4.6

6.1 5.8 5.2

6.1 6.1 5.5

6.1 6.1 6.1

7.0 7.0 7.0

8.4 8.4 8.4

Table 6 Conductor 6 Conductor Clearances

Span Lengths Neglec ting the influence upon the span length of such local local conditions as the necessity for the following the configuration con figuration of roads, canals cana ls or railways, it is interest in teresting ing to to note that there is one definite value value for span length which will give the m inimum inim um overall overall cost of the line. As the length of span increases, increa ses, the number num ber of insulators and supports decreases dec reases resulting in decrease in cost but at the same tim timee the height of the support will go up to allow for more sag and the length of the cross arms must be inc reased to take take up increased spacing, this will cause increase in cost. Moreover, the insulators constitute the weakest part of transmission line l ine and reduction redu ction in number num ber of towers per km km with the use u se of longer l onger span increases the reliability reliabili ty of the line. Thus, it is not possible to give any hard and fast rule as to the best span length to be adopted, and the only way to determine it is to calculate the total cost per km for several several different di fferent span lengths, leng ths, and plot the results to get the most economical econom ical span length. Many times, time s, it happens that the c onductor onduc tor size size determined determined from electrical calculations comes out rather small, it is possible to reduc e the the total total c ost of line by using a thic ker and stronger conductor, and increasing the span length. Sometimes Sometim es it is not feasibl feasiblee to determine determ ine the height of the line support and span length based on line cost alone as lightning hazards increase greatly with the increase in height of c onductors above above the ground. T he usual spans are: a. Wooden Poles: 40 – 50 m b. Steel Steel T ubular Poles: 50 – 80 m c. RCC Poles: 80 – 200 m d. Steel Towers: T owers: 200 – 400 m and above For river river – crossings exceptionally excep tionally long spans up to 800 m or so hav h avee been satisfactorily employed. empl oyed.

where S is sag in meters and V is li ne voltage voltage in kV. kV. Electrical Enginee ring Department | Engr. Engr. Gerard Francesco Francesco DG. Apoli nario

23

Electrical Transmission and Distribution System and Design Sag and a nd T ension (Stress) Analysis Analysis Overhead lines are supported on mechanical structures consisting con sisting of com ponents like insulators, cross arms, poles poles or towers etc. The T he strength of these com ponents must be such that there is no mech m echanical anical failure failure of line, even under the worst weather conditions. Analysis Analysis of sag and tension of c onductor onduc tor is an im portant portant consideration in overhead transmission as well as distribution line li ne design. desig n. The continuity continuity and quality of electric electric service depend largely on whether the conductors have been properly installed. install ed. T hus, it must be determined determin ed in advance the amount amoun t of sag sag and tension to be given to the conductor at a given temperature, maximum wind, and possible ice loading. To specify the tension to be used in stringing the line c onductors, the values values of sag and tension in sum mer and a nd winter win ter conditions conditions must be known. know n. Excessiv Excessive tension may m ay cause mechanical failure of o f the conduct c onductor or itself itself because bec ause conductor cond uctor tension tension contributes c ontributes to to the mec hanica hanicall load on structure angles in the line and at dead ends. The main factors in the design and stringing of conductors on the supports supports 1. Conductor load per unit l ength 2. Conduc tor tension (less than 50% of o f its ultimate tensile le strength even even when there is 12.7 mm m m radial co ating of of 2 ice ic e and a wind pressure of the order 380 N/m ) 3. Span, that is, distance di stance between supports 4. Temperature For determining determi ning the c onductor onductor load, the factors fac tors that need to to be considered are: 1. Weight of conduc tor itself 2. Weight of ice or snow clinging to conductor 3. Wind blowing against conductor con ductor The maximum effective weight of the conductor is the vector sum of the vertical weight of the conductor and horizontal wind pressure. It is extremely important to include inc lude the most adverse condition. c ondition. From the the design point of view, it is considered that the wind is blowing at right angles to the line and to act against the projected area of the c onductor, onduc tor, which inc ludes ludes the projected area of ice or snow that may be c linging lingin g to to it. From the practic al point of view, ec onomic onom ic design dictates dictates the following: 1. Sag of conductor should be minimum to refrain from extra pole height 2. Sufficient clearance cl earance above ground level. level. 3. To avoid providing excessive horizontal spacing between conductors to prevent those swinging together in mid span. Sag of the conduc c onductor tor decreases decreases because its tension pull the the conduc con ductor tor up. But at the sam e tim time, e, tension elongates the

conduc con ductor tor from elastic stretching which tends to relieve relieve tension and sag increases. T he elastic property property of wire is m easured by its modul m odul us of elasticity.

TA kg/m σ =  e = modul usstrofeselasticity

where = stress per unit area T = conductor conduc tor tension tension A = actual cross section of conduc con ductor tor Elongation of the c onductor due to to the tension is

Elongation is high if modulus of elasticity is low. Thus, a small c hange in length of c onductor causes large effect effect on sag and tension tension of conduc con ductor. tor. Sag and stresses in conductors are dependent on the following things: 1. Initial tension put on them when they are clamped in place. 2. Weight of the conductor cond uctor themselves. 3. Ice or sleet c hanging on them. them. 4. Wind pressure. Stress depends on sag; any span can be used provided the the poles or towers are high enough and strong enough. The matter is merely one of extending the catenary in both directions. direc tions. Cost of towers sharply i ncrease ncrea se with height and and loading. Example: A galvanized steel tower mem ber has original length of 22 cm and c ross – sectional area area 13 cm 2. With working axial, tensile load of 125 kN, the change in length was 0.2 mm. Calculate: a. Stress b. Strain c. Modulus Modul us of elasticity elastic ity d. Percent elongation e. If ultimate u ltimate stress stress is 110000 11000 0 N/mm 2, determine the factor fac tor of safety. s afety. Sag in Overhead Lines While erecting an ov o verhead line, li ne, it is very important imp ortant that conduc con ductors tors are under safe tension. If the conductors cond uctors ar a re too too much stretched between supports in a bid to save conduc con ductor tor material, the the stress in the conduc co nductor tor may reach reach unsafe value value and in certain c ases the conductor may m ay break break due to excessive tension. To permit safe tension in the conduc con ductors, tors, they are not fully stretched but c an have have a dip or sag. SAG – the difference in level between points of supports and the lowest point in the conduc c onductor. tor.


24


Figure 51 Conductor 51 Conductor Suspended in Equal Supports

Figure 51 shows a conductor suspended between equal level supports suppo rts A and B. the the conductor co nductor is not fully ful ly stretched but can ca n have have a dip. T he lowest point on the conduc co nductor tor is O and the sag is S. T he following points can be c onsidered: onsidered: 1. When a conduc con ductor tor is suspended between supports supports at the same level takes the shape of a catenary. catenary. However, if the sag is i s very very small compa c ompared red with the span, then the sag curve is said to be parabola. parabola . 2. The tension at any point on the conductor acts tangentially. Thus, tension T O at the lowest pt. pt. O ac ts horizontally horizontally as shown in Figure 52. 3. The horizontal component of tension is constant throughout the the length l ength of wire. 4. T he tension at supports is approxim ately equal to the the horizontal horizontal tension acting ac ting at any point of the wire. T hus, hus, if T is the tension at pt. B, then T = T O.

length of span, etc. remain remai n the same. Working tensile tensile strength of the conductor con ductor is determined determi ned by multiplyi mul tiplying ng the ultimate stress and area of cross section and dividing by a factor of safety. 4. T emperature. All All metallic m etallic bodies expand with the rise rise in temperature and, therefore, the length of the conduc con ductor tor increases increases with the rise in temperature, and and so does the the sag. T wo conditions conditions are considered in making sag sag – tension analysis: 1. At At Minimum Mi nimum Temperature. Temperature. T he lowest sag and maximum tension in conductor section occurs when the temperature is minimum and wind maximum. T ension on the conductor c onductor should not exceed the the breaking strength of the c onductor onduc tor divided divided by a factor of safety of 2.5. 2. At At Maxim um T empera empe rature. ture. On the other hand, maxim um sag oc curs when temperature is maximu maximum m and there is no wind win d pressure.

S = wTL

where w = weight of conductor in kg/m; L = length of the span in meters and and T = tension tension in the c onductor in kg kg Calculation of Sag In an overhead li ne, the the sag should shoul d be so adjusted that that the tension in the conduc con duc tors is within safe limits. The Th e tensi tension on is gov g overned erned by c onduc tor weight, weight, effect effect of wind, w ind, ice ic e loading loading and temperature temp erature variations. variations. It is a standard practice prac tice to keep keep conductor tension less than 50 % of its ultimate tensile strength. Thus, the minimum safety factor in respect of conductor conduc tor tension tension should be 2. a. CATENARY CABLE (high sag) - the unit weight weig ht is uniformly uniform ly distributed along the cabl e - for span of more mo re than than 300 meters me ters

Figure 52 Conductor 52 Conductor Tension Tensi on

T he factors affecting affecting the sag in an overhead overhead line are: 1. Weight of the Conduc C onductor. tor. T his affects affects the sag directly direc tly.. Heavier the conductor, greater will be the sag. In locations where ice formation takes place on the conductor conduc tor,, this will also increase in the sag. 2. Length of the Span. Sag is directly direc tly proportional to the the square of the span length. Hence other conditions, such as type type of c onductor, working tension, tension, temperature etc., remaining the same a section with longer span will have have muc h greater sag. 3. Working Tensile Strength. The sag is inversely proportional to the working tensile strength of conductor if other conditions such as temperature,

Figure 53 Free 53 Free body diagram of portion portion cable

Consider Figure 53, If T O is the tension at the lowest point (point O) on the wire and where whe re the the c urve urve is horizontal and T the tension at point B (distance l from the lowest point O), the portion OB (length l) is in equilibrium under the action of three forces, namely name ly T O, T and the weight of the wire wi re of length l acting vertically downward through a center of gravity, wl.


25

Electrical Transmission and Distribution System and Design So T wl TO

            y =  1        1  ≅  wxTO  12 = 2 T = TO  wlwl w x  T =  TO  TO sinh TO  T = TO11 sinh wxTO = TO cosh wxTO

If the fourth and higher hi gher order terms are neglec ted,

From Figure Fig ure 55 the T ension T at point B is given by Figure 54 The 54 The Force Triangle

The above three forces can be represented by a triangle shown in Figure 54 and fr from om this triangle

tan θ = wTOl ddxy  t a n θ = dldl = dxddldlx dy   = 11 wl  dx = 1 t tan θ = 1  TO dx =  1 dldlwl TO  x = ± x = TwO sinh− TwlO  C T = TO cosh 2TwLO   w L w L y ≅ 2TO2 = 8TO x =  s i=nht−an θ =  == sin×h  sinhl =  = ssininhh       =  sinh 2 =  2  48  ⋯  y = TO cosh   D D = ≅ TO × wL  TO × wL ≅ L  wL w 2TO w 48TO 2 48TO     w L y = TwO cosh wxTO  TwO = TwO cosh wxTO  1 =LL  24TO T = T cos h O            w x w x w x wx wx wx     = T 1    ⋯ ≅ T 1        O O    wx T T T 2T 24T 2T O O O O O O cosh TO = 11  2  4  6  ⋯ From the above above triangle, we c an say dl = T, dx = T O and dy dy = wl. Thus, T hus,

and

or

Simplifying,

Figure 56

If the line is supported between two two points A and B at the same level and the length of the span is L (Figure (F igure 56), then then

Integrating Integrating both sides, sid es, we get

at the supports

From initial conditions when x = 0 and l = 0, we get C = 0, thus,

T he sag sag ymax is i s the valu valuee of y at A or B and is given by

or

and

or

Again Again

Length of line in a half span

Integrating Integrating both sides, sid es, we get

Neglecting terms of order exceeding cube, we have

From initial conditions when x = 0 and y = 0, we get thus,

T his is the equation of the curve called cal led the catenary. catenary. T he function cosh is the hy hyperbolic perbolic c osine and is such that

or length of line in full span length

Tension,


26


    w L w L T = TO 1 8TO  = TO  8TO = TO  wy T   = TO  wy  Sag,S = w8TL wL Total lengtngthof wire = LL  24TO

If x = L/2, then

The maximum dip (sag) is represented by the value of y at either supports sup ports A and an d B. At At support B, set x = L/2 and y = S

 L Ssag = w wL2T2 S = 8T

T he maximum maxim um tension tension occurs when x = L/2 and is giv gi ven as Approximation: Neglecting

Where: S = m aximum sag L = length of the span w = weight per unit length of conductor cond uctor T = tension tension in the conduc tor T o get the approximate approximate length length of cable c able in the span,

, being very very small sm all c ompare ompa redd with unity, unity, T

becomes nearlyequal to T O and

b. PARABOLIC CABLE (Approximate Solution) (low sag) - for span of 300 m or less - the unit weight is assumed uniformly - distributed along the horizontal span l. - almost accurate for sag/span ratio less than 10%

L =yL  =8yS3L  wL  T = T =  TO   2 

T ension at the the point poi nt of supports, supports,

When the supports are at unequal levels (Parabolic Cable)

Figure 57 Parabolic 57 Parabolic Cable

- assuming the curvature is so small that the

curve length is equal to its horiz ho rizontal ontal projec tion ion

(OP≈ x). 2 forces forc es are acting on portion OP.

1. The weight wx of conductor acting at a distance x/2 from O. 2. T he tension tension T acting at O. O. Getting the mom ents at pt. O. O.

TM=OT =O 0 T = wx wx2x y = 2T

where: y = sag at any point y x = horizontal horizontal distance distanc e from the lowest point of the cable

Figure 58 Unequal 58 Unequal Level

Let: L = span length w = weight per unit length of conductor condu ctor T = tension tension in the conduc tor h = differ di fferenc encee in levels levels between two supports x1 = distance of support at lower level (horizontal) x2 = distance distanc e of support at higher level level (horizontal) (horizontal) S2 = vertical distanc e (sag) (sag) at point poin t O from tower 2 S1 = vertical vertical distanc e (sag) at point p oint O from tower 1 At At x = x1 and y = S1

SS ==    

Sag, and


27


x  x = L → 1    S  S == wxx xxx  x hh = = Sx 2T Sx or  = x  x → 2 x =xL=xx  2hT = L x  2hT wL/wL L   hT wL 2x = LL  2hT2xhT/wL   =   2 wL x =xLL=xx  2hT = LxL x  2hT wL wL  2x = L  x= L  hT  2 wL

Also, Also,

But,

Equate 1 and 2 If

multiply multiply by ½

If

multiply mul tiply by ½

Examples: 1. A transmission line conduc con ductor tor has been suspended suspended freely from two towers and has taken the form of a catenary that that has c = 487.68 m. T he span between the the two towers is 152 m, and weight of the conductor is 1160 kg/km. Calculate Calcu late the the following: a. Length of the conductor b. Sag c. Maximum and minimum value of conductor tension using catenary method d. Approxim Approximate ate value value of tension by using parabolic parabolic method. 2. A 132 – kV transmission line has the following data: weight of conductor c onductor = 680 kg/km ultimate strength = 3100 kg length of span = 260 m safety factor = 2 Calculate the height above ground at which the conductor should be supported. Ground clearance required is 10 meters. me ters. 3. A transmission line conduc con ductor tor at a river crossing is supported from two towers at heights of 70 m above water level. T he horizontal horizontal distance distanc e between towers is is 300 m. m . If the tension in conductor condu ctor is 1,500 kg, find the clearance cl earance at a point midway between the the towers. The The 2 size of conductor is 0.9 cm . Density of conductor material is 8.9 g/cm 3 and suspension length of the string is 2 meters. m eters. 4. An An overhead overhead transmission line at a river river crossing is supported from two towers at heights of 40 m and 90

m above water wa ter level, level, the horizontal horizontal distance betw be twee eenn the towers being 400 m. If the maximum allowable tension is 2000 kg, find the clearance between the conductor and water at a point midway between the towers. The weight weigh t of the conduc c onductor tor is 1 kg/m. 5. The towers of height 30 m and 90 m respectively support a transm ission line conduc tor at water water crossing. T he horizontal distance between the towers towers is 500 5 00 m. if the tension in the conduc co nductor tor is 1600 kg, find find the minimum clearance of the conductor and water clearance cl earance midway m idway the supports. Weight Weight of conduc co nducttor is 1.5 kg/m. Bases of the towers can c an be at water level. 6. An An overhead overhead transmission line at a river river crossing is supported from from two towers at heights of 50 m and 100 m above the water level, the horizontal distance between the towers being 400 m. If the maximum allowable allowab le tension is 1,800 kg and the conduct con ductor or weigh weighss 1 kg/m, find the clearance cl earance between the the conductor cond uctor and and water at a point mid – way between the towers. 7. A conduc c onductor tor is strung across a river, river, being supported supported at the two ends at heights of 20 m and 16 m respectively, from the bed of the river. The distance between the supports is 375 m and the weight of the conduc con ductor tor = 1.2 kg/m. IfIf the clearance clearan ce of the conduc cond ucttor from the river bed be 9 m, find the horizontal horizontal tension in the c onductor. onduc tor. Assume a parabolic configur co nfiguration ation and and that there there is no wind w ind or ice loading. loading . 8. A transmission line conduc con ductor tor at a river river crossing is supported from two towers at heights of 20 m and 60 m above water level. T he horizontal horizontal distance between between the towers is 300 m . IfIf the tension in the conducto con ductorr is is 1800 kg and the conductor weighs 1.0 kg per meter, find the clearance between the conductor and the water level at a point mid – way wa y between the towers. Use approxim ate method. method. 9. A transmission line li ne over over a hillside hil lside where the gradient gradient is 1:20, is supported by two 22 m high towers with 300 300 m between between them. them. T he lowest conductor is fixed 2 m below the top of each eac h tower. tower. Find the clearanc c learanc e of the the conductor from the ground. Given that conductor weighs 1 kg/m and the all allowable owable tension is 1500 kg. 10. A transmission tower on a level ground gives a minimu min imum m clearance of 8 meters me ters for its lowest c onduct onductor or with a sag of 10 m for a span of 300 m. If the same tower is to be used over a slope of 1 in 15, find the minimu min imum m ground clearance c learance obtained for the same span, same conductor c onductor and same weather conditions. 11. A transmission line conduc con ductor tor is supported from from two towers at heights of 70 m above water level. The horizontal horizontal distance distanc e between the towers is 300 m. If the tension in the c onduc tors is 1500 kg, find the c learan learance ce


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Electrical Transmission and Distribution System and Design at a point mid – way between the towers. T he size size of 2 the conductor is 0.9 cm and density of conductor material is 8 ·9 gm/cm 3. 12. An An overhead overhead line has a span of 260 m; the weight of the line li ne conductor cond uctor is 0.68 kg kg per meter run. Calc ulat ulate the maxim um sag in the the line. lin e. T he maxim um allowa allowabl ble tension in the line is 1550 kg. 13. A transmission line li ne has a span span of 150 m between the the level supports. The conductor has a cross – sectional area of 2 cm 2. The ultimate strength is 5000 kg/cm 2. T he specific speci fic gravity gravity of the m aterial is 8.9 gm/cm 3. If the wind pressure is 1.5 kg/m length of conductor, calcul cal culat atee the sag at the c enter of the the conduc con ductor tor if factor factor of safety is 5. 14. Two towers of height 40 m and 30 m respectively support a transmission line conductor at water crossing. T he horizontal distance between the towers towers is 300 3 00 m. If the tension in the c onductor onduc tor is 1590 kg, find find the clearance of the conductor at a point mid-way between the supports. sup ports. Weight Weight of c onductor onduc tor is 0.8 kg/m. Bases of the towers can be at the water wa ter level. Effect Effect of Ice and Wind Loadin Load ing g In areas where it becomes too cold in winter, there is a possibility possibili ty of formation of an ic e coating on the line conductors. The formation of an ice coating on a line conductor has a twofold effect – increase in weight and effective diameter of the conductor. In this condition the weight of conductor, together with weight of ice acts vertically erticall y downwards. Thus, T hus, the the total vertical weight wei ght acti ac ting ng on the conductor per meter length is w C + w i where w C is the weight of conduc c onductor tor in kg per meter length and w i is the weight of ice coating per meter length, w C is known w i is determined as follows: Let the diameter of conductor be d meters and radial thickness of ic e coating coa ting be t meter, as illustrated in Figure Figure 59. The overall diameter of ice covered conductor, as obvious from Figure 59, becom es equal to to (d + 2t) meter.

