chapter 2:transmission lines - ~ my edu blog~ · pdf filevoltage rate under 20 kv categorised...

Chapter 2:Transmission Lines ET302

7

S te sa nK u a sa

P e n g h a n ta r a n

P e n g a g ih a n

P e n g g u n a

Figure 2.1 Transmission system in overhead line

2.1 Transmission system

Transmission system was one large system linking generation station tousers through distribution system. Transmission system supply largeamount of energy of generator station to burden centres. Distribution systemalso will supply energy from transmission system and distribute to majorsubstation and small substation to the consumer.

Electricity can be supplied and distributed whether in alternating current(A.C) or direct current (DC). In practical, 3 phase 3 line system applies intransmission system meanwhile for 3 phase 4 line AC system applies indistribution system. Figure 2.1 show transmission system in overhead line.

Important considerations in transmission line operation are referring tovoltage and lost power fall which occurred on-line and also transmission lineefficiency. Components such as R resistance, L inductance and Ccapacitance found in transmission line influence that circumstances.

2.2 Short Line

Transmission line which possess long less than 60 kms and operating involtage rate under 20 kV categorised as short line in transmission system.Refering to short distance and low operation voltage, then as a result of thison-line fitness is also small then capacitance effect can be neglected in thisline system, such short line performance is depends on resistance andinductance found in transmission line. In real transmission line, resistanceand inductance found all along that transmission line. But in line case shortamount of obstruction and inductance lumped at one place or part.

Power station

Transmission

Distribution

Consumer


Page 18

2.3 Resistance and Inductance in line

Transmission line for electric circuit usually having a few parameter such asresistance, inductance and capacitance. This parameter is not same allalong transmission line which affect to voltage regulation and alsotransmission line efficiency. This effect also depends on some transmissionline length.

2.3.1 Series Resistance Conductor

Several factors should be taken into consideration such as line length,diameter of the line, material and environmental temperature cross sectionalarea.

Current flow in oppositional direction in its and this state is known asresistance. R resistance in ohm formed in this transmission line refering tolong and diameter a conductor and can be stated as ;

R =a ……(equation: 2.1)

Where is resistance conductor.

Resistance of the conductor ()is not only depends to material used byconductor but also depends to environmental temperature. Seriesresistance’s value found in line can be pointed through following equation:-

If 1 and 2 is resistance value which correlated with temperature value t1and t2 then,

2 = 1[1 + (t1 - t2)] ……(equation: 2.2)

Where is coefficient for material's temperature used to design conductor.Temperature coefficient’s value for resistance also not fixed but depends onbeginning temperature. Temperature coefficient for resistance rendered as;

= 0 / (1 + 0t1) ……(equation: 2.3)

where 0 is a temperature coefficient for resistance when 0C

Through equations which indicated can be concluded that resistance occurin transmission conductor are continuously started of transmission line untilarrived to distribution line to users.


Page 19

2.3.2 Inductance

Conductor in transmission line system is not only have resistance eveninductance also exist on that line. If we refer to Figure 2.2(a), shows twoconductor single phase, note in separatist part both that conductor markingby distance, D. When current flow through this two conductor at any time-itwill flow in opposite direction and further create field around both conductor.This pole will be permanently giving pressure between one by another likethose portrayed on Figure 2.2(b). (see arrow of field movement).

Figure 2.2: Magnetic force vein induced between two conductor

Both conductor in figure 2.2(a) and 2.2(b), will form rectangle loop for eachround through live flux current flow result in both conductor. When fluxresultant this chain to that loop it will produce inductance. Althoughseparatist distance between this conductor is large about 1 metres to 10metres, because existence this compact flux number will form larger coil andinfluence inductance existence.

Inductance existence in conductor for each round per metre ( when γ ≤ D )provable through equation as follows;

L loge

4iD

henry / meter ……(equation: 2.4)

where,μ = permeability absolute medium.μi = permeability absolute conductor material.

γ D

+ -

γ D

- +

(a) (b)


Page 20

2.4 Short Line circuit diagram

Short transmission line could be identified briefly with draw a single linecircuit. Some components found on this line could be used to makecalculation to determine voltage efficiency and fall which occurred in shortline transmission system. Figure 2.3 shows a short line single line circuit.

Referring to figure 2.3, a few parameters could be identified namely,

Vsn - Voltage in transmission endVm - Voltage in recipient endI - Load current in R tailingR - Loop resistance ()X - Loop inductance ()

Short Line Vector Diagram

A

O

C

I

A

B

G H

D F

Vsn

Vm

sn

m

Vm kos m

IX sin m

Im kos m

Vm kos m

IR

IX

Figure 2.4 Short line vector diagram

Vsn Vm Load

R XI

I

Figure 2.3 Single line circuit of short line

Line

Neutral


Page 21

A vector diagram for short line having relation with parameters found in shortline individual line figure. In this case vector diagram follow can be drawn likeFigure 2.2.4.

