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DISTRIBUTION PROTECTION OVERVIEW etouches
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Table of Contents,IEEE Device Designations 3,System Overview 4. Distribution 5 30MVA Transformer Protection damage curve 9. Relay Overcurrent Curves 20,Symmetrical Components 24. Distribution Fuse Protection 39,Coordinating Time Intervals 43. Transformer Relay Settings using Electromechanical Relays 52. Transformer Differential Protection 67,Smart Grid Challenges Solutions 70. Moscow 13 8kV Feeder Coordination Example 76, IEEE Device Designations commonly used in Distribution Protection.
Avista sometimes adds letters to these such as F for feeders T for transformers B for bus and BF for. breaker failure , 2 Time delay relay 52 a Circuit breaker auxiliary switch closed. 27 Undervoltage relay when the breaker is closed , 43 Manual transfer or selective device 52 b Circuit breaker auxiliary switch closed. We use these for cutting in and out when the breaker is open . instantaneous overcurrent relays 59 Overvoltage relay . reclosing relays etc 62 Time Delay relay, 50 or 50P Instantaneous overcurrent 63 Sudden pressure relay . phase relay 79 AC Reclosing relay , 50N or 50G Instantaneous overcurrent 81 Frequency relay . ground or neutral relay 86 Lock out relay which has several contacts . 50Q Instantaneous Negative Sequence Avista uses 86T for a transformer lockout . overcurrent relay 86B for a bus lockout etc , 51 or 51P Time delay overcurrent phase 87 Differential relay .
relay 94 Auxiliary tripping relay ,51N or 51G Time delay overcurrent. ground or neutral relay ,51Q Time delay Negative Sequence. overcurrent relay ,52 AC circuit breaker , System Overview Distribution Protection. Objective , Protect people company personnel and the public and equipment by the proper. application of overcurrent protective devices ,Devices include .
Relays operating to trip open circuit breakers or circuit switchers and or fuses blowing. for the occurrence of electrical faults on the distribution system . Design tools used ,1 Transformer and conductor damage curves . 2 Time current coordination curves TCC s fuse curves and relay overcurrent. elements based on symmetrical components of fault current . Documentation , 1 One line diagrams and Schematics with standardized device designations as defined. by the IEEE Institute of Electrical and Electronics Engineers keeps everyone on the. same page in understanding how the system works ,2 TCC s. System Overview Inside the, Substation Fence, 115 kV SYSTEM. 12 16 20 MVA HIGH LEAD LOW, 115 13 8kV DELTA WYE, 13 8 kV BUS.
Feeder relay 500A 13 8 kV, FDR 515 PT 1 13 8 kV FDR. 3 5158 512, 556 ACSR SLG 5346,Transformer PT 2A PT 2B 500 A FDR. 4 ACSR 2 ACSR, 556 ACSR 3 3699 1 PHASE, SLG 3060 4 ACSR. 2 ACSR PT 3B PT 3A PT 4, 2 0 ACSR 3 1210, PT 3C SLG 877. 3 3453, SLG 2762, PT 5 LINE RECLOSER P584 FUSE, 4 ACSR 3 558.
3 1907, SLG 1492 SLG 463, 1 PHASE PT 6A PT 6B WYE WYE. 4 ACSR 4 ACSR PT 8, PT 6C 65T, System Overview Outside the. Substation Fence, 115 kV SYSTEM,Midline 12 16 20 MVA HIGH LEAD LOW. 115 13 8kV DELTA WYE, 13 8 kV BUS,relay 500A 13 8 kV. FDR 515 PT 1 13 8 kV FDR, 3 5158 512, 556 ACSR SLG 5346.
PT 2A PT 2B 500 A FDR, 4 ACSR 2 ACSR, 556 ACSR 3 3699 1 PHASE. SLG 3060 4 ACSR, 2 ACSR PT 3B PT 3A PT 4, 2 0 ACSR 3 1210. PT 3C SLG 877, 3 3453, SLG 2762, PT 5 LINE RECLOSER P584 FUSE. 4 ACSR 3 558, 3 1907, SLG 1492 SLG 463, 1 PHASE PT 6A PT 6B WYE WYE. 4 ACSR 4 ACSR PT 8, PT 6C 65T, 115 kV SYSTEM,System Overview MOSCOW.
Each device has at least one curve plotted with current and time values XFMR. 12 16 20 MVA HIGH LEAD LOW, on the Time Coordination Curve 115 13 8kV DELTA WYE. 13 8 kV BUS, 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. 1000 5 1000, 500A 13 8 kV, 700 700 FDR 515 PT 1 13 8 kV FDR. 4 3 5158 512, 400 556 ACSR SLG 5346, 300 6 300 PT 2A PT 2B 500 A FDR. A Conductor damage curve k 0 06710 A 556000 0 cmils. Conductor AAC 4 ACSR 2 ACSR, 200 FEEDER 252 SMALLEST CONDUCTOR TO PROTECT 200.
