OFFICE: CENTER OF ELECTRICAL ENERGY SYSTEM (CEES)FACULTY ELECTRICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA 81310 UTM JOHOR BAHRUJOHOR DARUL TAKZIM

ROOM NO: P07-412Tel: 07-5536262

WELCOME

• POWER SYSTEM ENGINEERING• SEE 4423

• LECTURER: ASSOC PROF MD SHAH MAJID• ROOM NO: P08-409/P07-417(CEES)• TEL: 07 -5535295/5536262 (CEES)• E-mail: mdshah@fke.utm.my, mdshah@ieee.org

SEE 4423POWER SYSTEM ENGINEERING

SEMESTER II SESSION 2009/2010

Assoc Prof MD SHAH MAJID

4

SEE 4423POWER SYSTEM ENGINEERING

SEMESTER 1 SESSION 2009/2010

CHAPTER 1

INTRODUCTION TO PROTECTION SYSTEM

Introduction:Need for Protective Systems

• Power system consists of; generators, transformers, transmission and distribution lines, etc…..etc.

– Sistem kuasa mengandungi: penjana, pengubah, talian penghantaran dan pengagihan dsbnya

• Short circuits and other abnormal conditions often occur on a power system– Litar pintas dan keadaan tidak normal kerap berlaku pada sistem kuasa

• Heavy current associated with short circuits is likely to cause damage to equipments

– Arus yg tinggi ketika litar pintas mungkin menyebabkan kerosakan pada peralatan jika geganti perlindungan dan pemutus litar tidak di bekalkan untuk perlindungan bagi setiap seksyen sistem kuasa

Need for Protective Systems

• Short circuit is a faults (power engineer)

• Failure of conducting path due to a break in a conductor is a type of fault

• Fault occurs – automatic protective device is needed to isolate the faulty element as quickly as possible to keep the healthy section of the system in normal operation

• Fault must be cleared within fraction of a second

• Uncleared short circuits may cause total failure of the system

• A protective scheme includes transducers, protective relays and circuit breakers to isolate the faulty section of the system from the healthy section

• Protection is needed not only against short circuit but also against any abnormal condition– Overspeed of generators and motors– Overvoltage– Underfrequency– Loss of excitation– Overheating of stator or rotor of an alternator– Etc…etc…etc

• Protection is a pre-requisite for an effective and reliable system

• A protective relay does not anticipate or prevent the occurrence of fault, rather it takes action only after a fault has occurred – except Buchholz relay; a gas actuated relay

Nature and causes of Faults

• Faults are caused either by;

– Insulation failure –– Conducting path failure

• results in short circuit

• Faults on transmission and distribution lines are caused by overvoltage– Lightning and switching surges– External conducting objects falling on overhead line.

• Birds also may cause faults on overhead line if their bodies touch one of the phase and earth wire

• If conductors are broken – failure of conducting path and the conductor becomes open-circuited– If broken conductor fall to the ground, results in short circuit

• Joint failure on cables and overhead lines also cause a failure of the conducting path

• Opening of one or two of the three phases makes the system unbalanced– Set-up harmonics

• Other causes of faults in o/h line– Direct lightning strokes– Aircraft– Snakes– Ice and snow loading– Abnormal loading– Storm– Earthquakes– Creepers– Etc…..etc……etc……etc

• Cables, transformer generators and other equipment:– Failure of solid insulation due to aging– Heat– Moisture– Overvoltage– Mechanical damage– Accidental with earth– Flashover due to overvoltages– Etc…etc…..etc

Types of faults

• Symmetrical faults

• Unsymmetrical faults

• Symmetrical faults– Kerosakan tiga fasa atau tiga fasa ke bumi

• Unsymmetrical faults – Single line to earth, line-to-line, double line to

earth, open-circuited phases

• Faults can interrupt power system in several ways:

– Heavy current to flow

• Effect : overheating of power system component

– Fault is a short circuit and exist as electrical arc and liquid e.g air

• Effect: equipment faulty and fire

• Fault – can increase/decrease voltage system outside its acceptable range

– Can cause unstable three phase system, improper operation of three phase equipments

– Prevent power flow

– Can cause system to be unstable and collapse

• Faults incur MONEY.

