DISTRIBUTION PROTECTION OVERVIEW - WSU … PROTECTION OVERVIEW ... Transformer Relay Settings using Electromechanical Relays ……… ... Transformer Differential Protection

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DISTRIBUTION PROTECTION OVERVIEW

Kevin Damron & Mike DiedeschAvista Utilities

Presented March 17th, 2014

At the 31st AnnualHands-On Relay School

Washington State UniversityPullman, Washington

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Table of Contents

System Overview ………………………………………………………....… 3

Distribution (3-30MVA ) Transformer Protection ………………….…..…... 14

Relay Overcurrent Curves …………………………………………..……. 22

Symmetrical Components ……………………………………………..…….. 8

Distribution Fuse Protection ………………………………………….……. 24

Conductors ………………..………..………………………………………. 28

Electromechanical Relays used on Distribution Feeders…………….…….… 31

Coordinating Time Intervals …………………………………………….… 36

Transformer Relay Settings using Electromechanical Relays ………….…… 48

IEEE Device Designations …………………………………………….…….. 51

Transformer & Feeder Protection using Microprocessor Relays ………….… 52

Transformer Differential Protection ……………………………………….… 61

Moscow 13.8kV Feeder Coordination Example ……………………………. 64

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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

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR

1 PHASE#4 ACSR #4 ACSR

#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

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System Overview – Inside the Substation Fence

Transformer relays

Feeder relay

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XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

System Overview – Outside the Substation Fence

Midline recloser relay

XFMR 12/16/20 MVA115/13.8kV

115 kV SYSTEM

13.8 kV BUS500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3Ø = 5158SLG = 5346

3Ø = 3453SLG = 2762

PT-2BPT-2A

3Ø = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3Ø = 1907SLG = 1492

3Ø = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172MOSCOW

1 PHASE#4 ACSR

3Ø = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3Ø = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 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)

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TIME-CURRENT CURVES @ Voltage 13.8 kVHORS 2010 By JDH

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

STATION VCB 252 PROTECTION

MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR

MAXIMUM MIDLINE FUSE

Fault I=5665.9 A

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1. IDR A-777 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096sIa= 679.8A (5.7 sec A) T= 0.78s H=8.33

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2. IDR 252 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072sIa= 5665.9A (35.4 sec A) T= 0.30s

3. IDR 252 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s3Io= 0.0A (0.0 sec A) T=9999s

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4. 252 140T FUSE stn S&C Link140TTotal clear.Ia= 5665.9A T= 0.09s

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5f

5. Phase unit of recloser MID LINE OCR Fast: ME-341-B Mult=0.2 Slow: ME-305-A Add=1000. Ia= 5665.9A T(Fast)= 0.03s

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6. T FUSE S&C Link 50TMinimum melt.Ia= 5665.9A T= 0.01s

A

A. Conductor damage curve. k=0.06710 A=556000.0 cmilsConductor AACFEEDER 252 SMALLEST CONDUCTOR TO PROTECT

B

B. Transf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.2 percent.IDAHO RD 12/16/20 MVA XFMR

FAULT DESCRIPTION:Bus Fault on: 0 IDR 252 13.8 kV 3LG

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System Overview –Each 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

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MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1


2



4


5

5f


6


A


B



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System Overview –

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.

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Transformer Protection - Damage Curve – ANSI/IEEE C57.109-1985

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)

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T IME-CURRENT CURVES @ Voltage By

For No.

Comment Date

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1. M15CTR=

A

A. T ransf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.2 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.

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Copper Conductor Damage Curves ACSR Conductor Damage Curves (2/0 damage at 1500A @ 100sec.) (2/0 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)

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TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor ACSR Conductor Damage Curves No.Comment Date 12/13/05

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1. M15-515 Phase INST INST TD=1.000CTR=160 Pickup=7.A No inst. TP@5=0.048s

A

A. Conductor damage curve. k=0.08620 A=355107.0 cmilsConductor ACSR336.4 ACSR

B

B. Conductor damage curve. k=0.08620 A=167800.0 cmilsConductor ACSR AWG Size 4/04/0 ACSR

C

C. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 2/02/0 ACSR

D

D. Conductor damage curve. k=0.08620 A=83690.0 cmilsConductor ACSR AWG Size 1/01/0 ACSR

E

E. Conductor damage curve. k=0.08620 A=52630.0 cmilsConductor ACSR AWG Size 2#2 ACSR

FF. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWG Size 4#4 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

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TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Copper Conductor Damage Curves No.Comment Date 12/13/05

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1. M15-515 Phase INST INST TD=1.000CTR=160 Pickup=7.A No inst. TP@5=0.048s

A

A. Conductor damage curve. k=0.14040 A=105500.0 cmilsConductor Copper (bare) AWG Size 2/02/0 Copper

B

B. Conductor damage curve. k=0.14040 A=83690.0 cmilsConductor Copper (bare) AWG Size 1/01/0 Copper

C

C. Conductor damage curve. k=0.14040 A=52630.0 cmilsConductor Copper (bare) AWG Size 2#2 Copper

D

D. Conductor damage curve. k=0.14040 A=33100.0 cmilsConductor Copper (bare) AWG Size 4#4 Copper

EE. Conductor damage curve. k=0.14040 A=20820.0 cmilsConductor Copper (bare) AWG Size 6#6 Copper

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Conductor Rating556 730

336.4 5304/0 3402/0 2701/0 230#2 180#4 140

ACSR Ampacity Ratings

Conductor Rating

2/0 3601/0 310#2 230#4 170#6 120

CopperAmpacity Ratings

Conductor Ampacities

Conductor at 25°C ambient taken from the Westinghouse Transmission & Distribution book.

