DISTRIBUTION PROTECTION OVERVIEW - etouches Protection - Damage Curve – ANSI/IEEE C57.109 -1985 The main damage curve line shows only the thermal effect from transformer throughfault

DISTRIBUTION PROTECTION OVERVIEW

Kevin Damron amp Calvin HowardAvista Utilities

Presented March 12th 2018

At the 35th AnnualHands-On Relay School

Washington State UniversityPullman Washington

Table of Contents

IEEE Device Designations helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 3

System Overview helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 4

Distribution (5-30MVA ) Transformer Protection damage curve helliphelliphellip 9

Relay Overcurrent Curves helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 20

Symmetrical Components helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 24

Distribution Fuse Protection helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 39

Coordinating Time Intervals helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 43

Transformer Relay Settings using Electromechanical Relays helliphelliphelliphelliphelliphellip 52

Transformer Differential Protection helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 67

Smart Grid Challenges amp Solutionshelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 70

Moscow 138kV Feeder Coordination Example helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip 76

2 ndash Time delay relay27 ndash Undervoltage relay43 ndash Manual transfer or selective device

We use these for cutting in and out instantaneous overcurrent relays reclosing relays etc

50 (or 50P) ndash Instantaneous overcurrent phase relay

50N (or 50G) ndash Instantaneous overcurrentground (or neutral) relay

50Q ndash Instantaneous Negative Sequence overcurrent relay

51 (or 51P) ndash Time delay overcurrent phase relay

51N (or 51G) ndash Time delay overcurrent ground (or neutral) relay

51Q ndash Time delay Negative Sequence overcurrent relay

52 ndash AC circuit breaker

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

52a ndash Circuit breaker auxiliary switch closedwhen the breaker is closed

52b ndash Circuit breaker auxiliary switch closed when the breaker is open

59 ndash Overvoltage relay62 ndash Time Delay relay63 ndash Sudden pressure relay79 ndash AC Reclosing relay81 ndash Frequency relay86 ndash Lock out relay which has several contacts

Avista uses 86T for a transformer lockout 86B for a bus lockout etc

87 ndash Differential relay94 ndash Auxiliary tripping relay

System Overview - Distribution Protection

ObjectiveProtect people (company personnel and the public) and equipment by the proper application of overcurrent protective devices

Devices includeRelays operating to trip (open) circuit breakers or circuit switchers andor fuses blowing for the occurrence of electrical faults on the distribution system

Design tools used1 ndash Transformer and conductor damage curves2 - Time-current coordination curves (TCCrsquos) fuse curves and relay overcurrent elements based on symmetrical components of fault current

Documentation1 - One-line diagrams and Schematics with standardized device designations as defined by the IEEE (Institute of Electrical and Electronics Engineers) ndash keeps everyone on the same page in understanding how the system works 2 - TCCrsquos

XFMR 121620 MVA

115138kV

115 kV SYSTEM

138 kV BUS500A 138 kV

FDR 515

DELTAWYE

LINE RECLOSER P584

3Oslash = 5158SLG = 5346

3Oslash = 3453SLG = 2762

PT-2BPT-2A

3Oslash = 3699SLG = 3060

PT-3B PT-3A PT-4

4 ACSR

1 PHASE4 ACSR 4 ACSR

4 ACSR

2 ACSR4 ACSR

20 ACSR2 ACSR

3Oslash = 1907SLG = 1492

3Oslash = 322SLG = 271

3-250 KVAWYEWYE

PT-865T

A-172MOSCOW

1 PHASE4 ACSR

PT-6BPT-6A

HIGH LEAD LOW

3Oslash = 1210SLG = 877

556 ACSR

138 kV FDR 512

500 A FDR

System Overview ndash Inside the Substation Fence

Transformer relays

Feeder relay

XFMR 121620 MVA

115138kV

115 kV SYSTEM

138 kV BUS500A 138 kV

FDR 515

DELTAWYE

LINE RECLOSER P584

3Oslash = 5158SLG = 5346

3Oslash = 3453SLG = 2762

PT-2BPT-2A

3Oslash = 3699SLG = 3060

PT-3B PT-3A PT-4

4 ACSR

2 ACSR4 ACSR

20 ACSR2 ACSR

3Oslash = 1907SLG = 1492

3-250 KVAWYEWYE

PT-865T

A-172MOSCOW

1 PHASE4 ACSR

PT-6BPT-6A

HIGH LEAD LOW

3Oslash = 1210SLG = 877

556 ACSR

138 kV FDR 512

500 A FDR

System Overview ndash Outside the Substation Fence

Midline recloser relay

XFMR 121620 MVA

115138kV

115 kV SYSTEM

138 kV BUS500A 138 kV

FDR 515

DELTAWYE

LINE RECLOSER P584

3Oslash = 5158SLG = 5346

3Oslash = 3453SLG = 2762

PT-2BPT-2A

3Oslash = 3699SLG = 3060

PT-3B PT-3A PT-4

4 ACSR

2 ACSR4 ACSR

20 ACSR2 ACSR

3Oslash = 1907SLG = 1492

3-250 KVAWYEWYE

PT-865T

A-172MOSCOW

1 PHASE4 ACSR

PT-6BPT-6A

HIGH LEAD LOW

3Oslash = 1210SLG = 877

556 ACSR

138 kV FDR 512

500 A FDR

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 kVHORS 2010 By JDHFor Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007baseolr NoComment At Sub 3LG=5667A SLG=5863A L-L=4909A Date 11-25-2008

TRANSFORMER PROTECTION

STATION VCB 252 PROTECTION

MAXIMUM FEEDER FUSE

HI-SIDE CTS

MIDLINE OCR

MAXIMUM MIDLINE FUSE

Fault I=56659 A

1 IDR A-777 51P 351 SEL-VI TD=1200CTR=6005 Pickup=2A No inst TP5=03096sIa= 6798A (57 sec A) T= 078s H=833

2 IDR 252 51P 351S SEL-EI TD=1500CTR=8005 Pickup=6A No inst TP5=04072sIa= 56659A (354 sec A) T= 030s

3 IDR 252 50G 351S INST TD=1000CTR=8005 Pickup=3A No inst TP5=0048s3Io= 00A (00 sec A) T=9999s

4 252 140T FUSE stn SampC Link140TTotal clearIa= 56659A T= 009s

5 Phase unit of recloser MID LINE OCR Fast ME-341-B Mult=02 Slow ME-305-A Add=1000 Ia= 56659A T(Fast)= 003s

6 T FUSE SampC Link 50TMinimum meltIa= 56659A T= 001s

A Conductor damage curve k=006710 A=5560000 cmilsConductor AACFEEDER 252 SMALLEST CONDUCTOR TO PROTECT

B Transf damage curve 1200 MVA Category 3Base I=50200 A Z= 82 percentIDAHO RD 121620 MVA XFMR

