21252735 Protection System

PROTECTIVE RELAYS(Setting, Testing and Commissioning)

Prepared by

Er.R.KRISHNAN,B.E

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ContentsContentsContentsContents ::::

1.1.1.1. InstrumentInstrumentInstrumentInstrument transformertransformertransformertransformer forforforforprotectionprotectionprotectionprotection 1111

2.2.2.2. SymmetricalSymmetricalSymmetricalSymmetrical faultfaultfaultfault calculationcalculationcalculationcalculation 15151515

3.3.3.3. OvercurrentOvercurrentOvercurrentOvercurrent relaysrelaysrelaysrelays 37373737

4.4.4.4. DifferentialDifferentialDifferentialDifferential relaysrelaysrelaysrelays 61616161

5.5.5.5. GeneratorGeneratorGeneratorGeneratorprotectionprotectionprotectionprotection relaysrelaysrelaysrelays 69696969

6.6.6.6. DistanceDistanceDistanceDistance relaysrelaysrelaysrelays 103103103103

7777 MotorMotorMotorMotorprotectionprotectionprotectionprotection 121121121121

8.8.8.8. CommissioningCommissioningCommissioningCommissioning teststeststeststests 129129129129

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Refrences:Refrences:Refrences:Refrences:

1. ‘The art and science of protective relaying’C. Russel Mason – John Willy ( USA)

2. ‘Protective relays application guide ’GEC Alsthom

3. ‘Protection of industrial power system’T. Davies – Pergman press – oxford

4. ‘Protective relaying – Principle and application’J. Lew is Blackburn,Inc - MarcelDekker Inc.- New York and Basel

5. ‘Protective Relaying – Theory and Applications’Walter A. Elmore – ABB T&D company Inc.

6. ‘Elements of power system analysis ’W.D. Stevenson _ Mc Graw Hill Book Co

7. ‘Power system protection’S.P.Patra, S.K.Basu, S. Choudhuri – Oxfors 7 IBH Publishing Co

8 ‘Power system protection– Static relays’T.S.M. Rao – Tata Mc Graw Hill publish ing Co

9. ‘Protective relaying ’Chernobrovov – MIR Publisher

10. ‘Protection Techniques in Electrical Systems’Helmut Ungrad; Wiliband winkler; Andrzej Wiszniewski- Marcel Decker - New york - Basel

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1.Instrument1.Instrument1.Instrument1.Instrument TransformersTransformersTransformersTransformers forforforfor ProtectionProtectionProtectionProtection

CurrentCurrentCurrentCurrent transformers:transformers:transformers:transformers:Protective relays of a-c type are actuated by current and voltage

supplied by current transformers and voltage transformers. Thetransformers provide insulation against the high voltage of the powercircuit and also supplying the relays with quantities proportional to thatof the power circuit, but sufficiently reduced in magnitude, so that thecurrent can be carried by small cross sectional area of cables associatedwith panel wiring and the relays can be made small and inexpensive.

Construction :CTs are usually designed, so that the primary winding is the

line conductor, which passes through an iron ring which carries thesecondary winding. Most of the CTs are of this type and are known asbar-primary or ring wound CTs.Design :

CTs conform to normal transformer e.m.f equation where theaverage induced voltage is proportional to the product of the rate ofchange of flux and the number of turns. The normal criterion for thedesign of CT is to limit the flux to a value where the core saturationstarts, known as knee-point flux. The current that produces thissaturation flux is the maximum magnetizing current. The magnetizingcurrent and consequently the flux changes from zero to maximum per ¼cycle and there fore the rate of change of flux is

φ − 0⎯⎯⎯ = 4φ webers/cycle

¼or at a frequency of f cycles

4φf webers/secgiving an average induced voltage of

Vav = 4φfN where N is the number of turnsor in r.m.s value the knee point voltage is

V = 4.444φfN as Vrms = 1.11VavAlso the flux φ = B×A where B is the flux density in tesla, and A is

the core area in m2

So the Knee point voltage V = 4.444BAfN

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INSTRUMENT TRANSFORMER FOR PROTECTION

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The knee point voltage can be easily estimated if the fluxdensity of the steel at knee-point and approximate dimension of the coreare known.

For example a CT ratio of 300/1 with a core area of 40×30mmand a maximum flux of 1.5 tesla ,

40 30V = 4.44 × 1.5 × ⎯⎯ ×⎯⎯× 300 ×

50 = 118.8 V1000 1000

Burden :The load on a CT is called the burden. The burden is usually

expressed either as a VA (volt-ampere) or as an impedance. In theformer case the VA is taken to be at the CT nominal secondary current.For example a 5VA burden on a 1Amp CT would have an impedance of5ohms.

5VA⎯⎯⎯ = 5V

1A5V

Impedance = ⎯⎯⎯ = 5Ω1A

or on a 5A CT,5VA

⎯⎯⎯ = 1V5A1V

Impedance = ⎯⎯⎯ = 0.2Ω5A

All burdens are connected in series and the increase in impedanceincrease the burden on the CT. a CT is unloaded if the secondarywinding is short circuited. Under this condition the VA burden is zerobecause the voltage is zero. The errors of CT depends on the phase angleof the burden as well as its impedance.

Operation :The equivalent circuit of a ring type CT is shown in fig 1.1. Ie

is the magnetizing current Rct is secondary winding resistance and Rb &Xb are burden resistance and reactance. In a CT the primary AT mustequal to the secondary AT and the magnetizing AT

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N1I1 = N2( I2 + Ie)In practice Ie is small compared to I2 and is therefore ignored in

all CT calculations with exception of those concerned with ratio andphase angle error.

fig 1.1

The magnetizing current depends on the voltage V2 which inturn depends on the product of the secondary current and the impedanceof the burden plus CT secondary winding resistance.

V2 = I2 ( Rct + Rb + jX b )

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fig 1.2The vector diagram of the CT is shown in the fig1.2. The

difference between the I1 and I2 is the ratio error and θ is the phaseangle error.

The magnetizing current Ie lags V2 by 90° . It can be seen that if the burden was wholly resistance then the ratioerror would be minimum. Whereas if the burden was wholly reactivethen the ratio error would be maximum and the phase angle errorminimum.

fig 1.3The magnetizing characteristic of a CT is shown in fig1.3.

Since Ie small compared to the secondary current up to the knee-point ofthe magnetizing characteristic, the ratio and phase angle error will besmall in this region. This means the primary / secondary currentrelationship will be maintained in this region.

i.e. the product I2 ( Rct + Rb + jX b ) is Vke.g. Rct = 1Ω, Rb + jX b = 7 + j0 Ω, and Vk

= 150Vthen linearity is repeated up to a secondary current of

V2 150I2 = ⎯⎯⎯⎯⎯⎯⎯ = ⎯⎯⎯ = 18.75 A

√( Rct + Rb + jX b ) 1 + 7If linearity is to maintain up to 20 times CT rating the total impedanceshould not exceed

150

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( Rct + Rb + jX b ) = ⎯⎯ = 7.5Ω20

Effect on open circuited CT :If the burden impedance (Rb + jX b ) is very high, then the

voltage calculated from the equation Rct + Rb + jX b would be very large, will above knee point value and Ie would becomesignif icantly large and I2 would be reduced in the ampere-turn balanceequation

N1I1 = N2( I2 + Ie)Since Ie = 0 in an open circuited CT, all the input amp-turns will be usedas magnetizing amp-turns and drive the CT into saturation. It can beseen from the magnetizing characteristic that greatly increasedmagnetizing current will not cause much increase in the average voltage.However the change in flux from zero to knee-point value is notaccomplished in 1/4th cycle but in perhaps 1/100th

of this time. Thus the rate of change of flux and therefore, the inducedvoltage during this period would be 100 times the knee-point voltage.Insulation can be damaged due to this high short duration voltage andover heating caused by the great increase of iron losses.Specification of CT :

The CTs are usually specif ied in terms of rated burden,accuracyclass and accuracy limit as follows:

15 VA, class 5P10where 15 is the rated burden in VA

5P is the composite error at rated accuracy limit and10 is the accuracy limit factor.

The representation of the above CT means 15 VA burden, and the errorwill be within 5% at 10 times rated current.Standard values of rated burden are 5, 7.5, 10, 20, and 30 VA.Accuracy limit values are usually 5P and 10P.standard accuracy limit factors are 5, 10, 15, 20, and 30.

General protective purpose CTs are frequently specif ied interms of knee-point voltage, magnetizing current at knee-point or atsome other point and secondary resistance. The knee-point voltage isdefined as the point on the magnetizing curve of the CT at which an

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increase of 10% secondary voltage would increase the magnetizingcurrent by 50%. Such CTs are known as class PS or class X CTs.

Bearing in mind the high value of secondary current which aprotective CT may be required to deliver, it is desirable to make thesecondary winding resistance as low as practicable to limit copper lossand therefore heating.Application :

while selecting a CT for a particular application, the connectedburden, the mode of operation and the variation of impedance over therange of relay setting should be considered, rather than the range ofdevices which are connected.

For example, if a relay with burden 3VA and setting range of50% to 200% of nominal current is set at 50%, then

the relay setting is 0.5Avoltage across the coil at this current, V = 3VA / 0.5A = 6Vthe impedance of the relay Z = 6V / 0.5A = 12Ω

If the same relay is set at 200%, thenthe relay setting is 2Avoltage across the coil at this current, V = 3VA / 2A = 1.5Vthe impedance of the relay Z = 1.5V /2A = 0.75A

If the characteristic of the relay is to be maintained up to 20 times therelay setting, then the knee-point voltage should not be less than

20×6 = 120V for 50% setting20×1.5 = 30V for 200% setting

The lowest setting must taken into account while specifying the CT. i.e.Vk of 120V. However there is an alleviating factor that the relay will getsaturated magnetically at 20 times its setting and the impedance will bereduced. The reduction for over current relays is about half theimpedance at setting, which means that in the above case the knee-pointvoltage of 60V would be satisfactory.

In many cases the CT associated with over current protectionmust cater to earth fault relays also. If the earth fault relay having aminimum setting of 20%,

the voltage at relay setting = 3VA /0.2A = 15V andimpedance = 15V / 0.2A = 75Ω

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If the characteristic of the relay is to be maintained up to 10 times therelay, the knee-point voltage should be greater than 10×15 = 150V orallow ing saturation 75V.

In this case the s ize is determined by earth fault relay. Asuitable CT would be 7.5VA, 5P10. This would produce a voltage of7.5V at rated current when connected to 7.5 ohm burden and would haveonly 5% error at 10 times rated current. i.e.at a voltage of 10×7.5= 75V.

As rough guide the knee-point voltage is the product of the VArating and the ALF divided by the rated secondary current. So for as7.5VA, class 5P10, 5A CT

7.5 × 10the knee-point voltage Vk = ⎯⎯⎯⎯ = 15V

5If more than one relay is to be connected to one set of CTs the

total burden must be considered. It is usually sufficiently accurate to addseries burden impedance arithmetically.Effect magnetizing current on relay setting :

The overall setting of the protection system is affected by themagnetizing current of the CTs. The effect may not be signif icant inover current relays, whereas in earth fault relays it will have some effecton the overall setting. In differential protection system the effect ofmagnetizing current is considerable where a large number oftransformers are connected together. For example, a bus-bar protectionscheme.

The primary operating current (POC) of a protective system isthe sum of the relay setting current and the magnetizing current of allthe connected CTs at the voltage across the relay at setting multiplied bythe CT ratio.IllustrationIllustrationIllustrationIllustration 1111:

A 2000/5 bar primary CT has 20 cm2 of iron and a secondarywinding resistance of 0.32 ohm. The maximum current for which the CThas to operate is 40 KA, 50Hz. The relay burden 2ohm.

Given B max for stally 1.0 teslaBmax for cross 1.48 tesla

Determine the suitable core material without saturation.40,000

Secondary current = ⎯⎯⎯ = 100A400

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Assuming no saturationVk = 100 ( 2 + 0.32 ) = 232 voltsVk = 4.44BAf N

232B = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

4.44×1×20 ×10−4×50 ×400

= 1.36 teslaBmax for cross is 1.48 tesla. So more suitable core material

without saturation is CROSS.

IllustrationIllustrationIllustrationIllustration 2222 :A 100/5 bar primary CT supplies to a 3VA over current relay

set at 10% with a stalloy core. The maximum dimension of the CThousing on the circuit breaker are 22 cm dia and 19cm deep. It isrequired to cater for a current of 10 times the relay setting.Find 1) Knee-point voltage

2) Cross sectional area of the core

3) CT dimension.

Secondary current to operate the relay = 0.5 AVA 3

Volts to operate the relay = ⎯⎯⎯ = ⎯⎯ = 6 volts.Setting 0.5

Vk, the knee-point voltage = 10×6 = 60 voltsVk = 4.44 BAfN

60A = ⎯⎯⎯⎯⎯⎯⎯ = 0.0135 m2

4.44×1.0×50 ×20

i.e. 0.0135×104 = 135 cm2

So the core dimension may be 11.25cm × 12cm.

IllustrationIllustrationIllustrationIllustration 3333 :Three 100/5 bar primary CT with secondary resistance of

0.08Ω are connected to cater to three phase fault protection set to125%

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and earth fault relays set to 40%. The relay burden at setting is 3VA forphase fault relays and 2.4 VA for earth fault relays. What is themaximum primary current to operate (POC) both phase and earth faultrelays?

3 3O/C Relay impedance = ⎯⎯⎯⎯ = ⎯⎯⎯ = 0.077Ω

(5×1.25)2 6.252

2.4 2.4E/F Relay impedance = ⎯⎯⎯⎯ = ⎯⎯⎯ = 0.4Ω

(5×0.4)2 22

Phase fault relay setting :Total imped ance = Rct + Rrelay

= 0.08 + 0.077 = 0.157ΩVolts from CT = 0.157 × 6.25 = 0.98 volts.

Assuming the magnetizing current from the magnetizing characteristiccurve at 0.98 volts is 0.25Amps.

Primary current = N( Is + Imag)= 20 ( 6.25 + 0.25) = 130 Amps.

Earth fault relays setting :Voltage across the earth fault relay = 2 ×0.6 = 1.2 Volts

Assuming the magnetizing current from the magnetizing characteristiccurve at 1.28 volts is 0.3Amps.Now the energized CT must supply exciting current to the other twoCTs.Current through Rph relay +exciting current to Yph & B ph = 2 + 0.3 + 0.3 = 2.6Amps.Volts from CT = 2.6 ( 0.077 + 0.08 ) + 1.2

= 1.608 voltsAssuming the magnetizing current from the magnetizing characteristiccurve at 1.608 volts is 0.41Amps.Total secondary current = 0.41 + 2.6 = 3.01Amps.

100Primary fault current = ⎯⎯ × 3.01 = 60.2 Amps.

5

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fig 1.4IllustrationIllustrationIllustrationIllustration 4444 :

An earthed 132 KV transformer is protected by REF protectionusing four CTs and an earth fault relay with stabilizing resistor set at40%. The system fault level is 3000MVA and loop lead impedance fromCT to relay is 2 ohms. The ratio of the CT is 500/1A and Rct is 0.7ohms. Find the primary fault setting and the minimum stabilizingresistor value. Assume relay burden at 1VA.

3500E/F current = ⎯⎯⎯⎯ = 15.3KA

√3 × 13215300

CT sec. Current = ⎯⎯⎯ = 30.6A500

Relay voltage setting = 30.6 (0.7 + 2) = 82.6 volts.Assuming the magnetizing current from the magnetizing characteristiccurve at 82.6 volts is 0.0246Amps.Primary fault setting = 500 (4×0.0246) + 0.4

= 250 Amps.Stabilizing resister = ( Vk //// Is ) - (VA / Is2 )

82.6 1= ⎯⎯

− ⎯⎯0.4 0.42

= 250.25Ω

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VoltageVoltageVoltageVoltage transformerstransformerstransformerstransformers ::::The voltage transformer for use with protection schemes has to

fulfill only one condition, which is the secondary output voltage of thevoltage transformer must be an accurate representation of the primaryvoltage in both magnitude and phase angle. To meet this requirementvoltage transformers are designed to operate at very low flux densities.Magnetizing current at low flux densities will be very low, and thereforethe ratio and phase angle errors are also will be small. To achieve thisthe core area for a given output is larger than that of power transformer,which increases the over all s ize of the voltage transformer. In addition afive limbed construction is used instead of three one in powertransformers in order to reduce the magnetic interference betweenphases. The condition of magnetic interference doesn’t arise when threesingle phase units are used as is common in EHV system, since eachphase unit will have a core with a closed magnetic circuit.

AccuracyAccuracyAccuracyAccuracy ::::Ratio error : The error in the secondary voltage due to incorrect ratio iscommonly known as the ratio error and is expressed as a percentage asfollows:

(KnVs − Vp)⎯⎯⎯⎯⎯⎯ ×100%

Vpwhere Kn is the nominal ratio (rated

primary/secondaryvoltageVs is the actual secondary terminal voltageVp is the actual primary terminal voltage

A small turns compensation is usually be employed in V.T, so that theerror will be positive for low burden and negative for high burden.Phase angle error : The phase angle error is the phase differencebetween the reversed secondary and the primary voltage vector. It ispositive when the reverse secondary voltage leads the primary vectorand negative when it lags the primary vector.

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Limits of error and phase difference for different classes of metering andprotective voltage transformers are given in table below:

Table 1

Limits of error for measuring voltage transformer.Table 2

Limits of error for protective voltage transformer.

VoltageVoltageVoltageVoltage factorfactorfactorfactor ::::the voltage factor Vf is the upper limit of operating voltage,

expressed in per unit of rated voltage. Earth faults cause a displacementof system neutral, particularly in the case of unearthed or impedanceearthed systems, resulting rise in voltage on the unearthed phases. Theserise in voltage are important for correct relay operation and ability of thevoltage transformer withstanding under such condition.

Voltage factors with permissible duration for different type ofconnection and system earthing are shown in the table below:

Table 3

Accuracyclass

0.8 to 1.2 times rated voltage0.25 to1.0 times rated burden at 0.8pf

Ratio error (%) Phase difference (minutes)0.1 ± 0.1 ± 50.2 ± 0.2

± 100.5 ± 0.5 ± 201.0 ± 1.0 ± 403.0 ± 3.0

---

Accuracyclass

0.5to Vf (voltage factor) times rated voltage0.25 to1.0 times rated burden at 0.8pf

Ratio error (%) Phase difference (minutes)3P ± 3 ± 1205P ± 6 ± 240

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ProtectionProtectionProtectionProtection ofofofof voltagevoltagevoltagevoltage transformertransformertransformertransformer :voltage transformer are generally protected by HRC fuses on

the primary side for voltage up to 66KV. Above 66KV the HRC fusesdon’t have sufficient interrupting capacity. As the voltage transformersare designed to operate at a low flux density their impedance is low andtherefore a secondary side short circuit will produce a fault current ofmany times rated current. Hence the secondary side of the voltagetransformer are usually protected by fuses or miniature circuitbreakers.

ResidualResidualResidualResidual connectionconnectionconnectionconnection ofofofof voltagevoltagevoltagevoltage transformertransformertransformertransformer :It is important that a voltage of correct magnitude and phase

angle presented to the directional earth-fault relays and earth faultelements of impedance relays. As an earth-fault can be in any one of thethree phase, it is not possible to derive a voltage in a conventionalmanner. The residual or broken-delta connection of voltage transformersas shown in fig 1.5 is the solution to the above problem. Under 3φbalanced condition the sum of the three voltages connected in brokendelta is zero. If one voltage absent or reduced because of earth-fault on

Voltagefactor Vf

Rating

Method of primary connection andSystem earthing

1.2 continuous

Between lines in any networkBetween transformer star-point and earth inany network.

