Chapter 3
Modeling Of a Synchronous Machine
In Chapter 2 we have discussed the 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)
3.1
(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 a and 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 a and a it can be observed that the mmf due to the conductors
3.2
a and 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 threephase instantaneous ac voltages and currents in the stator windings are represented as
va , vb , vc and ia , ib , ic . 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 v fd with a field current
i fd . 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.  s  P m / 2 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  s  P m / 2 . 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 qaxis 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 this type of damper winding, when currents circulating through them,
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 voltage v1d
3.3
and current i1d , is considered whose mmf is along d-axis. Two damper windings
represented as 1q, 2q , with a voltage
v1q , v2 q and current i1q , i2 q 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
Number
1
of
windings
2
in
d-axis
3
1
2
Model
Model
Model
1.0
1.1
1.2
Model
Model
2.1
2.2

 

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 daxis and two damper windings along q-axis. Model 2.2 is widely used in many
industry grade transient stability simulations software.
3.4
3.1.1 Stator and rotor winding voltage equations
Applying KVL at the stator windings the following equations can be written
va  rs ia 
d a
dt
(3.1)
vb   rs ib 
d b
dt
(3.2)
vc  rs ic 
d c
dt
(3.3)
where, rs 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
v fd  rfd i fd 
(3.4)
dt
d 1d
dt
v1d  r1d i1d 
v1q  r1q i1q 
d  fd
(3.5)
d 1q
v2 q  r2 q i2 q 
(3.6)
dt
d 2 q
(3.7)
dt
Where, rfd , r1d , r1q , r2 q and  fd , 1d , 1q , 2 q are the rotor field, 1d, 1q and 2q winding
resistances and flux linkages, respectively.
3.5
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:
i fd 
 a 
 Laa Lab Lac  ia   Lafd La1d La1q La 2 q   
 i1d 
      L L L  i    L L
L
L


b
ba
bb
bc
b
bfd
b
d
b
q
b
q
1
1
2
 

 
i 
 c 
 Lca Lcb Lcc  ic   Lcfd Lc1d Lc1q Lc 2 q   1q 



  
 i2 q 
iabc
abc
Lss

Lsr
(3.8)
abc   Lss iabc  Lsr irotor
(3.9)
irotor
In equation (3.8) the diagonal elements of the matrix Lss represent the self
inductance of a, b, c windings and off-diagonal elements represent the mutual
inductance among a, b, c phases. The matrix Lsr represents the mutual inductance
between the stator and rotor windings. A similar expression for flux linkage of the
rotor windings can be written as
0  i fd 
 L fdfd L fd 1d 0
  fd 
 L fda L fdb L fdc 
i
 
 

 a 
0  i1d 
 1d     L1da L1db L1dc  i    L1dfd L1d 1d 0
 1q 
 L1qa L1qb L1qc   b   0
0 L1q1q L1q 2 q  i1q 
 
 

 ic  

 i2 q 
 2 q 
 L2 qa L2 qb L2 qc  
0
0
L
L
2 q1q
2q 2q  

 iabc 
 

rotor
Lrs
Lrr
(3.10)
irotor
………
rotor   Lrs iabc  Lrr irotor
(3.11)
In the matrix Lrr , the mutual inductance between the d-axis windings ( fd ,1d )
and the q-axis windings (1q, 2q ) is zero as the flux due to these windings are in
quadrature. The elements of inductance matrices Lss , Lsr , Lrr , and Lrs 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.
3.6
It can be observed from Fig. 3.1 at  s  0 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  s  90 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  s  0 and  s  180 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
3.7
s  
Pa  Pavg  Pp cos(2 s )
(3.12)
mmf f a  N a ia of a  a winding, where N a is number of turns of phase-a winding,
can be split into two components that is d -axis and q -axis component.
f ad  f a cos( s )
(3.13)
f aq   f a sin( s )
(3.14)
Now these two mmf f ad , f aq complete their path through two air gaps and hence two
permeance Pd , Pq (permeance along d and q axis). The flux along d -axis is f ad Pd and
along q -axis is f aq Pq . These fluxes along the mmf axis of a  a is given as
aa  f ad Pd cos( s )  f aq Pq sin( s )
 f a Pd cos 2 ( s )  f a Pq sin 2 ( s )
 P (1  cos(2 s )) Pq (1  cos(2 s )) 
 N a ia  d


2
2


 P  Pq ( Pd  Pq )

cos(2 s ) 
 N a ia  d

2
 2

if the flux linkage of phase-a due to current ia is represented as aa then
laa 
aa
ia

 Pd  Pq
N aaa
 N a2 
ia
 2
P  Pq

2 d
  Na 

 2

 cos(2 s )

(3.15)
if we assume that there is a leakage inductance lls because of flux of phase-a linking
itself then, (3.15) can be written as laa  laao  laap cos(2 s ) .
 Pd  Pq
laao  lls  N a2 
 2
P  Pq

2 d
 , laap  N a 

 2
Where,

.

Similarly we can derive an expression for the mutual inductance between aphase and b-phase. The mmf of b-phase can be split into two components along d-
3.8
axis and q-axis. The mmf along these axes can be multiplied with the permeance
along d-axis and q-axis then effective flux linkage along a  a mmf axis can be
found.
ba  fbd Pd cos( s )  fbq Pq sin( s )
 f b Pd cos( s ) cos( s  120)  fb Pq sin( s ) sin( s  120)
Pq
P

 N b ib  d (cos(2 s  120)  cos(120))  (cos(120)  cos(2 s  120)) 
2
 2

 P  Pq ( Pd  Pq )

 N b ib   d

cos(2 s  120) 
4
2


(3.16)
if the flux linkage of phase-a due to current ib represented as ba then
 Pd  Pq ( Pd  Pq )


cos(2 s  120) 
4
2


ba  N aba  N a N bib  
(3.17)
if the number of turns of a and b phase are same, that is N a  N b . Also, assuming
leakage inductance lls is same for all the phases, (3.17) can be approximately written
as,
1
lab   laao  laap cos(2 s  120)
2
(3.18)
Similarly it can be proved that mutual inductance between phase-a and phase-c will be
1
lac   laao  laap cos(2 s  120)
2
and
that
1
lbc   laao  laap cos(2 s  180)
2
Hence,
3.9
between
phase-b
and
phase-c
is
1

 laao  laap cos(2 s  120)
lls  laao  laap cos(2 s )
2

1
Lss    laao  laap cos(2 s  120) lls  laao  laap cos(2 s  120)
 2
 1
  laao  laap cos(2 s  120)  1 laao  laap cos(2 s  180)
2
 2
1

 laao  laap cos(2 s  120) 
2

1
 laao  laap cos(2 s  180) 

