distribution overview ppt

8/11/2019 Distribution Overview PPT

1/83

1

DISTRIBUTION PROTECTION OVERVIEW

Kevin Damron & Mike Diedesch

Avista Utilities

Presented

March 17th, 2014

At the 31st Annual

Hands-On Relay School

Washington State University

Pullman, Washington


2/83

2

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.. 31Coordinating 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


3/83

3

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 blowingfor the occurrence of electrical faults on the distribution system.

Design tools used:

1 Transformer and conductor damage curves,

2 - Time-current coordination curves (TCCs), 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 - TCCs


4/83

XFMR

12/16/20

MVA

115/13.8kV

115 kV SYSTEM

13.8 kV BUS

500A, 13.8 kVFDR #515

DELTA/WYE

LINE RECLOSER P584

3 = 5158SLG = 5346

3 = 3453SLG = 2762

PT-2BPT-2A

3 = 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

3 = 1907SLG = 1492

3 = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172

MOSCOW

1 PHASE#4 ACSR

3 = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3 = 1210SLG = 877

PT-1

PT-5

PT-7

FUSE

556 ACSR

556 ACSR

13.8 kV FDR #512

500 A FDR

4

System Overview Inside the

Substation Fence

Transformer

relays

Feeder relay


5/83

5

XFMR

12/16/20MVA

115/13.8kV

115 kV SYSTEM

13.8 kV BUS

500A, 13.8 kV

FDR #515

DELTA/WYE

LINE RECLOSER P584

3 = 5158SLG = 5346

3 = 3453SLG = 2762

PT-2BPT-2A

3 = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3 = 1907SLG = 1492

3 = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172

MOSCOW

1 PHASE#4 ACSR

3 = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3 = 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


6/83

XFMR

12/16/20 MVA

115/13.8kV

115 kV SYSTEM

13.8 kV BUS

500A, 13.8 kVFDR #515

DELTA/WYE

LINE RECLOSER P584

3 = 5158SLG = 5346

3 = 3453SLG = 2762

PT-2BPT-2A

3 = 3699SLG = 3060

PT-3B PT-3A PT-4

#4 ACSR


#4 ACSR

#2 ACSR#4 ACSR

2/0 ACSR#2 ACSR

3 = 1907SLG = 1492

3 = 322SLG = 271

3-250 KVAWYE/WYE

PT-865T

A-172

MOSCOW

1 PHASE#4 ACSR

3 = 558SLG = 463

PT-3C

PT-6BPT-6A

PT-6C

HIGH LEAD LOW

3 = 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)

SECOND

S

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 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

1

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

2

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

4

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

5

5f

5. Phase unit of recloser MID LINE OCRFast: ME-341-B Mult=0.2

Slow: ME-305-A Add=1000.Ia= 5665.9A T(Fast )= 0.03s

6

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

6

System Overview Each device has at least one curve plotted with current and time values

on the Time Coordination Curve.


7/83

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)

SECOND

S

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1

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

2

2. IDR 252 51P 351S SEL-EI TD=1.500

CTR=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

4


5

5f


Slow: ME-305-A Add=1000.Ia= 5665.9A T(Fast )= 0.03s

6


A


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


7

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 theleft on the TCC correspond to those

devices farther from the substation

where the fault current is less.


8/83

8

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)

SECONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 By

For No.

Comment Date

1

1. M1CTR=

A

A. T ransf. dam age 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 additionalthermal 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.


9/83

9

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)

SECONDS

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

1000

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

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 DLH

For ACSR Conductor Damage Curves No.

Comment Date 12/13/05

1

1. M15-515 Phase INST INST TD=1.000CTR=160 Pickup=7.ANo 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 AWGSize 4/04/0 ACSR

C

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

D

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

E

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

FF. Conductor damage curve. k=0.08620 A=33100.0 cmilsConductor ACSR AWGSize 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)

SEC

ONDS

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

1000

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

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


For Copper Conductor Damage Curves No.

