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THE ELECTRICAL DISTRIBUTION
SYSTEM OF A 69KV TO 12KV MODEL
A Project
Presented to the faculty of the Department of Electrical and Electronic Engineering
California State University, Sacramento
Submitted in partial satisfaction of
the requirements for the degree of
MASTER OF SCIENCE
in
Electrical and Electronic Engineering
by
Julio Gutierrez
Emilio Nuñez
FALL
2012
Sample Copyright page
[l]
© 2012[Year of graduation]
Julio Gutierrez
Emilio Nuñez
ALL RIGHTS RESERVED
ii
[Thesis/Project Approval Page]
THE ELECTRICAL DISTRIBUTION
SYSTEM OF A 69KV TO 12KV MODEL
A Project
by
Julio Gutierrez
Emilio Nuñez
Approved by:
__________________________________, Committee Chair
Turan Gönen, Ph.D.
__________________________________, Second Reader
Salah Yousif, Ph.D.
____________________________
Date
[Thesis/Project Format Approval Page]
iii
Student: Julio Gutierrez
Emilio Nuñez
I certify that these students have met the requirements for format contained in the
University format manual, and that this project is suitable for shelving in the Library and
credit is to be awarded for the project.
__________________________, Graduate Coordinator
B. Preetham. Kumar, Ph.D.
r
Department of Electrical and Electronic Engineering
iv
___________________
Date
Abstract
of
THE ELECTRICAL DISTRIBUTION
SYSTEM OF A 69KV TO 12KV MODEL
by
Julio Gutierrez
Emilio Nuñez
A reliable and efficient power system benefits not only the power utility company, but
the consumer as well. This report provides clarification to concepts and calculations in
the analysis of an electrical distribution system involving a 69KV to 12KV substation.
The project will analyze different kinds of faults at different locations around the
substation in order to see how the system will react during abnormal operating
conditions.
A fundamental part of achieving a reliable power system is the integration of protective
devices. Different protective devices will be discussed in the report. Analyzing the
potential fault conditions will provide the information needed to define the sizing and
settings for these devices. A MATLAB script will be developed to assess the
calculations.
_______________________, Committee Chair
Turan Gönen, Ph.D.
_______________________
Date
v
DEDICATION
[Optional]
I dedicate my work to all those close to me for inspiring me.
vi
[This Table of Contents covers many possible headings. Pages are optional.
Use only the headings that
TABLE OF CONTENTS
Page
Dedication .................................................................................................................... vi
List of Tables ............................................................................................................. xii
List of Figures ........................................................................................................... xiii
Chapter
1. INTRODUCTION………………………………………………………………... 1
2. LITERATURE SURVEY ........................................................................................3
2.1. Introduction .......................................................................................................3
2.2. Distribution Substation .....................................................................................3
2.3. Transmission Line.............................................................................................4
2.4. Bundle Conductors ...........................................................................................5
2.5. Selecting Material Grounding Conductors .......................................................6
2.6. Heat Impulse .....................................................................................................6
2.7. Sizing Conductors .............................................................................................7
2.8. Substation Transformer.....................................................................................8
2.9. Over-Current Relay Protection of Transmission Lines ................................... 8
2.10. Faults .............................................................................................................. 9
2.11. Single Line to Ground Fault ........................................................................ 10
2.12. Line to Line Fault .........................................................................................11
vii
2.13. Double Line-to-Ground Fault .......................................................................11
2.14. Three Phase Fault..........................................................................................11
2.15. Electromechanical vs. Solid-State Relays ................................................... 13
2.16. Relay Selection .............................................................................................14
2.17. Overcurrent Relay With Or Without Time Delay ........................................15
2.18. Current Independent Overcurrent Relay ...................................................... 17
2.19. Current Dependent Overcurrent Relay ........................................................ 17
2.20. Moving Operating Characteristic Along The X-Axis.................................. 18
2.21. Moving Operating Characteristic Along The Y-Axis.................................. 19
2.22. Reclosing ..................................................................................................... 19
2.23. Differential Relay ........................................................................................ 20
2.23.1. Applications Of Differential Protection .............................................. 20
2.24. Distance Relays............................................................................................ 23
2.24.1. Distance Measurement ........................................................................ 23
2.24.2. Hardware Components Of Distance Relays ....................................... 26
2.25. Directional Pilot Relaying ........................................................................... 27
2.25.1. Directional Comparison ...................................................................... 27
2.25.2. Information Transfer Between Ends ................................................... 28
2.26. Maintenance And Testing Of Relays ........................................................... 31
2.26.1. Installation Or Commissioning Tests.................................................. 31
2.26.2. Periodic Tests ...................................................................................... 33
viii
3. MATHEMATICAL MODEL ............................................................................... 34
3.1. Introduction .................................................................................................... 34
3.2. Fault Currents................................................................................................. 34
3.2.1. Deriving The Substransient Current ................................................... 37
3.2.2. Deriving The Transient Current .......................................................... 38
3.3. Heat Impulse .................................................................................................. 39
3.4. Sizing Of The Conductor ............................................................................... 44
3.5. The Analysis Of The Fault Currents .............................................................. 47
3.5.1. Single Line To Ground Fault Current ................................................. 49
3.5.2. Line To Line Fault Current ................................................................. 50
3.5.3. Double Line To Ground Fault Current ............................................... 51
3.5.4. Three Phase Fault Current ...................................................................53
3.6. Operating Characteristic Of The Differential Relay .......................................54
4. APPLICATION OF THE MATHEMATICAL MODEL ..................................... 57
4.1. Subtransient Fault Current Calculations For The Secondary 12kv Side ....... 57
4.2. Transient Current ........................................................................................... 59
4.3. Heat Impulse .................................................................................................. 61
4.4. Conductor Sizing ........................................................................................... 62
4.5. Fault Current At Transient Stage For Overhead Feeders............................... 64
4.6. Fault At The End Of The Line For Overcurrent Relay.................................. 68
4.6.1. Three Phase Fault Calculation .............................................................68
ix
4.6.2. Line To Line Fault ...............................................................................68
4.7. End Of The Line Fault For Overhead Ground Fault Relay ........................... 69
4.7.1. Single Line To Ground Fault .............................................................. 69
4.7.2. Double Line To Ground Fault............................................................. 71
4.8. Fault At The 12kv Bus For The Overcurrent Relay ...................................... 72
4.8.1. Three Phase Fault Calculation .............................................................72
4.8.2. Line To Line Fault ...............................................................................73
4.9. Fault At The 12kv Bus Overhead Ground Overcurrent Relay ...................... 73
4.9.1. Single Line To Ground Fault .............................................................. 73
4.9.2. Double Line To Ground Fault............................................................. 74
4.10. Fault Currents For The Underground Line..................................................76
4.10.1. Impedance Of A Six Mile Long Underground Transposed Line ....... 77
4.11. Fault At The End Of The Line For Overcurrent Relay ...............................80
4.11.1. Three Phase Fault Calculation .............................................................80
4.11.2. Line To Line Fault .............................................................................. 80
4.12. Fault At The End Of The Line For Underground Relay ............................ 81
4.12.1. Single Line To Ground Fault .............................................................. 81
4.12.2. Double Line To Ground Fault............................................................. 83
4.13. Fault At The 12kv Bus For The Underground Line ................................... 84
4.13.1. Three Phase Fault Calculation ............................................................ 84
4.13.2. Line To Line Fault .............................................................................. 85
x
4.14. Fault At The 12kv Bus For The Underground Relay ..................................85
4.14.1. Single Line To Ground Fault ...............................................................85
4.14.2. Double Line To Ground Fault..............................................................87
4.15. Summary Of Distribution Line Faults .........................................................89
4.15.1. Overhead Distribution Line Fault Summary........................................89
4.15.2. Underground Distribution Line Fault Summary ..................................89
4.16. Overhead And Underground Distribution Line Relay Settings ..................90
4.16.1. Overhead Line Protection ....................................................................90
4.16.2. Ground Fault Protection .......................................................................91
4.16.3. Power Transformer Protection ............................................................ 92
4.17. Protection Scheme For The Entire System ..................................................95
4.17.1. Protection Relays, Line And Ground Relays .......................................95
4.17.2. Differential Relay Installation..............................................................97
4.17.3. Relay Characteristics Summary .......................................................... 98
4.17.4. Diagram For The Control Circuit ....................................................... 99
4.17.5. Overall Protection Scheme ............................................................... 101
5. CONCLUSION ................................................................................................... 102
Appendix A. Conductor Characteristics ................................................................... 104
Appendix B. MATLAB Code ................................................................................... 105
Bibliography ............................................................................................................. 112
xi
LIST OF TABLES
Tables
Page
1.
Table 3.1 Data provided from Equation 3.50 used to minimize ...………........46
2.
Table 4.1 Overhead distribution line fault summary ………………………....89
3.
Table 4.2 Underground distribution line fault summary ..................... ……….89
4.
Table 4.3 Relay characteristics………….……… .. …………………………. 98
xii
LIST OF FIGURES
Figures
Page
1.
Figure 2.1 General representation of a single line-to-ground fault…...………12
2.
Figure 2.2 General representation of a line-to-line fault……………………...12
3.
Figure 2.3 General representation of a double line-to-ground..……………....12
4.
Figure 2.4 General representation of a three phase fault…...……….………...13
5.
Figure 2.5 An induction relay having one static coil and one shaded pole…...16
6.
Figure 2.6 Operating characteristic of a current independent overcurrent relay.
……...…………………………………………………………………………17
7.
Figure 2.7 Operating characteristic of a current dependent overcurrent relay.
..........…...………. ………………………………………………………........18
8.
Figure 2.8 Operating characteristics of the excitation coil along the x-axis.
.........…...……….…….....…………………………………………………….18
9.
Figure 2.9 Operating characteristics of an induction relay along the y-axis.
……..…………………………………………………………………...……..19
10.
Figure 2.10 Basic differential protection of a three-winding transformer...…..21
11.
Figure 2.11 Restricted earth protection of power transformer …...………….. 22
12.
Figure 2.12 Unbiased differential protection of a bus …...…………………... 22
13.
Figure 2.13 Unbiased differential protection characteristics for a bus..……….25
14.
Figure 2.14 Principle connections of an impedance distance relay…...……….26
15.
Figure 2.15 Basic principle of directional pilot relaying…...………. ………...28
xiii
16.
Figure 2.16 Basic circuits of DC pilot schemes…...………. ………………...29
17.
Figure 2.17 DC communication circuit. …...………………………………....30
18.
Figure 2.18 Primary injection test circuit…...………………………………...32
19.
Figure 3.1 One line diagram of power system model…...…………………....47
20.
Figure 3.2 Sequence networks …...…………………………………………..49
21.
Figure 3.3 Sequence network for a single line-to-ground fault …...………....49
22.
Figure 3.4 Sequence network for a line-to-line fault …...………. …………..51
23.
Figure 3.5 Sequence network for a double line-to-ground fault …...………..53
24.
Figure 3.6 Sequence network for a three phase fault…...……………………54
25.
Figure 3.7 Differential Relay…...……………………………………………55
26.
Figure 4.1 Transmission line protection relays………………………………95
27.
Figure 4.2 The two internal relays inside each relay…………………………96
28.
Figure 4.3 Differential relay installation……………………………………..97
29.
Figure 4.4 Control circuit for the breaker…………………………………....99
30.
Figure 4.5 Control circuit continued………………………………………..100
xiv
1
Chapter 1
INTRODUCTION
The modes of distribution operations of an electric power system are diverse and
sometimes complicated specially the coordination of its protective devises. Electricity is
one of the fundamental resources of a modern industrialized society, which provides
commodities not available without it. Modern societies have been accustomed to receive
their electric power instantly, at exactly the required amount and at the correct voltage
and frequency with minimum interruptions.
This remarkable performance is achieved
through careful planning, design, installation and operation of a very complex network
[2]. A network composes of a variety of components such as generators, step up/down
transformers, and a web network of transmission and distribution lines. This intricate
system supplies a variety of customers, such as, residential, commercial, industrial,
agricultural and special loads. The electric power received by the customer may appear
to be working in steady state without any interruption, and infinite in capacity. Yet, the
power system is subjected to constant disturbances created by random load changes, by
faults created by natural causes, and sometimes just equipment or operator error [2].
Even with all the random disturbances the system appears to be operating in a steady
state mainly due to a couple of factors. The first is the sheer size of a specific power
system in comparison to certain loads or generation equipment. The second factor is the
fast reactions of the protective relaying system, which is the objective of this project.
2
The rapid response to correct any disturbance is an essential part of relaying in power
systems. The relay response must be quick and as accurate as possible in order to
minimize power interruption to the customers. In order to accomplish this tremendous
task at hand the analysis of different faults at different locations will be analyzed. The
purpose of this project is to study a 69 kV to 12 kV system for an overhead and
underground operation. Calculations for faults at the end of a six mile long line, and at
the 12 kV bus will be conducted for both systems as well. The fault analysis is conducted
for four different scenarios; fault calculations for single line to ground, double line to
ground, line to line, and three phase faults. The results will be compared to find which
disturbance is the largest by tabulating the results and from the results of each disturbance
settings for the protective equipment will be defined to counteract the problem.
Equation Chapter (Next) Section 1
Equation Chapter (Next) Section 1
Equation Section (Next)
3
Chapter 2
LITERATURE SURVEY