2

lls  laao  laap cos(2 s  120) 

…………. (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 qaxis 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,
lafd cos( s )

Lsr  lbfd cos( s  120)
l cos(  120)
s
 cfd
la1d cos( s )
 la1q sin( s )
 la 2 q sin( s )


lb1d cos( s  120)  lb1q sin( s  120)  lb 2 q sin( s  120) 
lc1d cos( s  120)  lc1q sin( s  120)  lc 2 q sin( s  120) 
…………………
(3.20)
Now let,
lafd  lbfd  lcfd  lsfd
la1d  lb1d  lc1d  ls1d
la1q  lb1q  lc1q  ls1q
la 2 q  lb 2 q  lc 2 q  ls 2 q
then (3.20) can be written in a simplified form as
lsfd cos( s )

Lsr  lsfd cos( s  120)

lsfd cos( s  120)
ls1d cos( s )
 ls1q sin( s )
 ls 2 q sin( s )


ls1d cos( s  120)  ls1q sin( s  120)  ls 2 q sin( s  120) 
ls1d cos( s  120)  ls1q sin( s  120)  ls 2 q sin( s  120) 
...
3.10
…….. (3.21)
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 dq 0
transformation, is as given below:
Tdqo
 cos( s ) cos( s  120) cos( s  120) 
 k1   sin( s )  sin( s  120)  sin( s  120) 
 k2

k2
k2
The choice of k1 , k2 is arbitrary. In standard practice however a value of
(3.22)
k1  2 / 3
and k2  1 / 2 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 k1  2 / 3 and
k2  1 / 2 . 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:
3.11
ia  I m cos(s t )
(3.23)
ib  I m cos(s t  120)
(3.24)
ic  I m cos(s t  120)
(3.25)
Now let us apply the Park’s transformation on the stator currents given in (3.23) to
(3.25)
id 
ia   I m cos(s t   s ) 
 
  

iq   Tdqo ib    I m sin(s t   s ) 
i 
ic  

0
0
(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    s t
(3.27)
Substituting (3.27) in (3.26) lead to
id   I m cos( ) 
  

iq     I m sin( ) 
i  

0
0
(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
ia , ib , ic can be transformed to dc quantities id , iq , io . It can also be observed that for a
balanced system
I m  id2  iq2
(3.29)
3.12
1
I 0  (ia  ib  ic )  0
3
(3.30)
The reason for choosing k1  2 / 3 and k2  1 / 2 can be understood from (3.28) to
(3.30). Due this choice the peak value of the time independent currents id , iq 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 id , iq in (3.28) produce the same mmf along d  axis and
q  axis as suggested by Blondel two-reaction theory. The currents id , iq 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 id , iq can act as
real and imaginary parts of a current phasor.
We can now convert all three-phase ac quantities into dc quantities through dq 0
transformation. Just like (3.28) we can represent three-phase stator voltages and flux
linkage in terms of dq 0 components as
 vd 
vd 
va 
va 
 

 
 
1 
 vq   Tdqo vb  or vb   Tdq 0 vq 
v 
v 
vc 
vc 
 0
 0
(3.31)
 d 
d 
 a 
 a 
 





1 
 q   Tdqo  b  or  b   Tdq 0 q 
 
 
 c 
 c 
 0
 0
(3.32)
where,
3.13
1
Tdqo
 cos( s )
  cos( s  120)
 cos( s  120)
 sin( s )
1
 sin( s  120) 1
 sin( s  120) 1




(3.33)
Now in order to check the effect of dq 0 transformation of voltages and current on the
instantaneous power we can find the relation between the complex power in terms of
abc components and dq 0 as
 va 
S   vb 
 vc 
T
T
 vd  
ia  

 i   T 1  v  
 b   dqo  q  
v  
ic  
 o 

T

id  
 1   
 Tdqo iq  

i  
 o 

vd 
id  vd 
 
 
1 T
1  
 vq  Tdqo  (Tdqo ) iq   vq 
v 
i   v 
 o
o  o
T
3
2

0

0



0 0
 id 
3  
0 iq
2  
 
0 3  io 

(3.34)
or
3
S  va ia  vb ib  vc ic  (vd id  vq iq  2vo io )
2
Hence,
the
transformation
(3.35)
is
not
power
invariant
that
is va ia  vbib  vc ic  (vd id  vq iq  vo io ) . The readers can verify that by choosing
k1  2 / 3 and k2  1/ 2 we can get a power invariant transformation i.e.
va ia  vbib  vc ic  (vd id  vq iq  vo io )
Now applying dq0 transformation to equations (3.1)-(3.3), we can get
3.14
 vd 
id 
d 
 rs 0 0 
d 1  
 


1  
 vq   Tdqo 0 rs 0  Tdqo iq   Tdqo dt Tdqo q 
v 
i 
 
0 0 rs 
 o
o
 o
d 
d 
 rs 0 0  id 
 
  d 1



1 d 
  0 rs 0  iq   Tdqo q  Tdqo  TdqoTdqo q 
dt
dt
 
 
0 0 rs  io 
 o
 o
 d 
 rs 0 0  id  0   0  d 
  
  d  



  0 rs 0  iq    0 0  q   q 
dt
 
0 0 rs  io  0 0 0  o 
 o 
(3.36)
Hence, we can write the stator and rotor equations in terms of dq 0 as
vd   rs id  q 
vq   rs iq  d 
vo  rs io 
d d
dt
(3.37)
d q
(3.38)
dt
d o
dt
v fd  r fd i fd 
v1d  r1d i1d 
v1q  r1q i1q 
(3.39)
d  fd
dt
(3.40)
d 1d
dt
(3.41)
d 1q
v2 q  r2 q i2 q 
(3.42)
dt
d 2 q
(3.43)
dt
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
d d d q
,
are called as transformer
dt
dt
induced voltages and these are induced due to the variation of the flux with respect to
3.15
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 dq 0 transformation is not
required for rotor side parameters as they are already defined either along d or q - axis.
Applying dq 0 transformation to flux linkage equation (3.9) and (3.11), we get
i fd 
 d 
id 
 
 
i1d 
1  
 q   Tdqo LssTdqo iq   Tdqo Lsr i 
 
i 
 1q 
 0
0
i2 q 
(3.44)
Substituting equation (3.19) and (3.21) along with equation (3.1) in (3.44) will lead to,
after simplification,
3
 2 (laao  laap )
 d 

 

 q     0
 
0
 0


lsfd

 0
0

Let
ld 
0
0
3
(laao  laap )
2
0
ls1d
0
0
ls1q
0
0
0
lls


 id 
 i 
 q
 i0 


(3.45)
i fd 
0  
 i1d 
ls 2 q   
i1q
0   
i2 q 
3
3
laao  laap  and lq   laao  laap 

