CONTENTS (CHAPTER-WISE)

Chapter 1

REPRESENTATION OF POWER SYSTEMS

• One Line Diagram

• Table of Symbols
• Sample Example System
• Impedance Diagram
• Reactance Diagram
• Examples

Chapter 2

SYMMETRICAL THREE PHASE FAULTS

• Transients on a Transmission line

• Short Circuit of Unloaded Syn. Machine
• Short Circuit Reactances
• Sort Circuit Current Oscillogram
• Short Circuit of a Loaded machine
• Examples

Chapter 3

SYMMETRICAL COMPONENTS

• What are Symmetrical Components?

• Resolution of components, Neutral Shift
• Phase shift in Y-Δ Transformer Banks
• Power in terms of Symmetrical Components
• Sequence Imps. & Sequence Networks
• Sequence Networks of system elements
• Examples

2
 Chapter 4

UNSYMMETRICAL FAULTS

• Review of Symmetrical Components

• Preamble to Unsymmetrical Fault Analysis
• L-G, L-L, L-L-G & 3-Phase Faults
• Faults on Power Systems
• Effect of Fault Impedance
• Open Conductor Faults
• Examples

Chapter 5
CIRCUIT BREAKERS:
Requirement of a circuit breaker, difference between an isolator and circuit
breaker, Basic principles of operation of a circuit breaker, Phenomena of arc,
properties of arc, initiation and maintenance of arc, arc interruption theories -
Slepian’s theory and energy balance theory.
Re-striking voltage, recovery voltage, Rate of rise of Re-striking voltage, AC circuit
breaking, current chopping, capacitance switching, resistance switching. DC circuit
breakers. Rating of Circuit breakers
Air Circuit breakers – oil Circuit breakers - SF6 breaker – Vacuum circuit breakers -
principle of operation and constructional details. Advantages and disadvantages of
different types of Circuit breakers.

PREREQUISITE SUBJECTS:

1. DC and Synchronous Machines

2. Transmission and distribution
3. Transformers and induction machines

3
ELECTRICAL POWER SYSTEMS-

An Introduction

Energy in electrical form, apart from being clean, can be generated (converted from other
natural forms) centrally in bulk; can be easily controlled; transmitted efficiently; and it is
easily and efficiently adaptable to other forms of energy for various industrial and
domestic applications. It is therefore a coveted form of energy and is an essential
ingredient for the industrial and all-round development of any country.

The generation of electrical energy (by converting other naturally available forms of
energy), controlling of electrical energy, transmission of energy over long distances to
different load centers, and distribution and utilization of electrical energy together is
called an electrical power system.

The subsystem that generates electrical energy is called generation subsystem or

generating plants (stations). It consists of generating units (consisting of turbine-
alternator sets) including the necessary accessories. Speed governors for the prime
movers (turbines; exciters and voltage regulators for generators, and step-up transformers
also form part of the generating plants.

The subsystem that transmits the electrical energy over long distances (from generating
plants to main load centers) is called transmission subsystem. It consists of transmission
lines, regulating transformers and static/rotating VAR units (which are used to control
active/reactive powers).

The sub system that distributes of energy from load centers to individual consumer points
along with end energy converting devices such as motors, resistances etc., is called
distribution subsystems. It consists of feeders, step-down transformers, and individual
consumer connections along with the terminal energy converting electrical equipment
such as motors, resistors etc.

Electrical energy cannot be stored economically and the electric utility can exercise little
control over the load demand (power) at any time. The power system must, therefore, be
capable of matching the output from the generators to demand at any time at specified
voltage and frequency.

With the constant increase in the electrical energy demand, more and more generating
units, the transmission lines and distribution network along with the necessary controlling
and protective circuits make the power system a large complex system. It is considered as
one of the largest man-made systems. Hence highly trained engineers are needed to
develop and implement the advances of technology for planning, operation and control of
power systems.

4
The objective of the Course

The objective this course is to present methods of analysis with respect to the operation
and control of power systems. Planning and expansion, operation and control of a power
system require modeling (representation of the system suitable for analysis), load flow
studies, fault calculations, protective schemes, and stability studies. In addition, there are
more advanced issues such as economic operation which involve special algorithms for
secured and economical operation of power systems.

Load flow analysis is the determination of the voltage, current, real and reactive
powers at various points in the power network under normal operating conditions.

A fault in a power network is any failure which interferes with the normal
operation of the system. Fault calculations or Fault analysis consist of
determining the fault currents for various types of faults at various points of the
network.

Faults can be very destructive to power systems. System protection schemes are
therefore be evolved and implemented for the reliability and safety of power
systems.

The economic operation requires power systems to be operated at such conditions

which will ensure minimum cost of operation meeting all the conditions.

5
 CHAPTER 1

REPRESENTATION OF POWER SYSTEMS

[CONTENTS: One line diagram, impedance diagram, reactance diagram, per unit
quantities, per unit impedance diagram, formation of bus admittance &
impedance matrices, examples]

1.1 One Line Diagram

In practice, electric power systems are very complex and their size is unwieldy. It is very
difficult to represent all the components of the system on a single frame. The
complexities could be in terms of various types of protective devices, machines
(transformers, generators, motors, etc.), their connections (star, delta, etc.), etc. Hence,
for the purpose of power system analysis, a simple single phase equivalent circuit is
developed called, the one line diagram (OLD) or the single line diagram (SLD). An SLD
is thus, the concise form of representing a given power system. It is to be noted that a
given SLD will contain only such data that are relevant to the system analysis/study
under consideration. For example, the details of protective devices need not be shown for
load flow analysis nor it is necessary to show the details of shunt values for stability
studies.

Symbols used for SLD

Various symbols are used to represent the different parameters and machines as single
phase equivalents on the SLD,. Some of the important symbols used are as listed in the
table of Figure 1.

6
Example system
Consider for illustration purpose, a sample example power system and data as under:
Generator 1: 30 MVA, 10.5 KV, X”= 1.6 ohms, Generator 2: 15 MVA, 6.6 KV, X”=
1.2 ohms, Generator 3: 25 MVA, 6.6 KV, X”= 0.56 ohms, Transformer 1 (3-phase):
15 MVA, 33/11 KV, X=15.2 ohms/phase on HT side, Transformer 2 (3-phase): 15
MVA, 33/6.2 KV, X=16.0 ohms/phase on HT side, Transmission Line: 20.5 ohms per
phase, Load A: 15 MW, 11 KV, 0.9 PF (lag); and Load B: 40 MW, 6.6 KV, 0.85 PF
(lag). The corresponding SLD incorporating the standard symbols can be shown as in
figure 2.

7
It is observed here, that the generators are specified in 3-phase MVA, L-L voltage and
per phase Y-equivalent impedance, transformers are specified in 3-phase MVA, L-L
voltage transformation ratio and per phase Y-equivalent impedance on any one side and
the loads are specified in 3-phase MW, L-L voltage and power factor.

1.2 Impedance Diagram

The impedance diagram on single-phase basis for use under balanced conditions can be
easily drawn from the SLD. The following assumptions are made in obtaining the
impedance diagrams.

Assumptions:
1. The single phase transformer equivalents are shown as ideals with impedances on
appropriate side (LV/HV),
2. The magnetizing reactances of transformers are negligible,
3. The generators are represented as constant voltage sources with series resistance or
reactance,
4. The transmission lines are approximated by their equivalent -Models,
5. The loads are assumed to be passive and are represented by a series branch of
resistance or reactance and
6. Since the balanced conditions are assumed, the neutral grounding impedances do not
appear in the impedance diagram.

Example system
As per the list of assumptions as above and with reference to the system of figure 2, the
impedance diagram can be obtained as shown in figure 3.

8
1.3 Reactance Diagram

With some more additional and simplifying assumptions, the impedance diagram can be
simplified further to obtain the corresponding reactance diagram. The following are the
assumptions made.
Additional assumptions:
 The resistance is often omitted during the fault analysis. This causes a very
negligible error since, resistances are negligible
 Loads are Omitted
 Transmission line capacitances are ineffective &
 Magnetizing currents of transformers are neglected.

Example system
as per the assumptions given above and with reference to the system of figure 2 and
figure 3, the reactance diagram can be obtained as shown in figure 4.

Note: These impedance & reactance diagrams are also refered as the Positive Sequence
Diagrams/ Networks.

1.4 Per Unit Quantities

during the power system analysis, it is a usual practice to represent current, voltage,
impedance, power, etc., of an electric power system in per unit or percentage of the base
or reference value of the respective quantities. The numerical per unit (pu) value of any
quantity is its ratio to a chosen base value of the same dimension. Thus a pu value is a
normalized quantity with respect to the chosen base value.
Definition: Per Unit value of a given quantity is the ratio of the actual value in any
given unit to the base value in the same unit. The percent value is 100 times the pu value.
Both the pu and percentage methods are simpler than the use of actual values. Further,
the main advantage in using the pu system of computations is that the result that comes
out of the sum, product, quotient, etc. of two or more pu values is expressed in per unit
itself.

9
In an electrical power system, the parameters of interest include the current, voltage,
complex power (VA), impedance and the phase angle. Of these, the phase angle is
dimensionless and the other four quantities can be described by knowing any two of
them. Thus clearly, an arbitrary choice of any two base values will evidently fix the other
base values.
Normally the nominal voltage of lines and equipment is known along with the complex
power rating in MVA. Hence, in practice, the base values are chosen for complex power
(MVA) and line voltage (KV). The chosen base MVA is the same for all the parts of the
system. However, the base voltage is chosen with reference to a particular section of the
system and the other base voltages (with reference to the other sections of the systems,
these sections caused by the presence of the transformers) are then related to the chosen
one by the turns-ratio of the connecting transformer.
If Ib is the base current in kilo amperes and Vb, the base voltage in kilovolts, then the base
MVA is, Sb = (VbIb). Then the base values of current & impedance are given by
Base current (kA), Ib = MVAb/KVb
= Sb/Vb (1.1)
Base impedance, Zb = (Vb/Ib)
= (KVb2 / MVAb) (1.2)
Hence the per unit impedance is given by
Zpu = Zohms/Zb
= Zohms (MVAb/KVb2) (1.3)
In 3-phase systems, KVb is the line-to-line value & MVAb is the 3-phase MVA. [1-phase
MVA = (1/3) 3-phase MVA].

Changing the base of a given pu value:

It is observed from equation (3) that the pu value of impedance is proportional directly to
the base MVA and inversely to the square of the base KV. If Zpunew is the pu impedance
required to be calculated on a new set of base values: MVAbnew & KVbnew from the
already given per unit impedance Zpuold, specified on the old set of base values,
MVAbold & KVbold , then we have

Zpunew = Zpuold (MVAb /MVAb ) (KVb /KVb ) (1.4)

new old old new 2

On the other hand, the change of base can also be done by first converting the given pu
impedance to its ohmic value and then calculating its pu value on the new set of base
values.

Merits and Demerits of pu System

Following are the advantages and disadvantages of adopting the pu system of
computations in electric power systems:

Merits:

10
  The pu value is the same for both 1-phase and & 3-phase systems
 The pu value once expressed on a proper base, will be the same when refereed to
either side of the transformer. Thus the presence of transformer is totally
eliminated
 The variation of values is in a smaller range 9nearby unity). Hence the errors
involved in pu computations are very less.
 Usually the nameplate ratings will be marked in pu on the base of the name plate
ratings, etc.

Demerits:
 If proper bases are not chosen, then the resulting pu values may be highly absurd
(such as 5.8 pu, -18.9 pu, etc.). This may cause confusion to the user. However,
this problem can be avoided by selecting the base MVA near the high-rated
equipment and a convenient base KV in any section of the system.

1.5 pu Impedance / Reactance Diagram

for a given power system with all its data with regard to the generators, transformers,
transmission lines, loads, etc., it is possible to obtain the corresponding impedance or
reactance diagram as explained above. If the parametric values are shown in pu on the
properly selected base values of the system, then the diagram is refered as the per unit
impedance or reactance diagram. In forming a pu diagram, the following are the
procedural steps involved:
1. Obtain the one line diagram based on the given data
2. Choose a common base MVA for the system
3. Choose a base KV in any one section (Sections formed by transformers)
4. Find the base KV of all the sections present
5. Find pu values of all the parameters: R,X, Z, E, etc.
6. Draw the pu impedance/ reactance diagram.

