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Power System Study & Relay Co-ordination

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1.SYNOPSISA Power System consists of various electrical components such as Generating units,

transformers (Power and Distribution), transmission lines, isolators, circuit breakers, bus bars,cables, relays, instrument transformers, distribution feeders, and various types of loads.

Faults may occur in any part of power system such as short circuit & earth fault. Faults may

be of the following types-Single Line to Ground (SLG), Double Line to Ground (DLG), Line to

Line (LL), three phase short circuit etc. This results in flow of heavy fault current through the

system. Fault level also depends on the fault impedance which depends on the location of fault

referred from the source side. To calculate fault level at various points in the power system, fault

analysis is necessary.

The protection system operates and isolates the faulty section. The operation of the protection

system should be fast and selective i.e. it should isolate only the faulty section in the shortest

possible time causing minimum disturbance to the system. Also, if main protection fails to

operate, there should be a backup protection for which proper relay co-ordination is necessary.

The scope of the project is

1. To study and analyze a part of power system in Tata Motors Ltd. with respect to fault

analysis at different fault locations and verifying the short time current withstand capacity of the

protective devices placed in the system.

2. To study and ensure the Relay Co-ordination in the part of power system.

3. To analyze the effect of addition of 54.35 MW captive generation plant on the fault level.

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2. INTRODUCTION

A power system consists of Generating Units, Transformers (Power and Distribution) andTransmission lines. The system must be protected against flow of heavy short circuit currents

which cause permanent damage to the equipments if not cleared within sustainable time. For this

purpose circuit breakers and protective relaying is provided to disconnect the faulty section of

the system. Switchgear and protection devices are installed at each voltage level for normal

routine switching, control and monitoring and automatic switching during abnormal conditions

like short circuits, over current etc.

Short circuit currents in ac system are determined mainly by reactance of generators,

transformers and lines up to the point of fault in case of phase to phase faults. In case of circuit

breakers, their rupturing capacities are based on symmetrical short circuit current which is the

most severe amongst all other types of fault currents. So we are going to consider three phase

symmetrical short circuit fault for fault analysis of the power system.[1]

The power system in Tata Motors Ltd. consists of supply from MSEDCL and its own Captive

Power Plant. The Captive Power Plant is run during load shedding exercise to maintain

continuous supply. Captive power is supplied by OLD-DG House and MAN-DG House which in

all consists of 11generators. Out of these 11 generators- 8 generators are of OLD-DG House, of

which 6 are of 2.5 MW and remaining 2 are of 2.2 MW, 3 generators are of MAN-DG House

each of 11.65 MW each. Thus the Captive Power Plant in all contributes 54.35 MW.

For short circuit analysis we assume three phase short circuit faults at different voltage levels

and locations. Three phase short circuit faults are most severe faults and give pessimistic results.

As per IS Standards 10% overvoltage is permitted during normal condition. So while calculating

fault levels we consider that the faults occur at 10% overvoltage, to avoid any risk to the

protective gear as well as the equipment.

Faults in a particular section are cleared by its protective switchgear .i.e. its primary

protective switchgear, failing which the back-up protection should act. The back-up protection

should not act prior to primary protection but also within sustainable time. For achieving this,

proper relay co-ordination is necessary.

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3. POWER SYSTEM

3.1 Single Line Diagram:-

A complete diagram of power system representing all the three phases becomes too

complicated and cumbersome for a system of practical size. Hence it is much more practical to

represent a power system by means of simple symbols for each component. This representation

is called a single line diagram.

In this diagram, generator and load are represented by a circle, transformer is represented by a

primary and secondary coil, circuit breaker is represented by a square, transmission line by single

horizontal line and bus-bar by single vertical line. Generator and transformer connections such as

star, delta and neutral earthing are indicated by a symbol drawn by the side of their

representation. Ratings of the generators, transformers and motor are mentioned below the

diagram.[2]

Consider a power system shown in the figure:-

Fig. 1 Typical Single Line Diagram of Power System

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3.2 Working of Power System:-

3.2.1Normal Condition:-

Power is generated by generator, acting as a source, within the range of 11kV to 25 kV. This

voltage is stepped-up with the help of step-up transformer (T 1) up to 66 kV to 765 kV or higher

to reduce transmission losses. This power is transmitted over the transmission lines .This voltage

is stepped down to a level, as desired by the load, with help of step-down transformer (T 2).[1]

3.2.2 Abnormal Condition:-

Abnormal condition or fault is nothing but a defect in electrical circuit of an electrical

equipment due to which current is diverted from intended path. As the fault impedance is low,

fault currents are relatively high. During faults, power flow is diverted towards the fault and the

supply to the neighboring zone is affected. In order to isolate the faulty section from the healthy

part and to maintain continuity of supply, circuit breakers are employed in power system.[1]

3.3Need for Protection of Power System:-

Modern power systems are growing with more generators, transformers and large network in

the systems. For system protection, a high degree of reliability is required. In order to protect the

system from damage, due to fault currents and/or abnormal voltages caused by faults, need for

reliable protective devices, such as relays and circuit breakers arises. The most common

electrical hazard against which protection is needed is the short circuit. Also protection is

required against overloads, over-voltage, under-voltage, open-phase, power swings, under and

over-frequency, instability etc.[2]

Faults:-

A fault in electrical equipment is defined as a defect in the electrical circuit due to which

current is diverted from intended path. The fault impedance is low so fault currents are relatively

high. In an electrical power system comprising of generator, transformer, transmission lines, load

and switchgear, faults are inevitable.

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Causes:-

Sr.

No.

Equipment Cause of Fault Percent of Total

Fault

1. Alternators (Generators) --Stator Faults

--Rotor Faults

--Faults in associated equipments

--Faults in protective system

6 - 8

2. Transformers --Insulation Failure

--Faults in tap changer

--Faults in bushing

--Faults in protection circuit

--Overloading, over-voltage

1012

3. Current Transformers

&

Potential Transformers

--Over-voltages

--Insulation failures

--Breaking of conductors

--Wrong connections

1520

4. Switchgear --Insulation failure

--Mechanical defect--Leakage of air/oil/gas

--Inadequate ratings

--Lack of maintenance

10 - 12

Table1. Faults and Percent Contribution

Effects:-

The nature of faults simply implies any abnormal condition which causes a reduction in basic

insulation strength between conductors. Faults in certain important equipment can affect stability

of power system. During fault voltages of three phases become unbalanced. For example, if a

fault occurs in a motor, the motor winding is likely to get damaged. Further if motor is not

disconnected quickly, excessive fault currents can cause damage to starting equipment, supply

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connections etc. A fault in bus zone of power station can cause tripping of all generator units in

power station.[1]

To avoid these severe effects due to faults we need to protect the power system with the help

of switchgear protective devices viz. relays, circuit breakers, lightning arrestors etc.

3.4 Types of Fault:-

3.4.1 Symmetrical Faults:-

A fault involving all three phases is known as a symmetrical (balanced) fault.

Types of symmetrical faults:-

A) All three phases to ground (L-L-L-G)B) All three phases short circuited (L-L-L)This type of fault occurs infrequently. It is an important type of fault with simple calculations

and pessimistic answers .This type of fault imposes the most severe duty on circuit breakers and

therefore used in the determination of circuit breaker ratings.

3.4.2 Unsymmetrical Faults:-

A fault involving one or two phases is known as an unsymmetrical (unbalanced) fault.

Types of unsymmetrical faults:-1) Single phase to ground (L-G)

2) Phase to phase (L-L)

3) Two phases to ground (L-L-G)

4) Phase to phase and third phase to ground

These types of faults occur usually in the power system.[1]

3.5 Protection:-

Practical power system consists of large number of generators, transformers and load

connected in a complex network. No part of the power system can be left unprotected. For a

system to operate a high degree of reliability is required. In order to protect the system (lines and

equipments) from damage due to fault currents and/or abnormal voltages caused by faults, the

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need for reliable protective devices, such as relays and circuit breakers arises. Choice of

protection depends upon type and rating of protected equipment, its importance, location,

probable abnormal conditions, costs etc.