FAULT DESCRIPTIONBus Fault on 0 IDR 252 138 kV 3LG

System Overview ndashEach device has at least one curve plotted with current and time values on the Time Coordination Curve

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 7CURRENT (A)

SECONDS

400500

MAXIMUM FEEDER FUSE

HI-SIDE CTS

MIDLINE OCR

Fault I=56659 A

System Overview ndash

Damage Curves- transformer- conductor

So what curve goes where

What are the types of curves

Protective Curves- relay- fuse

Damage curves are at the top and to the right of the TCC

Protective curves lowest and to the left on the TCC correspond to those devices farther from the substation where the fault current is less

Transformer Protection - Damage Curve ndash ANSIIEEE C57109-1985

The main damage curve line shows only the thermal effect from transformer through-fault currents It is graphed from data entered below (MVA Base Amps Z ) 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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage ByFor NoComment Date

1 M15 CTR=1

A Transf damage curve 1200 MVA Category 3Base I=50200 A Z= 82 percent

The dog leg on the curve is added to allow for additional thermal and mechanical damage from (typically more than 5) through-faults over the life of a transformer serving overhead feeders

Dog leg curve - 10 times base current at 2 seconds

Main curve - 25 times base current at 2 seconds

Time at 50 of the maximum per-unit through fault current = 8 seconds

Avista has Category III size (5-30MVA) Distribution Transformers in service per the above standard

Copper Conductor Damage Curves ACSR Conductor Damage Curves (20 damage at 1500A 100sec) (20 damage at 900A 100sec)

Conductor Damage Curves

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 7CURRENT (A)

SECONDS

30405070

300400500700

345710

30405070100

3004005007001000

TIME-CURRENT CURVES Voltage 138 kV By DLHFor ACSR Conductor Damage Curves NoComment Date 121305

1 M15-515 Phase INST INST TD=1000CTR=160 Pickup=7A No inst TP5=0048s

A Conductor damage curve k=008620 A=3551070 cmilsConductor ACSR3364 ACSR

B Conductor damage curve k=008620 A=1678000 cmilsConductor ACSR AWG Size 4040 ACSR

C Conductor damage curve k=008620 A=1055000 cmilsConductor ACSR AWG Size 2020 ACSR

D Conductor damage curve k=008620 A=836900 cmilsConductor ACSR AWG Size 1010 ACSR

E Conductor damage curve k=008620 A=526300 cmilsConductor ACSR AWG Size 22 ACSR

FF Conductor damage curve k=008620 A=331000 cmilsConductor ACSR AWG Size 44 ACSR

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 7CURRENT (A)

SECONDS

30405070

300400500700

345710

30405070100

3004005007001000

TIME-CURRENT CURVES Voltage 138 kV By DLHFor Copper Conductor Damage Curves NoComment Date 121305

A Conductor damage curve k=014040 A=1055000 cmilsConductor Copper (bare) AWG Size 2020 Copper

B Conductor damage curve k=014040 A=836900 cmilsConductor Copper (bare) AWG Size 1010 Copper

C Conductor damage curve k=014040 A=526300 cmilsConductor Copper (bare) AWG Size 22 Copper

D Conductor damage curve k=014040 A=331000 cmilsConductor Copper (bare) AWG Size 44 Copper

EE Conductor damage curve k=014040 A=208200 cmilsConductor Copper (bare) AWG Size 66 Copper

Conductor Rating556 7303364 53040 34020 27010 2302 1804 140

ACSR Ampacity Ratings

Conductor Rating

20 36010 3102 2304 1706 120

CopperAmpacity Ratings

Conductor Ampacities

Conductor at 25degC ambient taken from the Westinghouse Transmission amp Distribution book

Comparing a 140T fuse versus a4 ACSR Damage curve The 140T wonrsquot protect the conductor below about 550 amps where the curves cross

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

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 kV4 ACSR amp 140T By ProtectionFor Aspen File HORS M15 EXPolr NoComment Date 306

1 Moscow 515 Kear 140T Kearney 140TTotal clear

A Conductor damage curve k=008620 A=331000 cmilsConductor ACSR AWG Size 4

Transformer Protection using 115 kV Fuses

Used at smaller substations up to 75 MVA transformer due to low cost of protectionOther advantages are- Low maintenance- Panel house amp station battery not required

There are also several disadvantages to using fuses however which arebull Low interrupting rating from 1200A (for some older models) up to 10000A at 115 kV By contrast a circuit switcher can have a rating of 25KAIC and our breakers have normally 40KAIC bull The fuses we generally use are rated to blow within 5 minutes at twice their nameplate rating Thus a 65 amp fuse will blow at 130 amps This compromises the amount of overload we can carry in an emergency and still provide good sensitivity for faults

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 KVROK 451R2 By JDHFor ROK FDR 451 BUS FAULTS 3LG=3580A SLG=3782A L-L=3100A NoComment ASPEN FILE ROK NEW XFMR (2005 BASE)OLR Date 2-13-07

STATION TRANSFORMER PROTECTION

FEEDER VCR PROTECTION

MAXIMUM FEEDER FUSE

1 LAT RP4211 50N CO-11 INST TD=1000CTR=5005 Pickup=2A No inst TP5=0048s

2 SMD-2B 65E VERY SLOW 176-19-065Minimum melt H=833

3 LAT RP4211 50P CO-11 INST TD=1000CTR=5005 Pickup=35A No inst TP5=0048s

4 LAT RP4211 51P CO-11 CO-11 TD=2000CTR=5005 Pickup=3A No inst TP5=05043s

5 LAT RP4211 51N CO-11 CO-11 TD=4000CTR=5005 Pickup=2A No inst TP5=10192s

6 LAT RP4211 FUSE SampC Link 65TTotal clear

A Transf damage curve 750 MVA Category 3Base I=31378 A Z= 73 percentRockford 138kV - ROCKFORD115 115kV 1 T

bull 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)

bull The sensitivity to detect lo-side SLG faults isnrsquot as good as using a relay on a circuit switcher or breaker This is because we use DELTAWYE connected transformers so the phase current on the 115 kV is reduced by the radic3 as opposed to a three phase fault

bull Some fuses can be damaged and then blow later at some high load point

bull 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 138 kV become 50 of normal

bull No indication of faulted zone (transformer bus or feeder)

Transformer Protection using 115 kV Fuses - continued

05 PU05 PU

Transformer Protection using a Circuit SwitcherShowing Avistarsquos present standard using Microprocessor relays

138kV 115kV

Transformer Protection using a Circuit SwitcherShowing Avistarsquos old standard using Electromechanical relays

Transformer Protection using a Circuit Switcher

Some advantages to this over fuses arebull Higher interruptingbull Relays can be set to operate faster

and with better sensitivity than fuses

bull Three phase operationbull Provide better coordination with

downstream devices

Some disadvantages would bebull Higher costbull Higher maintenancebull Requires a substation battery panel

house and relayingbull Transformer requires CTrsquos

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 kVIDR FEEDER 251 By JDHFor Idaho Rd Feeder 251 in Idaho Rd PHASE 1 2007baseolr NoComment At Sub 3LG=5667A SLG=5863A L-L=4909A Date 11-25-2008