16
II EXAMPLES ON PER UNIT ANALYSIS:

Problem #1:
Two generators rated 10 MVA, 13.2 KV and 15 MVA, 13.2 KV are connected in parallel
to a bus bar. They feed supply to 2 motors of inputs 8 MVA and 12 MVA respectively.
The operating voltage of motors is 12.5 KV. Assuming the base quantities as 50 MVA,
13.8 KV, draw the per unit reactance diagram. The percentage reactance for generators is
15% and that for motors is 20%.

Solution:
The one line diagram with the data is obtained as shown in figure P1(a).

Selection of base quantities: 50 MVA, 13.8 KV (Given)

Calculation of pu values:
XG1 = j 0.15 (50/10) (13.2/13.8)2 = j 0.6862 pu.
XG2 = j 0.15 (50/15) (13.2/13.8)2 = j 0.4574 pu.
Xm1 = j 0.2 (50/8) (12.5/13.8)2 = j 1.0256 pu.
Xm2 = j 0.2 (50/12) (12.5/13.8)2 = j 0.6837 pu.
Eg1 = Eg2 = (13.2/13.8) = 0.9565 00 pu
Em1 = Em2 = (12.5/13.8) = 0.9058 00 pu
Thus the pu reactance diagram can be drawn as shown in figure P1(b).

17
Problem #2:
Draw the per unit reactance diagram for the system shown in figure below. Choose a base
of 11 KV, 100 MVA in the generator circuit.

Solution:

18
The one line diagram with the data is considered as shown in figure.

Selection of base quantities:

100 MVA, 11 KV in the generator circuit(Given); the voltage bases in other sections are:
11 (115/11.5) = 110 KV in the transmission line circuit and 110 (6.6/11.5) = 6.31 KV in
the motor circuit.

Calculation of pu values:
XG = j 0.1 pu, Xm = j 0.2 (100/90) (6.6/6.31)2 = j 0.243 pu.
Xt1 =Xt2 = j 0.1 (100/50) (11.5/11)2 = j 0.2185 pu.
Xt3 =Xt4 = j 0.1 (100/50) (6.6/6.31)2 = j 0.219 pu.
Xlines = j 20 (100/1102) = j 0.1652 pu.
Eg = 1.000 pu, Em = (6.6/6.31) = 1.04500 pu
Thus the pu reactance diagram can be drawn as shown in figure P2(b).

Problem #3:
A 30 MVA, 13.8 KV, 3-phase generator has a sub transient reactance of 15%. The
generator supplies 2 motors through a step-up transformer - transmission line – step-
down transformer arrangement. The motors have rated inputs of 20 MVA and 10 MVA at
12.8 KV with 20% sub transient reactance each. The 3-phase transformers are rated at 35
MVA, 13.2 KV- /115 KV-Y with 10 % leakage reactance. The line reactance is 80
ohms. Draw the equivalent per unit reactance diagram by selecting the generator ratings
as base values in the generator circuit.

19
Solution:
The one line diagram with the data is obtained as shown in figure P3(a).

Selection of base quantities:

30 MVA, 13.8 KV in the generator circuit(Given);

The voltage bases in other sections are:

13.8 (115/13.2) = 120.23 KV in the transmission line circuit and
120.23 (13.26/115) = 13.8 KV in the motor circuit.

Calculation of pu values:
XG = j 0.15 pu.
Xm1 = j 0.2 (30/20) (12.8/13.8)2 = j 0.516 pu.
Xm2 = j 0.2 (30/10) (12.8/13.8)2 = j 0.2581 pu.
Xt1 =Xt2 = j 0.1 (30/35) (13.2/13.8)2 = j 0.0784 pu.
Xline = j 80 (30/120.232) = j 0.17 pu.

Eg = 1.000 pu; Em1 = Em2 = (6.6/6.31) = 0.9300 pu

Thus the pu reactance diagram can be drawn as shown in figure P3(b).

20
Problem #4:
A 33 MVA, 13.8 KV, 3-phase generator has a sub transient reactance of 0.5%. The
generator supplies a motor through a step-up transformer - transmission line – step-down
transformer arrangement. The motor has rated input of 25 MVA at 6.6 KV with 25% sub
transient reactance. Draw the equivalent per unit impedance diagram by selecting 25
MVA (3), 6.6 KV (LL) as base values in the motor circuit, given the transformer and
transmission line data as under:
Step up transformer bank: three single phase units, connected –Y, each rated 10 MVA,
13.2/6.6 KV with 7.7 % leakage reactance and 0.5 % leakage resistance;
Transmission line: 75 KM long with a positive sequence reactance of 0.8 ohm/ KM and a
resistance of 0.2 ohm/ KM; and
Step down transformer bank: three single phase units, connected –Y, each rated 8.33
MVA, 110/3.98 KV with 8% leakage reactance and 0.8 % leakage resistance;

Solution:
The one line diagram with the data is obtained as shown in figure P4(a).

21
3-phase ratings of transformers:

T1: 3(10) = 30 MVA, 13.2/ 66.43 KV = 13.2/ 115 KV, X = 0.077, R = 0.005 pu.
T2: 3(8.33) = 25 MVA, 110/ 3.983 KV = 110/ 6.8936 KV, X = 0.08, R = 0.008 pu.

Selection of base quantities:

25 MVA, 6.6 KV in the motor circuit (Given); the voltage bases in other sections are: 6.6
(110/6.8936) = 105.316 KV in the transmission line circuit and 105.316 (13.2/115) =
12.09 KV in the generator circuit.

Calculation of pu values:
Xm = j 0.25 pu; Em = 1.000 pu.
XG = j 0.005 (25/33) (13.8/12.09)2 = j 0.005 pu; Eg = 13.8/12.09 = 1.41400 pu.
Zt1 = 0.005 + j 0.077 (25/30) (13.2/12.09)2 = 0.005 + j 0.0765 pu. (ref. to LV side)
Zt2 = 0.008 + j 0.08 (25/25) (110/105.316)2 = 0.0087 + j 0.0873 pu. (ref. to HV side)
Zline = 75 (0.2+j 0.8) (25/ 105.3162) = 0.0338 + j 0.1351 pu.

Thus the pu reactance diagram can be drawn as shown in figure P4(b).

22
1.8 Exercises for Practice

Problems

1. Determine the reactances of the three generators rated as follows on a common base of
200 MVA, 35 KV: Generator 1: 100 MVA, 33 KV, sub transient reactance of 10%;
Generator 2: 150 MVA, 32 KV, sub transient reactance of 8% and Generator 3: 110
MVA, 30 KV, sub transient reactance of 12%.
[Answers: XG1 = j 0.1778, Xg2 = j 0.089, Xg3 = j 0.16 all in per unit]

2. A 100 MVA, 33 KV, 3-phase generator has a sub transient reactance of 15%. The
generator supplies 3 motors through a step-up transformer - transmission line – step-
down transformer arrangement. The motors have rated inputs of 30 MVA, 20 MVA and
50 MVA, at 30 KV with 20% sub transient reactance each. The 3-phase transformers are
rated at 100 MVA, 32 KV- /110 KV-Y with 8 % leakage reactance. The line has a
reactance of 50 ohms. By selecting the generator ratings as base values in the generator
circuit, determine the base values in all the other parts of the system. Hence evaluate the
corresponding pu values and draw the equivalent per unit reactance diagram.
[Answers: XG = j 0.15, Xm1 = j 0.551, Xm2 = j 0.826, Xm3 = j 0.331, Eg1=1.0 0 , Em1 = Em2
0

= Em3 = 0.9100, Xt1 = Xt2 = j 0.0775 and Xline = j 0.39 all in per unit]

3. A 80 MVA, 10 KV, 3-phase generator has a sub transient reactance of 10%. The
generator supplies a motor through a step-up transformer - transmission line – step-down
transformer arrangement. The motor has rated input of 95 MVA, 6.3 KV with 15% sub
transient reactance. The step-up 3-phase transformer is rated at 90 MVA, 11 KV-Y /110
KV-Y with 10% leakage reactance. The 3-phase step-down transformer consists of three
single phase Y- connected transformers, each rated at 33.33 MVA, 68/6.6 KV with 10%
leakage reactance. The line has a reactance of 20 ohms. By selecting the 11 KV, 100
MVA as base values in the generator circuit, determine the base values in all the other
parts of the system. Hence evaluate the corresponding pu values and draw the equivalent
per unit reactance diagram.
[Answers: XG = j 1.103, Xm = j 0.165, Eg1=0.9100, Em= 1.02200, Xt1 = j 0.11, Xt2 = j
0.114 and Xline = j 0.17 all in per unit]

4. For the three-phase system shown below, draw an impedance diagram expressing all
impedances in per unit on a common base of 20 MVA, 2600 V on the HV side of the
transformer. Using this impedance diagram, find the HV and LV currents.

23
 [Answers: Sb = 20 MVA; Vb=2.6 KV (HV) and 0.2427 KV (LV); Vt=1.000, Xt = j 0.107,
Zcable = 0.136 +j 0.204 and Zload = 5.66 + j 2.26, I = 0.158 all in per unit, I
(hv)= 0.7 A and I (lv) = 7.5 A]

Objective type questions

1. Under no load conditions the current in a transmission line is due to.

a) Corona effects
b) Capacitance of the line
c) Back flow from earth
d) None of the above
2. In the short transmission line which of the following is used?
a)  - Model
b) T – Model
c) Both (a) and (b)
d) None of the above
3. In the short transmission line which of the following is neglected?
a) I2 R loss
b) Shunt admittance
c) Series impedance
d) All of the above
4. Which of the following loss in a transformer is zero even at full load?
a) Eddy current
b) Hysteresis
c) Core loss
d) Friction loss
5. The transmission line conductors are transposed to
a) Balance the current
b) Obtain different losses
c) Obtain same line drops
d) Balance the voltage

[Ans.: 1(b), 2(a), 3(b), 4(d), 5(c)]

24
 CHAPTER 2

SYMMETRICAL THREE PHASE FAULTS

[CONTENTS: Preamble, transients on a transmission line, short circuit of an unloaded

synchronous machine- short circuit currents and reactances, short circuit of a loaded
machine, selection of circuit breaker ratings, examples]

2.1 Preamble

in practice, any disturbance in the normal working conditions is termed as a FAULT. The
effect of fault is to load the device electrically by many times greater than its normal
rating and thus damage the equipment involved. Hence all the equipment in the fault line
should be protected from being overloaded. In general, overloading involves the increase
of current up to 10-15 times the rated value. In a few cases, like the opening or closing of
a circuit breaker, the transient voltages also may overload the equipment and damage
them.

In order to protect the equipment during faults, fast acting circuit breakers are put in the
lines. To design the rating of these circuit breakers or an auxiliary device, the fault
current has to be predicted. By considering the equivalent per unit reactance diagrams,
the various faults can be analyzed to determine the fault parameters. This helps in the
protection and maintenance of the equipment.

Faults can be symmetrical or unsymmetrical faults. In symmetrical faults, the fault

quantity rises to several times the rated value equally in all the three phases. For example,
a 3-phase fault - a dead short circuit of all the three lines not involving the ground. On the
other hand, the unsymmetrical faults may have the connected fault quantities in a random
way. However, such unsymmetrical faults can be analyzed by using the Symmetrical
Components. Further, the neutrals of the machines and equipment may or may not be
grounded or the fault may occur through fault impedance. The three-phase fault involving
ground is the most severe fault among the various faults encountered in electric power
systems.

2.2 Transients on a transmission line

Now, let us Consider a transmission line of resistance R and inductance L supplied by an

ac source of voltage v, such that v = Vm sin (t+) as shown in figure 1. Consider the
short circuit transient on this transmission line. In order to analyze this symmetrical 3-
phase fault, the following assumptions are made:
 The supply is a constant voltage source,
 The short circuit occurs when the line is unloaded and

25
  The line capacitance is negligible.

Figure 1. Short Circuit Transients on an Unloaded Line.