B Transf damage curve 12 00 MVA Category 3, Base I 502 00 A Z 8 2 percent . 100 IDAHO RD 12 16 20 MVA XFMR 100 556 ACSR 3 3699 1 PHASE. 70 70, SLG 3060 4 ACSR, 50 50, 40 40, 30 30 2 ACSR PT 3B PT 3A PT 4. TRANSFORMER PROTECTION, HI SIDE CT S 2 0 ACSR 3 1210. 20 1 IDR A 777 51P 351 SEL VI TD 1 200 20, CTR 600 5 Pickup 2 A No inst TP 5 0 3096s PT 3C SLG 877. Ia 679 8A 5 7 sec A T 0 78s H 8 33,S 10 10 3 3453.
O 7 2 7, SLG 2762, STATION VCB 252 PROTECTION,D 5 5 PT 5 LINE RECLOSER P584 FUSE. 2 IDR 252 51P 351S SEL EI TD 1 500,S 4 CTR 800 5 Pickup 6 A No inst TP 5 0 4072s 4. Ia 5665 9A 35 4 sec A T 0 30s, 3 3, 3 IDR 252 50G 351S INST TD 1 000. 2 CTR 800 5 Pickup 3 A No inst TP 5 0 048s, 5f B 2 4 ACSR 3 558. 3Io 0 0A 0 0 sec A T 9999s, 3 1907, MAXIMUM FEEDER FUSE.
SLG 1492 SLG 463, 1 4 252 140T FUSE stn S C Link140T. Total clear FAULT DESCRIPTION , 1 3 250 KVA, Ia 5665 9A T 0 09s. MIDLINE OCR, Bus Fault on 0 IDR 252 13 8 kV 3LG, 7 1 PHASE PT 6A PT 6B WYE WYE. 5 5 Phase unit of recloser MID LINE OCR 5 4 ACSR 4 ACSR PT 8. 4 Fast ME 341 B Mult 0 2, Slow ME 305 A Add 1000 . PT 6C 65T, 3 Ia 5665 9A T Fast 0 03s 3, MAXIMUM MIDLINE FUSE.
6 T FUSE S C Link 50T, Minimum melt 3 322, Ia 5665 9A T 0 01s. 1 1 A SLG 271, 07 07, 05 05, 04 04, 03 03, 02 02, Fault I 5665 9 A. 01 01, 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. CURRENT A , HORS 2010 TIME CURRENT CURVES Voltage 13 8 kV By JDH 7. For Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base olr No . Comment At Sub 3LG 5667A SLG 5863A L L 4909A Date 11 25 2008. System Overview , 10 2 3 4 5 7 100, 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7.
What are the types of curves , 700 700, Damage Curves . 6 300, A Conductor damage curve k 0 06710 A 556000 0 cmils. Conductor AAC, FEEDER 252 SMALLEST CONDUCTOR TO PROTECT 200. transformer, B Transf damage curve 12 00 MVA Category 3. Base I 502 00 A Z 8 2 percent , IDAHO RD 12 16 20 MVA XFMR 100.
conductor, 70 70, Protective Curves , TRANSFORMER PROTECTION. HI SIDE CT S, 1 IDR A 777 51P 351 SEL VI TD 1 200. CTR 600 5 Pickup 2 A No inst TP 5 0 3096s, Ia 679 8A 5 7 sec A T 0 78s H 8 33. O 7 2 7, STATION VCB 252 PROTECTION,D 5 5, 2 IDR 252 51P 351S SEL EI TD 1 500. S 4 CTR 800 5 Pickup 6 A No inst TP 5 0 4072s 4, Ia 5665 9A 35 4 sec A T 0 30s.
3 3, 3 IDR 252 50G 351S INST TD 1 000, CTR 800 5 Pickup 3 A No inst TP 5 0 048s. 3Io 0 0A 0 0 sec A T 9999s, 5f B 2 So what curve goes where . MAXIMUM FEEDER FUSE, 4 252 140T FUSE stn S C Link140T. Total clear , Ia 5665 9A T 0 09s, FAULT DESCRIPTION . Bus Fault on 0 IDR 252, 13 8 kV 3LG, Damage curves are at the top and to.