Figure 59 Ice 59 Ice and Wind Loading

π       = d 2  2 t  d 4 = π4 4dt4dt4t = πtd  t m

Volume of ice c oating per per meter length length of conduc tor

T he density of ice ic e is approxim ately ately 920 kg/m 3, so the weight w eight of ic e coating per meter length,

w= 2890.= 920×πt3td dt ktg/m kg

Due to weight of ice deposits on the line, and the wind pressure, the mechanical stress increases in the conduc con ductor tor and, therefore, the the line must be designed to withstand these stresses and tension. Under this conditio con dition, n, the weight of the conductor, together with weight of ice Figure 60 The 60 The Force Triangle acts vertically downwards while the wind loading w w acts horizontally, as shown in Figure 60. T otal weight of conductor per unit length

w = w  w  w wMaxim=ump×Sagd = 2wtL 8T θ = cos−www  w L  Vertical sagag = 8T cosθ

ww = Wind force in kg per meter length ww = Wind pressure per m 2 of projec ted area × projec projec ted area per meter length

When ice ic e and wind are acting ac ting simultaneously, simultaneously, the the lowest l owest point of the conduc c onduc tor does not remain vertic vertically ally down but away from from it at an angle θ given by the expression

T he maximum sag will not be vertical vertical but will be slant sag and vertic vertical al sag will wi ll be obtained by multiply multipl ying the slant slan t ssag ag

with cos θ

Effect of Temperatur e Change Change Sag and a nd stress vary vary with temperature tem perature because bec ause of the thermal expansion and contraction of the conductor. Temperature rise of conductor increase the length of conductor, and hence sag increases and tension decreases. dec reases. A temperature fall causes cau ses opposite effect. Maximum stress occurs at the lowest temperature, when the line has h as contracted co ntracted and is also possibly po ssiblycovered with ice ice and sleet. If conduc con ductor tor stress is c onstant while the the temperature temperature changes, the the change in length of the conductor conduc tor is where

∆l = l α∆t  ∆t = t  t ∆l= l  l

t0 = initial temperature l0 = conduc c onductor tor length at initial temperature l1 = c onductor onduc tor length at t1


29


α

= coefficient co efficient of linear expansion expansion of conduc con ductor tor /°C If the temperature tem perature is constant co nstant while the the conduc con ductor tor stress stress changes cha nges (loading), the change in length of the con ductor duc tor is

∆l∆T ==T lMA ∆TT

where T 0 = initial tension tension of c onductor M = modulus m odulus of elasticity elastic ity of c onductor A = actual metal cross section section of conduc con ductor tor Examples: 1. A transmission transmission line has a span of 200 meters between between level supports. The conductor has a cross – sectional area of 1.29 cm 2, weighs 1,170 kg/km and has a breaking stress stress of 4,218 kg/c m 2. Calc ulate the sag for a factor of safety of 5 allowing allowi ng a wind pressure of 122 2 kg per m of projec ted area. What is the vertic vertical al sag? 2. A transmission transmission line has a span of 150 m between level level supports. The conductor has a cross – sectional area of 2 cm 2. The tension in i n the conductor condu ctor is 2000 kg. IfIf the the specific gravity of the conductor material is 9.9g/cm 3 and wind win d pressure is 1.5kg/m length, l ength, calculat cal culatee the sag. sag. What is the vertical sag? 3. A transmission transmission line has a span of 150 meters between between supports, the supports being at the same level. The conductor conduc tor has a cross – sectional area of 2 c m 2. The ultimate strength is 5,000 kg/cm 2. T he spec ific gravity gravity of the m aterial is 8.9. If the wind pressure is 1.5 kg/m length of the c onductor, onduc tor, calculat cal culatee the sag sag at the c enter enter of the conduc con ductor tor if factor factor of safety is 5. 4. T he effective diameter of a line l ine is 1.96 cm and it weighs weighs 90 kg per 100 – meter length. What would be the additional loading due to ice of radial radi al thickness 1.25 1.25 cm 2 and a horizontal wind pressure of 30 kg/m of projecte pro jectedd area? Also, find the total weight per meter run of the line. Density of ic e is 920 kg/m 3. 5. T wo towers are supported at the same elev el evation. ation. It has a sag of 8 m when subjected to a tension T when a wind load loa d of 0.5 kg/m is considered. con sidered. The sag incre in creas ases es by 1.423 m maintaining mai ntaining the the same tension. tension . Determin Determinee the weight of the conductor. con ductor. 6. A transmission transmission line has a span of 275 m between level level supports. The T he conductor cond uctor has an effective effective diam eter of of 1.96 c m and weighs wei ghs 0.865 kg/m. Its Its ultimate ultima te strengt strengthh is 8060 kg. If the conductor has ice coating of radial thickness 1.27 cm and is subjec su bjected ted to to a wind pressure pressure 2 of 3.9 gm/cm of projected area, calculate sag for a safety factor factor of 2. Weight of 1 c c of ice is 0.91 0.91 gm. g m. 7. An An overhead overhead transmission line conduc con ductor tor having a parabolic configuration weighs 1.925 kg per meter of length. The T he area of cross – section of the conduc co nducto torr is 2.2 cm 2 and the ultimate ultim ate strength strength is 8000 kg/cm 2. The

supports are 600 m apart having 15 m difference of levels. Calculate the sag from the taller of the two supports which must be allowed so that the factor of safety shall be 5. Assume Assume that ice ic e load is 1 kg per meter run and there is no wind win d pressure. 8. A transmission transmission line has a span of 214 meters. T he line line 2 conductor conduc tor has a cross c ross – section of 3.225 cm and has h as 2 an ultimate breaking strength of 2,540 kg/cm . If the line is covered with ice and provides a combined copper and ice load of 1.125 kg/m while the wind pressure is 1.5 kg/m run. a. Calculate the maximum sag produced. Take a factor of safety of 3 b. Determine the vertic vertical al sag 9. A transmission transmission line has a span of 150 m between level level supports. The cross – sectional sec tional area of the conduct co nductor or 2 is 1.25 cm and weighs we ighs 100 1 00 kg per pe r 100 m. The breakin breakingg 2 stress is 4220 kg/cm . Calculat Calcu latee the factor of safety if the sag of the l ine is 3.5 m. m . Assume Assume a maxim um wind nd pressure of 100 kg per square meter. 10. A transmission line has a span of 250 m between between supports, the supports being at the same level. The conductor conduc tor has a c ross – sectional area area of 1.29 c m 2. The ultimate strength is 4220 kg/cm 2 and factor of safety is 2. T he wind pressure is 40 kg/cm 2. Calc ulate the height of the conductor con ductor above above ground lev l evel el at which which it should shoul d be supported if a minim m inimum um clearance of 7m 7m is to be kept between the ground and the the conductor. cond uctor. 11. A transmission transmission line has a span of 150 m between level level supports. The conductor has a cross – sectional area of 2 cm 2. The ultimate strength is 5000 kg/cm 2. The specific speci fic gravityof the the m aterial is 8.9 gm/cm 3. If the wind pressure is 1.5 kg/m length of the c onductor, onduc tor, calcu cal cula latte the sag if factor of safety is 5. Location of Line T he routing of a transmi ssion or distribution lines requires requires thorough inv in vestigations and for selecting selec ting the m ost ost desirabl desirable and practical route, route, following points should be c onsidered: onsidered: 1. Cost of construction 2. Cost of easements 3. Cost of clearing 4. Cost of maintenance Stringing Chart Under standard conditions, the sag is required to be determined for worst probable c onditions and the minimum minimum ground clearance clea rance is to be maintained mai ntained for these condition cond itionss. At At the tim timee of erection, the severe severe conditions con ditions do not prevai prevail,l, the temperature tem perature is usually higher, the designer, therefore, therefore, should know the sag to be allowed and the tension in the line to be allowed, allowe d, so that under no condition con dition there should be any danger to to the line.


30

Electrical Transmission and Distribution System and Design Stringing chart is helpful in knowing the sag and tension at any temperature. This chart gives the data for sag to be allowed and the tension to be allowed at a given temperature. For preparation of stringing or sag chart ch art first first calcul cal culat atee the the sag and tension on the conductor under the worst conditions, maximum wind pressure and minimum temperature, assuming a suitable factor of safety in fixing the maximum working tension for the conductor. Now evaluate evaluate the sag and tension for a series of tem perature in steps within the working range of temperatures. tem peratures. T he equation for determining determining stringing stringing chart of a line is derived as below:

Various values of f 2 and S2 are calculated repeatedly for different temperatures. Now the graph of tension vs temperature and sag vs temperature can be plotted, as shown in Figure 61. This graph is plotted for a fixed span and is called stringing chart. This stringing chart is very useful while erecting the transmission line conductors for adjusting sag and tension properly properl y.

 S =     T = T y = O       l =L  We know that

;

;

and span lengt l engthh



Let w1, f 1, l 1, S1 and t1 be the load per unit length, the stress, the span length, sag and temperature at the maxim um load load conditions (with the ice and wind and low temperature usually – 5.5°); w 2, f 2, l2, S2 and t2 be the values under stringing conditions, a is the area of cross section of the conductor, is the coefficient c oefficient of linear expansion and E is the modulus m odulus of elasticity.

α

l = L   = L    = L     w L  l = L  24af  α α  −  l ≅  −  L l = l  fl =E fL  Lt  t  α L                 L 24awLf  = LL 24awLf   f E f  Lt  tαl f  f   f      t  tαE= αE= 

Span length,

T he span length at maximum maxim um load condition is

T he temperature temperature rise from t 1 to t2 c auses an increase in the the span length of l 1 (t1 – t2) which is practically equal to L (t1 stress from f 1 to f 2 c auses a decrease in the the – t2). T he fall in stress length of T he new length length l 2 is thus given by

But

or

T he equation is a cubic cu bic one and can be solved graphic ally ly or analy anal ytically. tical ly. From this equation erection tension T 1 = f 1a can be determined determined such suc h that tension tension T 2 = f 2a under worst probable c onditions will not exceed the safe limi lim i t of tension tension.. After After determining determini ng f 2, the corresponding sag can be determined from the equation

S = w8f La

Figure 61 Tension 61 Tension vs Temperature and Sag vs Temperature Graph

Example: An An overhead overhead line having a span of 250 m is to be erected erected at a temperature of 40°C in still air c onditions. ItIt is desir d esired ed that a factor of safety of 2 should be maintained under bad weather conditions co nditions when whe n the temperature is 10°C and the the wind load is 378 N/m 2 of projected proj ected area. The T he data for for the ACSR ACSR conduc con ductor tor used used for the line is: diameter: diam eter: 1.95 cm, 2 area: 2.25 cm , weight = 8.31 N/m, breaking breaki ng load is 77,900 77,900 -6 N, linear coefficient is 18.44 x 10 /°C, Young’s modulus: 91.4 x 10 3 N/mm 2. Find the sag and tension under erecti erec tion on conditions. Sag Template For c orrect design and economy, the location loc ation of structures structures on the profile with a template tem plate is very essential. Sag tem plat plate is a convenient devic devicee used in the design of a transmiss transmi ssio ionn line to determine d etermine the loc ation and height of structure. structure. Sag Sag template can be relied upon to to provide the following: followin g: 1. Economic Econo mic layout layout 2. Minimu Min imum m errors in design and layout 3. Proper grading of structures 4. Prevention Prevention of excessiv exc essivee insulator swing Generally Generall y, two two types of towers are used: 1. T he standard standard or straight run or intermediate intermed iate tower tower 2. T he angle or anchor or tension tower The straight run towers are used for straight runs and normal conditions. The angle towers are designed to withstand heavy heavy loading loadin g as compared co mpared to standard towers because angle towers are used at angles, terminals and


31

Electrical Transmission and Distribution System and Design other points where a large unbalanc ed pull may be thrown on the supports. sup ports. For standard towers, for norm al or average average spans, the sag sag and the nature of o f the c urve urve (Catenary or Parabola) that the the line conductor will occupy under expected loading conditions con ditions is ev e valuated and plotted on template. temp late. Temp T empla latte will also al so show the required requi red minimum mi nimum ground clearance cl earance by plotting a curv c urvee parallel to the conduc tor shape curve. curve. For For the standard tower and sam e height, the tower footing line line can also be plotted on the template. Tower footing line is used for locating the position of towers and minimum ground clearance cl earance is maintained maintaine d throughout. throughout.

It is very tedious to make m ake calc ulations of sag sag and tension for each span individually and then to make adjustment while erecting the transmission line. In the erection of a transmission line the conduct co nductors ors are run out through snatch snatch blocks bloc ks attached attached to the support arms equally equall y tensioned tensioned at each end of a section of five or six blocks. When the conduc con ductors tors are clampe c lampedd to suspension suspen sion insulator strings, strings, the the equal tension is maintained by insulator swing. When the conduc con ductors tors are bound to pin type type insulators, insula tors, the flexibility flexibi lity of the support ensure equal tension. It is often c onvenient to make sag and tension calc c alculati ulations ons in terms of a hypothetical equivalent span, span , this tensions being applied app lied to each eac h span within the section sec tion of overhea overheadd line between the tensioning points. poin ts. If there are n spans spa ns of length L 1, L2, L3, etc. which are to be given an equivalent span Le, then the strung length of the equivalent e quivalent line must must be the same as that of the individual spans. This may be expressed as

     ∑ w L w L  n∑ L =LnL  24T = L  24T  L = ∑+ = +∑∑+⋯ L =  +++⋯

Since Figure 11 Sag 11 Sag Template Used for Locating Towers

Figure 62 shows show s the sag template templa te used for locating loc ating towers towers. In fact, there are no clear – cut guidelines guide lines for locating loc ating the the tower positions and several other alternatives may be examined. exami ned. Ground c learance learance depends on voltage level and it gives the span length and ground g round clearance cle arance at different different voltage levels: Minimum Span Length Voltage Level Ground (m) Clearance (m) 0.4 kV 80 4.6 11 kV 100 4.6 33 kV 150 – 200 5.2 66 kV 200 – 300 6.3 132 kV 350 – 360 6.3 220 kV 360 – 380 7.0 400 kV 400 8.8 Table 7 Ground 7 Ground Clearanc e for Different Different Voltage Voltage Lev el

Equivalent Span It may not be possible possib le to have a section sec tion of transmission line line consisting of successive spans of equal lengths because the location of the towers depends upon the profile of the land along which the transmission line is to be laid. Sometimes, the towers are forced to be located to give spans of different lengths so that minimum min imum interferen in terference ce is caused cau sed with the use of land. When When the succ essive essive spans spans are of unequal lengths changes in tension in load or temperature temperature will c ause unequal c hanges hanges in tension in the different spans.

So or

If the line tension T 0 is determined for the equivalent span, the sag for the individual in dividual spans m ay be calculated using the the approximate approxim ate value value of span. NOTE: T NOTE: The he method of sag template templa te for for locating lo cating the towers towers should not be used for long spans as well as where the slope of the profile is very steep. In suc h cases, c ases, actu ac tual al calcul cal culat ations ions for sag and tension should be made. Aeolian Aeol ian Vibration (Resonant (Resonant Vib Vibration) ration) Overhead conductors will subject to normal swinging in wind and apart a part from that, may subjec t to vibration vibration known as Aeolian Aeolian vibrations vibrations or resonant vibrations. Aeolian vibrations vibrations have low amplitude ampl itude (20 mm m m to 50 mm) m m) and high hi gh frequen frequencie ciess (5 – 100 Hz). T hese are caused cau sed by the vortex vortex phenomeno phen omenonn in the low l ow wind speed (5 – 20 km/hr). km /hr).

f = 50 du

where: u = wind veloc veloc ity (km/hr) (km/hr) dc = diameter of conductor (mm) (mm) T he length of a loop (half wave length) len gth) depends on tension tension T and conductor conduc tor weight weight w and is given by

λ = 2f1   wT

T hese vibrations vibrations are very com mon to to all c onductors onduc tors and and are always present. Since the vibrations are small in


32

Electrical Transmission and Distribution System and Design magnitude, magn itude, these these are less harmful. harm ful. T he ACSR ACSR conduct cond uctor or has high hi gh diameter diam eter to to weight ratio and is subjec t to to fatigue by these vibrations. A stock bridge dam per is used used to minim m inimize ize these these vibrations. vibrations. It com prises of two masses m asses at the the 63 Stock Bridge Damper end of a short length of Figure 63 Stock stranded steel cable cab le suspended from the conductor co nductor about about midway between two point (nodes) of the vibrations. Movement of the damper is caused by the vibration and energy is absorbed by the inter in ter – strand friction in the steel steel cable. The length of a typical damper is about 60 cm and weighs about 5 kg or more. Galloping Gal loping or Dancin Danc ing g of Conductors T hese vibrations vibrations are low frequenc y (0.25 (0.25 – 2 Hz) and high amplitude (up to 6 m) and are generally caused by asymm asymm etrical layer of ice ic e formation. formation. T his vibration vibration is self – excited exci ted ty type. When the ice i ce c oated conductor is acted upon upon by a light drift wind particularly particu larly where the ground slopes at right angles to the transmission line vibration is initiated because wind travels up the slope and appears to get underneath the the conductor. co nductor. T he stranding of c onduct onduc tors ors significantly contributes to these vibrations. These vibrations may m ay cause c ause flashover flashover between conduc tors. It is difficult difficul t to to prevent prevent these vibrations. These The se vibrations vibrations may cause cau se flashover between c onductors. onductors. It is difficul t to prevent prevent these vibra vibrations tions but horizontal hor izontal configura con figuration tion of transm transm issio issionn line can be used to reduce the impact of galloping or dancing of c onductors. INSULATORS necessary - provides insulation between the line conductors and supports. And And thus, prevent prevent any leakage current from conduc con ductor tor to to earth. - are mounted on the cross arms and an d the line li ne c onductor onductorss 64 Parts of Insulator are attached to the Figure 64 Parts insulators to provide necessary clearances between conduc con ductors tors and metalwork. - prevent short circuiting between the different phase conduc con ductors tors and provide provide necessary mechanical suppor support for the the line c onductors. - one of the most important and vulnerable links in transmission and distribution practice and, therefore, proper selection is of utmost importance for the succ essful operation of overhead overhead transmission and and distribution system. system.

1. 2. 3. 4. 5. 6.

7.

1.

2.

Characteristics Chara cteristics of Insulators Ins ulators High mech m echanical anical strength strength to withstand conduc con ductor tor load load,, wind load and ice loading i f any. any. High insulation in sulation resistance to avoid avoid leakage c urrent to to earth. T he insulator insulator material should be non – ferrous, free from from impurities and crac ks. High relative permittivity permittivity so that dielec tric strength is high. High ratio of puncture punc ture strength strength to flashover. Ability Ability to withstand large temperature variations, it should not crac k when subjec ted to high temperatur temperatures during summer and low temperatures during winter. T he dielectric di electric strength should remain unaffected unde underr different conditions cond itions of temperature temperature and pressure. T he material used should not be porous and should be imperv imp ervious ious to fluids and gases in the atmosphere. atm osphere. Insulator Materials Porcelain - is produced by firing at a high temperature, a mixture of Kaolin, Kaol in, feldspar and quartz qua rtz.. - most commonly used material for overhead line insulators. ech anically stronger than than glass. - is m echanically - gives less trouble from leakage, and is less susceptible to temperature variations and its surface is not affected by dirt deposits. de posits. homog eneous as glass thus its satisfactory - is not homogeneous performance performanc e in service depends d epends to a consider consid erab able le extent on its preservation whic h is only onl y of the order of 25 mic rons in thickness. thickness. - fault cannot be detected easily since it is not transparent. - tension is usually weak and does not withstand tensile stresses stresses exceeding exc eeding 5 kg/mm 2. strength of a - dielec tric strength and compressive strength mec hanically sound porc elain are about 6.5 kV/mm. - com plicated design for an operating voltage Glass - cheaper than porcelain - high resistivity resistivity and dielectric diel ectric strength (14 kV/mm kV/mm of thickness of the the material) m aterial) - have simpler design and even one piece design can be used for an a n operating voltage - quite homogeneous material and can withstand higher compressive stresses as compared to porcelain - lower coefficient of thermal expansion which minimi min imizzes the strains due to temperature chang c hanges es - transparent in nature


33

Electrical Transmission and Distribution System and Design moisture moi sture more readily readi ly condenses cond enses on its surface surface and facilitates the accumulation of dirt deposits thus giving high surface surface leakage - in large sizes, great mass of material combined with irregular shape may result in internal strain after after cooling c ooling - can ca n be used in 25 kV under und er ordinary atmospheric atmospheric conditions co nditions and 50 kV in dry atmosphere Steatite - naturally occurring magnesium silicate, usually found combined with oxides in varying proportions. - has muc h higher tensile and an d bending stress than than porcelain - can be used at tension towers or when transmission line takes a sharp turn Special artificial artificial material al - used in insul ators for low voltages - can ca n be easily molded m olded into any shape without any internal stress metall ic fittings fittings c an also be firmly embedded embe dded in in the the - metallic material without wi thout diffic difficulty ulty - they deteriorate de teriorate rapidly in bad clima cl imatic tic conditio conditions ns and on being subjected to flashover their carbonized carbonized surface surface form form a conducting c onducting path Factors Factor s involve invo lved d in Insulat In sulator or Design Required to withstand both m echanical ech anical and elec trical ical stresses. Surface leakage path must have sufficiently high resistance to avoid any c urrent leakage to earth. Design must be suc h that that the stress developed developed owing owing to contraction and expansion in any part of the insulator does not lead to any a ny defect. defec t. In case of electrical breakdown due to flashover, the insulator continues c ontinues to ac t in its proper c apacityafter the the event event unless frac fractured tured by the heat of the arc, but after a puncture, it is permanently damaged due to excessive heat. Provide sufficient thickness of the porcelain in the insulator to resist puncture punc ture by the com bined effect effect of of the line l ine voltage and any probable proba ble transient voltage rise rise on the line. The ratio of puncture strength to flash over voltage, termed as factor of safety, safety, must be high to provide provide a good margin for the protection of insulators from complete com plete failure. It is desirable that porcelain may not come in direct contact with a hard metal screw thread. Normally, cem ent is used between metal and the porcelain porce lain.. The cement used must not cause any fracture by expansion or contraction. co ntraction. -

3.

4.

1. 2. 3.

4.

5.

6.

7.

