54
3.1 Chapter 3 Modeling Of a Synchronous Machine In Chapter 2 we have discussed about small-signal and transient stability of a synchronous machine, connected to an infinite bus, represented by a classical model. We have seen that there are several disadvantages of representing a synchronous generator by a classical model. In this Chapter we will look at detailed modeling of a synchronous machine. 3.1 Representation of Synchronous Machine Dynamics While modeling a synchronous machine, different ways of representation, conventions and notations are followed in the available literature. Hence, at the outset the notations and conventions used for representing a synchronous machine should be clear. In this course, IEEE standard (1110-1991) “IEEE guide to synchronous machine modeling” has been followed for representing the synchronous machine. (a)

Chapter 3nptel.ac.in/courses/108106026/chapter3.pdf · dq d q aa s fP fP fP fP P P Ni PP P P Ni

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3.1

Chapter 3 Modeling Of a Synchronous Machine

In Chapter 2 we have discussed about small-signal and transient stability of a

synchronous machine, connected to an infinite bus, represented by a classical model.

We have seen that there are several disadvantages of representing a synchronous

generator by a classical model. In this Chapter we will look at detailed modeling of a

synchronous machine.

3.1 Representation of Synchronous Machine Dynamics

While modeling a synchronous machine, different ways of representation,

conventions and notations are followed in the available literature. Hence, at the outset

the notations and conventions used for representing a synchronous machine should be

clear. In this course, IEEE standard (1110-1991) “IEEE guide to synchronous

machine modeling” has been followed for representing the synchronous machine.

(a)

3.2

(b)

Fig. 3.1: Synchronous machine (a) sectional view (b) Stator and rotor windings with

mmf along the respective axis.

The conventions and notations used along with their significance will be

explained in this Chapter. To model and mathematically represent a synchronous

machine first all the windings that need to be included in the model should be

identified. Consider sectional view of the synchronous machine shown in Fig. 3.1 (a).

The synchronous machine in Fig. 3.1 is a two pole salient machine. A general model

with n poles will be dealt latter in this Chapter.

The conductors anda a represent the sectional view of one turn of a-phase

stator winding. The dot in the conductor “a” represents current coming out of the

conductor and represents in to the conductor. By applying right hand thumb rule at

conductor anda a it can be observed that the mmf due to the conductors

3.3

anda a lie along axis marked A-axis. Similarly, the mmf due to ,b b and ,c c lie

along B and C axis, respectively. As an electrical circuit the stator can be represented

as three windings corresponding to three-phases, as shown in Fig. 3.1 (b). The three-

phase instantaneous ac voltages and currents in the stator windings are represented as

, ,a b cv v v and , ,a b ci i i . According to the generator convention, currents out of the

stator windings are considered as positive where as currents into the rotor windings

are considered as positive.

The rotor field is excited by a dc voltage represented as fdv with a field current

fdi . The mmf generated by the rotor field excitation lies normal to the pole surface,

along the direct axis or d-axis. The d-axis of the rotor is at angle m with respect to

the stator a a mmf axis that is A-axis. Angle m is in mechanical radians and in

case of two poles machine the electrical and mechanical angle are one and the same.

But in case of multiple poles the electrical angle is related to the mechanical angle

through the number of poles i.e. / 2s mP where P is the number of poles and s

is the rotor angle in electrical radians. In case of two poles machine the electrical and

mechanical rotor angle will be same as is the case for the synchronous machine shown

in Fig. 3.1. The analysis holds true for multiple pole machine as well but with the

additional condition that / 2s mP . For rest of the Chapters we will be expressing

the angle in terms of electrical radians unless specified other wise. The axis in

quadrature (leading or lagging by 90 ) with respect to the d-axis is called as

quadrature axis or q-axis. The q-axis can either be represented as leading d-axis or

lagging d-axis. Both the conventions are followed in the literature. However, here q-

axis is taken as leading d-axis according to the IEEE 1110-1991 standard.

Representing damper windings needs clarification. The damper windings are

copper bars placed usually in the slots of the pole face. The ends of the copper bars

are shorted forming a closed path for the currents to flow. The magnetic field

generated by these damper windings, due to currents circulating through these

windings, will be along the d-axis. However, the rotor core itself may act as closed

path for induced currents during non-synchronous operations. Hence, to properly

account for the action of the damper windings and damping effect of rotor core three

damper windings are considered. One damper winding represented as 1d , with a

3.4

voltage 1dv and current 1di , is considered whose mmf is along d-axis. Two damper

windings represented as 1 , 2q q , with a voltage 1 2,q qv v and current 1 2,q qi i are

considered whose mmf is along q-axis. In the d-axis and q-axis rotor windings the

current in to the winding is considered as positive.

For very accurate representation of synchronous machine, even more number

of damper windings may be considered along d and q axis. According to the number

of windings considered along each axis a model number is give as following [1]

Table 3.1: Classifications of synchronous machine model based on number of

windings in each axis

Number of windings in q-axis

0 1 2 3

Model

1.0

Model

1.1

Model

1.2

Model

2.1

Model

2.2

Number

of

windings

in

d-axis

1

2

3 Model

3.3

The first number in the model number given in Table 3.1 represents number of

windings in d axis and second number represents number of windings in q axis.

There should be at least one winding, field winding, in the d axis. Hence, the first

model 1.0 means that rotor is represented by one field winding, zero d-axis damper

winding and zero q-axis damper windings. From the view point of complexity, in the

representation of many windings along d axis and q axis, the maximum number

of winding that can be represented along any axis is fixed at 3. The model which is

shown in Fig. 3.1 (b) is 2.2 that is one field winding, one damper winding along d-

axis and two damper windings along q-axis. Model 2.2 is widely used in many

industry grade transient stability simulation softwares.

3.5

3.1.1 Stator and rotor winding voltage equations

Applying KVL at the stator windings the following equations can be written

aa s a

dv r i

dt

(3.1)

bb s b

dv r i

dt

(3.2)

cc s c

dv r i

dt

(3.3)

where, sr is the stator resistance and is assumed to be same in all the three

phases. The flux linkages in a, b, and c phases are represented as , ,a b c . The rate

of change of flux linkages in phase a, b and c lead to an induced emf (electro-motive

force) which is equal to the terminal phase voltage plus the drop in the stator

resistance (since we are using generator convention), as can be seen from equations

(3.1) to (3.3). Now applying KVL at the d and q axis rotor windings will give the

following expressions

fdfd fd fd

dv r i

dt

(3.4)

11 1 1

dd d d

dv r i

dt

(3.5)

11 1 1

qq q q

dv r i

dt

(3.6)

22 2 2

qq q q

dv r i

dt

(3.7)

Where, 1 1 2, , ,fd d q qr r r r and 1 1 2, , ,fd d q q are the rotor field, 1d, 1q and 2q winding

resistances and flux linkages, respectively.

3.6

3.1.2 Stator and rotor windings flux linkage equations

The flux linkages of different windings can be expressed in terms of current

through the windings and inductance of the windings as:

1 1 2

1

1 1 21

1 1 22

abc abcsssr

rotor

fd

afd a d a q a qa aa ab ac ad

b ba bb bc b bfd b d b q b qq

c ca cb cc c cfd c d c q c qqiL L

i

iL L L LL L L i

iL L L i L L L L

iL L L i L L L L

i

(3.8)

abc ss abc sr rotorL i L i (3.9)

In equation (3.8) the diagonal elements of the matrix ssL represent the self

inductance of a, b, c windings and off-diagonal elements represent the mutual

inductance among a, b, c phases. The matrix srL represents the mutual inductance

between the stator and rotor windings. A similar expression for flux linkage of the

rotor windings can be written as

1

1 1 11 1 1 1

1 1 1 1 1 1 1 2

2 2 2 2 2 1 2 2

0 0

0 0

0 0

0 0abc

rotor rs rr

fdfd fd dfd fda fdb fdc

adfd d dd da db dc

bq qa qb qc q q q q

c

q qa qb qc q q q qi

L L

L LL L Li

L LL L Li

L L L L Li

L L L L L

1

1

2

rotor

fd

d

q

q

i

i

i

i

i

(3.10)

………

rotor rs abc rr rotorL i L i (3.11)

In the matrix rrL , the mutual inductance between the d-axis windings ( ,1fd d )

and the q-axis windings (1 ,2q q ) is zero as the flux due to these windings are in

quadrature. The elements of inductance matrices , , , andss sr rr rsL L L L are dependent on

the angle s . The dependency of stator and rotor inductances on the angle s can be

understood from the way in which the air gap between the stator and rotor varies with

respect to time.