• If fault is isolated as quickly as possible and accurate –less money required

• Cost of protective equipment is 5% of the total cost of the system

• PROTECTION DOES NOT MEAN PREVENTION

– A protective relay does not anticipate or prevent the occurrence of fault, rather it takes action only after a fault has occurred – except Buchholz relay; a gas actuated relay

Functions of Protective System

• Fast and automatically opens the faulty section in the power system

• Increase system reliability and security – only affected area will be isolated and maintain the healthy line– Ensure consumers receive continuity of supply

• Delay in isolating the fault – System unstable, loss of synchronism, total failure of the system– Fire

Requirements of Protective System

• The basic requirements of a protective system are as follows:

– Discrimination/selectivity– Sensitivity– Reliable – Stability– Speed

(Discrimination/Selectivity)

• Ability to select either to operate or not – Keupayaan untuk memilih sama ada bekerja atau tidak

• Select to isolate the faulty section only (the rest normal condition) – Memilih untuk mengasingkan bahagian rosak sahaja (yang lain berkeadaan

normal)

• The nearest circuit breaker will trip – Pemutus litar yang hampir sahaja terbelantik (trip)

SD1SD2 SD3

SD1 SD2SD3

Radial protection system

Sensitivity

• Relay should operate when the magnitude of the current exceeds the preset value

• This value is called pick up current

• Should not operate when current is below its pick-up value

• Should be sufficiently sensitive to operate when the operatinfg current just exceeds the pick-up value – relate with minimum operating current

Reliability

• A protective system must operate reliably when a fault occurs in its zone of protection

• Failure may due to its protective element system; CT, PT, CB, relay etc….

• Reliability of protective system 95%

Stability

• A protective system should remain stable even when a large current is flowing through its protective zone due to an external fault

• Concerned circuit breaker is supposed to clear the fault

SD1 SD2SD3

Radial system protection

YesNo

Speed-fast operation

• Isolate fault as quickly as possible (at shortest time possible )

• Isolate disturbances before loss of synchronism and plant stop operation – time should not exceed critical clearing time

• Speed balanced with economy – Cost of protective equipment should be relevant to the cost of the protected

zone

• Avoid fire to equipments, interruption of supply to consumer, voltage drop operating time; 1 cycle, ½ cycle are also available; distribution system > 1 cycle

Operating time

• Total time to accomplish the isolation

• Calculated at the instance of fault until the given trip signal

• Must be low for the sake of plant and equipment safety

• Delay time – for discrimination

Protection economy

• Protection system can be designed as: – Simple and cheap

– Complex and expensive

• Baesd on:– The cost of fault

– Safety level requirement

Protection economics…

• Cost of fault– Cost of damage toward the plant

– Cost loss of revenue due to cut-out supply

– Cost of customers confidence

• Higher the cost of fault, more expensive the protection system

• Higher the plant kVA, more complex protection system required

Protection system at customer level

• Simple fuse – protect equipment and certain circuits

• Fius ringkas melindungi alat-alat dan litar tertentu

• Miniature circuit breaker- MCB

Protection system at distribution level

• Fuse and switch fuse

• Automatic reclosing circuit breaker– For rural area

– Transient fault is corrected by self clearing

Protection system at transmission level

• Technical consideration outweighs the economy

• 275kV dan 400kV System require a protection system which is:– very reliable– Full discrimation– High speed

• Expensive and complex

• Provide back-up protection

Zone of Protection

• Power system is divided into several zones

• Each zone of protection is provided with 2 types of protection:

