Review of Symmetrical Components

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Oct 13, 2013 (3 years and 9 months ago)

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1
ECE525
Lecture 16
Symmetrical
Components

Review of basics

Review

of

basics

Sequence Equivalents

Fault Analysis
Symmetrical Components
Fall 2012
ECE525
Lecture 16
References

Areva NPAG: Chapter 4 (analysis) Chapter 5
(equipment models)
(equipment

models)

J.L. Blackburn, Protective Relaying: Principles and
Applications,2nd Edition,Marcel-Dekker, 1998.
Chapter 4

P.M. Anderson, Analysis of Faulted Power Systems,
IEEE Press 1995
Symmetrical Components
Fall 2012
IEEE

Press

1995

J.L. Blackburn, Symmetrical Components for Power
Systems Engineering, Marcel-Dekker, 1993.
2
ECE525
Lecture 16
History

Fortescue 1918

Fortescue
,
1918
» Unbalanced n-phase system can be broken
down into sets of balanced n-phase systems
» Add using superposition
Symmetrical Components
Fall 2012
ECE525
Lecture 16
What is it?

Method for analysis of multiphase systems
d b l d diti
un
d
er un
b
a
l
ance
d
con
diti
ons
» Steady-state conditions
» Phasor Analysis
» Visualization

䅬汯睳 a f慳a e晦楣楥湴 睡w to 慮慬祺a
Symmetrical Components
Fall 2012

䅬汯睳

a

晡獴

敦晩捩敮e

睡w



慮慬祺a

畮扡污湣敤u捯湤楴楯渠楮⁲敡氠瑩浥

Relays set to look at specific components
3
ECE525
Lecture 16
Basic Equations

De-couple voltage or current into
3 b l d 3 h t
3

b
a
l
ance
d

3
p
h
ase se
t
s
» Positive phase sequence
(ABCABC…): V
1
» Negative phase sequence
(ACBACB…): V
2
Symmetrical Components
Fall 2012
» Zero phase sequence (A=B=C), V
0
» All RMS phasors
» Per phase analysis
ECE525
Lecture 16
Per Phase Symmetrical
Component Equations
a = 1 @ 120
0
V
A
= V
A0
+ V
A1
+ V
A2
V
B
= V
B0
+ V
B1
+ V
B2
= V
A0
+ a
2
V
A1
+ aV
A2
Symmetrical Components
Fall 2012
V
C
= V
C0
+ V
C1
+ V
C2
= V
A0
+ aV
A1
+ a
2
V
A2
4
ECE525
Lecture 16
Analysis Equations
V
A0
=V
0
=(V
A
+ V
B
+ V
C
)/3
V
A1
=V
1
=(V
A
+ aV
B
+ a
2
V
C
)/3
V
A
2
=V
2
=(V
A
+a
2
V
B
+ aV
C
)/3
Symmetrical Components
Fall 2012

Same expressions for current

Can express in Matrix form too
ECE525
Lecture 16
Analysis Equations
V 0


1 1 1


V







1 1
3
1
1
3
2




Va0




1 1
2
1
1




Va0




V
a
0
Va1
Va2








1
3
1
1
a
1
a
2
a
2
a
4










V
a
Vb
Vc









Symmetrical Components
Fall 2012
Vb
Vc






1
1
a
3
1

a
3 2

a
3
2

a










Va1
Va2







1
1
a
2
a
1
a
1
a
2








Va1
Va2







5
ECE525
Lecture 16
Circuit Analysis

Represent power system in per phase
sequence networks

Each network contains voltage and
impedance elements for the sequence

Reduce network to Thevenin Equivalent
i h h
Symmetrical Components
Fall 2012
i
n eac
h
p
h
ase sequence

All sources generally positive sequence
ECE525
Lecture 16
Basic Sequence
Networks
Zero Sequence
Positive Sequence
Negative Sequence
jX0
-
Va
0
+
Zero

Sequence
Positive

Sequence
Ea
-
Va
1
+
+
jX2
-
Va
2
+
Negative

Sequence
Symmetrical Components
Fall 2012
jX0
+
Ia
0
jX1
+
Ia
1
jX2
+
Ia
2
6
ECE525
Lecture 16
Basic Sequence
Networks