Thus the line can be modeled by a lumped R-L series circuit. Let the short circuit take
place at t=0. The parameter,  controls the instant of short circuit on the voltage wave.
From basic circuit theory, it is observed that the current after short circuit is composed of
the two parts as under: i =is +it, Where, is is the steady state current and it is the transient
current. These component currents are determined as follows.
Consider, v = Vm sin (t+)
= iR + L (di/dt) (2.1)
and i = Im sin (t+-) (22.)
Where Vm = 2V; Im = 2I; Zmag = [R2+(L)2]= tan-1(L/R) (2.3)
Thus is = [Vm/Z] sin (t+-) (2.4)
Consider the performance equation of the circuit of figure 1 under circuit as:
iR + L (di/dt) = 0
i.e., (R/L + d/dt)i = 0 (2.5)
In order to solve the equation (5), consider the complementary function part of the
solution as: CF = C1 e(-t/) (2.6)
Where  (= L/R) is the time constant and C1 is a constant given by the value of steady
state current at t = 0. Thus we have,
C1 = -is(0)
= - [Vm/Z] sin (-)
= [Vm/Z] sin (-) (2.7)
Similarly the expression for the transient part is given by:
it = -is(0) e(-t/)
= [Vm/Z] sin (-) e(-R/L)t (2.8)
Thus the total current under short circuit is given by the solution of equation (1) as
[combining equations (4) and (8)],

26
 i =is +it
= [2V/Z] sin (t+-) + [2V/Z] sin (-) e(-R/L)t (2.9)
Thus, is is the sinusoidal steady state current called as the symmetrical short circuit
current and it is the unidirectional value called as the DC off-set current. This causes the
total current to be unsymmetrical till the transient decays, as clearly shown in figure 2.

Figure 2. Plot of Symmetrical short circuit current, i(t).

The maximum momentary current, imm thus corresponds to the first peak. Hence, if the
decay in the transient current during this short interval of time is neglected, then we have
(sum of the two peak values);
imm = [2V/Z] sin (-) + [2V/Z] (2.10)
now, since the resistance of the transmission line is very small, the impedance angle ,
can be taken to be approximately equal to 90 0. Hence, we have
imm = [2V/Z] cos  + [2V/Z] (2.11)

27
This value is maximum when the value of  is equal to zero. This value corresponds to
the short circuiting instant of the voltage wave when it is passing through zero. Thus the
final expression for the maximum momentary current is obtained as:
imm = 2 [2V/Z] (2.12)
Thus it is observed that the maximum momentary current is twice the maximum value of
symmetrical short circuit current. This is refered as the doubling effect of the short circuit
current during the symmetrical fault on a transmission line.

2.3 Short circuit of an unloaded synchronous machine

2.3.1 Short Circuit Reactances

Under steady state short circuit conditions, the armature reaction in synchronous
generator produces a demagnetizing effect. This effect can be modeled as a reactance, Xa
in series with the induced emf and the leakage reactance, X l of the machine as shown in
figure 3. Thus the equivalent reactance is given by:
Xd = Xa +Xl (2.13)
Where Xd is called as the direct axis synchronous reactance of the synchronous machine.
Consider now a sudden three-phase short circuit of the synchronous generator on no-load.
The machine experiences a transient in all the 3 phases, finally ending up in steady state
conditions.

Figure 3. Steady State Short Circuit Model

Immediately after the short circuit, the symmetrical short circuit current is limited only by
the leakage reactance of the machine. However, to encounter the demagnetization of the
armature short circuit current, current appears in field and damper windings, assisting the
rotor field winding to sustain the air-gap flux. Thus during the initial part of the short
circuit, there is mutual coupling between stator, rotor and damper windings and hence the
corresponding equivalent circuit would be as shown in figure 4. Thus the equivalent
reactance is given by:
Xd” = Xl +[1/Xa + 1/Xf + 1/Xdw]-1 (2.14)

28
Where Xd” is called as the sub-transient reactance of the synchronous machine. Here, the
equivalent resistance of the damper winding is more than that of the rotor field winding.
Hence, the time constant of the damper field winding is smaller. Thus the damper field
effects and the eddy currents disappear after a few cycles.

Figure 4. Model during Sub-transient Period of Short Circuit

In other words, Xdw gets open circuited from the model of Figure 5 to yield the model as
shown in figure 4. Thus the equivalent reactance is given by:
Xd’ = Xl +[1/Xa + 1/Xf ]-1 (2.15)
Where Xd’ is called as the transient reactance of the synchronous machine.
Subsequently, Xf also gets open circuited depending on the field winding time constant
and yields back the steady state model of figure 3.

Figure 5. Model during transient Period of Short Circuit

Thus the machine offers a time varying reactance during short circuit and this value of
reactance varies from initial stage to final one such that: Xd  Xd’  Xd’

2.3.2 Short Circuit Current Oscillogram

Consider the oscillogram of short circuit current of a synchronous machine upon the
occurrence of a fault as shown in figure 6. The symmetrical short circuit current can be
divided into three zones: the initial sub transient period, the middle transient period and
finally the steady state period. The corresponding reactances, Xd,” Xd’ and Xd
respectively, are offered by the synchronous machine during these time periods.

29
 Figure 6. SC current Oscillogram of Armature Current.

The currents and reactances during the three zones of period are related as under in terms
of the intercepts on the oscillogram (oa, ob and oc are the y-intercepts as indicated in
figure 6):
RMS value of the steady state current = I = [oa/2] = [Eg/Xd]
RMS value of the transient current = I’ = [ob/2] = [Eg/Xd’]
RMS value of the sub transient current = I = [oc/2] = [Eg/Xd”] (2.16)

2.4 short circuit of a loaded machine

In the analysis of section 2.3 above, it has been assumed that the machine operates at no
load prior to the occurrence of the fault. On similar lines, the analysis of the fault
occurring on a loaded machine can also be considered.
Figure 7 gives the circuit model of a synchronous generator operating under steady state
conditions supplying a load current Il to the bus at a terminal voltage Vt. Eg is the induced
emf under the loaded conditions and Xd is the direct axis synchronous reactance of the
generator.

Figure 7. Circuit models for a fault on a loaded machine.

30
Also shown in figure 7, are the circuit models to be used for short circuit current
calculations when a fault occurs at the terminals of the generator, for sub-transient current
and transient current values. The induced emf values used in these models are given by
the expressions as under:

Eg = Vt + j ILXd = Voltage behind syn. reactance

Eg’= Vt + j ILXd’ = Voltage behind transient reactance
Eg“= Vt + j ILXd” = Voltage behind subtr. Reactance (2.17)

The synchronous motors will also have the terminal emf values and reactances. However,
then the current direction is reversed. During short circuit studies, they can be replaced by
circuit models similar to those shown in figure 7 above, except that the voltages are given
by the relations as under:

Em = Vt - j ILXd = Voltage behind syn. reactance

Em’= Vt - j ILXd’ = Voltage behind transient reactance
Em“= Vt - j ILXd” = Voltage behind subtr. Reactance (2.18)

The circuit models shown above for the synchronous machines are also very useful while
dealing with the short circuit of an interconnected system.

2.5 Selection of circuit breaker ratings

For selection of circuit breakers, the maximum momentary current is considered
corresponding to its maximum possible value. Later, the current to be interrupted is
usually taken as symmetrical short circuit current multiplied by an empirical factor in
order to account for the DC off-set current. A value of 1.6 is usually selected as the
multiplying factor.
Normally, both the generator and motor reactances are used to determine the momentary
current flowing on occurrence of a short circuit. The interrupting capacity of a circuit
breaker is decided by Xd” for the generators and Xd’ for the motors.

2.6 Examples

Problem #1: A transmission line of inductance 0.1 H and resistance 5  is suddenly

short circuited at t = 0, at the far end of a transmission line and is supplied by an ac
source of voltage v = 100 sin (100t+150). Write the expression for the short circuit
current, i(t). Find the approximate value of the first current maximum for the given
values of  and . What is this value for =0, and =900? What should be the instant of
short circuit so that the DC offset current is (i)zero and (ii)maximum?

31
Solution:

Figure P1.

Consider the expression for voltage applied to the transmission system given by
v = Vm sin(t+) = 100 sin (100t+150)
Thus we get: Vm = 100 volts; f = 50 Hz and  = 150.
Consider the impedance of the circuit given by:
Z = R + jL = 5 + j (100) (0.1) = 5 + j 31.416 ohms.
Thus we have: Zmag=31.8113 Ohms; =80.9570 and =L/R=0.1/5=0.02 seconds.
The short circuit current is given by:
i(t) = [Vm/Z] sin (t+-) + [Vm/Z] sin (-) e-(R/L)t
= [100/31.8113] [sin (100t+150-80.9570) + sin(80.9570-150) e-(t/0.02)]
= 3.1435 sin(314.16 t – 65.96) +2.871 e–50t
Thus we have:
i) imm = 3.1435 + 2.871 e–50t
where t is the time instant of maximum of symmetrical short circuit current. This instant
occurs at (314.16 tc – 65.960) = 900 ; Solving we get, t = 0.00867 seconds so that imm = 5
Amps.
ii) imm = 2Vm/Z = 6.287 A; for =0, and =900 (Also, imm = 2 (3.1435) = 6.287 A)
iii) DC offset current = [Vm/Z] sin (-) e-(R/L)t
= zero, if (-) = zero, i.e.,  = , or  = 80.9570
= maximum if (-) = 900, i.e.,  =  - 900, or  = - 9.0430.

Problem #2: A 25 MVA, 11 KV, 20% generator is connected through a step-up

transformer- T1 (25 MVA, 11/66 KV, 10%), transmission line (15% reactance on a base
of 25 MVA, 66 KV) and step-down transformer-T2 (25 MVA, 66/6.6 KV, 10%) to a bus
that supplies 3 identical motors in parallel (all motors rated: 5 MVA, 6.6 KV, 25%). A
circuit breaker-A is used near the primary of the transformer T1 and breaker-B is used
near the motor M3. Find the symmetrical currents to be interrupted by circuit breakers A
and B for a fault at a point P, near the circuit breaker B.

32
Solution:
Consider the SLD with the data given in the problem statement. The base values are
selected as under:

Figure P2(a)
Selection of bases:
Sb = 25 MVA (common); Vb = 11 KV (Gen. circuit)- chosen so that then Vb = 66 KV
(line circuit) and Vb = 6.6 KV (Motor circuit).
Pu values:
Xg=j0.2 pu, Xt1=Xt2=j0.1 pu; Xm1=Xm2=Xm3=j0.25(25/5)=j1.25 pu; Xline=j0.15 pu.
Since the system is operating at no load, all the voltages before fault are 1 pu.
Considering the pu reactance diagram with the faults at P, we have:

Figure P2(b)

Current to be interrupted by circuit breaker A = 1.0 /j[0.2+0.1+0.15+0.1]

= - j 1.818 pu = - j 1.818 (25/[3(11)]) = - j 1.818 (1.312) KA = 2.386 KA
And Current to be interrupted by breaker B = 1/j1.25 = - j 0.8 pu
= - j0.8 (25/[3(6.6)]) = - j0.8 (2.187) KA = 1.75 KA.

33
Problem #3: Two synchronous motors are connected to a large system bus through a
short line. The ratings of the various components are: Motors(each)= 1 MVA, 440 volts,
0.1 pu reactance; line of 0.05 ohm reactance and the short circuit MVA at the bus of the
large system is 8 at 440 volts. Calculate the symmetrical short circuit current fed into a
three-phase fault at the motor bus when the motors are operating at 400 volts.