MAXIMUM FEEDER FUSE

HI-SIDE CTS

LO-SIDE CTS

MINIMUM FAULT TO DETECT3LG=2460A SLG=1833A L-L=2130A

1 IDR A-777 51P 351 SEL-VI TD=1200CTR=6005 Pickup=2A No inst TP5=03096s H=833

2 IDR A-777 51G 351 SEL-VI TD=1000CTR=6005 Pickup=1A No inst TP5=0258s H=833

3 IDR A-777 51N 351 SEL-EI TD=5700CTR=12005 Pickup=3A No inst TP5=15473s

4 IDR A-777 51Q 351 SEL-EI TD=4200CTR=6005 Pickup=13A No inst TP5=11401s H=833

5 IDR A-777 51Q 587 W2 SEL-EI TD=4300CTR=12005 Pickup=54A No inst TP5=11672s

6 IDR 251 50P 351S INST TD=1000CTR=8005 Pickup=7A No inst TP5=0048s

7 IDR 251 51P 351S SEL-EI TD=1500CTR=8005 Pickup=6A No inst TP5=04072s

8 IDR 251 50G 351S INST TD=1000CTR=8005 Pickup=3A No inst TP5=0048s

9 IDR 251 51G 351S SEL-EI TD=3900CTR=8005 Pickup=3A No inst TP5=10587s

10 IDR 251 51Q 351S SEL-EI TD=3500CTR=8005 Pickup=52A No inst TP5=09501s

11 251 140T FUSE stn SampC Link140TTotal clear

A Conductor damage curve k=008620 A=1331000 cmilsConductor ACSR AWG Size 20FEEDER 251 SMALLEST CONDUCTOR TO PROTECT

Transformer Protection using a Breaker

This is very similar to using a circuit switcher with a couple of advantages such asbull Higher interrupting ndash 40kAIC for the one shown belowbull Somewhat faster tripping than a circuit switcher (3 cycles vs 6 ndash 8 cycles)bull Possibly less maintenance than a circuit switcherbull The CTrsquos would be located on the breaker so it would interrupt faults on the bus section up to

the transformer plus the transformer high side bushings

SEL Various Relay Overcurrent Curves Extremely Inverse ndash steepestVery InverseInverseModerately InverseShort Time Inverse ndash least steep

The five curves shown here have the same pickup settings but different time dial settings

These are basically the same as various E-M relays

Avista uses mostly extremely inverse on feeders to match the fuse curves

Relay Overcurrent Curves10 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 7CURRENT (A)

SECONDS

400500

1 EXTREMELY INVERSE SEL-EI TD=15000CTR=8005 Pickup=6A No inst TP5=40718s

2 INVERSE SEL-I TD=2000CTR=8005 Pickup=6A No inst TP5=08558s

3 MODERATELY INVERSE SEL-2xx-MI TD=1000CTR=8005 Pickup=6A No inst TP5=0324s

4 SHORT TIME INVERSE SEL-STI TD=1000CTR=8005 Pickup=6A No inst TP5=01072s

5 5 VERY INVERSE SEL-VI TD=6000CTR=8005 Pickup=6A No inst TP5=15478s

Time dial - at what time delay will the relay operate to trip the breaker- the larger time dial means more time delay- also known as ldquoleverrdquo from the electromechanical relay days- Instantaneous elements have a ldquotime dialrdquo of 1 and operate at 005 seconds- Instantaneous curves are shown as a flat horizontal line starting at the left at the pickup value and plotted at 005 seconds

Relay Overcurrent Curves - Pickups amp Time Dials Pickup - the current at which the relay will operate to trip the breaker- also known as ldquotaprdquo from the electromechanical relay days-expressed in terms of the ratio of the current transformer (CTR) that the relay is connected to - eg a relay with a CTR of 120 and a pickup (or tap) of 4 will operate to trip the breaker at 480 amps

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 7CURRENT (A)

SECONDS

400500

MAXIMUM FEEDER FUSE

HI-SIDE CTS

LO-SIDE CTS

Same Time Dial = 9Right curve picks up at 960 AmpsLeft curve picks up at 320Amps

Relay Overcurrent Curves - Example of Different Pickup amp Time Dial Settings

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 7CURRENT (A)

SECONDS

400500

1 EXT INV 1 SEL-EI TD=15000CTR=8005 Pickup=6A No inst TP5=40718s

2 2 ~EXT INV 1 SEL-EI TD=12000CTR=8005 Pickup=6A No inst TP5=32574s 3

3 ~EXT INV 2 SEL-EI TD=9000CTR=8005 Pickup=6A No inst TP5=24431s

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 7CURRENT (A)

SECONDS

400500

Same Pickup = 960 AmpsTop curve Time Dial = 15Bottom curve Time Dial = 2

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 kVINT 12F1 By JDHFor Indian Trail feeder 12F1 in INT 2007baseolr NoComment AT SUB 3LG= 5719A LL=4951A SLG= 5880A Date 1-29-08

STATION VCB 12F1 PROTECTION

1 1 INT A-742 51P 351 SEL-VI TD=10000CTR=6005 Pickup=08A No inst TP5=25797s H=833

2 INT 12F1 51P 351S SEL-VI TD=1800CTR=8005 Pickup=3A No inst TP5=04643s

3 INT 12F1 51G 351S SEL-VI TD=0900CTR=8005 Pickup=3A No inst TP5=02322s

4 INT 12F1 51Q 351S SEL-VI TD=0500CTR=8005 Pickup=3A No inst TP5=0129s

5 5 INT A-742 51G 351 SEL-VI TD=5000CTR=6005 Pickup=08A No inst TP5=12898s H=833 6

6 INT A-742 51Q 351 SEL-VI TD=3000CTR=6005 Pickup=08A No inst TP5=07739s H=833

A Transf damage curve 1200 MVA Category 3Base I=50200 A Z= 84 percentINDIAN TRAIL 121620 MVA XFMR

B Conductor damage curve k=006710 A=5560000 cmilsConductor AAC556 kCMil AAC

Microprocessor relays have different types of curves based on the type of fault current being measured

51P ndash phase time overcurrent

51N or 51G ndash ground time overcurrent

51Q ndash negative sequence time overcurrent

Transformer relay curves

Feeder relay curves51P ndash phase time overcurrent

This brings us to a brief discussion of

Symmetrical Components

Symmetrical Components for Power Systems Engineering J Lewis Blackburn

There are three sets of independent components in a three-phase system positive negative and zero for both current and voltage Positive sequence voltages (Figure 1) are supplied by generators within the system and are always present A second set of balanced phasors are also equal in magnitude and displaced 120 degrees apart but display a counter-clockwise rotation sequence of A-C-B (Figure 2) which represents a negative sequence The final set of balanced phasors is equal in magnitude and in phase with each other however since there is no rotation sequence (Figure 3) this is known as a zero sequence

