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 1 CHAPTER 3 BASICS OF PROTECTIVE RELAYING This chapter of the teaching material introduces the basics of power system relaying. It places a  particular emphasis on overcurrent, impedance and differential protection. A set of Simulink models and exercises has been developed and included with the software to illustrate the following co ncepts: A single-phase model of a simple power system is developed using the Power System Blo ckset contained within Simulink. Circuit breakers (CBs), voltage transformers (VTs), and current transformers (CTs) are all modeled as ideal elements. Appropriate relays are modeled using their generic description. The protective equipment (CBs, VTs, CTs and relays) is connected together to allow for a closed-loop simulation. In this setup, the relay trip signal s are fed back to the CBs. The configuration of the models may be changed by using the cut-and-paste and drag-and-drop operations on the elements in the power system. Additionally, the parameters may be changed by double clicking the icon corresponding to the element of interest. The student assignments for this class require the following activities: Perform certain power system simulations of faults and observe how a given protection  pri nc i ple (overcurrent, impedance and differential) works. Determine correct settings for the relays in a given power system. Verify by simulation that the relays operate as expected. Model malfunctioning of the protective equipment and verify the back-up protection functions. This material has been used at the undergraduate level (ELEN459). 3.1 OVERCURRENT RELAYING PRINCIPLE 3.1.1 Introduction This exercise shows how overcurrent relaying principle and overcurrent definite-time relays can  be used to protect a radial electric power network. The following issues are explained in the introducti on and co vered in the Matlab models, simulations, examples and problem assignments: Rules for protecting a network using overcurrent relays. Requirements for instrumentation (number and locations of the instrument transformers) and switchi ng apparatus (number and locations of the circuit breakers). Analysis of the normal load conditions for selecting the instrument transformers and setting the relays. Analysis of the fault conditio ns for selecting the instrument tr ansformers and setting the relays. Setting and coordinating the relays. Simulation of the radial network protected with overcurrent relays. Checking the relay operation and coordination including protective equipment failures (incorrect relay and circuit breaker operation).
Transcript

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CHAPTER 3 BASICS OF PROTECTIVE RELAYING

This chapter of the teaching material introduces the basics of power system relaying. It places a particular emphasis on overcurrent, impedance and differential protection. A set of Simulink

models and exercises has been developed and included with the software to illustrate the

following concepts:• A single-phase model of a simple power system is developed using the Power System

Blockset contained within Simulink.• Circuit breakers (CBs), voltage transformers (VTs), and current transformers (CTs) are all

modeled as ideal elements.• Appropriate relays are modeled using their generic description.

• The protective equipment (CBs, VTs, CTs and relays) is connected together to allow for aclosed-loop simulation. In this setup, the relay trip signals are fed back to the CBs.

The configuration of the models may be changed by using the cut-and-paste and drag-and-dropoperations on the elements in the power system. Additionally, the parameters may be changed by

double clicking the icon corresponding to the element of interest.The student assignments for this class require the following activities:

• Perform certain power system simulations of faults and observe how a given protection principle (overcurrent, impedance and differential) works.

• Determine correct settings for the relays in a given power system.

• Verify by simulation that the relays operate as expected.• Model malfunctioning of the protective equipment and verify the back-up protection

functions.

This material has been used at the undergraduate level (ELEN459).

3.1 OVERCURRENT RELAYING PRINCIPLE

3.1.1 Introduction

This exercise shows how overcurrent relaying principle and overcurrent definite-time relays can be used to protect a radial electric power network. The following issues are explained in theintroduction and covered in the Matlab models, simulations, examples and problem assignments:

• Rules for protecting a network using overcurrent relays.• Requirements for instrumentation (number and locations of the instrument transformers) and

switching apparatus (number and locations of the circuit breakers).

• Analysis of the normal load conditions for selecting the instrument transformers and settingthe relays.

• Analysis of the fault conditions for selecting the instrument transformers and setting therelays.

• Setting and coordinating the relays.• Simulation of the radial network protected with overcurrent relays.

• Checking the relay operation and coordination including protective equipment failures(incorrect relay and circuit breaker operation).