Refering to Figure 2.4, a parameter identified as;

OA Transmission end voltage , VmOI Load Current, IAB effect of resistence drop at the line, IRBC effect of reactance drop at the line, IXOC Receiver end voltage, Vsn

Refering to this parameter we could find end voltage transmision and furtherdetermine his power factor. Try the following solution:-

OC 22 )()( BCFBDFOD 22 )sin()( IXVIRosV RRRR

and

S

RRS V

IRVOCOF

coscos

In fact, from this vector diagram, we will be able to determine VS and powerfactor, we also can determine regulation percent for a short line by referringto the parameter. Regulation percent voltage refering to Figure 2.4 can bewritten as.

Regulation Percent = 100sincos

xVIXIR

S

RR ……(Equation: 2.5)

2.5 Regulation Voltage Change with Load Power Factor

In this condition, voltage drop are same in magnitude and phase but phaserelationship with end voltage receiver and end voltage transmission arechanged.

A Voltage drop at end voltage receiver with increas of load for inductive case(power factor lagging) and increase with capacitance load (power factorleading). End voltage receiver not only depends to load but also in powerfactor. Regulation voltage changing at end voltage transmission for differentpower factor could be describe through locust figure such as Figure 2.4.

Refering to Figure 2.5, AO vector show end voltage receiver (VR) at loadstate and OX line with angle ΦR, is load power factor with cos ΦR.Besidethat OX line also showed current (I) load phase. AB line drew parallel to OXline having IR resistance's drop lR and vertical line BC drawn vertical to OXline having inductance’s drop IX.


Page 22

Referring to figure 2.5, ABC is impedance triangle and CA is impedancedrop total IZ line. While OC shows end voltage receiver phase (VS) anddifference between VS and VR or ( OC - OA ) is voltage drop on-line and arealso known as regulation end transmission.

This regulation change able clearly seen if we see maximum regulasi in Bpoint and empty regulasi in S point by referring to locust figure (Figure 2.5)through following equation :-.

Regulation = IR cos θR + IX sin θRRegulation wiil be maximum when d ( regulation ) / d θ = 0IR (-sin θ ) + IX ( cos θ ) = 0

atau

Tan θ = X / R

O

VR

VS

IX

IR

C

BA

M

N

H F

E G

K

Q

SPVR

VS

O’ IR D

IX

ΦR

Φ

Φ

I

Transmission End Voltage, VS

Receiver End Voltage VR

Figure 2.5 Vm and Vsn Locus

X

Powerfactorleading

Powerfactorlagging


Page 23

2.6 Regulation per Unit

When load at end receiver get power supply then will happen voltage dropeffect from resistance and inductance in conductor. Hence voltage value atend receiver in Vm usually less compared end voltage transmission in Vsn.Different voltage drop in end receiver and end transmission stated as endtransmission voltage percent known as regulation.

Regulation per unit definable as voltage change in end receiver part whenfull load in halted, this will make voltage in transmission end equal toreceiver end. This situation can be made to appear in form of similarity asfollows:

Regulation Percent = 100xVmVmVsn ……(Equation: 2.6)

Where is Vsn is transmission end voltage and Vm receiver end voltage.Asknown regulation help to maintain voltage value in load terminal by settinglimit ( 5% error voltage) by using suitable control equipment.

2.7 Transmission Efficiency

When load given supply through transmission line will happendisappearance in conductor because of resistance and power effect sent inload transmission end line less than power supplied in end transmission.Transmission line efficiency found as power ratio received to power transmitor can be may be written as;

Transmission Efficiency = 100Re x

PowerTransmitPowerceived

= 100xsPowerLosserOutputPowe

rOutputPowe

T = 100coscosIm x

snVsnIsnmVm ……(Equations: 2.7)

Where is Vm, Im and cos m is voltage, current and receiver end powerfactor while Vsn, Isn dan cos sn is voltage, current and transmission endpower factor.


Page 24

2.8 Regulation per Unit And Efficiency Calculation

Voltage Regulation and transmission line visible with clearer through simplecalculations examples like in this part.

A transmission line a capable phase power 1,100 kW to factory with avoltage 11 kV in power factor 0.8 lagging. This line have a resistance 2 andinductance coil 3 Get;

i). Voltage at transmission end.ii). Regulation Percent.iii). Transmission Line Efficiency

Given;Resistance, R = 2Inductance, X = 3Power , P = 1,100 kwPower factor m = 0.8 (mengekor)Receiver end voltage, Vm = 11,000 V

Load Current, I =mVm

xPcos

000,1

I =8.0000,11

000,1100,1xx

I = 125 A

i). Value of transmission end voltage

known, cos m = 0.8so that sin m = 0.6

Vsn = 22 )sin()cos( IXmVmIRmVm

Vsn = 22 )31256.0000,11()21258.0000,11( xxxx

Vsn = 11,426 V

Example 2.1:

Solution :


Page 25

ii). Regulation Voltage Unit.