2.1 INTRODUCTION
The purpose of this chapter is to introduce and describe important terminology and
fundamental concepts involved with distribution and transmission design, and the fault
analysis associated with it. Some basic theories and concepts are covered in more detail
as questions and issues regarding certain aspects of the project are introduced and must
be addressed during the design and analysis process.
2.2 DISTRIBUTION SUBSTATION
By definition, each distribution substation consists of at least one power transformer.
Distribution substations are used for transforming transmission voltages above 50 kV to
distribution voltages at 50 kV and below. Usually, substation power transformers are
installed for three-phase operations. A bank of three single-phase transformers or a single
three-phase transformer may be used. Distribution substations are sized and placed so
that the load can be served cost effectively by reducing feeder losses and construction
costs, while taking into account service reliability. To assure that the growing demand for
electricity is met utilities assess the consequences of different proposed alternatives and
scenarios, and the impact they would have on the rest of the system [3].
4
A great deal of additional equipment is installed with each distribution station
transformer in order to produce a certain level of reliability to the distribution system for
the customer. For example, protection equipment, such as, circuit breakers, fuses,
disconnect switches, and monitoring equipment are installed so that abnormal conditions
can be detected and the transformers can be manipulated. This is so the transformer can
be removed from service for routine maintenance, or shut down when voltages or
currents fluctuate enough to set off a safety trip operation.
The scope of this project is focused on the system model and the electrical fault analysis
that is done for a distribution system involving a 69 kV to 12 kV substation without
defining specific load support requirements.
2.3 TRANSMISSION LINE
It is important that a substation has the proper transmission line built. The purpose of the
transmission line is to transfer electric energy from the generating units to the loads,
which needs the power to run their machines. Transmission lines also interconnect
neighboring utilities, which permits not only economic dispatch of power within regions
during normal conditions, but also the transfer of power between regions during
emergency [4].
Transmission planning is a very important step for utility companies. The purpose of the
transmission planning is to anticipate the needs of future loads. Like the wonderful state
of California, utility companies have to decide their transmission system to meet the
5
demands of businesses and homes. In general, transmission lines have two primary
functions: first, to transmit electrical energy from the generators to the loads within a
single utility and second, to provide paths for electrical energy to flow between utilities
[8]. These conductors used for transmission lines also have different designs each with
different properties that needs to be considered as well.
2.4 BUNDLE CONDUCTORS
Instead of using one single conductor per phase, utility companies use several conductors
of the same size. This bundle of conductors will form a transmission line. This method
will cost a little more compared to using just one single conductor, but in the long run it
will save a lot of money. By having two or more conductors per phase in close proximity
compared with the spacing between phases, the voltage gradient at the conductor surface
will be significantly reduced [8]. The advantages of using bundled conductors instead of
single conductors per phase is that first it reduces line inductive reactance; second, is a
reduction in voltage gradient; third, is that it will produce an increased corona critical
voltage and, therefore, less corona power loss, audible noise, and radio interference;
fourth, it will allow more power to be carried per unit mass of the conductor; and fifth,
the amplitude and duration of high-frequency vibrations may be reduced. The
disadvantages of bundled conductors include first, increased wind and ice loading;
second, suspension is more complicated and duplex or quadruple insulator strings may be
6
required; third, the tendency to gallop is increased; fourth, increased cost; fifth, increased
clearance requirements of the structure; and sixth, increased charging in kilovolt amperes.
2.5 SELCTING MATERIAL GROUNDING CONDUCTORS
When engineers design the sizing of conductors, they often choose to have it in copper or
in aluminum. Copper conductor is resistant to underground corrosion, since copper is
cathodic with respect to other materials buried in its vicinity [1]. Because of this factor,
copper conductors are used for underground systems. There are several procedures that
are done to minimize the corrosion process. One, insulate the affected metals with plastic
tape and or asphalt compound; two, arrange the crossing of copper conductors and other
metals at right angles as closely as possible; three, use cathodic protection of affected
metals in the area, or use nonmetallic pipes [1][5]. Aluminum is the other metal that
engineers would use for a conductor. Because aluminum conductors are very corrosive
they are usually not used for underground systems. Corrosive aluminum conductors are
not conductive, so most of the time for the purpose of saving money aluminum is used
only for overhead cable.
2.6 HEAT IMPULSE
When a fault hits a line, it will generate a large increase in heat than under normal
circumstances. The heat that is generated by the fault is called the heat impulse. Because
of the possibilities of a fault and future load growth, the thermal capacity of the conductor
7
chosen for a system is set with a larger thermal limit than what is needed. This will
determine the size of the conductor. If the conductor is not made to handle the
temperature it will damage the conductor. In this case, it will cost utility companies a lot
of money to repair it. For this reason, the conductors must be made in a way where it can
handle the worst case of fault current that will hit the transmission line under the worst
conditions.
2.7 SIZING CONDUCTORS
Transmission lines have the capability of transmitting power, which is restricted to the
thermal loading and stability limits. The temperature of the conductor reflects the
thermal loading limit of the line. The real power loss of the line is increased as the
temperature of the conductors rise. The thermal limit is specified to be at 75% of the
current carrying capacity given for a conductor at a temperature of 50° Celsius. This
temperature of 50° Celsius is based on a 25° Celsius air temperature with a 25° Celsius
rise in conductor temperature.
The sizing of a conductor is important. If the conductor is made to be small for the
system this can result in a breakdown of the transmission line when a large fault hits the
system. If the conductor is made too big, then it will cost the utility companies a lot of
money. Ultimately, the utility customer will have to absorb the cost of these mistakes.
Thus, the sizing of conductor has to be made just right to be prepared to handle the
largest fault and not be made too expensive at the same time. In order to find the current
8
carrying capacity, the power of the line and the rated phase voltage of the selected line
should be known.
2.8 SUBSTATION TRANSFORMER
Between a transmission line and distribution line is a transformer. The objective of a
transformer is to either raise or lower the voltages of a three-phase distribution system.
There a several methods to connect a three-phase distribution, which are wye-wye, deltadelta, wye-delta, and delta-wye connections. For a substation transformer connection the
most common types are delta-delta and delta-wye solidly grounded. The delta-wye
connection is very suitable for high-voltage transmission systems because they supply a
stable neutral point that is grounded, which prevents the possibility of oscillation. This
also means that because the primary side and the secondary side are not connected under
the same formation a phase shift of 30° does for voltage. For the focus of this project,
analysis will be done using a delta-wye solidly grounded connection.
2.9 OVERCURRENT RELAY PROTECTION OF POWER TRANSMISSION LINES
Relay protection must be provided to the power transmission lines, whether they as
designed to operate as underground or overhead power systems. There is a significant
amount of problems that may emerge. Some of the eminent problems are, break down of
insulation, physical damage to the conductors, human errors as well as natural elements,
and overall these problems may cause a sudden excessive current change along the power
9
line. In order to protect both people and the power system hardware from these excessive
changes of current, an adequate relay protection to the system must be provided.
2.10 FAULTS
A fault is an irregular flow of current that bypasses the normal load in a power system.
The frequency of fault occurrences on the overhead transmission lines depends mainly
upon weather conditions of a given geographic region. In general, the overhead power
transmission lines are more often found in faulted conditions than an underground
transmission lines. When a fault occurs the system can experience an open or short
circuit. Open circuits for the most part do not impact or damage the system normally,
whereas, a short circuit fault can result in damaging different elements in the system. The
following is a list of fault occurrence: single phase-to-ground faults, phase-to-phase
faults, two phase-to-ground faults and three phase faults, while their magnitude of danger
to the power system is described as follows. Phase-to-ground amounts to 70% of faults,
line-to-ground faults 15%, double line-to-ground faults 10%, and three phase faults
amounts to 5%. This statistical distribution of fault occurrence is not the same for every
voltage level, especially for higher voltages [3]. The case of three-phase fault on a 400kV
overhead power transmission line is very rare, but it is more common in an underground
power transmission line.
10
For a given power system both maximum and minimum fault currents are calculated. In
order to calculate fault currents the positive-, negative-, and zero-sequence Thevenin
impedances of the system must be identified.
The overhead transmission lines are one of the most vulnerable points in power systems.
The majority of fault occurrences in overhead transmission lines are attributed to wind,
trees, animals, cranes vandalism, airplanes, excessive ice loading, and vehicles colliding
with the poles. For the focus of this project only short circuit faults will be analyzed.
2.11 SINGLE LINE TO GROUND FAULT
A line to ground fault occurs when a single conductor becomes damaged physically and
finds an alternate path to ground. This causes the system to become unbalanced, and
because it only takes only one of the three conductors to create this kind of fault 70% of
all faults is typically single line to ground faults. Under the right conditions a single line
to ground fault can produce a worst outcome than a three phase fault when the generators
involved have a good connection to ground or that connection to ground has a low
impedance, and the location where the fault occurs is the wye side of a delta wye
grounded transformer. The phase that has contact to ground is given an impedance [1].
11
2.12 LINE TO LINE FAULT
A line to line fault occurs when a short between two circuits occur. There are a number of
reasons that could have happen. For example, bad insulation, an animal shorting the lines
with its own body, and various other manmade possible scenarios that could have
occurred as well [1] [5].
2.13 DOUBLE LINE TO GROUND FAULT
Double line-to-ground faults are not very likely. Two of the three conductors are
physically damaged and share a path to ground. This kind of fault makes the system
unbalanced with each damaged phase having their own separate fault impedance and a
common ground impedance [1] [5].
2.14 THREE PHASE FAULT
When all three phases of a transmission line are short-circuited and have the same path to
ground a three-phase fault has occurred. During a three-phase fault the system is
completely balanced. There are no negative- or zero-sequence currents. This leaves all
three phases with an equal separation of 120° from each phase. Three-phase faults are the
less likely to occur [1] [5].
12
F
a
b
c
+
Vaf
-
Ibf = 0
Zf
Icf = 0
Iaf
n
Figure 2.1 General representation of a single line-to-ground fault [1] [5].
a
F
b
c
Iaf=0
Ibf
Icf = -Ibf
Zf
Figure 2.2 General representation of a line-to-line fault [1] [5].
F
a
b
c
Iaf = 0
Icf
Ibf
Zf
Zf
Zg
n
Ibf + Icf
N
Figure 2.3 General representation of a double line-to-ground fault [1] [5].
13
F
a
b
c
Zf
Zf
Zg
n
Icf
Ibf
Iaf
Zf
Iaf +Ibf + Icf = 3Ia0
N
Figure 2.4 General representation of a three phase fault [1] [5].
2.15 ELECTROMECHANICAL VS. SOLID-STATE RELAYS
Relays today are either electromechanical or solid-state relays. For electromechanical
relays the contacts are closed or open by a magnetic force. Solid-state relays do not have
contacts and the switching is done electronically. The decision to use one over the other
is defined by the electrical requirements, cost constraints, and life expectancy expected of
the device. The use and demand of Solid-state relays over time has increased greatly, but
electromechanical relays remain common. Many of the functions done by heavy duty
equipment require the switching capabilities of electromechanical relays. The basic
mechanisms for these types of relays is a pivot arm that is designed for translatory
movement, a rotating pivot arm, and electromagnetic relays with an attractive or
repealing plunger. The induction electromagnetic relay consists of one excitation coil and
two short-circuited copper rings. These two types are usually instantaneously activated,
but a time delay may also be introduced by adding a classical clock mechanism. Solid
14
State relays switch the current using electronic devices, such as silicon controlled
rectifiers, that are non-moving.
Each relay has both their advantages and disadvantages. Solid State relays do not have to
energize a coil or open contacts, so less voltage is required to turn them off or on.
Because there are no physical parts to move they also turn on and turn off faster. This
also means that Solid State Relays are not subject to arcing and do not wear out.
However, when any part of the Solid State Relay becomes defective the entire relay must
be replaced, whereas on an Electromechanical Relays usually only the contacts have to be
replaced. Because of how a Solid State Relays is built residual electrical resistance and/or
current leakage occurs whether the relay is left open or closed. The small voltage drops
created are not usually a problem. But, for Electromechanical Relays because of their
relatively large distance between contacts, which acts as a form of insulation
Electromechanical Relays, provide a cleaner on or off condition [2]. For the purpose of
this project a mechanical relay will be used.
2.16 RELAY SELECTION
Selection of the type of mechanical relay and number of relays for protection purposes of
transmission lines depends upon the type of power system grounding conditions, i.e.,
whether a power transmission line belongs to the effectively grounded or isolated system.
This notion is defined by the ratio of the zero-sequence reactance, X0, and positivesequence reactance, X1, as well as the ration between the zero-sequence resistance, R0,
15
and positive-sequence reactance. The above classifications work well in the case of
power transmission lines belonging to medium and very high voltage levels. However, at
the distribution voltage levels, that classification is more specifically defined by
recognizing three distinctly different power systems classified on the basis of their neutral
points status with respect to ground. Overall there are three types of power systems and
they are described as follows:
1. A power system, which has all of its neutrals connected directly to the ground.
2. A power system, which has all of its neutrals connected to ground through some
kind of ground-fault-current limiting resistance, reactance, and impedance.
3. A power system which has no neutrals connected to ground
2.17 OVERCURRENT RELAY WITH OR WITHOUT TIME DELAY
Protective relays such as the overcurrent relay are devices used to monitor abnormalities
within a power system. These relays obtain input signals from current and potential
transformers, they compare the faulted current to the rated current and send the signal to
the over current-relay. The over-current relay falls in the category of induction
electromagnetic relays. The relay used in this study consists of one excitation coil and
two short-circuited copper rings as shown in Figure 2.5.
This relay consists of a laminated magnetic core, where both poles are split into two
sections. These sections are called teeth, and one of each tooth in the north/south poles
contains a small copper ring. These short circuited rings are magnetically affected, and
16
such technique is known as magnetic shading of poles. The gap in the magnetic core is
for the introduction of a rotating disc, where such disc is usually made up of aluminum.
The purpose of the split on the north and south poles is to have the common flux Φ to
create Φ1 and Φ2, which are used to reduce the vibrations and move the aluminum disc.
The contacts K1 and K2 allow for the introduction of time delay, where contact K2 can be
easily slipped along the calibrated scale. The distance between contacts can be easily
adjusted, and thanks to this hardware design, the relay actuating time is adjusted.
K3
IF
Restraining
Electromagnet
Disc
K4
IF
8
Magnetic Core
6
K1
K2
4
2
1
S
Disc
Figure 2.5 An induction relay having one static coil and one shaded pole [2].
17
2.18 CURRENT INDEPENDENT OVERCURRENT RELAY
In Figure 2.6 shows the operating characteristic of a current independent overcurrent
relay. In this type of relays the time delay, tp, is not dependent on upon the value of the
