Electrical System Protection_E-Portfolio

ARSALAN AHMED USMANI

Unit Chair: Dr Aman Than Oo Lecture Teacher: Dr Mohammad Arif

1 E-portfolio Electrical System Protection

Executive Summary

In the course of electrical system protection of the masters of engineering program, the

student are assessed based upon their learning throughout the course either through the

lecture classes, assignments, projects or lab experiments. These learning is presented by the

student in the format of e-portfolio.

E-portfolio is a collection of documents and evidences of the learning outcomes that a

student achieve in the subject of his study, it is to demonstrate that the student has

achieved appropriate knowledge and understanding for the unit he has taken.

This e-portfolio ha got two parts, the first one will have five learning outcome; the first one

will be show that the student has applied advance technical knowledge and the use of

instrument and computer applications, the second one will be to demonstrate the

application of advance systematic approaches for the use of electrical protection equipment,

third one will be to show that the student has applied problem solving and advance

research principles, the forth learning outcome will demonstrate has use innovative and

advance solution and the fifth learning outcome the student has communicated with the

engineering team and community.

The second part of the e-portfolio will contain the two projects that are conducted in our

course of electrical system protection. One of the project is to study faults in the

transmission line at different buses in the network and the other project is related to the

study of the protection applied on the 66KV to 22KV feeder and substation and to analyse

we whether the network is redundant or not.

The evidences of the learning outcome are shown in the appendices at the end of the report,

that include the evidence of the lab report, lectures, online learning and some of the pages

from the textbook.


Table of Contents

1 Introduction .................................................................................................................................... 3

2 Learning outcomes .......................................................................................................................... 4

2.1 The learning outcome 1 is: ...................................................................................................... 4





3 PROJECT REPORT........................................................................................................................... 10

3.1 Project 1 ................................................................................................................................ 10

3.1.1 Abstract: ........................................................................................................................ 10

3.1.2 Introduction: ................................................................................................................. 10

3.1.3 Fault current: ................................................................................................................ 35

3.1.4 Conclusion: .................................................................................................................... 36

3.2 Project 2 ................................................................................................................................ 37

3.2.1 Abstract: ........................................................................................................................ 37

3.2.2 Introduction: ................................................................................................................. 37

3.2.3 Redundancy in network ................................................................................................ 38

3.2.4 Protection schemes ....................................................................................................... 43

3.2.5 Conclusion: .................................................................................................................... 57

4 Conclusion ..................................................................................................................................... 58

5 References .................................................................................................................................... 59

5.1 Books ..................................................................................................................................... 60

6 Appendix ....................................................................................................................................... 62

6.1 Appendix 1: Applying advanced technical knowledge:......................................................... 62

6.2 Appendix 2: Applying advanced systematic approaches to conduct and manage electrical

systems protection equipment ......................................................................................................... 90

6.3 Appendix 3: Apply problem-solving and advanced research principles ............................... 95

6.4 Appendix 4: Apply innovative and advanced solutions as part of electrical systems

protection ....................................................................................................................................... 108

6.5 Appendix 5: Communication with engineering teams and community to investigate and

present advanced electrical protection systems ............................................................................ 111


1 INTRODUCTION

Electrical system protection is one of the most vital part if the electric power system. It is

necessary to have proper protection schemes designed for the electric network and its

component, so that they can function properly and in fault condition be isolated from the

rest of the network.

The objective of a protection scheme is to keep the power system stable by isolating only

the components that are under fault, whilst leaving as much of the network as possible still

in operation.

Since the current electric grid is aging, electrical engineer are finding better solutions for

controlling the electric network, therefore the advent of microprocessor based relay has

made the relay operation more precise yet it has become more complex. Electrical engineer

needs to learn know the advance technology to understand the changing electric grid.

Therefore this electric system protection course is designed for the students to get the most

advance knowledge about the protection of grid.

This postgraduate engineering course was for eleven weeks and was dived into different

topic shown in the table below

Class topics Week Lab experiment Week

Fundamental of Protection

Systems

1 Lab Induction 1

2 2

Current Transformers 3 Observe Current Transformer (CT)

characteristics

3

Overcurrent Transformer 4 Protection of a three phase Induction

Motor

4

5 5

Transformer Protection 6 Observation of Faults in a Three Phase

Power Transformer

5

6

Distance Protection 7 Protection of a Three Phase Power

Transformer

7

8 8

Motor Protection 9 Protection of Synchronous Generator

Protection

9

10

Grounding Principles 10

Feeder Protection 11


2 LEARNING OUTCOMES

2.1 THE LEARNING OUTCOME 1 IS: This learning outcome specifies in applying advance technical knowledge and the use of

computer assisted applications to complete protection of the electrical equipment was

effectively accomplished while undertaking the course of electrical protection system.

The lab experiments that we conducted throughout the course has enable us to apply the

most latest and advance technical knowledge that we have learned in the course to

practically implemented. Electrical labs were having installed with advance trainer which

made it possible for the students to effectively and successfully implement our theoretical

knowledge.

The application of our knowledge gained in the lectures were practically applied in the labs.

One of the lab was over current protection of a three phase Induction motor, the induction

motor is the most essential element of industry or any manufacturing plant, almost 70% of

the electricity consumption in the manufacturing plant accounts to load of induction motor.

In that lab experiment we created fault and used the current transformer to detect the fault

that was initiated by the fault module in the workstation. The fault module was placed to

create phase fault and ground fault at the terminals of induction motor. The over current

was measured by current transformer and the over current relay was able to sense it by

indicating it by the tripping indicator on the relay further the relay was able to send signal to

the control relay.

The other lab experiment we did was over current protection of three phase power

transformer, transformers step up or steps down the voltages required for the transmission

of electricity they are the most reliable yet most expensive and critical part of the

transmission line, hence the proper protection of this element of the transmission line is

essential. In this experiment we initiated fault at the load side of the transformer, the fault

includes phase to phase and phase to earth fault. The fault currents are measured using the

current transformer and are detected by the overcurrent relay and AC/DC sensitive relay.

The relay then clears the fault using the control relay.

Another lab experiment was the differential protection of synchronous generator,

Synchronous generator are commonly used in the electrical power system to convert

mechanical energy into electrical energy, one such practical use of these generators are in

the wind turbines to control the full speed of the wind, they can be used as prime mover for

various other applications where electricity is generated. Protection of synchronous

generator is therefore highly required and in this experiment we have seen the use of two

sets of identical current transformers connected across the synchronous generator and the

current leaving and the current entering the synchronous generator is measured through

them, any significant difference of current can operate the relay, we created the fault using

universal fault module and the relay detected it and clears it using the control relay.


The connection between the host computer and the EMS-workstation was developed using

the data acquisition unit in the module through the USB connection going from data

acquisition unit to the host computer. The connection allows the use of computer to

monitor and record the voltages and current value of the circuit using the software LVDAC-

EMS, the software also allows the use the data to be transferred to the excel files and the

data could also be viewed in the simulation format as well. The software was also able to

connect with the prime mover module of the synchronous generator to control the speed of

the generator using the host computer.

[Appendix 1: Shows the implementation of the experiment.]


2.2 THE LEARNING OUTCOME 2 IS:

This learning outcome wants the student to demonstrate the systematic approach to

manage and conduct the use of electric system protection equipment.

This learning outcome was achieved by having the lab experiment arranged in the

systematic and sequentially way so as the student gets familiar with the lab equipment

before the student use them to monitor the actual protection. Every lab experiment started

off with the risk assessment of all the modules that will be used in the lab.

After connecting the circuit according to the lab guide, the power connection was not

switched ON until the connections were verified by the instructor, so as to keep both the

student and the equipment safe.

The protection relay are set to respond in the quickest time by setting the relay manually,

while conducting the lab many of the modules were used in the work station namely;

Transmission grid “A”, Universal fault module, interconnection module, data acquisition

module, power supply module, faultable Transformer (FT), current Transformers (CTs), and

resistive Load.

The relay station consisted of three phase overcurrent relay, AC/DC Sensitive overcurrent

relay and the control relays.

Appendix 2 gives the evidence of operating instrument for protection of three phase

transformer, synchronous generator and three phase induction motor against faults.



This learning outcome specifies that the student should apply problem solving and advance

research principle as a part of design and development of electrical system protection study.

The lectures that were given in the classroom and the presentations that were presented

teach the design consideration of the protection network in the transmission, and

distribution lines. Some of the examples of the problem solving skills were gained during the

lecture were;

Selection of fuse rating on the radial distribution line having different load current and therefore the coordination of the fuse,

Calculating fault current using current transformer at a particular ratio and specification,

Calculating the fault current on the radial transmission line and indicating the relay that will detect the fault current at a particular places on the line,

Calculating the pickup current off the differential relay across the transformer,

Calculating the pickup current of the over current relay once it is placed with the transformer while taking into consideration of the overload capability of the transformer,

Calculating the impedance setting of the three zones of distance protection mho relay in the power system and determining which relay will trip under the certain current rating.

The prove of this is shown in the Appendix 3


This learning outcome specifies that the student needs to demonstrate the innovative and

advance solutions in performing the electrical system protection tasks,

During our lab experiment of over current protection of three phase induction motor, we

connected the current transformer module with the ratio 0.5:5 A which made the circuit

highly sensitive to any small rise in current. When we started the induction motor, the

inrush current of the motor was high enough for the overcurrent relay to detect that rise in

current as a fault and the tripping indicator on the relay turned ON. Our group along with

the lab instructor was able to figure out that the current transformer ratio was too less

hence we changed the CT module with the ratio of 2.5:5 A to make the circuit stable.