2
2
.
Define,
lmd = ld - lls
lmq = lq - lls , then the following equations can be written.
3.16
and
d  (lls  lmd )id  lsfd i fd  ls1d i1d
(3.46)
q  (lls  lmq )iq  ls1q i1q  ls 2 q i2 q
(3.47)
o  lls io
(3.48)
3
2
(3.49)
3
2
(3.50)
3
2
(3.51)
3
2
(3.52)
 fd   lsfd id  l fdfd i fd  l fd 1d i1d
1d   lsfd id  l1dfd i fd  l1d 1d i1d
1q   ls1q iq  l1q1q i1q  l1q 2 q i2 q
2 q   ls 2 q iq  l2 q1q i1q  l2 q 2 q i2 q
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 dq 0 are same as the peak
values of abc components and the rated MVA expressed in terms of dq0 components
has an additional factor 3 / 2 involved. Hence, the stator base quantities in terms of
dq 0 components can be expressed as given in (3.54).
3.17
Vbase  V rms (l  n) kV,


Sbase  Rated MVA  3Vbase I base


Sbase
kA,
I base 

3Vbase


V

Z base  base ,
I base


base  s  2 f elec.rad, f  50 or 60 Hz 

2

mbase  base mech.rad

P

Z base

Lbase 
H
base


V
1

base  base Wb-turn, tbase 
s

base
base
Vdqo _ base  2 Vbase kV
3
Sbase  Vdqo _ base I dqo _ base MVA
2
S
2 Sbase
2 Sbase
I dqo _ base 

 2 base
3 Vdqo _ base 3 2 Vbase
3Vbase
Z dqo _ base 
Vdqo _ base
Ldqo _ base 
Z dqo _ base
dqo _ base 
Vdqo _ base
I dqo _ base
base
base

H
Wb-turn





 2 I base kA 











(3.53)
(3.54)
So, far we have defined stator base quantities in terms of abc and
dq 0 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
can take the same power base, Sbase , 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:
3.18
Sbase  V fd _ base I fd _ base


lmd

I fd _ base 
I dqo _ base

lsfd

Sbase
3 lsfd


V fd _ base 
Vdqo _ base 
I fd _ base 2 lmd


V fd _ base

 fd _ base 

base


V fd _ base
Z fd _ base 

I fd _ base


V fd _ base

L fd _ base 
base I fd _ base

(3.55)
Similarly we can define the base quantities of damper windings along d-axis as
following:
l
3 ls1d

Vdqo _ base , I1d _ base  md I dqo _ base 
I1d _ base 2 lmd
ls1d


V1d _ base
V1d _ base

, L1d _ base 


I1d _ base
base I1d _ base

Sbase  V1d _ base I1d _ base , V1d _ base 
V1d _ base
1d _ base 
, Z1d _ base
base
Sbase

(3.56)
The base quantities of windings along q-axis are given as:
Sbase  V1q _ base I1q _ base  V2 q _ base I 2 q _ base
I1q _ base 
lmq
l s1 q
I dqo _ base , I 2 q _ base 
lmq
ls 2 q
I dqo _ base
V1q _ base 
3 ls1q
3 ls 2 q
Vdqo _ base , V2 q _ base 
Vdqo _ base
2 lmq
2 lmq
1q _ base 
V1q _ base
Z 2 q _ base 
V2 q _ base
base
, 2 q _ base 
I 2 q _ base
, L1q _ base 
V2 q _ base
base
, Z1q _ base 
V1q _ base
base I1q _ base
V1q _ base
I1q _ base
, L2 q _ base 
Let,
3.19
,
V2 q _ base
base I 2 q _ base

















(3.57)


Vdqo _ base
Vdqo _ base
Vdqo _ base


iq
id
i0

, Iq 
, I0 
Id 
I dqo _ base
I dqo _ base
I dqo _ base


v fd
v1q
v1d

V fd 
, V1d 
, V1q 
,

V fd _ base
V1d _ base
V1q _ base

v2 q
i fd

i1d
V2 q 
, I fd 
, I1d 
,

V2 q _ base
I fd _ base
I1d _ base


i2 q
i1q
d

, I 2q 
, d 
,
I1q 

I 2 q _ base
dqo _ base
I1q _ base

q
 fd

1d
q 
, fd 
, 1d 
,

dqo _ base
 fd _ base
1d _ base




rs

 1q  1q , 2 q  2 q , Rs 
,
Z dqo _ base
1q _ base
2 q _ base


rfd
r1q
r2 q 
r1d
R fd 
, R1d 
, R1q 
, R2 q 
Z fd _ base
Z1d _ base
Z1q _ base
Z 2 q _ base 
Vd 
vd
, Vq 
vq
, V0 
v0
,
(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
Vd   Rs I d 

1 d d
q 
base
base dt
(3.59)
Vq   Rs I q 

1 d q
d 
base
base dt
(3.60)
Vo   Rs I o 
1 d 0
base dt
(3.61)
V fd  R fd I fd 
1
d fd
base dt
(3.62)
3.20
V1d  R1d I1d 
V1q  R1q I1q 
base
d 1d
dt
1
d 1q
1
(3.63)
(3.64)
base dt
V2 q  R2 q I 2 q 
1
d 2 q
(3.65)
base dt
The instantaneous stator power in dq 0 , 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
S
d d
3 
  rs id  q 
dt
2 
d q 

d o


i


r
i




 iq  2  rs io 
d
s
q
d

dt 
dt



 
 io 
 
 3  d 
d q

d  3
3
   d id 
iq  2 o io     d iq  q id    rs id 2  rs iq 2  2rs io 2   (3.66)
dt
dt  2
2
 2  dt

For a balanced system, v0 , i0 are zero hence equation (3.66) can be written as
 3  d 
d  3

3
S    d id  q iq     d iq  q id    rs id 2  rs iq 2  
dt  2
2
 2  dt

(3.67)
It can be observed from equation (3.67) that there are three terms in the equation. The
first term,
d q 
3  d d
id 
iq  , corresponds to the rate of change of magnetic energy

dt 
2  dt
in the stator coils. The second term,
gap. The third term,
3
  d iq  q id  , is the power transferred over air
2
3
rs id 2  rs iq 2  , is stator copper losses. The power transferred

2
over the air gap appears as torque, at the shaft speed, either opposing the motion of
3.21
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
tem  te       d iq  q id 
P  2
3P
or te 
 d iq  qid  N.m
22
(3.68)
Here, te 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
Sbase
1 3


