DISTRIBUTION PROTECTION OVERVIEW
Mike Diedesch & Kevin Damron– Avista Utilities
Presented
March 12, 2011
At the 29th Annual
HANDS-ON Relay School
Washington State University
Pullman, Washington
1
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
2
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
3
System Overview – Inside the
Substation Fence
115 kV SYSTEM
MOSCOW
A-172
XFMR 12/16/20 MVA
115/13.8kV
HIGH LEAD LOW
DELTA/WYE
13.8 kV BUS
500A, 13.8 kV
FDR #515
Feeder relay
Transformer
relays
#4 ACSR
556 ACSR
PT-2A
PT-1
PT-2B
PT-3B
500 A FDR
#2 ACSR
556 ACSR 3Ø = 3699
SLG = 3060
#2 ACSR
1 PHASE
#4 ACSR
PT-3A
2/0 ACSR
PT-3C
PT-4
3Ø = 1210
SLG = 877
3Ø = 3453
SLG = 2762
LINE RECLOSER P584
PT-5
#4 ACSR
1 PHASE
#4 ACSR
13.8 kV FDR
#512
3Ø = 5158
SLG = 5346
3Ø = 558
SLG = 463
3Ø = 1907
SLG = 1492
PT-6A
FUSE
3-250 KVA
WYE/WYE
PT-6B
#4 ACSR
PT-6C
#4 ACSR
PT-8
65T
3Ø = 322
SLG = 271
4
PT-7
System Overview – Outside the
Substation Fence
115 kV SYSTEM
MOSCOW
A-172
XFMR 12/16/20 MVA
115/13.8kV
Midline
recloser
relay
HIGH LEAD LOW
DELTA/WYE
13.8 kV BUS
500A, 13.8 kV
FDR #515
#4 ACSR
556 ACSR
PT-2A
PT-1
PT-2B
PT-3B
500 A FDR
#2 ACSR
556 ACSR 3Ø = 3699
SLG = 3060
#2 ACSR
1 PHASE
#4 ACSR
PT-3A
2/0 ACSR
PT-3C
PT-4
3Ø = 1210
SLG = 877
3Ø = 3453
SLG = 2762
LINE RECLOSER P584
PT-5
#4 ACSR
1 PHASE
#4 ACSR
13.8 kV FDR
#512
3Ø = 5158
SLG = 5346
3Ø = 558
SLG = 463
3Ø = 1907
SLG = 1492
PT-6A
FUSE
3-250 KVA
WYE/WYE
PT-6B
#4 ACSR
PT-6C
#4 ACSR
PT-8
65T
3Ø = 322
SLG = 271
PT-7
5
115 kV SYSTEM
MOSCOW
System Overview –
A-172
XFMR 12/16/20 MVA
115/13.8kV
Each device has at least one curve plotted with current and time values
on the Time Coordination Curve.
HIGH LEAD LOW
DELTA/WYE
13.8 kV BUS
1000
10
2
3
4
5
7
100
2
5
3
700
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
4
500
500
400
400
6
300
200
300
A. Conductor damage curve. k=0.06710 A=556000.0 cmils
Conductor AAC
200
FEEDER 252 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
70
70
50
50
40
40
30
30
TRANSFORMER PROTECTION
HI-SIDE CT'S
20
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
Ia= 679.8A (5.7 sec A) T= 0.78s H=8.33
S
E
C
O
N
D
S
500A, 13.8 kV
FDR #515
1000
#4 ACSR
556 ACSR
PT-2A
#2 ACSR
PT-3B
2
STATION VCB 252 PROTECTION
5
PT-3C
PT-4
3Ø = 1210
SLG = 877
3Ø = 3453
SLG = 2762
LINE RECLOSER P584
PT-5
5
FUSE
4
1
3
3
3. IDR 252 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
3Io= 0.0A (0.0 sec A) T=9999s
2
B
5f
#4 ACSR
2
MAXIMUM FEEDER FUSE
4. 252 140T FUSE stn S&C Link140T
Total clear.
Ia= 5665.9A T= 0.09s
MIDLINE OCR
1
.7
.5
1
FAULT DESCRIPTION:
Bus Fault on:
0 IDR 252
13.8 kV 3LG
.7
.5
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
.4
.3
.2
1 PHASE
#4 ACSR
2/0 ACSR
7
2. IDR 252 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
Ia= 5665.9A (35.4 sec A) T= 0.30s
4
500 A FDR
#2 ACSR
PT-3A
10
7
PT-2B
13.8 kV FDR
#512
3Ø = 5158
SLG = 5346
556 ACSR 3Ø = 3699
SLG = 3060
20
10
PT-1
1 PHASE
#4 ACSR
PT-6A
#4 ACSR
PT-6C
.3
.2
6. T FUSE S&C Link 50T
Minimum melt.
Ia= 5665.9A T= 0.01s
.1
.1 A
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
3-250 KVA
WYE/WYE
PT-6B
.4
MAXIMUM MIDLINE FUSE
3Ø = 558
SLG = 463
3Ø = 1907
SLG = 1492
#4 ACSR
PT-8
65T
3Ø = 322
SLG = 271
PT-7
Fault I=5665.9 A
.01
10
2
HORS 2010
For
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
JDH
No.
Date 11-25-2008
7
.01
6
System Overview –
1000
10
2
3
4
5
7
100
2
5
3
700
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
500
400
400
6
300
200
300
A. Conductor damage curve. k=0.06710 A=556000.0 cmils
Conductor AAC
200
FEEDER 252 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
70
70
50
50
40
40
30
30
20
TRANSFORMER PROTECTION
HI-SIDE CT'S
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
Ia= 679.8A (5.7 sec A) T= 0.78s H=8.33
20
10
10
7
2
STATION VCB 252 PROTECTION
5
Protective Curves:
- relay
- fuse
5
4
1
3
3
3. IDR 252 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
3Io= 0.0A (0.0 sec A) T=9999s
2
Damage Curves:
- transformer
- conductor
7
2. IDR 252 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
Ia= 5665.9A (35.4 sec A) T= 0.30s
4
What are the types of curves?
700
4
500
S
E
C
O
N
D
S
1000
B
5f
2
So, what curve goes where?
MAXIMUM FEEDER FUSE
4. 252 140T FUSE stn S&C Link140T
Total clear.
Ia= 5665.9A T= 0.09s
MIDLINE OCR
1
.7
.5
13.8 kV 3LG
.7
.5
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
.4
.3
.2
1
FAULT DESCRIPTION:
Bus Fault on:
0 IDR 252
.4
Damage curves are at the top and to
the right of the TCC.
.3
MAXIMUM MIDLINE FUSE
.2
6. T FUSE S&C Link 50T
Minimum melt.
Ia= 5665.9A T= 0.01s
.1
.1 A
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
Protective curves lowest and to the
left on the TCC correspond to those
devices farther from the substation
where the fault current is less.
Fault I=5665.9 A
.01
10
2
HORS 2010
For
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
JDH
No.
Date 11-25-2008
7
.01
7
Transformer Protection - Damage Curve – ANSI/IEEE C57.109-1985
Avista has Category III size (5-30MVA) Distribution Transformers in service per the above standard.
The main damage curve line shows only the thermal effect
from transformer through-fault currents. It is graphed
from data entered below (MVA, Base Amps, %Z ):
1000
10
2
3
4
5
7
100
2
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
1000
700
700
500
500
400
400
300
300
200
200
A. T ransf. damage c urve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
100
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.
Time at 50% of the maximum per-unit
through fault current = 8 seconds.
100
70
70
50
50
40
40
30
30
20
20
S 10
E
C
O 7
N
D 5
S
4
10
7
5
4
3
3
A
2
1
1
.7
.7
.5
.5
.4
.4
.3
.3
.2
.2
.1
.1
.07
.07
.05
Dog leg curve - 10 times base current at 2 seconds.
.05
1
.04
.04
.03
.03
.02
.02
.01
Main curve - 25 times base current at 2 seconds.
2
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
T IME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
No.
Comment
Date
.01
1. M15
CT R=
By
For
7
8
Conductor Damage Curves
Copper Conductor Damage Curves
(2/0 damage at 1500A @ 100sec.)
10
2
3 4 5
7
100
2
3 4 5
7
1000
2
3 4 5
7
10000
ACSR Conductor Damage Curves
(2/0 damage at 900A @ 100sec.)
2
3 4 5
7
10
2
3 4 5
7
100
2
3 4 5
7
1000
2
3 4 5
7
10000
2
3 4 5
7
1000
1000
1000
700
700
500
400
300
500
400
300
200
200
200
200
100
100
100
100
70
70
70
70
50
40
30
50
40
30
50
40
30
50
40
30
20
20
20
20
S 10
E
C 7
O 5
N 4
D
3
S
10
S 10
E
C 7
O 5
N 4
D
3
S
1000
1. M15-515 Phase INST INST TD=1.000
CTR=160 Pickup=7.A No inst. TP@5=0.048s
700
500
400
300
7
A. Conductor damage curve. k=0.14040 A=105500.0 cmils
Conductor Copper (bare) AWG Size 2/0
2/0 Copper
5
4
3
B. Conductor damage curve. k=0.14040 A=83690.0 cmils
Conductor Copper (bare) AWG Size 1/0
1/0 Copper
2
1
2
1
C. Conductor damage curve. k=0.14040 A=52630.0 cmils
Conductor Copper (bare) AWG Size 2
#2 Copper
.7
.5
.4
.3
.7
.5
.4
.3
D. Conductor damage curve. k=0.14040 A=33100.0 cmils
Conductor Copper (bare) AWG Size 4
#4 Copper
.2
.2
E. Conductor damage curve. k=0.14040 A=20820.0 cmils
.1 Conductor Copper (bare) AWG Size 6
#6 Copper
.07
.05
.04
.03
E
D
C
.1
B A
.07
.05
.04
.03
1
.02
.02
10
2
3 4 5
7
100
2
3 4 5 7 1000
CURRENT (A)
2
3 4 5
7
10000
2
3 4 5
7
.01
Comment
Copper Conductor Damage Curves
By
DLH
No.