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Overcurrent relaying

Figure 3.1 shows a radial network consisting of an equivalent system source (it typically

represents a transformer connected to the transmission network) and two lines connected betweenthree busbars. A number of loads are supplied from the busbars. This network is of the "radial" type

because it is supplied from one source and does not contain any loops.

Overcurrent Relaying Principle

VTVT 3VTVT2VTVT 1

V3V2

V1

TRIP3

TRIP2

TRIP1

Afault

B

System A

iTR

Overcurrent

Relay3i TR

Overcurrent

Relay2

i TR

Overcurrent

Relay1

Load4 Load3Load2Load1

Afault

B

Line2

Afault

B

Line1

I3

I2

I1

Fault

EA

LK

iCT

CT3

K

L

iCT

CT 2

K

L

iCT

CT1

CB

CB3

CB

CB2

CB

CB1

Bus3Bus2Bus1

Figure 3.1 A sample radial network with overcurrent relays (taken from MERIT2000)

Radial networks can be protected against faults using the overcurrent relaying principle. Toexplain the principle, consider a power system element on a radial network that is located at thelargest distance from the supplying source. This would correspond to loads 3 and 4 on the system

shown in Figure 3.1. If a fault were to occur in this element, a fault current that is significantlylarger than the normal load current would flow. This situation would cause an instantaneousovercurrent relay (OR) to operate.

An instantaneous overcurrent relay (OR) is a device that measures the magnitude of a current

and compares it against a threshold value. If the current is higher than the threshold (for example,a fault current), then the relay operates by sending a signal to the circuit breaker (CB) to open thecircuit (trip) and disconnect the faulted element from the rest of the system. For example, if fault

F-3 were to occur on the system of Figure 3.1, OR-3 would operate and remove load 3 from therest of the power system. This action would allow the network to continue supplying power to

loads 1, 2, and 4.

Note the following:

• In order to protect a given element, a current transformer (CT) needs to be used to measurethe current. The CT should be installed at the element's terminal that is closest to the

supplying source.

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• A CB must also be installed at the terminal closest to the supplying source to disconnect theelement in the case of a fault.

• An overcurrent relay must measure the current magnitude and compare it against thethreshold.

• The fault clearing time corresponds to the sum of the operating time of the relay and the

operating time of the CB.

The same methodology that was discussed and used earlier for load 3 may also be used to protectline 2. This is accomplished by placing CB-2, CT-2, and OR-2 on the left terminal of line 2. The

setup is shown in Figure 3.1. However, OR-2 cannot distinguish between fault F-3 and fault F-2(notice that when fault F-3 occurs, OR-2 must not operate and that when fault F-2 occurs, OR-2must operate). Therefore, a certain time delay is applied to make OR-2 wait for the slowest relay

protecting the lines and loads connected to busbar 3.

The ORs with fixed delays are called definite-time overcurrent relays. If the fault prolongs beyond the longest clearing time of the elements connected to busbar 3, then the relay OR-2should operate. In this situation, there are two possible reasons why OR-2 would operate. Either

the fault is actually on line 2 or the fault is downstream from line 2 and OR-3. Therefore, it isimportant that the ORs constituting the protection system apply certain time delays in order to

ensure selectivity of operation.

Similarly, in order to protect the line 1, CB-1, CT-1 and OR-1 should be installed between the

line and the supplying source. OR-1 must be coordinated with the relay downstream (in this caseOR-2) to ensure selectivity of operation.

The definite-time overcurrent principle is illustrated in Figure 3.2. The figure shows the timedelays of the relays in the protection system of the network.

OR-1

CB1

BUS-1

OR-2

CB2

BUS-2

OR-3

CB3

BUS-3

tOR-3

tOR-2

tOR-1

time

distance

> tCB-2

> tCB-3

Figure 3.2 Illustration of the definite -time overcurrent protection principle

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The protection system of Figure 3.2 operates as follows:• If there is a fault on one of the elements, some or all relays upstream between the fault and

the supplying system will pick-up (activate) and begin timing-out. However, only the relayclosest to the fault is supposed to trip since it has the shortest delay. After the relay trips, the

remaining upstream relays will re-set and none will trip. The relays installed downstream

from the fault will not pick-up because there is no fault current flowing there. The part of thenetwork downstream from the operating relay (including fault) will be cut-off. The upstream

part will operate normally after the fault is cleared.• If the relay or its corresponding breaker fails to operate, the fault will be cleared by the next

upstream relay. In this particular situation, the upstream relay would be functioning as back-up protection. If this relay fails, the next relay upstream operates. In such a case, moreelements will get tripped than originally desired. Ho wever, bear in mind that even though the

equipment malfunctioned the action is still optimal under this fault condition.