Solution:
Consider the SLD with the data given in the problem statement. The base values are
selected as under:

Figure P3.
Sb = 1 MVA; Vb = 0.44 KV (common)- chosen so that Xm(each)=j0.1 pu, Em = 1.000,
Xline=j0.05 (1/0.442) = j 0.258 pu and Xlarge-system -= (1/8) = j 0.125 pu.
Thus the prefault voltage at the motor bus; Vt = 0.4/0.44 = 0.90900,
Short circuit current fed to the fault at motor bus (If = YV);
If = [0.125 + 0.258]-1 + 2.0 }0.909 = [20.55 pu] [1000/(3(0.4))]
= 20.55 (1.312) KA = 26.966 KA.

Problem #4: A generator-transformer unit is connected to a line through a circuit

breaker. The unit ratings are: Gen.: 10 MVA, 6.6 KV, Xd” = 0.1 pu, Xd’ = 0.2 pu and Xd
= 0.8 pu; and Transformer: 10 MVA, 6.9/33 KV, Xl = 0.08 pu; The system is operating
on no-load at a line voltage of 30 KV, when a three-phase fault occurs on the line just
beyond the circuit breaker. Determine the following:
(i) Initial symmetrical RMS current in the breaker,
(ii) Maximum possible DC off-set current in the breaker,
(iii) Momentary current rating of the breaker,
(iv) Current to be interrupted by the breaker and the interrupting KVA and
(v) Sustained short circuit current in the breaker.
Solution:

34
Consider the base values selected as 10 MVA, 6.6 KV (in the generator circuit) and
6.6(33/6.9) = 31.56 KV(in the transformer circuit). Thus the base current is:
Ib = 10 / [3(31.56)] = 0.183 KA
The pu values are: Xd” = 0.1 pu, Xd’ = 0.2 pu and Xd = 0.8 pu; and XTr = 0.08 (6.9/6.6)2
= 0.0874 pu; Vt = (30/31.6) = 0.9500 pu.
Initial symmetrical RMS current = 0.9500 / [0.1 + 0.0874] = 5.069 pu = 0.9277 KA;
Maximum possible DC off-set current = 2 (0.9277) = 1.312 KA;
Momentary current rating = 1.6(0.9277) = 1.4843 KA; (assuming 60% allowance)
Current to be interrupted by the breaker (5 Cycles) = 1.1(0.9277) = 1.0205 KA;
Interrupting MVA = 3(30) (1.0205) = 53.03 MVA;
Sustained short circuit current in the breaker = 0.9500 (0.183) / [0.8 + 0.0874]
= 0.1959 KA.

2.7 Exercises for Practice

PROBLEMS
1. The one line diagram for a radial system network consists of two generators, rated 10
MVA, 15% and 10 MVA, 12.5 % respectively and connected in parallel to a bus bar A at
11 KV. Supply from bus A is fed to bus B (at 33 KV) through a transformer T 1 (rated: 10
MVA, 10%) and OH line (30 KM long). A transformer T2 (rated: 5 MVA, 8%) is used in
between bus B (at 33 KV) and bus C (at 6.6 KV). The length of cable running from the
bus C up to the point of fault, F is 3 KM. Determine the current and line voltage at 11 kV
bus A under fault conditions, when a fault occurs at the point F, given that Zcable = 0.135
+ j 0.08 ohm/ kM and ZOH-line = 0.27 + j 0.36 ohm/kM. [Answer: 9.62 kV at the 11 kV
bus]
2. A generator (rated: 25MVA, 12. KV, 10%) supplies power to a motor (rated: 20 MVA,
3.8 KV, 10%) through a step-up transformer (rated:25 MVA, 11/33 KV, 8%),
transmission line (of reactance 20 ohms) and a step-down transformer (rated:20 MVA,
33/3.3 KV, 10%). Write the pu reactance diagram. The system is loaded such that the
motor is drawing 15 MW at 0.9 leading power factor, the motor terminal voltage being
3.1 KV. Find the sub-transient current in the generator and motor for a fault at the
generator bus. [Answer: Ig” = 9.337 KA; Im” = 6.9 KA]
3. A synchronous generator feeds bus 1 and a power network feed bus 2 of a system.
Buses 1 and 2 are connected through a transformer and a line. Per unit reactances of the
components are: Generator(bus-1):0.25; Transformer:0.12 and Line:0.28. The power
network is represented by a generator with an unknown reactance in series. With the
generator on no-load and with 1.0 pu voltage at each bus, a three phase fault occurring on
bus-1 causes a current of 5 pu to flow into the fault. Determine the equivalent reactance
of the power network. [ Answer: X = 0.6 pu]
4. A synchronous generateor, rated 500 KVA, 440 Volts, 0.1 pu sub-transient reactance is
supplying a passive load of 400 KW, at 0.8 power factor (lag). Calculate the initial
symmetrical RMS current for a three-phase fault at the generator terminals.
[Answer: Sb=0.5 MVA; Vb=0.44 KV; load=0.8–36.90; Ib=0.656 KA; If=6.97 KA]

35
OBJECTIVE TYPE QUESTIONS

1. When a 1-phase supply is across a 1-phase winding, the nature of the

magnetic field produced is
a) Constant in magnitude and direction
b) Constant in magnitude and rotating at synchronous speed
c) Pulsating in nature
d) Rotating in nature
2. The damper windings are used in alternators to
a) Reduce eddy current loss
b) Reduce hunting
c) Make rotor dynamically balanced
d) Reduce armature reaction
3. The neutral path impedance Zn is used in the equivalent sequence network
models as
a) Zn2
b) Zn
c) 3 Zn
d) An ineffective value
4. An infinite bus-bar should maintain
a) Constant frequency and Constant voltage
b) Infinite frequency and Infinite voltage
c) Constant frequency and Variable voltage
d) Variable frequency and Variable voltage
5. Voltages under extra high voltage are
a) 1KV & above
b) 11KV & above
c) 132 KV & above
d) 330 KV & above
[Ans.: 1(c), 2(b), 3(c), 4(a), 5(d)]

36
 CHAPTER 3: SYMMETRICAL COMPONENTS

[CONTENTS: Introduction, The a operator, Power in terms of symmetrical components, Phase shift in Y-
Δ transformer banks, Unsymmetrical series impedances, Sequence impedances, Sequence
networks, Sequence networks of an unloaded generator, Sequence networks of elements,
Sequence networks of power system]

3.1 INTRODUCTION

Power systems are large and complex three-phase systems. In the normal operating
conditions, these systems are in balanced condition and hence can be represented as an
equivalent single phase system. However, a fault can cause the system to become
unbalanced. Specifically, the unsymmetrical faults: open circuit, LG, LL, and LLG faults
cause the system to become unsymmetrical. The single-phase equivalent system method
of analysis (using SLD and the reactance diagram) cannot be applied to such
unsymmetrical systems. Now the question is how to analyze power systems under
unsymmetrical conditions? There are two methods available for such an analysis:
Kirchhoff’s laws method and Symmetrical components method.

The method of symmetrical components developed by C.L. Fortescue in 1918 is a

powerful technique for analyzing unbalanced three phase systems. Fortescue defined a
linear transformation from phase components to a new set of components called
symmetrical components. This transformation represents an unbalanced three-phase
system by a set of three balanced three-phase systems. The symmetrical component
method is a modeling technique that permits systematic analysis and design of three-
phase systems. Decoupling a complex three-phase network into three simpler networks
reveals complicated phenomena in more simplistic terms.

Consider a set of three-phase unbalanced voltages designated as Va, Vb, and Vc.
According to Fortescue theorem, these phase voltages can be resolved into following
three sets of components.

1. Positive-sequence components, consisting of three phasors equal in magnitude,

displaced from each other by 1200 in phase, and having the same phase sequence as
the original phasors, designated as Va1, Vb1, and Vc1
2. Negative-sequence components, consisting of three phasors equal in magnitude,
displaced from each other by 1200 in phase, and having the phase sequence opposite
to that of the original phasors, designated as Va2, Vb2, and Vc2
3. Zero-sequence components, consisting of three phasors equal in magnitude, and with
zero phase displacement from each other, designated as V a0, Vb0, and Vc0
Since each of the original unbalanced phasors is the sum of its components, the
original phasors expressed in terns of their components are

Va = Va1 + Va2 + Va0

Vb = Vb1 + Vb2 + Vb0
Vc = Vc1 + Vc2 + Vc0 (3.1)

37
The synthesis of a set of three unbalanced phasors from the three sets of symmetrical
components is shown in Figure1.

Figure 3.1 Graphical addition of symmetrical components

To obtain unbalanced phasors.

3.2 THE OPERATOR ‘a’

The relation between the symmetrical components reveals that the phase displacement
among them is either 1200 or 00. Using this relationship, only three independent
components is sufficient to determine all the nine components. For this purpose an
operator which rotates a given phasor by 1200 in the positive direction (counterclockwise)
is very useful. The letter ‘a’ is used to designate such a complex operator of unit
magnitude with an angle of 1200. It is defined by

a = 1120 0 = -0.5 + j 0.866 (3.2)

38
If the operator ‘a’ is applied to a phasor twice in succession, the phasor is rotated through
2400. Similarly, three successive applications of ‘a’ rotate the phasor through 3600.

To reduce the number of unknown quantities, let the symmetrical components of

Vb and Vc can be expressed as product of some function of the operator a and a
component of Va. Thus,

Vb1 = a 2 Va1 Vb2 = a Va2 Vb0 = Va0

Vc1 = a Va1 Vc2 = a 2 Va2 Vc0 = Va0
Using these relations the unbalanced phasors can be written as

Va = Va0 + Va1 + Va2

Vb = Va0 + a 2Va1 + a Va2
Vc = Va0 + a Va1 + a 2Va2 (3.3)

In matrix form,

va  1 1 1 va 0 
v   1 a 2 a   v  (3.4)
 b   a1 
 vc 1 a a2 va 2

va   v a 0  1 1 1 
 
Let Vp  vb  ;
 Vs   va1  ; A  1 a 2 a  (3.5)
 vc va 2 1 a a2

The inverse of A matrix is


1 1 1  
A  3 1 a a 2 
1  1 
(3.6)
 
1 a 2 a 
With these definitions, the above relations can be written as

Vp = A Vs; Vs = A-1Vp (3.7)

Thus the symmetrical components of Va, Vb and Vc are given by

Va0 = 1/3 (Va + Vb + Vc)

Va1 = 1/3 (Va + a Vb + a 2Vc)
Va2 = 1/3 (Va + a 2Vb + a Vc) (3.8)

Since the sum of three balanced voltages is zero, the zero-sequence component voltage in
a balanced three-phase system is always zero. Further, the sum of line voltages of even an
unbalanced three-phase system is zero and hence the corresponding zero-sequence
component of line voltages.

39
NUMERICAL EXAMPLES

Example 1 : The line currents in a 3-ph 4 –wire system are Ia = 100<300; Ib = 50<3000;
Ic = 30<1800. Find the symmetrical components and the neutral current.

Solution:
Ia0 = 1/3(Ia + Ib + Ic) = 27.29 < 4.70 A
Ia1 = 1/3(Ia + a Ib + a2Ic) = 57.98 < 43.30 A
Ia2 = 1/3(Ia + a2 Ib + a Ic) = 18.96 < 24.90 A
In = Ia + Ib + Ic = 3 Ia0 = 81.87 <4.70 A

Example 2: The sequence component voltages of phase voltages of a 3-ph system are:
Va0 = 100 <00 V; Va1 = 223.6 < -26.60 V ; Va2 = 100 <1800 V. Determine the phase
voltages.

Solution:
Va = Va0 + Va1 + Va2 = 223.6 <-26.60 V
Vb = Va0 + a2Va1 + a Va2 = 213 < -99.90 V
Vc = Va0 + a Va1 + a2 Va2 = 338.6 < 66.20 V

Example 3: The two seq. components and the corresponding phase voltage of a 3-ph
system are Va0 =1<-600 V; Va1=2<00 V ; & Va = 3 <00 V. Determine the other phase
voltages.

Solution:
Va = Va0 + Va1 + Va2
Va2 = Va – Va0 – Va1 = 1 <600 V
Vb = Va0 + a2Va1 + a Va2 = 3 < -1200 V
Vc = Va0 + a Va1 + a2 Va2 = 0 V

Example 4: Determine the sequence components if Ia =10<60 0 A; Ib =10<-600 A; Ic =

10 <1800 A.