Another solution we find out while making our project is that when we were making our

circuit in the Matlab program Simulink, we took the transmission line as the pi module in

our program, which made the voltage and current simulation of the circuit out of phase

after some time and even the magnitude decreased and kept decreasing after that, so we

replace the module with the inductance one.

Appendix 4: provide the evidence for this particular situation.



This learning outcome specifies that the student needs to communicate with engineering

teams and the community to investigate and present advanced electrical protection systems.

The projects that were assigned and the lab courses that were conducted both requires the

students to form a group, the students were divided into several different groups so that

they could effectively communicate and contribute to the understanding of the electrical

protection system.

Before the commencement of the lab experiment the student were required to make risk

assessment of all the modules they would be using in the experiment, the planning and

group teamwork was necessary for all the group members to complete the lab experiment

within the timeframe. The groups consult the lab supervisor for any sort of help they

required or for verifying of the connections before they switch on the main supply to

measure the required readings

The projects that were given to us as part of the course were also completed forming a

group, the group regularly consulted the unit teacher for his assistance if the group faces

any problem to find the solution to the particular problem.

Both the project and the labs enhanced the communication and teamwork skills for the

individual student and the assistance from the unit teacher helped the student in becoming

the active member of the group.

[Appendix 5: Provides the pictures of the lab experiments conducted in the team, under the

guidance of lab supervisor.]


Project reports

Group members:

Arsalan Ahmed Usmani

Balaji


3 PROJECT REPORT

3.1 PROJECT 1

3.1.1 Abstract:

In this project, the different types of faults are simulated in the given network by using power

system block set model. The faults are simulated at different regions of the network. In those

regions fault currents are observed with different fault resistance.

3.1.2 Introduction:

In electrical power systems consists of various complex interacting elements, there always exists the

possibility of disturbances and faults [1]. The cause of electric power system faults is insulation

breakdown. This breakdown can be due to various reasons like, Lightning ionizing air, Wires blowing

together in the wind, Animals or plants coming in contact with the wires. Different types of faults

can interrupt the healthy operation of the power system. Some of the major electrical faults are

phase faults include phase-to-phase fault (fault, when two lines come into contact), phase-to-ground

fault (fault, when one line come into contact with ground), and three phase faults (fault, when three

lines come into contact). The fault current will vary for different types of faults. The faults are

simulated using simulation tool Matlab.

1. BUS DATA:

The faults are applied to the 66kV sub-transmission network shown in the Figure1. Bus A is

connected to Transmission substation. Bus B and C are connected to Bus A, so that all the Bus A, B,

and C are supplied with 66kV. Substation B and C are step-down to 22kV by using step-down

transformers. The Bus length, type, impedance and ratings of transmission and step-down

transformers are given below.

Figure 1. 66kV sub-transmission network


Data 1. Bus Details

2. FAULT DETAILS:

The faults are applied to 22kV Bus at substation B and 66kV Bus at substation C.

SUBSTATION B:

Faults: Three phase fault, Phase-to-phase fault, Phase-to-earth fault

Fault currents should be given at 22kV side.

SUBSTATION C:

Faults: Three phase fault, Phase-to-earth fault for the cases of zero

fault resistance (0Ω) and a fault resistance of 10Ω.

Fault currents should be given at 66kV side.


3. MATLAB:

Matlab is a software package which can be used to perform analysis and solve mathematical and

engineering problems [2]. The simulation is done in Simulink, which is an add-on product to Matlab.

Simulink is a block diagram environment for multi-domain simulation and model-based design [3]. It

provides an interactive, graphical environment for modelling, simulating, and analyzing of dynamic

systems [2]. In this project, faults are simulated at two different locations, substation B and

substation C. The faults are three-phase fault, phase-to-phase fault, and phase-to-earth fault.

BLOCK SET MODELS:

Block set models used for substation B and C are almost same. The block set models are three phase

source, three phase mutual inductance (Z1-Z0), three phase transformer (two winding), scope, three

phase VI measurement, power GUI, ground, and single phase series RLC block .

Three-phase source:

Figure 2. Three phase source

The Three-Phase Source block uses a balanced three-phase voltage source with an internal R-L

impedance. The three voltage sources are connected in Y with a ground connection [4]. The block

diagram of three phase source is shown in the above Figure2. Parameters of this block are

mentioned in below Table1. From the data, the source voltage is given as 66kV and their phase-to-

phase rms voltage is 66000/√2 = 46.66kV

Table 1. Parameters of Three-phase source

S. No. Parameters Value

1. Phase-to-phase rms voltage 46,669V

2. Phase angle of phase A 0°

3. Frequency 50Hz

4. Internal connection Yg

5. 3-phase short circuit level 1500MVA

6. X/R ratio 0.57


Three-phase mutual inductance:

Figure 3. Three-phase mutual inductance

The Three-Phase Mutual Inductance Z1-Z0 block implements a three-phase balanced inductive and

resistive impedance with mutual coupling between phases [5]. The block diagram of three phase

source is shown in the above Figure3. Three blocks of three-phase mutual inductance are used for

this project. Each block represents connection between two buses like Bus A to B, Bus B to C, and

Bus A to C.

Calculated the resistance and inductance from the data. For all the line section, the impedances are

same, whereas the length of the line is different. The impedance is Z1=0.19+j0.35Ω/km and Z0 = 0.33

+ j1.68 Ω/km, which is in the format of Z=R+jX (plus (+) sign indicates the reactance is inductive

reactance). By using the formula, L= X/ (2πf), the inductance is calculated. Parameters of the Bus A

to B block are mentioned in below Table2. Parameters of the Bus B to C block are mentioned in

below Table3. Parameters of the Bus A to C block are mentioned in below Table4.

Table 2. Parameters of Three phase mutual inductance of Bus A to B

S. No. Parameters Value [R L]

1. Positive-sequence parameters [1.9 0.0111]

2. Zero-sequence parameters [3.3 0.0534]

Table 3. Parameters of Three phase mutual inductance of Bus B to C




Table 4. Parameters of Three phase mutual inductance of Bus A to C





Initially, pi-transmission line block had been used instead of three-phase mutual inductance. Due to

improper sine wave, pi-transmission block had been removed. The improper sine wave is caused by

capacitance of the line, whose data is not given. So that, three phase mutual inductance block is

used in the place of pi section. The block diagram of three phase pi section line is given below in

Figure4.

Figure 4. Block Diagram of pi section line

Three-phase transformer:

Figure 5. Three phase Transformer (Two winding)

The Three-phase transformer block is changed to step-down transformer by changing the

parameters of winding1 and winding 2. This block implements a three-phase transformer using three

single phase transformers [6]. From the given data, the ratio of the transformer is given as 66000 to

22000V Dy1, which means the value of 66kV is step-down to 22kV and Dy1 means the primary

winding is in delta configuration, secondary winding in star configuration and “1” represents the

low voltage winding should lag the high voltage winding by 30°. The block diagram of three phase

transformer is shown in the above Figure4. Parameters of this block are mentioned in below Table5.

Table 5. Parameters of step down transformer

S. No. Parameters Value

1. Nominal power and frequency 30MVA and 50Hz

2. Winding 1

[Voltage Resistance Inductance]

[46.6kV 1.5 0.10]

3. Winding 2

[Voltage Resistance Inductance]

[15.5kV 1.5 0.10]

4. Magnetization Resistance 3.24MΩ

5. Magnetization Inductance 8597.9H


Scope:

Figure 6. Block diagram of scope

The Scope block displays inputs signals with respect to simulation time [8]. This block is

used to observe the waveform of voltage and current with respect to time. The number of

inputs to the scope can be changed by modifying the axes setting in parameters. If the

inputs are voltage and current, the waveform of both voltage and current can be observed in

the same waveform with respect to time. So that, lagging of voltage and current can be

easily observed. The block diagram of scope is shown in Figure6.

Three-phase VI measurement:

Figure 7. Block diagram of three-phase VI measurement

The Three-Phase V-I Measurement block is used to measure instantaneous three-phase voltages and

currents in a circuit. When connected in series with three-phase elements, it returns the three

phase-to-ground voltages and currents [9]. The block diagram of three phase VI measurement is

shown in Figure7.

Ground:

Figure 8. Block diagram of ground

The Ground block implements a connection to the ground [10]. This block is used to create a phase-

to-earth fault in some cases and used to connect to R load to act as the return path of the circuit or

close the circuit. This block is used in substation C, in which fault resistance should be given to three

phase fault and phase-to-earth fault. The fault resistance is connected between the phases and

ground. The block diagram of ground is shown in Figure8.


Series RLC branch:

Figure 9.Block diagram of series RLC branch

The Series RLC Branch block implements a single resistor, inductor, or capacitor [11]. In this project,

load is taken as R load, whose values are assumed to 1kΩ. The block diagram of series RLC branch is

shown in the above Figure9. This block is used also as a fault resistance at substation C. In substation

C, different fault resistance is used and the corresponding fault currents are observed. The series RLC

branch is changed to resistor by modifying the branch parameter to R, which means resistor. As this

load value changes corresponding fault current value will change.

Initially, three phase RLC load is used as load, in which the low voltage winding and high voltage

winding of step-down transformer are in same phase, there is no 30° lag in between them which

was created in three phase transformer block. The assumption of parameters in RLC load contains

voltage, frequency, active power, inductive reactive power, and capacitive reactive power, which are

too many. So that, RLC branch is used instead of RLC load, which has less assumption and more

importantly the low voltage winding of three phase transformer get 30°lag from the high voltage

winding of three phase transformer.