fault current IF. The operating characteristic of the current independent overcurrent relay
is a straight line, where the actuating time or pick-up time (tp) is present.
t (s)
Time Delay
Operating Zone
Legend:
t=time
I=current
F=Fault
p=pick-up value
n=rated value
Pick up Time
tp
Ip = 1.2 In
Fault Current
IF (A)
Figure 2.6 Operating characteristic of a current independent overcurrent relay [2].
2.19 CURRENT DEPENDENT OVERCURRENT RELAY
The fault clearing time of the current dependent relays is a function of the fault current
magnitude, which is similar to an exponential decay function. For the induction relays
containing one static coil and magnetically shaded pole, the torque is directly
proportional to the square of its excitation current. A graphical representation of the
dependent overcurrent relay-operating characteristic is shown in Figure 2.7. Note that a
great variety of characteristics are available depending on the relay manufactures.
18
t (s)
Time Delay
3.5
0.5
Operating Zone
Minimum Pick up Time
Fault Current
IF (A)
Figure 2.7 Operating characteristic of a current dependent over-current relay [2].
2.20 MOVING OPERATING CHARACTERISTIC ALONG THE X-AXIS
The technique of moving the characteristic curves along the x-axis is accomplished by
changing the actuating flux, Φ, caused by a fault current, IF, flowing through a number of
turns. Changing the number of turns on the excitation coil can also do this, and this is
depicted in Figure 2.8.
t (s)
Time Delay
3.5
ts 0.5
Boundary Time
Fault Current
IF (A)
Figure 2.8 Operating characteristics of the excitation coil along the x-axis [2] [3].
19
2.21 MOVING OPERATING CHARACTERISTIC ALONG THE Y-AXIS
Changing the distance between the movable contact K1 and stationary contact K2 causes
the operating characteristic of an overcurrent relay to move along the y-axis. Both of
these contacts are shown in Figure 2.9. The distance between the contacts is provided in a
scale form, which is calibrated either in seconds or in terms of some other time
coefficient. The operating characteristics must be provided by the manufactures.
t (s)
G
R4
R3
R2
Legend:
R = relay
I = current
n = rated value
t = time
a = actuating
G = generator
F = fault
ta4
Time Delay
R1
ta3
ta2
ta1
In
2In
Fault Current
IF (A)
Figure 2.9 Operating characteristics of an induction relay along the y-axis [2] [3].
2.22 RECLOSING
The circuit recloser is an over-current protective device that automatically trips and
recloses a number of times to clear temporary faults or isolate permanent faults. The
reclose time can be set to different operations sequences such as two instantaneous trip
operations follow by time delays. The purpose of the instantaneous tripping is to
20
supplement the inverse time relays, by providing faster tripping operation times for
maximum reclose in faulted conditions. Multi-short reclosing schemes are applied to the
automatic reclosure to control the lock out of the 12kV primary feeder circuit breaker. In
reality reclosure attempts are made at 3 and 18 seconds are initiated following a
protective trip on all over current relays. The initial reclosure attempt at three seconds is
the minimum adjustable time delay. The second reclosure at eighteen seconds is a
cumulative effect of this time delay in addition to the minimum reclosure time interval of
15 seconds. On the other hand, reclosing schemes is not desirable in underground faults
because they are permanent as they are usually line-to-ground.
2.23 DIFFERENTIAL RELAY
According to Kirchhoff’s law the vector sum of all the currents entering into a circuit
should be equal to zero, unless a fault current is added to the path, which is not included
into the vector sum. This is how a differential relay works. If the secondary of the current
transformers that are connected to the protected circuits are paralleled with each other
and the relay protection, than no current should flow into the relay unless a fault current
enters into the system. The relay will provide an additional shunt path for the fault current
to flow through. This method is called the Unit Protection [2].
2.23.1 APPLICATIONS OF DIFFERENTIAL PROTECTION
One of the applications of differential protection is to generator protection Figure 2.10.
21
Differential Relays
Primaries of
auxiliary transformers supplying
amplitude comparators
Y
Y
Y
Relay
Restraining
Operating
Figure 2.10 Basic differential protection of a three-winding transformer [2].
The currents at the end of each phase of the stator windings will be compared through the
current transforms’, which are as close as possible to be identical. The secondary currents
should be balance during normal loads and fault currents, thus the relay should receive no
other currents from the system other than that of the fault current in the stator windings.
The same setup is used to protect for the overall system of a transformer. Except that in a
delta-wye transformer, the secondary side is connected wye-delta. The reason for this
type of setup is to make up the difference for the delta-wye transposition. To prevent the
magnetizing inrush current from operating the relay the transformer is first energized
from one side. Figure 2.11 shows a simple method of protecting the transformer windings
of against earth faults.
22
Power Transformer
Io
Io
Io
Ires
Residual
Overcurrent
relay
Restricted earth relay
In
Figure 2.11 Restricted earth protection of power transformer [2].
In bus protection, stability can be maintained by using high impedance relay. By
paralleling the current transformers’ at the bus, the relays can be very reliable. Shown in
Figure 2.12 the relay is connected to trip all the breakers. Only one phase shown.
Sources
Bus
Breakers
C.T’s
Feeders
Figure 2.12 Unbiased differential protection of a bus [5].
Relay
23
2.24 DISTANCE RELAYS
Distance relay protection of power transmission lines is very important, because it is the
fundamental for normal transmission lines through a power network. All the different
kinds of relays such as over current relays, ground fault current relays, thermal relays,
and other are used to back up the distance relays. Distance relay is the most important
relays of the power transmission line protection [1].
2.24.1 DISTANCE MEASUREMENT
The most of the reliable and positive type of protection relay is comparing the current
entering the circuit with the current leaving the circuit. This kind of principle is not
economical on the transmission lines and feeders the length, voltage or arrangement of
the line. So, in a distance relay, instead of comparing the local current line with current at
the far end of the line, the relays compares the local voltage with local current in the
corresponding phase, or convenient components of them.
For a fault at the far end of the line, the local relay voltage will be the IZ drop of the line.
It follows that the current to voltage ratio for a fault at the far end will be V/I=Z, where Z
is the impedance of the line, Figure 2.13. For an internal fault the protected section of the
line is V/I<Z. For a fault beyond the next section, V/I=Z. Since Z is proportional to the
line length between the relay and the fault it is also a measure of the distance to the fault;
hence the term distance relay [1].
24
In order to measure the same distance on all faults involving more than one phase as
shown in Figure 2.13 the distance relay will compare the potential between the two
faulted phases with the vectorial distance of their currents. As shown in Figure 2.13 (b)
𝑉𝑎
𝑏
for a b-c fault the relay measures𝐼𝑎−𝐼𝑏
= 𝑍1 , where Ib is on the wrong directional
therefore the minus sign. Similarly, for phase to ground faults as shown in Figure 2.13 (c)
𝑉 −𝑛
the phase c-to-ground the relay measures 𝐼 𝑐−𝐼 = 𝑍1 .
𝑐
𝑛
But since the current in the ground return path is difficult to get to, the relay is given the
equivalent current, which is a function of the current transforms residual and the phase c
relay measures𝐼
𝑉𝑐
𝑐 −𝐾𝐼𝑟𝑒𝑠
= 𝑍1 .
25
ZS
ZL
Block
EG
Trip
IZL
VRelay
I
(a)
+
Vab
-
a
b
Internal
Fault
Relay
Setting
Ia
Ib
Fault
c
(b)
a
b
+
Vcn
-
c
n
Ic
In
(c)
Figure 2.13 Unbiased differential protection characteristics for a bus [2].
External
Fault
26
2.24.2 HARDWARE COMPONENTS OF DISTANCE RELAYS
In order for distance relay to function properly, it should contain at least the following
components: one excitation organ, one directional organ, one or more measuring organs,
one or more time organ, and one executive organ, as shown in Figure 2.14.
A
B
C
Legend:
EO = excitation organ
DO = directional organ
MO = memi organ
TO = time organ
EXO = executive organ
+
CB
EXO
Power Transmission Line
-
EO
DO
M01
+
I>
CT
If
F
W
Z1<
T01
T02
+
Z2<
T03
+
T1
+
T2
Vf
PT
M02
+
T3
-
-
Legend:
CB= circuit breaker
I= current
F= fault
V= voltage
CT= current transformer
PT= potential transformer
T= timer
I>= overcurrent
W= directional relay
Z<= impedance relay
Figure 2.14 Principle connections of hardware components of an impedance distance relay [2].
27
2.25 DIRECTIONAL PILOT RELAYING
A unit form of protection used when it is important to clear the faults at the same time at
both ends of the protected section of the line. For unit protection it is important to
exchange information at the same time about the faults conditions at the ends of the
protected section of the line by either a pilot-wire or a carrier channel. There are two
basic principles that are used (a) to compare the direction of power flow at the ends and
(b) continuously compare the instantaneous phase relation of the currents at the two ends.
2.25.1 DIRECTIONAL COMPARISON
As shown in the directional comparison pilot schemes, the direction of power flow is
compared by means of the relative position where the directional relays at the two ends of
the protected section contacts. This kind of protection uses the fact that during an external
fault, the power will flow into the protected section at one end and out at the other. But
during an internal fault, the power can flow inwards at both ends.
Directional relays at the protected section at each end are connected to block tripping
when the power flow caused by the fault reaches the protected line at the bus bar as
shown in Figure 2.15. As it should be interconnecting these directional relays through a
pilot wire or a carrier channel, can compare the position of their contacts and thus the
location of the fault would be determined. As shown in Figure 2.15 (b) an external fault
has been accrued and will cause the directional relay at the end nearest the fault to
prevent tripping at both ends of the protected section. For the load current as shown in
28
Figure 2.15 (a) will have the same as an external fault for the; the relay at the load will
block the tripping. On the other hand, on the internal fault as shown in Figure 2.15 (c) the
tripping will not be prevented or blocked because the power will flow from the bus into
the line at both ends not from the line to the bus.
A
B
(a)
LOAD
(Open if volt restraint used)
A
B
(b)
External
Fault
A
(c)
Internal
Fault
B
Figure 2.15 Basic principle of directional pilot relaying. (a) Normal conditions. (b) External fault.
(c) Internal fault [4] [5].
2.25.2 INFORMATION TRANFER BETWEEN ENDS
The communication circuit can transfer the information from one end to the other end.
There are several ways of transferring the information between two ends. Two methods
that have been used is a pair of pilot wires, and the second one is a carrier channel using
the power lines themselves, or a v.h.f Radio signal transmitted directly between the line
29
terminals. For short lines, pilot wires were generally used and cheaper, and the carrier
channel was used for long lines because it is more economical. There are some examples
of each of the above communication circuits as shown in Figure 2.16 to Figure 2.18. On
Figure 2.16 are the basic circuits of dc pilot schemes (a) Series pilot scheme. (b) pilot
scheme are shown.
B
CB
CB
CB
A
Protected
T
D
C
Line section
D
T
D
Pilot
Pilot
(a)
D
D
D
B
B
F
F
B
F
T
T
A
C
CB
CB
T
B
CB
(b)
Figure 2.16 Basic circuits of DC pilot schemes. (a) Series pilot scheme. (b) Shunt pilot scheme
[2].
30
On Figure 2.17 (a) Basic dc connection of carrier blocking scheme and (b) Use of 3-zone
distance relay for carrier blocking. Finally, Figure 2.17(a) distance reaches settings for
carrier blocking scheme (b) Mho characteristics for carrier blocking scheme. Other forms
of communication circuits that are used as well and in other kinds of applications for
power systems are fiber optic cables, and the communication circuits used by telephone
companies.
Protected Line
+
Line Trap
Coupling
Capacitor
фTX
фT
Transmitter
Receiver
C
фB
GTX
GT
GTX
R
H
R
фTX
Receiver Relay
GB
-
(a)
+
(фT) Y2
Y1
GT
R
R
OY3
T2
T3
o
GTX
Trip
фTX
-
(b)
Figure 2.17 DC communication circuit. (a) Basic DC connections of carrier blocking scheme. (b)
Use of 3-zone distance relay for carrier blocking [2].
31
2.26 MAINTENCE AND TESTING OF RELAYS
The protective relay is usually used to protect very expensive equipment. In order to
make these devices function normally and smoothly they should be taken care of
properly. But when neglected they may become inoperative and could become a hazard
in themselves. In general, the protective relays and their trip circuits should be
periodically checked in order to ensure that they would always be ready to operate
normally. There are very good types of test or relay protective:
(a) Installation or commissioning Tests
(b) Periodic tests to check the calibration and condition of the relay.
(c) More frequent tests of a simple nature cause movement of the parts, and to
check the continuity of the trip circuit.
2.26.1 INSTALLATION OR COMMISSIONING TESTS
The first thing should be to examine for damage in transit. Care should be taken to not
bend any light parts when removing packing pieces, such as disc wedges. Other
important precautions are (a) to prevent handling contact surfaces; (b) to dust the cover
before removing it; (c) to see the packing pieces are removed and the armatures move
freely, (d) to avoid permanent magnets with ferrous objects such as screwdrivers.
Each relay unit should be given a mechanical inspection to see that the armature moves
freely and that the contacts have the necessary travel and wipe to ensure reliable
operation. Also check the manufacturers setting, if given, in the instruction book. An
32
inspection light and dental mirror should be used to see that the magnetic gaps are clean
before the relay is left in service. In order to check the current- transformers, voltagetransformers and wiring associated with the relays it is usual also to make overall tests
from the primary circuit. The primary current is usually supplied by a test transformer of
about 5kVA supplied from a low voltage lighting or power source, such as 240 volts, 30
amperes source, and tapped for various voltages (say 1 to 10 volts) necessary to give line
currents up to 1000A depending on the impedance of the circuit; this current is sufficient
to check the polarity of the connections but not to simulate fault currents, the latter being
done in the secondary injection tests to check the relay characteristics as in Figure 2.18.
The secondary wiring can be further checked if necessary by a low reading ohmmeter or
by the ringing method using a bell and battery [4] [6].
Primary Circuit
230 v A.C.
Relay
A
11500/230 Volts
5Kva
Figure 2.18 Primary injection test circuit [6].
C.T
33
2.26.2 PERIODIC TESTS
Since the installation tests, if the wiring has not been changed, it is not necessary to
recheck the polarity of the current-transformers. Also, the current-transformers can be
checked from their secondary or tertiary. However, the frequency of these periodic tests
depends on many factors. In clean, dry surrounding once a year is enough, or in even
every three years in the case of modern relay with a high torque/friction ratio, especially
if the tripping contacts are relieved by seal-in relay.
Equation Chapter (Next) Section 1
34
Chapter 3
MATHEMATICAL MODEL
3.1 INTRODUCTION
The objective of this chapter is to develop an understanding of the equations used during
the fault analysis of the system. Deriving these equations will provide a better
understanding of the power system during these abnormal conditions.
3.2 FAULT CURRENTS
In deriving the fault current occasionally the precise time varying fault current is not
required. If this is the case, the following simplification can be made. First, the time
constant is very small in comparison to the other time constants. Due to the relatively
small value of the angle θ, the fault current components associated with a transient
process in the equivalent windings located along the q-axis are ignored. Second, in order
to compensate the fault current for this decrease, it is assumed that cos θ = 1 and
𝛼0 = 0, 𝑖. 𝑒. , cos(𝜃 − 𝛼0 ) = 1. Finally, the magnetic symmetry is assumed along the d
and q axis, i.e., 𝑋𝑑" = 𝑋𝑑′ , which is really true for the cylindrical rotor generators. The last
simplification eliminates the components of the fault currents containing twofold
frequency. After this simplification, the induced electric magnetic field or EMF projected
along the quadrature axis, 𝐸𝑞 , is defined by Equation 3.1.
Eq  V cos   X D I D (V )
(3.1)
35
where Equation 3.2
I D  I sin     ( A)
(3.2)
is the load current preceding the fault and projected along the d axis. When a generator
enters the faulted condition and the no load operating condition is formed, the
corresponding subtransient, transient, and steady-state short circuit current components
are defined by
I 0" 
V
 A ,
X d"
(3.3)
and
(3.4)
In the case that the generator enters the faulted condition from any load operating
conditions carrying any current, I, prior to the fault, the last three components of the fault
current are modified as follows
I "  I 0"  I d ( A),
(3.5)
(3.6)
and
(3.7)
When the above simplifications are presented, the expression of a three-phase fault
current in phase A, 𝑖𝐴 (𝑡), can be described by Equation 3.5.
iA  t   id  t   i pd  t   ia  t  A ,
(3.8)
36
where 𝑖𝑑 (𝑡) represents the first, while the current 𝑖𝑝𝑑 (𝑡) represents the second term.
When the 𝑖𝑑 (𝑡)component is combined with Equation 3.8, Equation 3.9 is produced [1].
 V