Following are the conditions for which protection is required:-

A. Short CircuitB. OverloadC. Under-voltage and Over-voltageD. Open PhaseE. Unbalanced Phase CurrentsF. Reversal of PowerG. Under-frequency and Over-frequencyH. Over-temperatureI. Power SwingsJ. Instability

The occurrence of short circuits may lead to heavy disturbances in normal operation (damage

to equipment, impermissible drop in voltage etc). The protective scheme is designed to

disconnect or isolate the faulty section from the system without any delay.

Main functions of protection are to detect the presence of faults and their locations. The

protective devices should initiate the action for quick removal from service of any element in

case of short circuit or in abnormal condition which may hamper the effective operation of the

rest of the system.[2]

The components usually used for protection of system are relays, circuit breakers, isolators,

instrument transformers etc.

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Basic circuit diagram of protective scheme:-

1- Circuit Breaker

2- Relay

3- Trip Coil of Circuit Breaker

4- Trip Circuit

5- Battery

6- Relay Contacts

7- Potential Transformer

8- Current Transformer

9- Auxiliary Switch Contacts

Fig.2 Basic Circuit Diagram of Protective Scheme [1]

3.5.1 Working of Protective Scheme:-

Figure-2 shows basic connections of circuit breaker control for the opening operation.

The protected circuit X is shown by dashed line. When a fault occurs in the protected circuit the

relay (2) connected to CT and PT actuates and closes its contacts (6).

Current flows from battery (5) in the trip circuit (4). As the trip coil of circuit breaker (3)

is energized, the circuit breaker operating mechanism is actuated and it operates for the opening

operation.

Thus the fault is sensed and the trip circuit is actuated by the relay and the faulty part is

isolated.[1]

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3.6 Protective Relaying:-

The protective relaying of a power system is planned along with the system design.

Protective relaying senses the abnormal condition in a part of power system and gives an alarmor isolates that part from healthy system. Protective relaying is a team work of CT, PT,

protective relays, time delay relays, trip circuits, circuit breakers etc. Protective relaying plays an

important role in minimizing the faults and also in minimizing the damage in the event of

faults.[3]

3.6.1 Functions of protective relaying:-

A) To sound an alarm or to close the trip circuit of a circuit breaker so as to disconnect acomponent during an abnormal condition in the component.

B) To disconnect the abnormally operating part so as to prevent subsequent faults. For e.g.Overload protection of a machine not only protects the machine but also prevents

insulation failure.

C) To disconnect the faulty part quickly so as to minimize the damage to the faulty part.For example - If machine is disconnected immediately after a winding fault, only a few

coils may need replacement. But if the fault is sustained, the entire winding may get

damaged and machine may be beyond repairs.

D) To localize the effect of fault by disconnecting the faulty part from healthy part, causingleast disturbance to the healthy system.

E) To disconnect the faulty part quickly so as to improve system stability, service continuityand system performance. Transient stability can be improved by means of improved

protective relaying.[1]

3.6.2 Desirable qualities of protective relaying:-

A) Selectivity, Discrimination F) StabilityB) Sensitivity, Power consumption G) System SecurityC) ReliabilityD) AdequatenessE) Speed, Time

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3.7 Primary and Back-up Protection:-

For attaining higher reliability, quick action and improvements in operating flexibility of the

protection schemes, separate elements of a power system , in addition to main or primary

protection , are provided with a back-up and auxiliary protection.

First in line of defense is main protection which ensures quick action and selective clearing of

faults within the boundary of the circuit section or the element it protects. Main protection is

essentially provided as a rule.

Back up protection gives back up to the main protection, when the main protection fails to

operate or is cut out for repairs etc.

Failure of the main protection may be due to any of the following reasons:-

A) D.C supply to the tripping circuit failsB) Current or voltage supply to the relay failsC) Tripping mechanism of the circuit breaker failsD) Circuit breaker fails to operateE) Main protective relay fails

Back up protection may be provided either on the same circuit breakers which will be opened by

the main protection or may use different circuit breakers. Usually, more than the faulty section is

isolated when the back up protection operates. Very often the main protection of a circuit acts asback up protection for the adjacent circuit. Back up protection is provided where main protection

of the adjacent circuit fails to back up the given circuit. For simplification, back up protection

can have a lower sensitivity factor and be operative over a limited back up zone i.e. be operative

for only part of the protected circuit.

Methods of back up protection can be classified as follows:-

A) Relay Back-upB) Breaker Back-upC) Remote Back-upD) Centrally Co-ordinated Back-up

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Back-up protection by Time Grading principle:-

Fig.3 Primary and Back-up Protection by Time Grading Principle

In this, current is measured at various points along the current path, for e.g., at source,

intermediate locations, consumers end. The tripping time at these locations are graded in such a

way that the circuit breaker nearest to the faulty section operates first, giving primary protection.

The circuit breaker at the previous section operates only as a back-up.

In Fig.3 the tripping time at sections C, B and A are graded such that for a fault beyond C,

breaker at C operates as a primary protection. Relays at A and B also may start operating but

they are provided with enough time lags so that breaker at B operates only if breaker at C does

not.

Thus, for a fault beyond C, breaker at C will operate after 0.1 second. If it fails to operate, the

breaker at B will operate after 0.6 second (Back-up for C) and if the breaker at B also fails to

operate, breaker at A will operate after 1 second (Back-up for B and C).[1]

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3.8 Instrument Transformers:-

The values of voltage or current in a power circuit are too high to permit convenient direct

connection of measuring instruments or relays, so coupling is made through instrumenttransformers. These instrument transformers are required to produce scale down replica of the

input quantity to the accuracy expected for particular operation. The performance of instrument

transformers during and following large instantaneous changes in input quantity is important, in

that this quantity may depart from sinusoidal waveform. This deviation may persist for an

appreciable period. The resulting effect on instrument performance is usually negligible,

although for precision metering, a persistent change in accuracy of transformer may be

significant.

As many protective systems are required to operate in a time shorter than period of transient

disturbance in the output of instrument transformers following a system fault. The errors in

instrument transformers may abnormally delay the operation of protection or cause unnecessary

operation. So the functioning of these transformers must be examined analytically.

If the primary is energized while the secondary winding is open circuited and transformer will

become, an iron-cored inductor and will present relatively high impedance. A current will flow

and voltage drop will develop across the winding in proportion to its impedance. The current will

be entirely expended in magnetizing the core. The voltage drop in primary winding is because of

the e.m.f induced due to flux and e.m.f induced in secondary winding. If the circuit of the

secondary winding is closed through impedance, a proportionate current will flow; this current

produces m.m.f which opposes the flux. The tendency of the flux to be reduced by

demagnetizing force combined with corresponding reduction in primary back e.m.f causes

increase in primary current. If primary winding is lossless and the applied voltage is constant, the

flux will be maintained at initial value and increase in primary m.m.f would be identical to that

of secondary winding.[3]

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3.8.1 Current Transformers:-

Current Transformers are connected in AC power circuits for indication and metering

purposes and for protective relays.

Current transformer basically consists of an iron-core on which primary and one or two

secondary windings are wound. Primary winding of CT is connected in series with load and

carries actual power system current (normal and fault) while secondary is connected to

measuring circuit or relay. Primary winding is usually single turn winding and number of turns

on the secondary winding depend upon the current to be carried by power circuit. Larger the

current, more is the number of turns on secondary.

Ratio of primary current to secondary current is known as transformation ratio of CT. Current

ratio of CT is usually high. Secondary current ratings are of order of 5 A, 1A and 0.1A. Primary

current ratings vary from 10A to 3000A or more. The current in secondary winding of CT is

governed by current flowing in the primary winding of CT and not by load impedance on

secondary.

3.8.2 Potential Transformers:-

Potential transformers are employed for voltages above 380V to feed potential coils of

indicating, measuring instruments and protective relays.