Power GUI:

Figure 10. Block diagram of Power GUI

The powerGui block contains four methods to solve the circuit, which are continuous, ideal, discrete

and phasor, in which continuous method is used for simulation. The powergui block is necessary for

simulation of any Simulink model containing SimPower systems blocks. It also gives you access to

various graphical user interface (GUI) tools and functions for the steady-state analysis of SimPower

systems models and the analysis of simulation results [12]. The steady-state value of voltage and

current are observed under the steady-state analysis. So that, fault currents and phase shifts are

easily observed.


FAULTS AT SUBSTATION B:

In this section, faults are applied to 22kV Bus at substation B, which means, the secondary of the

step-down transformer in substation B. Totally, four models are simulated in this section, which are

no fault model, three phase fault model, phase-to-phase model, and phase-to-earth model.

No fault model:

The circuit is drawn as shown in the Figure1 in simulink. This model shows the normal operation of

the circuit, which means no fault is applied to the circuit. The circuit diagram of the no fault model is

shown below in Figure11. The steady-state values of voltage and current are observed and shown in

Figure12. In the Figure12, the first and second marked values are primary and secondary values of

Bus B transformer, which have almost 30° lag. In that figure, the below marked values are the

normal current of 22kV transformer.

The waveform of step-down transformer of the BUS B is shown in Figure13. In that waveform, the

first and second inputs are primary and secondary voltages of step-down transformer respectively.

The third and fourth inputs are primary and secondary currents of step-down transformer

respectively. The waveforms of voltage and current are in sinusoidal. The phase shift of voltage and

current are not accurately shown in waveform.

Figure 11. Circuit diagram for No fault


Figure 12. Steady state values


Figure 13. Waveform of step-down transformer

Three-phase fault:

In this model, three phase fault is simulated. In the secondary of the step-down transformer at Bus B,

all the three phases are shorted to ground. The circuit diagram of three-phase fault shown in the

Figure14. The steady-state values of voltage and current are observed and shown in the Figure15. In

that figure, the below marked values are the three phase fault current of 22kV transformer of

substation B.

The waveform of three-phase fault at step-down transformer of the BUS B is shown in Figure16. In

that waveform, the first and second inputs are primary and secondary voltages of step-down

transformer respectively. The third and fourth inputs are primary and secondary currents of step-

down transformer respectively. The waveform of secondary voltage of the step-down transformer is

zero due to the three-phase fault and the secondary current of all three phases are increased due to

the fault.


Figure 14.Circuit diagram for three phase fault

Figure 15. Steady state values of three-phase fault


Figure 16. Waveform of three-phase fault

Phase-to-phase fault:

In this model, phase-to-phase fault is simulated. In the secondary of the step-down transformer at

Bus B, the second and third phases are shorted. The circuit diagram of phase-to-phase fault shown in

the Figure17. The steady-state values of voltage and current are observed and shown in the Figure18.

In that figure, the below marked values are the phase-to-phase fault current of 22kV transformer.

The waveform of phase-to-phase fault at step-down transformer of the BUS B is shown in Figure19.

In that waveform, the first and second inputs are primary and secondary voltages of step-down

transformer respectively. The third and fourth inputs are primary and secondary currents of step-

down transformer respectively. The waveform of the secondary voltage of step-down transformer of

two phases are reduced due to the phase-to-phase fault and the fault current of two phases are

increased due to the fault.


Figure 17.Figure. Circuit diagram of phase-to-phase fault

Figure 18.Steady state value of phase-phase fault


Figure 19. Waveform of phase-phase fault

Phase-to-earth fault:

In this model, phase-to-earth fault is simulated. In the secondary of the step-down

transformer at Bus B, the third phase is shorted to ground. The circuit diagram of phase-earth fault

shown in the Figure20. The steady-state values of voltage and current are observed and shown in

the Figure21. In that figure, the below marked values are the phase-to-earth fault current of 22kV

transformer. The waveform of phase-to-phase fault at step-down transformer of the BUS B is shown

in Figure22. In that waveform, the first and second inputs are primary and secondary voltages of

step-down transformer respectively. The third and fourth inputs are primary and secondary currents

of step-down transformer respectively. The waveform of the secondary voltage of step-down

transformer of third phase is reduced to zero due to the phase-to-earth fault and the fault current of

one phase is increased compared to other two phases, which is due to the fault. But the value of the

fault current is low compared to the other faults.


Figure20. Circuit diagram of phase-to-earth fault

Figure 21.Steady-state of phase-earth fault


Figure 22. Waveform of phase-to-earth fault

FAULTS AT SUBSTATION C:

In this section, faults are applied to 66kV Bus at substation C, which means, the primary of the step-

down transformer in substation C. The fault current is measured over the Bus A-B line. Totally, five

models are simulated in this section, which are no fault model, three phase fault model with fault

resistance of 10Ω, three phase fault model with fault resistance of 0Ω, phase-to-earth model with

fault resistance of 10Ω, and phase-to-earth model with fault resistance of 0Ω.

No Fault Model:

This model shows the normal operation of the circuit, which means no fault is applied to the circuit.

The circuit diagram of the no fault model is shown below in Figure23. The steady-state values of

voltage and current are observed and shown in Figure24. In the Figure24, the first and second

marked values are primary and secondary values of Bus C transformer, which have almost 30° lag. In

that figure, the below marked values are the normal current of BUS A and BUS B. The waveform of

voltage and current in BUS A and B are shown in the Figure25.


Figure 23. Circuit diagram of no fault

Figure 24. Steady state of no fault


Figure 25. Waveform of no phase fault

Three Fault Model with 10Ω:

In this model, three phase fault is simulated. In the primary of the step-down transformer at

Bus C, all the three phases are shorted to ground with fault resistance of 10Ω. The circuit diagram of

three-phase fault with fault resistance of 10Ω shown in the Figure26. The steady-state values of

voltage and current are observed and shown in the Figure27. In that figure, the below marked values

are the fault current of third phase at Bus A, Bus B and Bus C, which clearly shows that fault current

is flowing in Bus C and Bus A but not in Bus B.

The waveform of three-phase fault with resistance of 10Ω at Bus C is shown in Figure28. In that

waveform, the first and second inputs are the voltages of Bus A and Bus B respectively. The third and

fourth inputs are the currents of Bus A and Bus B respectively. The waveform of the current at Bus A

is high due to the three phase fault, whereas the current at Bus B is normal. The fault is created at

Bus C, so that, fault current is flowing in Bus C and Bus A. The fault current at Bus A is small

compared to fault current at Bus C, which is clearly shown in the Figure27.


Figure 26. Circuit diagram of three phase fault with fault resistance of 10Ω

Figure 27. Steady state of three phase fault with 10Ω


Figure 28. Waveform of three phase fault with 10Ω

Three Fault Model with 0Ω:

In this model, three phase fault is simulated. In the primary of the step-down transformer at

Bus C, all the three phases are shorted and ground directly, which is considered to be the fault

resistance of 0Ω. The circuit diagram of three-phase fault with fault resistance of 0Ω shown in the

Figure29. The steady-state values of voltage and current are observed and shown in the Figure30. In

that figure, the below marked values are the fault current of third phase at Bus A, Bus B and Bus C,

which clearly shows that fault current is flowing in Bus C and Bus A but not in Bus B. The fault

current is same in both fault resistance of 0Ω and 10Ω for three phase fault.




is high due to the three phase fault, whereas the current at Bus B is normal. The fault is created at

Bus C, so that, fault current is flowing in Bus C and Bus A. The fault current at Bus A is small

compared to fault current at Bus C, which is clearly shown in the Figure30.


Figure 29.Circuit diagram of three phase fault with fault resistance of 0Ω

Figure 30. Steady state of three phase fault with 0Ω


Figure 31. Waveform of three phase fault with 0Ω

Phase-earth Fault Model with 10Ω:

In this model, phase-to-earth fault is simulated. In the primary of the step-down transformer

at Bus C, third phase is shorted to ground with fault resistance of 10Ω. The circuit diagram of phase-

earth fault with fault resistance of 10Ω shown in the Figure32. The steady-state values of voltage

and current are observed and shown in the Figure33. In that figure, the below marked values are the

fault current of third phase at Bus A, Bus B and Bus C, which clearly shows that fault current is

flowing in Bus C and Bus A but not in Bus B.



fourth inputs are the currents of Bus A and Bus B respectively. The waveform of the current in Bus A,

only one phase is shorted. So that, fault current is flowing only in one phase due to that other

waveforms looks almost zero because of the normal current. The fault is created at Bus C, so that,

fault current is flowing in Bus C and Bus A.


Figure 32.Circuit diagram of phase-earth fault with fault resistance of 10Ω

Figure 33. Steady state of phase-earth fault with 10Ω


Figure 34. Waveform of phase-earth fault with 10Ω

Phase-earth Fault Model with 0Ω:

In this model, phase-to-earth fault is simulated. In the primary of the step-down transformer

at Bus C, third phase is shorted directly to the ground, which means the fault resistance is 0Ω. The

circuit diagram of phase-earth fault with fault resistance of 0Ω shown in the Figure35. The steady-

state values of voltage and current are observed and shown in the Figure36. In that figure, the below

marked values are the fault current of third phase at Bus A, Bus B and Bus C, which clearly shows

that fault current is flowing in Bus C and Bus A but not in Bus B and the fault current is flowing only

in third phase, which is due to phase-to-earth fault. The fault current is slightly higher in Bus C and A

compared to the fault current with 10Ω fault resistance, which is because of the fault resistance. The

fault resistance is zero in this model, so that fault current will be high in this model.




is high due to the phase-earth fault, whereas the current at Bus B is normal. The fault is created at

Bus C, so that, fault current is flowing in Bus C and Bus A, not in Bus A to B. The voltage of Bus B in

fault phase reduced slightly.