1.2 continuous

1.5 30s -do-1.2 contin

uousBetween line and earth in an non effectivelyearthed neutral system with automatic earthfault tripping.

1.9 30s -do-1.2 contin

uousBetween line and earth in an isolated neutralsystem without automatic E/F tripping or inan resonant earthed system withoutautomatic E/F tripping.

1.9 8 hrs -do-

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fig 1.5that phase, then the difference between the normal voltage and thatvoltage is delivered to the relay.

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It necessary for the primary winding neutral of the voltagetransformer to be earthed for the zero sequence exciting current to flow.If the primary winding is not earthed, an broken-delta winding maydevelop a voltage. This broken-delta voltage so developed is not relatedto any residual in the primary system, but entirely third harmonic.

Voltage transformers are usually provided with a normal starconnected secondary winding and a broken-delta connected tertiarywinding. Alternatively the residual voltage can be obtained by using astar/broken-delta connected auxiliary Voltage transformer connectedfrom the secondary of the main voltage transformer. For this conditionto be successful, the main voltage transformer must fulfill therequirements for handling zero sequence voltage i.e. it must be of fivelimbed construction, have an earthed primary neutral and rated forsuitable voltage factor. The star point of the main voltage transformersecondary winding and the auxiliary voltage transformer primarywinding must be interconnected to complete the zero sequence circuit.

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2.Symmetrical2.Symmetrical2.Symmetrical2.Symmetrical FaultFaultFaultFault CalculationCalculationCalculationCalculation

The analysis of three phase balanced fault condition consiststhree stages. They are

a) representation of the given power system with itsfaultconditions by equivalent positive sequence network.

b) the solution of the network in terms of its common base ofvoltage, current and impedance.

c) conversion of the resulting common base values to actualvalues.

The system positive sequence network is the equivalent singlephase representation of the complete power system. In this network eachcomponent items of plant is represented by its equivalent positivesequence circuit using per-unit values to a commonMVA base.Over head lines and cables: Over head lines and cables are representedby nominal π circuit. The shunt arm impedance are usually large incomparison with the series arm impedance, and representation by theseries arm alone is usually sufficiently accurate for most practicalpurposes.Transformers & synchronous machines: Transformers and synchronousmachine impedances are predominantly reactive with high X/R ratios inthe order of 10 to 20 times. It is therefore usually sufficiently accurate toignore the resistive component of the impedance and to assume all theimpedance are purely reactive.Loads: load imped ances are large in value in comparison with seriesimpedance of the power system plant and they have only a small effecton the fault current under short circuit condition. Loads are thereforeignored in the short circuit calculation.Transformer tap position: For the purpose of fault calculations, it issufficient to ignore the actual tap position and to assume all thetransformers to be operating in nominal-ratio tap-position. The errorintroduced by this assumption is neglig ibly small in so far as the totalfault current is concerned.Equivalent source: The representation of a complex power systemnetwork can usually be simplif ied considerably by the use of anequivalent generator to represent the whole or parts of a given network.

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2.SYMMETRICAL FAULT CALCULATION

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Thus a complete network, as seen from any given point, may berepresented, using Thevnin`s theorem, as a single driving voltage inseries with a single impedance. This equivalent generator can beobtained with sufficient accuracy from an estimated knowledge of thesystem three phase fault level at the point in question, the pre-fault valueof the voltage at this point being assumed equal to the nominal ratedvalue.Treatment of complex quantities: The impedance of positive sequencenetwork are all complex impedances and must ,therefore be representedby R+jX form. In many cases resistance component of the impedanceare small compared with reactance components, and in such cases it isoften sufficient to treat imped ance as pure reactance, thus ensuring aconsiderable simplif ication in computation. The use of such a purereactance form of representation, it should be noted , results in a short-circuit current slightly greater than the true value.Plant impedance values: The impedance values employed in anyparticular fault calculation should, where ever possible, be the knownvalues appropriate to the particular item of the plant concerned. Whereprecise actual values are not known, however, it may be permissible touse typical values appropriate to similar plant of similar load andvoltage.Neutral earthing: Neutral earthing arrangements have no effect onbalanced three phase load condition or short-circuit condition, andtherefore disregarded in the derivation of the system positive sequencenetwork.Per-Per-Per-Per-unitunitunitunit notationnotationnotationnotationofofofof impedanceimpedanceimpedanceimpedance:

on perusal of a power system, it can be seen that there areseveral voltage level in a system. The common practice is to refer plantMVA in terms of per-unit or percentage values and the transmission lineand cable constants in ohm/km. Bef ore any system fault levelcalculation, the system parameters must be referred to common basequantities, and represented as a unif ied system of impedances in eitherohmic, percentageor per-unit values.

The base quantities are power and voltage. Normally they aregiven in terms of three phase power in MVA and line voltage in KV.The base impedance resulting from these base quantities is :

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(KVbase)2Zba = ⎯⎯⎯⎯⎯ ohm 2.1MVAbase

The per-unit value of any impedance in a system is the ratio of theactual impedance to base impedance. So,

ZohmZp= ⎯⎯⎯ 2.2Z base

substituting eqn. 2.1 in eqn 2.2MVAbaseZpu= Z ohm × ⎯⎯⎯⎯ 2.3(KVbase )2

Having chosen the base quantities of suitable magnitude, allsystem impedance may be converted to the base quantities by theequation given below:

New baseMVAZpu( New baseMVA) = Zpu(Given base MVA) × ⎯⎯⎯⎯⎯⎯⎯ 2.4

Given base MVA

The fault MVA of the system isFault MVA = √3 × KV × KA(f ault)

= 3 × (KV/√3) × KA(f ault)KV KV

= 3 × ⎯⎯× ⎯⎯⎯⎯√3 √3 × Z F.(ohm)

(KV)2= ⎯⎯⎯⎯ 2.5

ZF.(ohm)Substituting the value of fault impedance in ohms in terms of basequantities and per-unit values

(KV)2 ×MVAbaseFault MVA = ⎯⎯⎯⎯⎯⎯⎯⎯

ZF.pu × (KV)2MVA base= ⎯⎯⎯⎯ZF.pu

The source impedance is merely a value which represents theimpedance between the system under consideration and the source. Thisis determined from the fault level at the incoming bus bar.

MVAbaseThe source impedance Zs = ⎯⎯⎯⎯ 2.6MVAfault

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The elements of the power system are usually specif ied as follows:a) generator and transformer - percentage impedance on

rating.b) feeders and interconnectors - actual impedance per phase.c) reactors - voltage drop at rated current.

To convert these per - unit values on a common base the equations 2.3and 2.4 can be written as

Z% MVAbaseZpu = ⎯⎯ ×⎯⎯⎯⎯ 2.7

100 MVAratingMVAbaseZpu = Z ohm × ⎯⎯⎯⎯ 2.8(KVrated)2

VR MVAbaseZpu = ⎯⎯×⎯⎯⎯⎯ 2.9IR (KVrated)2

PlantPlantPlantPlantparametersparametersparametersparameters :Synchronous reactance Xd :It is a measure of the steady state stabilityof the machine; the smaller its value the more stable is the machine. Thevalue 1/Xd approximate to the short circuit ratio (SCR). The onlydifference is the SCR takes saturation into account whereas Xd isderived from the air-gap line.Transient reactance Xd` : The transient reactance covers the behavior ofa machine during the 0.1 to 3 seconds after disturbance. This generallycorresponds to the speed of change in a system, and is usually employedin studies of transient stability.Sub transient reactance Xd`` : The sub transient reactance determinesthe in itial current peaks follow ing a disturbance and in the case ofsudden fault is of importance for selecting the rupturing capacity of theassociated circuit breakers.Two winding transformers : A transformer may be replaced in a powersystem by an equivalent T network in which the cross member is theshort circuit impedance, and the column the excitation imped ance. Asthe excitation impedance is many times higher than the short circuitimpedance it is neglected in fault studies. The per-unit impedance of atransformer is the same regardless of whether it is determined ohmicvalues referred to the HT or LT side of the transformer. The per-unit

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impedance of transformers are usually marked on the name plate on thebase determined by the ratings. For three single phase transformersconnected as a three phase unit, the three phase rating are determinedfrom the single phase rating of each individua l transformer. The per-unitimpedance of the three phase unit is the same as that for each individualtransformer.Three winding transformer: Both the primary and secondary winding ofa two winding transformer the same MVA rating, but all the threewindings of a three winding transformer may have different MVA rating.The per-unit or percentage impedance of each windings are given onthe rating of its own windings. The per-unit impedance of all thewindings in the impedance diagram must be expressed on the sameMVA base.

Neglect ing the excitation impedance, the equivalent circuit of athree winding transformer may be represented by a star impedance asshown in fig 2.1wherep,s,t are primary secondary and tertiary windingsrespectively. The common star point is fictious and unrelated to theneutral of the system.

fig 2.1The impedance of any of these branches can be determined by

considering the short circuit impedance between pairs of windings withthe third winding open.Thus

Zps = Zp + ZsZts = Zt + Zs

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6

Zpt = Zp + Zt

from which Zp = ½(Zps + Zpt − Zst)Zs = ½ (Zps + Zts − Zpt)Zt = ½ (Zpt + Zts − Zps)

IllustrationIllustrationIllustrationIllustration 1:1:1:1:Fig 2.2 shows a power system with all power system

components.

In order to calculate the fault level at different locations firstconvert all plant impedance values to per-unit values on a common base.

Assume a common base of 10MVA.Z% MVAbaseZg = ⎯⎯ ×⎯⎯⎯⎯100 MVArating25 10

= ⎯⎯×⎯⎯ = 0.1 p.u100 25

Interconnector betweenbus-1 and bus-2MVAbaseZL 1-2 = Zohm × ⎯⎯⎯⎯(KVrated)210

= 0.05 × ⎯⎯ =0.0042 p.u(11)2

Transformer8 10

ZT = ⎯⎯×⎯⎯ = 0.1333 p.u100 6

Interconnector betweenbus-3 and bus-4

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7

10ZL 3-4 = 0.008× ⎯⎯ = 0.00735p.u

(3.3)2

VR MVAbaseZ pu = ⎯⎯×⎯⎯⎯⎯IR (KVrated)2

43.7 10= ⎯⎯ ×⎯⎯⎯ = 0.05p.u

800 (3.3)2The per-unit values to common base are shown in fig 2.3 below.

10Fault MVA for a fault at bus-1 = ⎯⎯⎯ = 100MVA

0.110

Fault MVA for a fault at bus-2 = ⎯⎯⎯⎯⎯⎯ = 96MVA0.1+ 0.0042

10Fault MVA for a fault at bus-3 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 42MVA

0.1 + 0.0042 + 0.1333

10Fault MVA for a fault at bus-4 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

0.1+ 0.0042+0.1333+0.00735

= 40.84 MVA

10Fault MVA for a fault at bus-5 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

0.1+ 0.0042+0.1333+0.00735+0.05

= 33.9MVA

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8

The fault currents at various points for a fault at bus-5 iscalculated as follows:

33.9Fault current at bus-5 = ⎯⎯⎯⎯ = 5.93 KA.

√3 × 3.3

33.9Fault current at 11KV interconnector = ⎯⎯⎯⎯ =1.78KA.

√3 × 11

fig 2.4

IllustrationIllustrationIllustrationIllustration 2:2:2:2:

fig 2.5

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9

Assume a common base of 10MVA10

Source impedance Zs = ⎯⎯ = 0.04 p.u250

There are two interconnectors each of two 400mm2 cable in parallel.Assume the reactance per phase of 11KV three core cable = 80µΩ/ m.The reactance of each interconnector = 0.5 × 800 × 80 × 10-6

= 0.032Ω

10ZL = 0.032× ⎯⎯ = 0.002644p.u

1126 10

XT = ⎯⎯×⎯⎯ = 0.3p.u10 2

The interconnecting cable to the 415volt bus is of six single core.i.e. 2 × 1000mm2

Assume the reactance /phase of 415volt cable = 100µΩ/ m.The reactance of 415volt cable = 0.5 × 30 × 100 × 10-6

= 0.0015Ω10

Z l = 0.0015× ⎯⎯⎯ = 0.0871p.u(0.415)2

fig 2.6 fig 2.7

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10

An impedance diagram as shown in fig 2.6 can be drawn withthe above calculated values to a common base. From this diagram thefault level at any particular point can be determined. In morecomplicated arrangement it may be necessary to redraw the impedancediagram after combining the impedance of various parts of the system tosimplif y the calculation. Some times it may be necessary to make morethan one redraw, before the calculation is complete.

From the simplif ied impedance diagram fig 2.7 the fault at11KV substation B is

10⎯⎯⎯⎯⎯⎯ = 242 MVA0.04+ 0.00132

Fault at 415V substation C is10

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 42.58 MVA0.4 + 0.001132 + 0.19355

IllustrationIllustrationIllustrationIllustration 3:3:3:3:A 110KV sub-station A with a three phase fault level of 1158

MVA feeds radially to another 110KV sub-station B, 30KM away fromsub-station A. Two 110KV / 33KV, 10MVA transformers are connectedbetween 110KV and 33KV bus. Calculate the fault level at 33KV busand 110KV bus at sub-station B. The conductor used for 110KVtransmission line is panther conductor.

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11

fig 2.8Assume baseMVA = 100 MVA.

Base MVASource impedance of the grid SS Z s = ⎯⎯⎯⎯⎯

Fault MVA100

= ⎯⎯⎯ = 0.018635p.u1158

R&X values of panther conductor:R = 0.1547Ω/KM, jX = 0.3975Ω/KMIgnore resistive value which will make little error in overall calculation.X = 0.3975 × 30 = 11.925Ω

MVAbaseXp.u = Xohm × ⎯⎯⎯⎯⎯(KVbase)2100

= 11.925 × ⎯⎯⎯ = 0.09855p.u1102

Z% MVAbaseZTp.u = ⎯⎯×⎯⎯⎯⎯⎯100 MVArating

10 100= ⎯⎯ ×⎯⎯ = 1p.u

100 10

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12

The impedance diagram is drawn as shown fig 2.9 and the simplif ieddiagram is shown in fig 2.10.

MVAbaseFault MVA at sub-station B 110KV bus = ⎯⎯⎯⎯⎯Zf

100= ⎯⎯⎯⎯⎯⎯⎯ = 540MVA

0.8635 + 0.09855

100Fault MVA at sub-station B 33Kvbus = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

0.8635 + 0.09855 + 0.5

= 146MVA


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13

fig 2.11The part of a power system shown in fig 2.11 is a power station

with two incoming supplies and an inter bus-bar reactor which isswitched in when the generators are in operation to keep the fault levelto 260MVA which is the rupturing capacity of the switchgear.Assume a common base of 10MVA.

24 10ZG = ⎯⎯×⎯⎯⎯ = 0.32p.u

100 (6/0.8)10.5 10

ZT = ⎯⎯×⎯⎯⎯ = 0.32p.u100 15MVAbase 10

ZS = ⎯⎯⎯⎯ = ⎯⎯⎯ = 0.004p.uMVAfault 2500VR MVAbase

ZR = ⎯⎯⎯ × ⎯⎯⎯⎯IR (KV)2

436 10= ⎯⎯⎯ × ⎯⎯⎯ = 0.045p.u

800 112

The impedance diagram with the per-unit impedance drawn to acommon base as shown in fig 2.12.

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14

fig 2.12there is no simple series or parallel combination which can beeliminated and so delta-star conversion must be made to 0.07, 0.07, and0.045 impedances.The equation for conversion is

Z 1Z2Za = ⎯⎯⎯⎯Z1 + Z 2 + Z3Z 2Z3Zb = ⎯⎯⎯⎯⎯

Z1 + Z 2 + Z3Z3Z 1

Zc = ⎯⎯⎯⎯⎯Z1 + Z 2 + Z3

0.07 × 0.07Za = ⎯⎯⎯⎯⎯⎯⎯⎯ = 0.0265

0.07+ 0.07+ 0.045

0.045 × 0.07Zb = Zc = ⎯⎯⎯⎯⎯⎯⎯⎯ = 0.017

0.7 + 0.07 + 0.045The modified imped ance diagram and the impedance diagrams fordifferent stages of reduction are shown in fig 2.13(a) to fig 2.13(f).

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15

fig 2.13

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16

10The fault level for a for a fault at bus-3 is = ⎯⎯⎯ = 253.47MVA

0.03945The various stages of determining the current distribution are elaboratedin fig 2.14 (a) to fig 2.14(f) and fig 2.14(g) shows the fin al currentdistr ibution for a fault in bus-3.

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17

fig 2.14

IllustrationIllustrationIllustrationIllustration 5:5:5:5:Application of bus impedance matrix in fault calculation.

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18

fig 2.151) Assume a common base of 200MVA.2) Convert all the impedances to a common base.

18.2 200ZG = ⎯⎯×⎯⎯⎯ = 0.164p.u

100 22214 200

ZT1 = ⎯⎯ ×⎯⎯⎯ = 0.1333p.u100 210

15 200ZT2 = ⎯⎯×⎯⎯⎯ = 0.25p.u

100 120200

ZL = 24.8× ⎯⎯⎯ = 0.0656p.u2752

3) Draw the impedance diagram.

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19

fig 2.164) Draw the admittance diagram.

fig 2.17

5) Form the admittance matrix (Y bus).

6) Invert the admittance matrix to get the bus impedance matrix(Zbus).

The fault current for a three phase fault on bus-k isVf

If = ⎯⎯Zkk

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20

The fault current for a three phase fault on bus-5 isVf 1.0

If = ⎯⎯ = ⎯⎯ = 2.918p.uZ 55 0.455

Bus voltage at bus-n for a for a bus fault at bus-k isZ nkVn = Vf − ⎯⎯ VfZ kk

= Vf − Z nk If (since If = Vf/Zkk)Bus voltages at bus 3 & 4 are

V3 = Vf − Z35 × If = 1.0 − 0.33 × 2.198 = 0.2747p.uV4 = Vf − Z 45 × If = 1.0 − 0.33 × 2.198 = 0.2747p.uV2 = Vf − Z25 × If = 1.0 − 0.297 × 2.198 = 0.3473p.uV1 = Vf − Z 15 × If = 1.0 − 0..164× 2.198 = 0.6395p.u

Currents to faults arefrom bus-n = Vn Ynkfrom bus-3 = 0.2747 × 4 = 1.099p.ufrom bus-4 = 0.2747 × 4 = 1.099p.ufrom bus-1 to bus-2 = ( V1 − V2 ) Y12

= (0.6395 − 3473) × 7.5 = 2.198p.uSo the fault MVA at bus-1 = 200 × 2.198 = 4396MVA

439.6Fault current = ⎯⎯⎯⎯ = 0.9229KA

√3 × 275Similarly the fault MVA supplied from bus-1 tobus-2 is

= 200 × 2.198 = 439.6MVA439.6

Fault current = ⎯⎯⎯⎯ = 15.38KA√3 × 16.5

The current flow through other branches can be calculated a similar wayand are shown in fig 2.18

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21

fig 2.18From the same short circuitmatrix i.e. Z-bus, similar information can befound for faults on any of the other buses.


The three phase rating of a three winding transformers are:Primary Y - connected 66KV, 30MVAsecondary Y - connected 13.2KV, 20MVATertiary ∆ - connected 2.3KV, 10MVA

Neglect ing resistance the leakage impedance are:Zps = 7% on 30MVA 66KV base.Zpt = 9% on 30MVA 66KV base.Zst = 8% on 20MVA 13.2KV base.