Each network connects fromreference bus

Each

network

connects

from

reference

bus

(top) to fault bus (bottom)

Impedance will differ between sequences

Zero sequence will also include ground
im
p
edance
Symmetrical Components
Fall 2012
p

Connect them as appropriate for different
fault types
ECE525
Lecture 16
Basic Sequence
Networks

Impedance will differ between sequences

Impedance

will

differ

between

sequences

Zero sequence will also include ground
impedance

Connect them as appropriate for different
fault t
yp
es
Symmetrical Components
Fall 2012
yp
7
ECE525
Lecture 16
Fault Analysis

Fault Types:
» Single line to ground
» Line to line
» Double line to ground
» Three phase (positive sequence unless
unbalanced fault im
p
edances
)
Symmetrical Components
Fall 2012
p )
» Phase open
» Phase open and line to ground
» Simultaneous faults
ECE525
Lecture 16
Fault Detection For
Protection Purposes

Basic fault analysis techniques calculate

Basic

fault

analysis

techniques

calculate

currents/voltages at fault location
» ABC or symmetrical components

Need fault signature as seen at the relay
location
Symmetrical Components
Fall 2012
» Rough location of fault for correct response
8
ECE525
Lecture 16
Single Line to Ground
Connections

Constraints at fault location (phase A):
» V
A
=0
» I
B
=I
C
0

Therefore:
» V
A
= V
0
+ V
1
+ V
2
=0
I
(
I
+ 0 + 0)/3 d
Symmetrical Components
Fall 2012
»
I
0
=
(
I
A
+

0

+

0)/3
an
d
»I
1
= I
2
=(I
A
+ 0 + 0)/3 so
» I
0
= I
1
= I
2
=(I
A
)/3
ECE525
Lecture 16
SLG Connections

To Satisfy these

To

Satisfy

these

constraints we connect
the three networks in
series

Connect reference
point of one network to
N0
F0
N1
F1
Symmetrical Components
Fall 2012
point

of

one

network

to

fault location of next
one
N2
F2
9
ECE525
Lecture 16
SLG Faults

Solve for
I
can calculate
I

Solve

for

I
0
,
can

calculate

I
A

Next solve for V
1
, V
2
, and V
0
and
calculate phase voltage

Note that in general jX
o
will be replaced
with Z
0
=
j
X
0
+ 3Z
gr
+ 3Z
f
Symmetrical Components
Fall 2012
0
j
0
gr

f
» The factor of 3 results since 0 sequence is
the same to all 3 phases
ECE525
Lecture 16
Phase to Phase Faults

Two phases (often B and C for analysis)
shorted togethe
r
» Not shorted to ground
» V
B
=V
C
» I
B
= -I
C

䙲潭 獹浭整物捡s 捯浰潮敮c 敱畡瑩潮猺
Symmetrical Components
Fall 2012

䙲潭

獹浭整物捡s

捯浰潮敮c

敱畡瑩潮猺
» I
0
=0
» I
1
= -I
2
and V
1
=V
2
10
ECE525
Lecture 16
Phase to Phase Faults

Fault impedance
(if )
(if
any
)
appears
between the
networks

No Zero
Sequence
Network
N1
F1
N2
F2
Symmetrical Components
Fall 2012
Network

V
B
=I
B
*Z
f
+ V
C
ECE525
Lecture 16
Double Line to Ground

Two phases shorted together and to ground

Two

phases

shorted

together

and

to

ground

Could have several impedances
» I
A
 0
» V
B
= (Z
f
+ Z
gr
)*I
B
+ Z
f
*I
f
» V
c
= (Z
f
+ Z
gr
)*I
C
+ Z
f
*I
f
Symmetrical Components
Fall 2012
N0
F0
N1
F1
N2
F2
11
ECE525
Lecture 16
Fault Detection For
Protection Purposes

