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Relay hardware is becoming even more standardised, to the point at

which versions of a relay may differ only by the software they contain.

This accurate prediction in the preface to the Third Edition of the ProtectiveRelay Application Guide (PRAG), 1987, has been followed by the rapiddevelopment of integrated protection and control devices. The change intechnology, together with significant changes in Utility, Industrial andCommercial organisations, has resulted in new emphasis on Secondary SystemsEngineering.

In addition to the traditional role of protection & control, secondary systems

are now required to provide true added value to organisations.

When utilised to its maximum, not only can the integration of protection &control functionality deliver the required reduction in life-time cost of capital,but the advanced features available (Quality of Supply, disturbance recordingand plant monitoring) enable system and plant performance to be improved,increasing system availability.

The evolution of all secondary connected devices to form digital controlsystems continues to greatly increase access to all information available withinthe substation, resulting in new methodologies for asset management.

In order to provide the modern practising substation engineer with referencematerial, the Network Protection& Automation Guide provides a substantially

revised and expanded edition of PRAG incorporating new chapters on all levelsof network automation. The first part of the book deals with the fundamentals,basic technology, fault calculations and the models of power system plant,including the transient response and saturation problems that affectinstrument transformers.

The typical data provided on power system plant has been updated andsignificantly expanded following research that showed its popularity.

The book then provides detailed analysis on the application of protectionsystems. This includes a new Chapter on the protection of a.c. electrifiedrailways. Existing chapters on distance, busbar and generator protection havebeen completely revised to take account of new developments, includingimprovements due to numerical protection techniques and the applicationproblems of embedded generation. The Chapter on relay testing andcommissioning has been completely updated to reflect modern techniques.Finally, new Chapters covering the fields of power system measurements,power quality, and substation and distribution automation are found, to reflectthe importance of these fields for the modern Power System Engineer.

The intention is to make NPAG the standard reference work in its subject area- while still helping the student and young engineer new to the field. We trustthat you find this book invaluable and assure you that any comments will becarefully noted ready for the next edition.

1 Introduction

N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e 3

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Introduction 3.1

Vector algebra 3.2

M anipulation of complex quantit ies 3.3

Circuit quantit ies and conventions 3.4

Impedance notation 3.5

Basic circuit laws, 3.6theorems and network reduction

References 3.7

3 F u n d a m e n t a l T h e o r y

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N e t w o r k P r o t e c t i o n & A u t o m a t i o n G u i d e 1 7

3.1 INTRODUCTION

The Protection Engineer is concerned with limiting theeffects of disturbances in a power system. Thesedisturbances, if allowed to persist, may damage plantand interrupt t he supply of electric energy. They aredescribed as faults (short and open circuits) or powerswings, and result from natural hazards (for instancelightning), plant failure or human error.

To facil it ate rapid removal of a disturbance from a power

system, the system is divided into 'protection zones'.Relays monitor the system quantities (current, voltage)appearing in these zones; if a fault occurs inside a zone,the relays operate to isolate t he zone from the remainderof the power system.

The operating characteristic of a relay depends on theenergizing quantities fed to it such as current or voltage,or various combinations of these two quantities, and onthe manner in which the relay is designed to respond tothis inf ormation. For example, a directional relaycharacteristic would be obtained by designing the relayto compare the phase angle between voltage and current

at the relaying point . An impedance-measuringcharacteristic, on the other hand, would be obtained bydesigning the relay to divide voltage by current. Manyother more complex relay characteristics may beobtained by supplying various combinations of currentand voltage to the relay. Relays may also be designed torespond to other system quantities such as frequency,power, etc.

In order to apply protection relays, it is usually necessaryto know the limiting values of current and voltage, andtheir relative phase displacement at the relay location,for various types of short circuit and their position in the

system. This normally requires some system analysis forfaults occurring at various points in the system.

The main components that make up a power system aregenerating sources, transmission and distributionnetworks, and loads. Many t ransmission and distributi oncircuits radiate from key points in the system and thesecircuit s are control led by circuit breakers. For thepurpose of analysis, the power system is treated as anetwork of circuit elements contained in branchesradiating from nodes to form closed loops or meshes.The system variables are current and voltage, and in

3 Fundamental T heor y

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3

Fundamen

talTheory

1 8

steady state analysis, they are regarded as time varyingquant it ies at a single and constant frequency. Thenetwork parameters are impedance and admittance;these are assumed to be linear, bilateral (independent ofcurrent direction) and constant f or a constant f requency.

3.2 VECTOR ALGEBRA

A vector represents a quantity in both magnitude anddirection. In Figure 3.1 the vector OP has a magnit ude|Z| at an angle with the reference axis OX.

Figure 3.1

It may be resolved into two components at right anglesto each other, in this case x and y. The magnitude orscalar value of vectorZis known as the modulus |Z|, andthe angle is the argument, or amplitude, and is writtenas arg.Z. The conventional method of expressing a vectorZ

is to write simply |Z|.

This form completely specifies a vector for graphicalrepresentation or conversion into other forms.

For vectors to be useful, they must be expressedalgebraically. In Figure 3.1, the vector

Z is the resultant

of vectorially adding its components x and y;algebraically this vector may be written as:

Z= x + jy Equation 3.1

where the operator j indicates that the component y isperpendicular to component x. In electricalnomenclature, the axis OC is the 'real' or 'in-phase' axis,and the vertical axis OY is called the 'imaginary' or'quadrature' axis. The operator j rotates a vector anti-clockwise through 90. If a vector is made to rotate ant i-clockwise through 180, then t he operator j hasperformed its function twice, and since the vector hasreversed its sense, then:

j x j or j2 = -1

whence j = -1

The representation of a vector quantity algebraically interms of i ts rectangular co-ordinates is called a 'complexquantity'. Therefore,x + jy is a complex quantity and isthe rectangular form of the vector |Z| where:

Equation 3.2

From Equations 3.1 and 3.2:Z= |Z| (cos + jsin ) Equation 3.3

and since cos and sin may be expressed inexponential form by the identities:

it follows thatZmay also be written as:

Z= |Z| ej Equation 3.4

Therefore, a vector quantity may also be representedtrigonometrically and exponentially.

3.3 MANIPULATIONOF COM PLEX QUAN TITIES

Complex quantities may be represented in any of thefour co-ordinate systems given below:

a. Polar Z

b. Rectangular x + jy

c. Trigonomet ric |Z| (cos + jsin )

d. Exponential |Z| ej

The modulus |Z| and the argument are together knownas 'polar co-ordinates', and x and y are described as'cartesian co-ordinat es'. Conversion between co-ordinate systems is easily achieved. As the operator jobeys the ordinary laws of algebra, complex quanti t ies inrectangular form can be manipulated algebraically, ascan be seen by the following:

Z1 +

Z2 = (x1+x2) + j(y1+y2) Equation 3.5

Z1 -

Z2 = (x1-x2) + j(y1-y2) Equation 3.6

(see Figure 3.2)

cos

= e ej j

2

sin

= e e

j

j j

2

Z x y

yx

x Z

y Z

= +( )

=

=

=

2 2

1

tan

cos

sin

Figure 3.1: Vector OP

0

Y

X

P

|Z |

y

x

q

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3

Fundamen

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Equat ion 3.7

3.3.1 Complex variables

Some complex quanti t ies are variable wi th, for example,time; when manipulating such variables in differentialequations it is expedient to write the complex quantityin exponential form.

When dealing with such functions it is important toappreciate that the quantity contains real and imaginarycomponents. If i t is required to investigate only one

component of the complex variable, separation intocomponents must be carried out after the mathematicaloperation has taken place.

Example: Determine the rate of change of the realcomponent of a vector |Z|wtwith time.

|Z|wt= |Z| (coswt+ jsinwt)

= |Z| ejwt

The real component of the vector is |Z|coswt.

Differentiating|Z| e jwtwith respect to time:

=jw|Z| (coswt+ jsinwt)

Separating into real and imaginary components:

Thus, the rate of change of the real component of avector |Z|wt is:

- |Z| w sinwt

d

dt Z e Z w wt jw wt jwt( )= +( )sin cos

d

dtZ e jw Z e

jwt jwt

=

Z Z Z Z

Z

Z

Z

Z

1 2 1 2 1 2

1

2

1

21 2

= +

=

3.3.2 Complex Numbers

A complex number may be defined as a constant thatrepresents the real and imaginary components of aphysical quantit y. The impedance parameter of anelectric circuit is a complex number having real andimaginary components, which are described as resistance

and reactance respectively.Confusion often arises between vectors and complexnumbers. A vector, as previously defined, may be acomplex number. In this context, it is simply a physicalquantity of constant magnitude acting in a constantdirection. A complex number, which, being a physicalquantity relating stimulus and response in a givenoperation, is known as a 'complex operator' . In thiscontext, it is distinguished from a vector by the fact thatit has no direction of its own.

Because complex numbers assume a passive role in anycalculation, the form taken by the variables in the

problem determines the method of representing them.

3.3.3 Mathematical Operators

Mathematical operators are complex numbers that areused to move a vector through a given angle withoutchanging the magnitude or character of the vector. Anoperator is not a physical quantity; it is dimensionless.