3.7

It can be observed from Fig. 3.1 at 0s the rotor d-axis is aligned along the

magnetic flux axis of a a winding. The air gap that has to be traversed by the

magnetic flux produced by the a a is minimum, that is twice the air gap length

between the pole face and stator along the d -axis, in this position and hence the

permeance is maximum. As the angle s increases, the air gap that has to be traversed

by the magnetic field of the a a starts increasing. At 90s the air gap is

maximum, that is twice the air gap length from stator to the rotor along q -axis, and

permeance is minimum. The variation of the permeance with respect to angle s is

shown in Fig. 3.2.

Fig. 3.2: Variation of permeance with respect to angle s

It can be observed from Fig. 3.2 that there is a positive average permeance

which is constant and there is double frequency oscillating component. This is

because for a two pole machine when the rotor rotates 180 there will be two

permeance peaks at 0s and 180s which indicates that the permeance is

varying at double the rotor speed for a two pole machine. In view of this observation

the permeance of the path that a a magnetic flux has to pass can be written as

s

3.8

cos(2 )a avg p sP P P (3.12)

mmf a a af N i of a a winding, where aN is number of turns of phase-a winding,

can be split into two components that is d -axis and q -axis component.

cos( )ad a sf f (3.13)

sin( )aq a sf f (3.14)

Now these two mmf ,ad aqf f complete their path through two air gaps and hence two

permeance ,d qP P (permeance along d and q axis). The flux along d -axis is ad df P and

along q -axis is aq qf P . These fluxes along the mmf axis of a a is given as

2 2

cos( ) sin( )

cos ( ) sin ( )

(1 cos(2 ))(1 cos(2 ))

2 2

( )cos(2 )

2 2

aa ad d s aq q s

a d s a q s

q sd sa a

d q d qa a s

f P f P

f P f P

PPN i

P P P PN i

if the flux linkage of phase-a due to current ai is represented as aa then

2 2 cos(2 )2 2

d q d qaa a aaaa a a s

a a

P P P PNl N N

i i

(3.15)

Now, (3.15) can be written as cos(2 )aa aao aap sl l l .

Where, 2 2,2 2

d q d qaao a aap a

P P P Pl N l N

.

Similarly we can derive an expression for the mutual inductance between a-

phase and b-phase. The mmf of b-phase can be split into two components along d-

axis and q-axis. The mmf along these axes can be multiplied with the permeance

3.9

along d-axis and q-axis then effective flux linkage along a a mmf axis can be

found.

cos( ) sin( )

cos( )cos( 120 ) sin( )sin( 120 )

(cos(2 120 ) cos(120 )) (cos(120 ) cos(2 120 ))2 2

( )cos(2 120 )

4 2

ba bd d s bq q s

b d s s b q s s

qdb b s s

d q d qb b s

f P f P

f P f P

PPN i

P P P PN i

(3.16)

if the flux linkage of phase-a due to current bi represented as ba then

( )cos(2 120 )

4 2d q d q

ba a ba a b b s

P P P PN N N i

(3.17)

if the number of turns of a and b phase are same, that is a bN N . Also, assuming

leakage inductance lsl is same for all the phases, (3.17) can be approximately written

as,

1cos(2 120 )

2ab aao aap sl l l (3.18)

Similarly it can be proved that mutual inductance between phase-a and phase-c will be

1cos(2 120 )

2ac aao aap sl l l and that between phase-b and phase-c is

1cos(2 180 )

2bc aao aap sl l l

3.10

Hence,

1 1cos(2 ) cos(2 60 ) cos(2 60 )

2 21 1

cos(2 60 ) cos(2 120 ) cos(2 180 )2 21 1

cos(2 60 ) cos(2 180 ) co2 2

aao aap s aao aap s aao aap s

ss aao aap s aao aap s aao aap s

aao aap s aao aap s aao aap

l l l l l l

L l l l l l l

l l l l l l

s(2 120 )s

…………. (3.19)

The mutual inductance between the rotor windings and the stator windings is

straight forward as the air gap that has to be traversed by the d-axis winding and q-

axis windings mmf to link the stator windings is fixed. The rotor windings flux along

the mmf axis of a a will vary only according to the angle s .

Hence,

1 1 2

1 1 2

1 1 2

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

cos( 120 ) cos( 120 ) sin( 120 ) sin( 120 )

cos( 120 ) cos( 120 ) sin( 120 ) sin( 120 )

afd s a d s a q s a q s

sr bfd s b d s b q s b q s

cfd s c d s c q s c q s

l l l l

L l l l l

l l l l

………………… (3.20)

Now let,

1 1 1 1

1 1 1 1

2 2 2 2

afd bfd cfd sfd

a d b d c d s d

a q b q c q s q

a q b q c q s q

l l l l

l l l l

l l l l

l l l l

then (3.20) can be written in a simplified form as

1 1 2

1 1 2

1 1 2

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

cos( 120 ) cos( 120 ) sin( 120 ) sin( 120 )

cos( 120 ) cos( 120 ) sin( 120 ) sin( 120 )

sfd s s d s s q s s q s

sr sfd s s d s s q s s q s

sfd s s d s s q s s q s

l l l l

L l l l l

l l l l

... …….. (3.21)

3.11

The complexity of analyzing the system of equations, (3.1)-(3.12) along with

(3.19)-(3.20), arises from the fact that inductances are a function of the angle s and

they vary with respect to time. R. H. Park [2]-[3] has proposed a method of changing

the time varying ac quantities into time independent quantities through a

transformation. This transformation is called as Park’s transformation. Next we will

discuss about the Park’s transformation.

3.2 Synchronous Machine Dynamics in Synchronous Reference Frame

The synchronous machine dynamics depend on the rotor angle with respect to

the a a mmf axis, which varies with time. Due to this time varying nature of the

parameters of the synchronous machine it becomes very difficult to analyze the

system. R. H. Park has proposed a method to transform time varying ac quantities into

time invariant quantities [2], [3]. Park’s transformation is explained below.

Parks transformation

The transformation matrix proposed by R. H. Park [3], also called as 0dq

transformation, is as given below:

1

2 2 2

cos( ) cos( 120) cos( 120)

sin( ) sin( 120) sin( 120)s s s

dqo s s sT k

k k k

(3.22)

The choice of 1 2,k k is arbitrary. In standard practice however a value of

1 2 / 3k

and 2 1 / 2k are used. In this course the same value mentioned above are used. There

is an alternative choice which is also used. The alternative choice is 1 2 / 3k and

2 1 / 2k . We will discuss the effect of these choices on the modeling later.

Let us apply Park’s transformation to the three-phase stator currents of a synchronous

generator given below:

3.12

cos( )a m si I t (3.23)

cos( 120 )b m si I t (3.24)

cos( 120 )c m si I t (3.25)

Now let us apply the Park’s transformation on the stator currents given in (3.23) to

(3.25)

0

cos( )

sin( )

0

d a m s s

q dqo b m s s

c

i i I t

i T i I t

ii

(3.26)

But s is the rotor angle with respect to the stationary a a mmf axis, in electrical

radians. Instead of taking s with respect to a stationary reference if we take a

synchronously rotating reference, same speed as the rotor then, we can express as

s st (3.27)

Substituting (3.27) in (3.26) lead to

0

cos( )

sin( )

0

d m

q m

i I

i I

i

(3.28)

Which mean that if the rotor, rotating at synchronous speed, has a fixed angle

difference of with respect to a synchronously rotating reference then ac quantities

, ,a b ci i i can be transformed to dc quantities , ,d q oi i i . It can also be observed that for a

balanced system

2 2m d qI i i (3.29)

3.13

0

1( ) 0

3 a b cI i i i (3.30)

The reason for choosing 1 2 / 3k and

2 1 / 2k can be understood from (3.28) to

(3.30). Due this choice the peak value of the time independent currents ,d qi i is equal

to the peak value of the stator current that is there is a direct correlation between the

stator currents and the Park’s transformed time independent currents. The physical

meaning of this transformation can be understood from Blondel two-reaction theory

[2]. Blondel two-reaction theory says that the traveling wave of mmf, created in the

air gap of the synchronous machine due to the combined effect of three-phase stator

mmfs, can be split into two sinusoidal components in such a way that the peak of one

component is always aligned along d axis and the peak of other component aligned

along q axis. The currents ,d qi i in (3.28) produce the same mmf along d axis and

q axis as suggested by Blondel two-reaction theory. The currents ,d qi i can also be

understood as the currents through two fictitious windings, rotating at synchronous

speed, producing same mmf as that of the fixed stator windings along d axis and

q axis, respectively. In fact from (3.29) it can also be observed that ,d qi i can act as

real and imaginary parts of a current phasor.

