1. 5.1
Chapter 5
Representation of Synchronous Machines for Stability studies
The equations (3.149) – (3.157) and (3.161) - (3.162) of Chapter 3 representing
the synchronous machine dynamics, equations (4.27) – (4.29), (4.54) - (4.55) and
(4.65) given in Chapter 4 representing the dynamics of generator exciter, steam
turbine and speed governor describes the complete behaviour of a steam generating
system. However, the q -axis of the synchronous generator has to be located with
respect to some reference in order to define the rotor angle. The synchronous
generator terminal conditions like complex voltage, real and reactive power decide the
rotor angle. The synchronous generator terminal conditions in turn depend on the
interaction of the synchronous generator with the network conditions. While
representing a synchronous machine in a system, the representation should include the
dynamic behaviour of the generator as well as its interaction with the network. The
effect of synchronous machine interaction with the rest of the network can be
understood through the steady state conditions.
5.1 Steady State Condition
In the steady state condition the rotor rotates at synchronous speed. Since rotor
rotates at synchronous speed the damper winding currents 1 1 2, ,d q qI I I are zero. Since
the synchronous generator is in steady state, left hand side of equations (3.149) to
(3.157) will become zero. Hence, from equations (3.149) to (3.157) the following
expression can be written
d s d qV R I (5.1)
q s q dV R I (5.2)
' '
( )fd q d d dE E X X I (5.3)
' '
1 ( )d q d ls dE X X I (5.4)
' '
( )d q q qE X X I (5.5)
' '
2 ( )q d q ls qE X X I (5.6)
2. 5.2
Taking the d-axis flux linkage equations in the steady state we get
( )d d d md fdX I X I (5.7)
( )fd md d fd fdX I X I (5.8)
Eliminating the filed current fdI from equations (5.7) and (5.8), equation (5.7) can be
written
2
' '
( )md md
d d d fd d d q
fd fd
X X
X I X I E
X X
(5.9)
Similarly,
' '
q q q dX I E (5.10)
Substituting equation (5.9) and (5.10) in equations (5.1) and (5.2) lead to
' '
d s d q q dV R I X I E (5.11)
' '
q s q d d qV R I X I E (5.12)
Let the three-phase terminal voltage be given in per unit as
cos( )
cos( 120 )
cos( 120 )
ta m s
tb m s
tc m s
v V t
v V t
v V t
(5.13)
by applying Park’s transformation defined in Chapter 3, we get
cos( )
sin( )
d m s s
q m s s
V V t
V V t
(5.14)
3. 5.3
Here, s is the angle between the d -axis and the phase-a mmf axis. Let
0s st where 0 is the initial angle between the d -axis and the phase-a mmf
axis at time 0t , also noting that in per unit the RMS value of the voltage tV and
peak value of the voltage mV are one and the same, equation (5.14) can be written
as
0
0
cos( )
sin( )
d t
q t
V V
V V
(5.15)
From (5.15) it can be observed that though ,d qV V are scalar quantities, due the
trigonometric nature of the equation the terminal voltage tV can be represented as a
ac phasor in dq axis. Just like an ac phasor in real and imaginary axis rotating at
synchronous speed in electrical radians the dq axis also rotates at the speed of the
rotor which is synchronous speed in steady state. Hence, the terminal voltage can be
represented as a phasor in dq axis as
0t d qV V jV (5.16)
Where, 2 2
t d qV V V and angle 0 is the angle between the terminal
voltage phasor and d axis as shown in Fig. 5.1. Instead of taking the angle between
the d axis and terminal voltage phasor let us take the angle between the terminal
voltage phasor and the q axis represented as 090 . Then we get
sin( ) cos( )d q t tV jV V j V (5.17)
4. 5.4
Fig. 5.1: Phasor diagram of terminal voltage in dq axis
Here, tV is a complex quantity in dq axis. Similarly, the stator current tI lagging
the terminal voltage tV by an angle , can be expressed as a phasor in dq axis as
sin( ) cos( )d q t tI jI I j I (5.18)
Substituting equations (5.11) and (5.12) in equation (5.17), the following expression
can be written
' ' ' '
t s d q q d s q d d qV R I X I E j R I X I E (5.19)
Rearranging terms in (5.19), the following expression can be derived as
' ' ' '
' ' ' '
0
' ' ' '
( ) ( ) ( )
fd
t s d q q d s q d d q
s d q q d q q d q q q d d d q
E
s q t d q q q q d d d d d q
s q t q d d fd
V R I X I E j R I X I E
R I jI jX I jI jX I jI X I E j X I E
R jX I E X X I j X X I X X I E
R jX I j X X I E
…………………………………………………………………………… (5.20)
0
d-axis
q-axis
tV
5. 5.5
Noting that md
fd fd md fd
fd
X
E V X I
R
, we can write equation (5.20) as
md fd d q d t s q tj X I X X I V R jX I (5.21)
let, the internal voltage of the synchronous generator be represented as
q md fd d q dE X I X X I , or q md fd d q dE j X I X X I then
q t s q tE V R jX I (5.22)
qE is defined as the internal voltage behind the synchronous reactance qX . Equation
(5.22) can be represented as an equivalent electrical circuit as shown in Fig. 5.2.