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Comparing a 140T fuse versus a#4 ACSR Damage curve.

The 140T won’t protect the conductor below about 550 amps where the curves cross.

Conductor Protection Graph

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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)

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TIME-CURRENT CURVES @ Voltage 13.8 kV#4 ACSR & 140T By Protection

For Aspen File: HORS M15 EXP.olr No.

Comment Date 3/06

1

1. Moscow 515 Kear 140T Kearney 140TTotal clear.

A

A. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWG Size 4

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Transformer Protection using 115 kV Fuses

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Transformer Protection using 115 kV Fuses

Used at smaller substations up to 7.5 MVA transformer due to low cost of protection.Other advantages are:- Low maintenance- Panel house & station battery not required

There are also several disadvantages to using fuses however which are: Low interrupting rating from 1,200A (for some older models) up to 10,000A at 115 kV. By contrast a circuit switcher can have a rating of 25KAIC and our breakers have normally 40KAIC. 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)

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TIME-CURRENT CURVES @ Voltage 13.8 KVROK 451R2 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

STATION TRANSFORMER PROTECTION

FEEDER VCR PROTECTION

MAXIMUM FEEDER FUSE

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1. LAT RP4211 50N CO-11 INST TD=1.000CTR=500/5 Pickup=2.A No inst. TP@5=0.048s

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2. SMD-2B 65E VERY SLOW 176-19-065Minimum melt. H=8.33

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3. LAT RP4211 50P CO-11 INST TD=1.000CTR=500/5 Pickup=3.5A No inst. TP@5=0.048s

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4. LAT RP4211 51P CO-11 CO-11 TD=2.000CTR=500/5 Pickup=3.A No inst. TP@5=0.5043s

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5. LAT RP4211 51N CO-11 CO-11 TD=4.000CTR=500/5 Pickup=2.A No inst. TP@5=1.0192s

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6. LAT RP4211 FUSE S&C Link 65TTotal clear.

A

A. Transf. damage curve. 7.50 MVA. Category 3Base I=313.78 A. Z= 7.3 percent.Rockford 13.8kV - ROCKFORD115 115.kV 1 T

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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.

No indication of faulted zone (transformer, bus or feeder).

Transformer Protection using 115 kV Fuses - continued

A

B

C

a

b

c

1.0 PU

0.5 PU0.5 PU

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Transformer Protection using a Circuit SwitcherShowing Avista’s present standard using Microprocessor relays.

13.8kV 115kV

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Transformer Protection using a Circuit SwitcherShowing Avista’s old standard using Electromechanical relays.

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Transformer Protection using a Circuit Switcher

Some advantages to this over fuses are: Higher interrupting. Relays can be set to operate faster

and with better sensitivity than fuses.

Three phase operation. Provide better coordination with

downstream devices.

Some disadvantages would be: Higher cost. Higher maintenance. Requires a substation battery, panel

house and relaying. Transformer requires CT’s.

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)

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TIME-CURRENT CURVES @ Voltage 13.8 kVIDR FEEDER 251 By JDH





MAXIMUM FEEDER FUSE

HI-SIDE CT'S

LO-SIDE CT'S

MINIMUM FAULT TO DETECT:3LG=2460A, SLG=1833A, L-L=2130A

1

1. IDR A-777 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s H=8.33

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2. IDR A-777 51G 351 SEL-VI TD=1.000CTR=600/5 Pickup=1.A No inst. TP@5=0.258s H=8.33

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3. IDR A-777 51N 351 SEL-EI TD=5.700CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s

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4. IDR A-777 51Q 351 SEL-EI TD=4.200CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s H=8.33

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5. IDR A-777 51Q 587 W2 SEL-EI TD=4.300CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s

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6. IDR 251 50P 351S INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

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7. IDR 251 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s

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8. IDR 251 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

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9. IDR 251 51G 351S SEL-EI TD=3.900CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s

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10. IDR 251 51Q 351S SEL-EI TD=3.500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

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11. 251 140T FUSE stn S&C Link140TTotal clear.

A

A. Conductor damage curve. k=0.08620 A=133100.0 cmilsConductor ACSR AWG Size 2/0FEEDER 251 SMALLEST CONDUCTOR TO PROTECT

B


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Transformer Protection using a Breaker

This is very similar to using a circuit switcher with a couple of advantages such as: Higher interrupting – 40kAIC for the one shown below. Somewhat faster tripping than a circuit switcher (3 cycles vs. 6 – 8 cycles). Possibly less maintenance than a circuit switcher. The CT’s 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.

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SEL Various Relay Overcurrent Curves. Extremely Inverse – steepestVery InverseInverseModerately InverseShort Time Inverse – 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)

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TIME-CURRENT CURVES @ Voltage By

For No.