We can now convert all three-phase ac quantities into dc quantities through 0dq

transformation. Just like (3.28) we can represent three-phase stator voltages and flux

linkage in terms of 0dq components as

10

0 0

d a a d

q dqo b b dq q

c c

v v v v

v T v or v T v

v vv v

(3.31)

10

0 0

d a a d

q dqo b b dq q

c c

T or T

(3.32)

where,

3.14

1

cos( ) sin( ) 1

cos( 120) sin( 120) 1

cos( 120) sin( 120) 1dqo

s s

s s

s s

T

(3.33)

Now in order to check the effect of 0dq transformation of voltages and current on the

instantaneous power we can find the relation between the complex power in terms of

abc components and 0dq as

1 1

1 1

30 0

23

( ) 0 02

0 0 3

TT

a a d d

b b dqo q dqo q

c c o o

T T

d d d dT

q dqo dqo q q

o o o

v i v i

S v i T v T i

v i v i

v i v i

v T T i v i

v i v

q

oi

(3.34)

or

3( 2 )

2a a b b c c d d q q o oS v i v i v i v i v i v i (3.35)

Hence, the transformation is not power invariant that

is ( )a a b b c c d d q q o ov i v i v i v i v i v i . The readers can verify that by choosing

1 2 / 3k and 2 1/ 2k we can get a power invariant transformation i.e.

( )a a b b c c d d q q o ov i v i v i v i v i v i

Now applying 0dq transformation to equations (3.1)-(3.3), we can get

3.15

1 1

1 1

0 0

0 0

0 0

0 0

0 0

0 0

d s d d

q dqo s dqo q dqo dqo q

so o o

s d d d

s q dqo dqo q dqo dqo q

s o o o

v r id

v T r T i T Tdt

rv i

r id d

r i T T T Tdt dt

r i

0 0 0 0

0 0 0 0

0 0 00 0

s d d d

s q q q

s o o o

r id

r idt

r i

(3.36)

Hence, we can write the stator and rotor equations in terms of 0dq as

dd s d q

dv r i

dt

(3.37)

qq s q d

dv r i

dt

(3.38)

oo s o

dv r i

dt

(3.39)

fdfd fd fd

dv r i

dt

(3.40)

11 1 1

dd d d

dv r i

dt

(3.41)

11 1 1

qq q q

dv r i

dt

(3.42)

22 2 2

qq q q

dv r i

dt

(3.43)

In equation (3.37) and (3.38), the terms ,q d are called as speed induced

voltages in the stator and these induced voltages are due to the variation of the flux

with respect to space. Similarly, the terms dd

dt

, qd

dt

are called as transformer

3.16

induced voltages and these are induced due to the variation of the flux with respect to

time. In steady state the transformer induced voltages will be zero and only speed

induced voltages will be present. It has to be observed that 0dq transformation is not

required for rotor side parameters as they are already defined either along ord q - axis.

Applying 0dq transformation to flux linkage equation (3.9) and (3.11), we get

11

1

0 02

fd

d dd

q dqo ss dqo q dqo srq

q

ii

iT L T i T L

ii

i

(3.44)

Substituting equation (3.19) and (3.21) along with equation (3.1) in (3.44) will lead to,

after simplification,

0 0

11

1 21

2

3( ) 0 0

23

0 ( ) 02

0 0 0

0 0

0 0

0 0 0 0

aao aap

d d

q aao aap q

fdsfd s d

ds q s q

q

q

l li

l l i

i

il l

il l

i

i

(3.45)

Let 3 3

2 2d aao aap q aao aapl l l and l l l . Define, md d lsl l l= - and mq q lsl l l= - .

Here, ,d ql l are inductances along d, q-axis corresponding to the total flux

linkage. However, there will be leakage flux in the slots, at winding ends

and zigzag paths between stator and rotor tooth faces. To account for this

leakage a leakage inductance lsl is considered and is assumed to be same

3.17

along d, q -axis. The mutual inductance along d, q –axis is represented as

,md mql l . With these assumptions the following equations can be written.

1 1( )d ls md d sfd fd s d dl l i l i l i (3.46)

1 1 2 2( )q ls mq q s q q s q ql l i l i l i (3.47)

o ls ol i (3.48)

1 1

3

2fd sfd d fdfd fd fd d dl i l i l i (3.49)

1 1 1 1 1

3

2d sfd d dfd fd d d dl i l i l i (3.50)

1 1 1 1 1 1 2 2

3

2q s q q q q q q q ql i l i l i (3.51)

2 2 2 1 1 2 2 2

3

2q s q q q q q q q ql i l i l i (3.52)

It can be observed from (3.46) and (3.52) that the mutual inductances, though

now are not a function of time, are not same. In order to make all the mutual

inductances between the windings equal, for sake of reducing complexity in analysis,

a proper per unit system can be chosen. Hence, let us look at per unit representation

of the synchronous machine model developed so far.

3.3 Per Unit Representation

Let stator line to neutral rms voltage be taken as the base voltage and the

generator three-phase apparent power be taken as the base power. With these two

bases defined the other base quantities can be defined as given in equation (3.53).

These are the base quantities in stator reference frame. It can be observed from

equation (3.28) that the voltage and current in terms of 0dq are same as the peak

values of abc components and the rated MVA expressed in terms of 0dq components

3.18

has an additional factor 3 / 2 involved. Hence, the stator base quantities in terms of

0dq components can be expressed as given in (3.54).

( ) kV,

Rated MVA 3

kA,3

,

2 elec.rad/s, 50 60 Hz

2mech.rad/s

H

1Wb-turn, s/rad

base rms

base base base

basebase

base

basebase

base

base s

mbase base

basebase

base

basebase base

base base

V V l n

S V I

SI

V

VZ

I

f f or

PZ

L

Vt

(3.53)

_

_ _

__

__

_

__

__

2 kV

3MVA

22 2

2 2 kA3 3 32

H

Wb-turn

dqo base base

base dqo base dqo base

base base basedqo base base

dqo base basebase

dqo basedqo base

dqo base

dqo basedqo base

base

dqo basedqo base

base

V V

S V I

S S SI I

V VV

VZ

I

ZL

V

(3.54)

So, far we have defined stator base quantities in terms of abc and

0dq components. Now we can define rotor base quantities. The rotor base quantities

should be chosen in such a way that the mutual inductances between the stator and

rotor windings and between rotor windings themselves should be same. To do this we

3.19

can take the same power base, baseS , as the power base for the rotor windings also but

define current bases in a such a way that the mutual inductances among different

windings become same. The field winding base quantities are shown below:

_ _

_ _

_ __

__

__

_

__

_

3

2

base fd base fd base

mdfd base dqo base

sfd

sfdbasefd base dqo base

fd base md

fd basefd base

base

fd basefd base

fd base

fd basefd base

base fd base

S V I

lI I

l

lSV V

I l

V

VZ

I

VL

I

(3.55)

Similarly we can define the base quantities of damper windings along d-axis as

following:

11 _ 1 _ 1 _ _ 1 _ _

1 _ 1

1 _ 1 _ 1 _1 _ 1 _ 1 _

1 _ 1 _

3, ,

2

, ,

base s d mdbase d base d base d base dqo base d base dqo base

d base md s d

d base d base d based base d base d base

base d base base d base

S l lS V I V V I I

I l l

V V VZ L

I I

(3.56)

The base quantities of windings along q-axis are given as:

3.20

1 _ 1 _ 2 _ 2 _

1 _ _ 2 _ _1 2

1 21 _ _ 2 _ _

1 _ 2 _1 _ 2 _ 1 _

,

3 3,

2 2

, ,

base q base q base q base q base

mq mqq base dqo base q base dqo base

s q s q

s q s qq base dqo base q base dqo base

mq mq

q base q baseq base q base q base

base base

S V I V I

l lI I I I

l l

l lV V V V

l l

V VZ

1 _

1 _

2 _ 1 _ 2 _2 _ 1 _ 2 _

2 _ 1 _ 2 _

,

, ,

q base

q base

q base q base q baseq base q base q base

q base base q base base q base

V

I

V V VZ L L

I I I

(3.57)