Find the per-unit impedance of the star connected equivalent circuit fora base 30MVA, 66KV in the primary circuit.

Zps and Z pt were measured in the primary circuit and arealready expressed on the proper base for the equivalent circuit. Nochange of voltage required for Zst. The required change in MVA base ismade as follows.

30Zst = 8 %× ⎯⎯ = 12 %

20So

Zp = ½ (j0.07 + j0.09 - j0.12) = j 0.02p.u

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22

Zs = ½ (j0.07 + j0.12 - j0.09) = j 0.05p.uZ t = ½ (j0.09 + j0.12 - j0.07) = j 0.07p.u

fig 2.19 shows the star connected equivalent circuit.

fig 2.19

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3.Over3.Over3.Over3.Over currentcurrentcurrentcurrent relaysrelaysrelaysrelaysGradingGradingGradingGradingmarginmarginmarginmargin :-

The time interval between the operation of two adjacent relaysdepends on1) the circuit breaker operating time2) overshoot time of the relay, i.e. disc movement after the removal of

the current3) error due to variation in the ideal characteristic curve4) contact gap, i.e. final margin on completion of operation.

A margin of 0.5s is normal grading margin. With faster moderncircuit breakers and lower over shoot times 0.4s is reasonable.

Operating characteristic of over current relays :-

characteristic operating time1

RI curves t = ⎯⎯⎯⎯⎯⎯ x K0.339- 0.236/ I

0.14Normal inverse t =⎯⎯⎯⎯⎯⎯ x K

I0.02 - 113.5

very inverse t =⎯⎯⎯⎯⎯⎯ x KI - 180

Extremely inverse t =⎯⎯⎯⎯⎯⎯ x KI2 -1120

Long time stand by earth fault t = ⎯⎯⎯⎯⎯⎯ x KI - 1

Logarithmic inverse t = 5.8 - 1.35 log n (I / I n)where t = relay operating time

K = scale constant or TMS according to curveI = multiple of set current Is or PSM

Page 41 / 153

3.OVER CURRENT RELAYS

2

fig 3.1

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3

IllustrationIllustrationIllustrationIllustration onononon overoveroverover currentcurrentcurrentcurrent relayrelayrelayrelay calculationcalculationcalculationcalculation1111 :-

fig 3.2Calculate the maximum and minimum load currents bus. Provide

current transformers of ratio as per system load requirements. Set therelay with an equivalent primary current well above the maximum loadcurrent. The relay setting must be well below the minimum fault current.Complete the data’s in a table as given below:

Table 3.1

Over current relays are intended to provide a discriminativeprotection against system faults and they do not give precise overloadprotection

location

Totalimpedancein ohms

Faultcurrent inAmps

Totalload

current

CT.Ratio

Relaycurrentsetting

Min Max Max Min percent

primary

current

A 0.81 1.62 7840 3920 500 400/5 150 600B 1.41 2.22 4504 2860 350 400/5 125 500C 2.36 3.17 2694 2003 175 200/5 100 200D 4.56 5.37 1393 1183 75 100/3 100 100

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4

Starting with the relay in substation D farther from the powersource, the relay plug setting multiplier (PSM) is calculated from theknowledge of the maximum fault current flowing from this point and therelay settingSubstation DCT ratio 100/5ARelay used Normal inverseCurrent setting 100% i.e. 100AMax fault level at substation D 1393ATherefore relay PSM 1393/100=13.93

From the normal inverse characteristics of the relay theoperating time of the relay at 13.93 times the relay plug setting and 1.0time multiplier setting (TMS) is 2.6 seconds. There is no relay follow ingrelay at D. Still a small delay is required for better discrimination.Further for electromagnetic relays the contact travel of the relay at Dshould not be unduly small so as to avoid the possibility of tripping dueto mechanical shock. So a TMS of 0.05 is a wise option.Hence the actual tripping of the relay at D is

0.05x2.6 = 0.13 sec’sA grading margin of 0.5 sec is adopted in this example, so that

the relay at substation C should have an operating time for fault atsubstation D.Substation C:-CT ratio 200/5ARelay used normal inverseCurrent setting 100% i.e. 200AMax fault level grading level C with D is 1393ATherefore relay PSM is 1393/200=6.965

From the normal inverse characteristic of the relay theoperating time of the relay at 6.95 times the relay plug setting and 1.0TMS is 3.6 sec.Required relay discriminating time = 0.13 + 0.5 = 0.63 secTherefore the required relay TMS = 0.63 / 3.6= 0.175

The calculations now proceeded for relay Cwith a close-upfaults at substation C.The maximum fault current for a fault

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5

just outside substation C bus bars =2691ATherefore relay PSM =2691/200 = 13.45

Now from the normal inverse characteristic of the relay theoperating time of the relay at 13.45 times the relay plug setting and 1.0TMS is 2.6 sec. The TMS previously calculated for the relay C is 0.175.This gives an actual relay operating time for a close-up fault atmaximum fault level.

ie 0.175x 2.6=0.455 sec.Table 3.2

Tc :- Relay operating time from standard curve for the given PSMTa :- relay actual operating time

The grading of the remaining relays are proceeded in similarway as illustrated above and tabulated.

Finally the discriminating curves of relay at substation A,B,Cand D are plotted on log-log sheet (fig…). This can be done using atemplate of relay operating characteristic at TMS of 1.0. From the plot itcan be seen that a grading margin of 0.5 sec has been achieved.

ProcedureProcedureProcedureProcedure forforforfor plottingplottingplottingplotting discriminatingdiscriminatingdiscriminatingdiscriminating curvecurvecurvecurve :-For this purpose it is necessary to prepare a transparent

template of the ‘time current curve’ of the relay on a log - log papercorresponding to TSM = 1. The template is so made that the origin ofthe log - log paper corresponds to 1 - second of the curve along the Y -axis and 100% plug setting for 1 - multiple P.S along the X - axis as infig 3.2(a).

Relay settingPSM Tc TMS Ta

Relay at D Fault at D 13.93 2.6 0.05 0.13Relay at C Fault at D 6.965 3.6 0.175 0.13+ 0.5 = 0.63Relay at C Fault at C 13.45 2.6 0.175 0.455Relay at B Fault at C 5.382 4.1 0.233 0.455 + 0.5 = 0.955Relay at B Fault at B 9 3.15 0.233 0.734Relay at A Fault at B 7.5 3.45 0.358 0.734 + 0.5 = 1.234Relay at A Fault at A 13.06 2.65 0.358 0.95

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6

fig 3.2(a)After having made the template the values of the current (in

amps) are plotted on a log -log paper along the X - axis and the time ofoperating of the relay ( in seconds) along the Y - axis.

For plotting the characteristic of the relay - D in the aboveillustration, the origin of the template is kept at 100 amps and 0.05 secsand the curve of the template is transferred on the log - log paper.Similarly for plotting the characteristic of the relay - C the orgin of thetemplate should be kept at 200 amps and 0.175 secs and the transfered tothe log - log paper. Repeat the procedure for B and A relays with theorigin of the of template kept at ( 500A, 0.233 s ) and ( 600A, 0.358 s )respectively.

DiscriminationDiscriminationDiscriminationDiscrimination ofofofof overoveroverover currentcurrentcurrentcurrent relayrelayrelayrelay’’’’ssss bybybyby bothbothbothboth timetimetimetimeandandandand currentcurrentcurrentcurrent :-Relay co-ordination in the case of discrimination by time alone

has the disadvantage of, more severe faults are cleared in the longestoperating time. Discrimination by current can be applied only whenthere is appreciable impedance between two circuit breakers. With theuse of inverse characteristic the time of operation is inverselyproportional to the fault current and the actual characteristic is afunction of time and current setting.

IllustrationIllustrationIllustrationIllustration 2222 :-

fig 3It is necessary to convert all the system imped ance to a

common base.Common base used :10 MVA

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7

percentage imped ance of 4 MVA transfer MVAbaseto common base : =Z% x ⎯⎯⎯⎯MVArated10

= 7 x ⎯ = 17.5%4

Percentage impedance of cable between MVAbaseD &E to a common base of 10 MVA : = Z ohm × ⎯⎯⎯⎯ × 100

( KV )210

= 0.04 x ⎯ x 100 = 0.33%112

Percentage impedance of cable between 10C&D to a common base of 10 MVA : = 0.24 x⎯ x 100 =1.98%

112percentage imped ance of 30 MVA 10transformer : = 22.5 x⎯ =7.5%

3010

Percentage impedance of 132KV line = 6.2 x⎯⎯ x100 = 0.36%1322

Base MVAPercentage impedance of 132KV source =⎯⎯⎯⎯⎯ x 100

Fault MVA10

= ⎯⎯ x 100 =0.29%3500

Table 1

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8

Table 2

A voltage base of 3.3KV has been chosen and the operatingcharacteristic of the highest rated fuse, that is 200A on the outgoing3.3KV circuit is plotted on a log-log sheet. The grading of the overcurrent relays at various substations of the radial system is carried out asdetailed below. A grading margin of 0.4 is adopted in this example

Relay location D:-

Relay used :- Extremely inverse characteristicBase MVA

Relaylocation

Total impedance inZpu

Fault current inAmps at 3.3KV base

Totalloadcurrentat

3.3KVbase

CT.Ratio

CT. Ratioto thebase of33kv

Relay currentsetting

Min Max Max Min percent

primarycurrentat 3.3kvbase

D 0.1013 0.2795 17268 6257 700 250/5 833/5 100 833C 0.0815 0.1013 21466 17268 1500 500/5 1666.6/5 100 1666.6B 0.0065 0.0815 269160 21466 6000 150/1 6000/1 100 6000A 0.0029 0.0065 603292 269160 20000 500/1 20000/1 100 20000

PSM Tc TMS TCRelay at D Fault at D 20.7 0.2 0.05 0.1Relay at C Fault at D 10.36 0.8 0.625 0.1+ 0.4 = 0.5Relay at C Fault at C 12.88 0.47 0.625 0.29Relay at B Fault at C 3.58 6 0.115 0.29+ 0.4 = 0.69Relay at B fault at B 44.8 0.2 0.115 0.023Relay at A Fault at B 13.458 0.43 1 0.023 + 0.4 =0.423Relay at A Fault at A 30.1 0.2 1 0.2

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9

Fault level up to 3.3KV bus = ⎯⎯⎯⎯⎯Σ Z pu

10

= ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

0.175+0.0033+0.0198+0.075+0.0036+0.002910

= ⎯⎯⎯⎯ = 35.76 MVA0.2796

Fault MVA 35.76Fault current = ⎯⎯⎯⎯⎯ = ⎯⎯⎯⎯

√3 x base KV √3 x 3.3= 6257Amps at 3.3KV

or 35.76=⎯⎯⎯ = 1877Amps at 11KV√3 x 11

10Fault level close to bus D = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

0.0198+0.075+0.0036+0.002910

= ⎯⎯⎯ = 98.7MVA0.101398.7MVA

Full load current at 3.3KV base =⎯⎯⎯⎯⎯ = 1.7268KAmps√3 x3.3KV

4MVAMaximum load current at 3.3Kv base = ⎯⎯⎯⎯ = 0.7KA

√3 x3.3KVCT ratio adopted 250/5Acorresponding CT ratio to the base of 3.3KV = 250 x (11/3.3) / 5

=833.3/5ASo a relay of 100% P.S (plug setting) corresponds to a current

of 833.3Amps(3.3KVbase).The calculations for the remaining relay locations are

proceeded in similar way as illustrated above and tabulated

Calculations were made for all the relay locations for the TMS(time lever setting) in the same way as explained in the previousexample and tabulated in Table 2

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10

The discriminating curves were plotted for all the relays at substations A, B, C, D &E on a log-log sheet (fig…).

ILLUSTRATIONILLUSTRATIONILLUSTRATIONILLUSTRATION 3333Coordination of over current relays for an industrial system :-

fig 3.4The relays used for protection are :i) for motor protection (relay G) a thermal relay with high set unit for

instantaneous over current elementii) for transformer protection (relay H) over current relay with very

inverse characteristic and high over current elementiii) for feeder protection (relay J) a normal inverse characteristiciv) for generator a voltage controlled normal inverse characteristic

with dual characteristic

For convenience a common voltage base of 6.6KV chosen andscheme for relay co-ordination redrawn with this base.

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11

fig 3.5Fuse characteristic :- The operating characteristic of the fuse should beplotted on a log-log sheet with a suitable scale for maximum fault levelson a common base voltage of 6.6KV.Table 1

Induction motor :-Rating of the motor =100KW.Full load current =139 Amps at 415V base

ie =8.75 Amps at 6.6KV baseMotor starting current ( 6 times full load current) =6 x 139 = 834 Amps

ie 52.48Amps at 6.6KV baseStarting time = 10 secsThe relay may be set for 100% setting corresponding to 9.43Amps at6.6KV base ie. 13% overload.

The thermal characteristic of the relay is

Table 2

Operating current at Operating Time415V Base 6.6KV Base

795 50 40955 60 131160 70 61270 80 31590 100 0.92700 170 0.1

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12

Plot the above characteristic on the same log-log sheet. Theinstantaneous unit may be set at 1.3 times the starting current ie 1.3 x840 = 1090 amps at 415 V base or 68.25amps at 6.6 KV base. Plot theseinstantaneous characteristic also.

Transformer protection:

Relay used …. Very inverse time over current plus instantaneouselement.

CT ratio ……… 75/5A.750 x 103

full load current of transformer = ⎯⎯⎯⎯⎯ = 65.6Amps√3 x6.6 x 103

The relay current setting ……. = 100% ie 75Amps.This setting will provide adequate margin and better co-ordination withthe fuse

10 x 106Max fault level =⎯⎯⎯⎯⎯ = 874.7Amp

√3 x 6.6 x103Relay PSM = 847.7/75 = 11.6

Relay operating time at 11.6 times plug setting and 1.0 TMS is1.41secs. The grading margin between relay and fuse will be

t’ = 0.4t + 0.15 secwere t = nominal operating time.So the grading margin = 0.4 x 0.01 + 0.15 =0.154 (say 0.16)

multiples of current operating current operating time415vbase

6.6KVbase

Hot Cold

1.25 187.5 11.79 1400 5501.5 225 14.15 700 2402 300 18.85 300 1003 450 28.3 105 354 600 37.5 55 185 750 47.2 33 116 900 56.6 23 7.5

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13

Plot the operating characteristic on the log-log sheet to thecommon 6.6KV voltage base using a template of the ver inversecharacteristic.The instantaneous element must be set above the maximum throughfault current.

Fault MVA of the transformer = MVA/Zpu = 0.75 / 0.07 =10.714MVA

Fault MVA 10.714Through fault current = ⎯⎯⎯⎯⎯ = ⎯⎯⎯ = 0.93725KA

√3 x KV √3 x 6.6The relay must be set at 1.3 x937.25 ie 1218Amps.Plot the operating characteristic on the log-log sheet.

Feeder protection:

Relay used = normal inverse characteristic.CT Ratio = 200 / 5

The current setting on this relay should be based on themaximum load of the 6.6 KV bus, plus a suitable margin for over loadon the transformer. Since no load values are specified the requirementmay assumed as 100%Maximum fault for grading = 1218ampsRelay PSM = 1218 / 200 = 6.1Relay operating time for6.1 times PS and 1.0 TMS is =3.8 secs

To grade relay J with relay H the operating time of relay H at1218amps must be added to the fixed grading margin of 0.5sec.

ie 0.19 + 0.5 = 0.69sec.So required TMS =0.69 / 3.8 = 0.18Plot the operating characteristic on the log-log sheet.

Generator :Relay used ……. The relay used is with a duel characteristic

which changes its pick-up value to 40% of its nominal plug setting and

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14

the operating characteristic changes from a long IDMT characteristic tostandard IDMT characteristic.

CT Ratio = 500 / 5 AUnder close-up fault condition the voltage at the bus bars dorps

below the setting of the under voltage units, the over current relaychanges its characteristic from the over load curve to the fault curve.Then the effective setting becomes (0.4 x 500 )Amps ie 200Amps whichhappens to be the current setting of the 6.6KV feeder relay J with whichit has to grade. So a setting of 100% can be used.

The possible fault MVA of the 5MVA generator isMVA Rating 5⎯⎯⎯⎯⎯ = ⎯ x 100 =33.33MVA

Xd` 15

33.33So the fault current =⎯⎯⎯ = 2.916 KA

√3 x 6.6Relay PSM = 2916 / 200 = 14.57Relay operating time for 14.57 times PS and 1.0 TMS is 2.5 secs.Relay JJJJ operating time at 2916 Amps is 0.38 secs.Grading margin 0.5 sec.Hence operating time on fault curve should be

0.38 + 0.5 = 0.88 secsSo TMS = 0.88 / 2.5= 0.352

Plot the operating characteristic of the fault curveand over loadcurveon the log-log sheet.

ExamplesExamplesExamplesExamples ofofofof earthearthearthearth relayrelayrelayrelay connectionsconnectionsconnectionsconnections ::::

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Earth fault relays protecting over head lines are normallyconnected to three CTs in a residual current circuit. (fig 3.6).

fig 3.6Instead of three separate CTs one open coreCT (Core Balance

CT) can be used in networks three cable (Fig 3.7).

fig 3.7

In case of non-directional earth fault relays a capacitor acrossthe CBCT terminal will increase the sensitivity of the relay.( Fig 3.8)

The sensitivity of the relay can be increased by running the cableseveral times through the CBCT. The sensitivity will increase twice fortwo turns. (Fig3.9)

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16

fig 3.8 fig 3.9

The sensitivity of the relay may be increased by mounting twoCBCTs on the cable and CT terminal connected in parallel (fig 3.10)

It is possible two parallel cables through same CBCT (fig3.11). However if the cables cannot be arranged symmetrically ,there isa possibility of undue current to relay. This may operate the relayunnecessarily for high short circuit currents.

fig 3.10 fig 3.11

In case of two or more cables operating in parallel and fittedwith one CBCT for each cable, it is possible to connect all the CTs inparallel and connected to the same relay (fig 3.12).

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17

fig 3.12

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4.Differential4.Differential4.Differential4.Differential relayrelayrelayrelay

CTCTCTCT connectionconnectionconnectionconnection forforforfor differentialdifferentialdifferentialdifferential relay:relay:relay:relay:The basic requirement of differential relay connection must satisfy thefollow ing conditions.1) The relay must not operate for external faults.2) The relay must operate for severe internal faults .

fig 4.1Connect the CTs of star winding of power transformer in delta

and CTs of delta winding in star. Assume the ratio of power transformerand CTs as 1:1, so that the current magnitude are equal. Once the properconnections are arrived the actual ratio can be taken into account.Assume arbitrarily the current flow in the transformer, but observe therequirement imposed by the polarity marks that the current flows inopposite directions in the winding on the same core. The completedconnections for the two winding transformer for differential relaying isshown in fig 4.1

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4.DIFFERENTIAL RELAY

2

PhasePhasePhasePhase andandandand ratioratioratioratio compensationcompensationcompensationcompensation ofofofof thethethethe biasedbiasedbiasedbiased differentialdifferentialdifferentialdifferential protectionprotectionprotectionprotection forforforforpowerpowerpowerpower transformerstransformerstransformerstransformers usingusingusingusing interposinginterposinginterposinginterposing currentcurrentcurrentcurrent transformerstransformerstransformerstransformers ::::

In the case of power transformers there may be a phase rotationbetween primary and secondary, e.g. group of connection yd5 or yd11,which has to be compensated before the currents measured in primaryand secondary can be compared by differential relay. Should the ratiosof the current transformers primary and secondary not compared to ratedcurrents, the amplitude must also be compensated, so that differencebetween them at the relay becomes zero under normal load conditions.Finally the zero sequence component must be eliminated on windingwith grounded star-points.

