Chapter 8: Symmetrical Components

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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

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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:

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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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Zero Sequence

Positive Sequence

Negative Sequence

a

b

c

a

c

b

Va

Vb

Vc

where

a = 1/120 = (-1 + j 3)/2

a2 = 1/240 = 1/-120

a3 = 1/360 = 1/0

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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 +ZccZabZbcZca)/3

Z01 = Z20 =

(Zaa+a2Zbb+aZccaZabZbca

2Zca)/3

Z02 = Z10 =

(Zaa+aZbb+a2Zcca

2ZabZbcaZca)/3

Z12 = (Zaa+a2Zbb+aZcc+2aZab+2Zbc+2a

2Zca)/3

Z21 = (Zaa+aZbb+a2Zcc+2a

2Zab+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

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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

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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

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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

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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

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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

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Three-winding (three-phase) transformers

ZERO SEQUENCE

Zh Zx

Zt

0

POSITIVE AND NEGATIVE SEQUENCE

Zh

Zt

Zx

0

h x

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%

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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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

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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.