Busbar Protection (1)

download Busbar Protection (1)

of 90

description

busbar protection

Transcript of Busbar Protection (1)

  • 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 - RL = 1.5 ( 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

  • Relation between rectangular and polar form

    R = P cos Polar form

    Rectangular form Z = R + jX

    X = P sin

    P

    P = R2 + X2 = tan-1 X/R

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

    impedance:

    Z = R + jX

  • Transmission line protection

    R-X diagram

    R

    jX

    Z1=R1+jX1 Z2=R2+jX2

    Load area

    P1 1 P2 2

    12

  • 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 p , z0 = 25.4 + j101 pBC : z1= z2 = 5.8 + j16.9 pBD : z1= z2 = 8 + j25 p

    C

    A B21 D

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

    = 4.91 71.5 s BC : z1= z2 = 0.92 + j2.7 sBD : z1= z2 = 1.28 + j4 s

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

  • R

    jX

    A

    B

    71.5

    zone1

    zone2

    zone3

  • Earth fault compensate

    Kn = ( z0 z1 ) / 3z1

    = 2.51 + j11.51 = 11.78 77.64.67 + 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 relays zone, Its 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. Its possible that all the zones are not trip!. Memory feature is now used to make high speed trip instead. Its 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