POWER TRANFORMER PROTECTION
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
Edgar Polanco
SPRING 2017
ii
© 2017
Edgar Polanco
ALL RIGHTS RESERVED
iii
POWER TRANSFORMER PROTECTION
A Project
by
Edgar Polanco
Approved by:
__________________________________, Committee Chair Tracy Toups, Ph.D.
__________________________________, Second Reader Preetham Kumar, Ph.D.
____________________________ Date
iv
Student: Edgar Polanco
I certify that this student has 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 ___________________ Preetham Kumar, Ph.D. Date
Department of Electrical and Electronic Engineering
v
Abstract
of
POWER TRANSFORMER PROTECTION
by
Edgar Polanco
The project focuses on the relay protection of two power transformers, a 30MVA and a
40MVA respectively. Specifically, the project focuses on the design and the application
of the 51, 50, 50G and 87 protection schemes for the two power transformers. The design
results were used to write a test procedure and a test plan using AcSELerator and Test
Universe Software for the four different relays.
Approved by:
__________________________________, Committee Chair Tracy Toups, Ph.D.
_______________________ Date
vi
ACKNOWLEDGEMENTS
I want to thank professor Tracy Toups for his support and guidance throughout this project.
vii
TABLE OF CONTENTS
Page
Acknowledgements ............................................................................................................ vi
List of Tables ..................................................................................................................... xi
List of Figures ................................................................................................................... xii
Chapter
1. INTRODUCTION .......................................................................................................... 1
2. BACKGROUND OF THE STUDY............................................................................... 4
2.1 High Voltge Transformer Faults ......................................................................... 4
2.2 Main Types of Transformer External Faults ..................................................... 5
2.3 Transformer External Short Circuits .................................................................. 6
2.4 Over Voltage and Frequency Reduction Effects .............................................. 7
2.5 Transformer Work Cycles in Adverse Conditions ........................................... 7
2.6 Main Types of Transfomer External Faults ...................................................... 8
2.7 Three-Phase Faults ............................................................................................... 9
2.8 Line-to-Line Faults ............................................................................................. 10
2.9 Double Line-to-Ground Faults .......................................................................... 10
2.10 Single Line-to-Ground Faults ........................................................................... 10
2.11 Protection ............................................................................................................. 11
2.12 Instantaneous and Time Overcurrent Protection Relays (50/51).................. 14
viii
2.13 Ground Fault Protection Relays (50N) ............................................................ 14
2.14 Differential Protection Relays (87) .................................................................. 15
2.15 Circuit Breakers .................................................................................................. 15
2.16 Current Transformer Requirements .................................................................. 17
3. CALCULATIONS AND DRAWINGS ....................................................................... 20
3.1 30 MVA Transformer Ratings and Loads ....................................................... 20
3.2 T1 Primary, Secondary and Ground Rated Currents ..................................... 21
3.3 Determining The Current Transformer (CT) Ratios ...................................... 22
3.4 CT Secondary Currents ...................................................................................... 23
3.5 Relaying CT Performance Calculations ......................................................... 24
3.6 CT Burden Voltage (VB) for Differential CT For T1 ................................... 26
3.7 T1 Relay Differential Tap Calculations ........................................................... 28
3.8 T1 Relay Differential Pickup (Dial) Calculations .......................................... 30
3.9 Calculation of Phase Angle Correction (WXCTC) ........................................ 31
3.10 Restraint Slope Settings (SLP) ......................................................................... 31
3.11 The 30 MVA Transformer Slopes in The SEL 787 Relay ............................ 33
3.12 40 MVA Transformer (T2) Calculations ......................................................... 37
3.13 T2 Primary, Secondary and Ground Rated Currents ..................................... 38
3.14 Determining the Current Transformer (CT) Ratios ........................................ 39
3.15 T2 CT Secondary Currents ............................................................................... 40
3.16 Relay and CT Performance Calculations ......................................................... 41
ix
3.17 CT Burden Voltage (VB) For Differential CTs For T2 ................................. 43
3.18 T2 Relay Differential Tap Calculations ........................................................... 44
3.19 T2 Relay Differential Pickup (Dial) Calculations .......................................... 45
3.20 Calculation of Phase Angle Correction (WXCTC) ........................................ 46
3.21 Restraint Slope Percentage Settings (SLP) ..................................................... 47
3.22 The 40 MVA Transformer Slopes in The SEL 787 Relay ............................ 47
3.23 Relay Settings ...................................................................................................... 49
3.24 Drawings .............................................................................................................. 51
4. TEST PROCEDURE AND PLAN .............................................................................. 56
4.1 Test Procedure ..................................................................................................... 56
4.2 Relay Settings ...................................................................................................... 56
4.3 Wiring Connection Procedure ........................................................................... 57
4.4 Wiring Diagram .................................................................................................. 58
4.5 Programming The SEL 787 Relay.................................................................... 59
4.6 Setting The 50/51 And 50 Reays in Test Universe ........................................ 68
4.7 Setting The 87 Element in Test Universe ........................................................ 74
4.8 Test Plan ............................................................................................................... 85
4.9 Programming and Configuring The SEL Relay ............................................. 87
4.10 Programming and Configuring The Omicron ................................................. 90
4.11 Manual “Inverse Time Over Current” Relay Test (51) ................................. 91
4.12 Testing the 51, 50 And 50G Elements ............................................................. 93
x
4.13 Manual “Instantaneous Over Current” Relay Test (50)................................. 93
4.14 Manual “Ground Fault” Relay Test (50G) ...................................................... 94
4.15 Programming The Differential Relay in Omicron.......................................... 95
4.16 Testing The 87 Element ..................................................................................... 97
4.17 Point-By-Point Test of the 87 Relay ................................................................ 97
4.18 Automated Multiple Point Test of the 87 Relay ............................................. 98
5. CONCLUSION ............................................................................................................ 99
6. References .................................................................................................................. 103
xi
LIST OF TABLES Tables Page
1. 30 MVA Transformer Data ………………..................…………...……………. 20
2. 30 MVA Transformer Relay Settings…….………………..……..……...………36
3. 30 MVA Transformer Relay and CT Data Settings …………..………………....37
4. 40 MVA Transformer Data…………………………..……………………...…...38
5. 40 MVA Transformer Relay Settings……….……………………….……….….50
6. 40 MVA Transformer Relay and CT Data Settings ……………...….……...…...51
7. 30 MVA Relay and CT Data Settings and Ratings…………..….…………….....57
8. 30 MVA Transformer Relays Pick-up Currents…..……....……..………………58
9. 51 Relay Test Form……………………………………….....…………..…….…93
10. 50 Relay Test Form...…………………….....…….....………………………...…95
11. 50G Relay Test Form……….......………...…..…………..…………………...…96
12. 87 Relay Test Form…….…………………………………………………..……99
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LIST OF FIGURES Figures Page
1. 30 MVA Transformer……………………………………………………………20
2. Differential Operating Slope Zones…………………………………………….32
3. Slopes 1 and 2 of 30MVA Transformer………………………………..……..…36
4. 40 MVA Transformer……………………….…………….……………..………38
5. Slopes 1 and 2 of 40MVA Transformer…...…..……………..………..………...50
6. Single Line Diagram….........………………………...………………….……….52
7. 30 MVA 30 MVA Transformer 51,50 and 50N CT Connections…………….....53
8. 30 MVA Transformer Differential CT Connections…………...………….....….54
9. 40 MVA Transformer 51, 50 and 50N CT Connections………......…………….55
10. 40 MVA Transformer Differential CT connections…………..............................56
11. SEL-787 and OMICRON CMC-356 Wiring Diagram……..................................59
12. AcSELerator Software Access Tab…………………………………….……..…60
13. AcSELerator Settings Access Tab…………………………………….……..…..60
14. Communication Parameters……………………….…………………………..…61
15. Setting Editor Selection………………………….………………………………62
16. AcSELerator Transformer Configuration……………………………….…….…63
17. Phase Overcurrent Relay Configuration…….……………………………..….…64
18. Phase Overcurrent Relay Configuration………………….…….……….…….…65
19. Ground Overcurrent Relay Configuration…..……….…….………………….…65
xiii
20. Differential Relay Configuration……………………………………………...…66
21. Trip and Logic Configuration………………………………………………...….67
22. Circuit Breaker Trip Contacts…………………………………………………....68
23. Test Universe Access Tab…...…………………………………………………...69
24. QuickCMC Access Module…………………………………………………...…69
25. Hardware Configuration Test Setup…………………………………………..…70
26. Hardware Configuration Access…………………………………………………71
27. Output Configuration Currents…………………………………………………..72
28. Testing the Overcurrent Elements…………………………………………….…73
29. Test Results Example…………………………………………………………….74
30. Universe Access Tab…………………………………………………………..…75
31. Differential Module Access Tab…………………………………………..…….75
32. Test Object Access Tab………………………………………………………….76
33. Differential Access Screen………………………………………………………77
34. Differential Protection Configuration……………………………………………78
35. Differential CT Configuration…………………………………………………...79
36. Protection Device Configuration……………………………………………..….80
37. Slope 1 Characteristic Definition…………………………………………..….…81
38. Slope 2 Characteristic Definition…………………………………………..….…82
39. Operating Characteristic Diagram……………………………………………….83
40. Non-Trip Test View…………………………………………………………….. 84
xiv
41. Trip Test View………………………………………………………………..….85
42. TCC Curve from [12]……………………………………………………………87
1
CHAPTER I
INTRODUCTION Power transformers are the most expensive and important equipment in substations. They
are also instrumental in output voltages at generating stations and vital in all cases for the
normal operation of the electric power system. They are considered very reliable equipment
and are used to step-up or step-down transmitted voltages. While in operation, the windings
and the core of the transformer are subjected to different forces during normal operation
such as [11]:
• vibration,
• expansion and contraction due to thermal cycling
• warming due to the magnetic flux
• forces due to fault currents
• and heating due to overloading or improper cooling
When a failure in a transformer occurs, the damage is usually severe. In some cases and
depending on the size, the transformer has to be transported for repairs, which can take a
considerable amount of time.
Transformers are continuously subjected to electrical and mechanical forces which cause
deterioration and can lead to failure of the same. Failures in power transformers can cause
both internal and external damages which will make the unit incapable to perform its
functions, both electrically and mechanically. The root causes that lead to failures in
transformers can be grouped as follows [11]:
2
• Windings Failures: Some of the reasons that lead to this type of failure are
deterioration of the insulation, manufacturing defects, overheating, mechanical stress and
vibration.
• Terminals and vacuum changer: Failures due to improper assembly, damage during
transportation, excessive vibration or improper design.
• Bushings failures: The causes for this type of failure include vandalism, pollution
and animals coming in contact with them.
• On Load Tap Changer (OLTC): These failures may be due to malfunction
mechanism, contact problem, and insulation contamination.
• Other faults: Other causes of failures include: insulation failures, current
transformer failures and fault.
To operate a transmission system with a transformer out of service is always very
complicated. Frequently, the impact of a transformer out of service is more serious than
that of a transmission line [9,10].
This project begins with defining some of the common fault conditions that affect power
transformers followed by a discussion of various types of relay devices used to protect
against transformer damage. Theoretical and mathematical calculations will further narrow
the scope of this project to determine the relay protection settings necessary for the
protection of a single power transformer bank. Differential, instantaneous, time over
current and transformer ground fault protection relays will be addressed in the scope of this
project. Rated currents will be used to determine the current transformers (CT) ratios for
the relays and the secondary CT burden voltage and voltage saturation will be calculated.
3
The taps and pickup currents will be computed using guidelines prescribed in the
Schweitzer SEL-787 instructions manual. Finally, the slope percentage will be calculated
to address the differential CT mismatch between the primary and secondary sides of the
power transformers [12].
4
CHARPTER II
BACKGROUND OF THE STUDY 2.1 HIGH VOLTGE TRANSFORMER FAULTS
Power transformers are subjected to internal and external faults, which if not detected and
corrected may developed into severe faults and lead to prolong outrages. Transformers in
continuous operation are exposed to all incidents of overload and short circuits that can
occur both on the high and low voltage sides of the transformer. A burned transformer may
be the result of one or more faults, which can range from thermal or electrical overload to
failures in the cooling system, or manufacturing windings defects [1-5].
External transformer faults are faults that physically occur outside the transformer. From
the perspective of the life of the transformer, these types of failures are as important as
those produced internally, and if not properly cleared the conditions that cause these faults
will result in a reduction of the life of the transformer, which can lead to continuous
undesired malfunctions and even destruction over a long period of time.
Internal transformer faults occur due to failures on the transformer terminals, the windings
or the insulation. Winding faults are the most difficult to detect, because in their early
stages, the faults involve only a few turns and are virtually impossible to detect. When the
fault spreads to a larger number of turns, it is possible to detect it from the reverse
component of the current. Statistically, most internal faults that occur within the windings
are ground faults or faults between winding turns. The severity of this type of fault depends
on the design and the type of grounding and connection to the neutral point [1-9].
5
The faults between phases within three phase transformer tanks are unlikely, but possible,
and impossible in single phase transformer banks. The main causes of faults between
phases are arcs in the bushing and equipment failures while performing tap changes under
load. There are certain types of internal faults called Incipient, but do not pose immediate
danger to the transformer. However, if undetected they can become faults or large
magnitude. The main faults within this group are faults in the core of the transformer, as a
result of insulation damage, and failures in the oil due to losses or blockages in the
circulation of the same. In both cases overheating will occur. [6-8].