Note the following:

• To protect a radial network using ORs, the CBs, CTs and relays should be installed betweeneach power system element and the supplying system.

• The closer a fault is to the source, the higher the fault current and the longer the clearingtime. The combination of these two problems is a major disadvantage of this overcurrent protection. To improve the protection and shorten the average clearing time, relay may be

designed in such a way that their delay time depends on the fault current magnitude i.e. faultlocation.

Analysis of the load and fault conditions for selecting the protective equipment

• The load and fault conditions must be analyzed in order to select the CTs and CBs and to set

the relays.• The power flow and short-circuit calculations are very easy for radial networks and are not

discussed.

One needs to consider maximal and minimal fault current for each line. The maximal faultcurrent is used to select the CBs and the CTs and the minimal fault current is used to check

the sensitivity of the protection system.

Selecting the CTs

The following parameters of a CT should be considered from the protective relaying standpoint:• The rated primary current should be higher than the load current during normal load

conditions or acceptable overloads.

• The rated secondary current should match the rated input current of used relays (typically 5Aor 1A).

• A CT should not saturate under the maximum fault current flowing through the CT in a givennetwork (i.e. for bolted faults just downstream from the CT).

Selecting the CBs

The following parameters of a CB should be considered:• The maximal current that can be interrupted by a CB must be higher than the maximal fault

current flowing through this CB in a given network (i.e. for faults just downstream from theCB).

• The longest operating time of a CB must be known for proper setting of the correspondingovercurrent relay.

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Selecting the relays

The following parameters of an overcurrent relay should be considered:

• The rated input current should match the rated secondary current of the corresponding CT.• The setting ranges (pick-up current and time delay) should match the needs of a given relay

application (location).

Setting the relays

There are two settings of the definite-time overcurrent relay: the pick-up current and the timedelay. The setting process starts from the relay that is most distant from the source and progresses upstream relay by relay.

Pick-up current

The pick-up current should be set as low as possible but high enough to avoid picking up during

overload, transient and switching conditions. Typically, the pick-up current is set at 120-150% of the maximal load current. It is important remember that the relay is set based on the secondary

ampere values. Therefore, the load current must be converted into secondary qua ntities bydividing by the CT ratio.

Time delay

The time delay is set according to the following rule:

time delay = maximal time delay of the relays downstream + maximal CB time +security margin

The security margin depends on the accuracy of the installed relays and the variation of the CBoperating time. Typically, it is in the range of 0.1-0.3 sec.

Both the pick-up current and the time delay are set in certain steps. When coordinating the

relays, the actual time dials should be considered (resulting from the existing steps) and not thevalues calculated prior to setting the physical relays.

Sensitivity check

After all the relays have been set, a sensitivity check should be performed. This is accomplished by considering faults at the end of each line and checking if the relay protecting the line (primary

protection) and at least one relay upstream (back-up protection) will pick-up for a minimum faultcurrent.

All of the described engineering activities will be illustrated by an example in the followingsections.

3.1.2 Teaching Goals and Objectives

After completing this computer exercise the students will learn how to protect a radial network using definite-time overcurrent relays. Particularly, the following issues are re-enforced: loadflow and short-circuit calculations, selecting the protective equipment, setting and coordinating

overcurrent relays, relay sensitivity check, analysis of the network operation under variety of conditions including faults and equipment mal-operations.

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3.1.3 Available Software

Description

The system shown in Figure 3.1 has been implemented as a Matlab model with the use of the

Power System Blockset for modeling the network and SIMULINK for modeling the relays. In

addition to the elements shown in Figure 3.1, the voltage transformers (VTs) are added tofacilitate voltage measurement. A number of scopes have also been connected to display the

most critical signals in the model such as currents, voltages and trip signals. For simplicity, themodel is a single-phase model. Therefore, the only fault type considered at this time is the single

line-to-ground fault.