Figure 35.Circuit diagram of phase-earth fault with fault resistance of 0Ω

Figure 36. Steady state of phase-earth fault with 0Ω


Figure 37. Waveform of phase-earth fault with 0Ω

3.1.3 Fault current:

Substation B:

For the load resistance of 1000Ω, the fault currents in three phases are tabulated below in Table6.

Table 6. FAULT CURRENT VALUES OF SUBSTATION B

S. No. Fault Type Phase A Phase B Phase C

1. No Fault 1.27A 1.27A 1.27A

2. Three phase Fault 374.18A 374.18A 374.18A

3. Phase-to-phase Fault 1.27A 324.68A 323.42A

4. Phase-to-earth Fault 2.19A 2.21A 3.81A


Substation C:

For the load resistance of 1000Ω, the fault currents in three buses are tabulated below in Table7. For

the three phase fault, the fault currents in all the phases in Bus A, B and C are same. For the phase-

to-earth fault, the fault current is only in third phase.

Table 7. FAULT CURRENT VALUES OF SUBSTATION C

S. No. Fault Type BUS A BUS B BUS C

1. No Fault 0.63A 0.46A 0.46A

2. Three phase Fault_0Ω 1069.53A 0.33A 2406.17A

3. Three phase Fault_10Ω 1069.53A 0.33A 2406.17A

4. Phase-to-earth Fault_0Ω 568.87A 0.42A 1279.60A

5. Phase-to-earth Fault_10Ω 490.11 0.42A 1102.17A

3.1.4 Conclusion:

The fault currents at both substation B and C are tabulated above in Table 6 and 7

respectively. This fault is simulated for the resistive load of 1000Ω and it is taken as the balance load,

in order to apply the balance fault. It is difficult to calculate the fault current under unbalanced

condition. As the load changes, the fault currents will change. In substation B, the normal current is

1.27A (no fault condition), which means the normal operation of the circuit. For three-phase fault,

the current is increased enormously in all three phases and for phase-phase fault, the current is

increased in faulted phases (like phase B and phase C). The phase-to-earth current is very low, when

compared to other faults but it is two times higher than the normal operation of the circuit, which is

still a threat to the system. In theoretically, the phase-to-earth fault current might be higher than

three-phase fault current, which is because of the fault impedance. The formula for the three-phase

fault is

[12], in which Vll is a line-line voltage and Zsc is the total impedance of

the circuit. The formulae for phase-to-phase and phase-to-earth faults are

and , where Zo is the ground impedance. So that, for different

impedance, the fault current will change and depend on the ground impedance, the phase-to-earth

fault current might be higher than the three-phase fault current.

From the Table7, it shows that, at substation C, the fault current is flowing in Bus A and C

but not in Bus B. The fault is applied at Bus C due to that, fault current of Bus C is higher than Bus A.

The fault current is flowing through Bus A to Bus C. There is no fault current flowing through Bus A to

Bus B. The fault applied at Bus C, did not affect the current of Bus B, whereas it affected the current

of Bus A, which are due to the length of the line and source path of Bus C through Bus A is shorter

compared to Bus B. The length from Bus A to C is shorter compare to the Bus B to C. But still, there is

a slight voltage fluctuation in Bus B due to the fault. Three-phase fault currents of 10Ω and 0Ω

resistors are same. The phase-to-earth fault current of 0Ω resistor is slightly higher than the phase-

to-earth fault current of 10Ω resistor. So that, the fault current can be reduced by applying the fault

resistance. The increase in fault resistance reduces the fault current but the fault resistance should

be lower than the impedance of the load, otherwise the fault current will flow through the load.


3.2 PROJECT 2

3.2.1 Abstract:

Electrical system protection is the essential for the equipments in the electricity grid to have

uninterrupted power supplied to the consumer. In this project our aim is to study the redundancy of

the network and the protection schemes installed at different section of the substation.

3.2.2 Introduction:

After looking at the single line diagram given in the project we are to write a report on the;

1. Redundancy in grid network shown by the SLD.

2. Protection schemes that are used installed at different section of the substation network.

First we need to research, how does the distribution network shown in the SLD is made redundant.

Consumer such as business, industries and even home users are really concerned about the

reliability of the network, to achieve the reliable network we need to have a redundant grid.

Achieving redundancy in the network is possible by having mesh topologies to form the grid network;

In case of any line failures the power supply could be rerouted so as to maintain the reliability of the

network. Therefore such topologies used for forming the network are useful in mitigating the

chances of outages for the consumer while the utility could repair the damaged and deactivated line.


3.2.3 Redundancy in network

Below is the single line diagram of 66/11kV Berserker Sub-station in Rockhampto, we need to

analyse whether the network is redundant or not.


66KV feeder redundancy

We can see from the picture below that there are two 66KV sources suppling to bus 1 and 2 and

they are also interconnected so as to provide back up for if one the line is disconnected due to any

fault.


Step down transformer (66KV/11KV) Redundancy

Two sources and two buses therefore there are two 66KV/11KV step down transformer and they are

linked together by the bus section 1 and 2. If one fails then the other could supply to the load. Being

same and connected made it possible.


11 KV bus network redundancy

The picture below shows the distribution network that has 16 of the 11KV distribution feeder that

goes to the end load of which many are spare that can be used for further load. Notice that all the

three buses are connected to each other and there is interconnecting buses 2 as well which

distributes the load.


In the picture below, we can see after some time when the single line diagram was revised there are

two new loads attached to the network where there were spare before. In the pitchure below two

distribution feeder are connected to each other by the change over board that makes the

distribution network even more redundant, as one 11KV feeder can supply electricity if there is fault

at the other feeder.


3.2.4 Protection schemes

The line diagram of Berserker Sub-station 66KV/11KV protection design


Legends


66 KV feeder protection

As both the feeder in the diagram are of same 66 KV rating therefore the protection of both are

similar therefore we will study only one of them.


66 KV Feeder protection

J11- M (protection relay multifunction Main)

Relay

function

number

Description Purpose of use

ITS Protection

intertrip send

function

Once the fault is detected this function send signal to the

backup relay

ITR Protection

intertrip

receive

function

This function on the relay is essential for the coordination of

the relays and it coordinates with the back up relay

p Metering in

protection

relay

Metering is essential in relays as it calculates the current in

realtime

03 Circuit

breaker fail

function

It is a relay that operates in response to the position of a

number of other devices (or to a number of predetermined

conditions) in an equipment, to allow an operating sequence to

proceed, or to stop, or to provide a check of the position of

these devices or of these conditions for any purpose [13].

21 Distance

function

It is a relay that functions when the circuit admittance,

impedance, or reactance increases or decreases beyond

predetermined limits.

87L Line current

differential

function

It is a protective relay that functions on a percentage or phase

angle or other quantitative difference of two currents or of

some other electrical quantities.

J11- B (protection relay multifunction backup), it doesn’t have 21 but has got 79 and 67/67 N

additional

67/67 N Directional

overcurrent

and earth

fault function

Phase-to-phase short-circuit protection, with selective tripping

according to fault current direction. It comprises a phase

overcurrent function associated with direction detection, and

picks up if the phase overcurrent function in the chosen

direction (line or busbar) is activated for at least one of the 3

phases.[14]

79 Auto reclose

function

Automation device used to limit down time after tripping due

to transient or semipermanent faults on overhead lines. The

recloser orders automatic reclosing of the breaking device after

the time delay required to restore the insulation has elapsed.

Recloser operation is easy to adapt for different operating

modes by parameter setting[15].


66 KV bus protection


66 KV bus protection

-J87- M (protection relay – Bus differential Main)

Relay function

number


J87 Bus current

differential

function

It is a protective relay that functions on a

percentage or phase angle or other quantitative

difference of two currents or of some other

electrical quantities.

-J03 (protection relay – Bus CHECKING OR INTERLOCKING RELAY)

99 Trip circuit

supervision

function

This function on the relay trips the circuit in case

of over flux at the bus

03 Circuit breaker

fail function

It is a relay that operates in response to the

position of a number of other devices (or to a

number of predetermined conditions) in an

equipment, to allow an operating sequence to

proceed, or to stop, or to provide a check of the

position of these devices or of these conditions for

any purpose[16].

-J87- B (protection relay – Bus differential Backup)

J87 Bus current

differential

function

It is a protective relay that functions on a

percentage or phase angle or other quantitative

difference of two currents or of some other

electrical quantities.


Step down transformer (66KV/11KV) protection


Step down transformer (66KV/11KV) protection

J11- M (protection relay multifunction Main)

Relay function

number


87 current

differential

function


angle or other quantitative difference of two currents or of some

other electrical quantities.

87G Transformer

differential relay

This relay is now also known as 87T because it is only use for

transformer whereas 87G is use for differential protection of

generator. This relay provides high speed phase and ground

protection of three phase transformer[17].

50/50N Instantaneous

over current

and earth fault

function

It is a relay that functions instantaneously on an excessive value

of current or on an excessive rate of current rise, thus indicating

a fault in the apparatus or circuit being protected

51/51N Inverse time

overcurrent and

earth fault

function

It is a relay with either a definite or inverse time characteristic

that functions when the current in an a-c circuit exceed a

predetermined value.

HV 03 Distance

function

It is a relay that operates in response to the position of a number

of other devices mainly as the name suggest it for high voltage

(or to a number of predetermined conditions) in an equipment,

to allow an operating sequence to proceed, or to stop, or to

provide a check of the position of these devices or of these

conditions for any purpose.