2.2 MAIN TYPES OF TRANSFORMER EXTERNAL FAULTS
External Overload
Electrical overload is the main reason for premature aging of a transformer. From a thermal
point of view, the overload occurs when the condition of thermal equilibrium in the
machine reaches temperature such that it causes degradation of dielectric insulated
conductors or sheets in the magnetic core of the transformer. The overload conditions that
involve parameters of a different nature are:
• Level of electric charge
• Environmental conditions such as temperature, humidity and altitude above sea
level.
• Continuous and non-continuous operating conditions
From a transformer protection stand point, the overload condition usually does not require
taking the transformer out of service. This allows actions dedicated to reducing the
6
conditions that produce this overload to take place while maintaining continuity of service.
Among the measures aimed at reducing the level of overloading of the machine include
load shedding and improving cooling conditions of the transformer [1,5,7].
Defects in the Cooling System
Defects in the cooling system can cause the transformer to overheat. The cooling system
failures are due to poor maintenance and improper operation of the radiators fans resulting
in transformer temperature rise. Such failures are not common, but it is worth mentioning
that proper and scheduled maintenance would help insure proper functionality of the
cooling system [1].
2.3 TRANSFORMER EXTERNAL SHORT CIRCUITS
The external transformer short circuit conditions are more severe types of faults a
transformer may be subjected to. From an electrical perspective, the intensity of a
symmetrical fault in a network is limited only by the short circuit power of the network. If
a short circuit occurs in the primary or secondary voltage side of the transformer, the short
circuit current at that point is obtained as a combination of the short circuit currents of the
network and the short circuit current of the power transformer. When a short circuit occurs,
in addition to the thermal effect produced by the intensity of the high current in the
windings, a high electro-dynamic force may develop between conductors and can cause
mechanical damages to the transformer [1,7,11,13].
7
2.4 OVER VOLTAGE AND FREQUENCY REDUCTION EFFECTS
When an AC voltage is applied to the magnetic circuit of a transformer, a flow of current
is produced and its value is proportional to the ratio of the voltage values and its frequency.
However, when this flow rate reaches the saturation zone in the B-H curve of the material
used for the construction of the transformer magnetic circuit, the current consumption
increases which increases transformer losses. The increase in losses are due to saturation
and can occur because of both: overvoltage and a reduction of the frequency of the supply
voltage [11].
2.5 TRANSFORMER WORK CYCLES IN ADVERSE CONDITIONS
From a conceptual standpoint, a reduction of life of a dielectric material is produced by
temperature increases, regardless of whether this is caused by heat dissipation due to the
Joule effect in the conductors, or by an increased temperature due to failure in the cooling
system.
Under certain operating conditions, such as emergencies or very adverse environmental
conditions, it is necessary that the transformer operates for a specified period of time above
its nominal ratings. This situation may be permissible if an average load level, generally
less than nominal, occurs for periods of 24-hour intervals [7-13].
8
2.6 MAIN TYPES OF TRANSFOMER EXTERNAL FAULTS
Short Circuits between Coils in the Same Phase
This is the most difficult type of fault to identify because in its early stages, when the fault
involves only a few turns, it is virtually impossible to detect, especially in the case of high
voltage transformers with a high number of turns. However, when the fault spreads to a
larger number of turns, it is possible to detect starting with the reverse component of
intensity [11].
Faults between Coils at Different Stages
Faults between coil turns of different phases can be detected by a differential protection
placed at the entry and exit of each transformer winding [11]
Faults between Phases in the Transformer
Insulation faults between a phase and its housing due to deterioration of dielectric materials
can cause the circulation of fault currents through solidly grounded or impedance
grounding transformers, depending on the type of grounding of the transformer. For
distribution transformers systems equipped with isolated neutrals, the intensity of current
flow is restricted to the existing parasitic loads.
Faults in the Wiring and the Insulation
These types of faults are caused by faulty and loose connections at the cable interface with
the isolator input and transformer output. In places with severe weather conditions, it
occurs with some regularity the event of flashover and surges in insulators, at the input and
9
output, due to accumulated dirt, especially in coastal environments where there are higher
concentrations salt, mists and fog. Such phenomena are also typical in facilities with high
concentrations of dust in the environment, such as the case of steel and lumber mills and
cement plants with dedicated substations. In many cases, the solution is periodic cleaning
of insulators, dielectrics and protection coatings that repel dirt. If the faults are caused by
bad connections, which are characterized by increased temperature, it is possible to detect
them by the use of infrared thermography detection –also known as IR scanning.
Sometimes, the faults can be caused by animals electrocuted to get close to the machine
for warmth [7-11].
Other External Faults
In areas of high frequency lightning and especially when power lines come into the
substation aerially, there is a risk of a lightning strike hitting any of these lines and
damaging the transformer. In order to protect the transformer from damage, as a result of
overcurrent of atmospheric origin, and due to possible inadequate electrical distribution
lines maneuvers, it is common practice the installation of lightening arrester on the high
voltage input as close as possible of the transformer terminals [7,11].
2.7 THREE-PHASE FAULTS
This type of fault consists of three phases coming in contact with each other directly or
through a low impedance value. An example of three-phase faults is the fall of a
10
transmission tower where there would be a high probability that the three-phase lines would
touch each other. This type of fault is the most serious in the system, producing the highest
currents. Therefore, the faulted system must be quickly identified and removed from the
source of system failure by the action of the protection system in the shortest possible time.
The mathematical analysis is the simplest because all the three phases are faulted in the
same way and become a symmetrical system of short-circuit currents [9,10,11,13].
2.8 LINE-TO-LINE FAULTS
This type of fault involves any two phases coming in contact with each other. Example of
this type include a tree limb falling and causing two overhead lines to rob each other, small
animals touching two lines simultaneously, released metallic balloons drifting into the air
and ending in between two overhead power lines and failures in insulated cables. This type
of faults produces an unbalanced system of currents with different intensities in all three
phases [9,10,11,13].
2.9 DOUBLE LINE-TO-GROUND FAULTS
In this type of fault, two phases come in contact with each other and with the ground. This
type of fault is statistically the least frequent [9,10,11,13].
2.10 SINGLE LINE-TO-GROUND FAULTS
Single line to ground faults occurs when any of the three phases comes in contact with the
ground. Statistically, these type of faults are the most predominate in the system. An
11
average of 80% of the faults that occur in overhead power lines are single-line-to-ground.
The resulting short circuit currents of this type depend on the fault impedance and
grounding connections of the transformer in service [9,10,11,13].
Of the four types of faults described herein, only three-phase faults produce a system of
symmetrical intensities in all three phases. To calculate the circulating currents: three-
phase, line-to-line, double-line-to-ground and single-line-to-ground symmetrical
components can be used. Three-phase faults not only produce the most critical state of the
faulted system, but are the simplest to study because they are symmetrical.
2.11 PROTECTION
Transformers have to be protected from both internal and external faults with sensitive and
selective devices that can quickly and effectively interrupt the flow of energy of the fault
location in order to reduce the transformer damage. Relays and circuit breakers are standard
equipment used for transformer protection. For different short-circuit overloads, relays
should provide actuation times as a function of the load, so that its response curve fits the
curve provided by the transformer manufacturer. There are a series of well-developed and
established protection schemes that can detect overloaded circuit conditions. Such schemes
are used to program protection devices to interrupt the faults or to perform an alarm
indicating the zone where and when the fault occurred. The protection elements are
installed on the main circuit or on the transformer itself depending on the type of scheme
to perform and the type of maneuver to be executed. Among the many protection schemes
12
that can be performed, the most common are overload and short circuit protection which
can be grouped as direct and indirect types.
a. Direct protection types are electromechanical. They are installed on the main circuit
and support all incidents that may arise in this circuit, causing it to open its own operating
switch before the circuit breaker does and operate on true RMS quantities.
b. Indirect type are electronic type. This system consist of sensors installed in the main
circuit providing a proportional signal to a relay acting on the firing element or alarm. This
type has to have minimum electrical protections like any other electrical system and must
follow the instructions of the manufacturers. They are fast, accurate and more versatile.
In practice, the complete protective system consists of several independent relay schemes
arranged to monitor various aspects of transformer operation. Wherever possible, the
individual schemes are arranged so that in addition to performing their primary function,
the coverage will include at least one parameter of another relay scheme. This ensures some
measure of security in the event of component failure. The major criteria for the selection
and application of relays are sensitivity, selectivity, speed and reliability [11,13].
Sensitivity
The relay equipment must be sufficiently sensitive so that it will operate reliably, when
required, under the fault condition that produces the least operating tendency [11,13].
13
Selectivity
The relay must be able to select between those conditions for which prompt action is
required and those for which no operation, or time delay operation is required. This ensures
that the proper circuit breakers are tripped and healthy equipment is not removed from
service needlessly [11,13].
Speed The relay must operate at the required speed corresponding with the application. Such
considerations are to minimize damage to faulted equipment, prevent thermal ratings from
being exceeded during fault conditions or to remove equipment from service when
operation in an abnormal condition may cause damage or interfere with the operation of
the remainder of the power system [11,13].
Reliability The relay equipment must have high reliability and shall operate within its calibration
setting after long periods of inactivity. Relay types should not be sensitive to dust, vibration
or electrical noise. Application of the relays should be such that neither normal loads nor
fault currents will impose conditions which may lead to thermal deterioration of the
elements [11,13].
14
2.12 INSTANTANEOUS AND TIME OVERCURRENT PROTECTION RELAYS (50/51)
This type of protection is applied to the transformer in the form of inverse time overcurrent
relays connected to current transformers located on the transformer HV bushings. This
scheme not only provides protection for transformer internal faults and LV secondary
connections but will sense sustained overloads and faults occurring on the LV of the
transformer. These overloads and faults are outside the zone of the differential relays. With
the equipment ratings, operation on the high side terminal faults is unlikely due to current
transformer saturation but reliable protection is achievable elsewhere. This type of
protection can be used on the high side of the transformer as backup to the differential
protection scheme [11,15].
2.13 GROUND FAULT PROTECTION RELAYS (50N)
This relay function is used on failure of one or more than one phase to ground phases, and
on flashover. Directional overcurrent relays are the most used for ground fault protection,
with zero sequence current and/or voltage being the most common polarizing sources. The
50N would detect ground faults within the wye winding of the transformer and in the phase
conductors between the current transformer and the windings, when an external source of
zero sequence current is available. The CT ratio and relay setting may be matched to the
rating of the transformer. As the only component of system current which can flow in this
CT circuit is zero sequence, the scheme is immune from operation for any system fault and
does not require coordination within the system phase or ground fault areas. [11].
15
2.14 DIFFERENTIAL PROTECTION RELAYS (87)
Differential protection of the main transformer requires comparison of LV and HV
currents, entering and leaving the transformer. Microprocessor based relays are equipped
with second a harmonic restraint feature to prevent unwanted operations during rapid
voltage recovery following the clearance of a close in-system fault. The operating zone of
this scheme is restricted to the primary equipment encompassed by the high and low side
current transformers and thus the scheme requires no current or time coordination with
other systems. Due to the high slope characteristic, the sensitivity is dependent upon the
through-fault current component and a significant portion of the winding may not be fully
protected. Two three-phase sets of current transformers are commonly used, one set on the
high voltage side and one set on the low voltage connections. Ratios are selected to ensure
that a secondary current match is achieved between the HV and LV current transformers
within the tap setting range of the relay used, allowing for the phase correction required
across the main transformer [15].
2.15 CIRCUIT BREAKERS
A circuit breaker is a device intended to open or close the continuity of an electric circuit
under normal or faulted conditions. The circuit breaker has to be capable of interrupting
electric currents of different magnitudes and power factors base of its ratings. The rating
of circuit breaker can range from a few amps and volts to several thousands of amps and
volts. For high power application, the circuit breaker has to be designed and tested to ensure
16
the electrical interruption of a faulted circuit, and have to be able to dissipate the energy
produced by the electric arc between its contacts without getting damage.
There are different types of circuit breakers and they all depend on their medium such as
air, oil, sulfur hexafluoride (SF6) gas or vacuum to lengthens, cool and quenches and arc
of high magnitude to due to an abnormal condition. Some circuit breakers are designed
with a high automatic reset speed. Since most faults are temporary and self-clearing, power
restoration is based on the idea that if a circuit de-energizes for a short period of time, it is
likely that, whatever the cause of the fault, the arc ionized in the fault is very quickly
disintegrated and dissipated. In power transformer protection there are two main types of
circuit breaker used: dead tank and live tank circuit breakers.
Dead Tank Circuit Breakers
This type of circuit breaker has a tank enclosing the medium where the arc is interrupted.
It connects to live circuits outside of the tank via bushings [9,10,13].
Live Tank Circuit Breakers
This type of circuit breaker if is live. The environment where the arc interrupted is under
high voltage. It takes less space and it is more economical than dead tank circuit breakers.
The operating mechanisms can be pneumatic, hydraulic or mechanical. Pneumatically
operated store energy for operation in compressed air and therefore require the use of
compressors. Hydraulically operated mechanisms store energy for operation in pressurized
oil and thus require the use of pumps. Mechanically operated store energy for operation in
17
springs, therefore motors are required for spring’s loading. Some circuit breakers operated
faster than others, but the range is between 0 and 5 cycles [11,13].
2.16 CURRENT TRANSFORMER REQUIREMENTS
Selection and application of current transformers (CTs) for large power transformers
protective systems involves: 1) full load continuous operation and 2) maximum fault
conditions. These two factors generally oppose one another and a compromise has to be
made within the limits executed by each application. The current transformer parameters
for protection of the power transformer have to be set with a correct ratio, level of burden
and accuracy at full load and saturation voltage [2,15].