 Vdq 0 _ base I dq 0 _ base 
2
2

base
base  2
P
P
3 P Vdq 0 _ base
3P
I dq 0 _ base 

 dq 0 _ base I dq 0 _ base
2 2 base
22
Tbase 
(3.69)
Dividing the electrical torque given in (3.68) by the base torque defined in (3.69)
leads to
3P
 d iq  qid 
te
Te 
 22
  d I q  q I d 
Tbase 3 P 
dq 0 _ base I dq 0 _ base
22
(3.70)
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
d  (lls  lmd )id  lsfd i fd  ls1d i1d
(3.71)
3.22
3
2
(3.72)
3
2
(3.73)
 fd   lsfd id  l fdfd i fd  l fd 1d i1d
1d   lsfd id  l1dfd i fd  l1d 1d i1d
From equation (3.58) we know that
Id 
id
I dqo _ base
, I fd 
i fd
I fd _ base
, I1d 
i1d
(3.74)
I1d _ base
We can express equation (3.74) in a different form as
id  I d I dqo _ base , i fd  I fd I fd _ base , i1d  I1d _ base I1d
(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
d
dqo _ base
 fd
 fd _ base
1d
1d _ base

(lls  lmd )( I d I dqo _ base )
dqo _ base

lsfd I fd I fd _ base
dqo _ base

ls1d I1d _ base I1d
dqo _ base
(3.76)

3 lsfd ( I d I dqo _ base ) l fdfd I fd I fd _ base l fd 1d I1d _ base I1d


 fd _ base
 fd _ base
 fd _ base
2
(3.77)

3 lsfd ( I d I dqo _ base ) l1dfd I fd I fd _ base l1d 1d I1d _ base I1d


1d _ base
1d _ base
1d _ base
2
(3.78)
It can be observed from equation (3.54)-(3.56) that dqo _ base  Vdqo _ base / base ,
 fd _ base  V fd _ base / base , 1d _ base  V1d _ base / base . Substituting these and also using
equation (3.58) in equation (3.76)-(3.78), we get
d 
base (lls  lmd )( I d I dqo _ base ) baselsfd I fd I fd _ base basels1d I1d _ base I1d


Vdqo _ base
Vdqo _ base
Vdqo _ base
3.23
(3.79)
 fd  
3 baselsfd ( I d I dqo _ base ) basel fdfd I fd I fd _ base basel fd 1d I1d _ base I1d


2
V fd _ base
V fd _ base
V fd _ base
(3.80)
 1d  
3 baselsfd ( I d I dqo _ base ) basel1dfd I fd I fd _ base basel1d 1d I1d _ base I1d


2
V1d _ base
V1d _ base
V1d _ base
(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
I fd _ base 
S
lmd
3 lsfd
Vdqo _ base
I dqo _ base , V fd _ base  base 
I fd _ base 2 lmd
lsfd
(3.82)
I1d _ base 
lmd
S
3l
I dqo _ base , V1d _ base  base  s1d Vdqo _ base
ls1d
I1d _ base 2 lmd
(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, lmd .
For example if we substitute
the field current base given in (3.82) in (3.79), the mutual inductance term lsfd gets
cancelled and replaced by lmd . We can also express the flux linkage equation of q axis windings in per units as
base (lls  lmq )( I q I dqo _ base ) basels1q I1q I1q _ base basels 2 q I 2 q _ base I 2 q


Vdqo _ base
Vdqo _ base
Vdqo _ base
(3.84)
 1q  
3 basels1q ( I q I dqo _ base ) basel1q1q I1q I1q _ base basel1q 2 q I 2 q I 2 q _ base


2
V1q _ base
V1q _ base
V1q _ base
(3.85)
 2q  
3 basels 2 q ( I q I dqo _ base ) basel2 q1q I1q I1q _ base basel2 q 2 q I 2 q I 2 q _ base


2
V2 q _ base
V2 q _ base
V2 q _ base
(3.86)
q 
3.24
Let us define some more quantities to further simplify equations (3.79) to (3.81) and
(3.84) to (3.86)
X ls 
xls

Z dq 0 _ base
X md 
X mq 
X fd 
X 1d 
X 1q 
X 2q 
xmd
xmq
Z dq 0 _ base
Z fd _ base
x1q1q
Z1q _ base
x2 q 2 q
Z 2 q _ base

Z dq 0 _ base
baselmq
Z dq 0 _ base
basel fdfd


Z fd _ base
baselmd I dq 0 _ base
Vdq 0 _ base
baselmq I dq 0 _ base
Vdq 0 _ base
basel fdfd I fd _ base
V fd _ base


basel1q1q
Z1q _ base

basel2 q 2 q
Z 2 q _ base
basel fd 1d I1d _ base
V fd _ base

basel1q1q I1q _ base
V1q _ base

basel2 q 2 q I 2 q _ base
V2 q _ base
basel fd 1d lsfd
V fd _ base ls1d
(3.88)
(3.89)
(3.90)
(3.91)
(3.92)
(3.93)
I fd _ base
(3.94)
basel fd 1d lsfd
Z fd _ base ls1d
X 1q 2 q 


baselmd
(3.87)
Vdq 0 _ base
 l I
x1d 1d
 l
 base 1d 1d  base 1d 1d 1d _ base
Z1d _ base
Z1d _ base
V1d _ base
X fd 1d 


basells I dq 0 _ base

Z dq 0 _ base

Z dq 0 _ base
x fdfd
basells
basel1q 2 q I 2 q _ base
V1q _ base

basel1q 2 q ls1q
V1q _ base ls 2 q
I1q _ base
(3.95)
basel1q 2 q lsfd
Z1q _ base ls1d
Let,
X d  X ls  X md , X q  X ls  X mq , X lfd  X fd  X md ,
X l1d  X 1d  X md , X l1q  X 1q  X mq , X l 2 q  X 2 q  X mq ,
3.25
(3.96)
The reactance X d , X q 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 K d 
L fd 1d
Lmd
, Kq 
L1q 2 q
Lmq
, we get
 d  X d ( I d )  X md I fd  X md I1d
(3.97)
 fd  X md ( I d )  X fd I fd  K d X md I1d
(3.98)
 1d  X md ( I d )  K d X md I fd  X 1d I1d
(3.99)
 q  X q ( I q )  X mq I1q  X mq I 2 q
(3.100)
 1q  X mq ( I q )  X 1q I1q  K q X mq I 2 q
(3.101)
 2 q  X mq ( I q )  K q X mq I1q  X 2 q I 2 q
(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 K d is assumed to be unity. Similarly, all mutual inductances/reactances along qaxis are equal if K q is assumed to be unity. In fact in actual synchronous generators
the values of K d and K q 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
Vd   Rs I d 

