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Lecture 19
Symmetrical Components
Read: Chapter 8.1 - 8.6 ; 8.8
Homework 6 – will be posted on the course website
Dr. Lei Wu
Department of Electrical and Computer Engineering
EE 333
POWER SYSTEMS ENGINEERING
Outline
Symmetric Components
Zero
Positive
Negative
Sequence Diagram of Systems Components
System Analysis Using Sequence Diagram
2
0 0 0a b cI I I= =2
1 1 1 1b a c aI I I Iα α= =2
2 2 2 2b a c aI I I Iα α= =
02
1
22
1 1 1
1
1
a a
p b a s
c a
I I
I I
I I
α α
α α
= = =
I A I
0 0
1 1
2 2
3 0 0
0 0
0 0
y n
s s s y
y
Z ZV I
V Z I
V IZ
+ = = =
V Z I
The system is decoupled
0 0( 3 )y nV Z Z I= + 1 1yV Z I= 2 2yV Z I=
Sequence Diagrams of Impedance Loads, cont’d
3
Each sequence network is uncoupled with the other two
This separation underlines the advantage of symmetrical
components
Neutral impedance does not appear in positive and negative
sequence networks
Positive and negative currents do not flow in the neutral
impedance
Sequence Diagrams of Impedance Loads
4
0 0( 3 )y nV Z Z I= + 1 1yV Z I= 2 2yV Z I=
Sequence Diagrams of Impedance Loads, cont’d
For a delta-connected balanced 3Φ load
Delta-connected balanced 3Φ load is equivalent to wye-connected
balanced 3Φ load with infinite neutral to ground impedance
3YZ
Z ∆=
5
01 2
1
22
1 1 1 01
13
34.111
ag
bg
cg
VV
V V
V V
α α
α α
−
100∠ ° 18.63∠63.43° = = 75∠180° = 67.32∠ − 26.45° 50∠90° ∠22.99°
A
00 2.446
3y n
VI
Z Z= = ∠ − 3.37°
+
11 13.46
y
VI
Z= = ∠ − 79.58° 2
2 6.822y
VI
Z= = ∠ − 30.14°
Sequence Diagrams of Impedance Loads Example
6
Sequence Diagrams of Impedance Loads Example
02
1
22
1 1 1 2.446
1 13.46
6.822 8.531
a
b
c
I I
I I
II
α α
α α
∠ − 3.37° 19.97∠ − 57.34° = = ∠ − 79.58° = 15.16∠132.42° ∠ − 30.14° ∠37.45°
A
7
For a symmetrical series impedance, which means
, we can obtain
Sequence Diagrams for a General 3Φ Load
ag aaa ab ac
bg ab bb bc b
ac bc cc ccg
V IZ Z Z
V Z Z Z I
Z Z Z IV
=
1
2 0 0
0 0
0 0
aa ab
s p aa ab
aa ab
Z Z
Z Z
Z Z
−+
= = −
−
Z A Z A
aa bb cc ab ac bcZ Z Z and Z Z Z= = = =
8
However, if the loads are not symmetrical, then
is not diagonal. The sequence networks are coupled and the
voltage drop across any one sequence network depends on all
three sequence currents.
The sequence diagrams can be directly applied for series
impedances and transmission lines, which usually have the same
impedance value in three phases.
sZ
Sequence Diagrams for a General 3Φ Load
9
Sequence diagrams for generators
Balanced 3Φ positive sequence generators only produce positive
sequence voltages; therefore only the positive sequence network
has a voltage source.
01 2
1
22
1 1 1 01
13
01
ag
bg
cg
VV V
V V V V
VV V
α α
α α
−
∠ ° 0 = = ∠ −120° = ∠0° ∠120°
A
10
Sequence diagrams for generators
Balanced 3Φ negative sequence generators only produce negative
sequence voltages; therefore only the negative sequence network
has a voltage source.
01 2
1
22
1 1 1 01
1 03
1
ag
bg
cg
VV V
V V V
V VV V
α α
α α
−
∠ ° 0 = = ∠120° = ∠ −120° ∠0°
A
11
Sequence diagrams for generators, cont’d
During a fault Z1 ≈ Z2 ≈ Xd”. The zero sequence impedance is
usually substantially smaller. The value of Zn depends on whether the generator is grounded.
Synchronous motors have the same sequence networks as synchronous generators, except the sequence currents are referenced into the networks.
12
Sequence diagrams for generators, example
0
1
2
277 0 5
260 120 15
10295 115
10 40 30 40
, , ?
ag
bg
cg
Y
a b c
V Z j
V Z j
Z jV
Z Z
I I I∆
∠ = Ω
= ∠ − = Ω = Ω∠
= ∠ Ω = ∠ Ω
13
Get the sequence networks
Sequence diagrams for generators, example
( )0
11
2
15.91 62.11
277.10 1.77 // 3 5 40
9.22 216.59
ag
bg Y Y
cg
VV
V A V Z Z Z
V V
−∆
∠ ′= = ∠ − = = ∠ Ω ∠
14
Calculate sequence currents
Calculate phasor currents
Sequence diagrams for generators, example
0
1
2
46.94 50.05
49.68 70.49
47.98 166.94
a
b
c
I I
I A I
II
∠ = = ∠ − ∠
0 0I = ( )1
1 // 3
48.19 48.83
l Y
VI
Z Z Z∆=
+
= ∠ −
( )2
2 // 3
1.60 169.53
l Y
VI
Z Z Z∆=
+
= ∠
15
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