Comment Date 12/13/05

1

1. M15-515 Phase INST INST TD=1.000CTR=160 Pickup=7.ANo inst. TP@5=0.048s

A

A. Conductor damage curve. k=0.14040 A=105500.0 cmils

Conductor Copper (bare) AWGSize 2/02/0 Copper

B

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

C

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

D

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

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


10/83

10

Conductor Rating556 730

336.4 530

4/0 340

2/0 270

1/0 230

#2 180

#4 140

ACSR

Ampacity Ratings

Conductor Rating

2/0 360

1/0 310

#2 230

#4 170

#6 120

Copper

Ampacity Ratings

Conductor Ampacities

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


11/83

11

Comparing a 140T fuse versus a

#4 ACSR Damage curve.

The 140T wont 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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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#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


12/83

12

Transformer Protection using

115 kV Fuses


13/83

13

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 blowwithin 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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 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

1

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

2

2. SMD-2B 65E VERY SLOW 176-19-065

Minimum melt.H=8.33

3

3. LAT RP4211 50P CO-11 INST TD=1. 000CTR=500/5 Pickup=3.5A No inst. TP@5=0.048s

4

4. LAT RP4211 51P CO-11 CO-11 TD=2. 000CTR=500/5 Pickup=3.A No inst. TP@5=0.5043s

5

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

6

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


14/83

14

The fuse time current characteristic (TCC) is fixed (although you can buy a standard, slowor very slow speed ratio which are different inverse curves).

The sensitivity to detect lo-side SLG faults isnt as good as using a relay on a circuitswitcher 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


15/83

15

Transformer Protection using a Circuit SwitcherShowing Avistas present standard using Microprocessor relays.

13.8kV 115kV


16/83

16

Transformer Protection using a Circuit SwitcherShowing Avistas old standard using Electromechanical relays.


17/83


18/83

18

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 CTs 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.


19/83

19

SEL Various Relay Overcurrent Curves.Extremely Inverse steepest

Very Inverse

Inverse

Moderately Inverse

Short 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-Mrelays.

Avista uses mostly extremely inverse on

feeders to match the fuse curves.

Relay Overcurrent 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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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


For No.

Comment Date

1

1. EXTREMELY INVERSE SEL-EI TD=15.000

CTR=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

3

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

4

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

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


20/83

20

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

amps10 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

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 kVIDR FEEDER 251 By JDH





MAXIMUM FEEDERFUSE

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=0.3096sH=8.33

2

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

3

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

4

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

5

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

6

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

7

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

8

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

9

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

10

10. IDR251 51Q 351S SEL-EI TD=3.500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

11

11. 251 140T FUSE st n 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

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


21/83

21

Same Time Dial = 9

Right curve picks up at 960 Amps

Left 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)

SE

CONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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


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

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

3


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)

SE

CONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 By

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

2. ~EXT INV 1 SEL-EI TD=9.000

CTR=800/5 Pickup=5.A No inst. TP@5=2.4431s3


4


5

5. ~EXT INV 4 SEL-EI TD=9.000


Same Pickup = 960 Amps

Top curve Time Dial = 15

Bottom curve Time Dial = 2


22/83

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

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 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


STAT ION VCB 12F1 PROTECTION

11. INT A-742 51P 351 SEL-VI TD=10.000CTR=600/5 Pickup=0.8A No inst. TP@5=2.5797sH=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

55. INT A-742 51G 351 SEL-VI TD=5.000

CTR=600/5 Pickup=0.8A No inst. TP@5=1.2898sH=8.33

6

6. INT A-742 51Q 351 SEL-VI TD=3.000CTR=600/5 Pickup=0.8A No inst. TP@5=0.7739sH=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 offault current being measured:

51P phase time overcurrent

51N or 51G ground time overcurrent

51Q negative sequence time overcurrent

Transformer relay curves

Feeder relay curves

51P phase time overcurrent

51N or 51G ground time overcurrent

51Q negative sequence time overcurrent

This brings us to a brief discussion of:


23/83

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.


