Lecture 8: Symmetrical Components

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Transcript Lecture 8: Symmetrical Components

Lesson 8
Symmetrical Components
8 Symmetrical Components
Notes on Power System Analysis
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Symmetrical Components
• Due to C. L. Fortescue (1918): a set of n
unbalanced phasors in an n-phase
system can be resolved into n balanced
phasors by a linear transformation
– The n sets are called 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:
8 Symmetrical Components
Notes on Power System Analysis
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Symmetrical component definition
• Three-phase voltages Va, Vb, and Vc
(not necessarily balanced, with phase
sequence a-b-c) can be resolved into
three sets of sequence components:
Zero sequence Va0=Vb0=Vc0
Positive sequence Va1, Vb1, Vc1 balanced
with phase sequence a-b-c
Negative sequence Va2, Vb2, Vc2 balanced
with phase sequence c-b-a
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Z e ro S e q u e n ce
Vc
c
Va
a
P o sitiv e S eq u e n ce
Vb
b
a
c
N eg ativ e S e q u e n ce
b
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Va
Vb
Vc
=
1
1
1
V0
1
a2
a
V1
1
a
a2
V2
where
a = 1/120° = (-1 + j 3)/2
a2 = 1/240° = 1/-120°
a3 = 1/360° = 1/0 °
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1
1
1
A=
1
a2
a
1
a
a2
V0
Va
Vb
Vc
Vp =
Vp = A Vs
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Vs =
V1
V2
Vs = A-1 Vp
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1 1 1
A-1 = (1/3) 1 a a2
1 a2 a
• We used voltages for example, but the
result applies to current or any other
phasor quantity
Vp = A Vs
Vs = A-1 Vp
Ip = A Is
Is = A-1 Ip
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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
These are the phase a symmetrical (or
sequence) components. The other phases
follow since the sequences are balanced.
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Notes on Power System Analysis
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Sequence networks
– A balanced Y-connected load has
three impedances Zy connected line to
neutral
and one impedance Zn connected neutral
to ground
a
Zy
b
Zy
c
Zn
g
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Sequence networks
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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a
b
c
Zy
Vp = Zp Ip
Zy
Zy
Zn
g
n
Vp = AVs = Zp Ip = ZpAIs
AVs = ZpAIs
Vs = (A-1ZpA) Is
Vs = Zs Is where
Zs = A-1ZpA
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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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a
V0
g
I0
Zy
3 Zn
Zerosequence
network
n
a
V1
n
I1
Zy
Positivesequence
network
I2
a
V2
n
Zy
Negativesequence
network
Sequence networks for Y-connected
load impedances
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a
V0
g
I0
ZD/3
Zerosequence
network
n
a
V1
n
I1
ZD/3
Positivesequence
network
I2
a
V2
n
ZD/3
Negativesequence
network
Sequence networks for D-connected load
impedances.
Note that these are equivalent Y circuits.
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Remarks
– Positive-sequence impedance is equal to
negative-sequence impedance for
symmetrical impedance loads and lines
– Rotating machines can have different
positive and negative sequence impedances
– 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
Zp =
Zs=A-1ZpA
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=
Zaa
Zab
Zca
Zab
Zbb
Zbc
Zca
Zbc
Zcc
Z0
Z10
Z20
Z01
Z1
Z21
Notes on Power System Analysis
Z02
Z12
Z2
16
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
Zp =
Zs =
8 Symmetrical Components
Zaa
Zab
Zab
Zab
Zaa
Zab
Zab
Zab
Zaa
Z0
0
0
0
Z1
0
0
0
Z2
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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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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
Sp = VpT Ip* = VsT ATA* Is*
1
ATA* =
1
1
1 1 1
1 a2
a
1 a a2 = 0 3 0
1
a2
1 a2 a
a
3 0 0
0 0 3
Sp = 3 VsT 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.
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Sequence networks for power apparatus
• Slides that follow show sequence
networks for generators, loads, and
transformers
• Pay attention to zero-sequence
networks, as all three phase currents
are equal in magnitude and phase
angle
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N
E
V1
I1
Positive
Z1
N
V2
I2
Zn
Negative
Z2
Y generator
G
3Zn
N
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I0
Z0
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V0
Zero
24
N
V1 I
1
Positive
Z
Ungrounded Y load
N
V2 I
2
Negative
Z
G
V0
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I0
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Z
Zero
N
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Zero-sequence networks for loads
G
V0
I0
Z
3Zn
N
Y-connected load
grounded through Zn
G
V0
8 Symmetrical Components
Z
D-connected load
ungrounded
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Y-Y transformer
H1
A
X1
a
B
b
C
c
N
n
ZN
8 Symmetrical Components
Zn
Zeq+3(ZN+Zn)
A
a
Va0
VA0 I0
g
Zero-sequence
network (per unit)
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Y-Y transformer
H1
A
X1
B
a
b
C
c
N
n
ZN
8 Symmetrical Components
Zn
A
Zeq
VA1 I1
a
Va1
n
Positive-sequence
network (per unit)
Negative sequence
is same network
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D-Y transformer
A
H1
X1
a
B
b
C
c
n
Zn
8 Symmetrical Components
A
VA0
Zeq+3Zn
I0
a
Va0
g
Zero-sequence
network (per unit)
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D-Y transformer
A
H1
X1
a
B
b
C
c
n
Zn
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Zeq
A
VA1 I1
a
Va1
n
Positive-sequence
network (per unit)
Delta side leads wye
side by 30 degrees
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D-Y transformer
A
H1
X1
a
B
b
C
c
n
Zn
8 Symmetrical Components
Zeq
A
VA2 I2
a
Va2
n
Negative-sequence
network (per unit)
Delta side lags wye
side by 30 degrees
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Three-winding (three-phase)
transformers Y-Y-D
H and X in grounded Y and T in delta
Zero sequence
Positive and negative
Neutral
Ground
ZT
H
ZH
ZX
X H
X
ZT
T
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ZX
ZH
T
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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%
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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
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