The primary winding of PT is connected directly to power circuits. To the secondary winding

various indicating metering instruments and relays are connected. Primary has large number of

turns while secondary has a much smaller number of turns. The primaries of PT are rated from

400 volts to several thousand volts and the secondary always for 110 volts. The ratio of rated

primary voltage to rated secondary voltage is known as transformation ratio.[3]

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4. RELAYS AND CIRCUIT BREAKERS

4.1 Relays

In a modern power system to have a normal operation of the system, without electrical

failure and damage to the equipment, there are two alternatives available- one is to design the

system so that faults do not occur; second is to take steps to protect the equipments from the ill-

effects of faults. As it is impossible to eliminate faults, the latter alternative is the only

alternative. Protective relay functions as a sensing device in a protection scheme, it senses the

fault, then determines its location and finally, sends tripping signal to the circuit breaker and the

circuit breaker disconnects the faulty part. So, the protective relay is the brain behind the

protection scheme and plays a vital role. Lesser the time required for clearing the fault, lesser is

the damage incurred.

To achieve all the above mentioned objectives, proper care should be taken in designing

and selecting an appropriate relay which must be reliable, efficient and fast in operation. The

relay should be sensitive enough to distinguish between normal and abnormal (faulty)

conditions.

4.1.1 Classification of Relays:

A) Attracted ArmatureB) Moving CoilC) InductionD) ThermalE) Motor OperatedF) MechanicalG) Magnetic AmplifierH) ThermionicI) SemiconductorJ) Photo-electric

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4.1.2 Induction Relay:-

The basic principle of induction motors is applied to relays designed to operate on the

induction principle. The moving conductor is placed in the two magnetic fields, displaced both in

time and phase, and produces the required torque. The two fields are derived from a single

quantity by energizing two electromagnets with the required phase shift. Another arrangement

can be that of energizing two magnets by separate sources. In both the cases, the torque

generated is given by:

T= K12 sin (i)

Where T= torque

1, 2 = flux produced in the two electromagnets and = angle between 1 and 2

Principle of Working:-

Fig. 4 Induction Relay

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In this arrangement, a U-shaped electromagnet and E-shaped electromagnet are used with a

disc free to rotate in between. A phase displacement between the fluxes produced by two

magnets is obtained by energizing two circuits whose outputs are relatively displaced in-phase.

The E-shaped electromagnet carries two windings; the primary and the secondary. The

primary winding carries relay current I1 while secondary winding is connected to the U-shaped

electromagnet. The primary current induces e.m.f in secondary and so circulates I2 in it. The flux

2 induced in U-shaped magnet by current in secondary winding of E-shaped magnet will lag

behind flux1 by an angle . The current generates a flux across the air gap which passes

through an aluminum disc placed in the air gap. This phase difference will develop a driving

torque on the disc given by equation (i).

This design is generally applied to over-current and over-voltage relays. The restraining force

is achieved by a spiral spring, the force of which must be overcome by the driving torque before

any operation can begin; this determines the setting or minimum operating current of the relay.

The disc is further controlled by a permanent magnet which produces an eddy current braking

torque, this torque being proportional to the speed at which the disc rotates.

Braking:-

It is important that the motion of the disc shall be limited to correct amount, proportional to

energizing current and its duration; that is, due to kinetic energy, after current cessation must be

as small as possible. The energy is minimized by keeping the disc weight low, using aluminum

as constructional material. In addition the operating and braking torques are made high so that

stored energy is quickly dissipated.

Plug Settings:-

One of the windings of the upper electromagnet is connected to secondary of CT in the line to

be protected and is tapped at intervals. The tappings are connected to a plug setting bridge by

which the number of turns in use can be adjusted, by giving the desired current settings. The plug

bridge is usually arranged to give seven sections of tappings to give over-current range from

50% to 200% in steps of 25%. If the relay is required to response for earth fault, steps are

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arranged to give a range from 10% to 70% or 20% to 80% in steps of 10%. The value assigned

to each tap are expressed in terms of full-load rating of CT with which the relay is associated and

presents a value above which the disc commences to rotate and finally closes the trip circuit.

Thus pick-up current equals the rated secondary current of CT multiplied by current setting.

For e.g.:- Suppose that an over-current relay having a setting of 50% is connected to a supply

circuit through CT of 500/5 A. The rated secondary current of CT is 5A and therefore pick-up

value will be 1.5 X 5 = 7.5 A. It means that with above current setting, the relay will actually

operate for a relay current equal to or greater than 7.5 A. Similarly for current settings of 50, 100

and 200%, the relay will operate for relay currents of 2.5 A, 5 A and 10 A respectively. The taps

are selected by inserting a single pin plug in appropriate position on a plug setting bridge.

When the pin is withdrawn for the purpose of changing the setting value, relay automaticallyadopts higher setting, thus the CTs secondary is not open circuited.

Time Setting Multiplier:-

In order to apply the relay in the power system it is necessary to be able to modify the time

scale of time-current characteristic. This can be achieved by control of amount of disc

movement, since operating time is proportional to such movement at any given current value.

The spring torque varies over angle of disc travel, so that the disc speed would vary and the timecharacteristic would change in shape for different values of d. To avoid this, disc is given a non-

circular shape, so that the radius measured through electromagnet pole increases as the disc is

rotated from start to contact make position. The increase in radius provided disc currents with a

wider path and hence causes the driving torque to increase, the amount of this change is

proportional to spring rate. This compensation makes possible the calibration of time adjustment

as a multiplier for use with a single characteristic curve, over a wide range.

The time multiplier setting (TMS) decides arc length through which disc travels, by reducing

length of travel, operating time is reduced. TMS is calibrated from 0 to 1 in steps of 0.05. These

figures do not represent actual operating time. These are multipliers to be used to convert the

time known from the relay nameplate curve (time-PSM curve) into the actual operating time.

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For e.g.:- If time setting is 0.2 and operating time obtained from time-PSM curve of relay is 5

seconds, then actual operating time of relay will be equal to 0.2 X 5 = 1 second.

When the relay picks-up the spring unwinds and the disc rotates to close the contacts. The

time multiplier settings are used to wind and unwind the spring. If more time of operation of

relay is needed, the spring is wound more. More the driving torque, lesser will be the time

required to operate. So the relay has inverse-time characteristics.

A set of typical time-current characteristics of an over-current induction disc relay is shown in

fig4. The horizontal scale is marked in terms of plug setting multiplier. It represents the number

of times the relay current is in excess of current setting. The vertical scale is marked in terms of

the time required for relay operation. The abscissa is taken as multiple of pick-up value so that

the same curves can be used for any value of pick-up i.e. if the curves are known for pick-up

value of 5A then the characteristics remain same for 2.5 A, 6.25 A, 7.5 A, 10 A or any other

pick-up value.

These curves are normally plotted on log-log graph papers as shown in fig 4. The advantage

of this is that if the characteristic for one particular pick-up value and one time multiplier setting

is known then characteristics can be obtained for any other pick-up value and time multiplier

setting.

The curves are used to estimate not only the operating time of relay for a given multiple of

pick-up value and time multiplier setting but also it is possible to know how far the relay moving

contacts would have travelled towards fixed contacts within any time interval. This method is

also useful in finding out whether the relay will pick-up and how long it will take for the relay

operation when the actuating activity is changing during the in-rush current period of starting a

motor.[1]

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Fig. 5 Current-Time Characteristics of an Over-Current Induction Disc Relay

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Fig. 6 CDG Over-Current Induction Disc Relay (Inverse) Type

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4.1.3Static relays-

A static relay refers to a relay in which there is no armature or other moving element and

response is developed by electronic, magnetic or other components without mechanical motion.

The solid state components used are transistors, diodes, resistors, capacitors, and thyristors etc.

Measurement is carried out by static circuits consisting of comparators, level detectors, filters

etc., while in a conventional electro-magnetic relay it is done by comparing operating torque (or

force) with restraining torque (or force).Static relays can be arranged to respond to electrical

inputs. However, other types of inputs such as heat, light, magnetic field, travelling waves etc.,

can be suitably converted into equivalent analogue or digital signals and then supplied to the

static relay.