id  t   2 
 I d  cos  2  IV 0  I d  cos  2 IV cos  A
 Xd

(3.9)
The current 𝑖𝑝𝑑 (𝑡) represents the second terms. When 𝑖𝑑 (𝑡) and 𝑖𝑝𝑑 (𝑡) are summed
together, they will produce 𝑖𝑝 (𝑡), Equation 3.10
i p  t   id  t   i pd  t   2I p  t  cos  A ,
(3.10)
I p (t )  ( I "  I ' )e t Td  ( I '  IV 0 )e t Td  IV ( A).
(3.11)
where
"
'
The a periodic, or DC component, 𝑖𝑎 (𝑡), which is provided as the last term, is
approximated by Equation 3.12 and Equation 3.13.
ia  t   

 1
2  1
1 
1 
V  "  "  cos  2  0      "  "  cos 0     e  t /Ta  A (3.12)
X

2  X d X q 

 d Xq 
ia  t   
2 1
1   t /T
V  t /Ta
e
 2 I 0"e t /Ta  A  .
 "  "  e a  2
2  Xd Xq 
X2
(3.13)
In conclusion, the approximated expression of a three-phase fault current in phase
A, 𝑖𝑎 (𝑡), is:
ia  t   i p  t   ia  t   2 I p  t  cos  ia  t 
 2  I "  I '  et /Td   I '  IV 0  et /Td  IV  cos  2 I 0"e t /Ta  A  .


"
'
(3.14)
37
3.2.1 DERIVING THE SUBTRANSIENT CURRENT
The subtransient current can be expressed in terms of the power angle, θ, and the load
angle, φ, as follows:


I
I "  I 0"  I sin       1  " sin      I 0"
 I0

 IX "
 V
 1  d sin      "  A  .
V

 Xd
(3.15)
Where V and I are the operating parameters, and 𝑋𝑑" is the machine design parameter,
measured in ohms (Ω) and calculated from:
X d"  xd"
Vn

In
(3.16)
For typical operating conditions of a synchronous generator, the parameters below can
help further simplify the equation:
V  Vn
(3.17)
I  In
(3.18)
cos   0.1  0.2
(3.19)
  25  40
(3.20)
xd"  0.1  0.2 pu
(3.21)
The value of the expression provided within the brackets may fall within the 1.08-1.12
range. The smaller value corresponds to salient pole synchronous generators and the large
values correspond to the cylindrical rotor synchronous generators. Therefore, the
38
practical formula for computing the subtransient component of the fault current can be
defined by Equation 3.22.
I "  1.1
Vn
 A
X d"
(3.22)
3.2.2 DERIVING THE TRANSIENT CURRENT
Similar to the calculations for deriving the subtransient current, the transient current is
done almost the same way. The formula earlier can be expressed in terms of the power
angle, θ, and the load angle, φ, as follows:


I
I '  I 0'  I sin      1  ' sin      I 0'
 I0

(3.23)

 V
IX '
 1  d sin      '  A
V

 Xd
For typical operating conditions of a synchronous generator, like below:
V  Vn
(3.24)
I  In
(3.25)
cos   0.1  0.2
(3.26)
  25  40
(3.27)
xd"  0.13  0.3 pu
(3.28)
The expression from Equation 3.23 within the brackets may fall within the 1.13-1.22
range. Therefore, a practical formula for the copulation of the transient component of the
fault current, 𝐼 ′ , can be written as:
39
I '  1.175
Vn
V
 1.2 n'  A 
'
Xd
Xd
(3.29)
3.3 HEAT IMPULSE
The time varying short circuit current produces the heat impulse in a conductor carrying
the current. Mathematically, it can be defined by:

H   iF2  t  dt  kA2 s 
(3.30)
0
where τ is the duration time of a short circuit. The heat impulse is identical with a Joule’s
heat in a conductor carrying the fault current. Since the heat produced by the fault will
increase the conductor temperature very fast, it is conveniently called the “heat impulse.”
The notion of a thermic current is introduced as a quantitative measure of the heat
impulse created by the short circuit current. Mathematically, it can be defined in a similar
fashion to the RMS definition of an AC current:
I 
1


 i  t  dt  A
2
F
(3.31)
0
where τ is the duration of a short circuit.
The relationship between the thermic current, 𝐼𝜏 , and the heat impulse, H, in the
connection to the formula earlier can be given by:
H  I2 (kA2 s )
or
(3.32)
40
I 
H

(A)
(3.33)
By definition the heat impulse, H, is expressed in the terms of fault current, 𝑖𝐹 (𝑡), and the
fault duration time, τ. In the first approximation, the fault current, 𝑖𝐹 (𝑡), can be expressed
in terms of the “quasi” periodic current component, 𝑖𝑝 (𝑡), and the aperiodic components,
𝑖𝑎 (𝑡). Therefore, the heat impulse, H, can be expressed by:

H    I p  t   ia  t   dt
2
0

I

2
P

 t  dt  2 I P  t  ia  t  dt  
0
0
i  t  dt  kA s 
2
a
(3.34)
2
0
Since the second integral, containing a twofold product, 2 , is far smaller in magnitude
than the other two integral terms, it may be neglected. Therefore, the expression of the
heat impulse, H, becomes:

H  I
0

2
P
 t  dt   ia2  t  dt  H P  H a
 kA s 
2
(3.35)
0
where

H P   I P2 (t )dt  kA2 s 
(3.36)
0
Is the heat impulse of a “quasi” periodic current, 𝐼𝑝 (𝑡),

H a   I a2 (t )dt  kA2 s 
0
is the heat impulse of the aperiodic current, 𝑖𝑎 (𝑡), and τ is the fault duration time.
(3.37)
41
The aperiodic current heat impulse can be described in a general form. Equation 3.38 will
express that:

Ha  

2 I 0" sin c

2


e 2 /Ta dt  1  e 2 /Ta Ta  I 0" sin c 
2
 kA s 
2
(3.38)
0
Since the impedance of the short circuit loop is predominantly inductive reactive, 𝜑𝑐 =
90°, it becomes:


H a  1  e 2 /Ta Ta I 0"2  kA2 s 
(3.39)
The duration of the short circuit, τ, for the faults taking place electrically “father away”
from a generator, is usually this:
3
2
  Ta  s 
(3.40)
because the aperiodic time constant, 𝑇𝑎 , in those cases is significantly smaller. Therefore,
the heat impulse given earlier may be approximated by:
H a  Ta I 0"2  kA2 s 
(3.41)
where 𝑇𝑎 is the aperiodic time constant of the short circuit loop. On the other hand, the
“quasi” periodic current heat impulse can be defined by:

HP  
0
2




 2 I P cos   2  c   dt





  I dt  
2
P
0
0



I cos 2    c  dt  kA2 s 
2


2
P
(3.42)
42
Since the value of the second integral term is somewhat smaller than the first, it may be
neglected. Therefore, by substituting the previously derived expression for 𝑖𝑝 (𝑡) into the
equation above, the periodic current heat impulse becomes:

HP   I

2
P
 t  dt  
0
0
 I "  I '  e  t /Td   I '  IV  e t /Td  IV  dt  kA2 s 


"
"
2
(3.43)
When the short circuit duration time, τ, is:

Td'
5
s
(3.44)
,
the expression for the periodic current heat impulse, 𝐻𝑝 , becomes:

H P   [( I "  I ')e
0
 t Td"
 I ' ]2 dt  I '2 (   ) (kAs)
(3.45)
where
2



"
"
 I"  T"
 I" 
Δ   '  1 d 1  e 2t /Td  2  '  1 Td" 1  e t /Td
I
 2
I

 s
(3.46)
is the corrected time interval.
The periodic current heat impulse, 𝐻𝑝 , can be calculated by the means of the transient
current component according to the equation above, if the fault duration time, τ, is
corrected by Δτ defined above, which takes care about the subtransient period.
The first term of a corrected time interval, Δτ, is very small. Even in the cases of short
circuit taking place close to the generator site, it may have the maximum value of 0.1𝑇𝑑" ,
which is negligible. The second term of a corrected time interval, Δτ, may have maximum
43
value equal to 𝑇𝑑" , being usually less than 0.5𝑠. Therefore, in the second approximation,
the periodic current heat impulse, 𝐻𝑝 , can be computed from:
H p  I '2  kA2 s 
(3.47)
For the short circuits taking place at the generator terminals, the periodic current heat
impulse, 𝐻𝑝 , is usually:
H P  I '2   Td"   kA2 s 
(3.48)
For the short circuits taking place in the power transmission network away from the
generators, the ratio (𝐼 " ⁄𝐼 ′ ) is close to one and Δτ may be again, in spite of the 𝑇𝑑"
increase. Beside this fact, the faults are lasting relatively longer in the power network and
τ is significantly longer then Δτ.
Sometimes the faults are permitted to last longer than 𝑇𝑑" ⁄5. the periodic current heat
impulse in those cases is computed below:

2
'
T"
H P  I d    I '  IV  et /Td  IV  dt  kA2 s 

5 T' 
'2
(3.49)
d
Which is valid when 𝑇𝑑" ⁄5 > 3𝑇𝑑" .The second term may not easily be approximated due
to the very large range within which the ratio 𝐼 ′ ⁄𝐼𝑉 may fall over, due to the different
fault locations, and due to the different time durations of the faults. The formula is
usually used in its original form.
44
In the distribution power networks having no generation, the currents 𝐼 ′ and 𝐼𝑉 are
numerically very close to each other. Because of this fact, the periodic current heat
impulse is computed from:
H P  IV2  I '2  kA2 s 
(3.50)
This expression for the heat impulse is used whenever the short circuit currents are
cleared within the time intervals

Tdo'
5
s
(3.51)
′
since the time constant 𝑇𝑑𝑜
does represent the upper limit of the time constant 𝑇𝑑′ . If this
is not the case, than the periodic heat impulse current are valid.
3.4 SIZING OF THE CONDUCTOR
By introducing the notions of boundary permitted temperature at the short term
overloading 𝜃𝑏𝑝 = 𝜈𝑏𝑝 − 𝜈𝑎 and permanently permitted temperature 𝜃𝑝𝑝 = 𝜈𝑝𝑝 − 𝜈𝑎 ,
where 𝜈𝑎 is ambient temperature, the heat impulse can obtain boundary value defined as
the following:
45
mc
H
k s R20
mc
H
k s R20




0
bp


pp
d
 kA s 
1     20 
2
d
1       20 


mc
In 1   ( bp  a  20)   In 1    pp  a  20  
k s R20
(3.52)
1   bp  20 
 qlc
In
 k s  1/ q  1    pp  20 
 q 2lc 1   bp  20 