– Primary protection

– Backup protection

GENERATOR PROTECTION

CIRCUIT BREAKER

HV SWITCHGEAR PROTECTION

TRANSFORMER PROTECTION

EHV SWITCHGEAR PROTECTION

TRANSMISSION LINE PROTECTION

EHV SWITCHGEAR PROTECTION

ZONE OF PROTECTION

Protection zone

• Region or area encompass a protection system

SD1 SD2SD3

Radial protection system

CB1CB2

CB3

Protection zone CB3

Main/primary protection

• In general primary protection is provided for each transmission line segment, major piece of equipment and switchgear

• If fault occurs, it is the duty of the primary protective scheme to clear the fault

• First line of defence - Responsible to isolate fault as quickly as possible

• Fails, back-up protection clears the fault

Backup Protection

• Act when primary protection fails – second line of defence

• Usually several back up protection scheme will act to control the power system

• Longer time delay

Back-up protection

SD1 SD2SD3

Radial protection system

Main protection

Zone of protection

• Zone of protection usually overlap

Fault detection

• Current magnitude

• Current in abnormal path (earth)

• Current balance (current out#current in)• Voltage balance

• Changes in impedance

• Protective system damage(Buchholz), power flow direction, temeprature & pressure

Protective system components

1. Circuit breaker

2. Transducer – current transformer (CT), voltage transformer (VT), potential transformer (PT)

3. Communication links– Pilot wire

– Radio link

– Overlapping signal, power line carrier (PLC)

Protective system components

4. Relay – Electromagnetic

– Static – improve reliability,versatile, fast response (1/4 cycle)

– Microprocessor- VLSI teknologi – current interest to power engineers; adv:attractive flexibility due to programmable approach, can provide protection at low cost and compete with conventional relays

5. Fuse

Transducer

Serves as a sensor to detect abnormal system conditions and to transform the high values of short

circuit current and voltage to a lower values

43

Protective relays

• Process the signals provided by the transducers which may be in the form of current, voltage or a combination of current and voltage

44

Circuit breaker

• Mechanical device used to energize and interrupt an electric circuit

45

Ex. Protective system: power flow balance

R1 R2

1. PL 1. P.L2. CT, PT

3

4. Relay

Classifications of relay based on its function

• Overcurrent relay

• Undervoltage relay

• Impedance relay

• Under frequency relay

• Directional relay etc….etc….etc

Classification of protective scheme

• Overcurrent protection

• Distance protection

• Current carrier protection

• Differential protection

Transformer in a protective system

• CT assume ideal• Normal rating:

– 1:1, 2:1. 2.5:1, 4:1, 5:1– 20:1, 40:1, 100:1, 200:1, 300:1,

600:1– 1000:1, 2000:1, 4500:1

• Secondary coil is connected to a “sensitive device” eg voltmeter

• Low Stray impedance so that voltage drop small

VpVs

Ip

Is

Feeder conductor

VT

CT

Current transformer (CT)

• Normal rating – 1A (Europe)– 5A (USA)

• Example :– 50:5, 100:5, 150:5, 200:5, 250:5– 450:5, 500:5,……1000:5, ……, 6000:

• Secondary coil is connected to a “current sensing device” of zero impedance

• Shunt impedance high so that I0 low.

(current transformer CT)

• Working principle similar to voltage transformer

• Supply by current source

• Primary winding connected in series with power circuit – Carries full load current

• Transformer impedance (referred to secondary and can be neglected)

• Secondary coil connected to load (burden)

(current transformer CT)

• Secondary winding feeds the protective system

• Current is reduced but must be almost the same with the power system current

• Secondary rated current 5 A or 1 A

• Fault current 10 – 20 X rated current. Transformer must be capable to operate reliably/accurately at this value

• Burden usually usually small – flux do not saturate

Current transformer construction

Primary winding

(1 turn)- from power system

Core

Power system current

Secondary winding wound on the core

Bar Type

Current transformer construction

Wound type

design

• CT – similar to a normal transformer emf equation

• Average induced voltage = product of No. of turn and rate of change of flux magnet; eaverage

= N(dΦ/dt)

• Knee point voltage(rms) = 4.44 BAfN

Burden

• Defined as the load connected across its secondary CT – express in VA (VA taken as nominal secondary

current of CT) or– Impedance (at the rated secondary current at a

given power factor usually 07. lagging

• Increase in impedance – increase burden• CT unloaded if secondary winding is short

circuited

Example

• 5VA burden at 1A transformer, gives 5 Ohms impedance.