Date 12/13/05
500
400
300
10
7
5
4
3
B. Conductor damage curve. k=0.08620 A=167800.0 cmils
Conductor ACSR AWG Size 4/0
4/0 ACSR
2
C. Conductor damage curve. k=0.08620 A=105500.0 cmils
Conductor ACSR AWG Size 2/0
2/0 ACSR
1
.7
1
.7
D. Conductor damage curve. k=0.08620 A=83690.0 cmils
Conductor ACSR AWG Size 1/0
1/0 ACSR
.5
.4
.3
.5
.4
.3
E. Conductor damage curve. k=0.08620 A=52630.0 cmils
Conductor ACSR AWG Size 2
#2 ACSR
.2
.2
F. Conductor damage curve. k=0.08620 A=33100.0 cmils
Conductor ACSR AWG Size 4
#4 ACSR
.1
.07
.05
.04
.03
F
E
DC
A.1
B
.07
.05
.04
.03
1
.02
.02
10
2
3 4 5
7
100
2
3 4 5 7 1000
CURRENT (A)
2
3 4 5
7
TIME-CURRENT CURVES @Voltage 13.8 kV
TIME-CURRENT CURVES @Voltage 13.8 kV
For
700
A. Conductor damage curve. k=0.08620 A=355107.0 cmils
Conductor ACSR
336.4 ACSR
2
.01
.01
1. M15-515 Phase INST INST TD=1.000
CTR=160 Pickup=7.A No inst. TP@5=0.048s
For
Comment
ACSR Conductor Damage Curves
10000
2
3 4 5
By
7
.01
DLH
No.
Date 12/13/05
9
Conductor Ampacities
Conductor at 25°C ambient taken from the Westinghouse Transmission & Distribution book.
Copper
Ampacity Ratings
ACSR
Ampacity Ratings
Conductor
556
336.4
4/0
2/0
1/0
#2
#4
Rating
730
530
340
270
230
180
140
Conductor
Rating
2/0
1/0
#2
#4
#6
360
310
230
170
120
10
Conductor Protection Graph
1000
10
2
3
4
5
7
100
2
700
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
1
500
500
400
400
300
300
200
200
100
100
1. Moscow 515 Kear 140T Kearney 140T
Total clear.
70
70
50
50
40
30
20
20
10
10
7
7
5
5
4
4
3
3
2
2
1
1
.7
.7
.5
.5
.4
.4
.3
.3
.2
.2
A
.1
.1
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
#4 ACSR & 140T
For
Comment
Aspen File: HORS M15 EXP.olr
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
3
@ Voltage 13.8 kV
4
5
7
10000
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.
40
A. Conductor damage curve. k=0.08620 A=33100.0 cmils
Conductor ACSR AWG Size 4
30
S
E
C
O
N
D
S
1000
2
3
By
4
5
Protection
No.
Date 3/06
7
.01
11
Transformer Protection using
115 kV Fuses
12
1000
Transformer Protection using
115 kV Fuses
10
2
3
4
5
7
100
2
3
4
700
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
2
500
500
400
400
300
300
6
A. Transf. damage curve. 7.50 MVA. Category 3
Base I=313.78 A. Z= 7.3 percent.
Rockford
13.8kV - ROCKFORD115 115.kV 1 T
200
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
100
70
70
50
5
40
40
30
30
STATION TRANSFORMER PROTECTION
20
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.
200
100
50
S
E
C
O
N
D
S
1000
4
20
2. SMD-2B 65E VERY SLOW 176-19-065
Minimum melt.
H=8.33
FEEDER VCR PROTECTION
10
10
4. LAT RP4211 51P CO-11 CO-11 TD=2.000
CTR=500/5 Pickup=3.A No inst. TP@5=0.5043s
7
5
7
5
5. LAT RP4211 51N CO-11 CO-11 TD=4.000
CTR=500/5 Pickup=2.A No inst. TP@5=1.0192s
4
3
3. LAT RP4211 50P CO-11 INST TD=1.000
CTR=500/5 Pickup=3.5A No inst. TP@5=0.048s
2
1. LAT RP4211 50N CO-11 INST TD=1.000
CTR=500/5 Pickup=2.A No inst. TP@5=0.048s
4
3
A
2
MAXIMUM FEEDER FUSE
1
1
6. LAT RP4211 FUSE S&C Link 65T
Total clear.
.7
.7
.5
.5
.4
.4
.3
.3
.2
.2
.1
.1
.07
.07
.05
1
.05
3
.04
.04
.03
.03
.02
.02
.01
10
2
ROK 451R2
For
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
ROK FDR 451, BUS FAULTS: 3LG=3580A, SLG=3782A, L-L=3100A
Comment ASPEN FILE: ROK NEW XFMR (2005 BASE).OLR
2
3
@ Voltage 13.8 KV
4
5
7
10000
2
3
By
4
5
JDH
No.
Date 2-13-07
13
7
.01
Transformer Protection using 115 kV Fuses - continued
• The fuse time current characteristic (TCC) is fixed (although you can buy a standard, slow
or very slow speed ratio which are different inverse curves).
• The sensitivity to detect lo-side SLG faults isn’t as good as using a relay on a circuit
switcher or breaker. This is because we use DELTA/WYE connected transformers so the
phase current on the 115 kV is reduced by the √3 as opposed to a three phase fault.
• Some fuses can be damaged and then blow later at some high load point.
• When only one 115 kV fuse blows, it subjects the customer to low distribution voltages.
For example the phase to neutral distribution voltages on two phases on the 13.8 kV become
50% of normal.
A
a
1.0 PU
B
0.5 PU
0.5 PU
C
b
c
• No indication of faulted zone (transformer, bus or feeder).
14
Transformer Protection using a Circuit Switcher
Showing Avista’s present standard using Microprocessor relays.
13.8kV
115kV
15
Transformer Protection using a Circuit Switcher
Showing Avista’s old standard using Electromechanical relays.
16
Transformer Protection using a Circuit Switcher
1000
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.
10
2
3
4
5
7
100
2
3
700
4
5
7
1000
3
4
5
7
10000
2
3
4
5
7
500
500
400
400
200
300
A. Conductor damage curve. k=0.08620 A=133100.0 cmils
Conductor ACSR AWG Size 2/0
200
FEEDER 251 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
300
70
70
50
50
40
40
30
20
S
E
C
O
N
D
S
1000
700
11
10
7
5
4
TRANSFORMER PROTECTION
HI-SIDE CT'S
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
H=8.33
2. IDR A-777 51G 351 SEL-VI TD=1.000
CTR=600/5 Pickup=1.A No inst. TP@5=0.258s
H=8.33
4. IDR A-777 51Q 351 SEL-EI TD=4.200
CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s
H=8.33
LO-SIDE CT'S
3. IDR A-777 51N 351 SEL-EI TD=5.700
CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s
30
3
5
4
9
20
10
10
7
7
5
4
1
2
3
Some disadvantages would be:
•
Higher cost.
•
Higher maintenance.
•
Requires a substation battery, panel
house and relaying.
•
Transformer requires CT’s.
2
3
5. IDR A-777 51Q 587 W2 SEL-EI TD=4.300
CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s
2
B
2
STATION VCB 251 PROTECTION
1
1
7. IDR 251 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
.7
.7
6. IDR 251 50P 351S INST TD=1.000
CTR=800/5 Pickup=7.A No inst. TP@5=0.048s
.5
.5
.4
.4
9. IDR 251 51G 351S SEL-EI TD=3.900
CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s
.3
.2
.3
.2
8. IDR 251 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
10. IDR 251 51Q 351S SEL-EI TD=3.500
CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s
.1
A
.1
.07
.07
.05
.04
MAXIMUM FEEDER FUSE
MINIMUM FAULT TO DETECT:
3LG=2460A, SLG=1833A, L-L=2130A
8
.05
6
.04
11. 251 140T FUSE stn S&C Link140T
Total clear.
.03
.03
.02
.02
.01
10
2
3
IDR FEEDER 251
For
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 251 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
7
.01
JDH
No.
Date 11-25-2008
17
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.
18
Relay Overcurrent Curves
1000
10
2
3
4
5
7
100
2
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
500
500
400
400
300
300
200
200
100
100
1
70
50
70
50
1. EXTREMELY INVERSE SEL-EI TD=15.000
CTR=800/5 Pickup=6.A No inst. TP@5=4.0718s
40
40
30
20
5
5. VERY INVERSE SEL-VI TD=6.000
CTR=800/5 Pickup=6.A No inst. TP@5=1.5478s
2
10
10
7
7
5
5
2. INVERSE SEL-I TD=2.000
CTR=800/5 Pickup=6.A No inst. TP@5=0.8558s
4
The five curves shown here have the same
pickup settings , but different time dial
settings.