MIDLINE OCR, 5 Phase unit of recloser MID LINE OCR. Fast ME 341 B Mult 0 2, the right of the TCC , Slow ME 305 A Add 1000 . 3 Ia 5665 9A T Fast 0 03s 3, MAXIMUM MIDLINE FUSE. 2 2, 6 T FUSE S C Link 50T, Minimum melt , Ia 5665 9A T 0 01s. Protective curves lowest and to the, 07 07 left on the TCC correspond to those.
05 05, devices farther from the substation, 02 02. where the fault current is less , Fault I 5665 9 A. 01 01, 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. CURRENT A , HORS 2010 TIME CURRENT CURVES Voltage 13 8 kV By JDH 8. For Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base olr No . Comment At Sub 3LG 5667A SLG 5863A L L 4909A Date 11 25 2008. Transformer Protection Damage Curve ANSI IEEE C57 109 1985. Avista has Category III size 5 30MVA Distribution Transformers in service per the above standard . The main damage curve line shows only the thermal effect. from transformer through fault currents It is graphed. 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. from data entered below MVA Base Amps Z 1000, 500 500.
400 400, 300 300, 200 200, A T ransf damage c urve 12 00 MVA Category 3. Base I 502 00 A Z 8 2 percent , 100 100, 70 70, 50 50. 40 40, 30 30, 20 20, S 10 10, O 7 7, D 5 5, The dog leg on the curve is added to allow for additional S 4. thermal and mechanical damage from typically more 2 A 2. than 5 through faults over the life of a transformer 1. serving overhead feeders 5, 3 3, 2 2,Time at 50 of the maximum per unit. 1 1,through fault current 8 seconds 07 07, 05 1 05.
04 04, 03 03, Dog leg curve 10 times base current at 2 seconds 02 02. 01 01, 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. Main curve 25 times base current at 2 seconds For, CURRENT A . T IME CURRENT CURVES Voltage By, 1 M15 P584 GND I. CT R 100 Pickup 3, Comment Date, Conductor Damage Curves.
Copper Conductor Damage Curves ACSR Conductor Damage Curves. 2 0 damage at 1500A 100sec 2 0 damage at 900A 100sec . 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. 1000 1000 1000 1000, 700 1 M15 515 Phase INST INST TD 1 000 700 700 1 M15 515 Phase INST INST TD 1 000 700. CTR 160 Pickup 7 A No inst TP 5 0 048s CTR 160 Pickup 7 A No inst TP 5 0 048s. 500 500 500 500, 400 400 400 400, 300 300 300 300. 200 200 200 200, 100 100 100 100, 70 70 70 70, 50 50 50 50. 40 40 40 40, 30 30 30 30, 20 20 20 20, A Conductor damage curve k 0 08620 A 355107 0 cmils. S 10 10 S 10 10, E Conductor ACSR,E 336 4 ACSR,C 7 7 C 7 7.
A Conductor damage curve k 0 14040 A 105500 0 cmils. O 5 O 5 5, Conductor Copper bare AWG Size 2 0 5 N 4. N 4 B Conductor damage curve k 0 08620 A 167800 0 cmils 4. 2 0 Copper 4 D,D Conductor ACSR AWG Size 4 0, S 3 3. S 3 3 4 0 ACSR, B Conductor damage curve k 0 14040 A 83690 0 cmils 2 2. 2 Conductor Copper bare AWG Size 1 0 2 C Conductor damage curve k 0 08620 A 105500 0 cmils. 1 0 Copper Conductor ACSR AWG Size 2 0, 1 1, 1 1, C Conductor damage curve k 0 14040 A 52630 0 cmils. 7 D Conductor damage curve k 0 08620 A 83690 0 cmils 7. 7 Conductor Copper bare AWG Size 2 7, Conductor ACSR AWG Size 1 0.
2 Copper 5 5, 5 5 1 0 ACSR, 4 4, 4 4, D Conductor damage curve k 0 14040 A 33100 0 cmils 3 E Conductor damage curve k 0 08620 A 52630 0 cmils 3. 3 3, Conductor Copper bare AWG Size 4 Conductor ACSR AWG Size 2. 4 Copper 2 2, 2 2 2 ACSR, E Conductor damage curve k 0 14040 A 20820 0 cmils 1 F Conductor damage curve k 0 08620 A 33100 0 cmils F E D C B A 1. 1 Conductor Copper bare AWG Size 6 E D C B A 1 Conductor ACSR AWG Size 4. 6 Copper 07 4 ACSR 07, 07 07, 05 1 05, 05 1 05 04 04. 04 04, 03 03, 03 03, 02 02, 02 02, 01 01, 01 01 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7.