Current Transformer Ratio
Three basic considerations apply in the selection of the CT ratio: load current, relay and
thermal limit, and saturation at maximum fault. The most severe condition will control the
ratio used. For applications involving differential protection other factors may control
ratios in addition to those described herein [2,11,13,15].
CT Load Current
The maximum continuous full load current must be determined for each current
transformer location. This value should be derived for the highest value condition such as
minimum system voltage and maximum equipment rating. This will set the lower limit of
the available ratios [2,11,13,15].
18
Relay Thermal Limits
Transformer ratio should be selected such that the maximum fault current seen by the relay
for the required clearance time does not exceed the relay thermal limit. It is usual for this
purpose to use the clearance time for the backup relaying plus breaker time at the maximum
fault level. An indication of the minimum current transformer ratio, which may be used to
remain below the relay thermal limit, can be obtained by calculating the maximum
permissible relay amperes for the appropriate clearing time applicable [2,11,13,15].
CT Saturation
Current transformer performance under fault current magnitudes is determined, from the
saturation standpoint, by the secondary burden. This includes the CT internal resistance,
lead resistance and relay burden. Selection of a CT ratio have a profound effect on the
terminal voltage to be expected during fault current flow. Assuming that the external
burden of the current transformer is constant, varying the CT ratio has two effects. First,
the lower the CT ratio the higher the secondary current and the higher the terminal voltage
required. Second, the lower the CT ratio, the lower the internal resistance which tends to
lower the terminal voltage. Combining these two, unless the internal resistance is much
larger than the external value, it may generally be assumed that the lower the ratio, the
higher will be the terminal voltage. The expected terminal voltage should be calculated for
the maximum fault current and a ratio selected to give a value which is below a
commercially available current transformer saturation voltage [11,12,13].
19
Current Transformer Level of Burden
The total full load burden, in VA, should be calculated for each current transformer
involved to insure that the CT VA rating is not exceeded. In the case of three-phase sets of
current transformers, the largest phase burden of the set should be used assuming that a
balanced system load will be used. CT lead burden should be calculated based on lead
resistance, actual full load secondary current and manufacturer’s published values of
equipment burden at a nominal unsaturated 5 amp load. If the results of this approach give
excessively high values or come close to available ratings, vectorial addition should be
used instead. Current transformer burdens should be summed vectorially to determine the
Transformer Correction Factor [11,13].
Current Transformer Accuracy
Accuracy class voltage is used to determine the accuracy class of the current transformer.
A maximum ratio error limit is required for up to twenty times full load current (20 x 5
Amps) into the secondary burden. If this voltage is somewhat below the transformer
saturation voltage, then the current transformer would produce this voltage at various
combinations of current and burden with predictable accuracy. Saturation voltage is the
maximum voltage to which the current transformer may be driven without a virtually
complete loss of functionality [11,13].
20
CHAPTER III
CALCULATIONS AND DRAWINGS
3.1 30 MVA TRANSFORMER RATINGS AND LOADS
15 MVA
GND
T1: 30MVA69/13.8 kVZ = 8%
15 MVA
13.8 kV
69 kV
15 MVA
GND
T1: 30MVA69/13.8 kVZ = 8%
15 MVA
13.8 kV
69 kV
Figure 1: 30 MVA Transformer
Table 1: 30 MVA Transformer Data
ITEM MVA VOLTAGE RATING IMPEDANCE
T1 30 69/13.8 kV 8% Load 1 15 13.8 kV NA Load 2 15 13.8 kV NA
21
3.2 T1 PRIMARY, SECONDARY AND GROUND RATED CURRENTS
T1 Primary Rated Current
IRate-PrimaryT1 = MVA
√3 𝑥𝑥 KV (3.1)
IRate-PrimaryT1 = 30 x 106
√3 x 69 x 103 = 251.02 (A).
This is the rated current on the primary side of the 30 MVA power transformers.
T1 Secondary Rated Current
IRated-SecondaryT1 = MVA
√3 𝑥𝑥 KV (3.2)
IRated-SecondaryT1 = 30 𝑥𝑥106
√3 𝑥𝑥13.8 𝑥𝑥 103 = 1,255.11 (A).
This is the rated current on the secondary side of the 30 MVA power transformers. The
effect of the transformer and system impedances are neglected in both in the primary and
secondary rated currents calculations.
T1 Ground Rated Current
IRated-GndT1 = MVALoad/3√3 x KV
(3.3)
IRated-GndT1 = 15 x 106 3⁄
√3 x 13.8 x 103 = 209.18 (A).
This is the single-phase current from of the ground of the 30 MVA transformer through
one of the two 15 MVA loads and back to the ground of the transformer ground. Since the
loads are similar, only one ground (IgroundT1) is calculated.
22
3.3 DETERMINING THE CURRENT TRANSFORMER (CT) RATIOS
The computed primary and secondary rated currents are used to determine the CT ratios
for the differential relay.
nCT1 = CT ratio of the primary to secondary CT currents. This CT is connected on the
primary side of the 30 MVA power transformer.
nCT1 = I1nI2n
(3.4)
Where I1n = primary rated current going through the CT and I2n = secondary current seen
by the CT winding leads.
nCT1 = 2505
= 50 , where 50 is the primary to secondary CT current ratio.
nCT2 = CT ratio of the primary to secondary CT currents. This CT is connected on the
secondary side of the 30 MVA power transformer.
nCT2 = I1nI2n
(3.5)
Where I1n = primary rated current going through the CT and I2n = secondary current seen
by the CT winding leads.
nCT2 = 12005
= 240, where 240 is the primary to secondary CT current ratio.
nCT3 = CT ratio of the primary to secondary CT currents. This CT is connected on the
23
wye-ground of the 30 MVA power transformer.
nCT3 = I1nI2n
(3.6)
Where I1n = primary rated current going through the CT and I2n = secondary current seen
by the CT winding leads.
nCT3 = 2505
= 50, where 50 is the primary to secondary CT current ratio.
3.4 CT SECONDARY CURRENTS
These currents are computed by using the primary and secondary calculated currents on
the primary and secondary side of the 30 MVA, and the current ratios of the current
transformer.
Primary Side of T1
Computing the secondary current the 87, and 50/51 relays will see from the primary rated
current of the 30 MVA transformer and the selected CT turn ration.
IHigh-Sec = IRate-PrimaryT1 x 1
nCT1 (3.7)
IHigh-Sec = 251.022 x
5250
= 5.02 (A) Secondary Side of T1
Computing the secondary current the 87 relay will see from the secondary rated current
of the 30 MVA transformer and the selected CT turn ratios.
24
IHigh-Sec = IRated-SecondaryT1 x 1
nCT2 (3.8)
ILow-Sec = 1,255.11 x 5
1200 = 5.02 (A)
Ground Side Wye-GND of T1
Computing the secondary current the 50N relay will see through the solidly ground of the
30 MVA transformer and the selected CT turn ratios.
IGround1-Sec = IRated-Gnd1x 1
nCT3 (3.9)
IGround1-Sec = 209.18 x 5250
= 4.18 (A)
3.5 RELAYING CT PERFORMANCE CALCULATIONS
To determine the maximum fault current through each set of differential CTs, the 87 relay
for a transformer through fault on each of the buses served by the CT1 and CT2
transformers, an infinite bus (conservative approach) on the high voltage side of the
transformer was assumed. This assumption ignores the system impedance. The maximum
current that the relays and CTs will see during a three-phase through fault is calculated as
follows:
30 MVA Transformer Primary Through Fault Current
IThrough-primaryT1 = MVA
√3 𝑥𝑥 𝐾𝐾𝐾𝐾𝐾𝐾𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃−𝑃𝑃 x %Z (3.10)
IThrough-primaryT1 = 30 𝑥𝑥106
√3∗ 69𝑥𝑥103 ∗0.08 = 3,137.77 (A).
25
30 MVA Transformer Secondary Through Fault Current
IThrough-primaryT1 = MVA
√3 𝑥𝑥 𝐾𝐾𝐾𝐾𝐾𝐾𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃−𝑃𝑃 x %Z (3.11)
IThrough-SecondaryT1 = 30 𝑥𝑥106
√3∗ 13.8𝑥𝑥103 ∗0.08 = 15,688.9 (A).
The 3-phase short circuit current through the.3
CTs on the high voltage (69kV) bushings of the CT1 and CT2 transformers is computed as
follows:
H Winding (W1 ): 250/5 Series CT, 250/5 tap, C400 accuracy class, wye-connected
W1 x IPrimaryT1 = IThrough-primaryT1 x 1
nCT1 (3.12)
W1*IPrimaryT1 = 3,137.77 x 5250
= 62.75 (A)
X Winding (W2); 1200/5 Series CT, 1200/5 tap, C400 accuracy class, delta-connected
W2 x ISecondaryT1 = IThrough-SecondaryT1 x 1
nCT2 (3.13)
W2 x ISecondaryT1 = 15,688.9 x 5
1200 = 65.27 (A)
Ground Winding (Wye-gnd); 250/5 Series CT, 250/5 tap, C400 accuracy class
Wye-gnd x IGnroundT1-Sec = IGnroundT1-Sec x 1
nCT2 (3.14)
Wye-gnd x IGnroundT1-Sec = 209.18 x 5250
= 4.18 (A) A guideline regarding relay currents is that the relay current during fault conditions should
not exceed 20 x the nominal CT secondary current of 5A. By inspection of the relay
26
currents above, none of the relay currents for a 3-phase fault exceed 100A (20 x 5 Amps),
and therefore this guideline is satisfied.
3.6 CT BURDEN VOLTAGE (VB) FOR DIFFERENTIAL CT FOR T1
The burden voltage is the maximum voltage that can develop at the secondary terminals
of the protection CTs. Therefore, this voltage must be calculated to ensure the correct
selection of the CTs.
VB = IF3 (RS + RL + RR) (3.15) Where: IF3 = W1x IPrimaryT1 = Relay current for 3-phase fault conditions on the high side IF3 = W2xISecondaryT1 = Relay current for 3-phase fault conditions on the low side IF3 = Wye-gndx IGoundT1 = Relay current for solidly conditions through one of the loads RS = CT secondary winding resistance + CT lead resistance in ohms RL = CT circuit two-way lead resistance to relay in ohms RR = Differential relay resistance in ohms
RS Calculation: Manufactured published data [11] of 0.211 Ω for a C400 CT secondary winding
resistance + CT lead resistance was used.
RL Calculation For this calculation a distance of 225 feet is assumed for the CT secondary terminals to
the relay panel. The CT wiring circuits consist of 1 #10AWG conductor per phase from is
1.21 ohms per the CTs on the HV side of the CTs to the corresponding Protection Relay
Panels. Per NEC 2014, Table 9, #1 10AWG conductor 1000 feet at 750C.
RL =1.21 Ω x 225
1000 Ω x 2 = 0.558 (Ω) (3.15.1)
The equation is multiplied by 2 to account for the two way wire circuit.
27
RR Calculation
The SEL 787 relay has published burdens of 0.27VA@ 5A input, and 2.51VA@ 15A input.
The 2.51VA is used to compute the RR since it represents the burden at higher than
nominal relay input; i.e., the burden
RR = VAI2
= 2.51VA152
= 0.011 (Ω) (3.15.2) This value will be used in each of the CT burden voltage calculations. High Side of 30 MVA Transformer: CT1 VB3 = W1*IPrimaryT1 x (Rs + RL + RR) (3.16) CT1 VB3 = 62.75 x (0.211 + 0.558 + 0.011) CT1 VB3 = 48.95 (V) Low Side of 30 MVA Transformer: CT2 VB3 = W2 x ISecondaryT1 (Rs + RL + RR) (3.17) CT2 VB3 = 62.27 x (0.211 + 0.558 + 0.011) CT2 VB3 = 48.57 (V) Ground Side of 30 MVA Transformer: CT3 VBGND = Wye-gnd x IGnroundT1-Sec x (Rs + RL + RR) (3.18) CT3 VBGND = 52.29 x (0.211 + 0.558 + 0.011) CT3 VBGND = 40.78 (V) Section 3 of the SEL-787 instruction manual indicates CT performance under through fault
conditions will be satisfactory if the burden voltage of each CT circuit is less than ½ of
the CT class C (accuracy class) voltage rating: VB3 < (0.5) (C400 rating). The C400 CT
secondary is rated for 400V.
CT1 Winding 1, T1 (H): CT1 VB3 = 48.95 (V) < 0.5 (400 V) (3.19) CT1 VB3 = 48.95 (V) < 200 (V)
28
CT1 Winding 1, T1 (X): CT2 VB3 = 48.27 (V) < 0.5 (400 V) (3.20) CT2 VB3 = 48.27 (V) < 200 (V) Ground Side of T1 CT3 VBGND = 48.78 (V) < 0.5 (400 V) (3.21) CT3 VBGND = 48.78 (V) < 200 (V) The maximum voltage that can develop on the secondary of the CT with less than a 10%
error of the selected the CT C400 is 400 (V). Based on the above calculations the C400 CT
will perform adequately under through fault conditions. In all three cases, which are
applicable to the 50/51, 87 and 50N relays, the calculations were found to be less than 400
(V). Therefore, the C400 can be used in all for four cases. For simplicity and to keep
uniformity, one class C, CT C400, was selected for all four relays schemes rather than CTs
of different ratios.
