ECE 576 – Power System Dynamics and Stability

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

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Lecture 6: Synchronous Machine Modeling

Special Guest: TA Soobae Kim

Announcements

• Read Chapter 3

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Synchronous Machine Modeling

• Electric machines are used to convert mechanical energy into electrical energy (generators) and from electrical energy into mechanical energy (motors)–Many devices can operate in either mode, but are usually

customized for one or the other

• Vast majority of electricity is generated using synchronous generators and some is consumed using synchronous motors, so that is where we'll start

• Much literature on subject, and sometimes overly confusing with the use of different conventions and nominclature

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3 bal. windings (a,b,c) – stator

Field winding (fd) on rotor

Damper in “d” axis(1d) on rotor

2 dampers in “q” axis(1q, 2q) on rotor

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Synchronous Machine Modeling

Dq0 Reference Frame

• Stator is stationary and rotor is rotating at synchronous speed

• Rotor values need to be transformed to fixed reference frame for analysis

• This is done using Park's transformation into what is known as the dq0 reference frame (direct, quadrature, zero)

• Convention used here is the q-axis leads the d-axis (which is the IEEE standard)–Others (such as Anderson and Fouad) use a q-axis lagging

convention

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11 1 1

11 1 1

22 2 2

fdfd fd fd

dd d d

qq q q

qq q q

dv i r

dtd

v i rdt

dv i r

dtd

v i rdt

Kirchhoff’s Voltage Law, Ohm’s Law, Faraday’s Law, Newton’s Second Law

aa a s

bb b s

cc c s

dv i r

dtd

v i rdt

dv i r

dt

shaft 2

2m e f

d

dt Pd

J T T TP dt

Stator Rotor Shaft

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Fundamental Laws

1

or ,d a

q dqo b

co

a d

b dqo q

c o

v v

v T v i

vv

v v

v T v

v v

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Dq0 transformations

2 2sin sin sin

2 2 3 2 3

2 2 2cos cos cos

3 2 2 3 2 3

1 1 1

2 2 2

shaft shaft shaft

dqo shaft shaft shaft

P P P

P P PT

with the inverse,

13

22

cos3

22

sin

13

22

cos3

22

sin

12

cos2

sin

1

shaftshaft

shaftshaft

shaftshaft

dqo

PP

PP

PP

T

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Dq0 transformations

abcabc Li

(symmetric)

dqodqo

dqoabc

iTLT

iTLTT

1

1

Not symmetric if T is not power invariant.

Example: If the magnetic circuit is assumed to be linear

Note: This transformation is not power invariant. This means that some unusual things will happen when we use it.

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Dq0 transformations

11 1 1

11 1 1

22 2 2

fdfd fd fd

dd d d

qq q q

qq q q

dv r i

dtd

v r idt

dv r i

dtd

v r idt

2

2

shaft

m e f

d

dt Pd

J T T TP dt

dd s d q

qq s q d

oo s o

dv r i

dtd

v r idt

dv r i

dt

Stator Rotor Shaft

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Transformed System

Electrical system:

2

(voltage)

(power)

dv iR

dtd

vi i R idt

Mechanical system:

2

2(torque)

2 2 2 2(power)

m e f

m e f

dJ T T T

P dt

dJ T T T

P dt P P P

Electrical & Mechanical Relationships

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Derive Torque

• Torque is derived by looking at the overall energy balance in the system

• Three systems: electrical, mechanical and the coupling magnetic field–Electrical system losses in form of resistance–Mechanical system losses in the form of friction

• Coupling field is assumed to be lossless, hence we can track how energy moves between the electrical and mechanical systems

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13

Energy Conversion

ooqqddccbbaa iviviviviviv 323

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Look at the instantaneous power:

dt

di

dt

di

dtd

idt

di

dtd

idt

di

dtd

iP

iririririiirP

iv

ivivivivivivP

qq

qq

dd

fdfd

cc

bb

aa

electtrans

qqqqddfdfdcbaselectlost

qq

qqddfdfdccbbaaelectin

22

11

11

222

211

211

2222

22

1111

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Change to Conservation of Power

222

211

211

2222

2211

11

323

23

323

23

qqqq

ddfdfdosqsdselectlost

qqqq

ddfdfdooqqddelectin

irir

iririririrP

iviv

ivivivivivP

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With the Transformed Variables

dt

di

dt

di

dtd

idt

di

dtd

idt

di

idt

dPdt

dii

dt

dPP

qq

qq

dd

fdfd

oo

qq

qdshaftd

ddqshaft

electtrans

22

11

113

23

223

23

223

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With the Transformed Variables

fdW

dt

PTe

2 dtd

ai dt

d abi dt

d b

ci dt

d cfdi

dt

d fddi1 dt

d d1

qi1 dt

d q1qi2

dt

d q2

This requires the lossless coupling field assumption

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Change in Coupling Field Energy

For independent states , a, b, c, fd, 1d, 1q, 2q

fdW

dt fW

dtd f

a

W

dt

d a f

b

W

dt

d b

f

c

W

dt

d c f

fd

W

dt

d fd

1

f

d

W

dt

d d1

1

f

q

W

dt

d q1

2

f

q

W

dt

d q2

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Change in Coupling Field Energy

2 f fe a

a

W WT i

P

etc.