The symbol j, which has been compounded withquadrature components of complex quantities, is anoperator that rotates a quantity anti-clockwise through

90. Another useful operator is one which moves avector anti-clockwise through 120, commonlyrepresented by the symbol a.

Operators are distinguished by one further feature; theyare the roots of uni ty. Using De Moivre's theorem, thenth root of unity is given by solving the expression:

11/n = (cos2m + jsin2m)1/n

where m is any integer. Hence:

where m has values 1, 2, 3, ... (n-1)

From the above expressionj is found to be the 4th rootand a the 3rd root of unity, as they have four and threedistinct values respectively. Table 3.1 gives some usefulfunctions of the a operator.

12 2

1/ cos sinnm

nj

m

n= +

Figure 3.2: Addit ion of vectors

0

Y

X

y1

y2

x2x1

|Z1|

|Z2|

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1=1+ j0 = ej0

1+ a + a2= 0

Table 3.1: Properties of the a operator

3 .4 CIRCUIT QUANTITIESAND CONVENTIONS

Circuit analysis may be described as the study of theresponse of a circuit to an imposed condition, forexample a short circuit . The circuit variables are currentand voltage. Conventionally, current flow result s fromthe application of a driving voltage, but there iscomplete duality between the variables and either maybe regarded as the cause of the other.

When a circuit exists, there is an interchange of energy;a circuit may be described as being made up of 'sources'and 'sinks' for energy. The parts of a circuit are describedas elements; a 'source' may be regarded as an 'active'element and a 'sink' as a 'passive' element. Some circuitelements are dissipative, that is, they are continuoussinks for energy, for example resistance. Other circuitelements may be alternately sources and sinks, forexample capacitance and inductance. The elements of acircuit are connected together to form a network havingnodes (terminals or junctions) and branches (seriesgroups of elements) that form closed loops (meshes).

In steady state a.c. circuit theory, the ability of a circuit

to accept a current flow resulting from a given drivingvoltage is called the impedance of the circuit . Sincecurrent and voltage are duals the impedance parametermust also have a dual, called admittance.

3.4.1 Circuit Variables

As current and voltage are sinusoidal functions of time,varying at a single and constant frequency, they areregarded as rotating vectors and can be drawn as planvectors (that is, vectors defined by two co-ordinates) ona vector diagram.

ja a=

2

3

a a j =2 3

1 32 =a j a

1 3 2 =a j a

a j ej2

4

31

2

3

2= =

a j ej

= + =12

3

2

2

3

For example, the instantaneous value, e, of a voltagevarying sinusoidally with time is:

e=Emsin(wt+) Equation 3.8

where:

Em is the maximum amplitude of the waveform;

=2f, the angular velocity, is the argument defining the amplitude of thevoltage at a time t=0

At t=0, the actual value of the voltage isEmsin . So ifEm is regarded as the modulus of a vector, whose

argument is , thenEmsin is the imaginary componentof the vector |Em|. Figure 3.3 il lustrates this quantityas a vector and as a sinusoidal function of time.

Figure 3.3

The current resulting from applying a voltage to a circuitdepends upon the circuit impedance. If the voltage is asinusoidal function at a given frequency and theimpedance is constant the current will also varyharmonically at the same frequency, so it can be shownon the same vector diagram as the voltage vector, and isgiven by the equation

Equation 3.9

where:

Equation 3.10

From Equations 3.9 and 3.10 it can be seen that theangular displacement between the current and voltagevectors and the current magnitude |Im|=|Em|/ |Z| isdependent upon the impedance

Z . In complex form the

impedance may be writtenZ=R+jX. The 'real

component', R, is the circuit resistance, and the

Z R X

X LC

XR

= +

=

=

2 2

1

1

tan

iE

Zwt

m= + ( )sin 3

Fundamen

talTheory

2 0

Figure 3.3: Representat ionof a sinusoidal funct ion

Y

X' X0

Y'

e

t= 0

t

|Em| Em

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'imaginary component',X, is the circui t reactance. Whenthe circuit reactance is inductive (that is, wL>1/wC), thecurrent 'lags' the voltage by an angle , and when it iscapacitive (that is, 1/wC>wL) it 'leads' the voltage by anangle .

When drawing vector diagrams, one vector is chosen as

the 'reference vector' and all other vectors are drawnrelative to the reference vector in terms of magnitudeand angle. The circuit impedance |Z| is a complexoperator and is distinguished from a vector only by thefact that it has no direction of it s own. A furt herconvention is that sinusoidally varying quantities aredescribed by their 'effective' or 'root mean square' (r.m.s.)values; these are usually written using the relevantsymbol without a suffix.

Thus:

Equat ion 3.11

The 'root mean square' value is that value which has thesame heating effect as a direct current quantity of thatvalue in the same circuit, and this definition applies tonon-sinusoidal as well as sinusoidal quantities.

3.4.2 Sign Conventions

In describing the electrical state of a circuit, it is oftennecessary to refer to the 'potential difference' existingbetween two points in the circuit . Since wherever such

a potential difference exists, current will flow and energywill either be transferred or absorbed, it is obviouslynecessary to define a potential difference in more exactterms. For this reason, the terms volt age rise and voltagedrop are used to define more accurately the nature of thepotential difference.

Voltage rise is a rise in potential measured in thedirecti on of current flow between two points in a circuit .Voltage drop is the converse. A circuit element wi th avolt age rise across it acts as a source of energy. A circuitelement with a voltage drop across it acts as a sink ofenergy. Volt age sources are usually active circuit

elements, while sinks are usually passive circuitelements. The posit ive direction of energy flow is fromsources to sinks.

Kirchhoff's first law states that the sum of the drivingvoltages must equal the sum of the passive voltages in aclosed loop. This is il lustrated by the fundamentalequation of an electric circuit :

Equat ion 3.12

where the terms on the left hand side of t he equation arevolt age drops across the circuit elements. Expressed in

iRLdi

dt Cidt e+ + =

1

I I

E E

m

m

=

=

2

2

steady state terms Equation 3.12 may be written:

Equation 3.13

and this is known as the equated-voltage equation [3.1].

It is the equation most usually adopted in electricalnetwork calculations, since it equates the driving

voltages, which are known, to the passive voltages,which are functions of the currents to be calculated.

In describing circuits and drawing vector diagrams, forformal analysis or calculations, it is necessary to adopt anotation which defines the positive direction of assumedcurrent flow, and establishes the direction in whichposit ive volt age drops and volt age rises act. Twomethods are available; one, the double suffix method, isused for symbolic analysis, the other, the single suffix ordiagrammatic method, is used for numericalcalculations.

In the double suffix method the positive direction ofcurrent f low is assumed to be from node a to node b andthe current is designated Iab . With the diagrammaticmethod, an arrow indicates the direction of current flow.

The voltage rises are positive when acting in thedirection of current f low. It can be seen from Figure 3.4that

E1 and

Ean are positive voltage rises and

E2 and

Ebn are negative volt age rises. In the diagrammat icmethod their direction of action is simply indicated by anarrow, whereas in t he double suff ix method,

Ean and

Ebn

indicate that there is a potential rise in directionsna and nb.

Figure 3.4 Methods or representing a circuit

E I Z =

3

Fundamen

talTheory

(a) Diagrammatic

(b) Double suffi x

a b

n

abI

Z3

Z2Z1

E1

Zan

Zab

Ean

Zbn

Ebn

E2

E1-E2=(Z1+Z2+Z3)I

Ean-Ebn=(Zan+Zab+Zbn)Iab

I

Figure 3.4 Methods of representing a circuit

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Voltage drops are also positive when acting in thedirection of current f low. From Figure 3.4(a) it can beseen that (

Z1+

Z2+

Z3)

I is the total voltage drop in the

loop in the direction of current flow, and must equate tothe total volt age rise

E1-

E2. In Figure 3.4(b), the volt age

drop between nodes a and b designatedVab indicates

that point b is at a lower potential t han a, and is positivewhen current flows from a to b. Conversely Vba is anegative volt age drop.

Symbolically:Vab =

Van -

Vbn

Vba =

Vbn -

Van Equation 3.14

where n is a common reference point.

3.4.3 Power

The product of the potential difference across and the

current through a branch of a circuit is a measure of therate at which energy is exchanged between that branchand the remainder of the circuit . If the potentialdifference is a positive voltage drop, the branch ispassive and absorbs energy. Conversely, if the potent ialdifference is a positive voltage rise, the branch is activeand supplies energy.

The rate at which energy is exchanged is known aspower, and by convention, the power is positive whenenergy is being absorbed and negative when beingsupplied. With a.c. circuit s the power alternates, so, toobtain a rate at which energy is supplied or absorbed, itis necessary to take the average power over one wholecycle.If e=Emsin(wt+) and i=Imsin(wt+-), then the powerequation is:

p=ei=P[1-cos2(wt+)]+Qsin2(wt+)

Equation 3.15where:

P=|E||I|cos and

Q=|E||I|sin

From Equation 3.15 it can be seen that the quantity P

varies from 0 to 2P and quantit y Q varies from -Q to +Qin one cycle, and that the waveform is of twice theperiodic frequency of the current voltage waveform.