Let,

00

_ _ _

00

_ _ _

111 1

_ 1 _ 1 _

2 12 1

2 _ _ 1 _

11

1 _

, , ,

, ,

, , ,

, , ,

qdd q

dqo base dqo base dqo base

qdd q

dqo base dqo base dqo base

fd qdfd d q

fd base d base q base

q fd dq fd d

q base fd base d base

qq

q base

vv vV V V

V V V

ii iI I I

I I I

v vvV V V

V V V

v i iV I I

V I I

iI

I

22

2 _ _

11

_ _ 1 _

1 21 2

1 _ 2 _ _

1 211 1 2

_ 1 _ 1 _ 2 _

, , ,

, , ,

, , ,

, , ,

q dq d

q base dqo base

q fd dq fd d

dqo base fd base d base

q q sq q s

q base q base dqo base

fd q qdfd d q q

fd base d base q base q base

iI

I

rR

Z

r r rrR R R R

Z Z Z Z

(3.58)

3.3.1 Stator and rotor winding voltage equations in per units

The left hand side parameters in equation (3.58) are nothing but synchronous

generator parameters in per units. Now equations (3.37)-(3.43) can be expressed in

per units, with base quantities defined by (3.53)-(3.57) and per unit synchronous

machine parameters represented as given in (3.58), as

3.21

1 dd s d q

base base

dV R I

dt

(3.59)

1 qq s q d

base base

dV R I

dt

(3.60)

01o s o

base

dV R I

dt

(3.61)

1 fdfd fd fd

base

dV R I

dt

(3.62)

11 1 1

1 dd d d

base

dV R I

dt

(3.63)

11 1 1

1 qq q q

base

dV R I

dt

(3.64)

22 2 2

1 qq q q

base

dV R I

dt

(3.65)

The instantaneous stator power in 0dq , given in (3.35), can also be expressed in per

units. Before expressing instantaneous stator power in per unit there is an important

aspect which can be understood from instantaneous stator power when expressed in

terms of stator voltage equations, given in (3.37) to (3.39). Hence, by substituting

equations (3.37) to (3.39) in (3.35), and rearranging the terms we get

2 2 2

32

2

3 3 32 2

2 2 2

qd os d q d s q d q s o o

qd od q o d q q d s d s q s o

dd dS r i i r i i r i i

dt dt dt

dd di i i i i r i r i r i

dt dt dt

(3.66)

For a balanced system, 0 0,v i are zero hence equation (3.66) can be written as

2 23 3 3

2 2 2qd

d q d q q d s d s q

ddS i i i i r i r i

dt dt

(3.67)

3.22

It can be observed from equation (3.67) that there are three terms in the equation. The

first term, 3

2qd

d q

ddi i

dt dt

, corresponds to the rate of change of magnetic energy

in the stator coils. The second term, 3

2 d q q di i , is the power transferred over air

gap. The third term, 2 23

2 s d s qr i r i , is stator copper losses. The power transferred

over the air gap appears as torque, at the shaft speed, either opposing the motion of

the rotor in case of generator or aiding the motion of the rotor in case of motor. Hence,

we can find the electrical torque, at shaft speed from the power transferred over the air

gap as

2 3

2

3N.m

2 2

e m e d q q d

e d q q d

t t i iP

Por t i i

(3.68)

Here, et is the electrical torque at the shaft with shaft speed m in mech.rad/s with

P number of poles. The electrical torque can now be expressed in per units. Before

that we need to define the electrical torque base at shaft speed and is given as

0_ 0 _

0_0 _ 0_ 0 _

1 32 2 2

3 3

2 2 2 2

basebase dq base dq base

base base

dq basedq base dq base dq base

base

ST V I

P PVP P

I I

(3.69)

Dividing the electrical torque given in (3.68) by the base torque defined in (3.69)

leads to

0_ 0 _

32 2

32 2

d q q de

e d q q dbase

dq base dq base

Pi it

T I IPT I

(3.70)

3.23

3.3.2 Stator and rotor flux linkage equations in per units

In section 3.3.1 the per unit representation of stator and rotor winding voltage

equations, in synchronous reference frame, were explained. Here, the per unit

representation of the flux linkage equations will be explained. Let us start with the

windings along d-axis

1 1( )d ls md d sfd fd s d dl l i l i l i (3.71)

1 1

3

2fd sfd d fdfd fd fd d dl i l i l i (3.72)

1 1 1 1 1

3

2d sfd d dfd fd d d dl i l i l i (3.73)

From equation (3.58) we know that

_

dd

dqo base

iI

I , 1

1_ 1 _

,fd dfd d

fd base d base

i iI I

I I (3.74)

We can express equation (3.74) in a different form as

_d d dqo basei I I , _ 1 1 _ 1,fd fd fd base d d base di I I i I I (3.75)

Since, we need to represent equation (3.71) to (3.73) in per units we need to divide

them by their respective base quantities and substituting equation (3.75) we get

_ _ 1 1 _ 1

_ _ _ _

( )( )ls md d dqo base sfd fd fd base s d d base dd

dqo base dqo base dqo base dqo base

l l I I l I I l I I

(3.76)

_ _ 1 1 _ 1

_ _ _ _

( )3

2fd sfd d dqo base fdfd fd fd base fd d d base d

fd base fd base fd base fd base

l I I l I I l I I

(3.77)

3.24

_ 1 _ 1 1 1 _ 11

1 _ 1 _ 1 _ 1 _

( )3

2sfd d dqo base dfd fd fd base d d d base dd

d base d base d base d base

l I I l I I l I I

(3.78)

It can be observed from equation (3.54)-(3.56) that _ _ /dqo base dqo base baseV ,

_ _ /fd base fd base baseV , 1 _ 1 _ /d base d base baseV . Substituting these and also using

equation (3.58) in equation (3.76)-(3.78), we get

_ _ 1 1 _ 1

_ _ _

( )( )base ls md d dqo base base sfd fd fd base base s d d base dd

dqo base dqo base dqo base

l l I I l I I l I I

V V V

(3.79)

_ _ 1 1 _ 1

_ _ _

( )3

2base sfd d dqo base base fdfd fd fd base base fd d d base d

fdfd base fd base fd base

l I I l I I l I I

V V V

(3.80)

_ 1 _ 1 1 1 _ 11

1 _ 1 _ 1 _

( )3

2base sfd d dqo base base dfd fd fd base base d d d base d

dd base d base d base

l I I l I I l I I

V V V

(3.81)

In order to simplify equations (3.79) to (3.81) for ease of analysis, we have defined

base quantities of rotor windings, given in (3.55) to (3.57), as

_ _md

fd base dqo basesfd

lI I

l , _ _

_

3

2sfdbase

fd base dqo basefd base md

lSV V

I l (3.82)

1 _ _1

mdd base dqo base

s d

lI I

l , 1

1 _ _1 _

3

2base s d

d base dqo based base md

S lV V

I l (3.83)

It is important to understand the choice of base quantities for rotor windings [4].

The rotor winding base quantities are chosen such that most of the mutual inductance

terms in equation (3.79) to (3.81) become same, mdl . For example if we substitute

the field current base given in (3.82) in (3.79), the mutual inductance term sfdl gets

cancelled and replaced by mdl . We can also express the flux linkage equation of q -

axis windings in per units as

3.25

_ 1 1 1 _ 2 2 _ 2

_ _ _

( )( )base ls mq q dqo base base s q q q base base s q q base qq

dqo base dqo base dqo base

l l I I l I I l I I

V V V

(3.84)

1 _ 1 1 1 1 _ 1 2 2 2 _1

1 _ 1 _ 1 _

( )3

2base s q q dqo base base q q q q base base q q q q base

qq base q base q base

l I I l I I l I I

V V V

(3.85)

2 _ 2 1 1 1 _ 2 2 2 2 _2

2 _ 2 _ 2 _

( )3

2base s q q dqo base base q q q q base base q q q q base

qq base q base q base

l I I l I I l I I

V V V

(3.86)

Let us define some more quantities to further simplify equations (3.79) to (3.81) and

(3.84) to (3.86)

0_

0 _ 0 _ 0_

base ls dq basels base lsls

dq base dq base dq base

l Ix lX

Z Z V

(3.87)

0_

0_ 0_ 0_

base md dq basemd base mdmd

dq base dq base dq base

l Ix lX

Z Z V

(3.88)

0_

0_ 0 _ 0_

mq base mq base mq dq basemq

dq base dq base dq base

x l l IX

Z Z V

(3.89)

_

_ _ _

fdfd base fdfd base fdfd fd basefd

fd base fd base fd base

x l l IX

Z Z V

(3.90)

1 1 1 _1 1 1 11

1 _ 1 _ 1 _

base d d d based d base d dd

d base d base d base

l Ix lX

Z Z V

(3.91)

1 1 1 1 1 1 1 _1

1 _ 1 _ 1 _

q q base q q base q q q baseq

q base q base q base

x l l IX

Z Z V

(3.92)

2 2 2 2 2 2 2 _2

2 _ 2 _ 2 _

q q base q q base q q q baseq

q base q base q base

x l l IX

Z Z V

(3.93)