The following models are used:

Voltage Source

Double click on the icon to open a property dialog and change the magnitude, phase and

frequency.

Lines and System Impedance

Double click on the icon to open a property dialog and change the resistance, reactance and faultlocation. Notice that a third terminal has been created that divides the line. The fault distance is

calculated from terminal A of the line. To place a fault, connect a fault icon to the third terminal.

Fault

Double click on the icon to change the fault resistance and inception time. Connect the icon'sonly terminal to the place where the fault should occur.

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Load

Double click on the icon to change the rated voltage, active and reactive powers, and frequency.

Current Transformer

The transformer should be connected between the K and L terminals. The current signal is

available at the "i" terminal. Double click on the icon to change the rated primary and secondarycurrents. The transformation ratio is defined as the ratio between the primary and secondary

currents.

Voltage Transformer

The transformer should be connected to a busbar. The voltage signal is available at the only

output terminal. Double click on the icon to change the rated primary and secondary voltages.The transformation ratio is defined as the ratio between the primary and secondary voltages.

Circuit Breaker

Double click on the icon to change the CB’s operating (delay) time. The additional input is for the trip signal. If the control signal is high into this input, the CB opens after its operating timeand at a point when the current waveform is zero.

Scope

This element is used for displaying signal waveforms. A double-click will open a displaywindow. The toolbar in the display window contains buttons that allow the user to zoom in and

out on different sections of the waveform.

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Display

This element is used to display numerical values of a signal at discrete points in time. The block does not need to be opened; it simply displays the numerical values on the icon.

Definite-Time Overcurrent Relay

The input terminal "i" should be fed with the current and the output terminal "TR" should be fed-

back to the CB. A double click on the icon will open up a new window that displays thefunctional model of the relay (see Figure 3.3). The model has the following characteristics:

• The waveform of the current is processed into a phasor.• The phasor magnitude is compared with the threshold A0. A double click on the A0 block

allows the user to change the pick-up current.• The results of the comparison are integrated to reflect the time delay. The integrator's output

is compared with the time threshold. A double click on the integrator block allows the user to

change the delay.• A hysteresis is used to model the relay operation. Once tripped, the relay does not reset by

itself.

Figure 3.3 Definite -time overcurrent relay model

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The modern relays are constructed in one of three technologies:

• Electro-mechanical relays use traditional relays whose contacts are physically closed or opened by a magnetic field that is produced by the flowing current. A definite-timeovercurrent relay can be implemented as a single element or as a combination of an

instantaneous overcurrent relay and a mechanical timer.

• Static (electronic) relays convert the current signal into a low energy voltage signal. The

voltage signal is processed by internal circuits that consist of diodes, transistors, operationalamplifiers, etc. They typically use rectifying circuits and electronic comparators. They also

have electronic timers that use a type of RC circuit to time-out the delays.

• Digital relays sample the input signal, convert the samples to a digital form, calculate the

phasor magnitude, and compare it against a threshold value. These operations are performedin real-time using digital algorithms. The delays use a stable clock as a reference. The clock allows the relay to time-out very precisely.

The relay in included computer exercises is a functional model that is only able to simulate the

basic functions of an overcurrent definite-time relay. For example, it can sense the magnitude,compare it to a threshold value and time-out a certain time delay. Different functions of the blocks in these models are indicated by different colors (for example, the displays are yellow and

the instrument transformers are green).

Default Data:

The model was developed with the following data:

• System voltage: 158 kV (114 % of 138 kV)

• System impedance: 1 + j10 Ω

Line 1 impedance: 2 + j20Ω

, fault location 0.5• Line 2 impedance: 2 + j20 Ω, fault location 0.5• Load 1: 138 kV, 100 MW, 30 MVAr

• Load 2: 138 kV, 100 MW, 30 MVAr • Load 3: 138 kV, 50 MW, 15 MVAr • Load 4: 138 kV, 50 MW, 15 MVAr

• Fault: 0 Ω, inception time 20 msec.• CB-1 operating time: 15 msec

• CB-2 operating time: 20 msec• CB-3 operating time: 25 msec

• CT-1: 1600/1 A•

CT-2: 1000/1 A• CT-3: 500/1 A

• OR-1: pick-up 1.134 A, delay 90 msec• OR-2: pick-up 0.907 A, delay 50 msec

• OR-3: pick-up 0.907 A, delay 5 msec.