The average value of the power exchanged in one cycleis a constant, equal to quantity P, and as this quantity isthe product of the voltage and the component of currentwhich is 'in phase' with the voltage it is known as the'real' or 'active' power.

The average value of quantity Q is zero when taken overa cycle, suggest ing t hat energy is stored in one half -cycleand returned to the circuit in the remaining half-cycle.Q is the product of voltage and the quadrature

component of current, and is known as 'reactive power'.

As P and Q are constants which specify the powerexchange in a given circuit, and are products of thecurrent and voltage vectors, then if

S is the vector

productE

Iit follows that with

Eas the reference vector

and as the angle betweenEand

I:

S = P + jQ Equation 3.16

The quantityS is described as the 'apparent power', and

is the term used in establishing the rating of a circuit.S has units of VA.

3.4.4 Single- Phase and Polyphase Systems

A system is single or polyphase depending upon whetherthe sources feeding it are single or polyphase. A sourceis single or polyphase according to whether there are oneor several driving voltages associated wi th it . For

example, a three-phase source is a source containingthree alternating driving voltages that are assumed toreach a maximum in phase order, A, B, C. Each phasedriving voltage is associated with a phase branch of thesystem network as shown in Figure 3.5(a).

If a polyphase system has balanced voltages, that is,equal in magnitude and reaching a maximum at equallydisplaced time intervals, and the phase branchimpedances are identical, it is called a 'balanced' system.It will become 'unbalanced' if any of the abovecondit ions are not sati sf ied. Calculat ions using abalanced polyphase system are simplified, as it is only

necessary to solve for a single phase, the solution for theremaining phases being obtained by symmetry.

The power system is normally operated as a three-phase,balanced, system. For t his reason the phase volt ages areequal in magnitude and can be represented by threevectors spaced 120or 2/3 radians apart, as shown inFigure 3.5(b).

3

Fundamen

talTheory

2 2

Figure 3.5: Three-phase systems

(a) Three- phase system

BC B'C'

N N'

EanEcn Ebn

A 'A

Phase branches

Directionof rotation

(b) Balanced system of vectors

120

120

120

Ea

Ec=aEa Eb=a2Ea

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From Figure 3.6 it can be seen that the base voltages inthe three circuits are related by the turns ratios of theintervening transformers. Care is required as thenominal transformation ratios of the transformersquoted may be different from the turns ratios- e.g. a110/33kV (nominal) transformer may have a turns ratioof 110/34.5kV. Therefore, the rule for hand calculations

is: 'to refer an impedance in ohms from one circuit toanother multiply the given impedance by the square ofthe turns ratio (open circuit voltage ratio) of theintervening transformer'.

Where power system simulation software is used, thesoftware normally has calculation routines built in toadjust transformer parameters to take account ofdifferences between the nominal primary and secondaryvoltages and turns ratios. In this case, the choice of basevoltages may be more conveniently made as the nominalvoltages of each section of the power system. Thisapproach avoids confusion when per unit or percentvalues are used in calculations in translating the finalresults into volts, amps, etc.

For example, in Figure 3.7, generators G1 and G2 have asub-transient reactance of 26% on 66.6MVA rating at11kV, and transformers T1 and T2 a voltage ratio of11/145kV and an impedance of 12.5% on 75MVA.Choosing 100MVA as base MVA and 132kV as basevoltage, find the percentage impedances to new basequantities.

a. Generator reactances to new bases are:

b. Transformer reactances to new bases are:

NOTE: The base voltages of the generator and circuits

are 11kV and 145kV respectively, that is, the turns

ratio of the transformer. The corresponding per unit

values can be found by dividing by 100, and the ohmic

value can be found by using Equation 3.19.

Figure 3.7

12 5100

75

145

13220 1

2

2. . %

( )

( )=

26 10066 6

11132

0 27

2

2 ( )

( )=

.. %

3.6 BASIC CIRCUIT LAWS,

THEOREMS AND NETWORK REDUCTION

Most practical power system problems are solved byusing steady state analytical methods. The assumptionsmade are that the circuit parameters are linear andbilateral and constant for constant frequency circuit

variables. In some problems, described as initial valueproblems, it is necessary to study the behaviour of acircuit in the transient state. Such problems can besolved using operational methods. Again, in otherproblems, which fortunately are few in number, theassumption of linear, bilateral circuit parameters is nolonger valid. These problems are solved using advancedmathematical techniques that are beyond the scope ofthis book.

3.6.1 Circuit Laws

In linear, bilateral circuits, three basic network lawsapply, regardless of the state of the circuit, at anyparticular instant of time. These laws are the branch,

junction and mesh laws, due to Ohm and Kirchhoff, andare stated below, using steady state a.c. nomenclature.

3.6.1.1 Branch law

The currentI in a given branch of impedance

Z is

proportional to the potential differenceV appearing

across the branch, that is,V=

I

Z.

3.6.1.2 Junction law

The algebraic sum of all currents entering any junction

(or node) in a network is zero, that is:

3.6.1.3 Mesh law

The algebraic sum of all the driving voltages in anyclosed path (or mesh) in a network is equal to thealgebraic sum of all the passive voltages (products of theimpedances and the currents) in the componentsbranches, that is:

Alternatively, the total change in potential around aclosed loop is zero.

3.6.2 Circuit Theorems

From the above network laws, many theorems have beenderived for the rationalisation of networks, either toreach a quick, simple, solution to a problem or torepresent a complicated circuit by an equivalent. Thesetheorems are divided into two classes: those concernedwith the general properties of networks and those

E Z I=

I= 0

3

FundamentalTheory

2 4

Figure 3.7: Section of a power system

G1

T1

T2

G2

132kVoverheadlines

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concerned with network reduction.

Of the many theorems that exist, the three mostimportant are given. These are: the SuperpositionTheorem, Thvenin's Theorem and Kennelly's Star/DeltaTheorem.

3.6.2.1 Superposition Theorem

(general network theorem)

The resultant current that flows in any branch of anetwork due to the simultaneous action of severaldriving voltages is equal to the algebraic sum of thecomponent currents due to each driving voltage actingalone with the remainder short-circuited.

3.6.2.2 Thvenin's Theorem(active network reduction theorem)

Any active network that may be viewed from twoterminals can be replaced by a single driving voltageacting in series with a single impedance. The driving

voltage is the open-circuit voltage between the twoterminals and the impedance is the impedance of thenetwork viewed from the terminals with all sourcesshort-circuited.

3.6.2.3 Kennelly's Star/Delta Theorem(passive network reduction theorem)

Any three-terminal network can be replaced by a delta orstar impedance equivalent without disturbing theexternal network. The formulae relating the replacementof a delta network by the equivalent star network is asfollows (Figure 3.8):

Zco = Z13 Z23 / (Z12 + Z13 + Z23)and so on.

Figure 3.8: Star/Delta network reduction

The impedance of a delta network corresponding to andreplacing any star network is:

Z

12=

Z

ao+

Z

bo+

Zao

Zbo

Zco

and so on.

3.6.3 Network Reduction

The aim of network reduction is to reduce a system to asimple equivalent while retaining the identity of thatpart of the system to be studied.

For example, consider the system shown in Figure 3.9.The network has two sources E and E, a line AOB

shunted by an impedance, which may be regarded as thereduction of a further network connected betweenA andB, and a load connected between O andN. The object ofthe reduction is to study the effect of opening a breakeratA or B during normal system operations, or of a faultatA or B. Thus the identity of nodesA and B must beretained together with the sources, but the branch ONcan be eliminated, simplifying the study. Proceeding,A,B, N, forms a star branch and can therefore be convertedto an equivalent delta.

Figure 3.9

= 51 ohms

=30.6 ohms

= 1.2 ohms (since ZNO>>> ZAOZBO)

Figure 3.10

Z Z ZZ Z

Z AN AO BO

AO BO

NO

= + +

= + +

0 45 18 85

0 45 18 85

0 75. .. .

.

Z Z ZZ Z

ZBN BO NO

BO NO

AO

= + +

= + +

0 75 18 85 0 75 18 85

0 45. .

. .

.

Z Z ZZ Z

Z AN AO NO

AO NO

BO

= + +

3

FundamentalTheory

Figure 3.8: Star-Delta network transformation

c

Zao Zbo Z12

Z23Z13

Oa b 1 2

3

(a) Star network (b) Delta network

Zco

Figure 3.9: Typical power system network

E' E''

N

0

A B1.6

0.75 0.45

18.85

2.55

0.4

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The network is now reduced as shown in Figure 3.10.

By applying Thvenin's theorem to the active loops, thesecan be replaced by a single driving voltage in series withan impedance as shown in Figure 3.11.

Figure 3.11

The network shown in Figure 3.9 is now reduced to thatshown in Figure 3.12 with the nodesA and B retainingtheir identity. Further, the load impedance has beencompletely eliminated.

The network shown in Figure 3.12 may now be used to

study system disturbances, for example power swingswith and without faults.