Basic fault analysis techniques calculate

Basic

fault

analysis

techniques

calculate

currents/voltages at fault location
» ABC or symmetrical components

Need fault signature as seen at the relay
location
Symmetrical Components
Fall 2012
» Rough location of fault for correct response
ECE525
Lecture 16
What Type of Fault?
-25
0
25
VA
VA
VB
VC
IA
IB
IC
-25
0
25
-25
0
25
2500
0
2500
VBVCIA
Symmetrical Components
Fall 2012
-
2500
-2500
0
2500
-2500
0
2500
1
2
3
4
5
6
7
8
9
10
11
IBIC
C
y
cles
12
ECE525
Lecture 16
What Type of Fault?
0
10000
IA
IA
IB
IC
IR
-10000
-10000
0
10000
0
10000
IB
I
C
Symmetrical Components
Fall 2012
-10000
-10000
0
10000
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
ECE525
Lecture 16
What Type of Fault?
0
10000
A
IA
IB
IC
-10000
-10000
0
10000
I
A
IB
Symmetrical Components
Fall 2012
-10000
0
10000
1
2
3
4
5
6
7
8
9
10
11
IC
C
y
cles
13
ECE525
Lecture 16
What Type of Fault?
0
5000
IA
IA
IB
IC
IR
-5000
-5000
0
5000
0
5000
IB
I
C
Symmetrical Components
Fall 2012
-5000
-2500
0
2500
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
ECE525
Lecture 16
What Type of Fault?
0
100
IA
IA
IB
IC
IR
-100
-100
0
100
0
100
IB
I
C
Symmetrical Components
Fall 2012
-100
-200
-0
200
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
14
ECE525
Lecture 16
What Type of Fault?
0
200
IA
IA
IB
IC
IR
-200
-200
0
200
0
200
IB
I
C
Symmetrical Components
Fall 2012
-200
-500
0
500
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
ECE525
Lecture 16
What Type of Fault?
0
250
IA
IA
IB
IC
IR
-250
-250
0
250
0
250
IB
I
C
Symmetrical Components
Fall 2012
-250
-100
0
100
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
15
ECE525
Lecture 16
Three Phase Fault, Right?
-25
0
25
VA
VA
VB
VC
IA
IB
IC
-25
0
25
-25
0
25
-2500
0
2500
VBVCIA
Symmetrical Components
Fall 2012
-2500
-2500
0
2500
-2500
0
2500
1
2
3
4
5
6
7
8
9
10
11
IBIC
C
y
cles
ECE525
Lecture 16
A Symmetrical Component
View of an Three-Phase Fault
45
9
0
135
A-Ground Fault
Component Magnitude Angle
Ia0 7.6 175
Ia1 2790 -64
Ia2 110 75.8
Va0 0 0
Va1
18 8
0
0
45
135
180
V1
Symmetrical Components
Fall 2012
Va1
18
.
8
0
225
27
0
315
I1
16
ECE525
Lecture 16
A to Ground Fault, Okay?
0
10000
IA
IA
IB
IC
IR
-10000
-10000
0
10000
0
10000
IB
I
C
Symmetrical Components
Fall 2012
-10000
-10000
0
10000
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
ECE525
Lecture 16
A Symmetrical Component View of
an A-Phase to Ground Fault
45
90
135
A-Ground Fault
Component Magnitude Angle
Ia0 7340 -79
Ia1 6447 -79
Ia2 6539 -79
Va0 46 204
0
45
135
180
V0
V1
V2
Symmetrical Components
Fall 2012
V
a1 123 0
225
270
315
I0
I1
I2
17
ECE525
Lecture 16
Single Line to Ground
Fault