LV 03 Line current

differential

function

It is a relay that operates in response to the position of a number

of other devices mainly as the name suggest it for low voltage

(or to a number of predetermined conditions) in an equipment,

to allow an operating sequence to proceed, or to stop, or to

provide a check of the position of these devices or of these


J11- B (protection relay multifunction backup), same functions as the Main

-J51G Neutral earth

fault



predetermined value and the fault current may be due to earth

fault



-J11 (protection relay multifunction)

Relay function

number


p Metering in

protection relay

Metering is essential in relays as it calculates the current in real-

time

03 Circuit breaker

fail function

It is a relay that operates in response to the position of a number of

other devices (or to a number of predetermined conditions) in an

equipment, to allow an operating sequence to proceed, or to stop,

or to provide a check of the position of these devices or of these


81 Under

frequency

function

It is a relay that functions on a predetermined value of frequency

(either under or over or on normal system frequency) or rate of

change of frequency.

79 Auto reclose

function

Automation device used to limit down time after tripping due to

transient or semipermanent faults on overhead lines. The recloser

orders automatic reclosing of the breaking device after the time

delay required to restore the insulation has elapsed. Recloser

operation is easy to adapt for different operating modes by

parameter setting.

99 Trip circuit

supervision

function

This function on the relay trips the circuit in case of over flux at the

bus

51NS Time sensitive

earth fault

function

This relay operates at a fixed time set in the relay and this relay

function could detect high impendence breakdown to earth[18].


over current

and earth fault

function

It is a relay that functions instantaneously on an excessive value of

current or on an excessive rate of current rise, thus indicating a

fault in the apparatus or circuit being protected

51/51N Inverse time

overcurrent and

earth fault

function

It is a relay with either a definite or inverse time characteristic that

functions when the current in an a-c circuit exceed a predetermined

value.


Capacitor bank protection


Capacitor bank protection

-J60.1 (protective rlay- capacitor bank unbalance)

Relay function

number


59 Under/

overvoltage

function

It is a relay that functions on a given value of over-voltage.

60 Voltage or

current balance

relay

It iis a relay that operates on a given difference in voltage, or

current input or output, or two circuits.

37 Under current

or under power

relay

It is a relay that functions when the current or power flow

decreases below a predetermined value.

68 Close/ reclose

inhabit function

It is a relay that initiates a pilot signal for blocking of tripping on

external faults in a transmission line or in other apparatus under

predetermined condition, or cooperates with other devices to

block tripping or to block re-closing on an out-of-step condition

or on power savings.


11 KV Bus protection



-J51- B (protection relay sum bus overcurrent)

Relay function

number


99 Trip circuit

supervision

function

This function on the relay trips the circuit in case of over flux at

the bus


over current

and earth fault

function

It is a relay that functions instantaneously on an excessive value

of current or on an excessive rate of current rise, thus indicating

a fault in the apparatus or circuit being protected

51/51N Inverse time

overcurrent and

earth fault

function



predetermined value.

-J87- M (protection relay – Bus differential Main)

J87 Bus current

differential

function


angle or other quantitative difference of two currents or of

some other electrical quantities.


3.2.5 Conclusion:

In this project we investigated the network diagram of the distribution substation and its

protection schemes on the different part of the network, as we have analysed all the equipments

that are part of the network is being protected by different characteristics of relay.

Some part of the network has got two relays installed one is the main and the other is the backup

so as to improve the reliability of the network where as the dependability of the network

increases as we have redundant network.

By doing the project we come across the practical use of protection relays in the substation that

has given us good knowledge about the protection schemes installed for different components.

We have notice that their need to be mant spare connections available in the distribution network

so as to compensate the load in future that can be seen as two new load were added within small

period of time.


4 CONCLUSION

It conclusion this e portfolio is about the electrical power protection unit that we have taken as apart

of post graduate engineering course. The e portfolio has been written to show all the topics that

were covered in the class and the lab experiments that were conducted at electrical lab.

There are also two projects that were to be completed as a part of e portfolio. The projects show the

understanding that the student have acquired about the subject during his studies. The projects

were completed with the help of unit teacher Mohammad Arif.

In this e portfolio there are five learning outcomes that were to be completed in the unit and the

justification for the learning outcome in the unit is in the appendix part of this e portfolio.

At the end I would like to thank the teaching staff especially unit teacher Mohammad Arif for his

valuable time that he has spent in teaching the subject and in helping the students understand the

subject.


5 REFERENCES

[1]. www.science.smith.edu/~jcardell/Courses/.../MatlabSimulinkTutorial.ppt. (n.d.).

[2] myelectrical. (n.d.). Retrieved from http://myelectrical.com/notes/entryid/192/fault-calculations-

introduction

[3] subhra jana. (2013). IEEE Explore

Mathworks. (n.d.). Retrieved from

http://www.mathworks.com.au/help/physmod/sps/powersys/ref/powergui.html?searchHighlight=c

ontinuous+powergui

[5] Mathworks. (n.d.). Retrieved from

http://www.mathworks.com.au/help/physmod/sps/powersys/ref/ground.html?searchHighlight=gro

und


http://www.mathworks.com.au/help/physmod/sps/powersys/ref/threephasevimeasurement.html?

searchHighlight=three+phase+measurement


http://www.mathworks.com.au/help/simulink/slref/scope.html?searchHighlight=scope


http://www.mathworks.com.au/help/physmod/sps/powersys/ref/threephasetransformertwowindin

gs.html


http://www.mathworks.com.au/help/physmod/sps/powersys/ref/threephasesource.html

[10] Mathworks. (n.d.). Retrieved from http://www.mathworks.com.au/products/simulink/

[11] Mathwork. (n.d.). Retrieved from

http://www.mathworks.com.au/help/physmod/sps/powersys/ref/seriesrlcbranch.html

[12] Mathwork. (n.d.). Retrieved from

http://www.mathworks.com.au/help/physmod/sps/powersys/ref/threephasemutualinductancez1z0

.html

[13] http://electrical-engineering-portal.com/ansi-codes-device-designation-numbers [14] Electrical engineering portal: http://electrical-engineering-portal.com/protection-relay-ansi-standards#ANSI 67 - Directional phase overcurrent [15]Electrical engineering portal: http://electrical-engineering-portal.com/protection-relay-ansi-standards#ANSI 67 - Directional phase overcurrent [16] http://electrical-engineering-portal.com/ansi-codes-device-designation-numbers

http://electrical-engineering-portal.com/ansi-codes-device-designation-numbers

http://electrical-engineering-portal.com/ansi-codes-device-designation-numbers


[17] ABB: http://www05.abb.com/global/scot/scot229.nsf/veritydisplay/744c18d45bda740bc1256ea8004bb325/$file/db41-359s%20%20%20%20%2087t.pdf [18] http://www.woodbeam.co.za/docs/2770-proddoc-20114729.pdf

5.1 BOOKS

[1] Mason, C. Russell. "The Art and Science of Protective Relaying". General Electric.

Retrieved 2009-01-26

[2] Stanley H. Horowitz and Arun G. Phadke, POWER SYSTEMRELAYING”, Third Edition,

[3] Muhammad Arif, Lecture notes , Deakin university

[4] Blackburn, J.L, A Domin, T.J, Protective Relaying: Principles and Applications, Third

Edition, 2006

http://www.geindustrial.com/pm/notes/artsci/artsci.pdf

http://en.wikipedia.org/wiki/General_Electric


Appendices


6 APPENDIX

6.1 APPENDIX 1: APPLYING ADVANCED TECHNICAL KNOWLEDGE:

Lab exercise 2: Over current protection of three-phase induction

motors

Objective of this experiment

To learn, to protect an induction motor from an over-current fault in the electric system

network. When I have completed this exercise, I will be familiar with over current

protection of three-phase induction motors.

Equipment used for this experiment

The list below shows the equipment required for this exercise in the EMS workstation and

Protective Relaying control station

1. Power Supply

2. Universal Fault Module (UFM)

3. Four-pole Squirrel-Cage Induction Motor

4. Prime Mover/Dynamometer

5. Transmission Grid “A”

6. Current Transformers (CTs)

7. Data Acquisition unit with Ammeter & Voltmeter

8. Three phase overcurrent relay

9. Connecting leads

Setting up Equipment:

1. Ensure that the Protective Relaying Control Station is connected to a power source.

Make sure the DC Power Supply of the Protective Relaying Control Station is turned off.

Make sure that all fault switches on the Three-Phase over Current Relay are set to the O (off)

position then install it in the Protective Relaying Control Station.


2. Make the following settings on the Universal Fault Module:

TD1 time delay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... ~1 s

SST1 time interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...~3 s

SST2 time interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....~1o s

3. Install the Interconnection Module (if available), Power Supply, Universal Fault Module,

Four-Pole Squirrel-Cage Induction Motor, Prime Mover/Dynamometer, Transmission Grid

"A", Current Transformers, AC Ammeter, and AC Voltmeter in the EMS Workstation.

Mechanically couple the Four-Pole Squirrel-Cage Induction Motor to the Prime Mover/

Dynamometer using the timing belt.

Make sure the Power Supply is turned off and its voltage control knob is set to the 0 position.

Connect the Power Supply to the power outlets.

On the Current Transformers (CTs) module, make sure that all switches are set to the 1

(close) position to short-circuit the secondaries of the current transformers.

4. Connect the Prime Mover/Dynamometer module to the 240 V - AC Power Supply. On the

Power Supply, turn on the 24-V AC power source and connect to the Data acquisition unit.

5. If cables are available, connect the Interconnection Module installed in the EMS

Workstation to the Interconnection Panel of the Protective Relaying Control Station using

the supplied cables. Connect the equipment as shown in Figures 2 and 3.