1 d d
q 
base
base dt
(3.103)
Vq   Rs I q 

1 d q
d 
base
base dt
(3.104)
3.26
Vo   Rs I o 
1 d 0
base dt
(3.105)
Rotor voltage equations
1
V fd  R fd I fd 
V1d  R1d I1d 
V1q  R1q I1q 
d fd
base dt
(3.106)
base
d 1d
dt
(3.107)
1
d 1q
1
(3.108)
base dt
V2 q  R2 q I 2 q 
1
d 2 q
(3.109)
base dt
Stator flux linkage equations
 d  X d ( I d )  X md I fd  X md I1d
(3.110)
 fd  X md ( I d )  X fd I fd  X md I1d
(3.111)
 1d  X md ( I d )  X md I fd  X 1d I1d
(3.112)
Rotor flux linkage equations
 q  X q ( I q )  X mq I1q  X mq I 2 q
(3.113)
 1q  X mq ( I q )  X 1q I1q  X mq I 2 q
(3.114)
 2 q  X mq ( I q )  X mq I1q  X 2 q I 2 q
(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
3.27
3.4.1 Sub-transient and transient reactance
X d"  X ls 
1
1
1
1


X md X lfd X l1d
(3.116)
X d" 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 I d flows then immediately
after the voltage is applied ( t  0 ) the flux linkages  fd and  1d will be zero. Hence
we can write equations (3.110) to (3.112) as
 d  X d ( I d )  X md I fd  X md I1d
(3.117)
 fd  0  X md ( I d )  X fd I fd  X md I1d
(3.118)
 1d  0  X md ( I d )  X md I fd  X 1d I1d
(3.119)
Solving for I fd , I1d in term of I d , also considering equation (3.96), we get
I fd 
I1d 
X md X lfd
X md X l1d
Id
 X md X l1d  X lfd X l1d
X md X lfd
X md X lfd  X md X l1d  X lfd X l1d
(3.120)
(3.121)
Id
Substituting equation (3.120) and (3.121) in (3.117), we get
2

X md
( X lfd  X l1d )
d   Xd 

X md X lfd  X md X l1d  X lfd X l1d


 I d

Since, X d  X ls  X md , equation (3.122) can be further simplified as
3.28
(3.122)

 d    X ls 

X md X lfd X l1d

 I d

X md X lfd  X md X l1d  X lfd X l1d





1

 I d   X d" I d
  X ls 
1
1
1 




X md X lfd X l1d 

(3.123)
Similarly the quadrature axis sub-transient reactance can be expressed as
X q"  X ls 
1
1
1
1


X mq X l1q X l 2 q
(3.124)
The direct axis transient reactance is defined as
X d'  X ls 
1
1
1

X md X lfd
(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
transients settle down faster as compared to field winding transients hence I1d can be
assumed to be zero. With this assumption, by solving equations (3.117) and (3.118),
we get
2



X md X lfd
X md
I d    X ls 



X md  X lfd 
X md  X lfd






1
 I d   X d' I d
   X ls 
1
1





X
X
md
lfd 

d   Xd 
3.29

 I d

(3.126)
Similarly, quadrature axis transient reactance can be defined as
X q'  X ls 
1
1
1

X mq X l1q
(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 I d is zero. At time t  0 the flux
linkage  1d is zero. Substituting these assumption in (3.107) and (3.112) we get
I fd  
X 1d
I1d
X md
(3.128)
dI fd
dI 
1 
 X 1d 1d   R1d I1d  V1d  0
 X md
base 
dt
dt 
(3.129)
or
1  dI fd 
1 X 1d dI1d R1d
I



base  dt 
base X md dt X md 1d
(3.130)
Substituting equations (3.130) and (3.129) in (3.106) the following expression can be
obtained
V fd   R fd
2
X1d
1  X md  X fd X 1d
I1d 

X md
base 
X md
 dI
 1d
 dt

(3.131)
Substituting equation (3.111), (3.130) in (3.106) a first order differential equation in
terms of I1d with a forcing function V fd , can be obtained as
3.30
 X fd dI fd X md dI1d 


  R fd I fd  V fd
 base dt base dt 
 X fd X 1d dI1d X fd R1d
X dI  R fd X 1d
 

I1d  md 1d  
I  V fd
base dt  X md 1d
X md
 base X md dt
2
1  X md  X fd X 1d


base 
X md
(3.132)
 dI
X fd R1d  R fd X 1d
I1d  V fd
 1d 
 dt
X
md

2
1  X md  X fd X1d


base  X fd R1d  X1d R fd
 dI
X md
V fd
 1d  I1d 
 dt
X
R

R
X
fd
1
d
fd
1
d

In a practical generator R1d  R fd hence X1d R fd can be neglected as compared to
X fd R1d in equation (3.132). With this assumption and further simplification we get


X md X lfd
 X ls 
base R1d 
X md  X lfd
1
 dI
X md
V fd
 1d  I1d 
 dt
X fd R1d

(3.133)
Hence, the open circuit direct axis sub-transient time constant, in seconds, is given as

X md X lfd
1
Tdo" 
 X l1d 
base R1d 
X md  X lfd

1

 base R1d



1
 X l1d 
1
1



X md X lfd







(3.134)
Similarly, the open circuit quadrature axis sub-transient time constant, in seconds, can
be express as


1
1
"
 X l 2q 
Tqo 
1
1
base R1q 


X mq X l1q







(3.135)
3.31
for finding the open circuit direct axis transient time constant, along with the
assumption taken for computing Tdo" , the current in the damper winding I1d is
assumed to be zero then from equation (3.106) and (3.111)
 fd  X fd I fd
X fd
dI fd
base R fd dt
(3.136)
 I fd 
1
V fd
R fd
(3.137)
Hence, the open circuit direct axis transient time constant, in seconds, is defined as
Tdo' 
X fd
(3.138)
base R fd
Similarly the open circuit quadrature axis transient time constant, in seconds is
defined as
Tqo' 
X 1q
(3.139)
base R1q
Also, defining new variables as
Eq' 
X
X md
X
 fd , E fd  md V fd , Ed'   mq  1q
X fd
R fd
X 1q
(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
 I fd   X fd X md   fd   X fd X md   X md 
 
  
 
 Id
 I1d   X md X 1d   1d   X md X 1d   X md 
3.32
(3.141)
2
 X lfd X l1d  X lfd X md  X l1d X md , then equation (3.141) can be
let,   X fd X 1d  X md
expressed as


1
2
X 1d fd  X md 1d   X 1d X md  X md
 Id

1
2
I1d 
 X md fd  X fd 1d   X fd X md  X md
 Id

I fd 


(3.142)
The coefficients of  fd , 1d , I d can be expressed in terms of sub-transient, transient
and steady state reactances as given below
2
 X md