There are eight such “reciprocity conditions for this model.

These are key conditions – i.e. the first one gives an expression for the torque in terms of the coupling field energy.

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Equate the Coefficients

shaft

fW

3

2 2 d q q d eP

i i T

dd

f iW

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3, , 3

2f f

q oq o

W Wi i

fdfd

f iW

1 1 2

1 1 2

, , ,f f fd q q

d q q

W W Wi i i

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Equate the Coefficients

These are key conditions – i.e. the first one gives an expression for the torque in terms of the coupling field energy.

Coupling Field Energy

• The coupling field energy is calculated using a path independent integration–For integral to be path independent, the partial derivatives of

all integrands with respect to the other states must be equal

• Since integration is path independent, choose a convenient path–Start with a de-energized system so all variables are zero– Integrate shaft position while other variables are zero, hence

no energy

– Integrate sources in sequence with shaft at final shaft value

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3For example,

2fdd

fd d

ii

3 ˆ ˆ ˆ2 2

3 3ˆ ˆ ˆ32 2

1 21ˆ ˆ ˆ ˆ

1 1 1 1 2 2

1 1 2

shaft PoW W i i df f d q q d shaftoshaft

qd oi d i d i dd d q q o o

o o od q o

fd q qdi d i d i d i dfd fd d d q q q q

o o o ofd d q q

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Do the Integration

Torque

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• Assume: iq, id, io, ifd, i1d, i1q, i2q are independent of shaft (current/flux linkage relationship is independent of shaft)

• Then Wf will be independent of shaft as well

• Since we have

dqqde iiP

T 22

3

fd q q d e

shaft

W 3 Pi i T 0

2 2

11 1 1

fdfd fd fd

dd d d

dr i v

dtd

r i vdt

2 shaft sP

t

ds d q d

qs q d q

os o o

dr i v

dtd

r i vdt

dr i v

dt

11 1 1

22 2 2

qq q q

qq q q

dr i v

dtd

r i vdt

2 3

2 2

s

m d q q d f

d

dtd P

J T i i Tp dt

Define Unscaled Variables

s is the ratedsynchronous speedd plays an important role!

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Convert to Per Unit

• As with power flow, values are usually expressed in per unit, here on the machine power rating

• Two common sign conventions for current: motor has positive currents into machine, generator has positive out of the machine

• Modify the flux linkage current relationship to account for the non power invariant “dqo” transformation

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BaseBaseBase PIV

, , ,

, ,

, ,

a b ca b c

BABC BABC BABC

a b ca b c

BABC BABC BABC

a b ca b c

BABC BABC BABC

v v vV V V

V V V

i i iI I I

I I I

where VBABC is rated RMS line-to-neutral stator voltage and

,3

B BABCBABC BABC

BABC B

P VI

V

Convert to Per Unit

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, , ,

, ,

, ,

qd od q o

BDQ BDQ BDQ

qd od q o

BDQ BDQ BDQ

qd od q o

BDQ BDQ BDQ

vv vV V V

V V V

ii iI I I

I I I

where VBDQ is rated peak line-to-neutral stator voltage and

2,

3BDQB

BDQ BDQBDQ B

VPI

V

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Convert to Per Unit

1 211 1 2

1 1 2

1 211 1 2

1 1 2

1 211 1 2

1 1 2

, , ,

, , ,

, , ,

fd q qdfd d q q

BFD B D B Q B Q

fd q qdfd d q q

BFD B D B Q B Q

fd q qdfd d q q

BFD B D B Q B Q

v v vvV V V V

V V V V

i i iiI I I I

I I I I

Convert to Per Unit

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Hence the variables are just normalizedflux linkages

Where the rotor circuit base voltages are

11

1 21 2

, ,

,

B BBFD B D

BFD B D

B BB Q B Q

B Q B Q

P PV V

I I

P PV V

I I

And the rotor circuit base flux linkages are

11

1 21 2

, ,

,

BFD B DBFD B D

B B

B Q B QB Q B Q

B B

V V

V V

Convert to Per Unit

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11

1

21 2

1 2

11

1

1 21 2

1 2

, , ,

1, ,

, , ,

,

fds ds fd d

BDQ BFD B D

qq q

B Q B Q

BDQ BFD B DBDQ BFD B D

BDQ BFD B D

B Q B QB Q B Q

B Q B Q

rr rR R R

Z Z Z

rr qR R

Z Z

V V VZ Z Z

I I I

V VZ Z

I I

Convert to Per Unit

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Convert to Per Unit

• Almost done with the per unit conversions! Finally define inertia constants and torque