Note that the time data are re-scaled to speed up the simulation. The actual values would be

approximately 10 times higher.

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Model Activation

The model is stored in the file “ee460_3.mdl ”. To use the model, copy the file into Matlab’s

default working directory (typically Matlab/BIN), run Matlab, and type "ee460_3".

The following hints are useful when working with the model:

• Double-click on an element to change its parameters or to see its components.• Double-click on a scope to see the plot.

• Use the zooming buttons in plot’s display window to zoom in or out.• To disconnect two blocks, click on the connecting line and press the DEL key.• To connect two elements together, draw a line that connects their terminals.

3.1.4 Numerical Example

This subsection presents calculations for selecting the protective equipment, setting the relaysand checking the relays' sensitivity.

Load flow calculations

The load current is calculated as (apparent power)/voltage. Using the default data from the modelthe currents become:• Load 4 draws 0.378 kA.

• Load 3 draws 0.378 kA.• Load 2 draws 0.756 kA.• Load 1 draws 0.756 kA.

• The load current through CT-3 is 0.378 kA (Load 3).• The load current through CT-2 is 0.756 kA (Loads 4 and 3).

• The load current through CT-1 is 1.512 kA (Loads 4, 3 and 2).

Selecting the CTs

Assume that the CTs have a 1A secondary current rating and primary current ratings: 500, 1000,1500, 1600, 2000, 2500 A.• CT-3: select 500 A (the closest value higher than 378 A). With this setting, the secondary

load current at CT-3 becomes 378/500 x 1 = 0.756 A.• CT-2: select 1000 A (the closest value higher than 756 A). With this setting, the secondary

load current at CT-2 becomes 756/1000 x 1 = 0.756 A.• CT-1: select 1600 A (the closest value higher than 1512 A). With this setting, the secondary

load current at CT-1 becomes 1512/1600 x 1 = 0.945 A.

Short-circuit calculations

• The fault current for faults F-3 and F-2 (Figure 3.1) is equal to (system voltage)/(system

impedance + line 1 impedance + line 2 impedance).• The fault current for fault F-1 (Figure 3.1) is equal to (system voltage)/(system impedance +

line 1 impedance).• The fault current for fault F0 (Figure 3.1) is equal to (system voltage)/(system impedance).• Using the default data from the model the following currents are found:

• Faults F-3 and F-2: 3.14kA (6.28 A secondary at CT-3, 3.14 A secondary atCT-2 and 1.963 A secondary at CT-1)

Fault F-1: 5.24kA.Fault F-0: 15.72kA

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Setting the relays

The relay OR-3 is set to operate without any delay. It is assumed that the relay operates in 5

msec. It is also assumed that this is the slowest relay of all the relays protecting the elements branching from bus 3. The pick-up current of OR-3 should be set at 1.2 x load current = 1.2 x

0.756 A = 0.907 A.

The pick-up current of OR-2 should be set at 1.2 x load current: 1.2 x 0.756 A = 0.907 A (Note

that the load current is higher for OR-3, but the secondary values are the same due to thedifferences in the ratios of CT-2 and CT-3). With a 20 msec security margin, the time delay of

OR-2 should be 5 msec (delay of OR-3) + 25 msec (operating time of CB-3) + 20 msec (securitymargin) = 50 msec.

The pick-up current of OR-1 should be set at 1.2 x load current = 1.2 x 0.945 A = 1.134 A. Witha 20 msec security margin and OR-2 to be the slowest relay at bus 2, the time delay of OR-1

should be 50 msec (delay of OR-2) + 20 msec (operating time of CB-2) + 20 msec (securitymargin) = 90 msec.

Sensitivity check

Fault F-3 is the most distant fault in the system. Therefore, it produces the least amount of

current. It is necessary to check and see if the relays will pick-up during this fault.• OR-3 will pick-up because 6.28 A (F-3 fault current at CT-3) > 0.907 A (setting of OR-3).• OR-2 will pick-up because 3.14 A (F-3 fault current at CT-2) > 0.907 A (setting of OR-2).

