Busbar Protection
Transcript of Busbar Protection
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