Flashover – an arc oc curs between the line c onductor onductor and and earth and the discharge jumps across the air gaps in its path. Puncture – the discharge occ occ urs from from conductor to to pin through the the body of o f the insulator. Types of Insulato Insul ators rs 1. Pin Type Insul ator - used for transmission and and distribution of electric power at voltage up to 33 kV. Figur e 65 Pin Type Insulators - modern pin type insulators are very very reliable reliabl e and inherent crac ks in porcelain porcel ain are very rare and never oc cur with with toughened glass insulators. of modern - life porcelain insulators is relatively long (expected about 50 years) - is designed to be mounted on pin which in turn is sec ured to the cross arm of the pole pol e - for lower voltages, one piece piec e type type of insulators i nsulators are are Figure 66 One 66 One Piece Pin Type Insulator used. - for higher voltages, stronger pin type insulators are used which consists of two or three pieces of porcelain cemented cem ented together together Advantages a. it is cheaper, since one piece pin insulator insulator can do do work of two suspension insulators b. requires shorter pole to giv gi ve the sam e conduct con ductor or clearance cl earance above the ground since the pin pi n insulat insulator raises the conductor above the cross arm while suspension insulator i nsulator suspends it bel ow the c ross oss arm c. it’s used above 50 kV is unecon un economical omical since sinc e it become bec ome very bulky and cumbersome c umbersome when designed on higher hi gher voltages voltages

Insulators Figure 68 66 Figure 67 33 67 33 k V Pin Type Insulators 68 66 kV Pin Type Insulator


34

Electrical Transmission and Distribution System and Design 2. Suspension Type Insulator – consists co nsists of number of porcelain porcel ain discs disc s connected connected in series by metal ink in the form of string. Advantages a. Suspension type insulators insu lators are chea c heape perr than pin type insulator for voltages beyond 50 kV. b. Each unit of disc of suspension type Figure 69 Suspension 69 Suspension Insulators insulator is designed for low voltages, usually usua lly 11 kV and can ca n be used by connecting them in series, the number depending dependi ng upon the working voltage. voltage. c. If any an y insulator insul ator is damaged, dama ged, the whole string does not become useless, damaged disc can be replaced. d.It d. It gives more flexibility to the line and mec hanical stresses are are reduced in this arrangement. The connection at the cross arms is such that the insulator string is free to swing in any direction, and thus takes up a position where it experiences experienc es only a pure p ure tensile stress e.When e. When used in conjunction with steel supporting structures, Figure 70 Suspension 70 Suspension Insulator Parts has the advantage of rendering the c onductor less liable liabl e to be affected affected by lightning disturbances. At every point of the support the wire is hung below the earthed c ros ross arm, thus enabling the tower to function as a lightning rod. f. Additional Additional insulation insulation required for the raise of voltage due to increased demand c an be obtained by adding one or m ore discs to the string. 71 Yoked Insulator g. In case of long spans Figure 71 Yoked Strings (river or valley crossings) where heavy conductor load is to be sustained, two – disc insulator strings strings can be yoked.

T he disadvantage of suspension type type insulators is that large spacing between conductors are required than with pin type type insulators due to to large amplitude ampli tude of the swing of the conductors, but this is not a serious disadvantage. Types of Suspension Suspen sion Insul Ins ulators ators 1. Hewlett or Interlinking Type of Suspension Insulators - is one of the earliest design - each disc consists only of one piece of porcelain, the central bulbous portion of which is provided with two curved tunnels lying in planes at right angles to each 72 Hewlett 72 Hewlett or Interlinking other. T he short Figure Ty pe Suspension Insulators Insulators steel strips forming the connection con nection between indiv indi vidual idual discs are threaded through these tunnels and thus loop through each other, being separated by a layer of porcelain which is totally in compression. - the main m ain advantages are: a. simple in design b. high mechanical strength since the porcelain in between the two tunnels is under compression com pression only c. no risk of breakage owing to the difference in expansion expansi on or c ontraction on of of the connecting links and the insulating materials d. no risk of interruption to the service in case ca se the the porcelain porce lain between the links get acci ac cident dentally ally broken, since the links keep the other units uni ts held together. - are more liable to puncture than any other type of suspension insulators, owing to the high electrostatic stress in the material between the links. 2. Cemented – Cap Type Suspension Suspen sion Insulators Insulators - is the most commonly used type and consists of a single disc – shaped piece of porcelain grooved on the under surface to Figure 73 Cemented 73 Cemented - Cap increase the surface Ty pe Suspension Insulators Insulators


35


-

-

-

-

leakage path, and to a metal cap at the top, and to a metal m etal pin underneath. very uniform distribution distrib ution of the electrostat elec trostatic ic stress in the m aterial between the connect c onnecting ing link is obtained in such insulators insulators main mai n drawback drawbac k has been that coefficients coeffic ients of cubical expression of the three materials – porcelain, porce lain, cem ent and steel are different di fferent and and no provision is made for their expansion and and the sudden temperature tem perature change c hangess occurr oc currin ingg in servic servicee are sufficient suffici ent to to set up internal stress stress whic h ultimat ultim ately ely crack the porc elain elain leading to to electrical failure. the cement itself is subject to volumetric changes depending on its moisture content, has often materially m aterially assisted assisted in the proc ess of of the failure of the insulator one way of i mproving the the design is by way of substituting the cem enting of the pin by purely

-

transmission lines, line s, strain strain insulator consists co nsists of an assembly of suspension insulator. insula tor. The disc of strain insulator is used in vertical plane.

mechanical mec hanical fixing, such as the “spring – ring”.

3. Core and Lin L inkk Type Suspension Insu Insulators lators - combin co mbines es the advantages advantages of the two previous types types and overcome overcom e their disadvantages eac h insulator disc is - each symm symmetrical etrically ly placed placed and it conforms to the electrost elec trostatic atic lines lines of force, thus avoiding placing materials material s of different permittivities in series. Figure 73 Core 73 Core and Link Type Suspension Insulators - metalwork consists of pressed steel spiders, spid ers, the the legs of which whic h are are fastened into the porc elain by an alloy having having approximately approxim ately the same sam e coefficie coe fficient nt of c ubical cal expansion as the porcelain. Thus, high mechanical stresses on the porcelain, whether due to sudden temperature variations or to the emp e mploymen loymentt of cement ce ment are are eliminated. - it allows al lows discs disc s to be formed out of quite thick thick porcelain porce lain thereby allowing allo wing the disc to be one o ne piec e only. only. - has high punc ture strength 3. Strain I nsulators nsulators - it is i s used for dead end of the line or c orner or sharp sharp curv cu rve, e, the line is subject subje ct to a greater tension. - For low voltage lines, li nes, shackle shackl e insulator is used as strain insulator. However, for high voltage

Figure 74 Strain 74 Strain Insulators

4. Post Insulator Insul atorss - are employed for supporting bus – bars and isolating switches, etc. - is like pin type type insulator but has a metal bases with with metal cap so that more than one unit can be mounted in series. - is mostly m ostly a solid core c ore insulator made of resin - porcelain is also used 5. Shackle Insulators Insulators - in early days, shackle insulator was used as strain insulator. - frequently used for low voltage distribution Figure 75 Shackle 75 Shackle Insulators lines. - the wet flash – over and dry flash – over are 10 kV and 25 kV - puncture punc ture voltage voltage is about abo ut 35 kV - operating voltage voltage is 1 kV - its weight, transverse mechanical load and total creepage creepa ge distance are 0.5 kg, 1150 kg and 63 mm respectively.


36

Electrical Transmission and Distribution System and Design -

-

the tapered hole in the shackle insulator distributes the load l oad more m ore evenly and reduces reduc es the the possibility possibili ty of breakage when heavily loaded. can ca n be directly direc tly fixed to the the pole with a bolt or to the cross arm are bell mounted to prevent water being held in contact co ntact with the spindle. is used at all positions, either intermediate, terminal or angle. Where the angle exceeds 60° deviation they are generally used in conjunction with shackle shac kle straps. straps.

Figure 76 Shackle 76 Shackle Type Insulators Fitted to Poles

6. Stay Insulato Insul ators rs - are of egg shape, also called cal led as guy insulators insul ators

Figure 77 Stay 77 Stay Insulators

-

-

consists co nsists of porcelain porce lain piece pierced with two holes at right angles a ngles to to each eac h other through through whic h two ends of the guy g uy wires are looped. T his keeps the the porcelain between two holes ho les at right angles to each other Figure 78 Cross 78 Cross Section of Stay Insulators through whic h two two ends of the the guy wires are looped. are provided at a height of about 3 m from the ground level

the size size used depends upon the tensile tensil e strength strength of stay wire Failure Failu re of Insula I nsulators tors 1. Cracking Cracki ng of Insulators - very common co mmon in case of pin type type insulators and and cemented – cap type type suspension suspensi on insulators. - occ urs due to unequal expansion of steel, steel, porcelain and cement during the varying conditions of cold and heat and dryness and dampness. - develops high stresses in the porcelain near the joint and it results results in tension tension failure. failure. - can be avoided to some extent by using elastic cushions cushi ons between the the shells. 2. Porosity of Material - due to under – firing or other c auses, always leads to failure after a comparatively short period of service. - the pores usually absorb moisture from the atmosphere or the cement, thereby decreasing the insulation resistivity resistivity of the m aterial. - gives rise to leakage current flowing through the the porcelain, resulting in a gradual rise in the temperature until porcelain porcelai n is punctured. - can be avoided by gl azing azing the insulator, insul ator, to some extent. 3. Improper Vitrification Vitrification - another cause of puncture punc ture of the material - can be avoided by carrying out suitable routine tests during manufacture. manu facture. 4. Flash – Over - most common com mon c ause ause of insulator failure failure - causes unequal expansion of the porcelain thereby shattering the insulator with big cracks and causes cau ses interruption interruption of the supply supp ly - can be avoided avoided by providing providing arcing arc ing horns or rings which whic h take up the the arc and divert it away from the insulator. 5. Mec hanical Stresses Stresses - very rare becau be cause se defective pieces are weeded weeded out in the routine factory test - occ urs if the com pressiv pressivee strength is quite high yet the tensile strength is not adequate and the insulator is always weak in tension and usually fails bec ause of it. 6. Short Circuits - birdage means me ans the short short circui c ircuiting ting of conducto conduc torr to to earth through the the large birds or similar simi lar objects. - can be avoided by providing providing bird guards near nea r the insulator on the cross arm, by increasing the -


37

Electrical Transmission and Distribution System and Design clearance cl earance of the the c onductor onduc tor from from earthed parts or or using suspension insulators instead of pin type, where the clearance between the earthed structure and the conduc c onduc tor is very large to warrant any birdage. 7. Deposition of Dust - if the insulator material is not properly glazed, the water will wil l stick over over it resulting into deposition de position of dust over it which is partially conducting and reduces reduc es flash flash – over over distance. di stance. - the deposits of dust and like matter m atter (salt, cement, cement, dust, etc.) on the interior surfaces surface s can cause cause much much havoc under the condition co ndition of fog and mist. - can ca n be easily avoided by cleaning cl eaning the insulator insulators periodically. Testing of Insulato In sulators rs An An insulator should have good mec hanical strength to withstand the load conditions, large dielectric strength to withstand the normal norma l operating voltage and over over voltages, but it should be free from any pores or voids that may lead to breakdown. Three distinct tests namely flash – over, performance and routine tests tests are to be c onducted onduc ted.. 1. Flash – Over Tests a. Power Frequency Frequenc y Dry Flash – Over Test - is performed by gradually increasing the voltage between the electrodes of the insulator mounted in a way it is to be used till the breakdown of surrounding air with formation formation of a sustained sustained arc occurs. oc curs. - is repeated 5 or 10 times. - for 5 readings, the mean is taken as the correct flash – over voltage - for 10 readings, comm onlyused method, the the median is found by arranging 10 values in ascending order and taking the mean of 5 th and 6th reading. The voltage voltage obtained is known as 50% dry flash – over voltage. voltage . - can ca n also be performed by apply appl ying the same voltage to 10 insulators together. T he spacin spacingg between the insulators should be at least 4 times the the length of each. eac h. - the voltage voltage till which whi ch five five insulators have have flashed – over is called the 50% power frequency dry flash – over voltage. voltage . voltage may be applied appl ied c onstantly and the the - the voltage voltage which causes flash – over after one minute is called one minute dry flash – over voltage. voltage . - the voltage voltage should not be less than the given standard.

b. Power Frequenc y Wet Flash – Over Test - the insulator is mounted in a way it is to be used and the voltage is applied gradually gradual ly but in addition to this the insulator insula tor is sprayed sprayed with with water. - the standard precipitation conditions are 3 mm per minute at an angle of 45 , the water having a resistance of 10000 Ω per cm 3 at normal atmospheric temperature and pressure. pressure . IfIf the resistiv resisti vity of o f pure water is more, it may be reduced by adding small quantities of a mineral salt or an inorganic salt. - voltage should not be less than the given standard. c. Impulse Frequency Frequen cy Flash – Over T est - the generator ge nerator developing developing lightning voltage is used. It develops de velops very very high hig h voltag voltages es at frequency of several hundred kHz. Such a voltage is i s applied appl ied to the insulator and spark – over voltage is noted. °

Impulse ratio = SparkImoverpulpulse volspatargegkeatoverpowervolftragegequee ncyncy

minimum mini mum acceptable acceptable flash – over vol voltage tagess are are given in standards. 2. Performance Performanc e T ests ests a. Puncture Test - is performed to determine determ ine puncture punc ture voltage. voltage. - the insulator is totally immersed in an insulating oil and is subjected to the voltage being gradually increased till the puncture takes plac e. The voltage at whic wh ichh conductio conductionn begins is called the puncture the puncture voltage. voltage . - The puncture voltage for string should be at least 30% higher than the power frequency wet flash – over voltage and pin insulators should not be less than that given in the standards. - as an acceptance test the voltage can be rapidly increased to the value given in standards for for pin insulators and 1.3 times the wet flash – over voltage for string insulators. No conductio conduc tionn should take place. b. Porosity Test - sample sam ple of every batc h of freshly freshl y fired fired insulators are taken. The specimens are broken into pieces and immersed in a 1% solution of fuchsine dye in alcohol under pressure of 150 kg/cm 2. After sufficien suffic ientt time (the product of test duration in hours and pressure in kg/cm 2 should not be less than -


38

Electrical Transmission and Distribution System and Design 1800) the pieces are removed from the testing pot and examined. - the slightest sli ghtest degree of porosity is indicated by a deep de ep penetration of the dye into the body of specimen. c. T emperatur emperaturee Cycle T est - the insulators are quickly and completely immersed imm ersed in a water – bath maintained mai ntained at the the temperature of 70°C above that of the cold water and left submerged for a period of T minutes. min utes. They Th ey then then are withdrawn and an d quickly quickly and completely immersed without being plac ed in an intermediate container, c ontainer, in a bat ba th of cold water for the same period of T minutes. This heating and cooling cycle is repeated 4 – 5 times and the insulators are then dried to observe any cracks in the glazing. , where M is the mass of the insulator in kg. d. Electromechanical Test - performed with suspension type type insulators insulators only and in this test the insulators are subjected to a power – frequency voltage equal to 75 % of o f their dry flash flash – over voltage and simultaneously sim ultaneously to a tensile load equal to ½ of the spec ified minimum failin failingg load. load. - tensile load is then steadily increased at a rate per m inute of about 20% of the specif speci fied ied minimum failing load, until puncture or breakage occurs. occ urs. - puncture punc ture or breakage should not be less than the specified m inimum failing failing load. e. Mechani Mec hanical cal T est - performed on pin type insul ators to determine determine their ultim ate mechanica mec hanicall strength. - the insulator is m ounted on a rigidly fixed pin capable ca pable of o f withstanding, without apprec iabl iable deformation, the loads to whic h it is subjected subjected during test. - it is subjected to a load equal to one half of the specified minimum failing load applied applied perpendicular to its axis in the plane of the side groove by means of a wire rope encircling the groove. The load is then increased inc reased at a rate rate per pe r minute of about abou t 20% 20% of the specified minimum failing load until breakage occurs. occ urs. - breakage should not oc cur at a load less than the specified m inimum failing failing load.

T = 15 0.7M

3. Routine Test T estss a. Proof Load or Mechanical Test - all types of insulators are assembled assembl ed and the thenn subjected to to tensile load whic h is 20 – 25% more than the normal tensile load which the insulators are likely to encounter in actual operation. The load loa d is applied appli ed for a duration duration of 1 minute m inute only. b. Corrosion Corrosion T est i nsulators with its fittings fittings is suspended in - the insulators a c opper sulphate solution at 15.6°C for for one minute. Then, the insulator is removed, wiped, cleaned and again put in copper sulphate solution. T his is repeated four tim es. es. Now, an exam ination there should not be any deposition of m etal over over it. c. High Voltage Voltage T est inverted and are put into - the pin insul ators are inverted water up to the neck. The water is also put into the spindle hole. Then high voltage is applied for 5 minutes. There should be no damage damag e to the the insulator after carry ca rrying ing out this test. POT ENTIA ENTIALL DISTRIBUTIO DIST RIBUTION N OVER SUSPENSION SUSPENSION INSULA INS ULAT T OR ST RING For overhead overhead lines l ines operating at high voltages (33 kV and above) use of number n umber of disc s connected conne cted in series, throug throughh metal links, is made. made . T he whole unit formed by c onnectin onnectingg several several discs d iscs in series is known as string of insulators. The insulators. The line conductor is secured to the bottom disc of the string and the top disc i s connec ted to the the cross arm of the pole or tower, as illustrated in the Figure. Fi gure. The Th e number of discs connec con nected ted in series in an insulator string depends upon the the line operating operating voltage voltage and can be seen in T able 8. Operating Voltage in kV 11 33 66 132 220 400

Number of Discs in Suspension Tension or dead Assembly – end Assembly 1 1 2 3 5 6 9 10 14 15 21 22

Table 8 Minim 8 Minim um N umber of Insulator Insulator Discs Required for Transmission Transmission Lines

T he number num ber of disc indicat indi cated ed in Table 8 is the usual numbe numberr used. However, in the case of transmission lines line s operatin operatingg at 66 kV or m ore, one disc less than the num ber indicated indicated in Table 8 is used on about 8 suspension structures near the substation. T his is accomplished acc omplished so that in the event event of a lightning lig htning surge appearing on the line, the insulator string string


39

Electrical Transmission and Distribution System and Design will flash – ov o ver and prevent the surge from travellin travellingg to the substation thus thus safeguarding safeguardin g the equipment equipme nt there. there. T he following points p oints may be noted regarding the potential potential distribution over over a string of suspension insulators: insula tors: 1. T he voltage voltage im pressed on each eac h string string of suspension suspension does not distribute itself uniformly ac ross the individual individual disc due to the presenc e of shunt capacitance. capa citance. 2. The disc nearest to the conductor has maximum voltage ac ross it. it. As we move towards the c ross arms arms the voltage voltage across each disc goes on decreasing. 3. The unit nearest to the conductor is under maximum electrical stress and is likely to be punctured. Therefore, means must be provided to equalize the potential across each unit. 4. If the voltage imp i mpressed ressed ac ross the string were dc , then hen the voltage across each unit would be the same. It is because bec ause insulator capacitances are ineffectiv ineffecti ve in dc. Each type of suspension insulator forms a capacitor capac itor C which is known as mutual capacitance or self – capacitance. capacitance. If there were mutual capacit cap acitance ance alone, then the c harging ng current has been the same through the entire disc and consequently, the voltage across each unit would hav ha ve been the same. However, in actual practice,

capacitance ‘C’ also

Important Impo rtant points in solving so lving String Efficiency Problem Problems 1. Maxim um voltage voltage appears ac ross the the disc nearest n earest the the conductor. 2. The charging current is given by the equation .

 = 

 =

3. The voltage across the string is equal to the phase voltage. 4. Line voltage voltage = √3 x voltage across the string T he voltage voltage distribution di stribution across different nits of an insulato i nsulatorr string and string efficiency can be mathematically determined with the help of an equivalent circuit of the insulator string string as shown in the the Figure 80.

Figure 79 Arrangement 79 Arrangement of Insulator Strings

exist between metal me tal fittings of each di scs and tower or eart earth which is also called shunt capacitance. capacitance . Due to shunt capaci cap acitance tance,, charging cha rging current is not the same through all all the disc and an d the voltage voltage across each eac h disc would not be the the same. T he ratio of voltage voltage across the whole string to the produc p roductt of the number of discs and the voltage across the disc nearest to the c onductor onduc tor is called as String Efficienc Effici ency. y.

 =     vol  t⋯age acr =ossoss th =e string  =   = N×voltagege acorss disc nearestest to conductor

where: N = number num ber of disc in the the string String Efficiency is an important consideration since it decides the potential distribution along the string. The greater the string efficiency , the more uniform the voltage vol tage distribution distribution .