Solution:
Ia0 = 1/3(Ia + Ib + Ic) =0A
Ia1 = 1/3(Ia + a Ib + a2Ic) = 10<600 A
Ia2 = 1/3(Ia + a2 Ib + a Ic) = 0 A
Observation: If the phasors are balanced, two sequence components will be zero.
0
Example 5: Determine the sequence components if Va = 100 <30 V; Vb = 100
<1500 V & Vc = 100 <-900 V.

Solution:
Va0 = 1/3(Va + Vb + Vc) =0V
Va1 = 1/3(Va + a Vb + a2Vc) = 0 V
0
Va2 = 1/3(Va + a2 Vb + a Vc) = 100<30 V
Observation: If the phasors are balanced, two sequence components will be zero.

40
Example 6: The line b of a 3-ph line feeding a balanced Y-load with neutral grounded is
open resulting in line currents: Ia = 10<00 A & Ic = 10<1200 A. Determine the sequence
current components.

Solution:
Ib = 0 A.
Ia0 = 1/3(Ia + Ib + Ic) = 3.33<600 A
Ia1 = 1/3(Ia + a Ib + a2Ic) = 6.66<00 A
Ia2 = 1/3(Ia + a2 Ib + a Ic) = 3.33<-600 A

Example 7: One conductor of a 3-ph line feeding a balanced delta-load is open.

Assuming that line c is open, if current in line a is 10<00 A , determine the sequence
components of the line currents.

Solution:
Ic = 0 A; Ia = 10<00 A.  Ib = 10<1200 A
Ia0 = 1/3(Ia + Ib + Ic) = 0A
Ia1 = 1/3(Ia + a Ib + a2Ic) = 5.78<-300 A
0
Ia2 = 1/3(Ia + a2 Ib + a Ic) = 5.78< 30 A
Note: The zero-sequence components of line currents of a delta load (3-ph 3-wire) system
are zero.

3.3 POWER IN TERMS OF SYMMETRICAL COMPONENTS

The power in a three-phase system can be expressed in terms of symmetrical components

of the associated voltages and currents. The power flowing into a three-phase system
through three lines a, b and c is

S = P + j Q = Va Ia* + Vb Ib* + Vc Ic * (3.9)

where Va , Vb and Vc are voltages to neutral at the terminals and Ia , Ib, and Ic are the
currents flowing into the system in the three lines. In matrix form

I a  Va   I a 

* T *

S  va
   V   I 
v  Ib
   b   b 
vb c

Ic Vc I c
Thus
S = [A V]T [AI] *

Using the reversal rule of the matrix algebra

S = V T AT A * I *

Noting that AT = A and a and a 2 are conjugates,

41
 I 
*

1 1 1 1 1 1   a0 
S  va 0 
va 2  1 a 2 a 1 a a  I
 
  a1 
2
va1
  
1 a a 2  1 a 2 a 
I a2
or, since A T A* is equal to 3U where U is 3x3 unit matrix

 I a0 
*

S  3 va 0  I a1 
 
va1 va 2

I a2
Thus the complex three-phase power is given by

S = Va Ia* + Vb Ib* + Vc Ic * = 3 Va0 Ia0 + 3 Va1 Ia1 + 3 Va2 Ia2 (3.10)

Here, 3Va0Ia0, 3Va1Ia1 and 3Va2Ia2 correspond to the three-phase power delivered to the
zero-sequence system, positive-sequence system, and negative-sequence system,
respectively. Thus, the total three-phase power in the unbalanced system is equal to the
sum of the power delivered to the three sequence systems representing the three-phase
system.

3.4 PHASE SHIFT OF COMPONENTS IN Y-Δ TRANSFORMER BANKS

The dot convention is used to designate the terminals of transformers. The dots are placed
at one end of each of the winding on the same iron core of a transformer to indicate that
the currents flowing from the dotted terminal to the unmarked terminal of each winding
produces an mmf acting in the same direction in the magnetic circuit. In that case, the
voltage drops from dotted terminal to unmarked terminal in each side of the windings are
in phase.

The HT terminals of three-phase transformers are marked as H1, H2 and H3 and the
corresponding LT side terminals are marked X1, X2 and X3. In Y-Y or Δ-Δ transformers,
the markings are such that voltages to neutral from terminals H1, H2, and H3 are in phase
with the voltages to neutral from terminals X1, X2, and X3, respectively. But, there will
be a phase shift (of 300) between the corresponding quantities of the primary and
secondary sides of a star-delta (or delta-star) transformer. The standard for connection
and designation of transformer banks is as follows:
1. The HT side terminals are marked as H1, H2 and H3 and the corresponding LT side
terminals are marked X1, X2 and X3.
2. The phases in the HT side are marked in uppercase letters as A, B, and C. Thus for
the sequence abc, A is connected to H1, B to H2 and C to H3. Similarly, the phases in
the LT side are marked in lowercase letters as a, b and c.
3. The standard for designating the terminals H1 and X1 on transformer banks requires
that the positive-sequence voltage drop from H1 to neutral lead the positive sequence
voltage drop from X1 to neutral by 300 regardless of the type of connection in the HT

42
 and LT sides. Similarly, the voltage drops from H2 to neutral and H3 to neutral lead
their corresponding values, X2 to neutral and X3 to neutral by 30 0.

Figure 3.2 Wiring diagram and voltage phasors of a Y-Δ transformer

With Y connection on HT side.

Consider a Y- Δ transformer as shown in Figure a. The HT side terminals H1, H2, and
H3 are connected to phases A, B, and C, respectively and the phase sequence is ABC.
The windings that are drawn in parallel directions are those linked magnetically (by being
wound on the same core). In Figure a winding AN is the phase on the Y-side which is
linked magnetically with the phase winding bc on the Δ side. For the location of the dots
on the windings VAN is in phase with Vbc. Following the standards for the phase shift, the
phasor diagrams for the sequence components of voltages are shown in Figure b. The
sequence component of VAN1 is represented as VA1 (leaving subscript ‘N’ for convenience
and all other voltages to neutral are similarly represented. The phasor diagram reveals
that VA1 leads Vb1 by 300. This will enable to designate the terminal to which b is
connected as X1. Inspection of the positive-sequence and negative-sequence phasor
diagrams revels that Va1 leads VA1 by 900 and Va2 lags VA2 by 900.

From the dot convention and the current directions assumed in Figure a, the phasor
diagram for the sequence components of currents can be drawn as shown in Figure c.
Since the direction specified for IA in Figure a is away from the dot in the winding and the
direction of Ibc is also away from the dot in its winding, IA and Ibc are 1800 out of phase.
Hence the phase relation between the Y and Δ currents is as shown in Figure c. From this
diagram, it can be seen that Ia1 leads IA1 by 900 and Ia2 lags IA2 by 900. Summarizing
these relations between the symmetrical components on the two sides of the transformer
gives:

43
 Figure 3.3 Current phasors of Y-Δ transformer with Y connection on HT side.

Va1 = +j VA1 Ia1 = +j IA1

Va2 = -j VA2 Ia1 = -j IA2 (3.11)
Where each voltage and current is expressed in per unit. Although, these relations are
obtained for Y- Δ transformer with Y connection in the HT side, they are valid even
when the HT side is connected in Δ and the LT side in Y.

NUMERICAL EXAMPLES

Example 8: Three identical resistors are Y-connected to the LT Y-side of a delta-star

transformer. The voltages at the resistor loads are |Vab| = 0.8 pu., |Vbc|=1.2 pu., and
|Vca|=1.0 pu. Assume that the neutral of the load is not connected to the neutral of the
transformer secondary. Find the line voltages on the HT side of the transformer.

Solution:
0
Assuming an angle of 180 for Vca, find the angles of other voltages

Vab = 0.8<82.80 pu
Vbc = 1.2<-41.40 pu
Vca = 1.0<1800 pu

The symmetrical components of line voltages are

Vab0 = 1/3 (Vab +Vbc + Vca) = 0

Vab1 = 1/3 (Vab +aVbc + a2Vca) = 0.985<73.60 V
Vab1 = 1/3 (Vab +a2Vbc + aVca) = 0.235<220.30 V
0 0
Since Van1 = Vab1<-30 and Van2 = Vab2<30
Van1 = 0.985<73.60-300
= 0.985<43.60 pu (L-L base)
Van2 = 0.235<220.30+300
= 0.235<250.30 pu(L-L base)

Since each resistor is of 1.0<0 pu. Impedance,

Ian1 = (Van1/Z) = 0.985<43.60 pu.

44
Ian2 = (Van2/Z) = 0.235<250.30 pu.

The directions are +ve for currents from supply toward the delta primary and away from
the Y-side toward the load. The HT side line to neutral voltages are

VA1 = - j Va1 = 0.985<-46.40

VA2 = +j Va2 = 0.235<-19.70
VA = VA1 +VA2 = 1.2<-41.30 pu.
VB1 = a2VA1 and VB2 = a VA2
0
VB = VB1 + VB2 = 1<180 pu.
VC1 = a VA1 and VC2 = a2VA2
VC = VC1 + VC2 = 0.8<82.90 pu.

The HT side line voltages are

VAB = VA – VB = 2.06<-22.60 pu. (L-N base)

= (1/3) VAB = 1.19<-22.60 pu. (L-L base)
VBC = VB – Vc = 1.355<215.80 pu. (L-N base)
= (1/3) VBC = 0.782<215.80 pu. (L-L base)
VCA = VC – VA = 1.78<116.90 pu. (L-N base)
= (1/3) VCA = 1.028<116.90 pu. (L-L base)

3.5 UNSYMMETRICAL IMPEDANCES

Figure 3.4 Portion of three-phase system representing three

unequal series impedances.

Consider the network shown in Figure. Assuming that there is no mutual impedance
between the impedances Za, Zb, and Zc, the voltage drops Vaa’, vbb’, and Vcc’ can be
expressed in matrix form as

45
 Vaa '   Z a 0 0  I a 
     
Vbb '    0 Zb 0 (3.12)
 I b 
Vcc ' 0 0 Zc Ic

And in terms of symmetrical components of voltage and current as

Vaa '0   Z a 0 0   I a 0 
A Vaa '1    0 Zb 0  A I a1 

 (3.13)
Vaa '2 0 0 Zc  I a 2

If the three impedances are equal ( i.e., if Za = Zb = Zc), Eq reduces to

Vaa’1 = Za Ia1; Vaa’2 = Za Ia2; Vaa’0 = Za Ia0 (3.14)

Thus, the symmetrical components of unbalanced currents flowing in balanced series

impedances (or in a balanced Y load) produce voltage drops of like sequence only.
However, if the impedances are unequal or if there exists mutual coupling, then voltage
drop of any one sequence is dependent on the currents of all the sequences.

46
 Figure 3.5 Sequence impedances of a Y-connected load.

NUMERICAL EXAMPLES

Example 9: A Y-connected source with phase voltages Vag = 277<00, Vbg = 260<-1200
and Vcg = 295<1150 is applied to a balanced Δ load of 30<400 Ω/phase through a line of
impedance 1<850 Ω. The neutral of the source is solidly grounded. Draw the sequence
networks of the system and find source currents.

Solution:
Va0 = 15.91<62.110 V
Va1 = 277.1<-1.70 V
Va2 = 9.22<216.70 V
Y eq. of Δ load = 10<400 Ω/phase
Zline = 1<850 Ω.
Zneutral = 0

Ia0 = 0<00 A
Ia1 = 25.82<-45.60 A
Ia2 = 0.86<172.80 A

Ia = 25.15<-46.80 A
Ib = 25.71<196.40 A
Ic = 26.62<73.80 A

47
3.6 SEQUENCE IMPEDANCES AND SEQUENCE NETWORKS

The impedance of a circuit to positive-sequence currents alone is called the impedance to

positive-sequence current or simply positive-sequence impedance, which is generally
denoted as Z1. Similarly, the impedance of a circuit to negative-sequence currents alone
is called the impedance to negative-sequence current or simply negative-sequence

48
impedance, which is generally denoted as Z2. The impedance of a circuit to zero-
sequence currents alone is called the impedance to zero-sequence current or simply zero-
sequence impedance, which is generally denoted as Z0. In the analysis of an
unsymmetrical fault on a symmetrical system, the symmetrical components of the
unbalanced currents that are flowing are determined. Since in a balanced system, the
components currents of one sequence cause voltage drops of like sequence only and are
independent of currents of other sequences, currents of any one sequence may be
considered to flow in an independent network composed of the generated voltages, if any,
and impedances to the current of that sequence only.