3.26

1 1 _ 11 _

_ _ 1

1

_ 1

base fd d d base base fd d sfdfd d fd base

fd base fd base s d

base fd d sfd

fd base s d

l I l lX I

V V l

l l

Z l

(3.94)

1 2 2 _ 1 2 11 2 1 _

1 _ 1 _ 2

1 2

1 _ 1

base q q q base base q q s qq q q base

q base q base s q

base q q sfd

q base s d

l I l lX I

V V l

l l

Z l

(3.95)

Let,

1 1 1 1 2 2

, , ,

, , ,

d ls md q ls mq lfd fd md

l d d md l q q mq l q q mq

X X X X X X X X X

X X X X X X X X X

(3.96)

The reactance ,d qX X are called as direct axis and quadrature axis synchronous

reactance. Substituting equations (3.87) to (3.96) in equations (3.79) to (3.81), (3.84)

to (3.86) and taking 1 1 2,fd d q qd q

md mq

L LK K

L L , we get

1( )d d d md fd md dX I X I X I (3.97)

1( )fd md d fd fd d md dX I X I K X I (3.98)

1 1 1( )d md d d md fd d dX I K X I X I (3.99)

1 2( )q q q mq q mq qX I X I X I (3.100)

1 1 1 2( )q mq q q q q mq qX I X I K X I (3.101)

2 1 2 2( )q mq q q mq q q qX I K X I X I (3.102)

We can observe from equations (3.97) to (3.102) the flux linkage equations in

per unit system are less complex than the flux linkage equations in actual values. The

second advantage is that all the mutual inductances/reactances along d-axis are equal

if dK is assumed to be unity. Similarly, all mutual inductances/reactances along q-

3.27

axis are equal if qK is assumed to be unity. In fact in actual synchronous generators

the values of dK and qK are very near to unity.

The per unit representation of the voltage equation and flux linkage equations of a

synchronous generator are summarized below:

Stator voltage equations

1 dd s d q

base base

dV R I

dt

(3.103)

1 qq s q d

base base

dV R I

dt

(3.104)

01o s o

base

dV R I

dt

(3.105)

Rotor voltage equations

1 fdfd fd fd

base

dV R I

dt

(3.106)

11 1 1

1 dd d d

base

dV R I

dt

(3.107)

11 1 1

1 qq q q

base

dV R I

dt

(3.108)

22 2 2

1 qq q q

base

dV R I

dt

(3.109)

Stator flux linkage equations

1( )d d d md fd md dX I X I X I (3.110)

1( )fd md d fd fd md dX I X I X I (3.111)

1 1 1( )d md d md fd d dX I X I X I (3.112)

Rotor flux linkage equations

3.28

1 2( )q q q mq q mq qX I X I X I (3.113)

1 1 1 2( )q mq q q q mq qX I X I X I (3.114)

2 1 2 2( )q mq q mq q q qX I X I X I (3.115)

3.4 Synchronous Machine Parameters

It is customary to represent the voltage equation and flux linkage equations in terms

of sub-transient and transient reactances, open circuit sub-transient and transient time

constants. Parameters of the industry grade synchronous generators are given in terms

of above parameters. Let us define the parameters

3.4.1 Sub-transient and transient reactance

"

1

11 1 1d ls

md lfd l d

X X

X X X

(3.116)

"dX is called as the direct axis sub-transient reactance. The expression given in

(3.116) can be derived. Suppose a voltage is applied at the stator terminals with all

other rotor circuits short circuited such that only current dI flows then immediately

after the voltage is applied ( 0t ) the flux linkages fd and 1d will be zero. Hence

we can write equations (3.110) to (3.112) as

1( )d d d md fd md dX I X I X I (3.117)

10 ( )fd md d fd fd md dX I X I X I (3.118)

1 1 10 ( )d md d md fd d dX I X I X I (3.119)

Solving for 1,fd dI I in term of dI , also considering equation (3.96), we get

3.29

1

1 1

md l dfd d

md lfd md l d lfd l d

X XI I

X X X X X X

(3.120)

11 1

md lfdd d

md lfd md l d lfd l d

X XI I

X X X X X X

(3.121)

Substituting equation (3.120) and (3.121) in (3.117), we get

21

1 1

( )md lfd l dd d d

md lfd md l d lfd l d

X X XX I

X X X X X X

(3.122)

Since, d ls mdX X X , equation (3.122) can be further simplified as

1

1 1

"

1

11 1 1

md lfd l dd ls d

md lfd md l d lfd l d

ls d d d

md lfd l d

X X XX I

X X X X X X

X I X I

X X X

(3.123)

Similarly the quadrature axis sub-transient reactance can be expressed as

"

1 2

11 1 1q ls

mq l q l q

X X

X X X

(3.124)

The direct axis transient reactance is defined as

' 11 1d ls

md lfd

X X

X X

(3.125)

For deriving equation (3.125) the same logic applied for finding sub-transient

reactance can be used with an additional assumption that the damper winding

3.30

transients settle down faster as compared to field winding transients hence 1dI can be

assumed to be zero. With this assumption, by solving equations (3.117) and (3.118),

we get

2

'11 1

md lfdmdd d d ls d

md lfd md lfd

ls d d d

md lfd

X XXX I X I

X X X X

X I X I

X X

(3.126)

Similarly, quadrature axis transient reactance can be defined as

'

1

11 1q ls

mq l q

X X

X X

(3.127)

3.4.2 Open circuit sub-transient and transient time constants

For finding the sub-transient open circuit time constant a voltage is applied to the field

winding with the stator terminals open circuited so dI is zero. At time 0t the flux

linkage 1d is zero. Substituting these assumption in (3.107) and (3.112) we get

11

dfd d

md

XI I

X (3.128)

11 1 1 1

10fd d

md d d d dbase

dI dIX X R I V

dt dt

(3.129)

or

1 1 11

1 1fd d d dd

base base md md

dI X dI RI

dt X dt X

(3.130)

3.31

Substituting equations (3.130) and (3.129) in (3.106) the following expression can be

obtained

211 1

11 md fd dd d

fd fd dmd base md

X X XX dIV R I

X X dt

(3.131)

Substituting equation (3.111), (3.130) in (3.106) a first order differential equation in

terms of 1dI with a forcing function fdV , can be obtained as

1

1 1 11 11 1

21 1 11

1

21

1

1

fd fd md dfd fd fd

base base

fd d fd d fd dd md dd d fd

base md md base md

md fd d fd d fd ddd fd

base md md

md fd d

base fd

X dI X dIR I V

dt dt

X X X R R XdI X dII I V

X dt X dt X

X X X X R R XdII V

X dt X

X X X

X R

1

11 1 1 1

d mdd fd

d d fd fd d fd d

dI XI V

X R dt X R R X

(3.132)

In a practical generator 1d fdR R hence 1d fdX R can be neglected as compared to

1fd dX R in equation (3.132). With this assumption and further simplification we get

11 1

1 1

1 md lfd d mdl d d fd

base d md lfd fd d

X X dI XX I V

R X X dt X R

(3.133)

Hence, the open circuit direct axis sub-transient time constant, in seconds, is given as

"1 1

1 1

1 1 11 1

md lfddo l d l d

base d md lfd base d

md lfd

X XT X X

R X X RX X

(3.134)

Similarly, the open circuit quadrature axis sub-transient time constant, in seconds, can

be express as

"2

1

1

1 11 1qo l q

base q

mq l q

T XR

X X

(3.135)

3.32

for finding the open circuit direct axis transient time constant, along with the

assumption taken for computing "doT , the current in the damper winding 1dI is

assumed to be zero then from equation (3.106) and (3.111)

fd fd fdX I (3.136)

1fd fdfd fd

base fd fd

X dII V

R dt R

(3.137)

Hence, the open circuit direct axis transient time constant, in seconds, is defined as

' fddo

base fd

XT

R (3.138)

Similarly the open circuit quadrature axis transient time constant, in seconds is

defined as

1'

1

qqo

base q

XT

R (3.139)

Also, defining new variables as

' '1

1

, , mqmd mdq fd fd fd d q

fd fd q

XX XE E V E

X R X (3.140)

We can express equations (3.103) to (3.115) using the sub-transient and transient

reactances and open circuit time constant along with equation (3.140). The rotor

currents along d -axis can be eliminated from equation (3.111) and (3.112) as

1 1

1 1 1 1

fd fd md fd fd md mdd

mdd md d d md d

I X X X X XI

XI X X X X

(3.141)