Formerly, the compensation of group of connection and mainCT ratios was performed outside the differential relay by appropriateconnections of the main CT secondaries or of interposing CTs insertedbetween the main CTs inserted between the main CTs and thedifferential relay as shown in the fig 4.2

fig 4.2The main CTs are Y-connected while three single phase inter

posing CTs are connected with the same vector group of connection asthe power transformer. Typical interposing CTs have several tapings foradjusting their ratio KICT to the composite ratio of the power transformer

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KT and the main CTs KS1and KS2. The follow ing relationship fordetermining the ratio of KICT is determined from fig 4.2

I2N / √3 . KS2KICT = ⎯⎯⎯⎯⎯

I1N / KS1KS1

= ⎯⎯⎯⎯⎯√3 .KT . KS2

where I1N and I2N are primary and secondary ratedcurrent of the transformer.

Inter posing CTs are also used in the case Y-connectedprimaries and secondaries. e.g. group of connection Yy0, to compensatedifference of ratio. Since the ICTs are also connected in Yy0 , the term√3 in the above equation is omitted while determining the ratio of KICT.

Phase and ratio compensation is essential in the case of threewinding transformer, because there is always a phase-shift between thecurrent of the three windings and the ratios of the three groups of mainCTs have to be adjusted to each other. The principle of compensationfor a Ydd three winding transformer with three groups of interposingCTs ICT1, ICT2 and ICT3 is shown in fig 4.3 in a simplif ied form.

fig4.3The ratios of the interposing CTs in the three legs ‘n’ can be determinedusing the generally applicable formula

SN 1

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KICT = ⎯⎯⎯⎯⎯ × ⎯√3. UN. KSN m

whereKSN = main CT ratios in legs n =1,2,3……….m = factordepending on ICT circuit

m = INP/√3 incase of Y/∆m = INP in case of Y/Y, INP being the main CT

nominal current i.e. the primary current ofICT

Sn = rated transformer power.Un = rated voltage of the respective winding.

IllustrationIllustrationIllustrationIllustration 1111

fig 4.4Matching transformer required =3.75 / 4.625Assume the secondary turns of matching transformer as 25 turns.The number of turns on the primary side Tp = ( Is / Ip ) x Ts

= ( 4.625 / 3.75 ) x25= 30.83 ≅ 31 turns.

Current imbalance due to the use of 31 turns limited to30.83 - 31

Error = ⎯⎯⎯⎯ x 100% = - 0.55%30.683

Inter posing CT of ratio 4.625 / 5 could also have been used inconjunction with the 200/5 CTs. In such case ICT used in conjunctionwith the 1000/5 CT should be 3.75/5A.

In omitting to provide a second ICT the effective setting wouldbe increased by the ratio (5 /4.625) x 100% i.e. 108.1%.

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5

This would give a setting of 1.081 x20% i.e. 21.62% on the20% plug setting. The slight increase in effective setting would beacceptable and thus a second ICT may be considered unnecessary.

IllustrationIllustrationIllustrationIllustration 2222

Three phase transformer , 30MVA , 11 / 66KV delta - star

fig 4.511KVWinding:

30x106Normal current at 11KV =⎯⎯⎯⎯⎯ = 1574 .6 Amps

√3 x 11 x 103Because the 11KV winding is Delta connected, the associated

current transformer will be star connected and under rated loadcondition the current per pilot phase will be

Is = (1574.6 /1600) x 1A =0.984 Amps.This current sufficiently close to relay rated current (1A).

6.6 KV winding :30 x 106

Normal current at 6.6KV = ⎯⎯⎯⎯⎯ = 262.43Amps.√3 x 6.6 x 103

To provide appropriate phase shif t correction the currenttransformer associated with the star winding of the main transformer

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6

should be connected in delta. However these CTs are connected in star.So the necessary phase correction may be carried out by providing a stardelta connected matching interposing current transformer. The outputcurrent per phase pilot of the 300 / 1A CT is given by

Ip = (262.43 /300) x 1A =0.875Amps.This should be adjusted by he interposing CT, so that

0.984Amps flows into the relay. If three single phase CTs are used, thenthe ratio should be

Is /√3 0.984 / √3⎯⎯ = ⎯⎯⎯⎯Ip 0.875

Assume the secondary of the ICT uses 215 turns, then the primary turnsrequired is given by

Is / √3 0.984 x 215Tp = ⎯⎯⎯ x Ts = ⎯⎯⎯⎯ = 139.6 turns ≅ 140 turns

Ip √3 x 0.875

IllustrationIllustrationIllustrationIllustration 3333Three winding transformer :400MVA /100 MVA / 300 MVA, 500 / 13.45 / 138KV Star - Delta

fig 4.6The current transformer ratios are chosen as a function of the windingvoltage and power rating of the particular winding with which they areassociated.

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500KVwinding :Based on the 400MVA the rated current is given by

400 x 106In = ⎯⎯⎯⎯⎯ = 462Amps

√3 x 500 x 103Secondary current from 500 / 5A CT

Is = (462 / 500) x 5 = 4.62Amps.The 500 / 5 A star connected CTs are associated with the 500KV starwinding, and thus the transition to delta connected secondary must bemade by means of an interposing CT.Assuming the secondary of the ICT be 43 turns.

Is /√3 5 x 43Then primary turns Tp = ⎯⎯ x Ts = ⎯⎯⎯ = 26.86or say 27 turns

Ip √3 x 4.62138KVwinding :

During external fault conditions the line CT out put currentmust be related to the primary input MVA.

So based on 400MVA, the corresponding current400 x 106

In = ⎯⎯⎯⎯⎯⎯ = 1674Amps.√3 x 138 x 103

secondary current of 138KV CT isIs = (1674 /1200) x 5 = 6.975 AmpsThe 1200 / 5A star connected CTs are associated with the

138KV star winding of the transformer. Hence necessary transition todelta connection must be made by means of an ICT.Assuming the secondary of the ICT be 43 turns,

Is /√3 5 x 43Then the primary turns Tp =⎯⎯ x Ts = ⎯⎯⎯ = 17.79, say 18 turns

Ip √3 x6.975

13.45KV winding :

Based on 400MVA, the corresponding current400 x 106

In = ⎯⎯⎯⎯⎯⎯ = 17170.2Amps.√3 x 13.45x 103

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secondary current of 138KV CT isIs = (17170.2 /5000) x 5 = 17.17 AmpsIn this case the transformer winding of the 13.45 KV side is

delta connected and the associated CTs are star connected. So there is nophase angle compensation required. ICT may be star connected and tapsshould be selected to reduce the 17.17A to 5A.Assuming the secondary of the ICT be 43 turns,

Is 5 x 43Then the primary turns Tp =⎯ x Ts = ⎯⎯ = 12.52or say 13 turns

Ip 17.17

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5.Generator5.Generator5.Generator5.Generator ProtectionProtectionProtectionProtection RelaysRelaysRelaysRelays

SelectionSelectionSelectionSelection ofofofof GeneratorGeneratorGeneratorGenerator neutralneutralneutralneutral groundinggroundinggroundinggrounding transformertransformertransformertransformer andandandand loadingloadingloadingloadingresistor:resistor:resistor:resistor:IllustrationIllustrationIllustrationIllustration

Generator voltage (between phases)…………….13.8KVFrequency ………………………………………. 50HzZero sequence capacitance per phase of thesystem components :a) Generator………………………………… C1 = 0.170µFb) Generator surge capacitor……………… .C2 = 0.250µFc) Generator leads (bus ducts )…………… C3 = 0.063µFd) Step-up transformer…………………….. C4 = 0.0035µFe) Unit transformer and miscellaneous……C5 = 0.0135µF

Total capacitance (C1+C2+C3+C4+C5……..) C = 0.5µFTotal residual capacitance 3C = 3 x 0.5 = 1.5µF

106 106Residual capacitive impedance = ⎯⎯ = ⎯⎯⎯⎯⎯ = 2123 ohms.

3ωC 3 x 314 x 0.5

If the effective earthing resistance is made equal to the totalcapacitive impedance of 2123 ohms, then with a generator terminal faultat nominal voltage, the neutral current is

13800 /( √3 x 2123) = 3.75Amps.The actual fault current will contain equal resistive and capacitvecomponents and will therefore be

3.75 + j3.75 = 5.3 AmpsEarthing transformer:

The primary knee point voltage should not be less than 1.3 x13.8KV ie 18KV. The applied voltage during an earth fault is normally13.8/√3 ie 8KV and with field forcing condition may be 18/√3 ie10.4KV. Hence in either case standard 11KV insulation will besatisfactory.

The maximum earth fault current at field forcing condition willincrease in proportion to the voltage rise and will therefore be

(10.8/8) x 3.75 =5 Amps

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5.GENERATOR PROTECTION RELAYS

2

Hence the maximum load ing under such condition is 5 x 10.4ie 52 KVA. A 30 seconds rating for maximum duty is adequate and isusually specif ied. Experience with practical distribution transformer hasshown that an overload of six times the continuous rating can beapplied for this time. The rating is however based on the maximumoutput and voltage which the transformer can produce.So the necessary continuous rating is (5 / 6) x 18 = 15 KVA.

The value may be rounded up to he nearest standard size, andso the rating of 15KVAmay be specif ied.Transformer secondary rating:

The rated secondary voltage may be any value and should bechosen to give a suitable secondary current. The secondary knee pointvoltage of 250V making the ratio of 18000 /250 Volts will give amaximum secondary current at normal generator voltage of (18000 /250) x3.75 ie 270AmpsEarthing resistor :

The equivalent resistor should be equal to the earth faultcapacitance of 2123 ohms.Therefore the secondary resistance = 2123 x (250 /18000)2

= 0.41 ohms.This is the total resistance required. The total transformer windingresistance in terms of the secondary should be deducted from this valueto obtain the value of load ing resistor.

A standard 15KVA transformer can be expected to have acopper loss of 450 watts. In terms of the secondary winding theresistance will be

W 450RT = ⎯⎯ = ⎯⎯ = 0.125 ohm

Is2 602Therefore the load ing resistor should be (0.41 - 0.125) i.e. 0.285 ohmand rated to carry

1.3 x 3.75 x 18000 / 250 = 350Amps for 30 seconds.

CTCTCTCT requirementsrequirementsrequirementsrequirements forforforfor relaysrelaysrelaysrelays usedusedusedused inininin balancebalancebalancebalance typetypetypetype protectiveprotectiveprotectiveprotective schemesschemesschemesschemes ::::

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The knee point voltage referred in the follow ing table offormulae is defined as the point on the magnetizing curve at which a10% increase in exciting voltage produces 50% increase in excitingcurrent.

Where If - Equivalent secondary current of maximum fault currentIs - Effective fault setting in secondary amperesIr - Relay operating currentn - Number of CT groupsq - Number of incoming circuitsRs - CT secondary resistanceRl - Max lead resistance between relay circuits and any CT

SelectionSelectionSelectionSelection ofofofof stabilizingstabilizingstabilizingstabilizing resistorsresistorsresistorsresistors forforforfor balancedbalancedbalancedbalanced typetypetypetype protectiveprotectiveprotectiveprotective schemesschemesschemesschemes:

In balance type protective schemes, spill current in relaycircuits can cause indiscriminate operation. To avoid such unwantedoperation stabilizing resistor is connected in series with the current relay.

The value of the stabilizing resister is chosen to ensure thatunder maximum steady state through fault conditions there isinsufficient voltage developed across the bus wires to cause a spillcurrent equal to the relay operating current.

In calculating the value of the stabilizing resister, the follow ingassumptions are made.

1) one set of CT is completely saturated.2) The whole of primary fault current is perfectly

transformed by the remaining CTs.

Application MIN Knee pointvoltage

MAX. excitingcurrent at V/2 volts

REF with 3 CTs 2 If (Rs +Rl ) (Is - Ir) / 3REF with 4 CTs 2 If (Rs +Rl ) (Is - Ir) / 4Bus differential 2 If (Rs +Rl ) (Is - Ir) / nResidual check 2 If (Rs +Rl ) (Is - Ir) / 3q

Machinedifferential

2 If (Rs +Rl ) (Is - Ir) / 2

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3) The maximum loop lead burden between the relay and CTsis used.

With one set of CTs saturated the maximum voltage developedacross the stabilizing resister and relay coil in series, is

If ( Rs + Rl ) / N voltswhere If is the maximum fault current

Rs is the CT internal resistanceRl is the maximum lead resistanceN is the CT ratio

For stability, the current through the relay coil at this voltagemust be insufficient to cause relay operation. A stabilizing resister ischosen which will just allow setting current to flow through the coil.Example 1 :-

Maximum voltage calculated by the above method….50 volts.Setting on the relay …..0.1 AmpsBurden of the relay at the adopted setting ….1 VARelay impedance at setting current = VA / Is2

= 1.0 / (0.1)2 = 100 ohmRelay circuit impedance (Stabilizing resister and relay) = Vk//// Is

= 50 / 0.1 = 500 ohmTherefore Stabilizing resister = ( Vk //// Is ) - (VA / Is2 )

= 500 - 100 = 400 ohm.

Example 2 :-

Machine rating 247 MVA, 15.75 KV.Sub transient reactance Xd`` = 21.4%CT ratio 10000 / 5Amps, RCT = 1.5 ohm

1 1Maximum sub transient through fault current = ⎯⎯ = ⎯⎯

Xd`` 0.214= 4.67 pu

247Generator base current = ⎯⎯⎯⎯ = 9.0 kA

√3 x 15.75Therefore If = 4.67x 9.06 = 42.3 kA

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secondary fault current = 42300 x ( 5 / 1000 ) = 21.15AmpsRelay voltage = If ( RCT + Rl )

= 21.15 ( 1.5 + 0.5 ) =42.3 voltsRelay setting = 0.25 Amps ( i.e. 5% of 5 Amps )Stabilizing resistor = ( Vk //// Is ) - (VA / Is2 )

42.3 1=⎯⎯ - ⎯⎯⎯ (assume relay burden 1VA)

0.25 (0.25)2=169.2 - 16 = 153 2 or say 160 ohms

SettingSettingSettingSetting ofofofof restrictedrestrictedrestrictedrestricted earthearthearthearth faultfaultfaultfault relayrelayrelayrelay ::::

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fig 5.1

Data required :1) CT Ratio……………………. ………………… 200 / 1 A2) magnetizing characteristic of CT…..Graph (fig), class X3) CT secondary winding resistance…………….. 3 ohm4) Lead resistance (loop)……………………………3 ohm5) Rating of transformer ……………………………. 30

MVA6) Ratio of transformer …………………………… 132 / 11

kV7) Required primary fault setting………………… 25%8) power transformer impedance ……………….. .. 9.5 %9) Min CT knee point voltage …………………….. 126 V

1 1If = ⎯⎯⎯⎯⎯ x 100 = ⎯⎯ x 100 = 10.5 %

% impedance 9.5%Minimum knee point voltage Vk = 2 If ( Rct + Rl )

= 2 x 10.5 x (3+3) = 126V25 30 MVA

Required primary fault setting = ⎯ x ⎯⎯⎯⎯ = 33Amp

100 √3 x 132 kVThere fore secondary fault setting = 33 x ( 1/200 ) =0.165

Amp

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For actual setting all currents through shunt path must be subtracted.Shunt path current = number CT x their magnetizing current.Thus actual setting = 0.165 - 3 x ImetrosilIn R.E.F application where through fault current is limited Imetrosil maybe ignored. Magnetizing current must be considered at relay setting Vswhich is now calculated.

From the magnetization curve in fig … the knee point voltageVk is 300 volts.A stabilizing voltage within the range of Vk/2 to Vk/4 is

normal.So Vs may be chosen as 150 volts.The value of Imag at Vs = 150 is 0.015 Amps.So setting in secondary amps = 0.15 - 3 x 0.015

= 0.12 AmpsThe stabilizing resistance (series voltage dropping resistance)

for a stabilizing setting of 150V canbe calculated as follows :Vstability - Vrelay

Rs = ⎯⎯⎯⎯⎯⎯Is

Vs Vr Vs Vr x Is= ⎯ - ⎯ = ⎯ - ⎯⎯⎯Is Is Is Is x IsVs VA

= ⎯ - ⎯⎯ (where VA is relay burden)Is Is2150 0.5

= ⎯⎯ - ⎯⎯0.12 (0.12)2

= 1250 - 34.7 = 1215 ohmsThe nearest standard value of 1250 ohms may be used.The 0.5 sec rating resistor Rs =Is2 R (where Ir = Vf / Rs )Vf is the peak value developed across the resistor Rs under internalcondition.

Vf = 4√(Vk3 x Rs x Ifs) x 1.3

where Ifs = IRMS of secondary fault currentVf = 4√(3003 x 1250 x 10.5) x 1.3 = 1003 voltsIR = 1003 / 1250 = 0.8 Amps for 0.5 sec.

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So , I2Rs = (0.8)2 x 1250 = 800Watts for 0.5 seccontinuos rating of the metrosil is given by

4 If⎯ x ⎯ x Vk where N is CT ratioπ N4 2100

=⎯ x ⎯⎯ x 300 = 4010 jou lesπ 200

which is well within the maximum rating of 3`` metrosil which is 43000jou les.

SettingSettingSettingSetting ofofofof fieldfieldfieldfield failurefailurefailurefailure relays:relays:relays:relays:

The general practice is to use an offset setting equal to half themachine transient reactance (Xd /2) and circle diameter equal tosynchronous reactance of the machine Xs for rotor angle upto 90° andwhen the machine cannot be operated at lead ing power factors.

For machines which can be operated at rotor angles upto 120°,the offset are modified to three quarters of the machine transientreactance (3Xd /4) and circle diameter equal to half the synchronousreactance (Xs/2).

IllustrationIllustrationIllustrationIllustration 1111:voltage ………………………………… 11 kV, 3 phase, 50

Hzoutput ………………………………… 30 MVA, 0.8 pfmachine transient reactance Xd` …… 19%machine synchronous reactance……. 200%CT ratio ………………………………… 1500/5 AVT ratio ………………………………… 11000/110 volts

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fig 5.2Offset setting:Machine transient reactance in secondary ohms = Xd% x (kV)2 x CTR

100 MVA PTR= 19 x 112 x (1500/5)100 30 (11000/110)

= 2.3 ohmRequired setting Xd /2 = 2.3/2 =1.15 ohm.

Circle diameter setting:Machine synchronous reactance in = Xs% x (kV)2 x CTR

secondary ohms 100 MVA PTR= 200 x 112 x (1500/5)100 30 (11000/110)

= 24.2 ohmRequired circle diameter Xs = 24.2 ohm

IllustrationIllustrationIllustrationIllustration 2222 ::::Afield failure relay with K1 (setting for diameter of the

impedance circle) and K2 setting for offset of the impedance circle )expressed as a percentage of nominal imped ance is used to protect asalternator with the follow ing characteristics:In = 2100 amps, Xs = 200%, Xd` = 30%Nominal primary current of CT = 2500AmpsFind the pickup current at two levels of the characteristic with nominalvoltage and phase angle of +90° for a relay with Vn =110V and In = 5A.

