API Curriculum Module on Fault Calculations

COLLABORATIVE POWER ENGINEERING CENTRES OF EXCELLENCE

FAULT CALCULATIONS IN

POWER SYSTEMS

PROF. ARINDAM GHOSH

School of Engineering Systems

Queensland University of Technology

Brisbane, Queensland

Course Notes

COURSE NOTES: FAULT CALCULATIONS IN POWER SYSTEMS Ghosh, A.

2010 UGSP7_Fault Calculations_Course Notes_Final 2010 i

© Queensland University of Technology, 2010

No part of this material may be reproduced without permission. For enquiries and requests to access

resources, contact the API University Representative on the Australian Power Institute website:

www.api.edu.au - Collaborative Power Engineering Centres of Excellence.

http://www.api.edu.au/


2010 UGSP7_Fault Calculations_Course Notes_Final 2010 ii

Table of Contents

Introduction: Setting the Scene 1

Assumed Knowledge 1

Objectives 1

Teaching and Learning Approaches 2

Module Resources 2

Course Notes 2

Presentation Packs 2

Activities 2

Acknowledgment 2

Author/s and Contact Details 3

1. Fault in an AC Circuit 5

2. Synchronous Machine Model 7

2.1 Short Circuit on an Unloaded Synchronous Generator 12

3. Symmetrical Faults in a Power System 15

3.1. Calculation of Fault Current using Impedance Diagram 15

3.2. Calculation of Fault Current using Zbus Matrix 17

4. Symmetrical Components 21

4.1. Symmetrical Component Transformation 21

4.2. Real and Reactive Power 25

4.3. Orthogonal Transformation 26

5. Sequence Circuits for Loads 29

5.1. Sequence Circuit for a Y-Connected Load 29

5.2. Sequence Circuit for a -Connected Load 30

6. Sequence Circuits for Synchronous Generator 35

7. Sequence Circuits for Symmetrical Transmission Line 37

8. Sequence Circuits for Transformers 41


2010 UGSP7_Fault Calculations_Course Notes_Final 2010 iii

8.1. Y-Y Connected Transformer 41

8.2 - Connected Transformer 43

8.3 Y- Connected Transformer 44

9. Sequence Networks 47

10. Unsymmetrical Faults 49

10.1 Single-Line-to-Ground Fault 49

10.2 Line-to-Line Fault 52

10.3 Double-Line-to-Ground Fault 55

10.4 Fault Current Computation using Sequence Networks 59

Conclusions 68

References 68

INTRODUCTION AND OVERVIEW

2010 UGSP7_Fault Calculations_Course Notes_Final 2010 1

INTRODUCTION: SETTING THE SCENE

Short circuits occur in power system due to various reasons like, equipment failure, lightning strikes,

falling of branches or trees on the transmission lines, switching surges, insulation failures and other

electrical or mechanical causes. All these are collectively called faults in power systems. A fault

usually results in high current flowing through the lines and if adequate protection is not taken, may

result in damages in the power apparatus. In this module, we shall discuss the effects of both

symmetrical faults and unsymmetrical on the system. Here the term symmetrical fault refers to those

conditions in which all three phases of a power system are grounded at the same point. For this

reason the symmetrical faults sometimes are also called three-line-to-ground (3LG) faults. The

unsymmetrical faults are of three types. These are:

Single-line-to-ground (1LG) fault

Line-to-line (LL) fault

Double-line-to-ground (2LG) fault

We shall first discuss the behavior of electrical circuits and synchronous generators under faulted

condition. We shall proceed to calculate fault currents in a network under balanced fault condition.

Following this, we shall discuss symmetrical components and how a power network can be modeled

for unbalanced operation. Finally we shall discuss unsymmetrical fault current calculations.

ASSUMED KNOWLEDGE

It is assumed that you will have a basic knowledge of:

Basic three-phase circuits and phasor calculations

Synchronous generators

Power systems representation and impedance diagrams

Ybus formulation

OBJECTIVES

On completion of this module you should be able to:

1. Compute fault currents for symmetrical faults.

2. Compute fault currents for unsymmetrical faults.

3. Compute and analyze circuits in terms of symmetrical components.

4. Use concepts learned in designing protective devices.



TEACHING AND LEARNING APPROACHES

This module is best taught through a combination of:

Group lectures presented by the instructor using the PowerPoint Slides which are part of the module material,

Group lecture discussions coordinated by the instructor,

Individual study of the notes provided to review lectures and learn details required to perform activities,

Performing individually fault calculations using calculators, and

Performing individually simulation studies to analyze different types of faults.

A set of presentation packs and activities are available in addition to the course notes as a suite of

resources for this module. These are listed below and can be downloaded by visiting www.api.edu.au

– Collaborative Power Engineering Centres of Excellence.

MODULE RESOURCES

COURSE NOTES

Fault Calculations in Power Systems

PRESENTATION PACKS

Symmetrical Faults (2 hrs)

Symmetrical Faults and Symmetrical Components (2 hrs)

Sequences Circuits of Loads, Generator and Line (2 hrs)

Sequence Circuits of Transformers and Sequence Networks (2 hrs)

Unsymmetrical Fault Current Calculations. (2 hrs)

Fault Current Calculation Examples (2 hrs)

ACTIVITIES

Tutorial: Fault Calculations - Problem Set (3 hrs)

Practical: Overcurrent Protection Coordination (1 hr)

Fault Calculations - PSCAD Simulation Exercise (3 hrs)

ACKNOWLEDGMENT

The author wishes to thank Dr. Manjula Dewadasa for his help in preparing the simulation example.

Also, this material has been developed with the support of API.

http://www.api.edu.au/



AUTHOR/S AND CONTACT DETAILS

Prof. Arindam Ghosh

Professor in Power Engineering

School of Engineering Systems

Queensland University of Technology

Brisbane, Queensland

Email: [email protected]

mailto:[email protected]

COURSE NOTES


1. FAULT IN AN AC CIRCUIT

Consider the single-phase circuit of Fig. 1 where Vs = 240 V (rms), the system frequency is 50 Hz, R =

0.864 , L = 11 mH (L = 3.46 ) and the load is R-L comprising of an 8.64 resistor and a 49.5 mH

inductor (L = 15.55 ). The current phasor when the switch S is open is given by

43.6329.1110.1005.501.19504.9

240j

jI A

This means that the current has a peak value of (11.29×2 =) 15.97 A.

We shall now use the switch S to simulate a short circuit across the load. When the switch is closed,

the fault current is given by

98.7523.672965301646.3864.0

240.j .

jI A

This implies that the fault current has a peak value of (67.23×2 =) 95.17 A. The current (i) waveform

is shown in Fig. 2. Even though the steady state fault current is about 6 times the un-faulted load

current, the peak of the transient current is around 125 A, i.e., over 10 times higher than the nominal

current. The peak current depends on the time of inception of the fault that is unpredictable.

We shall now discuss model of synchronous machine and how it can be represented in fault studies.

Fig. 1. A single-phase circuit in which a source supplies a load through a source impedance.



Fig. 2. The current waveform of the circuit of Fig. 5 before and after the closing of the switch S.



2. SYNCHRONOUS MACHINE MODEL

The schematic diagram of a synchronous generator is shown in Fig. 3. This contains three stator

windings that are spatially distributed. It is assumed that the windings are wye-connected. The

winding currents are denoted by ia, ib and ic. The rotor contains the field winding the current through

which is denoted by if. The field winding is aligned with the so-called direct (d) axis. We also define a

quadrature (q) axis that leads the d-axis by 90°. The angle between the d-axis and the a-phase of the

stator winding is denoted by θd.

Fig. 3. Schematic diagram of a synchronous generator.

Let the self-inductance of the stator windings be denoted by Laa, Lbb, Lcc such that

ccbbaas LLLL (1)

and the mutual inductance between the windings be denoted as

cabcabs LLLM (2)

The mutual inductances between the field coil and the stator windings vary as a function of θd and are

given by

dfaf ML cos (3)

120cos dfbf ML (4)



120cos dfcf ML (5)

The self-inductance of the field coil is denoted by Lff.