• OR-1 will pick-up because 1.963 A (F-3 fault current at CT-1) > 1.134 A (setting of OR-1).

Based on the sensitivity check, the protection system should work correctly. OR-2 provides back-up for OR-3 and OR-1 provides back-up for both OR-2 and OR-3.

Prediction of average clearing times• Faults on load 3 (5 msec (relay) + 25 msec (breaker) = 30 msec.

• Faults on line 2 (50 msec (relay) + 20 msec (breaker) = 70 msec.• Faults on line 1 (90 msec (relay) + 15 msec (breaker) = 105 msec.

Simulations

This subsection describes simulations performed using the available software. The exercises are

designed to validate the calculations from the previous subsection and to investigate theoperation of the protection system.

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Exercise 1

Disconnect the fault and run a simulation for the undisturbed system. Observe the currents and

voltages. Check if they match your calculations.

Figure 3.4 shows the current through load 3. The magnitude of the current reads 0.685 A

(secondary) instead of 0.756 A (secondary). The difference is due to the fact that the voltage at busbar 3 differs from the rated value. The load model maintains the power. Therefore, if the

voltage is not at the rated value (90.6V secondary instead of 100V secondary - see Figure 3.5),then the current will not be at the rated value either.

Figure 3.6 shows the current at CT-2. The magnitude of the current reads 0.685 A (secondary)instead of 0.756 A (secondary). The difference is due to the same reason discussed above. A

similar relation holds true for the current at CT-1. Although, load 1 operates at a voltage abovethe rated value - 105.5 V secondary (see figure 3.7).

The simulation validates the calculations. The differences result from the fact that the rated

voltage was assumed to exist when the load currents were calculated. The simulated values arecertainly more accurate. However, for protective relaying studies, the accuracy of thecalculations presented earlier is sufficient.

Figure 3.4 Load 3 current Figure 3.5 Voltage at bus 3

Figure 3.6 Load current at CT-2 Figure 3.7 Voltage at bus 1

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Exercise 2

Place the fault at F-3 on the system and disconnect all the relays so that the fault is not cleared.

Observe the fault current. Check if it matches your calculations.

Figure 3.8 displays the fault current. The current shows significant dc component decaying after

about 4 cycles. The steady-state magnitude reads 5.82 A (secondary at CT-3), while thecalculated value is 6.28 A. The difference results from the load current superimposed on the fault

current. Our short-circuit calculations neglected the load current according to the common andreasonable approach.

Figure 3.8 F-3 fault current (secondary A at CT-3)

Exercise 3

Reconnect the trip signals of the relays. Observe the system and explain its behavior. What is the fault clearing time?

Figure 3.9 shows the current at CT-3 and Figure 3.10 shows the voltage at bus 3. The relayoperates after 6 ms and the fault is cleared by OR-3 34 ms after the fault. The voltage drops to

zero during the fault and then raises up once the fault is cleared.

Figure 3.11 shows the current at CT-2. The current jumps from the normal load value to the faultvalue and then drops after the fault is cleared by OR-3. The post-fault magnitude is lower because the faulted load 3 is disconnected. Notice that relay OR-2 performs properly and does

not trip. The voltage at bus 2 behaves similarly to the voltage at the bus 3.

Figure 3.12 shows the current flowing through CT-1. The current at CT-2 is measured by OR-1.

A jump from the normal load value to the fault value and a drop after the fault is cleared by OR-3 can be observed. The post-fault magnitude is lower because the faulted load 3 is disconnected.

The relay OR-1 does not trip (correct behavior). The voltage at the bus 1 behaves similarly to the

voltage at the bus 3. However, the voltage drop is much lower because the fault is more distant(see Figure 3.13).

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Figure 3.9 Current at CT-3 (secondary A) Figure 3.10 Voltage at bus 3 (secondary V)

Figure 3.11 Current at CT-2 (secondary A) Figure 3.12 Current at CT-1 (secondary A)

Figure 3.13 Voltage at bus 1 (secondary V)

Exercise 4

Place the fault in the middle of the line 2. Observe the system and explain its behavior. What is

the fault clearing time?