Comment Date

1

1. EXTREMELY INVERSE SEL-EI TD=15.000CTR=800/5 Pickup=6.A No inst. TP@5=4.0718s

2

2. INVERSE SEL-I TD=2.000CTR=800/5 Pickup=6.A No inst. TP@5=0.8558s

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3. MODERATELY INVERSE SEL-2xx-MI TD=1.000CTR=800/5 Pickup=6.A No inst. TP@5=0.324s

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4. SHORT TIME INVERSE SEL-STI TD=1.000CTR=800/5 Pickup=6.A No inst. TP@5=0.1072s

5 5. VERY INVERSE SEL-VI TD=6.000CTR=800/5 Pickup=6.A No inst. TP@5=1.5478s

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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 “lever” from the electromechanical relay days- Instantaneous elements have a “time dial” of 1 and operate at 0.05 seconds.- Instantaneous curves are shown as a flat horizontal line starting at the left at the pickup value and plotted at 0.05 seconds.

Relay Overcurrent Curves - Pickups & Time Dials Pickup - the current at which the relay will operate to trip the breaker.- also known as “tap” from the electromechanical relay days-expressed in terms of the ratio of the current transformer (CTR) that the relay is connected to, - e.g., 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)

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.01

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1






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

LO-SIDE CT'S


1


2


3


4


5


6


7


8


9


10


11


A


B


21

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

Relay Overcurrent Curves - Example of Different Pickup & 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

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For No.

Comment Date

1

1. EXT INV 1 SEL-EI TD=15.000CTR=800/5 Pickup=6.A No inst. TP@5=4.0718s

2 2. ~EXT INV 1 SEL-EI TD=12.000CTR=800/5 Pickup=6.A No inst. TP@5=3.2574s

3

3. ~EXT INV 2 SEL-EI TD=9.000CTR=800/5 Pickup=6.A No inst. TP@5=2.4431s

4


5


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

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For No.Comment Date

1

1. EXT INV 1 SEL-EI TD=9.000CTR=800/5 Pickup=6.A No inst. TP@5=2.4431s

2


3


4


5


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

22

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

2

3

45

7

10

20

30

4050

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100

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400500

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1

TIME-CURRENT CURVES @ Voltage 13.8 kVINT 12F1 By JDH

For Indian Trail feeder 12F1 in INT 2007base.olr No.

Comment 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=10.000CTR=600/5 Pickup=0.8A No inst. TP@5=2.5797s H=8.33

2

2. INT 12F1 51P 351S SEL-VI TD=1.800CTR=800/5 Pickup=3.A No inst. TP@5=0.4643s

3

3. INT 12F1 51G 351S SEL-VI TD=0.900CTR=800/5 Pickup=3.A No inst. TP@5=0.2322s

4

4. INT 12F1 51Q 351S SEL-VI TD=0.500CTR=800/5 Pickup=3.A No inst. TP@5=0.129s

5 5. INT A-742 51G 351 SEL-VI TD=5.000CTR=600/5 Pickup=0.8A No inst. TP@5=1.2898s H=8.33

6

6. INT A-742 51Q 351 SEL-VI TD=3.000CTR=600/5 Pickup=0.8A No inst. TP@5=0.7739s H=8.33

A

A. Transf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.4 percent.INDIAN TRAIL 12/16/20 MVA XFMR

B

B. Conductor damage curve. k=0.06710 A=556000.0 cmilsConductor AAC556 kCMil AAC

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

51P – phase time overcurrent

51N or 51G – ground time overcurrent

51Q – negative sequence time overcurrent

Transformer relay curves

Feeder relay curves51P – phase time overcurrent

51N or 51G – ground time overcurrent

51Q – negative sequence time overcurrent

This brings us to a brief discussion of:

23

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.

24

Symmetrical Components

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

Symmetrical Components Notation:Positive Sequence current = I + = I1Negative Sequence current = I - = I2Zero Sequence current = I 0 = I0

Phase Current notation:IA – High side AmpsIa – Low side Amps

8.33= 115/13.8 = transformer Voltage (turns) ratioPhasor diagram from Fault-study Software.

Examples of three 13.8kV faults showing the current distribution through a Delta-Wye high-lead-low transformer bank :

25

Symmetrical Components – Positive Sequence, 3LG 13.8 kV FaultYou have only positive sequence voltage and current since the system is balanced.

A

B

C

a

b

c

RR

IA = 619 -88 Ia = 5158 -118IB = 619 152 Ib = 5158 122IC = 619 32 Ic = 5158 2Phase current = Sequence currentThat is; Ia = I+.

Phase currents and voltages for the 115kV side.

IA = Ia / 8.33 = 5158A / 8.33IA = 619A

26

Symmetrical Components – Negative Sequence, L-L 13.8 kV Fault

115kV side sequence currents and voltages

13.8kV side sequence currents and voltages

A

B

C

a

b

c

RR

IA = 309 -28 Ia = 0 0IB = 619 152 Ib = 4467 152IC = 309 -28 Ic = 4467 -28

3LG 13.8kV fault = 5158A,Ib=Ic=4467 A, 4467/5158 = 86.6%=3/2IB3LG = IBLL = 619AIA & IC = IB or Ia & Ic = Ib I2 = the phase current/3 = 4467/3 = 2579

Digital relays 50Q/51Q elements set using 3I2.3I2 = Ib x 3 = 4467 x 1.732 = 7737,

27

Symmetrical Components – Zero Sequence, 1LG 13.8 kV Fault

115kV side sequence currents and voltages

A

B

C

a

b

c

RR

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

5346/(8.33*3) = 370 amps. So the high side phase current is the 3 less as compared to the 3Ø fault.