The flux linkage equations are then given by

fafcbsasfafccababaaaa iLiiMiLiLiLiLiL (6)

fbfcasbsb iLiiMiL (7)

fafbascsc iLiiMiL (8)

fffccfbbfaaff iLiLiLiL (9)

For balanced operation we have

0 cba iii

Hence, the flux linkage equations for the stator windings (6) to (8) can be modified as

fafassa iLiML (10)

fbfbssb iLiML (11)

fcfcssc iLiML (12)

For steady state operation we can assume

constant ff Ii (13)

Also assuming that the rotor rotates at synchronous speed ωs, we obtain the following two equations

dt

d ds

(14)



0dsd t (15)

where θd0 is the initial position of the field winding with respect to the phase-a of the stator winding

at time t = 0. The mutual inductance of the field winding with all the three stator windings will vary as

a function of θd, i.e.,

0cos dsfaf tML (16)

120cos 0dsfbf tML (17)

120cos 0dsfcf tML (18)

Substituting (13) and (16) to (18) in (10) to (12) we get

0cos dsffassa tIMiML (19)

120cos 0dsffbssb tIMiML (20)

120cos 0dsffcssc tIMiML (21)

Since we assume balanced operation, we need to treat only one phase. Let the armature resistance of

the generator be R. The generator terminal voltage is given by

dt

dRi a

aa

(22)

where the negative sign is used for generating mode of operation in which the current leaves the

terminal. Substituting (19) in (22) we get

0sin dssffa

ssaa tIMdt

diMLRi

(23)

The last term of (23) is the internal emf ea that is given by

0sin||2 dsia tEe (24)

where the rms magnitude |Ei| is proportional to the field current



2

ffs

i

IME

(25)

Since θd0 is the position of the d-axis at time t = 0, we define the position of the q-axis at that instant

as

900d (26)

Therefore (15) can be rewritten as

90 tsd (27)

Substituting (26) in (24) we get

tEe sia cos||2 (28)

Replacing the last term of (23) by the internal emf ea, we get

aa

ssaa edt

diMLRi

(29)

The equivalent circuit of a synchronous generator is shown in Fig. 4. Let the current ia lag the internal

emf ea by θa. The stator currents are then

asaa tIi cos||2 (30)

120cos||2 asab tIi (31)

120cos||2 asac tIi (32)



Fig. 4. Three-phase equivalent circuit of a synchronous generator.

The single-phase equivalent circuit of the generator is shown in Fig. 5. The phase angle θa between ea

and ia is rather difficult to measure under load as ea is the no load voltage. To avoid this, we define

the phase angle between va and ia to be θ. We assume that ea leads va by δ. Therefore we can write

a (33)

Then the voltages and currents shown in Fig. 5 are given as

tV saa cos2 (34)

)cos(2 tEe sia (35)

)cos(2 tIi saa (36)

Equations (34) to (36) imply that

aaiaaa IIEEVV and ,0 (37)

From Fig. 5, we define the synchronous impedance as

sssdd MLjRjXRZ (38)

Then the terminal voltage equation can be written as



adaa IjXREV (39)

In general, the resistance R is negligible compared to the d-axis reactance. Therefore the machine can

be represented by the classical voltage behind a reactance model. For a cylindrical rotor machine, the

d-axis reactance is equal to the q-axis reactance. This is usually called the synchronous reactance Xs.

Therefore (39) can be modified as

adaasaa IjXEIjXEV (40)

Fig. 5. Single-phase equivalent circuit of a synchronous generator.

2.1 SHORT CIRCUIT ON AN UNLOADED SYNCHRONOUS GENERATOR

Fig. 6 shows a typical response of the armature current when a three-phase symmetrical short circuit

occurs at the terminals of an unloaded synchronous generator. It is assumed that there is no dc offset

in the armature current. From a high initial value, the magnitude of the current decreases

exponentially. This is due to the fact that the machine reactance changes due to the effect of

armature reaction. When a short circuit occurs, the flux linking the stator and the rotor cannot

change instantaneously due to the eddy currents flowing in the rotor and damper circuits. These eddy

currents oppose the change. The reactance due to the armature reaction is small during the initial

phase. The eddy current in the damper circuit decays first, followed by the decay of the eddy current

in the field circuit. Then the armature reaction gets to establish.

Fig. 6. Armature current of a synchronous generator as a short circuit occurs at its terminals.



The instantaneous expression for the fault current shown in Fig. 6 is given by

t

Xe

XXe

XXVti s

d

Tt

dd

Tt

dd

tfdd sin

111112

(41)

where Vt is the magnitude of the terminal voltage, and

dX is the direct axis subtransient reactance

dX is the direct axis transient reactance

dX is the direct axis synchronous reactance

with ddd XXX . The time constants are

dT is the direct axis subtransient time constant

dT is the direct axis transient time constant

In the expression of (41), we have neglected the effect of the armature resistance. Let us assume that

the fault occurs at time t = 0. Then the rms value of the initial current is

d

tff

X

VII

0

(42)

which is called the subtransient fault current. The duration of the subtransient current is dictated by

the time constant Td. As the time progresses and Td < t < Td, the first exponential term of (41) will

start decaying and will eventually vanish. However since t is still nearly equal to zero, we have the

following rms value of the current

d

tf

X

VI

(43)

This is called the transient fault current. Now as the time progress further and the second exponential

term also decays, we get the following rms value of the current for the sinusoidal steady state

d

tf

X

VI

(44)

In addition to the ac, the fault currents will also contain the dc offset. Note that a symmetrical fault

occurs when three different phases are in three different locations in the ac cycle. Therefore the dc

offsets in the three phases are different. The maximum value of the dc offset is given by

ATt

fdc eIi 2max

(45)

where TA is the armature time constant.



3. SYMMETRICAL FAULTS IN A POWER SYSTEM

3.1. CALCULATION OF FAULT CURRENT USING IMPEDANCE DIAGRAM

Let us first illustrate the calculation of the fault current using the impedance diagram with the help of

the following examples.

Example 1

Consider the power system of Fig. 7 in which a synchronous generator supplies a

synchronous motor. The motor is operating at rated voltage and rated MVA while

drawing a load current at a power factor of 0.9 (lagging) when a three phase

symmetrical short circuit occurs at its terminals. We shall calculate the fault current

that flow from both the generator and the motor.

We shall choose a base of 50 MVA, 20 kV in the circuit of the generator. Then the

motor synchronous reactance is given by

4.025

502.0 mX per unit

Also the base impedance in the circuit of the transmission line is

12.8750

662

baseZ

Fig.7. A generator supplying a motor load though a transmission line.

Therefore the impedance of the transmission line is

1148.012.87

10jjX line per unit.

The impedance diagram for the circuit is shown in Fig. 8 in which the switch S indicates

the fault.



Example 1

Fig. 8. Impedance diagram of the circuit of Fig. 7.

Before the fault, when the switch S is open, the motor draws a load at rated voltage

and rated MVA with 0.9 lagging power factor. Therefore

4359.09.09.0cos1 1 jIL per unit

Let us assume the voltage across the terminal A and B is 1.0 per unit when the fault

occurs. Then the subtransient voltages of the motor and the generator are

36.08256.04.00.1 jIjE Lm per unit

4633.02244.15148.00.1 jIjE Lg per unit

Hence, the subtransient fault currents, when the switch S closes, fed by the motor and

the generator are

0641.29.04.0

jj

EI m

m

per unit

3784.29.05148.0

jj

EI

g

g

per unit

and the total current flowing to the fault is

4425.4jIII mgf per unit

Note that the base current in the circuit of the motor is

8.1603183

1050 3

baseI A

Therefore while the load current was 1603.8 A, the fault current is 7124.7 A.

We shall now solve the above problem differently.

Example 2

The Thevenin impedance at the circuit between the terminals A and B of the circuit of

Fig. 8 is the parallel combination of the impedances j0.4 and j0.5148. This is then given

as

2251.05148.04.0

5148.04.0jjZth

per unit

Since voltage at the motor terminals before the fault is 1.0 per unit, the fault current is



Example 2

4425.40.1

jZ

Ith

f per unit

If we neglect the pre-fault current flowing through the circuit, then fault current fed by

the motor and the generator can be determined using the current divider principle,

i.e.,

5.25148.09148.0

0 jjj

II

f

m

per unit

9425.14.09148.0

0 jjj

II

f

g

per unit

If, on the other hand, the pre-fault current is not neglected, then the fault current

supplied by the motor and the generator are

0641.29.00 jIII Lmm per unit

3784.29.00 jIII Lgg per unit

3.2. CALCULATION OF FAULT CURRENT USING ZBU S MATRIX

Consider the circuit of Fig. 9 (a), the impedance diagram of which is drawn in Fig. 9 (b). We assume

that a symmetrical fault has occurred in bus-4 such that it is now connected to the reference bus. Let

us assume that the pre-fault voltage at this bus is Vf. To denote that bus-4 is short circuit, we add two

voltage sources Vf and Vf together in series between bus-4 and the reference bus. This is shown in

Fig. 9 (c). Also note that the subtransient fault current If flows from bus-4 to the reference bus. This

implies that a current that is equal to If is injected into bus-4. This current, which is due to the

source Vf will flow through the various branches of the network and will cause a change in the bus

voltages. Assuming that the two sources and Vf are short circuited, Vf is the only source left in the

network that injects a current If into bus-4. The voltages of the different nodes that are caused by

the voltage Vf and the current If are then given by



(a) (b)

(c)

Fig.9. (a) Single-line diagram of a simple power network, (b) its impedance diagram and (c) the equivalent network depicting

a symmetrical fault at bus-4.

f

bus

f I

Z

V

V

V

V

0

0

0

3

2

1

(46)

where the prefix indicates the changes in the bus voltages due to the current If.

