back to the basics event analysis using symmetrical...
TRANSCRIPT
1
© 2018 SEL
Back to the Basics – Event Analysis
Using Symmetrical Components
Amanvir Sudan
Schweitzer Engineering Laboratories, Inc.
• Motivation
• Overview
▪ Introduction
▪ Symmetrical components refresher
▪ Event analysis using symmetrical components
▪ Case studies
▪ Conclusions
Back to Basics
2
• Short-circuit calculations
• Protective elements (67G, 59Q, 49, and so on)
• Impedance-based fault location
• Fault identification
• Event analysis
IntroductionSymmetrical Components Usage
Symmetrical Components Refresher
Ia0
Ib0
Ic0
Zero-sequence
currents
Ia2
Ib2 Ic2
Negative-sequence
currents
Ia1
Ic1 Ib1
Positive-sequence
currents
A a0 a1 a2
B b0 b1 b2
C c0 c1 c2
I I I I
I I I I
I I I I
3
Symmetrical Components
Refresher
Z2
V2
Z1I1
V1
Z0
V0
I2
I03RF
E10°
I0 = I1 = I2
Symmetrical Components Refresher
I1 I2 = –I1
Z1 Z2
V1 V2
E10°
I1
Z1 Z0
V1 V2
Z2
3RF
I0I2
V0
E10°
4
• Captured fault oscillography
• Relay hard-coded algorithm
• User-set relay settings
▪ Numeric
▪ Logic
General Event Report Analysis
• Captured fault oscillography
• Anatomy of power system
• Relay location in power system
Event Report Analysis Using Symmetrical Components
5
1. Using system one-line diagram, build positive-,
negative-, and zero-sequence network diagrams
2. Connect networks based on known or expected
fault location and type
3. Using basic circuit theory, reduce connected
networks to calculate sequence currents at
relay location
Symmetrical Components-Based Approach
4. Calculate phase currents at relay location and verify
they match phase current waveforms recorded in
event report (similar procedure applies for voltages)
5. If calculated phase currents do not match event
report phase currents, change expected fault location
and type, and repeat procedure from Step 2
Symmetrical Components-Based Approach
6
• Explain nature of fault waveforms that may not
resemble classic waveforms
• Estimate fault location
• Determine or verify impedance values of various
power system elements
Symmetrical Components-Based Approach Can Be Used to Accomplish Several Tasks
Case Study 1
69 kV
51
87T
Feeder 1
Feeder 2
B-CØ
fault
12.47 kV
System
equivalent
source
Main
W1 W251
7
87T Relay Fault Oscillography
0
–
–
20
4069 kV-side
currents
12.47 kV-side
currents
IAW1
IBW1
ICW1
IAW2
IBW2
ICW2
0
–
–
20
40
–
Time (ms)
0 10 20 30 40 50 60 70
Seco
nda
ry A
mp
ere
s
Sequence Network Connection
V2
I1
V1
I2 = –I1
E10°I1
I2
V1
V2
SYS1Z XFMR
1Z
SYS2Z XFMR
2Z
8
Sequence Currents
0
1 1
2 2
I 0
I ' I • 3
I ' I • 3
A 1 2I I • 3 I • 3
A 1 1
A 1
I I • 3 I • 3
I I 90
Sequence Voltages
XFMR
1 1 1 1
XFMR
2 2 2 2
V ' (V I Z ) • 3
V ' (V I Z ) • 3
9
Phase Currents and Voltages
XFMRA 1 1 1
A 1
V 3V 0 I Z 90
I I 90
XFMRB 1 1
B 1
V 2I Z 270
I 2I 180
XFMRC 1 1 1
C 1
V 3V 180 I Z 90
I I 90
Phasor Diagram
IA
ICIB
VAVC
VB
10
Case Study 2
500 kV
87T
F4