Digital relays ground elements set using 3I0. 13.8kV side sequence currents and voltages

IA = 370 -118 Ia=3I0=5346 -118IB = 0 0 Ib = 0 0IC = 370 62 Ic = 0 0Ia = 3Va/(Z1 + Z2 + Z0)

28

Symmetrical Components - Summary of 13.8 kV Faults

3LG – positive sequence current

SLG – zero sequence current

L-L – negative sequence current

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

5158 / 8.33 = 619 Ia = 5158 = I1

5346/(8.33x3) = 370 Ia = 5346 = 3I0

If you have a Delta-Wye transformer bank, and you know the voltage ratio and secondary phase current values for 13.8kV 3LG (5158) and SLG (5346) faults, you can find the rest:

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

5158 x 3/2 = 4467 Ib x 3 = 7737 = 3I2 4467/(8.33x 3) = 309309 x 2 = 619

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

29

Symmetrical Components - Summary of 13.8 kV Faults

You’ve heard of a Line-to-Line fault, how about an “Antler-to-Antler” fault?

30

Electromechanical Relays used on Distribution Feeders

Avista’s standard distribution relay package (until the mid-1990’s) included the following:

3 phase 50/51 CO-11 relays,

1 reclosing relay

1 ground 50/51 CO-11 relay

.

31

Objectives: Protect the feeder conductor Detect as low a fault current as possible (PU = 50% EOL fault amps) 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. Carry normal maximum load (phase overcurrent only). Pickup the feeder in a cold load condition ( 2 times maximum normal load) or pickup ½ of the next feeder load·

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

2

3

4

5

7

10

20

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100

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300

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2

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1

TIME-CURRENT CURVES @ Voltage 13.8 KV By

For No.

Comment Date


FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

1


2

2. LAT421 50P CO-9 INST TD=1.000CTR=600/5 Pickup=4.6A No inst. TP@5=0.048s

3

3. LAT421 51P CO-9 CO-9 TD=1.900CTR=600/5 Pickup=4.A No inst. TP@5=0.4618s

4

4. LAT421 50N CO-9 INST TD=1.000CTR=600/5 Pickup=3.A No inst. TP@5=0.048s

5

5. LAT421 51N CO-9 CO-9 TD=3.000CTR=600/5 Pickup=3.A No inst. TP@5=0.7228s

6

6. LAT421 FUSE S&C Link100TTotal clear.

A

A. Transf. damage curve. 7.50 MVA. Category 3Base I=313.78 A. Z= 6.9 percent.Latah Jct 13.8kV - LATAHJCT115 115.kV T

B

B. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 1/0


32

13 kV BUS

MIN FAULT3LG = 2000 A1LG = 1000 A

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


Pickup setting criteria of 2/1 ratio of “end of line” fault duty / pickup- ensures that the relay will “see” the fault and operate when needed.

33

13 kV BUS

FAULT at MIDLINE3LG = 2000 A1LG = 1000 A

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

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

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


34


13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A

SECTION LOAD = 500 A

35


13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A N.O.




36

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.

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.

Lockout only for faults within the protected zone. That is; won’t lockout for faults beyond fuses, line reclosers etc.

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

37

13 kV BUS

RECLOSING SEQUENCE

Closed

Open0.5" 12" LOCKOUT

RESET = 120"(INST Blocked during Reset Time)


Reclosing – Sequence shown for a permanent fault

38

Distribution Fusing – Fuse Protection/Saving 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: Trip and clear the fault at the station (or line recloser) by the instantaneous trip before the fuse is damaged for a lateral fault. Reclose the breaker. That way if the fault were temporary the feeder is completely re-energized and back to normal. 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 customers. 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

2

3

4

5

7

10

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1

TIME-CURRENT CURVES @ Voltage 13.8 kVM15 515 51N to 140T By Protection

For Aspen File: HORS M15 EXP.olr No.

Comment Date 3/06

STATION FEEDER RELAYING

MAXIMUM FEEDER FUSE

1

1. M15 515 GND TIME CO-11 TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0192s

2

2. Moscow 515 Kear 140T Kearney 140TTotal clear.

3

3. M15 50N CO-11 INST TD=1.000CTR=160 Pickup=3.A No inst. TP@5=0.048s

39

Shows the maximum fault current for which S&C type T fuses can still be protected by a recloser/breaker instantaneous trip for temporary faults (minimum melt curve at 0.1 seconds):6T – 120 amps

8T – 160 amps10T – 225 amps12T – 300 amps15T – 390 amps20T – 500 amps25T – 640 amps30T – 800 amps40T – 1040 amps50T – 1300 amps65T – 1650 amps80T – 2050 amps100T – 2650 amps140T – 3500 amps200T – 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

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For No.

Comment Date

1

1. BR1 S&C 140T S&C Link140TMinimum melt.

2

2. ENDTAP S&C 40T S&C Link 40TMinimum melt.

3

3. M15 515 S&C 80T S&C Link 80TMinimum melt.

Distribution Fusing – Fuse Protection for Temporary Faults

40

NOTE: These values were taken from the S&C 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 it’s rating prior to the fault. This means there was prior heating of that fuse so it doesn’t 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

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For No.