Figure 80 Equivalent 80 Equivalent Circuit for 4 String Insulator Unit

 iωCV ωCVV I= = =ωCVVI1k I = I  i    ωCV = ωCV  ωC  V  V       V = Vk  V1  k = V1  3K  K I = I  i   ωCV = ωCV  ωCV  V  V

Applying Kirchhoff’s current law to node A, A, we w e get

Applying Kirchhoff’s current law to node B, we get

Applying Kirchhoff’s current law to node C, we get


40


V = Vk  Vk 5KV 1Kk = V1  6K  VV = V=4V 10K10VK 6KV6K  VK

Finally, Final ly, voltage between line conductor con ductor and earth

Similar Simi lar derivation can c an be had for string insulators in sulators consisting ing of any number num ber of units. When the number of insulators in the string is large it becomes laborious to work out the voltage distribution across each unit, for such c ases standard standard formula may be used. In general case if there there are n units in the string, V is the maxim um voltage voltage across ac ross the string, string, V1, V2, V3, … Vn denote the voltage voltage across the insulators starting starting from top, C is i s the capacitance between the links and KC be the shunt capacitance between the links and earth, the voltage distribution across the the mth m th unit (counted (c ounted from from top) is given as

1 1 2si n h Kcosh K cosh  m   K K  √ √ 2 2 V = V sinh(n√ K) 1 1 2si n h Kcosh K cosh  n   K K  √ √ 2 2 V = V sinh(n√ K)

and potential potential adjacent to the line conduct c onductor or

When the insulators are wet the value of mutual capaci cap acitance tance C increas inc reases es while C 1 remains rema ins constant con stant (except (except for the unit nearest the cross arm) so the value of K decreases, dec reases, more uniform potential di stribution is obtained obtained and the string string efficienc effici encyy increases. inc reases. T he value value of K varies and depends upon the leng th of the insulator string. string. T he larger the the number numbe r of insulator discs disc s in a string, the longer will be the string. T he longer the string, the greater must be the horizontal spacing between the insulator disc and the support sup port (pole or tower) to make an allowance allowa nce of conduc tor swing. The greater the horiz hori zontal ontal spacing between the insulator string and the support, the lesser the shunt capac c apacitance itance and vic vicee versa. versa. T hus, the value of K is low for loner strings and high for shorter strings. In prac tice, K varies from 0.1 to 0.1667. Examples: 1. In a 33 kV, kV, overhead overhead line, there are three units in the string of insulators. insul ators. If the capac ca pacitance itance between between eac eachh insulator pin and earth is 11% of self – c apacitance apac itance of each eac h insulator, find: find: a. the distribution of voltage voltage over 3 insulators b. string efficiency 2. For a string insulator with four discs, the capacitance of the disc di sc is i s ten times the c apacit apac itance ance between the the pin pin and earth. Calculate the voltage across each disc

when used on a 66 – kV line. li ne. Also, calc c alculat ulatee the string string efficiency. 3. A 3 – phase transmission line is being supported by three disc insulators. The potentials across top unit (near to the tower) and m iddle unit are 8 kV and 11 kV respectively. respectively. Calc ulate: a. the ratio of capacitance cap acitance between pin and earth to the self – capacitanc capac itancee of each unit b. the line lin e voltage voltage c. string efficiency 4. Each line of a 3 – phase system is suspended by a string of 3 similar insulators. If the voltage across the line unit uni t is 17.5 kV, kV, calcul cal culat atee the line to neutral voltage. voltage. Assume Assume that the shunt capaci capa citance tance between each th insulator and earth is 1/8 of the capacitance of the insulator itself. Also, Also, find the string efficiency. effici ency. 5. Each line of a 3 phase, 33 kV system is suspended by a string of 3 identica i denticall insulator discs. disc s. T he capacitan capacitance ce of eac h disc is 9 tim es the capac itance to ground. ground. Find nd voltage distribution d istribution ac ross each eac h insulator and the strin stringg efficiency. 6. An An insulator insul ator string consists of three three units, each eac h having having a safe working voltage of 15 kV. The ratio of self – capacitance to shunt capacitance of each unit is 8:1. Find the maximum maxi mum safe working voltage voltage of the string. Also, Also, find the string string efficienc y. 7. A string of 4 insulators has a self – capaci cap acitan tance ce equa equall to 10 times the pin to earth capac ca pacitanc itance. e. Find: a. the voltage distribution across various units expressed as a perc entage of total total voltage across across the string b. string efficiency 8. A string string of 5 insulators is connected conne cted across a 100 – kV line. If the capacitance of each disc to earth is 0.1 of the capaci capa citance tance of the insulator, calcul cal culate: ate: a. the distribution of voltage voltage on the insulator discs b. the string string efficienc effici encyy 9. T he three bus bar c onductors onduc tors in an outdoor substati substation on are supported by units of post type insulators. Each unit consists of a stack of 3 pin type insulators fixed one on the top of the other. The voltage across the lowest insulator is 13.1 kV and that across the next unit un it is 11 kV. Find the bus bar voltage voltage of the station. 10. A string string of 4 insulators is connected conne cted across a 100 – kV line. If the capac ca pacititance ance of each disc to earth is 0.1 of the capaci ca pacitance tance of the insulator. Calc ulate: a. T he Distribution of voltage on the insulator disc b. T he string efficienc efficienc y 11. Each conductor of a 3 – phase high voltage transmission line is suspended by a string of 4 suspension type type disc insulators. If the potential potential


41

Electrical Transmission and Distribution System and Design difference across the second unit from top is 13.2 kV and across the third from top i s 18 kV, kV, determine the the voltage between conductors. cond uctors. 12. A string of four insulators has a self – capacitance equal to 5 times pin to earth capaci cap acitan tance. ce. Find: a. the voltage distribution across various units as a percentage perc entage of total voltage voltage across the string b. string efficiency 13. A three – phase overhead transmission line is being supported by three – disc suspension insulators; the potentials across the the first and an d second insulator from from the top are 8 kV and 11 kV respec tively. tively. Calculate: Calc ulate: a. the line lin e voltage voltage b. the ratio of o f c apacit apac itance ance between pin and earth to self – capacitanc capac itancee of each unit c. the string efficiency 14. A 3 – phase p hase overhead transmi transmission ssion line is i s supported supported on 4 – disc suspension suspensio n insulators. T he voltage voltage ac ross ross the second and third discs are 13.2 kV and 18 kV respectively. Calculate the line voltage and mention the nearest standard voltage. voltage. 15. In a 3 phase, overhead overhead system, system, each eac h line is suspende suspendedd by a string of o f 3 insulators. The Th e voltage voltage across the top unit (near the tower) and m iddle unit are 10 kV and 11 kV respectively. respectively. Calculate: Calcu late: a. the ratio of shunt c apacitanc apac itancee to self – c apacit apac itan ance ce of each eac h insulator b. the string efficiency c. line voltage 16. Each line of a 3 – phase system is suspended by a string of 3 similar insulators. If the voltage across the line unit is 17.5 kV, calculate cal culate the line to neutral voltage voltage and string efficienc y. Assume Assume that shunt shun t capacit ca pacitan ance ce between each insulator and earthed metal work of tower to be 1/10 th of the capaci capa citance tance of the insulator. 17. T he three bus bar conduc con ductors tors in an outdoor substation on are supplied by units of post insulators. Each unit consists con sists of a stack of 3 pin insulators in sulators fixed one on the top of the other. The voltage across the lowest insulator is 8.45 kV and that ac ross the next is 7.25 kV. kV. Find the bus bar voltage voltage of the station. 18. A string of suspension insulators consists con sists of three three units. The T he c apacitance between between each link l ink pin and eart earth is 1/6 of the self – capacitance of each unit. If the maximum voltage per unit is not to exceed 35 kV, determine the maximum voltage that the string can withstand. Also, Also, calcu cal culat latee the string string efficienc y. 19. A string string of 4 insulators has self – c apacit apac itance ance equal to to 4 times the pin – to – earth capaci capa citan tance. ce. Calculate: Calc ulate: a. the voltage distribution across various units as a percentage perc entage of total voltage voltage ac ross the the string

b. string efficiency 20. A string of four suspension insulators is connect con nected ed across a 285 – kV line. li ne. T he self self – c apacitance apac itance of each each unit is equal to 5 times pin to earth capacitance. Calculate: a. the potential difference difference ac ross each unit b. the string string efficienc effici encyy Methods Meth ods in Impro I mproving ving Strin String g Efficiency T he m aximum voltage appears across the insulator neares nearestt to the line c onductor onduc tor decreases dec reases progressively progressively as the cros c rosss arm is approached. approac hed. If the the insulation i nsulation of the highest stressed stressed insulator breaks down or flashover takes place, the breakdown breakdown of other units will take plac e in succession. succ ession. 1. By using Insulators with Larger Discs or by providing each i nsulato nsulatorr unit un it with with metal cap - One method is to design the units such that the mutual capacit ca pacitance ance is i s muc m uchh greater than the shunt shunt capaci cap acitance tance.. T his can ca n be achieved ac hieved by using insulators with larger discs or providing each insulator unit with a metal cap. - The ratio K can be made 1/6 to 1/10 by this method. 2. By using u sing longe l ongerr cross arms - The value of string efficiency depends upon the value of K, (K is the fraction of the self – capacitance in the string), thus to determine the shunt capac c apacitance, itance, C e = KC. T he lesser the value value of K, the greater the string efficiency. K can be reduced by reducing the value of the shunt capacitance, to decrease it, the distance of conduc con ductor tor from from tower must be inc reased and thus thus larger cross arms must arms must be used. of - T he lim itation of cost and mec hanical strength of line supports do not all ow the cross arms to be too long and it has been found that in prac tice it is not possible to obtain the value of K less than 0.1. 3. By grading insul in sulato ators rs - In this method, insulators of different di fferent dimensio dime nsions ns are so chosen so that each has different capacitance. The insulators are capacitance graded, they are assembled in the string in such a way that the top unit has minimum capacitance and increasing to the bottom limit. - It can be shown that by this method complete equality of voltage across the units of insulator string can be obtained but this method needs many different sized insulators. This involves maintaining spares of all varieties of insulator discs which is contrary to the tendency of standardization. used in practice below 200 kV. - T his method is not used


42

Electrical Transmission and Distribution System and Design Consider a 4 – unit string. Let C be the capac ca pacititance ance of the top unit and let the capacitances of other units be C 2, C3 and C4 as shown in Figure 81.

Applying Kirchhoff’s current law to node 1 in Figure Fi gure 82, we

I  i ′ == iI i i ii = ωC =iV= ωKCV i = 2ωCV i= 2ωKCV i = ωC×2V= 2ωC KC C =i2 i = ωC×V= ×V= ωCωC C  = 2KC i C= ωKC= −V i = n  1ωCωCV C = − C = npKCp p get

Similarly, Also Also

The potential causing current units leaving the the top one) So Thus,

is 2V (voltage across two

T he potential potential causing causin g current is i s V So Thus,

C = KC I = I  i   ωCC =CvK=KωCωCC=v Cω1CvKK I = I  i    ωC v = ωC ω C vωC 2v    C = C  2KC = C1  K 2KC = C1 3K I = I  i     3v ωC v = ωC ω C vωC   C = C  3KC = C1 3 3K 3KC = C1 6K Figure 81 Grading 81 Grading Insulators

Assume Assume

Applying Kirchhoff’s current law to node A, A, we get

Apply Applying

Kirchhoff’s Kirchho ff’s current cu rrent law to node B, we get

Applying Kirchhoff’s current law to node C, we get

4. By Static Shielding - T he potential potential across each unit in a string string can be equaliz equali zed by using a guard g uard ring which is a m etal etal ring electrically connected connec ted to the conductor and surrounding bottom. The guard ring ring introduces capacitance between metal fittings and the line conductor. The guard ring is Figure 82 Insulator 82 Insulator with Guard contoured in such a Ring way that the shunt capacitance currents are equal to the line capacitance currents.

In general, if there are n units uni ts and or

similarly,

or the capac c apacitance itance of the the pth metal link to the line is given as

Arcing Horn and Grading Grading Ring In the event of flash – over the insulator is crac ked or broken up due due to the heat of the arc. Grading ring, in addition to equalization equa lization of voltage voltage distribution across the insulator units, when used in conjunction with arcing horn fixed at the top end of the string serves the purpose of 83 Insulator 83 Insulator String with Arcing arcing arci ng shield and protects FigureHorn and Grading Ring the insulator string from flash – over whenever overvoltage (under normal or abnormal abnorma l condition) con dition) appears between between the tower structure structure and the line lin e conduc co nductor. tor. T hey are designed to keep the arc away from the insulator string until it is interrupted by the device protecting the line. The arrangement of the arcing horns on a 7 – unit un it string string of suspension insulators i nsulators is shown shown in Figure 83. The combination of the arcing horn and grading ring provides path through the air medium and


43

Electrical Transmission and Distribution System and Design discharges disc harges the energy contained in the abnormal a bnormal voltage voltage and thus saves saves the insulator string. Examples: 1. A string of 6 suspension suspension insulators is to be graded to obtain uniform distribution of voltage across the string. If the pin to earth capacitance are all equal to C and the mutual capacitance of the top insulator is 10 C, find the mutual capacitance of each unit in terms of C (Figure (Fi gure 84). 2. The self – capacitance of each unit in a string of three suspension insulators is C. T he shunting c apacitance of of the connec c onnecting ting metal work of of Figure 84 each eac h insulator to earth earth is 0.15 C while whi le for line it is 0.1 C. Calculate: a. the voltage ac ross each insulator insul ator as a percentag percentagee of the line voltage to earth ea rth b. string efficiency 3. Each of three insulators forming a string has self –

capacitance of “C” farad. The shunt capacitance of

each cap of insulator is 0.25C to earth and 0.15C to line. Calculate the voltage distribution across each insulator as a percentage of line voltage to earth and the string effic efficienc ienc y. 4. It is required to grade a string having seven suspension insulators. If the pin to earth capacitance are all equal to C, determine the line to pin capaci cap acitance tance that that would give g ive the the same voltage across each insulator of the string. 5. A string of 8 suspension suspension insulators is to be fitted with a grading ring. If the pin pi n to eart e arthh capacitance are all equal to C, Figur e 85 find the values of line to pin capacitances that would give a uniform voltage distribution over the string (Figure 85). 6. Define the string efficiency. Calculate its value for a string of 3 insulator units used if the capacitance of each unit to Figur e 86 earth and line be 20% and 5%

of the self – capaci cap acitance tance of the the unit. Derive the formula used (Figure 86). 7. Each line of a 3 – phase system is suspended by a string of 3 identical insulators of self – c apacita apac itanc ncee C farad. T he shunt capacitance c apacitance of connecting connecting metal me tal wor work of each insulator is 0.2 C to earth and 0.1 C to line. Calculate Calc ulate the the string efficiency efficien cy of the system system if a guar gua rd ring inc reases the capacit capa citance ance to the line of metal wor w orkk of the lowest insulator insul ator to 0.3 C. 8. In a transmission line each conductor is at 20 kV and is supported by a string of 3 suspension insulators. The air capacitance between each cap – pin junction and the tower is 1/5 of Figur e 87 the capacitance of each insulator unit. A guard ring, effective only onl y over over the line – end insulator unit is fitted so that the voltages in the two units nearest the the line li ne – end are equal (Figure 87). 87). a. Calculate Calc ulate the voltage on the line – end unit b. Calculate Calc ulate the the value value of capac c apacitance itance Cx required 9. Each of the three insulato insulators rs forming a string has a self – capacitance of C farad. The shunt capacitance of each eac h insulator insula tor is 0.2C to earth and 0.1C to line. line . A guard ring inc reases the capacitance of line of the metal wor w orkk of the lowest insulator to 0.3C. Calculate the string efficienc y of the arrangement: a. with the guard ring b. without guard ring Nominal Voltage (kV)

69 138 230 345 345 500 500 765 Nominal Voltage (kV)

69 138 230

Phase Conductors Number of conductors per Bundle

Aluminum Cross – Section Area per Conductor (ACSR) (kcmil) 1 1 300 – 700 1 400 -1000 1 2000 – 2500 2 800 – 2200 2 2000 – 2500 3 900 – 1500 4 900 – 1300 Suspension Insulator String

Number of Strings per phase

1 1 1


Number of Standard Insulator Discs per Suspension String 4 to 6 8 to 11 12 to 21

Bundle Spacing (cm)

45.7 45.6 45.7 45.7

Minimum Clearance Clearance Phase – Phase – to – to – Phase Ground (m) (m)

4 to 5 6 to 9 6 to 9 6 to 9 9 to 11 9 to 11 13.7 Shield Wires

7.6 to 11 7.6 to 11 9 to 14 9 to 14 12.2

Type

Number

Diameter (cm)

Steel Steel Steel or ACSR

0,1 or 2 0,1 or 2 1 or 2

1.1 to 1.5

44

Electrical Transmission and Distribution System and Design 345

1

18 to 21

Alumoweld

2

345

1 and 2

18 to 21

Alumoweld

2

500

2 and 4

24 to 27

Alumoweld

2

500

2 and 4

24 to 27

Alumoweld

2

2 and 4

30 to 35

Alumoweld

2

765

0.87 to 1.5 0.87 to 1.5 0.98 to 1.5 0.98 to 1.5 0.98

Table 9 Ty pical Transmission Line Characteristics Characteristics

Corona - Occurs when an alternating potential difference is applied ac ross two two c onductors whose spac ing is large large as compared to their diameter, there is no apparent change in the condition of the atmospheric air surrounding the wires if voltage is low. However, when when the potential drop is inc i ncreased, reased, then then a point is reached reached when a faint luminous glow of bluish color appears along the lengths of c onductors and at the same time time a hissing sound is heard. This phenomenon is called “visual corona ”. - Is always accompanied by the production of ozone which whic h is readily detected because bec ause of its charac terist teristic odor. If the potential difference is further increas inc reased ed,, then the glow gl ow and hissing hissin g both increases inc reases in intensity till till a spark – over over between the c onductors takes place due due to the breakdown of air insulation. - T he whole phenomenon phenomenon – the hissing hissi ng noise, no ise, the viol violen entt glow, and the production of ozone gas is known as corona. ph enomenon is very muc h evident in transmissio transmissionn - T his phenomenon lines of 100 kV and above. co nductors tors are smooth and polished, polished , the the corona corona - If the conduc glow is uniform along their length but if there is any roughness, they they will w ill be picked pic ked up by relatively brighter brighter illumination. - In the case of conductors with spacing shorter as compared to their diameters (interaxial difference is less than 15 times the diameter of the conductor), sparking may take plac e without any visibl visiblee glow. gl ow. - If the potential difference between wires is direct d irect instead of alternating, there is a difference in the appearance of the two wires. The positive wire has a smooth glow about it whereas the glow about the negative wire is spotty. - Corona is accompanied by a loss of energy which increases inc reases very very rapidly once the visual c ritical voltage is exceeded. excee ded. Power loss, due to corona, co rona, heavily depends depends upon weather condition – during humid and moist clim ate corona loss is muc h increased. increased. - The energy loss accompanied by the phenomenon, called the corona, is dissipated in the form of light, heat, sound and chemi c hemica call action.

In c ase of ac system, system, the current due to corona co rona is non si nusoidal curr cu rren entt – sinusoidal. In practice, this non – sinusoidal and the non – sinusoidal voltage drop c aused byit may be more m ore important than than the power loss. Effects Effe cts of Corona Coron a 1. A violet violet glow g low is observed observed around the conductor. 2. It produc es a hissing noise. 3. It produces produ ces ozone which c an be readily detected by its characteristic cha racteristic odor. 4. T he glow is maxim um over over rough and dirty surfaces of the conductor. c onductor. 5. It is accompanied by a power loss – the wattmeter connected connec ted in the electric c ircuit will show a reading. 6. T he c harging current under corona c orona condition condition incre i ncreas ases es because bec ause the corona induce ind ucess harmonic curr c urrents. ents. Factors Factor s Affecting Corona 1. Atmosphere. Atmosphere. As As c orona is formed due to ioniz ioni zation of air surrounding the conductors, con ductors, therefore, it is affecte a ffectedd by the physical state of atmosphere. In the stormy weather, the number num ber of ions is more than norm al and and as such corona occurs at much less voltage as com pared with fair weather. 2. Conductor size. size . The Th e corona co rona effect effect depends upon the the shape and an d conditions c onditions of the conduct co nductors. ors. T he roughand and irregular surface will gi ve rise to more corona c orona becaus because unevenness of the surface decreases the value of breakdown voltage. Thus, a stranded conductor has irregular surface and henc e gives gives rise to more corona that a solid conductor. cond uctor. 3. Spacing between conductors. conductors . If the spacing between the conduc con ductors tors is made very very large as compared to their diameters, there may not be any corona coron a effect. It is becau b ecause se larger distance distanc e between between conduc con ductors tors reduces reduce s the elec trostatic trostatic stresses stresses at the conduc con ductor tor surface, thus avoiding avoiding corona c orona formation. 4. Line voltag vol tagee . T he line voltage greatly affects corona c orona.. If it is low, there is no change in the condition of air surrounding surrounding the the conductor conduc torss and hence no corona is formed. However, if the line lin e voltage voltage has such a value that electrostatic el ectrostatic stresses developed at the c onductor onductor surface make m ake the air around aroun d the conduc c onductor tor conducti conducting ng,, then corona c orona is formed. Tripl e Harmonic Current Currentss Due to Corona Coron a Corona forms when the voltage voltage of a conductor cond uctor passes the the disruptive disruptive c ritical voltage and disappears when the voltage descends desce nds through the the same value. value. T his occ urs on each conduc con ductor tor every every half cycle, and develops develops a pulsation in the voltage wave having three times the generator frequency. frequency. In an earthed system, system, this triple frequency voltage voltage caus cau ses a triple frequenc y current cu rrent to flow through the capaci capa citan tance ce of of the system system to earth and bac k through the earthed earthed neutral. -


45

Electrical Transmission and Distribution System and Design This effect is accentuated by the fact that the effective capaci cap acitance tance of the conductors condu ctors pulsates at triple frequency due to the inc rease and decrease of effective effective diameter diam eter of of conduc con ductors tors caused by the c orona. Because of c orona triple triple frequency currents flow through the ground in c ase of a non non e arthed system, system, the neutral has a voltage to earth of triple – earthed frequency. Advantages Advan tages of Corona 1. Due to corona formation, the air surrounding the conductor becomes conducting and hence virtual diameter diam eter of the conduc co nductor tor is inc reased. The incre in creas ased ed diameter reduces the electrostatic stresses between the conductors. 2. Corona reduc es the effects of transients produced produc ed by surges. Disadvantages Disad vantages of Corona 1. Corona is accompanied by a loss of energy. This affects the transmission effici ency of the line. li ne. 2. There is a definite dissipation of power although it is not so important except under abnormal weather conditions like storms etc. 3. Corrosion due to ozone formation. 4. T he current c urrent drawn by the line due to corona co rona losses is non – sinusoidal in character, hence it causes non – sinusoidal drop in the line which may cause some interference with neighboring neighbori ng commun com munication ication c ircui ircuits due to electromagnetic and electrostatic induction. Such a shape of corona current tends to introduce a large third harmon h armonic ic component. However, However, it has been found that corona c orona works as a safety valve valve for surges. 5. Particularly intense corona effects are observed at a working voltage voltage of 35 kV or o r higher. Hence, Henc e, designs designs must be made m ade to avoid avoid any corona co rona on the bus b us bars of substations rated rated for 35 kV and higher voltages voltages durin d uringg their normal n ormal operation. Corona disc harge around bus bus bars is extremely undesirable because the intense ionization of the air reduces its dielectric strength, makes it easier for the flashover to occur in the insulators and between phases particularly when the surfaces concerned are dirty or soiled with other deposits. The ozone produced due to corona discharge aggressively attacks the metallic components in the substations and switchgear, covering them with oxides. Moreover, the crackling sound of the c orona discharge disc harge in a substation masks masks other sounds like light crac kling noise due to arcing in a loose c ontact, the sound of an impen i mpending ding breakdo breakdown wn or creepage discharge in the equipment, the rattling noise due to the loosening of steel in a transformer core etc. The timely detection of such sounds is very important imp ortant if any serious serio us breakdown is to be avoided.