The single-phase equivalent circuit consisting of the impedances to currents of any one
sequence only is called the sequence network of that particular sequence. Thus, the
sequence network corresponding to positive-sequence current is called the positive-
sequence network. Similarly, the sequence network corresponding to negative-sequence
current is called negative-sequence network, and that corresponding to zero-sequence
current is called zero-sequence network. The sequence networks are interconnected in a
particular way to represent various unsymmetrical fault conditions. Therefore, to
calculate the effect of a fault by the method of symmetrical components, it is required to
determine the sequence networks.

3.7 SEQUENCE NETWORKS OF UNLOADED GENERATOR

Consider an unloaded generator which is grounded through a reactor as shown in Figure.

When a fault occurs, unbalanced currents depending on the type of fault will flow
through the lines. These currents can be resolved into their symmetrical components. To
draw the sequence networks of this generator, the component voltages/currents,
component impedances are to be determined. The generated voltages are of positive-
sequence only as the generators are designed to supply balanced three-phase voltages.
Hence, positive-sequence network is composed of an emf in series with the positive-
sequence impedance. The generated emf in this network is the no-load terminal voltage to
neutral, which is also equal to the transient and subtransient voltages as the generator is
not loaded. The reactance in this network is the subtransient, transient, or synchronous
reactance, depending on the condition of study.

Figure 3.6 Circuit of an unloaded generator grounded through reactance.

49
The negative- and zero-sequence networks are composed of only the respective sequence
impedances as there is no corresponding sequence emf. The reference bus for the
positive- and negative-sequence networks is the neutral of the generator.

The current flowing in the impedance Zn between neutral and ground is 3Ia0 as shown in
Fig. 3.6. Thus the zero-sequence voltage drop from point a to the ground, is given by: (-
Ia0Zg0 – 3Ia0Zn), where Zg0 is the zero-sequence impedance of the generator. Thus the
zero-sequence network, which is single-phase equivalent circuit assumed to carry only
one phase, must have an zero-sequence impedance of Zo = (Zg0 +3Zn).

From the sequence networks, the voltage drops from point a to reference bus (or ground)
are given by

Va1 = Ea - Ia1Z1
Va2 = - Ia2Z2
Va0 = - Ia0 Z0 (3.15)

Figure 3.7 Sequence current paths in a generator and

The corresponding sequence networks.

50
Eq. 3.15 applicable to any unloaded generator are valid for loaded generator under steady
state conditions. These relations are also applicable for transient or subtransient
conditions of a loaded generator if Eg’ or Eg” is substituted for Ea.

3.8 SEQUENCE IMPEDANCE OF CIRCUIT ELEMENTS

For obtaining the sequence networks, the component voltages/ currents and the
component impedances of all the elements of the network are to be determined. The usual
elements of a power system are: passive loads, rotating machines (generators/ motors),
transmission lines and transformers. The positive- and negative-sequence impedances of
linear, symmetrical, static circuits are identical (because the impedance of such circuits is
independent of phase order provided the applied voltages are balanced).

The sequence impedances of rotating machines will generally differ from one another.
This is due to the different conditions that exists when the sequence currents flows. The
flux due to negative-sequence currents rotates at double the speed of rotor while that the
positive-sequence currents is stationary with respect to the rotor. The resultant flux due to
zero-sequence currents is ideally zero as these flux components adds up to zero, and
hence the zero-sequence reactance is only due to the leakage flux. Thus, the zero-
sequence impedance of these machines is smaller than positive- and negative-sequence
impedances.

The positive- and negative-sequence impedances of a transmission line are identical,

while the zero-sequence impedance differs from these. The positive- and negative-
sequence impedances are identical as the transposed transmission lines are balanced
linear circuits. The zero-sequence impedance is higher due to magnetic field set up by the
zero-sequence currents is very different from that of the positive- or negative-sequence
currents ( because of no phase difference). The zero-sequence reactance is generally 2 to
3.5 times greater than the positive- sequence reactance. It is customary to take all the
sequence impedances of a transformer to be identical, although the zero-sequence
impedance slightly differs with respect to the other two.

3.9 SEQUENCE NETWORKS OF POWER SYSTEMS

In the method of symmetrical components, to calculate the effect of a fault on a power

system, the sequence networks are developed corresponding to the fault condition. These
networks are then interconnected depending on the type of fault. The resulting network is
then analyzed to find the fault current and other parameters.

Positive- and Negative-Sequence Networks: The positive-sequence network is obtained

by determining all the positive-sequence voltages and positive-sequence impedances of
individual elements, and connecting them according to the SLD. All the generated emfs
are positive-sequence voltages. Hence all the per unit reactance/impedance diagrams
obtained in the earlier chapters are positive-sequence networks. The negative-sequence
generated emfs are not present. Hence, the negative-sequence network for a power
system is obtained by omitting all the generated emfs (short circuiting emf sources) and

51
replacing all impedances by negative-sequence impedances from the positive-sequence
networks.

Since all the neutral points of a symmetrical three-phase system are at the same potential
when balanced currents are flowing, the neutral of a symmetrical three-phase system is
the logical reference point. It is therefore taken as the reference bus for the positive- and
negative-sequence networks. Impedances connected between the neutral of the machine
and ground is not a part of either the positive- or negative- sequence networks because
neither positive- nor negative-sequence currents can flow in such impedances.

Zero-Sequence Networks: The zero-sequence components are the same both in

magnitude and in phase. Thus, it is equivalent to a single-phase system and hence, zero-
sequence currents will flow only if a return path exists. The reference point for this
network is the ground (Since zero-sequence currents are flowing, the ground is not
necessarily at the same point at all points and the reference bus of zero-sequence network
does not represent a ground of uniform potential. The return path is conductor of zero
impedance, which is the reference bus of the zero-sequence network.).

If a circuit is Y-connected, with no connection from the neutral to ground or to another

neutral point in the circuit, no zero-sequence currents can flow, and hence the impedance
to zero-sequence current is infinite. This is represented by an open circuit between the
neutral of the Y-connected circuit and the reference bus, as shown in Fig. 3.8a. If the
neutral of the Y-connected circuit is grounded through zero impedance, a zero-impedance
path (short circuit) is connected between the neutral point and the reference bus, as
shown in Fig. 3.8b. If an impedance Zn is connected between the neutral and the ground
of a Y-connected circuit, an impedance of 3Zn must be connected between the neutral
and the reference bus (because, all the three zero-sequence currents (3Ia0) flows through
this impedance to cause a voltage drop of 3Ia0 Z0 ), as shown in Fig. 3.8c.

A Δ-connected circuit can provide no return path; its impedance to zero-sequence line
currents is therefore infinite. Thus, the zero-sequence network is open at the Δ-connected
circuit, as shown in Fig.3.9 However zero-sequence currents can circulate inside the Δ-
connected circuit.

The zero-sequence equivalent circuits of three-phase transformers deserve special

attention. The different possible combinations of the primary and the secondary windings
in Y and Δ alter the zero-sequence network. The five possible connections of two-
winding transformers and their equivalent zero-sequence networks are shown in Fig.3.10.
The networks are drawn remembering that there will be no primary current when there is
no secondary current, neglecting the no-load component. The arrows on the connection
diagram show the possible paths for the zero-sequence current. Absence of an arrow
indicates that the connection is such that zero-sequence currents cannot flow. The letters
P and Q identify the corresponding points on the connection diagram and equivalent
circuit:

52
 Figure 3.8 Zero-sequence equivalent networks of Y-connected load

Figure 3.9 Zero-sequence equivalent networks of Δ -connected load

1. Case 1: Y-Y Bank with one neutral grounded: If either one of the neutrals of a Y-Y
bank is ungrounded, zero-sequence current cannot flow in either winding ( as the
absence of a path through one winding prevents current in the other). An open circuit
exists for zero-sequence current between two parts of the system connected by the
transformer bank.
2. Case 2: Y-Y Bank with both neutral grounded: In this case, a path through
transformer exists for the zero-sequence current. Hence zero-sequence current can
flow in both sides of the transformer provided there is complete outside closed path
for it to flow. Hence the points on the two sides of the transformer are connected by
the zer0-sequence impedance of the transformer.

53
 Figure 3.10 Zero-sequence equivalent networks of three-phase
transformer banks for various combinations.

3. Case 3: Y- Δ Bank with grounded Y: In this case, there is path for zero-sequence
current to ground through the Y as the corresponding induced current can circulate in
the Δ. The equivalent circuit must provide for a path from lines on the Y side through
zero-sequence impedance of the transformer to the reference bus. However, an open
circuit must exist between line and the reference bus on the Δ side. If there is an
impedance Zn between neutral and ground, then the zero-sequence impedance must
include 3Zn along with zero-sequence impedance of the transformer.

54
4. Case 4: Y- Δ Bank with ungrounded Y: In this case, there is no path for zero-
sequence current. The zero-sequence impedance is infinite and is shown by an open
circuit.
5. Case 5: Δ-Δ Bank: In this case, there is no return path for zero-sequence current. The
zero-sequence current cannot flow in lines although it can circulate in the Δ windings.
6. The zero-sequence equivalent circuits determined for the individual parts separately
are connected according to the SLD to form the complete zero-sequence network.

Procedure to draw the sequence networks

The sequence networks are three separate networks which are the single-phase equivalent
of the corresponding symmetrical sequence systems. These networks can be drawn as
follows:

1. For the given condition (steady state, transient, or subtransient), draw the reactance
diagram (selecting proper base values and converting all the per unit values to the
selected base, if necessary). This will correspond to the positive-sequence network.

2. Determine the per unit negative-sequence impedances of all elements (if the values of
negative sequence is not given to any element, it can approximately be taken as equal
to the positive-sequence impedance). Draw the negative-sequence network by
replacing all emf sources by short circuit and all impedances by corresponding
negative-sequence impedances in the positive-sequence network.
3. Determine the per unit zero-sequence impedances of all the elements and draw the
zero-sequence network corresponding to the grounding conditions of different
elements.

NUMERICAL EXAMPLES

Example 10: For the power system shown in the SLD, draw the sequence networks.

55
EXERCISE PROBLEM: For the power system shown in the SLD, draw the sequence
networks.

56
 CHAPTER 4: UNSYMMETRICAL FAULTS

[CONTENTS: Preamble, L-G, L-L, L-L-G and 3-phase faults on an unloaded alternator without and with
fault impedance, faults on a power system without and with fault impedance, open
conductor faults in power systems, examples]

4.1 PREAMBLE

The unsymmetrical faults will have faulty parameters at random. They can be analyzed
by using the symmetrical components. The standard types of unsymmetrical faults
considered for analysis include the following (in the order of their severity):

 Line–to–Ground (L-G) Fault

 Line–to–Line (L-L) Fault
 Double Line–to–Ground (L-L-G)Fault and
 Three-Phase–to–Ground (LLL-G) Fault.

Further the neutrals of various equipment may be grounded or isolated, the faults can
occur at any general point F of the given system, the faults can be through a fault
impedance, etc. Of the various types of faults as above, the 3- fault involving the ground
is the most severe one. Here the analysis is considered in two stages as under: (i) Fault at
the terminals of a Conventional (Unloaded) Generator and (ii) Faults at any point F, of a
given Electric Power System (EPS).