• 5VA/1A = 5V• impedance = 5V/1A = 5Ω

• or

• At 5A CT• 5VA/5A• impedance = 1V/5A = 0.2 Ω

Example

• CT ratio 300/1, core area 40x30 mm

• Voltage at knee point?

• V = 4.44 x 0.0018 x 300 x 50 = 120 V

• 1.5 x 40 x 30 x10-6 = 0.0018 Wb ( B for sheet steel, 1.5 tesla at knee point)

CT open circuit

• If burden high, Es high– exceed Vkp (knee point voltage)• Io high ; I2 less

• Limited value when the secondary CT open circuit; I2 = 0;• Then N1I1 = N2 (I2+Io) = N2Io

• This will drive the CT to saturation level• dΦ/dt = 100 x Vkp (induced = 100 Vkp )• Cause insulation failure and overheated

Current transformer equivalent circuit

Load (burden) R

x r

I0’

Ic’Im’

Ip’

Es

Is

Vs

(nominal ratio) Kn

npp KII / ' = nKII / 0 '0 =

Ip = primary current

I0 = primary excitation current (no load)

Ip’

Current transformer equivalent circuit

Load (burden) R

x r

I0’

Ic’Im’Es

Is

Vs

θ

Ip’

Is

Ic’

Im’

I0’

Es

φ

'0

' III ps −=

'0 ' III sp +=

relationship Es and φ (flux)

)2

sin(N s

πωωφ += te m

dtdNesφ =

tm ωφφ sin =

te ms ωωφ cosN =

N=no of turn in secondary

°∠= 902

N msE ωφ

(Nominal Turn Ratio)• Determine from the given current ratio

– Eg.1000/5 (bar type)

1200rationominalKn ===

s

p

II

Actual Ratio• Actual ratio value Ip and Is

s

pt

s

p

II

K

II

′×=

′×= ratiowindingratioactual

• Winding ratio= secondary coil turn if transformer is of bar type l– Kt = N2/1

• Ideally

• Practically

1=′

s

p

II

1⟩′

s

p

II

Actual turn≥ nominal ratio

CT error

• Current error or ratio error

• Phase error

• Composite error

Ratio error or current error

100%ratio actualration actual-ratio nominal

error ratio ×=

Nominal ratio= Kn

Turn ratio= Kt

No compensation Kn=Kt

s

pt I

IK

′×=ratioactual

Cont’d

'

'n

'

'

n

K

Kerrorratio

pt

pts

s

pt

s

pt

IK

IKI

II

K

II

K

−=

−=

%100error ratio '

'

×−

=p

ps

I

II

difference in magnitude between Ip’ and Is

Kn=Kt

(no compensation)

Phase ErrorDifference in angle between Ip’ and Is

Phase error= θ

Ip’

Is

I0’

φIm’

θα

Es

Compensating winding

• Secondary coil winding is reduced (1 or 2 turn) to compensate current error due by I0’.