Dielectric Strength of Air T he value of potential gradient at which whi ch complete complete disruptio disruptionn of air occurs, oc curs, is called the disruptive strength or o r dielectric strength of strength of air. T he breakdown strength of air at a barometric pressure of of 760 mm, mm , and temperature of 25°C is 30 kV/cm kV/cm (maximum) (maximum) or 21.1 kV/cm (rms) and is denoted by . The value of dielec tric strength strength of air depends upon the density of air – is proportional p roportional to density of air over over a wide range and thus directly direc tly proportional proportional to the barom etric pressure, and inversely proportional to the absolute temperature. temp erature. T hus, hus, breakdown strength strength of at a barometric barom etric pressure of b cm of merc ury and temperature of t °C °C becomes where

g

g

δ  δ = b760×10 ×27327325  = 273t 3.92b

Disruptive Cri tical tical Voltage Let us c onsider the two – wire line shown sho wn in Figur Fi guree 88, where r is the radius of the line conductor con ductorss and and d is i s the distance distanc e between between their centers, where d is very large c ompared om pared to r. If a positive charge of q coulombs per meter of conduc con ductor tor length length is giv gi ven Figure 88 Two 88 Two - wire Line to c onductor A, then a negative charge of q coulombs per meter of conductor length will be induc ed on conductor B. B. Consider point P at x meters m eters from c onductor A. A. Electric Elec tric field intensity i ntensityat point P due to c harge on c onductor or A acting towards B as shown.

=   = − E =   − =    −  − − V = ∫ qEdx = ∫ 2πεq 1x  dd1xrdx V = 2πεlog x  qlogogd d xr r = πεlog r d  r ≅ d V = πεq log dr Electric Elec tric field intensity in tensity at point poi nt P due to induc ed char c harge ge on on conductor conduc tor B acting ac ting towards towards B as shown.

Resultant electric fiel d intensity at point P,

Potential Drop between c onductors A and B,

Now, since sinc e r is very very small as compared co mpared to d, and, therefore,


46

Electrical Transmission and Distribution System and Design Now gradient gradi ent at any point x from the c enter of the conduct condu ctor or A is given given by

E = 2πεq 1x  d 1 x = 2πεq  ∙ xdxd E = lπεogVdr ∙ 2πε1  ∙ xddx x = 2loVg dr ∙ xdd x E =  ∙ −  V = Line  to √ Li3 ne Voltage = √ V3       =  =  = log dr ≅ logdr

Substituting Substituting for q, we hav ha ve

where V is voltage between two conductors. c onductors. and

where V’ V’ is the line l ine to neutral voltage voltage of the system. system. In case c ase of three phase system, system,

From the expression for the potential gradient for a given transmission system, the potential gradient increases i ncreases as x decreases, dec reases, the the potential gradient is maxim um when x = r, the surface of the c onductor, onduc tor, and this value value is given by

where r is the radius radiu s of c onductors onduc tors in cm, d is the spacing spacing

in cm and V’ is the value of voltage of the conductors to

neutral, both g max and V being expressed in rms values. values. When the disruptive gradient of air is reached at the conductor conduc tor surface surface

g = rloVg dr V = grlog dr V = gδmrlog dr d V = 2.3gδmrlog r V = 21.1mδrlogdr d V = 21.1mδr×2.3log r

or disruptive critic al voltage, voltage,

In practice, corrections must be applied to the above formula of air density and surface conditions of the conduc con ductor tor and thus the comple co mplete te formula becomes

Substituting Substituting the value of g 0,

T he irregularity factor m 0 depends on the shape of cross – section of the wire and an d the state state of its surface. surface . Its Its value is unity for a smooth sm ooth wire of one strand of c ircular ircu lar section section and and less than uni ty for wires roughened due to weathering as shown in Table T able 10.

Polished wires Weathered wires 7 – strand cables, cab les, concentric con centric lay Cables with m ore than 7 – strands

1.0 0.93 to 0.98 0.83 to 0.87 0.80to 0.80to 0.85

Table 10 Irregularity Irregularity Factor

Examples: 1. A single phase overhead overhead line has two conduc c onduc tors of diameter diam eter 1 cm with w ith a spacing spaci ng of 1 m between c enters. enters. If the disruptive c ritical voltage for air is 21.1 kV/cm kV/cm,, for what value value of the line voltage voltage will corona co rona commence. 2. A 3 – phase line has conductors 2 cm in diameter spaced space d equilaterally equil aterally1 m apart. If the di electric elec tric strengt strength of air ai r is 30 kV (max) per pe r cm , find the disruptive critical critical voltage for the line. li ne. T ake air density factor and irregularity factor . 3. A 3 – phase, 220 22 0 kV, kV, 50 Hz transmission line consist c onsists of 1.2 c m radius conductors c onductors spaced 2 m at the corners corners of an equilateral triangle. Calculate the disruptive critic al voltage voltage between the lines. Irregularity factor fac tor = 0.96; temperature at 20°C. Barom etric pressure 72.2 72.2 cm of mercury. Dielectric strength of air = 21.1 kV(rms)/cm kV(rms)/cm.. Define Define the disruptiv di sruptivee c ritical voltage. 4. A 132 – kV line l ine with wi th 1.956 cm diameter diam eter conductor conductors is built so that corona takes place if the line voltage exceeds 210 kV (RMS). If the value of potential gradient at whic h ionization ionization occu oc curs rs can be taken as 30 kV per cm, c m, find the spaci ng between the conductors. 5. Find the disruptive critical voltage for a transmission line having: conductor conduc tor spacing spacing = 1 m conductor conduc tor (str (stranded) anded) radius = 1 c m barometric barometric pressure = 76 cm of Hg temperature = 40 ⁰C Air Air breakdown potential gradient (at 76 cm of Hg and at 25⁰C) = 21.1 kV (rms)/cm (rms)/cm.. 6. A 3 – phase overhead transmission line operates at 132 kV, kV, 50 Hz. Hz. T he conductors condu ctors are arranged in a 3m 3m delta configuration. configu ration. What What is the minim m inimum um diameter diameter of of conductor that can be used for no corona under fair weather conditions? c onditions? Assume Assume air density de nsity fac factor tor of 0.95 0.95 and irregularity factor of 0.85. 7. Taking the dielectric strength of air to be 30 kV/cm, calcul cal culat atee the the disruptive disruptive critic al voltage voltage for for a 3 – phase line with conductors of 1 cm radius and spaced symm symmetrical etrically ly 4 m apart. apa rt. 8. A conductor c onductor with 2.5 cm diameter is passed through a porcelain bushing having having and internal and external diameters of 3 cm and 9 c m respectiv respec tively. ely. The voltage between the conduc c onductor tor and an earthed clam clampp surrounding the porcelain is 20 kV rms. Determine


δ = 0. 9 52 m = 0.9

ε = 4

47

Electrical Transmission and Distribution System and Design whether corona will wil l be present in the air spac e around around the conductor. c onductor. Visual Critical Voltage In the case c ase of parallel wires, it is found that that visual visual corona does not begin at the voltage at which whic h the disruptive disruptive gradient of air , but a higher h igher voltage , called the visual critical voltage. voltage . Visual Critical Cri tical Voltage is defined as the minimu min imum m phase to neutral voltage at which glow appears all along the line conductors. Thus, when corona begins, the potential gradient at the conductor surface is higher than the disruptive disruptive gradient g radient . Contrary to what wh at might migh t be expect expec ted, ed, , the apparent strength of air is not constant but depends depends on the size size of the conduc con ductors, tors, air being apparently appa rently stronger stronger at the surface of sm aller conductors cond uctors than larger ones. Peek states that the disruptive critic al voltage voltage must be so exceeded that the stress is greater than the breakdown value up to a distance of cm from the conduc tor. or. T his visual visual c orona will occ ur when the breakdown value value is attained at the distance distanc e from from the axis, a xis, instead of at a t the distance r. this requires req uires that that the voltage to neutral be times the disruptive critical voltage. Thus, the

g

V V

g g g

0.3√ δrδr r0. 3 δr δ r √ 1  √. V = mgδr 1  0 √ 0.3δrδ.3r logDr D V = 21.1mδr 1  √ δrδr×2.3log r

visual c ritical voltage is

where m v is another an other irregularity factor hav h aving ing a value of 1.0 for smooth conduc tors and 0.98 – 0.93 for rough rou gh conduct c onductor or exposed to atmospheric severities severities and 0.72 for loc al coron coronaa on stranded conduc con ductors. tors. Because Becau se of irregular surface of the conductor, con ductor, the corona corona does not start simultaneously on the whole surface but it takes place at different points of the conductor which are pointed and this is known as local corona. For corona. For this m v = 0.72. for dec ided or general corona along the length of the conductor conduc tor m v may be taken taken as a s 0.82. Examples: 1. Find the disruptive critic al and visual corona voltage of a grid line lin e operating at 132 kV. conductor conduc tor diameter diameter = 1.9 cm cm conductor conduc tor spacing spacing = 3.81 m temperature = 44 ⁰C barometric barometric pressure pressure = 73.7 cm conduc con ductor tor surface factor: fine weather = 0.8 rough weather = 0.66. 2. Calculate the critical disruptive voltage and critical voltages for local and general corona on a three – phase overhead transmission line, c onsisting of three three

stranded copper c onductors spaced 2.5 m apart at the corners corne rs of an a n equilateral equil ateral triangle. triangle. Air Air temperature and and pressure are 21°C and 73.6 cm Hg respectively. The conductor diameter, irregularity factor and surface factors are 10.4 m m, 0.85 and 0.7, 0.8 respec tively. tively. 3. Find the disruptive critical voltage and visual corona voltage (local (loc al corona co rona as well as general c orona) for for a 3 – phase 220 kV line c onsisting of 22.26 mm m m diamet diameter conductors spaced in a 6 m delta configuration. The following data can be assumed: Temperature 25°C, pressure 73 cm of mercury, surface factor 0.84, irregularity factor for local corona 0.72, irregularity factor for for general (decided) (deci ded) co rona 0.82. 4. Find the disruptive critical and visual corona voltages voltages of a grid line li ne operating at 132 kV. kV. T he following data data is is given: conductor cond uctor diam eter eter 1.9 cm, cm , conductor spacin spacingg 3.81 c m, temperature 44°C, barometric barom etric pressure 73. 73.7 cm , conductor surface factor: fine weather 0.8, 0.8, rough weather 0.66. Corona Power Formation Form ation of corona is always al ways accompanied acc ompanied by dissipation dissipation of energy. T his loss will have some effect on efficiency efficien cy of of the line but bu t will not have any an y appreciable apprec iable effect on the line regulation. T his loss is i s affected both by atmospheric and line condi c ondition tion.. Soon after the critical voltage is reached, the corona loss increases inc reases as the square of the excess exce ss voltage. voltage. Under the fair – weather conditions, cond itions, the followi following ng empir emp irica icall formula for c orona power loss has been ascertained a scertained due to to Peek

kw 2 44 r  P = δ f 2 25 D(V  VV) ×10− phaskme V

where Vph is voltage to neutral in kV, is the disruptive critic al voltage to to neutral in kV and f is the supply frequency frequency in Hz. H z. Under the stormy weather conditions, is taken to be 0.8 times its fair – weather value and power loss due d ue to coron c oronaa is given by

kw 2 44 r  P = δ f  25 D (V  0.8V) ×10− phaskme

As As a matter of fact, with perfectly perfec tly smooth smoo th and cylindrical conduc con ductors tors no loss occurs occ urs until the the visual c ritical voltage is reached. It then follows the quadratic law for higher voltages. The empirical relation is derived by Peek has certain limitations and holds good only under certain conditions con ditions (the supply frequenc frequenc y lies between 25 and 120 Hz; Hz; the c onductor onduc tor radius is greater greater than 2.5 mm and ratio ratio



exceed e xceedss 1.8). Also, Also, a small error in m 0, the irregularity


48

Electrical Transmission and Distribution System and Design factor, will lead to wrong results when using the above formula. When the ratio



is less le ss than 1.8 Peterson’s formul formulaa is to

be applied a pplied for determining determini ng the corona losses and is given given as

P = 21×10log− fd( V)×K phaskmkwe  r

where k is a fac tor which whic h varies varies with the ratio

VV

in T able 11.

K

0.6 0.012

0.8

1.0

1.2

1.4

1.6



as given

1.8

2.0

2.2

0.018 0.05 0.08 0.30 1.0 3.5 Table 11 K 11 K factor for for C orona Power Loss

6.0

8.0

Examples: 1. A 3 phase, 220 kV, kV, 50 Hz transmission line consists of 1.5 cm radius conductor spaced 2 meters apart in equilateral triangular formation. formation. If the temperature tem perature is 40º C and atmospheric atmosphe ric pressure is 76 cm, c m, calculate calculate the the corona loss per per km of the line. T ake . 2. A 3 – phase pha se 220 kV 50 Hz transmission transmi ssion line consis c onsistts of 2 cm radius conductor spaced 2.5 m apart in equilateral triangular formation. formation. If the temperature tem perature is 20°C and atmospheric pressure 75 cm, m 0 = 0.8, determine determine the corona loss per km of line. 3. A certain 3 phase equilateral transmission line li ne has a total corona c orona loss of 53 kW at 106 kV and a loss of 98 kW at 110.9 110 .9 kV. What is the disruptive di sruptive c ritic al voltage? voltage? What is the corona coron a loss at 113 kV? kV? 4. Determine (i) the critical disruptive voltage (ii) the visual critical voltage; and (iii) the corona loss under foul weather conditions condi tions for 3 – phase ph ase line 160 km long, conductor diameter 1.036 cm; 2.44 m delta spacing. Air Air temperature 26.6°C, corresponding correspond ing to an approximate approxim ate barometric pressure of 73.15 cm of mercury, operating voltage 110 kV at 50 Hz, surface irregularity factor 0.85. Assume a value of m v = 0.72. Disruptive voltage under foul weather = 0.8 x fair weather value. 5. A 3 phase, 50 Hz, Hz, 220 kV transmission transmission line consist con sistss of conductor conduc torss of 1.2 1.2 cm radius spaced 2 meters at the corners of an equilateral equil ateral triangle. Calc ulate the the cor c oron onaa power loss per km of the line at a temperature tempe rature of 20 ⁰C and barometric ba rometric pressure of 72.2 72.2 c m. Take T ake the surface surface factors of the conduc c onductor tor as 0.96. 0.96. 6. Determine the corona characteristics of a 3 – phase line 160 km long. Conduc tor diameter 1.036 1.036 cm, c m, 2.44 44 m delta spacing, air temperature 26.67°C, altitude

 = 0.85

2440 m, m , c orresponding to an approxim ate barometr barometric pressure of 73.15 cm, operating voltage of 110 kV at 50 Hz. H z. 7. Estimate the corona co rona loss for a three phase, 110 kV, 50 Hz, Hz, 150 km long transmission line c onsisting of three three conduc con ductors tors each of 10 mm diameter and spaced spaced 2·5 2·5 m apart in an equilateral triangle formation. The temperature of air is 30ºC and the atmospheric pressure is 750 m m of o f mercury merc ury.. Take T ake irregularity facto factorr as 0.85. Ionization of air may m ay be assumed to take plac placee at a m aximum voltage gradient of 30 kV/cm kV/cm . 8. A 3 phase, 220 kV, kV, 50 Hz transmission line consists consists of 1.2 cm c m radius conductors c onductors spaced 2 m at the c orners of of an equilateral triangle. Calculate the corona loss per km of the line. The condition of the wire is smoothly weathered and the weather is fair wi th temperature of 20 ºC and barometric pressure of 72.2 cm of Hg. Factors Factor s affecting Corona Loss 1. Effect of System Frequency – Corona loss varies directly direc tly as the system system frequenc y. 2. Effect of System Voltage – T he electric field field in the the space around the conductor conduc torss depends mainly in the potential difference between the c onductors. onduc tors. Greater Greater the potential difference, greater the electric field and therefore, greater is the power loss due to corona. corona . In the region regi on near the disruptive disruptive critic al voltage, the rate of increase in power loss with the increase in system voltage is small but when Vph is large as compared with corona coron a loss increases inc reases at a very very fast rate with the inc rease in system system voltage. 3. Effect of Conductivity Cond uctivity of Air – – The T he conduct c onductivit ivityy of air depends on the number numb er of ions per unit volume of o f air, the size size and charge per ion. Both factors vary vary with altitude and atmospheric conditions. During rain and thunderstorms, ion content increases and therefore, atmosphere becomes more conducting. High conduc con ductiv tivity ity leads to greater corona loss. 4. Effect Effect of Density of Air – l oss inc reases wit with – Corona loss the decrease dec rease in the the density of air. T he corona loss of a transmission line lin e passing through a hilly area is higher higher than that of a similar line in planes due to reduced value of at high altitudes. 5. Effect of Conductor Radius – The electric field intensity decreases with the increase in radius of conductor. Hence with conductors of large radius, electric elec tric field field intensity in tensity decreases dec reases resulting in lower corona coron a power loss. 6. Effect of Conductor Cond uctor Surface – T he potential p otential gradient gradient at the surface of a stranded conductor is greater than than that for the equiv equi valent solid sol id conduc con ductor. tor. So, breakdown breakdown

V

δ


49

Electrical Transmission and Distribution System and Design voltage is low and corona loss is more for stranded conductors. - Roughness of the surface of the conductor condu ctor causes causes a field of distortion and gives rise to high hig h potential potential gradient causing higher corona loss. 7. Effect of Atmospheric Conditions – Corona loss, particularly particul arly local c orona discharges, increas inc reases es in rain and bad atmospheric c onditions, tions, such as fog, sleet and snowstorms. The effect is more pronounced if the conductor is of larger diameter because on smaller diameter diam eter conductors, the the radius of a rai n drop may not be very m uch smaller sm aller than the the radius of the c onductor onductor itself. 8. Effect Effect of Load Current – The T he heating of the conduct c onductor or due to flow of load current through it has an indirect reducing effect on the corona loss. Heating of conduc con ductor tor prevents prevents deposition of dew or snow on the surface of the conductor condu ctor and reduces corona loss. - During rains, the heating of conductors has no effect on the corona loss but, after the rain it accel ac celera erates tes the drying drying of the conductor cond uctor surface. surface. The time duration for which drops remain on the surface is reduced and so the corona loss is reduced. l ong transmission transm ission lines passing through through routes es - For long of varying altitudes, the average value of corona loss is i s determined determi ned by determining determini ng the corona loss loss per km at several points and then taking out an average. 9. Bundling of Conductors – A bundled conductor consists of two or more parallel sub – conductors at spacing of several diameters. These group of conductors form the phase conductors. Thus, the effective diameter of the bundled conductor is much larger than that of the equivalent eq uivalent single conducto co nductorr. The value of o f large diameter dia meter will reduce the corona loss. Methods Meth ods of Reducing Corona Effect Effect 1. By increasing conductor size . By inc reasing reasing conduc con ductor tor size, size, the the voltage at whic h corona occurs occurs is is raised and hence corona effects are considerably reduced. This is one of the reasons that ACSR conduc con ductors tors which have have a larger c ross – sec tional area area is used in transmission lines. 2. By i ncreasin ncreas ing g conductor spacing spacing . Byinc reasing reasi ng the the spacing between conductors, the voltage at which corona occ urs is raised raised and henc e corona effects can can be elimi el iminated. nated. However, However, spacing spac ing cannot ca nnot be increased increased too much otherwise the cost of supporting structure (bigger cross arms and supports) may increase to a considerable con siderable extent.