Three phase (3LG) fault - Positive sequence currents ndash for setting phase elements in relays Phase -Phase (L-L) fault - Negative sequence currents ndash for setting negative sequence elements in relaysSingle phase (1LG or SLG) fault - Zero sequence currents ndash for setting ground elements in relays

Symmetrical Components NotationPositive Sequence current = I + = I1Negative Sequence current = I - = I2Zero Sequence current = I 0 = I0

Phase Current notationIA ndash High side AmpsIa ndash Low side Amps

833= 115138 = transformer Voltage (turns) ratioPhasor diagram from Fault-study Software

Examples of three 138kV faults showing the current distribution through a Delta-Wye high-lead-low transformer bank

Symmetrical Components ndash Positive Sequence 3LG 138 kV FaultYou have only positive sequence voltage and current since the system is balanced

IA = 619 ang-88deg Ia = 5158 ang-118degIB = 619 ang 152deg Ib = 5158 ang 122degIC = 619 ang 32deg Ic = 5158 ang 2degPhase current = Sequence currentThat is Ia = I+

Phase currents and voltages for the 115kV side

IA = Ia 833 = 5158A 833IA = 619A

Symmetrical Components ndash Negative Sequence L-L 138 kV Fault

115kV side sequence currents and voltages

IA = 309 ang-28deg Ia = 0 ang 0degIB = 619 ang 152deg Ib = 4467 ang 152degIC = 309 ang-28deg Ic = 4467 ang-28deg

3LG 138kV fault = 5158AIb=Ic=4467 A 44675158 = 866=radic32IB3LG = IBLL = 619AIA amp IC = IB or Ia amp Ic = Ib I2 = the phase currentradic3 = 4467radic3 = 2579

Digital relays 50Q51Q elements set using 3I23I2 = Ib x radic3 = 4467 x 1732 = 7737

Symmetrical Components ndash Zero Sequence 1LG 138 kV Fault

3I0 is the sum of the 3 phase currents and since Ib amp Ic= 0 then 3I0 = Ia This means the phase and ground overcurrent relays on the feeder breaker see the same amount of current

5346(833radic3) = 370 amps So the high side phase current is the radic3 less as compared to the 3Oslash fault

Digital relays ground elements set using 3I0 138kV side sequence currents and voltages

IA = 370 ang-118deg Ia=3I0=5346 ang-118degIB = 0 ang 0deg Ib = 0 ang 0degIC = 370 ang 62deg Ic = 0 ang 0degIa = 3Va(Z1 + Z2 + Z0)

Symmetrical Components - Summary of 138 kV Faults

3LG ndash positive sequence current

SLG ndash zero sequence current

L-L ndash negative sequence current

IA = 619 Ia = 5158IB = 619 Ib = 5158IC = 619 Ic = 5158

5158 833 = 619 Ia = 5158 = I1

5346(833xradic3) = 370 Ia = 5346 = 3I0

If you have a Delta-Wye transformer bank and you know the voltage ratio and secondary phase current values for 138kV 3LG (5158) and SLG (5346) faults you can find the rest

IA = 309 Ia = 0IB = 619 Ib = 4467IC = 309 Ic = 4467

5158 x radic32 = 4467 Ib x radic3 = 7737 = 3I2 4467(833x radic3) = 309309 x 2 = 619

IA = 370 Ia=5346IB = 0 Ib = 0IC = 370 Ic = 0

Yoursquove heard of a Line-to-Line fault how about an ldquoAntler-to-Antlerrdquo fault

Electromechanical Relays used on Distribution Feeders

Avistarsquos standard distribution relay package (until the mid-1990rsquos) included the following

3 phase 5051 CO-11 relays

1 reclosing relay

1 ground 5051 CO-11 relay

Objectivesminus Protect the feeder conductorminus Detect as low a fault current as possible (PU = 50 EOL fault amps) minus Other than 51P set pickup and time dial as low as possible and still have minimum Coordinating Time Interval (CTI) to next device CTI is minimum time between operation of adjacent devices minus Carry normal maximum load (phase overcurrent only)minus Pickup the feeder in a cold load condition (cong 2 times maximum normal load) or pickup frac12 of the next feeder loadmiddot

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 KV ByFor NoComment Date

FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

2 LAT421 50P CO-9 INST TD=1000CTR=6005 Pickup=46A No inst TP5=0048s

3 LAT421 51P CO-9 CO-9 TD=1900CTR=6005 Pickup=4A No inst TP5=04618s

4 LAT421 50N CO-9 INST TD=1000CTR=6005 Pickup=3A No inst TP5=0048s

5 LAT421 51N CO-9 CO-9 TD=3000CTR=6005 Pickup=3A No inst TP5=07228s

6 LAT421 FUSE SampC Link100TTotal clear

A Transf damage curve 750 MVA Category 3Base I=31378 A Z= 69 percentLatah Jct 138kV - LATAHJCT115 115kV T

B Conductor damage curve k=008620 A=1055000 cmilsConductor ACSR AWG Size 10

13 kV BUS

MIN FAULT3LG = 2000 A1LG = 1000 A

FEEDER SETTINGS51P = 2000 2 = 1000 A51N = 1000 2 = 500 A

Pickup setting criteria of 21 ratio of ldquoend of linerdquo fault duty pickup- ensures that the relay will ldquoseerdquo the fault and operate when needed

13 kV BUS

FAULT at MIDLINE3LG = 2000 A1LG = 1000 A

MIN FAULT at END OF LINE3LG = 1000 A1LG = 500 A

MIDLINE SETTINGS51P = 1000 2 = 500 A51N = 500 2 = 250 A

13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A

SECTION LOAD = 500 A

13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A NO

bull Most overhead feeders also use reclosing capability to automatically re-energize the feeder for temporary faults Most distribution reclosing relays have the capability of providing up to three or four recloses

-- Avista uses either one fast or one fast and one time delayed reclose to lockout

bull The reclosing relay also provides a reset time generally adjustable from about 10 seconds to three minutes This means if we run through the reclosing sequence and trip again within the reset time the reclosing relay will lockout and the breaker will have to be closed by manual means The time to reset from the lockout position is 3 to 6 seconds for EM reclosing relays

-- Avista uses reset times ranging from 90 to 180 seconds

bull Lockout only for faults within the protected zone That is wonrsquot lockout for faults beyond fuses line reclosers etc

bull Most distribution reclosing relays also have the capability of blocking instantaneous tripping

-- Avista normally blocks the INST tripping after the first trip to provide for a Fuse Protecting Scheme

Reclosing

13 kV BUS

RECLOSING SEQUENCE

Closed

Open05 12 LOCKOUT

RESET = 120(INST Blocked during Reset Time)