From the fourth row of (14) we can write

ff IZV 44 (47)

Combining (46) and (47) we get

3,2,1,44

44 iV

Z

ZIZV f

ifii

(48)

We further assume that the system is unloaded before the fault occurs and that the magnitude and

phase angles of all the generator internal emfs are the same. Then there will be no current circulating

anywhere in the network and the bus voltages of all the nodes before the fault will be same and

equal to Vf. Then the new altered bus voltages due to the fault will be given from (48) by

4,,1,144

4

iV

Z

ZVVV f

iifi

(49)



Example 3

Let us consider following parameters, in per unit, for the impedance diagram of Fig. 9

(b)

Z11 = Z22 = j0.25, Z12 = j0.2, Z13 = j0.25, Z23 = Z34 = j0.4 and Z24 = j0.5

Then Ybus is then given as

5.45.220

5.295.24

25.25.135

04513

jYbus per unit

Inverting the Ybus matrix, we get the Zbus matrix as

3926019740136701133.0

19740256501236012640

13670123601531009690

0.1133126400969015310

...

....

....

...

jZbus per unit

Let us now assume that the internal voltages of both the generators are equal to

1.00. Then the current injected in both bus-1 and 2 will be given by 1.0/j0.25 = j4.0

per unit. Now the altered bus voltages for a symmetrical fault in bus-4 are given from

(49) as

7114.03926.0

1133.011 V per unit

6518.03926.0

1367.012 V per unit

4972.03926.0

1974.013 V per unit

03926.0

3926.014 V per unit

Also since the Thevenin impedance looking into the network at bus-4 is Z44, the

subtransient fault current flowing from bus-4 is

5471.23926.0

1j

jI f per unit



4. SYMMETRICAL COMPONENTS

An unbalanced three-phase system can be resolved into three balanced systems in the sinusoidal

steady state. This method of resolving an unbalanced system into three balanced phasor system has

been proposed by C. L. Fortescue. This method is called resolving symmetrical components of the

original phasors or simply symmetrical components. In this section we shall discuss symmetrical

components transformation and then will present how unbalanced components like Y- or -

connected loads, transformers, generators and transmission lines can be resolved into symmetrical

components. We can then combine all these components together to form what are called sequence

networks.

A system of three unbalanced phasors can be resolved in the following three symmetrical

components:

Positive Sequence: A balanced three-phase system with the same phase sequence as the original sequence.

Negative sequence: A balanced three-phase system with the opposite phase sequence as the original sequence.

Zero Sequence: Three phasors that are equal in magnitude and phase.

Fig. 10 depicts a set of three unbalanced phasors that are resolved into the three sequence

components mentioned above. In this the original set of three phasors are denoted by Va, Vb and Vc,

while their positive, negative and zero sequence components are denoted by the subscripts 1, 2 and 0

respectively. This implies that the positive, negative and zero sequence components of phase-a are

denoted by Va1, Va2 and Va0 respectively. Note that just like the voltage phasors given in Fig. 10, we

can also resolve three unbalanced current phasors into three symmetrical components.

Fig. 10. Representation of (a) an unbalanced network, its (b) positive sequence, (c) negative sequence and (d) zero

sequence.

4.1. SYMMETRICAL COMPONENT TRANSFORMATION

Before we discuss the symmetrical component transformation, let us first define the a-operator as

2

3

2

10120 jea j (50)



Note that for the above operator the following relations hold

on so and

1

2

3

2

1

22403606005

1203604804

3603

2402

000

000

0

0

aeeea

aeeea

ea

ajea

jjj

jjj

j

j

(51)

Also note that we have

02

3

2

1

2

3

2

111 2 jjaa

(52)

Using the a-operator we can write from Fig. 10 (b)

111

2

1 and acab aVVVaV (53)

Similarly from Fig. 10 (c) we get

2

2

222 and acab VaVaVV (54)

Finally from Fig. 10 (d) we get

000 cba VVV (55)

The symmetrical component transformation matrix is given by

c

b

a

a

a

a

V

V

V

aa

aa

V

V

V

2

2

2

1

0

1

1

111

3

1

(56)

Defining the vectors Va012 and Vabc as

c

b

a

abc

a

a

a

a

V

V

V

V

V

V

V

V ,

2

1

0

012

we can write (56) as

abca CVV 012 (57)

where C is the symmetrical component transformation matrix and is given by



aa

aaC2

2

1

1

111

3

1

(58)

The original phasor components can also be obtained from the inverse symmetrical component

transformation, i.e.,

012

1

aabc VCV (59)

Inverting the matrix C given in (58) and combining with (59) we get

2

1

0

1

2

1

0

2

2

1

1

111

a

a

a

a

a

a

c

b

a

V

V

V

C

V

V

V

aa

aa

V

V

V

(60)

From (60) we can write

210 aaaa VVVV (61)

21021

2

0 bbbaaab VVVaVVaVV (62)

2102

2

10 cccaaac VVVVaaVVV (63)

Finally, if we define a set of unbalanced current phasors as Iabc and their symmetrical components as

Ia012, we can then define

012

1

012

aabc

abca

ICI

CII

(64)

Example 4

Let us consider a set of balanced voltages given in per unit by

1200.1 and 1200.1 ,0.1 cba VVV

These imply

aVaV cb and 2

Then from (56) we get



Example 4

013

1 2

0 aaVa

pu 0.113

1 33

1 aaVa

013

1 24

2 aaVa

We then see that for a balanced system the zero and negative sequence voltages are

zero. Also the positive sequence voltage is the same as the original system, i.e.,

ccbbaa VVVVVV 111 and ,

All the quantities given in this example are in per unit.

Example 5

Let us now consider the following set of three unbalanced voltages

1209.0 and 1102.1 ,0.1 cba VVV

Resolving those using (56), we have

72.1470873.0

87.30296.1

16.681250.0

0466.00738.0

0695.00273.1

1161.00465.0

1209.0

1102.1

0.1

1

1

111

3

1

2

2

2

1

0

j

j

j

aa

aa

V

V

V

a

a

a

Therefore we have

16.68125.0000 cba VVV

87.1230296.1,13.1160296.1 11 cb VV

72.270973.0,72.2670873.0 22 cb VV

Furthermore note that

0.1210 aaaa VVVV

1102.1210 bbbb VVVV

1209.0210 cccc VVVV



4.2. REAL AND REACTIVE POWER

The three-phase power in the original unbalanced system is given by

abc

T

abcccbbaaabcabc IVIVIVIVjQP ***

(65)

where I is the complex conjugate of the vector I. Now from (59) and (64) we get

012

1

012 a

TT

aabcabc ICCVjQP (66)

Noting that

100

010

001

31CC T

we can write (66) as

2211003 aaaaaaabcabc IVIVIVjQP (67)

Thus the complex power is three times the summation of the complex power of the three

phase sequences.