Voltage

Voltage
» Negative and zero sequence 180 out of
phase with positive sequence

Current
» All sequence are in phase
Symmetrical Components
Fall 2012
ECE525
Lecture 16
A to B Fault, Easy?
0
10000
A
IA
IB
IC
-10000
-10000
0
10000
I
A
IB
Symmetrical Components
Fall 2012
-10000
0
10000
1
2
3
4
5
6
7
8
9
10
11
IC
C
y
cles
18
ECE525
Lecture 16
A Phase Symmetrical Component
View of an A to B Phase Fault
45
9
0
135
A-B Fault
Component Magnitude Angle
Ia0 3 -102
Ia1 5993 -81
Ia2 5961 -16
Va0 1 45
0
45
135
180
I2
V1
Symmetrical Components
Fall 2012
V
a1 99 0
225
27
0
315
I1
V2
ECE525
Lecture 16
C Phase Symmetrical Component
View of an A to B Phase Fault
45
9
0
135
I2
Component Magnitude Angle
Ic0 3 138
Ic1 5993 279
Ic2 5961 104
Vc0 1 -75
Vc1 99 0
0
45
135
180
V1
V2
Symmetrical Components
Fall 2012
V
c2 95 2.5
225
27
0
315
I1
19
ECE525
Lecture 16
Line to Line Fault

Voltage

Voltage
» Negative in phase with positive sequence

Current
» Negative sequence 180 out of phase with
positive sequence
Symmetrical Components
Fall 2012
ECE525
Lecture 16
B to C to Ground
0
5000
IA
IA
IB
IC
IR
-5000
-5000
0
5000
0
5000
IB
I
C
Symmetrical Components
Fall 2012
-5000
-2500
0
2500
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
20
ECE525
Lecture 16
A Symmetrical Component View
of a B to C to Ground Fault
45
9
0
135
Component Magnitude Angle
Ia0 748 97
Ia1 2925 -75
Ia2 1754 101
Va0 8 351
Va1 101 0
0
45
135
180
I0
I2
V0
V1
V2
Symmetrical Components
Fall 2012
V
a2 18 348
225
27
0
315
I1
ECE525
Lecture 16
Line to Line to Ground
Fault

Voltage

Voltage
» Negative and zero in phase with positive
sequence

Current
» Negative and zero sequence 180 out of
Symmetrical Components
Fall 2012
phase with positive sequence
21
ECE525
Lecture 16
Again, What Type of Fault?
0
100
IA
IA
IB
IC
IR
-100
-100
0
100
0
100
IB
I
C
Symmetrical Components
Fall 2012
-100
-200
-0
200
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
ECE525
Lecture 16
C Symmetrical Component
View of a C-Phase Open Fault
45
9
0
135
Component Magnitude Angle
Ic0 69 184
Ic1 101 4
Ic2 32 183
Vc0 0 162
Vc1 79 0
0
45
135
180
I0
I1
I2
V1
V2
Symmetrical Components
Fall 2012
V
c2 5 90
225
27
0
315
22
ECE525
Lecture 16
One Phase Open (Series)
Faults

Voltage

Voltage
» No zero sequence voltage
» Negative 90 out of phase with positive sequence

Current
» Negative and zero sequence 180 out of phase
with positive sequence
Symmetrical Components
Fall 2012
with

positive

sequence
ECE525
Lecture 16
What About This One?
0
200
IA
IA
IB
IC
IR
-200
-200
0
200
0
200
IB
I
C
Symmetrical Components
Fall 2012
-200
-500
0
500
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
23
ECE525
Lecture 16
Ground Fault with Reverse Load
45
9
0
135
Ic0 164 -22
Ic1 89 -113
Ic2 41 -6
Vc0 4 -123
Vc1 38 0
Vc2
6
130
0
45
135
180
I0
I2
V0
V1
V2
Symmetrical Components
Fall 2012
Vc2
6
-
130
225
27
0
315
I0
I1
ECE525
Lecture 16
Finally, The Last One!
0
250
IA
IA
IB
IC
IR
-250
-250
0
250
0
250
IB
I
C
Symmetrical Components
Fall 2012
-250
-100
0
100
1
2
3
4
5
6
7
8
9
10
11
I
IR
C
y
cles
24
ECE525
Lecture 16
Fault on Distribution System with
Delta – Wye Transformer
45
9
0
135
Component Magnitude Angle
Ic0 45 40
Ic1 153 -4
Ic2 132 180
Vc0 0.5 331
Vc1 40 0
0
45
135
180
I0
I1
I2
V0
V1
V2
Symmetrical Components
Fall 2012
Vc2 0.5 93
225
27
0
315