  X md
1  X md
1  ( X d  X d' )( X d'  X d" )  '
 fd  

1





1 
 Eq

fd 
 X md 
( X d'  X ls ) 2
   X fd

 X fd

 X md 
2
 X md

( X d  X d' )( X d'  X d" ) ' X md
 fd 
, Eq 


'
2

X fd
( X d  X ls )
 

'
2
1
1 X md
1 ( X d  X d )( X d'  X d" )
 1d
X md 1d 
 1d 
( X d'  X ls ) 2
X md 
X md

X X
1
X 1d fd  fd 1d
X md 

1
1  X lfd X l1d X md 
1  ( X d  X d' )( X d"  X ls ) 
2
 X 1d X md  X md  I d  X    I d  X  ( X '  X )  I d


lfd 
md 
d
ls

'
X lfd X l1d X md


( X d  X ls )
"
,


X
X
X
 X lfd 

md
d
ls

( X d  X d' )


 ( X d'  X d" ) '
X X
1
X md fd  fd  md  fd  
E
 ( X '  X )2 q

  X fd
d
ls

 X fd

( X d'  X d" )
X
, Eq'  md  fd 


'
2

( X d  X ls )
X fd
 

'
"
(Xd  Xd )
1
 1d
X fd 1d 
( X d'  X ls ) 2

X X
1
2
X fd X md  X md
I d   lfd md



 

( X d'  X d" )

I
Id
 d
( X d'  X ls )

Replacing the coefficient in (3.142) with new coefficients, equation (3.142) can be
expressed as
3.33
I fd 

1
X md
 ' ( X d  X d' )( X d'  X d" )  ' ( X d'  X ls )( X d"  X ls )

I d  1d  
 Eq 
 Eq 
'
2
'
"
( X d  X ls )
(Xd  Xd )



1
Eq'  ( X d  X d' )  I d  I1d  

X md
(3.143)
I1d 
( X d'  X d" )
  Eq'  ( X d'  X ls ) I d  1d 
( X d'  X ls ) 2
(3.144)
Substituting equation (3.143) and (3.44) in equation (3.110) the stator d -axis flux can
be expressed as
 d  X d ( I d )  X md I fd  X md I1d
 Eq'  X d' I d   X d'  X ls  I1d
  X d" I d 
(3.145)
( X d"  X ls ) ' ( X d'  X d" )
Eq  '
 1d
( X d'  X ls )
( X d  X ls )
In a similar way the q -axis flux linkage equation along with current I1q , I 2 q can be
expressed as
 q   X q" I q 
I1q 
I 2q 
( X q"  X ls )
( X q'  X ls )

Ed' 
( X q'  X q" )
( X q'  X ls )
1
 Ed'  ( X q  X q' )  I q  I 2 q 
X mq
( X q'  X q" )
( X  X ls )
'
q
2
E
'
d
 2q

(3.146)
(3.147)
 ( X q'  X ls ) I q  2 q 
(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
the network electrically, if its dynamics are considered then the entire network
3.34
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:
Tdo" 

X md X lfd
 X l1d 
base R1d 
X md  X lfd
1

1

 base R1d

 X lfd X l1d  X lfd X md  X l1d X md

X fd



 1 
  
1
1
1




 
base R1d  X fd  base R1d  X fd  base R1d


  


1
1

Tqo" 

'
base R2 q ( X q  X q" )

 ( X q'  X ls ) 2



1

'
 ( X d  X d" )

'
2
 ( X d  X ls )
















1 d d

 Vd  Rs I d 

base dt
base q
(3.149)
1 d q

 Vq  Rs I q 

base dt
base d
(3.150)
1 d 0
 Vo  Rs I o
base dt
(3.151)
Tdo'


( X '  X d" )
  Eq'  ( X d  X d' )  I d  ' d
 Eq'  ( X d'  X ls ) I d  1d   E fd
2
dt
( X d  X ls )


dEq'


(3.152)
3.35
Tdo"
d 1d
 Eq'  ( X d'  X ls ) I d  1d
dt
Tqo'

( X q'  X q" )
dEd'
'
'
Ed'  ( X q'  X ls ) I q  2 q
  Ed  ( X q  X q )  I q  '
2

dt
(
X
X
)

q
ls

"
Tqo
(3.153)

d 2 q
dt

 

  Ed'  ( X q'  X ls ) I q  2 q
 d   X d" I d 
 q   X q" I q 
(3.155)
( X d"  X ls ) ' ( X d'  X d" )
 1d
Eq  '
( X d'  X ls )
( X d  X ls )
( X q"  X ls )
( X q'  X ls )
Ed' 
( X q'  X q" )
( X q'  X ls )
(3.154)
 2q
(3.156)
(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  d s
J
 t m  te
 
2
 P  dt
(3.158)
Where, tm , te are the mechanical input torque from the prime mover and the electrical
output torque. J kg.m 2 is the inertia constant of the rotating machine. Now defining
machine inertia constant as
H
1  2

 J  base 
2  P

2
Sbase



MJ/MVA .
Replace J with the defined machine inertia constant in equation (3.158), then
HSbase d 2 s
 t m  te
2 2 dt 2
base
P
(3.159)
Rearranging we can express (3.159) as
3.36
tm
te
t
t
2 H d 2 s


 m  e  Tm  Te
2
Sbase
Sbase
base dt
Tbase Tbase
2
2
base
base
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
d d s

 base    base
dt
dt
(3.161)
2 H d 2
2 H d

 Tm  Te  D   base 
2
base dt
base 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 airgap flux linkage.
3. Hence, most of the effect of saturation is on the mutual inductances
X md , X mq .
3.37
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
Vt   d  X md I fd ,  q  0
(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
 t   d2  q2   d  Vt
(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 I fag produces 1.0 pu terminal voltage but with
saturation to produce the same terminal voltage the filed current required is I fnl which
is higher than I fag 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 I fsc . Because of the armature reaction in case of short circuit the field current I fsc
will correspond to an internal voltage which will lie on the linear portion of the open
circuit characteristics.
3.38
Vt ( pu )
Air-gap line
1.0
I sc ( pu )
1.0
I fag
I fnl
I fsc
If
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 X s _ unsat then we can write the following expression
KI fsc  1.0 X s _ unsat
(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
KI fag  1.0
(3.166)
Hence,
X s _ unsat 
I fsc
(3.167)
I fsag
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
3.39
produced 1.0 pu terminal voltage without saturation or from air-gap line. On similar
reasoning we can define the saturated synchronous reactance as
X s _ sat 
I fsc
(3.168)
I fsnl
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
SCR 
I fsnl
I fsc