Example 6

Let us consider the voltages given in Example 5. Let us further assume that

these voltages are line-to-neutral voltages and they supply a balanced Y-

connected load whose per phase impedance is ZY = 0.2 + j0.8 per unit. Then the

per unit currents in the three phases are

96.752127.1Y

aa

Z

VI pu

04.1744552.1Y

bb

Z

VI pu

04.440914.1Y

cc

Z

VI pu

Then the real and reactive power consumed by the load is given by

pu 9559.0

96.75cos0914.19.04552.12.12127.10.1

abcP



Example 6

pu 8235.3

96.75sin0914.19.04552.12.12127.10.1

abcQ

Now using the transformation (64) we get

75.711058.0

10.722486.1

12.1441516.0

1005.00331.0

1881.13839.0

0889.01229.0

2

1

0

j

j

j

I

I

I

a

a

a

pu

From the results given in Example 5 and from the above values we can compute

the zero sequence complex power as

0552.00138.03 0000 jIVjQP aa pu

The positive sequence complex power is

7415.39354.03 1111 jIVjQP aa pu

Finally the negative sequence complex power is

0269.00067.03 2222 jIVjQP aa pu

Adding the three complex powers together we get the total complex power

consumed by the load as

8235.39559.0210210 jQQQjPPPjQP abcabc pu

4.3. ORTHOGONAL TRANSFORMATION

Instead of the transformation matrix given in (58), let us instead use the transformation

matrix

aa

aaC2

2

1

1

111

3

1

(68)

The inverse of the above matrix is

2

21

1

1

111

3

1

aa

aaC

(69)



Note from (68) and (69) that C1 = (CT). We can therefore state C(CT) = I3, where I3 is (33)

identity matrix. Therefore the transformation matrices given in (68) and (69) are orthogonal.

Now since

100

010

0011 CCCC TT

we can write from (66)

221100 aaaaaaabcabc IVIVIVjQP (70)

This implies that real and reactive powers are directly obtained by the polar multiplication

voltage and current of the symmetrical component voltage and conjugate of currents. The

factor 3 is absent using this transformation, unlike that given in (67).

We shall now discuss how different elements of a power system are represented in terms of their

sequence components. In fact we shall show that each element is represented by three equivalent

circuits, one for each symmetrical component sequence.



5. SEQUENCE CIRCUITS FOR LOADS

In this section we shall construct sequence circuits for both Y and -connected loads separately.

5.1. SEQUENCE CIRCUIT FOR A Y-CONNECTED LOAD

Consider the balanced Y-connected load that is shown in Fig. 11. The neutral point (n) of the windings

are grounded through an impedance Zn. The load in each phase is denoted by ZY. Let us consider

phase-a of the load. The voltage between line and ground is denoted by Va, the line-to-neutral

voltage is denoted by Van and voltage between the neutral and ground is denoted by Vn. The neutral

current is then

02221110 33 acbacbaa

cban

IIIIIIII

IIII

(71)

Therefore there will not be any positive or negative sequence current flowing out of the neutral

point.

Fig. 11. Schematic diagram of a balanced Y-connected load.

The voltage drop between the neutral and ground is

03 ann IZV (72)

Now

03 anannana IZVVVV (73)

We can write similar expression for the other two phases. We can therefore write

1

1

1

3 0an

c

b

a

Y

n

n

n

cn

bn

an

c

b

a

IZ

I

I

I

Z

V

V

V

V

V

V

V

V

V

(74)

Pre-multiplying both sides of the above equation by the matrix C and using (58) we get



1

1

1

3 0012012 CIZIZV anaYa

(75)

Now since

0

0

1

1

1

1

C

We get from (75)

0

03

0

2

1

0

2

1

0 a

n

a

a

a

Y

a

a

a I

Z

I

I

I

Z

V

V

V

(76)

We then find that the zero, positive and negative sequence voltages only depend on their respective

sequence component currents. The sequence component equivalent circuits are shown in Fig. 12.

While the positive and negative sequence impedances are both equal to ZY, the zero sequence

impedance is equal to

nY ZZZ 30 (77)

If the neutral is grounded directly (i.e., Zn = 0), then Z0 = ZY. On the other hand, if the neutral is kept

floating (i.e., Zn = ), then there will not be any zero sequence current flowing in the circuit at all.

Fig. 12. Sequence circuits of Y-connected load: (a) positive, (b) negative and (c) zero sequence.

5.2. SEQUENCE CIRCUIT FOR A -CONNECTED LOAD

Consider the balanced -connected load shown in Fig. 13 in which the load in each phase is denoted

by Z. The line-to-line voltages are given by

caca

bcbc

abab

IZV

IZV

IZV

(78)

Adding these three voltages we get



cabcabcabcab IIIZVVV (79)

Denoting the zero sequence component Vab, Vbc and Vca as Vab0 and that of Iab, Ibc and Ica as Iab0 we can

rewrite (79) as

00 abab IZV (80)

Fig. 13. Schematic diagram of a balanced -connected load.

Again since

0 accbbacabcab VVVVVVVVV

we find from (80) that Vab0 = Iab0 = 0. Hence a -connected load with no mutual coupling does not

have any zero sequence circulating current. Note that the positive and negative sequence impedance

for this load will be equal to Z.

Example 7

Consider the circuit shown in Fig. 14 in which a -connected load is connected in

parallel with a Y-connected load. The neutral point of the Y-connected load is

grounded through an impedance. Applying Kirchoff’s current law at the point P in the

circuit we get

Y

ncba

Y

Y

nacabaa

Z

VVV

ZV

ZZ

Z

VV

Z

VV

Z

VVI

112

The above expression can be written in terms of the vector Vabc as

Y

nabca

Y

aZ

VV

ZV

ZZI

111113



Example 7

Fig. 14. Parallel connection of balanced and Y-connected loads.

Since the load is balanced we can write

1

1

1

111

111

111113

Y

nabcabc

Y

abcZ

VV

ZV

ZZI

Pre-multiplying both sides of the above expression by the transformation matrix C we

get

1

1

1

111

111

111113

012

1

012012 CZ

VVCC

ZV

ZZI

Y

naa

Y

a

Now since

000

000

003

111

111

1111CC

we get

0

0

1313

0012012

Y

naa

Y

aZ

VV

ZV

ZZI

Separating the three components, we can write from the above equation

11

13a

Y

a VZZ

I

22

13a

Y

a VZZ

I



Example 7

n

Y

a

Y

a VZ

VZ

I11

00

Suppose now if we convert the -connected load into an equivalent Y, then the

composite load will be a parallel combination of two Y-connected circuits one with

an impedance of ZY and the other with an impedance of Z/3. Therefore the positive

and the negative sequence impedances are given by the parallel combination of these

two impedances. The positive and negative sequence impedance is then given by

3

3

ZZ

ZZ

Y

Y

Now refer to Fig. 14. The voltage Vn is given by

03 aYncYbYaYnn IZIIIZV

From Fig. 14 we can also write Ia = Ia + IaY. Therefore

cYbYaYbccaabbccaab

cYbYaYcbacba

IIIIIIIII

IIIIIIIII

This implies that Ia0 = IaY0 and hence Vn = 3ZnIa0. We can then rewrite the zero sequence

current expression as

nY

aa

ZZ

VI

3

00

It can be seen that the Z term is absent from the zero sequence impedance.



6. SEQUENCE CIRCUITS FOR SYNCHRONOUS GENERATOR

The three-phase equivalent circuit of a synchronous generator is shown in Fig. 15, where the neutral

point grounded through a reactor with impedance Zn. The neutral current is then given by

cban IIII (81)

Fig. 15. Equivalent circuit of a synchronous generator with grounded neutral.

The derivation of Section 1 assumes balanced operation which implies Ia + Ib + Ic = 0. As per (81) this

assumption is not valid any more. Therefore with respect to this figure we can write for phase-a

voltage as

ancbasass

ancbsasan

EIIIMjIMjLjR

EIIMjILjRV

(82)

Similar expressions can also be written for the other two phases. We therefore have

cn

bn

an

c

b

a

s

c

b

a

ss

cn

bn

an

E

E

E

I

I

I

Mj

I

I

I

MLjR

V

V

V

111

111

111

(83)

Pre-multiplying both sides of (83) by the transformation matrix C we get

2

1

0

2

1

0

1

2

1

0

2

1

0

111

111

111

an

an

an

a

a

a

s

a

a

a

ss

an

an

an

E

E

E

I

I

I

CCMj

I

I

I

MLjR

V

V

V

(84)



Since the synchronous generator is operated to supply only balanced voltages we can assume that

Ean0 = Ean2 = 0 and Ean1 = Ean. We can therefore modify (84) as

0

0

000

000

003

2

1

0

2

1

0

2

1

0

an

a

a

a

s

a

a

a

ss

an

an

an

E

I

I

I

Mj

I

I

I

MLjR

V

V

V

(85)

We can separate the terms of (85) as

0000 2 agassan IZIMLjRV (86)

1111 aananassan IZEEIMLjRV (87)

2222 aassan IZIMLjRV (88)

Furthermore we have seen for a Y-connected load that Va1 = Van1, Va2 = Van2 since the neutral current

does not affect these voltages. However Va0 = Van0 + Vn. Also we know that Vn = 3ZnIa0. We

can therefore rewrite (86) as

00000 3 aanga IZIZZV (89)

The sequence diagrams for a synchronous generator are shown in Fig. 16.

