PROTECTION 2SUBSTATION

SOMPOL C.

Busbar Protection

Busbar ProtectionBus arrangement

1. Radial bus2. Main and transfer3. Double breaker double bus4. Ring bus5. Breaker and a half

Busbar ProtectionMain Bus

DisconnectSwitch

Circuit Breaker

Circuit

Radial bus

Busbar Protection1. Radial busAdvantages

Lowest costSmall land area requiredEasy to expandSimple to operateSimple protective relay

Busbar Protection1. Radial busDisadvantages

Low reliabilityLow flexibility of operation for maintenanceBus fault and failure of breaker requires substationbe removed from service

Busbar Protection

DisconnectSwitch

Circuit Breaker

Circuits

Main Bus

Transfer Bus

TransferCircuit

Breaker(N.O.)

Circuits

N.O

.

N.O

.

N.O

. TransferSwitch

Main and transfer

Busbar Protection2. Main and transferAdvantages

Small land area requiredEasy to expandIncreased flexibility of operation over radial busAny breaker can be removed from service without an outage

Busbar Protection 2. Main and transfer Disadvantages Increased cost over radial bus Increased complexity of operation over radial bus Increased complexity of protection over radial bus Low reliability

DisconnectSwitch

Circuit Breaker

Circuits

Bus No. 1

Bus No. 2

Circuits

Circuit Breaker

Busbar Protection

Double breaker double bus

Busbar Protection3. Double breaker double busAdvantages

Very high reliabilityVery flexibility operation Any breaker can be removed from service without an outage

Busbar Protection 3. Double breaker double bus Disadvantages High cost Large land area required Complex protective relaying and control

Circuit Breaker

Source

Load

Source

Load

Line DisconnectSwitch

DisconnectSwitch

Busbar Protection

Ring bus

Busbar Protection4. Ring busAdvantages

High reliabilityFlexibility operationLow costAny breaker can be removed from service without outageExpandable to breaker and a half configuration

Busbar Protection 4. Ring bus Disadvantages Complex protective relaying and control Failed breaker during fault caused outage of one additional circuit

Line DisconnectSwitch

Main Bus No. 1

Main Bus No. 2

Circuit Breaker

Disconnect Switch

Circuits

Circuits

Busbar Protection

Breaker and a half

Busbar Protection5. Breaker and a halfAdvantages

Very high reliabilityVery flexibility operation Any breaker can be removed from service without an outage

Busbar Protection 5. Breaker and a half Disadvantages Large land area required High cost Complex protective relaying and control

Busbar Protection

Approximate per unit cost Reliability

Radial 1 5Main and transfer 1.2 4Ring bus 1.25 3Breaker and a half 1.45 2Double breaker double bus 1.75 1

Busbar Protection

Radial bus

Busbar Protection

Main and transfer

Busbar Protection

Double breaker double bus

Busbar Protection

Breaker and a half

- What kind of bus arrangement in single line diagram 1 and 2 ?

- Where is the zone of protection of 87B1 and 87B2 ?

Practice 0

Busbar Protection

Criteria of Bus Differential relay (87B) Check the difference current between

the current flow in and out of the protected bus ( vector summation at relay = 0 )

Busbar Protection

Bus differential has 2 types1. High impedance2. Low impedance

Busbar ProtectionHigh impedance bus differential

1. Every bay must use same class and CT ratio2. Suitable for non switching substation3. Easy to expand4. Easy to use

Because of fault current at bus bar is very high, so some CT may saturate and make 87B misoperation on external fault…..

Assume one CT saturate on external fault

saturateVoltage = 0 at 87B

CT saturation- Equivalent circuit ( saturate )

E

Ip/n Is

Im

Lm

IfE

saturatesaturateRct

Ip/n Is

EIm = ∞Lm = 0

Rct

Rct

Equivalent circuit = Rct

Assume one CT saturate on external fault

Rct 87B

Voltage > 0 at 87B

Ex Calculation of 87BData- 3 phase fault current at bus = 25000 A ( If - 3phase )- 1 phase fault current at bus = 23600 A ( If - 1phase )- CT ratio 2000/5 ( N )- Rct = 1.2 Ω- RL = 1.5Ω ( lead resistance between relay and CT )- Relay setting range ; 175, 225, 275, 325 v- Vk = 800 v

Vs3 >= ( If / N )* ( Rct + RL ) ; 3 phase fault

Vs1 >= ( If / N )* ( Rct + 2RL ) ; 1phase fault

Setting of 87B ( Vs )

Vs1 = 249.6 v Vs3 = 169.8 v

So set Vs = 325 v;( Vk >= 2Vs )

- From single line diagram 2, if- 3phase fault =17000 A- 1phase fault = 13000 AWhat is the setting of Vs ?