Comment Date

1

1. T FUSE S&C Link100TMinimum melt.

2


Distribution Fusing – Fuse to Fuse Coordination

41

Typical continuous and 8 hour emergency rating of the S&C 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

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1

T IME-CURRENT CURVES @ Voltage By

For No.

Comment Date

1

1. BR1 S&C 140T S&C Link140TMinimum melt.

Distribution Fusing – S&C T-Fuse Current Ratings

General Rule: Fuse Blows at 2X Rating in 5 Minutes

42

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

DEVICES: CTI (Sec.)

Relay – Fuse Total Clear 0.2Relay – Series Trip Recloser 0.4Relay – Relayed Line Recloser 0.3Lo Side Xfmr Relay – Feeder Relay 0.4Hi Side Xfmr Relay – Feeder Relay 0.4Xfmr Fuse Min Melt – Feeder Relay 0.4

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

2

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4

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7

10

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2

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1






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1


2



4


5

5f


6


A


B



43


DEVICES: CTI (Sec.)


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

2

3

4

5

7

10

20

30

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50

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100

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300

400

500

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1000

2

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.01

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1






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1


2



4


5

5f


6


A


B



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

2

3

4

5

7

10

20

30

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100

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300

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1000

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.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kVRAT 231 MID C390R By JDH

For Huetter feeder 142 (2008) Rat feeder 231 (2009+) Midline Recloser C390R No.

Comment Saved in: HUE142-RAT231 MID C390R 2007base.olr 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

1. C390R MID 51P CO-9 CO-9 TD=2.500CTR=300/5 Pickup=5.A No inst. TP@5=0.6037s

2

2. C390R MID 50P CO-9 INST TD=1.000CTR=300/5 Pickup=6.A No inst. TP@5=0.048s

3 3. C390R MID 50N CO-9 INST TD=1.000CTR=300/5 Pickup=5.A No inst. TP@5=0.048s

4

4. C390R MID 51N CO-9 CO-9 TD=2.500CTR=300/5 Pickup=5.A No inst. TP@5=0.6037s

5

5. HUE 142 50P 251 INST TD=1.000CTR=160 Pickup=7.A No inst. TP@5=0.048s

6

6. HUE 142 51P 251 SEL-EI TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.4072s

7

7. HUE ML C270R 50P INST TD=1.000CTR=400/5 Pickup=8.1A No inst. TP@5=0.048s

8

8. HUE ML C270R 51P CO-11 TD=1.500CTR=400/5 Pickup=7.A No inst. TP@5=0.3766s

9

9. HUE 142 50N 251 INST TD=1.000CTR=160 Pickup=3.A No inst. TP@5=0.048s

10

10. HUE 142 51N 251 SEL-EI TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0858s

11

11. HUE 142 51Q 251 SEL-EI TD=4.000CTR=160 Pickup=5.2A No inst. TP@5=1.0858s

12

12. HUE ML C270R 50N INST TD=1.000CTR=400/5 Pickup=3.A No inst. TP@5=0.048s

13

13. HUE ML C270R 51N CO-11 TD=7.000CTR=400/5 Pickup=3.A No inst. TP@5=1.8052s

14

14. RAT 231 50N CO-11 INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

15

15. RAT 231 51N CO-11 CO-11 TD=5.000CTR=800/5 Pickup=3.A No inst. TP@5=1.3192s

16

16. RAT 231 50P CO-11 INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

17

17. HUE 142 FUSE S&C Link 65TTotal clear.

18

18. RAT 231 FUSE S&C Link100TTotal clear.

19

19. RAT 231 51P CO-11 CO-11 TD=1.800CTR=800/5 Pickup=6.A No inst. TP@5=0.4533s

A

A. Conductor damage curve. k=0.08620 A=133100.0 cmilsConductor ACSR AWG Size 2/0MIDLINE C390 FED FROM RAT 231 - MINIMUM CONDUCTOR TO PROTECT

B

B. Conductor damage curve. k=0.06710 A=556000.0 cmilsConductor AACMIDLINE C390 FED FROM HUE 142 - MINIMUM CONDUCTOR TO PROTECT

44


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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

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10

20

30

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50

70

100

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300

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500

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1000

.01

.02

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.05

.07

.1

.2

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.7

1

.01

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.05

.07

.1

.2

.3

.4

.5

.7

1


For Idaho Rd Feeder 251 in Idaho Rd PHASE 1 2007base.olr No.Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A Date 11-25-2008



MAXIMUM FEEDER FUSE

HI-SIDE CT'S

LO-SIDE CT'S


1


2


3


4


5


6


7


8


9


10


11


A


B


45


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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

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50

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100

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.01

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.1

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1

.01

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.1

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.7

1

TIME-CURRENT CURVES @ Voltage 13.8 KV By

For No.