Fig. 16. Sequence circuits of synchronous generator: (a) positive, (b) negative and (c) zero sequence.



7. SEQUENCE CIRCUITS FOR SYMMETRICAL TRANSMISSION LINE

The schematic diagram of a transmission line is shown in Fig. 17. In this diagram the self impedance of

the three phases are denoted by Zaa, Zbb and Zcc while that of the neutral wire is denoted by Znn. Let us

assume that the self impedances of the conductors to be the same, i.e.,

ccbbaa ZZZ

Since the transmission line is assumed to be symmetric, we further assume that the mutual

inductances between the conductors are the same and so are the mutual inductances between the

conductors and the neutral, i.e.,

cabcab ZZZ

cnbnan ZZZ

The directions of the currents flowing through the lines are indicated in Fig. 17 and the voltages

between the different conductors are as indicated.

Fig. 17. Lumped parameter representation of a symmetrical transmission line.

Applying Kirchoff’s voltage law we get

nnnaaannnaaaan VVVVVVV (90)

Again

nancbabaaaaa IZIIZIZV (91)

cbaanannnn IIIZIZV (92)

Substituting (91) and (02) in (90) we get

nnnancbanabaanaanaan IZZIIZZIZZVV (93)



Since the neutral provides a return path for the currents Ia, Ib and Ic, we can write

cban IIII (94)

Therefore substituting (94) in (93) we get the following equation for phase-a of the circuit

cbannnabaannnaanaan IIZZZIZZZVV 22 (95)

Denoting

annnabmannnaas ZZZZZZZ 2 Zand 2

(7.46) can be rewritten as

cbmasnaan IIZIZVV (96)

Since (96) does not explicitly include the neutral conductor we can define the voltage drop across the

phase-a conductor as

naanaa VVV (97)

Combining (96) and (97) we get

cbmasaa IIZIZV (98)

Similar expression can also be written for the other two phases. We therefore get

c

b

a

smm

msm

mms

cc

bb

aa

I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

(99)

Pre-multiplying both sides of (99) by the transformation matrix C we get

012

1

012 a

smm

msm

mms

aa IC

ZZZ

ZZZ

ZZZ

CV

(100)

Now

msmsms

msmsms

msmsms

smm

msm

mms

smm

msm

mms

ZaZaZaaZZZ

ZaaZZaZaZZ

ZZZZZZ

aa

aa

ZZZ

ZZZ

ZZZ

C

ZZZ

ZZZ

ZZZ

112

112

2

1

1

111

22

22

2

21



Hence

ms

ms

ms

msmsms

msmsms

msmsms

smm

msm

mms

ZZ

ZZ

ZZ

ZaZaZaaZZZ

ZaaZZaZaZZ

ZZZZZZ

aa

aaC

ZZZ

ZZZ

ZZZ

C

3300

0330

0063

3

1

112

112

2

1

1

111

3

1

22

22

2

21

Therefore from (100) we get

2

1

0

2

1

0 2

a

a

a

ms

ms

ms

aa

aa

aa

I

I

I

ZZ

ZZ

ZZ

V

V

V

(101)

The positive, negative and zero sequence equivalent circuits of the transmission line are shown in Fig.

18 where the sequence impedances are

abaams ZZZZZZ 21

annnabaams ZZZZZZZ 63220

Fig. 18. Sequence circuits of symmetrical transmission line: (a) positive, (b) negative and (c) zero sequence.



8. SEQUENCE CIRCUITS FOR TRANSFORMERS

In this section we shall discuss the sequence circuits of transformers. As we have seen earlier that the

sequence circuits are different for Y- and -connected loads, the sequence circuits are also different

for Y and connected transformers. We shall therefore treat different transformer connections

separately.

8.1. Y-Y CONNECTED TRANSFORMER

Fig. 19 shows the schematic diagram of a Y-Y connected transformer in which both the neutrals are

grounded. The primary and secondary side quantities are denoted by subscripts in uppercase letters

and lowercase letters respectively. The turns ratio of the transformer is given by = N1:N2.

The voltage of phase-a of the primary side is

03 ANANNANA IZVVVV

Fig. 19. Schematic diagram of a grounded neutral Y-Y connected transformer.

Expanding VA and VAN in terms of their positive, negative and zero sequence components, the above

equation can be rewritten as

0210210 3 ANANANANAAA IZVVVVVV (102)

Noting that the direction of the neutral current In is opposite to that of IN, we can write an equation

similar to that of (102) for the secondary side as

0210210 3 ananananaaa IZVVVVVV (103)

Now since the turns ratio of the transformer is = N1:N2 we can write

AN

an

an

AN VV

V

V

N

N

2

1

AaaA IIININ 21



Substituting in (103) we get

0210210 31

AnANANANaaa IZVVVVVV

Multiplying both sides of the above equation by results in

0

2

210210 3 AnANANANaaa IZVVVVVV (104)

Finally combining (102) with (104) we get

0

2

210210 3 AnNAAAaaa IZZVVVVVV (105)

Separating out the positive, negative and zero sequence components we can write

11

2

11 Aaa VV

N

NV

(106)

22

2

12 Aaa VV

N

NV

(107)

0

2

2100

2

10 3 AnNAaa IZNNZVV

N

NV

(108)

From (106) and (107) we see that the positive and negative sequence relations are the same as that

we have used for representing transformer circuits used in impedance diagram of Fig. 9. Hence the

positive and negative sequence impedances are the same as the transformer leakage impedance Z.

The zero sequence equivalent circuit is shown in Fig. 20. The total zero sequence impedance is given

by

nN ZNNZZZ2

210 33 (109)

Fig. 20. Zero sequence equivalent circuit of grounded neutral Y-Y connected transformer.



The zero sequence diagram of the grounded neutral Y-Y connected transformer is shown in Fig. 21 (a)

in which the impedance Z0 is as given in (109). If both the neutrals are solidly grounded, i.e., Zn = ZN =

0, then Z0 is equal to Z. The single line diagram is still the same as that shown in Fig. 21 (a). If however

one of the two neutrals or both neutrals are ungrounded, then we have either Zn = or ZN = or

both. The zero sequence diagram is then as shown in Fig. 21 (b) where the value of Z0 will depend on

the neutral which is kept ungrounded.

Fig. 21. Zero sequence diagram of (a) grounded neutral and (b) ungrounded neutral Y-Y connected transformer.

8.2 - CONNECTED TRANSFORMER

The schematic diagram of a - connected transformer is shown in Fig. 22. From this circuit we have

21210210 ABABBBBAAA

BAAB

VVVVVVVV

VVV

(110)

Fig. 22. Schematic diagram of a - connected transformer.

Again

ababAB VVN

NV

2

1


2121 ababABABAB VVVVV (111)

The sequence components of the line-to-line voltage VAB can be written in terms of the sequence

components of the line-to-neutral voltage as

303 11 ANAB VV (112)



303 22 ANAB VV (113)

Therefore combining (111)-(113) we get

303303303303 2121 ananANAN VVVV (114)

Hence we get

2211 and anANanAN VVVV (115)

Thus the positive and negative sequence equivalent circuits are represented by a series impedance

that is equal to the leakage impedance of the transformer. Since the -connected winding does not

provide any path for the zero sequence current to flow we have

000 aA II

However the zero sequence current can sometimes circulate within the windings. We can then

draw the zero sequence equivalent circuit as shown in Fig. 23.

Fig. 23. Zero sequence diagram of - connected transformer.

8.3 Y- CONNECTED TRANSFORMER

The schematic diagram of a Y- connected transformer is shown in Fig. 24. It is assumed that the Y-

connected side is grounded with the impedance ZN. Even though the zero sequence current in the

primary Y-connected side has a path to the ground, the zero sequence current flowing in the -

connected secondary winding has no path to flow in the line. Hence we have Ia0 = 0. However the

circulating zero sequence current in the winding magnetically balances the zero sequence current

of the primary winding.



Fig. 24. Schematic diagram of a Y- connected transformer.