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Reactance at the relay terminals areXalt x In CT x Vn altXset = ⎯⎯⎯⎯⎯⎯⎯⎯⎯

In alt Vn PTXs = 200 x 2500 = 238%

2100Xd` = 30 x 2500 = 35.7 ≅ 36%

2100Suitable setting are

K1 = Xs = 238%K2 = Xd /2 = 36/2 = 18%

fig 5.3Relay nominal voltage Vn = 110VRelay nominal current In = 5ARelay nominal impedance Zn = Vn / (√3In )

=110 / (√3 x 5 ) = 12.7 ohms.Under this condition with nominal test voltage 110volts the relay wouldpickup at

IA = U x 100 = 110 x 100 = 48.1AZn K2 12.7 x 18

IB = U x 100 = 110 x 100 = 3.38AZn K2 12.7 x (238+18)

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IllustrationIllustrationIllustrationIllustration 3333 ::::PGEN = 200MW PT Ratio = 15.75kV /110VUN.GEN = 15.85kV CT Ratio = 10000 /5 AIN.GEN = 8625 Amps Un.Relay = 100VXd = 1.9808pu In.Relay = 5AXd` = 0.2428puXn = 0.2pu

fig 5.4In.Relay x IN.CT.Prix UN.VT.Sec x UN.GEN

Xsec (pu) = Xpri (pu) x⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯IN.CT.Sec x IN.GEN x Un.Relay x UN.VT.PriCTR x UN.GEN x In. Relay= Xpri (pu) x ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯PTR x IN.GEN x Un.Relay

UN.Relay 100Zn.Relay = ⎯⎯⎯⎯ = ⎯⎯ = 11.55 ohm

√3 x IN.Relay √3 x 5CTR = 2000, PTR =143.18

a) Setting impedance element for dynamic stability characteristic:1 0.2428x 2000 x 15750 x 5

Z1A = Z2A = ⎯ x ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ x 100%2 143.18 x 8625 x 100

= - 15.48% ≅ 16%2000 x 15750 x 5

Z1B = 1.9808 x⎯⎯⎯⎯⎯⎯⎯⎯ x 100%143.18x 8625 x 100

= - 252.6 ≅ 253%Z2B = Z 2B / 2 = 253/2 ≅ 126%

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fig 5.5

b) Setting impedance element for steady state and dynamic stabilitycharacteristic:

fig 5.6CTR x UN.GEN x In. Relay

Z1A = XN x ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯PTR x IN.GEN x Un.Relay2000 x 15750 x 5

= 0.2 x ⎯⎯⎯⎯⎯⎯⎯⎯ x100%143.15x 8625 x 100

= + 25.5 ≅ 26%2000 x 15750 x 5

Z1B = 1.1 x 19808 x ⎯⎯⎯⎯⎯⎯⎯⎯ x100% + Z1A

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143.15 x 8625 x 100= 277.8 + 26 = 303.888 ≅ 304%

Z 2A = - Xd` / 2 = 16% (refer part A of this example )Z 2B = 1.1 x Xd - (Xd /2)

= 277.8 - 16 ≅ 262%Tripping time

tZ1 = 1.0s ( set between 0.5 to 3sec)tZ1 = 0.25s ( set between0.5 to 3sec)

Setting of under voltage elementThe under voltage protection element is usually set to the

critical system voltage at which the generator loss of field jeopardizesthe system stability. The limit is normally 80% of the generator voltage.

UN.Gen 15750U = 0.8 x ⎯⎯⎯ = 0.8 x ⎯⎯⎯ = 88%

PTR 143.18

SettingSettingSettingSetting ofofofof voltagevoltagevoltagevoltage relays:relays:relays:relays:IllustrationIllustrationIllustrationIllustration 1111 ::::

An under voltage relay is fed from a set voltage transformers ofrated secondary Un.VT of 110volts. The relay is expected to operatewhen the voltage goes down to the level of 85% of the rated V.T.The rated voltage of the relay Un.relay = 100volts.The relay setting will be

Un.VT 110U</ %Un = ⎯⎯⎯ x 85% i.e. ⎯⎯⎯ x 85% = 93.5%

Un.relay 100

IllustrationIllustrationIllustrationIllustration 2222 ::::

An over voltage relay is to be used as a residual earth faultrelay, connected to an open delta connection of the secondary of a set ofvoltage transformers. The relay is to operate for a residual voltage of20% of the fully developed residual voltage.

Rated voltage of the relay Un.relay = 100voltsVoltage transformer ratio 6600 / 110

√3 / √3

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14

The relay setting will beUn.VT 110

U</ %Un = ⎯⎯⎯ x 20% i.e. ⎯⎯⎯ x 20% = 22.2%Un.relay 100

SettingSettingSettingSetting ofofofof voltagevoltagevoltagevoltage relaysrelaysrelaysrelays connectedconnectedconnectedconnected forforforfor earthearthearthearth fault:fault:fault:fault:

IllustrationIllustrationIllustrationIllustration 3:3:3:3:A voltage relay is to be set at 10% of the phase to neutral

voltage. Normal Voltage Vn is 6600volts and voltage transformer ratio6600/110volts.a) If the relay is supplied from a voltage transformer connected

between the neutral point and ground, the relay measures the zerosequence voltage V0 of the neutral.The voltage 10%Vn, brought to the secondary of the transformergives

10 6600 110 11V = ⎯⎯ x ⎯⎯ x ⎯⎯ = ⎯⎯ = 6.35 ≅ 7volts.

100 √3 6600 √3The operating level of the relay should be set at 7volts.

b) If the relay is supplied from an open delta, it measures the sum ofthe three phases to neutral voltage, which is by definition equal tothree times the zero sequence voltage V0 of the network.

The relay is supplied by an open delta of three phase voltagetransformerwhose ratios are

6600 / 110√3 / √3

The voltage 10%Vn brought to the secondary of any one of thevoltage transformer is equal to

10 6600 110/√3 11V = ⎯⎯ x ⎯⎯ x ⎯⎯⎯ = ⎯⎯ = 6.35 volts

100 √3 6600/√3 √3In fact the voltage applied to the relay is equal to three times

this calculated value.So relay voltage is ⇒ 3 x (11/√3) ≅ 19voltsSo the operating level of the relay should be set at 19volts or the nearesttap available.

SettingSettingSettingSetting ofofofof powerpowerpowerpower relays:relays:relays:relays:

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IllustrationIllustrationIllustrationIllustration 1:1:1:1:Active power of the generator PGEN = 185 KVA x 0.8 =1500KWNominal current of the relay IN = 5ANominal voltage of the relay VN = 100VCT ratio = 150/5APT ratio = 10KV/100VRelay to be set 3% reverse power.Reference power of the relay PR = √3 x UN x IN x CTR x PTR

= √3 x 5 x 100 x 30 x 100= 2598KW

If the relay should pick up at a reverse power of greater than 3%(referred to the active power of the generator) the adjustment value onthe relay canbe calculated as

PGEN 1500KWAdjustment Value % = ⎯⎯ x %P = ⎯⎯⎯⎯ x 3% = 2%

PR 2598KWHence the relay should be set at 2%.


Active power of the generator PGEN = 210 MWNominal current of the relay IN = 5ANominal voltage of the relay VN = 100VCT ratio = 10000/5APT ratio = 15.75KV/110VRelay to be set 1% reverse power.Reference power of the relay PR = √3 x UN x IN x CTR x PTR

= √3 x 100 x 5 x 2000 x 143.18= 248MW

PGEN 210MWAdjustment Value % = ⎯⎯ x %P = ⎯⎯⎯⎯ x 1% = 0.846%

PR 248MWSince the relay is fed from voltage transformers of 110V, the actualsetting for 1% reverse power is

U⎯ x 0.846 = 0.93 ≅ 1%UN


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16

Active power of the generator PGEN = 500KW, 400V, 50HzNominal current of the relay IN = 5ANominal voltage of the relay VN = 230VCT ratio = 1000/5ASingle phase connection, PT not required.Relay to be set 5% reverse power.Reference power of the relay PR = UN x IN x CTR x PTR

= 230 x 5 x 200 x 1= 230KW

PG / phase = 500/3 =166.66KWPGEN 166.66

So adjustment Value % =⎯⎯ x %P = ⎯⎯⎯⎯ x 5% = 3.62%PR 230

SettingSettingSettingSetting ofofofof NegativeNegativeNegativeNegative phasephasephasephase sequencesequencesequencesequence relaysrelaysrelaysrelays ::::

Negative phase sequence currents in stator of a generator due tounbalanced loads or faults induce double frequency eddy currents in therotor. These currents if allowed to persists, can cause series over heatingand the purpose of the negative phase sequence relay is to disconnect themachine before such excess temperature is reached. The timecharacteristic of the machine in order to avoid tripping.IllustrationIllustrationIllustrationIllustration 1111 ::::Negative sequence current withstand rating I2S of the generator = 10%I22t of the generator = 7.5characteristic of negative sequence relay(CTN31of English Electric Co)

I22t = K1 x K3

K1 selectable between 1 to 10Setting:

Select a plug board setting I2S equal to nearest below thegenerator continuous negative phase sequence current withstand rating

i.e. I2S = 10%Choose appropriate value of K3 from the table of relay characteristic

I2S 7.5% 10% 15% 20% 30%K3 1 1.78 4 7.1 16

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17

i.e. K3 = 1.78Choose the time multiplier setting K1 from the formula

I22t = K = K1 x K3i.e. K1 x K3 = 7.5

K1 = 7.5 / 1.8 = 4.21Select the nearest lower tap value of K1 available

i.e. K1 = 3.9Comparison of relay operating time and generator withstand time at fivetimes relay setting current :Generator withstand time tm = K / I22t = 7.5 / (5 × 0.1)2 = 30 secRelay operating time from graph of relay characteristic when K1is 1 andat 5 × setting current is 7secSo relay operating time tr = K1 × (operating whenK1 = 1)

i.e. tr = 3.9 × 7 = 27.3sec.


Generator parametersNominal current 800ACT Ratio 1000/5AContinuously permissible unbalance current I2S 40%I22t of generator 60s

Relay particularsNominal current In 5ASetting range I2t 0.02 to 0.5 × In ( trip )

I2w 0.02 to 0.5× In (warning )Time selectable T (DEF) 1 to 200sec

T (INV) 300 to 3600sec

Tripping characteristic (inverse)T

t = ⎯⎯⎯⎯(I2 / I2s)2 − 1

where t = tripping timeT = thermal time constant

Normal CT secondary current = 800 × (5/1000) = 4 Amps

Continuously permissible ⎫

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18

negative current related ⎬ I2s = 4 × ( 40/100 ) = 1.6Ampsto secondary side of CT ⎭I2s setting = ( I2s secondary ) ÷ Ini.e. = 1.6 / 5 = 0.32Time constant T for selection of tripping characteristic is

T = I22t / I2s = 60 /(0.4)2 = 375 or say 360secFor warning I2wmay be set at a lower value of 35%.

35% × IN.G 0.35 × 800I2w setting = ⎯⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯ = 0.28

CTR × In (1000/5) × 5tw may set at a fixed time of 5sec.


Generator parametersNominal current 9060ACT Ratio 10000/5AContinuously permissible unbalance current I2S 5%I22t of generator 8s

Relay particularsNominal current In 5A (TypeRARIB of ASEA)

⎛ In ⎞2operating characteristic t = ⎜ ⎯ ⎟ × K

⎝ I2 ⎠whereK is I22t constant of generator.

Relay setting current I2s = 9060 × (5/10000) = 4.53Ai.e. (4.53/ 5) × In = 0.906 InK = 8 on the relay.Continuous negative sequence withstand characteristic of 5%implies that maximum difference between the line current isabout 10%. So set the trip relay at 8% and alarm relay at 4%.

SettingSettingSettingSetting ofofofof distancedistancedistancedistance backupbackupbackupbackup protectionprotectionprotectionprotection relaysrelaysrelaysrelays forforforfor generatorsgeneratorsgeneratorsgenerators ::::

The over current relays have the disadvantage that, ongenerators the fault current reduces sharply, as the generator reactancechanges from sub-transient to transient to synchronous and is generallyinsufficient for reliable operation of the over current relays whichoperate after a time delay. The voltage control feature also becomes

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19

ineffective when AVR’s are used in conjunction with the generator,because they tend to maintain normal voltage at generator terminals.Therefore distance type relays are generally preferred.

IllustrationIllustrationIllustrationIllustration 1111 ::::System details

Rating of generator 210MW/247MVA, 15.75KV.Number of machine operating in parallel 2Length of the longest line 185 KMLine impedance /KM 0.335 ohmRating of transformer 250 MVA, Ratio 15.75/230KV% impedance 14 %

Taking under-reaching effect into consideration the impedance settingrequired to cover the entire length is

ZT + NZLwhere ZT is the impedance of the transformer and ZL is the

impedance of the longest line both in primary ohms on HV basis. N isthe number of generator in parallel.

(KV)2ZT = Zpu × ⎯⎯⎯

MVA14 2302

= ⎯ × ⎯⎯ = 29.62 ohms100 250

ZL = 185 × 0.335 = 61.98 ohmsZ T + ZL = 29.62 + 61.98 = 153.58ohms

Impedance expresses at the ⎫generator voltage level ⎭ = Z × ( VL / VH )2

= 153.58 × (15.75 / 230 )2= 0.72 ohm

CTRSetting in secondary ohms = Z × ⎯⎯

PTR10000 110

= 0.72 × ⎯⎯ × ⎯⎯5 15750

=10.057 ohmsThe relays are supplied with delta voltages and delta currents

and connected for phase fault measurements. Therefore it will have asetting of 1.15 times (derivation of which is given separately) actual

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20

setting i.e. If the actual setting is 3 to12 ohms the effective setting willbe 3.45 to13.2ohms.So the setting required on the relay = 10.057 / 1.15

= 8.745ohms.

Alternate criteria for setting :

Criteria ⇒ Set at 70% of the load imped ance.KV2 2002

Z load = ⎯⎯⎯ = ⎯⎯ = 1.004Ω/PhMVA 247

70Primary setting = 1.004× ⎯⎯ = 0.703Ω/ Ph

100CTR

Secondary setting = Z × ⎯⎯⎯PTR

10000 110= 0.703 × ⎯⎯⎯ × ⎯⎯⎯ = 9.82Ω/ Ph

5 15750Applying the effect of delta voltages and delta currents

9.82Relay setting = ⎯⎯⎯ = 8.539 ≅ 8.5Ω/Ph

1.15

SettingSettingSettingSetting ofofofof transformertransformertransformertransformer overoveroverover excitationexcitationexcitationexcitation relays:relays:relays:relays:

Generator step up transformers need protection against the riskof damage which may be caused when the transformers are operated at aflux density higher than the design values. When the power transformersare over excited the leakage flux increases and this results in heavyhysteresis and eddy current loss in non laminated parts of thetransformers. If the temperature rise due to this losses is excessive, theinsulation may be damaged and flashover may occur. These conditionsare more likely to occur when the unit is on open circuit with generatorfield energized, and machine speed is increasing, or decreasing fromsynchronous speed. Over excitation relays are designed to prevent overfluxing in generator step-up transformers with V/f characteristics.

IllustrationIllustrationIllustrationIllustration ::::

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21

fig 5.7For this specif ic example a relay (Type RATUB of asea make)

is considered.The characteristic of the relay is

Kt = 0.8+ 0.18 × ⎯⎯⎯

( m − 1)2( V/f )

where m = ⎯⎯ and(V/f)set

K = the adjustable time constant.Transformer working flux density = 1.6 Tesla (assumed )Transformer saturation flux density = 1.9 Tesla (assumed )Flux ratio = 1.9 /1.6 = 1.185 ≅ 1.1 (say).The maximum permitted excitation corresponds to

110⎯⎯ × 1.1 = 2.42V/Hz on the relay side50

So V/f start level setting =2.4 V/Hz.Calculate appropriate K factor for 150% excitation.

( V/f ) 1.5150% excitation corresponds to ⎯⎯ = ⎯ = 1.3636

(V/f)set 1.1

fig 5.8From the excitation characteristic of step up transformer (see graph)time for 150% excitation is 5.5secs.

t − 0.8 ⎛ V/f ⎞2K = ⎯⎯⎯ × ⎜ ⎯ − 1⎟

0.18 ⎝ V/fset ⎠

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22

5.5 − 0.8= ⎯⎯⎯⎯ × ( 1.3636− 1 )2 ≅ 3.4

0.18Choose K = 3 and check the operating time for different points on theexcitation curve of the transformer and plot it on the same sheet.

0.18 × 3t = 0.8 + ⎯⎯⎯⎯⎯

⎛ V/f ⎞2⎜ ⎯ − 1 ⎟⎝ V/fset ⎠

This plot should be below and very close to the excitationcharacteristic of the transformer. Otherwise choose a different K value.

The relay should be checked for operation by applying avoltage equal to

relay setting × (test frequency / normal frequency)× normalvoltageof the relay

i.e. 110volt relay with setting of 1.1 at49.5 Hz will operate at1.1× (49.5 /50) × 110 = 119.79 vols.

GeneratorGeneratorGeneratorGeneratorpolepolepolepole slippingslippingslippingslipping protectionprotectionprotectionprotection :It is not a simple operation to establish the relay setting in

relation to the machine and system parameters, since the setting dependson the machine rate of slip which is a function of the machine inertia,machine load, type of fault causing the pole slipping condition, thelength of time it taken for fault to be cleared. In general the longer thefault clearance time, the higher the in itial rate of slip. It has beenestablished by tests, that for a tandem generator the average rate of slipduring the first half slip cycle will usually be in the range of 250 to 400degrees/sec, while for cross compound units the average in itial rate ofslip will be 400 to 800 degrees/sec. However for both type of generatorthe average rate of slip during the reminder of the slip cycle will fall onthe range of 1200 to 1600 degrees/sec.

It is generally recommended that under pole slipping conditionthe machine be disconnected from the system without any delay afterthe power swing has passed on an angle of 270 degrees, which ensuresthat the voltage across the opening poles of the generator breaker are notmore than 90 degrees.

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23

Fast removal from the power system of a large generatorrunning out of step avoids the risk of loss of auxiliar ies to othergenerators which remains instep and also minimizes the disturbances toother system loads.

The general recommendation for a the setting of pole slippingrelay is as follows :

1) The angle of both the directional and the blinder unit be setat as close as possible to the system impedance angle.

2) That the blinder unit be set to operate an angle of 90° tothe total system impedance along the power swing locus.

3) That the time unit be set in accordance with expressionθ1 − θ

t = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯slip rate in degrees/sec

where t = time in secθ1 = 270°θ = Approximately 180°(angle subtended by the

unit at the point along which the powerswing locus crosses it with the twoends ofthe total system impedance line.)

4) That the over current unit be set at approximately 115% ofthe generator full load rating.

Determination of pole slipping relay setting :1) From the site data determine the follow ing with reference

to machine voltagea) Transformer reactance XT.b) Generator reactance XG (This is obtained from sub

transient reactance. ( 90% of Xd`` is effective for poleslipping.)

c) Source impedance Zs.2) Plot the above on R - X diagram to determine the line

impedance A - C.3) Determine the power swing locus. This is perpendicular

bisector of line A - B.4) Insert the directional and blinder characteristic on the

diagram. Angle to be as near the line angle as possible.5) Directional line through origin passes through point P on

the power swing locus.

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24

6) To determine the offset measure along the R - axis towhere the blinder intersects ( This give offset in primaryohms.)