3.33

let, 21 1 1fd d md lfd l d lfd md l d mdX X X X X X X X X , then equation (3.141) can be

expressed as

21 1 1

21 1

1

1

fd d fd md d d md md d

d md fd fd d fd md md d

I X X X X X I

I X X X X X I

(3.142)

The coefficients of 1, ,fd d dI can be expressed in terms of sub-transient, transient

and steady state reactances as given below

2 ' ' "1 '

1 ' 2

2 ' ' "'

' 2

2

1 1

( )( )1 1 11 1

( )

( )( ),

( )

(1 1 1

fd d md md md d d d dd fd fd fd q

md fd md fd md d ls

md d d d d mdq fd

d ls fd

md dmd d d

md md

X X X X X X X X XX E

X X X X X X X

X X X X X XE

X X X

X X XX

X X

' ' "

1' 2

' "12

1 '

'1 "

'

)( )

( )

( )( )1 1 1

( )

( ),

( )

d d dd

d ls

lfd l d md d d d lsd md md d d d

lfd md d ls

lfd l d mdd lslfd md d ls

d d

X X

X X

X X X X X X XX X X I I I

X X X X

X X XX XX X X X

X X

' "'

' 2

' "'

' 2

' "

1 1' 2

' "2

'

( )1

( )

( ),

( )

( )1

( )

( )1

( )

fd md d dmd fd fd q

fd d ls

fd d d mdq fd

d ls fd

d dfd d d

d ls

lfd md d dfd md md d d d

d ls

X X X XX E

X X X

X X X XE

X X X

X XX

X X

X X X XX X X I I I

X X

Replacing the coefficient in (3.142) with new coefficients, equation (3.142) can be

expressed as

3.34

' ' " ' "' '

1' 2 ' "

' '1

( )( ) ( )( )1

( ) ( )

1( )

d d d d d ls d lsfd q q d d

md d ls d d

q d d d dmd

X X X X X X X XI E E I

X X X X X

E X X I IX

(3.143)

' "

' '1 1' 2

( )( )

( )d d

d q d ls d dd ls

X XI E X X I

X X

(3.144)

Substituting equation (3.143) and (3.44) in equation (3.110) the stator d -axis flux can

be expressed as

1

' ' '1

" ' "" '

1' '

( )

( ) ( )

( ) ( )

d d d md fd md d

q d d d ls d

d ls d dd d q d

d ls d ls

X I X I X I

E X I X X I

X X X XX I E

X X X X

(3.145)

In a similar way the q -axis flux linkage equation along with current 1 2,q qI I can be

expressed as

" ' "" '

2' '

( ) ( )

( ) ( )q ls q q

q q q d qq ls q ls

X X X XX I E

X X X X

(3.146)

' '1 2

1( )q d q q q q

mq

I E X X I IX

(3.147)

' "

' '2 2' 2

( )( )

( )q q

q d q ls q qq ls

X XI E X X I

X X

(3.148)

So far the flux linkage equations were expressed in terms of sub-transient,

transient and steady state reactances. In order to express the voltage equations in

terms of these new parameters some simplification can be considered. In the equations

(3.103) and (3.104), the stator voltage equations, the rate of change of stator fluxing

linkage along ,d q -axis can be neglected. Since, the stator is connected to the rest of

3.35

the network electrically, if its dynamics are considered then the entire network

dynamics i.e. transformers, transmission lines, load etc have to be considered which

will increase the computational burden immensely. Also the stator as well as network

transients are much faster as compared to the rotor dynamics and hence as compared

to rotor dynamics the stator and network transients can be considered as instantaneous

changes. Second assumption is, in the speed induced voltage terms, ,d q the

speed is assumed to be synchronous speed that is base . This assumption does not

mean the speed is constant but the effect of speed variation on the induced voltage is

insignificant and one more advantage is that this assumption counter balances the first

assumption thereby nullifying effect of both assumptions. With this assumptions and

substituting (3.143) to (3.148) in equations (3.106) to (3.109), the following

expression can be obtained.

Note:

1 1"1

1 1

' "1 1 1

' 2

1 1

1 1 1 1 1

( )

( )

md lfd lfd l d lfd md l d mddo l d

base d md lfd base d fd

fdbase d fd base d base d d d

d ls

X X X X X X X XT X

R X X R X

XR X R R X X

X X

"' "

2

' 2

1 1

( )

( )

qobase q q q

q ls

TR X X

X X

1 dd s d q

base base

dV R I

dt

(3.149)

1 qq s q d

base base

dV R I

dt

(3.150)

01o s o

base

dV R I

dt

(3.151)

' ' "

' ' ' ' '1' 2

( )( ) ( )

( )q d d

do q d d d q d ls d d fdd ls

dE X XT E X X I E X X I E

dt X X

(3.152)

3.36

" ' '11( )d

do q d ls d dd

T E X X Idt

(3.153)

' "'

' ' ' ' '2' 2

( )( ) ( )

( )q qd

qo d q q q d q ls q qq ls

X XdET E X X I E X X I

dt X X

(3.154)

2" ' '2( )q

qo d q ls q q

dT E X X I

dt

(3.155)

" ' "" '

1' '

( ) ( )

( ) ( )d ls d d

d d d q dd ls d ls

X X X XX I E

X X X X

(3.156)

" ' "" '

2' '

( ) ( )

( ) ( )q ls q q

q q q d qq ls q ls

X X X XX I E

X X X X

(3.157)

Apart from equations (3.149) to (3.157), equation of motion or swing equation also

needs to be considered for synchronous machine. The equation of motion is given as,

given in Chapter 2, equation (2.4),

2

2

2 sm e

dJ t t

P dt

(3.158)

Where, ,m et t are the mechanical input torque from the prime mover and the electrical

output torque. 2.J kg m is the inertia constant of the rotating machine. Now defining

machine inertia constant as

21 2

2MJ/MVA

base

base

JP

HS

.

Replace J with the defined machine inertia constant in equation (3.158), then

2

222

base sm e

base

HS dt t

dtP

(3.159)

Rearranging we can express (3.159) as

3.37

2

2

2

2 2

s m e m em e

base basebase base base

base base

d t t t tHT T

S Sdt T T

P P

(3.160)

The rotor angle s is with respect to fixed stator reference frame but instead it can be

expressed with respect to synchronously rotating reference frame, as given in equation

(3.27) s baset , also including the damping torque as given in equation (2.32)

Chapter 2 in (3.160) lead to

sbase base

dd

dt dt

(3.161)

2

2

2 2m e base

base base

H d H dT T D

dt dt

(3.162)

Equations (3.149) to (3.157), along with equations of motion (3.161) to (3.162),

describe the behavior of a synchronous machine.

3.5 Effect of Saturation on Synchronous Machine Modeling

The modeling of synchronous machine discussed so far assumed that there is no

saturation either in the stator or rotor core. Before considering the effect of saturation

some assumptions are made [5], they are

1. The leakage inductances of stator and rotor have negligible effect due to

saturation of the stator and rotor core as the leakage flux path is mostly

through the air.

2. The leakage fluxes of stator and rotor have negligible effect on saturation

of the stator and rotor core and the saturation mostly depends on the air-

gap flux linkage.

3. Hence, most of the effect of saturation is on the mutual inductances

,md mqX X .

3.38

4. The effect of saturation on mutual inductances can be computed through

the open circuit characteristics of the synchronous generator and the same

effect can be considered to be applicable even when the generator is

loaded.

5. There will be no coupling inductances between the d-axis and q-axis

windings due to the nonlinearity introduced by the saturation.

In order to discuss the effect of saturation the open circuit characteristic should be

understood. From equation (3.110) and (3.113) it can be observed that at no load

open circuit condition the following expression hold

, 0t d md fd qV X I (3.163)

It can be observed that in per unit the stator flux linkage and the stator terminal

voltage in no load condition are same. A stator flux linkage can be defined in case of

open circuit as

2 2t d q d tV (3.164)

It can be understood from equations (3.163) and (3.164) that the terminal

voltage or the stator flux linkage is directly proportional to the field current in case

saturation is not considered as can be seen from the air-gap line shown in Fig. 3.3.

With saturation of stator and rotor core considered the relation between the terminal

voltage and the field current during open circuit is no longer linear.

In case of no saturation a field current fagI produces 1.0pu terminal voltage but with

saturation to produce the same terminal voltage the field current required is fnlI which

is higher than fagI as can be seen from Fig. 3.3. In the same figure the short circuit

characteristics are also given. The field current to produce 1.0 pu short circuit current

is fscI . Because of the armature reaction in case of short circuit the field current fscI

will correspond to an internal voltage which will lie on the linear portion of the open

circuit characteristics.