Figure 3.12

Most reduction problems follow the same pattern as theexample above. The rules to apply in practical networkreduction are:

a. decide on the nature of the disturbance ordisturbances to be studied

b. decide on the information required, for example

the branch currents in the network for a fault at aparticular location

c. reduce all passive sections of the network notdirectly involved with the section underexamination

d. reduce all active meshes to a simple equivalent,that is, to a simple source in series with a singleimpedance

With the widespread availability of computer-basedpower system simulation software, it is now usual to usesuch software on a routine basis for network calculations

without significant network reduction taking place.However, the network reduction techniques given aboveare still valid, as there will be occasions where suchsoftware is not immediately available and a handcalculation must be carried out.

In certain circuits, for example parallel lines on the sametowers, there is mutual coupling between branches.Correct circuit reduction must take account of thiscoupling.

Figure 3.13

Three cases are of interest. These are:

a. two branches connected together at their nodes

b. two branches connected together at one node only

c. two branches that remain unconnected

3

FundamentalTheory

2 6

Figure 3.10: Reduction usingstar/delta transform

E'

A

51 30.6

0.4

2.5

1.21.6

N

B

E''

Figure 3.12: Reduction of typical

power system network

N

A B

1.2

2.5

1.55

0.97E'

0.39

0.99E''

Figure 3.11: Reduction of active meshes:Thvenin's Theorem

E'

A

N

(a) Reduction of left active mesh

N

A

(b) Reduction of right active mesh

E''

N

B B

N

E''31

30.630.6

31

0.4 x 30.6

52.6

1.6 x 51

E'52.6

5151

1.6

0.4

Figure 3.13: Reduction of two brancheswith mutual coupling

(a) Actual circuit

IP Q

P Q

(b) Equivalent when ZaaZbb

(c) Equivalent when Zaa=Zbb

P Q

21Z= (Zaa+Zbb)

Zaa

Zbb

Z=ZaaZbb-Z

2ab

Zaa+Zbb-2Zab

Zab

Ia

Ib

I

I

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Considering each case in turn:

a. consider the circuit shown in Figure 3.13(a). Theapplication of a voltage Vbetween the terminals Pand Q gives:

V= IaZaa + IbZab

V= IaZab + IbZbbwhere Ia and Ib are the currents in branches a andb, respectively and I= Ia + Ib , the total currententering at terminal Pand leaving at terminal Q.

Solving for Ia and Ib :

from which

and

so that the equivalent impedance of the originalcircuit is:

Equation 3.21

(Figure 3.13(b)), and, if the branch impedances are

equal, the usual case, then:

Equation 3.22

(Figure 3.13(c)).

b. consider the circuit in Figure 3.14(a).

Z Z Zaa ab= +( )1

2

ZV

I

Z Z Z

Z Z Z

aa bb ab

aa bb ab

= =

+

2

2

I I IV Z Z Z

Z Z Za b

aa bb ab

aa bb ab

= + =+ ( )

22

IZ Z V

Z Z Z

b

aa ab

aa bb ab

=( )

2

IZ Z V

Z Z Za

bb ab

aa bb ab

=( )

2

The assumption is made that an equivalent starnetwork can replace the network shown. Frominspection with one terminal isolated in turn and avoltage Vimpressed across the remaining terminalsit can be seen that:

Za+Zc=Zaa

Zb+Zc=Zbb

Za+Zb=Zaa+Zbb-2Zab

Solving these equations gives:

Equation 3.23

-see Figure 3.14(b).

c. consider the four-terminal network given in Figure3.15(a), in which the branches 11' and 22' areelectrically separate except for a mutual link. Theequations defining the network are:

V1=Z11I1+Z12I2

V2=Z21I1+Z22I2

I1=Y11V1+Y12V2

I2=Y21V1+Y22V2

where Z12=Z21 and Y12=Y21 , if the network isassumed to be reciprocal. Further, by solving the

above equations it can be shown that:

Equation 3.24

There are three independent coefficients, namelyZ12, Z11, Z22, so the original circuit may bereplaced by an equivalent mesh containing fourexternal terminals, each terminal being connected

to the other three by branch impedances as shownin Figure 3.15(b).

Y Z

Y Z

Y Z

Z Z Z

11 22

22 11

12 12

11 22 122

=

=

=

=

Z Z Z

Z Z Z

Z Z

a aa ab

b bb ab

c ab

=

=

=

3

FundamentalTheory

Figure 3.14: Reduction of mutually-coupled branches

with a common terminal

A

C

B

(b) Equivalent circuit

B

C

A

(a) Actual circuit

Zaa

Zbb

Zab

Za=Zaa-Zab

Zb=Zbb-Zab

Zc=Zab

Figure 3.15 : Equivalent circuits forfour terminal network with mutual coupling

(a) Actual circuit

2

1

2'

1'

2'

1'


2

1Z11

Z22

Z11

Z22

Z12 Z12 Z12 Z12Z21

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defining the equivalent mesh in Figure 3.15(b), andinserting radial branches having impedances equalto Z11 and Z22 in terminals 1 and 2, results inFigure 3.15(d).

3.7 REFERENCES

3.1 Power System Analysis. J. R. Mortlock andM. W. Humphrey Davies. Chapman & Hall.

3.2 Equivalent Circuits I. Frank M. Starr, Proc. A.I.E.E.Vol. 51. 1932, pp. 287-298.

3

FundamentalTheory

2 8

Figure 3.15: Equivalent circuits for

four terminal network with mutual coupling

2'

1'

(d) Equivalent circuit

11

C 2

(c) Equivalent with allnodes commonedexcept 1

Z11 Z12

Z11

Z12

Z12

Z12

-Z12 -Z12Z12

In order to evaluate the branches of the equivalentmesh let all points of entry of the actual circuit becommoned except node 1 of circuit 1, as shown inFigure 3.15(c). Then all impressed voltages exceptV1 will be zero and:

I1 = Y11V1

I2 = Y12V1

If the same conditions are applied to the equivalentmesh, then:

I1 = V1Z11

I2 = -V1/Z12 = -V1/Z12

These relations follow from the fact that the branchconnecting nodes 1 and 1' carries current I1 andthe branches connecting nodes 1 and 2' and 1 and2 carry current I2. This must be true since branchesbetween pairs of commoned nodes can carry no

current.By considering each node in turn with theremainder commoned, the following relationshipsare found:

Z11 = 1/Y11

Z22 = 1/Y22

Z12 = -1/Y12

Z12 = Z12 = -Z21 = -Z12

Hence:

Z11 = Z11Z22-Z212_______________Z22

Z22 = Z11Z22-Z212_______________Z11

Z12 = Z11Z22-Z212_______________Z12 Equation 3.25

A similar but equally rigorous equivalent circuit isshown in Figure 3.15(d). This circuit [3.2] followsfrom the fact that the self-impedance of any circuitis independent of all other circuits. Therefore, itneed not appear in any of the mutual branches if itis lumped as a radial branch at the terminals. So

putting Z11 and Z22 equal to zero in Equation 3.25,

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Introduction 4.1

Three phase fault calculations 4.2

Symmetrical component analysis 4.3of a three-phase network

Equations and network connections 4.4for various types of faults

Current and voltage distribution 4.5in a system due to a fault

Effect of system earthing 4.6on zero sequence quantities

References 4.7

4 F a u l t C a l c u l a t i o n s

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4.1 INTRODUCTION

A power system is normally treated as a balancedsymmetrical three-phase network. When a fault occurs,the symmetry is normally upset, resulting in unbalancedcurrents and voltages appearing in the network. The onlyexception is the three-phase fault, which, because itinvolves all three phases equally at the same location, isdescribed as a symmetrical fault. By using symmetricalcomponent analysis and replacing the normal system

sources by a source at the fault location, it is possible toanalyse these fault conditions.

For the correct application of protection equipment, it isessential to know the fault current distributionthroughout the system and the voltages in differentparts of the system due to the fault. Further, boundaryvalues of current at any relaying point must be known ifthe fault is to be cleared with discrimination. Theinformation normally required for each kind of fault ateach relaying point is:

i. maximum fault current

ii. minimum fault currentiii. maximum through fault current

To obtain the above information, the limits of stablegeneration and possible operating conditions, includingthe method of system earthing, must be known. Faultsare always assumed to be through zero fault impedance.

4.2 THREE-PHASE FAULT CALCULATIONS

Three-phase faults are unique in that they are balanced,that is, symmetrical in the three phases, and can becalculated from the single-phase impedance diagramand the operating conditions existing prior to the fault.

A fault condition is a sudden abnormal alteration to thenormal circuit arrangement. The circuit quantities,current and voltage, will alter, and the circuit will passthrough a transient state to a steady state. In thetransient state, the initial magnitude of the fault currentwill depend upon the point on the voltage wave at whichthe fault occurs. The decay of the transient condition,until it merges into steady state, is a function of theparameters of the circuit elements. The transient currentmay be regarded as a d.c. exponential current

4 Fault Calculations

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4

Fault

Calc

ulations

Let this current be 1.0 per unit. It is now necessary tofind the fault current distribution in the various branchesof the network and in particular the current flowing fromA toXon the assumption that a relay atXis to detectthe fault condition. The equivalent impedances viewedfrom either side of the fault are shown in Figure 4.4(a).