Voltage oltag e Limitation Limitati onss of Lines Lin es T he basis for the design of a transmission li ne is essentially essentially financial, the most economical line being the most acc eptable. It is because bec ause power loss due to corona c orona is of no great im portance. ItIt is considered con sidered satisfactory to design a line for operation at a voltage just below the disruptive critical voltage for fair weather (taking ). It is economic eco nomical al to have have a small c orona loss in bad weather, for a fraction of the year (storms are experienc ed at intervals) intervals) rather than have large conductors and heavy supporting structures to to avoid avoid c orona entirely. entirely. Moreover, corona is considered beneficial because of its inherent advantages. advantages. T RANS RANSMISSION MISSION LINE PA PARAMETERS (R, L, C) Series Resistance Resistance – is the most m ost important cause of power loss in a transmission line.

δ=1

R = ρ Al

where R – Resistance of the the line in Ω l – Length of the line in meter A – Cross – sectional sec tional area area of conductor in m2 ρ – resistivity resistivity of o f conductor cond uctor in Ω – m Copper – 1.77x10 -8 Ω–m @ 20⁰C or 10.66 Ω–cmil/ft Aluminum – 2.38x10 -8 Ω–m @ 20⁰C or 17 Ω–cmil/ft 1. In a single singl e phase 2 wire dc d c line, the total total resistance is equal to double the resistance of either conductor known as loop resistance. 2. In case ca se of a 3 – phase transmission transmi ssion line, li ne, the resistan resistance ce per phase is the resistance of one conduc con ductor. tor. Conversions 1 inch = inch = 1000 1000 m ils 1 cmil =

 sq.mil

Quantity Symbol SI Units English Resistivity ρ Ω – m Ω – cmil/ft Length l m ft Cross Sectional Sectional A Cmil Area Resistance Rdc Ω Ω Table 12 Comparisons of SI and Englis h Units for Calculating Calculating Conductor Resistance



Skin Effect – the tendency of the alternating current to concen con centrate trate near the the surface of a conductor. The skin effect depends on the followi ng factors: 1. Nature of the material m aterial 2. Diameter Diam eter of the wire 3. Frequency Figure 89 Skin 89 Skin Effect 4. Shape of the wire Note: Skin Note: Skin effect is negli gible when whe n the supply frequenc y is less than 50 Hz and the c onductor onduc tor diameter diameter is less than than 1 cm.


50


∝ρ  T=1T t = ρ T  t   ρ = ρ(1 ∝ t  t)

T he skin effec effectt is muc h smaller with stranded stranded conduc tors ors than with wi th solid conductors. cond uctors. It increases inc reases with the increase of cross section, section , permeability and supply frequency.

∝ ∝ ∝

Figure 90 Stranded 90 Stranded Conductor

In practice, stranded conductors are invariably used for transmission and distribution lines and hollow conductors for solid bus – bars. T his is done to overcom overcom e the the adverse adverse effects of skin effect. T o get the number numb er of strands, use the formula:

number number of strands = 3x  3x  1 R = kR     +√+ k=

x = the numbe n umberr of layers including includi ng the the single singl e center strands strands. Due to to skin effect e ffect



where F = 0.105d 2f – for copper F = 0.0063d 2f – for aluminum d = diam eter in inches f = frequenc y in c ps or Hz Material Copper Annealed Hard Drawn Aluminum Hard Drawn Brass

%conductivity

Ω – m

Ω–

100 %

x 10-8 1.72

cmil/ft 10. 37

Temperature Con stant °C 234.5

97.3 %

1.77

10. 66

241.5

61 %

2.83

17.00

228.1

20 – 27 %

6.4 – 38.51 480 8.4 Iron 17.2 % 10 60 180 Silv er 18 % 1.59 9.6 243 Steel 2 – 14 % 12 – 72 – 180 – 980 88 530 Table 13 % 13 % Conductiv ity, y, resistivity vity and temperature temperature constants of conductor metals ρ @ 20°C

RR = TT  tt R = R∝(=1 ∝∝ t  t) 1 t ∝

where: R2 – resistance at temperature 2 R1 – resistance at temperature 1 – temperature coefficien c oefficientt of resistance in temperatur temperature 1 in °C resistancee in temperatur temperature – T emperature coefficient of resistanc t in °C coefficient of resistance in 0°C – T emperature coefficient t2 – temperature 2 in °C t1 – temperature 1 in °C speci fic resistance at temperature 2 ρ2 – specific speci fic resistance at temperature 1 ρ2 – specific T he increase in resistance resistance due to Spiraling For stranded conductors, alternate layers of strands are spiraled in opposite oppo site directions to hold the strands together. together. T he strands strands make 1% or 2% longer lo nger than the actual ac tual length. length. T hus, dc resistance resistance will increase. increase. 1 % inc % inc rease in resistanc resistancee for 3 strands 2 % increase in resistance for concentrically stranded conductor conduc torss (4 or more). Examples: 1. A copper conduc tor has its specific speci fic resistanc resistancee of 1.6 of 1.6 x -6 10 Ω – cm at 0 ⁰C and a resistance temperature coefficient of 1/254.4 per ⁰C at 20 ⁰C. Determine Determ ine the following: a. T he specific resistance @ 60 ⁰C b. T he temperature coef co efficient ficient of resistance resistance at 60 ⁰C 2. T able of electrical characteristics characteristics shows of all Alumi Aluminum num Merigold Merig old stranded conductor cond uctor list a dc resistance of 0.01558 Ω/1000 ft. at 20 ⁰C and ac con ductor or resistance of 0.0956 Ω/mile at 50 ⁰C. T he conduct has 61 strands and its size is 1.113 x 10 6 c mil. mil . Verify Verify the dc resistance and find the ratio of ac to dc resistance. resistance. Let speci fic resistance be equal to 17 Ω – cmil/ft. 3. An An aluminum alum inum c onductor onductor is com posed of 37 strands strands each eac h having a diameter dia meter of 0.312 cm and has a length of 5.3 inches. inc hes. Compute for the the equivalent eq uivalent MCM cros cross sectional area and for the dc resistance of this conductor. Proximity Effect T he inductance induc tance and, therefore, the current distribution in a conductor is also affected by the presence of other conduc con ductors tors in its vicinity. ic inity. This Thi s effect effect is known as proximit as proximityy effect.

T he resistance resistance is also dependent upon the temperature temperature


51

Electrical Transmission and Distribution System and Design Effec Effects ts of Proximity Effect Affects Affects the current distribution and results in: 1. an inc rease in the the resistance resistance of the c onductor 2. decrease dec rease of self – reactance reac tance of the conduc tor T he proximity effect is dependent on the following: a. Conduc tor Size Size b. Frequenc y of the Supply c. Resistivity Resistivity of the material d. Relative Permeability Permeabil ity of the material T he phenomen pheno menon on is m ore noticeable for large conducto c onductorrs, high frequencies and closer proximity. The magnitude of the effect, at normal supply frequencies, frequenci es, in the case c ase of the wide spac ing of c onductors required for for overhead overhead transmission lines, is so small that it can be ignored. However, the the effect effec t is noticeable noticea ble in case ca se of cables ca bles where where the spacing between the the conduc tors is small. small. In the case of stranded conductors, each wire traverses alternately weaker and stronger portions of the magnetic magnetic field caused c aused by the external current carrying conduc tor. or. T hus, the the average average value of the field fiel d along the path of any wire remains remai ns the same, and if the currents in the c onductor onductor follow the paths of the individual wires, the effect is substantially eliminated. elimi nated. Line Inductance Inductance - is the property by whic h a circ uit opposes chang cha nges es in the value of a varying varying current flowing through it. - causes cau ses opposition only to vary varying ing currents. - does not cause any opposition to steady or direct current. - In case of transmission and distribution lines, the current flowing is varying varying or alternating current, the effect of induc in ductance, tance, in addition to that of resistance, is therefore to be considered. co nsidered. - T he opposition to the flow flow of varying varying current owing to induc tance is viewed viewed as a voltage voltage drop. It is well known fact that a current carrying conductor is surrounded by conc entric circ c ircles les of magnetic lines. l ines. In case case of AC system system this field set up around the c onductor onduc tor is not not constant but changing and links with the same conductor as well as with other conductors. con ductors. Due to these flux linkages, linkages, the line li ne possesses inductance, induc tance, defined defined as the flux linkages linkages per unit uni t current. Thus, Thu s, for determination d etermination of inductanc e of of a circui ci rcuit,t, determination of flux linkages li nkages is essential. Mathematically, inductance is defined as flux linkages/ampere.

where:

ψ

 = 

– flux linkages in Wb – turns I – c urrent in amperes

Besides resistanc e, the transmission transm ission line has inductance inductance as as well as capacitance. The resistance, inductance and

capaci cap acitance tance are termed as the “parameters” “param eters” of the line li ne and are uniformly uniform ly distributed along the entire length le ngth of the line. For single phase line the parameters are usually represented on loop inductance i nductance basis and for 3 – phase pha se line line on per conductor basis. Flux Linkages o f a Conductor Conductor A long straight cylindric al con ductor carrying carrying a current is surrounded by a magnetic m agnetic field. T he magnetic magnetic lines of for force ce will exist inside the conductor as well as outside the conductor. Both fluxes contribute to the inductance of the conductor. Flux Linkages of a Conductor Due to Internal Flux. Consider a long straight cylindrical conductor of radius r meters and carrying a current of I amperes, as shown in Figure 91.

Figure 91 Internal 91 Internal Flux of a Conductor

In overhead lines, it may m ay be assumed without wi thout appreciab appreciable le error that the current c urrent is uniformly distributed. distributed. T he current inside of a line of force force radius x,

I = πrI ×πx = Irx Currsetantnce = Irx × 2πx1 = 2πrI  AT/m H = 2π×di B = μμH Wb⁄m or T μ = 4π×10− B = 2πrμI Wb⁄m d∅ = B×I×dx= 2πrμx Idx webers I  I

Field Fiel d strength strength inside the c onductor onduc tor at a distance di stance x from the center,

Flux density, density,

where and µ is i s the relativ rela tivee permeability permeab ilityof the medium m edium and for for non non magn etic materials µ=1. – magnetic

Now the flux through a c ylindric al shell of radial thic knes knesss dx and axial length l ength one meter, But this flux links with only onl y the current lying within tthe he c ircle ircle of radius x, with


or

52


  μ x x μ I    dψ= dψ = 2πr×dx× r ×I=I = 2πr dx weber weber turns   μ I   ψ = ∫ 2πr μI dx Wb Wb  turns/m rns/m ψ = 8π Wbturns/m − μ μ μ ×10   L = = Henry

L = μ2πμ ln dd = 2×10− lnDr (H⁄m)

Linkages of the shell,

T hus, inductance due to external external flux,

T otal flux linkages from center of the the conductor con ductor and up to to the surface of the conduc tor,

Inductance of a Single – Phase Two – Wire Wire Li ne Consider a single – phase line consisting of two parallel conduc con ductors tors A and B of radii r 1 and r 2 spaced space d d meters apart (d being very very large compared com pared to r 1 and r 2). Conduc C onductors tors A and B c arry the same current in magnitude m agnitude but opposite in in directions, direc tions, as one forms the return path for the other. The The induc tance of each conductor conductor is due to internal flux linka li nkage gess and external flux linkages li nkages and the following points poi nts are to to be be noted regarding external fflux lux linkages: l inkages: 1. A line of flux flux produced produc ed due to current in conduc co nductor tor A at a distance equal to or greater than (d + r 2) from the center cen ter of conductor conduc tor A links with a zero – net current, cu rrent, as as the c urrent flowing in the two conductors cond uctors A and B are are equal in magnitude m agnitude but opposit oppositee in directions. 2. Flux linkages at a distanc e (d – r 2) link l ink with a current c urrent I and those between (d – r 2) and (d + r 2) link with a current varying from I to zero. As As a simplify simpl ifying ing assumption, assumption, it can be assumed that that all the flux produc ed by current cu rrent in conduc co nductor tor Alinks all the curre c urrent nt up to the center cen ter of conductor condu ctor B and that the flux beyond the center cen ter of the conduc co nductor tor B does not link any a ny current.

T hus, inductance due to internal flux, flux,

8π 8 m

Flux Linkages of a Conductor Due to External Flux. Consider two points 1 and 2 distant d 1 and d 2 from the center cen ter of the c onductor. onduc tor. Since the flux paths paths are c oncent onc entrric circles around the conductor, whole of the flux between points 1 and an d 2 lies l ies within the c oncentric onc entric cylindrical cylindrical surface surfacess passing through these these points poi nts 1 and 2.

Figure 92 External 92 External Flux of a Conductor

T he field strength at any distance distanc e x from from the center cen ter of the conduc con ductor tor (x > r),

H = 2πxI AT/m B = μH = 2πrμI Wb⁄m d∅ = 2πxμI webers per meted∅r dψ = 2πxμI dx Wbturns per meter  ψ = ∫ 2πxμI dxd= 2πμ Ilnxdd ψ = 2×10−I ln d Wbturns/m

Flux density, density,

Figure 93 Single 93 Single Phase, Two Wire Conductor

Now flux linkages per meter is equal to since flux external to conduc co nductor tor links all the current in the conductor conductor once and only once

T he above above assumption assump tion simplifies the calcu cal culations lations and results obtained are quite accurate specially when d is much greater than r 1 and r 2, as is usually the case in overhead overhead lines. l ines. Based on the above assumption flux linkages of conduct con ductor or A due to to external flux flux can be determin determined ed by substituting d 2 = d and d 1 = r 1. T hus, flux flux linkages of conductor c onductor A due to external flux,

T otal flux linkages linkages between between points 1 and 2

Flux linkages of c onductor onduc tor A due to to internal in ternal flux, flux,

So, the flux through a cy c ylindric al shell of radial thicknes thic knesss dx dx and axial length len gth one meter,

ψ = 2×10−I ln rd Wbturns/m ψ = 12 I×10− Wbb  turns/m rns/m d −

T otal flux flux linkages of conduc tor A

ψ = ψ  ψ = 0.5  2ln rI×10 Wbb  turns/m ns/m


53


L = 2×1070.25  2l2dln rd1HH/m L = 2×107ln r114 m μ L = 1×10−μ  2μ ln D H

T otal inductance induc tance of conductor con ductor A, A,

If

is not equal to 1 then,

2 rm L = 2 =×100.77788ln d  H r1′ m L = 2×107ln rd2′ mH L = L  L = 2×107ln rd1′  ln rd2′ L = 4×107ln dr′ H/m r−

The product is known as Geom Geometric etric M ean Radiu Radiuss (GMR) of the c onductor onduc tor is equal to 0.7788 times the radius of the conductor . Let it be represented by r’ 1 where

T he current in each conductor sets sets up a certain flux due its own current. c urrent. The sum of all these these fluxes is the total flux of the system system and the total flux linkages li nkages of any one condu conduct ctor or is the sum of its linkages with all the individual in dividual fluxes set set up by the conduc con ductors tors of the the system. system. Now let us determine the flux linkages linkage s of conductor c onductor 1 due due to current c urrent I1 c arried by the conduc tor itself and flux linkages linkages to c onductor onduc tor 1 due to c urrents carried carrie d by other conduc tors ors

(2, 3, …, n)

The flux linkages of conductor 1 due to its own current I 1 (internal and external), up to to point P

ψ = 2×10−I ln dr′ Wbt Wbturns/m ψ = 2×10−Iln dd Wbturns/m

T he flux linkages of c onductor 1 due to current in c onductor or 2

similarly, simi larly, induc tance of conductor B,

Loop inductance induc tance of the line,

If r 1’ = r 2’ = r’, the loop inductance induc tance of the line is given as T he idea of replaci ng the original c onductor onductor of radii r by by a fictitious conductor of radii r’ is quite attractive because streamlined equations for inductance can be developed without bogging down in ac counting coun ting for the internal flux. Flux Linkages of One Conductor in a Group of Conductors

Consider a group of o f parallel conduct c onductors ors 1, 2, 3 … n carrying carrying currents I1, I2, I3 … In respectively, respectively, as illustrated ill ustrated in Figur F iguree 94. Let it be assumed that the the sum of the c urrents in various various conduc con ductors tors is zero. zero.

Flux due to conduc co nductor tor 2 that that lies between conductors con ductors 2 and and 1 does not link conductor 1 and therefore the distances involved are d 2p and d12. T hus, the the expression for flux linkages of conduc c onductor tor 1 due to to currents in all conductor conduc torss can be written written as

ψ = 2×10− I ln dr′  I ln dd  Iln dd  ⋯  I ln dd Wb  turns/m

To account for the total flux linkages to conductor 1, the point P must mu st approach approac h infinityand in this condi tion Then,

d ≅ d ≅ d … d ≅ d     lim I  I  I  ⋯ Ilnd = 0 →∞

This simplifies the equation thus, the equation for the flux linkages to to conductor conduc tor 1 becomes

ψ = 2×10− Iln r1′  I ln d1  I ln d1  ⋯ I ln d1 Wb  turns/m

Inductance of Composite Conductor Lines – Self and Mutual GMDs Consider a single – phase line consisting of two parallel conductors A and B, conductor A consisting of x and conduc con ductor tor B of y strands, strands, as illustrated in Figure 95. Let the conduc c onductor tor A and an d B carry currents c urrents I and – I respectively (since conductors of a 2 – wire line carry the same current cu rrent but in opposite directions).

Figure 94 Cross 94 Cross - Sectional View of Group of n Conductors and Distances of Conductors from a Rem ote Point P

T heoretically, heoretically, the flux due to a conductor con ductor extends from from the center cen ter of the conduc con ductor tor right right up to infinity but let us assume assume that the flux linkages l inkages extend up to a rem ote point P and the respective distanc distances es are as marked m arked in Figure 94.

Figur e 95 Stranded and Parallel Conductors


54

Electrical Transmission and Distribution System and Design Assuming Assuming uniform current c urrent density in both the conduc c onducttors ors the current c urrent carried by each eac h strand of conductor c onductor A will be I/x while that carried by each strand strand of c onductor B will be – I/y. T he flux linkages of strand 1 in c onductor A will be

ψ = 2×10− xI ln r1′  ln d1  lnId1 1⋯  ln d11 1 1 − y ln r′  ln d  ln d  ⋯ ln d Wb  2×10 turns/m   −      

L = ψxI = 2×10 ln  r′d ,d,d , d, d , ,……, d, d H/m  d, d, d, … , d  − L = 2×10 ln r′ ,d, d , … , d H/m L = L  L xL  ⋯ L (d , d , d ,… ,d )(d , d , d , … , d )…d , d , d ,… ,d 

Inductanc e of strand 1 of conduc con ductor tor A

Similarly, Simi larly, the expression for induc tance of strand 2 can ca n be be written as

T hus, we see see that the different strands of a conduc con ductor tor hav have different inductances T herefore, inductance of conduc con ductor tor A, A,

             − L = 2×10 ln d, d , d , … , d( d, d , d , … , d )… d , d, d ,… ,d

In the above expression, the numerator of argument of ln is called the GMD GMD (often called mutual GMD) GMD ) between conduc con ductors tors A and B and the denominator denom inator of argument ln is called GM GMR R (often called self GMD). GMD). GMD (Geometric ( Geometric Mean Distance) Distance ) and GMR (Geometric ( Geometric Mean Radius) Radius ) are denoted by D m and DS respectively.

L = 2×10−ln DD mH L = 2×10−ln DD mH L = L  L D = D = D    D L = 4×10− lnD mH

similarly, simi larly, induc tance of conductor B,

Loop inductance is

2. A single – phase line has two parallel conductors conductors 2 meters apart. apart. T he diameter of each c onductor is 1.2 1.2 cm. cm . Calculate the loop inductance inductance per km of the line. 3. A single – phase line has two parallel conductors 1 meter apart. The radius of each conductor is 0.5 cm. Calculate the loop inductance per km of the line. 4. A single – phase transmission line has two parallel conductors 3 m apart. The radius of each conductor being 1 cm. Calculate the loop inductance per km length of the line if the material of the condu ctor is: a. Copper b. Steel with relative relative permeabi pe rmeability lityof 100. 5. Find the loop induc tance per km km of a single – phase overhead transmission line when conductors have relative relative permeability permeab ility of: a. 1 b. 100 Each conductor c onductor has a diameter diam eter of 1 c m and they are spaced 5 m apart. apart. 6. One circuit of a single – phase transmission line is composed of three solid 0.25 cm radius wires. The return circuit circ uit is com posed of two 0.5 cm radius wires. wires. Find the induc tance tance of the the complete com plete line in H/m. 7. T wo conductors conductors of a single – phase line, each of 1 cm diameter, are arranged in a vertical plane with one conductor mounted 1 m above the other. A second identical identic al line is mounted mou nted at the the same height as the first and spaced spac ed horizontally 0.25 m apart apa rt from from it. T he two two upper and the two lower conduc con ductors tors are connected connected in parallel. Determine the inductance per km of the resulting resulting double circuit circ uit line. 8. A 20 – km single phase p hase transmission line l ine having having 0.823 0.823 cm diameter diam eter has two line conduc c onductors tors separated separated by 1.5 meter. The conductor has a resistance of 0.311 ohm per kilometer. Find the loop impedance of this line at 50 Hz. H z. 9. Calculate Calc ulate the GMR of a 6/3 mm Al, 1/3 mm Steel ACS ACSR R conductor.

If conduc con ductors tors A and B are are identic al

Examples: 1. What is the inductance per loop meter of two parallel conductor conduc torss of a single – phase system if each has a diameter diam eter of 1 cm and their axes are 5 cm apart when when conduc con ductors tors have have a relativ rel ativee permeability perm eability of (a) unity un ity and (b) 100? T he relative relative permeability permea bility of the surrounding ng medium is unity in both cases. End effects may be neglec ted and the current c urrent may be assumed unifor un iforml mlyy distributed over over cross – section sec tion of the the wires.