Operating Principle:-

Fig. 7Block Diagram of a Static Relay

Static Relays consists of Input Stage and Output Stage.

Input Stage:-

The input is derived from line CT and PT. The output of these CT and PT are not suitable for

static components so they are brought down to suitable level by auxiliary CT and PT. Then aux

CT output is given to rectifier, which rectifies the input. This is then smoothened by

smoothening circuit to remove the ripple. The smoothened and ripple free output is given to

comparator. The comparator compares the inputs and generates error signal which is given to

level detector. The level detector determines its input level with respect to its predetermined

setting and gives output only if input is greater than threshold value. The output of level detector

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is then amplified by amplifier to strengthen the weak signal. The amplifier output is then given to

output device. Time delay can be introduced between two level detectors if needed.

Output Stage:-

Output stage of static relays consists of either permanent magnet moving coil relay (PMMC)

or thyristors in series with trip coil of circuit breaker.

4.1.4 Microprocessor Based Relays & Numerical Relays:-

Fig. 8 Microprocessor based Relay

The increased growth of power system both in size and complexity has brought about the

need for fast and reliable relays to protect major equipments and to maintain system stability.

The conventional protective relays are either electromagnetic or static type. Electromagnetic

relays suffer from high burden on instrument transformers, high operating time, contact problems

etc. Though static relays have certain advantages such as compactness, low burden, less

maintenance and high speed over electromagnetic relays but they do suffer from inflexibility,

inadaptability to changing operating conditions of system and its complexity.

With the development of economically powerful and sophisticated microprocessors, there is a

growing interest in developing microprocessor-based relays which are more flexible as they are

programmable and they are very much superior to static relays and conventional electromagnetic

relays.

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Operation:-

A current signal from CT is converted into proportional voltage signal using I to V converter.

The ac voltage proportional to load current is converted into dc using precision rectifier and is

given to multiplexer (MUX) which accepts more than one input and gives one output.

Microprocessor sends command signal to the multiplexer to switch on desired channel to accept

rectified voltage proportional to current in a desired circuit. Output of MUX is fed to analog to

digital converter (ADC) to obtain signal in digital form. Microprocessor then sends a signal ADC

for start of conversion (SOC), examines whether the conversion is completed and on receipt of

end of conversion (EOC) from ADC, receives the data in digital form. The microprocessor then

compares the data with pick-up value. If the input is greater than pick-up value the

microprocessor send a trip signal to circuit breaker of the desired circuit.

Incase of instantaneous over current relay there is no intentional time delay and circuit

breaker trips instantly. Incase of normal inverse, very inverse, extremely inverse and long inverse

over current relay the inverse current-time characteristics are stored in the memory of

microprocessor in tabular form called as look-up table.

4.1.5 Merits of Microprocessor based, Numerical & Static relays

A) Flexibility- A variety of protection functions can be accomplished with suitablemodifications in the software only either with the same hardware or with slight

modifications in the hardware.

B) Reliability- A significant improvement in the relay reliability is obtained because the useof fewer components results in less interconnections and reduced component failures.

C) Obtaining different types of relay characteristics- given the system requirements, it ispossible to provide better 0matching of protection characteristics since these

characteristics are stored in the memory of the microprocessor.

D) Digital communication- The microprocessor based relay furnishes easy interface withdigital communication equipments.

E) Modular frame- The relay hardware consists of standard modules resulting in ease ofservice.

F) Low burden- The microprocessor based relays impose minimum burden on theinstrument transformers.

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G) Sensitivity - Greater sensitivity and high pickup ratio.H) Speed- With static relays, tripping time of cycle or even less can be obtained.I) Resetting is also fast.

However static relays suffer from some limitations as follows:-

Limitations of static relays:-

A) Auxiliary voltage requirement.B) Electrostatic Discharges-These charges are developed by rubbing of two insulating

components. Even small discharges can damage the components which would normally

withstand 100 V.

C) Voltage transients-Static relays are sensitive to voltage transients which are caused byoperation of breaker and isolator in the primary circuit of CTs and PTs.

D) Serious overvoltage is also caused by breaking of control circuit, relay contacts etc. Suchvoltage spikes of small duration can damage the semiconductor components and also

cause mal operation of relays.

E) Temperature dependence of static relays- The characteristics of semiconductor devicesare affected by ambient temperature.

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4.2 Circuit Breaker:-

Circuit Breakers are automatic switches which can interrupt fault currents. During normal

operating conditions the circuit breaker is in closed position. During abnormal or faulty

conditions, relays sense the fault and close the trip circuit of circuit breaker. There after the

circuit breaker opens. So circuit breaker is the device which actually isolates the faulty part and

is final output device of protective scheme. On opening of circuit breaker contacts an arc is

drawn out between them. This arc can be extinguished using different media like SF6 gas,

vacuum etc.

4.2.1 SF6 Circuit Breaker:-

Sulphur hexafluoride (SF6) is an inert, heavy gas with good dielectric strength and arc

extinguishing properties. The dielectric strength of SF6 gas is greater than that of atmospheric air

and it increases with pressure. SF6 widely used in electrical equipment like high voltage metal

enclosed cables, high voltage metal clad switchgear, bushings, circuit-breakers, current

transformers etc.

Single Puffer Action:-

Initially the interrupter is in fully closed position. The moving cylinder (1) is coupled withmovable contact (2) against fixed piston (5) and there is a relative motion between moving

cylinder and fixed piston and gas is compressed in cavity (6). This trapped gas is released

through nozzle (4) during arc extinction process. During travel of moving contact and moving

cylinder the gas puffs over arc and reduces arc diameter by axial convection and radial

dissipation. At current zero arc diameter becomes too small to arc gets extinguished. The puffing

action continues for sometime even after arc extinction and contact space is filled with cool,

fresh gas. Due to electro negativity of gas it regains its dielectric strength rapidly after final

current zero.

Several types of SF6 circuit breakers have been developed for rated voltages from 3.6 to 760

kV. SF6 gas insulated metal-clad switchgear comprises factory assembled metal-clad, substation

equipment like circuit breaker, isolators, earth switches, bus-bars etc. These are filled with SF6

gas. Such substations are compact and are being favored in densely populated urban areas.

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Fig. 9 Single Puffer Action of SF6 Circuit Breaker

4.2.2 Vacuum Circuit Breaker:-

Fig. 10 Vacuum Interrupter

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Vacuum is used as dielectric material. When two contacts are separated in vacuum module

arc is drawn between them. An intensively hotspot is created at the instant of contact separation

from which metal vapors shoot off, constituting plasma. The amount of vapor in plasma is

proportional to the vapor emission from electrodes, hence to the arc current. As contacts separate

contact space is filled with vapor of positive ions liberated from contact material. During

decreasing mode of current the rate of vapor emission reduces and amount of plasma tends to

zero. After natural current zero the remaining metal vapor condenses and medium regains the

dielectric strength rapidly and thus striking of arc is prevented.

Vacuum interrupter is rated up to 36 kV and beyond which two interrupters are required.[1]

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5. PROTECTION OF POWER SYSTEM COMPONENTS

5.1 Generator Protection:-

Protection of generator is complex and elaborate because of following reasons:-

A) Generator is costly equipment and one of the major links in power system.B) Generator is not single equipment but is associated with the unit-transformers, auxiliary

transformers, station bus-bars, excitation system, prime-mover, voltage regulating

equipment, cooling system etc. Therefore the protection of generator is to be coordinated

with the associated equipment.

C) The generator capacity has sharply risen in recent years from 30MW to 500MW with theresult that loss of even a single machine may cause overloading of associated machines in

the system and eventual system instability.

The basic function of protection applied to generators is, therefore, to reduce the outage

period to a minimum by rapid discriminative clearance of faults. Unlike other apparatus, opening

of a breaker to isolate the faulty generator is not sufficient to prevent further damage, since

generator will continue to supply power to a stator winding fault until its field excitation is

suppressed. It is, therefore, necessary that the field is opened, fuel supply to the prime-mover is

stopped and sometimes braking application also becomes imperative.