7) Time settingθ1 − θ

t = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯slip rate in degrees/sec

Angle θ1 is fixed at 270°. To determine the angle θ thenfrom the point where the directional characteristiccrosses the power swing locus, link with point A and B onthe system impedance line and measure the anglesubtended this will be around 180°. For almost all typesof machines the average rate of slip after the first halfcycle will be between1200 and 1600 degrees/sec.

IllustrationIllustrationIllustrationIllustration ::::Generator rating 247MVA, 15.75KVSub transient reactance Xd`` = 21.4%Generator transformer rating 250MVA, (240/√3) /15.75KVPercentage impedance 14%Fault level of the 240 KV bus 5000 MVA.

fig 5.9

KV2Generator sub transient reactance in ohms = Xd``pu × ⎯⎯⎯

MVA21.4 15.752

= ⎯⎯ × ⎯⎯⎯100 247

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25

= 0.2149ΩFor pole slip XG = 0.9 × X

= 0.9 × 0.2149= 01934Ω

KV2Transformer reactance in ohms XT = Xpu × ⎯⎯⎯

MVA14 15.752

= ⎯⎯ × ⎯⎯⎯100 250

= 0.1389ΩFault MVA

Source impedance = ⎯⎯⎯⎯⎯⎯KV2

5000= ⎯⎯⎯

15.752= 0.0496Ω

Assume system angle = 80°Plot the values on the graph sheet.

fig 5.10

From the graph off -set = 0.185Ω

CTROff - set in secondary ohms = Primary ohm × ⎯⎯⎯

PTR

10000 110= 0.185 × ⎯⎯⎯ × ⎯⎯

5 15750= 2.584Ω

θ1 − θTime setting = ⎯⎯⎯⎯⎯

Rate of slipFrom the graph θ = 182°Assume the rate of slip applicable to machine is 1600 degrees/sec

So 270 − 182t = ⎯⎯⎯⎯

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26

1600= 55 m . sec

Over current relay setting = 115% of FLC= 4.525× 0.115= 5.2 amps

Setting of pole slipping relays with lens characteristic :

Relay type ZPT 408, BBC make.

fig5.11KV2 15.752

Load impedance = ⎯⎯⎯ = ⎯⎯⎯ = 1.004ΩMVA 247

Assume minimum load resistance RL.MIN = 85%ZA = XT + XN

= 0.1389 + 0.0496 = 0.1885ΩZB = Xd`

= 0.2149ΩZC = 0.9 XT

= 0.9 × 0.1389 = 0.125Ω

Relay setting in secondary ohmsCTR

ZA.sec = ZA × ⎯⎯PTR

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27

10000 110= 0.1885 × ⎯⎯⎯×⎯⎯⎯

5 15750= 0.1885 × 13.968 = 2.63Ω

ZB = 0. 2149 × 13.968 = 3ΩZC = 0.125× 13.968 = 1.746Ω

⎡ RL.MIN ⎤* = 180° − 2 × tan –1 ⎢1.54 × ⎯⎯⎯ ⎥

⎣ ZA +ZB ⎦⎡ 0.85 × 1.004 ⎤

= 180° − 2 × tan -1 ⎢1.54 × ⎯⎯⎯⎯⎯⎯ ⎥⎣ 0.1885 + 2149 ⎦

= 180° − 2× tan-1(3.258) = 34°( If α is less than 90° set at 90°)

setting available is 90° to 150°So set at α = 90°Set φ = 85° ( 5° higher than system angle )operation time of the relay 50 ms (fixed )

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28Page 93 / 153

6.Distance6.Distance6.Distance6.Distance relaysrelaysrelaysrelaysGeneralGeneralGeneralGeneral PrinciplePrinciplePrinciplePrinciple :

Distance protection monitors the imped ance of the protectedunit as seen from the relay location. The impedance is high when thereis no fault on the protected unit and low when there is a fault. Distanceprotection is used as a primarily protection for transmission lines, andas a backup protection for large generators, power transformer and auto-transformer.

The important advantage of distance protection in comparisonwith graded over current protection is its extremely fast operation forfaults in the first zone of protection, which is usually between 80 to 90%of the protected line. Thus faults close to the source can be tripped justquickly as faults at the end of the line. From fig 6.1 it can also be seenthat each of the distance relays RZ1 to RZ3 has several impedance zones

fig 6.1with different operating times. The second and higher zones provideback-up protection for other section of line, should either theirprotection or circuit breaker fails to clear the fault. This means thatassuming a fault on L2( fault location F2) distance relay RZ1would tripcircuit breaker-1 after the time t1II should the relay RZ2 fail to clear thefault in its first time step (operating time t2I). A fault at F3 (last 10 to15% of the line) on the other hand would be detected by distance relay

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6.Distance relays

2

RZ2 in its second Zone and tripped after the time t2II. Should this nottake place, RZ1 would trip circuit breaker-1 until the time t3III hasexpired, because the fault lies in its third impedance zone.

It follows from this explanation that discrimination isachieved in the case of distance protection by giving tripping priority tothe relay which measures the lowest impedance, i.e. the shortest distanceto the fault. It was also explained, that the first zone cannot be set tocover the full length of the line. In order to ensure proper discriminationunder all conditions it has to be set short and as a result fau8lts in thelast 10 to 15% are detected in the second Zone. This margin has to beobserved, because if the first zone were to be set to 100% of the line,faults at the beginning of the next section of the line might be detectedby mistake due to inaccuracies of the distance measurement and lineimpedance data. This apparent disadvantage is overcome by installing acommunication channel between the relay at two end of the line or, ifauto-reclosure is used, by an appropriate auto-reclosure log ic.

MeasuringMeasuringMeasuringMeasuring principle:principle:principle:principle:A simplif ied illustration of the principle of distance

measurement is given in fig 6.2 a & 6.2 b.It assumes that only the positive-sequence component of the

line impedance is measured and the zone of protectioncorresponds to

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6.Distance relays

3

Protected circuitfig 6.2 a

UL = IF.ZLAF UM =IF .ZMUL` = UL / KNU

Measuring principleImpedance characteristic

fig 6.2 bthe first zone of any of the relays e.g. RZ1, as shown in the fig 6.1. Thismeans that all faults occurring between station-A and zone limit G are in

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6.Distance relays

4

the operating area of the distance relay characteristic, and fault outsidethis area are not detected in the first zone. It is the relays job to decidewhether the fault impedance lies within or outside the characteristic.One possibility of doing this is to compare the absolute values of twovariables, one which is the voltage drop UL across the impedance of thefault loop Z LAF and the other is the voltage drop UM across what isreferred to as the replica impedance ZM connected to the secondary ofthe CT. The reflected impedance of the replica on the primary side ofthe current transformer corresponds to the line impedance Z1I i.e. theimpedance of the first zone. The two variables at the input of theamplitude comparator M are thus given by the relationship

UL IK . ZLAFUL` = ⎯⎯ = ⎯⎯⎯⎯

1KNUKNU

IK . ZMUM = IK` . ZM = ⎯⎯⎯⎯

2KNI

where IK = primary fault currentIK` = secondary value of the

primary faultcurrent IK i.e. IK`

= IK / KNIKNI = nominal CT ratioKNU = nominal V.T ratioZ LAF = impedance of the line loop

betweensection A and

fault location FZM = replica impedance, in this

case ZM = Z1IAs mentioned above , the absolute values of the voltages UL`

and UM are compared and therefore for a limit of the first zone (G infig6.2b)

| UL` | = | UM |3

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6.Distance relays

5

This means that the signals applied to the amplitude comparator M arebalanced and the condition for tripping is not fulfilled. If both sides ofequation 1 and 2 are divided by IK and IK` equation 3 becomes

ZL` = ZM = Z1IThe graphical representation of the characteristic in the R/X plane is acircle with its center at the origin of the system coordinates ( fig 6.2b).This is also the location of the beginning of the protected line. Since theratio between X and R of the line is known, the impedance of the linebetween A and B can be represented by the impedance vector ZAB if acorresponding scalar value is assumed. The operating area of theprotection lies inside the circular characteristic and the restrained areaoutside. Thus tripping condition is expressed by the follow ing inequality.

|ZM| > |ZL |4which in terms of voltage means |UL| < |UM| and is the case assumedin fig 6.2b. The fault location F lies in the first zone and therefore |ZLAF|is less than Z LAG = ZM. Since the same fault current flows through bothimpedances, the voltage |UL | is also less than UM and protection trips.For faults beyond the limit of the first zone, |Z L | > |ZM| and thereforealso |UL | > |UM| and the protection restrains.

The consideration up to the present were concerned with theapplication of distance protection to radial lines. If fault energy can besupplied from both ends, distance relays have to be installed at bothterminal stations as shown in fig 6.3.

Assuming that the relay should only trip for faults in theforward direction, the two relays must have operating characteristicwhich are capable of making a clear directional decision. Aprotection

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6.Distance relays

6

fig 6.3with an impedance characteristic according to fig 6.2b cannot fulfill thisrequirement, because the impedance vector seen by the relays RZI andRZ2 for the fault locations F1 and F3 are also in the operating area. Forrespective relays, these faults are in the third quadrant of the R/X planeas shown for RZ1 in fig c. to prevent a distance relay from operating forfaults in the reverse direction, its characteristics in the R/X plane mustnot encroach on the third quadrant. The desired mode of operation is

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6.Distance relays

7

achieved by relays having a “mho” characteristic (mho - inverse of ohm)as shown in fig 6.3. A characteristic of this kind can be obtained usingboth amplitude and phase comparator. In the case of phase comparatorthe relevant input signals are

S1 = UL and S2 = UM − ULfrom which can be derived

S1 = Z L and S2 = ZM − Z LThe parameter mentioned by the phase comparator is the phase

angle β between the input signals. The limit values of β = 90°, for thisangle describes the circular operating characteristic i.e. β < 90° forfaults outside the first zone and β > 90° for faults inside the first zone.

Apart from the circular mho characteristic shown above toillustrate the basic principle of distance protection, operatingcharacteristic which shapes more closely suited to the conditions inpracticeare achieved by solid - state distance relays.

SettingSettingSettingSetting recommendationrecommendationrecommendationrecommendation :Positive sequence impedance:-

The line impedance converted to the secondary side of theinstrument transformerwith the follow ing formula

CT RatioZsec = Z pri × ⎯⎯⎯⎯

PT RatioCTpri

VTsec = Z pri ×⎯⎯ × ⎯⎯

CTsec VTsecDue to errors in relays, current transformers, voltage

transformers and inaccuracies inline data an under reaching zone 1reach is normally set at 80% of the calculated line impedance. For thesame reason an over reaching zone 2 reach is set to cover the protectedline plus 50% of the adjacent shortest line or 120% of the protected linewhichever is greater. The zone 2 setting should never exceed 80% of theimpedance corresponding to the protected line plus the first zone reachof the shortest adjacent line and the impedance corresponding to theprotected line plus the impedance of the maximum number oftransformers in parallel on the bus at the remote end of thee protectedline. The zone 3 reach is set to cover 120% of the protected line plus the

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6.Distance relays

8

longest adjacent line. It should at least two times the zone 1 setting. Thezone 3reverse function can also be used for backup protection of the busbars behind the relay and is typically at 25% of zone 1 setting.Zero sequence compensation :-

The measuring loop at single phase to ground faults consists oftwo impedances, the positive sequence impedance the positive sequenceimpedance Z1and the zero sequence Z0 . To measure the impedance uptofault point correctly, a compensation factor known as zero sequencecompensation or residual compensation must be applied.

Z0 − Z 1This compensation is calculated as Kn = ⎯⎯⎯⎯

3Z1Fault resistance:-

The single phase to ground fault is of great importance indistance protection as normally more than 70% of the faults ontransmission line are single phase to ground. The fault resistantcomposed of two components, the arc resistance and the tower footingresistance. The arc resistance as per Warington formula is

28707 × lRarc = ⎯⎯⎯⎯

I1.4where I = the actual fault current

l =length of the arc in meters.“l’’ is approximately 2 -3 times the arc foot spacing. The tower

footing resistance must be calculated or measured for the specif ic casesas the variation of this parameter is very large.

Distance relay cannot be used to detect very high resistiveground faults as the reach is limited by the load impedance and loadtransfer.Margin between resistive reach and load impedance in the ase ofquadrilateral characteristic :-

the maximum permissible resistive reach for any zone shouldbe checked to ensure that sufficient setting margin between the relayboundary and minimum load impedance. The minimum load impedanceis calculated as

(KV)2Z load = ⎯⎯ = ohm/phase.

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6.Distance relays

9

MVASince the safety margin is required to avoid load encroachment

under three phase fault conditions and to avoid healthy phase relayoperation under combined three phase load and ground faults, both thephase to phase and phase to ground characteristic should be considered.

The load impedance is a function of minimum operationvoltage and maximum load current under emergency conditions.

UminZ load.min = ⎯⎯⎯⎯ = ohm /phase

√3 × ImaxTo avoid load encroachment the resistive reach should be set

less than 80% of the minimum load imped ance.Power swing blocking :-

power swings are variations in power flow which occur whenthe voltage of generator at different points of the power system sliprelative to each other to cater for changes of load magnitude anddirection or as a result of fault and their subsequent clearance.

fig 6.4The results of a power swing may cause the impedance

presented to the distance relay to move away from the normal load areaand into the relay characteristic. In the case transient power swing it isimportant that the distance relay should not trip and should allow thepower system to return to a stable condition. For this reason an optionalpower swing blocking relay is provided in distance protection schemes.Since the power swing is a balanced three phase condition a single phase

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6.Distance relays

10

relay with off-set mho characteristic is provided to encircle the zone 3characteristic. The time taken for the power swing to cross thecharacteristic is measured. If this time is longer than the set time of thepower swing blocking relay, then the power swing has occurred andtripping is blocked if the time is less than the set time, then the faultoccurred and normal tripping is allowed.

When circular characteristic such as mho or off-set mho typesare used, the power swing blocking characteristic should have adiameter of atleast 1.3 times the diameter of the zone 3 characteristic.Normally this is set at 150% of the zone 3 setting in the forward reachand 25%of the forward reach in the reverse reach.

Zone time setting :- Normally no time delay will be provided for zone 1tripping. Zone 2 delay should be set to discriminate with primaryprotection of the next line section including circuit breaker trip time.Generally a zone 2 time delay setting of 0.2 to 0.4 is satisfactory and thezone 3 time delay should be twice that of zone 2time setting.


fig 6.5Length of the protected line180KMLength of the adjacent shortest line 220KMlength of the adjacent longest line 380KMPositive sequence impedance0.0264 + j0.3289 ohm/KM

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6.Distance relays

11

i.e.0.33∠85.42° ohm/KMZero sequence impedance0.08887 + j1.1094 ohm/KM

i.e.1.113∠85.42° ohm/KMLine angle85.42°Relay angle75°CT ratio1000 /1 APT ratio400,000/110V

CTR1000 110Transformation ratio = ⎯⎯ = ⎯⎯ × ⎯⎯⎯= 0.275

PTR1 400,000Primary impedance of the

protected line = 0.33 × 180= 59.4 ohmSecondary impedance of the

protected line = Pri.Impedance × Transformation ratio

= 59.4 × 0.275 = 16.335ohm

Sec. Impedance along the Pri.Imprelay angle =

⎯⎯⎯⎯⎯

Cos(θL− θR)

16.335

= ⎯⎯⎯⎯⎯⎯ = 16.61ohmCos(85.42 − 75 )Primary impedance of the

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6.Distance relays

12

adjacent shortest line = 0.33 × 220= 72.6 ohmSecondary impedance of the

adjacent shortest line = 72.6 × 0.275= 19.965 ohmSec. Impedance along the 19.965

relay angle =⎯⎯⎯⎯⎯ = 20.299 ohm

Cos(85.42 − 75)Primary impedance of the

adjacent longest line = 0.33 × 380= 125.4 ohmSecondary impedance of the

adjacent longest line = 125.4 × 0.275= 34.485 ohmSec. Impedance along the 34.485

relay angle =⎯⎯⎯⎯⎯ = 35.063 ohmCos(85.42 − 75)Zone 1 reach = 80%of the protected line

= 0.8 ×16.61 = 13.287ohmZone 2 reach = 120% of theprotected line

orprotected

line + 50%of the adjacentshortest

line whichever is greater= 1.2 ×

16.61 or (16.61+ 0.5 × 20.299)= 19.932

or 26.7998 ohmZone 2 reach adopted = 26.7998ohmZone 3 forward reach = 120% of (protected line +longest

Page 105 / 153

6.Distance relays

13

adjacent line)= 1.2

( 16.61 + 35.0632 )= 62

ohmZone 3 reverse reach = 25% of Zone 1 reach

=0.25 × 13.287 =3.3217ohmStarter setting = 125%of Zone 3setting

i.e. Forward reach = ( 0.25 ×Z3F ) + Z3F=

( 0.25 × 62 ) + 62 = 77.5 ohmReverse reach = ( 0.25 × Z 3F ) + Z3R

=(0.25× 62 ) + 3.3217 = 18.82ohmPower swing blocking relay setting = 150%of the Zone 3 setting

i.e. Forward reach = ( 0.5 ×Z3F ) +Z3F

= ( 0.5 × 62 ) + 62 = 93 ohmReverse reach = ( 0.5 × Z 3F ) +

Z3R=

(0.5× 62 ) + 3.3217 = 34.32ohm

Z0 − Z 1Neutral compensation = ⎯⎯⎯⎯ ×1003ZI

1.113 − 0.33

= ⎯⎯⎯⎯⎯ × 100 =79.09%

3 ×0.33Time setting : Zone 2 time = 0.4 sec

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6.Distance relays

14

Zone 3 time = 0.8sec

IllustrationIllustrationIllustrationIllustration 2222 :::: (reactance relay)

fig 6.6Length of the protected line70KMLength of the adjacent shortest line 30KMlength of the adjacent longest line 55KMPositive sequence impedance 0.016+ j0.4 ohm/KMZero sequence impedance0.04 + j1.0 ohm/KMLine angle tan −1(0.4/0.16)67.3°Relay angle75° (power swing

blocking relay )CT ratio300 /1 APT ratio110,000/110V


PTR1 110,000

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6.Distance relays

15

Primary impedance of theprotected line = (0.16 + j0.4) × 70 =

11.2 + j28 ohmSecondary impedance of the

protected line = Pri. Impedance ×Transformation ratio

= 11.2 + j28ohm × 0.3 = 3.36 + j8.4 ohmPrimary impedance of the

adjacent shortest line = (0.16+ j0.4) × 30 =4.8 + j12 ohmSecondary impedance of the

adjacent shortest line = (4.8 + j12) × 0.3 =1.44 + j3.6 ohmPrimary impedance of the

adjacent longest line = (0.16 + j0.4) × 55= 8.8 + j22 ohmSecondary impedance of the

adjacent longest line = (8.8 + j22) × 0.3 =2.64 + j6.6 ohmZone 1 reach (reactance only) = 80% of the protected line

=0.8 × 8.4 = 6.72 ohmZone 2 reach (reactance only) = 120% of the protected line

or

protected line + 50%of the adjacentshortest

line whichever is greater=

1.2 × 8.4 or (8.4+ 0.5 × 3.6 )=

10.08 or 10.2 ohmZone 2 reach adopted = 10.2 ohmZone 3 forward reach = 120% of (protectedline + longest

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6.Distance relays

16

adjacent line)=

1.2 ( 8.4 + 6.6 )=

18 ohmZone 3 reverse reach = No reverse reach setting inreactance relaysMho starting unit :

