Energy Conversion Lab

SYMMETRICAL COMPONENTS Symmetrical components allow phase quantities

of voltage and current to be replaced by three separated balanced symmetrical components

Consider three phase balanced components

where

oooa

oac

ao

ab

ao

aa

aaawhereaIII

IaIIIII

3601 ,2401 ,1201

120

240

0

32

111

1211

111

∠=∠=∠=

=∠=

=∠=

=∠=

Ia1

Ib1

Ic1

Ia2

Ic2

Ib2

Ia0

Ib0

Ic0

Energy Conversion Lab

SYMMETRICAL COMPONENTS

Define the operator a 1+a+a2 = 0 where a=1∠120o

The order of phasor abc: positive phase sequence acb: negative phase sequence abc (positive) sequence acb (negative) sequence

zero sequence

2222

222

222

240

120

0

ao

ac

ao

ab

ao

aa

IaIIaIII

III

=∠=

=∠=

=∠=

111

1211

111

120

240

0

ao

ac

ao

ab

ao

aa

aIIIIaII

III

=∠=

=∠=

=∠=

000cba III ==

Energy Conversion Lab

SYMMETRICAL COMPONENTS Consider three phase unbalanced currents Ia, Ib, Ic

the symmetrical component of the currents

the matrix form of the abc currents in term of symmetrical component

the symmetrical component in term of the three phase current

2210210

2120210

210

aaacccc

aaabbbb

aaaa

IaaIIIIIIaIIaIIIII

IIII

++=++=

++=++=

++=

012

2

1

0

2

2

1 11 1 1

aabc

a

a

a

c

b

a

AIIIII

aaaa

III

===>

=

*11012

31 AAwhereIAI abc

a == −−

Energy Conversion Lab

SYMMETRICAL COMPONENTS Consider three phase unbalanced currents Ia, Ib, Ic

the symmetrical component in term of the three phase current

Zero sequence current one-third of the sum of the phase currents in a three phase system with ungrounded neutral, the

zero sequence current can’t exist (KCL) if neutral is grounded, zero-sequence current flow

between neutral and ground

( )

( )

( )cbaa

cbaa

cbaa

aIIaII

IaaIII

IIII

++=

++=

++=

22

21

0

313131

Energy Conversion Lab

SYMMETRICAL COMPONENTS Consider three phase unbalanced voltages Va, Vb, Vc

the symmetrical component of the currents

the matrix form of the abc currents in term of symmetrical component

the symmetrical component in term of the three phase current

2210210

2120210

210

aaacccc

aaabbbb

aaaa

VaaVVVVVVaVVaVVVVV

VVVV

++=++=

++=++=

++=

012

2

1

0

2

2

1 11 1 1

aabc

a

a

a

c

b

a

AVVVVV

aaaa

VVV

===>

=

*11012

31 AAwhereVAV abc

a == −−

Three Phase Transformations Transformation is used to decouple variables

with time-varying coefficients and refer all variables to a common reference frame

Transformation to decouple abc phase variables [f012]=[T012][fabc]

where

The symmetrical transformation is applicable to steady-state vectors or instantaneousquantities

[ ]

=

aaaaT

1 1

1 1 1

31

2

2012

ojea 12013

2

∠==π

Energy Conversion Lab

SYMMETRICAL COMPONENTS Consider three phase unbalanced currents Va, Vb, Vc

the symmetrical component in term of the three phase current

Three phase complex power in terms of symmetrical components S3φ = Vabc

T Iabc* = (AVa

012)T(AIa012)*

since AT=A, ATA*=3 S3φ = 3(Va

012)T(Ia012)* = 3 Va

0 Ia0* + 3 Va

1 Ia1* + 3 Va

2 Ia2*

total unbalanced power can be obtained from the sum of the symmetrical component powers

( )

( )