Overloading of a generator is caused either due to partial breakdown of winding insulation or

due to excessive load on the power supply system. Over current protection for alternators is not

considered necessary; since modern generators are capable of withstanding complete short-

circuit at their terminals for sufficient time without much overheating and damage. On

occurrence of such faults, the generator can be disconnected from the system manually.

In case an overload protection is provided for generators, such a protection might disconnect

generator from system due to momentary troubles outside the power station or temporary

overload on system and thus interfere with continuity of supply.[2]

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5.2 Transformer Protection:-

Power transformers are static devices, totally enclosed and usually oil immersed, and

therefore, chances of fault occurrence on them are very rare. But the consequences of even a rarefault may be very serious unless the transformer is quickly disconnected from the system. Hence

automatic protection of transformers against possible faults is essential and of utmost

importance.

The faults occurring in power transformers are open-circuit faults (an open-circuit in one

phase of a three phase transformer), earth faults, phase-to-phase faults, inter-turn faults and

overheating from overloading or from some internal cause such as core-heating.

Interphase (phase-to-phase) short-circuits are most frequent on leads of three phase

transformers, while the interphase short-circuits within the winding are less frequent. Earth faults

and inter-turn faults have the highest probability on the power transformers. Winding short-

circuits, also called the internal faults, generally result from failure of insulation due to

temperature rise or deterioration of transformer oil.

An open-circuit in one phase of a three phase transformer may cause undesirable heating but

this condition is relatively harmless and so no relay protection is required against open-circuits.

On the occurrence of such a fault, the transformer can be disconnected manually from the

system.

The choice of a protective device for a transformer depends upon several factors such as:-

A) Type of transformer i.e., distribution or power transformerB) Size of transformerC) Type of coolingD) System where used i.e., its electrical location in the networkE) Importance of service for which it is required.

For distribution transformers employed in rural areas, the normal practice is to use the fuses

for its protection against external faults but for urban distribution network, where discrimination

is absolutely necessary, fuses will not serve the purpose.

For power transformers, the protection is to be provided usually against dangerous overloads

and excessive temperature rise. Dangerous overloads may be due to external faults or the internal

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ones. External faults, however, are cleared by the relay system outside the transformer within the

shortest possible time in order to avoid any danger to the transformer due to these faults.

Differential protection is the most important type of protection used for protection against

internal phase-to-phase and phase-to-earth faults. The other protection systems employed for

protection of transformers against internal faults are Buchholz protection, core-balance leakage

protection, combined leakage and overload protection, restricted earth-fault protection.[2]

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6. POWER SYSTEM ANALYSIS

6.1 Analysis OF Power System:-

The power system at TATA Motors Ltd, Pimpri consists of a 220 kV switchyard. This 220

kV voltage supply from the MSEDCL is then stepped down to 22 kV in the same switchyard

with help of three step down transformers. The transformers are of 220/22 kV and of 30, 30 and

40 MVA each. For calculation purposes we have considered the transformers to be of 30 MVA.

It consists of four Main Receiving Stations (MRS) of which New MRS forms the ring main

pattern. The remaining three MRS are connected to load through transformers of 2MVA or 1.5

MVA and ratio of step down of 22 kV /415 V.

Ring Main is an electrical distribution scheme in which outgoing feeders are connected to the

main supply through a ring circuit. Here the main supply after stepping down to 22 kV is fed to

the New MRS. From the New MRS it is fed to MRS-I, MRS-II, MRS-III each having two

sections. Also MRS-I and MRS-III are connected via MRS-II to New MRS, this being an

alternate connection, seldom used. The OLD-DG feeds power to MRS-I 22 kV bus. The MAN-

DG feeds power to New MRS. This completes the ring main system. This ensures uninterrupted

power supply to the plant even on occurrence of fault.

The generator houses contribute 54.35 MW to the system. According to the load shedding

schedule, the deficit power is generated by required number of generator units.

The company has maximum connected load of 55 MW. The Generator Houses contribute

54.35 MW to the plant, ensuring continuity of supply to the plant up to 99 percent.

Short circuit analysis is done for MRS-II without considering the effect of addition of OLD-

DG House and MAN-DG House on the fault level. Similarly, analysis is done initially

considering the effect of addition of only the OLD-DG House and then both OLD-DG House

and MAN-DG House. For both these cases fault levels are calculated. We then compare the fault

levels for all the cases. If there is an increase in fault level in any of the case then we have to

ensure that the switchgear protective devices already installed should be capable of protecting

the equipments for these increased fault levels. If they are not found suitable then we have to

suggest suitable switchgear protective devices.

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We are going to co-ordinate the relays of MRS-II up to the 22 kV level. Beyond which the

timings are specified by the MSEDCL. Depending upon the fault levels at various assumed fault

locations in MRS-I, the relays in MRS-II are to be co-ordinated. Co-ordination is done in such a

manner that no art of MRS-II is left unprotected. Also a fault in any part of MRS-II, it should be

cleared in minimum time. This is done so as to reduce the damage to protective gear and the

equipments and also to maintain system stability.

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7. SHORT CIRCUIT ANALYSIS

7.1 Steps Involved in Short Circuit Analysis:-

For short circuit analysis we consider three phase short circuit as it is the most severe fault

amongst all the faults. We are going to assume three phase short circuit on various locations

from 415V to 22kV level i.e. the New MRS incomer. The impedances of generators,

transformers, cables and motors are contributing to the change in fault level at different

locations.

Here we first calculate the fault levels for MRS-II without effect of addition of OLD-DG

House and MAN-DG House for 22 kV. Then we consider that all the eight generators of OLD-

DG House are run along with main supply. We calculate the contribution of the eight generators

to fault level at 22 kV and add it to the earlier fault level. We consider that all the generators are

run simultaneously to consider the worst case on occurrence of fault.

Secondly, we consider the fault level contribution of the three generators of MAN-DG

connected at 22 kV. The fault level contributions of all the three generators are then added to

fault level at MRS-II along with OLD-DG House. Usually all the three generators are not run at

the same time however we consider this for worst case calculations.

For calculating the contribution of generators to the fault level at 22 kV we require the short

circuit ratio of the generators. From the short circuit ratio we calculate the transient reactance and

subtransient reactance of the generator. Induction motor and synchronous motor contribute to

fault level as they act as generator for a short period due to inertia of connected load. Generally

we consider motors of rating greater than 30 hp for fault level contribution at 22 kV. As it is only

for few cycles its effect is not reflected on the New MRS 22 kV bus.

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Formulae used for calculations of short circuit analysis.

Z pu =

Fault MVA =

Fault Current =

Base MVA = 30 MVA

Base Voltage = 22kV

7.2 Calculations: - (Without considering OLD-DG House and MAN-DG House)

Following calculations for 22 kV remain same for every MRSII substation:-

Fig. 11 Impedance Diagram for Faults on 22 kV bus on MRS-II

%Z x Base MVA

Transformer Rating

Base MVA

Z(pu)T

Fault MVA

3 x Voltage

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For Fault Fa:-

Z(pu)T =0.039+0.0045+0.00090.1x(0.039+0.0045+0.0009)

=0.04440.00444

We consider 10% negative tolerance as per IEC Standards (1)

So,

Z(pu)T =0.03996 pu

Fault MVA = 30 / (0.0996)

=750.75

Fault Current = 750.75 / (3 x22000)

=19.70 kA

For Fault Fb:-

Z(pu)T =0.039+0.0045 - 0.1x(0.039+0.0045)

=0.04350.00435 = 0.03915 pu from(1)

Fault MVA = 30 / (0.03915)

=766.28

Fault Current = 766.28 / (3 x 22000)

=20.10 kA

For Fault Fc:-

Z(pu)T =0.039 - 0.0039

=0.0351 pu from(1)

Fault MVA = 30 / (0.0351)

=854.70

Fault Current = 854.70 / (3 x 22000)

=22.43 kA

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For Fault F2:-

Zpu = 0.0654 x 30/2

=0.981 pu

Z(pu)T =0.981+0.039+0.0045+0.0009

=1.02540.10254 from (1)