Figure 4.3

Figure 4.4

The currents from Figure 4.4(a) are as follows:

From the right:

From the left:

There is a parallel branch to the right ofA

1 21

2 760 437

.

..= p.u.

1 55

2 760 563

.

..= p.u.

Therefore, current in 2.5 ohm branch

and the current in 1.2 ohm branch

Total current enteringXfrom the left, that is, fromA toX, is 0.437 + 0.183 = 0.62p.u. and from B to X is0.38p.u. The equivalent network as viewed from therelay is as shown in Figure 4.4(b). The impedances oneither side are:

0.68/0.62= 1.1 ohmsand

0.68/0.38= 1.79ohms

The circuit of Figure 4.4 (b) has been included becausethe Protection Engineer is interested in these equivalentparameters when applying certain types of protectionrelay.

4.3 SYMMETRICAL COMPONENT ANALYSISOF A THREE-PHASE NETWORK

The Protection Engineer is interested in a wider variety offaults than just a three-phase fault. The most commonfault is a single-phase to earth fault, which, in LVsystems, can produce a higher fault current than a three-

phase fault. Similarly, because protection is expected tooperate correctly for all types of fault, it may benecessary to consider the fault currents due to manydifferent types of fault. Since the three-phase fault isunique in being a balanced fault, a method of analysisthat is applicable to unbalanced faults is required. It canbe shown [4.2] that, by applying the 'Principle ofSuperposition', any general three-phase system ofvectors may be replaced by three sets of balanced(symmetrical) vectors; two sets are three-phase buthaving opposite phase rotation and one set is co-phasal.These vector sets are described as the positive, negative

and zero sequence sets respectively.The equations between phase and sequence voltages aregiven below:

Equation 4.1

E E E E

E a E aE E

E aE a E E

a

b

c

= + +

= + +

= + +

1 2 0

21 2 0

12

2 0

=

=

2 5 0 563

3 7 0 38

. .

. . p.u.

= =1 2 0 5633 7

0 183. .

.. p.u.

Figure 4.3: Network with fault at node A

N

A

V

B

A

X

(b) Typical physical arrangement of nodeA with a fault shown atX

(a) Three - phase fault diagram for a fault at nodeA

BusbarCircuit breaker

1.55

1.2

2.5

0.39

Figure 4.4: Impedances viewed from fault

N

V

A

N

V

X

1.55 1.21

1.791.1

(a) Impedance viewed from node A

(b) Equivalent impedances viewed from node X

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Equation 4.2

where all quantities are referred to the reference phaseA. A similar set of equations can be written for phaseand sequence currents. Figure 4.5 illustrates theresolution of a system of unbalanced vectors.

Figure 4.5

When a fault occurs in a power system, the phaseimpedances are no longer identical (except in the case of

three-phase faults) and the resulting currents andvoltages are unbalanced, the point of greatest unbalancebeing at the fault point. It has been shown in Chapter 3that the fault may be studied by short-circuiting allnormal driving voltages in the system and replacing thefault connection by a source whose driving voltage isequal to the pre-fault voltage at the fault point. Hence,the system impedances remain symmetrical, viewed fromthe fault, and the fault point may now be regarded as thepoint of injection of unbalanced voltages and currentsinto the system.

This is a most important approach in defining the fault

conditions since it allows the system to be representedby sequence networks [4.3] using the method ofsymmetrical components.

4.3.1 Positive Sequence Network

During normal balanced system conditions, only positivesequence currents and voltages can exist in the system,and therefore the normal system impedance network is apositive sequence network.

When a fault occurs in a power system, the current in the

E E aE a E

E E a E aE

E E E E

a b c

a b c

a b c

12

22

0

1

3

1

3

1

3

= + +( )

= + +( )

= + +( )

fault branch changes from 0 toI and the positive

sequence voltage across the branch changes fromVto

V1;

replacing the fault branch by a source equal to the changein voltage and short-circuiting all normal driving voltagesin the system results in a current

I flowing into the

system, and:

Equation 4.3

whereZ1 is the positive sequence impedance of the

system viewed from the fault. As before the fault nocurrent was flowing from the fault into the system, itfollows that

I1 , the fault current flowing from the

system into the fault must equal - I . Therefore:

V1 =

V-

I1

Z1 Equation 4.4

is the relationship between positive sequence currentsand voltages in the fault branch during a fault.

In Figure 4.6, which represents a simple system, thevoltage drops

I1

Z1 and

I1

Z1 are equal to (

V-

V1 )

where the currentsI1 and

I1 enter the fault from the

left and right respectively and impedancesZ1 and

Z1

are the total system impedances viewed from either sideof the fault branch. The voltage

Vis equal to the open-

circuit voltage in the system, and it has been shown thatV

E

E (see Section 3.7). So the positive sequence

voltages in the system due to the fault are greatest at thesource, as shown in the gradient diagram, Figure 4.6(b).

Figure 4.6

4.3.2 Negative Sequence Network

If only positive sequence quantities appear in a powersystem under normal conditions, then negative sequencequantities can only exist during an unbalanced fault.

If no negative sequence quantities are present in the

IV V

Z=

)( 11

4

Fault

Calc

ulations

3 4

Figure 4.6: Fault at F:Positive sequence diagrams

(a) System diagramN

F

X

N

X

F

N'

(b) Gradient diagram

ZS1 Z'1

Z'1

Z'1

Z''1

I'1

I'1

I'1

I1

V1

V1

V'1+

I'1Z

'1

V

I''1

E' E'

Figure 4.5: Resolution of a systemof unbalanced vectors

a2E2

a2E1

aE1

aE2

Eo

Eo

Eo

E1

E2 Ea

Eb

Ec

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fault branch prior to the fault, then, when a fault occurs,the change in voltage is

V2 , and the resulting current

I2

flowing from the network into the fault is:

Equation 4.5

The impedances in the negative sequence network aregenerally the same as those in the positive sequencenetwork. In machines

Z1

Z2 , but the difference is

generally ignored, particularly in large networks.

The negative sequence diagrams, shown in Figure 4.7, aresimilar to the positive sequence diagrams, with twoimportant differences; no driving voltages exist beforethe fault and the negative sequence voltage

V2 is

greatest at the fault point.

Figure 4.7

4.3.3 Zero Sequence Network

The zero sequence current and voltage relationshipsduring a fault condition are the same as those in thenegative sequence network. Hence:

V0 = -

I0

Z0 Equation 4.6

Also, the zero sequence diagram is that of Figure 4.7,

substitutingI0 for

I2 , and so on.

The currents and voltages in the zero sequence networkare co-phasal, that is, all the same phase. For zerosequence currents to flow in a system there must be areturn connection through either a neutral conductor orthe general mass of earth. Note must be taken of thisfact when determining zero sequence equivalent circuits.Further, in general

Z1

Z0 and the value of

Z0varies

according to the type of plant, the winding arrangementand the method of earthing.

IV

Z2

2

2

=

4.4 EQUATIONS AND NETWORK CONNECTIONSFOR VARIOUS TYPES OF FAULTS

The most important types of faults are as follows:

a. single-phase to earth

b. phase to phase

c. phase-phase-earthd. three-phase (with or without earth)

The above faults are described as single shunt faultsbecause they occur at one location and involve aconnection between one phase and another or to earth.

In addition, the Protection Engineer often studies twoother types of fault:

e. single-phase open circuit

f. cross-country fault

By determining the currents and voltages at the fault

point, it is possible to define the fault and connect thesequence networks to represent the fault condition.From the initial equations and the network diagram, thenature of the fault currents and voltages in differentbranches of the system can be determined.

For shunt faults of zero impedance, and neglecting loadcurrent, the equations defining each fault (using phase-neutral values) can be written down as follows:

a. Single-phase-earth (A-E)

Equation 4.7

b. Phase-phase (B-C)

Equation 4.8

c. Phase-phase-earth (B-C-E)

Equation 4.9

d. Three-phase (A-B-C or A-B-C-E)

Equation 4.10

It should be noted from the above that for any type offault there are three equations that define the faultconditions.

I I I

V V

V V

a b c

a b

b c

+ + =

=

=

0

I

V

V

a

b

c

=

==

0

0

0

I

I I

V V

a

b c

b c

=

=

=

0

I

I

V

b

c

a

=

==

0

0

0

4

Fault

Calc

ulations

Figure 4.7: Fault at F:Negative sequence diagram

(a) Negative sequence networkN

F

X

F

X

N

(b) Gradient diagram

ZS1

Z'1

Z'1

Z''1

I'2

I2

V2

V2

V2+ I'2Z'1

I''2

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When there is a fault impedance, this must be taken intoaccount when writing down the equations. For example,with a single-phase-earth fault through fault impedanceZf, Equations 4.7 are re-written:

Equation 4.11

Figure 4.8

4.4.1 Single-phase-earth Fault (A-E)

Consider a fault defined by Equations 4.7 and by Figure4.8(a). Converting Equations 4.7 into sequencequantities by using Equations 4.1 and 4.2, then:

Equation 4.12

V1 = - (

V2 +

V0 ) Equation 4.13Substituting for

V1 ,

V2 and

V0 in Equation 4.13 from

Equations 4.4, 4.5 and 4.6:V-

I1

Z1 =

I2

Z2+

I0

Z0

but, from Equation 4.12,I1 =

I2 =

I0 , therefore:

V=

I1 (

Z1 +

Z2+

Z3 ) Equation 4.14

The constraints imposed by Equations 4.12 and 4.14indicate that the equivalent circuit for the fault isobtained by connecting the sequence networks in series,as shown in Figure 4.8(b).