Reclosing ndash Sequence shown for a permanent fault

Distribution Fusing ndash Fuse ProtectionSaving Scheme

Fault on fused lateral on an overhead feeder- Station or midline 51 element back up fuse- Station or midline 50 element protects fuse

During fault bull Trip and clear the fault at the station (or line recloser) by the instantaneous trip before the fuse is damaged for a lateral fault bull Reclose the breaker That way if the fault were temporary the feeder is completely re-energized and back to normal bull During the reclose the reclosing relay has to block the instantaneous trip from tripping again That way if the fault still exists you force the time delay trip and the fuse will blow before you trip the feeder again thus isolating the fault and re-energizing most of the customersbull Of course if the fault were on the main trunk the breaker will trip to lockout

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 kVM15 515 51N to 140T By ProtectionFor Aspen File HORS M15 EXPolr NoComment Date 306

STATION FEEDER RELAYING

MAXIMUM FEEDER FUSE

1 M15 515 GND TIME CO-11 TD=4000CTR=160 Pickup=3A No inst TP5=10192s

3 M15 50N CO-11 INST TD=1000CTR=160 Pickup=3A No inst TP5=0048s

Shows the maximum fault current for which SampC type T fuses can still be protected by a recloserbreaker instantaneous trip for temporary faults (minimum melt curve at 01 seconds)6T ndash 120 amps8T ndash 160 amps10T ndash 225 amps12T ndash 300 amps15T ndash 390 amps20T ndash 500 amps25T ndash 640 amps30T ndash 800 amps40T ndash 1040 amps50T ndash 1300 amps65T ndash 1650 amps80T ndash 2050 amps100T ndash 2650 amps140T ndash 3500 amps200T ndash 5500 amps

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 7CURRENT (A)

SECONDS

400500

1 BR1 SampC 140T SampC Link140TMinimum melt

2 ENDTAP SampC 40T SampC Link 40TMinimum melt

3 M15 515 SampC 80T SampC Link 80TMinimum melt

Distribution Fusing ndash Fuse Protection for Temporary Faults

NOTE These values were taken from the SampC data bulletin 350-170 of March 28 1988 based on no preloading and then preloading of the source side fuse link Preloading is defined as the source side fuse carrying load amps equal to itrsquos rating prior to the fault This means there was prior heating of that fuse so it doesnrsquot take as long to blow for a given fault

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 7CURRENT (A)

SECONDS

400500

1 T FUSE SampC Link100TMinimum melt

Distribution Fusing ndash Fuse to Fuse Coordination

Typical continuous and 8 hour emergency rating of the SampC T rated silver fuse links plus the 140T and 200T

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 7CURRENT (A)

SECONDS

400500

Distribution Fusing ndash SampC T-Fuse Current Ratings

General Rule Fuse Blows at 2X Rating in 5 Minutes

Typical Coordinating Time Intervals (CTI) that Avista generally uses between protective devicesOther utilities may use different times

DEVICES CTI (Sec)

Relay ndash Fuse Total Clear 02Relay ndash Series Trip Recloser 04Relay ndash Relayed Line Recloser 03Lo Side Xfmr Relay ndash Feeder Relay 04Hi Side Xfmr Relay ndash Feeder Relay 04Xfmr Fuse Min Melt ndash Feeder Relay 04

Coordinating Time Intervals10 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 7CURRENT (A)

SECONDS

400500

MAXIMUM FEEDER FUSE

HI-SIDE CTS

MIDLINE OCR

Fault I=56659 A

DEVICES CTI (Sec)

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 7CURRENT (A)

SECONDS

400500

MAXIMUM FEEDER FUSE

HI-SIDE CTS

MIDLINE OCR

Fault I=56659 A

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 7CURRENT (A)

SECONDS

400500

TIME-CURRENT CURVES Voltage 138 kVRAT 231 MID C390R By JDHFor Huetter feeder 142 (2008) Rat feeder 231 (2009+) Midline Recloser C390R NoComment Saved in HUE142-RAT231 MID C390R 2007baseolr Date 2-25-08

HUETTER FDR 142 PROTECTION

RATHDRUM FDR 231 PROTECTION

HUE 142 MID C270R PROTECTION

MIDLINE 390R PROTECTION

CALLED HUE 142 LINE IN POWERBASERECLOSE IS 1SEC 12SEC LO 120 SEC RESET

RECLOSE IS 1SEC 12 SEC LO 180 SEC RESET

LARGEST DOWNSTREAM FUSE FROM 390R MIDLINE

1 C390R MID 51P CO-9 CO-9 TD=2500CTR=3005 Pickup=5A No inst TP5=06037s

2 C390R MID 50P CO-9 INST TD=1000CTR=3005 Pickup=6A No inst TP5=0048s

3 3 C390R MID 50N CO-9 INST TD=1000CTR=3005 Pickup=5A No inst TP5=0048s

4 C390R MID 51N CO-9 CO-9 TD=2500CTR=3005 Pickup=5A No inst TP5=06037s

5 HUE 142 50P 251 INST TD=1000CTR=160 Pickup=7A No inst TP5=0048s

6 HUE 142 51P 251 SEL-EI TD=1500CTR=160 Pickup=6A No inst TP5=04072s

7 HUE ML C270R 50P INST TD=1000CTR=4005 Pickup=81A No inst TP5=0048s

8 HUE ML C270R 51P CO-11 TD=1500CTR=4005 Pickup=7A No inst TP5=03766s

9 HUE 142 50N 251 INST TD=1000CTR=160 Pickup=3A No inst TP5=0048s

10 HUE 142 51N 251 SEL-EI TD=4000CTR=160 Pickup=3A No inst TP5=10858s

11 HUE 142 51Q 251 SEL-EI TD=4000CTR=160 Pickup=52A No inst TP5=10858s

12 HUE ML C270R 50N INST TD=1000CTR=4005 Pickup=3A No inst TP5=0048s

13 HUE ML C270R 51N CO-11 TD=7000CTR=4005 Pickup=3A No inst TP5=18052s

14 RAT 231 50N CO-11 INST TD=1000CTR=8005 Pickup=3A No inst TP5=0048s

15 RAT 231 51N CO-11 CO-11 TD=5000CTR=8005 Pickup=3A No inst TP5=13192s

16 RAT 231 50P CO-11 INST TD=1000CTR=8005 Pickup=7A No inst TP5=0048s

17 HUE 142 FUSE SampC Link 65TTotal clear

18 RAT 231 FUSE SampC Link100TTotal clear

19 RAT 231 51P CO-11 CO-11 TD=1800CTR=8005 Pickup=6A No inst TP5=04533s

A Conductor damage curve k=008620 A=1331000 cmilsConductor ACSR AWG Size 20MIDLINE C390 FED FROM RAT 231 - MINIMUM CONDUCTOR TO PROTECT

B Conductor damage curve k=006710 A=5560000 cmilsConductor AACMIDLINE C390 FED FROM HUE 142 - MINIMUM CONDUCTOR TO PROTECT

DEVICES CTI (Sec)