Vsn = 11,426 VVm = 11,000 V

Regulation Percent = 100xVmVmVsn

= 100000,11

000,11426,11 x

= 3.873 %iii). Transmission Line Efficiency.

Power Losses = I2 R = (125)2 x 2= 31,250 atau 31.25 kw

Transmission Efficiency = 100Re

xPowerceivingPoweronTransmissi

T = 1100 x 1000 x 100(1100 x 1000) + 31250

T = 97.24%

A transmission line three phase 11 kV owns resistance 1.5 and inductance4 for each phase. Calculate regulation percent and efficiency line percentif total end receiver load, 5000 kVA in power factor 0.8 lagging and voltagesupplied until last distance was 11 kV.

Resistance for each conductor R = 1.5Inductance for each conductor, X = 4.0

phase voltage at end of receiver, Vm =3

000,11 = 6,351 V

Transmission Load = 5,000 kVA

Load power factor , kos m = 0.8 (lagging)

Line Current, I =Vmx

xkVAinPower3

100

262.4 amp

Transmission end voltage for each phase,

=351,63

000,1000,5xx =

262.43 A

Example2.2:

Solution :


Page 26

Vsn = Vm + IR cosm + IX sin m= 6,351+(262.43x1.5x0.8)+(262.43x4x0.6)= 7,295.8 V

Transmission end voltage VSL = 3 x 7,295.8 = 12,637V

So that,

Regulation Voltage Percent = 100xVmVmVsn

= 100000,11

000,11637,12 x

= 14.88 %and

Transmission line efficiency = 100xLossesPowerPowerOutput

PowerOutput

T = 5000k x1005000k + 103280.64

T = 97.98%

MEDIUM AND LONG TRANSMISSION LINE

2.9 Introduction of Medium Line and Long Line

Besides short line, medium line and long line also in transmission lines. Ascompared both this line with short line, effect of loss caused resistance andinductance in conductor were more numerous and large. With existencelarge lost power on-line medium and long thus it also also influencetransmission line efficiency. As such voltage regulation should be done alsois large to overcome loss in acceptance end transmission line.

2.9.1 Medium Line and Long Line

Transmission line which possess line length between 60 to 150 kms and linevoltage between 20 kV to 100 kV classified as medium line. If for line shortcapacitance effect deserted, in medium line capacitance effect taken.

Transmission line which possess long exceeding 150 kms and voltage reachmore 100 kV it classified as long line. As known each line influenced byresistance, capacitance, inductance and conductivity. Due to this fromcalculation aspect loss found in long line was enormous compared short andmedium line.


Page 27

2.9.2 Medium Line and Long Line circuit.

Medium Line Circuit

Referring to figure 2.6, a few parameters could be identified,

Vsn Transmission end voltageVm Receiver end voltageIsn Transmission end currentIm Receiver end currentIc Capacitance currentR Loop Capacitance ()X Loop Inductance ()C Capacitance (farad)

Actually exist three way that can use to determine medium line individual linefigure such as end condenser methods, T method and method . Figure 2.6is method end condenser. This method crumple up capacitance in load endpart. If in short line capacitance deserted, in medium line also capacitancetaken into account this is because availability addition to voltage and lengthvalue line.

Due to this regulation calculation in medium line will participate considercapacitance and leak reactants in line and its can be explained depends onvalue of receiving voltage.

Refering to circuit in Figure 2.6, we would note line current (Isn) is total loadcurrent (Im) and also current recharging (Ic) for capacitance.Written as;

Isn = Im + Ic ……(Equations: 2.8)

If we write equation for current recharging for capacitance, Ic = jwCVm andIm load current = Im (cost θm - jsin θm)

Vsn Vm load

R XISn

Figure 2.6 Medium line individual circuit

C

line

Neutral

Im

IC


Page 28

We will find out Isn = Im cost θm - jIm Sin θm j, as such with this equationis found;

Voltage drop in line = Isn ( R JX ) from this equation we will know voltagevalue found in transmission end as follows;

Vsn = Vm + Isn (R + jX). ……(Equations: 2.9)

Through this equation we can see how power flow occurring in line deliverysystem medium and further we be able to size his voltage regulation.

Long Line Circuit

A transmission line who sent out electricity power release heat due toconductor resistance. So long line act as a barrier. Transmission line also actas a inductance because every conductor are surrounded by a magneticfield in transmission line length. Long transmission line also acted as a itscapacitor because conductor act as pelitic capacitor. Resistance, inductanceand capacitance as a transmission line are dispensed uniform in line length,with magnetic field around conductor along with electric field created bydifferent between that field.

We could imagine that a transmission line contain thousands resistor,inductor and basis capacitor like those demonstrated at figure 2.7.