3.39

Fig. 3.3: Open circuit characteristics of a synchronous generator

If we assume that the stator resistance is negligible and taking the unsaturated

synchronous reactance as _s unsatX then we can write the following expression

_1.0fsc s unsatKI X (3.165)

Where, K is the slope of the air-gap line. Equation (3.165) is written on the basis that

during a short circuit the terminal voltage is zero and the internal voltage is fully

applied across the synchronous reactance and hence multiplying the synchronous

reactance with the short circuit current gives the internal voltage. Similarly, for

producing 1.0 pu terminal voltage during no load an expression can be written as

1.0fagKI (3.166)

Hence,

_fsc

s unsatfsag

IX

I (3.167)

The synchronous reactance unsaturated can now be defined as the ratio of field

current required to produce rated short circuit current to the field current required to

fscIfnlIfagI fI

1.0

( )tV pu

1.0

Air-gap line

( )scI pu

3.40

produced 1.0 pu terminal voltage without saturation or from air-gap line. On similar

reasoning we can define the saturated synchronous reactance as

_fsc

s satfsnl

IX

I (3.168)

Short circuit ratio (SCR) is defined as the field current required to produced the rated

terminal voltage at rated speed at no load considering the effect of saturation to the

field current required to produce the rated short circuit current

_

1fsnl

fsc s sat

ISCR

I X (3.169)

Now in order to estimate the effect of saturation on the synchronous reactance Fig. 3.4

can be used.

Fig. 3.4: Open circuit characteristics to compute the effect of saturation

fnlIfagI fI

flux linkage

t

Air-gap line _t ag

J

lin

3.41

From equation (3.163) and (3.164) and the assumption made that the saturation does

not effect the leakage inductance the saturated and unsaturated d-axis inductance can

be written as

_tagt

md unsatfag fnl

XI I

(3.170)

_t

md satfnl

XI

(3.171)

Dividing equation (3.171) by (3.170) leads to

_ _ _t t

md sat md unsat md unsattag t J

X X X

(3.172)

From Fig. 3.4 it can be observed that if t is below or equal to lin then there is no

nonlinearity and hence J is zero so from equation (3.172) the saturated and

unsaturated mutual inductances are same. If t is greater than lin then J can be

approximated by the exponential mathematical expression given in equation (3.173)

sat t linBJ satA e (3.173)

In case of cylindrical rotor it can be assumed that the effect of saturation is same

along both the d and q axis hence

_ _t

mq sat mq unsatt J

X X

(3.174)

In case of salient pole rotor since the air-gap along the q-axis is significantly more we

can assume that the effect of saturation is negligible. Equation (3.173) and (3.174) are

defined with respect no load condition but it can be extended to loaded condition as

well. This can be done as following

3.42

1

1 2

( )

( )

d ls md d md fd md d ls d md

q ls mq q mq q mq q ls q mq

X X I X I X I X I

X X I X I X I X I

(3.175)

Substituting equation (3.103) and (3.104), in steady state, in equation (3.175) gives

md d ls d q s q ls d

mq q ls q d s d ls q

X I V R I X I

X I V R I X I

(3.176)

or

md mq q s q ls d d s d ls q

d q s ls d q

j V R I X I j V R I X I

j V jV R jX I jI

(3.177)

Now if we define an internal voltage behind the leakage reactance as

l t s ls tE V R jX I (3.178)

Then we can write observe from equation (3.177) and (3.178) that

2 2t md mq lE (3.179)

The flux linkage computed in equation (3.179) can be used in equation (3.172) and

(3.174) to compute the saturated mutual inductances along d and q axis. The saturated

reactances _ _,md sat mq satX X have to be used to compute the sub-transient and transient

reactances, and time constants. The saturated values of sub-transient and transient

reactances and time constants should be used in equations (3.149) to (3.157).

3.6 Estimation of Synchronous Machine Parameters through

Operational Impedance

The synchronous machine parameters, sub-transient and transient inductances

and time constants, should be found out through some experiments. An operational

3.43

impedance can be defined which can be computed through test and then this

operational impedance can be related to the synchronous machine parameters. This

can be done through standstill frequency response [6], [7]. Before going to the details

of the test let us first define operational impedance.

The stator equations given in equations (3.103) to (3.105) can be written in the

Laplace domain (with d

dtreplaced by 0s j ) gives

rd s d q d

base base

sV R I

(3.180)

rq s q d q

base base

sV R I

(3.181)

0o s obase

sV R I

(3.182)

If the flux linkage can be defined as

( ) ( )d dop d op fdX s I G s V (3.183)

( )q doq qX s I (3.184)

0 ( )oop oX s I (3.185)

where, ( )dopX s , ( )doqX s are the operational reactances along d and q axis. Now

suppose a test is done by applying a variable frequency stator voltage with the rotor at

stand still such that only the d – axis voltage and current will be present, due to the

position of the rotor with respect to the stator, then substituting equation (3.183) in

equation (3.180) with 0r will lead to

( ) ( )d s dop d op fdbase base

s sV R X s I G s V

(3.186)

3.44

If in the rotor the field is shorted then fdV is zero and hence with ,d dV I known from

the test at different frequencies the operational d – axis reactance ( )dopX s can be

found. Similarly, we can repeat the test by applying a test voltage to the stator with

rotor at standstill and positioned in such a way with respect to the rotor that only q

axis voltage and current are present then substituting equation (3.184) in (3.181), with

0r lead to equation (3.187) from which the operational reactance ( )doqX s can be

found for different frequencies.

( )q s qop qbase

sV R X s I

(3.187)

Theoretically we can express ( )dopX s , ( )doqX s in terms of the sub-transient and

transient inductances and time constants. This can be done as following, with the

damper windings voltages 1 0dV , by expressing voltage equations of the d-axis

windings in Laplace domain as

fd fd fd fdbase

sR I V

(3.188)

1 1 1d d dbase

sR I

(3.189)

Substituting rotor flux linkage equations in equations (3.188) and (3.189) and

elimination of rotor current lead to

3.45

2

1 1

2

1 1

1 1

2

1 1

md d d mdbase base base

fd d

d d fd fd mdbase base base

d dbase

d d fd fd mdbase base base

s s sX R X X

I Is s s

R X R X X

sR X

s s sR X R X X

fdV

(3.190)

1

1 1

mdd d fd

based d

base

XsI I I

sR X

(3.191)

Substituting equations (3.190) and (3.191) in the d-axis stator flux linkage equations

lead to equation (3.183) where

21 1

2

1 1

2

( )md d d md fd fd

base base base basedop d

d d fd fd mdbase base base

s s s sX R X X R X

X s Xs s s

R X R X X

(3.192)

1 1

2

1 1

( )md d d md

base baseop

d d fd fd mdbase base base

s sX R X X

G ss s s

R X R X X

(3.193)

Equation (3.192) and (3.193) can also be expressed in factored form in terms of time

constants as

' "

' "

' "

1 1( )

1 1

1( )

1 1

d d

dop d

do do

kdop o

do do

sT sTX s X

sT sT

sTG s G

sT sT

(3.194)

3.46

Where, ' ",do doT T are the open circuit transient and sub-transient time constant as given

in equations (3.185) and (3.186). ' ",d dT T are called the short circuit transient and sub-

transient time constants and given as

"1

1

'

1

1

1

1

,

md lfd lsd l d

base d md ls lfd ls md lfd

md lsd lfd

base fd md ls

md l do kd

fd d

X X XT X

R X X X X X X

X XT X

R X X

X XG T

R R

(3.195)

In a similar way the q-axis operational reactance can also be defined as

' "

' "

1 1( )

1 1

q q

qop q

qo qo

sT sTX s X

sT sT

(3.196)

The open circuit time constant ' ",qo qoT T are already defined. The short circuit time

constant are given as

1"2

2 1 1

'1

1

1

1

mq l q lsq l q

base q mq ls l q ls mq l q

mq lsq l q

base q mq ls

X X XT X

R X X X X X X

X XT X

R X X

(3.197)

It can be observed equations (3.94) and (3.96) that for 0s

( ) ,

( )

dop d

qop q

X s X

X s X

(3.198)

As s , like immediately after a disturbance the effect on stator flux linkage,

3.47

' "

"

' "

' "

"

' "

( ) ,

( )

d d

dop d d

do do

q q

qop q q

qo qo

T TX s X X

T T

T TX s X X

T T

(3.199)

In case the damper windings are absent then as s

'

'

'

'

'

'

( ) ,

( )

d

dop d d

do

q

qop q q

qo

TX s X X

T

TX s X X

T

(3.200)

In a synchronous generator ' ' " "do d do dT T T T and ' ' " "

qo q qo qT T T T . Hence, if the

gain variation of the transfer function ( )dopX s and ( )qopX s with respect to frequency

variation is plotted then the corner frequencies will correspond to the time constants

' ' " ", , ,do d do dT T T T and ' ' " ", , ,qo q qo qT T T T with 'doT , '

qoT corresponding to lowest corner

frequency and "dT , "

qT corresponding to highest corner frequency. With this all the

synchronous generator parameters can be found out. But how to apply a test voltage

such that only d-axis or q-axis, voltage and current will be present to find the

operational impedance along either d-axis or q-axis has to be explained.