Coordinating Time Intervals

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 7CURRENT (A)

SECONDS

400500

MAXIMUM FEEDER FUSE

HI-SIDE CTS

LO-SIDE CTS

DEVICES CTI (Sec)

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 7CURRENT (A)

SECONDS

400500

MAXIMUM FEEDER FUSE

DEVICES CTI (Sec)

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 7CURRENT (A)

SECONDS

30405070

300400500700

345710

30405070100

3004005007001000

TIME-CURRENT CURVES Voltage 138 kV By DLHFor M15 A-172 CS with fdr 515 with Kearney 140T Fuse NoComment Three Phase Fault Date 121205

1 Moscow 515 Kear 140T Kearney 140TMinimum meltI= 51581A T= 010s

2 M15-515 Phase Time CO-11 TD=1500CTR=160 Pickup=6A No inst TP5=03766sI= 51581A (322 sec A) T= 033s

3 M15 A-172 Phase CO- 9 TD=1500CTR=120 Pickup=2A Inst=1200A TP5=03605sI= 6190A (52 sec A) T= 103s H=833

FAULT DESCRIPTIONClose-In Fault on 0 MoscowCity2 138kV - 0 BUS1 TAP 138kV 1L 3LG

3LG Fault Coordination Example

Top ndash Ckt Swr w Phase E-M relay

----- 04 sec ------

Middle - E-M Phase relay for a 500 Amp Feeder

----- 02 sec ------

Bottom ndash 140T Feeder Fuse (Total Clear)

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 7CURRENT (A)

SECONDS

30405070

300400500700

345710

30405070100

3004005007001000

TIME-CURRENT CURVES Voltage 138 kV By DLHFor Moscow CS with fdr 515 with 140T fuses NoComment Single Line to Ground Fault Date 121205

2 M15 515 GND TIME CO-11 TD=4000CTR=160 Pickup=3A No inst TP5=10192sI= 53465A (334 sec A) T= 029s

4 4 M15 A-172 GND TIME CO-11 TD=4000CTR=240 Pickup=4A No inst TP5=10192sI= 53465A (223 sec A) T= 083s

FAULT DESCRIPTIONClose-In Fault on 0 MoscowCity2 138kV - 0 BUS1 TAP 138kV 1L 1LG Type=A

SLG Fault Coordination Example

Top ndash Ckt Swr w E-M Phase relay

----- 04 sec ------

Middle - E-M relays (Phase amp Ground) for a 500 Amp Feeder

----- 02 sec ------

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 7CURRENT (A)

SECONDS

30405070

300400500700

345710

30405070100

3004005007001000

TIME-CURRENT CURVES Voltage 138 kV By DLHFor Moscow CS A-172 with fdr 515 with 140T Fuse NoComment Line to Line Fault Date

FAULT DESCRIPTIONClose-In Fault on 0 MoscowCity2 138kV - 0 BUS1 TAP 138kV 1L LL Type=B-C

L-L Fault Coordination Example

Top ndash Ckt Swr w E-M relay

----- 04 sec ------

Middle - E-M (Phase) relay for a 500 Amp Feeder

----- 02 sec ------

The Final Product 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 7CURRENT (A)

SECONDS

TIME-CURRENT CURVES Voltage 138 kVINT 12F2 amp MID 466R By JDHFor Indian Trail feeder 12F2 and 12F2 Midline 466R in INT 2007baseolr NoComment AT SUB 3LG= 5719A LL=4951A SLG= 5880A Date 1-29-08

EAST BRANCH - FEEDER MAXIMUM FUSE

FAULT DUTY AT 140T - WHEN FED FROM NORTH (NO PT)

12F2 MIDLINE 466R PROTECTION

WEST OR NORTH BRANCH - FEEDER MAXIMUM FUSE

FAULT DUTY AT 150E 3LG=2318 LL=2008 SLG=1684

3LG=1932 LL=1673 SLG=1388

FAULT DUTY AT 140T - WHEN FED FROM WEST MID 466R 3LG=2093 LL=1813 SLG=1520

9amp15

11amp4

1 INT A-742 51P 351 SEL-VI TD=1200CTR=6005 Pickup=2A No inst TP5=03096s H=833

2 INT A-742 51G 351 SEL-VI TD=1000CTR=6005 Pickup=1A No inst TP5=0258s H=833

3 INT A-742 51N 351 SEL-EI TD=5600CTR=12005 Pickup=3A No inst TP5=15201s

4 INT A-742 51Q 351 SEL-EI TD=4200CTR=6005 Pickup=13A No inst TP5=11401s H=833

5 INT F2 150E FUSE SMU-20_150ETotal clear

6 INT 12F2 50P 351S INST TD=1000CTR=8005 Pickup=7A No inst TP5=0048s

7 INT 12F2 51P 351S SEL-EI TD=1500CTR=8005 Pickup=6A No inst TP5=04072s

8 INT 12F2 50G 351S INST TD=1000CTR=8005 Pickup=3A No inst TP5=0048s

9 INT 12F2 51G 351S SEL-EI TD=3900CTR=8005 Pickup=3A No inst TP5=10587s

10 INT 12F2 51Q 351S SEL-EI TD=3500CTR=8005 Pickup=52A No inst TP5=09501s

11 INT A-742 51Q 587 W2 SEL-EI TD=4300CTR=12005 Pickup=54A No inst TP5=11672s

12 F2 MID 466R 50P 351R INST TD=1000CTR=5001 Pickup=168A No inst TP5=0048s

13 F2 MID 466R 51P 351R SEL-EI TD=2300CTR=5001 Pickup=144A No inst TP5=06243s

14 F2 MID 466R 50G 351R INST TD=1000CTR=5001 Pickup=089A No inst TP5=0048s

15 F2 MID 466R 51G 351R SEL-EI TD=4200CTR=5001 Pickup=089A No inst TP5=11401s

16 F2 MID 466R 51Q 351R SEL-EI TD=3800CTR=5001 Pickup=154A No inst TP5=10315s

17 INT F2 140T FUSE stn SampC Link140TTotal clear

B Conductor damage curve k=014040 A=526300 cmilsConductor Copper (bare) AWG Size 2NORTH BRANCH SMALLEST TRUNK CONDUCTOR

C Conductor damage curve k=004704 A=3500000 cmilsCable XLPE 90C250CEAST BRANCH SMALLEST TRUNK CONDUCTOR - 350 CN15

D Conductor damage curve k=008620 A=1055000 cmilsConductor ACSR AWG Size 20WEST BRANCH SMALLEST TRUNK CONDUCTOR

An example of a completed 138kV Feeder Coordination Study with 20 Time-Current Curves representing

Instantaneous amp Time-Delay Curves for the - Transformer high side protection- Transformer low side protection- Station feeder breaker protection- Midline feeder breaker protection

Two Fuses

Transformer Damage Curve

Three Conductor Damage Curves

Criteria (for outdoor bus arrangement not switchgear)