6. Make the following settings:

On the Prime Mover I Dynamometer (set in brake mode)

MODE switch . . . . . . . . . . . . . . . . . . . . . . . . . . . . DYNamometer

LOAD CONTROL MODE switch . . . . . . . . . . . . . . . . . . MANual

MANUAL LOAD CONTROL knob . . . . . . . . . . . . . . . . MINimum

DISPLAY switch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TORQUE

On Transmission Grid "A" set

Switch S1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 (open)

On the Universal Fault Module set

INITIATE FAULT button . . . . . . . . . . . . . . . . . released position


FAULT DURATION switch . . . . . . . . . . . . . . . . . . . . . . 0.05 - 5 s


Figure 2. Connection diagram of the equipment in the EMS Workstation

Figure 3: Connection diagram of the equipment in the Protective Relaying Control Station

7. Set the current set point of the Three-Phase Over Current Relay to approximately 450% of

the nominal full-load current of the three-phase induction motor, taking into account the

transformation ratio of the current transformers.

Set the time delay of the Three-Phase Over Current Relay to 0 s.

8. Turn on the DC Power Supply of the Protective Relaying Control Station.

On Transmission Grid "A", set switch S2 to the 0 (open) positions to open contactor CR2.

This will prevent operation of the over-current protection system and allow the operation of

the Three-Phase Over-current Relay to be observed.

9. Turn on the Power Supply while observing the motor currents indicated by the AC

Ammeter (I1, I2, and I3). The induction motor should start rotating.

On the Prime Mover/Dynamometer, set the MANUAL LOAD CONTROL knob so that the

mechanical load torque (indicated on the module display) is equal to 1.0 N-m (9.0 lbf-in),

which is the nominal full-load torque of the motor.

10. Turn on the Power Supply while observing the motor currents (I1, I2, I3) and the tripping

indicator (red LED) on the Three-Phase over Current Relay. The induction motor should start

rotating.


Turn off the Power Supply.

EXPERIMENTAL OUTPUT AND MY OBSERVATION:

11. Repeat the previous step a few times.

Is the over current protection system stable when the induction motor is starting?

Yes No

At first it wasn’t stable and the over current relay was indicating a fault because of the

inrush current at starting of the induction motor which is required to accelerate the motor

and the load to the specific speed, however after changing the current transformer module

from the ratio of 0.5: 5 A to 2.5: 5 A ratio made the relay not to operate at the starting of

the motor.

Test - bypassing Protective Relays

12. Turn on the Power Supply.

On the Universal Fault Module, depress the INITIATE FAULT button to produce a fault at the

terminals of the induction motor. While doing this, observe the circuit currents (I1, I2, I3)

and the tripping indicator on the Three-Phase Over-Current Relay.

Describe what has happened.

The nominal current in each phase is about 0.25 A but as we initiate a fault in the line of the

induction motor by the universal fault module, the current in the faulty phase 1 and phase 2

increases drastically to about 2.5 A which is detected by the current transformer 1 and 2.

The ratio of CT is 2.5:5.0 A, therefore the fault signal of 5 A is passed on to over current relay,

which sense the fault in the system and tripping indicator indicate the fault. It clears the

fault within small period of time.

Time E(V) I1 (A) I2 (A) I3 (A)

1 397.8 0.266 0.256 0.279

2 397.9 0.266 0.256 0.279

3 398.1 0.265 0.257 0.279

4 333.2 2.232 2.53 0.363

5 332.7 2.229 2.527 0.364

6 397.8 0.265 0.255 0.279

7 398.1 0.265 0.256 0.279

8 398 0.265 0.256 0.279

9 397.8 0.265 0.256 0.279


On the Universal Fault Module, place the INITIATE FAULT button in the released position.

Test - with Protective Relays

13. On Transmission Grid "A", set switch S2 to the 1 (close) position to close contractor CR2.

This will allow operation of the over current protection system.

On the Universal Fault Module, depress the INITIATE FAULT button to produce a fault at the

terminals of the induction motor. While doing this, observe the circuit currents (I1, I2, I3)

and the tripping indicator on the Three-Phase Over-Current Relay.

Describe what has happened.

As the switch S2 was open it prevented the operation of over current protection system, but

as we close the switch S2 and then initiates a fault using universal fault module the fault

current passes through the current transformer to the over current relay which then signals

the control relay to clear the fault, now the control relay CR1 starts to blink to have fault

cleared. If instead of CR1 there would have been an isolator or circuit breaker then it would

have operated and disconnected the circuit.

0

0.5

1

1.5

2

2.5

3

1 2 3 4 5 6 7 8 9 10

Current when fault is initiated

I1 (A) I2 (A) I3 (A)

10 397.6 0.265 0.256 0.279


Time E(V) I1 (A) I2 (A) I3 (A)

1 398.8 0.262 0.255 0.281

2 398.7 0.262 0.255 0.28

3 398.9 0.263 0.255 0.28

4 399.1 0.263 0.256 0.281

5 399 0.262 0.255 0.281

6 399.3 0.263 0.256 0.28

7 396.2 2.692 2.946 0.285

8 398.7 0.263 0.255 0.28

9 399.1 0.263 0.256 0.28

10 399.2 0.263 0.256 0.28

Has the fault been cleared by the over current protection system?

Yes No

Does the over current protection system provide fast, effective protection against faults at

the induction motor terminals?

Yes No

On the Universal Fault Module, place the INITIATE FAULT button in the released position.

14. Turn off the Power Supply and turn off all modules at the EMS workstation.

Turn off the DC Power Supply of the Protective Relaying Control Station.

Remove all leads and cables.

0

0.5

1

1.5

2

2.5

3

3.5

1 2 3 4 5 6 7 8 9 10

Axi

s Ti

tle

Current during fault

I1 (A)

I2 (A)

I3 (A)


LEARNING OUTCOMES:

In this experiment, I get to know about the protection system of the three phase induction

motor. Three phase induction motor is the critical component of many plants and

industries therefore any fault in these can result in unplanned downtime and can prove

very costly.

The main faults in induction motor can be mechanical, electrical or environmental. Our

focus in this experiment was to identify and take possible actions if any line to line or line

to earth fault occurs in the induction motor.

At first we connected the current transformers with the ratio 0.5:5 A which made the

circuit highly sensitive to any small rise in current that is why when we started the

induction motor the overcurrent relay senses the inrush current, which is 7 to 9 the

nominal current and starts to indicate the fault, so therefore we change the CTs with the

ratio of 2.5:5 A.

First we setup the EMS work station so as to make over current relay sense the fault when

the fault is generated using the universal fault module. The overcurrent relay setting is set

to 45% because the nominal current is 0.25 A and when it passes through the CTs with

ratio of 2.5:5 A, then the current becomes 0.5A and the current set point of the Three-

Phase Over Current Relay is 450% of the nominal full-load current of the three-phase

induction motor therefore it is 2.25A and which is 45% of the current setting in the over

current relay.

As we initiate the fault in step 12 the over current relay’s tripping indicator starts to

indicate the fault and when we initiate a fault in step 13 after the switch S2 is close, we

find the relay sense the fault and further sends the signal to control relay.

This experiment also shows that the three phase over current relay has got two settings in

it, one is current setting (that work on the principle that the fault current reduces as the

fault is further away from the protection scheme) and the other is time delay hence we

can calibrate both of them in such a way so as to have a backup protection in presence of

the main protection scheme.


Lab exercise 4: Over current protection of three phase transformer

Objective of this experiment

The objective of this experiment is to observe the faults in a three phase Power Transformer

and this will show why transformer protection is important and to observe the over current

protection system. Usually the protection that is used for the transformer is differential,

restricted earth fault and over current protection. We are using over current protection

relays in this experiment only.

In the first part of the exercise, I will set up the equipment in the EMS Workstation and the

Protective Relaying Control Station.

In the second part of the exercise, I will connect the equipment as shown in Figures. In this

circuit, I will initiate fault at the load side of the transformer to observe fault current for

phase-to-phase fault and whether the relay sense the fault and cleared the fault using

control relays . Again I will initiate fault at the load side of the circuit to observe phase-to-

earth fault and whether the fault is sense by the over current relay and hence cleared by the

control relay. In this part of the experiment I will use three phase overcurrent relay, AC/DC

sensitive relay and over current relay, in the next experiment I will use these relays to clear

the faults.

When the fault will be initiated the current transformer will detect the over current and

further it’s connected to three phase over current relay, this initiate the trip mechanism of

over current relay which is further connected to control relay CR1, the trip current goes to

CR1. Therefore the contact CR1-C closes to note the fault and starts to light up. The contact

CR-1B opens to open the contractor CR1, this separates the transformer from the sources.

In this experiment we will short one of the phase of the load and will observe with happen

with the protection system. We will also initiate the fault in the primary side of the

transformer and observe what the protection system is going to do.


EQUIPMENT USED FOR THIS EXPERIMENT

The list below shows the equipment required for this exercise in the EMS workstation and

Protective Relaying control station

1. Power Supply

2. Universal Fault Module (UFM)

3. Faultable Transformer (FT)

4. Transmission Grid A

5. Current Transformers (CTs)

6. Resistive Load (2 sets; for 533 ohm load on each phase)

7. Data Acquisition unit with Ammeter & Voltmeter

8. Interconnection module

9. Three phase overcurrent relay

10. AC/DC Sensitive overcurrent relay

11. Connecting leads


SETTING UP EQUIPMENT:

Connect the equipment according to the diagram shown below, make sure the power

supply is off before making the connections and once they are made recheck them with the

instructor.

Using the two resistor module to implement the resistance required. Initially turning the

fault switches on transformer to 0.

Set the time delay of the Three-Phase over Current Relay to approximately 5 s.

Adjust the current set point and hysteresis of the AC/DC Current Sensitive Relay to 100 mA

and 5%, respectively.