In
 k s  20 1    pp  20 
where q is the cross-section of a conductor, γ and 𝜌20 is its specific weight and resistivity
at 20℃, repectively. Thus, the formula in Equation 3.52 provides the means of
determining the boundary heat impulse for the given cross-section of a conductor and
permitted temperature.
From Equation 3.52, it is possible to determine the minimum cross-section, 𝑞𝑚𝑖𝑛 , of a
conductor needed to absorb the entire heat boundary impulse, 𝐻𝑏 , under condition not to
step over the boundary permitted temperature, 𝜃𝑏𝑝 , i.e.,
qmin 
1
1   bp  20 
c
ln
 20 1    pp  20 
ks H b
 mm 
2
(3.53)
Typically data for the quantities appearing in the formula in Equation 3.53 are provided
in Table 3.1, which covers only copper and aluminum, because they are exclusively used
46
in substations and power plant design so often. In Table 3.1 the characteristics of both
aluminum and copper has been listed.
ρ20 (Ωmm/m2) α (1/°C) ρ
γ
υa(°C) υpp(°C) υbp(°C)
3
(J/kg°C) (kg/m )
0.0286*10-6
0.004
908.5
2.7*103
35
65(70)
180
Cu 0.0178*10-6
0.004
387.3
8.9*103
35
65(70)
200
Al
Table 3.1 Data provided from Equation 3.50 used to minimize 𝑞𝑚𝑖𝑛 .
When these data are plugged into the equation above, the minimum cross-section for a
conductor can be computed as the following:
qmin cu  7.4 ks H b  mm 2 
(3.54)
qmin Al  12 ks H b  mm 2 
(3.55)
47
3.5 THE ANALYSIS OF THE FAULT CURRENTS
Our substation configuration is:
69kv
12kv
Y
Network
ZA1=0
.
ZA1=RA1+Jxa1
ZA2=RA2+jXA2
TL
S=20 MVA
XT0, XT1, XT2
FAULT
Z10=R10+jX10
Z11=R11+jX11
Z12=R12+jX12
Figure 3.1 One line diagram of power system model.
For the analysis of the fault current, the assumption is that there is no load condition.
Since, no load will result in a smaller fault current, because it is desirous for the relay to
pick up the smallest fault current. For the impedances above, the zero subscript refers to
the zero sequence; the number one subscript refers to the positive sequence; and the
number two subscript refers to the negative sequence.
For the analysis of fault currents, only consider the no load condition, which will result in
smaller values of fault currents. This is the case, because it is ideal for the protection
relays to detect the minimum fault current. If the protection relays can detect the
minimum fault current, any larger fault currents of course can be detected by the
protection relays, which will result in a safer protection. To analyze fault currents, the
main approach is to use the sequence networks formed during faulted conditions. After
establishing the fault sequence networks, the fault sequence currents are calculated. Then,
48
the phase fault currents can be computed by adding individual sequence current
components together. At that end, the sequence network needs to be first established as
seen from the fault location, which is at the end of the transmission line. Thus, the first
task is to refer all the impedances to the secondary side of the transformer. Note that for
the Delta-Wye connection of the transformer, the fault cannot see the zero sequence
impedance of the active network, so the zero sequence impedance of the active network
is ignored in the calculation.
The impedances referred to the secondary side of the transformer are:
Z A1S  Z A1   LV / HV  Ω
(3.56)
Z A2 S  Z A2   LV / HV  Ω
(3.57)
X T 0 S  X T 0   LowVoltage2  /  S3   Ω 
(3.58)
X T 1S  X T 1   LowVoltage2  /  S3   Ω 
(3.59)
X T 2 S  X T 2   LowVoltage2  /  S3   Ω 
(3.60)
2
2
Thus, the total zero sequence impedance 𝑍0 seen at the fault is:
Z0  X T 0 S  Z10
(3.61)
Likewise, the total positive sequence impedance 𝑍1 seen at the fault is:
Z1  Z A1S  X T 1S  Z11
(3.62)
And the total negative sequence 𝑍2 seen at the fault is:
Z 2  Z A2 S  X T 2 S  Z12
(3.63)
49
F0
F1
F2
IA0
IA1
IA2
Z1
Z0
+
-
Z2
Vf=12000 0.577 0
N1
N0
(a)
N2
(b)
(c)
Figure 3.2 Sequence networks: (a) The zero sequence network. (b) The positive sequence
network. (c) The negative sequence network [1][5].
3.5.1 SINGLE LINE TO GROUND FAULT CURRENT
The general representation of the single line to ground fault is represented in Figure 3.3.
N0
+
-
Z0
IA0
N1
Vf
Z1
Z2
IA1
N2
IA2
Figure 3.3 Sequence network for a single line-to-ground fault [1] [5].
Assume the fault occurs in phase a, and fault impedances are zero. As seen, the fault
occurs when there is a short circuit between one phase to ground. The interconnection of
50
the sequence networks is a series connection as shown in Figure 3.3. As seen, the fault
sequence currents are:
I a 0  I a1  I a 2 
Vf
3   Z0  Z1  Z 2 

120000
3   Z 0  Z1  Z 2 
(3.64)
Therefore, the phase fault currents are:
I af  I a 0  I a1  I a 2  3I a 0
(3.65)
Ibf  I a 0  a 2 I a1  aI a 2  0
(3.66)
I cf  I a 0  aI a1  a 2 I a 2  0
(3.67)
3.5.2 LINE TO LINE FAULT CURRENT
The general representation of the line to line fault is represented in Figure 3.4. The line to
line fault occurs when there is a short circuit between two phases. Assume the fault is
between phase a and phase b, and all fault impedances are zero. The interconnection of
the sequence network is represented in Figure 3.4. As seen, the sequence fault currents
are:
Ia0  0
I a1   I a 2 
Vf
Z1  Z 2

(3.68)
120000
3  Z1  Z 2 
(3.69)
Therefore, the fault phase currents are:
Ibf  I cf  a2 I a1  aI a 2 ,
where
(3.70)
51
a 2  1240,
(3.71)
a  1120
(3.72)
Iaf  Ia0  Ia1  Ia2  0
(3.73)
F0
F1,2
IA1
IA0
IA2
Z1
Z0
+
-
Z2
Vf
N1
N0
(a)
N2
(b)
Figure 3.4 Sequence network for a line-to-line fault [1][5].
3.5.3 DOUBLE LINE TO GROUND FAULT CURRENT
The general representation of the double line to ground fault is represented in Figure 3.5.
As seen, the fault occurs when there is a short circuit between two phases and ground.
Assume the fault occurs with phase b and phase c, and the fault impedances are zero. The
interconnection of the sequence networks is represented in Figure 3.5. From the
interconnection, the fault sequence currents are:
I a1 
120000


Z2  
3  Z1   Z 0 

Z0  Z2  


(3.74)
52
 Z0 
Ia2   
  I a1
 Z0  Z2 
(3.75)
 Z2 
Ia0   
  I a1
 Z0  Z2 
(3.76)
Therefore, the phase fault currents are:
I af  0
(3.77)
Ibf  I a 0  a2 I a1  aI a 2
(3.78)
Icf  I a 0  aI a1  a2 I a 2
(3.79)
a  1120
(3.80)
a 2  1240  1  120
(3.81)
Where
The total fault current flowing to the neutral line or ground is:
I n  I bf  I cf  3I a 0
(3.82)
53
IA1
IA0
Z1
Z0
+
-
Z2
Vf
N1
N0
IA2
N2
Figure 3.5 Sequence network for a double line-to-ground fault [1] [5].
3.5.4 THREE PHASE FAULT CURRENT
The general three-phase fault condition representation is shown in Figure 3.6. As seen,
three-phase fault occurs when three-phases are shorted together and are shorted to
ground. Assume all the fault impedances are zero, and there are direct contacts between
the lines and ground as shown in the representation. As seen, three-phase fault is a
balanced fault since all fault impedances are zero. For this type of fault, each sequence
network is short circuited over its own fault impedance and isolated from each other.
Since only the positive sequence network has its own voltage source, therefore:
I a1 
120000
3  Z1
(3.83)
54
The fault current in each phase is:
I af  I a1
(3.84)
I bf  I a1 1240
(3.85)
I cf  I a1 1120
(3.86)
Thus, the total current flowing to the neutral or ground is:
I n  I af  I bf  I cf  0
(3.87)
F0
F1
F2
IA0
IA1
IA2
Z1
Z0
Z2
+
N1
N0
(a)
(b)
N2
(c)
Figure 3.6 Sequence network for a three phase fault [1][5].
3.6 OPERATING CHARACTERISTIC OF THE DIFFERENTIAL RELAY
By using the percentage differential relay in this project, the power transformer is
protected. In Figure 3.7 it illustrates the schematic diagram and the characteristics of the
percentage differential relay.
55
Trip
Restrain
Restrain
Protected Zone
87
(a)
I1'
I2'
If’
I1
I1
R
I2
R
Op
I2
I1-I2
(b)
Positive-torque
region
Operating
Characteristics
Negative-torque
region
(c)
Figure 3.7 Differential Relay (a) general schematic diagram, (b) schematic diagram during a fault,
and (c) operating characteristics.
56
According to the schematic diagram normally the CT secondary currents merely circulate
between CT’s and no current flows through the relay operating winding.
Therefore:
I1  I 2  0
(3.88)
In case of the fault occurs outside the protection zone, no current flows through the
operating winding. On the other hand, when the fault occurs within the protection zone,
the fault current flowing through the operating winding will be:
I1  I 2  I f
(3.89)
If it exceeds the predetermined setting of the differential relay, the relay will be activated,
tripping the circuit breakers. Because of these characteristics of the differential relay, it
can be used to protect the power transformer. Both circuit breaker and the current
transformer are present on both sides of the power transformer.
Equation Chapter 4 Section 1
57
Chapter 4
APPLICATION OF THE MATHEMATICAL MODEL
4.1 SUBTRANSIENT FAULT CURRENT FOR THE SECONDARY 12KV SIDE
In order to make reference to the primary or secondary side of the system the sub fix 69
and 12 are used respectively. The assumption that the subtransient fault current on the
primary side is 10,800 amps is made in order to analyze the system. This assumption is
used to calculate the subtransient fault current for the secondary side. The value for the
assumed current has to be large enough to simulate the worst case scenario in a faulted
system, which is a three phase fault. Now with this assumption the subtransient reactance
for the primary side can be calculated by means of the following formula.
I"69 =
1.1*V69
"
3 * X 69
,
(4.1)
Where
I"69  10800 A
(4.2)
V69  69 kV
(4.3)
Now solving for X "69
10800( A) 
1.1* 69 *103
"
3 * X 69
(4.4)
"
X 69

3
1.1* 69 *10
 4.06()
3 *10800
58
The next step is to refer the subtransient reactance from the primary side to the secondary
side. This is done by using the subtransient reactance from the primary side and referring
it to the secondary side by means of the transformer turn ratio, hence 1/n2.
2
V 
X  12   X 12" ()
 V69 
"
69
2
 12 *103 
X 12"  4.06 * 
 0.1228().
3 
 69 *10 
(4.5)
The subtransient reactance from the primary side was referred to the secondary side, now
the same procedure is conducted for the subtransient reactance of the transformer. For
this step another assumption is made for the value of the transformers positive sequence
reactance and is as follows.
X trf"  x1
V122
S
12*10 
 0.12*
(4.6)
3 2
X trf"
20*106
 0.864()
Now that the reactance for the transformer has being calculated, the next step is to add
the secondary subtransient and transformers reactance. This result yields the total
subtransient reactance on the secondary side.
X T" 1  X 12"  X trf" ()
(4.7)
X T" 1  0.1228  0.864  0.9868
With the information obtained the subtransient fault current can now be calculated from
the data above.
59
1.1V12 1.1*12 *103
I 

 7722.9667 A
3 X T" 1
3 * 0.9868
"
12
(4.8)
4.2 TRANSIENT CURRENT
The procedure for calculating the transient current is the done the same way that the
subtransient current was calculated. The fault current needs to be brought from the
primary side to the secondary side as for the subtransient current. For this stage of the
calculations another assumption for the value of the transient current for the primary side
is made. The value for the primary side transient current is made to be 10,780 amps and
the zero sequence reactance for the transformer is 0.10. With the following given values
I'69  10780 A
(4.9)
V69  69*103V
(4.10)
Solving for X '69 is as follows
I 69' 
1.2V69
3 X 69'
10780 
X '69 
1.2 * 69 *103
3 X 69'
(4.11)
1.2 * 69 *103
 4.43
3 *10780
Now the transient reactance from the primary side will be referred to the primary side
following the same steps as that to the subtransient reactance.
60
2
V 
X  12   X 12' ()
 V69 
'
69
2
 12*10 
X 12'  4.43* 
 0.1340().
3 
 69*10 
3
(4.12)
The transient reactance for the transformer is in calculated based on the previews
assumption that the zero sequence reactance is 0.10. With the assumption made the
calculations are as follows.
X trf'  x0
V122
S
12*10 
 0.10*
(4.13)
3 2
X trf'
20*106
 0.72()
Then the total transient reactance is the summation of the transient reactance of the
secondary side and the transient reactance of the transformer and it is as follows.
X T' 2  X 12'  X trf' ()
(4.14)
X T' 2  0.1340  0.72  0.8540 
The transient current on the secondary side can now be calculated by means of the
following formula.
I12' 
1.2V12 1.2 *12 *103

 9735.1801 A
3 X T' 2
3 * 0.8540
(4.15)
61
4.3 HEAT IMPULSE
The calculations done for the subtransient current in the previews section are utilized to
calculate the subtransient resistance. The equivalent resistance for the system is the
addition of the subtransient resistance and the resistance of the transformer. The
resistance of the transformer is assumed to be at least ten percent of the previously
calculated subtransient reactance. From equation 4.8 where current
"
I12
 7722.9667 A
(4.16)
The resistance is calculated as follows
I12" 
V12
3R12"
(4.17)
"
R 12

12 *103
 0.8971 
3 * 7722.9667
The addition of the subtransient resistance and 10% of the subtransient reactance yields
the equivalent resistance of the secondary side. The equivalent resistance is required in
order to calculate the periodic time constant ω = 2πf, which is used to calculate the heat
impulse. The heat impulse is the heat created by the subtransient current in the secondary
side.
"
"
Re  RTrf
 R12
()
(4.18)
Re  (10%) * (0.864)  0.8971  0.9835 
Ta 
X 12"
0.1228
( s) 
 3.3119 *104 sec
 Re
377 *0.9835
(4.19)
62
H a  Ta I12"2 (kA2 s )  3.3119 *104 * (7722.9667) 2  19.7536 kA2 .s
(4.20)
The heat impulse for transient fault current is represented by the multiplication of the
transient fault current and the fault duration time. The fault duration time τ is the length
of time of the fault duration. For the purpose of this study the fault duration is considered
to be 0.7 seconds and it is described as follows.
H P   * I12' 2 (kA2 s )  0.7 * (9735.1801)2  66341.6121 kA 2 s
(4.21)
Now that the heat impulse for the transient and subtransient fault current calculations are
done, the heat impulse boundary. The heat impulse boundary is heat produced by the
total fault current on the line. Hence by just adding the heat impulse of the transient and
subtransient fault current the boundary is obtained.
Hb  Ha  H p (kA2 s)  19.4216  66341.6121  66361.3657 kA2 s
(4.22)
4.4 CONDUCTOR SIZING
The values previously calculated are used to calculate the size of the conductor, with the
characteristics of copper and aluminum. The skin effect is taken into consideration which
is represented by ks, which in terms is the current that flows around the conductor [1]. In
appendix A.1 and A.2 the characteristics of both aluminum and copper are listed. When
the data is plugged into equation 3.51, the minimum cross-section for a conductor can be
computed in a simplified form as shown in Equation 4.23.
63
qmin cu  7.4 k s H b ( mm 2 )
qmin cu  7.4 1* 66361.3657  1906 mm 2
2
 0.0254 
6
2
1 cmil =  
  506.7 *10 mm
 2 
1 cmil =
(4.23)
1906
 3772 kcmil
506.7 *106
3772
 627 kcmil
6
Now for the aluminum conductor which is used for the overhead system. The same steps
as above are used to calculate the size of the conductor by using Equation 3.52.
qmin al  12 k s H b ( mm 2 )
qmin al  12 1* 66361.3657  3091 mm 2
2
 0.0254 
6
2
1 cmil =  
  506.7 *10 mm
 2 
1 cmil =
3091
 6100 kcmil
506.7 *106
6100
 677 kcmil
9
(4.24)
64
4.5 FAULT CURRENT AT TRANSIENT STAGE FOR OVERHEAD FEEDERS
The most common type of overhead conductors used in the industry are: all Aluminum
conductors (AAC), all Aluminum alloy conductors (AAAC), Aluminum conductor steelreinforced (ACSR), and copper conductors (Cu) [1]. The overhead conductor for this
study is ACSR, and it will be assumed that the line is transposed. For the underground
conductor the material is copper.
The line consists of 666 kcmil, operating at a temperature of 50C at 60 cycles and the
distances between the conductors are as follows:
DAB  26"  2.1670 ft
(4.25)
DBC  18" 44"  62"  5.1670 ft
(4.26)
DAC  26" 18" 44"  7.3333 ft
(4.27)
Then to calculate the positive/negative and zero impedance the following equations are
used [1].

