SUBSTATIONPROTECTION 2 SOMPOL C. Busbar Protection Busbar Protection Bus arrangement 1. Radial bus 2. Main and transfer 3. Double breaker double bus 4. Ring bus 5. Breaker and a half Busbar Protection Main Bus Disconnect Switch Circuit Breaker Circuit Radial bus . Radial bus Advantages Lowest cost Small land area required Easy to expand Simple to operate Simple protective relay .Busbar Protection 1. Busbar Protection 1. Radial bus Disadvantages Low reliability Low flexibility of operation for maintenance Bus fault and failure of breaker requires substation be removed from service . O.O. N.Busbar Protection Main Bus Disconnect Switch Transfer Circuit Breaker (N. Circuit Breaker Transfer Switch Transfer Bus Circuits Circuits Main and transfer .) N. N.O.O. Busbar Protection 2. Main and transfer Advantages Small land area required Easy to expand Increased flexibility of operation over radial bus Any breaker can be removed from service without an outage . 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 .Busbar Protection 2. 1 Disconnect Switch Circuit Breaker Circuit Breaker Circuits Bus No. 2 Double breaker double bus .Busbar Protection Circuits Bus No. Double breaker double bus Advantages Very high reliability Very flexibility operation Any breaker can be removed from service without an outage .Busbar Protection 3. Busbar Protection 3. Double breaker double bus Disadvantages High cost Large land area required Complex protective relaying and control . Busbar Protection Load Line Disconnect Switch Source Disconnect Switch Circuit Breaker Source Load Ring bus . Ring bus Advantages High reliability Flexibility operation Low cost Any breaker can be removed from service without outage Expandable to breaker and a half configuration .Busbar Protection 4. Busbar Protection 4. Ring bus Disadvantages Complex protective relaying and control Failed breaker during fault caused outage of one additional circuit . 1 Line Disconnect Switch Disconnect Switch Circuit Breaker Circuits Breaker and a half Main Bus No.Busbar Protection Circuits Main Bus No. 2 . Breaker and a half Advantages Very high reliability Very flexibility operation Any breaker can be removed from service without an outage .Busbar Protection 5. 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 1 5 Main and transfer 1.25 3 Breaker and a half 1.75 1 Radial .2 4 Ring bus 1.45 2 Double breaker double bus 1. Busbar Protection Radial bus . Busbar Protection Main and transfer . Busbar Protection Double breaker double bus . Busbar Protection Breaker and a half . Practice 0 .What kind of bus arrangement in single line diagram 1 and 2 ? .Where is the zone of protection of 87B1 and 87B2 ? . 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 types 1. High impedance 2. Low impedance . Suitable for non switching substation 3.Busbar Protection High impedance bus differential 1. Easy to use . Every bay must use same class and CT ratio 2. Easy to expand 4. .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 saturate Voltage = 0 at 87B . Equivalent circuit ( saturate ) Ip/n Is Rct saturate Im E Lm Ip/n E Is Rct Lm = 0 Im = ∞ Rct E If Equivalent circuit = Rct .CT saturation . Assume one CT saturate on external fault Rct 87B Voltage > 0 at 87B . 175.Vk = 800 v .1phase ) .CT ratio 2000/5 ( N ) . 325 v .Ex Calculation of 87B Data .5 Ω ( lead resistance between relay and CT ) .3 phase fault current at bus = 25000 A ( If .Rct = 1.Relay setting range . 275.RL = 1.1 phase fault current at bus = 23600 A ( If . 225.3phase ) .2 Ω . Setting of 87B ( Vs ) Vs1 >= ( If / N )* ( Rct + 2RL ) ; 1phase fault Vs3 >= ( If / N )* ( Rct + RL ) ; 3 phase fault Vs1 = 249.6 v Vs3 = 169.8 v So set Vs = 325 v; ( Vk >= 2Vs ) Practice 1 - From single line diagram 2, if - 3phase fault =17000 A - 1phase fault = 13000 A What is the setting of Vs ? Busbar Protection Low impedance bus differential 1. Can use difference CT ratio for each bay 2. Suitable for switching substation 3. Not easy to expand Busbar Protection Function 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. Practice 2 From single line diagram 2 - Which circuit breaker should be tripped if 87B1 operated? - What is the operating time of 87B? Transmission line Protection . Transmission line protection Since the impedance of a transmission line is proportional to its length. Such a relay is called distance relay ( 21 ). . for distance measurement it is appropriate to use a relay capable of measuring the impedance of a line up to a predetermined point. 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 Relay point ZS Vs Z line Ir Vr Zr Z load Impedance seen by relay Zr = Zline + Zload . Transmission line protection Operating condition : Basic operation of distance relay . 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 Rectangular form Polar form R = P cos Ɵ X = P sin Ɵ Z = R + jX PƟ P Ω 2 2 =√ R + X -1 Ɵ = tan X/R . Transmission line protection We use R-X diagram to represent the line impedance: Z = R + jX Ω . Transmission line protection jX P1 Ɵ1 Z1=R1+jX1 P2 Ɵ2 Z2=R2+jX2 Ɵ2 Ɵ1 R Load area R-X diagram . Transmission line protection jX R Plain impedance . Transmission line protection Plain impedance has no direction ! . Transmission line protection Plain impedance with direction . Transmission line protection jX Mho R . Transmission line protection jX R Offset mho . Transmission line protection Quadrilateral . zone1. zone2.3 difference tripping time T3A T2 Z1A A Z2A A B Z1C Z3C Z2C T3C T2 Z1B Z1B T2B Z3A T2C B C .Transmission line protection Stepped distance protection . zone3 .3 zone of protection. delay trip T2 Zone3 = 100 % line+120 % next line delay trip T3 . instantaneous trip Zone2 = 120 % line .Transmission line protection Example criteria Zone1 = 85 % line . Base voltage =115 kV .CT ratio = 800/5 .Ex Calculation of 21 Data .Conductor type : 477 MCM AAC – 590 A .PT ratio 115/115 kV/V .Base MVA = 100 MVA . 