( )cbaa

cbaa

cbaa

aVVaVV

VaaVVV

VVVV

++=

++=

++=

22

21

0

313131

Sequence Impedances of Y-connected Loads Consider a three phase balanced load with self and

mutual elements (Fig. 10.4 PSA-Saddat)

a s n m n m n a

b m n s n m n b

m n m n s n cc

V Z Z Z Z Z Z IV Z Z Z Z Z Z I

Z Z Z Z Z Z IV

+ + + = + + + + + +

s n m n m nabc

m n s n m n

m n m n s n

Z Z Z Z Z ZZ Z Z Z Z Z Z

Z Z Z Z Z Z

+ + + = + + + + + +

Vabc = Zabc Iabc

Zabc

Aaaaa =

2

2

1 11 1 1

Sequence Impedances of Y-connected Loads Consider a three phase balanced load with self

and mutual elements in the above figure voltage equation: Vabc = Zabc Iabc

use transformation: AVa012=ZabcAIa

012

Va012=Z012Ia

012, where Z012 = A-1ZabcA Z012 in case of the above figure

impedances of nonzero terms appears in principle diagonal

for a balanced load, three sequence impedances are independent

current of each phase sequence produces voltage drops of the same phase sequence only

−−

++=

ms

ms

mns

ZZZZ

ZZZZ

0 0 0 0 0 0 23

012 Aaaaa =

2

2

1 11 1 1

Park Transformation Park transformation to decouple

three-phase quantities into two-phase variables (generator notation) [fdq0]=[Tdq0(θd)][fabc] generator notation, θq = θd + π/2

relationship between qd and abc quantities, positive d-axis is along with magnetic

field winding axis positive q-axis is along with internal

voltage ωLaf if internal voltage leads magnetic field by

90 degree (generating)

[ ]

+

−−

+

−

=

21

21

21

32in-

32in- sin

32cos

32cos cos

32)(0

πθπθθ

πθπθθ

θ ddd

ddd

ddq ssT

c

θdq

d

a

b ω=ωs

ω=0

[ ]

+

+

−

−=−

1 3

2in- 3

2cos

1 3

2in- 3

2cos

1 sin- cos

)( 10

πθπθ

πθπθ

θθ

θ

dd

dd

dd

ddq

s

sT

Park Transformation Park transformation (motor notation)

[fdq0]=[Tdq0(θd)][fabc] motor notation, θq = θd - π/2

relationship between qd and abc quantities, positive d-axis is along with magnetic

field winding axis positive q-axis is along with negative of

the internal voltage ωLaf if (induced voltage – motoring)

d-axis is referred from a-axis

[ ]

+

−

+

−

=

21

21

21

32in

32in sin

32cos

32cos cos

32)(0

πθπθθ

πθπθθ

θ ddd

ddd

ddq ssT

θd

d

q

a

b

c

ω=ωs

ω=0

Park Transformation Park transformation to decouple

abc phase variables [fqd0]=[Tqd0(θq)][fabc] generator notation, θq = θd + π/2

relationship between qd and abc quantities, q-axis is along with internal voltage d-axis is along with the magnetic

field q-axis is referred from a-axis

[ ]

+

−

+

−

=

21

21

21

32in

32in sin

32cos

32cos cos

32)(0

πθπθθ

πθπθθ

θ qqq

qqq

qqd ssT

θq

d

q

a

b

c

ω=ωs

ω=0

[ ]

+

+

−

−=−

1 3

2in 3

2cos

1 3

2in 3

2cos

1 sin- cos

)( 10

πθπθ

πθπθ

θθ

θ

qq

qq

qq

qqd

s

sT

Transformation Between abc and qd0 Starting from positive sequence vector

Let , the second row can be cancelled, the above matrix can be reformed in terms of real part and imaginary part

=

c

b

a

iii

aaaa

iii

31

31

31

1 1

2

2

0

2

1

( )

