POWER TRANFORMER PROTECTION A Project Presented to …

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

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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

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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

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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

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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

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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]:

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• 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.


secondary side of the 30 MVA power transformer.

nCT2 = I1nI2n

(3.5)



nCT2 = 12005

= 240, where 240 is the primary to secondary CT current ratio.


23

wye-ground of the 30 MVA power transformer.

nCT3 = I1nI2n

(3.6)



nCT3 = 2505


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


√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 pickup 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



transformer. The ratios are computed as follow:


primary side of the 40 MVA power transformer.

nCT1 = I1nI2n

(3.30)



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


40

3.15 T2 CT SECONDARY CURRENTS



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


√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


√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



= 331.69∗1.2

70 = 5.69 (A)


= 3346.96 ∗ 1.270

= 57.38


= 3346.96

70 = 47.81 (A)

The pickup 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



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


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


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



57

Table 8: 30 MVA Transformer Relays Pick-up Currents



= 25102∗1.2

50 = 6.02 (A)


= 3,137.02 ∗ 1.250

= 75.3 (A)


= 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

78

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

79

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

80

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

81

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

82

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

83

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

84

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

85

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]

87

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


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


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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

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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.

http://www.bookdepository.com/author/J-Lewis-Blackburn

http://www.bookdepository.com/author/Thomas-J-Domin

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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”

http://www.amazon.com/s/ref=dp_byline_sr_book_1?ie=UTF8&text=Mark+W.+Earley&search-alias=books&field-author=Mark+W.+Earley&sort=relevancerank

http://www.amazon.com/s/ref=dp_byline_sr_book_2?ie=UTF8&text=Christopher+D.+Coache&search-alias=books&field-author=Christopher+D.+Coache&sort=relevancerank

http://www.amazon.com/s/ref=dp_byline_sr_book_3?ie=UTF8&text=Mark+Cloutier&search-alias=books&field-author=Mark+Cloutier&sort=relevancerank

http://www.amazon.com/s/ref=dp_byline_sr_book_4?ie=UTF8&text=Gil+Moniz&search-alias=books&field-author=Gil+Moniz&sort=relevancerank

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