4.4.2 Phase-phase Fault (B-C)

From Equation 4.8 and using Equations 4.1 and 4.2:I1 = -

I2 Equation 4.15

I0 = 0V1 =

V2 Equation 4.16

From network Equations 4.4 and 4.5, Equation 4.16 canbe re-written:

V-

I1

Z1 =

I2

Z2+

I0

Z0

I I I I o a1 21

3= = =

I

I

V I Z

b

c

a a f

=

==

0

0

V-

I1

Z1 =

I2

Z2

and substituting forI2 from Equation 4.15:

V=

I1 (

Z1 +

Z2) Equation 4.17

The constraints imposed by Equations 4.15 and 4.17indicate that there is no zero sequence network

connection in the equivalent circuit and that the positiveand negative sequence networks are connected inparallel. Figure 4.9 shows the defining and equivalentcircuits satisfying the above equations.

Figure 4.9

4.4.3 Phase-phase-earth Fault (B-C-E)

Again, from Equation 4.9 and Equations 4.1 and 4.2:I1 = -(

I2 +

Io ) Equation 4.18

andV1 =

V2 =

V0 Equation 4.19

Substituting forV

2and

V

0using network Equations 4.5

and 4.6:I2

Z2 =

I0

Z0

thus, using Equation 4.18:

Equation 4.20

Equation 4.21

Now equatingV1 and

V2 and using Equation 4.4 gives:

V-

I1

Z1 = -

I2

Z2or

V=

I1

Z1 -

I2

Z2

Substituting forI2 from Equation 4.21:

or

Equation 4.22I V

Z Z

Z Z Z Z Z Z 1

0 2

1 0 1 2 0 2

=+ )(

+ +

V ZZ Z

Z ZI= +

+

1

0 2

0 21

IZ I

Z Z2

0 1

0 2

= +

IZ I

Z Z0

2 1

0 2

= +

4

Fault

Calc

ulations

3 6

Figure 4.8: Single-phase-earth fault at F

(a) Definition of fault

F

C

B

A


Ia

Ib

VaF1

N1

N2 N0

F2 F0Vb

Vc

Ic

Ib =0

Ic =0

Va=0

V

Z1 Z2 Z0

Figure 4.9: Phase-Phase fault at F


F

C

B

A


Ia

Ia =0

Ib =-Ic

Vb=-Vc

Ib

Ic

F1

N1

N2 N0

F2 F0Vb

Va

Vc

V

Z1 Z2 Z0

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From the above equations it follows that connecting thethree sequence networks in parallel as shown in Figure4.10(b) may represent a phase-phase-earth fault.

Figure 4.10 Phase-phase-earth fault

4.4.4 Three-phase Fault (A-B-C or A-B-C-E)

Assuming that the fault includes earth, then, fromEquations 4.10 and 4.1, 4.2, it follows that:

Equation 4.23

andI0 = 0 Equation 4.24

SubstitutingV2 = 0 in Equation 4.5 gives:

I2 = 0 Equation 4.25

and substitutingV1 = 0 in Equation 4.4:

0 =V1 -

I1

Z1

orV=

I1

Z1 Equation 4.26

Further, since from Equation 4.24Io = 0, it follows from

Equation 4.6 thatVo is zero when

Zo is finite. The

equivalent sequence connections for a three-phase faultare shown in Figure 4.11.

Figure 4.11

4.4.5 Single-phase Open Circuit Fault

The single-phase open circuit fault is showndiagrammatically in Figure 4.12(a). At the fault point,the boundary conditions are:

Equation 4.27

I

V V

a

b c

=

= =

0

0

V V

V V

a0

1 2 0

=

= =

Hence, from Equations 4.2,

V0 = 1/3 Va

V1 = 1/3 Va

V2 = 1/3 Va

and therefore:

Equation 4.28

From Equations 4.28, it can be concluded that thesequence networks are connected in parallel, as shown inFigure 4.12(b).

4.4.6 Cross-country Faults

A cross-country fault is one where there are two faultsaffecting the same circuit, but in different locations and

possibly involving different phases. Figure 4.13(a)illustrates this.

The constraints expressed in terms of sequencequantities are as follows:

a) At point F

Equation 4.29

Therefore:

Equation 4.30

b) At point F

Equation 4.31

and therefore:

Ib1 = Ib2= Ib0 Equation 4.32

To solve, it is necessary to convert the currents andvoltages at point F to the sequence currents in thesame phase as those at point F. From Equation 4.32,

I I

V

' '

'

a c

b

= =

=

0

0

I I I

V V V

a a a

a a a

1 2 0

1 2 0 0

= =

+ + =

I I

V

b c

a

+ =

=

0

0

V V V V

I I I I

a

a

1 2 0

1 2 0

1 3

0

= = =

= + + =

4

Fault

Calc

ulations

Figure 4.10: Phase-phase-earth fault at F


F

C

B

A


F1 F2 F0

N0N2

N1

Va

Vb

Vc

Ib

Ic

Ia

Ia=0

Vb=0

Vc=0

Z1 Z2 Z0

V

Figure 4.11: Three-phase-earth fault at F

(a) Definition of fault (b) Equivalent circuit

F

C

B

A F1 F2 F0

N0N2

N1Ia

Ia+Ib+Ic=0

IcIb

Vb

Vc

Va

Va+Vb+Vc=0

Z0Z2Z1

V

Figure 4.12: Open circuit on phase A

Va Va'a

bc

1

Ib

IcVc

N1 N2 N0

I1 P1

Q1

2

I2 P2

Q2

0

I0 P0

Q0

Vb'

Vc'

P Q

(a) Circuit diagram


+veSequenceNetwork

-veSequenceNetwork

ZeroSequenceNetwork

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4

Fault

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ulations

3 8

(a) `A' phase to ground at F and `B' phase to ground at F'

a-e b-e


F'F

Ia1F1 F'1

N1 N'1

I'a1

V'a1Va1

Ia2F2 F'2

N2 N'2

I'a2

V'a2Va2

Ia0F0 F'0

N0 N'0

I'a0

V'a0aV'a0

aI'a0

a2V'a2

a2I'a2

1a2

1a

Va0

Figure 4.13: Cross - country fault - phase A to phase B

a2Ia1 = aIa2 = Ia0

or

Ia1 = a2Ia2 = aIa0 Equation 4.33

and, for the voltages

Vb1 + Vb2+Vb0 = 0

Converting:

a2Va1 + aVa2+Va0 = 0

or

Va1 + a2Va2+ aVa0 = 0 Equation 4.34

The fault constraints involve phase shifted sequencequantities. To construct the appropriate sequencenetworks, it is necessary to introduce phase-shiftingtransformers to couple the sequence networks. This

is shown in Figure 4.13(b).

4.5 CURRENT AND VOLTAGE DISTRIBUTIONIN A SYSTEM DUE TO A FAULT

Practical fault calculations involve the examination ofthe effect of a fault in branches of network other than

the faulted branch, so that protection can be appliedcorrectly to isolate the section of the system directlyinvolved in the fault. It is therefore not enough tocalculate the fault current in the fault itself; the faultcurrent distribution must also be established. Further,abnormal voltage stresses may appear in a systembecause of a fault, and these may affect the operation ofthe protection. Knowledge of current and voltagedistribution in a network due to a fault is essential forthe application of protection.

The approach to network fault studies for assessing theapplication of protection equipment may be summarised as

follows:

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a. from the network diagram and accompanying data,assess the limits of stable generation and possibleoperating conditions for the system

NOTE: When full information is not availableassumptions may have to be made

b. with faults assumed to occur at each relaying point

in turn, maximum and minimum fault currents arecalculated for each type of fault

NOTE: The fault is assumed to be through zeroimpedance

c. by calculating the current distribution in thenetwork for faults applied at different points in thenetwork (from (b) above) the maximum throughfault currents at each relaying point areestablished for each type of fault

d. at this stage more or less definite ideas on the typeof protection to be applied are formed. Furthercalculations for establishing voltage variation atthe relaying point, or the stability limit of thesystem with a fault on it, are now carried out inorder to determine the class of protectionnecessary, such as high or low speed, unit or non-unit, etc.

4.5.1 Current Distribution

The phase current in any branch of a network isdetermined from the sequence current distribution in theequivalent circuit of the fault. The sequence currents areexpressed in per unit terms of the sequence current inthe fault branch.

In power system calculations, the positive and negativesequence impedances are normally equal. Thus, thedivision of sequence currents in the two networks willalso be identical.

The impedance values and configuration of the zerosequence network are usually different from those of thepositive and negative sequence networks, so the zerosequence current distribution is calculated separately.