1
(3.169)
X s _ sat
Now in order to estimate the effect of saturation on the synchronous reactance Fig. 3.4
can be used.
Air-gap line
 t _ ag
flux linkage
J
t
 lin
I fag
I fnl
If
Fig. 3.4: Open circuit characteristics to compute the effect of saturation
3.40
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
t
X md _ unsat 
X md _ sat 
I fag

 tag
(3.170)
I fnl
t
(3.171)
I fnl
Dividing equation (3.171) by (3.170) leads to
X md _ sat 
t
t
X md _ unsat 
X
 tag
 t  J md _ unsat
(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)
 J  Asat e B
sat
 t  lin 
(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
X mq _ sat 
t
X
 t  J mq _ unsat
(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
3.41
 d   X ls  X md  ( I d )  X md I fd  X md I1d   X ls I d  md
 q   X ls  X mq  ( I q )  X mq I1q  X mq I 2 q   X ls I q  mq
(3.175)
Substituting equation (3.103) and (3.104), in steady state, in equation (3.175) gives
 md   d  X ls I d  Vq  Rs I q  X ls I d
(3.176)
 mq   q  X ls I q  Vd  Rs I d  X ls I q
or
 md  j mq  Vq  Rs I q  X ls I d  j  Vd  Rs I d  X ls I q 

  j Vd  jVq    Rs  jX ls   I d  jI q 

(3.177)
Now if we define an internal voltage behind the leakage reactance as
El  Vt   Rs  jX ls  I t
(3.178)
Then we can write observe from equation (3.177) and (3.178) that
2
2
 t   md
 mq
 El
(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 X md _ sat , X mq _ sat 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
3.42
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
replaced by s  j0 ) gives
dt
Vd   Rs I d 
r
s
q 

base
base d
(3.180)
Vq   Rs I q 
r
s
d 

base
base q
(3.181)
Vo   Rs I o 
s
base
0
(3.182)
If the flux linkage can be defined as
 d   X dop ( s) I d  Gop ( s)V fd
(3.183)
 q   X doq ( s ) I q
(3.184)
 0   X oop ( s) I o
(3.185)
where, X dop ( s ) , X doq ( 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 r  0 will lead to


s
s
Vd    Rs 
X dop ( s )  I d 
 G (s)V fd 
base
base op


3.43
(3.186)
If in the rotor the field is shorted then V fd is zero and hence with Vd , I d known from
the test at different frequencies the operational d – axis reactance X dop ( 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
r  0 lead to equation (3.187) from which the operational reactance X doq ( s ) can be
found for different frequencies.


s
Vq    Rs 
X qop ( s )  I q
base


(3.187)
Theoretically we can express X dop ( s ) , X doq ( s ) in terms of the sub-transient and
transient inductances and time constants. This can be done as following, with the
damper windings voltages V1d  0 , by expressing voltage equations of the d-axis
windings in Laplace domain as
s
base
s
base
 fd   R fd I fd  V fd
(3.188)
 1d   R1d I1d
(3.189)
Substituting rotor flux linkage equations in equations (3.188) and (3.189) and
elimination of rotor current lead to
3.44
I fd
2



 s
 s

s


X md  R1d 
X 1d 
X md 

base
base  base





I

2  d






s
s
s
 R 

  1d  X 1d   R fd   X fd     X md  
base
base

  base
 





s


R
X

 1d

base 1d 



V

2  fd






s
s
 R  s X
X fd   
X md  
1d   R fd 
  1d 
base
base

  base
 

I1d 
X md
s
base 
s
 R1d 
base


X 1d 

I
d
(3.190)
 I fd 
(3.191)
Substituting equations (3.190) and (3.191) in the d-axis stator flux linkage equations
lead to equation (3.183) where
 s

s
s
s
2 

X md
X 1d  2
X md  R fd 
X fd  
 R1d 
base
base
base



X dop ( s )  X d   base
2



  s

s
s


X 1d   R fd 
X fd   
X md 
 R1d 





base
base


  base



(3.192)




s
s


X md  R1d 
X 1d 
X md 
base
base




Gop ( s )  
2 
  R  s X  R  s X    s X  
 

1d   fd
  1d 
base fd   base md  
base


(3.193)
Equation (3.192) and (3.193) can also be expressed in factored form in terms of time
constants as
X dop ( s )  X d
Gop ( s )  Go
1  sT 1  sT 
1  sT 1  sT 
'
d
"
d
'
do
"
do
(3.194)
1  sTkd 
1  sT 1  sT 
'
do
"
do
3.45
Where, Tdo' , Tdo" are the open circuit transient and sub-transient time constant as given
in equations (3.185) and (3.186). Td' , Td" are called the short circuit transient and subtransient time constants and given as

X md X lfd X ls
 X l1d 
X md X ls  X lfd X ls  X md X lfd
base R1d 

X md X ls 
1
Td' 
 X lfd 

X md  X ls 
base R fd 
X
X
Go  md , Tkd  l1d
R fd
R1d
Td" 
1




(3.195)
In a similar way the q-axis operational reactance can also be defined as
X qop ( s )  X q
1  sT 1  sT 
1  sT 1  sT 
'
q
"
q
'
qo
"
qo
(3.196)
The open circuit time constant Tqo' , Tqo" are already defined. The short circuit time
constant are given as
Tq" 