Connection diagram of the equipment in the EMS works station is below:


Connection diagram of the protective relaying control station is below:


RESULTS AND OBSERVATION

Turning on the DC power supply, to turn the protection relaying control station and setting the

transmission system switch S2 to 0, to allow only protective relay to work and not the whole control

system.

Then we turn ON the main power supply of the EMS station and turning the knob of voltage to 100%.

At no fault condition the current and line to line voltages are:

I1 = 0.409 A E1 = 389.6 V I2 = 0.392 A E2 = 356.6 V I3 = 0.008 A

Now we initiate a fault by pressing the initiate fault button this short circuits one phase of the

three phase load connected to the faultable transformer, we the observe the line current and line to

line voltage and we also observes tripping indicator that is the red light on the overcurrent relay and

AC/DC current sensitive relay.

11 = 1.088 A E1 = 382.5V 12 = 1.084A E2 = 314.3 V 13 = 0.010 A

When one phase of three phase load is short circuited, the current in the primary and the secondary winding of that phase increases almost 3 times of the nominal current, and the voltage is also dropped in secondary side of the transformer and that is indicated by the three phase over current relay and the AC/DC current sensitive relay.

Now we introduce other fault by switching the fault switch to FS1, and transformer T1 to the 1

position to insert an earth fault near the middle of the primary winding of transformer T1. While

doing this, we observe the circuit currents and voltages and also observe the tripping indicator that

is the red light on 3 phase over current and AC/DC current sensitive relay.

I1 = 0.540A E1 =389.2 V I2 = 0.412A E2 =355.6 V I3 = 0.940A

We can observe that when we introduce earth fault in the primary winding of transformer the

current in the ground going from transformer increases almost 10 times of the nominal

current, and that is shown by the ammeter at I3. Therefore the transformer primary current

also increases as there is more current drawn by the ground. These changes in current is

also sensed by the over current relay and AC/DC current sensitive relay by red light

indicating it.


Now we set the switch S2 on the transmission grid module to close position to allow the use of

protection system.

We initiate the fault by the fault button on the EMS station to short circuit one of the phase of the

three phase load connected to the fault able transformer. We measure current and line to line

voltage and also observe the tripping indicator that is the red light on 3 phase over current and

AC/DC current sensitive relay.

I1 = 1.108A E1 =390.4 V I2 = 1.106A E2 =317.6 V I3 = 0.011A

When one phase of three phase load is short circuited, the current in the primary and the secondary winding of that phase increases almost 3 times of the nominal current, and the voltage is also dropped in secondary side of the transformer and that is indicated by the three phase over current relay and the AC/DC current sensitive relay. The relay then further removes the power supply through the control relay and hence the current becomes zero later on.

On Control Relays 1 of the Protective Relaying Control Station, press the RESET button of control

relay CR1 to reset the protection system.

Now we introduce other fault by switching the fault switch to FS1, and transformer T1 to the 1

position to insert an earth fault near the middle of the primary winding of transformer T1. While

doing this, we observe the circuit currents and voltages and also observe the tripping indicator that

is the red light on 3 phase over current and AC/DC current sensitive relay.

I1 = 0.577A E1 =391.3V I2 = 0.268A E2 =357.0 V I3 = 1.031A We can observe that when we introduce earth fault in the primary winding of transformer the current in the ground going from transformer increases almost 10 times of the nominal current, and that is shown by the ammeter at I3. Therefore the transformer primary current also increases as there is more current drawn by the ground. These changes in current is also sensed by the over current relay and AC/DC current sensitive relay by red light indicating it.The relay then further removes the power supply through the control relay and hence the current becomes zero later on. On the Faultable Transformers, set fault switch FS1 of transformer T1 to the 0 position to remove

the fault. Turn OFF the supply


DISCUSSION AND LEARNING OUTCOMES

The graphs of the voltage E1 and current I1 of the primary side of the transformer are shown below:

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

1

9

17

25

33

41

49

57

65

73

81

89

97

10

5

11

3

12

1

12

9

13

7

14

5

15

3

16

1

16

9

17

7

18

5

19

3

20

1

M7-I1

0

50

100

150

200

250

300

350

400

450

1

9

17

25

33

41

49

57

65

73

81

89

97

10

5

11

3

12

1

12

9

13

7

14

5

15

3

16

1

16

9

17

7

18

5

19

3

20

1

M1-E1


The graphs of the voltage E2 and current I2 of the secondary side of the transformer are shown

below:

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

1

9

17

25

33

41

49

57

65

73

81

89

97

10

5

11

3

12

1

12

9

13

7

14

5

15

3

16

1

16

9

17

7

18

5

19

3

20

1

M8-I2

0

50

100

150

200

250

300

350

400

1

9

17

25

33

41

49

57

65

73

81

89

97

10

5

11

3

12

1

12

9

13

7

14

5

15

3

16

1

16

9

17

7

18

5

19

3

20

1

M2-E2


The graph of current I3 is shown below:

The graph of I3 clearly shows that when the earth fault near the middle of the primary

winding of transformer T1 is initiated by switching the fault switch FS 1 and transformer T1

to 1 position, the high current flows that is shown in the graph of I3.

When one of the 3 phase is shorted at the load side then there is also high current at the

primary and the secondary side of the transformer this can be shown by the graph of I1

which is current at the primary and graph of I2 which is current at secondary of the fault able

transformer.

The voltage and current is also zero several times in the graphs above that is because of the

control relay that disconnects the supply to the faultable transformer, hence there is zero

voltage and current at the primary side and the secondary side that can be clearly seen in

the graph.

this experiment clearly shows that the transformer should be protected by the short circuit

fault and phase to earth fault.

Right settings of the over current relay helps to detect the fault in the shortest possible time

and clear the fault as it occurs.

Transformer faults i.e. short circuits are the result of internal electrical faults, the most

common one being the phase to earth faults. Somewhat less common are the turn-to-turn

faults. Thus the transformer need to have proper protection as they are the most expensive

and critical part of the transmission system.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1

9

17

25

33

41

49

57

65

73

81

89

97

10

5

11

3

12

1

12

9

13

7

14

5

15

3

16

1

16

9

17

7

18

5

19

3

20

1

M9-I3


Lab exercise 5: Differential protection of three-phase induction

motors


the LVEMS- Workstation


6.2 APPENDIX 2: APPLYING ADVANCED SYSTEMATIC APPROACHES TO CONDUCT AND MANAGE

ELECTRICAL SYSTEMS PROTECTION EQUIPMENT

Comprehensive instructions given on the EMS-workstation about the use of workstation in

the electrical lab.


Risk assessment done before ever lab;


6.3 APPENDIX 3: APPLY PROBLEM-SOLVING AND ADVANCED RESEARCH PRINCIPLES Selection of fuse rating on the radial distribution line having different load current and

therefore the coordination of the fuse

School of Engineering, Deakin University, Waurn Ponds Campus, Geelong, VIC 3220 Australia

Tutorials

Question 4Consider the radial distribution line shown in Figure 1, where customers are served all along the length of the feeders. Fuse A is the main feeder protection, and Fuses B and C are installed on lateral feeders to limit the outage due to remote faults, for example, for faults beyond B or C.

51

The maximum and minimum

available fault currents, in

amperes, at each location are

shown in the boxes. Also shown

is the normal load current

flowing through each fuse. Check

the coordination of the fuses.

Select fuse ratings for A, B, and C

that will coordinate properly. Figure 1


Tutorials

Question-4

Table-1: Coordination b/w EEI-NEMA Type T Fuse Links

52



Tutorials

Question-4

53

Table 2: Continuous Current Carrying Capacity of EEI-NEMA Fuse Links


Tutorials

Answer Q-4

As a first trial, lets consider Fuse C be a 15T fuse.

The load current is 21 A, but the 15T is capable of 23 A, according to Table 2.

Therefore this fuse is of adequate rating, although there is little room for load growth.

From Table 1, for T links, we see that the 15T will coordinate with the 25T fuse at location B for currents up to 730 A. but the maximum fault current (at C) is 1550 A.

Therefore, we select the 30T fuse for location B.

54



Tutorials

Answer Q-4

The 30T can carry 45 A continuously and, from Table 1, 30T will coordinate with the 15T protecting fuse up to 1700 A. This is a good choice.

The 30T must coordinate with A for fault currents up to 1800 A.

To carry the load current at A, we must select the 80T fuse, which can carry 120 A (from Table 2).

The 80T will coordinate with the 30T for fault currents up to 5000 A, and this system has only 1800 A available.

Thus, the workable solution is 80T at A, 30T at B, and 15T at C.

If you wish to allow for a greater load growth at C, depending on the nature of the load served and its likelihood for growth. This would require a larger fuse at C, which will then require that all fuse selections be reconsidered.

55

Calculating the fault current using current transformer at a particular ratio


TutorialsQuestion-2An overcurrent relay is set to operate at 8A is connected to a CT with a ratio of 100/5A. It is given that CT impedance Z2 = 0.082 ohms.

Find whether the relay will detect the fault current of 200A on the primary side for the following burdens and also calculate the CT errors.

Zb = 0.8 ohms

Zb = 3.0 ohms

45



Tutorials

Answer Q-2 (a)

(a) Zb = 0.8 ohms

Considering the equivalent circuit of Question 1

E2 = I2 (Zb + Z2) = 8(0.8 + 0.082) = 7.056V

From the CT characteristics chart, the corresponding Ie for E2 = 0.4A

Hence, the corresponding primary current I1`= I2 + Ie = 8 + 0.4 = 8.4 A

The corresponding primary current of CT I1 = I1`* CT ration = 8.4*(100/5) = 168 A

So, the relay will operate for a primary fault current of 200A.