3.7 T1 RELAY DIFFERENTIAL TAP CALCULATIONS
A tap is the minimum amount of current the relay needs to operate provided by the
secondary of the current transformer. The Schweitzer SEL-787 instruction manual
guideline was used to calculate the optimum input current taps based on the following
equation.
TAPn = MVA
√3∗ VWDGn ∗CTRn x C (3.22)
Where: C = 1, if WnCT setting = Y (wye connected CT's) C = √3, if WnCT setting= D (delta connected CT's) MVA = Maximum Power Transformer capacity Setting (must be the same for all TAPn Calculations) VWDGn =Winding line-to-line voltage setting, in kV CTRn = Current Transformer ratio setting.
29
The relay calculates the tap settings and incorporates these settings provided the
following criteria are met: ·
1. The TAP settings are within the range of 0.1 x In and 31 .0 x In (where In is the
relay input nominal current rating, which is 5 amperes).
2. The ratio of TAPmax/TAPmin ≤ 7.5
T1 Taps Values
The calculated tap values for each winding of input of Transformer 1 are as calculated
follows:
TAP(l)(H ) = 30 x 106
√3 x 69 x 103 x 250 5⁄ x 1
TAP(l)(H ) = 5.02 (A) Delta Connected CTs
TAP(2)(X ) = 30 x 106
√3 x 13.8 𝑥𝑥 103 x 1200 5⁄ x √3
TAP(2)(X ) = 9.06 (A)
To verify if the relay will accept these tap settings, a comparison is made to the setting
criteria noted above. For criterion #1, each tap setting must be in the range of 0.1 x In to
31 x In (0.1 x 5 = 0.5, and 31*In = 5.0 x 31 = 155 amperes). Both taps settings calculated,
5.02 Amps and, 5.23 Amps, meet this criterion.
For criterion #2, the ratio of TAPmax/TA Pmin ≤ 7 .5
30
TAPmaxTAP min
≤ 7.5 (3.23) 5.235.02
= 1.04 ≤ 7.5
Therefore, the CT ratios and calculated tap settings are in compliance with the relay setting
criteria, and the relay will automatically incorporate these optimum tap settings for internal
computations if the transformer MVA data is entered in the Configuration Settings list.
3.8 T1 RELAY DIFFERENTIAL PICKUP (DIAL) CALCULATIONS
The operating current pickup or dial is set to provide the maximum sensitivity for
differential protection while maintaining secure operation of the protected system. Under
balanced load and through fault conditions, the phasor sum of the currents entering the
relay should sum to near zero. For faults within the zone of protection, the phasor sum of
currents will exceed the pickup setting and will cause the relay to signal circuit breaker
tripping. The SEL schweitzer relays 787 instruction manuals suggests a pickup value of
0.3 amperes, provided the following criterion is met:
Pick up min ≥ 0.1 x InTAP min
(3.24)
0.3 ≥ 0.1 x 5TAP min
0.3 ≥ 0.1 x 55.02
= 0.099
The criterion is met and the 0.3 can be employed.
31
3.9 CALCULATION OF PHASE ANGLE CORRECTION (WXCTC)
The SEL schweitzer 787 relay is capable of performing input current phase compensation
for each winding when this function is enabled. This feature allows the relay to be used
with any combination of delta or wye-connected CTs. For both CT applications, the power
transformers are connected Delta on the 69kV H winding and wye-grounded on the 13.8
kV X winding. The CTs providing input to the relay W1 terminals (Transformer H
winding) are connected in Wye. The CTs providing input to the relay W2 terminals
(Transformer X winding) are connected in delta. Phase compensation setting is started with
the W1 relay input from the wye connected CTs. This input, which is selected as the
reference point from which the phase compensation settings of other two windings will be
determined, and will be set to zero (no phase shift). The W2 relay input currents from the
(13.8 kV system) wye connected CTs will lag the W1 currents by 30° due to the power
transformer phase shift. In summary, the relay windings phase angle compensation set
points are: W1 CTC = 0, for 0° and W2 CTC = 1, for 30°
3.10 RESTRAINT SLOPE SETTINGS (SLP)
When CTs saturate, the secondary waveform signature current gets distorted. If the CTs
saturate, they would not operate properly and the relay would not trip. The purpose of the
slope is to compensate for CT mismatch between the differential CTs at CT saturation..
Microprocessor based relays are equipped with two differential slopes to remedy the
differential CT mismatch at CT saturation. The slope is a boundary that defines the
operating from the non-operating zone of the relay. The relay region of operation is above
32
the slope boundary, Figure 2. The lower the slope (Slope 1), the more sensitive the relay is
and the more prone to nuisance tripping. The higher the slope (Slope 2), the less sensitive
and less prone to nuisance tripping the relay would be [2].
Figure 2: Differential Operating Slope Zones
Slope 2 = �IprimT1�−|IsecT1|�IprimT1�+|IsecT1| X 100 (3.25)
Slope 2 = |3,137.77|−|15,688.9||3,137.77|+|15,688.9| X 100
Slope 2 = 45.2% Therefore, the slope for the 30 MVA power transformer would be set at 45% to
compensate for the differential CT’s mismatch.
33
3.11 THE 30 MVA TRANSFORMER SLOPES IN THE SEL 787 RELAY
The 87 function has two slopes, 1 and 2. Slope 1 starts at the origin and end at 6 units of
tap inside the relay, x = 6, on the x-axis. Slope 2 starts where slope 1 ends, at x = 6. If the
unit of tap inside the relay is set at 5A secondary (the CT secondary set at 5.A), the 6 per
unit of tap is 6 x 5 = 30A secondary. This mean that at 30A secondary the 87 differential
function will start plotting on slope 2. The SEL-87 function has a setting called RS1. The
RS1 setting is the breakpoint at which slope 1 transitions to slope 2. This value is set by
default at 6 units of tap inside the relay, but can be changed to fit the design needs.
Slope 1
Slope 1 starts at the origin, (0,0) and it travels to the RS1 point setting (x2, y2). The value
of x2 = 6. Using the slope intercept formula the value of y at x = 6 can be found as follow.
y = mx + b (3.26)
Given:
b = 0, because slope 1 goes through the origin
m = 25%, this is the first default slope of the SEL-787 relay
x2 = 6, the RS1 default setting
y2 = 0.25(6) + 0
y2 = 1.5
Slope 1 will start at the origin (0,0) and ends at (6,1.5). This indicates that slope 1 is
anchored at the point (0,0) and stops at (6,1.5). This slope would have the following line
equation, y = 0.25x
34
Slope 2
Slope 2 starts where slope 1 ends at (6, 1.5). There is a point at which Slope 2 crosses the
unrestrained setting. The unrestrained setting is set by default at 10 multiples of tap. The
unrestrained setting can be represented as a horizontal line at x = 10. Once slope 2 crosses
the unrestrained horizontal line at x = 10, slope 2 is no longer used. For slope 2, slope
formula can be used to find the y value of the second point
Given:
b = ?, it has to be calculated m = 45.2%, Calculated slope x2 = 6, the RS1 default setting (x1 = 6, y1 = 1.5) (x2 = 10, y2 = ?)
Slope 2 = y2− y1 x2− 1
= = y2− 1.510−6
0.6 = y2− 1.5
4
0.6 (4) = y2 – 1.5 2.4 + 1.5 = y2 y2 = 3.9
thus, the beginning point for slope 2 is (6,1.5) and the stopping point is at (10, 3.9).
If the unit of tap inside the relay is set at 5A secondary, the calculated 3.9 per unit of tap is
3.9*5 = 19.5A secondary. The relay would shut off slope 2 at this point and will transition
to the instantaneous element. By entering the values for restrain (O87P), unrestraint
35
(U87P) and the restrain current slope (IRS1), the relay would use the slope formula to
calculate the value of y1 = 1.5 for Slope 1, and the value of y2 = 3.9 for lope 2. The y-
intercept for slope 2 is calculated using the point (10, 3.9) and he equation (3.26):
y = mx + b
3.9 = 0.6*(10) + b
3.9-.6*(10) = b
b = - 2.1
thus, the calculated equation of slope 2 would be y = 0.6x – 2.1 This indicates that slope 2
is anchored at the point (0, -1.2) and travel through the points (6, 1.5) and (10, 3.9)
Figure 3: Slopes 1 and 2 of 30MVA Transformer
36
Table 2: 30 MVA Transformer Relays Settings
ELEMENT PICK-UP CURRENT SETTINGS
51 IPrimary−Rated ∗ 1.2CT Ratio
= 25102∗1.2
50 = 6.02 (A)
50 IThrough− Primary−Rated ∗ 1.2CT Ratio
= 3,137.02 ∗ 1.250
= 75.3 (A)
50G IThrough− Primary−Rated CT Ratio
= 3,137.02
50 = 62.75 (A)
The pick- up setting were tabulated in Table 2 for easy and convenient reading to program
the relay in ASELerator. The 1.2 multiplier used in the pickup current for the 50 and 51
relays accounts for current imbalance between phases. This multiplier was not used for the
50G relay because this relay is only monitoring current on the ground connection of the
Wye transformer winding.
Table 3: 30 MVA Transformer Relay and CT Data Settings
CT RATIO SLOPE 1 SLOPE 2 TAP DIFF CURRENTS
CTR1 250/5 CTR2 1200/5 CTRN 250/5 Point 1 (0,0) (6, 1.5) Point 2 (6, 1.5) (10, 3.9) Slope % 25% 60% TAP 1 5.02 TAP 2 9.06 Idiff > 0.3 Idiff >> 10
Table 3 shows the data computed in chapter two for convenient and easy reading to
program the relay in ASELerator and Test Universe Software. A 25% slope was used for
37
Slope 1. While the calculated slope 2 was 45.2%, a 60% percent slope was used to
compute the rest of the parameters.
3.12 40 MVA TRANSFORMER (T2) CALCULATIONS
20 MVA
GND
T2: 40MVA69/13.8 kVZ = 10%
20 MVA
13.8 kV
69 kV
20 MVA
GND
T2: 40MVA69/13.8 kVZ = 10%
20 MVA
13.8 kV
69 kV
Figure 4: 40 MVA Transformer
Table 4: 40 MVA Transformer Data
ITEM MVA VOLTAGE RATING IMPEDANCE
T2 40 69/13.8 kV 10% Load 1 20 13.8 kV NA Load 2 20 13.8 kV NA
38
3.13 T2 PRIMARY, SECONDARY AND GROUND RATED CURRENTS
T2 Primary Rated Current
IRate-PrimaryT2 = MVA
√3 𝑥𝑥 KV (3.27)
IRate-PrimaryT2 = 40 𝑥𝑥 106
√3 𝑥𝑥 69 𝑥𝑥 103 = 331.69 (A).
This is the rated current on the primary side of the 40 MVA power transformers. T2 Secondary Rated Current
IRated-SecondaryT2 = MVA
√3 𝑥𝑥 KV (3.28)
IRated-SecondaryT2 = 40 𝑥𝑥106
√3 𝑥𝑥13.8 𝑥𝑥 103 = 1,673.48 (A).
This is the rated current on the secondary side of the 40 MVA power transformers. T2 Transformer Ground Current
IRated-GndT2 = MVALoad/3√3 x KV
(3.29)
IRated-GndT2 = 20 x 106 3⁄
√3 x 13.8 x 103 = 278.91 (A).
This is the single phase current from of the ground of the 40 MVA transformer through
one of the two 20 MVA loads and back to the ground of the transformer ground. Since the
loads are similar, only one ground (IgroundT2) is calculated.
39
3.14 DETERMINING THE CURRENT TRANSFORMER (CT) RATIOS
These currents are computed by using the primary and secondary calculated currents on
the primary and secondary side of the 40 MVA, and the current ratios of the current
transformer. The ratios are computed as follow:
nCT1 = CT ratio of the primary to secondary CT currents. This CT is connected on the
primary side of the 40 MVA power transformer.
nCT1 = I1nI2n
(3.30)
Where I1n = primary rated current going through the CT and I2n = secondary current seen
by the CT winding leads.
nCT1 = 3505
= 70 , where 70 is the primary to secondary CT current ratio. nCT2 = CT ratio of the primary to secondary CT currents. This CT is connected on the secondary side of the 30 MVA power transformer
nCT2 = I1nI2n
(3.31)
nCT2 = 16005
= 320, where 320 is the primary to secondary CT current ratio. nCT3 = CT ratio of the primary to secondary CT currents. This CT is connected on the secondary side of the 30 MVA power transformer
nCT3 = I1nI2n
(3.32)
nCT3 = 2505
= 50, where 70 is the primary to secondary CT current ratio.
40
3.15 T2 CT SECONDARY CURRENTS
These currents are computed by using the primary and secondary calculated currents on
the primary and secondary side of the 40 MVA, and the current ratios of the current
transformer.
Primary Side of T2
Computing the secondary current the 87, and 50/51 relays will see from the primary rated
current of the 40 MVA transformer and the selected CT turn ration.
IHigh-Sec = IRate-PrimaryT2 x 1
nCT1 (3.33)
IHigh-Sec = 331.69 x
5350
= 4.74 (A)
Secondary Side of T2
Computing the secondary current the relay from the will from the secondary of the
current transformer based on the computed secondary rated current.