Zone 3 reach in reactance = 18 ohmLine angle = 67.3°

18Zone 3 reach in impedance = ⎯⎯⎯ = 19.51ohmsin67.3°

6.72Zone 3 reach in impedance = ⎯⎯⎯ = 7.285ohm

sin67.3°Starter setting (impedance) = 125%of Zone 3setting

i.e. Forward reach = ( 0.25 ×Z 3F ) + Z 3F=

( 0.25 ×19.51 ) + 19.51= 24.38ohmReverse reach = ( 0.25 × Z 3F ) + 0.25

Z1F=

(0.25× 19.51) + ( 0.25 ×7.285)= 6.7

ohmPower swing blocking relay setting = 150%of the Zone 3 setting

i.e. Forward reach = ( 0.5×Z 3F ) + Z3F=

( 0.5 × 19.51) + 19.51= 29.265ohm

or say 30 ohm

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6.Distance relays

17

Reverse reach = ( 0.5 × Z 3F ) + Z3R=

(0.5× 19.51) + 7.285 = 13.397ohm

or say 14 ohm

X0 − X1Neutral compensation = ⎯⎯⎯⎯ ×1003XI

1 − 0.4

= ⎯⎯⎯⎯ × 100 = 50%

3 ×0.4

IllustrationIllustrationIllustrationIllustration 3333 :::: (Quadrilateral characteristic relay)

fig 6.7Length of the protected line135KMLength of the adjacent shortest line 16KMlength of the adjacent longest line 140KMPositive sequence impedance 0.08 +j0.4125 ohm/KMZero sequence impedance0.263 + j1.51 ohm/KM

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6.Distance relays

18

Line angle tan −1(0.4125/0.08)79°Max.load transfer250MVACT ratio800 /1 APT ratio230,000/110V


PTR1 230,000Primary impedance of the

protected line = (0.08 +j0.4125) × 135

=10.8 + j55.68 ohmSecondary impedance of the

protected line = Pri. Impedance ×Transformation ratio

= 10.8 + j55.68 × 0.3826

= 4.13 + j21.3 ohmPrimary impedance of the

adjacent shortest line = (0.08+ j0.4125) × 16=

1.28 + j6.6 ohmSecondary impedance of the

adjacent shortest line = (1.28+ j6.6) × 0.3826=

0.4889+ j2.525 ohmPrimary impedance of the

adjacent longest line = 0.08 + j0.4125) ×140

=11.2 + j57.75 ohm

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6.Distance relays

19

Secondary impedance of theadjacent longest line = (11.2 + j57.75) ×

0.3826=

4.285 + j22.09 ohm

28709 ×lArc resistance = ⎯⎯⎯

I1.4assume arc length = 10 meters and minimum fault current 1KA

28709 ×10Rarc = ⎯⎯⎯⎯ =18.1 ohm or

say 20ohm10001.4

Reactance reach settingZone 1 reach (reactance only) = 80% of the protected line

= 0.8 × 21.3 = 17.04 ohm

Zone 2 reach (reactance only) = 120% of the protected line

or

protected line + 50%of the adjacentshortest

line whichever is greater= 1.2

× 21.3 or (21.3+ 0.5 × 2.525)=

25.56 or 22.5625 ohmZone 2 reach adopted = 25.56ohmZone 3 forward reach = 120% of (protected line+ longest

adjacent line)=

1.2 ( 21.3 + 22.09 )

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6.Distance relays

20

=43.39 ohmMaximum transformer at station B is 200MVA

Per unit impedance Zpu = 0.14pu(average)

MVA 200Fault MVA = ⎯⎯⎯ = ⎯⎯ = 1428MVA

Zpu 0.14(KV)2

2302Fault impedance = ⎯⎯⎯⎯⎯ = ⎯⎯ = 37.07 ohm

MVA(fault) 1428The over reach of 4.26 ohm (i.e. 25.56 − 21.3) of zone 2 at

station A is much smaller than the above impedance. Hence the settingchosen for zone 2 is O.K.

( KV)2 2302Z load = ⎯⎯⎯⎯⎯ = ⎯⎯ = 211.6 ohm

MVA(load) 250

X0 − X1Zero sequence compensation factor = ⎯⎯⎯

3X1

1.51− 0.4125

= ⎯⎯⎯⎯⎯ =0.8863 × 0.4125Resistive reach

Rfn = 20 ohm/loop for ground faultRf = 3 ohm/phase for phase to phase faultRfn = 20 × 0.3826 = 7.652 ohm (secondary )Rf = 3 × 0.3826 = 1.148 ohm (secondary)

Maximum setting of resistive reach isZ load (MIN)

Rmax < 0.8 × ⎯⎯⎯⎯⎯1.2

80.96< 0.8 × ⎯⎯⎯ = 54 ohm

1.2Required setting

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6.Distance relays

21

XRset = Rf + ⎯⎯⎯ for ph-ph faults

tan φR

X KN RNset = Rf + ⎯⎯⎯ +⎯⎯⎯for ph-ph faultstan φR tan φN

where φR = tan −1(X/R) = tan −1(0.4125/0.08) = 79°and

X0 − X11.51 − 0.4125

φN = tan −1 ⎯⎯⎯ = tan −1

⎯⎯⎯⎯⎯⎯ = 80.53°R0 − R10.263 - 0.08

Zone 1

17.04Rset = 1.148+ ⎯⎯⎯ = 4.46 ohm

tan79

17.04 0.886×17.04RNset = 7.652 + ⎯⎯⎯ +

⎯⎯⎯⎯⎯

tan 79 tan 80.53= 7.652 + 3.312 +

2.52 = 13.48ohmZone 2

25.56Rset = 1.148 + ⎯⎯⎯

= 6.116ohmtan 79

25.56 0.886×25.56RNset = 7.652 + ⎯⎯⎯ +

⎯⎯⎯⎯⎯⎯

tan 79 tan 80.53= 16.4 ohm

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6.Distance relays

22

Zone 343.39

Rset = 1.148 + ⎯⎯⎯ =9.582 ohmtan 79

43.39 0.886×43.39RNset = 7.652 + ⎯⎯⎯ +

⎯⎯⎯⎯⎯

tan 79 tan 80.53= 22.5

ohmAs the required reach 22.5 ohm is much less than the minimum

load impedance, there is no risk for load encroachment.

Page 115 / 153

7.Motor7.Motor7.Motor7.Motor ProtectionProtectionProtectionProtection

SettingSettingSettingSetting ofofofof motormotormotormotorprotectionprotectionprotectionprotection relaysrelaysrelaysrelays ::::

IllustrationIllustrationIllustrationIllustration ::::Motor rating 1700KWVoltage 6.6 KVFull load current 148.7 AmpsCT ratio 200/1 AmpsStarting time 7 secStarting current 6 × FLCStalling current 6 × FLClocked rotor withstand time Hot 10sec

Cold 19secThermal characteristic

Thermal setting :CT sec

Is = FLC × ⎯⎯⎯CT pri

1= 148.7× ⎯⎯ = 0.7435 or say 0.76

400Thermal curve selection (τ ) :

The characteristic of modern thermal relays an exponentialcharacteristic up to two times Ieq (equivalent thermal current ) and anadiabatic characteristic beyond two times Ieq.The operating time of relay below 2×Ieq is given by

⎡ Ieq 2 − K3(IL)2 ⎤t = τ log e ⎢ ⎯⎯⎯⎯⎯ ⎥⎣ Ieq 2 − (K Is)2 ⎦

whereτ is the heating time constantIL equivalent pre fault load currentIs thermal current setting of the relayK 1.03 (minimum operating current )K3 constant used to provide different

× rated current 1.4 2 3 4 6Hot time 400 200 90 48 19Cold time 390 170 70 31 10

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7.Motor protection

2

hot/cold ratioIeq equivalent current of over load defined by

the equation Ieq = (I12 − K2 I22)½

During cold start pre fault load current IL = 0⎡ Ieq 2 ⎤

t = τ log e ⎢ ⎯⎯⎯⎯⎯ ⎥⎣ Ieq 2 − (K Is)2 ⎦

dividing the equation by Is2⎡ ( Ieq / Is ) 2 ⎤

t = τ log e ⎢ ⎯⎯⎯⎯⎯ ⎥⎣ (Ieq / Is)2 − K2 ⎦⎡ Ieq.th 2 ⎤

t = τ log e ⎢ ⎯⎯⎯⎯⎯ ⎥ (1)⎣ Ieq.th 2 − K2 ⎦

where Ieq.th = Ieq / Is which is the equivalent loadcurrent expressed in terms thermal setting Is.

SoIeq × FLC × CTsec

Ieq.th = ⎯⎯⎯⎯⎯⎯⎯⎯Is × CTpri

The operating time of the relay above 2 × Ieq is given byK2τ

t = ⎯⎯Ieq 2

K2τt = ⎯⎯ (2)

Ieq.th 2Using the equation (1) and (2) calculate τ at three different points, say1.4, 3, and 6 (locked rotor current )of the cold withstand characteristic ofthe motor.Ieq.th at 1.4 times Ieq is given by

1.4× 148.7× 1Ieq.th = ⎯⎯⎯⎯⎯⎯ = 1.37

0.76 × 200tc

τ1.4 = ⎯⎯⎯⎯⎯⎯⎯⎯⎡ Ieq.th 2 ⎤

log e ⎢ ⎯⎯⎯⎯⎯ ⎥⎣ Ieq.th 2 − K2 ⎦

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7.Motor protection

3

400τ1.4 = ⎯⎯⎯⎯⎯⎯⎯⎯

⎡ 1.372 ⎤log e ⎢ ⎯⎯⎯⎯⎯ ⎥

⎣1.37 2 − 1.032⎦= 479.78 sec or say 7.996 minutes

Ieq.th at 3 times Ieq is given by3 × 148.7 × 1

Ieq.th = ⎯⎯⎯⎯⎯⎯ = 2.9350.76 × 200

tc × Ieq.th2τ3 = ⎯⎯⎯⎯K2

90 × 2.9352τ3 = ⎯⎯⎯⎯ =730.7 sec

1.032or say 12.18 minutes.

Ieq.th at 6 times Ieq is given by6 × 148.7 × 1

Ieq.th = ⎯⎯⎯⎯⎯⎯ = 5.870.76 × 200

tc × Ieq.th2τ6 = ⎯⎯⎯⎯K2

19 × 5.872τ6 = ⎯⎯⎯⎯ = 617 sec

1.032or say 10.28 minutes.

The τ selected for the motor should be lower than all the above threevalues. Since τ1.4 is only 7.996 select a τ of 7. So the relay operatingtime at 1.4 times Ieq and a τ of 7 is given by

⎡ 1.372 ⎤7 × 60 × log e ⎢ ⎯⎯⎯⎯⎯ ⎥ = 350 sec

⎣1.37 2 − 1.032⎦The value of τ selected should be such that the gap between the

motor withstand time and relay operating time at any multiple of thesetting current is greater than 15%of the motor withstand time or 4 secwhichever is greater.

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7.Motor protection

4

Since the gap between motor withstand time and relayoperating time doesn’t meet the above requirement when τ is 7, thevalue of τ should be selected to the next lower value.

i.e. a τ of 6Hot/Cold ratio selection :

The value of hot/cold ratio should be such that the relay hotcharacteristic comes below the motor hot characteristic and the gapbetween the hot withstand time and relay hot operating time at anymultiple of the setting current is greater than 15% of the motorwithstand time or 4 sec whichever is greater.

Select a hot /cold ratio if 0.67. The relay hot /cold operatingtime, motor hot /cold operating time etc. at various multiples of Ieq.thare shown below in the table. If the above criteria is not met choose adifferent hot/cold ratio.

Instantaneous short circuit protection :starting current CTsec

Isc = 1.3 × ⎯⎯⎯⎯⎯⎯⎯×⎯⎯⎯thermal setting CTpri6 × 147.8 1

= 1.3× ⎯⎯⎯⎯ ×⎯⎯0.76 200

= 7.63 Amps.If the relay has the provision of doubling the set range of Isc

during staring, then Isc should be set at (7.63 / 2) i.e. 3.82 Amps.

τ Ieq Ieq.th motorwithstand time

relay operatingtime

gap

7 1.4 1.37 400 350 50

Sl.No Multiples of Issetting

Motor withstandtime

Relay operatingtime

Ieq Ieq.th cold hot cold hot1 1.4 1.37 400 390 300.14 201.092 2.0 1.96 200 170 116.83 78.273 2.03 1.98 198 168 96.84 64.884 3 2.93 90 70 44.34 29.75 4 3.913 48 31 24.44 16.716 6 3.87 19 10 11.08 7.43

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7.Motor protection

5

Unbalance current setting :For definite time mode set at less than

1 starting current CTsecI2 = ⎯×⎯⎯⎯⎯⎯⎯⎯×⎯⎯⎯

3 thermal setting CTpriIf more than 1 set 1.00

1 6 × 147.8 1= ⎯×⎯⎯⎯⎯×⎯⎯

3 0.76 200= 1.96 Amps.So set at 1.00

Time setting :a) For circuit breaker control motor set at 100 m. sec.b) For contactor control motor with backup fuse set at 1.2

times fuse operating time at maximum contactor rating anda minimum of 0.5 sec.

c) The element can be set at inverse mode if I2 during startingis more than100%. For inverse time mode adopt 30 %setting.

Locked rotor protection :Current setting:

full load current CTsecIstl = 2 × ⎯⎯⎯⎯⎯⎯⎯×⎯⎯⎯

thermal setting CTpri147.8 1

= 2 × ⎯⎯ ×⎯⎯0.76 200

= 1.956Amps.Set at 2

Time setting :a) for starting time less than hot stall withstand time, set less

than hot locked rotor withstand time and more than startingtime.

b) for starting time more than hot stall withstand time, usespeed switch and protection set at 0.5 × hot stall withstandtime.

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7.Motor protection

6

c) for high inertia motors like ID fan motors etc. time shouldbe 25% of starting time.

For the given motor time setting = starting time + 2 seci.e. tstl = 7 + 2 = 9sec

This is more than hot stall withstand time. So adopt a setting of7 sec only.

For slip ring motorscurrent setting

starting current CTsecIstl = 1.25× ⎯⎯⎯⎯⎯⎯⎯×⎯⎯⎯

thermal setting CTpritime settingtstl = 0.5 × hot stall withstand time.

Prolonged start protection :current setting

starting current at 80% voltage CTsecIst = 0.8× ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ×⎯⎯⎯

thermal setting CTpriIf the starting current at reduced voltage is not available from

the manufacturer, assumestarting current at 80% voltage = 0.8 × starting current at 100% voltageSo

0.8 × starting current CTsecIst = 0.8× ⎯⎯⎯⎯⎯⎯⎯⎯⎯×⎯⎯⎯

thermal setting CTpri0.8 × 6 × 148.7 1

Ist = 0.8× ⎯⎯⎯⎯⎯⎯× ⎯⎯0.76 200

= 3.76 AmpsFor slip ring motors

full load current CTsecIst = 1.2× ⎯⎯⎯⎯⎯⎯⎯×⎯⎯⎯

thermal setting CTpriTime setting

set at 1.2 × starting timeor

max. starting time + 5sec whichever is greater.For the given motor set at tst = 7 + 5 =12 sec.

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7.Motor protection

7

Earth fault setting :For residual and solid ly earthed system

CTsecIe/f = 0.1 × full load current × ⎯⎯⎯

CTpriFor insulated system

CTsecIe/f = 0.8 × full load current × ⎯⎯⎯

CTpriAssuming this is a resistance earthed system, current setting

1Ie/f = 0.1 × 148.7× ⎯⎯

200= 0.0743Amps. I.e. 74 mA

This may be set at 40 mA

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7.Motor protection

8Page 123 / 153

8.Commissioning8.Commissioning8.Commissioning8.Commissioning TestsTestsTestsTestsCTCTCTCT polaritypolaritypolaritypolarity teststeststeststests:

In many protection schemes the relative polarity of CT is moreimportant and therefore the correct polarity must be ensured before theyare connected. The normal current flow convention of CT is when theprimary flows from P1 toP2, the secondary current flows from S1to S2in the external circuit as shown in the fig 8.1

fig 8.1fig 8.2

A simple way to check the polarity is flick test which uses abattery, center zero ammeter, and a push button connected as shown inthe fig8.2. When the push button is pressed the ammeter makes apositive flick and a negative flick when the push button is released, ifthe assumed polarity is correct.

CTCTCTCT ratioratioratioratio checkcheckcheckcheck :The polarity and ratio check can be carried out by the primary

inject ion tests described below:The circuit for the test is shown in the fig 8.3. A short

circuit is placed across the phases on one side of the CT’s and singlephase inject ion is carried out on the other side. One ammeter A2 isprovided in the phase side of the CT circuit and another ammeter A3 isprovided in the residual circuit of the CT. current is injected through theprimary conductor and measured on test set ammeter A1. The secondary

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8.Commissioning Tests

2

current is measured on the ammeter A2. The ratio of the value on A1 tothat on A2 should approximate to the ratio marked on the CT nameplate.

fig 8.3

The ammeter A3 in the residual circuitwill read few milliamps if the CTpolarity is correct. The reading A3 twice that of A2 shows wrongpolarity. Single phase inject ion should be carried out for each pair ofphases.

If an earth fault relay with low setting is provided in theresidual circuit its operating should be temporarily short circuited duringthe test to avoid overheating.

PolarityPolarityPolarityPolaritycheckcheckcheckcheck usingusingusingusing seriesseriesseriesseries injectioninjectioninjectioninjection ::::The test circuit is as shown in the fig8.4 & 8.5 below.

Inject current approximate to rated current. The ammeter A3will read three times A2 if the polarity is correct. If the reading ofammeter A3 is equal to A2,the relative polarity of one of the CT iswrong.

Page 125 / 153


3

Test circuit for bar primary CTsfig 8.4

Test circuit for ring type CTsfig 8.5

CTCTCTCT magnetizingmagnetizingmagnetizingmagnetizing curve:curve:curve:curve:

Page 126 / 153


4

A variable voltage supply is connected across the secondaryterminals of the CT as shown in the fig 8.6. If the knee point voltage ishigher than local mains supply use a step up interposing transformer.

Fig 8.6The voltmeter is connected in such away that the ammeter

doesn’t read the voltmeter current. It will be useful to find roughly thevoltage at which saturation starts by increasing the voltage until there isa large increase in current for a small change in voltage. From this it canbe decided at what values to take readings to give sufficient point to plotthe curve. Initially take readings in large steps (say 20 or 30 volts) andsmaller steps (say 10 or 5 volts) when saturation starts. Tabulate andplot the readings.

The knee point voltage is the voltage at which an increase of10% voltage will result in 50% increase in magnetizing current .From the above reading

For 100vols Ie = 46 mAFor (100 +10 ) % i.e. 100 volts Ie = 65mA

65 − 46i.e. increase = ⎯⎯⎯ × 100 = 41%

46For 100vols Ie = 65 mAFor (110 +10 ) % i.e. 110 volts Ie = 105 mA

105 − 65

Voltage 0 20 40 60 80 90 100 110 120current in mA 0 1 16 23 31 37 46 65 105

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i.e. increase = ⎯⎯⎯ × 100 = 61%65

Hence knee point voltage is between100 and 110 volts.VoltageVoltageVoltageVoltage transformertransformertransformertransformer polaritypolaritypolaritypolarity checkcheckcheckcheck ::::

The polarity of the voltage transformer can be checked with thetest described for current transformer. While checking, the batteryshould be connected to the primary winding and the polarity windingconnected to the secondary winding. If the VT is of capacitor type thenthe polarity of the transformer at the bottom of the capacitor should bechecked.