Refering to Figure 2.7, a few parameter could be identified in among them;

Vsn Transmission end voltageVm Receiver end voltageIsn Transmission end currentIm Receiver end current

Figure 2.7 Long line individual circuit

Vsn

ISn

LoadVsn

Im

B/n G/n

R/n X/n


Page 29

Apart from that several statements can be done refering to same circuit,i). Line are made up a few constant parameter namely resistance,

inductance, capacitance and conductivity all along long transmissionline distribution.

ii). Resistance (R) and inductive reactance (X) is element on-line serialtransmission

iii). Capacitance string (B) and leaking conductivity (G) are element shunt,leaking conductivity also caused on-line lost power. It derives ofinsulation leak or corona effect in conductor.

iv) Leakage current which flowed through shunt graduate is maximum intransmission end line and lesser continuously when headed for receiverend and finally become zero after arriving to receiver end of line.

2.10 Voltage Effects on Trasmission Efficiency.

Transmission line efficiency is not only influenced by several constant occurin transmission line namely resistance, inductance and capacitance likealready discussed before. However line efficiency also influenced by voltagevalue are brought by something corona line and effect in that line. In this partthe effects will discuss briefly.

2.10.1 High Voltage Transmission ( 132 kV / 275 kV)

Usually withdrawn electricity by one generator station is about 33 kV, 22 kVand 11 kV. Of that generator pass energy to a few transformer through spanline on by various extension way such as ring system, radial system, networksystem etc. The generated voltage is step up to the wanted value such as132 kV, 275 kV or 500 kV. This voltage purpose of raising was to reduceexpenditure in large cable size usage because cable size used is based onenormity current which flowed. At the same time, transmission line efficiencycan be increased.

The situation explicable with double power (W) are sent through threephase transmission system which possess line voltage (E) and cost powerfactor kos , produce equations as follows

Current LinekosE

WI3

…(Equations: 2.10)

Let say :ι = Length of conductor line = current density=Resistance of conductor material A = Conductor cross section


Page 30

line loss may be written as:

kosEWRI 33 2 …………(Equations: 2.11)

Line efficiency transmission line:

inputoutput

kosE

31 …(Equations: 2.12)

Voltage drop for each line:= IR = ι ……(Equations: 2.13)

Copper volume:

= 3 ι AkosEW3

……(Equations: 2.14)

Refering to this equation some assumptions could be made among them:-

Equation (2.11), giving picture that lost power are proportional inverse withE, also inversely proportional with power factor cost .

Equation (2.12), show that efficiency increased transmission line by addingtotal voltage in line and power factor.

Equation (2.13), resistance decrease at every line are fixed ( when and ιassumed fixed. Voltage regulation can be repaired because fall percentagevoltage could be reduced with increasing value E.

Copper volume of necessity for transmission line are proportional inversewith a voltage and system power hereby copper need dwindling.

Can be concluded of this equation altogether, when voltage and powersystem value enhanced then this result would give efficiency in transmissionline by conductor material saving also can be done and further reducedelivery cost with small cable size in long line.

Due to this line capacity advancing by the increase of transmission linevoltage. Undeniable that cost for transmission line and terminal equipmentalso rising with the increase of transmission line voltage but overall cost isproportional with delivery voltage. Moreover it will save cost in keeping withlost power reduction which occurred in transmission line. Effect of that, thedelivery total cost decrease with the increase of transmission line voltage.


Page 31

2.10.1.1 Corona

Corona was electrical discharge emerge around overhead line conductor,due to air flow where would disturb radio waves and creating lost power. Inlow voltage there was no change which occurred can be influenced by aircondition around conductor. However when potential different and graduallyincrease, at one level, glow luminance (luminous glow) weak purple color willrise with hissing sound. This phenomenon is known as virtual corona andparticipated by gas production which has identified through system featurehis smell. All situations such as hissing sound, purple radiation and gasproduction smelling known as corona.

If conductor are homogeneous and smooth and similar state fixed all alongconductor, in other circumstances rugged parts will issue brightness. If rangein conductor not too big compared to diameter, arc may be might stand in forbefore glow luminance seem. This happened in keeping with statementwhere range in small conductor do not have time quite enabling glowluminance happened.

In the case of a positive conductor DC system has a uniform warmth andbrightness of the rising of the negative division. For AU system, according tothe corona current is not sinusoidal. Corona, accompanied by loss of power,losing is caused by light, heat, noise and chemical reactions. Corona existsat small transmission line side effects of them;

i). Result in lost power that is during uncertain weather condition.ii). Found voltage drop which not sinusoidal with current not corona

sinusoidal, this situation cause little trouble of communication circuit asa result of electromagnetic and electrostatic induction.

iii). Harmonic wave distortion which many especially in third hormonic, thatexists in transmission line.Corona formation produce ozone gas with chemical reaction inconductor and creating corrosion

Corona effect in transmission line like those above-mentioned of courseundeniable especially for long transmission linei). conductor diameter (for example with ACSR).ii). Use more from one conductor for each phase, namely use bouquet

conductor (bundle conductor)iii). Add more range in conductor hereby stress by magnetic static could be

reduced and thereby corona effect also could be reduced.