Figure 3.14 shows the current at CT-2. Relay OR-2 operates 52 ms after the fault occurs. The

fault is cleared after additional 24 ms. The voltage at bus 2 drops during the fault and thenrebuilds once the fault is cleared (see Figure 3.15). Relay OR-3 does not operate because it does

not measure any fault current (see Figure 3.16). Figure 3.17 shows that the current at CT-1 jumpsfrom the normal load value to the fault value and then drops after the fault is cleared by OR-2.

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The post-fault magnitude is lower because the loads 3 and 4 are disconnected. Relay OR-1 doesnot trip (correct).

Figure 3.14 Fault Current at CT-2 (secondary A) Figure 3.15 Voltage at bus 2 (secondary V)

Figure 3.16 Fault Current at CT-3 (secondary A) Figure 3.17. Fault Current at CT-1 (secondary A)

The protection system operates correctly. When line 2 suffers a fault, it is disconnected by OR-2.Since loads 3 and 4 can only be fed by line 2, they are left out of service due to the fault. The restof the system is left in service. This includes line 1 and loads 1 and 2.

Exercise 5

Place the fault in the middle of the line 1. Observe the system and explain its behavior. What isthe fault clearing time?

Figure 3.18 shows the current at CT-1. OR-1 operates 92 ms after the fault occurs. The fault iscleared after additional 109 ms. The voltage at bus 1 drops during the fault and then rebuilds

once the fault is cleared (see Figure 3.19). The downstream relays OR-2 and OR-3 do not operate because they do not measure any fault current.

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Figure 3.18 Fault Current at CT-1 (secondary A) Figure 3.19 Voltage at bus 1 (secondary V)

The protection system operates correctly. When line 1 suffers a fault, it is disconnected by OR-1.Since loads 2, 3 and 4 can only be fed by the line 1, they are left out of service due to the fault.The rest of the system (load 1) is left in service.

Exercise 6

Place the fault at the load 3. Disconnect OR-3 from the CB-3 to simulate relay/breaker mal-operation. Observe the system and explain its behavior. What is the fault clearing time?

Relay OR-3 operates after 6 ms (compare to exercise 3). However, the fault should be cleared 25ms after the relay operation. The reason this does not happen is because the trip signal is not

relayed to the circuit breaker (broken wire) or the CB-3 fails to operate and the fault is notcleared. This is an example of a fault situation combined with a malfunction of the

protection/switching equipment.

In order to provide system protection in such cases, there needs to be at least one extra level of

back-up protection. In the case of the overcurrent relaying, the next relay upstream acts as a back-up element. In this situation, relay OR-2 picks-up and operates 52 ms later. Breaker CB-2

then opens 76 ms after the fault. As a result of this sequence of events, line 2 gets removed fromservice and load 4 becomes disconnected. However, since there is a fault at load 3 and CB-3 failsto trip, the only way to clear this fault is to operate CB-2. This action is optimal even though

more loads are left out of service.

The relay upstream (OR-1) also picks-up but does not trip because the fault gets cleared by OR-2

and CB-2. However, if CB-2 fails as well, OR-1 will trip taking out both the lines 1 and 2.

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Figure 3.20 The fault current at CT-3 (sec. A) Figure 3.21 The fault current at CT-1 (sec. A)

Exercise 7 Place the fault in the middle of the line 2. Disconnect OR-2 from the CB-2 to simulaterelay/breaker mal-operation. Observe the system and explain its behavior. What is the fault clearing time?

As in the previous exercise, the primary relay for line 2 (OR-2) operates but CB-2 fails to open

the circuit and the fault is not cleared. Therefore, the back-up relay upstream (OR-1) picks-up.Since the fault lasts beyond the time delay of OR-1, the relay trips 93 ms after the fault and breaks the supply through line 1 109 ms after the fault (see Figure 3.22). As a result of this

sequence of events, line 1 gets removed from service and the load 2 is disconnected. However,since there is a fault on the line 2 and the CB-2 fails to trip, the only way to clear this fault is to

operate CB-1. The action is optimal even though more loads are left out of service.

Figure 3.22 Fault current at CT-1 (secondary A)

Exercise 8

Place faults at different locations with different fault resistances and check if the protection

system works correctly. Include mal-operation of the relays and/or breakers.


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