=0.923 pu

Fault MVA = 30 / (0.923)

=32.50

Fault Current =32.5/(3 x 415)

=45.21 kA

For Fault F3:-

Zpu = 0.0626 x 30/2

=0.939 pu

Z(pu)T =0.939+0.039+0.0045+0.0009

=0.98340.09834

=0.88506 pu from(1)

Fault MVA = 30 / (0.88506) = 33.896

Fault Current = 33.896 / (3 x 415)

=47.15 kA

For Fault F4:-

Zpu = 0.0541 x 30/1.5

=1.082 pu

Z(pu)T =1.082+0.039+0.0045+0.0009

=1.12640.11264 from(1)

=1.01376 pu

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Fault MVA = 30 / (1.01376)

=29.6

Fault Current = 29.6 / (3 x 433)

=39.45 kA

For Fault F5:-

Zpu = 0.048 x 30/1.5

=0.96 pu

Z(pu)T =0.96+0.039+0.0045+0.0009

=1.00440.10044 from(1)

=0.90396 pu

Fault MVA = 30 / (0.90396)

=33.19

Fault Current = 33.19 / (3 x 433)

=44.25 kA

For Fault F6:-

Zpu = 0.0619 x 30/2

=0.9285 pu

Z(pu)T =0.9285+0.039+0.0045+0.0009

=0.97290.09729 from(1)

=0.87561 pu

Fault MVA = 30 / (0.87561)

=34.26

Fault Current = 34.26 / (3 x 433)

=45.68 kA

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B) 8-12-15-21-29-39

Fig. 13 Impedance Diagram of Substation B

For Fault F1:-

Zpu = 0.048 x 30/1.5

=0.96 pu

Z(pu)T =0.96+0.039+0.0045+0.0009

=1.00440.10044 from (1)

=0.904 pu

Fault MVA = 30 / (0.904)

=33.18


=44.25 kA

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For Fault F2:-

Zpu = 0.0568 x 30/1.5

=1.136 pu

Z(pu)T =1.136+0.039+0.0045+0.0009

=1.18040.11804 from (1)

=0.1.0623 pu

Fault MVA = 30 / (1.0623)

=28.23


=37.65 kA

For Fault F3:-

Zpu = 0.0483 x 30/1.5

=0.966 pu

Z(pu)T =0.966+0.039+0.0045+0.0009

=1.01040.10104 from (1)

=0.909 pu

Fault MVA = 30 / (0.909)

=33.00


=43.98 kA

For Fault F4:-

Zpu = 0.0587 x 30/1.5

=1.174 pu

Z(pu)T =1.174+0.039+0.0045+0.0009

=1.21840.12184 from (1)

=1.096 pu

Fault MVA = 30 / (1.096)

=27.35

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=36.47 kA

For Fault F5:-

Zpu = 0.0627 x 30/2

=0.9405 pu

Z(pu)T =0.9405+0.039+0.0045+0.0009

=0.98490.09849 from (1)

=0.8864 pu

Fault MVA = 30 / (0.8864)

=33.84


=47.08 kA

For Fault F6:-

Zpu = 0.0514 x 30/1.5

=1.028 pu

Z(pu)T =1.028+0.039+0.0045+0.0009

=1.07240.10724 from (1)

=0.9652 pu

Fault MVA = 30 / (0.9652)

=31.08

Fault Current =31.08(3 x 433)

=41.44 Ka

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C) 41-43-44-55-56-57-58-61

Fig.14 Impedance Diagram of Substation C

For Fault F1:-

Zpu = 0.0626 x 30/2

=0.939 pu

Z(pu)T =0.939+0.039+0.0045+0.0009

=0.9834 pu

Z(pu)T =0.9834 - 0.09834 from (1)

=0.88506 pu

Fault MVA = 30 / (0.88506)

=33.896

Fault Current = 33.896 / (3 x 415)

=47.15 kA

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For Fault F2:-

Zpu = 0.0625 x 30/2

=0.9375 pu

Z(pu)T =0.9375+0.039+0.0045+0.0009

=0.98190.09819 from (1)

=0.8837 pu

Fault MVA = 30 / (0.8837)

=33.94


=45.26 kA

For Fault F3:-

Zpu = 0.0625 x 30/2

=0.9375 pu

Z(pu)T =0.9375+0.039+0.0045+0.0009

=0.98190.09819 from (1)

=0.8837 pu

Fault MVA = 30 / (0.8837)

=33.94


=45.26 kA

For Fault F4:-

Zpu = 0.0627 x 30/2

=0.9405 pu

Z(pu)T =0.9405+0.039+0.0045+0.0009

=0.98490.09849 from (1)

=0.8864 pu

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Fault MVA = 30 / (0.8864)

=33.84


=47.08 kA

For Fault F5:-

Zpu = 0.051 x 30/2

=0.765 pu

Z(pu)T =0.765+0.039+0.0045+0.0009

=0.80940.08094 from (1)

=0.7284 pu

Fault MVA = 30 / (0.7284)

=41.18


=57.29 kA

For Fault F6:-

Zpu = 0.051 x 30/2

=0.765 pu

Z(pu)T =0.765+0.039+0.0045+0.0009

=0.80940.08094 from (1)

=0.7284 pu

Fault MVA = 30 / (0.7284)

=41.18


=57.29 kA

For Fault F7:-

Zpu = 0.051 x 30/2

=0.765 pu

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Z(pu)T =0.765+0.039+0.0045+0.0009

=0.80940.08094 from (1)

=0.7284 pu

Fault MVA = 30 / (0.7284)

=41.18


=57.29 kA

For Fault F8:-

Zpu = 0.0653 x 30/2

=0.98 pu

Z(pu)T =0.98+0.039+0.0045+0.0009

=1.02390.10239 from (1)

=0.9215 pu

Fault MVA = 30 / (0.9215)

=32.55


=45.30 kA

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D) 51-52-53-54

Fig. 15 Impedance Diagram of Substation D

For Fault F1:-

Zpu = 0.0652 x 30/2

=0.978 pu

Z(pu)T =0.978+0.039+0.0045+0.0009

=1.02240.10244 from (1)

=0.9201 pu

Fault MVA = 30 / (0.9201)

=32.60


=45.35 kA

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For Fault F2:-

Zpu = 0.0661 x 30/2

=0.9915 pu

Z(pu)T =0.9915+0.039+0.0045+0.0009

=1.03590.10359 from (1)

=0.9323 pu

Fault MVA = 30 / (0.9323)

=32.17


=44.76 kA

For Fault F3:-

Zpu = 0.0649 x 30/2

=0.9735 pu

Z(pu)T =0.9735+0.039+0.0045+0.0009

=1.01790.10179 from (1)

=0.9161 pu

Fault MVA = 30 / (0.9161)

=32.74


=45.55 kA

For Fault F4:-

Zpu = 0.0658 x 30/2

=0.987 pu

Z(pu)T =0.987+0.039+0.0045+0.0009

=1.03140.10314 from (1)

=0.9282 pu

Fault MVA = 30 / (0.9282)

=32.31

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=44.96 kA

E) 32-59

Fig. 16 Impedance Diagram of Substation E

For Fault F1:-

Zpu = 0.0578 x 30/1.5

=1.156 pu

Z(pu)T =1.156+0.039+0.0045+0.0009

=1.20040.12004 from (1)

=1.08 pu

Fault MVA = 30 / (1.08)

=27.76


=37.02 kA

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For Fault F2:-

Zpu = 0.0654 x 30/2

=0.981 pu

Z(pu)T =0.981+0.039+0.0045+0.0009

=1.02540.10254 from (1)

=0.923 pu

Fault MVA = 30 / (0.923)

=32.50


=45.21 kA

F) 33-34-35-37

Fig. 17 Impedance Diagram of Substation F

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For Fault F1:-

Zpu = 0.0646 x 30/2

=0.969 pu

Z(pu)T =0.969+0.039+0.0045+0.0009

=1.01340.10134 from (1)