If Co and C1 are described as the zero and positivesequence distribution factors then the actual current ina sequence branch is given by multiplying the actualcurrent in the sequence fault branch by the appropriatedistribution factor. For this reason, if

I1 ,

I2 and

I0 are

sequence currents in an arbitrary branch of a networkdue to a fault at some point in the network, then thephase currents in that branch may be expressed in termsof the distribution constants and the sequence currentsin the fault. These are given below for the variouscommon shunt faults, using Equation 4.1 and theappropriate fault equations:

a. single-phase-earth (A-E)

Equation 4.35

b. phase-phase (B-C)

Equation 4.36

c. phase-phase-earth (B-C-E)

Equation 4.37

d. three-phase (A-B-C or A-B-C-E)

Equation 4.38

As an example of current distribution technique, considerthe system in Figure 4.14(a). The equivalent sequencenetworks are given in Figures 4.14(b) and (c), togetherwith typical values of impedances. A fault is assumed atA and it is desired to find the currents in branch OB dueto the fault. In each network, the distribution factors aregiven for each branch, with the current in the faultbranch taken as 1.0p.u. From the diagram, the zerosequence distribution factor Co in branch OB is 0.112and the positive sequence factor C1 is 0.373. For anearth fault atA the phase currents in branch OB fromEquation 4.35 are:

Ia = (0.746+ 0.112)

I0

= 0.858I0

andIb =

Ic = -(0.373 + 0.112)

I0

= -0.261I0

By using network reduction methods and assuming thatall impedances are reactive, it can be shown thatZ1 =

Z0 =j0.68ohms.

Therefore, from Equation 4.14, the current in fault

branch IV

a =

0 68.

I C I

I a C I

I aC I

'

'

'

a

b

c

=

=

=

1 1

21 1

1 1

I C C I

I a a C

Z

Z a C C I

I a a C Z

ZaC C I

'

'

'

a

b

c

= ( )

= ( )

= ( ) +

1 0 0

2

1

0

1

2

1 0 0

21

0

11 0 0

I

I a a C I

I a a C I

'

'

'

a

b

c

=

= ( )

= ( )

0

21 1

21 1

I C C I

I C C I

I C C I

'

'

'

a

b

c

= +( )

= ( )

= ( )

2 1 0 0

1 0 0

1 0 0

4

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ulations

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Assuming that |V| = 63.5volts, then:

If

Vis taken as the reference vector, then:Ia = 26.8 -90AIb = Ic =8.15 -90A

The vector diagram for the above fault condition isshown in Figure 4.15.

Figure 4.15

I Ix

Aa013

63 5

3 0 68 31 2= = =

.

..

4.5.2 Voltage Distribution

The voltage distribution in any branch of a network isdetermined from the sequence voltage distribution. Asshown by Equations 4.4, 4.5 and 4.6 and the gradientdiagrams, Figures 4.6(b) and 4.7(b), the positivesequence voltage is a minimum at the fault, whereas the

zero and negative sequence voltages are a maximum.Thus, the sequence voltages in any part of the systemmay be given generally as:

Equation 4.39

Using the above equation, the fault voltages at bus B inthe previous example can be found.

From the positive sequence distribution diagram Figure4.8(c):

From the zero sequence distribution diagram Figure4.8(b):

For earth faults, at the faultI1 =

I2 =

I0 =j31.2A, when

|V| = 63.5volts and is taken as the reference vector.Further,

Z1 =

Z0 =j0.68ohms.

Hence:V1 = 63.5 - (0.216x 31.2)

= 56.760 volts

V2 = 6.74180 voltsV0 = 2.25180 volts

and, using Equations 4.1:Va =

V1 +

V2 +

V0

= 56.76-(6.74 + 2.25)Va = 47.80Vb = a

2V1 + a

V2 +

V0

= 56.76a2-(6.74a + 2.25)Vb = 61.5-116.4 volts

= [ ]I Z j0 0 0 608.

V I Z j '0 = ( ) + ( ){ }[ ]0 0 0 165 2 6 0 112 1 6 . . . .

V V I Z j '2 = [ ]1 1 0 464.

V V I Z j '1 = ( ) + ( ){ }[1 1 0 395 0 75 0 373 0 45 . . . .

V V I Z C Z

V I Z C Z

V I Z C Z

n

n

n

n

n

n

n

n

n

1 1 1 1

1

1

2 2 1 1

1

1

0 0 0 0

1

0

' =

=

=

'

'

4

Fault

Calc

ulations

4 0

Figure 4.15: Vector diagram: Fault currents

and voltages in branch OB due to P-E fault at bus A

V'c =61.5-116.4

V'a =47.8-0

V'b =61.5-116.4

V=63.5-0

I'b =I'c =8.15-90

I'a =26.8-90

A

Power system

B

Fault

Load

O

(a) Single line diagram

A0

B

0.165 0.112

0.08

0.053

0.755 0.1921.0

(b) Zero sequence network

j7.5j0.4

j0.4

j0.9j2.6 j1.6

j1.6j0.75 j0.45

j4.8

j2.5

j18.850.3731.0 0.395

(c) Positive and negative sequence networks

0.422

0.022

0.556

A0

0.183

B

Figure 4.14: Typical power system

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Vc = a

V1 + a

2V2 +

V0

= 56.75a -(6.74a2 + 2.25)Vc = 61.5116.4 volts

These voltages are shown on the vector diagram, Figure4.15.

4.6 EFFECT OF SYSTEM EARTHINGON ZERO SEQUENCE QUANTIT IES

It has been shown previously that zero sequence currentsflow in the earth path during earth faults, and it followsthat the nature of these currents will be influenced bythe method of earthing. Because these quantities areunique in their association with earth faults they can beutilised in protection, provided their measurement andcharacter are understood for all practical systemconditions.

4.6.1 Residual Current and Voltage

Residual currents and voltages depend for their existenceon two factors:

a. a system connection to earth at two or more points

b. a potential difference between the earthed pointsresulting in a current flow in the earth paths

Under normal system operation there is a capacitancebetween the phases and between phase and earth; thesecapacitances may be regarded as being symmetrical anddistributed uniformly through the system. So even when

(a) above is satisfied, if the driving voltages aresymmetrical the vector sum of the currents will equateto zero and no current will flow between any two earthpoints in the system. When a fault to earth occurs in asystem an unbalance results in condition (b) beingsatisfied. From the definitions given above it followsthat residual currents and voltages are the vector sum ofphase currents and phase voltages respectively.

Hence:

Equation 4.40

Also, from Equations 4.2:

Equation 4.41

It should be further noted that:

Equation 4.42

V V V

V V V

V V V

ae an ne

be bn ne

ce cn ne

= +

= +

= +

I I

V V

R

R

=

=

3

3

0

0

I I I I

V V V V

R a b c

R ae be ce

= + +

= + +

and

and sinceVbn = a

2Van ,

Vcn =a

Van then:

VR = 3

Vne Equation 4.43

whereVcn - neutral displacement voltage.

Measurements of residual quantities are made usingcurrent and voltage transformer connections as shown in

Figure 4.16. If relays are connected into the circuits inplace of the ammeter and voltmeter, it follows that earthfaults in the system can be detected.

4.6.2 SystemZ0 /

Z1 Ratio

The systemZ0/

Z1 ratio is defined as the ratio of zero

sequence and positive sequence impedances viewed from

the fault; it is a variable ratio, dependent upon themethod of earthing, fault position and system operatingarrangement.

When assessing the distribution of residual quantitiesthrough a system, it is convenient to use the fault pointas the reference as it is the point of injection ofunbalanced quantities into the system. The residualvoltage is measured in relation to the normal phase-neutral system voltage and the residual current iscompared with the three-phase fault current at the faultpoint. It can be shown [4.4/4.5] that the character ofthese quantities can be expressed in terms of the system

Z0/

Z1 ratio.The positive sequence impedance of a system is mainlyreactive, whereas the zero sequence impedance beingaffected by the method of earthing may contain bothresistive and reactive components of comparablemagnitude. Thus the express of the

Z0 /

Z1 ratio

approximates to:

Equation 4.44

Expressing the residual current in terms of the three-

phase current andZ0/

Z1 ratio:

Z

Z

X

Xj

R

X

0

1

0

1

0

1

=

4

Fault

Calc

ulations

(a) Residual current

C

B

A

(b) Residual voltage

Vae

Ia

Ib

Ic

VceVbe

A

V

Figure 4.16: Measurement of residual quantities

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a. Single-phase-earth (A-E)

whereK=

Z0/

Z1

Thus:

Equation 4.45

b. Phase-phase-earth (B-C-E)

Hence:

Therefore:

Equation 4.46

Similarly, the residual voltages are found by multiplyingEquations 4.45 and 4.46 by -

K

V.

a. Single-phase-each (A-E)

Equation 4.47

b. Phase-phase-earth (B-C-E)

Equation 4.48

The curves in Figure 4.17 illustrate the variation of theabove residual quantities with the

Z0 /

Z1 ratio. The

residual current in any part of the system can beobtained by multiplying the current from the curve bythe appropriate zero sequence distribution factor.Similarly, the residual voltage is calculated bysubtracting from the voltage curve three times the zerosequence voltage drop between the measuring point inthe system and the fault.