4
3
3
2
2
3
3. MODERATELY INVERSE SEL-2xx-MI TD=1.000
CTR=800/5 Pickup=6.A No inst. TP@5=0.324s
1
1
These are basically the same as various E-M
relays.
.7
.7
.5
.5
4
.4
.4
.3
.3
.2
.2
4. SHORT TIME INVERSE SEL-STI TD=1.000
CTR=800/5 Pickup=6.A No inst. TP@5=0.1072s
.1
.1
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
.01
SEL Various Relay Overcurrent Curves.
Extremely Inverse – steepest
Very Inverse
Inverse
Moderately Inverse
Short Time Inverse – least steep
30
20
S
E
C
O
N
D
S
1000
700
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
7
Avista uses mostly extremely inverse on
feeders to match the fuse curves.
.01
By
For
No.
Comment
Date
19
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
1000
10
2
3
4
5
7
100
2
3
700
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.
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
500
500
400
400
300
200
300
A. Conductor damage curve. k=0.08620 A=133100.0 cmils
Conductor ACSR AWG Size 2/0
200
FEEDER 251 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
70
70
50
50
40
30
20
S
E
C
O
N
D
S
1000
700
11
10
7
5
4
40
TRANSFORMER PROTECTION
HI-SIDE CT'S
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
H=8.33
2. IDR A-777 51G 351 SEL-VI TD=1.000
CTR=600/5 Pickup=1.A No inst. TP@5=0.258s
H=8.33
4. IDR A-777 51Q 351 SEL-EI TD=4.200
CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s
H=8.33
LO-SIDE CT'S
3. IDR A-777 51N 351 SEL-EI TD=5.700
CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s
30
3
5
4
9
20
10
10
7
7
5
4
1
2
3
3
5. IDR A-777 51Q 587 W2 SEL-EI TD=4.300
CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s
2
B
2
STATION VCB 251 PROTECTION
1
1
7. IDR 251 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
.7
.7
6. IDR 251 50P 351S INST TD=1.000
CTR=800/5 Pickup=7.A No inst. TP@5=0.048s
.5
.5
.4
.4
9. IDR 251 51G 351S SEL-EI TD=3.900
CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s
.3
.2
.3
.2
8. IDR 251 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
10. IDR 251 51Q 351S SEL-EI TD=3.500
CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s
.1
A
.1
.07
.05
.04
.07
MAXIMUM FEEDER FUSE
MINIMUM FAULT TO DETECT:
3LG=2460A, SLG=1833A, L-L=2130A
8
.05
6
.04
11. 251 140T FUSE stn S&C Link140T
Total clear.
.03
.03
.02
.01
.02
10
2
3
IDR FEEDER 251
For
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 251 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
7
.01
JDH
No.
Date 11-25-2008
20
Relay Overcurrent Curves - Example of Different Pickup & Time Dial Settings
Same Time Dial = 9
Right curve picks up at 960 Amps
Left curve picks up at 320Amps
1000
10
2
3
4
5
7
100
2
3
4
5
7
1000
2
3
4
5
7
10000
2
Same Pickup = 960 Amps
Top curve Time Dial = 15
Bottom curve Time Dial = 2
3
4
5
7
10
2
3
4
5
100
2
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
1000
700
700
700
500
500
500
500
400
400
400
400
300
300
300
300
200
200
200
200
100
100
70
70
50
50
40
40
30
30
20
20
1. EXT INV 1 SEL-EI TD=9.000
CTR=800/5 Pickup=6.A No inst. TP@5=2.4431s
100
70
2. ~EXT INV 1 SEL-EI TD=9.000
CTR=800/5 Pickup=5.A No inst. TP@5=2.4431s
50
5
40
30
4
3
2 1
3. ~EXT INV 2 SEL-EI TD=9.000
CTR=800/5 Pickup=4.A No inst. TP@5=2.4431s
20
1. EXT INV 1 SEL-EI TD=15.000
CTR=800/5 Pickup=6.A No inst. TP@5=4.0718s
10
7
5. ~EXT INV 4 SEL-EI TD=9.000
CTR=800/5 Pickup=2.A No inst. TP@5=2.4431s
5
70
2
2. ~EXT INV 1 SEL-EI TD=12.000
CTR=800/5 Pickup=6.A No inst. TP@5=3.2574s
50
3
40
30
4
3. ~EXT INV 2 SEL-EI TD=9.000
CTR=800/5 Pickup=6.A No inst. TP@5=2.4431s
20
4. ~EXT INV 3 SEL-EI TD=6.000
CTR=800/5 Pickup=6.A No inst. TP@5=1.6287s
10
7
1000
100
1
4. ~EXT INV 3 SEL-EI TD=9.000
CTR=800/5 Pickup=3.A No inst. TP@5=2.4431s
S
E
C
O
N
D
S
7
1000
700
5
S
E
C
O
N
D
S
10
10
5
7
7
5. ~EXT INV 4 SEL-EI TD=2.000
CTR=800/5 Pickup=6.A No inst. TP@5=0.5429s
5
5
4
4
4
4
3
3
3
3
2
2
2
2
1
1
1
1
.7
.7
.7
.7
.5
.5
.5
.5
.4
.4
.4
.4
.3
.3
.3
.3
.2
.2
.2
.2
.1
.1
.1
.1
.07
.07
.07
.07
.05
.05
.05
.05
.04
.04
.04
.04
.03
.03
.03
.03
.02
.02
.02
.02
.01
.01
.01
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
7
10
By
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
7
.01
By
For
No.
For
No.
Comment
Date
Comment
Date
21
1000
10
2
3
4
5
7
100
2
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
1000
700
A. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.4 percent.
INDIAN TRAIL 12/16/20 MVA XFMR
500
400
500
400
300
300
B. Conductor damage curve. k=0.06710 A=556000.0 cmils
Conductor AAC
556 kCMil AAC
200
200
100
100
70
70
50
50
TRANSFORMER PROTECTION
40
40
1
1. INT A-742 51P 351 SEL-VI TD=10.000
CTR=600/5 Pickup=0.8A No inst. TP@5=2.5797s
H=8.33
30
30
20
20
Transformer relay curves
5
5. INT A-742 51G 351 SEL-VI TD=5.000
CTR=600/5 Pickup=0.8A No inst. TP@5=1.2898s
H=8.33
S 10
E
C 7
O
N
D 5
S
4
10
6
7
6. INT A-742 51Q 351 SEL-VI TD=3.000
CTR=600/5 Pickup=0.8A No inst. TP@5=0.7739s
H=8.33
2
51P – phase time overcurrent
51N or 51G – ground time overcurrent
5
4
3
2
Microprocessor relays have different
types of curves based on the type of
fault current being measured:
3
3
STATION VCB 12F1 PROTECTION
2. INT 12F1 51P 351S SEL-VI TD=1.800
CTR=800/5 Pickup=3.A No inst. TP@5=0.4643s
A
51Q – negative sequence time overcurrent
2
4
1
1
.7
.7
3. INT 12F1 51G 351S SEL-VI TD=0.900
CTR=800/5 Pickup=3.A No inst. TP@5=0.2322s
.5
.5
.4
.4
.3
.3
4. INT 12F1 51Q 351S SEL-VI TD=0.500
CTR=800/5 Pickup=3.A No inst. TP@5=0.129s
.2
Feeder relay curves
51P – phase time overcurrent
.2
51N or 51G – ground time overcurrent
.1
.1 B
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
.01
10
INT 12F1
For
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Indian Trail feeder 12F1 in INT 2007base.olr
Comment AT SUB: 3LG= 5719A, LL=4951A, SLG= 5880A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
7
51Q – negative sequence time overcurrent
This brings us to a brief discussion of:
.01
JDH
No.
Date 1-29-08
22
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.
23
Symmetrical Components
Examples of three 13.8kV faults showing the current distribution through a
Delta-Wye high-lead-low transformer bank :
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
Phasor diagram from Fault-study Software.
8.33= 115/13.8 = transformer Voltage (turns)
ratio
24
Symmetrical Components – Positive Sequence, 3LG 13.8 kV Fault
You have only positive sequence voltage and current since the system is balanced.
A
a
R
R
B
b
C
c
IA = 619 ∠-88°
Ia = 5158 ∠-118°
IB = 619 ∠ 152°
Ib = 5158 ∠ 122°
IC = 619 ∠ 32°
Ic = 5158 ∠ 2°
Phase current = Sequence current
That is; Ia = I+.
Phase currents and voltages for the 115kV side.