CT error = Ie/I1`*100% = (0.4/8.4)*100 = 4.76%

This is a reasonable error.

46


TutorialsAnswer Q-2 (b)

(b) Zb = 3.0 ohms

Considering the equivalent circuit of Question 1

E2 = I2 (Zb + Z2) = 8(3.0 + 0.082) = 24.656V

From the CT characteristics chart, the corresponding Ie for E2 = 30.0A (approximately)

Hence, the corresponding primary current I1`= I2 + Ie = 8 + 30.0 = 38.0 A

The corresponding primary current of CT I1 = I1`* CT ration = 38.0*(100/5) = 760 A

So, the relay will not operate for a primary fault current of 200A.

CT error = Ie/I1`*100% = (30.0/38.0)*100 = 78.95%

This is very high error.47


Calculating the fault current on the radial transmission line and indicating the relay that will

detect the fault current at a particular places on the line.


Tutorials

Question-3

Consider a simple 3 phase radial system as shown in Figure. Calculate fault currents at different fault locations and indicate which relays will operate for which faults. Consider F4 and F3 is very close to each other.

38


Tutorials

Answer Q-3 (1/8)

Grading by current relies on the fact that the fault current along the length decreases as the distance from the source to the fault location increases.

The relays controlling various circuit breakers need to be set to operate at suitable values such that only the relay nearest to the fault will trip its breaker.

Note, we are treating all impedances as scalars. This is often done if the impedances are basically inductive. This often applies in transmission and larger distribution MV systems.

39



Tutorials

Answer Q-3 (2/8)

For a three phase source (as shown in figure),

the fault current level at F4 is

)( 1

/

4

LS

f

FZZ

VI

40

Where

Vf/Ø = Source phase to ground voltage

ZS = Source impedance

ZL1 = Line impedance


Tutorials

Answer Q-3 (3/8)

ZL1 = 0.12x2 = 0.24 ohm

So,

VkV

V f 63503

11/

ohmMVA

kVZS 484.0

250

)11()( 22

kAIF 77.8)24.0484.0(

63504

41



Tutorials

Answer Q-3 (4/8)

The relay R3 controlling the CB if set to operate somewhat below 8.77kA to protect the whole cable from A to B. “Somewhat below” is a vague and unsatisfactory term. In practice we will consider how the source impedance (fault level) might change and the accuracy of the relays and CTs to determine how far below 8.77kA we would need to set the relay to get reliable operation.

The relay would not be able to distinguish b/w faults F4 and F3, as the distance b/w the fault locations is very small. Hence the corresponding changes from IF4 to IF3 would be negligible.

42


Tutorials

Answer Q-3 (5/8)

For fault current at F2:

ZL1 = 0.12x2 = 0.24 ohm

ZL2 = 0.02x2 = 0.04 ohm

So kAIF 31.8

)04.024.0484.0(

63502

)( 21

/

2

LLS

f

FZZZ

VI

VkV

V f 63503

11/

ohmMVA

kVZS 484.0

250

)11()( 22

43



Tutorials

Answer Q-3 (6/8)

Relay R2 should operate for faults “somewhat below” 8.31kA.

It will be practically difficult to grade R3 and R2 for a fault at F2 because the currents are similar - 8.7kA and 8.3kA. Recall the accuracy of a 10P CT is 10% i.e. 0.8kA.

For the fault current at F1:

Current IF1 is)( 21

/

1

TLLS

f

FZZZZ

VI

44


Tutorials

Answer Q-3 (7/8)

ZL1 = 0.12x2 = 0.24 ohm

ZL2 = 0.02x2 = o.04 ohm

(this is a 7% reactance on a 4MVA base)

So,

VkV

V f 63503

11/

ohmMVA

kVZS 484.0

250

)11()( 22

ohmxxMVA

kVZ T 12.207.0

4

)11(%7

)( 22

kAIF 20.2)12.204.024.0484.0(

63501

45



Tutorials

Answer Q-3 (8/8)

The relay R1, if set to operate reliably at 2.20kA would protect equipment downstream of the transformer for faults at F1.

R2 could be set to trip above 2.20kA (and below 8.31kA) and would protect the cable from B to C and the primary side of the transformer. It no longer provides back up to F1 for faults at F1. This is a weakness for current grading.

46

Calculating the pickup current off the differential relay across the transformer,


Tutorials

Question-5

A single phase transformer is rated at 69/110kV, 20MVA to be protected by a differential relay.

a) What are the ideal ratios of the CTs if they have 5A secondary windings?

b) The primary CT is 300:5. The secondary CT is 200:5. For these ratios what is the differential current at full load?

c) The differential relay has a constraint winding that means the relay needs a certain difference current to pick up. Calculate what the differential current in part (b) is in per-unit terms. Round this up to the nearest 2%.This will be the constraint setting.

d) The transformer has a 1% magnetizing current. At zero transformer load what is the differential current in the relay? Set a pickup current that is rounded up to the nearest 2%. Draw the pickup region of the relay on a i1s i2s plane (in per unit terms).

56



Tutorials

AnswerQ-5 (a)

The current in the primary for the rated load :

=(20x1000)/69 = 289.8A

Current in the secondary for the rated load :

=(20x1000)/110 = 181.8A

The perfect ratio for CT is 289.8:5 and 181.8: 5.

57


Tutorials

Answer Q-5 (b)

We can select CT ratio at the primary side as 300:5 and at the secondary side 200:5

For this selected CT ratios, CT at primary side of large Transformer will produce current at the secondary of the CT as:

= (289.8x5)/300 = 4.83A

Similarly, CT at the secondary side of the large Transformer will produce current at the secondary of the CT as:

= (181.8x5)/200 = 4.54A

So, the differential current is (4.83 - 4.54) = 0.29A

58



Tutorials

Answer Q-5 (c)

Differential current is 0.29 A, which is (0.29/4.83) = 6.004% of the primary side CT current at rated load.

We would need to set the differential current at 8% to have a margin of safety in through faults. We could reduce this by having a CT with taps on the nominal 5A winding. In an electronic relay we could set a scaling factor.

59


Tutorials

Answer Q-5 (d)

As mentioned Transformer magnetising current is 1% i.e at the primary side magnetising current is 2.89A which results magnetising current of 0.0483A on the primary side of CT. This is 1.6% of the current so we set the pickup current at 2%. Relay will trip if the differential current is more that 8% as mentioned in part c.

A graphical representation of the trip and operate region is shown below.

60


Calculating the impedance setting of the three zones of distance protection mho relay in the

power system and determining which relay will trip under the certain current rating.


TutorialsQuestion-3

Three-zone mho relays are used for the protection of the power system as shown in Figure 2. Positive sequence impedances (in Ohm) are given. The rated voltage at Bus-1 is 500kV and the CT and VT ratios are 1500:5 and 4500:1 respectively. In a three zone scheme for B3, Zone-1 protects 80% of line 1-2, Zone-2 protects 120% of line 1-2 and Zone-3 protects 100% of line 1-2 and 120% of line 2-3.

(a) Determine the actual impedance settings of 3 zones.

(b) The maximum current through the line 1-2 during emergency loading condition is 1400A at 0.9p.f. Lagging. Will any of the relays trip during this condition?

19

Figure 2. Three-zone mho relays are used for the protection of the power system


Tutorials

Answer Q-3 (a)

Impedance settings for phase ‘a’ to ground relay is given by Z = Vag/Ia

To express the same quantity for the secondary side of the Relay, as:

Hence, we can write the secondary side impedance settings as:

20

ratioVT

VV

ag

Secondaryag

)( ratioCT

II a

Secondarya

)(

ratioVT

ratioCTZx

ratioVT

ratioCTx

I

V

ratioCT

IratioVT

V

Za

ag

a

ag

.

.

.

.

151/4500

5/1500 ZZxZ



Tutorials

Answer Q-3 (a)

The line impedances given are:

Z12 = 6+j60

Z23 = 5+j50

So, the actual relay setting for relay at B3 as:

Zone 1 setting: 0.8 * Z12/15 = 0.8(6+j60)/15= (0.32 + j3.2) = 3.216 84.29oΩ

Zone 2 setting: 1.2 * Z12/15 =1.2(6+j60)/15 = (0.48 + j4.8) = 4.824 84.29oΩ

Zone 3 setting: (Z12 + 1.2*Z23)/15 = ((6+j60) + 1.2(5+j50))/15 = (0.80+j8.0) = 8.04

84.29oΩ

21


Tutorials

Answer Q-3(b)

The bus voltage at Bus-1 is 500kV and the maximum current for an emergency loading condition is 1400A with 09.p.f lagging.

So, maximum current = 1400 -25.84o A

Hence, the impedance seen by the secondary of the relay as:

Since this impedance is higher than the Zone-3 impedance settings, so the impedance during the emergency loading condition is outside the trip setting of any of the 3 zones. Therefore none of the relays will trip.

22

84.25)9.0(cos 1

84.2575.1384.251400315

1000500

15 xx

xZ


6.4 APPENDIX 4: APPLY INNOVATIVE AND ADVANCED SOLUTIONS AS PART OF ELECTRICAL SYSTEMS

PROTECTION

Both of them are current transformer module but with different ratings


The current transformers that were changed because of the ratio


The waveform of the pi model of the transmission line, because of the capacitance it becomes

distorted after sometime

The prime mover setting of the generator


6.5 APPENDIX 5: COMMUNICATION WITH ENGINEERING TEAMS AND COMMUNITY TO

INVESTIGATE AND PRESENT ADVANCED ELECTRICAL PROTECTION SYSTEMS

Electrical System Protection_E-Portfolio

Documents

Transcript of Electrical System Protection_E-Portfolio