F3
F1
34.5 kV
13.8 kV
F2
Solar 1
Solar 2System
equivalent
source
W1W2
W3
Load
Load
ac
dc
ac
dc
87T Relay Fault Oscillography
IAW1
IBW1
ICW10
–
0
0.4500 kV-side
currents
34.5 kV-side
currents
0
0.08
–
–
0.2
560Time (ms)
580 600 620 64051N1
IAW2
IBW2
ICW2
IAW3
IBW3
ICW3
660
Seco
nda
ry A
mp
ere
s
11
Feeder Relay Oscillography
0
–
100
Feeder relay
oscillography
560
Time (ms)
2:03:01.602466239 AM 640
–
0
–
20
27C1
IA
IB
IC
VA_kV
VB_kV
VC_kV
VA_kV.Mag
VB_kV.Mag
VC_kV.Mag
C-phase undervoltage
element picks up for
about 3 cycles
Prim
ary
Am
pere
sK
ilovolts
Case Study 2Fault Possibilities
F3, F4 F2 F1
F3, F4 F2 F1
SOURCE1E
SOURCE1Z
SOURCE2Z
SOURCE0Z
T5001Z
T5002Z
T5000Z
T34.51Z
T34.52Z
T34.50Z
T13.81Z
T13.82Z
T13.80Z
LOAD1Z
LOAD2Z
LOAD0Z
12
Case Study 2Sequence Network
Connection
F3, F4
F3, F4
I2
I0
I1 0
I2
0
I0 = I1 = I2
SOURCE1E
SOURCE1Z
SOURCE2Z
SOURCE0Z
T5001Z
T5002Z
T5000Z
T34.51Z
T34.52Z
T34.50Z
T13.81Z
T13.82Z
T13.80ZSOURCE
0IXFMR0I
SOURCE0I
XFMR0I
Case Study 3
System equivalent source
87T
High-side breaker
230 kV
W1 IBW4W3
W2
Capacitor
bank
OPEN
34.5 kV
50/51
PV1 PV2 PV3 PV4
1Ø
A-GØ
fault
Solar 1 Solar 2 Solar 3 Solar 4
ac
dc
ac
dc
ac
dc
ac
dc
13
PV2 50/51 Relay Fault Oscillography
0
40000
20000
Time (ms)11:56:17.216145052 PM 250
0
–
40
51G
VA_kV
VB_kV
VC_kV
IA
IB
IC
IAMag
IB.Mag
IC.Mag
150
–
–
51P67G167P1
Prim
ary
Am
pere
s
0
–
20
Time (ms)11:56:17.220916667 PM 250
0
–
100
87R
150
TRIP187R287R387BL
230 kV-side currents
and neutral CT current
34.5 kV-side
currentsSeco
nda
ry A
mp
ere
s
IAW1
IBW1
ICW1
IBW4
IAW2
IBW2
ICW2
87T Relay Fault Oscillography
14
Case Study 387T Relay Secondary
Circuit Wiring
ICW
2
A-G fault
IFAULT
PV2
CT
IAW2
ZCBAZCBA ABCZ
87T
Relay
IAW
1
IBW
1
ICW
1
IAW
2
IBW
2
ICW
3
IAW
3
IBW
3
ICW
4
IAW
4
IBW
4
Mis
sin
g c
on
nectio
n
IAW2N CBA
N
From PV1
CT
From PV3
CT
From PV4
CT
From 230 kV
high-side
breaker CT
From
capacitor
bank CT
A
From
neutral
CT dot
polarity
Case Study 3Sequence Network
Connection
I1
I1
I2
I0
I2
I0
I0
SOURCE1E
2300I
SOURCE1Z
SOURCE2Z
SOURCE0Z
T2300I T13.8
0I
T2301Z T34.5
1ZT13.81Z
T2302Z T34.5
2ZT13.82Z
T2300Z T34.5
0ZT13.80Z
15
Case Study 3
Sequence Current
Calculations
T230 T13.8
0 1 2 0 0I = I = I = I + I
T2300 0
T13.80
T13.8 T2300 0
I • I
Z
Z Z
m
m =
T13.8
0 0I • I m
T2300
2 T2301
2 T2302
1 1 1 IIAW1
IBW1 1 I
ICW1 1 I
02
0
20
1 1 1IAW •I
IBW1 1 I
ICW1 I1
m
0
0
0
IAW ( )• I
IBW ( )• I
ICW ( )• I
m
m
m
1IBW ICW •IAW
2
m
m
IAW1 = 21.9 0°
IBW1 = 7.45 180°
ICW1 = 6.97 180°
23
1
0.25
m
m
m
• Performance report card
• Hold valuable information about power system
ConclusionsEvent Reports
16
• Provide holistic view of power system
• Explain non-classic fault waveforms at relay location
• Estimate fault location
• Determine or validate power system element
impedance, or build relationships between fault
oscillography and element impedance
ConclusionsEvent Report Analysis Using
Symmetrical Components
Questions?