IA = Ia / 8.33 = 5158A / 8.33
IA = 619A
25
Symmetrical Components – Negative Sequence, L-L 13.8 kV Fault
A
a
R
R
B
b
C
c
IA = 309 ∠-28°
IB = 619 ∠ 152°
IC = 309 ∠-28°
Ia = 0 ∠ 0°
Ib = 4467 ∠ 152°
Ic = 4467 ∠-28°
115kV side sequence currents and voltages
3LG 13.8kV fault = 5158A,
Ib=Ic=4467 A, 4467/5158 = 86.6%=√3/2
IB3LG = 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,
13.8kV side sequence currents and voltages
26
Symmetrical Components – Zero Sequence, 1LG 13.8 kV Fault
A
a
R
R
B
b
C
c
IA = 370 ∠-118°
Ia=3I0=5346 ∠-118°
IB = 0 ∠ 0°
Ib = 0 ∠ 0°
IC = 370 ∠ 62°
Ic = 0 ∠ 0°
Ia = 3Va/(Z1 + Z2 + Z0)
115kV side sequence currents and voltages
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
27
Symmetrical Components - Summary of 13.8 kV Faults
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:
3LG – positive sequence current
IA = 619 Ia = 5158
IB = 619 Ib = 5158
IC = 619 Ic = 5158
5158 / 8.33 = 619
L-L – negative sequence current
IA = 309 Ia = 0
5158 x √3/2 = 4467
IB = 619 Ib = 4467 4467/(8.33x √3) = 309
IC = 309 Ic = 4467 309 x 2 = 619
Ia = 5158 = I1
Ib x √3 = 7737 = 3I2
SLG – zero sequence current
IA = 370 Ia=5346
IB = 0 Ib = 0
IC = 370 Ic = 0
5346/(8.33x√3) = 370
Ia = 5346 = 3I0
28
Symmetrical Components - Summary of 13.8 kV Faults
You’ve heard of a Line-to-Line fault, how about an “Antler-to-Antler” fault?
29
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
.
30
1000
Electromechanical Relays
used on Distribution Feeders
10
2
3
4
5
7
100
2
3
4
5
7
700
1000
2
3
4
5
7
10000
2
3
4
5
7
700
1
500
500
400
400
300
300
6
200
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·
200
100
100
A. Transf. damage curve. 7.50 MVA. Category 3
Base I=313.78 A. Z= 6.9 percent.
Latah Jct
13.8kV - LATAHJCT115 115.kV T
70
70
50
50
40
40
B. Conductor damage curve. k=0.08620 A=105500.0 cmils
Conductor ACSR AWG Size 1/0
30
30
5
TRANSFORMER PROTECTION
20
20
1. SMD-2B 65E VERY SLOW 176-19-065
Minimum melt.
H=8.33
S
E
C
O
N
D
S
1000
10
3
10
FEEDER VCB PROTECTION
3. LAT421 51P CO-9 CO-9 TD=1.900
CTR=600/5 Pickup=4.A No inst. TP@5=0.4618s
7
5
7
5
5. LAT421 51N CO-9 CO-9 TD=3.000
CTR=600/5 Pickup=3.A No inst. TP@5=0.7228s
4
3
4
3
2. LAT421 50P CO-9 INST TD=1.000
CTR=600/5 Pickup=4.6A No inst. TP@5=0.048s
A
2
2
4. LAT421 50N CO-9 INST TD=1.000
CTR=600/5 Pickup=3.A No inst. TP@5=0.048s
1
1
MAXIMUM FEEDER FUSE
.7
.7
6. LAT421 FUSE S&C Link100T
Total clear.
.5
.5
.4
.4
.3
.3
.2
.2
B
.1
.1
.07
.07
.05
4
.05
2
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
3
@ Voltage 13.8 KV
4
5
7
10000
2
3
4
5
By
For
No.
Comment
Date
31
7
.01
Electromechanical Relays
used on Distribution Feeders
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.
FEEDER SETTINGS
51P = 2000 / 2 = 1000 A
51N = 1000 / 2 = 500 A
13 kV BUS
MIN FAULT
3LG = 2000 A
1LG = 1000 A
32
Electromechanical Relays
used on Distribution Feeders
FEEDER SETTINGS
51P = 2000 / 2 = 1000 A
51N = 1000 / 2 = 500 A
MIDLINE SETTINGS
51P = 1000 / 2 = 500 A
51N = 500 / 2 = 250 A
13 kV BUS
FAULT at MIDLINE
3LG = 2000 A
1LG = 1000 A
MIN FAULT at END OF LINE
3LG = 1000 A
1LG = 500 A
33
Electromechanical Relays
used on Distribution Feeders
500A FEEDER SETTINGS
51P = 960 A
51N = 480 A
13 kV BUS
SECTION
LOAD = 500 A
34
Electromechanical Relays
used on Distribution Feeders
500A FEEDER SETTINGS
51P = 960 A
51N = 480 A
N.O.
13 kV BUS
SECTION
LOAD = 500 A
SECTION
LOAD = 250 A
SECTION
LOAD = 250 A
35
Electromechanical Relays
used on Distribution Feeders
Reclosing
•
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.
36
Electromechanical Relays
used on Distribution Feeders
Reclosing – Sequence shown for a permanent fault
13 kV BUS
RECLOSING SEQUENCE
Closed
Open
0.5"
12"
LOCKOUT
RESET = 120"
(INST Blocked during Reset Time)
37
Distribution Fusing – Fuse Protection/Saving Scheme
1000
10
2
3
4
5
7
100
2
3
700
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.
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
500
500
400
400
300
300
200
200
100
100
70
70
50
S
E
C
O
N
D
S
1000
700
2
50
1
40
40
30
30
20
20
10
10
7
7
5
5
4
4
3
3
2
2
STATION FEEDER RELAYING
1
1
1. M15 515 GND TIME CO-11 TD=4.000
CTR=160 Pickup=3.A No inst. TP@5=1.0192s
.7
.7
3. M15 50N CO-11 INST TD=1.000
CTR=160 Pickup=3.A No inst. TP@5=0.048s
.5
.5
.4
.4
MAXIMUM FEEDER FUSE
.3
.3
2. Moscow 515 Kear 140T Kearney 140T
Total clear.
.2
.2
.1
.1
.07
.07
.05
.05
3
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
M15 515 51N to 140T
For
Comment
Aspen File: HORS M15 EXP.olr
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
Protection
No.
Date 3/06
38
7
.01
Distribution Fusing – Fuse Protection for Temporary Faults
1000
10
2
3
4
5
7
100
2
700
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
1
500
500
400
400
2
300
S
E
C
O
N
D
S
1000
3
300
200
200
100
100
70
70
50
50
40
40
30
30
20
20
1. BR1 S&C 140T S&C Link140T
Minimum melt.
10
10
7
7
5
5
3. M15 515 S&C 80T S&C Link 80T
Minimum melt.
4
4
3
3
2
2
2. ENDTAP S&C 40T S&C Link 40T
Minimum melt.
1
1
.7
.7
.5
.5
.4
.4
.3
.3
.2
.2
.1
.1
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
7
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 amps
10T – 225 amps
12T – 300 amps
15T – 390 amps
20T – 500 amps
25T – 640 amps
30T – 800 amps
40T – 1040 amps
50T – 1300 amps
65T – 1650 amps
80T – 2050 amps
100T – 2650 amps
140T – 3500 amps
200T – 5500 amps
.01
By
For
No.
Comment
Date
39
Distribution Fusing – Fuse to Fuse Coordination
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.
1000
10
2
3
4
5
7
100
2
700
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
500
500
400
400
300
300
1
200
200
100
100
1. T FUSE S&C Link100T
Minimum melt.
70
70
2. 251 140T FUSE stn S&C Link140T
Total clear.
50
S
E
C
O
N
D
S
1000
700
2
50
40
40
30
30
20
20
10
10
7
7
5
5
4
4
3
3
2
2
1
1
.7
.7
.5
.5
.4
.4
.3
.3
.2
.2
.1
.1
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
7
.01
By
For
No.
Comment
Date
40
Distribution Fusing – S&C T-Fuse Current Ratings
Typical continuous and 8 hour emergency rating of the S&C T rated silver fuse links plus the 140T
and 200T.
1000
10
2
3
4
5
7
100
2
700
3
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
500
500
400
400
300
300
200
200
100
100
70
70
50
50
40
40
30
30
20
20
1. BR1 S&C 140T S&C Link140T
Minimum melt.
S 10
E
C
O 7
N
D 5
S
4
General Rule: Fuse Blows at 2X Rating in 5 Minutes
1000
700
1
10
7
5
4
3
3
2
2
1
1
.7
.7
.5
.5
.4
.4
.3
.3
.2
.2
.1
.1
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
T IME-CURRENT CURVES
2
@ Voltage
3
4
5
7
10000
2
3
4
5
7
By
For
No.
Comment
Date
41
.01
1000
Coordinating Time Intervals
10
2
3
4
5
7
100
2
5
3
700
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
4
500
500
400
400
6
200
300
A. Conductor damage curve. k=0.06710 A=556000.0 cmils
Conductor AAC
200
FEEDER 252 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
300
70
70
50
50
40
40
30
20
Typical Coordinating Time Intervals (CTI)
that Avista generally uses between protective
devices.
Other utilities may use different times.
S
E
C
O
N
D
S
1000
30
TRANSFORMER PROTECTION
HI-SIDE CT'S
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
Ia= 679.8A (5.7 sec A) T= 0.78s H=8.33
20
10
10
7
2
STATION VCB 252 PROTECTION
5
7
5
2. IDR 252 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
Ia= 5665.9A (35.4 sec A) T= 0.30s
4
4
1
3
3
3. IDR 252 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
3Io= 0.0A (0.0 sec A) T=9999s
2
B
5f
2
MAXIMUM FEEDER FUSE
4. 252 140T FUSE stn S&C Link140T
Total clear.