X mq X l1q X ls
 X l 2q 
base R2 q 
X mq X ls  X l1q X ls  X mq X l1q
1

X mq X ls
1
Tq' 
 X l1q 
base R1q 
X mq  X ls








(3.197)
It can be observed equations (3.94) and (3.96) that for s  0
X dop ( s )  X d ,
X qop ( s )  X q
(3.198)
As s   , like immediately after a disturbance the effect on stator flux linkage,
3.46
X dop ( s )  X d
X qop ( s )  X q
T T   X ,
T T 
T T   X
T T 
'
d
"
d
'
do
"
do
'
q
"
q
'
qo
"
qo
"
d
"
q
(3.199)
In case the damper windings are absent then as s  
X dop ( s )  X d
X qop ( s )  X q
T   X ,
T 
T   X
T 
'
d
'
do
'
q
'
qo
'
d
(3.200)
'
q
In a synchronous generator Tdo'  Td'  Tdo"  Td" and Tqo'  Tq'  Tqo"  Tq" . Hence, if the
gain variation of the transfer function X dop ( s ) and X qop ( s ) with respect to frequency
variation is plotted then the corner frequencies will correspond to the time constants
Tdo' , Td' , Tdo" , Td" and Tqo' , Tq' , Tqo" , Tq" with Tdo' , Tqo' corresponding to lowest corner
frequency and Td" , Tq" 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, vtest , 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
vtest  vb  vc , va  0, vb  vc
(3.201)
itest  ib  ic , ia  0, ib  ic
3.47
3
3
3
3
3
vb 
vc 
vtest , id 
ib 
ic  3 itest
2
2
2
2
2
1
1
1
1
vq  vb  vc  0, iq  ib  ic  0
2
2
2
2
vd 
(3.202)
let the line-line test voltage and test current be defined as vtest  3 2 Vt cos ot   o 
and itest  3 2 I t cos ot  o  then
3
3 2 Vt cos ot   o  ,
2
id  3 3 2 I t cos o t  o 
vd 
(3.203)
In per units equation (3.203) can be written as
3
Vt _ pu cos ot   o  ,
2
id  3 I t _ pu cos ot  o 
vd 
(3.204)
Hence, the d-axis operational impedance as a function of the variable frequency o is
given as
Z dop o  
Vt _ pu
I t _ pu
2
Vd
Id
(3.205)
 1
Vd 

1 Vt _ pu
  Rs  j o X dop   Z dop o  
base
Id 
2 I t _ pu
 2
X dop

  1 Vt _ pu
  j base 
 Rs 

o  2 I t _ pu

(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
3.48
1
1
1
1
vd   vb  vc  0, id   ib  ic  0
2
2
2
2
3
3
3
3
3
vq 
vb 
vc 
vtest , iq 
ib 
ic  3 itest
2
2
2
2
2
(3.207)
Hence, the operational reactance of q-axis can be written as

 1

1 Vt _ pu
  Rs  j o X qop   Z qop o  
base
Iq 
2 I t _ pu
 2
Vq
X qop

  1 Vt _ pu
  j base 
 Rs 

o  2 I t _ pu

3.49
(3.208)
Example Problems
E1. A three-phase 300 MVA, 20 kV, 0.9 pf, 50 Hz, 2 pole synchronous generator has
the following parameters.
laa = 4.675 + 0.0534 cos (2q ) mH
æ
pö
lab = -2.3375 - 0.0534 cos çç2q + ÷÷÷ mH
çè
3ø
lls = 0.5792 mH
lafd = 67.2 cos (2q ) mH
l fd = 1084.08 mH
rs = 0.0014 W
r fd = 0.0635 W
Define the base quantities and express all the generator parameters in per units in dq 0
reference frame.
Sol:
The d-axis and q-axis inductances are given as
3
ld = ´(4.675 + 0.0534) = 7.0926 mH
2
(E1.1)
3
lq = ´(4.675 - 0.0534) = 6.9324 mH
2
(E1.2)
lmd = ld - lls = 6.5134 mH
(E1.3)
lmq = lq - lls = 6.3532 mH
(E1.4)
3.50
The base quantities are defined as
üï
ïï
ïï
ïï
ïï
ïï
S
ïï
Ibase = base = 8.665 kA,
ïï
3Vbase
ïï
ïï
wbase = 2´ p ´50 = 314.159 rad / s
ïï
ïï
Vdq 0 _ base = 2Vbase = 16.32 kV ,
ïï
ïï
I dq 0 _ base = 2 Ibase = 12.25 kA,
ïï
ïï
Vdq 0 _ base
Z dq 0 _ base =
= 1.3317 W, ïï
ïï
I dq 0 _ base
ïý
ïï
Z dq 0 _ base
Ldq 0 _ base =
= 4.2389 mH , ïï
ïï
wbase
ïï
lmd
ï
I fd _ base =
I dq 0 _ base = 1.1873 kA,ïï
ïï
lafd
ïï
ïï
Sbase
= 252.66 kV
V fd _ base =
ïï
I fd _ base
ïï
ïï
V fd _ base
ïï
= 212.86 W,
Z fd _ base =
ïï
I fd _ base
ïï
ïï
Z fd _ base
ï
= 677.55 mH , ïï
L fd _ base =
ïïþ
wbase
20
= 11.54 kV ,
3
Sbase = 300 MVA,
Vbase =
(E1.5)
The per unit generator parameters can now be computed as
Ld =
ld
Ldq 0 _ base
Lmd =
Lls =
= 1.673 pu,
lmd
Ldq 0 _ base
lls
Ldq 0 _ base
Lq =
r fd
Z fd _ base
Ldq 0 _ base
= 1.5365 pu, Lmq =
= 0.1366 pu, L fd =
Llfd = L fd - Lmd = 0.0635 pu , Rs =
R fd =
lq
= 1.6354 pu,
lmq
Ldq 0 _ base
l fd
L fd _ base
rs
Z dq 0 _ base
= 0.0002983 pu ,
3.51
= 1.4987 pu,
= 1.6 pu ,
= 0.00105 pu,
(E1.6)
The transient d-axis inductance and open circuit time constant can be calculated as
L'd = Lls +
Td' 0 =
Lmd Llfd
Lmd + Llfd
= 0.1975 pu ,
(E1.7)
L'd
= 2.1074 s
wbase R fd
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
X d = 1.508, X q = 1.489, X md = 1.371
X mq = 1.352, X ls = 0.1366, X fd = 1.6
X lfd = 0.229, Td' 0 = 8 s, X q' = 0.65
Tq' 0 = 1.0 s, X d" = 0.23, Td"0 = 0.03 s
With these parameters defined find the following generator parameters:
R fd , R1d , R1q , X d' , X l1q , X1q , X l1d , X1d
Sol:
The direct axis transient reactance is given in (3.125), substituting the given
parameters in (3.125) lead to
X d'  X ls 
1
1
1

X md X lfd
 0.1366 
1
1
1

1.371 0.229
 0.3328
The direct axis transient open circuit time constant is given as Tdo' 
(E2.1)
X fd
base R fd
which the per unit field winding resistance can be found as given in (E2. 2)
3.52
, from
R fd 
X fd
 T
'
base do

0.3328
 0.0001324
314.159  8
(E2.2)
From equation (3.127) the following expression can be written for finding the leakage
inductance of the quadrature axis winding 1q
X mq
X mq
X l 1q 
X q'  X ls
 0.8277
(E2.3)
1
Similarly the rest of the parameters can be computed as following
X1q = X l1q + X mq = 2.179
R1q =
X q'
wbaseTq' 0
X l1d =
(E2.4)
= 0.006935
(E2.5)
X md + X lfd
= 0.6775
X md X lfd
-1
X d" - X ls
(E2.6)
X1d = X l1d + X md = 2.0485
R1d =
X d"
wbaseTd"0
(E2.7)
= 0.0927
(E2.8)
3.53
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.
3.54