Figure 96 Cross 96 Cross - Section of a 7 Strand Conductors


55

Electrical Transmission and Distribution System and Design 10. In a single – phase line , conductors conduc tors a and a′ in parallel form one circuit while conductors b and b′ in parallel form the return path. Calc ulate the total induc tance tance of of the line li ne per km if c urrent is equally shared by the two parallel c onductors. Conductor diameter in 2.0 c m.

Figure 97 Single 97 Single Phase Parallel Conductors

Inductance of Thr ee – Phase Overhead Lines With Unsymmetrical Spacing. Consider Spacing. Consider a 3 – phase line with conductors cond uctors A, A, B and C; each ea ch of radius r meters. Let the the spacing between between them be d 1, d2 and d3 and the current flowing through them be I A, IB and IC respectively.

L = 2×10− ln r1  ln  dd  j√3ln  dd H/m L = 2×10− ln r1  ln  dd  j√3l√3 ln dd H/m

Similarly,

and

Thus, Thu s, we see that when the conductors cond uctors of a 3 – phase transmission transmission line are not equidistant equidistant from each other, the flux flux linkages and inductances inductances of various va rious phases are ar e differe different nt which wh ich causes unequal voltage drop in the three phases and transfer of power powe r between between phases pha ses due to mutual mutual inductances even eve n if the current in the conductors are balanced. The unbalancing effect because b ecause of irregular irre gular spacing spacing of conductors is avoided avo ided by transposition of conductors, as shown in Figure 99. In practice, the conductors are so transposed that each of the three possible arrangements of conductors exist one hird of the total total length of the line. – third

Figure 99 Transposition 99 Transposition of a 3 - Phase Line

Figure 98 3 98 3 - phase Transmission Line with Unsymmetrical Spacing

T he effects effects of transposition transposition are as follows: a. Each conductor has the same average inductance, which is given as

L = L  L3 d dLd L = 2×10− ln r′ H/m L = 2×10− lnr′ H/m

T he flux linkages of c onductor onduc tor A due to its i ts own current c urrent I A and other conductors c onductors IB and IC

ψ = 2×10−I ln r1  I ln d1  I ln d1 Wbb  turns/m ns/m ψ = 2×10−I ln r1  I ln d1  Iln d1 Wbb  turns/m ns/m ψ = 2×10−Iln r1  I ln d1  I ln d1 Wbturns/m

Similarly, and

I = I = I = I Ij0. 8=66I I = I0.5 j0.866 I = I0.0. 5  If the system system is balanc ed, (say) (say) in magnitude Taking I A A as a reference phasor, the currents are represented, in symbolic form as ; and

b. If c onductors onduc tors are equispaced (let the spacing be equal equal to d), the inductance of each conductor will be the same

For stranded stranded conductor conduc tor r’ will be replaced by D S (self c.

GMD) When the conductors of three – phase transmission line are in the same plane, as shown in Figure 100.

ψ ψ = 2×10−Iln 1  ln d d  j√3ln  d Wbb  turns/ns/m

Substituting Substituting these values of IB and IC in the expression for for we get



and

r    

d

L = 2×10− ln r1  ln  dd  j√3ln  dd H/m


Figure 100

56

Electrical Transmission and Distribution System and Design In this position d 1 = d2 = d and d3 = 2d Thus,

L = 2×10− ln rd  12−ln22 d j0.0.866 ln 2 H/m L− = 2×10 l n H /m  r L = 2×10 ln rd  12 ln22  j0.0.866 ln2  H/m

d. When the conductors are at the corner of a right – angled triangle as shown in Figure Fi gure 101. In this position d1 = d2 and d 3 =

√ 2

d

Figur e 101

L = 2×10− ln rd  12 ln−√2 √2 d j0.0.866 ln √2 H/m L−  =d2×10 1 ln r H/m L = 2×10 ln r  2 ln√2 j0.866ln√2  H/m Thus,

Examples: 1. Find the induc tance per km of a 3 – phase transmiss transm issio ionn line using 1.24 cm diam eter conductor conduc torss when these these are placed at the corners of an equilateral triangle of each side 2 m. 2. Find the inductance per km per phase of a 3 – phase overhead transmission line using 2 cm diameter conduc con ductor tor when these are placed plac ed at the the corners c orners of an an equilateral triangle of side 4 meters. 3. T he three conductors of of a 3 – phase pha se line are arranged arranged at the corners corne rs of a triangle of sides 2 m, 2.5 m and 4.5 m. Calculate the inductance per km of the line when the c onductors onduc tors are regularly transposed. The Th e diame di ametter of each c onductor is 1.24 1.24 cm. 4. Calculate the inductance of each conductor in a 3 phase, 3 wire system when the conductors are arranged in a horizontal horizontal plane with wi th spacing spac ing such suc h that that D31 = 4 m; D 12 = D23 = 2m. T he conductors are are transposed and have a diameter di ameter of 2.5 c m. 5. T he three conductors of of a 3 – phase pha se line are arranged arranged at the c orners of a triangle of sides 4, 5 and 6 m eters. eters.

Calculate Calc ulate inductance per km of each conduct con ductor or when when conduc con ductors tors are regularly transposed. transposed. T he diameter of of each line conductor is 2 cm. c m. 6. The three conductors of 3 phase overhead line are arranged in a horizontal plane with a spacing of 4 m between adjacent conductors. The diameter of each conductor conduc tor is 2 cm. c m. Determine the inductance per per km per phase of the line if the lines are transposed. 7. A three – phase transmission line has its conductors arranged in a triangle so that two of the distances distance s is 25 25 ft. and an d the the third distance is 42 ft. Determine the induc tance and the inductiv induc tivee reactance reactanc e per phase per mile. mil e. Assuming Assuming D S of c onductors onduc tors is 0.0284 ft. ft. 8. Determine the inductance per km of a 3 – phase transmission line using 20 mm diameter conductors when conductors are at the corners of a triangle with spacing spaci ng of 4, 5 and 6 meters. Conductors are regular regul arly ly transposed. 9. Determine the induc tance of a 3 – phase symmetrical line whose c onductors are are placed at the the c orners of an equilateral equila teral triangle triangle of sides 1 meter. The T he diameter of of each c onductor onductor is 20 mm. 10. A three – phase overhead line is designed with an equilateral spac ing of 3.5 m with a c onduc tor diamet diameter of 1.2 cm. If the line is constructed with horizontal spacing spaci ng with suitably transposed conduc tors, find find spacing between adjacent conductors which would would give the same sam e value value of induc tance as in the equilater equi lateral al arrangement. 11. A three phase, 50 Hz, Hz, 30 km long line has four 4/0 wires (1.5 cm c m diameter) diam eter) spaced space d horizontally horizontally 2 m apart in a plane. The Th e wires are carrying carrying c urrents I a, Ib, Ic, and the fourth wire wi re is neutral that carries ca rries zero zero current. The The phase currents cu rrents are:

II == 20 30j 2 4 A 20  j 2 6 A I = 5050  j50 A

T he line is i s untransposed. untransposed. a. Find the flux linkages linkag es of the the neutral wire. b. Find the voltage induc ed in the neutral wire. c. Find the voltage drop in each of the three – phase wires. Inductan Indu ctance ce of Three Th ree – Phase Lines Li nes with more mo re than one one circuit It is usual practic e to to run run 3 – phase pha se transmi transmission ssion lines with more than one circ c ircuit uit in parallel on the same tower becaus becausee it gives greater reliability and a higher transmission capacity. If such circuits are so widely separated that the mutual induc i nductance tance between them becomes be comes negligible, the the


57

Electrical Transmission and Distribution System and Design induc tance of the equivalent equivalent single circuit ci rcuit would be half h alf of each of the individual circuits considered alone. But in actual practice practic e the separation is not very very wide and the mutual inductance is not negligible. GMD method is used for determination of induc tance per phase by consider co nsiderin ingg the various conduc co nductors tors connected in parallel as strands of of one composite com posite conductor. conductor. It is desirable to have a configuration that provides minimu min imum m induc tance to have maximum transmission capacity. This is possible only with low GMD and high GMR. Therefore, the practice is to have the individual conductors of a phase widely wid ely separated to to Figure 102 Arran 102 Arrangeme gement nt of Conductors in a Double Circuit 3 provide high GMR and the – Phase Line distance between the phases small to give low GM D. Thus, Thu s, in the case of a double circu c ircuitit in vertical formation the arrangem ent of conductors condu ctors woul would be as illustrated i llustrated in Figure 100. Inductance of 3 – Phase Double Circuit Line with Symmetrical Symmetrical Spacing Consider a 3 – Phase Double Circuit Circu it connec ted in Parall Parallel el – Conduc tors A, A, B, C formi fo rming ng one circuit c ircuit and conductors co nductors A’, A’, B’, C’ forming the other o ne, as illustrated in Figure 103 (conductors (cond uctors symm symm etrically etrically spaced). space d).

 3d 3 d √ − L = 2×10 ln 2dr H/m

Inductanc e of conductor A, A,

Similarly, inductance of remaining conductors can be worked out, which will be the same sam e as L A. T his is becau bec ausse the conductors of different phases are symmetrically placed. Since Sinc e conductors co nductors are electrically in parallel, parall el, inductan induc tance ce of each phase

L = 1×10− ln √ 2dr3d3d H/m

Inductance of 3 – Phase Double Circuit Line with Unsymm Unsymm etrical Spacing but Transp Tran sposed osed Now consider a 3 – phase double circuit connected in

parallel c onductors A, A, B and C forming formi ng one circuit ci rcuit and A’, B’ and an d C’ forming the other one, as illustrated in Figure 102.

Since the conductors are thoroughly transposed, the conduc con ductor tor situations situations in the transposition cycle wo uld be, as illustrated in F igures 104(1), 104(2) and 104(3).

Figure 104

ψ = 2×10− I ln r1  ln 4d1 1  d 1  I ln d1  ln d1 d  I ln 2d  ln d

Flux linkages with conduc tor A in position (1)

Figur e 103 103 3 – Phase Double Circuit rcuit

ψ = 2×10−I ln r1  ln 2d11   I1 ln1d  ln √3d1  − I l1n √3d  ln d 1 ψ = 2×10 −I ln 2dr 1 I  I ln1√3d ψ = 2×10 I l√3dn 2dr   I ln √3d ψ = 2×10−Iln 2dr  Wb turns/m

Flux linkages of phase A conductors cond uctors

Similarly, Simi larly, flux linkages with conductor c onductor A in position (2) and and (3)

ψ = 2×10−I ln r1  l 1n d1  I l1n d1  ln d1  d  I1ln d  l1n d  d 1 1 ψ = 2×10− I ln   ln 2 2  Iln  ln 


r   14d4d1  d2 1 2d d  I ln d  lnd  d

58


ψ 13= ψ2  ψ3213  ψ Wb  turnsurns per meter ψ = 2×107IAln 2r′d4d41dd211  dd22213 Wb

Av Average flux linkages linkag es with conduc con ductor tor A

L =

 

123d1 d21  d22 13 2×107 ln r′ 4d21  d22 13 H/m 1 1 2 2 6 1 2 d d  d 1 1 2 7 6 2×10 ln 2 r 4d21  d22 H/m+ + 1 1 d 1 2×107 ln 26 r 2 H/m

and inductance

Inductance of each phase,

L = 12 L = L=

     

If the distance d 2 is too large as compared to d 1, would tend to be uni ty and induc in ductance tance per phase,

Examples: 1. Figure 105 shows the spacing spaci ng of a double circui c ircuit 3 phase overhead line. The phase sequence is ABC and the line is completely transposed. T he conduct con ductor or radius in 1.3 cm. Find the Figur e 105 inductance per phase per kilometer. 2. Find the inductance per phase per km of double circuit 3 phase line shown in Figure 106. The conduc con ductors tors are transposed transposed and are of radius 0.75 cm each. eac h. The T he phase sequence sequence Figur e 106 is ABC. 3. A three three – phase double dou ble circui ci rcuitt line consists of 300,000 300,000 cmil cm il 26/6 26/6 ACSR Ostrich Ostrich conductor conduc torss arranged arranged as shown in Figure Fi gure 107. Find the 60 Hz inductiv ind uctivee reactance in ohm per mile per phase. Assume Assume D S = 0.0229 ft.

4. Calculate the inductance per phase per meter for a three phase double circuit line whose phase conduc con ductors tors have a radius radiu s of 5.3 cm with w ith the the horiz hori zontal ontal conduc con ductor tor arrangement as shown in Figure Fi gure 108.

Figur e 108

Bundled Bund led Conductors T he demand of elec tric power is increasing increa sing throughout the the world and in many m any countries, it is doubling every five five to eight years. T he power stations are usually usual lylocated loc ated far away frfrom the load centers. Thus, transmission of large amounts of power over long distances can be accomplished most economic eco nomically ally only by using extra high voltages (or simply EHV) EHV) is nece n ecessary ssary.. Voltages Voltages m ore than 230 kV fall in i n this category. An An increase inc rease in transmission voltage voltage results in reduction reduc tion of electrical elec trical losses, inc rease in transmission on efficienc y, improvement of voltage regul ation and reductio reductionn in conduc co nductor tor material requirement. requirement. At At voltages above 300 kV, kV, corona causes cau ses a significa signi ficant nt power loss l oss and interfe i nterferrence ence with com munication munication circuit c ircuits, s, if round single singl e conductor con ductor per per phase is used. Instead of going go ing for a hollow holl ow conductor it is preferable to use more than one conduc cond uctor tor per phase which which is called c alled the bundling of conductors. co nductors. Lines of 400 kV and higher voltages invariably used bundled conductors. A bundled conduct con ductor or is a conduct con ductor or made up of two or more more conduc con ductors, tors, called the sub – conductors, c onductors, per phase in clos close proximity proximi ty com pared with the spacing spac ing between phase pha sess. The The basic difference difference between a composi com posite te conductor and a bundled conductor is that sub conductors of a bundled conductor are separated to each other by a constant distance from 0.2 m to 0.6 m depending upon designed voltage and surrounding c onditions throughout the length length of the line l ine with the help of spac ers whereas the wires of a composite conductor touch each other. The bundled conductors have filter material or air space inside so that the overall overall diameter di ameter is increased. inc reased. The use of bundled conductors per phase reduces the voltage gradient near the line and an d thus, reduces reduc es the possibilities possibili ties of the corona coron a discharge. Although Although bundled conduc con ductors tors are used on EHV transmission lines primarily reduce corona loss and radio interference, they have several other advantages over single conductors cond uctors such as: 1. Transmit power with reduced losses, thereby giving increased inc reased transmission effic efficiency. iency. 2. Have higher c harging c urrent which whic h helps in im provin provingg the power factor. 3. Since GMD or GMR is increased, the inductance per phase is reduced.

Figur e 107


59

Electrical Transmission and Distribution System and Design 4. Have comparatively lower surge impedance with a corresponding correspond ing increase inc rease in the maxim um power transfer capabili tyas seen in Table T able 14. Number Numb er of sub – c onduct ondu ctor orss 1 2 4 8 Relative power transfer 1.0 1.3 1.6 1.7 Table 14 Relativ 14 Relativ e Power Transfer Transfer VS Sub – Conductors

It is i s to be noted n oted that there is a little to be gained by using using more than four sub – conductors per phase, two or three sub – conductors per phase are sufficient for most of the EHV lines. For two strand bu ndle

Figur e 112

2. The bundled conductors of a line are arranged as shown. Each conductor has a radius r. Find the expression of the mean geom etric radius in terms terms of r.

Figure 113

 = D×dD DL =2×10 − ln D

Figure 109 2 109 2 Strand Bundle Conductor

For three strand b undle

DL = 2×10 =  − Dl×dn D D


For four strands

DL = =2×101.09−  Dln×dD D

3. A bundled and transposed 3 phase transmission line has a conductor arrangement shown. The identical conductors have a radius of 0.74 cm. The spacing between phase conductors is 30 cm. Determine the line reactance reac tance per phase per mile at 60 Hz. Hz. 4. Find the inductive reactance of a 3 – phase p hase bundled bundled conductor line with 2 conductors per phase with spacing spaci ng of 40 c m. Phase to to phase separation is 7 m in horizontal horizontal configuration. c onfiguration. All All c onduc tors are ACSR ACSR with with diameter diam eter of 3.5 3.5 c m. com pare the above above value value with that conduc tor line. of an ‘equivalent’ ‘equ ivalent’ single – conductor 5. Determine the induc in ductan tance ce per km per phase pha se of a single circui ci rcuitt 460 kV line using two bundle conduc tors per phase as shown in i n Figure 113. 11 3. The diameter di ameter of each each conductor conduc tor is 5 cm.


Figure 114

Examples: 1. A single – phase transmission line uses bundled arrangement arrangeme nt shown. Line A has its c onductors onduc tors situat situated at the corners of an equilateral equ ilateral triangle. All All conduc con ducttors ors are identical identic al and eac h has a radius of 2 cm. Determine Determine the following: a. The inductance of each line and the total inductance in m H per km. b. For operation of the line at a frequency of 60 Hz. Hz. Determine Determine the total line reac tance in ohm/mile.

Shunt Capac itance of T ransmissio ransmissionn Lines As As any two conduc co nductors tors of an overhead overhead transmission line l ine//s separated by air, which acts as an insulator, therefore, capacitance exists between any two overhead line conductors. The capacitance between the conductors is defined as the charged per unit potential difference. The capaci cap acitance tance is uniformly distributed along the whole length of the line and may be regarded as a uniform series capacitor connected between the conductors or a shunt capaci cap acitance tance between between any conductor conduc tor and the the earth. When When an alternating voltage voltage is impressed i mpressed on a transmission transm ission line, line, the c harge at any point inc reases and decreases with wi th the the increase and decrease of the instantaneous value of the


60

Electrical Transmission and Distribution System and Design flow of voltage between the the conductors con ductors at the point. poi nt. Thus Th us,, a charging cha rging current c urrent flow to the the line even if i t is open c ircuit irc uited ed.. It affects the voltage drop along the line as well as the efficienc y and the power factor factor of the line. l ine.

Capacicitance = Q/V

where Q – Charge in coulombs V – Elec tric potential in volts volts C – Capacitance in Farad If the c apacitanc apac itancee of an overhead line is high, high , the line draw drawss more charging cha rging current, whic h com c ompensat pensates es or cancels c ancels the the lagging component of load current. Hence, the resultant current flowing in the line is reduced. The reduction in resultant current flowing in the line results in: 1. reduction of line losses and so increase of transmission efficiency efficien cy 2. reduction in voltage drop or improvement of voltage regulation The other advantages of a transmission line having high capacitance are increased load capacity and improved power factor. Charging Current:

I = X = 2πfCV

Electric Potential T he electric potential at a point poin t due to a charge cha rge is the work done in bringing bringi ng a unit positiv p ositivee c harge from infinity to to that point. Potential Charge of a Single Singl e Conductor Consider a long l ong straight straight cylindric al conductor A of radius r meters and having a charge q coulombs per meter of its length. T he electric elec tric field field intensity at distance distanc e x from from the center of the conductor cond uctor is given by the formula:

E = 2πεqεx V/m

free space (8.85 x 10 ) ε0 = permittivity of free -12

V = 2πεq ln 1r V

Potential Charge at a Conductor in a Group of Charged Conductors Consider a group of long straight c onductors onduc tors A, A, B, C, D, E, c oulombs bs per per …, F having charges q 1, q2, q3, q4, q5, …, qf coulom meter length respectively (Figure 116).

Figure 116

∞ = ∫ 2πεq dx/x volts = ∫∞ 2πεqdx/x volts ∞ = ∫ 2πεqdx/x volts ∞ = ∫ 2πεq dx/x volts V = ∫∞ 2πεq dx/x  ∫∞ ∞2πεq qdx/x  ∫∞ 2πεq  dx/x  ⋯  ∫ 2πεdx/x q  q  q  ⋯ q = 0.0.     1 1 1 V = {q ln  q ln  ⋯⋯  q ln 1 }

Potential at A due to to its own charge, c harge, q1

Potential at A due to to charge cha rge q2

Potential at A due to to charge cha rge q3

Potential at A due to to charge cha rge qn

Overall potential difference between conductor c onductor A and infinite neutral plane.

Simplify Simpl ifying, ing, assuming balanced bala nced load conditions,

εE ==1 q V/m 2πεx q ∞   dx 2πε  V = ∫ x V Figur e 115

T aking air as medium,

As As x approaches approac hes infinity, infinity, the value of E approaches approac hes zero, zero, thus, the potential difference between conductor co nductor A and the neutral wire is:

 2πε 

r  d

 d−

Capacitance Capac itance of Single Phase Overhead Overhead Line Consider a single – phase pha se overhead overhead line with two paralle pa rallell conduc con ductors, tors, each eac h of radius r meters m eters placed plac ed at a distanc e of of d m eters in air. It is i s assumed that the the distance d between between the conductors is large in comparison to the radii of the conductors. Therefore, the density of charge on either conduc con ductor tor will be practically unaff unaffected ected by the the charge cha rge on the the conductor and will be uniform throughout the length. A uniformly distributed charge on a conductor cond uctor acts ac ts as thoug thoughh


61

Electrical Transmission and Distribution System and Design it is conc c oncentrated entrated on the conduc tor axis. T herefore, for for our present analysis it is assumed that the charge +Q coulombs ombs on c onductor A and -Q coulombs on c onductor B are are concen con centrated trated at the centers of the two conductors cond uctors which which ar are separated from from each eac h other by d meters. m eters.