Consider now the symmetrical component relational equations derived from the three
sequence networks corresponding to a given unsymmetrical system as a function of
sequence impedances and the positive sequence voltage source in the form as under:

Va0 = - Ia0Z0
Va1 = Ea - Ia1Z1
Va2 = - Ia2Z2 (4.1)

These equations are refered as the sequence equations. In matrix Form the sequence
equations can be considered as:

Va0 0 Z0 0 0 Ia0
Va1 = Ea – 0 Z1 0 Ia1
Va2 0 0 0 Z2 Ia2 (4.2)

This equation is used along with the equations i.e., conditions under fault (c.u.f.), derived
to describe the fault under consideration, to determine the sequence current Ia1 and hence
the fault current If, in terms of Ea and the sequence impedances, Z1, Z2 and Z0. Thus
during unsymmetrical fault analysis of any given type of fault, two sets of equations as
follows are considered for solving them simultaneously to get the required fault
parameters:
 Equations for the conditions under fault (c.u.f.)

57
  Equations for the sequence components (sequence equations) as per (4.2) above.
4.2 SINGLE LINE TO GROUND FAULT ON A CONVENTIONAL (UNLOADED)
GENERATOR

Figure 4.1 LG Fault on a Conventional Generator

A conventional generator is one that produces only the balanced voltages. Let Ea, nd Ec
be the internally generated voltages and Zn be the neutral impedance. The fault is
assumed to be on the phase’a’ as shown in figure 4.1. Consider now the conditions under
fault as under:

c.u.f.:
Ib = 0; Ic = 0; and Va = 0. (4.3)
Now consider the symmetrical components of the current Ia with Ib=Ic=0, given by:
Ia0 1 1 1 Ia
Ia1 = (1/3) 1 a a2 0
Ia2 1 a2 a 0 (4.4)
Solving (4.4) we get,
Ia1 = Ia2 = Ia0 = (Ia/3) (4.5)

Further, using equation (4.5) in (4.2), we get,

Va0 0 Z0 0 0 Ia1
Va1 = Ea – 0 Z1 0 Ia1

58
 Va2 0 0 0 Z2 Ia1 (4.6)

Pre-multiplying equation (4.6) throughout by [1 1 1], we get,

Va1+Va2+Va0 = - Ia1Z0 + Ea – Ia1Z1 – Ia2 Z2
i.e., Va = Ea – Ia1 (Z1 + Z2 + Z0) = zero,

Or in other words,
Ia1 = [Ea/(Z1 + Z2 + Z0)] (4.7)

.
Figure 4.2 Connection of sequence networks for LG Fault
on phase a of a Conventional Generator

The equation (4.7) derived as above implies that the three sequence networks are
connected in series to simulate a LG fault, as shown in figure 4.2. Further we have the
following relations satisfied under the fault conditions:

1. Ia1 = Ia2 = Ia0 = (Ia/3) = [Ea/(Z1 + Z2 + Z0)]

2. Fault current If = Ia = 3Ia1 = [3Ea/(Z1 + Z2 + Z0)]
3. Va1 = Ea - Ia1Z1 = Ea(Z2+Z0)/(Z1+Z2+Z0)
4. Va2 = - Ea Z2/(Z1+Z2+Z0)
5. Va0 = - Ea Z0/(Z1+Z2+Z0)
6. Fault phase voltage Va = 0,
7. Sound phase voltages Vb = a2Va1+aVa2+Va0; Vc = aVa1+a2Va2+Va0
8. Fault phase power: VaIa* = 0, Sound pahse powers: VbIb* = 0, and VcIc* = 0,
9. If Zn = 0, then Z0 = Zg0,

59
 10. If Zn = , then Z0 = , i.e., the zero sequence network is open so that then,
If=Ia=0.

4.3 LINE TO LINE FAULT ON A CONVENTIONAL GENERATOR

Figure 4.3 LL Fault on a Conventional Generator

Consider a line to line fault between phase ‘b’ and phase ‘c’ as shown in figure 4.3, at the
terminals of a conventional generator, whose neutral is grounded through a reactance.
Consider now the conditions under fault as under:
c.u.f.:
Ia = 0; Ib = - Ic; and Vb = Vc (4.8)
Now consider the symmetrical components of the voltage V a with Vb=Vc, given by:
Va0 1 1 1 Va
Va1 = (1/3) 1 a a2 Vb
Va2 1 a2 a Vb (4.9)

Solving (4.4) we get,

Va1 = Va2 (4.10)
Further, consider the symmetrical components of current Ia with Ib=-Ic, and Ia=0; given
by:
Ia0 1 1 1 0
Ia1 = (1/3) 1 a a2 Ib
Ia2 1 a2 a -Ib (4.11)

60
Solving (4.11) we get,
Ia0 = 0; and Ia2 = -Ia1 (4.12)
Using equation (4.10) and (4.12) in (4.2), and since Va0 = 0 ( Ia0 being 0), we get,
0 0 Z0 0 0 0
Va1 = Ea – 0 Z1 0 Ia1
Va1 0 0 0 Z2 -Ia1 (4.13)
Pre-multiplying equation (4.13) throughout by [0 1 -1], we get,
Va1-Va1 = Ea – Ia1Z1 – Ia1Z2 = 0
Or in other words,
Ia1 = [Ea/(Z1 + Z2)] (4.14)

Figure 4.4 Connection of sequence networks for LL Fault

on phases b & c of a Conventional Generator

The equation (4.14) derived as above implies that the three sequence networks are
connected such that the zero sequence network is absent and only the positive and
negative sequence networks are connected in series-opposition to simulate the LL fault,
as shown in figure 4.4. Further we have the following relations satisfied under the fault
conditions:
1. Ia1 = - Ia2 = [Ea/(Z1 + Z2)] and Ia0 = 0,
2. Fault current If = Ib = - Ic = [3Ea/(Z1 + Z2)] (since Ib = (a2-a)Ia1 = 3Ia1)
3. Va1 = Ea - Ia1Z1 = Ea Z2/(Z1+Z2)
4. Va2 = Va1 = EaZ2/(Z1+Z2)
5. Va0 = 0,
6. Fault phase voltages;Vb = Vc = aVa1+a2Va2+Va0 = (a+a2)Va1 = - Va1
7. Sound phase voltage; Va = Va1+Va2+Va0 = 2Va1;
8. Fault phase powers are VbIb* and VcIc*,
*
9. Sound phase power: VaIa = 0,

61
 10. Since Ia0=0, the presence of absence of neutral impedance does not make any
difference in the analysis.

4.4 DOUBLE LINE TO GROUND FAULT ON A CONVENTIONAL

GENERATOR

Figure 4.5 LLG Fault on a Conventional Generator

Consider a double-line to ground fault at the terminals of a conventional unloaded

generator, whose neutral is grounded through a reactance, between phase ‘b’ and phase
‘c’ as shown in figure 4.5, Consider now the conditions under fault as under:
c.u.f.:
Ia = 0 and Vb = Vc = 0 (4.15)
Now consider the symmetrical components of the voltage with Vb=Vc=0, given by:
Va0 1 1 1 Va
Va1 = (1/3) 1 a a2 0
Va2 1 a2 a 0 (4.16)
Solving (4) we get,
Va1 = Va2 = Va0 = Va/3 (4.17)
Consider now the sequence equations (4.2) as under,
Va0 0 Z0 0 0 Ia0
Va1 = Ea – 0 Z1 0 Ia1
Va2 0 0 0 Z2 Ia2 (4.18)

Pre-multiplying equation (4.18) throughout by

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 1/Z0 0 0
-1
Z = 0 1/Z1 0
0 0 1/Z2 (4.19)
We get,
Va1 0 Z0 0 0 Ia0
-1 -1
Z Va1 = Z Ea – Z-1Z10 0 Ia1
Va1 0 0 0 Z2 Ia2 (4.20)

Using the identity: Va1= (Ea – Ia1Z1) in equation (4.19), pre-multiplying throughout by [1
1 1] and finally adding, we get,
Ea/Z0 - Ia1(Z1/Z0) + (Ea/Z1)- Ia1 + Ea/Z2 - Ia1(Z1/Z2) = (Ea/Z1) – (Ia0+Ia1+Ia2)
= (Ea/Z1) - Ia = (Ea/Z1) (4.21)
Since Ia = 0, solving the equation (4.21), we get,
Ia1 = { Ea/ [Z1 + Z2Z0/(Z2+Z0)] } (4.22)

Figure4.6 Connection of sequence networks for LLG Fault on

phases b and c of a Conventional Generator

The equation (4.22) derived as above implies that, to simulate the LLG fault, the three
sequence networks are connected such that the positive network is connected in series
with the parallel combination of the negative and zero sequence networks, as shown in
figure 4.6. Further we have the following relations satisfied under the fault conditions:
1. Ia1 = {Ea/ [Z1+Z2Z0/(Z2+Z0)]}; Ia2= -Ia1Z0/(Z2 + Z0) and Ia0 = -Ia1Z2/(Z2 + Z0),
2. Fault current If: Ia0=(1/3)(Ia+Ib+Ic) = (1/3)(Ib+Ic) = If/3, Hence If = 3Ia0
3. Ia = 0, Vb=Vc=0 and hence Va1=Va2=Va0=Va/3
4. Fault phase voltages;Vb = Vc = 0
5. Sound phase voltage; Va = Va1+Va2+Va0 = 3Va1;
6. Fault phase powers are VbIb* = 0, and VcIc* = 0, since Vb=Vc=0

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 7. Healthy phase power: VaIa* = 0, since Ia=0
8. If Z0=, (i.e., the ground is isolated), then Ia0=0, and hence the result is the same
as that of the LL fault [with Z0=, equation (4.22) yields equation (4.14)].

4.5 THREE PHASE TO GROUND FAULT ON A CONVENTIONAL

GENERATOR

Figure 4.7 Three phase ground Fault on a Conventional Generator

Consider a three phase to ground (LLLG) fault at the terminals of a conventional

unloaded generator, whose neutral is grounded through a reactance, between all its three
phases a, b and c, as shown in figure 4.7, Consider now the conditions under fault as
under:
c.u.f.:
Va = Vb = Vc = 0, Ia + Ib + Ic = 0 (4.23)
Now consider the symmetrical components of the voltage with V a=Vb=Vc= 0, given by:
Va0 1 1 1 0
Va1 = (1/3) 1 a a2 0
Va2 1 a2 a 0 (4.24)

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Solving (4.24) we get,
Va1 = Va2 = Va0 = 0 (4.25)
Thus we have
Va1 = Ea1 – Ia1Z1 (4.26)
So that after solving for Ia1 we, get,
Ia1 = [ Ea / Z1 ] (4.27)

Figure 4.8 Connection of sequence networks for 3-phase ground

Fault on phases b and c of a Conventional Generator

The equation (4.26) derived as above implies that, to simulate the 3-phase ground fault,
the three sequence networks are connected such that the negative and zero sequence
networks are absent and only the positive sequence network is present, as shown in figure
4.8. Further the fault current, If in case of a 3-phase ground fault is given by
If = Ia1= Ia = (Ea/Z1) (4.28)
It is to be noted that the presence of a neutral connection without or with a neutral
impedance, Zn will not alter the simulated conditions in case of a three phase to ground
fault.

4.6 UNSYMMETRICAL FAULTS ON POWER SYSTEMS

In all the analysis so far, only the fault at the terminals of an unloaded generator have
been considered. However, faults can also occur at any part of the system and hence the
power system fault at any general point is also quite important. The analysis of
unsymmetrical fault on power systems is done in a similar way as that followed thus far
for the case of a fault at the terminals of a generator. Here, instead of the sequence
impedances of the generator, each and every element is to be replaced by their
corresponding sequence impedances and the fault is analyzed by suitably connecting
them together to arrive at the Thevenin equivalent impedance if that given sequence.
Also, the internal voltage of the generators of the equivalent circuit for the positive

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sequence network is now Vf (and not Ea), the pre-fault voltage to neutral at the point of
fault (PoF) (ref. Figure 4.9).