The voltage in phase-a of both sides of the transformer is related by

ababAN VVN

NV

2

1

Also we know that

NANA VVV

We therefore have

0210

0210210

3

3

ANababab

ANANANANAAA

IZVVV

IZVVVVVV

(116)

Separating zero, positive and negative sequence components we can write

03 000 abANA VIZV (117)

303 111 aabA VVV (118)

303 222 aabA VVV (119)

The positive sequence equivalent circuit is shown in Fig. 25 (a). The negative sequence circuit is the

same as that of the positive sequence circuit except for the phase shift in the induced emf. This is

shown in Fig. 25 (b). The zero sequence equivalent circuit is shown in Fig. 25 (c) where Z0 = Z + 3ZN.

Note that the primary and the secondary sides are not connected and hence there is an open circuit

between them. However since the zero sequence current flows through primary windings, a return

path is provided through the ground. If however, the neutral in the primary side is not grounded, i.e.,



ZN = , then the zero sequence current cannot flow in the primary side as well. The sequence diagram

is then as shown in Fig. 25 (d) where Z0 = Z.

Fig. 25. Sequence diagram of a Y- connected transformer: (a) positive sequence, (b) negative sequence, (c) zero sequence

with grounded Y-connection and (d) zero sequence with ungrounded Y-connection.



9. SEQUENCE NETWORKS

The sequence circuits developed in the previous sections are combined to form the sequence

networks. The sequence networks for the positive, negative and zero sequences are formed

separately by combining the sequence circuits of all the individual elements. Certain assumptions are

made while forming the sequence networks. These are listed below.

Apart from synchronous machines, the network is made of static elements.

The voltage drop caused by the current in a particular sequence depends only on the impedance of that part of the network.

The positive and negative sequence impedances are equal for all static circuit components, while the zero sequence component need not be the same as them. Furthermore subtransient positive and negative sequence impedances of a synchronous machine are equal.

Voltage sources are connected to the positive sequence circuits of the rotating machines.

No positive or negative sequence current flows between neutral and ground.

Example 8

Let us consider the network shown in Fig 26. The values of the various reactances are

not important here and hence, are not given in this figure. However various points of

the circuit are denoted by the letters A to G. This has been done to identify the

impedances of various circuit elements. For example, the leakage reactance of the

transformer T1 is placed between the points A and B and that of transformer T2 is

placed between D and E.

Fig. 26. Single-line diagram of a 3-machine power system.

The positive sequence network is shown in Fig. 27. The negative sequence diagram,

shown in Fig. 28, is almost identical to the positive sequence diagram except that the

voltage sources are absent in this circuit. The zero sequence network is shown in Fig.

29. The neutral point of generator G1 is grounded. Hence a path from point A to the

ground is provided through the zero sequence reactance of the generator. The primary



Example 8

side of the transformer T1 is -connected and hence there is discontinuity in the circuit

after point A. Similar connections are also made for generator G2 and transformer T2.

The transmission line impedances are placed between the points BC, CD and CF.

The secondary side of transformer T3 is ungrounded and hence there is a break in the

circuit after the point F. However the primary side of T3 is grounded and so is the

neutral point of generator G3. Hence the zero sequence components of these two

apparatus are connected to the ground.

Fig.27. Positive sequence network of the power system of Fig. 26.

Fig. 28. Negative sequence network of the power system of Fig. 26.

Fig. 29. Zero sequence network of the power system of Fig. 26.



10. UNSYMMETRICAL FAULTS

The sequence circuits and the sequence networks developed in the previously will now be used for

finding out fault current during unsymmetrical faults. For the calculation of fault currents, we shall

make the following assumptions:

The power system is balanced before the fault occurs such that of the three sequence networks only the positive sequence network is active. Also as the fault occurs, the sequence networks are connected only through the fault location.

The fault current is negligible such that the pre-fault positive sequence voltages are same at all nodes and at the fault location.

All the network resistances and line charging capacitances are negligible.

All loads are passive except the rotating loads which are represented by synchronous machines.

Based on the assumptions stated above, the faulted network will be as shown in Fig. 30 where the

voltage at the faulted point will be denoted by Vf and current in the three faulted phases are Ifa, Ifb

and Ifc. We shall now discuss how the three sequence networks are connected when the three types

of faults discussed above occur.

Fig. 30. Representation of a faulted segment.

10.1 SINGLE-LINE-TO-GROUND FAULT

Let a 1LG fault has occurred at node k of a network. The faulted segment is then as shown in Fig. 31

where it is assumed that phase-a has touched the ground through an impedance Zf. Since the system

is unloaded before the occurrence of the fault we have

Fig. 31. Representation of 1LG fault.



0 fcfb II (117)

Also the phase-a voltage at the fault point is given by

fafka IZV (118)

From (117) we can write

0

0

1

1

111

3

1

2

2

012

fa

fa

I

aa

aaI

(119)

Solving (119) we get

3210

fa

fafafa

IIII

(120)

This implies that the three sequence currents are in series for the 1LG fault. Let us denote the zero,

positive and negative sequence Thevenin impedance at the faulted point as Zkk0, Zkk1 and Zkk2

respectively. Also since the Thevenin voltage at the faulted phase is Vf we get three sequence circuits

that are similar to the ones shown in Fig. 16. We can then write

222

111

000

fakkka

fakkfka

fakkka

IZV

IZVV

IZV

(121)

Then from (120) and (121) we can write

0210

210

fakkkkkkf

kakakaka

IZZZV

VVVV

(122)

Again since

0210 3 faffafafaffafka IZIIIZIZV

we get from (122)

fkkkkkk

f

faZZZZ

VI

3210

0

(123)

The Thevenin equivalent of the sequence network is shown in Fig. 32.



Fig. 32. Thevenin equivalent of a 1LG fault.

Example 9

A three-phase Y-connected synchronous generator is running unloaded with rated

voltage when a 1LG fault occurs at its terminals. The generator is rated 20 kV, 220

MVA, with subsynchronous reactance of 0.2 per unit. Assume that the subtransient

mutual reactance between the windings is 0.025 per unit. The neutral of the generator

is grounded through a 0.05 per unit reactance. The equivalent circuit of the generator

is shown in Fig. 33. We have to find out the negative and zero sequence reactances.

Fig. 33. Unloaded generator of Example 9.

Since the generator is unloaded the internal emfs are

1200.11200.10.1 cnbnan EEE

Since no current flows in phases b and c, once the fault occurs, we have from Fig. 33

0.4

05.02.0

1j

jI fa



Example 9

Then we also have

2.0 fann IXV

From Fig. 33 and (83) we get

72.1240536.1866.06.0025.0

72.1240536.1866.06.0025.0

0

jIjVEV

jIjVEV

V

fancnc

fanbnb

a

Therefore

3.0

7.0

4.0

72.1240536.1

72.1240536.1

0

012 CVa

From (87) we can write Z1 = j(Ls + Ms) = j0.225. Then from Fig. 16 we have

3333.1225.0

7.01

1

11 j

jZ

VEI aan

fa

Also note from (120) that

210 fafafa III

Therefore from Fig. 16 we get

15.015.03.030

00 jjZ

I

VZ n

a

ag

225.02

22 j

I

VZ

a

a

Comparing the above two values with (86) and (88) we find that Z0 indeed is equal to

j(Ls 2Ms) and Z2 is equal to j(Ls + Ms). Note that we can also calculate the fault

current from (123) as

3333.1

05.0315.0225.0225.0

10 j

jI fa

10.2 LINE-TO-LINE FAULT

The faulted segment for an L-L fault is shown in Fig. 34 where it is assumed that the fault has occurred

at node k of the network. In this the phases b and c got shorted through the impedance Zf. Since the

system is unloaded before the occurrence of the fault we have



0faI (124)

Also since phases b and c are shorted we have

fcfb II (125)

Therefore from (124) and (125) we have

fb

fb

fb

fbfa

Iaa

Iaa

I

ICI2

2

012

0

3

10

(126)

We can then summarize from (126)

21

0 0

fafa

fa

II

I

(127)

Therefore no zero sequence current is injected into the network at bus k and hence the zero

sequence remains a dead network for an L-L fault. The positive and negative sequence currents are

negative of each other.

Fig. 34. Representation of L-L fault.

