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Introduction 4.1 Three phase fault calculations 4.2 Symmetrical component analysis 4.3 of a three-phase network Equations and network connections 4.4 for various types of faults Current and voltage distribution 4.5 in a system due to a fault Effect of system earthing 4.6 on zero sequence quantities References 4.7 4 Fault Calculations

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Page 1: 4 Fault Calculations - fecime.org · Introduction 4.1 Three phase fault calculations 4.2 Symmetrical component analysis 4.3 of a three-phase network Equations and network connections

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

Page 2: 4 Fault Calculations - fecime.org · Introduction 4.1 Three phase fault calculations 4.2 Symmetrical component analysis 4.3 of a three-phase network Equations and network connections

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 1 •

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 systemsources 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 current

iii. 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 Calcul at ions

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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

• 4 •

Fau

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• 3 2 •

superimposed on the symmetrical steady state faultcurrent. In a.c. machines, owing to armature reaction,the machine reactances pass through 'sub transient' and'transient' stages before reaching their steady statesynchronous values. For this reason, the resultant faultcurrent during the transient period, from fault inceptionto steady state also depends on the location of the faultin the network relative to that of the rotating plant.

In a system containing many voltage sources, or havinga complex network arrangement, it is tedious to use thenormal system voltage sources to evaluate the faultcurrent in the faulty branch or to calculate the faultcurrent distribution in the system. A more practicalmethod [4.1] is to replace the system voltages by a singledriving voltage at the fault point. This driving voltage isthe voltage existing at the fault point before the faultoccurs.

Consider the circuit given in Figure 4.1 where the drivingvoltages are

—E and

—E’ , the impedances on either side of

fault point F are —Z1’ and

—Z1’’ , and the current through

point F before the fault occurs is —I .

Figure 4.1:

The voltage—V at F before fault inception is:

—V =

—E -

—I

—Z‘ =

—E’’ +

—I

—Z’’

After the fault the voltage—V is zero. Hence, the change

in voltage is -—V . Because of the fault, the change in the

current flowing into the network from F is:

and, since no current was flowing into the network fromF prior to the fault, the fault current flowing from thenetwork into the fault is:

By applying the principle of superposition, the loadcurrents circulating in the system prior to the fault may

I f I VZ Z

Z Z= − =

+( )∆ 1 1

1 1

' ''

' ' '

∆ I VZ

VZ Z

Z Z= − = −

+( )1

1 1

1 1

' ''

' ' '

be added to the currents circulating in the system due tothe fault, to give the total current in any branch of thesystem at the time of fault inception. However, in mostproblems, the load current is small in comparison to thefault current and is usually ignored.

In a practical power system, the system regulation issuch that the load voltage at any point in the system iswithin 10% of the declared open-circuit voltage at thatpoint. For this reason, it is usual to regard the pre-faultvoltage at the fault as being the open-circuit voltage,and this assumption is also made in a number of thestandards dealing with fault level calculations.

For an example of practical three-phase faultcalculations, consider a fault at A in Figure 3.9. With thenetwork reduced as shown in Figure 4.2, the load voltageat A before the fault occurs is:

Figure 4.2:

—V = 0.97

—E’ - 1.55

—I

For practical working conditions,—E’ ⟩⟩⟩ 1.55

—I and —

E’’ ⟩⟩⟩ 1.207—I . Hence

—E’ ≅ —

E’’≅ —V.

Replacing the driving voltages —E’ and

—E’’ by the load

voltage—V between A and N modifies the circuit as shown

in Figure 4.3(a).

The node A is the junction of three branches. In practice,the node would be a busbar, and the branches arefeeders radiating from the bus via circuit breakers, asshown in Figure 4.3(b). There are two possible locationsfor a fault at A; the busbar side of the breakers or theline side of the breakers. In this example, it is assumedthat the fault is at X, and it is required to calculate thecurrent flowing from the bus to X.

The network viewed from AN has a driving pointimpedance |Z1| = 0.68 ohms.

The current in the fault is.

VZ1

V E I= + ×+

+

0 99 1 2 2 52 5 1 2

0 39. . .. .

.' '

Figure 4.1: Network with fault at F

N

FZ '1 Z ''1

I

VE' E''Figure 4.2: Reduction of typical

power system network

N

1.55 ΩA B

2.5 Ω

1.2 Ω

0.39 Ω

0.99E ''0.97E '

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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 • 3 3 •

• 4 •F

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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 to X on the assumption that a relay at X is 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 of A

1 212 76

0 437..

.= p.u.

1 552 76

0 563..

.= p.u.