24/83

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 + = I1

Negative Sequence current = I - = I2

Zero Sequence current = I 0 = I0

Phase Current notation:

IA High side Amps

Ia 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 aDelta-Wye high-lead-low transformer bank :


25/83


26/83

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 0

IB = 619 152 Ib = 4467 152

IC = 309 -28 Ic = 4467 -28

3LG 13.8kV fault = 5158A,

Ib=Ic=4467 A, 4467/5158 = 86.6%=3/2IB3LG = IBLL = 619A

IA & 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,


27/83

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 phasecurrent is the 3 less as compared to the 3 fault.

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

IA = 370 -118 Ia=3I0=5346 -118

IB = 0 0 Ib = 0 0

IC = 370 62 Ic = 0 0

Ia = 3Va/(Z1 + Z2 + Z0)


28/83

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 = 5158

IB = 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 = 0

IB = 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 = 0

IC = 370 Ic = 0


29/83

29

Symmetrical Components - Summary of 13.8 kV Faults

Youve heard of a Line-to-Line fault, how about an Antler-to-Antler fault?


30/83

30

Electromechanical Relays

used on Distribution Feeders

Avistas standard distribution relay

package (until the mid-1990s) included

the following:

3 phase 50/51 CO-11 relays,

1 reclosing relay

1 ground 50/51 CO-11 relay

.


31/83

31

Objectives:

Protect the feeder conductor

Detect as low a fault current aspossible (PU = 50% EOL fault amps)

Other than 51P, set pickup and timedial as low as possible and still haveminimum Coordinating Time Interval

(CTI) to next device. CTI is minimum

time between operation of adjacent

devices.

Carry normal maximum load (phaseovercurrent only).

Pickup the feeder in a cold loadcondition ( 2 times maximum normalload) or pickup 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

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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

For No.

Comment Date


FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

1

1. SMD-2B 65E VERY SLOW 176-19-065Minimum melt.

H=8.33

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




32/83


33/83

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 A

1LG = 500 A

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




34/83

34



13 kV BUS

500A FEEDER SETTINGS51P = 960 A51N = 480 A

SECTIONLOAD = 500 A


35/83

35



13 kV BUS

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

SECTIONLOAD = 250 A

SECTIONLOAD = 500 A

SECTIONLOAD = 250 A


36/83

36

Most overhead feeders also use reclosing capability to automatically re-energize the feederfor 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 tothree 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. Thetime 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; wont lockout for faults beyondfuses, 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


37/83

37

13 kV BUS

RECLOSING SEQUENCE

Closed

Open

0.5" 12" LOCKOUT

RESET = 120"

(INST Blocked during Reset Time)



Reclosing Sequence shown for a permanent fault


38/83

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 linerecloser) by the instantaneous trip before the fuse

is damaged for a lateral fault. Reclose the breaker. That way if the fault weretemporary 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 thetime 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

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 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


39/83

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 amps

12T 300 amps

15T 390 amps

20T 500 amps

25T 640 amps

30T 800 amps40T 1040 amps

50T 1300 amps

65T 1650 amps

80T 2050 amps

100T 2650 amps

140T 3500 amps

200T 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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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


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


40/83

40

NOTE: These values were taken from the S&C data bulletin 350-170 of March 28, 1988 based on nopreloading and then preloading of the source side fuse link. Preloading is defined as the source side

fuse carrying load amps equal to its rating prior to the fault. This means there was prior heating of

that fuse so it doesnt 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)

SE

CONDS

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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 By

For No.

Comment Date

1

1. T FUSE S&C Link100TMinimum melt.

2

2. 251 140T FUSE stn S&C Link140TTotal clear.