Now from Fig. 34 we get the following expression for the voltage at the faulted point

fbfkckb IZVV (128)

Again

21

2

2

2

1

2

2211

210210

kakakaka

kckbkckb

kckckckbkbkbkckb

VVaaVaaVaa

VVVV

VVVVVVVV

(129)

Moreover since Ifa0 = Ifb0 = 0 and Ifa1 = Ifb2, we can write

1

2

21

2

21 fafbfafbfbfb IaaaIIaIII (130)



Therefore combining (128)-(130) we get

121 fafkaka IZVV (131)

Equations (128) and (131) indicate that the positive and negative sequence networks are in parallel.

The sequence network is then as shown in Fig. 35. From this network we get

fkkkk

f

fafaZZZ

VII

21

21

(132)

Fig. 35. Thevenin equivalent of an LL fault.

Example 10

Let us consider the same generator as given in Example 9. Assume that the generator is

unloaded when a bolted (Zf = 0) short circuit occurs between phases b and c. Then we

get from (125) Ifb = Ifc. Also since the generator is unloaded, we have Ifa = 0. Therefore

from (83) we get

0.1 anan EV

fbfbbnbn IjIjEV 225.01201225.0

fbfccncn IjIjEV 225.01201225.0

Also since Vbn = Vcn, we can combine the above two equations to get

849.345.0

12011201

jII fcfb

Then

2222.2

2222.2

0

849.3

849.3

0

012

j

jCI fa

We can also obtain the above equation from (132) as

2222.2225.0225.0

121 j

jjII fafa

Also since the neutral current In is zero, we can write Va = 1.0 and

5.0 bncb VVV



Example 10

Hence the sequence components of the line voltages are

5.0

5.0

0

5.0

5.0

0.1

012 CVa

Also note that

5.0225.00.1 11 faa IjV

5.0225.0 22 faa IjV

which are the same as obtained before.

10.3 DOUBLE-LINE-TO-GROUND FAULT

The faulted segment for a 2LG fault is shown in Fig. 36 where it is assumed that the fault has occurred

at node k of the network. In this the phases b and c got shorted through the impedance Zf to the

ground. Since the system is unloaded before the occurrence of the fault we have the same condition

as (124) for the phase-a current. Therefore

fcfbfa

fcfbfcfbfafa

III

IIIIII

0

0

3

3

1

3

1

(133)

Also the voltages of phases b and c are given by

03 fafcbfkckb IZIIZVV (134)

Therefore

kbka

kbka

kbka

kb

kb

ka

ka

VaaV

VaaV

VV

V

V

V

CV2

2

012

2

3

1

(135)

We thus get the following two equations from (135)

21 kaka VV (136)

kbkakakakbkaka VVVVVVV 223 2100 (137)



Substituting (134) and (136) in (137) and rearranging we get

Fig. 36. Representation of 2LG fault.

0021 3 fafkakaka IZVVV (138)

Also since Ifa = 0 we have

0210 fafafa III (139)

The Thevenin equivalent circuit for 2LG fault is shown in Fig. 37. From this figure we get

fkkkk

fkkkk

kk

f

fkk

l

kkkk

f

fa

ZZZ

ZZZZ

V

ZZZZ

VI

3

33

02

02

1

021

1

(140)

The zero and negative sequence currents can be obtained using the current divider principle as

fkkkk

kkfafa

ZZZ

ZII

302

210

(141)

fkkkk

fkk

fafaZZZ

ZZII

3

3

02

0

12

(142)



Fig. 37. Thevenin equivalent of a 2LG fault.

Example 11

Let us consider the same generator as given in Examples 9 and 10. Let us assume that

the generator is operating without any load when a bolted 2LG fault occurs in phases b

and c. The equivalent circuit for this fault is shown in Fig. 38. From this figure we can

write

fcfbnnbn IjIjVVE 025.02.01201

fbfcnncn IjIjVVE 025.02.01201

fcfbn IIjV 05.0

Combining the above three equations we can write the following vector-matrix form

1201

1201

25.0025.0

025.025.0

fc

fb

I

Ij

Solving the above equation we get

8182.1849.3

8182.1849.3

jI

jI

c

b

Hence

6162.1

8283.2

2121.1

8182.1849.3

8182.1849.3

0

012

j

j

j

j

jCi fa



Example 11

Fig. 38. Equivalent circuit of the generator in Fig. 8.4 for a 2LG fault in phases b and c.

Note that we can also obtain the above values using (140)-(142). Note from Example 9

that

0 and 3.005.0315.0,225.0 021 fZjjZjZZ

Then

8283.2

525.0

3.0225.0225.0

0.11 j

j

jjj

I fa

6162.1525.0

3.012 j

j

jII fafa

2121.1525.0

225.010 j

j

jII fafa

Now the sequence components of the voltages are

3636.0225.00.1 11 faa IjV

3636.0225.0 22 faa IjV

3636.03.0 00 faa IjV

Also note from Fig. 38 that

0909.10225.0 fcfbnana IIjVEV

and Vb = Vc = 0. Therefore



Example 11

3636.0

3636.0

3636.0

0

0

0909.1

012 CVa

which are the same as obtained before.

10.4 FAULT CURRENT COMPUTATION USING SEQUENCE NETWORKS

In this section we shall demonstrate the use of sequence networks in the calculation of fault currents

using sequence network through some examples.

Example 12

Consider the network shown in Fig. 39. The system parameters are given below:

Generator G: 50 MVA, 20 kV, X = X1 = X2 = 20%, X0 = 7.5%

Motor M: 40 MVA, 20 kV, X = X1 = X2 = 20%, X0 = 10%, Xn = 5%

Transformer T1: 50 MVA, 20 kV/110 kVY, X = 10%

Transformer T2: 50 MVA, 20 kV/110 kVY, X = 10%

Transmission line: X1 = X2 = 24.2 , X0 = 60.5

We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus-

2.

Fig. 39. Radial power system of Example 12.

Let us choose a base in the circuit of the generator. Then the per unit impedances of

the generator are:

075.0,2.0 021 GGG XXX

The per unit impedances of the two transformers are

1.021 TT XX

The MVA base of the motor is 40, while the base MVA of the total circuit is 50.

Therefore the per unit impedances of the motor are



Example 12

0625.040

5005.0,125.0

40

501.0,25.0

40

502.0 021 nMMM XXXX

For the transmission line

24250

1102

baseZ

Therefore

25.0242

5.60,1.0

242

2.24021 LLL XXX

Let us neglect the phase shift associated with the Y/ transformers. Then the positive,

negative and zero sequence networks are as shown in Figs. 40-42.

Fig. 40. Positive sequence network of the power system of Fig. 39.

Fig. 41. Negative sequence network of the power system of Fig. 39.

Fig. 42. Zero sequence network of the power system of Fig. 39.

From Figs. 40 and 41 we get the following Ybus matrix for both positive and negative

sequences



Example 12

141000

1020100

0102010

001015

21 jYY busbus

Inverting the above matrix we get the following Zbus matrix

1667.01333.01000.00667.0

1333.01867.01400.00933.0

1000.01400.01800.01200.0

0667.00933.01200.01467.0

21 jZZ busbus

Again from Fig. 42 we get the following Ybus matrix for the zero sequence

2.3000

01440

04140

0003333.13

0 jYbus

Inverting the above matrix we get

3125.0000

00778.00222.00

00222.00778.00

000075.0

0 jZbus

Hence for a fault in bus-2, we have the following Thevenin impedances

0778.0,18.0 021 jZjZZ

Alternatively we find from Figs. 40 and 42 that

18.045.03.021 jjjZZl

0778.035.01.00 jjjZl

(a) Single-Line-to-Ground Fault: Let a bolted 1LG fault occurs at bus-2 when the system

is unloaded with bus voltages being 1.0 per unit. Then from (123) we get

2841.2

0778.018.02

1210 j

jIII fafafa

per unit

Also from (120) we get

8524.63 0 jII fafa per unit

Furthermore Ifb = Ifc = 0. From (121) we get the sequence components of the voltages



Example 12

as

4111.018.0

5889.018.01

1777.00778.0

222

112

002

faa

faa

faa

IjV

IjV

IjV

Therefore the voltages at the faulted bus are

11.1079061.0

11.1079061.0

0

22

12

02

1

a

a

a

c

b

a

V

V

V

C

V

V

V

(b) Line-to-Line Fault: For a bolted LL fault, we can write from (132)

7778.218.02

121 j

jII fafa

per unit

Then the fault currents are

8113.4

8113.4

00

2

1

1

fa

fa

fc

fb

fa

I

IC

I

I

I

Finally the sequence components of bus-2 voltages are

5.018.0

5.018.01

0

222

112

02

faa

faa

a

IjV

IjV

V

Hence faulted bus voltages are

5.0

5.0

0.1

22

12

02

1

a

a

a

c

b

a

V

V

V

C

V

V

V

(c)Double-Line-to-Ground Fault: Let us assumes that a bolted 2LG fault occurs at bus-2.