1 Ωp .4 + j101 Ωp BC : z1= z2 = 5.Impedance data : AB : z1= z2 = 9.8 + j16.7 + j29.Length of line AB = 70 km. line BC = 30 km.9 Ωp BD : z1= z2 = 8 + j25 Ωp C 21 A B D . line BD = 50 km . z0 = 25.Data . 5° Ωs .65 Ωs .28 + j4 Ωs Setting Zone 1 = 85% line AB = 4.17 71.Multiply by 0.55 + j4.17 71. 0.16.92 + j2. so AB : z1= z2 = 1.5° Ωs .91 71. z0 = 4 + j16.7 Ωs BD : z1= z2 = 1.1 sec .5 sec Zone3 = 100% line AB + 120% line AB = 9.1 Ωs = 4.5° Ωs Zone2 = 120% line AB = 5 71.5° Ωs BC : z1= z2 = 0. 5° A zone1 R .jX zone3 zone2 B 71. 97 14.78 77.67 + j13.51 + j11.Earth fault compensate Kn = ( z0 – z1 ) / 3z1 = 2.799 6.6° 4.73 71.15° .5° = 0.51 = 11. Which circuit breaker should be tripped if 21 line 1 operated ( both primary and back up )? .Where is the zone of protection of 21? .Which circuit breaker should be reclosed? .Practice 3 From single line diagram 2 . CT. PT error .Relay error So zone1 should not set 100 % line .Transmission line protection Zone1 can over trip due to : .Impedance data and calculation error . Tele Protection . Too slow! To solve this problem ‘ communication system ’ is required. .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. Transmission line protection Teleprotection scheme 1. Permissive underreach transfer trip ( PUTT ) 2. Permissive overreach transfer trip ( POTT ) . Permissive underreach transfer trip ( PUTT ) .Transmission line protection 1.Send carrier by zone 1 .High speed trip ( by pass T2 ) when zone 2 start and carrier received . Permissive overreach transfer trip ( POTT ) .Send carrier by zone 2 .High speed trip ( bypass T2 ) when zone 2 start and carrier received .Transmission line protection 2. Transmission line protection Benefit of teleprotection .also initiate recloser as zone 1 .clear fault 100% line as fast as zone 1 .on more over trip . Transmission line protection Other functions in distance relay * Power swing blocking * Fuse failure * Switch onto fault ( SOTF ) . and sometime voltage and current in the system are disturb by fault.Transmission line protection Power Swing Blocking distance relay operate by detect impedance in its zone. Wrong operation! . It’s trip. System impedance also change and if impedance move into relay’s zone. By detect rate of change of the impedance. relay use PSB.Transmission line protection To prevent this situation. relay will know which one is fault which one is power swing and block itself to trip… . impedance will be zero also. Zero impedance means fault is very close to distance relay and should trip the transmission line.Transmission line protection Fuse failure Distance relay calculate impedance by the ratio of voltage to current. If voltage goes to zero. . Distance relay use ‘ zero sequence concept’ to protect itself from misoperation .Transmission line protection PT fuse blows can make distance relay see ‘zero impedance’ in spite of no fault in high voltage system. sometime ground are forgotten to remove from line. When CB is closed. it closed to fault.Transmission line protection Switch OnTo Fault ( SOTF ) For safty in transmission line maintenance. After finish the job. the line should be grounded for all 3 phases. . .Transmission line protection Distance relay use healthy voltage for reference. Memory feature is now used to make high speed trip instead. no voltage reference at all. It’s possible that all the zones are not trip!. It’s SOTF…. so when close into 3 phase fault . Helpful for temporary fault .Close circuit breaker after tripped by distance relay ( only trip by high speed zone ) .Transmission line protection Auto recloser relay ( 79 ) .Dead time and reclaim time should be set properly .Single or multi shots .Single or three poles . Trip & reclose Dead time . Only check voltage level for charge line function ( voltage check Function. and phase angle at both sides of circuit breaker ( sync.Transmission line protection Synchrocheck relay ( 25 ) .Supervise recloser relay befor close circuit breaker by check voltage level. BH-LH or LL-LB) . frequency. BH-LD or DL-LB ) . Function . 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’. . BF Feeder Primary – 51/51G.Example : Back up Protection Transformer Primary – 87K. BF . self protection. Back up – 51T. Back up – 51T. Back up – protection back up and others distance relay ( 230 kV ) .Back up Protection Transmission line Primary – zone1. Back up – zone2. BF Primary – protection primary. zone3 and others distance relay ( 115 kV ). BF . Back up Protection Bus bar Primary – 87B. BF . zone3 of others distance relay at remote end substation ( 115 kV ). Back up – zone2. 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. it is considered that the circuit breaker has failed to trip. If current is still flowing after preselected time delay. . .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. On this function by cut off switch ( BFCO) . Current detector operate ( 50BF) 3. Main protection operate 2. Breaker fail timer operate ( 62BF) 4. Another lockout relay. .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. is required. 86BF. which circuit breakers should be tripped? From single line diagram 1 .If circuit breaker 7052 fail to trip.If circuit breaker 80722 fail to trip.Practice 4 From single line diagram 2 . which circuit breakers should be tripped? . Question ? . Have a nice day! The end .