Therefore, current in 2.5 ohm branch

and the current in 1.2 ohm branch

Total current entering X from the left, that is, from A toX, is 0.437 + 0.183 = 0.62 p.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 = 1.79 ohms

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 ANALYSIS OF 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, negativeand 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 5633 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 node A with a fault shown at X

(a) Three - phase fault diagram for a fault at node A

BusbarCircuit breaker

1.55Ω

1.2Ω

2.5Ω

0.39Ω

Figure 4.4: Impedances viewed from fault

N

V

A

N

V

X

1.55Ω 1.21Ω

1.79Ω1.1Ω

(a) Impedance viewed from node A

(b) Equivalent impedances viewed from node X

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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

…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 ofthree-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 faultconditions 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

13

13

13

= + +( )= + +( )= + +( )

fault branch changes from 0 to —I and the positive

sequence voltage across the branch changes from —V to

—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 thesystem, and:

…Equation 4.3

where —Z1 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 currents —I1’ and

—I1’’ enter the fault from the

left and right respectively and impedances—Z1’ and

—Z1’’

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

—V is equal to the open-

circuit voltage in the system, and it has been shown that—V ≅ —

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= −

− )( 1

1

• 4 •

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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 '1∆Z '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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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 5 •

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 isgenerally 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,substituting

—I0 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 —Z0 varies

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

IVZ

22

2= −

4.4 EQUATIONS AND NETWORK CONNECTIONS FOR 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 circuitf. cross-country fault

By determining the currents and voltages at the faultpoint, 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 •F

ault

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s

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 ''1I '2

I2

V2

V2

V2 + I '2∆Z '1

I ''2

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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

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 impedance—Zf , 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 4.13

Substituting 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 = 0—V1 =

—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 Io a1 213

= = =

I

I

V I Z

b

c

a a f

=

=

=

0

0

—V -

—I1

—Z1 =

—I2

—Z2

and substituting for —I2 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 networkconnection 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

and—V1 =

—V2 =

—V0 …Equation 4.19

Substituting for —V2 and

—V0 using network Equations 4.5

and 4.6:—I2

—Z2 =

—I0

—Z0

thus, using Equation 4.18:

…Equation 4.20

…Equation 4.21

Now equating —V1 and

—V2 and using Equation 4.4 gives:

—V -

—I1

—Z1 = -

—I2

—Z2

or—V =

—I1

—Z1 -

—I2

—Z2

Substituting for —I2 from Equation 4.21:

or

…Equation 4.22I V

Z Z

Z Z Z Z Z Z1

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 •

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• 3 6 •

Figure 4.8: Single-phase-earth fault at F

(a) Definition of fault

F

C

B

A

(b) Equivalent circuit

Ia

Ib

Va F1

N1

N2 N0

F2 F0Vb

Vc

Ic

Ib =0Ic =0Va=0

V

Z1 Z2 Z0

Figure 4.9: Phase-Phase fault at F

(a) Definition of fault

F

C

B

A

(b) Equivalent circuit

Ia

Ia =0Ib =-IcVb=-Vc

IbIc

F1

N1

N2 N0

F2 F0Vb

Va

Vc

V

Z1 Z2 Z0

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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 • 3 7 •

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

and—I0 = 0 …Equation 4.24

Substituting —V2 = 0 in Equation 4.5 gives:

—I2 = 0 …Equation 4.25

and substituting —V1 = 0 in Equation 4.4:

0 =—V1 -

—I1

—Z1

or—V =

—I1

—Z1 …Equation 4.26

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

Equation 4.6 that —Vo 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 andpossibly 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:

I ’b1 = I ’b2 = I ’b0 …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 •F

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Figure 4.10: Phase-phase-earth fault at F

(a) Definition of fault

F

C

B

A

(b) Equivalent circuit

F1 F2 F0

N0N2

N1

Va

Vb

Vc

Ib

Ic

Ia

Ia=0Vb=0Vc=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

Ic Ib

Vb

Vc

Va

Va+Vb+Vc=0

Z0Z2Z1

V

Figure 4.12: Open circuit on phase A

Va Va'νa

νb

νc

ν1

Ib

IcVc

N1 N2 N0

I1 P1

Q1

ν2

I2 P2

Q2

ν0

I0 P0

Q0

Vb'

Vc'

P Q

(a) Circuit diagram

(b) Equivalent circuit

+veSequenceNetwork

-veSequenceNetwork

ZeroSequenceNetwork

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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

• 4 •

Fau

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• 3 8 •

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

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

a-e b-e

(b) Equivalent circuit

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

a2 I ’a1 = aI ’a2 = I ’a0

orI ’a1 = a2I ’a2 = aI ’a0 …Equation 4.33

and, for the voltages

V ’b1 + V ’b2 +V ’b0 = 0

Converting:

a2V ’a1 + aV ’a2 +V ’a0 = 0

or

V ’a1 + a2V ’a2 + aV ’a0 = 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. Thisis shown in Figure 4.13(b).