Page 32

OVERHEAD LINE INSULATOR

2.11 Principle and design and Overhead Line

Overhead line distribution system usually involving a few key componentnamely wireline, insulation, tower post etc. However in chapter, insulatorapplies in overhead line distribution system will be discussed. This Insulatorcould be identified is based on design.

2.11.1 Overhead Line Insulators

Conductor for distribution system overhead line is guaranteed his securitywith electric equipment assistance named insulator. This insulator thenwould not be have been leaks current to earth from conductor through thisequipment. Due to this insulator play important role succeed distributionoperand overhead line system. Figure 2.8 show either insulator form ofequipment.

A few important features and should be taken into consideration beforeinsulator applies in installation any overhead line distribution system amongthem;

i.). Physical strength: ability arrested suitable burden heavily somethingconductor.

ii). Having high insulation resistance to prevent current leak to earth.iii). High resilience ratio of breach leap result voltage.iv). Material used to develop insulator was the type that does not permeable

liquid or hollow and influenced by temperature changes.v). This process does not contain impurity and crack and impervious of

liquid substance and gas of aerospace.

Figure 2.8: Few insulator form overhead line


Page 33

Overhead line insulator common use including pin insulator, suspensioninsulator, tension insulator and shackle insulator . In this chapter, only pininsulator, suspension and insulator insulator tension will be discused .

Pin Insulator

Pin insulator designs by having steel pin could be installed in post bar tower.This insulator have screw in pin part steel while conductor placing at thisinsulator top and bound with use wire aluminum soft with a couple coil.Pottery part kept apart from steel division with a kind of soft metal (timbre).Design this pin insulator visible in figure 2.9(a) for low voltage a pin insulatorused is fair enough.

For transmission line high voltage also stronger and large pin insulatorutilized. Type pin insulator high voltage is different with insulator constructionlow voltage pin. Insulator construction high voltage pin comprising two ormore ceramic layer simultaneous cement. Use a insulator unit pin fairenough for delivery system 33 kV if exceeded over in this voltage proportiontwo or more insulator arrangement pin used. However inside pin insulatoruse overhead line delivery system is not economy for voltage exceeding 80kV. Figure 2.9(b) show schematic figure a pin insulator for overhead line.

(a) Design


Page 34

Figure 2.9 Pin Insulator

2.11.3 Suspension Insulator

Insulator installation suspension in dependent overhead line to voltagecapacity under by conductor in line singles. Capacity expansion voltage atone line will participate increase number insulator installation suspension inthat line. By transmission line and distribution as most handling exceeding33 kV then sistam become bigger and distance between increased line far.

This situation cause pin insulator have no capacity to keep this line system.Due to this to overcome this problem suspension insulator designs, insulatorexact figure pin can be observed in Figure 2.10(a)

Suspension Insulator suspend in different tower post bar with pin insulatorplacing at bar top. For type this insulator, conductor will be connected ininsulator lower part suspension. As such we can add range in tower bar withconductor through insulator addition suspension arranged by network.Insulator increasing number suspension at one line are referring to linevoltage capacity, weather condition, transmission line and size assemblytype suspension insulator used.

Steel pin

Tower Bar

Conductor

glazed porcelain

Binding Wire

Plumbed

Shield

(b) Schematic


Page 35

Because this insulator is fitted by network then replacement any insulatorcould be made without change insulator entire network. Figure 2.9(b) showschematic form a suspension insulator

Steel CapCement

Glass or Pottery

Socket

Steel PinBalls

Figure 2.10(b) Suspension Insulator schematic

Figure 2.10(a) Suspension Insulator


Page 36

2.11.4 Strain Insulator

At one overhead line state having high tension, for example at end or hairpinat transmission line. For tension its low voltage line use shackle insulator.Whereas for line tension insulator high voltage tension used. Usually ininsulator installation in transmission line two or more insulator used. Insulatordisk tension fitted with horizontally different with suspension insulator willfitted with vertically. Process of tension insulator are same with suspensioninsulator, however height size of suspension insulator are exceededtension insulator (Refer to Figure 2.11).

Advantages and Disadvantages Overhead Line Insulator

The advantages of suspension insulator compared pin insulator can beexplained as follows;

i). Suspension insulator is cheaper from his cost aspect for capacity linewhich exceeded 50 kV.

ii). For each unit of suspension insulator type will designs by refer at lowvoltage capacity around 11kV . When used in high voltage capacity thenenough with connect suspension insulator by serial, disk number useddepends on voltage value in line.

iii). If occur unanticipated damage in which points suspension insulator, justreplace damaged disk only and no need replace insulator entire networkthat suspension.

iv). Suspension insulator is more flexible assembled on in line. Insulatorconnection tension in tower bar is ease to turn to any direction.

v). Pin insulator suitable to be fitted in low post compared suspensioninsulator.