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( ),

, , ,

2

B

B s

fwm e BM ELEC FW B

B B BB

1 2J 2H2 PH M

S

TT T ST T T T

2T T TP

1

1

1

ds d q d

s s

qs q d q

s s

os o o

s

dR I V

dt

dR I V

dt

dR I V

dt

11 1 1

1

1

fdfd fd fd

s

dd d d

s

dR I V

dt

dR I V

dt

11 1 1

22 2

1

12

qq q q

s

qq q

s

dR I V

dt

dR I V

dt

2

s

M d q q d FWs

d

dtH d

T I I Tdt

Synchronous Machine Equations

32

32

cos2

32

cos2

cos2

32

cos2

32

cos2

cos2

isssc

isssb

isssa

vsssc

vsssb

vsssa

tII

tII

tII

tVV

tVV

tVV

Sinusoidal Steady-State

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Here we consider the applicationto balanced, sinusoidal conditions

0

2cos

2

2sin

2

o

vssshaftBDQ

BABCsq

vssshaftBDQ

BABCsd

V

tP

VVV

V

tP

VVV

V

2sin

2

2cos

2

0

s BABCd shaft s is

BDQ

s BABCq shaft s is

BDQ

o

I I PI t

I

I I PI t

I

I

Transforming to dq0

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issq

issd

vssq

vssd

II

II

VV

VV

cos

sin

cos

sin

/ 2

/ 2

jj vsV jV e V ed q s

jj isI jI e I ed q s

2 shaft sP

t

Simplifying Using d

• Recall that

• Hence

• These algebraic equations can be written as complex equations,

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The conclusion is if we know d, thenwe can easily relatethe phase to the dq values!

Summary So Far

• The model as developed so far has been derived using the following assumptions–The stator has three coils in a balanced configuration, spaced

120 electrical degrees apart–Rotor has four coils in a balanced configuration located 90

electrical degrees apart–Relationship between the flux linkages and currents must

reflect a conservative coupling field–The relationships between the flux linkages and currents must

be independent of shaft when expressed in the dq0 coordinate system

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Assuming a Linear Magnetic Circuit

• If the flux linkages are assumed to be a linear function of the currents then we can write

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1 1

1 1

2 2

a a

b bss shaft sr shaft

c c

fd fd

d drs shaft rr shaft

q q

q q

i

iL Li

i

iL L

i

i

The rotorself-inductancematrix Lrr is independentof qshaft

L12 = 0 L12 = + maximum

L12 = - maximum

Inductive Dependence on Shaft Angle

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Stator Inductances

• The self inductance for each stator winding has a portion that is due to the leakage flux which does not cross the air gap, Lls

• The other portion of the self inductance is due to flux crossing the air gap and can be modeled for phase a as

• Mutual inductance between the stator windings is modeled as

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cos( )A B shaftL L P

cos( )A B shaft

1L L P offset

2

The offset angleis either 2 /3 or-2 /3

1

1 11

1 1

2 2

d d

q q

o odqo srdqo ss dqo

fd fd

d drrrs dqo

q q

q q

i

i

iT LT L Ti

iLL T

i

i

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Conversion to dq0 for Angle Independence

1 1

1 1

1 1 1 1 1 1

3

23

2

d s md d sfd fd s d d

fd sfd d fdfd fd fd d d

d s d d fd d fd d d d

L L i L i L i

L i L i L i

L i L i L i

1 1 2 2

1 1 1 1 1 1 2 2

2 2 1 2 1 2 2 2

3

23

2

q s mq q s q q s q q

q s q q q q q q q q

q s q q q q q q q q

L L i L i L i

L i L i L i

L i L i L i

o s oL i

Conversion to dq0 for Angle Independence

41

,md A B

mq A B

3L L L

33

L L L3

Convert to Normalized at f = ws

• Convert to per unit, and assume frequency of ws

• Then define new per unit reactance variables

42

, ,

, ,

, ,

,

,

s mqs s s mds md mq

BDQ BDQ BDQ

s fdfd s fd 1d sfds 1d 1dfd 1d fd 1d

BFD B1D BFD s1d

s 1q1q s 2q2q s 1q2q s1q1q 2q 1q2q

B1Q B2Q B1Q s2q

fd fd md 1d 1d md

1q 1q mq 2q 2

LL LX X X

Z Z Z

L L LLX X X

Z Z Z L

L L L LX X X

Z Z Z L

X X X X X X

X X X X X

,

q mq

d s md q s mq

X

X X X X X X

1

1

1 1 1

1

d d d md fd md d

fd md d fd fd d md d

d md d d md fd d d

d

X I X I X I

X I X I c X I

X I c X I X I

c

1 2

1 1 1 2

2 1 2 2

1

q q q mq q mq q

q mq q q q q mq q

q mq q q mq q q q

q

X I X I X I

X I X I c X I

X I c X I X I

c

o s oX I

Normalized Equations

1 1 2,fd d q qd q

md mq

X Xc c

X X

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