Protect the Transformer from thermal damage- Refer to Damage Curves

Backup feeder protection (as much as possible)- Sensitivity is limited because load is higher

Coordinate with downstream devices (feeder relays)

Carry normal maximum load (phase only)

Pick up Cold Load after outages

Distribution Transformer Electromechanical Relays

Relays 3 High Side Phase Overcurrent (with time and instantaneous elements) 1 Low Side Ground Overcurrent (with time and instantaneous elements) Sudden Pressure Relay

Distribution Transformer Electromechanical Relays - Setting Criteria

Distribution Transformer amp Feeder Protectionwith Microprocessor Relays

Advantages

More precise TAP settings

More Relay Elements

Programmable Logic Buttons

Lower burden to CT

Event Reports

Communications

Coordinate with like elements (faster)

More Settings

Microprocessor Relay

Elements We Set 51P 50P 50P2 (FTB) 51G 50G (115kV) 51N (138kV) 50N1 (FTB)

FTB = Fast Trip Block (feeder relays must be Microprocessor)

Fast Trip Blocking

Distribution Transformer Relay Setting Criteria

Electromechanical Relays 51P Pickup (time overcurrent)

Donrsquot trip for load or cold load pickup (use 24 highest MVA rating)

Ends up being higher than the feeder phase element pickup

Example 121620 MVA unit would use 24205 = 240 amps (1940A low side)

51P Time Lever (time dial)

Coordinate with feeder relays for maximum fault (close in feeder fault)

CTI is 04 seconds

Worst coordination case Oslash-Oslash on low side (discrepancy due to deltawye conn)

Multiple feeder load can make the transformer relay operate faster

50P Pickup (instantaneous overcurrent)

Must not trip for feeder faults set at 170 of low side bus fault

Accounts for DC offset

Microprocessor Relays 51P - Phase Time Overcurrent Pickup

Set to 240 of nameplate (same as EM relay)

51P - Time Dial

Set to coordinate with feederrsquos fastest element for each fault

CTI is still 04 seconds

50P1 ndash Phase instantaneous 1 Pickup

Direct Trip

Set to 130 of max 138 kV fault (vs 170 with EM relay)

50P2 ndash Phase instantaneous pickup for Fast Trip Block scheme

Set above transformer inrush

4 Cycle time delay

Blocked if any feeder overcurrent elements are picked up

Phase Overcurrent Settings ndash Current measured on 115kV side

Electromechanical Relays 51N ndash Ground Time Overcurrent Pickup

Set to same sensitivity as feeder phase relay (in case the feeder ground relay is failed)

This setting is higher than the feeder ground so we lose some sensitivity for backup

51N - Time Lever (time dial)

Coordinate with feeder relays for ground faults

CTI is 04 seconds

50N - Pickup (instantaneous overcurrent)

DO NOT USE

Microprocessor Relays 51N - Ground Time Overcurrent Pickup

Set slightly higher than feeder ground pickup (about 13 times)

51N - Time Dial

Set to coordinate with feeder ground

CTI is 04 seconds

50N1 ndash Ground Instantaneous for Fast Trip Block

Set slightly above the feeder ground instantaneous pickup

Time Delay by 4 cycles

Blocked if Feeder overcurrent elements are picked up

Transformer Ground Overcurrent Settings - Current calculated from 115kV CTs

51G - Ground Time Overcurrent Pickup

Set very low (will not see low side ground faults due to transformer connection) Usually 120 Amps

51GTime Dial

Set very low

50G1 ndash Ground Instantaneous 1 Pickup

Set very low Usually 120 Amps

SymmetricalComponents

Inrush ndash The current seen when energizing a transformer

Need to account for inrush when using instantaneous elements (regular or FTB)

Inrush can be approximately 8 times the nameplate of a transformer

Note that the microprocessor relay only responds to the 60 HZ fundamental and that this fundamental portion of inrush current is cong 60 of the total So to calculate a setting we could use the 8 times rule of thumb along with the 60 value For a 121620 MVA transformer the expected inrush would be 812506 = 288 amps We set a little above this number (360 Amps)

Transformer inrush UNFILTEREDcurrent The peak current is about 1200 amps

Transformer inrush FILTERED current (filtered by digital filters to show basically only 60 HZ)

Microprocessor Feeder Relay Overcurrent Reclosing Fast Trip Block Output Breaker Failure Output

Elements We Set 51P 50P 51G 50G 51Q

Feeder Phase Overcurrent Settings

51P - Phase Time Overcurrent Pickup

Set above load and cold load (960 Amps for 500 Amp feeder)

Same as EM pickup

51P - Time Dial

Same as EM relay (coordinate with downstream protection with CTI)

Select a curve

50P ndash Phase instantaneous Pickup

Set the same as the 51P pickup (960 Amps)

Feeder Ground Overcurrent Settings

51G - Phase Time Overcurrent Pickup

Same as EM pickup which is 480 Amps

51G - Time Dial

Same as EM relay (coordinate with downstream protection)

Select a curve

50G ndash Phase instantaneous Pickup

Set the same as the 51G pickup

51G Time Dial

Select a curve

EXTERNAL FAULTTHE SECONDARY CURRENTS FLOW THROUGH BOTHRESTRAINT COILS IN THE SAME DIRECTION AND THENCIRCULATE BACK THROUGH THE CTS THEY DO NOTFLOW THROUGH THE OPERATE COIL

CURRENT FLOW THROUGH AN EM 87 DIFFERENTIAL RELAY FOR AN INTERNAL AND EXTERNAL FAULT

CT POLARITY MARK

Transformer Differential Protection ndash External Fault

INTERNAL FAULTTHE SECONDARY CURRENTS FLOW THROUGH BOTHRESTRAINT COILS IN OPPOSITE DIRECTIONS ADD AND THEN FLOW THROUGH THE OPERATE COIL ANDBACK TO THE RESPECTIVE CTS

CT POLARITY MARK

Transformer Differential Protection ndash Internal Fault

bull The differential relay is connected to both the high and low side transformer BCTrsquos

bull EM Differential RelaySince the distribution transformer is connected delta ndash wye the transformer CTrsquos have to be set wye ndash delta to compensate for the phase shift

Transformer Differential Protection

BACK-TO-BACK MIDLINE RECLOSERS COORDINATION

Feeder Breaker

Switch

MidlineRecloser

Feeder Breaker

Feeder BreakerSwitch Midline

Recloser NO TieMidlineRecloser

Feeder Breaker

Recloser Closed TieMidlineRecloser

Feeder Breaker

TRTR TRT-LO

T = TripR = RecloseLO = Lockout

RECLOSERSWITCH FUNCTIONALITY

Substation

FEEDER 1

FEEDER 2

FWDFWD

Substation

FEEDER 1

FEEDER 2

CLOSED

Now Z02R is aswitch 75

GRAB YOUR HANDOUTMOSCOW FEEDER 515 PROTECTION EXAMPLE

The Scenario

A hydraulic midline recloser on a 138kV feeder in Moscow ID is being replaced by a newer relayed recloser The recloser is P584 on feeder 515