10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7 CURRENT A . CURRENT A , TIME CURRENT CURVES Voltage 13 8 kV By DLH. TIME CURRENT CURVES Voltage 13 8 kV By DLH, For ACSR Conductor Damage Curves No . For Copper Conductor Damage Curves No , Comment Date 12 13 05. Comment Date 12 13 05 10, Conductor Ampacities, Conductor at 25 C ambient taken from the Westinghouse Transmission Distribution book . ACSR Copper, Ampacity Ratings Ampacity Ratings, Conductor Rating Conductor Rating.
556 730, 336 4 530 2 0 360, 4 0 340 1 0 310, 2 0 270 2 230. 1 0 230 4 170, 2 180 6 120, 4 140, Conductor Protection Graph. 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. 1000 1000, 700 700, Comparing a 140T fuse versus a. 300 300, 4 ACSR Damage curve , 200 200, The 140T won t protect the. 1 Moscow 515 Kear 140T Kearney 140T, Total clear .
100 conductor below about, 70 70, 50 50, 550 amps where the curves cross . 40 40, A Conductor damage curve k 0 08620 A 33100 0 cmils. 30 Conductor ACSR AWG Size 4 30, 20 20,S 10 10,O 7 7. D 5 5,S 4 4, 3 3, 2 2, 1 1, 7 7, 5 5, 4 4, 3 3, 2 2. 1 A 1, 07 07, 05 05, 04 04, 03 03, 02 02, 01 01, 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7.
CURRENT A , 4 ACSR 140T, For Aspen File HORS M15 EXP olr. TIME CURRENT CURVES Voltage 13 8 kV By, Protection. Comment Date 3 06, Transformer Protection using, 115 kV Fuses. 10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7. 1000 1000, Transformer Protection using 700, 500 500. 115 kV Fuses 400, 300 6, A Transf damage curve 7 50 MVA Category 3.
Base I 313 78 A Z 7 3 percent , 200 Rockford 13 8kV ROCKFORD115 115 kV 1 T 200. Used at smaller substations up to 7 5 MVA 100 100,transformer due to low cost of protection 70 70. Other advantages are 50, Low maintenance 30, STATION TRANSFORMER PROTECTION. 2 SMD 2B 65E VERY SLOW 176 19 065, Panel house station battery not required Minimum melt . S 10 FEEDER VCR PROTECTION 10, C 4 LAT RP4211 51P CO 11 CO 11 TD 2 000.
O 7 CTR 500 5 Pickup 3 A No inst TP 5 0 5043s 7, D 5 5 LAT RP4211 51N CO 11 CO 11 TD 4 000 5. There are also several disadvantages to S 4 CTR 500 5 Pickup 2 A No inst TP 5 1 0192s 4. 3 3 LAT RP4211 50P CO 11 INST TD 1 000 3,using fuses however which are 2. CTR 500 5 Pickup 3 5A No inst TP 5 0 048s, 1 LAT RP4211 50N CO 11 INST TD 1 000 A 2. Low interrupting rating from 1 200A for CTR 500 5 Pickup 2 A No inst TP 5 0 048s. MAXIMUM FEEDER FUSE,some older models up to 10 000A at 115 kV 1. 6 LAT RP4211 FUSE S C Link 65T, Total clear , By contrast a circuit switcher can have a rating 5 5.
4 4,of 25KAIC and our breakers have normally 3 3,40KAIC 2 2. The fuses we generally use are rated to blow 1 1,within 5 minutes at twice their nameplate 07 07. 05 1 3 05,rating Thus a 65 amp fuse will blow at 130 04 04. 03 03,amps This compromises the amount of 02 02,overload we can carry in an emergency and. still provide good sensitivity for faults 01, 10 2 3 4 5 7 100 2 3 4 5 7 1000.
CURRENT A , 2 3 4 5 7 10000 2 3 4 5 7, ROK 451R2 TIME CURRENT CURVES Voltage 13 8 KV By JDH. For ROK FDR 451 BUS FAULTS 3LG 3580A SLG 3782A L L 3100A No . Comment ASPEN FILE ROK NEW XFMR 2005 BASE OLR Date 2 13 07. Transformer Protection using 115 kV Fuses continued. The fuse time current characteristic TCC is fixed although you can buy a standard slow. or very slow speed ratio which are different inverse curves . The sensitivity to detect lo side SLG faults isn t as good as using a relay on a circuit. switcher or breaker This is because we use DELTA WYE connected transformers so the. phase current on the 115 kV is reduced by the 3 as opposed to a three phase fault . Some fuses can be damaged and then blow later at some high load point . When only one 115 kV fuse blows it subjects the customer to low distribution voltages . For example the phase to neutral distribution voltages on two phases on the 13 8 kV become. 50 of normal A a, B b, 0 5 PU 0 5 PU, C c, No indication of faulted zone transformer bus or feeder .


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