Practice 1

Busbar ProtectionLow impedance bus differential

1. Can use difference CT ratio for each bay2. Suitable for switching substation3. Not easy to expand

Busbar ProtectionFunction of bus differential

Trip all circuit breakers that connected to the fault bus via 86B ( bus differential lockout relay ) and interlock all circuit breakers also.

From single line diagram 2- Which circuit breaker should be tripped if 87B1

operated?- What is the operating time of 87B?

Practice 2

Transmission line Protection

Transmission line protection Since the impedance of a transmission line

is proportional to its length, for distance measurement it is appropriate to use a relay capable of measuring the impedance of a line up to a predetermined point. Such a relay is called distance relay ( 21 ).

Transmission line protection The basic principle of impedance measurement

( Z ) involves the comparison of the fault current ( I ) with the voltage ( V ) “ seen ” by the relay at the relaying point.

Zr = Vr / Ir

Transmission line protection

Impedance seen by relay Zr = Zline + Zload

Z load Zr Vr

Z line ZS

Vs

Ir

Relay point

Transmission line protection

Basic operation of distance relayOperating condition :

Transmission line protection Since the relay see current via CT and voltage

via VT, so actual impedance that relay seen is :

Z relay = Zr * CT ratio / PT ratio

Transmission line protection We use R-X diagram to represent the line

impedance:

Z = R + jX Ω

Relation between rectangular and polar form

R = P cos ƟPolar form

Rectangular form Z = R + jX Ω

X = P sin Ɵ

P Ɵ

P =√ R2 + X2

Ɵ = tan-1 X/R

Transmission line protection We use R-X diagram to represent the line

impedance:

Z = R + jX Ω

Transmission line protection

R-X diagram

R

jX

Z1=R1+jX1 Z2=R2+jX2

Load area

P1 Ɵ1 P2 Ɵ2

Ɵ1Ɵ2

Transmission line protection

Plain impedance

R

jX

Transmission line protection

Plain impedance has no direction !

Transmission line protection

Plain impedance with direction

Transmission line protection jX

Mho

R

Transmission line protection

Offset mho R

jX

Transmission line protection Quadrilateral

Transmission line protection Stepped distance protection - 3 zone of protection, zone1, zone2, zone3 - 3 difference tripping time

A B C Z1A

Z2A Z3A

Z1B

Z1B

Z1C Z2C

Z3C

T2A

T3A

T2C

T2B

T2B

T3C

Transmission line protection

Zone1 = 85 % line , instantaneous trip Zone2 = 120 % line , delay trip T2 Zone3 = 100 % line+120 % next line delay trip T3

Example criteria

Ex Calculation of 21Data- Base voltage =115 kV- Base MVA = 100 MVA- CT ratio = 800/5- PT ratio 115/115 kV/V- Conductor type : 477 MCM AAC – 590 A

Data- Length of line AB = 70 km, line BC = 30 km, line BD = 50 km- Impedance data :

AB : z1= z2 = 9.7 + j29.1 Ωp , z0 = 25.4 + j101 ΩpBC : z1= z2 = 5.8 + j16.9 ΩpBD : z1= z2 = 8 + j25 Ωp

C

A B21 D

Multiply by 0.16, soAB : z1= z2 = 1.55 + j4.65 Ωs , z0 = 4 + j16.1 Ωs

= 4.91 71.5° Ωs BC : z1= z2 = 0.92 + j2.7 ΩsBD : z1= z2 = 1.28 + j4 Ωs

Setting Zone 1 = 85% line AB = 4.17 71.5° Ωs Zone2 = 120% line AB = 5 71.5° Ωs , 0.5 secZone3 = 100% line AB + 120% line AB = 9.17 71.5° Ωs ,1 sec

R

jX

A

B

71.5°

zone1

zone2

zone3

Earth fault compensate

Kn = ( z0 – z1 ) / 3z1

= 2.51 + j11.51 = 11.78 77.6°4.67 + j13.97 14.73 71.5°

= 0.799 6.15°

From single line diagram 2- Which circuit breaker should be tripped if 21

line 1 operated ( both primary and back up )?- Which circuit breaker should be reclosed?- Where is the zone of protection of 21?