IHigh-Sec = IRated-SecondaryT2 x 1
nCT2 (3.34)
ILow-Sec = 1,673.48 x 5
1600 = 5.22 (A)
41
Ground Side Wye-GND of T2
Computing the secondary current the 50N relay will see through the solidly ground of the
40 MVA transformer and the selected CT turn ration
IGround1-Sec = IRated-GndT2 x 1
nCT3 (3.35)
IGround1-Sec = 278.91 x 5250
= 5.57 (A)
3.16 RELAY AND CT PERFORMANCE CALCULATIONS
To determine the maximum fault current through each set of differential CTs to the 87
relay for a transformer through fault on each of the buses served by the CT1 and CT2
transformers, an infinite bus (conservative approach) on the high voltage side of the
transformer was assumed. This assumption ignores the system. The maximum current the
relay and CTs will see, during a 3-phase through fault is calculated as follows:
Transformer 2 Primary Through Fault Current
IThrough-primaryT2 = MVA
√3 𝑥𝑥 𝐾𝐾𝐾𝐾𝐾𝐾𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃−𝑃𝑃 x %Z (3.36)
IThrough-primaryT2 = 40 𝑥𝑥106
√3∗ 69𝑥𝑥103 ∗0.10 = 3,346.96 (A).
Transformer 2 Secondary Through Fault Current
IThrough-primaryT2 = MVA
√3 𝑥𝑥 𝐾𝐾𝐾𝐾𝐾𝐾𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃−𝑃𝑃 x %Z (3.37)
IThrough-SecondaryT2 = 40 𝑥𝑥106
√3∗ 13.8𝑥𝑥103 ∗0.10 = 16,734.8 (A).
42
The 3-phase short circuit current through the CTs on the high voltage (69kV) bushings of
the CT1 and CT2 transformers is as follows:
H Winding (W1 ): 350/5 SR CTs, 350/5 tap, C400 accuracy class, wye-connected
W1 x IPrimaryT2 = IThrough-primaryT2 x 1
nCT1 (3.38)
W1 x IPrimaryT2 = 3,346.96 x 5350
= 47.81 (A)
X Winding (W2); 1600/5 SR CTs, 1600/5 tap, C400 accuracy class, delta-connected
W2 x ISecondaryT2 = IThrough-SecondaryT2 x 1
nCT2 (3.39)
W2 x ISecondaryT2 = 16,734.8 x 5
1,600 = 52.30 (A)
Ground Winding (Wye-gnd); 350/5 SR CTs, 250/5 tap, C400 accuracy class
Wye-gnd x IGnroundT2-Sec = IGnroundT2-Sec x 1
nCT3 (3.40)
Wye-gnd x IGnroundT2-Sec = 278.91 x 5250
= 5.58 (A) A guideline regarding relay currents is that the relay current during fault conditions should
not exceed 20 x the nominal CT secondary current of 5A. By inspection of the relay
currents above, none of the relay currents for a 3-phase fault exceed 100A, and therefore
this guideline is satisfied.
43
3.17 CT BURDEN VOLTAGE (VB) FOR DIFFERENTIAL CTS FOR T2
The burden voltage is the maximum voltage that can develop at the terminals of the
protection CTs. Therefore, this voltage must be calculated to ensure the correct selection
of the CTs
VB = IF3 (RS + RL + RR) (3.41)
Where: IF3 = W1* IPrimaryT2 = Relay current for 3-phase fault conditions on the high side IF3 = W2*ISecondaryT2 = Relay current for 3-phase fault conditions on the low side IF3 = Wye-gnd* IGoundT2 = Relay current for solidly conditions through one of the loads RS = CT secondary winding resistance + CT lead resistance in ohms RL = CT circuit two-way lead resistance to relay in ohms RR = Differential relay resistance in ohms
RL =1.21 Ω x 225
1000 Ω x 2 = 0.558 (Ω) (3.41.1)
RR = VAI2
= 2.51VA152
= 0.011 (Ω) (3.41.2) This value will be used in each of the CT burden voltage calculations. High Side of T1 : CT VB3 = W1*IPrimaryT2 x (Rs + RL + RR) (3.42) CT VB3 = 47.81 x (0.211 + 0.558 + 0.011) CT VB3 = 39.29 (V) Low Side Side of T1: CT VB3 = W2 x ISecondaryT2 (Rs + RL + RR) (3.43) CT VB3 = 52.3 x (0.211 + 0.558 + 0.011) CT VB3 = 40.79 (V) Ground Side of T1: CT VBGND = Wye-gnd x IGnroundT2-Sec x (Rs + RL + RR) (3.44) CT VBGND = 5.58 x (0.211 + 0.558 + 0.011) CT VBGND = 4.35 (V)
44
Section 3 of the SEL-787 instruction manual indicates CT performance under through fault
conditions will be satisfactory if the burden voltage of each CT circuit is less than ½ of
the CT C (accuracy class) voltage rating: VB3 < (0.5) (C400 rating). The C400 CT
secondary is rated for 400V.
CT1 Winding 1, T1 (H): CT1 VB3 = 39.29 (V) < 0.5 (400 V) (3.45) CT1 VB3 = 48.95 (V) < 200 (V) CT1 Winding 1, T1 (X): CT2 VB3 = 40.79 (V) < 0.5 (400 V) (3.46) CT2 VB3 = 40.79 (V) < 200 (V) Ground Side of T1 CT3 VBGND = 48.78 (V) < 0.5 (400 V) (3.47) CT3 VBGND = 4.35 (V) < 200 (V) The maximum voltage that can develop on the secondary of the CT without exceeding a
10% error of the selected the CT C400 is 400 (V). In all three cases, which are applicable
to the 50/51, 87 and 50N relays, the calculations were found to be less than 400 (V).
Therefore, the C400 can be used in all for four cases. For simplicity and to keep uniformity,
one class of CT C400 was selected rather than CTs of different ratings.
3.18 T2 RELAY DIFFERENTIAL TAP CALCULATIONS
The calculated tap values for each winding of input of T2 are as calculated follows:
Wye connected CTs
TAP(l)(H ) = 40 x 106
√3 x 69 𝑥𝑥 103 x 350 5⁄ x 1
TAP(l)(H ) = 4.78 (A)
45
Delta Connected CTs
TAP(2)(X ) = 40 x 106
√3 x 13.8 𝑥𝑥 103 x 1600 5⁄ x √3 (3.51)
TAP(2)(X ) = 9.06 (A)
To verify if the relay will accept these tap settings, a comparison is made to the setting
criteria noted above. For criterion #1, each tap setting must be in the range of 0.1 x In to
31 x In (0.1 x 5 = 0.5, and In x 5 = 5 x 31 = 155 amperes). Both taps settings calculated,
6.69 Amps and, 2.42 Amps, meet this criterion.
For criterion #2, the ratio of TAPmax/TA Pmin ≤ 7 .5
TAPmaxTAP min
≤ 7.5 (3.48) 6.692.242
= 1.97 ≤ 7.5
Therefore, the CT ratios and calculated tap settings are in compliance with the relay setting
criteria, and the relay will automatically incorporate these optimum tap settings for internal
computations if the transformer MVA data is entered in the Configuration Settings list.
3.19 T2 RELAY DIFFERENTIAL PICKUP (DIAL) CALCULATIONS
The operating current pickup is set to provide the maximum sensitivity for differential
protection while maintaining secure operation of the protected system. Under balanced
load and through fault conditions, the phasor sum of the currents entering the relay should
sum to near zero. For faults within the zone of protection, the phasor sum of currents will
exceed the pickup setting and will cause the relay to signal circuit breaker tripping. The
46
SEL schweitzer relays 787 instruction manuals suggests a pickup minimum value of 0.3
amperes, provided the following criterion is met:
Pick up min ≥ 0.1 x InTAP min
(3.49)
0.3 ≥ 0.1 x 5TAP min
0.3 ≥ 0.1 x 52.42
= 0.21
The criterion is met and the 0.3 ≥ can be employed.
3.20 CALCULATION OF PHASE ANGLE CORRECTION (WXCTC)
The SEL schweitzer 787 relay is capable of performing input current phase compensation
for each winding when this function is enabled. This feature allows the relay to be used
with any combination of delta or wye-connected CTs. For both CT applications, the power
transformers are connected Delta on the 69kV H winding and wye-grounded on the 13.8
kV X winding. The CTs providing input to the relay W1 terminals (Transformer H
winding) are connected in Wye. The CTs providing input to the relay W2 terminals
(Transformer X winding) are connected in delta. Phase compensation setting is started with
the W1 relay input from the wye connected CTs. This input, which is selected as the
reference point from which the phase compensation settings of other two windings will be
determined, and will be set to zero (no phase shift). The W2 relay input currents from the
(13.8 kV system) wye connected CTs will lag the W1 currents by 30° due to the power
transformer phase shift. In summary, the relay windings phase angle compensation set
points are: W1 CTC = 0, for 0° W2CTC = 1, for 30°
47
3.21 RESTRAINT SLOPE PERCENTAGE SETTINGS (SLP)
Slope 2 for the 40 MVA transformer is calculated using the same methodology as that of
30 MVA transformer.
Slope 2 = �IprimT2�−|IsecT2|�IprimT2�+|IsecT2| X 100 (3.50)
Slope 2 = |3,346.96|−|16,734.80||3,346.96|+|16,734.80| X 100
Slope 2 = 68.5%
This Slope for the 40 MVA transformer would be set at 70% .
3.22 THE 40 MVA TRANSFORMER SLOPES IN THE SEL 787 RELAY
The starting and ending points and the calculation of the equation of the line for Slope 1
for the 40 MVA transformer is the same as for the 30MVA transformer.
Slope 2
Like for the 30 MVA transfer, slope 2 of the 40 MVA transformer starts where slope 1
ends at (6, 1.5) and ends at the unrestrained setting set by default at 10 multiples of tap.
The y2 value for the second point wound be different and consequently the y-intercept.
Given:
b = ?, it has to be calculated m = 70% x2 = 6, the RS1 default setting (x1 = 6, y1 = 1.5) (x2 = 10, y2 = ?)
48
Slope 2 = y2− y1 x2− 1
= y2− 1.510− 6
0.70 = y2− 1.54
0.70*(4) = y2 – 1.5 0.70*(4) + 1.5 = y2
y2 = 4.3 thus, the beginning point for slope 2 is (6,1.5) and the stopping point is at (10,4.3). If the unit of tap inside the relay is set at 5A secondary, the calculated 4.3 per unit of tap is
4.3*5 = 32.5 secondary. The relay would shut off slope 2 at this point and will transition
to the instantaneous element. By entering the values for restrain (O87P), unrestraint
(U87P) and the restrain current slope (IRS1), the relay would use the equation of the line
formula, y = mx + b, to calculate the value of y = 1.5 for Slope 1, and the value of x2 = 10
for lope 2. The y-intercept for slope 2 is calculated using the point (10,4.3) and he equation
of the line y = mx + b as follow:
y = mx + b
4.3 = 0.70*(10) + b
4.3-.0.70*(10) = b
b = - 2.7
thus, the calculated equation of slope 2 would be y = 0.70x – 2.7 This indicates that slope
2 is anchored at the point (0, - 2.7) and travel through the points (6, 1.5) and (10, 4.3)
49
Figure 5: Slopes 1 and 2 of 40MVA Transformer
3.23 RELAY SETTINGS
Table 5: 40 MVA Transformer Relay Settings
ELEMENT PICK-UP CURRENT SETTINGS
51 IPrimary−Rated ∗ 1.2CT Ratio
= 331.69∗1.2
70 = 5.69 (A)
50 IThrough− Primary−Rated ∗ 1.2CT Ratio
= 3346.96 ∗ 1.270
= 57.38
50G IThrough− Primary−Rated CT Ratio
= 3346.96
70 = 47.81 (A)
The pick- up setting were tabulated in Table 5 for easy convenient easy reading to program
the relay in ASELerator. The 1.2 multiplier used in the pickup current for the 50 and 51
relays accounts for current imbalance between phases. This multiplier was not used for the
50
50G relay because this relay is only monitoring current on the ground connection of the
Wye transformer winding.
Table 6: 40 MVA Transformer Relay and CT Data Settings
CT RATIO SLOPE 1 SLOPE 2 TAP DIFF CURRENTS
CTR1 350/5 CTR2 1600/5 CTRN 250/5 Point 1 (0,0) (6, 1.5) Point 2 (6, 1.5) (10, 4.3) Slope % 25% 70% TAP 1 4.78 TAP 2 9.06 Idiff > 0.3 Idiff >> 10
Table 6 shows the data computed in chapter two for convenient and easy reading to
program the relay in ASELerator and in Test Universe.
51
3.24 DRAWINGS
GND
5287
52
50/51
50N
5287
52
50/51
50N
52
50/51
52
50/51
52
50/51
52
50/51
52
50/51
52
50/51
T1: 30MVA69/13.8 kVZ = 8%
T2: 40MVA69/13.8 kVZ = 10%
15 MVA 15 MVA 20 MVA 20 MVA
69 kV
13.8 kV 13.8 kVNO
Legend:50 = Tine overcurrent Relay51 = Instantaneous Relay50N = Ground Relay52 = Circuit Breaker 87 = Differential RelayGND = Ground T1 = Transformer 1T2 = Transformer 2
Figure 6: Single Line Diagram
52
CT = 250/5A
B
C
a
b
c
IC
IA
IbIb
Ic
Legend:CT = current transformer50 = Tine overcurrent relay51 = Instantaneous relay50N = Ground relayOP = Operating CoilGND = Ground
30 MVA Transformer
IB
Ic
50/51 Relay
CT = 250/5
CT = 250/5
IBIB
IAIA IaIa
OP 50NRelay
GND
CT = 250/5
OPOP
OPOP
OPOP
Phase C
Phase B
Phase A
Figure 7: 30 MVA Transformer 51,50 and 50N CT Connections
Figure 7 was created from a combination of generic examples in [12], and [13]. Such
Figure, reflects the specific application of the project depicted in Figure 6.