• Secondary coil current will be high, but the value will be minimised by the excitation current component

• This can minimised the current error

Example T1

• A current transformer 50 Hz has a secondary current of 3A,Sebuah• Secondary impedance circuit (0.6+j0.45),• Maximum flux 0.253 weber• Magnetisation current and core loss current (primary circuit) at this load

are 0.2 and 0.15 A respectively• Turn ratio1:10, assume no compensation• Determine no of turn, ratio error and phase error of transformer

Ip’

CT equivalent circuit

Load (burden)

= 0.6+j0.45

r

I0’

Ic’Im’Es

Is=3 ∠ αº

Vs

θ

Ip’

Is

Ic’

Im’

I0’

Es

φ '0

' III ps −=

'0 ' III sp +=

npp KII / ' = nKII / 0 '0 =

α

Is=3 ∠ αº

N1:N2 = 1:10

Im = 0.2 A

Ic = 0.15 A

°∠××

= 902

ωφmss

NE °+∠=°∠×°∠=

=

)87.36(25.287.3675.03

ααZIE ss

°=−=°=°+

13.5387.3690

90)87.36(α

α

Kt = 10 = Kn

Ip’

Is

I0’

φIm’

θα

Ic’

Es

∴ N1:N2 = 4:40

40502253.0

225.2

25.22

=××

×=

=××

π

ωφ

s

ms

N

N

Is=3 ∠ αº

Im = 0.2 A

Ic = 0.15 A

15.020.00 jI +=

015.002.0101)15.020.0(0

j

jI

+=

×+=′

°∠=°∠++=

+′=′

9975.52024.313.533015.0020.0

0

jIII sp

Ip’

Is

I0’φ

Im’

θ

Ic’

Es

%794.0

%100024.3

024.33errorratio

−=

×−

=′

′+=

p

ps

I

II

1325.09975.5213.53errorphase

=−=

Ex 1

• A current transformer has 15 turn on the primary and 75 turn on the secondary. The relay burden is (1.2+j2.6)Ω and secondary winding impedance of the transformer is (3.6+j0.4)Ω. If the magnitude of the secondary voltage is 26 volt and magnetization current referred to secondary is 0.24∠-45o amp. Determine the ratio error and phase error.

Ex 2

• A new non compensated CT with a cross sectional area of 40 cm2 has a nominal ratio of 2000/5. When 5A is supplied to a relay, the combination of load relay and secondary winding impedance is (10+j5.77) Ω. If the magnetisation amp-turn is 100 and core loss amp-turn is 50, determine the ratio error and phase error. Using the same data, calculate also the compensating turn to reduce the ratio error to a minimum.

Ex. 3

• A CT, 50 Hz has a primary winding of 10 turn and its cross sectional area is 8 cm2. Secondary current is 5A. The total load and secondary impedance is 0.8+j0.5 Ω. The excitation current required at the primary winding to establish a working flux with the secondary open is 0.6∠45o amp. If the flux density is 0.1878 wb/m2,determine the nominal ratio of the transformer. Determine also the ratio error and phase error of the transformer.

Composite error

• BS 3938:1973– (r.m.s.) value; difference between ideal secondary

current (i.e CT is ideal ; no excitation component) with actual secondary current.

• Encompass (current error, phase error and harmonic effect)

Accuracy limit current of a protection system CT

• Accuracy limit current (saturation current)– The maximum current a CT can sustain before

exceed its accuracy

– Specify either in term of primary current or secondary current

• Accuracy limit factor (saturation factor)– Ratio of accuracy limit primary current and

rated primary current

– Standard values are 5, 10, 15, 20 and 30

CT specification

• Expressed in– VA at rated current /class accuracy/accuracy limit factor

– Burden standard rated values

• 2.5,5,7.5,10,15 and 30 VA

– 2 accuracy class: 5P and 10P – gives composite error at rated accuracy limit 5% and 10% respectively

– Standard accuracy limit factor:

• 5, 10, 15, 20 and 30

Rating of CT

• Method to explain CT

– 10VA/5P/25

Accuracy limit current is 25 x rated current

Transformer works in good condition/reliably until 25 x its rated curentClass – error

do not exceed 5%

Rated burden

Ex 1

• CT 10VA/5P/15

• Calculate the accuracy limit factor when the burden is half of its initial value.