Practice 3

Transmission line protection Zone1 can over trip due to : - CT, PT error - Impedance data and calculation error - Relay error

So zone1 should not set 100 % line

Tele Protection

Transmission line protection Because zone 1 cannot clear fault at the end

of line ( 15 % ), fault must be cleared by zone 2 with delay time T2. Too slow!

To solve this problem ‘ communication system ’is required.

Transmission line protection Teleprotection scheme 1. Permissive underreach transfer trip ( PUTT )

2. Permissive overreach transfer trip ( POTT )

Transmission line protection

- Send carrier by zone 1 1. Permissive underreach transfer trip ( PUTT )

- High speed trip ( by pass T2 ) when zone 2 start and carrier received

Transmission line protection

- Send carrier by zone 2 2. Permissive overreach transfer trip ( POTT )

- High speed trip ( bypass T2 ) when zone 2 start and carrier received

Transmission line protection

- clear fault 100% line as fast as zone 1Benefit of teleprotection

- on more over trip - also initiate recloser as zone 1

Transmission line protection Other functions in distance relay * Power swing blocking * Fuse failure * Switch onto fault ( SOTF )

Transmission line protection Power Swing Blocking distance relay operate by detect impedance in its

zone, and sometime voltage and current in the system are disturb by fault. System impedance also change and if impedance move into relay’s zone, It’s trip. Wrong operation!

Transmission line protection

To prevent this situation, relay use PSB. By detect rate of change of the impedance, relay will know which one is fault which one is power swing and block itself to trip…

Transmission line protectionFuse failure

Distance relay calculate impedance by the ratio of voltage to current. If voltage goes to zero, impedance will be zero also. Zero impedance means fault is very close to distance relay and should trip the transmission line.

Transmission line protectionPT fuse blows can make distance relay see

‘zero impedance’ in spite of no fault in high voltage system. Distance relay use ‘ zero sequence concept’ to protect itself from misoperation

Transmission line protectionSwitch OnTo Fault ( SOTF )

For safty in transmission line maintenance, the line should be grounded for all 3 phases. Afterfinish the job, sometime ground are forgotten to remove from line. When CB is closed, it closed to fault.

Transmission line protectionDistance relay use healthy voltage for

reference, so when close into 3 phase fault , no voltage reference at all. It’s possible that all the zones are not trip!. Memory feature is now used to make high speed trip instead. It’s SOTF….

Transmission line protection Auto recloser relay ( 79 )

- Close circuit breaker after tripped by distance relay( only trip by high speed zone )

- Single or multi shots - Single or three poles - Dead time and reclaim time should be set properly - Helpful for temporary fault

Trip & reclose

Dead time

Transmission line protection Synchrocheck relay ( 25 )

- Supervise recloser relay befor close circuit breaker by check voltage level, frequency, and phase angle at both sides of circuit breaker ( sync. Function ,

BH-LH or LL-LB) - Only check voltage level for charge line function

( voltage check Function, BH-LD or DL-LB )

Back up Protection

Back up Protection In protection system, each equipment

should have 2 sets of protective relay. One set we call ‘primary protection’ and the another is call ‘back up protection’.

Back up ProtectionExample :TransformerPrimary – 87K, self protection, Back up – 51T, BFFeederPrimary – 51/51G, Back up – 51T, BF

Back up ProtectionTransmission linePrimary – zone1, Back up – zone2, zone3

and others distance relay ( 115 kV ), BFPrimary – protection primary, Back up –

protection back up and others distance relay ( 230 kV ) , BF

Back up ProtectionBus barPrimary – 87B, Back up – zone2, zone3 of

others distance relay at remote end substation ( 115 kV ), BF

Back up Protection Principle of Breaker failure Measure the duration of fault current from the

instance at which any relay operates to trip circuit breaker. If current is still flowing after preselected time delay, it is considered that the circuit breaker has failed to trip.

Back up Protection Principle of Breaker failure Normally breaker failure timer should less than

zone 2 timer of distance relay at remote end substation to limit the tripping area only in substation that breaker fail.

Back up Protection Element of Breaker failure 1. Main protection operate 2. Current detector operate ( 50BF) 3. Breaker fail timer operate ( 62BF) 4. On this function by cut off switch ( BFCO)

Back up Protection Function of Breaker failure When breaker fail to trip, the tripping and

interlocking of all other circuit breakers connected to the failed circuit breaker will be initiated. Another lockout relay, 86BF, is required.

From single line diagram 2- If circuit breaker 80722 fail to trip, which circuit

breakers should be tripped?

Practice 4

From single line diagram 1- If circuit breaker 7052 fail to trip, which circuit

breakers should be tripped?

Question ?

Have a nice day!

The end