Comment Date


FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

1


2

2. LAT421 50P CO-9 INST TD=1.000CTR=600/5 Pickup=4.6A No inst. TP@5=0.048s

3

3. LAT421 51P CO-9 CO-9 TD=1.900CTR=600/5 Pickup=4.A No inst. TP@5=0.4618s

4

4. LAT421 50N CO-9 INST TD=1.000CTR=600/5 Pickup=3.A No inst. TP@5=0.048s

5

5. LAT421 51N CO-9 CO-9 TD=3.000CTR=600/5 Pickup=3.A No inst. TP@5=0.7228s

6

6. LAT421 FUSE S&C Link100TTotal clear.

A

A. Transf. damage curve. 7.50 MVA. Category 3Base I=313.78 A. Z= 6.9 percent.Latah Jct 13.8kV - LATAHJCT115 115.kV T

B

B. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 1/0

46


DEVICES: CTI (Sec.)



47


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

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

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.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor M15 A-172 CS with fdr 515 with Kearney 140T Fuse No.Comment Three Phase Fault Date 12/12/05

1

1. Moscow 515 Kear 140T Kearney 140TMinimum melt.I= 5158.1A T= 0.10s

2

2. M15-515 Phase Time CO-11 TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.3766sI= 5158.1A (32.2 sec A) T= 0.33s

3

3. M15 A-172 Phase CO- 9 TD=1.500CTR=120 Pickup=2.A Inst=1200A TP@5=0.3605sI= 619.0A (5.2 sec A) T= 1.03s H=8.33

FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L 3LG

3LG Fault Coordination Example:

Top – Ckt Swr w/ Phase E-M relay

----- 0.4 sec. ------

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

----- 0.2 sec. ------

Bottom – 140T Feeder Fuse (Total Clear)

48


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

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

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1

.01

.02

.03

.04

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1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Moscow CS with fdr 515 with 140T fuses No.Comment Single Line to Ground Fault Date 12/12/05

1


2

2. M15 515 GND TIME CO-11 TD=4.000CTR=160 Pickup=3.A No inst. TP@5=1.0192sI= 5346.5A (33.4 sec A) T= 0.29s

3


4 4. M15 A-172 GND TIME CO-11 TD=4.000CTR=240 Pickup=4.A No inst. TP@5=1.0192sI= 5346.5A (22.3 sec A) T= 0.83s

FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L 1LG Type=A

SLG Fault Coordination Example:

Top – Ckt Swr w/ E-M Phase relay

----- 0.4 sec. ------

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

----- 0.2 sec. ------


49


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

2

3457

10

20

304050

70

100

200

300400500700

1000

2

3457

10

20

304050

70

100

200

300400500700

1000

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

.01

.02

.03

.04

.05

.07

.1

.2

.3

.4

.5

.7

1

TIME-CURRENT CURVES @ Voltage 13.8 kV By DLHFor Moscow CS A-172 with fdr 515 with 140T Fuse No.Comment Line to Line Fault Date

1


2


3

3. M15 A-172 Phase CO- 9 TD=1.500CTR=120 Pickup=2.A Inst=1200A TP@5=0.3605sI= 619.0A (5.2 sec A) T= 1.03s H=8.33

FAULT DESCRIPTION:Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L LL Type=B-C

L-L Fault Coordination Example:

Top – Ckt Swr w/ E-M relay

----- 0.4 sec. ------

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

----- 0.2 sec. ------


50

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

2

3

4

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7

10

20

30

40

50

70

100

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300

400

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1000

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.01

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1

TIME-CURRENT CURVES @ Voltage 13.8 kVINT 12F2 & MID 466R By JDHFor Indian Trail feeder 12F2 and 12F2 Midline 466R in INT 2007base.olr No.Comment AT SUB: 3LG= 5719A, LL=4951A, SLG= 5880A Date 1-29-08


STATION VCB 12F2 PROTECTION

EAST BRANCH - FEEDER MAXIMUM FUSE

FAULT DUTY AT 140T - WHEN FED FROM NORTH (N.O. 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