• Secondary resistance of CT 0.15 ohm

Solution

Es

r

Burden (R)

Burden impedance 10VA at rated current 5A

V

RI

IRIIVVA

2

=

××=×=

ohmR 4.0510 2 ==

Solution

Es

r

Burden (R)V

Total impedance referred to secondary winding

= 0.15 + 0.4 = 0.55 ohm

Accuracy limit current= Accuracy limit factor × Rated current

A75515 currentlimit accuracy

=×=

Accuracy limit factor= ratio accuracy limit current and rated current

VV

25.41750.55 emfsecondary

=×=

When burden is half

R = 0.2 ohm

Total secondary impedance= 0.15 + 0.2 = 0.35 ohm

Es

r

Burden (R=0.4)

Smaller the burden, the higher the accuracy Therefore :

burden has to be low possible

A8.11735.0/25.41 current limit accuracy new

==

56.235/8.117factorlimit accuracy new

==

CT Class X

• BS 3938 – CT for special application is known as Class “X’

• Expressed in turn ratio, knee point voltage, excitation current at the specified voltage and secondary winding resistance

– Application – scheme where the phase fault stability and/or accurate time grading required

• Transformer rating specify in term of e.m.f maximum that can be use by the transformer

• knee point– Point where an increase of 10% in secondary e.m.f require an addition of 50% excitation current

Definition of knee point

Excitation voltage

Excitation current

+10%

+50%

Knee point

Voltage transformer/potential transformer(VT/PT)

• Supply voltage lower than the sistem voltage

• Nominal sekunder voltage110V

• 2 types– Wound ( electromagnetic) type(> 132 kV non economical)– Capacitor type- CVT

• Element X and C consist of tuning circuit – to reduce the ratio and phase angle error secondary voltage

1/ωC2 = ωL pd 50 Hz; V2 = V’

Vp

C1

C2

L

V2V’ Vs

Rb

Xb

V’ = 12 kV, C1 = 2000pF

Talian

C2 L

V’V2 110 V

Protection techniques

• Protection scheme:– Arrangements of CT, VT, pilot wire and relay so

that the required operating characteristics can be obtained

• Each section of the power system network is monitored by several protection scheme.

Main/primary scheme usually consists of:

Unit Scheme

Backup scheme consists of:

Non Unit Scheme

Unit scheme

• ZONE OF PROTECTION IS WELL DEFINED

Protected zone

R

R

FAULT:Non similar current flow in the relay CB will trip

CT CT

relay

P

Q

Pilot wire

Difference current scheme: circulating current

Unit protection scheme

Protected zon

RFAULT:Current flow in circuit, result e1≠e2

CT

Differencec current scheme: voltage balance

R

Healthy and fault external zone (through fault):

No current in relay, e1=e2

e1 e2

Balance power direction

R1 R2

PL P.LCT, PT

Relay

Trip

Restrain

Restrain

Trip

R1 R2

Trip

Restrain

Restrain

Trip

Power flow entering the feeder (from point A) will cause relay (R1) close trip contact CB1

CB1 CB2

Power flow exit fron the feeder (at point B) will cause relay R2 rotate and close restrain contact of the further circuit breaker (remote) (CB1)

(relay R2 send restrain signal CB1)

A B

R1 R2

Trip

Restrain

Restrain

Trip

Power enter at A and exit at B, block trip will prevent the CB from opening.

CB1 CB2

A B

R1 R2

Trip

Restrain

Restrain

Trip

Power enter at A and at B (fault occurs), both CB open

CB1 CB2

A B

R1 R2

Trip

Restrain

Restrain

Trip

Power enter at A and no power at B, circuit breaker A, CB1 will open

CB1 CB2

A B

NON UNIT SCHEME

supervise/monitor one or several points in the power system

signal will be detected at the points

this signal is the input to the relay

section monitored/supervised is not clearly definedZONE OF PROTECTION IS NOT WELL DEFINED

2 examples of non unit scheme(1.) Over Current Protection

(2.) Distance Protection

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•Q & A