4.5 CURRENT AND VOLTAGE DISTRIBUTION IN A SYSTEM DUE TO A FAULT

Practical fault calculations involve the examination ofthe effect of a fault in branches of network other thanthe 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 asfollows:

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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 • 3 9 •

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 pointin 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 at A the phase currents in branch OB fromEquation 4.35 are:

—Ia = (0.746 + 0.112)

—I0

= 0.858—I0

and—I ’b =

—I ’c = -(0.373 + 0.112)

—I0

= -0.261—I0

By using network reduction methods and assuming thatall impedances are reactive, it can be shown that—Z1 =

—Z0 = j0.68 ohms.

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 CZZ

a C C I

I a a CZZ

aC C I

'

'

'

a

b

c

= − −( )

= −( ) − −

= −( ) − +

1 0 0

21

0

1

21 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 •F

ault

Cal

cula

tion

s

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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

Assuming that |—V | = 63.5 volts, then:

If —V is taken as the reference vector, then:

—I ’a = 26.8∠ -90° A—I ’b = I ’c =8.15∠ -90° A

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

Figure 4.15

I Ix

Aa013

63 53 0 68

31 2= = =..

.

4.5.2 Voltage DistributionThe 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 thezero 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 fault —I1 =

—I2 =

—I0 = j31.2A, when

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

—Z1 =

—Z0 = j0.68 ohms.

Hence:—V’1 = 63.5 - (0.216 x 31.2)

= 56.76 ∠ 0° volts—V’2 = 6.74 ∠ 180° volts—V’0 = 2.25 ∠ 180° volts

and, using Equations 4.1:—Va =

—V1 +

—V2 +

—V0

= 56.76 -(6.74 + 2.25) —V’a = 47.8 ∠ 0°—V’b = a2

—V’1 + a

—V’2 +

—V’0

= 56.76a2 -(6.74a + 2.25) —V’b = 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 11

1

2 2 1 11

1

0 0 0 01

0

' = − −

= − −

= − −

'

'

• 4 •

Fau

lt C

alcu

lati

ons

• 4 0 •

Figure 4.15: Vector diagram: Fault currentsand 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

j7.5Ωj0.4Ω

j0.4Ω

j0.9Ωj2.6Ω j1.6Ω

j1.6Ωj0.75Ω j0.45Ω

j4.8Ω

j2.5Ω

j18.85Ω0.3731.0 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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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 • 4 1 •

—V’c = a

—V’1 + a2

—V’2 +

—V’0

= 56.75a -(6.74a2 + 2.25) —V’c = 61.5 ∠ 116.4° 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 since —Vbn = a2

—Van ,

—Vcn =a

—Van then:

—VR = 3

—Vne …Equation 4.43

where —Vcn - neutral displacement voltage.

Measurements of residual quantities are made usingcurrent and voltage transformer connections as shown inFigure 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 System—Z0 /

—Z1 Ratio

The system —Z0 /

—Z1 ratio is defined as the ratio of zero

sequence and positive sequence impedances viewed fromthe 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 and

—Z0 /

—Z1 ratio:

ZZ

XX

jRX

0

1

0

1

0

1= −

• 4 •F

ault

Cal

cula

tion

s

(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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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

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

where —K =

—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.

V K

KVR =

+( )3

2 1

V K

KVR = −

+( )3

2

II K

R

3

32 1ϕ

= −+( )

IV Z

Z Z Z K

VZ

R = −+

= −+( )

3

2

3

2 11

1 0 12

1

IV Z Z

Z Z Z1

1 0

1 0 122

=+( )+

I IZ

Z ZIR = = −

+3

30

1

1 01

II K

R

3

3

2φ=

+( )

I VZ

31

φ =

I VZ Z K

VZ

R =+

=+( )

32

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 •

Fau

lt C

alcu

lati

ons

• 4 2 •

Figure 4.17: Variation of residual quantities at 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

VR

and

I Ras

mul

tiple

s of

V a

ndI 3

(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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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 • 4 3 •

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 = 0and

VR =—VbF +

—VcF

= -3 —Ean

At source:—VR = 3

—Vne = -3

—Ean

since—Ean +

—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 between—VR at source and

—VR at fault.

In practice, there is some leakage impedance betweenneutral and earth and a small residual current would bedetected at X if 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 inFigure 4.16, the residual current will be

—Ian , that is, the

neutral 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 istherefore 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 point X:—VR =

—VXn +

—Vbn +

—Vcn

• 4 •F

ault

Cal

cula

tion

s

Figure 4.19: Solid fault-resistance neutral

(a) Circuit diagram

C

B

AX 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

CBA

X

b

n

c

a

X

(b) Vector diagram

(c) Residual voltage at fault

F

F

(d) Residual voltage at relaying 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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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

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 and —ZL become greater. The ultimate value of

—VFn

will depend on the effectiveness of the earthing, and thisis 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 •

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