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Chapter 8: Symmetrical Components
1
ELCT 551: Power System Analysis & Design
Outlines
Definition of Symmetrical Components
Sequence Network: Load Impedance and Transmission Line Impedance
Sequence Network: Rotating Machines and Transformers
Power in Sequence Networks
2
1. Symmetrical Components
• Due to C. L. Fortescue (1918): a set of n unbalanced phasors in an n-phase system can be resolved into n sets of balanced phasors by a linear transformation – The n sets are called symmetrical
components
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Symmetrical Components
One of the n sets is a single-phase set and the others are n-phase balanced sets
Here n = 3 which gives the following case:
4
Definition of symmetrical components
• Three-phase voltages Va, Vb, and Vc (not necessarily balanced) can be resolved into three sets of sequence components:
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Definition of symmetrical components
Assume normal system phase sequence is abc
Zero sequence: Va0=Vb0=Vc0
Positive sequence Va1, Vb1, Vc1 balanced with phase sequence abc
Negative sequence Va2, Vb2, Vc2 balanced with phase sequence cba
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7
Zero Sequence
Positive Sequence
Negative Sequence
a
b
c
a
c
b
Va
Vb
Vc
where
a = 1/120º = (-1 + j 3)/2
a2 = 1/240º = 1/-120º
a3 = 1/360º = 1/0º
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Va
=
1 1 1 V0
Vb 1 a2 a V1
Vc 1 a a2 V2
Vp = A Vs Vs = A-1 Vp
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A =
1 1 1
1 a2 a
1 a a2
Vp =
Va
Vb
Vc
Vs =
V0
V1
V2
Vp = A Vs Vs = A-1 Vp
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A-1 = (1/3)
1 1 1
1 a a2
1 a2 a
Va = V0 + V1 + V2
Vb = V0 + a2V1 + aV2
Vc = V0 + aV1 + a2V2
V0 = (Va + Vb + Vc)/3
V1 = (Va + aVb + a2Vc)/3
V2 = (Va + a2Vb + aVc)/3
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V0 = (Va + Vb + Vc)/3
V1 = (Va + aVb + a2Vc)/3
V2 = (Va + a2Vb + aVc)/3
• These are the phase a symmetrical (or sequence) components.
• The other phases follow since the sequences are balanced.
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• We used voltages for example, but the result applies equally well to current or any other phasor quantity
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Vp = A Vs Vs = A-1 Vp
Ip = A Is Is = A-1 Ip
2. Sequence networks: Load Impedance and Transmission Line Impedance
• A balanced Y-connected load has three impedances Zy connected line to neutral and one impedance Zn connected neutral to ground
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Zy
Zy
g
c
b
a
Zn
Zy
Sequence networks
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Vag
=
Zy+Zn Zn Zn Ia
Vbg Zn Zy+Zn Zn Ib
Vcg Zn Zn Zy+Zn Ic
or in more compact notation Vp = Zp Ip
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Zy
n
Vp = Zp Ip
Vp = AVs = Zp Ip = ZpAIs
Vs = (A-1ZpA) Is
Vs = Zs Is where
Zs = A-1ZpA
Zy
Zy
g
c
b
a
Zn
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Zs =
Zy+3Zn 0 0
0 Zy 0
0 0 Zy
V0 = (Zy + 3Zn) I0 = Z0 I0
V1 = Zy I1 = Z1 I1
V2 = Zy I2 = Z2 I2
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Zy n
g
a
3 Zn V0
I0
Zero- sequence network
Zy
n
a
V1
I1
Positive- sequence network
Zy
n
a V2
I2
Negative- sequence network
Sequence networks for Y-connected load impedances
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ZD/3
n
a
V1
I1
Positive- sequence network
ZD/3
n
a V2
I2
Negative- sequence network
Sequence networks for D-connected load impedances. Note that these are equivalent Y circuits.