9&15

11&4

1

1. INT A-742 51P 351 SEL-VI TD=1.200CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s H=8.33

2

2. INT A-742 51G 351 SEL-VI TD=1.000CTR=600/5 Pickup=1.A No inst. TP@5=0.258s H=8.33

3

3. INT A-742 51N 351 SEL-EI TD=5.600CTR=1200/5 Pickup=3.A No inst. TP@5=1.5201s

4

4. INT A-742 51Q 351 SEL-EI TD=4.200CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s H=8.33

5

5. INT F2 150E FUSE SMU-20_150ETotal clear.

6

6. INT 12F2 50P 351S INST TD=1.000CTR=800/5 Pickup=7.A No inst. TP@5=0.048s

7

7. INT 12F2 51P 351S SEL-EI TD=1.500CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s

8

8. INT 12F2 50G 351S INST TD=1.000CTR=800/5 Pickup=3.A No inst. TP@5=0.048s

9

9. INT 12F2 51G 351S SEL-EI TD=3.900CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s

10

10. INT 12F2 51Q 351S SEL-EI TD=3.500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

11

11. INT A-742 51Q 587 W2 SEL-EI TD=4.300CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s

12

12. F2 MID 466R 50P 351R INST TD=1.000CTR=500/1 Pickup=1.68A No inst. TP@5=0.048s

13

13. F2 MID 466R 51P 351R SEL-EI TD=2.300CTR=500/1 Pickup=1.44A No inst. TP@5=0.6243s

14

14. F2 MID 466R 50G 351R INST TD=1.000CTR=500/1 Pickup=0.89A No inst. TP@5=0.048s

15

15. F2 MID 466R 51G 351R SEL-EI TD=4.200CTR=500/1 Pickup=0.89A No inst. TP@5=1.1401s

16

16. F2 MID 466R 51Q 351R SEL-EI TD=3.800CTR=500/1 Pickup=1.54A No inst. TP@5=1.0315s

17

17. INT F2 140T FUSE stn S&C Link140TTotal clear.

A

A. Transf. damage curve. 12.00 MVA. Category 3Base I=502.00 A. Z= 8.4 percent.INDIAN TRAIL 12/16/20 MVA XFMR

B

B. Conductor damage curve. k=0.14040 A=52630.0 cmilsConductor Copper (bare) AWG Size 2NORTH BRANCH SMALLEST TRUNK CONDUCTOR

C

C. Conductor damage curve. k=0.04704 A=350000.0 cmilsCable XLPE 90C/250CEAST BRANCH SMALLEST TRUNK CONDUCTOR - 350 CN15

D

D. Conductor damage curve. k=0.08620 A=105500.0 cmilsConductor ACSR AWG Size 2/0WEST BRANCH SMALLEST TRUNK CONDUCTOR

An example of a completed 13.8kV Feeder Coordination Study with 20 Time-Current Curves representing:

Instantaneous & 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

51

2 – Time delay relay.27 – Undervoltage relay.43 – Manual transfer or selective device.

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

50 (or 50P) – Instantaneous overcurrent phase relay.

50N (or 50G) – Instantaneous overcurrentground (or neutral) relay.

50Q – Instantaneous Negative Sequence overcurrent relay.

51 (or 51P) – Time delay overcurrent phase relay.

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

51Q – Time delay Negative Sequence overcurrent relay.

52 – 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.

52/a – Circuit breaker auxiliary switch closedwhen the breaker is closed.

52/b – Circuit breaker auxiliary switch closed when the breaker is open.

59 – Overvoltage relay.62 – Time Delay relay63 – Sudden pressure relay.79 – AC Reclosing relay.81 – Frequency relay.86 – Lock out relay which has several contacts.

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

87 – Differential relay.94 – Auxiliary tripping relay.

Mike’s Turn

52

53

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

54

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

55

Phase Overcurrent Settings – Current measured on 115kV side

P Pickup (time overcurrent)

Don’t trip for load or cold load pickup (use 2.4 * highest MVA rating)

Ends up being higher than the feeder phase element pickup

Example: 12/16/20 MVA unit would use 2.4*20*5 = 240 amps (1940A low side)

51P Time Lever (time dial)

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

CTI is 0.4 seconds

Worst coordination case: Ø-Ø on low side (discrepancy due to delta/wye 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


56

Ground Overcurrent Settings – Current measured on 13.8 kV side

N Pickup (time overcurrent)

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 0.4 seconds

50N Pickup (instantaneous overcurrent)

DO NOT USE!!!!


57

Distribution Transformer & 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

58


Elements We Set: 51P 50P 50P2 (FTB) 51G 50G (115kV) 51N (13.8kV) 50N1 (FTB)

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

59

Transformer Phase Overcurrent Settings - Current measured on 115kV Side

51P - Phase Time Overcurrent Pickup

Set to 240% of nameplate (same as EM relay)

51P Time Dial

Set to coordinate with feeder’s fastest element for each fault

CTI is still 0.4 seconds

50P1 – Phase instantaneous #1 Pickup

Direct Trip

Set to 130% of max 13.8 kV fault (vs. 170% with EM relay)

50P2 – 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


60

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 – Ground Instantaneous #1 Pickup

Set very low. Usually 120 Amps.


SymmetricalComponents!

61

Transformer Neutral Overcurrent Settings - Current measured by neutral CT or calculated from 13.8kV CTs.

51N - Ground Time Overcurrent Pickup

Set slightly higher than feeder ground pickup (about 1.3 times)

51NTime Dial

Set to coordinate with feeder ground

CTI is 0.4 seconds

50N1 – 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


62

Inrush – 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 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 12/16/20 MVA transformer, the expected inrush would be 8*12*5*0.6 = 288 amps. We set a little above this number (360 Amps).


63

Transformer inrush UNFILTEREDcurrent. The peak current is about 1800 amps.

Transformer inrush FILTERED current (filtered by digital filters to show basically only 60 HZ). Peak is about 700 amps.


64


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

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

65

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 – Phase instantaneous Pickup

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


66

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 – Phase instantaneous Pickup

Set the same as the 51G pickup


67

Feeder Negative Sequence Overcurrent Settings

51Q - Negative Sequence Time Overcurrent Pickup

Set with equivalent sensitivity as the ground element (not affected by load)

Coordinate with downstream protection

51Q Time Dial

Same as EM relay (coordinate with downstream protection)

Select a curve

50Q – Negative Sequence instantaneous Pickup

DON’T USE!!!!!

Contributions from motors during external faults could trip this


68

The differential relay is connected to both the high and low side transformer BCT’s.

EM Differential RelaySince the distribution transformer is connected delta – wye the transformer CT’s have to be set wye – delta to compensate for the phase shift.