Figure 2.11 : Tension Insulator Schematictegangan


Page 37

Disadvantages of suspension insulator compared pin insulator can beexplained as follows;

i) Suspension insulator no suitable to be fitted in low tower post.ii) Suspension insulator need high and strong post and this increase

cost delivery.iii). Damaged at pin insulator difficult to be detected as compared

suspension insulator.iv). Limited pin insulator capacity only in voltage below 80 kV.v). Suspension insulator need wide column between conductor compared

pin insulator.

Tests Conducted on Overhead Line Insulator

Overhead line insulator are important instrument to achieve electricitytransmission process and distribution at some area. Due to this suitable andsafe insulator selection should be taken into consideration, so that damagein line does not happen. Ensure insulators are used at well off overhead lineseveral tests should be done before its used or being marketed. Test shoulsbe done before used and marketed are:-

(a). Design test (Test emerge arc or flashover)(b). Performance test(c). Habit test

(a). Design test

Design test done to ensure electric performance and mechanical insulator toseveral trial condition such as test emerge dry arc, test emerge wet arc andtest emerge contamination arc. Insulator test of methods listed thiscommonly made to three insulator randomly chosen. This insulator will betested either fulfilling standard or on the other hand.

i. Test Emerge Dry Arc

Voltage emerge arc is voltage cause insulator surface breach insulate, allowcurrent flow through insulator face from conductor to supporter bar. Ainsulator fixing with a safe minimum voltage imposed at insulator. Thisminimum voltage is dependent to insulator type and size.

In tests emerge this dry arc, a net and dry insulator assembled on asupporter. A voltage with power frequency system imposed in that insulator.This voltage will increase by way staggered until minimum voltage forinsulator under test. This minimum voltage need to be given in insulatorwithin not less than 30 seconds . If emerge arc does not happen in thatperiod, insulator is good. Voltage raised again by way gradually untilflashover happened within 10 seconds. Lightning overvoltage beingrecorded. This process is repeated as much as four times. Average voltage


Page 38

of flashover could not less than minimum voltage of past dry lightning that isfixed.

ii. Test Emerge Wet Arc

This test equal to test (a) except under synthetic rain which possessresistance and temperature that is fixed. Angle and speed rate synthetic rainwater decline also in fix. In this time, insulator should arrested lowerminimum voltage of test (a) long 30 seconds at least without emerge arc.

iii. Test Emerge Contamination Arc

This test equal to test (b) except involving contamination with fog, salt,smoke, dust or chemical. Usually, lightning overvoltage is ½ value than tests(b).

(b). Performance Test

Performance test is another test used to determine the overhead line insulators aresafe (to meet required standards). Among the tests involved dalan performance testis;

i. Broken Test Insulate

Breakdown tests carried out on 3% per cent of the total insulation produced. To testthe dielectric insulation. Insulation may be able to withstand extreme lightningevents have suffered, but it must be replaced if it had been breakdown.For insulation design, flash past should occur at voltages less than voltagebreakdown. At the time of testing, the insulation shall be immersed in a cleaninsulating oil to prevent flash past. The value of the test voltage is raised slowly andthe insulation must be able to withstand 1.3 times the voltage dry flash past withoutbreakdown.

ii. Pulse Test.

Previous voltage lightning pulses obtained by using a signal pulse as in Figure 2.11.


Page 39

frequencypowereovervoltagLightingeOvervoltagPulseRatioPulse

iii. Mechanical Test

A insulator network suspend tested by one tension 1.2 times doublemaximum load usually and type insulator pin tested with moment bentokan2.5 times double maximum burden usually. After the test, dry lightningovervoltage test required again give ensure be provided change in lightningovervoltage.

iv. Temperature Test

The insulator is immersed in rotation in water barrels temperature 70degrees celsius and 7 degrees celsius. Overall immersion number is sixtimes, invariably take one hour long. The insulator then dried and flash testdry implemented.

v. Porosity Test

This is glaze test in pottery insulator. The insulator is weighed in dry statethen it immersed in water and under pressure long 24 hours. After that theinsulator is issued, his face dried. Difference between both the reading showthe pottery deep water result gelis imperfect.

(c). Habit Test

Habit test are involving high voltage and test test corrosion which conductedon all insulator. For test insulator erosion and rustiness usually divulge tocopper solution of sulphate in temperature 15.6 celcius in time a minute.After one minute insulator tested will in transferred and then polished and

Magnitude

LeadingWave

LaggingWave

TimeFigure 2.12: test wave form pulse


Page 40

cleaned and then disclosed again to copper solution of sulphate. Ianya donerepeatedly up to four times. After that checked to be sure there were no anyrustiness and metal cleave to insulator tested.

High voltage test committed against pin insulator, where itoverturned and placed inside water which hit keparas insulator neck. Wateralso placed in hole spindle. Then high voltage are supplied in 5 minutestime. After going through this test should good insulator will not sufferdamage.