 Deq  
Z1  Z 2   ra  j 0.1213ln 
 L
 Ds  


 De3
Z 0   ra  3re   j 0.1213ln 
 D  D2
eq
 s


  L
 
(4.28)
(4.29)
Where L is the length of the line and equals 6mi
De is the equivalent depth of earth return and it is described as follows [1].
1
  2
De  2160   ft
f 
(4.30)
65
If the frequency is substituted as (60Hz) and assume that the earth resistivity  = 50-m
then the above equation becomes.
De  278.85 50  1971.8012 ft
(4.31)
The resistance of Carson’s equivalent earth return conductors is given by

re  1.588 103 f  
 mi 
(4.32)
re  0.09528 / mi
(4.33)
Therefore at 60Hz
The resistance of one conductor in ohms per mile is ra, which the information of can be
obtained from Appendix A.1.
ra  .1601  / mi
(4.34)
Ds is the geometric mean radius of a single conductor in feet, which is also obtained from
Appendix A.1.
Ds  0.0337 ft
(4.35)
1
Deq   DAB  DBC  DAC  3  3 2.167  5.167  7.333  4.3464 ft
(4.36)

 4.3464  
Z1  Z 2  0.1601  j 0.1213ln 

 0.0337  

(4.37)

  0.1601  j 0.5895    6mi   0.9606  j 3.5370 
 mi 
66

 1971.80123  
Z 0   0.1601  3  0.09528   j 0.1213ln 
 6mi
2 
 0.0337  4.3464  

(4.38)
 2.6756  j16.8935 
The impedance of the line for the zero sequence is:
Z 0, line  2.6756  j16.8935 
(4.39)
The impedance of the line for the positive and negative sequence is:
Z1,line  Z 2,line  0.9606  j 3.5370 
(4.40)
For the substation configuration referrer to Figure 3.1.
The impedance for zero, positive and negative sequences of the transformer as obtained
from SMUD are:
X 0, trf  0.1 pu
(4.41)
X 1, trf  X 2, trf  0.12 pu
(4.42)
Since the reactance of the transformer T, is referred to 12 (kV) voltage level, it has the
value of
Z 0, trf  X 0  x
V2
120002
 0.1
 0.72 
ST
20 106
Z1, trf  Z 2, trf  X1  X 2  x
V2
120002
 0.12 
 0.8640 
ST
20 106
(4.43)
(4.44)
The impedance for the zero and positive/negative sequences of the network at the
primary side from SMUD are:
Z 0, p  0.0902  j1.7761 
(4.45)
67
Z1, p  Z 2, p  0.3524  j3.7035 
(4.46)
The impedance for the zero and positive and negative sequences of the network at the
secondary side are described below. Because the Delta Wye grounded connection, the
zero sequence will equal to zero in the secondary side.
Z0  0 
Z1, s (network )  Z 2, s  Z1or 2
(4.47)
1
nT21
2
 12 
  0.3524  j3.7035     0.0107  j 0.1120 
 69 
(4.48)
The total impedance of zero, positive, and negative sequences seen at the end of the
transmission line is:
Z 0,total  Z 0,trf  Z 0,line
(4.49)
  0  j 0.72    2.6756  j16.8935  2.6756  17.6135 
Z1  Z 2  Z1, s  Z1, trf  Z1, line
  0.0107  j 0.1120    0  j 0.8640    0.9606  j 3.5370 
Z1, total  Z 2, total  0.9713  j 4.5130 
(4.50)
68
4.6 FAULT AT THE END OF THE LINE FOR OVERCURRENT RELAY
4.6.1 THREE PHASE FAULT CALCULATION
Since only the positive sequence network is involved:
I F 1max 
I F 1max 
V
12000

3   Z1 
3   Z1, s  Z1, trf  Z1, line 
3
12000
 0.0107  j0.1120    j0.8640    0.9606  j3.5370 

(4.51)
 315.7755  j1467.2037  1500.8  77.85 A
4.6.2 LINE TO LINE FAULT
By assuming the fault happened between phase (b) and phase (c) therefore;
I af  0
I bf   I cf
(4.52)
and the sequence currents can be found as follows
Ia0  0
I a1   I a 2 
V
Z1  Z 2
(4.53)
69
I a1 

V
12000

3   Z1  Z 2 
3  Z 1,s  Z 2,s  Z1,trf  Z 2,trf  Z1,line  Z 2,line 
12000
(4.54)
3   2   0.0107  j 0.1120   2   0  j0.8640   2  .9606  j3.5370  
 157.8878  j 733.6019  750.4  77.85 A
I F  min 


3  90  I a1
(4.55)
I F  min  3   750.4  77.85 1  90   1300  167.85 A
4.7 END OF THE LINE FAULT FOR OVERHEAD GROUND FAULT RELAY
4.7.1 SINGLE LINE TO GROUND FAULT
I a 0  I a1  I a 2 
V
3   Z 0  Z1  Z 2 
I af  I a 0  I a1  I a 2  3I a 0  3I a1  3I a 2
(4.56)
(4.57)
70
I a1 

V
3   Z 0  Z1  Z 2 
3  Z 0,line  Z 0, trf
12000
 Z1,s  Z 2,s  Z1,trf  Z 2,trf  Z1,line  Z 2,line 
I a1 
12000
3  2  Z 1,s 2  Z1,trf  2  Z1,line  Z 0,line  Z 0,trf 
I a1 
12000
3   0.0214  j 0.224    j1.728  1.9212  j 7.0740   Z 0,line  Z 0,trf 
I a1 
12000
3  1.9426  j 9.026    2.6756  j16.8935   j 0.72  

12000
3   4.6182  j 25.9195
 46.1602  j 259.0724  263.1525  79.90 A
I fault  3I a1  789.4575  79.90 A
(4.58)
(4.59)
71
4.7.2 DOUBLE LINE TO GROUND FAULT
The sequence current can be found as follows:
I a1 
I a1 
I a1 
I a1 
12000


Z2  
3  Z 1   Z0 

Z0  Z2  


12000



Z 2(s trf line )
3   Z1,s  Z1,trf  Z1,line    Z 0( trf line ) 



Z 0( trf line )  Z 2( s trf line )  


12000
3[(.9713  j 4.5130)  (2.6756  j17.6135) 
(.9713  j 4.5130)
]
(3.6469  j 22.1265)
(4.60)
12000
3  1.6977  j8.1078
 171.4117  j818.6200  836.3735  78.17 ( A)
 Z0 
Ia2   
  I a1
Z

Z
2
 0


2.6756  j17.6135
Ia2   
   836.3735  78.17  (4.61)
  2.6756  j17.6135   0.9713  j 4.5130 
 144.4014  j 648.5762  664.4568102.55 A
72
 Z2 
Ia0   
  I a1
 Z0  Z2 


0.9713  j 4.5130
Ia0   
   836.37  78.17
  2.6756  j17.6135   0.9713  j 4.5130  

(4.62)
  27.0623  j170.0329  172.173099 A
Ibf  I a 0  a2 I a1  aI a 2
(4.63)
Icf  I a 0  aI a1  a2 I a 2
(4.64)
a  1120 & a 2  1240  1  120
(4.65)
Where
The total fault current flowing to the neutral is:
I n  I bf  I cf  3I a 0
I n  3  172.173099  516.519199
(4.66)
4.8 FAULT AT THE 12KV BUS FOR THE OVERCURRENT RELAY
4.8.1 THREE PHASE FAULT CALCULATION
I F 1max 

V
12000

3   Z1   Z1,s  Z1,trf 
12000
3  .0107  j 0.1120    j 0.8640  
 78.0465  j 7097.7225  7098.1423  89.37 A
(4.67)
73
4.8.2 LINE TO LINE FAULT
By assuming the fault happened between phase (b) and phase (c), therefore
I af  0
(4.68)
I bf   I cf
(4.69)
and the sequence currents can be found as follows:
Ia0  0
I a1   I a 2 
V

3   Z1  Z 2 
I a1 

(4.70)
V
Z1  Z 2
12000
3  Z 1,s  Z 2,s  Z1,trf  Z 2,trf 
12000
3   2   0.0107  j 0.1120   2   0  j 0.8640 
(4.71)

 38.9865  3548.8579  3549.072  89.37 A
I F min  3  I a1  90  3   3549.072  89.37   1  90 
(4.72)
 6147  179.37 A
4.9 FAULT AT THE 12KV BUS OVERHEAD GROUND OVERCURRENT RELAY
4.9.1
SINGLE LINE TO GROUND FAULT
I a 0  I a1  I a 2 
V
3   Z 0  Z1  Z 2 
(4.73)
74
I af  I a 0  I a1  I a 2  3I a 0  3I a1  3I a 2
I a1 
V
12000

3   Z 0  Z1  Z 2 
3  Z 1,s  Z 2,s  Z1,trf  Z 2,trf  Z 0,trf 
I a1 
12000
3  2  Z 1,s 2  Z1,trf  Z 0,trf 
(4.74)
(4.75)
I a1 
12000
3  2   0.0107  j 0.1120   2   j 0.8640    j 0.72  
 2592.807  89.54 A
I fault  3I a1  7778.422  89.54 A
(4.76)
4.9.2 DOUBLE LINE TO GROUND FAULT
By assuming the fault happened between phase (b) and phase (c) therefore
I af  0  I a 0  I a1  I a 2
(4.77)
I bf   I cf  3  I a1  90
(4.78)
The sequence current can be found as follows:
75
I a1 
12000


Z2  
3  Z 1   Z0 

Z 0  Z 2  


I a1 
12000



Z 2( s trf )
3   Z1,s  Z1,trf    Z 0,trf 
 


Z

Z
0,
trf
2(
s

trf
)



I a1 
(4.79)
12000



0.0107  j 0.976
3   0.0107  j 0.976     j 0.72  


 j 0.72    0.0107  j0.976   


 190.668  j 4974.8335  4978.486  89.48 A
 Z0 
Ia2   
  I a1
 Z0  Z2 


0  j 0.72
Ia2   
   4978.486  89.48
  0  j 0.72    0.0107  j 0.976  
 32.5135  j 2113.2168  2113.466198.88 A

(4.80)
76
 Z2 
Ia0   
  I a1
 Z0  Z2 


0.0107  j 0.976
Ia0   
   4978.486  89.48
0

j
0.72

0.0107

j
0.976







(4.81)
 23.3102  j 2864.9112  286590.47 A
Ibf  I a 0  a2 I a1  aI a 2
(4.82)
Icf  I a 0  aI a1  a2 I a 2
(4.83)
a  1120 & a 2  1240  1  120
(4.84)
where
The total fault current flowing to the neutral is:
I n  I bf  I cf  3I a 0
I n  3   286590.47  859590.47
(4.85)
4.10 FAULT CURRENTS FOR THE UNDERGROUND LINE
Recalling from section 9.5, the impedance for the zero positive and negative sequences of
the network at the primary side from SMUD are:
Z 0, p  0.0902  j1.7761 
(4.86)
Z1, p  Z 2, p  0.3524  j3.7035 
(4.87)
77
4.10.1 IMPEDANCE OF A SIX MILE LONG UNDERGROUND TRANSPOSED LINE
The construction material for the underground system is copper and the conductors are
equidistant from each other at 2.10 inches or 0.1475 feet apart. The conductor size for
the underground system is approximated to be 600 kcmils as calculated in section 4.4,
operating at a temperature of 50C at 60 cycles. For the sequence impedance the same
formulas as before are used to resolve.