VoltageVoltageVoltageVoltage transformertransformertransformertransformer ratioratioratioratio checkcheckcheckcheck ::::the check can be carried-out when the main circuit is first made

alive. The secondary winding voltage of the VT is compared withsecondary winding voltage of a VT already connected to the same bus-bars.

VoltageVoltageVoltageVoltage transformertransformertransformertransformer phasingphasingphasingphasing checkcheckcheckcheck ::::The secondary connection of a three phase voltage transformer

or a bank of three single phase transformers should be checked forcorrect polarity as detailed below:

fig 8.7

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With the main circuit alive the secondary voltage of the VTbetween phase to phase and phase to neutral should be measured forcorrect magnitude. If an existing proven VT is available in the sameprimary system and the secondary earthing is employed then the correctphasing can be proved by measuring the voltage between the respectivephases of both the VT’s. This voltage should be normally little or novoltage if the phasing is correct.

If the VT has a broken delta territory winding then a checkshould be made for the voltage across the broken terminals Vn and VL,when a rated balanced three phase voltage applied to the primary ofthe VT. The broken delta voltage should be below 5volts when ratedburden connected.

SecondarySecondarySecondarySecondary injectioninjectioninjectioninjection tests:tests:tests:tests:OverOverOverOver currentcurrentcurrentcurrent andandandand earthearthearthearth faultfaultfaultfault relays:relays:relays:relays:

fig 8.8The over current relays can be checked using an over current

test set. Alternatively a test setup as shown in the fig 8.8 may be used.While checking over current relays care should be taken to avoidexcessive currents flowing through the coil for a long time to avoidexcessive heating.

Instantaneous over current relays are checked for minimumcurrent required for operation (pick-up) of the relay and maximum

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current at which the relay resets (drop-off) for each current setting of therelay.

Definite time over current relays are first checked for pick-upand drop-off as detailed above with timer setting at zero. Then the timedelay should be checked at 1.3 times (approx.) the current setting.

For inverse time over current relays the follow ing checksshould be performed.1) Check that starting current ( i.e. the current at which the disc just

begins to move but does not completes its travel to close thecontact.) is within specif ied limits.

2) Check that the closing current ( i.e. the minimum current at whichat which the disc completes its travel to close its contact) iswithin its limits.

3) Check the operating time of the relay at 2,5,10 times the currentsetting (plug setting ). These times should be within the tolerancelimits. (For electronic and numerical relays the checking ofstarting and closing current does not arise.

DifferentialDifferentialDifferentialDifferential relaysrelaysrelaysrelays ::::The sensitivity of the differential relay can be checked using a

over current test set as described below:

fig 8.9 fig 8.10While checking unbiased differential relay the current setting

should be checked by slowly raising the current until the relay pickup.

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This should be checked for all the setting and finally for the requiredsetting.

The biased differential relay can be checked with the overcurrent test set by inject ing current simultaneously to the operating coiland the bias coil as shown in fig8.10 above, at different points of thebias characteristic by adjusting the rheostats R1and R2. Record theoperating currents for different values of bias current. The ratio of theoperating current A1 to the through current A2 is approximate to thevalue of the bias slope of the relay.

The relative value of the resistor R2 and R1 should roughly thesame as the ratio of the bias slope of the relay.

e.g. for bias slope of 30%, R2/R1 = 0.3To determine the position of the characteristic inject a current

of I2Amps through R2 and adjust R1 until the relay operates.Let the operating current be I1

I2 + ( I1 + I2 )I1Then the average bias IB =⎯⎯⎯⎯⎯ = I2 + ⎯

22

I1Then the bias slope of the relay is ⎯

IBRepeat the test for other values of currents to plot the curve.

NegativeNegativeNegativeNegative phasephasephasephase sequencesequencesequencesequence relaysrelaysrelaysrelays ::::

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fig 8.11The current setting of the negative phase sequence relay is

usually expressed in terms of negative phase sequence currents. Therelay can be checked with over current test set. If the relay is providedwith an external filter current inject ion should be made before the filter.

The relay can be checked for phase to earth fault or phase toearth fault simulation. If the relay is checked for phase to earth faultsimulation 1/3rd of the injected current is negative sequence current. Ifthe relay is checked for phase to phase fault simulation 1/√3 times of theinjected current is negative sequence current.

For phase to earth fault simulation of a relay rated for 1ampsand I2s setting of 10%

10I2s = ⎯⎯ × 1 = 0.1Amps

100In order to obtain 0.1Amps negative phase sequence current a

test current of 3 × 0.1 Amps should be injected into each phase in turn.Test current = 0.1 × 3 × 2 = 0.6 Amps.

The factor 2 is for two times the set current and thefactor 3 is for phase

to neutral simulation.For phase to phase fault simulation of a relay rated for 5amps

and I2s setting of 7.5%

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7.5I2s = ⎯⎯ × 5 = 0.375Amps

100In order to obtain 0.375Amps negative phase sequence current,

a test current of √3 × 0.375 Amps should be injected into each pair ofphase in turn.

Test current = 0.375 × √3 × 4 = 2.6 Amps.The factor 4 is for four times the set current and the

factor √3 is for phase to phasesimulation.

If the relay has an inverse time characteristic check operatingtime at different points of the curve.

DirectionalDirectionalDirectionalDirectional relaysrelaysrelaysrelays ::::

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fig 8.12The directional characteristic and maximum torque angle of

many relays can be checked with a phase shif ting transformer as shownin the fig8.12. The phase shif ting transformer permits the phase angle ofthe relay voltage to be varied with respect to the relay current. Adjustthe current flowing through the relay current coil to correspond to ratedcurrent and the applied voltage to the voltage coil corresponds to therated voltage. Rotate the phase shif ter till the phase angle meter readszero degree (unity PF). Check the relay is in operated position. If notreverse the voltage or current connection and check the operation of therelay at UPF. Rotate the phase shif ter in the clockwise direction until therelay contact opens and then rotate back in the anti clockwise directionand note the phase angle meter reading φ1,when the relay contact justclose. Continue to rotate in the anti clockwise direction till the relaycontact open again, then rotate backwards in clockwise direction andnote the phase angle meter reading φ2, when the contact just close. Themaximum torque angle line of the relay is given by the bisector of φ1 −φ2 and the angle between this line and the zero degree line (UPF)givesthe maximum torque angle of the relay which should be within ±4° ofthe declared maximum torque angle (MTA).

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fig 8.13

DistanceDistanceDistanceDistance relays:relays:relays:relays:Distance relay s are required to measure impedance accurately

over a wide range of current and voltages. The impedance measurementof these relays can be checked using a phase shif ting transformerexplained earlier for checking directional relays. These equipmentmeasures the impedance at static conditions i.e. the current increasedslowly or the voltage is decreased slowly until the relay operates. Whiletesting high speed distance relays, it is important to apply simulatedfault conditions suddenly, because during the fault condition the relayvoltage falls suddenly from nominal voltage to fault voltage and thecurrent from the load value to fault value accompanied by changes inphase angles. So in order to check the distance relays under dynamiccondition most of the manufacturers of relay have developed dynamictest set. The basic principle of operation of the dynamic test set is asdetailed below:

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fig 8.14Normal working voltage is applied to the relay through a

voltage auto transformer. When the fault contactor is closed the faultcurrent If is applied to the relay. At the same instant the voltage appliedto the relay collapses to the fault voltage VL. The magnitude of the faultvoltage VL depends on the ratio of the source impedance to lineimpedance (Zs / ZL). The voltage auto transformer has 10%and 1%tapings to allow the line impedance Z L to be matched to the relay ohmicsetting. By varying the source impedance Z S the relay ohmic setting canbe checked over a wide range of voltage and current magnitudes.

Usually the distance relay ohmic setting is checked at actualline angle. For this a choke and resistance with tapping of the lineimpedance ZL are provided. Angles other than line impedance anglesmay be chosen for checking the characteristic of mho or reactance relays .The points usually checked on the mho and reactance relaycharacteristic are shown in fig 8.15.

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fig 8.15

PrimaryPrimaryPrimaryPrimary injectioninjectioninjectioninjection teststeststeststests ::::OverOverOverOver currentcurrentcurrentcurrent andandandand earthearthearthearth faultfaultfaultfault relaysrelaysrelaysrelays ::::

The sensitivity of the over current relays can be checked byphase to phase inject ion. Phase to phase inject ion is explained earlier inCT ratio and polarity checks. For checking earth fault relays singlephase inject ion should be used as shown in fig8.16.

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DirectionalDirectionalDirectionalDirectional relaysrelaysrelaysrelays ::::The directional feature of the directional over current, earth

fault, distance and other wattmetric relays can be checked by the use ofload current.

The phase fault directional over current relays can be checkedby use of load current when the load current is appreciable and thedirection is in no doubt. The relay contact should close in the operatingdirection and open in the reverse direction. The direction of the currentto the relay can be reversed by cross connecting the voltage or currentleads to the relay through the test plug as shown in fig 8.17.

fig 8.17Earth fault over current relays are usually fed from broken delta

voltage of the three phase voltage transformer and residual current of themain current transformers. Under normal condition the relay is notenergized, so it is necessary to simulate the earth fault condition. Withone phase of the voltage transformer disconnected and short circuitedwhile the current transformers of the other two phases disconnectedand short circuited as shown in fig8.18, simulates a condition of earthfault on the phase from which voltage is disconnected. The relay shouldoperate when the load current flows in the operating direction.

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fig 8.18Mho relays canbe made to operate with load current flowing in

the operating direction by removing the restraint voltage from the relay.In most modern distance relays a convenient switch link or plug isprovided o the relay to facilitate this test. The removal of the restraintvoltage from the mho relay changes the mho circle characteristic into aplain directional characteristic with the same characteristic angle asshown in the fig8.19.

fig 8.19With lagg ing current flowing in the operating direction the relay will

operate a soon as the restraint voltage is removed. The relay should be

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then checked for non operation for load flow in the opposite direction.This can be done by inserting a test plug with crossover connection ofthe current transformer leads as shown in the fig 8.20 to reverse thecurrent flow to the relay.

fig 8.20

GeneratorGeneratorGeneratorGeneratordifferentialdifferentialdifferentialdifferential protectionprotectionprotectionprotection ::::

The sensitivity of this type of differential scheme can bechecked by primary inject ion test as shown in the fig 8.21.

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fig 8.21Primary current is passed through one of the main CT and

slowly raised until the relay operates. This gives the true operatingcurrent necessary in the primary to cause operation, that includes themagnetizing current of the CT shunting the relay. the voltage acrossthe relay coil and stabilizing resistance should also be measured at thetime of operation of the relay to check the voltage developed by themain CT that causes the relay operation. The sensitivity check should becarried out by inject ing through the CT primary of each phases in turn tocheck all units of the relay.

Another method of checking the sensitivity is by use ofmachine itself to supply the primary current by carefully controlling themachine excitation from a low value to the required primary current. Forthis the main circuit breaker is lef t open and a three phase short circuit isprovided in the generator link cubicle to simulate an internal fault asshown in fig 8.22.

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fig 8.22The machine is then run up to full speed and the excitation is raisedslowly until the three elements of the relay operates. The suitability ofthis type of checking depends on the type of excitation and degree ofcontrol available.

The stability of the generator differential protection can bechecked by providing a short circuit at the bus side of the generatorcircuit breaker as shown in fig8.23.

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fig 8.23The circuit breaker should be closed and excitation of the machineraised until full load current is circulated in the primary circuit. Theoperating coil of the relay should be short circuited and an ammeter oftwice the rated current of the CT should be used initially until thecorrect polarity of the CT is confirmed. Wrongly connected or opencircuitedCT will producehigher spill current reading in the ammeter. Ifthe ammeter reads very little current when circulating full load in theprimary circuit, the ammeter may replaced with a higher sensitive meterto measure the spill current which will in the order of milli-amperes.

The stability of motor and generator protection can be checkedby primary injection as detailed below :

fig 8.23(a)

Current injected through R & y phases as shown in fig 8.23(a)keeping the R & Y windings short circuited. While primary current isflowing, the relay should maintain stability and the reading of theammeters should read few milli-amps only. Wrongly connected oropen circuited CT will produce higher spill currents. It is advisable toshort circuit the relay coils and stabilizing resistors throughout the test.Repeat the test for Y & B and B & R phases to confirm the stability inall the phases.

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TransTransTransTransformerformerformerformer biasedbiasedbiasedbiased differentialdifferentialdifferentialdifferentialprotectionprotectionprotectionprotection :The sensitivity of the relay can be checked with a single phase

primary injection test set as described for generator differentialprotection. The test can also be carried out by inject ing current throughone CT to simulate earth fault or two CTs to simulate phase fault asshown in fig 8.24 & 8.25.

Earth fault simulation

fig 8.24

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Phase fault simulationfig 8.25

The stability of the transformer differential protection can bechecked by circulating full load current through the main CT’s primarywindings. The best source of power for this test is a generator ofadequate current rating. Full load current can be circulated through theby putting a shot circuit at one side of the transformer external to theprotection. Connect the terminals of the generator to the other side of thetransformer and slowly excite the generator at rated speed until full loadcurrent flows through the primary winding of the transformer.Approximately 12% of the transformer winding rated voltage will haveto be raised to circulate full load current.

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fig 8.26If there is no machine available to circulate full load current

through the transformer, connect the transformer to the local bus barsthat have voltage rating of approximately 5% to 12% of one of thewinding. The transformer short circuit impedance must be known inorder to calculate the bus-bar voltage required to circulate full loadcurrent. If both the above possibility is not available of protection can bechecked when the transformer is first put in load.

When the load current flowing through the transformerwindingthe protective relay should remain stable and the spill current in theoperating coil of the relay should be very small, provided the CT ratiosassociated with each winding of the transformer are chosen correctlyand the transformer tap changer is at the nominal tap. Measure the spillcurrent through the relay during the load test with the tap changer at itsmaximum and minimum tap. This spill current expressed as apercentage of load current indicates the minimum amount of bias ofthe relay required to maintain stability for through faults. If the relayoperates on load with the bias set correctly, the circuit diagram shouldbe checked for correct CT connection according to the vector group ofthe transformer being protected. The possible errors in diagram or actualwiring and the respective spill current are tabulated below.

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RestrictedRestrictedRestrictedRestricted earthearthearthearth faultfaultfaultfault relayrelayrelayrelay ::::The sensitivity of the restricted earth fault (REF) protection can

be checked by inject ing current using a single phase test set througheach of the main CT in turn as shown in fig 8.27.

Sensitivity checkfig 8.27

POSSIBLE ERRORS PILOTCURRENT

S

OPERATINGCURRENTS

R Y B R Y B1 One set of CT reversed I I I 2I 2I 2I2 Three pilots transposed I I I √3I √3I √3I3 Y&B pilots transposed I I I - √3I √3I4 One set of CT reversed and

three pilots transposedI I I I I I

5 One set of CT reversed andY&B pilots transposed

I I I 2I I I

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Stability checkfig 8.28

While carrying out this test measure the voltage developed across therelay coil and stabilizing resistance so as to check the voltage by themain CT to relay operation.

The stability of the REF protection can be checked by inject ingcurrent through the neutral CT and each phase CT in turn as shown infig8.28. If the protection is combined with differential protection theconnection to the differential relay should be short circuited. Whileprimary current is flowing the relay should maintain stability and thereading in ammeter connected in series with the relay coil should be fewmilliamperes only. It is advisable to short circuit the relay coil and thestabilizing through the test.

NegativeNegativeNegativeNegative phasephasephasephase sequencesequencesequencesequence relays:relays:relays:relays:The sensitivity of the NPS relay can be checked by injecting

current using a single phase test set through each pair of phases the inturn as shown in fig 8.29.

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fig 8.29NPS relays are usually calibrated in NPS currents only. To check therelay for phase to phase fault simulation, inject √3 times the settingcurrent i.e. 1/√3 times the injected current is the negative phasesequence current.

To test the relay at four times the set current for a NPS settingof 7.5% and a CT ratio of 1000/5A with phase to phase faultsimulation, inject a primary current of

7.5⎯⎯ ×1000 × √3 ×4 i.e. 520 Amps.

100For phase to earth fault simulation the same test current should be

7.5⎯⎯ ×1000 × 3 ×4 i.e. 900 Amps.

100The relay should also be checked for load test when it is first put onservice. When the load is balanced and phase sequence of the currentsupplied to the relay is correct, the relay should not operate. To confirmthe relay for operation on NPS current, twoof the main CT inputs to therelay should be transposed. This can be done by either using a test pluginserted into the relay or by reconnecting the leads too the terminals ofthe relay. in this case all the phase current entering the relay is NPScurrent and the relay will operate if the current is above the set current.

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Bus-barBus-barBus-barBus-bar protectionprotectionprotectionprotection ::::In bus-bar protection scheme all the current transformers are

connected to form a circulating current scheme. So the ratios of al theCTs should be same and connected with correct polarity. The polarityand ratio of the bus-bar protection scheme can be checked by choosingone CT circuit as reference after checking its ratio and polarity and useit to check the remaining CT circuits.

To check the polarity and ratio of the reference CT provide ashort in the bus-bars as shown in fig 8.30.

fig 8.30Inject current through R and Y phase of the bus-bars. The ratio

of the ammeter reading A1 to that of A2 should approximate to the ratiomarked on the CT name plate. The ammeter A3 connected in the neutraltest link will read a few milliampere if the CT polarity is correct.Reading A3 twice that of A2 showswrong polarity. Repeat the test withcurrent inject ion in Y and B phases bus-bars. During the test theoperating coil of the relay and stabilizing resistance should be shortcircuited to avoid overheating as they are not continuously rated. Aconvenient way to do this is by manually operating the supervisionrelay which in turn short circuit the bus- wires. If the scheme of

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duplicate bus bar type in the CTs are switched between the main andreserve bus wires by isolator auxiliary switches, then it is advisable tocarryout the test with bus bar isolators in both the position in order tocheck the wiring.

After checking the ratio and polarity of the reference CTs, theother CTs can be checked against the reference CT. This is carried outby providing a temporary short circuit on the test circuit and inject ing aprimary current in R & Y phases of the reference circuit as shown in thefig 8.31

fig 8.31The ammeter A2 provide the ratio check and A3 check the polarity asexplained earlier. Repeat the test with current inject ion in Y & B phases.If the scheme is a duplicate bus bar type the test should be carried outwith the bus bar isolators in both the positions to check the wiring.

The sensitivity of the bus bar protection relay can be checkedby passing single phase primary current through one CT only as shownin fig 8.32 and measuring the current necessary to cause operation of the

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fig8.32circulating current relay with maximum number of CTs in id le shunt.If the scheme provided with an overall check feature, the sensitivity ofthis can be checked in the same manner and at the same time. Thesensitivity check should be carried out for each discriminating zone. Avoltmeter should be connected across the relay coil and stabilizingresistor in order to check the correctness of value of stabilizing resistor.

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Table 1Relay location Total impedance

in Zpu

Fault current in Ampsat 3.3KV base

Total loadcurrent at3.3KVbase

CT.Ratio

CT. Ratioto thebase of33kv

Relay current setting

Min Max Max Min percent primarycurrent at3.3kv base

D 0.1013 0.2795 17268 6257 700 250/5 833/5 100 833C 0.0815 0.1013 21466 17268 1500 500/5 1666.6/5 100 1666.6B 0.0065 0.0815 269160 21466 6000 150/1 6000/1 100 6000A 0.0029 0.0065 603292 269160 20000 500/1 20000/1 100 20000

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Documents

21252735 Protection System