VK

KVR =

+( )3

2 1

VK

KVR =

+( )3

2

I

I K

R

3

3

2 1=

+( )

IV Z

Z Z Z K

V

ZR =

+=

+( )3

2

3

2 1

1

1 0 12

1

IV Z Z

Z Z Z1

1 0

1 0 122

=+( )+

I IZ

Z ZIR = =

+3

30

1

1 01

I

I K

R

3

3

2=

+( )

IV

Z3

1 =

IV

Z Z K

V

ZR =

+=

+( )3

2

3

21 0 1

4.6.3 Variation of Residual Quantities

The variation of residual quantities in a system due todifferent earth arrangements can be most readilyunderstood by using vector diagrams. Three exampleshave been chosen, namely solid fault-isolated neutral,solid fault-resistance neutral, and resistance fault-solidneutral. These are illustrated in Figures 4.18, 4.19 and4.20 respectively.

4

Fault

Calc

ulations

4 2

Figure 4.17: Variation of residual quantitiesat fault point

Residual voltage forSingle-Phase-Earth fault

Residual current forDouble-Phase-Earth fault

1 2 3 4 5

0.5

0

1.0

1.5

2.0

2.5

3.0

=Z0

Z1K

Residual voltage forDouble-Phase-Earth fault

Residual current forDouble-Phase-Earth fault

VRandIRasmultiplesofV

andI3

(a) Circuit diagram

C

B

A

(c) Residual voltage diagram

X

N

F

b

n

c

a(F)

(b) Vector diagram

Iab+Iac

Iab+Iac

-VcF=Eac

-VbF=Eab VR

VcFVbF

Iab

Iab

Iab

Iac

Iac

Iac

Figure 4.18: Solid fault-isolated neutral

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4.6.3.1 Solid fault-isolated neutral

From Figure 4.18 it can be seen that the capacitance toearth of the faulted phase is short circuited by the faultand the resulting unbalance causes capacitance currentsto flow into the fault, returning via sound phasesthrough sound phase capacitances to earth.

At the fault point:

VaF= 0

and

VR =VbF+

VcF

= -3Ean

At source:VR = 3

Vne = -3

Ean

sinceEan +

Ebn +

Ecn = 0

Thus, with an isolated neutral system, the residualvoltage is three times the normal phase-neutral voltageof the faulted phase and there is no variation betweenVR at source and

VR at fault.

In practice, there is some leakage impedance betweenneutral and earth and a small residual current would bedetected atXif a very sensitive relay were employed.

4.6.3.2 Solid fault-resistance neutral

Figure 4.19 shows that the capacitance of the faultedphase is short-circuited by the fault and the neutralcurrent combines with the sound phase capacitivecurrents to give

Ia in the faulted phase.

With a relay at X, residually connected as shown in

Figure 4.16, the residual current will be Ian , that is, theneutral earth loop current.Figure 4.19

At the fault point:VR =

VbF+

VcF since

VFe = 0

At source:VR =

VaX+

VbX+

VcX

From the residual voltage diagram it is clear that there islittle variation in the residual voltages at source and fault,as most residual voltage is dropped across the neutralresistor. The degree of variation in residual quantities is

therefore dependent on the neutral resistor value.

4.6.3.3 Resistance fault-solid neutral

Capacitance can be neglected because, since thecapacitance of the faulted phase is not short-circuited,the circulating capacitance currents will be negligible.

At the fault point:VR =

VFn +

Vbn +

Vcn

At relaying pointX:VR =

VXn +

Vbn +

Vcn

4

Fault

Calc

ulations

Figure 4.19: Solid fault-resistance neutral

(a) Circuit diagram

C

B

A

X F

b

n

c

a(F)

X

(b) Vector diagram

(at source)

(at fault)

(c) Residual voltage diagram

Ia

Ia

Iab

Iab Iab

Iac

Iac

Ian

ZL

-Vcf

-Vbf

Vbf

VcX VcF

VbX

VR

VR

VaX

-VbX

-VcX

-VXn

Iab

Ia

IabIan

Iac-IaZL

(a) Circuit diagram

C

B

A

X

b

n

c

a

X

(b) Vector diagram

(c) Residual voltageat fault

F

F

(d) Residual voltage atrelaying point

IF

IF

IF-IFZL

-IFZS

IF

ZS ZL

Vcn

Vcn

VR VR

Vcn

Vbn

Vbn Vbn

VbF

VFn

VFn

VXn

VXn

Van

VcF

Figure 4.20: Resistance fault-solid neutral

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From the residual voltage diagrams shown in Figure 4.20,it is apparent that the residual voltage is greatest at thefault and reduces towards the source. If the faultresistance approaches zero, that is, the fault becomessolid, then

VFn approaches zero and the voltage drops in

ZS andZL become greater. The ultimate value of

VFn

will depend on the effectiveness of the earthing, and this

is a function of the system Z0/Z1 ratio.

4.7 REFERENCES

4.1 Circuit Analysis of A.C. Power Systems,Volume I.Edith Clarke. John Wiley & Sons.

4.2 Method of Symmetrical Co-ordinates Applied tothe Solution of Polyphase Networks. C.L.Fortescue. Trans. A.I.E.E.,Vol. 37, Part II, 1918, pp1027-40.

4.3 Power System Analysis. J.R. Mortlock and M.W.

Humphrey Davies. Chapman and Hall.4.4 Neutral Groundings. R Willheim and M. Waters,Elsevier.

4.5 Fault Calculations. F.H.W. Lackey, Oliver & Boyd.

4

Fault

Calc

ulations

4 4

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Introduction 5.1

Synchronous machines 5.2

Armature reaction 5.3

Steady state theory 5.4

Salient pole rotor 5.5

Transient analysis 5.6

Asymmetry 5.7

Machine reactances 5.8

Negative sequence reactance 5.9

Zero sequence reactance 5.10

Direct and quadrature axis values 5.11

Effect of saturation on machine reactances 5.12

Transformers 5.13

Transformer positive sequence equivalent circuits 5.14

Transformer zero sequence equivalent circuits 5.15

Auto-transformers 5.16

Transformer impedances 5.17

Overhead lines and cables 5.18Calculation of series impedance 5.19

Calculation of shunt impedance 5.20

Overhead line circuits with or without earth wires 5.21

OHL equivalent circuits 5.22

Cable circuits 5.23

Overhead line and cable data 5.24

References 5.25

5 E q u i v a l e n t C i r c u i t s a n d P a r a m e t e r so f P o w e r S y s t e m P l a n t

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5.1 INTRODUCTION

Knowledge of the behaviour of the principal electricalsystem plant items under normal and fault conditions isa prerequisite for the proper application of protection.This chapter summarises basic synchronous machine,

transformer and transmission line theory and givesequivalent circuits and parameters so that a fault studycan be successfully completed before the selection andapplication of the protection systems described in laterchapters. Only what might be referred to as 'traditional'synchronous machine theory is covered, as that is all thatcalculations for fault level studies generally require.Readers interested in more advanced models ofsynchronous machines are referred to the numerouspapers on the subject, of which reference [5.1] is a goodstarting point.

Power system plant may be divided into two broad

groups - static and rotating.The modelling of static plant for fault level calculationsprovides few difficulties, as plant parameters generallydo not change during the period of interest followingfault inception. The problem in modelling rotating plantis that the parameters change depending on theresponse to a change in power system conditions.

5.2 SYNCHRONOUS MACHINES

There are two main types of synchronous machine:cylindrical rotor and salient pole. In general, the former

is confined to 2 and 4 pole turbine generators, whilesalient pole types are built with 4 poles upwards andinclude most classes of duty. Both classes of machineare similar in so far that each has a stator carrying athree-phase winding distributed over its inner periphery.Within the stator bore is carried the rotor which ismagnetised by a winding carrying d.c. current.

The essential difference between the two classes ofmachine lies in the rotor construction. The cylindricalrotor type has a uniformly cylindrical rotor that carriesits excitation winding distributed over a number of slots

5 Equivalen t Circuit s and Parameterso f Po we r Sy ste m Pl an t

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most common. Two-stroke diesel engines are oftenderivatives of marine designs with relatively large outputs(circa 30MW is possible) and may have running speeds ofthe order of 125rpm. This requires a generator with alarge number of poles (48 for a 125rpm, 50Hz generator)and consequently is of large diameter and short axiallength. This is a contrast to turbine-driven machines that

are of small diameter and long axial length.


5

EquivalentCircuitsandParametersofPowerS

ystemP

lant

4 8

around its periphery. This construction is unsuited tomulti-polar machines but it is very sound mechanically.Hence it is particularly well adapted for the highestspeed electrical machines and is universally employed for2 pole units, plus some 4 pole units.

The salient pole type has poles that are physically

separate, each carrying a concentrated excitationwinding. This type of construction is in many wayscomplementary to that of the cylindrical rotor and isemployed in machines having 4 poles or more. Except inspecial cases its use is exclusive in machines having morethan 6 pole

7750909 power network protection and automation guide

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