Ia= 5665.9A T= 0.09s
MIDLINE OCR
1
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 0.4
Xfmr Fuse Min Melt – Feeder Relay 0.4
.7
.5
13.8 kV 3LG
.7
.5
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
.4
.3
.2
1
FAULT DESCRIPTION:
Bus Fault on:
0 IDR 252
.4
.3
MAXIMUM MIDLINE FUSE
.2
6. T FUSE S&C Link 50T
Minimum melt.
Ia= 5665.9A T= 0.01s
.1
.1 A
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
Fault I=5665.9 A
.01
10
2
HORS 2010
For
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
7
JDH
No.
Date 11-25-2008
42
.01
1000
Coordinating Time Intervals
10
2
3
4
5
7
100
2
5
3
700
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
4
500
500
400
400
6
200
300
A. Conductor damage curve. k=0.06710 A=556000.0 cmils
Conductor AAC
200
FEEDER 252 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
300
70
70
50
50
40
40
30
20
Typical Coordinating Time Intervals (CTI)
that Avista generally uses between protective
devices.
Other utilities may use different times.
S
E
C
O
N
D
S
1000
30
TRANSFORMER PROTECTION
HI-SIDE CT'S
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
Ia= 679.8A (5.7 sec A) T= 0.78s H=8.33
20
10
10
7
2
STATION VCB 252 PROTECTION
5
7
5
2. IDR 252 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
Ia= 5665.9A (35.4 sec A) T= 0.30s
4
4
1
3
3
3. IDR 252 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
3Io= 0.0A (0.0 sec A) T=9999s
2
B
5f
2
MAXIMUM FEEDER FUSE
4. 252 140T FUSE stn S&C Link140T
Total clear.
Ia= 5665.9A T= 0.09s
MIDLINE OCR
1
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 0.4
Xfmr Fuse Min Melt – Feeder Relay 0.4
.7
.5
13.8 kV 3LG
.7
.5
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
.4
.3
.2
1
FAULT DESCRIPTION:
Bus Fault on:
0 IDR 252
.4
.3
MAXIMUM MIDLINE FUSE
.2
6. T FUSE S&C Link 50T
Minimum melt.
Ia= 5665.9A T= 0.01s
.1
.1 A
.07
.07
.05
.05
.04
.04
.03
.03
.02
.02
Fault I=5665.9 A
.01
10
2
HORS 2010
For
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 252 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
7
JDH
No.
Date 11-25-2008
43
.01
1000
10
2
3
4
5
7
100
2
3
4
5
700
Coordinating Time Intervals
7
1000
2
3
4
5
7
10000
2
3
4
5
7
500
400 RATHDRUM FDR 231 PROTECTION
19. RAT 231 51P CO-11 CO-11 TD=1.800
17
300
CTR=800/5 Pickup=6.A No inst. TP@5=0.4533s
500
400
B. Conductor damage curve. k=0.06710 A=556000.0 cmils
Conductor AAC
MIDLINE C390 FED FROM HUE 142 - MINIMUM CONDUCTOR TO PROTECT
18
300
16. RAT 231 50P CO-11 INST TD=1.000
CTR=800/5 Pickup=7.A No inst. TP@5=0.048s
200
200
15. RAT 231 51N CO-11 CO-11 TD=5.000
CTR=800/5 Pickup=3.A No inst. TP@5=1.3192s
100
100
13
14. RAT 231 50N CO-11 INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
70
1000
700
A. Conductor damage curve. k=0.08620 A=133100.0 cmils
Conductor ACSR AWG Size 2/0
MIDLINE C390 FED FROM RAT 231 - MINIMUM CONDUCTOR TO PROTECT
70
15
50
50
40 HUETTER FDR 142 PROTECTION
RECLOSE IS 1SEC, 12 SEC, LO, 180 SEC RESET
20
Typical Coordinating Time Intervals (CTI)
that Avista generally uses between protective
devices.
Other utilities may use different times.
S
E
C
O
N
D
S
1
4
10
8
19
11
20
10. HUE 142 51N 251 SEL-EI TD=4.000
CTR=160 Pickup=3.A No inst. TP@5=1.0858s
10
10
6
9. HUE 142 50N 251 INST TD=1.000
CTR=160 Pickup=3.A No inst. TP@5=0.048s
7
7
5
5
11. HUE 142 51Q 251 SEL-EI TD=4.000
CTR=160 Pickup=5.2A No inst. TP@5=1.0858s
4
4
3
3
HUE 142 MID C270R PROTECTION
CALLED "HUE 142 LINE" IN POWERBASE
2
RECLOSE IS 1SEC, 12SEC, LO, 120 SEC RESET
2
8. HUE ML C270R 51P CO-11 TD=1.500
CTR=400/5 Pickup=7.A No inst. TP@5=0.3766s
7. HUE ML C270R 50P INST TD=1.000
CTR=400/5 Pickup=8.1A No inst. TP@5=0.048s
1
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 0.4
Xfmr Fuse Min Melt – Feeder Relay 0.4
30
5. HUE 142 50P 251 INST TD=1.000
CTR=160 Pickup=7.A No inst. TP@5=0.048s
.7
DEVICES:
40
6. HUE 142 51P 251 SEL-EI TD=1.500
CTR=160 Pickup=6.A No inst. TP@5=0.4072s
30
1
.7
13. HUE ML C270R 51N CO-11 TD=7.000
CTR=400/5 Pickup=3.A No inst. TP@5=1.8052s
.5
.4
.5
.4
12. HUE ML C270R 50N INST TD=1.000
CTR=400/5 Pickup=3.A No inst. TP@5=0.048s
.3
.3
MIDLINE 390R PROTECTION
1. C390R MID 51P CO-9 CO-9 TD=2.500
CTR=300/5 Pickup=5.A No inst. TP@5=0.6037s
.2
.2
2. C390R MID 50P CO-9 INST TD=1.000
CTR=300/5 Pickup=6.A No inst. TP@5=0.048s
.1
A
.1 B
4. C390R MID 51N CO-9 CO-9 TD=2.500
CTR=300/5 Pickup=5.A No inst. TP@5=0.6037s
.07
.05
.07
12 3 2
3. C390R MID 50N CO-9 INST TD=1.000
CTR=300/5 Pickup=5.A No inst. TP@5=0.048s
.04
9
14
7
.05
5
16
.04
.03
.03
LARGEST DOWNSTREAM FUSE FROM 390R MIDLINE
17. HUE 142 FUSE S&C Link 65T
.02
Total clear.
.02
18. RAT 231 FUSE S&C Link100T
Total clear.
.01
10
2
3
RAT 231 MID C390R
For
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
2
TIME-CURRENT CURVES @ Voltage 13.8 kV
Huetter feeder 142 (2008) Rat feeder 231 (2009+) Midline Recloser C390R
Comment Saved in: HUE142-RAT231 MID C390R 2007base.olr
3
4
5
7
10000
2
3
By
4
5
JDH
No.
Date 2-25-08
44
7
.01
1000
10
2
3
4
5
7
100
2
3
700
4
5
7
1000
2
3
4
5
7
10000
2
3
4
5
7
700
11
500
500
Coordinating Time Intervals
200
300
A. Conductor damage curve. k=0.08620 A=133100.0 cmils
Conductor ACSR AWG Size 2/0
200
FEEDER 251 SMALLEST CONDUCTOR TO PROTECT
100
B. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.2 percent.
IDAHO RD 12/16/20 MVA XFMR
100
300
70
70
50
50
40
40
30
20
Typical Coordinating Time Intervals (CTI)
that Avista generally uses between protective
devices.
Other utilities may use different times.
400
400
S
E
C
O
N
D
S
10
7
5
4
TRANSFORMER PROTECTION
HI-SIDE CT'S
1. IDR A-777 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
H=8.33
2. IDR A-777 51G 351 SEL-VI TD=1.000
CTR=600/5 Pickup=1.A No inst. TP@5=0.258s
H=8.33
4. IDR A-777 51Q 351 SEL-EI TD=4.200
CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s
H=8.33
LO-SIDE CT'S
3. IDR A-777 51N 351 SEL-EI TD=5.700
CTR=1200/5 Pickup=3.A No inst. TP@5=1.5473s
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 0.4
Xfmr Fuse Min Melt – Feeder Relay 0.4
30
3
5
4
9
20
10
10
7
7
5
4
1
2
3
3
5. IDR A-777 51Q 587 W2 SEL-EI TD=4.300
CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s
2
B
2
STATION VCB 251 PROTECTION
1
1
7. IDR 251 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
.7
DEVICES:
1000
.7
6. IDR 251 50P 351S INST TD=1.000
CTR=800/5 Pickup=7.A No inst. TP@5=0.048s
.5
.5
.4
.4
9. IDR 251 51G 351S SEL-EI TD=3.900
CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s
.3
.2
.3
.2
8. IDR 251 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
10. IDR 251 51Q 351S SEL-EI TD=3.500
CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s
.1
A
.1
.07
.05
.04
.07
MAXIMUM FEEDER FUSE
MINIMUM FAULT TO DETECT:
3LG=2460A, SLG=1833A, L-L=2130A
8
.05
6
.04
11. 251 140T FUSE stn S&C Link140T
Total clear.