=0.9121 pu

Fault MVA = 30 / (0.931

=32.89


=45.76 kA

For Fault F2:-

Zpu = 0.0652 x 30/2

=0.978 pu

Z(pu)T =0.978+0.039+0.0045+0.0009

=1.02240.10244 from (1)

=0.9201 pu

Fault MVA = 30 / (0.9201)

=32.60


=45.35 kA

For Fault F3:-

Zpu = 0.0652 x 30/2

=0.978 pu

Z(pu)T =0.978+0.039+0.0045+0.0009

=1.02240.10244 from (1)

=0.9201 pu

Fault MVA = 30 / (0.9201)

=32.60

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=45.35 kA

For Fault F4:-

Zpu = 0.0629 x 30/2

=0.9435 pu

Z(pu)T =0.9435+0.039+0.0045+0.0009

=0.98790.09879 from (1)

=0.88911 pu

Fault MVA = 30 / (0.88911)

=33.74


=45 kA

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G) 17-18-28-40-49

Fig. 18 Impedance Diagram of Substation G

The compressor house consists of 6.6 kV bus to which motor load is connected. The motors

are of different ratings which are greater than 30 HP. Generally motors of 30 HP or more are

considered while calculating the fault levels. When a fault occurs the power to the motor is

interrupted and the motor continuous running due to the inertia of the connected load thus the

motors acts as generator and contributes to fault level.

There are 9 motors connected to the 6.6 kV bus.

The maximum running load = (4 X 522 X 6) + (2 X 900 X 6) + (2 X 315 X 6)

= 29000 kW

(3 X 6.6 X 0.86 X 0.94)

= 3138.10 A this is to be added to the fault level at F2 and F3.

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Maximum running load in ampere reflected on the primary of the transformer is

= 138.10 X 6.6

22

= 941.43 A = 0.94 kA this is to be added to the fault level at FA.

Therefore the new fault level at 22 kV bus at fault FA = 19.7 + 0.94 = 20.64 kA

For Fault F1:-

Zpu = 0.0638 x 30/2

=0.957 pu

Z(pu)T =0.957+0.039+0.0045+0.0009

=1.00140.10014 from (1)

=0.90126 pu

Fault MVA = 30 / (0.90126)

=33.28


=46.30 kA

For Fault F2:-

Zpu = 0.0748 x 30/6.26=0.3584 pu

Z(pu)T =0.3584+0.039+0.0045+0.0009

=0.40280.04028 from (1)

=0.3625 pu

Fault MVA = 30 / (0.3625)

=82.75


=7.23 + 3.13 = 10.37 kA

For Fault F3:-

Zpu = 0.0762 x 30/6.25

=0.3657 pu

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Z(pu)T =0.3657+0.039+0.0045+0.0009

=0.41010.04101 from (1)

=0.3691 pu

Fault MVA = 30 / (0.3691)

=81.36


=7.11 + 3.13 = 10.25 kA

For Fault F4:-

Zpu = 0.0673 x 30/2

=1.0095 pu

Z(pu)T =1.0095+0.039+0.0045+0.0009

=1.05390.10539 from (1)

=0.94851 pu

Fault MVA = 30 / (0.94851)

=31.62


=42.17 kA

For Fault F5:-

Zpu = 0.0625 x 30/2

=0.9375 pu

Z(pu)T =0.9375+0.039+0.0045+0.0009

=0.98190.09819 from (1)

=0.8837 pu

Fault MVA = 30 / (0.8837)

=33.94


=45.25 Ka

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7.3 Calculations considering OLD DG House:-

Old generator house consists of eight alternators of which six are of 2.5MW and two are of

2.2MW.It also consists of six transformers. Two are step down transformers of 0.2 MVA, 0.5

MVA and 6.6kV/415V which are nothing but station transformers. Remaining four are step up

transformers of 6.6kV/22Kv out of which two are of 8MVA and the other two are of 6.25 MVA.

For fault calculations, we consider only the transformers connected in parallel to the MRS-II.

Even though the generators are not in operation, the transformers have to be kept charged. So,

the equivalent impedance of the transformers has to be taken into consideration for fault

calculation. Here fault current contributed by all eight generators is equal to the fault current

contributed by a single generator multiplied by eight. Hence

Total fault current contributed = (Fault current contributed by single generator) x 8

In OLD-GEN House, all eight generators are generating power at 6.6 kV. Short circuit ratio

of all eight generators is 0.7. Hence synchronous reactance of all eight generators is given by:

Xd = 1 / 0.7

= 1.4285 pu

Also, subtransient reactance is given by:

Xd = 0.2 x 1.4285= 0.2857 pu

MVA rating for all eight generators is 3.125. Hence fault MVA for all eight generators is given

by:

Fault MVA = 3.125 / 0.2857

= 10.9375

Now fault current contributed by each generator at 6.6 kV is given by:

Fault Current = 10.9375 / (3 x 6.6)

= 0.96 kA

Consider any one generator. It is connected to another 6.6 kV bus through a cable of impedance

0.0009 pu. So total impedance is given by:

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Z = Xd + 0.0009

=0.2857 + 0.0009

= 0.2866 pu

Fault MVA = 3.125 / 0.2866

= 10.90

Fault Current = 10.90 / (3 x 6.6)

= 0.9538 kA

Now this generator, with cable of 0.0009 pu in series, is connected to 22kV bus through a step up

transformer of 6.6/22 kV. So total impedance is given by:

ZT = (Xd + 0.0009 +Z transformer)

= (0.2857 + 0.0009 + 0.1097)

= 0.3963 pu

Fault MVA = 3.125 / 0.3963

= 7.89


= 0.21 kA

Total fault current contributed by = (Fault current contributed by single generator) x 8

eight generators

= 0.21 x 8

= 1.68 kA

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For MRS-II:-

For Fault FA:-

Fault Current = 19.70 kA (Without OLD-DG House and MAN-DG House)

Fault current with OLD-DG House = 19.70 + 1.68

=21.38 kA

For Fault FB:-



=21.78 kA

For Fault FC:-



= 24.11 kA

7.4 Calculations considering OLD DG house and MAN DG House:-

MAN DG house consists of three alternators 11.65MW. It also consists of five transformers.

Two are step down transformers of 2 MVA and 11kV/415V which are nothing but stationtransformers. Remaining three are step up transformers of 11kV/22kV and 22MVA.

Even though the generators are not in operation, the transformers have to be kept charged. So,

the equivalent impedance of the transformers has to be taken into consideration for fault

calculation. Here fault current contributed by all three generators is equal to the fault current

contributed by a single generator multiplied by three. Hence

Total fault current contributed = (Fault current contributed by single generator) x 3

In OLD-GEN House, all three generators are generating power at 11 kV. Short circuit ratio of

all eight generators is 0.447. Hence synchronous reactance of all eight generators is given by:

Xd = 1 / 0.447

= 2.237 pu

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Also, subtransient reactance is given by:

Xd = 0.2 x 2.237

= 0.4474 pu

MVA rating for all eight generators is 3.125. Hence fault MVA for all eight generators is given

by:

Fault MVA = 14.65 / 0.4474

= 32.74

Now fault current contributed by each generator at 6.6 kV is given by:


= 1.72 kA

Consider any one generator. It is connected to another 6.6 kV bus through a cable of impedance

0.0009 pu. So total impedance is given by:

Z = Xd + 0.0009

=0.4474 + 0.0009

= 0.4484 pu

Fault MVA = 14.65 / 0.4484

= 32.74


= 1.72 kA

Now this generator, with cable of 0.0009 pu in series, is connected to 22kV bus through a step up

transformer of 11/22 kV. So total impedance is given by:

ZT = (Xd + 0.0009 +Z transformer)

= (0.4474 + 0.0009 + 0.00617)

= 0.4545 pu

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Fault MVA = 14.65 / 0.4545

= 32.24


= 0.85 kA

Total fault current contributed by = (Fault current contributed by single generator) x 3

eight generators = 0.85 x 3

= 2.54 kA

For MRS-II:-

For Fault FA:-


New Fault current with OLD-DG House and MAN-DG House = 19.70 + 1.68 + 2.54

=23.92 kA

For Fault FB:-



=24.32 kA

For Fault FC:-



= 26.65 kA

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8. RELAY CO-ORDINATION

8.1 Introduction to Relay Co-ordination:-

Relay co-ordination plays an important role in the protection of power system. For

proper protection, proper co-ordination of relays with appropriate relay settings is to be

done. Relay settings are done in such a way that proper co-ordination is achieved along

various series network. However the review of Co-ordination is always essential since various

additions / deletion of feeders and equipments will occur after the initial commissioning of

plants. As power can be received from generators of captive power plant, the analysis becomes

complex.