Distribution Fusing Fuse to Fuse Coordination


41/83

41

Typical continuous and 8 hour emergency rating of the S&C T rated silver fuse links plus the 140Tand 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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

1000

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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


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


42/83

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 0.2

Relay Series Trip Recloser 0.4

Relay Relayed Line Recloser 0.3

Lo Side Xfmr Relay Feeder Relay 0.4

Hi Side Xfmr Relay Feeder Relay 0.4Xfmr Fuse Min Melt Feeder Relay 0.4

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

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1

1. IDR A-777 51P 351 SEL-VI TD=1.200

CTR=600/5 Pickup=2.A No inst. TP@5=0.3096sIa= 679.8A (5.7 sec A) T= 0.78s H=8.33

2

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

4

4. 252 140T FUSE stn S&C Link140T

Total clear.Ia= 5665.9A T= 0.09s

5

5f


Slow: ME-305-A Add=1000.Ia= 5665.9A T(Fast)= 0.03s

6


A


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



43/83

43



devices.


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

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

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






MAXIMUM FEEDER FUSE

HI-SIDE CT'S

MIDLINE OCR


Fault I=5665.9 A

1

1. IDR A-777 51P 351 SEL-VI TD=1.200

CTR=600/5 Pickup=2.A No inst. TP@5=0.3096sIa= 679.8A (5.7 sec A) T= 0.78s H=8.33

2

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

4


5

5f


Slow: ME-305-A Add=1000.Ia= 5665.9A T(Fast)= 0.03s

6


A


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



44/83

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 71000 1000


45/83

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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

.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 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=0.3096sH=8.33

2

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

3

3. IDR A-777 51N 351 SEL-EI TD=5. 700CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s

4

4. IDR A-777 51Q 351 SEL-EI TD=4. 200CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401sH=8.33

5

5. IDR A-777 51Q 587 W2 SEL-EI TD=4. 300CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s

6

6. IDR 251 50P 351S INST TD=1.000


7

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

8

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

9

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

10

10. IDR 251 51Q 351S SEL-EI TD=3. 500CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s

11

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

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

45



devices.


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 71000 1000


46/83

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

2

3

4

5

7

10

20

30

40

50

70

100

200

300

400

500

700

.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

For No.

Comment Date


FEEDER VCB PROTECTION

MAXIMUM FEEDER FUSE

1

1. SMD-2B 65E VERY SLOW 176-19-065Minimum melt.

H=8.33

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.


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 71000 1000

1


47/83

47


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

70

100

200

300

400500

700

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

.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


For M15 A-172 CS with fdr 515 with Kearney140T Fuse No.

Comment Three Phase Fault Date 12/12/05

1. Moscow515 Kear 140T Kearney140TMinimummelt.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)

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 71000 1000

1


48/83

48


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

70

100

200

300

400500

700

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

.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


For MoscowCS with fdr 515 with 140T fuses No.

Comment Single Line to Ground Fault Date 12/12/05


2

2. M15 515 GND TIME CO-11 TD=4.000

CTR=160 Pickup=3.A No inst. TP@5=1.0192sI= 5346.5A (33.4 sec A) T= 0.29s

3

3. M15-515 Phase Time CO-11 TD=1.500CTR=160 Pickup=6.A No inst. TP@5=0.3766sI= 5346.4A (33.4 sec A) T= 0.31s

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. ------


Clear)

10 2 3 4 5 7 100 2 3 4 5 7 1000 2 3 4 5 7 10000 2 3 4 5 71000 1000

1


49/83

49


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

70

100

200

300

400500

700

2

3

45

7

10

20

30

4050

70

100

200

300

400500

700

.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


For MoscowCS A-172 with fdr 515 with 140T Fuse No.

Comment Line to Line Fault Date


2

2. M15-515 Phase Time CO-11 TD=1.500

CTR=160 Pickup=6.A No inst. TP@5=0.3766sI= 4467.0A (27.9 sec A) T= 0.43s

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. ------


Clear)


50/83

IEEE Device Designations commonly used in Distribution Protection


51/83

51

2 Time delay relay.

27 Undervoltage relay.

43 Manual transfer or selective device.

We use these for cutting in and outinstantaneous overcurrent relays,

reclosing relays etc.

50 (or 50P) Instantaneous overcurrent

phase relay.