Figur e 117

∞ ∞ Q ∫   ∫ V =  2πε dx/xQ  d 2πε Qdx/x V = 2πεln r ∞ ∞  Q V = ∫ 2πεdx/xQ ∫d 2πεQ dx/x V = 2πεln r VV = =VQ lnVd πε r

Potential at conductor c onductor A and Neutral Neu tral Plane, Plane ,

Potential at conductor c onductor B and Neutral Plane, Plane ,

Examples: 1. A single – phase transmission line has two parallel conductors 3 m apart, the radius of each conductor being 1 cm. Calculate the capacitance of the line per km. Giv Gi ven that ε0 = 8.854 x 10 -12 F/m. 2. A single – phase transmission line has two parallel conductors 1.5 meters apart, the diameter of each conductor being 0.5 cm. Calculate line to neutral capacitance capac itance for for a line 80 km long. 3. The length of transmission lines consisting of two identical identic al conduct c onductors ors with one cm radius is 50 miles. mi les. It is desired that the capacitive reactance to be at least needed 10000 Ω. Determine the separation distanc e needed between the lines to ac hieve this result @ 60 Hz. Hz. Capacitance of Three Phase Overhead Overhead Line Unsymmetrically Unsymmetrically Spaced Line For an untransposed unsymmetrical 3 – phase line the capacitances between conductor to neutral of the three conductors are different. Supposes that the line is, as shown in Figure F igure 98, and that voltages voltages V A, VB, VC are applied applied to the c onduc tors with the result that the the chargers c hargers per met m eter er length are q 1, q 2, q3 respectively. Potential at c onductor A and a nd Neutral Plane,

both potentials with respect to the same neutral plane. Since Sinc e unlike c harges attract attract each other, thus the potenti p otential al difference between conductor cond uctor is:

Capacitance between between conductors (for 1 phase, 2 wire line) l ine)

Similarly,

and

 =  =    Figur e 118

Capacitance between conductor and neutral (for 1 phase, 3 wire line)

Since q3=-(q1+q2),

Multiplying Mul tiplying by

and

C = 2C   C = C = C = 2lπεn dr / Figur e 119

∞ ∞ q  V = ∫ 2πεdx/x∞ q∫ 2πεq dx/x ∫    dx/x 2πε    1 1 1 1 V = 2πε q ln r  q ln d  q lnd V = 2πε1  q ln1r  q ln d1  q ln d1 V = 2πε1  q ln 1r  q ln d1  q ln d1 Vln =2πε1 q ln dr  q ln dd  d d d d 1 V = 2πεq ln r ln d  q ln d ln d Vln =2πε1 q ln dd  q ln dr   d d d d 1 V = 2πεq ln d ln d  q ln r ln d

Multiplying Mul tiplying by


62


V lnV  VlndlnV lnd   r q = lndr lndr  lndd lnddd d V d    l n  l n r V d C = 2πε lndr lndr lndd lndd F/m d V d    l n  l n C = 2πε lndr lnr dr Vln dddln dd F/m d r  VV ln dd l n C = 2πε lndr lndr  ln dd ln dd F/m V = V  V3  V

Simplifying

Capacitance Capac itance of conductor cond uctor A to to neutral,

Similarly, Simi larly, capaci cap acitance tance of conductor cond uctor B to to neutral,

and

Unsymmetrical Unsymmetrical Lime with Transposed Conductors If the charge per unit length is same in every part of the transposed cycl cyclee (Figure 120), 12 0), average value value of voltage voltage of conduc con ductor tor A, will be

Figur e 120

V = 2πε11  q ln 11r  q ln d11  q ln d11 V = 2πε q ln r  q ln d  q lnd V = 2πε1  q ln1r  q ln d1  q ln d1 V = 6πε1  q ln r1  q  q ln dd1d  1  V = 2πε q ln drdd C = l2πεn Dr F/m

Voltage of o f conduc co nductor tor Ain positions p ositions (1), (2) and (3) we have

Similar Simi lar expression for C BN and CCN can be obtained. Equilaterally Spaced Lines For the equilateral spacing d 1 = d2 = d3 = d

C = 2lnπεdr F/m

Figur e 121

Examples: 1. A 200 km, 3 phase transmission line has its conduct con ductors ors placed plac ed at the the corners of an equilateral equi lateral triangle of 2.5 2.5 m side. T he radius of each c onductor is 1 cm. Calc ulate: a. line to neutral capacitance cap acitance of the line b. charging c urrent per phase phase if the line is maintained m aintained at 66 kV, kV, 50 Hz. Hz. 2. A 3 – phase overhead transmission line has its conductors arranged at the corners of an equilateral triangle of 2 m side. Calc ulate the capacitance of eeac achh line conductor per km. Given that diameter of each conductor conduc tor is 1.25 1.25 cm. 3. Calculate Calc ulate the capac itance itance of a 100 km long l ong 3 phase, p hase, 50 Hz overhead transmission line consisting of 3 conductor conduc tors, s, each of diameter diam eter 2 c m and spaced spac ed 2.5 m at the corners of an equilateral triangle. 4. A 3 phase, 50 Hz, Hz, 132 kV overhead overhead line has conductors placed in a horizontal plane 4 m apart. Conductor diameter is 2 cm. If the line length is 100 km, calc c alculate ulate the c harging harging c urrent urrent per phase assum assumin ingg com plete transposition. transposition. 5. A 3 phase, 50 Hz, Hz, 66 kV overhead overhead line conductors condu ctors are are placed in a horizontal plane as shown in Figure 122. T he conductor diameter diameter is 1.25 cm. if the line length is 100 km, calc ulate: a. Capacitance per phase b. The charging current per phase, assuming com plete transposition of the line.

and

So, average average value of voltage of c onductor onduc tor A,

Simplify Simpl ifying, ing, since q 1+q2+q3 = 0

Capacitance Capac itance of conductor cond uctor A to to neutral,

Figure 122

6. The three conductors A, B and C of a 3 – phase line are arranged in a horizontal plane with wi th D AB = 2 m and DBC = 2.5 m. m . Find line – to – neutral n eutral capaci capa citance tance per per km if diameter of each conductor is 1.24 cm. cm . The conduc con ductors tors are transposed transposed at regular intervals. 7. T he three three conductors conduc tors of a 3 – phase pha se line are arranged arranged at the corners of a right angled isosceles triangle. If each equal side of this triangle triangle is 2 m, find line – to – neutral capac ca pacitanc itancee per km. T ake the diameter of eac eachh conduc con ductor tor as 1.24 1.24 cm . The c onductors are transpos transposed ed at regular regu lar intervals.


63

Electrical Transmission and Distribution System and Design 8. A 3 phase, 50 Hz, Hz, 132 kV overhead overhead line has conduc con ductors tors placed in a horizontal plane 4.56 m apart. apart. Conduc tor diameter is 22.4 mm. m m. IfIf the line length l ength is 100 100 km, calculate the charging current per phase, assuming c omplete transposition. transposition. 9. T hree conductor conduc torss of a 3 – phase overhead line are arranged in a horizontal plane pla ne 6 m apart. T he diameter er of eac h conductor is 1.24 c m. Find the the capac c apacitance itance of each eac h conductor to to neutral per 100 km km of o f the line. Capacitance of Double Circuit 3 – Phase Overhead Lines Normally Normal ly used c onductor onduc tor configurations configurations are of hexagonal hexagonal spacing and flat vertical spacing. It has been found that modified modi fied GMD GM D method holds good g ood for determination d etermination of of capacitance capac itance of transposed transposed double double circuit circ uit 3 – phase overhead overhead lines li nes with equilateral equi lateral spac ing and with flat vertica verticall spacing. spaci ng. ItIt is reasonable rea sonable to assume that the modified m odified GMD GMD method c an be used for determination determ ination of capacit c apacitance ance of of a line with any c onfiguration intermediate between these these two two configurations. In the case of calc c alculations ulations of induc tance, determinatio determinationn of self GMD G MD (or GM R) of conduc tor is necessary because bec ause of of internal flux linkages of the conductor. But in case of calcul cal culat ations ions of capacitance, capaci tance, sinc sincee all charges c harges reside on the the surface, actual radius of the conduc c onductor tor is used. Symmetrically Symmetrically Spaced Line Consider a 3 – phase double circuit ci rcuit connected in paralle parallell – conduc con ductors tors A, B and C forming formi ng one circ uit and conducto cond uctorrs A’, B’ and C’ forming forming another circuit (conductors symm symmetrical etrically ly spaced). space d). Let the charge c harge over over conductors c onductors A, A, B and C be q 1, q 2 and q3 coulom cou lombs bs per meter length and q 1 + q2 + q3 =0. Potential of conductor A with respect to neutral infinite plane (Figure 103).

V = ∫∞ 2πεq dx/x∞  ∫q∞ 2πεq dx/x∞ q  ∫√∞ 2πεq  dx/x  ∫∞ 2πεq  dx/x  ∫√ 2πε dx/x  ∫ 2πε dx/x q √ 3d3d q √ 3d3d

V = 2πε ln 2dr = 2πε ln 2r C = ln2πε√ 2r3d3d F/m C = C = C = C = ln2πε√ 2r3d3d F/m

T he capacitance of conductor A to neutral,

Similar expressions for C BN and C CN can be obtained and we have

This is because the conductors of different phases are symmetrical placed. T he capacitance per phase is

C = 2C = ln4πε√ 2r3d3d F/m

Flat Vertically Vertically Spaced Line Consider conduc c onductors tors arranged, as shown in Figure 104, corresponding to different position in the transposition positions.

Potential at conductor A with respect to ‘infinite’ neutral

V = ∫∞ 2πεq  dx/x ∞ ∫∞ 2πεqq dx/x  ∫∞∞2πεq dx/xq  ∫ ∞ + 2πεdx/x  ∫ + 2πε dx/x q   dx/x ∫   2πε   1 1 V = 2πε q ln r4d  d  q lndd1  d  q ln2d1d V =12πε1  q ln r1d1 q ln dd1  d1  q ln dd1 1d V = 2πε q ln r4d  d  q lnln2dd  q lndd  d V = V V3   V  V = 2πεq ln 2rd 4d d  dd V C = ln 2d 2πε d  d F/m r 4d  d

plane

Similarly,

Capacitance Capac itance of conductor A, A,

Similar expression for capacitance C BN and C CN can be obtained. Capacitance per phase will be double of C AN

C = ln 2d 4πε d  d F/m r 4d  d C =  F/m C = 2C F/m

Looking at the above expression the simplified simpl ified form form of the two cases is and Examples: 1. Six conductors of a double – c ircuit ircu it transmission lin linee are arranged in hexagonal hexag onal formation. D AC’ = DCA’ = 7 m and D BB’ = 9 m. T he diameter of each conductor c onductor is 2.5 cm. Find the capacitive reactance to neutral and the


64

Electrical Transmission and Distribution System and Design charging current per km per phase at 132 kV and 50 Hz, Hz, if i f the line is regularly regularl y transposed. 2. A 100 – km double circuit transmission line with 7 strand copper copp er conductors c onductors having having 5 mm diameter, diameter, has has 6 c onductors onduc tors arranged arranged in vertical vertical formation. forma tion. The line l ineis transposed at regular regul ar intervals. intervals. Find the capacit cap acitan ance ce per phase of the the line. line . d 1 = 8 m and d2 = 6 m. 3. Find the capacitance per phase per km of a double circuit 3 phase line shown in Figure 123. The conductors are transposed and are of radius 1 cm each. eac h. The phase sequence is ABC. ABC.

3. A 50 Hz, Hz, 3 – phase line (transposed) of an ACSR moose conductor (overall diameter = 31.8 mm) per phase has flat horizontal spacing of 10 m between adjacent conductors. Compare the inductive and capaci cap acitiv tivee reactances in ohm per km per phase of this line with that of a line (transposed) using a three conductor ACSR hyn x conductors (each having overall diameter = 19.6 mm) having 10 m spacing measured from the center ce nter of the bundles. The T he bundl bundle conduc con ductor tor in each phase are arranged in an equilate eq uilaterral triangle formation with spacing between the conductor conduc torss in the bundle as 40 cm . Effec Effectt of Earth on Capacitanc e of T ransmission ransmission Line So far, in determination of capacitance of transmission lines, the presence of earth was ignored. But it is not true; the presence of earth affec affects ts the elec tric field of a line li ne and and so the capaci cap acitance tance.. The effect of earth on c apacitanc apac itancee can can be modelled m odelled by method of images.

Figur e 123

Capacitance of Bundled Conductor Line Same principle as inductance of bundled conductor conduc tor line line except excep t that radius is use instead of GM R.

C = l2πεnDD F/m DD == √rd√ rd D = 1.09√r d

where: Dm = mutual GMD of the circuit DS = self GMD of the circuit For 2 bundle: For 3 bundle: For 4 bundle: Examples: 1. Find out the capacitance per km to neutral neu tral of the three – phase line as Figur e 124 shown in Figure 124. 124. T he lines are regularly transposed. transposed. T he radius of each each conductor conduc tor is r = 0.5 cm. 2. Figure 125 12 5 shows a compl c ompletely etely transposed 50 Hz, 250 250 km long three phase line has flat horizontal spacing with 10 m between adjac ent conductors. If the outside outside radius is 1.2 cm and the line voltage is 220 kV, determine the charging current, per phase and total reactive power in MVAr MVAr supplied by the line capac itor.

Figur e 125

Figur e 126 126 Electric Field of Two Long Parallel, Oppositely Charged Conductors

T he electric field field of too long, parallel conductors having having charge cha rge +q +q and -q per unit is such that it has ha s zero zero potential potential between the conductors, as shown in Figure 126. If a conduc con ducting ting sheet of infinite infinite dim ensions ensions is placed at the zer zeroo p lane, the electric el ectric field fiel d is not n ot distributed. Further Further,, – potential plane, if the conductor conduc tor carrying carrying charge – q is now removed, the electric elec tric field above the the c onduc ting sheet stay stayss intact, intac t, while that below belo w it disappears. Using these well – known facts in in reverse, the presenc presen c e of ground below bel ow a charged ch arged conduct conductor or can be replace repl acedd by a fictitious fic titious conduc tor having equal and opposite charge c harge and located l ocated far below the ground surface surface as the overhead conductor above it – such fictitious conduc con ductors tors is the mirror m irror image of the overhead c onduct onduc tor. or.


65

Electrical Transmission and Distribution System and Design This method of producing the same electric field as in the presence of earth ea rth is c alled the method of images imag es originally originally suggested by Lord Kelvin. Effect of Earth on the Capacitance of Single – Phase Transmission Transmission Line Considering the case of a single – phase overhead line,

assume conductors A’ and B’ as image conductors of

conduc con ductor tor Aand B respectively, respectively, as shown in i n figure 127. Let the height hei ght of conductors cond uctors be h meters above above the charge of +q c oulombs per meter length leng th and -q coulombs co ulombs per met meter length on c onductors onduc tors Aand B respectively. respectively.

the height of the conductor from the ground, therefore,

r 1   ≅ 

and for all al l practical practic al purposes the the effect effect of

earth earth on line capac itance itance can be neglected. Effect of Earth on the Capacitance of Three – Phase Transmission Transmission Line Figure 128 shows the c onductors of a 3 – phase line along along with im age conductors. conductors. T he line is assumed to be transposed and in the first part of the transposition cycle conductor A is in position 1, B is in position 2 and C is in position 3. Let the charges c harges on line conductors co nductors be q A, qB and qC while those those on image conduc tors tors be – q A, - qB and – qC respectively as shown in the figure.

Figur e 127 127 Single - Phase Transmission Line with Images

T he equation for the voltage voltage drop V AB as determined determ ined by two two

charged conductors A and B, and their images A’ and B’

  1 d r √4h  d  V = 2πε q ln r  q ln2hd  q ln 2h  q ln √4h  d 1  ℎ  ℎ         V = q l n  l n  q l n  l n    2πε  ℎ  ℎ     V = πεqln r√4h2hd  d  q ln   ln ℎℎ  C = πεd ln [r 1  4hd]  1  ℎ ℎ ℎ V = 2πε2πε q ln   ln  ℎℎℎ   r 1    q ln   ln ℎℎℎℎℎℎ 

may be written as follows:

tors of a 3 - Phas e Line w ith Image Charges Charges Figure 128 Conduc 128 Conductors

The equation for the three sections of transposition cycle can be written for the voltage drop V AB as determined by three charged ch arged c onductors onductors and their images imag es.. With conduct c onductor or in position posi tion 1, B in position 2 and C in position po sition 3, we have

Since q A = q = q B’ and qB = - q = q A’, we have

Capacitance Capac itance between conductors A and B,

Similarly, Simi larly, equations for V AB can ca n be written for the second second and third sections of the transposition cycle. If the fairly acc urate assumption assumption of constant co nstant charge per unit length of of conductor throughout the transmission cycle is made, the average average value of three sections of the c ycle cl e is given by

The above expression for capacitance reveals that the presence of earth modifies the radius of conductor r to

. T he effect effect of earth on the capaci capa citance tance of the the

system system is to increase inc rease it. However, However, normally normall y the distance of separation between the conductors is much smaller than

where:

D =  ddd


66

Electrical Transmission and Distribution System and Design T he equation for the average average value value of the voltage V AC can be determine determi ne din the the same sam e way. way. Using V AB + V AC = 3V AN and q A + qB + qC = 0, the c apacitanc apac itancee to to neutral is

C = ln Dr ln2πε  hhhhhh F/m

The above expression for capacitance reveals that the presence of earth increases the line li ne capacit c apacitance ance by a small small amount. However, the effect is very – very small because the height of the conductors is large as compared to the distance between them. Examples: 1. T he conductors in a single – phase pha se transmission line line are 6 m above ground. Each c onductor has a diameter diameter of 1.5 cm and the two conductors are spaced 3 m apart. Calculate Calc ulate the capacitan capa citance ce per km of the the line l ine (i) excluding ground effect and (ii) including the ground effect. 2. T he conductors conductors in a three three – phase transmission line are 6 m above ground. Each conductor con ductor has a diameter diameter of 1.5 c m and the two adjac ent conductors are spaced spaced 3 m apart. Calculate Calc ulate the capacitance per km of the line (i) exclu e xcluding ding ground effect and (ii) inc luding the groun groundd effect. 3. T he conduct c onductors ors in a single phase p hase 10 km transmiss transm issio ionn line are 6m above the the ground. g round. Each c onductor is of 1.5 1.5 cm diameter diam eter and the conductors are spaced 3m apart. If the supply voltage is 33 kV. Determine the following: a. T he capacit capac itance ance of the line including the effect of ground. b. T he capacitance c apacitance of the line negl ecting the the effect effec t of of ground. c. T he charging chargi ng current between conductors. Transposition Tran sposition of 3 – Phase Lines The inductance and capacitance of each phase will be different in case of conduc con ductors tors of three – phase line li ne being spaced irregularly. The apparent resistance of the conductors is also affected because transfer of power between the phases, which occur due to mutual induc tance. Thus, all the three line c onstants are affected affected by irregular irregula r spacing of the the c onduc tors in a 3 – phase line. Also, Also, due to unsymmetrical unsymmetrica l spac ing, the magnetic magn etic field field external to the conductors is not zero, thereby causing induc ed voltages in adjac ent electrical circuit ci rcuits, s, particular particul arly ly the telephone lines that may cause disturbances in the telephone lines. T he unbalancing effect because of irregular spacing of line line conduc con ductors tors can be av a voided by transposition of line line conductors. Transposition of line conductors means changing the position of the three phases on the line

supports twice over the total length leng th of the the line. li ne. In practi prac tice ce,, the line li ne conductors co nductors should be transposed that each of the the three possible arrangem ents ents of conductors condu ctors exist for one one – third of the total length of line. lin e. T his is illustrat illu strated ed in Figures Figures 120 and 129.

Figur e 129 129 Transposition of Three Phase Lines

An An unbalanced unbalan ced system, system, may be of conduc con ducto torrs, symm symmetrical etrically ly placed, plac ed, and cannot be represented by three simple inductances, without making transposition. The effect of unbalanced currents is neutralized in case of conduc con ductors tors being transposed regularly at intervals. intervals. The transposition of conductors also reduces the disturbances to the nearby communication circuits. In com posite line, the the line li ne c arrying arrying telephone line conduc c onducttors ors below the power line conductors, it is also necessary to transpose the telephone line conduc tors to keep down the disturbances. REFERENCES:

[1] Das, Da s, Debapriya (2006). Electrical Power System. System. New Age Age International Publishers [2] Glover, G lover, J. Dunc an, Sarma, Mulukutla S. and Ov O verbye, erbye, Thomas (2012). Power System: Analysis & Design. Design. CENGAGE Learning. [3] Gupta, J.B. (2012). Transmission and Distribution of Electrical Power. S.K. Power. S.K. Kataria & Sons [4] Mehta, V.K. V.K. and Mehta, Me hta, R. (2005). Principles of Power System. System. S.CHAND Publishing [4] Ramar, S. and Kuruseelan, S. (2013). Power System Analysis. Analysis. PHI Learning Private Private Limited Lim ited [5] Stevenson, William D. Jr. (1982). Elements of Power System Analysis. McGraw Analysis. McGraw – Hill Book Company Compan y [6] T heraja, B.L. B.L. and Theraja, T heraja, A.K. A.K. (2003). A Textbook of Electrical Technology . S. Chand Chan d & Company LT D.


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