Let a test voltage, testv , be applied across b-phase and c-phase with a-phase open and

the standstill rotor is positioned such that its d-axis lead the mmf axis of a-phase of

the stator by 90 so that s in parks transformation will be fixed at 90 since the rotor

is standstill, then

, 0,

, 0,test b c a b c

test b c a b c

v v v v v v

i i i i i i

(3.201)

3.48

2 3 3 2 3 2 3 3 2, 3

3 2 2 3 2 3 2 2 3

2 1 1 2 1 10, 0

3 2 2 3 2 2

d b c test d b c test

q b c q b c

v v v v i i i i

v v v i i i

(3.202)

let the line-line test voltage and test current be defined as 3 2 costest t o ov V t

and 2 costest t o oi I t then

2 33 2 cos ,

3 22

3 2 cos3

d t o o

d t o o

v V t

i I t

(3.203)

In per units equation (3.203) can be written as

_

_

2 3cos ,

3 22

3 cos3

d t pu o o

d t pu o o

v V t

i I t

(3.204)

Hence, the d-axis operational impedance as a function of the variable frequency o is

given as

_

_

2t pu ddop o

t pu d

V VZ

I I (3.205)

_

_

_

_

1 1

2 2

1

2

t pud os dop dop o

d base t pu

t pubasedop s

o t pu

VVR j X Z

I I

VX j R

I

(3.206)

Similarly for finding the q-axis operational reactance test voltage has to be applied

like before with the standstill rotor positioned such that the angle between d-axis of

the rotor and a-phase is zero then

3.49

2 1 1 2 1 10, 0

3 2 2 3 2 2

2 3 3 2 3 2 3 3 2, 3

3 2 2 3 2 3 2 2 3

d b c d b c

q b c test q b c test

v v v i i i

v v v v i i i i

(3.207)

Hence, the operational reactance of q-axis can be written as

_

_

_

_

1 1

2 2

1

2

q t puos qop qop o

q base t pu

t pubaseqop s

o t pu

V VR j X Z

I I

VX j R

I

(3.208)

3.50

Example Problems

E1. A three-phase 300 MVA, 20 kV, 0.9 pf, 50 Hz, 2 pole synchronous generator has

the following parameters.

( )

( )

4.675 0.0534cos 2

2.3375 0.0534cos 23

0.5792

67.2cos

1084.08

0.0014

0.0635

aa

ab

ls

afd

fd

s

fd

l mH

l mH

l mH

l mH

l mH

r

r

q

pq

q

= +

æ ö÷ç=- - + ÷ç ÷çè ø

=

=

=

= W

= W

Define the base quantities and express all the generator parameters in per units in 0dq

reference frame.

Sol:

The d-axis and q-axis inductances are given as

( )34.675 0.0534 7.0926

2dl mH= ´ + = (E1.1)

( )34.675 0.0534 6.9324

2ql mH= ´ - = (E1.2)

6.5134md d lsl l l mH= - = (E1.3)

6.3532mq q lsl l l mH= - = (E1.4)

3.51

The base quantities are defined as

0 _

0 _

0 _0 _

0 _

0 _0 _

2011.54 ,

3300 ,

8.665 ,3

2 50 314.159 /

2 16.32 ,

2 12.25 ,

1.3317 ,

4.2389

base

base

basebase

base

base

dq base base

dq base base

dq basedq base

dq base

dq basedq base

base

V kV

S MVA

SI kA

V

rad s

V V kV

I I kA

VZ

I

ZL

w p

w

= =

=

= =

= ´ ´ =

= =

= =

= = W

= =

_ 0 _

__

__

_

__

,

1.1873 ,

252.66

212.86 ,

677.55 ,

mdfd base dq base

afd

basefd base

fd base

fd basefd base

fd base

fd basefd base

base

mH

lI I kA

l

SV kV

I

VZ

I

ZL mH

w

üïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïïýïïïïïïïïïï= = ïïïïïïï= = ïïïïïïïï= = W ï

= =þ

ïïïïïïïïïï

(E1.5)

The per unit generator parameters can now be computed as

0 _ 0 _

0 _ 0 _

0 _ _

0 _

1.673 , 1.6354 ,

1.5365 , 1.4987 ,

0.1366 , 1.6 ,

0.0635 , 0.00105 ,

qdd q

dq base dq base

mqmdmd mq

dq base dq base

fdlsls fd

dq base fd base

slfd fd md s

dq base

ffd

llL pu L pu

L L

llL pu L pu

L L

llL pu L pu

L L

rL L L pu R pu

Z

rR

= = = =

= = = =

= = = =

= - = = =

=_

0.0002983 ,d

fd basepu

Z=

(E1.6)

3.52

The transient d-axis inductance and open circuit time constant can be calculated as

'

''0

0.1975 ,

2.1074

md lfdd ls

md lfd

dd

base fd

L LL L pu

L L

LT s

Rw

= + =+

= =

(E1.7)

It has to be noted here that in per units the inductance and inductance reactance are

same.

E2. A synchronous generator has the following parameters in per units

' '0

' " "0 0

1.508, 1.489, 1.371

1.352, 0.1366, 1.6

0.229, 8 , 0.65

1.0 , 0.23, 0.03 , 50

d q md

mq ls fd

lfd d q

q d d s

X X X

X X X

X T s X

T s X T s f Hz

= = =

= = =

= = =

= = = =

With these parameters defined find the following generator parameters:

'1 1 1 1 1 1, , , , , , ,fd d q d l q q l d dR R R X X X X X

Sol:

The direct axis transient reactance is given in (3.125), substituting the given

parameters in (3.125) lead to

' 1 10.1366 0.3328

1 1 1 11.371 0.229

d ls

md lfd

X X

X X

(E2.1)

The direct axis transient open circuit time constant is given as ' fddo

base fd

XT

R , from

which the per unit field winding resistance can be found as given in (E2. 2)

3.53

'

1.60.0006366

314.159 8fd

fdbase do

XR

T

(E2.2)

From equation (3.127) the following expression can be written for finding the leakage

inductance of the quadrature axis winding 1q

1

'

0.8277

1

mql q

mq

q ls

XX

X

X X

(E2.3)

Similarly the rest of the parameters can be computed as following

1 1 2.179q l q mqX X X= + = (E2.4)

11 '

0

0.006935q

qbase q

XR

Tw= = (E2.5)

1

"

0.1782md lfd

l dmd lfd

fdd ls

X XX

X XX

X X

= =-

-

(E2.6)

1 1 1.5494d l d mdX X X= + = (E2.7)

11 "

0

0.1643dd

base d

XR

Tw= = (E2.8)

3.54

References

1. IEEE Task Force, “Current usage and suggested practices in power system

stability simulations for synchronous machines”, IEEE Trans. On Energy

Conversion, Vol. EC-1, No. 1, 1986, pp. 77-93.

2. A. Blondel, “The two-reaction method for study of oscillatory phenomena in

coupled alternators,” Revue Generale de L Electricite, Vol. 13, pp.235-251,

February 1923, March 1923.

3. R. H. Park, “Two-reaction theory of synchronous machines-generalized

method of analysis-part I,” AIEE Trans., Vol. 48, pp.716-727, 1929; part II,

Vol. 52,pp. 352-355, 1933.

4. A.W. Rankin, “Per-Unit Impedances of Synchronous Machines,” AIEE Trans.,

Vol. 64, Part I, pp. 569-573, August 1945; Part II, pp. 839-845, December

1945.

5. G. Shackshaft and P. B. Henser, “Model of generator saturation for use in

power system studies”, Proc. IEE, Vol. 126, No. 8, pp. 759-763, 1979.

6. M. E. Coultes and W. Watson, “Synchronous machine models by standstill

frequency response tests,” IEEE Trans. Power Appar. Syst., PAS-100, 4, Apr.

1981, 1480-1489.

7. IEEE Std 115A, “IEEE trail used standard procedure for obtaining

synchronous machine parameters by standstill frequency response testing,”

Supplement to ANSI/IEEE std. 115-1983, IEEE, New York, 1984.