.03
.03
.02
.01
.02
10
2
3
IDR FEEDER 251
For
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
Idaho Rd Feeder 251 in Idaho Rd PHASE 1 2007base.olr
Comment At Sub: 3LG=5667A, SLG=5863A, L-L=4909A
2
3
@ Voltage 13.8 kV
4
5
7
10000
2
3
By
4
5
JDH
No.
Date 11-25-2008
45
7
.01
1000
10
2
3
4
5
7
100
2
3
4
5
7
700
Coordinating Time Intervals
1000
2
3
4
5
7
10000
2
3
4
5
7
700
1
500
500
400
400
300
300
6
200
200
100
100
A. Transf. damage curve. 7.50 MVA. Category 3
Base I=313.78 A. Z= 6.9 percent.
Latah Jct
13.8kV - LATAHJCT115 115.kV T
70
50
40
40
B. Conductor damage curve. k=0.08620 A=105500.0 cmils
Conductor ACSR AWG Size 1/0
30
5
TRANSFORMER PROTECTION
20
20
1. SMD-2B 65E VERY SLOW 176-19-065
Minimum melt.
H=8.33
Typical Coordinating Time Intervals (CTI)
that Avista generally uses between protective
devices.
Other utilities may use different times.
70
50
30
S
E
C
O
N
D
S
10
3
10
FEEDER VCB PROTECTION
3. LAT421 51P CO-9 CO-9 TD=1.900
CTR=600/5 Pickup=4.A No inst. TP@5=0.4618s
7
5
7
5
5. LAT421 51N CO-9 CO-9 TD=3.000
CTR=600/5 Pickup=3.A No inst. TP@5=0.7228s
4
3
4
3
2. LAT421 50P CO-9 INST TD=1.000
CTR=600/5 Pickup=4.6A No inst. TP@5=0.048s
A
2
2
4. LAT421 50N CO-9 INST TD=1.000
CTR=600/5 Pickup=3.A No inst. TP@5=0.048s
1
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 0.4
Xfmr Fuse Min Melt – Feeder Relay 0.4
1000
1
MAXIMUM FEEDER FUSE
.7
.7
6. LAT421 FUSE S&C Link100T
Total clear.
.5
.5
.4
.4
.3
.3
.2
.2
B
.1
.1
.07
.07
.05
4
.05
2
.04
.04
.03
.03
.02
.02
.01
10
2
3
4
5
7
100
2
3
4
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES
2
3
@ Voltage 13.8 KV
4
5
7
10000
2
3
4
5
By
For
No.
Comment
Date
46
7
.01
1000
10
2
3 4 5
7
100
2
1
3 4 5
7
1000
2
3 4 5
7
10000
2
3 4 5
7
Coordinating Time Intervals
500
400
300
1. Moscow515 Kear 140T Kearney140T
Minimummelt.
I= 5158.1A T= 0.10s
500
400
300
200
2. M15-515 Phase Time CO-11 TD=1.500
CTR=160 Pickup=6.A No inst. TP@5=0.3766s
I= 5158.1A (32.2 sec A) T= 0.33s
200
3. M15 A-172 Phase CO- 9 TD=1.500
CTR=120 Pickup=2.A Inst=1200A TP@5=0.3605s
I= 619.0A (5.2 sec A) T= 1.03s H=8.33
70
100
70
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)
1000
700
700
50
40
30
100
50
40
30
20
20
2
S 10
E
C 7
O 5
N 4
D
3
S
10
3
7
5
4
3
2
2
1
1
.7
.7
.5
.4
.3
.5
.4
.3
.2
.2
.1
.1
.07
.07
.05
.04
.03 FAULT DESCRIPTION:
Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L 3LG
.02
.05
.04
.03
.01
10
2
3 4 5
7
100
2
3 4 5 7 1000
CURRENT (A)
.02
2
3 4 5
7
TIME-CURRENT CURVES @Voltage 13.8 kV
For
M15 A-172 CS with fdr 515 with Kearney140T Fuse
Comment Three Phase Fault
10000
2
3 4 5
By
7
DLH
No.
Date 12/12/05
47
.01
1000
10
2
3 4 5
7
100
2
1
3 4 5
7
700
Coordinating Time Intervals
70
----- 0.4 sec. -----Middle - E-M relays (Phase &
Ground) for a 500 Amp Feeder
----- 0.2 sec. -----Bottom – 140T Feeder Fuse (Total
Clear)
3 4 5
7
10000
2
3 4 5
7
50
40
30
2
4
1000
700
500
400
300
2. M15 515 GND TIME CO-11 TD=4.000
CTR=160 Pickup=3.A No inst. TP@5=1.0192s
I= 5346.5A (33.4 sec A) T= 0.29s
3. M15-515 Phase Time CO-11 TD=1.500
CTR=160 Pickup=6.A No inst. TP@5=0.3766s
I= 5346.4A (33.4 sec A) T= 0.31s
100
Top – Ckt Swr w/ E-M Phase relay
2
1. Moscow 515 Kear 140T Kearney140T
Minimummelt.
I=5346.4A T= 0.09s
500
400
300
200
SLG Fault Coordination
Example:
1000
200
100
70
50
40
30
4. M15 A-172 GND TIME CO-11 TD=4.000
CTR=240 Pickup=4.A No inst. TP@5=1.0192s
I= 5346.5A (22.3 sec A) T= 0.83s
20
20
3
10
S 10
E
C 7
O 5
N 4
D
3
S
7
5
4
3
2
2
1
1
.7
.7
.5
.4
.3
.5
.4
.3
.2
.2
.1
.1
.07
.07
.05
.04
.03
.05
.04
.03
.02 FAULT DESCRIPTION:
Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L 1LG Type=A
.02
.01
10
2
3 4 5
7
100
2
3 4 5 7 1000
CURRENT (A)
2
3 4 5
7
TIME-CURRENT CURVES @Voltage 13.8 kV
For
MoscowCS with fdr 515 with 140T fuses
Comment Single Line to Ground Fault
10000
2
3 4 5
By
7
DLH
No.
Date 12/12/05
48
.01
1000
10
2
3 4 5
7
100
2
1
3 4 5
700
Coordinating Time Intervals
1000
2
3 4 5
7
10000
2
3 4 5
7
500
400
300
2. M15-515 Phase Time CO-11 TD=1.500
CTR=160 Pickup=6.A No inst. TP@5=0.3766s
I= 4467.0A (27.9 sec A) T= 0.43s
200
3. M15 A-172 Phase CO- 9 TD=1.500
CTR=120 Pickup=2.A Inst=1200A TP@5=0.3605s
I= 619.0A (5.2 sec A) T= 1.03s H=8.33
100
70
1000
700
1. Moscow515 Kear 140T Kearney140T
Minimummelt.
I= 4467.0A T= 0.12s
500
400
300
200
L-L Fault Coordination
Example:
7
100
70
50
40
30
50
40
30
20
20
2
Top – Ckt Swr w/ 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)
S 10
E
C 7
O 5
N 4
D
3
S
10
3
7
5
4
3
2
2
1
1
.7
.7
.5
.4
.3
.5
.4
.3
.2
.2
.1
.1
.07
.07
.05
.04
.03
.05
.04
.03
FAULT DESCRIPTION:
.02 Close-In Fault on: 0 MoscowCity#2 13.8kV - 0 BUS1 TAP 13.8kV 1L LL Type=B-C
.02
.01
10
2
3 4 5
7
100
2
3 4 5 7 1000
CURRENT (A)
2
3 4 5
7
TIME-CURRENT CURVES @Voltage 13.8 kV
For
MoscowCS A-172 with fdr 515 with 140T Fuse
Comment Line to Line Fault
10000
2
3 4 5
By
DLH
No.
Date
49
7
.01
The Final Product
1000
10
2
3
4
5
7
100
2
700
3
17
4
5
7
1000
2
3
An example of a completed 13.8kV
Feeder Coordination Study with 20
Time-Current Curves representing:
1. INT A-742 51P 351 SEL-VI TD=1.200
CTR=600/5 Pickup=2.A No inst. TP@5=0.3096s
H=8.33
200
2. INT A-742 51G 351 SEL-VI TD=1.000
CTR=600/5 Pickup=1.A No inst. TP@5=0.258s
H=8.33
Two Fuses
40
5
7
B. Conductor damage curve. k=0.14040 A=52630.0 cmils
Conductor Copper (bare) AWG Size 2
NORTH BRANCH SMALLEST TRUNK CONDUCTOR
300
400
D. Conductor damage curve. k=0.08620 A=105500.0 cmils
Conductor ACSR AWG Size 2/0
WEST BRANCH SMALLEST TRUNK CONDUCTOR
100
70
50
40
30
7. INT 12F2 51P 351S SEL-EI TD=1.500
CTR=800/5 Pickup=6.A No inst. TP@5=0.4072s
20
6. INT 12F2 50P 351S INST TD=1.000
CTR=800/5 Pickup=7.A No inst. TP@5=0.048s
30
3
15
9 9&
15
11
11
4 &4
16
10
20
13
10
8. INT 12F2 50G 351S INST TD=1.000
CTR=800/5 Pickup=3.A No inst. TP@5=0.048s
7
10
7
7
10. INT 12F2 51Q 351S SEL-EI TD=3.500
CTR=800/5 Pickup=5.2A No inst. TP@5=0.9501s
5
5
4
4
1
2
12F2 MIDLINE 466R PROTECTION
3
3
13. F2 MID 466R 51P 351R SEL-EI TD=2.300
CTR=500/1 Pickup=1.44A No inst. TP@5=0.6243s
A
2
2
12. F2 MID 466R 50P 351R INST TD=1.000
CTR=500/1 Pickup=1.68A No inst. TP@5=0.048s
1
15. F2 MID 466R 51G 351R SEL-EI TD=4.200
CTR=500/1 Pickup=0.89A No inst. TP@5=1.1401s
1
.7
14. F2 MID 466R 50G 351R INST TD=1.000
CTR=500/1 Pickup=0.89A No inst. TP@5=0.048s
.7
.5
.5
16. F2 MID 466R 51Q 351R SEL-EI TD=3.800
CTR=500/1 Pickup=1.54A No inst. TP@5=1.0315s
.4
.3
.3
.2
.2
EAST BRANCH - FEEDER MAXIMUM FUSE
FAULT DUTY AT 150E; 3LG=2318, LL=2008, SLG=1684
5. INT F2 150E FUSE SMU-20_150E
Total clear.