=

c

b

a

iii

aaaa

ii

i

21

21

21

1 1

32 2

2

0

*( ) ( )**

12 23 iii ==

sd

sq jiii −=

=

c

b

as

d

sq

iii

aaaR

ii

i

21

21

21

)I(- )I(- 1)R( (a) 1

32 2

2

0

=

c

b

as

d

sq

iii

ii

i

21

21

21

23-

23- 1

21-

21- 1

32

0

[ ] [ ][ ]abcs

qdsqd iTi 00 =

Transformation Between abc and qd0

Balanced three-phase current in term of t

Using the qd0 transformation, iqd0 becomes

Scaled current space vector

( ) )3

4cos( ),3

2cos( ,cos φπωφπωφω +−=+−=+= tIitIitIi emcembema

0

)2

cos()sin(

)cos(

0 =

++=+−=

+=

i

tItIi

tIi

ememsd

emsq

πφωφω

φω

{ }tj

atjj

mtj

m

eemsd

sq

eee eIeeIeI

tjtIjiiiωωφφω

φωφω~2

)sin()cos()( ===

+++=−=+

quantityphasor theis which ,2

1~ φjma eIIwhere =

Transformation Between abc and qd0

Scaled current space vector

clearly for balanced three-phase current, iqs and ids

are orthogonal and they have the same peak value as the abc phase current

Ids peaks 90o ahead of iqs and the resultant current

I rotates counter-clockwise at a speed of ωe from initial position of φ to the a phase axis at t=0

)( φω +=−= tjm

sd

sq

eeIjiii

d

qa

b

c

sd

sq jiii −=

Transformation Between qd0 to arbitrary reference frame

New rotating qd axes with stationery qd axes

qd component space vector form

the above equation implies rotating stationery qd components backward by angle θ

synchronous rotating frame w.r.t. stationery frame

quantities in synchronous frame are constant relationship between syn. frame and peak value phasor of a phase

current

syn. frame quantities and peak value phasor quantities of phase a current are the same

( ) θjsd

sqdq ejiijii −−=−

−=

sd

sq

d

q

i

iii

θθθθ

cos sinsin cos

∫ +=t

dttt0

)0()()( θωθ

( )))0(sin())0(cos(

))0((

emem

tjsd

sq

ed

eq

jIIejiijii ee

θφθφ

θω

−+−=

−=− +−

( )( ) a

ed

eq

tja

tjed

eq

sd

sq

Ijii

eIejiijiii ee

~2

~2

=−∴

=−=−= ωω

Transformation Between abc and qd0 Full transformation from stationery qd frame to

arbitrary qd rotating frame full transformation form

In matrix notation, [iqd0] in terms of original abc currents [iabc]

Total instantaneous power into three phase circuit in arbitrary qd0 frame

no restriction on abc currents, could be balanced or unbalanced

−=

00 1 0 00 cos sin0 sin cos

ii

i

iii

sd

sq

d

q

θθθθ

( ) 0031

23 ivivivivivivP ddqqccbbaaabc ++=++=

]][[]][][[]][[][ 0000 abcqdabcs

qdsqdqd iTiTTiTi === θθ

Project 4-1Complex quantities in transformation

Transform the instantaneous three-phase ac current to space vectors in positive and negative-sequence in the spatial domain The abc currents are of the form

ia=10cos(2πt)ib=10cos(2πt-2π/3)ic=10cos(2πt+2π/3)

Using the following dq0 transformation matrix

[ ]

+

−−

+

−

=

21

21

21

32in-

32in- sin

32cos

32cos cos

32)(0

πθπθθ

πθπθθ

θ ddd

ddd

ddq ssT

Project 4-1Complex quantities in transformation

Show the two rotating space vectors id and iqcomponents corresponding to sinusoidal and complex phase currents run the dq0 sequence component in

stationary frame ωe=0 frame rotating frame ωe

-ωe frame 2 ωe frame 5 ωe frame

Sequence Current Space Vector Sequence space vector

Balanced three-phase current in term of t

Sequence current space vector

Resultant airgap mmf

for balanced 3 phase currents, Fs is a rotating space vector which has a sinusoidal spatial distribution around the airgap with speed of ωe