INSULATOR NETWORK OVERHEAD LINE

2.12 Introduction Insulator Network Overhead Line

After we learn a fewinsulator form overhead line andalso tests which operate oninsulator. We have obtaindescription a little bit insulatorneed for delivery systemoverhead line However alsoneed we know that insulator areused at overhead line namelysuspension insulator (Figure1.13), having different voltagedistribution within onenetwork.This potential differencewould cause networkinefficiency insulation happenedwhen occurence of interference(lightning) on overhead line.Although that kecepan this canbe repaired with a couple method which will in discuss further dalan thisinput.

2.12.1 Potential Distribution in Network Insulator

Overhead line which operates in high voltage capacity using somedisk number (suspension insulator) connected by serial. Connection thesedisks by serial overall known as network insulator. Each suspensioninsulator having metal installation own and each metal installation for per unitthis having relative fitness on metal installation to different units. For network

Figure 2.13 Insulator installation suspension inoverhead line


Page 41

this insulator mutual capacitance are referring between metal installationinsulator per unit suspension. Shunt fitness or air fitness also refering toeach metal installation between insulator unit suspension with tower post toearth.

Because availability faggot voltage in insulator network suspension whendihidupkan, cause unequal voltage division at every insulator will happen.

Potential difference voltage found in insulator network is different, forinsulator near network with conductor having voltage percentage value highcompared near insulator with tower post. This situation cause voltagedivision no linear in this network insulator. Situation explicable refer solutionin Figure 2.14 (a) and (b).

Refering to Figure 2.14, known;C : Mutual capacitanceC1 : Shunt fitness or air fitnessV1 : Voltage negotiate first suspension insulator unit (near to tower post)V2 : Voltage negotiate second suspension insulator unit.V3 : Voltage negotiate third suspension insulator unit (near to conductor)E : Voltage between conductor and earth.Take K = C1 / C or C1 = KC

C

C

C

C1

C1

C1

V1

V2

V3

B

A

I1

i2

i1

I2

i3I3

E

Figure b

I4

V1

V1 + V2 + V3

V1 + V2

V1

V2

V3

B

A

E

Figure a

C

Figure 2.14 Process (a)and equivalent circuit (b) for insulator network


Page 42

Use law kirchhoff in node A we find out:-

I2 = I1 + i1CV2 = CV1 + C1V1CV2 = CV1 + KCV1CV2 = C(V1 + KV1)

V2 = (V1 + KV1)V2 = V1(1 + K) ……………… Get V1

By using law kirchhoff in node B we find out:-

I3 = I2 + i2CV3 = CV2 + C1( V1 + V2) …………..…..Voltage negotiate C1 air's fitnessfrom tower post to insulator unit to two = ( V1+ V2) …… figure above andnoteCV3 = CV2 + KC( V1 + V2)CV3 = C[V2 + K ( V1 + V2)]

V3 = [V2 + K( V1 + V2)]V3 = [KV1 + V2(1 + K)]V3 = [KV1 + V1 (1 + K) (1 + K)]V3 = V1 [K + (1 + K) (1 + K)] ……………….Simplified.V3 = V1 (K + 1 + 2 K + K²)V3 = V1 (1 + 3 K + K²) …...…………..Get V1

Voltage between tower conductor and post (to earth) :-

E = V1 + V2 + V3E = V1 + V1(1 + K) + V1 (1 + 3 K + K²)E = V1 ( 3 + 4K + K2 )

From this equation is found :-

V1 = E / ( 3 + 4K + K2 ) …..{Equations: 2.16)

After getting V1 value further obtain V2 value and V3. Of in this retrieval wewill see bezaupaya voltage negotiate each this network insulator.

A network to four insulator used to hang up a 33kV conductor, threeoverhead line phase. Kapasitan air or shunt between every cap and towerwas one tenth (1/10) from each fitness unit. Calculate voltage hinder eachinsulator.

C1 = KC

know V2 = V1 (1 + K)

Example 2.3:

Solution :


Page 43

Given: E = 33kVK = C1/C = 1/10 = 0.1

With use equation were obtained of 8.1 we know;

V2 = V1(1 + K)V3 = V1 (1 + 3 K + K²)

For equation V4 voltage hinder fourth insulator, available with method usedjust like in 8.1 and found as;

V4 = V1 (1 + 6K + 5K2 + K3)