.1
B
D
C
.1
.07
Three Conductor Damage Curves
1000
500
STATION VCB 12F2 PROTECTION
.4
Transformer Damage Curve
4
200
9. INT 12F2 51G 351S SEL-EI TD=3.900
CTR=800/5 Pickup=3.A No inst. TP@5=1.0587s
S
E
C
O
N
D
S
3
11. INT A-742 51Q 587 W2 SEL-EI TD=4.300
CTR=1200/5 Pickup=5.4A No inst. TP@5=1.1672s
50
Instantaneous & Time-Delay Curves
for the:
- Transformer high side protection,
- Transformer low side protection,
- Station feeder breaker protection,
- Midline feeder breaker protection.
2
A. Transf. damage curve. 12.00 MVA. Category 3
Base I=502.00 A. Z= 8.4 percent.
INDIAN TRAIL 12/16/20 MVA XFMR
3. INT A-742 51N 351 SEL-EI TD=5.600
CTR=1200/5 Pickup=3.A No inst. TP@5=1.5201s
70
10000
C. Conductor damage curve. k=0.04704 A=350000.0 cmils
Cable XLPE 90C/250C
EAST BRANCH SMALLEST TRUNK CONDUCTOR - 350 CN15
4. INT A-742 51Q 351 SEL-EI TD=4.200
CTR=600/5 Pickup=1.3A No inst. TP@5=1.1401s
H=8.33
100
7
700
TRANSFORMER PROTECTION
300
5
5
500
400
4
.07
WEST OR NORTH BRANCH - FEEDER MAXIMUM FUSE
FAULT DUTY AT 140T - WHEN FED FROM WEST MID 466R
3LG=2093, LL=1813, SLG=1520
.05
.04
148
12
.05
6
FAULT DUTY AT 140T - WHEN FED FROM NORTH (N.O. PT)
3LG=1932, LL=1673, SLG=1388
.03
.04
.03
17. INT F2 140T FUSE stn S&C Link140T
Total clear.
.02
.01
.02
10
2
3
4
5
7
100
INT 12F2 & MID 466R
For
2
3
5
7
1000
CURRENT (A)
TIME-CURRENT CURVES @ Voltage 13.8 kV
Indian Trail feeder 12F2 and 12F2 Midline 466R in INT 2007base.olr
Comment AT SUB: 3LG= 5719A, LL=4951A, SLG= 5880A
4
2
3
4
5
7
10000
2
3
By
4
5
7
JDH
No.
Date 1-29-08
50
.01
IEEE Device Designations commonly used in Distribution Protection
Avista sometimes adds letters to these such as F for feeders, T for transformers, B for bus and BF for
breaker failure.
2 – Time delay relay.
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 overcurrent
ground (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.
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.
51
Mike’s Turn
52
Distribution Transformer
Electromechanical Relays
Criteria (for outdoor bus arrangement, not switchgear)
ƒ Protect the Transformer from thermal damage
- Refer to Damage Curves
ƒ Backup feeder protection (as much as possible)
- Sensitivity is limited because load is higher
ƒ Coordinate with downstream devices (feeder relays)
ƒ Carry normal maximum load (phase only)
ƒ Pick up Cold Load after outages
53
Distribution Transformer
Electromechanical Relays - Setting Criteria
Relays:
ƒ 3 High Side Phase Overcurrent (with time and instantaneous elements)
ƒ 1 Low Side Ground Overcurrent (with time and instantaneous elements)
ƒ Sudden Pressure Relay
54
Distribution Transformer
Electromechanical Relays - Setting Criteria
Phase Overcurrent Settings – Current measured on 115kV side
ƒ 51P Pickup (time overcurrent)
ƒ Don’t trip for load or cold load pickup (use 2.4 * highest MVA rating)
ƒ Ends up being higher than the feeder phase element pickup
ƒ Example: 12/16/20 MVA unit would use 2.4*20*5 = 240 amps (1940A low side)
ƒ51P Time Lever (time dial)
ƒ Coordinate with feeder relays for maximum fault (close in feeder fault)
ƒ CTI is 0.4 seconds
ƒ Worst coordination case: Ø-Ø on low side (discrepancy due to delta/wye conn.)
ƒ Multiple feeder load can make the transformer relay operate faster
ƒ50P Pickup (instantaneous overcurrent)
ƒ Must not trip for feeder faults Æ set at 170% of low side bus fault
ƒ Accounts for DC offset
55
Distribution Transformer
Electromechanical Relays - Setting Criteria
Ground Overcurrent Settings – Current measured on 13.8 kV side
ƒ 51N Pickup (time overcurrent)
ƒ Set to same sensitivity as feeder phase relay (in case the feeder ground relay is failed)
ƒ This setting is higher than the feeder ground, so we lose some sensitivity for backup
ƒ51N Time Lever (time dial)
ƒ Coordinate with feeder relays for ground faults
ƒ CTI is 0.4 seconds
ƒ50N Pickup (instantaneous overcurrent)
ƒ DO NOT USE!!!!
56
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Advantages:
ƒ More precise TAP settings
ƒ More Relay Elements
ƒ Programmable Logic / Buttons
ƒ Lower burden to CT
ƒ Event Reports!!!!!!!!!
ƒ Remote Communications
ƒ Coordinate with like elements
(faster)
ƒ More Settings
Microprocessor Relay
57
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Elements We Set:
ƒ 51P
ƒ 50P
ƒ 50P2 (FTB)
ƒ 51G
ƒ 50G (115kV)
ƒ 51N (13.8kV)
ƒ 50N1 (FTB)
FTB = Fast Trip Block (feeder relays must be Microprocessor)
58
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Transformer Phase Overcurrent Settings - Current measured on 115kV Side
ƒ
51P - Phase Time Overcurrent Pickup
ƒ
ƒ
ƒ
ƒ
Set to 240% of nameplate (same as EM relay)
51P Time Dial
ƒ
Set to coordinate with feeder’s fastest element for each fault
ƒ
CTI is still 0.4 seconds
50P1 – Phase instantaneous #1 Pickup
ƒ
Direct Trip
ƒ
Set to 130% of max 13.8 kV fault (vs. 170% with EM relay)
50P2 – Phase instantaneous pickup for Fast Trip Block scheme
ƒ
Set above transformer inrush
ƒ
4 Cycle time delay
ƒ
Blocked if any feeder overcurrent elements are picked up
59
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Transformer Ground Overcurrent Settings - Current calculated from 115kV CTs
ƒ
51G - Ground Time Overcurrent Pickup
ƒ
ƒ
51GTime Dial
ƒ
ƒ
Set very low (will not see low side ground faults due to transformer
connection). Usually 120 Amps.
Set very low
50G1 – Ground Instantaneous #1 Pickup
ƒ
Set very low. Usually 120 Amps.
Symmetrical
Components!
60
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Transformer Neutral Overcurrent Settings - Current measured by neutral CT or
calculated from 13.8kV CTs.
ƒ
51N - Ground Time Overcurrent Pickup
ƒ
ƒ
ƒ
Set slightly higher than feeder ground pickup (about 1.3 times)
51NTime Dial
ƒ
Set to coordinate with feeder ground
ƒ
CTI is 0.4 seconds
50N1 – Ground Instantaneous for Fast Trip Block
ƒ
Set slightly above the feeder ground instantaneous pickup
ƒ
Time Delay by 4 cycles
ƒ
Blocked if Feeder overcurrent elements are picked up
61
Distribution Transformer & Feeder Protection
with Microprocessor Relays
ƒ Inrush – The current seen when
energizing a transformer.
ƒ Need to account for inrush when using
instantaneous elements (regular or FTB).
ƒ 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).
62
Distribution Transformer & Feeder Protection
with Microprocessor Relays
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 about
700 amps.