53
CT = 250/5 CT = 1200/5A
B
C
a
b
c
IA
IB
IC
IA-IB
IB-IC
IC-IAIC-IA
IA-IB
IB-IC
Ia
IbIb
IcIC-IA
Ic-Ia
Ia-Ib
30 MVA Transformer
CT = 250/5
CT = 250/5
CT = 1200/5
CT = 1200/5
Ib-Ic
`
IBW1 ICW1IAW1IAW1 IBW2 IAW2ICW2ICW2
Differential Reay
GND
GND
Legend:CT = current transformerGND = Ground Differential Relay = SEL-787 IAW1= Phase A current, Winding 1, IBW1 = Phase B current, Winding 1 ICW1 = Phase C current, Winding , IAW2 = Phase a current, Winding 2IBW2 = Phase b current, Winding 2, ICW2 = Phase c current, Winding 2
Figure 8: 30 MVA Transformer Differential CT Connections
Figure 8 was created from a combination of generic examples in [12], [13] and [17]. Such
Figure, reflects the specific application of the project depicted in Figure 6.
54
CT = 350/5A
B
C
a
b
c
IC
IA
IbIb
Ic
Legend:CT = current transformer50 = Tine overcurrent relay51 = Instantaneous relay50N = Ground relayOP = Operating CoilGND = Ground
40 MVA Transformer
IB
Ic
50/51 Relay
CT = 350/5
CT = 350/5
IBIB
IAIA IaIa
OP 50NRelay
GND
CT = 250/5
OPOP
OPOP
OPOP
Phase C
Phase B
Phase A
Figure 9: 40 MVA Transformer51, 50 and 50N CT Connections
Figure 9 was created from a combination of generic examples in [12], and [13]. Such
Figure, reflects the specific application of the project depicted in Figure 6.
55
CT = 350/5 CT = 1600/5A
B
C
a
b
c
IA
IB
IC
IA-IB
IB-IC
IC-IAIC-IA
IA-IB
IB-IC
Ia
IbIb
IcIC-IA
Ic-Ia
Ia-Ib
40 MVA Transformer
CT = 350/5
CT = 350/5
CT = 1600/5
CT = 1600/5
Ib-Ic
`
IBW1 ICW1IAW1IAW1 IBW2 IAW2ICW2ICW2
Differential Reay
GND
GND
Legend:CT = current transformerGND = Ground Differential Relay = SEL-787 IAW1= Phase A current, Winding 1, IBW1 = Phase B current, Winding 1 ICW1 = Phase C current, Winding , IAW2 = Phase a current, Winding 2IBW2 = Phase b current, Winding 2, ICW2 = Phase c current, Winding 2
Figure 10: 40 MVA Transformer Differential CT connections
Figure 10 was created from a combination of generic examples in [12], [13] and [17].
Such Figure, reflects the specific application of the project depicted in Figure 6.
56
CHAPTER IV
TEST PROCEDURE AND PLAN 4.1 TEST PROCEDURE
A step-by-step test procedure was written using AcSELerator and Test Universe Software.
The objective of the procedure is to test the 51, 50, 50G and 87 relay and CT settings and
ratings computed and established in chapter two for the 30MVA transformer. In the
AcSELerator library, a 787 relay was selected to test the four relay elements. The data to
program this relay were aggregated from chapter two in tables 7 and 8 below. Using the
SEL-787 and the OMICRON CMC-356 manuals, an instructional connection procedure
and a wiring diagram were and included in this test procedure. The test procedure to test
the data computed for the 40 MVA transformer would be identical as for the 30 MVA
transformer. The data and parameters for the 40 MVA would be different can be found in
tables 4, 5 and 6 in chapter two.
4.2 RELAY SETTINGS
Table 7: 30 MVA Relay and CT Data Settings and Ratings
CT RATIO SLOPE 1 SLOPE 2 TAP DIFF CURRENTS
CTR1 250/5 CTR2 1200/5 CTRN 250/5 Point 1 (0,0) (6, 1.5) Point 2 (6, 1.5) (10, 3.9) Slope % 25% 60% TAP 1 5.02 TAP 2 9.06 Idiff > 0.3 Idiff >> 10
57
Table 8: 30 MVA Transformer Relays Pick-up Currents
ELEMENT PICK-UP CURRENT SETTINGS
51 IPrimary−Rated ∗ 1.2CT Ratio
= 25102∗1.2
50 = 6.02 (A)
50 IThrough− Primary−Rated ∗ 1.2CT Ratio
= 3,137.02 ∗ 1.250
= 75.3 (A)
50G IThrough− Primary−Rated CT Ratio
= 3,137.02
50 = 62.75 (A)
4.3 WIRING CONNECTION PROCEDURE
STEP PROCEDURE 1 Connect ports 1 and 2 from the Binary Input/output of the OMICRON to the
output terminals A03 and A04 of the relay, Figure 1. Through this connection, the relay signals the OMICRON to stop supplying current after a trip signal has occurred.
2 For the primary side of the transformer, connect the “CURRENT OUTPUT A” of the OMICRON to the “CURRENT INPUTS” of the relay to the as follow, Figure 1.
1. Port 1 of the “CURRENT OUTPUT A” to Z01 of the relay 2. Port 2 the of “CURRENT OUTPUT A” to Z03 of the relay 3. Port 3 the of “CURRENT OUTPUT A” to Z05 of relay 4. Port N the of “CURRENT OUTPUT A” to Z02, Z04 and Z06 of the relay
3 For the Secondary side of the transformer, connect the “CURRENT OUTPUT B” of the OMICRON to the “CURRENT INPUTS” of the relay to the as follow, Figure 1.
1. Port 1 of the “CURRENT OUTPUT B” to Z07 of the relay 2. Port 2 the of “CURRENT OUTPUT B” to Z09 of the relay 3. Port 3 the of “CURRENT OUTPUT B” to Z11 of the relay 4. Port N the of “CURRENT OUTPUT A” to Z08, Z09 and Z12 of the relay.
4 Connect the serial cable from Port 3 on the back of the relay to the USB port of the computer.
5 Connect the power cable on the back of the relay terminals, A01, A02, and ground.
58
4.4 WIRING DIAGRAM
A01
A02
A03
A04
A05
A06
A06
A07
A08
A09
A10
A11
A12
A12
A07
A08
A09
A10
A11
A12
`Port 1
+/H - /N
GND
Port 2TX RX Port 3
E01
E02
E03
E04
E05
E06
E06
E07
E08
E08
C09
C10
C11
C12
C13
C14
C14
C15
C16
C16
C09
C10
C11
C12
C13
C14
C15
C16
C01
C02
C03
C04
C05
C06
C06
C07
C08
C08
C01
C02
C03
C04
C05
C06
C07
C08
D09
D10
D11
D12
D13
D14
D14
D15
D16
D16
D09
D10
D11
D12
D13
D14
D15
D16D01
D02
D03
D04
D05
D06
D06
D07
D08
D08
D01
D02
D03
D04
D05
D06
D07
D08
E07
E08
E08
1 2 3 N
1 2 3 N
CURRENT OUTPUT A
CURRENT OUTPUT B
BINARY / ANALOG INPUT (CAT III / 300 V)
1 2 3 N 4 N
VOLTAGE OUTPUT AUXDC
1 2 3 4
BINARY OUTPUT ANALOG DC INPUT
0 ± 20 mA 0 ± 10V
OMICRON
CMC-356
SCHWEITZER
SEL-787120 VAC
1 2 3 4 5 6 7 8 9 10
I
O
POWER
Z07
Z08
Z09
Z10
Z11
Z12
Z01
Z02
Z03
Z04
Z05
Z06
Figure 11: SEL-787 and OMICRON CMC-356 Wiring Diagram
59
4.5 PROGRAMMING THE SEL-787 RELAY
6 Double click on the AcSeLerator Ikon on the desk top (Figure 12) 7 Click on new (Figure 13). 8 Select the SEL-787 under Device Family and Device Model, and 003 under
Version, and click OK (Figure 13).
Figure 12 Figure 13
Figures 12: AcSELerator Software Access Tab
:
Figures 13: AcSELerator Setting Access Tab
60
9 To establish communication with the SEL-787 relay click on Communication (Figure 14). The Communications Parameters in Figure 14 will appear.
• Select Serial. Under Device form the drop down menu select COM1: Communications Port, 19200. Click on Apply, and Ok
Figure 14: Communications Parameters
61
Figure 15: Setting Editor Selection
10 To configure the SEL-787 Reay, click on Group 1, and on Configuration. Under the Configuration scree enter the following data as shown in Figure 16. MVA = 30 Winding 1: W1CT = WYE, CTR1 = 50, VWDG1 = 69 Winding 2: W2CT = DELTA, CTR1 = 250, VWDG2 = 13.8 NEUTRAL CT RATIO: CTRN1 = 50
62
Figure 16: AcSELerator Transformer Configuration
63
11 To program the 51 element, click on Group 1, Set 1, Time Overcurrent Elements, Winding 1, and Phase Time Overcurrent. Under 51P1P enter 6.02 and under 51PC1 select U2 (Figure 17). The calculations for this data can be found in Table 2 of chapter 2.
Figure 17: Phase Overcurrent Relay Configuration
12 To program the 50 element, click on Group 1, Set 1, Overcurrent Elements, Winding 1, and Phase Overcurrent. Under 50P11P enter 75.30 and under 51P11D select 0.00, Figure 18. The calculations for this data can be found in Table 2 of chapter 2.
64
Figure 18: Phase Overcurrent Relay Configuration
13 To program the 50G element, click on Group 1, Set 1, Overcurrent Elements, Winding 2, and Neutral Overcurrent. Under Element 2, and under 5N12P type 62.75 and under 50N12D type 0.00. The calculations for this data can be found in Table 8 of chapter 2, Figure 19.
Figure 19: Ground Overcurrent Relay Configuration
65
14 To program the 87 element, Figure 20. • Click on Group 1, Set 1, and Transformer Differential Elements. • Under E87, select Y for yes form the drop down menu. • Under TAP1, Winding 1 type 5.02 • Under TAP2, Winding 2 Current Tap type 9.06 • Under O87P, select 0.30, this is the minimum pick up • Under SLP1, type 25 • Under SLP2, type 60 • Under IRS1, type 6.0. This is the default Restrain current in per unit of
tap • Under U87P, type 10. This is the default Unrestraint current in per unit
of tap
Figure 20: Differential Relay Configuration
66
15 Next, the transformer Trip and Close Logic equations need to be set (Figure 21).
• Click on Group 1, Set 1, and Trip and close logic. Because the transformer was configured before, the equation for the differential element, under TRXFMR Trip XFMR Equation (SELLogic) is set automatically by the program as 87R OR 87U.
• The equation for the 51, 50 and 50G have to be entered under Trip 1 Equation (SELogic) as 50P11T OR 51P1T OR 50G11T
Figure 21: Trip and Logic Configuration
67
16 The circuit breaker trip contacts need to be program (Figure 22). In a real application, this output would be wired directly to the circuit breaker.
• Click on Group 1, Set 1, Logic 1, Outputs, and Slot A. Slot A has three output contacts and any of them can be used. For this application Out102 was selected arbitrarily
• Under OUT102 select Y from the drop down menu • Under OUT102 (SELogic) type TRIPXFRM OR TRIP
17 Finally, the setting have to be send to the relay. Click on Sending Active Settings and On Terminal as shown in Figure 12. The AcSELertor screen can now be minimized to program the OMICRON 356.
Figure 22: Circuit Breaker Trip Contacts
68
4.6 SETTING THE 50/51 AND 50 REAYS IN TEST UNIVERSE
18 The OMICRON is a current generator and will be used to inject secondary current on the relay to test it. To program the OMICRON, click on Test Universe (Figure 23) on the desktop.
19 Once the main screen opens, click o QuikCMCm (Figure 24).
Fig 23 Fig 24
Figure 23: Test Universe Access Tab
Figure 24: QuickCMC Access Module
69
20 Under the Home tab, click on Hardware Configuration Test Setup (Figure 25) 21 Click on Details (Figure 25).
Figure 25: Hardware Configuration Test Setup
70
Figure 26: Hardware Configuration Access
71
22 Under voltage outputs select <not Used> (Figure 27). 23 Under Current Outputs select select 6 x32A, 430VA @25A, 25Vrms
(Figure 27). 24 Select not used under Voltage Outputs and under Current Outputs select 6
x32A, 430VA @25A, 25Vrms and click OK (Figure27). The 6 x32A, 430VA @25A, 25Vrms is the rating of each port of the of the Current Outputs A and B of the OMICRON.
Figure 27: Output Configuration Currents
72
25 Under Analog Outputs, enter the magnitude currents separated by of 120o phase shift, and a 60 Hz frequency for each current. Set the delay to 200 ms, the Step Ramp to 100 mA and choose IL1, IL2, IL3. To find where the different elements would trip, the current can be ram-up by clicking on the arrows below Auto Step, Figure 28.
Figure 28: Testing the Overcurrent Elements
73
26 Once the relay has tripped, a report can be obtained by clicking on Report View (to the right of the test Screen (Figure 29).