50N (or 50G) Instantaneous overcurrent

ground (or neutral) relay.

50Q Instantaneous Negative Sequence

overcurrent relay.

51 (or 51P) Time delay overcurrent phase

relay.

51N (or 51G) Time delay overcurrentground (or neutral) relay.

51Q Time delay Negative Sequence

overcurrent relay.

52 AC circuit breaker.

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 closed

when the breaker is closed.

52/b Circuit breaker auxiliary switch closed

when the breaker is open.

59 Overvoltage relay.

62 Time Delay relay

63 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.


52/83

Mikes Turn

52

Distribution Transformer


53/83

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 TransformerElectromechanical Relays



54/83

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


Electromechanical Relays - Setting Criteria


El t h i l R l S tti C it i


55/83

55

Phase Overcurrent Settings Current measured on 115kV side

P Pickup (time overcurrent)

Dont 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: - 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





56/83

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!!!!

Electromechanical Relays Setting Criteria

Distribution Transformer & Feeder Protection


57/83

57

with 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




58/83

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)



59/83

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 feeders 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




60/83

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.


Symmetrical

Components!



61/83

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



i i


62/83

62

Inrush The current seen when

energizing a transformer.

Need to account for inrush when usinginstantaneous 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).



ith Mi R l


63/83

63

Transformer inrush UNFILTERED

current. The peak current is about 1800

amps.

Transformer inrush FILTERED current

(filtered by digital filters to show

basically only 60 HZ). Peak is about700 amps.



ith Mi R l


64/83

64


Microprocessor Feeder Relay

Overcurrent

Reclosing

Fast Trip Block Output Breaker Failure Output

Elements We Set:

51P

50P

51G

50G

51Q




65/83

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)





66/83

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





67/83

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

DONT USE!!!!!

Contributions from motors during external faults could trip this


Transformer Differential Protection


68/83

68

The differential relay is connected to both the high and low side transformer BCTs.

EM Differential RelaySince the distribution transformer is connected delta wye the transformer CTs

have to be set wye delta to compensate for the phase shift.

Transformer Differential Protection External Fault


69/83

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 DIFFERENTIALRELAY FOR AN INTERNAL AND EXTERNAL FAULT

BCT'S

BCT'S

CT POLARITY MARK

Transformer Differential Protection Internal Fault


70/83

70

87R1

87

OP

87R2

PRI I

PRI I

INTERNAL FAULT

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

SEC I

SEC I

BCT'S

BCT'S

CT POLARITY MARK


71/83

71

GRAB YOUR HANDOUT!MOSCOW FEEDER 515 PROTECTION EXAMPLE

The Scenario


72/83

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.

MOSCOW FEEDER 515 PROTECTION EXAMPLE


73/83

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?)


74/83

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?


75/83

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 31.4amps. The Avista transformer fusing standard says to use a 65T on this

transformer so thats what well choose.

FUSE - POINT 6B:


76/83

76

What we know: 3 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 2,050amps for temporary faults and the 3 fault is 1,907 amps so we could choose an

80T or higher from that standpoint


77/83

77

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

POINT 5 Midline Recloser


78/83

78

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 someload growth by setting at 2*conductor rating. 300 Amp phase pickup setting. NOTE: we dont design

settings to protect for overload.

Conductor Protection: From the table, Avistas 300A feeder settings can protect down to #2 ACSR, so300A 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 6

1. The - 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 PU from that standpoint (used - because its 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 were 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.

TRUNK FUSE - POINT 3C:

What we know:


79/83

79

What we know: #4 ACSR 3 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 minimumline 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.


80/83

80

TRUNK FUSE Point 3C - Solutions?


81/83

81

TRUNK FUSE Point 3C - Solutions?

Move Recloser

Re-conductor

FEEDER BREAKER - POINT 1:


82/83

82

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 - 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 cant be saved by an instantaneous trip at the maximum fault duties.


83/83

83

Questions?

distribution overview ppt

Documents