So thatV2 = V1(1 + K)V2 = V1(1 +0.1)V2 = 1.1V1

V3 = V1 (1 + 3 K + K²)V3 = V1(1+ 3(0.1) +(0.1)2)V3 = 1.31V1

V4 = V1 (1 + 6K + 5K2 + K3)V4 = V1(1+ 6(0.1) + 5(0.1)2 + (0.1)3)V4 = 1.651V1

Voltage between tower conductor and post (to earth) :-

E = V1 + V2 + V3 + V4E = V1 + 1.1V1 + 1.31V1 + 1.651V1E = 5.061V1

and

E =3

33000 = 19050V

So that;V1 = E / 5.061V1 = 19050 / 5.061V1 = 3764.7V

By including V1 value in V2 ,V3 equation, and V4, from this equation isfound :-

V2 = 1.1V1V2 = 1.1(3764.7)V2 = 4141.2V

V3 = 1.31V1


Page 44

V3 = 1.31(3764.7)V3 = 4931.8V

V4 = 1.651V1V4 = 1.651(3764.7)V4 = 6215.5V

Refering to voltage values negotiate insulation per unit found V4 voltage'svalue obstruct fourth insulator in insulation network (near to conductor) washigh compared V1 voltage's value in near insulator with tower post namelyfirst insulator.

2.12.2 Network Efficiency

Because availability unequal voltage division at every network insulator areused at overhead line what when occurence of interrupt or ganguan lightningresult. Namely voltage negotiate near insulator with higher conductor anddecrease until to near insulator with tower post. Then will reduce efficiencyin insulation network used. Network efficiency also influenced by insulatornumber suspension applies in a network. Apart from that depend also tofitness ratio air (capacitance between tower unit and post) with mutualcapacitance (capacitance between unit) at one network. Network efficiencyfor use overhead line definable as;

Network Efficiency = x 100%

Or can be writen as,

Network Efficiency = %100xnVTE …..{Equations: 2.17)

Where;E = Voltan hinder networkn = number of insulator arranged by serial in network insulatorVT = Voltage Hinder Insulator near to conductor

A network to four insulators used to hang up a 33kV conductor, threeoverhead line phase. Kapasitan air or shunt between every cap and towerwas one tenth (1/10) from each fitness unit. Calculate insulator efficiency thisnetwork..

Voltage Hinder Network

n x Voltage Hinder Insulatornear to conductor

Example 2.4:

Solution :


Page 45

By using answer achieved of example 2.3 namely;

V4 = 6215.5V (Voltage Hinder Insulator near to conductor)Given;

E =3

33000 = 19053V

n = 4

So that;

Network efficiency = %100xnVTE

= 19053 x 100%4 x 6215.5

= 76.6%

2.12.3 Repairing Potential Distribution in Network Insulator

Although found potentisal inside voltage a insulator system networkwhich reduces insulator efficiency network for overhead line. This problemcould be overcome and repair by variety way among them extend tower bar,capacitance grading, shielding static and use guard ring. In this input onlytwo ways would discuss to improve efficiency in network insulator namelycross arm and guard ring

(a) Cross-arm

Cross arm mean increase bar distance serves to suspend networkinsulator of tower post. Through this method network efficiency can beincreased. If referring to example 2.3, to get network efficiency is obviouslynamely by adding bar distance from tower post will participate reduce Kvalue (capacitor ratio). When K value reduced namely lower of 0.1 then willincrease insulator efficiency network. However this method is on to high andlarge tower post only because for post small tower had no enough capacityto support long bar weight and also network insulator. Figure 2.15 showschematic for cross arm method

D

D = Bar length

Conductor

Tower Bar

Figure 2.15 Cross arm schematic


Page 46

(b) guard ring

Ring way obstruction can be done with use static shield. This staticshield assembled on end lower part insulator unit connected by using joiningof metal in suspension insulator and then connects to line conductor.

Guard ring which functions as curtain for per unit, reduce earth capacitanceand create capacitance between insulator line and cap. Capacitance value isbig in nearby unit part with guard ring and this will reduce voltage fallnegotiate insulator per unit network. Through this way same voltagedistribution negotiates per unit is impossible obtained in practically. Howeverit could be considered to increase decently possible network efficiency. Thisguard ring method visible like figure 2.16(a).

Figure 2.16 : Guard ring

Refering to figure 2.16(b), a guard ring were installed in conductor part

C

C

C

C1

C1

C1 Cx

I1

I2

I3

i1

i2

i3

Ix

Cy

Iy

V3

V2

V1

(b) Equivalent circuit

Obstruction Ring

Tower post

Conductor

Cz

IzObstruction Ring

Tower Post

(a) Construction

Arc Horn


Page 47

so that level voltage voltage be equivalent. In this situation capacitanccurrent i1, i2 and i3 become equal capacitance current conductor to Ix pin, Iyand Iz. as such voltage value insulation per unit is the same namely V1 = V2= V3 = V.

Supposing Cx, Cy and Cz is capacitance obstruction need to be samevoltage division;

Through solution with method used kirchhoff law at every node we willachieve;

At point A,

C1V = Cx3VC1V = Cx3V

Cx = C1 / 3

At point B,

C12V = Cy 2VC1V = Cy 2V

Cy = C1

At point C,

C13V = CzVC13V = CzV

Cz = 3C1

chapter 2:transmission lines - ~ my edu blog~ · pdf filevoltage rate under 20 kv categorised...

Documents