ZD/3 n
g
a
V0
I0
Zero- sequence network
• Remarks: –Positive-sequence impedance is equal to
negative-sequence impedance for any symmetrical impedance load (not a rotating machine)
–Rotating machines can have different positive and negative sequence impedances
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• Remarks: – Zero-sequence impedance is usually
different than the other two sequence impedances
– Zero-sequence current can circulate in a delta but the line current (at the terminals of the delta) is zero in that sequence
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• General case unsymmetrical impedances (Unbalanced Load Impedance)
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Zs=A-1ZpA =
Z0 Z01 Z02
Z10 Z1 Z12
Z20 Z21 Z2
Zp =
Zaa Zab Zca
Zab Zbb Zbc
Zca Zbc Zcc
Z0 = (Zaa+Zbb+Zcc+2Zab+2Zbc+2Zca)/3
Z1 = Z2 = (Zaa+Zbb +Zcc–Zab–Zbc–Zca)/3
Z01 = Z20 =
(Zaa+a2Zbb+aZcc–aZab–Zbc–a2Zca)/3
Z02 = Z10 =
(Zaa+aZbb+a2Zcc–a2Zab–Zbc–aZca)/3
Z12 = (Zaa+a2Zbb+aZcc+2aZab+2Zbc+2a2Zca)/3
Z21 = (Zaa+aZbb+a2Zcc+2a2Zab+2Zbc+2aZca)/3
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• Special case symmetrical impedances (Transmission Line and Balanced Load)
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Zs =
Z0 0 0
0 Z1 0
0 0 Z2
Zp =
Zaa Zab Zab
Zab Zaa Zab
Zab Zab Zaa
Z0 = Zaa + 2Zab
Z1 = Z2 = Zaa – Zab
Z01=Z20=Z02=Z10=Z12=Z21= 0 Vp = Zp Ip Vs = Zs Is
• This applies to impedance loads and to series impedances (the voltage is the drop across the series impedances)
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3. Sequence Network: Rotating Machines and Transformers
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Zn
Positive seq. V1
Z1
E
+
-
I1 +
-
Z2
I2 +
V2
-
Zo + 3 Zn
Io
Vo
+
-
Generator
Negative seq.
Zero seq.
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Ungrounded Y load (Motor)
+
0
-
Vo
+
-
+
V2
-
Z
Io
Z
I2
V1
+
-
Z
I1
N
G
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(a) GROUNDED WYE LOAD
(b) DELTA LOADVo
+
-
Vo
+
-
Z
Io
G
Z
Io
3 Zn
N
G
Y-Y transformer
29
A
B
C
N
H1 X1 a
b
c
n
Zn ZN
Zeq+3(ZN+Zn)
g
A
VA0 I0
Zero-sequence network (per unit)
Va0
a
Y-Y transformer
30
A
B
C
N
H1 X1 a
b
c
n
Zn ZN
Zeq
n
A
VA1 I1
Positive-sequence network (per unit) Negative sequence is same network
Va1
a
D-Y transformer
31
A
B
C
H1 X1 a
b
c
n
Zn
Zeq+3Zn
g
A
VA0 I0
Zero-sequence network (per unit)
Va0
a
D-Y transformer
32
A
B
C
H1 X1 a
b
c
n
Zn
Zeq
n
A
VA1 I1
Positive-sequence network (per unit)
Delta side leads wye side by 30 degrees
Va1
a
D-Y transformer
33
A
B
C
H1 X1 a
b
c
n
Zn
Zeq
n
A
VA2 I2
Negative-sequence network (per unit) Delta side lags wye side by 30 degrees
Va2
a
34
Three-winding (three-phase) transformers
ZERO SEQUENCE
Zh Zx
Zt
0
POSITIVE AND NEGATIVE SEQUENCE
Zh
Zt
Zx
0
h x
t
35
Three-winding transformer data: Windings Z Base MVA H-X 5.39% 150 H-T 6.44% 56.6 X-T 4.00% 56.6 Convert all Z's to the system base of 100 MVA: Zhx = 5.39% (100/150) = 3.59% Zht = 6.44% (100/56.6) = 11.38% Zxt = 4.00% (100/56.6) = 7.07%
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Calculate the equivalent circuit parameters: Solving: Zhx = Zh + Zx Zht = Zh + Zt Zxt = Zx +Zt Gives: Zh = (Zhx + Zht - Zxt)/2 = 3.95% Zx = (Zhx + Zxt - Zht)/2 = -0.359% Zt = (Zht + Zxt - Zhx)/2 = 7.43%
Typical relative sizes of sequence impedance values
• Balanced three-phase lines:
Z0 > Z1 = Z2
• Balanced three-phase transformers (usually):
Z1 = Z2 = Z0
• Rotating machines: Z1 Z2 > Z0
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Unbalanced Short Circuits
• Procedure: – Set up all three sequence networks
– Interconnect networks at point of the fault to simulate a short circuit
–Calculate the sequence I and V
– Transform to ABC currents and voltages
38
4. Power in sequence networks
Sp = Vag Ia* + Vbg Ib
* + Vcg Ic*
Sp = [Vag Vbg Vcg] [Ia* Ib
* Ic*]T
Sp = VpT
Ip*
= (AVs)T
(AIs)*
= VsT
ATA* Is
*
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Power in sequence networks
40
ATA* =
1 1 1 1 1 1
=
3 0 0
1 a2 a 1 a a2 0 3 0
1 a a2 1 a2 a 0 0 3
Sp = 3 VsT Is*
Sp = VpT Ip* = Vs
T ATA* Is*
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Sp = 3 (V0 I0* + V1 I1
* +V2 I2*) = 3 Ss
In words, the sum of the power calculated in the three sequence networks must be multiplied by 3 to obtain the total power.
This is an artifact of the constants in the transformation. Some authors divide A by 3 to produce a power-invariant transformation. Most of the industry uses the form that we do.