 De3
Z 0   ra  3re   j 0.1213ln 
 D  D2

eq
 s

 Deq
Z1  Z 2   ra  j 0.1213ln 
 Ds


  L
 
 
 L
 
(4.88)
(4.89)
Where L is the length of the line which equals six miles De is the equivalent depth of
earth return and ra is the resistance of a single conductor.
1
  2
De  2160   ft
f 
(4.90)
If the frequency is substituted as (60Hz) and assume that the earth resistivity  = 50-m
then the above equation becomes.
De  278.85 50  1971.8012 ft
(4.91)
The resistance of Carson’s equivalent earth return conductors at 60Hz is given as

re  1.588 103 f  
 mi 

re  0.09528  
 mi 
(4.92)
78
The resistance of a single conductor in ohms per mile is ra which is obtained from
appendix A.2 and it equals to: ra = 0.1095 ohms per mile. The geometric mean radius or
Ds at 60 cycles is 0.0285 feet and it is also displayed at the same appendix. The next step
is to calculate the equivalent distance Deq using the same equations used in section 4.5.
The zero, positive and negative sequence impedances are then calculated.
1
Deq   DAB  DBC  DAC  3  3 .1475  .1475  .1475  0.1475 ft
(4.93)
The sequence impedances are

 1971.80123  
Z 0   0.1095  3  0.09528   j 0.1213ln 
6
2 
 0.0285  0.1475  

Z 0  2.3720  j 21.9402 

 0.1475  
Z1  Z 2  0.1095  j 0.1213ln 
   6   0.6570  j1.1964  
 0.0285  

(4.94)
(4.95)
The impedance of the line for the zero sequence is:
Z 0, line  2.3720  j 21.9402 
(4.96)
The impedance of the line for the positive and negative sequence is:
Z1,line  Z 2,line  0.6570  j1.1964 
(4.97)
The impedance for zero, positive and negative sequences of the transformer as obtained
from SMUD are:
X 0, trf  0.1 pu
(4.98)
X 1, trf  X 2, trf  0.12 pu
(4.99)
79
Since the reactance of the transformer T, is referred to 12 (kV) voltage level, it has the
values of:
Z 0, trf
V2
120002
 X0  x
 0.1
 0.72 
ST
20 106
Z1, trf  Z 2, trf  X1  X 2  x
V2
120002
 0.12 
 0.8640 
ST
20 106
(4.100)
(4.101)
The impedance for the zero, positive, and negative sequences of the network at the
primary side from SMUD are:
Z 0, p  0.0902  j1.7761 
(4.102)
Z1, p  Z 2, p  0.3524  j3.7035 
(4.103)
The impedance for the zero and positive and negative sequences of the network at the
secondary side are described below. Because the Delta Wye grounded connection, the
zero sequence will equal to zero in the secondary side.
Z0  0 
Z1, s ( network )  Z 2, s  Z1or 2
(4.104)
1
ntrf2
(4.105)
2
 12 
  0.3524  j 3.7035    0.0107  j 0.1120 
 69 
The total impedance of zero, positive, and negative sequences seen at the end of the
transmission line is:
80
Z 0, total  Z 0, trf  Z 0, line
  0  j 0.72    2.3720  j 21.9402   2.3720  j 22.6602 
(4.106)
Z1  Z 2  Z1, s  Z1, trf  Z1, line
  0.0107  j 0.1120    0  j 0.8640    0.6570  j1.1964 
(4.107)
Z1, total  Z 2, total  0.6677  j 2.1724 
4.11 FAULT AT THE END OF THE LINE FOR OVERCURRENT RELAY
4.11.1 THREE PHASE FAULT CALCULATION
Since only the positive sequence network is involved:
I F 1max 
V
12000

3   Z1 
3   Z1, s  Z1, trf  Z1, line 
I F 1max 
12000
3  0.6677  j 2.1724 
 895.6111  j 2913.9217  3048.4518  72.90 A
4.11.2 LINE TO LINE FAULT
By assuming the fault happened between phase (b) and phase (c), therefore;
Iaf = 0 and Ibf = -Icf
Sequence currents are described as
(4.108)
81
Ia0  0
I a1   I a 2 

(4.109)
V
V

Z1  Z 2
3   Z1  Z 2 
12000
3  Z 1,s  Z 2,s  Z1,trf  Z 2,trf  Z1,line  Z 2,line 
(4.110)

12000
3  [2   0.6677  j 2.1724 ]
 447.8055  j1456.9609  1524.2260  72.90 A
I F  min 


3  90  I a1
(4.111)
I F  min  3  1524.2260  72.90 1  90   2640.0368  162.90 A
4.12 FAULT AT THE END OF THE LINE FOR UNDERGROUND RELAY
4.12.1 SINGLE LINE TO GROUND FAULT
I a 0  I a1  I a 2 
V
3   Z 0  Z1  Z 2 
I af  I a 0  I a1  I a 2  3I a 0  3I a1  3I a 2
(4.112)
(4.113)
82
I a1 

I a1 
V
3   Z 0  Z1  Z 2 
3  Z 0,line  Z 0, trf
12000
 Z1,s  Z 2,s  Z1,trf  Z 2,trf  Z1,line  Z 2,line 
12000
3  2  Z 1,s 2  Z1,trf  2  Z1,line  Z 0,line  Z 0,trf 

12000
3   0.0214  j 0.224    j1.728  1.3140  j 2.3928  Z 0,line  Z 0,trf 

12000
3  1.3354  j 4.3448   2.3720  j 21.9402    j 0.72  
I a1 
(4.114)
(4.115)
12000
3   3.7074  j 27.0050 
 34.5695  j 251.8067  254.1686  82.18 A
I fault  3I a1  762.5057  82.18 A
4.12.2 DOUBLE LINE TO GROUND FAULT
(4.116)
83
The sequence currents are calculated as follows:
12000


Z2  
3  Z 1   Z0 

Z0  Z2  


I a1 
12000
I a1 





Z 2(s trf line )
3   Z1,s  Z1,trf  Z1,line    Z 0( trf line ) 
 


Z

Z
0(
trf

line
)
2(
s

trf

line
)



12000



 0.6677  j 2.1724 
3  .6677  j 2.1724    Z 0( trf line ) 
 

2.372

j
22.6602

.6677

j
2.1724







12000
3 .6677  j 2.1724     2.3720  j 22.6602    0.089433  j 0.015941 

I a1 
12000
3  1.2411  j 4.1610 

(4.117)
 456.0103  j1528.9655  1595.5190  73.40 A
 Z0 
Ia2   
  I a1
Z

Z
0
2




2.3720  j 22.6602
(4.118)
 
  1595.519  73.40 
  2.3720  j 22.6602   .6677  j 2.1724  
 511.3131  j1360.1145  1453.0494107.60 A
84
 Z2 
Ia0   
  I a1
 Z0  Z2 


0.6677  j 2.1724
 
  1595.519  73.40
2.3720

j
22.6602

0.6677

j
2.1724







(4.119)
  16.4077  j144.0093  144.941096.50 A
Ibf  I a 0  a2 I a1  aI a 2
(4.120)
Icf  I a 0  aI a1  a2 I a 2
(4.121)
a  1120 & a 2  1240  1  120
(4.122)
Where
The total fault current flowing to the neutral is:
I n  I bf  I cf  3I a 0
(4.123)
I n  3  144.941096.50   434.823096.50
4.13 FAULT AT THE 12KV BUS FOR THE UNDERGROUND LINE
4.13.1 THREE PHASE FAULT CALCULATION
I F 1max 

V
12000

3   Z1   Z1,s  Z1,trf 
3 
12000
.0107  j 0.1120    j0.8640 

 78.0465  j 7097.7225  7098.1423  89.37 A
(4.124)
85
4.13.2 LINE TO LINE FAULT
By assuming the fault happened between phase (b) and phase (c), therefore
I af  0
(4.125)
I bf   I cf
(4.126)
Ia0  0
(4.127)
I a1   I a 2 
I a1 

V
Z1  Z 2
V
12000

3   Z1  Z 2 
3  Z 1, s  Z 2, s  Z1,trf  Z 2,trf
(4.128)

12000
3   2   0.0107  j 0.1120   2   0  j 0.8640 

(4.129)
 38.9865  3548.8579  3549.072  89.37 A
I F  min  3  I a1  90
 3   3549.072  89.37   1  90 
(4.130)
 6147  179.37 A
4.14 FAULT AT THE 12KV BUS FOR THE UNDERGROUND RELAY
4.14.1 SINGLE LINE TO GROUND FAULT
I a 0  I a1  I a 2 
V
3   Z 0  Z1  Z 2 
I af  I a 0  I a1  I a 2  3I a 0  3I a1  3I a 2
(4.131)
(4.132)
86
I a1 
V
12000

3   Z 0  Z1  Z 2 
3  Z 1,s  Z 2,s  Z1,trf  Z 2,trf  Z 0,trf 
I a1 
12000
3  2  Z 1,s 2  Z1,trf  Z 0,trf 
(4.133)
I a1 
12000
3  2   0.0107  j 0.1120   2   j 0.8640    j 0.72  
I a1  2592.807  89.54 A
I fault  3I a1  7778.422  89.54 A
(4.134)
87
4.14.2 DOUBLE LINE TO GROUND FAULT
By assuming the fault happened between phase (b) and phase (c), therefore
I af  0  I a 0  I a1  I a 2
(4.135)
I bf   I cf  3  I a1  90
(4.136)
The sequence current can be found as follows:
I a1 
12000


Z2  
3  Z 1   Z0 

Z0  Z2  


I a1 
12000



Z 2( s trf )
3   Z1,s  Z1,trf    Z 0,trf 



Z 0,trf  Z 2( s trf )  


(4.137)
I a1 
12000
  0.0107  j 0.976  



3

0.0107  j 0.976
   j 0.72    j 0.72    0.0107  j 0.976   


 190.668  j 4974.8335  4978.486  89.48 A
 Z0 
Ia2   
  I a1
Z

Z
2
 0


0  j 0.72
Ia2   
   4978.486  89.48
  0  j 0.72    0.0107  j 0.976  
 32.5135  j 2113.2168  2113.466198.88 A

(4.138)
88
 Z2 
Ia0   
  I a1
 Z0  Z2 


0.0107  j 0.976
Ia0   
   4978.486  89.48
  0  j 0.72    0.0107  j 0.976  

(4.139)
 23.3102  j 2864.9112  286590.47 A
Ibf  I a 0  a2 I a1  aI a 2
(4.140)
Icf  I a 0  aI a1  a2 I a 2
(4.141)
a  1120 & a 2  1240  1  120
(4.142)
where
The total fault current flowing to the neutral is:
I n  I bf  I cf  3I a 0
(4.143)
I n  3   286590.47  859590.47
(4.144)
89
4.15 SUMMARY OF DISTRIBUTION LINE FAULTS
4.15.1 OVERHEAD DISTRIBUTION LINE FAULT SUMMARY
Fault Type
End of the Line Fault (A) Fault at the 12kV Bus (A)
Single Line to Ground
789.4575
7778.422
Double Line to Ground 516.5191
8595
Line to Line
1300
6147
Three Phase
15000
7098.1423
Table 4.1 Overhead distribution line fault summary.
4.15.2 UNDERGROUND DISTRIBUTION LINE FAULT SUMMARY
Fault Type
End of the Line Fault (A) Fault at the 12kV Bus (A)
Single Line to Ground
762.5057
7778.422
Double Line to Ground 434.8230
8595
Line to Line
2640.0368
6147
Three Phase
3048.4518
7098.1423
Table 4.2 Underground distribution line fault summary.
90
4.16 OVERHEAD AND UNDERGROUND DISTRIBUTION LINE RELAY
SETTINGS
4.16.1 OVERHEAD LINE PROTECTION
The 20 MVA system as described in figure 4.1 can yield the rated current flowing in the
12kV side is described in the calculation below. The IEEE standard is that the rated
current can be either increased or decreased between 20%.

In  




  962.25 A
3 *12 kV 

20MVA

(4.145)
By taking the standard into consideration, the value for the rated current for this task is
1000 amps. By considering the secondary connecting leads been less than 100 meters
long, then the secondary rated current is In = 5 A. Thus the turn ration for the current
transformer CT protecting the line is described as follows.
n CT 
I1n 1000 A

 200
I 2n
5A
(4.146)
The relays in use are over current relays which are current dependent devises that are
equipped with instantaneous parts. The minimum fault currents are chosen as the
minimum pick up currents for the relays. Referring to the overhead distribution fault
summary from section 4.15.1 the double line to ground at the end of the line for both
overhead and underground are the minimum fault currents. In order to guarantee core
reliable operation for relay protection a 25% of the minimum pick up current is taken into
consideration. Then by taking the 25% the minimum pick up current for the overhead
91
system is 384 amps and for the underground system is 323 amps. The minimum pick up
current values are referred to the secondary side and the calculations are as shown.
1 
 5 
I P (OH )  I mp     384  
 1.92 A
1000 
 nct 
(4.147)
1 
 5 
I P (UG )  I mp     323  
 1.615 A
1000 
 nct 
(4.148)
For the instantaneous organs the three phase fault currents at the end of the line are used
as the pickup currents. As show in the table above the three phase fault currents are 1500
A and 3048.4518 A, for overhead and underground respectively. Then the calculations
for the secondary side of the current transformer for the pickup current of the
instantaneous organs for both systems are shown.
1
 5 
I P (OH )  I mp     1500  
 7.50 A
1000 
 nct 
(4.149)
1
 5 
I P (UG )  I mp     3048.4518  
 15.2422 A
1000 
 nct 
(4.150)
4.16.2 GROUND FAULT PROTECTION
The rated current flowing in the secondary side is:
 20 MVA 
In  
  962.25 A
 3 12 kV 
(4.151)
And therefore the maximum asymmetric possible load means that all the impedance on
the line concentrates on one phase only.
92
In 
962.25
 320.75 A
3
(4.152)
The value obtained above can be either increased or decreased by a number between
20%, which in this case is an even value of 300 amps was chosen. The secondary leads
are less than 100 meters long. Since the secondary rated current is 5 amps, then the turn
ratio for the ground current transformer protecting the line is shown below.
nct 
I1n 300 A

 60
I 2n
5A
(4.153)
Any current above 120 amps will cause interference with phone lines, and thus the
minimum pickup current is set at 120 amps. The minimum pickup current for the ground
over current relay at the secondary side for both systems is calculated as:
 1 
 5 
I P  I mp  
  120  
2A
 300 
 nct 
(4.154)
4.16.3 POWER TRANSFORMER PROTECTION
The protection of the power transformer against faults at the 12 kV bus and within the
power transformer a differential relay is utilized. The power transformer connection is of
a delta-wye grounded type, thus the compensation of a 30 degree phase shift between the
primary and secondary windings is enforced. In order to omit the phase shift on the
secondary side the connection between the differential relay and the current transformer
the connection is wye on the delta side of the power transformer. Then it is connected in
delta configuration on the wye connected section of the power transformer. In order to
93
determine the settings for the differential relay the turn ration of the current transformer is
used. The normal rated current on the high voltage side (69 kV) is calculated as:
 20 106VA 
I n ( HV )  
  167.35 A
3
 3  (69 10 ) V 
(4.155)
The 20% IEEE standard can be applied to the rated current calculated above, and thus an
even value of In = 150 amps is chosen. As mention before the secondary leads are less
than 100 meter long and that yields a secondary rated current of In = 5 amps, then the turn
ratio computation is depicted below.
I CT 
I p ,n
I s ,n

150 A
 30
5A
(4.156)
Thus, the normal rated current flowing into the secondary side of the current transformer
is the normal rated current on the high voltage side multiplied by the ration of the rated
currents.
I normal  167.35 
5A
 5.58 A
150 A
(4.157)
The next step is to calculate the normal rated current for the low voltage side (12 kV).
This is done in the same manner as calculations above for the high voltage side.
 20 106VA 
I n ( HV )  
  962.25 A
3
3

(12

10
)
V


(4.158)
Once aging by making use of the IEEE standard the chosen value for the current is 950
amps, thus In = 150 amps. The secondary rated current once again is 5 amps due to the
94
length of the connecting leads. With this values take into consideration the turn ratio is
described below.
I CT 
I p ,n
I s ,n