Figure 29: Test Results Example
74
4.7 SETTING THE 87 ELEMENT IN TEST UNIVERSE
27 To program the Test Universe to test the 87 element, click on Test Universe (Figure 30)
28 Click on Differential (Figure 31)
Figure 30 Figure 31
Figures 30: Universe Access Tab
Figures 31: Differential Module Access Tab
75
29 Under the File tab, click on Test Object (Figure 32) 30 Double click on Differential (Figure 32).
Figure 32: Test Object Access Tab
76
Figure 33: Differential Access Screen
77
31 Select Protected Object (Figure 34). • Under Protected Object select Transformer from the dropdown menu • Under Vector Group, select DY1. This indicates Delta configuration on
the primary side and Wye on the secondary side with a 300 phase shift • Under Nominal Values enter 69kV for the primary side and 13.8 kV for
the secondary side • Enter 30 MVA for the power on the primary and secondary sides • For the Vector Group enter D for primary side and Y1 for the
secondary side. For the Starpoint Grounding select Yes from the drop down menu on the secondary side of the transformer
Figure 34: Differential Protection Configuration
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32 Select CT (Figure 35). • Under CT Nominal Values enter the CT Primary and Secondary
Currents in this case is 250/5 for the primary side and 1200/5 for the secondary side
• Under Ground CT Nominal Values enter the ground CT ratings. In this case is 250/5
Figure 35: Differential CT Configuration
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33 Select Protection Device (Figure 36). • Set the Factor K1 to 1 • Under Diff Current Settings set Idiff > to 0.3 and Idiff > > to 10. These
are the Restrain and Unrestraint values also programed in the AcSELrator under Transformer Differential Element, see Figure 10
Figure 36: Protection Device Configuration
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34 Select Characteristic Definition (Figure 37). • Set the Start Point at (0.0, 0.0) and the end point at (6.0, 1.50). Based
on this two points, the slope of 0.25 will be automatically calculated. Ensure the two point as correct under the Defined Segments. The program will plot the graph to the right based on these two points. To see the graph on the Idiff-Ibias plain click on Add
Figure 37: Slope 1 Characteristic Definition
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35 Slope 2 is also programmed under the same Characteristic Definition (Figure 38)
• To set Slope 2, under Define Segments highlight the end point for Slope 1, click on Cut from here and then click on Add. The program will use this point as the starting point for Slope 2
• Under End point, enter the point (10, 3.9). The program will automatically compute the 0.6 Slope for Slope 2
• Click Add to see both Slopes on the Idiff-Ibias plain, click on Add. • Click on OK to exit this module and to enter the test module
Figure 38: Slope 2 Characteristic Definition
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36 After exiting the Characteristic Definition in Figure 38, the program will display the Operating Characteristic Diagram. The 87 relay will operate in the Trip Region, Figure 39. The trip region is above the plotted slopes and below the Maximum Torque Line. The solid blue ramps represent Slopes 1 and 2. The dash lines below and above the solid blue line represent the relay tolerance.
Figure 39: Operating Characteristic Diagram
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37 Figure 40 shows the 87 relay does not trip for current under the slope lines, the Non Trip region. This indicates currents outside the protected zone.
Figure 40 : Non-Trip Test View
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38 Figure 41 shows the 87 relay trips for differential currents in the Trip Region. The Trip Region above the Slope solid lines and under the Maximum Torque Line.
Figure 41: Trip Test View
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4.8 TEST PLAN
This test plan is intended to work in conjunction with the test procedure to test the relay
elements protecting power transformer. The job plan provides all the steps needed to
program the SEL relay and the OMICRON current generator and to test the 51, 50, 50G
and 87 elements. All the data to program the SEL relay and OMICRON were computed
in chapter 2.
In the absence of a project specific coordination study, the U2 TCC curve and the 0.5 Dial
(Figure 42) can used to explain the inverse time characteristic operation of the 51 relay and
the instantaneous operation of the 50 and 50G relay elements. Both the horizontal and
vertical axes in Figure 42 have a set of values ranging from 0.5 to one hundred in log-log
scale. On the horizontal-axis, these values represent a multiple of the Tap Current. In the
Figure, these values are referred to as Multiples of Pickup. On the vertical-axis, these
values represents the time in seconds that would take the relay to operate for any of these
Multiples of Pickup multiplied by the Tap Current (Multiples of Pickup x Tap Current).
In conclusion, the relay depends on these two paraments to trip:
• the Multiples of Pickup multiplied by the Tap Current on the horizontal axis of
Figure 42
• and the time, in seconds, that would take the relay to trip for any of these Multiples
of Pickup multiplied by the Tap Current
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Figure 42: TCC Curve from [12]
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4.9 PROGRAMMING AND CONFIGURING THE SEL RELAY
The equipment needed to connect, energize and configure the SEL relay and OMICRON
current generator are: the SEL-787 Relay, the Omicron CMC-356, flexible cables with safe
tip connectors, a 9 pin serial cable, power cables, a computer, AcSELerator and test
Universe Software.
1. Wire the SEL 787 relay with the OMICRON by using the Wring Connection
Procedure and the Wiring Diagram in the test procedure. Ensure all the connections
are correct before continuing.
2. Connect both the OMICRON and the SEL -787 to 120 VAC power source and turn
their power switches to the ON position. Verify that both unites are ON.
3. From the desk top, open AcSELerator. A new window titled “Settings” will appear.
4. From the Settings window, select Communication. A new window called
“Communications Parameter” will appear. Under the Active Connection Type,
select the Serial tab. Under Device, select COM1: Communications Port. Under
Data Speed select 19200, click on Apply and on OK
5. From the Settings window, click on New. A “Settings Editor Section – Settings
Database” window will appear. From the Device Family and the Device Model,
select SEL-787. Under Version, select 003. Click OK.
6. A new window called “Device Part Number” will appear. Enter the correct part
number of the SEL relay in order to activate all the modules needed to configure
the relay, click on OK.
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7. A new window called “SEL 787 Settings” will appear. From the left-hand side
of this window select: Group 1, Set 1, and then click on Configuration to access the
Configuration page to enter the power transformer and CT parameters.
8. Under the Configuration page:
• Under MVA Maximum Transfer Capacity (MVA) enter the transformer rated
30 MVA
• Under Winding 1 and below W1CT Winding 1 CT Connection, select WYE.
Under the CTR1 Winding 1 CT Ration enter 50. Under VWDG1 Winding 1
Line-to-Line Voltage (kV) enter 69 kV.
• Under Winding 2 and below Winding 2 CT Connection, select Delta. Under
CTR2 Winding 2 Phase CT Ratio, select the CT renter 240. Under the VWDG2
Winding 2 Line-to-Line Voltage (kV), enter 13.8 kV
9. Under CTRN1 Neutral (1N1) CT Ratio, enter 50. Under Group 1, Set 1, select
Time Over Current Element. Then, select Winding 1 and Phase Tine Overcurrent.
A new window on the right-hand side called “Phase Tine Overcurrent” will
appear. Under the 51P1P Winding 1 Max Phase Time Overcurrent Trip Level
(amps), enter 6.02 Amps. Under 51P1C Winding 1Max Phase TOC Curve
Selection, select U2.
10. Under Group 1, Set 1, select Over Current Elements. Then, select Winding 1 and
Phase Overcurrent. A new window will appear on the right-hand side called “Phase
Overcurrent”. Under Element 1 and under the 50P11P Winding 1 Phase Inst
Overcurrent Trip Level (amps), enter 75.30 Amps. Under 50P11D Winding 1
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Phase Inst Overcurrent Trip Delay (seconds), enter 0.00. Under 50P11TC Winding
1 Phase Inst Overcurrent Torque Control (SELogic), enter 1.
11. Under Group 1, Set 1, select Over Current Elements. Then, select Winding 2 and
Neutral Overcurrent. A new window will appear on the right-hand side called
“Neutral Inst Overcurrent”. Under Element 2 and under the 50P11P Winding 1
Phase Inst Overcurrent Trip Level (amps), enter 75.30 Amps. Under 50N12P
Neutral Inst Overcurrent Trip Level 1 (amps), enter 62.75 Amps. Under 50N12D
Neutral Inst Overcurrent Trip Level Delay (seconds) enter 0.0. Under 50N12TC
Neutral Inst Overcurrent Torque Control (Set Logic), enter 1.
12. Under Group 1, Set 1, select Transformer Differential Elements. A new window
will appear on the right-hand side called “Transformer Differential Elements”.
Under E87 Enable Transformer Differential Protection, select Y. Under TAP1
Winding 1 Current Tap (Auto Calculated), make sure it is set to 5.02 Amps. Under
TAP2 Winding 2 Current Tap (Auto Calculated) make sure it is 9.06 Amps. Under
O87P Restrain Element Operating Current PU (multiple of tap), enter 0.3. Under
SLP1 Restrain Slope 1 Percentage, enter 25. Under SLP2 Restrain Slope 2
Percentage enter 60. Under IRS1 Restrain Current Slope 1 limit enter 6.0. Under
U87P Unrestrained Element Current PU, enter 10.
13. Click on Group 1, Set 1, select Trip and Close Logic. A new window will appear
on the right-hand side called “Trip and Close Logic”. Under TRXFMR Trip
XFMR Equation (SELlogic) type 87R OR 87 U. Under Element 1 and under TR1
Trip 1 Equation (SELogic), type 50P11T OR 51P1T OR 50G11T.
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14. Click on Group 1, Set 1, Logic 1, Output and select Slot A. A new window will
appear on the right-hand side called “Slot A”. Under OUT102 and below
OUT102FS OUT102 Fail Safe, select Y. Under OUT102 (SELogic), select
TRXFMR OR TRIP.
4.10 PROGRAMMING AND CONFIGURING THE OMICRON
1. From the desk top, open Test Universe.
2. The Test Universe screen will appear. Under Test Modules, select QuickCMC
3. A new window will appear. Click on the Hardware Configuration tab.
4. Under General and under Test Set(s), click on Details
5. Under CMC356 (??????) Voltage Outputs, select <not used>. Under CMC356
(??????) Current Outputs, select 6 x 32A, 430 VA @25A, 25 Vrms. click on Ok,
Apply and on OK
6. On the left-hand, Under Test View QuickCMC1 and under Analog Outputs, set the
currents for the primary winding as follow:
• IL1 = 5.02, angle 0, 60 Hz
• IL1 = 5.02, angle -120, 60 Hz
• IL1 = 5.02, angle 120, 60 Hz
7. Under the same Test View QuickCMC1 and below the “On Trigger”, set the delay
to 200 ms.
8. Under the same Test View QuickCMC1 and under Step/Ramp, set:
• the Delay to 200 ms
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• set the Signal(s) to IL1, IL2, IL
• se the Magnitude to Quantity
• set the Size to 100 mA
4.11 MANUAL “INVERSE TIME OVER CURRENT” RELAY TEST (51)
As an example, Table 9 below shows a set of arbitrarily CT secondary calculated currents.
These are the currents the CT selected of 250:5 Amps would read on its secondary winding
leads. To calculate these currents, the computed primary rated current of 251.02 Amps, the
Multiples of Pickup from Figure 42 and the CT ratio of 50 (250/5 = 50) were used. The 51
relay would take a different amount of time to trip for these different currents, the higher
the current, the less time it would take to trip.
The CT secondary currents and their corresponding trip times were organized on the same
table in two separate columns for easy reading. The U2 TCC curve and the 0.5 Dial were
used to obtain the inverse time characteristic of the 51 relay as shown in Figure 42. A Trip
(Y/N) column was added to indicate whether or not the relay for these CT secondary
currents and times would trip.
The same three columns were copied and right of the filled columns and were left blank to
be filled with the actual tested data.
To begin testing of the element, increase the magnitude currents manually by clicking on
the upward arrow below the “Auto Step” tab. Use Table 9 to fill in the record the results
of the test
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Pick-Up = 5.02 (Amps)
Computed CT Secondary (A)
Estimated Trip (s)
Trip (Y/N)
Actual CT Secondary (A)
Actual Time (s)
Trip (Y/N)
5.02 No Trip N7.53 1.5 Y
15.06 0.35 Y22.59 0.25 Y35.14 0.15 Y
1) 1*251.02 /50 = 5.02 (A)2) 1.5*251.02 /50 = 7.53 (A)3) 3*251.02 /50 = 15.06 (A)4) 4.5*251.02 /50 = 22.59 (A)5) 7*251.02 /50 = 35.14 (A)
1) 5.02 * 50 = 5.02 (A)2) 7.53* 50 = 375(A)3) 15.06* 50 = 753 (A)4) 22.59 * 50 = 1125.5 (A)5) 35.14 * 50 = 1757 (A)
30 MVA TRASFORMER RELAY TEST FORM (51)
CT RATIO = 250:5 on the Primary Side of T1
The CT ratio = 50
3 x Pickup
1 x Pickup
Pick-Up
1.5 x Pickup
4 .5x Pickup7 x Pickup
51Using the U2 TCC Curve and the 0.5 Dial of the 787 SEL Manual
Computed CT Primary (A) = Computed CT Secondary (A) * CT RatioNOTE: The Primary current seen by the CT is only computed for refernce
Computed CT Secondary (A) = (Pick-up Multiple*I-primary rated currnet) / CT Ratio
NOTES: The Primary I-rated current was computed as 251.02 Amps, and the Tap was computed as 5.02 Amps. For this reason, the CT = 250/5 was selected.
Table 9: 51 Relay Test Form
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4.12 TESTING THE 51, 50 AND 50G ELEMENTS
The SEL 787 relay and OMICRON have been programed with all the calculated data from
chapter two and the manual test of the 51, 50 and 50G elements can begin.