63
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Microprocessor Feeder Relay
ƒ Overcurrent
ƒ Reclosing
ƒ Fast Trip Block Output
ƒ Breaker Failure Output
Elements We Set:
ƒ 51P
ƒ 50P
ƒ 51G
ƒ 50G
ƒ 51Q
64
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Feeder Phase Overcurrent Settings
ƒ
ƒ
ƒ
51P - Phase Time Overcurrent Pickup
ƒ
Set above load and cold load (960 Amps for 500 Amp feeder)
ƒ
Same as EM pickup
51P Time Dial
ƒ
Same as EM relay (coordinate with downstream protection with CTI)
ƒ
Select a curve
50P – Phase instantaneous Pickup
ƒ
Set the same as the 51P pickup (960 Amps)
65
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Feeder Ground Overcurrent Settings
ƒ
51G - Phase Time Overcurrent Pickup
ƒ
ƒ
ƒ
Same as EM pickup which is 480 Amps
51G Time Dial
ƒ
Same as EM relay (coordinate with downstream protection)
ƒ
Select a curve
50G – Phase instantaneous Pickup
ƒ
Set the same as the 51G pickup
66
Distribution Transformer & Feeder Protection
with Microprocessor Relays
Feeder Negative Sequence Overcurrent Settings
ƒ
ƒ
ƒ
51Q - Negative Sequence Time Overcurrent Pickup
ƒ
Set with equivalent sensitivity as the ground element (not affected by load)
ƒ
Coordinate with downstream protection
51Q Time Dial
ƒ
Same as EM relay (coordinate with downstream protection)
ƒ
Select a curve
50Q – Negative Sequence instantaneous Pickup
ƒ
DON’T USE!!!!!
ƒ
Contributions from motors during external faults could trip this
67
Transformer Differential Protection
•
The differential relay is connected to both the high and low side transformer BCT’s.
•
EM Differential Relay
Since the distribution transformer is connected delta – wye the transformer CT’s
have to be set wye – delta to compensate for the phase shift.
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Transformer Differential Protection – External Fault
PRI I
SEC I
CURRENT FLOW THROUGH AN E/M 87 DIFFERENTIAL
RELAY FOR AN INTERNAL AND EXTERNAL FAULT
CT POLARITY MARK
BCT'S
87
R1
EXTERNAL FAULT
THE SECONDARY CURRENTS FLOW THROUGH BOTH
RESTRAINT COILS IN THE SAME DIRECTION AND THEN
CIRCULATE BACK THROUGH THE CT'S. THEY DO NOT
FLOW THROUGH THE OPERATE COIL
87
OP
87
R2
BCT'S
PRI I
SEC I
69
Transformer Differential Protection – Internal Fault
PRI I
SEC I
CT POLARITY MARK
87
R1
BCT'S
87
OP
87
R2
INTERNAL FAULT
THE SECONDARY CURRENTS FLOW THROUGH BOTH
RESTRAINT COILS IN OPPOSITE DIRECTIONS, ADD
AND THEN FLOW THROUGH THE OPERATE COIL AND
BACK TO THE RESPECTIVE CT'S
BCT'S
SEC I
PRI I
70
HANDOUT
MOSCOW FEEDER 515 PROTECTION EXAMPLE
• Where do we start the protection design?
– Gather Information:
• The feeder rating is 500 Amps.
• The fault duty at the bus is 5,188 for a 3Ø fault and 5,346 for a SLG
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Where do we start (Cont’d)
• What other information do we have/need?
– This feeder already has a line recloser and the Field Engineer
wants to keep the same location. Reason for coordination was
replacing the older recloser with a newer relayed recloser.
– We then have to run feeder fault studies (normally done by a
Distribution Engineer) and look at the feeder configuration so we
know what fusing and relay settings to use.
– We then generally start near the end of the feeder and work back
towards the substation. A lot of times though we have to go back
and forth because our first choice of a fuse size may not work
correctly when we figure out the protection towards the substation
and we have to make compromises.
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How do we proceed?
• At each coordination point
– Temporary Faults - From Table 1
• What can a fuse be protected up to
– Fuse-to-Fuse Coordination - From Table 2
• What will coordinate with the downstream device
– Fuse Loading – From Table 3
• Are we above the fuse rating
– Conductor Protection - From Table 7
• minimum conductor that can be protected by the feeder settings or
fuse
– Fault Detection
• Can we detect the fault by our 2:1 margin
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Begin at the END
• NOTE: The actual Moscow feeder 515 does not look
exactly like this even though the substation and line
recloser settings we will arrive at are the actual settings.
•
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.
•
POINT 7: This is the end of a 3Ø lateral and the fault duty is: 3Ø = 322
amps and SLG = 271 amps
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POINT 6B:
• What we know:
– 3Ø lateral feeding to a 65T at the end
– #4 ACSR exists.
• Coordination:
– Temporary Faults - 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
– Fuse-to-Fuse Coordination – From Table 2, we see that we need an 80T, or larger,
fuse to coordinate with the downstream 65T fuses (at fault duties < 1400A
w/preload)
– 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
end fault with a 2:1 margin. The SLG is 463 so the maximum fuse we could use
would be: 463/2 (2:1 margin) = 231 amps so the largest fuse we could use is 231/2
(blows at twice the rating) = 115 so we could use a 100T or smaller fuse for this
criteria
– Loading – Assume the load at the end is all there is so no need to worry
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Point 6B - So what fuse size should we use?
• 80T
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POINT 5 – Midline Recloser:
•
What we know:
– 3Ø line of #4 ACSR feeding to point 6
– 3LG = 3453 A, 1LG = 2762 A
•
Coordination:
– Temporary Faults - The fault duties at the recloser are: 3Ø = 3,453 amps
and SLG = 2,762. We will assume we have a fused lateral just beyond
the line recloser and we want to protect it for temporary faults. From
Table 1 we see that we can protect a 100T fuse up to 2,650 and a 140T up
to 3,500 amps. Therefore, from this standpoint we could use a 140T fuse
beyond the recloser
– Conductor Protection - From Table 7 we see that #4 ACSR can be protected by a
100T or smaller fuse. That means that we won’t be able to set the line recloser to
coordinate with a 140T and still protect the conductor. Therefore, we have to
compromise and we will sacrifice the fuse protection for temporary faults in order
to gain protection for the conductor. NOTE: This normally isn’t too bad because
most faults are SLG and are somewhat less than the calculated value because of
fault resistance and distance to the fault out on the lateral
– Coordination – We will try and coordinate with a 100T fuse and still protect the #4
ACSR conductor. Once we determine the overcurrent pickup values, we will
choose a Time Lever to provide 0.2 seconds (see Table 6) coordination with the
77
fuse and then will recheck to determine if we are satisfied with the conductor
protection
POINT 5 (Cont’d):
1.
Fault Detection & Loading - The loading information we obtained from the
Distribution Group is 84 amps normal and cold load ≅ 2 times = 168 amps.
Therefore, we can set the phase PU at?
•
•
2.
3.
4.
200 amps which would carry the load and pick it up cold load.
However, #4 ACSR can carry around 140 amps (see Table 4) so we would probably
want to set about 300 amps to allow for load growth. NOTE: We do not set the phase
overcurrent to protect the conductor from normal or emergency loading
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)
SLG Fault Detection – 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. Again this is 300 amps so in this
case the ground PU will be set the same as the phase
Instantaneous Units – The phase instantaneous will be set at 350 A and the
ground instantaneous unit will be set at 300 A. Note that they can’t reach all the
way to point 7 so they can’t protect the point 6C fuse from temporary faults
occurring near the end of the lateral
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POINT 3C:
•
What we know:
–
–
•
The fault duty at point 3 is: 3Ø = 3,699 amps and the SLG is 3,060
#4 ACSR 3Ø trunk feeding to the line recloser with a SLG of 2,762 amps
Coordination:
1.
2.
3.
4.
Temporary Faults - From Table 1 we see that a 140T fuse can be protected up to
3,500 amps for temporary faults and the 3Ø fault is 3,699 amps. This again
appears to indicate we would want to use a 200T fuse
Coordination – We set the line recloser up to coordinate with a 100T fuse so we
wouldn’t be able to coordinate with the recloser by using a 140T so this would
indicate we need a 200T
Conductor Protection - From Table 7 we see that #4 ACSR requires a 100T fuse
or smaller to be protected. OOPS
The above coordination and conductor protection dilemma means that we can’t
solve this problem by fusing this lateral. We also can’t remove the fuse because
the substation setting of 960 A can’t protect the conductor either.
79
Point 3C - Solutions?
• Move Recloser
• Reconductor
80
POINT 1:
•
•
What we know:
–
This is a 500 amp feeder so must set to carry load. NOTE: The load is 325
amps and cold load 650
–
–
The fault duty at point 1 is: 3Ø = 5,158 amps and SLG = 5,346
We want a Fuse Protection Scheme.
Coordination:
1.
2.
3.
Temporary Faults – We have decided that a 140T fuse is the maximum fuse we
will use on the feeder even though it can’t be protected at the maximum fault
duties.
Coordination – We will coordinate with a 140T fuse at the maximum SLG fault
duty of 5,346 amps with 0.2 seconds coordination time per Table 6. We will
also coordinate with the line recloser settings with 0.3 seconds coordination time
again per Table 6
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 so no problem
since all laterals are fused.
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POINT 1 (Cont’d):
4.
Fault Detection & Loading – We will set the phase pickup to carry 500 amps
normally plus pick up cold load so will set at 960 amps.
•
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 so no problem
•
SLG Fault Detection – Again per our criteria we will set as low as possible
and still coordinate with the maximum downstream devices with the same
coordination times as above
•
The SLG at point 5 is 2,762 so our margin to detect that fault is: 2762/480 =
5.7:1 so no problem
•
Instantaneous Units – The phase instantaneous will be set at 1,120A and
the ground instantaneous unit will be set at 480 A
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Questions?
83