8 Symmetrical ComponentsNotes on Power System Analysis1 Lesson 8 Symmetrical Components.

Notes on Power System Analysis 18 Symmetrical Components

Lesson 8

Symmetrical Components


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:


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-cNegative sequence Va2, Vb2, Vc2 balanced

with phase sequence c-b-a


Zero Sequence

Positive Sequence

Negative Sequence

a

b

c

a

c

b

Va

Vb

Vc


wherea = 1/120° = (-1 + j 3)/2 a2 = 1/240° = 1/-120° a3 = 1/360° = 1/0 °

Va

=

1 1 1 V0

Vb 1 a2 a V1

Vc 1 a a2 V2


Vp = A Vs Vs = A-1 Vp

A =1 1 11 a2 a1 a a2

Vp =

Va

Vb

Vc

Vs =

V0

V1

V2


A-1 = (1/3)1 1 11 a a2

1 a2 a

Ip = A Is Is = A-1 Ip

• We used voltages for example, but the result applies to current or any other phasor quantity

Vp = A Vs Vs = A-1 Vp


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)/3These are the phase a symmetrical (or sequence) components. The other phases follow since the sequences are balanced.


Sequence networks– A balanced Y-connected load has three impedances Zy connected line to

neutral and one impedance Zn connected neutral to ground

Zy

Zy

g

cba

Zn


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


Zy

n

Vp = Zp Ip

Vp = AVs = Zp Ip = ZpAIs

AVs = ZpAIs

Vs = (A-1ZpA) Is

Vs = Zs Is where

Zs = A-1ZpA

Zy

Zy

g

c

ba

Zn


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


Zy n

g

a3 ZnV0

I0

Zero-sequencenetwork

Zy

n

aV1

I1

Positive-sequencenetwork

Zy

n

aV2

I2

Negative-sequencenetwork

Sequence networks for Y-connected load impedances


ZD/3

n

aV1

I1

Positive-sequencenetwork

ZD/3

n

aV2

I2

Negative-sequencenetwork

Sequence networks for D-connected load impedances.Note that these are equivalent Y circuits.

ZD/3 n

g

aV0

I0

Zero-sequencenetwork

Notes on Power System Analysis 15

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

8 Symmetrical Components


• General case unsymmetrical impedances

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


• Special case symmetrical impedances

Zs =

Z0 0 00 Z1 00 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= 0Vp = Zp Ip Vs = Zs Is

• This applies to impedance loads and to series impedances (the voltage is the drop across the series impedances)


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*


Power in sequence networks

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*


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.

Notes on Power System Analysis 23

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

8 Symmetrical Components


Y generator

Zero

I1V1

Z1

Z2

I2

Z0I0

V0

N

G

N

Negative

N

Positive

Zn

V2

3Zn

E


Ungrounded Y load

Zero

I1V1

Z

Z

I2V2

ZI0V0

N

G

NNegative

NPositive


Zero-sequence networks for loads

ZI0V0

N

G

3Zn

ZV0

G

Y-connected load grounded through Zn

D-connected load ungrounded


Y-Y transformer

A

B

C

N

H1 X1 a

b

c

nZnZN

Zeq+3(ZN+Zn)

g

AVA0 I0

Zero-sequencenetwork (per unit)

Va0

a


Y-Y transformer

A

B

C

N

H1 X1 a

b

c

nZnZN

Zeq

n

A

VA1 I1

Positive-sequencenetwork (per unit)Negative sequence

is same network

Va1

a


D-Y transformer

A

B

C

H1 X1 a

b

c

nZn

Zeq+3Zn

g

AVA0 I0

Zero-sequencenetwork (per unit)

Va0

a


D-Y transformer

A

B

C

H1 X1 a

b

c

nZn

Zeq

n

AVA1 I1

Positive-sequencenetwork (per unit)

Delta side leads wyeside by 30 degrees

Va1

a


D-Y transformer

A

B

C

H1 X1 a

b

c

nZn

Zeq

n

AVA2 I2

Negative-sequencenetwork (per unit)Delta side lags wyeside by 30 degrees

Va2

a

Notes on Power System Analysis8 Symmetrical Components 32

Three-winding (three-phase) transformers Y-Y-D

ZX

ZT

ZHH X

Ground

Zero sequence

ZXZHH X

Neutral

ZT

T

Positive and negative

T

H and X in grounded Y and T in delta


Three-winding transformer data:Windings Z Base MVAH-X 5.39% 150H-T 6.44% 56.6X-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%


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


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

8 Symmetrical ComponentsNotes on Power System Analysis1 Lesson 8 Symmetrical Components.

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