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

Notes on Power System Analysis 18 Symmetrical Components

Lesson 8

Symmetrical Components

Notes on Power System Analysis 28 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:

Notes on Power System Analysis 38 Symmetrical Components

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

Notes on Power System Analysis 48 Symmetrical Components

Zero Sequence

Positive Sequence

Negative Sequence

a

b

c

a

c

b

Va

Vb

Vc

Notes on Power System Analysis 58 Symmetrical Components

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

Notes on Power System Analysis 68 Symmetrical Components

Vp = A Vs Vs = A-1 Vp

A =1 1 11 a2 a1 a a2

Vp =

Va

Vb

Vc

Vs =

V0

V1

V2

Notes on Power System Analysis 78 Symmetrical Components

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

Notes on Power System Analysis 88 Symmetrical Components

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.

Notes on Power System Analysis 98 Symmetrical Components

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

Notes on Power System Analysis 108 Symmetrical Components

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

Notes on Power System Analysis 118 Symmetrical Components

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

Notes on Power System Analysis 128 Symmetrical Components

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

Notes on Power System Analysis 138 Symmetrical Components

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

Notes on Power System Analysis 148 Symmetrical Components

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

Notes on Power System Analysis 168 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

Notes on Power System Analysis 178 Symmetrical Components

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

Notes on Power System Analysis 188 Symmetrical Components

• Special case symmetrical impedances

Zs =

Z0 0 00 Z1 00 0 Z2

Zp =

Zaa Zab Zab

Zab Zaa Zab

Zab Zab Zaa

Notes on Power System Analysis 198 Symmetrical Components

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)

Notes on Power System Analysis 208 Symmetrical Components

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*

Notes on Power System Analysis 218 Symmetrical Components

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*

Notes on Power System Analysis 228 Symmetrical Components

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

Notes on Power System Analysis 248 Symmetrical Components

Y generator

Zero

I1V1

Z1

Z2

I2

Z0I0

V0

N

G

N

Negative

N

Positive

Zn

V2

3Zn

E

Notes on Power System Analysis 258 Symmetrical Components

Ungrounded Y load

Zero

I1V1

Z

Z

I2V2

ZI0V0

N

G

NNegative

NPositive

Notes on Power System Analysis 268 Symmetrical Components

Zero-sequence networks for loads

ZI0V0

N

G

3Zn

ZV0

G

Y-connected load grounded through Zn

D-connected load ungrounded

Notes on Power System Analysis 278 Symmetrical Components

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

Notes on Power System Analysis 288 Symmetrical Components

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

Notes on Power System Analysis 298 Symmetrical Components

D-Y transformer

A

B

C

H1 X1 a

b

c

nZn

Zeq+3Zn

g

AVA0 I0

Zero-sequencenetwork (per unit)

Va0

a

Notes on Power System Analysis 308 Symmetrical Components

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

Notes on Power System Analysis 318 Symmetrical Components

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

Notes on Power System Analysis 338 Symmetrical Components

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%

Notes on Power System Analysis 348 Symmetrical Components

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%

Notes on Power System Analysis 358 Symmetrical Components

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

Notes on Power System Analysis 368 Symmetrical Components

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