950 A
 190
5A
(4.159)
The normal rated current flowing in the secondary of the current transformer is,
I normal  I CT 
1
I CT
 962.25 
1
 5.06 A
190
(4.160)
As a result the normal rated current flowing through the differential relay, which is just
the difference between the two rated currents in the secondary side of the current
transformer.
I normal  I normal  5.58  5.06  0.52 A
(4.161)
Now the fault currents are as follows,
I fault 
512.55 A
 0.53
962.25 A
(4.162)
It is 0.53 times greater that the rated current on the secondary side of the power
transformer, therefore the minimum pickup current setting for the differential relay is the
value shown.
I min( P )  0.53  0.52  0.27 A
(4.163)
By using the three phase fault current obtained in section 4.15.1 which equals to
7098.1423 amps will help determine the pickup current settings for the instantaneous
relay.
95
I fault 
7098.1423
 7.38
962.25
(4.164)
Thus the rated current on the secondary side of the power transformer is 7.38 times
greater. Therefore the pickup current setting for the instantaneous relay is shown below.
I max( P )  7.38  0.52  3.838 A
(4.165)
4.17 PROTECTION SCHEME FOR THE ENTIRE SYSTEM
4.17.1 PROTECTION RELAYS, LINE AND GROUND RELAYS
The primary task is to protect the entire system, both the overhead and underground lines.
In order to protect the system the relays are placed in a couple of strategic locations. The
first strategic location is right after the circuit breaker. The second location is at the
neutral wire of the secondary side of the delta-wye connected transformer to protect
against ground faults. The connection configurations for the relays are depicted in
Figure 4.1
a
CB
b
CB
Legend
:ground point
:Relay
c
CB
:current transformer
RL: the relay protecting line
faults
RL
RL
RL
Figure 4.1 Transmission line protection relays.
RG
RG: the relay protecting ground
faults
96
The relays are made up of two internal relays connected in series, where one is the
instantaneous and the other the time delay. The instantaneous part is used to pick up the
maximum three phase fault current without a time delay. In the other hand the time delay
system is used to pick up the minimum fault currents with some time delay.
R1
Instantaneous
Relay
Figure 4.2 The two internal relays inside each relay.
Time-delayed
Relay
97
4.17.2 DIFFERENTIAL RELAY INSTALLATION
Y
Y
C
Circuit Breaker
B
a
Circuit Breaker
A
Legend
R : the restrain coil
Op : the operating windings,
which consist of 2 internal
winings: one instantaneous
winding and one time delay
winding.
R
R
Op
R
R
Op
R
R
Op
Figure 4.3 Differential relay installation.
b
c
98
4.17.3 RELAY CHARACTERISTICS SUMMARY
Relay location
Minimum pick up
Max pick up for Instantaneous
current
apparatus
A
A
Underground Lines
1.615
15.2422
Ground Fault Relays
2
2
Overhead Lines
1.92
7.50
Both sides of power
0.27
3.838
Transformer for
differential
Relay
Table 4.3 Relay characteristics.
99
4.17.4 DIAGRAM FOR THE CONTROL CIRCUIT
The circuit shown below represents the controls for opening and closing the circuit
breaker according to any specific time interval.
+
Fuse
RL
RL
B
RL
RG
RD
B
H
T1
H
T1(3s)
H
T1
C
O
M
T2
T2
T1
T3
M
Figure 4.4 Control circuit for the breaker.
In the above figure T1, T2, T3 are the timers and are assumed to have internal relays
which are activated instantaneously when the timers close their contacts after a time
delay.
100
From the diagram above Rl represents the contacts for the line relay, and Rg represents
the contacts for the ground relay which protects against ground fault currents in the
neutral wire. The last one Rd represents the contacts for the differential relay, which
protects the power transformer.
+
M
N
B
Legend
:normally open contact
N
:normally closed contact
T2(18s)
N
A
L
T2
T3
LED
Reset
L
O
M
L
T3(60s)
L
B
O
O
LED
Figure 4.5 Control circuit continued.
:Relay A
:light indicator
101
4.17.5 OVERALL PROTECTION SCHEME
The function of the controller shown in the previews circuit is to control the circuit
breaker when a fault occurs by reclosing. The breaker opens and closes at 3 seconds after
the first fault, if the fault persist the breaker will reopen. The period of time for the
second reclosing is 18 seconds .If the faults still persist the breaker will reopen once more
and remain open. The breaker remains in open mode until personnel remove the faulted
conditions. The breaker is in faulted conditions the current flows in the secondary side of
the transformer, activating relays R1 or Rg. When the relays are activated the contacts
close and energize relay B in the controller circuit. It is important to note that the
controller is fed by an independent DC source.
102
Chapter 5
CONCLUSION
It is important for the reader to understand that the procedure followed in this project is
not the only method in power systems field to perform a study similar to the one
presented. For engineers throughout the utility companies it is important to know how to
size a conductor for either overhead or underground. There is a more important and
indispensable practice in distribution systems, which is to know how to set up relay
protection.
In this study the first calculations done were to find the impedance in the secondary side
with the impedance from the primary side. With the calculations obtained the transient
and subtransient fault currents where acquired, which then supported the calculations for
the heat impulse. The sizing of the conductor was conducted with the values from the
heat impulse. The simplified version of the equation was used to find the size of the
conductor by using the characteristics of copper and aluminum. The skin effect which is
basically the current that flows around the conductor is used in the calculation, and since
there is no current flowing in the center of the conductor the value used for the skin effect
was one. The materials used for the conductors were aluminum and copper for the
overhead and underground conductors respectively. The size for the overhead was found
to be 666 kcmils and for the underground is 600 kcmils.
103
The next procedure performed in the study was to calculate the fault currents for the
transient and subtransient stage for the overhead system. With the spacing between the
conductors specified and the information from the table in appendix A.1 the sequence
impedance of the line could be found. Then with the system information provided by the
utility company such as the sequence impedance for the transformer, and the sequence
impedance for the primary side of the network the networks secondary side impedance
was obtained.
The sequence impedance of the line and secondary side were implemented into acquiring
the fault calculations. The fault calculation are needed to set up the relays, and thus
calculations for three phase, line to line, single line to ground, double line to ground were
conducted. The fault calculations were done for the overhead and underground systems
at the end of the line and at the 12 kV bus. The results were then tabulated into a table
summarizing the results showing were the largest fault currents are found for this
particular study with the given set up. The three phase fault for the overhead and
underground systems was the largest for the fault calculations at the end of the line. The
calculations at the 12 kV bus resulted as the double line to ground fault having the largest
amperage. With the faulted currents attained the settings of the relays were done for line
protection, ground protection and transformer protection and the characteristics of the
relays set up was tabulated for better understanding of the system. Then a simplified
controller for the systems was introduces to make the entire mechanism operates as
intended.
104
APPENDIX A
CONDUCTOR CHARACTERISTICS
A.1 ACSR CHARACTERISTICS
Aluminum
Circular
Mills
Strands
Layers
666600
54
3
Steel
Strand
Strands Diameter
[in]
7
0.1111
Strand
Diameter
[in]
0.1111
Frequency
[Hz]
Outside
Diameter
[in]
60
1.000
Geometric
Mean
Radius Ds
[ft]
Resistance
r
[Ω/mi]
at 50C
Inductive
Reactance xa
at 1ft spacings
[Ω/cond./mi]
Shunt Capacitive
Reactance 𝑥𝑎′
at 1ft spacings
[MΩmi/cond.]
0.0337
0.1601
0.412
0.0943
A.2 COPPER CHARACTERISTICS
Aluminum
Circular
Mills
Strands
Layers
600000
37
2
Strand
Diameter
[in]
0.1273
Frequency
[Hz]
Outside
Diameter
[in]
Weight
[lb/mi]
60
0.891
9781
Geometric
Mean
Radius Ds
[ft]
Resistance
r
[Ω/mi]
at 50C
Inductive
Reactance xa
at 1ft spacings
[Ω/cond./mi]
Shunt Capacitive
Reactance 𝑥𝑎′
at 1ft spacings
[MΩmi/cond.]
0.0285
0.1095
0.432
0.0977
Weight
[lb/mi]
4527
105
APPENDIX B
MATLAB CODE
B.1 HEAT IMPULSE CALCULATIONS
%----GIVEN VALUES-----------------------------%-Primary Side Voltage-----------------V69=69*10^3;
% Secondary Side Voltage
V12=12*10^3;
% TRF MVA Rating
S=20*10^6;
% Assumed Fault Current
I69FS=10800;
% Assumed TRF Positive Sequence Reactance
X1=0.12;
%-Primary side transient fault current--I69FT=10780;
% Zero Sequence Reactance
X0=0.1;
%-Duration of the Fault in Seconds------t=0.7;
% w=2*pi*frequency
w=377;
% Percentage of the Subtransient Resistance
RTRF=0.1*0.864;
%----------------------------------------------------%----Subtransient Fault Current For the Secondary Side
%----------------------------------------------------%----Subtransient Reactance Primary Side-----% From I69=(1.1*V69)/(X69*(3)^(0.5))
X69FS=(1.1*V69)/(I69FS*(3)^(0.5))
%----Substransient Reactance Secondary Side--X12FS=X69FS*(V12/V69)^2
106
%----Substransient Reactance of TRF----------XTRFFS=X1*(V12^2)/S
%----Total Reactance On The Secondary Side--XT1=X12FS+XTRFFS
%----Subtransient Fault Current -------------I12FS=1.1*V12/(XT1*(3)^(0.5))
%--------------------%----Transient Current
%--------------------%----Transient Reactance Primary Side-------X69FT=1.2*V69/(I69FT*(3)^(0.5))
%----Transient Reactance Secondary Side-----X12FT=X69FT*(V12/V69)^(2)
%----Transient Reactance of TRF-------------XTRFFT=X0*V12^2/S
%----Transient Reactance Secondary Side-----XT2=X12FT+XTRFFT
%----Transient Current Secondary Side-------I12FT=1.2*V12/(XT2*(3)^(0.5))
%---------------%----Heat Impulse
%---------------%----Subtransient resistance-----------------R12=V12/(I12FS*(3)^(0.5))
%----Equivalent Resistance-------------------RE=RTRF+R12
%----Heat Impulse----------------------------TA=X12FS/(w*RE)
HA=TA*(I12FS)^(2)
107
HP=t*(I12FT)^(2)
% Sum of Heat Impedance of Subtransient/Transient
HB=HA+HP
%---RESULTS-----------X69FS =
4.0575
X12FS =
0.1227
XTRFFS =
0.8640
XT1 =
0.9867
I12FS =
7.7236e+003
X69FT =
4.4346
X12FT =
0.1341
XTRFFT =
0.7200
108
XT2 =
0.8541
I12FT =
9.7337e+003
R12 =
0.8970
RE =
0.9834
TA =
3.3101e-004
HA =
1.9746e+004
HP =
6.6322e+007
HB =
6.6342e+007
109
B.2 END OF THE LINE FAULT CALCULATIONS
%------------------------------------------------------%----Fault at the end of the line for over current Relay
%------------------------------------------------------%----GIVEN VALUES---------------------------------% Secondary Side Voltage
V12=12000;
% Zero Sequence Impedances
Z0TRF=0.72*j;
Z0LINE=2.67560+16.8935*j
% Positive Sequence Impedances
Z1S=0.0107+0.1120*j;
Z1TRF=0+0.864*j;
Z1LINE=0.9606+3.5370*j;
% Negative Sequence Impedances
Z2S=0.0107+0.1120*j;
Z2TRF=0+0.864*j;
Z2LINE=0.9606+3.5370*j;
%----3 phase fault---------------------------------IF1MAX=V12/((Z1S+Z1TRF+Z1LINE)*(3)^0.5)
%----Line to Line Fault----------------------------Ia1_LL=V12/((Z1S+Z1TRF+Z1LINE+Z2S+Z2TRF+Z2LINE)*(3)^0.5)
IFMIN=Ia1_LL*(-j*(3)^(0.5))
%----Single Line to Ground Fault-------------------Ia1_LG=V12/((Z1S+Z1TRF+Z1LINE+Z2S+Z2TRF+Z2LINE+Z0TRF+Z0LINE)*3^0.5
)
IFAULT_LG=3*Ia1_LG
%----Double Line to Ground Fault-------------------Z0=Z0TRF+Z0LINE
Z1=Z1S+Z1TRF+Z1LINE
Z2=Z2S+Z2TRF+Z2LINE
Ia1_DLG=V12/((Z1+(Z0*(Z2/(Z0+Z2))))*(3)^(0.5))
Ia2_DLG=-(Z0/(Z0+Z2))*Ia1_DLG
Ia0_DLG=-(Z2/(Z0+Z2))*Ia1_DLG
% Total Fault Current Through In
In=3*Ia0_DLG
110
%----RESULTS--------------Z0LINE =
2.6756 +16.8935i
IF1MAX =
3.1578e+002 -1.4672e+003i
Ia1_LL =
1.5789e+002 -7.3360e+002i
IFMIN =
-1.2706e+003 -2.7347e+002i
Ia1_LG =
4.3770e+001 -2.5248e+002i
IFAULT_LG =
1.3131e+002 -7.5745e+002i
Z0 =
2.6756 +17.6135i
Z1 =
0.9713 + 4.5130i
111
Z2 =
0.9713 + 4.5130i
Ia1_DLG =
1.7141e+002 -8.1862e+002i
Ia2_DLG =
-1.4436e+002 +6.4858e+002i
Ia0_DLG =
-2.7052e+001 +1.7003e+002i
In =
-8.1156e+001 +5.1010e+002i
112
BIBLIOGRAPHY
[1] Gonen, T. 2009.Electrical power transmission system engineering analysis and
design, 2nd ed. Florida: CRC Press.
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[3] Gonen, T. 1986.Electrical power distribution system engineering. New York:
McGraw-Hill.
[4] Sadaat, H. 1999.Power system analysis. New York: McGraw-Hill.
[5] Sadaat, H. 2002.Power system analysis, 2nded. New York: McGraw-Hill.
[6] Granger, J. J. and W. D. Stevenson.1994. Power system analysis. New York:
McGraw-Hill.
[7] SMUD. 1995. UG/OH Conductor sequence impedance study.
[8] Gonen, T. 1988. Modern power system analysis. New York: McGraw-Hill.
[9] James W. Nilsson, and Susan A. Riedel. 2004. Electric Circuits, 7thed. New Jersey:
Prentice Hall.
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