Figure 4.9 Unsymmetrical faults in Power Systems

Thus, for all the cases of unsymmetrical fault analysis considered above, the sequence
equations are to be changed as under so as to account for these changes:

Va0 0 Z0 0 0 Ia0
Va1 = Vf – 0 Z1 0 Ia1
Va2 0 0 0 Z2 Ia2 (4.29)

(i) LG Fault at any point F of a given Power system

Let phase ‘a’ be on fault at F so that then, the c.u.f. would be:
Ib = 0; Ic = 0; and Va = 0.
Hence the derived conditions under fault would be:
Ia1 = Ia2 = Ia0 = (Ia/3)
Ia1 = [Vf / (Z1 + Z2 + Z0)] and
If = 3Ia1 (4.30)

(ii) LL Fault at any point F of a given Power system

Let phases ‘b’ and ‘c’ be on fault at F so that then, the c.u.f. would be:
Ia = 0; Ib = - Ic; and Vb = Vc
Hence the derived conditions under fault would be:
Va1 = Va2; Ia0 = 0; Ia2 = -Ia1
Ia1 = [Vf / (Z1 + Z2)] and
If = Ib = - Ic = [3 Vf / (Z1 + Z2)] (4.31)

(ii) LLG Fault at any point F of a given Power system

Let phases ‘b’ and ‘c’ be on fault at F so that then, the c.u.f. would be:
Ia = 0 and Vb = Vc = 0
Hence the derived conditions under fault would be:
Va1 = Va2 = Va0 = (Va/3)

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 Ia1 = {Vf / [Z1+Z2 Z0/(Z2+Z0)]}
Ia2= -Ia1Z0/(Z2 + Z2); Ia0 = -Ia1 Z2/(Z2 + Z2) and
If = 3Ia0 (4.32)

(ii) Three Phase Fault at any point F of a given Power system

Let all the 3 phases a, b and c be on fault at F so that then, the c.u.f. would be:
Va = Vb = Vc = 0, Ia + Ib + Ic = 0
Hence the derived conditions under fault would be:
Va1 = Va2 = Va0 = Va/3
Va0 = Va1 = Va2 = 0; Ia0 = Ia2 = 0,
Ia1 = [Vf /Z1] and If = Ia1=Ia (4.33)

4.7 OPEN CONDUCTOR FAULTS

Various types of power system faults occur in power systems such as the shunt type faults
(LG, LL, LLG, LLLG faults) and series type faults (open conductor and cross country
faults). While the symmetrical fault analysis is useful in determination of the rupturing
capacity of a given protective circuit breaker, the unsymmetrical fault analysis is useful in
the determination of relay setting, single phase switching and system stability studies.

When one or two of a three-phase circuit is open due to accidents, storms, etc., then
unbalance is created and the asymmetrical currents flow. Such types of faults that come
in series with the lines are refered as the open conductor faults. The open conductor faults
can be analyzed by using the sequence networks drawn for the system under
consideration as seen from the point of fault, F. These networks are then suitably
connected to simulate the given type of fault. The following are the cases required to be
analyzed (ref. fig.4.10).

Figure 4.10 Open conductor faults.

(i) Single Conductor Open Fault: consider the phase ‘a’ conductor open so that then the
conditions under fault are:
Ia = 0; Vbb’ = Vcc’ = 0
The derived conditions are:

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 Ia1 + Ia2 + Ia0 = 0 and
Vaa1’ = Vaa2’ = Vaa0’ = (Vaa’/3) (4.34)
These relations suggest a parallel combination of the three sequence networks as shown
in fig. 4.11.

Figure 4.11 Sequence network connection for 1-conductor open fault

It is observed that a single conductor fault is similar to a LLG fault at the fault point F of
the system considered.

(ii) Two Conductor Open Fault: consider the phases ‘b’ and ‘c’ under open condition so
that then the conditions under fault are:
Ib = Ic = 0; Vaa’ = 0
The derived conditions are:
Ia1 = Ia2 = Ia0 = Ia/3 and
Vaa1’ = Vaa2’ = Vaa0’ = 0 (4.35)
These relations suggest a series combination of the three sequence networks as shown in
fig. 4.12. It is observed that a double conductor fault is similar to a LG fault at the fault
point F of the system considered.

Figure 4.12 Sequence network connection for 2-conductor open fault.

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(iii) Three Conductor Open Fault: consider all the three phases a, b and c, of a 3-phase
system conductors be open. The conditions under fault are:
Ia + Ib + Ic = 0
The derived conditions are:
Ia1 = Ia2 = Ia0 = 0 and
Va0 = Va2 = 0 and Va1 = Vf (4.36)
These relations imply that the sequence networks are all open circuited. Hence, in a strict
analystical sense, this is not a fault at all!

4.8 FAULTS THROUGH IMPEDANCE

All the faults considered so far have comprised of a direct short circuit from one or two
lines to ground. The effect of impedance in the fault is found out by deriving equations
similar to those for faults through zero valued neutral impedance. The connections of the
hypothetical stubs for consideration of faults through fault impedance Zf are as shown in
figure 4.13.

Fig
ure 4.13 Stubs Connections for faults through fault impedance Zf.

(i) LG Fault at any point F of a given Power system through Zf

Let phase ‘a’ be on fault at F through Zf, so that then, the c.u.f. would be:
Ib = 0; Ic = 0; and Va = 0.
Hence the derived conditions under fault would be:
Ia1 = Ia2 = Ia0 = (Ia/3)
Ia1 = [Vf / (Z1 + Z2 + Z0+3Zf)] and
If = 3Ia1 (4.37)

(ii) LL Fault at any point F of a given Power system through Zf

Let phases ‘b’ and ‘c’ be on fault at F through Zf, so that then, the c.u.f. would be:
Ia = 0; Ib = - Ic; and Vb = Vc
Hence the derived conditions under fault would be:
Va1 = Va2; Ia0 = 0; Ia2 = -Ia1
Ia1 = [Vf / (Z1 + Z2+Zf)] and
If = Ib = - Ic = [3 Vf / (Z1 + Z2+Zf)] (4.38)

(iii) LLG Fault at any point F of a given Power system through Zf

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Let phases ‘b’ and ‘c’ be on fault at F through Zf,, so that then, the c.u.f. would be:
Ia = 0 and Vb = Vc = 0
Hence the derived conditions under fault would be:
Va1 = Va2 = Va0 = (Va/3)
Ia1 = {Vf / [Z1+Z2(Z0+3Zf)/(Z2+Z0+3Zf)]}
Ia2= -Ia1(Z0+3Zf)/(Z2+Z0+3Zf); Ia0 = -Ia1Z2/(Z2+(Z0+3Zf) and
If = 3Ia0 (4.39)

(iv) Three Phase Fault at any point F of a given Power system through Zf
Let all the 3 phases a, b and c be on fault at F, through Zf so that the c.u.f. would be: Va =
IaZf ; Hence the derived conditions under fault would be: Ia1 = [Vf /(Z1+Zf); The
connections of the sequence networks for all the above types of faults through Zf are as
shown in figure 4.14.

LG Fault LL Fault

LLG Fault 3-Ph. Fault

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 Figure 4.15 Sequence network connections for faults through impedance

4.9 EXAMPLES

Example-1: A three phase generator with constant terminal voltages gives the following
currents when under fault: 1400 A for a line-to-line fault and 2200 A for a line-to-ground
fault. If the positive sequence generated voltage to neutral is 2 ohms, find the reactances
of the negative and zero sequence currents.
Solution: Case a) Consider the conditions w.r.t. the LL fault:
Ia1 = [Ea1/(Z1 + Z2)]
If = Ib = - Ic = 3 Ia1
=3 Ea1 / (Z1 + Z2) or
(Z1 + Z2) = 3 Ea1 / If
i.e., 2 + Z2 = 3 [2000/1400]
Solving, we get, Z2 = 0.474 ohms.
Case b) Consider the conditions w.r.t. a LG fault:
Ia1 = [Ea1/(Z1 + Z2+Z0)]
If = 3 Ia1
= 3 Ea1 / (Z1 + Z2+Z0) or
(Z1 + Z2+Z0) = 3 Ea1 / If
i.e., 2 + 0.474 + Z0 = 3 [2000/2200]
Solving, we get, Z0 = 0.253 ohms.

Example-2: A dead fault occurs on one conductor of a 3-conductor cable supplied y a 10

MVA alternator with earhed neutral. The alternator has +ve, -ve and 0-sequence
components of impedances per phase respectively as: (0.5+j4.7), (0.2+j0.6) and (j0.43)
ohms. The corresponding LN values for the cable up to the point of fault are:
(0.36+j0.25), (0.36+j0.25) and (2.9+j0.95) ohms respectively. If the generator voltage at
no load (Ea1) is 6600 volts between the lines, determine the (i)Fault current, (ii)Sequence
components of currents in lines and (iii)Voltages of healthy phases.
Solution: There is LG fault on any one of the conductors. Consider the LG fault to be on
conductor in phase a. Thus the fault current is given by:
(i) Fault current: If = 3Ia0 = [3Ea/(Z1 + Z2 + Z0)]
= 3(6600/3)/ (4.32+j7.18)
= 1364.24 58.970.

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(ii) Sequence components of line currents:
Ia1 = Ia2 = Ia0 = Ia/3 = If/3 = 454.75 58.970.
(iii) Sound phase voltages:
Va1 = Ea - Ia1Z1 = Ea(Z2+Z0)/(Z1+Z2+Z0) = 1871.83 -26.170,
Va2 = - EaZ2/(Z1+Z2+Z0) = 462.91 177.60,
Va0 = - EaZ0/(Z1+Z2+Z0) = 1460.54 146.5 ,
0

Thus,
Sound phase voltages Vb = a2Va1+aVa2+Va0 = 2638.73 -165.80 Volts,
And Vc = aVa1+a2Va2+Va0 = 3236.35 110.80 Volts.

Example-3: A generator rated 11 kV, 20 MVA has reactances of X1=15%, X2=10% and
X0=20%. Find the reactances in ohms that are required to limit the fault current to 2 p.u.
when a a line to ground fault occurs. Repeat the analysis for a LLG fault also for a fault
current of 2 pu.
Solution: Case a: Consider the fault current expression for LG fault given by:
If = 3 Ia0
i.e., 2.0 = 3Ea / j[X1+X2+X0]
= 3(1.000) / j[0.15+0.1+0.2+3Xn]
Solving we get
3Xn = 2.1 pu
= 2.1 (Zb) ohms = 2.1 (112/20) = 2.1(6.05)
= 12.715 ohms.
Thus Xn = 4.235 ohms.

Case b: Consider the fault current expression for LLG fault given by:
If = 3Ia0 = 3 { -Ia1X2/(X2 + X0+3Xn)}= 2.0,
where, Ia1 = {Ea/ [X1+X2(X0+3Xn)/(X2+X0+3Xn)]}
Substituting and solving for Xn we get,
Xn = 0.078 pu
= 0.47 ohms.

Example-4: A three phase 50 MVA, 11 kV generator is subjected to the various faults

and the surrents so obtained in each fault are: 2000 A for a three phase fault; 1800 A for a
line-to-line fault and 2200 A for a line-to-ground fault. Find the sequence impedances of
the generator.
Solution: Case a) Consider the conditions w.r.t. the three phase fault:
If = Ia = Ia1 = Ea1/Z1
i.e., 2000 = 11000/ (3Z1)
Solving, we get, Z1 = 3.18 ohms (1.3 pu, with Zb = (112/50) = 2.42 ohms).

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Case b) Consider the conditions w.r.t. the LL fault:
Ia1 = [Ea1/(Z1 + Z2)]
If = Ib = - Ic = 3 Ia1
=3 Ea1 / (Z1 + Z2) or
(Z1 + Z2) = 3 Ea1 / If
i.e., 3.18 + Z2 = 3 (11000/3)/1800
Solving, we get, Z2 = 2.936 ohms = 1.213 pu.
Case c) Consider the conditions w.r.t. a LG fault:
Ia1 = [Ea1/(Z1 + Z2+Z0)]
If = 3 Ia1
= 3 Ea1 / (Z1 + Z2+Z0) or
(Z1 + Z2+Z0) = 3 Ea1 / If
i.e., 3.18+ 2.936 + Z0 = 3 (11000/3)/ 2200
Solving, we get, Z0 = 2.55 ohms = 1.054 pu.

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