Transformer Differential Protection

69

87R1

87OP

87R2

PRI I

PRI I

SEC I

SEC I

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

CURRENT FLOW THROUGH AN E/M 87 DIFFERENTIAL RELAY FOR AN INTERNAL AND EXTERNAL FAULT

BCT'S

BCT'S

CT POLARITY MARK

Transformer Differential Protection – External Fault

70

87R1

87OP

87R2

PRI I

PRI I

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

SEC I

SEC I

BCT'S

BCT'S

CT POLARITY MARK

Transformer Differential Protection – Internal Fault

71

GRAB YOUR HANDOUT!MOSCOW FEEDER 515 PROTECTION EXAMPLE

72

The Scenario

A hydraulic midline recloser on a 13.8kV 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 “fuse saving” scheme with one instantaneous trip. The field engineers decide if they want to enable or disable the instantaneous tripping.

73

MOSCOW FEEDER 515 PROTECTION EXAMPLE

• Where do we start the protection design?1) Gather Information:

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

74

How do we proceed?

• At each coordination point– Loading– Coordination

• What will coordinate with the downstream device• Are we above the fuse rating

– Conductor Protection• minimum conductor that can be protected by the feeder settings or

fuse

– Fault Detection• Can we detect the fault by our 2:1 margin

– Fuse Saving for Temporary faults• What can a fuse be protected up to?

75

Begin at the END

• 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 31.4 amps. The Avista transformer fusing standard says to use a 65T on this transformer so that’s what we’ll choose.

76

FUSE - POINT 6B:What we know:

– 3Ø lateral feeding to a 65T at the end– #4 ACSR conductor

Fuse Selection:– Loading – Assume the load is all downstream of point 8, so a 65T or higher will

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

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

line end fault with a 2:1 margin. The SLG is 463 so we calculate the max fuse as follows:

• 463/2 (2:1 margin) = 231 amps. “T” fuses blow at twice their rating so divide by 2 again, with a result of 115. Fuse must be 100T or less.

– Coordination – From Table 2, we see that we need an 80T or larger to coordinate with the downstream 65T fuse (at fault duties < 1400A w/preload)

– Fuse Saving - From Table 1 we see that an 80T fuse can be protected up to 2,050 amps for temporary faults and the 3Ø fault is 1,907 amps so we could choose an 80T or higher from that standpoint

77

Point 6B - So what fuse size should we use?

78

POINT 5 – Midline Recloser• What we know:

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

• Settings Considerations:– Loading: Cold load would be 84*2 = 168 A, but the conductor can carry 140A max. Size for some

load growth by setting at 2*conductor rating. 300 Amp phase pickup setting. NOTE: we don’t design settings to protect for overload.

– Conductor Protection: From the table, Avista’s 300A feeder settings can protect down to #2 ACSR, so 300A pickup is still acceptable. Review curves to confirm.

– Fault Detection: detect the minimum fault with a 2:1 margin. Here the min fault is at point 61. The Ø-Ø fault at point 6 is 0.866*1907 = 1,651 so our margin to detect that fault would be 1651/300

= 5.5:1 so no problem with the 300 amps Ø PU from that standpoint (used Ø-Ø because it’s the minimum multi phase fault)

2. The SLG at point 6 is 1,492 so we could set the ground up to 1492/2 = 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 we’re trying to use a 100T. Based on curves, this is 300 amps (same as phase, which is unusual).

– Coordination: Based on fault study, set the time dial to coordinate with a 100T fuse with 0.2 second CTI.

– 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.

79

TRUNK FUSE - POINT 3C:What we know:

– #4 ACSR 3Ø trunk – Load is 84 Amps– Downstream protection is the midline recloser

Fuse Selection:– Loading – Assume the load is the same as at the midline recloser. We set that at

300A, and a 200T can carry 295 A continuous.– Conductor Protection - From Table 7 we see that #4 ACSR can be protected by a

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

line end fault with a 2:1 margin. • SLG at midline is 2762 so even a 200T would provide enough sensitivity.

– Coordination – PROBLEM! We have coordinated the midline with a 100T, so the upstream fuse must coordinate with that.

– Fuse Saving: As before, some fuse saving is compromised due to higher fault duty.

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TRUNK FUSE Point 3C - Solutions?

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TRUNK FUSE Point 3C - Solutions?

• Move Recloser• Re-conductor

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FEEDER BREAKER - POINT 1:What we know:

– This is a 500 amp feeder design. The load is 325 amps and cold load 650 per field engineer.– 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 1/0 ACSR or higher. Laterals with smaller conductor must be fused.

Fault Detection: The Ø-Ø fault at point 5 is 0.866*3453 = 2990 so our margin to detect that fault would

be 2990/960 = 3.1:1 The SLG at point 5 is 2,762 so our margin to detect that fault is: 2762/480 = 5.7:1

Coordination: Use fault study to calculate settings to achieve 0.2 second CTI to fuses and 0.3 second CTI to the recloser.

Fuse Saving – We have decided that a 140T fuse is the maximum fuse we will use on the feeder even though it can’t be saved by an instantaneous trip at the maximum fault duties.

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Questions?

Documents

DISTRIBUTION PROTECTION OVERVIEW - WSU … PROTECTION OVERVIEW ... Transformer Relay Settings using Electromechanical Relays ……… ... Transformer Differential Protection