Then

0543.00778.018.0 jjjZl

eq

Thus from (140) we get the positive sequence current as

2676.418.0

11 j

ZjI

eq

fa

per unit



Example 12

The zero and negative sequence currents are then computed from (141) and (142) as

9797.2

0778.018.0

18.010 j

j

jII fafa

per unit

2879.1

0778.018.0

0778.012 j

j

jII fafa

per unit

Therefore the fault currents flowing in the line are

89.42657.6

11.137657.6

0

2

1

0

1

fa

fa

fa

fc

fb

fa

I

I

I

C

I

I

I

Furthermore the sequence components of bus-2 voltages are

2318.018.0

2318.018.01

2318.00778.0

222

112

002

faa

faa

faa

IjV

IjV

IjV

Therefore voltages at the faulted bus are

0

0

6954.0

22

12

02

1

a

a

a

c

b

a

V

V

V

C

V

V

V

Example 13

Let us now assume that a 2LG fault has occurred in bus-4 of the system of Example 12

instead of the one in bus-2. Therefore

3125.0,1667.0 021 jXjXX

Also we have

1087.03125.01667.0 jjjZl

eq

Hence

631.31667.0

11 j

ZjI

eq

fa

per unit



Example 13

Moreover

2631.1

3125.01667.0

1667.010 j

j

jII fafa

per unit

3678.2

3125.01667.0

3125.012 j

j

jII fafa

per unit

Therefore the fault currents flowing in the line are

04.205298.5

96.1595298.5

0

2

1

0

1

fa

fa

fa

fc

fb

fa

I

I

I

C

I

I

I

We shall now compute the currents contributed by the generator and the motor to the

fault. Let us denoted the current flowing to the fault from the generator side by Ig,

while that flowing from the motor by Im. Then from Fig. 40 using the current divider

principle, the positive sequence currents contributed by the two buses are

2103.175.0

25.011 j

j

jII faga per unit

4206.275.0

5.011 j

j

jII fama per unit

Similarly from Fig. 41, the negative sequence currents are given as

7893.075.0

25.022 j

j

jII faga per unit

5786.175.0

5.022 j

j

jII fama per unit

Finally notice from Fig. 42 that the zero sequence current flowing from the generator

to the fault is 0. Then we have

00 gaI

2631.10 jIma per unit

Therefore the fault currents flowing from the generator side are

93.67445.1

07.1737445.1

904210.0

2

1

0

1

ga

ga

ga

gc

gb

ga

I

I

I

C

I

I

I

and those flowing from the motor are



Example 13

93.258512.3

07.1548512.3

904210.0

2

1

0

1

ma

ma

ma

mc

mb

ma

I

I

I

C

I

I

I

It can be easily verified that adding Ig and Im we get If given above.

In the above two examples we have neglected the phase shifts of the Y/ transformers. However

according to the American standard, the positive sequence components of the high tension side lead

those of the low tension side by 30, while the negative sequence behavior is reverse of the positive

sequence behavior. Usually the high tension side of a Y/ transformer is Y-connected. Therefore as

we have seen in Fig. 25, the positive sequence component of Y side leads the positive sequence

component of the side by 30 while the negative sequence component of Y side lags that of the

side by 30. We shall now use this principle to compute the fault current for an unsymmetrical fault.

Example 14

Let us consider the same system as given in Example 12. Since the phase shift does not

alter the zero sequence, the circuit of Fig. 42 remains unchanged. The positive and the

negative sequence circuits must however include the respective phase shifts. These

circuits are redrawn as shown in Figs. 43 and 44.

Note from Figs. 43 and 44 that we have dropped the 3 vis-à-vis that of Fig. 25. This is

because the per unit impedances remain unchanged when referred to the either high

tension or low tension side of an ideal transformer. Therefore the per unit impedances

will also not be altered.

Fig. 43. Positive sequence network of the power system of Fig. 39 including transformer phase shift.



Example 14

Fig. 44. Negative sequence network of the power system of Fig. 39 including transformer phase shift.

Since the zero sequence remains unaltered, these currents will not change from those

computed in Example 12. Thus

00 gaI and 2631.10 jIma per unit

Now the positive sequence fault current from the generator Iga1, being on the Y-side of

the Y/ transformer will lead Ima1 by 30. Therefore

602103.13012103.11 jIga per unit


Finally the negative sequence current Iga2 will lag Ima2 by 30. Hence we have

607893.03017893.02 jjIga per unit


Therefore

04.200642.1

1809996.1

04.200642.1

2

1

0

1

ga

ga

ga

gc

gb

ga

I

I

I

C

I

I

I

Also the fault currents flowing from the motor remain unaltered. Also note that the

currents flowing into the fault remain unchanged. This implies that the phase shift of

the Y/ transformers does not affect the fault currents.



Example 15

Let us consider the same power system as shown in Fig. 26, the sequence diagrams of

which are given in Figs. 27 to 29. With respect to Fig. 26, let us define the system

parameters as:

Generator G1: 200 MVA, 20 kV, X = 20%, X0 = 10%



Transformer T1: 300 MVA, 220Y/22 kV, X = 10%

Transformer T2: Three single-phase units each rated 100 MVA, 130Y/25 kV, X

= 10%

Transformer T3: 300 MVA, 220/22 kV, X = 10%

Line B-C: X1 = X2 = 75 , X0 = 100

Line C-D: X1 = X2 = 75 , X0 = 100

Line C-F: X1 = X2 = 50 , X0 = 75

Let us choose the circuit of Generator 3 as the base, the base MVA for the circuit is

300. The per unit reactances are then computed as shown below.

Generator G1: 3.0200

3002.0 X , X0 = 0.15

Generator G2: 1312.022.22

182.0

2

X , X0 = 0.0656

Generator G3: 2.0X , X0 = 0.15

Transformer T1: 121.0200

2201.0

2

X

Transformer T2: 1266.022.22

251.0

2

X

Transformer T3: 121.020

221.0

2

X

Line B-C: 5625.033.133

7521 XX , 75.0

33.133

1000 X

Line C-D: 5625.033.133

7521 XX , 75.0

33.133

1000 X

Line C-F: 375.033.133

5021 XX , 5625.0

33.133

750 X



Example 15

Neglecting the phase shifts of Y/ connected transformers and assuming that the

system is unloaded, we shall find the fault current for a 1LG fault at bus-1 (point C of

Fig. 26).

From Figs. 27 and 28, we can obtain the positive and negative sequence Thevenin

impedance at point C as (verify)

X1 = X2 = j0.2723 per unit

Similarly from Fig. 29, the Thevenin equivalent of the zero sequence impedance is

X0 = j0.4369 per unit


0188.1

4369.02723.02

10 j

jI fa

per unit

Then the fault current is Ifa = 3Ifa0 = 3.0565 per unit.

CONCLUSIONS

In this module, the fundamentals of fault calculations are discussed. Throughout the course phasor

techniques are employed for fault calculation assuming that a power system is in its sinusoidal steady

state. This implies that if a fault is allowed to persist, the calculated fault currents will flow to the fault

and the various parts of the network. However, before that can happen, the fault will be interrupted

by protective devices. The rms currents, assuming that a fault is allowed to persist, gives us an

indication of what the maximum current that can flow in that part of the circuit. The instantaneous

current can be anywhere between 1.0 to 1.8 times the calculated phasor current. This gives an

indication of what the rating of the circuit breaker that will withstand this current be. We can also

determine the conductor sizes based on this calculations. Furthermore, the fault current levels are

used for coordination of overcurrent relays in power distribution systems. Thus we find that the

application of fault studies is numerous. Moreover, the symmetrical component analysis presented in

this module can be used to analyze any unbalanced circuit.

REFERENCES

For further reading

[1] J. D. Glover and M. S. Sarma, Power System Analysis and Design, 5th Ed., Cengage Learning,

Stamford, CT, 2012.

[2] J. J. Grainger and W. D. Stevenson, Power Stem Analysis, McGraw-Hill, New York, 1994.



[3] D. P. Kothari and I. J. Nagrath, Modern Power Systems Analysis, McGraw-Hill, Boston, 2008.

API Curriculum Module on Fault Calculations

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