Relay co-ordination can be done by selecting proper plug setting and time multiplication

setting of the relay, considering maximum fault current at the relay location. After selecting the

plug setting and time multiplier setting, the co-ordination can be checked graphically.

When plotting co-ordination curves, certain time intervals must be maintained between the

curves of various protective devices in order to ensure the correct sequential operation of the

devices when co-coordinating inverse time over current relays.

For a given fault current, the operating time of IDMT relay is jointly determined by its plug

and time multiplier settings. Thus this type of relay is most suitable for proper coordination.

Operating characteristics of this relay are usually given in the form of a curve with operating

current of plug setting multiplier along the X axis and operating time along Y axis.

Calculation of relay operating time:

In order to calculate the actual relay operating time, the following things must be known.

A) Time / PSM CurveB) Plug SettingC) Time SettingD) Fault CurrentE) Current Transformer Ratio

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The procedure for calculating the actual relay operating time is as follows:-

A) Convert the fault current into the relay coil current by using the currenttransformer ratio.

B) Express the relay current as a multiple of current setting, i.e. calculate thePSM.

C) From the Time/PSM curve of the relay, with the calculated PSM the corresponding timeof operation can be obtained.

D) Determine the actual time of operation by multiplying the above time of therelay by time-setting multiplier in use.

But here we are going to follow the procedure as follows:

A) Fault current is 19.70 kA. We assume PMS=100% i.e. 1 for all relays. For first level,we assume TMS=0.1.

B) Calculate Rated C.T. Secondary Current.C) Calculate Multiple of set current i.e. PSM.D) Calculate time of operation of relays only for first case.E) For the remaining levels, follow steps A) to C). For these levels, assume appropriate time

of operation of relays.

F) Calculate TMS for each level except first level using same formula as in step

8.2 Formulae used:-

t =

Where,

t= Operating time in sec TMS= Time Multiplier Setting

k ,, = Curve constants

I= Fault Current

I>= Set Current

k x TMS(I/I>) -1

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Fig. 19 Various Inverse Characteristics of Induction Disc Relays on Log Scale

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Table 2 Table of constants for various curves

Fig. 18 shows various inverse characteristics of induction disc relays. Characteristics are of

four types:

A) Standard or Normal InverseB) Very InverseC) Extremely InverseD) Long Inverse

For Normal inverse over current characteristics, the operation time is inversely proportional to

the applied current.

Very inverse over current characteristics are particularly suitable if there is a substantial

reduction of fault current. The characteristics of this relay are such that its operating time is

approximately doubled for a reduction in current from 7 to 4 times the relay current setting. This

permits the use of the same time multiplier setting for several relays in series.

For Extremely inverse over current characteristics, the operation time is approximately

inversely proportional to the square of the applied current.

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8.3 Calculations for MRS-I:-

Fig. 20 Relay Co-ordination for MRS-I

We select Normal Inverse Curve initially.

k=0.14

=0.02

=2.97

Plug Setting=100% i.e. 1

Fault Current I =19.70 kA

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Rated C.T. Secondary Current = Plug Setting x C.T. Secondary Current

PSM = Fault Current in C.T. Primary / (C.T. Transformation Ratio x Rated C.T.

Secondary Current)

1) C.T Ratio = 200/5TMS = 0.1


= 1 x 5

=5

Multiple of set current (PSM) = 19.7 kA/200 A

= 98.50

t1 = (0.14 x 0.1) / (98.5)0.02

- 1

= 0.15 sec

2) C.T Ratio = 800 / 5We assume co-ordination time as 0.15 sec.

t2 = 0.15 + 0.15

= 0.30 sec.


=1 x 5

= 5

Multiple of set current = 19.7 kA/800 A

= 24.63

TMS = 0.3 x ((24.63)0.02

- 1) / 0.14

= 0.14

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t 3 = 0.3 + 0.1

= 0.4 sec.


= 1 x 5

= 5

Multiple of set current = 19.7 kA / 1200 A

= 16.42

TMS = 0.4 x ((16.42)0.02

- 1) / 0.14

= 0.16


t3

= 0.4 + 0.2

= 0.6 sec.


= 1x 5

=5


= 16.42

TMS = 0.6 x ((16.42)0.02

- 1) / 0.14

= 0.25

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t 3 = 0.6 + 0.1

= 0.7 sec.


= 1 x 5

= 5


= 32.83

TMS = 0.7 x ((32.83)0.02

- 1) / 0.14

= 0.36


t 3 = 0.7 + 0.2

= 0.9 sec.


= 1 x 5

= 5


= 13.13

TMS = 0.9 x ((13.13)0.02

- 1) / 0.14

= 0.34

We know the actual time required for operation of relay will be the time of operation we have

assumed and time multiplier setting.

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9. CONCLUSION

1. Analysis of Power System

We studied a part of power system of Tata Motors Ltd., Pimpri and its single line

diagram. From the short circuit analysis carried out on the substations we calculated the faultlevels at 415 V to 22 kV levels. The fault current is inversely proportional to fault impedance up

to the location of fault and the voltage level. From incomer of New MRS to outgoing feeder of

MRS-II the fault impedance is increasing while the voltage remains constant resulting in

decrease in fault level. At low voltage side of distribution transformers the voltage level is

significantly lower than high voltage side as the transformation ratio is high. The effect of lower

voltage level is more than the effect of increase in fault impedance which causes the fault level to

rise considerably as compared to the 22 kV level.

2. Relay Co-ordination

We co-ordinated the over current relays from the outgoing feeder of MRS-I to incomer of

New MRS. The actual operating time for the relays at

Outgoing feeder of MRS-I is 0.15 sec

Incomer of MRS-I is 0.30 sec

Outgoing feeder of MAN-DG is 0.40 sec

Incomer of MAN-DG is 0.60 sec

Outgoing of New MRS is 0.70 sec

Incomer of New MRS is 0.90 sec

For a fault on outgoing feeder of MRS-I, where the fault level is 19.7 kA, the relay employed

at that location, i.e. the primary relay, should operate within 0.15 seconds. In case of failure of

primary relay there is a back-up relay provided at the incomer of MRS-I which is set to operate

within 0.30 sec. This relay is set to operate with a time interval so as to avoid its tripping earlier

than the primary relay. In a similar way the relays till incomer of New MRS are co-ordinated

such that the maximum time of operation is less than the sustainable time of the circuit breaker.

3. Analysis of effect of addition of captive power plant of 54.35 MW

The OLD-DG House and MAN-DG House contribute 1.68 kA and 2.54 kA respectively to

the fault level at incomer of New MRS. So the changed fault level at the incomer of New MRS is

26.65 kA.

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The short time rating of the circuit breaker at incomer of NEW MRS is 26.3 kA for 1 sec.

Therefore the sustainable time of this circuit breaker for26.65 kA is

26.32

X 1 = 26.652

X t

t = 0.97 sec.

New MRS incomer has the highest fault level amongst all the 22 kV buses. The selected

operating time for this circuit breaker for a fault of 19.7 kA at outgoing feeder of MRS-I is 0.9

sec. Therefore for the changed fault level of26.65 kA the sustainable time is

19.72 X 0.9 = 26.652X t

t=0.5 sec.

This time is less than sustainable time of the circuit breaker. Though the fault level is

maximum, fault will be cleared without any damage.

Hence, we conclude that there is no need to change either the switchgear protective devices or

the relay settings as they are capable of clearing the fault for increased fault levels.

report-15.06.10-1300737626

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