( ) )3

4cos( ),3

2cos( ,cos πωπωω −=−== tIitIitIi emcembema

( )*12

2

21

2323

ieIaiiaii

eIiaaiii

tjmcba

tjmcba

e

e

==++=

=++=

− ω

ω

( ) ( )tINeieiNF eamjj

saa ωθθθ −

=+= − cos

23

24sin

12sin

*)( 21 ii =

=

c

b

a

iii

aaaa

iii

31

31

31

1 1

2

2

0

2

1

Relation Between Space Vector And Phase Quantity

Current space vector and phase currents current space vector

peak value of phase current

relations between current space vector and phase expression

Balanced sequence current space vector

( ) ( )iiii23or

32

11 ==

tjm

eeIi ω

23

1 =

tjm

eeIi ω=

( )( ) 00

31

31

2323

0

*

12

2

21

=∗=++=

==++=

=++=

−

mcba

tjmcba

tjmcba

Iiiii

ieIaiiaii

eIiaaiii

e

e

ω

ω

qd0 Transformation to Series RL Consider a three phase balanced RL transmission line

with self and mutual elements]][[]][[][][ iLpiRVV Rs +=−

Where

=

csgs

bsgs

asgsabc

S

vvv

V ][

=

crgr

brgr

rabc

R

vvv

Varg

][

+

+

+

=

gcgg

ggbg

gggaabc

rrrrrrrrrrrr

R ][

−+−−+−−+

−−+−+−−+

−−+−−+−+

=

cgggcccgbgggbccgagggac

bgcgggbcbgggbbbgagggab

agcgggacagbgggabagggaaabc

LLLLLLLLLLLLLLLLLLLLLL

LLLLLLLLLLLL

222

2][

qd0 Transformation to Series RL Consider a three phase balanced line with self and mutual

elements in Fig. 5.17 voltage equation: Δ[Vabc]= [Rabc][iabc]+p [Labc][iabc] use transformation:

[Tqd0(θ)] -1Δ Vaqd0= [Rabc] [Tqd0(θ)]-1 [iqd0]+ [Labc] p [Tqd0(θ)]-1 [iqd0]

[Tqd0(θ)] [Tqd0(θ)]-1 Δ Vqd0= [Tqd0(θ)] [Rabc] [Tqd0(θ)]-1 [iqd0]+ [Tqd0(θ)] [Labc] p ( [Tqd0(θ)]-1 [iqd0] )

Δ Vqd0= [Rqd0] [iqd0]+ [Lqd0] p[iqd0]+ [Tqd0(θ)] [Labc] [iqd0] p [Tqd0(θ)]-1

[Rqd0]=[Tqd0(θ)] [Rabc] [Tqd0(θ)]-1,

[Lqd0]= [Tqd0(θ)] [Labc] [Tqd0(θ)]-1

Rqd0 and Lqd0 in case of Fig. 5.17

impedances of nonzero terms appears in principle diagonal

+−

−=

ms

ms

ms

rrrr

rrRqd

2 0 0 0 0 0 0

0

+−

−=

ms

ms

ms

LLLL

LLLqd

2 0 0 0 0 0 0

0

qd0 Transformation to Series RL

Transform from abc to qd0 equivalent circuit

speed voltage

speed voltage

dtdiLLirrv

dtd

iLLdtdiLLirrv

dtd

iLLdtdi

LLirrv

msms

qqms

dmsdmsd

qdms

qmsqmsq

000 )2()2(

)()()(

)()()(

+++=∆

−−−+−=∆

−+−+−=∆

θ

θ

Space Vector and Transformations Air gap mmf due to current ia(t)

Fa1=(Nsine/2) ia(t) cos(θa), Fa1 is centered about a-phase winding axis

space vector notation

Resultant airgap mmf by currents flowing into all three windings

aaaaa tiiwhereiNF θcos)( ,2sin

1 ==

( ) ( )

( ) ( ){ } ( )aaaa

aa

jjcba

jcba

j

j

c

j

bajj

c

j

baj

ccbbaacba

cbas

eieiNaiaiieaiaiieN

eieiieeieiieN

iiiNiiiNFFFF

θθθθ

ππθ

ππθ

θθθ

−−

−−−

+=+++++=

+++

++=

++=++=

++=

12sin22sin

34

32

34

32

sin

sinsin

111

44

4

coscoscos22

2cos

aa jj

aee θθ

θ−+

=

=

c

b

a

iii

aaaa

iii

31

31

31

1 1

2

2

0

2

1

Given a one-line diagram of a three-phase system as shown below, sketch the input-output relations between the qd0 component generator is represented by an equivalent voltage

source E behind a source inductance Lg

since it is a three-wire system, no zero sequence component. zero-sequence circuit is omitted

Project. 4-2 qd0 Transformation