The field engineer would like to replace the recloser in the existing location

Protection engineer must review the feeder protection and report back to the field engineer

NOTE Avista designs for a ldquofuse savingrdquo scheme with one instantaneous trip The field engineers decide if they want to enable or disable the instantaneous tripping

MOSCOW FEEDER 515 PROTECTION EXAMPLE

bull Where do we start the protection design1) Gather Information

bull Feeder Rating (expected load)bull Protective devices (Breakers line reclosers fuses)bull Fault Duty (at each protective device)bull Conductors to be protectedbull Project deadline (tomorrow)

How do we proceed

bull At each coordination pointndash Loadingndash Coordination

bull What will coordinate with the downstream devicebull Are we above the fuse rating

ndash Conductor Protectionbull minimum conductor that can be protected by the feeder settings or

ndash Fault Detectionbull Can we detect the fault by our 21 margin

ndash Fuse Saving for Temporary faultsbull What can a fuse be protected up to

Begin at the END

bull POINT 8 This is a customers load and we are using 3-250 KVA transformers to serve the load The full load of this size of bank is 314 amps The Avista transformer fusing standard says to use a 65T on this transformer so thatrsquos what wersquoll choose

FUSE - POINT 6BWhat we know

ndash 3Oslash lateral feeding to a 65T at the endndash 4 ACSR conductor

Fuse Selectionndash Loading ndash Assume the load is all downstream of point 8 so a 65T or higher will

still suffice ndash Conductor Protection - From Table 7 we see that 4 ACSR can be protected by a

100T or smaller fusendash Fault Detection - Under the Relay Setting Criteria we want to detect the minimum

line end fault with a 21 margin The SLG is 463 so we calculate the max fuse as follows

bull 4632 (21 margin) = 231 amps ldquoTrdquo fuses blow at twice their rating so divide by 2 again with a result of 115 Fuse must be 100T or less

ndash Coordination ndash From Table 2 we see that we need an 80T or larger to coordinate with the downstream 65T fuse (at fault duties lt 1400A wpreload)

ndash Fuse Saving - From Table 1 we see that an 80T fuse can be protected up to 2050 amps for temporary faults and the 3Oslash fault is 1907 amps so we could choose an 80T or higher from that standpoint

Point 6B - So what fuse size should we use

POINT 5 ndash Midline Recloserbull What we know

ndash Conductor to protect 4 ACSR (can carry 140A load)ndash Max fuse to coordinate with = 100T (FUTURE per distribution engineer)ndash Maximum load (per engineer) = 84A

bull Settings Considerationsndash Loading Cold load would be cong 842 = 168 A but the conductor can carry 140A max Size for some

load growth by setting at cong 2conductor rating 300 Amp phase pickup setting NOTE we donrsquot design settings to protect for overload

ndash Conductor Protection From Table 7 Avistarsquos 300A feeder settings can protect down to 2 ACSR so 300A pickup is still acceptable Review curves to confirm

ndash Fault Detection detect the minimum fault with a 21 margin Here the min fault is at point 61 The Oslash-Oslash fault at point 6 is 08661907 = 1651 so our margin to detect that fault would be 1651300

= 551 so no problem with the 300 amps Oslash PU from that standpoint (used Oslash-Oslash because itrsquos the minimum multi phase fault)

2 The SLG at point 6 is 1492 so we could set the ground up to 14922 = 745 amps and still detect the fault However our criteria says to set as low as possible and still coordinate with the largest downstream device and from above wersquore trying to use a 100T Based on curves this is 300 amps (same as phase which is unusual)

ndash Coordination Based on fault study set the time dial to coordinate with a 100T fuse with 02 second CTI

ndash Fuse Saving The fault duty at the recloser is 3453 3PH and 2762 1LG An instantaneous trip will protect the 80T fuse if the fault duty is less than 2050 Amps Some fuse protection is compromised but there is nothing we can do about it

TRUNK FUSE - POINT 3CWhat we know

ndash 4 ACSR 3Oslash trunk ndash Load is 84 Ampsndash Downstream protection is the midline recloser

Fuse Selectionndash Loading ndash Assume the load is the same as at the midline recloser We set that at

300A and a 200T can carry 295 A continuous (Table 3)ndash Conductor Protection - From Table 7 we see that 4 ACSR can be protected by a

100T or smaller fuse ndash Fault Detection - Under the Relay Setting Criteria we want to detect the minimum

line end fault with a 21 margin bull SLG at midline is 2762 so even a 200T would provide enough sensitivity

ndash Coordination ndash PROBLEM We have coordinated the midline with a 100T so the upstream fuse must coordinate with that

ndash Fuse Saving As before some fuse saving is compromised due to higher fault duty

TRUNK FUSE Point 3C - Solutions

bull Move Recloserbull Re-conductor

FEEDER BREAKER - POINT 1What we know

ndash This is a 500 amp feeder design The load is 325 amps and cold load 650 per field engineerndash Downstream protection is either the midline recloser or a 140T fuse

Settings Considerations Loading A standard 500 amp feeder phase pickup setting of 960 A will carry all normal

load and pick up cold load The ground pickup does not consider load and will be set at 480A

Conductor Protection The main trunk is 556 ACSR and from Table 7 a 500 amp feeder setting of 960 A can protect 10 ACSR or higher Laterals with smaller conductor must be fused

Fault Detection The Oslash-Oslash fault at point 5 is 08663453 = 2990 so our margin to detect that fault would

be 2990960 = 311 The SLG at point 5 is 2762 so our margin to detect that fault is 2762480 = 571

Coordination Use fault study to calculate settings to achieve 02 second CTI to fuses and 03 second CTI to the recloser

Fuse Saving ndash We have decided that a 140T fuse is the maximum fuse we will use on the feeder even though it canrsquot be saved by an instantaneous trip at the maximum fault duties

Questions

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Relay Overcurrent Curves

Relay Overcurrent Curves - Pickups amp Time Dials

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Symmetrical Components ndash Positive Sequence 3LG 138 kV Fault You have only positive sequence voltage and current since the system is balanced

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3LG Fault Coordination ExampleTop ndash Ckt Swr w Phase E-M relay----- 04 sec ------Middle - E-M Phase relay for a 500 Amp Feeder ----- 02 sec ------ Bottom ndash 140T Feeder Fuse (Total Clear)

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L-L Fault Coordination ExampleTop ndash Ckt Swr w E-M relay ----- 04 sec ------ Middle - E-M (Phase) relay for a 500 Amp Feeder ----- 02 sec ------ Bottom ndash 140T Feeder Fuse (Total Clear)

The Final Product

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The Scenario

How do we proceed

Begin at the END

FUSE - POINT 6B

POINT 5 ndash Midline Recloser

TRUNK FUSE - POINT 3C

FEEDER BREAKER - POINT 1