4.13 MANUAL “INSTANTANEOUS OVERCURRENT” RELAY TEST (50)
As a reference to test the 50 element, an example is shown on Table 10. The
recommendation in the SEL manual to set the instantaneous current as the primary through
fault current increased by 120% was used (Through Fault Current x 1.2). This is different
than the 150% to 200% of through fault current that IEEE recommends [2]. Due to the
lack of resources, a specific coordination study could not be conducted to determine the
best value for the instantaneous current. Instead, the SEL U2 TCC curve using the 0.5 Dial
was used to test this relay. Two columns were left blank: “Tested Instantaneous Fault
Current (A) and “Trip (Y/N)” to be filled with the actual tested magnitude of fault current
and to state whether or not the relay tripped.
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Computed CT Secondary (A)
Computed Intantaneous Current (A)
Trip (Y/N)
Tested Instantaneous Fault Current (A)
Trip (Y/N)
75.3 3,765.32 Y
CT Secondary Pickup @ Instantaneous Current = (3137.77 *1.2)/50 = 75.3 (Amps)
Instantaneous Current = 3,765.32
CT Secondary Current Pick-Up @ Instantaneous Current CT Secondary Pickup @ Instantaneous Current = (I-Though Pri * 120%)/CT Ratio
30 MVA TRASFORMER RELAY TEST FORM (50)
CT RATIO = 250:5 on the Primary Side of T1
Instantaneous Current = I-Though Pri * 120%Instantaneous Current = 3,137.77 *1.2
Using the U2 TCC Curve and the 0.5 Dial of the 787 SEL Manual
Computed Pickup Curent = 5.02 (Amps)
15 * Pick up
Pick-Up
Pickup = 5.02 (Amps)
NOTES: The through fault current was computed as 3,737.77
Table 10: 50 Relay Test Form
4.14 MANUAL “GROUND FAULT” RELAY TEST (50G)
As a reference to test 50G element, an example is shown on Table 11. Because this relay
has a s similar functionality as the 50 relay, the same principle as that of the 50 relay test
can be used to test the 50G relay. Since the 50G relay is monitoring the ground current on
the ground of the power transformer, there is no need to take into account a multiplier like
in the 50 relay where a 120% multiplier was used. The U2 TCC curve using the 0.5 Dial
was used to work the example on Table 12.
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Ground Fault Current (A)
Trip (Y/N)
Tested Fault Current (A)
Trip (Y/N)
3,737.77
Ground CT Current Pick-up = 3,137.77 / 50 = 62.76 (Amps)
30 MVA TRASFORMER RELAY TEST FORM (50G)
CT RATIO = 250:5 on the Primary Side of T1Using the U2 TCC Curve and the 0.5 Dial of the 787 SEL Manual
NOTES: The Primatry I-Though was used to test the 50G element Ground CT Current Pick-up Ground CT Current Pick-up = I-Though Primary / CT Ratio
3,737.77
Computed Through Fault Current (A)
Table 11: 50G Relay Test Form
4.15 PROGRAMMING THE DIFFERENTIAL RELAY IN OMICRON
1 From the desk top, open Test Universe.
2 The “Test Universe” screen will appear. Under Test Modules, select: Differential,
Advance Differential, and Diff Operating Characteristic
3 A new window will appear. Click on the Hardware Configuration tab.
4 Under General and below Test Set(s), click on Details.
5 Under CMC356 (??????) Voltage Outputs, select <not used>. Under CMC356
(??????) Current Outputs, select 6 x 32A, 430 VA @25A, 25 Vrms, click on Ok,
click on Apply and on OK.
6 On the let top left-had side, click on Test Object
7 The Test Object window will appear, double click on Differential.
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8 The differential Protection Parameters window will appear. Under the protected
Object Tab select Transformer. Under the Number of Windings, select 2. Under
Nominal Values and under the Primary column:
• For the primary Winding/ Leg Name, type Primary
• For Voltage, type 69 kV
• For Power, type 30 MVA
• For Vector Group, type D
For the secondary Winding/ Leg Name, type Secondary
• For Voltage, type 13.8 kV
• For Power, type 30 MVA
• For Vector Group, type Y1
• For the Secondary Winding. select Y
9 Click on the CT tab. Under CT Nominal Values. enter 250 Amps for the primary
current and 5 Amps for the secondary current. Under the secondary enter 1200
Amps for the primary current and 5 Amps for the secondary.
10 Click on the Protection Device Tab. For Factor K1, select 1. Under Diff Current
Settings set Idiff> to 0.3 and Idiff>> to 10.
11 Select the Characteristic Definition tab. Under the Start Point, for Ibias type 0.00
and for Idiff type 0.00. Under the End Point, for Ibias type 6.00 and for Idiff type
1.50. The program will automatically calculate Slope 1. Click on the Cut from
here and on click on the Add tabs.
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12 With the mouse, select the point under Define Segments. Click on the Cut from
here and on the Add tabs.
13 Under the End Point, for Ibias type 10.00 and for Idiff type 3.9. Click on the Add
tab. The program will automatically calculate Slope 2.
14 Click on the Harmonics tab. From the drop down menu, select 2. For the Ixf/Idiff
type 15.00%. For the Tol. relative type 3.00% and for the Tol. absolute type
1.00%. Click on the Ixf/Idiff to update the entries.
15 Repeat step 14 for the 4th and 5th harmonics. Click OK to access the Test View
Window.
16 Under Test View, select the Short Test tab.
17 Under Fault Type, select L1-L2-L3.
18 The program will display the two slopes previously programmed on the Ibias – Idiff
plain, under the Operating Characteristic Diagram
4.16 TESTING THE 87 ELEMENT
At this time the 87 relay has been programmed in both AcSELeartor and OMICRON and
is ready to be tested. Two tests were explored in this job plan: a Point-By-Point test and an
Automated Test.
4.17 POINT-BY-POINT TEST OF THE 87 RELAY
1 Click on the Trip Region of the Ibias – Idiff plain. Under the Operating
Characteristic Diagram, click on the Add tab. Under the Short Test and Test
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Points (to the left of the Ibias – Idiff plain). The program will display the values
associated with that test point and the time it takes for the relay to operate.
2 The previous step can be repeated for more points to be tested.
1 Points in the non-trip region can be selected to ensure the proper operation of the
relay. The results of any of the Point-By-Point test on Table 13 bellow.
4.18 AUTOMATED MULTIPLE POINT TESTOF THE 87 RELAY
1. To perform a automated multiple test point test, under the Test View window, click
on Test Search, and click on the Add Sweep Tab. An Add Multiples Window will
appear.
• For the Start Value, type the tap value, 5.02 Amps
• For the End Value, enter a value form 0 to 10
• For the Step Size select the Tap Value, 5.02 Amps
• Click on Add to Table
2 The program will run an automated test of several point. The Start Value, End
Value and Step Size can be changed depending on the desired test criteria. The
results of any of this test recorded on Table 12 bellow.
Idiff Point Ibias Point tactual tnominal
Table 12: 87 Relay Test Form
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CHAPTER V
CONCLUSION
This transformer protection project involved a model of two power transformers fed from
69 kV common bus. The rating of the transformers are 30 MVA with 8% impedance and
40 MVA with 10% impedance respectively. Each of these transformers stepped down the
69 kV voltage to 13.8 kV, feed their own loads and can tied on the low side via high voltage
switch. The power transformers’ rated and through fault currents were computed for each
transformers. The intend of the project is to protect both transformers with four different
relays schemes, the 51, 50, 50G and 87.
The 51 and 50 relays were used on the high side of the power transformers. The 51 to
protect the power transformers from external faults and the 50 from internal faults. The
50G relay was used on the ground side of wye power transformers to protect them from
ground faults. The 87 relay was to protect the power transformers from internal faults by
comparing the high and the low side rated currents of the power transformers. The
computed rated currents on the high side and low side of the power transformers were used
to size the CTs. The rated currents were multiplied by the inverse of the CTs selected turn
ratios to insure the correct the secondary current from the CTs would come to the relay.
Through fault currents were computed on the high and low voltage sides of both power
transformers and multiplied by the inverse turn ratios of the CTs selected to ensure the
guideline of 20 x the nominal CT secondary would not be exceeded.
To ensure the level of burden met the relay specifications, the CT’s burden voltage and
voltage saturation were computed by multiplying the 3-phase short circuit current through
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the CTs on the high, low and ground side of the power transformers. The total burden was
summed as: the CTs secondary winding, the lead resistance, the hard wire resistance form
the CT circuit secondary terminals to the relay panel, and the relay resistance. The
computed burden voltage in section three of the SEL-787 instruction manual was used to
insure the CT burden voltage was less than 0.5 of the CT voltage rating.
The differential relay tap setting were calculated using the same Schweitzer’s relay 787
instruction manual. The guidelines used ensured two conditions were met: 1) the tap values
had to be within the range of 0.1 x In and 31 .0 x In, and 2) the ratio of TAPmax/TAPmin
had to be ≤ 7.5.
The guidelines in the same section of the SEL 787 instruction manual were used to insure
the operating minimum current pick up was ≥ (0.1 x In)/(TAP min) in order to provide
the maximum sensitivity for differential protection while maintaining secure operation of
the power transformers. The pickup values were computed by following the instruction
manual recommendation of a minimum of 0.3 Amperes pickup value.
To address the CT differential mismatch to test the 87 element, a slope percentage was
computed by using a formula from differential relay a lab procedure. This formula consists
of the absolute difference of the primary and secondary power transformer’s currents
divided by the summation of the same quantities multiplied by 100. For the 30MVA
transformer, Slope 2 was computed as 45.2%. For the 40MVA transformer, Slope 2 was
computed as 68.5%. Slope 1 was used between the minimum pick-up and the Slope 2 break
point. Slope 1 provides security against nuisance tripping as a result of CT inaccuracy.
Class C CT accuracy is +/-10%, thus a 20% setting was used. Slope 2 was set above its
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own break point. Slope 2 provides security against nuisance tripping for CT saturation
cases [17].
A model was created in ETAP, however a coordination study was not possible because
the program does not have the library needed to conduct a coordination study to determine
if the setting and rating of the equipment selected were appropriate for this application. A
current time interval (CTI) of 3.5 seconds would have been used between relays to insure
proper coordination between the relay elements. The CTI would had been increase to 4
seconds if the initial 3.5 seconds proved poor time current coordination between the relays.
Finally, a test procedure and a test plan for the 30 MVA were written to test the data
computed in chapter two for this project model. The Schweitzer AcSELerator and the Test
Universe proprietary software programs were used to write the procedure. The data used
for this procedure was computed in chapter two, and organized in Tables 1, 2 and 3 in the
same chapter. Based on the instruction manual, the SEL 787 relay is capable of performing
input current phase compensation for each winding when this function is enabled
performing input current phase compensation for each winding when this function is
activated, i.e., the Delta winding, W1 CTC = 0, for 0° and Wye winding, W2 CTC = 1, for
30°. This commendation was implemented in the AcSELerator and Test Universe software
when writing the test procedure for 30 MVA transformer.
The same test procedure can be used for the 40 MVA transformer, but using the data
aggregated in Tables 4, 5 and 6 Chapter two need to be used. Because the test procedure is
102
lengthy and identical for both transformers, only the procedure for the 30 MVA was
written.
103
References
1. "Transformer Bank Protection." Gas and Electric Transmission Protection Bulletin 62e. June 1992
2. "An American National Standard." IEEE Guide for Protective Relay Application to Power Tans formers. June 1990.
3. Horowitz H. Stanley, Protective Relaying for Power System II. IEEE Inc: New York, 1992.
4. Hoebart Bernard, Power Transformer Handbook. Butterworth & Co. Ltd: France, 1997.
5. A.R. Van C. Warrington, Protective Relays: Their Theory and Practice. John Wiley & Sons, Inc.: New York, 1962. 380-412
6. Feinberg R., Modern Power Transformer Practice. The Macmillan Press Ltd: London,1979
7. Franklin A.C., The J&P Transformer Book. Butterworth & Co. Ltd: France, 1983.1997.
8. El-Hawary Mohamed E., Electric Power Systems: Design and Analysis. Reston Publishing Company, Inc.: Virginia 1983. 575-582
9. Turan Gonen, “Electrical Power Distribution System Engineering”, (Second Edition), CRC Press, Taylor & Francis Group, Boca Raton, FL, 2008.
10. Turan Gonen, “Electrical Power Transmission System Engineering: Analysis and Design”, (Second Edition), CRC Press, Taylor & Francis Group, Boca Raton, FL, 2011
11. Markovic, D. Miroslav. “Relay Protection of Power Systems.” TESLA Institute, 2006.
12. Instruction Manual for Schweitzer SEL-787-0 Current Differential and Overcurrent Relay (Manual P/N: SEL-787-0 dated 20050919).
13. J. Lewis Blackburn and Thomas J. Domin, “Protective Relaying : Principles and Applications” (Fourth Edition), CRC Press Inc., Taylor & Francis Group, Boca Raton, FL, 2014.
104
14. Mark W. Earley, Christopher D. Coache, Mark Cloutier, and Gil Moniz, “NFPA 70: National Electrical Code (NEC) 2014 Hand Book”, (13th Edition), National Fire Protection Association ,Quincy Massachusetts, Table 9.
15. The Art and Science of Protective Relaying, “http://www.gegridsolutions.com/multilin/notes/artsci/artsci.pdf”, Accessed on February 2016 through March 2017.
16. Tracy Toups and Prashanna Bhattarai, “Microprocessor Based Differential Relaying I” Lab procedure, 2014.
17. ANSI / IEEEC37.91, “Guide for Protective Relay Applications for Power Transformers”