Fig. 5.2: Electrical equivalent circuit of steady state synchronous generator
From equation (5.22) it can be observed that the internal voltage phasor qE is
aligned along the q axis as shown in Fig. 5.3. Since qE is aligned along q axis, it
is at angle from the terminal voltage tV . This angle is called as generator rotor
or load angle. The load angle can be computed if the real power output tP ,
reactive power output tQ and terminal voltage tV of the synchronous generator
are defined. Let us take terminal voltage as the reference and then locate the q axis
qX sR
tV
qE
tI
6. 5.6
at an angle with respect to the terminal voltage phasor. The following steps should
be followed
Fig. 5.3: Phasor diagram of the steady state synchronous generator
2 2
1
, cost t t
t
t t t
P Q P
I
V V I
(5.23)
0q t s q tE V R jX I (5.24)
1
cos sin
tan
cos sin
q t s t
q
t s t q t
X I R I
E
V R I X I
(5.25)
In case of steady state no load condition, the stator currents ,d qI I will become zero
and therefore the following expression holds
'
q q q fd t d md fdV E E E V X I (5.26)
'
0d d qV E (5.27)
dV dI
tV
d-axis
q-axis
tI
s tR I
q tjX I
qV
qI
qE
md fdX I
d q dX X I
7. 5.7
Hence, in the steady state no load case the internal voltage qE and the terminal
voltage tV are one and the same due to which the load angle becomes zero. As the
load increase the load angle also increases. It can also be observed that for 1 pu ,
internal voltage or terminal voltage in steady state no-load condition the field current
should be
1
fd
md
I
X
.
The load angle computed in equation (5.25) is with respect to the dq -axis of a
particular generator. In multi-machine systems there should be a common reference
instead of multiple individual generator references. Usually in a network with
multiple machines a slack bus or a reference bus is defined and all other bus angles in
the network are defined with respect to the common reference bus. Hence there
should be a conversion from dq -axis of individual generator to common reference in
real-imaginary (RI) axis or it is conversion from machine reference frame to common
network reference frame.
Fig. 5.4: Terminal voltage phasor in dq -axis and RI-axis
The load angle previously was defined as the angle by which the q-axis leads the
terminal voltage phasor but now it should be defined as the angle by which the q-axis
lead the common reference R-axis. In steady state both dq -axis and RI-axis rotate at
electrical synchronous speed hence the load angle in the new reference frame is fixed
for a given load condition. Even in case of transients since the effect of rotor speed
R-axis
q-axis
I-axis
tV
d-axis
RV
IV
qV dV
8. 5.8
variation on stator voltage is neglected the load angle can still be considered fixed.
From the phasor diagram shown in Fig. 5.4 the following expression can be derived
cos sin cos
sin cos sin
R t d q
I t d q
V V V V
V V V V
(5.28)
2
sin cos sin cos
( ) sin cos ( )
j
t R I d q q d
j
d q d q
V e V jV V V j V V
V jV j V jV e
(5.29)
2
( ) sin cos
j
d q t t tV jV V e V jV
(5.30)
Hence, when ever the stator voltage current expressed in dq -axis needs to be
transformed to network reference frame or RI-axis multiply the complex voltage or
current in dq -axis with the factor 2
j
e
where the load angle is the angle
between the q-axis of the individual generator with respect to the common reference
phasor R-axis.
To understand conversion from dq -axis to common reference let us take the
case of a single machine connected to an infinite bus with a voltage 0j
V e through a
transmission line with impedance T TR jX .
Fig. 5.5: Equivalent electrical circuit of single machine connected to an infinite bus
The generator terminal voltage is j
tV e
. Let the generator stator current be j
tI e
and the power factor angle is defined as . Both the angles and are
taken with respect to the infinite bus voltage phasor. Let the generator supply a
complex power P jQ to the infinite bus. From the equivalent circuit shown in Fig.
5.5 the following expressions can be written
0j
V e
qX sR
2
( )
j
j
d q tV jV e V e
2
j
qE e
2
( )
j
d qI jI e
TX TR
P jQ
9. 5.9
j
t j
t
P jQ
I e
V e
(5.31)
2 2 2
( ) ( )
j j j
q d q s q d q
j j
t s q t
E e V jV e R jX I jI e
V e R jX I e
(5.32)
Substituting q q md fd d q dE jE j X I X X I in equation (5.32) we get
2
j
j j j
q q t s q tjE e E e V e R jX I e
(5.33)
The load angle can be obtained from (5.33) and this angle is again with
respect to the infinite bus voltage phasor. Equation (5.33) can also be represented as a
phasor diagram as shown in Fig. 5.6. The network equations that is the relation
between the terminal voltage, stator current and the infinite bus voltage can also be
expressed in terms of dq -axis parameters as
02 2
2
( ) ( )
( ) ( )
j j
j
d q T T d q
j
d q T T d q
V jV e V e R jX I jI e or
V jV V e R jX I jI
(5.34)
From equation (5.34) separating real and imaginary terms we can get
sin
cos
d dT T
q qT T
V IR X V
V IX R V
(5.35)
The real and reactive power generated by the generator in terms of dq -axis
parameters is given as
*
*
2 2
( ) ( )
j j
j j
t t d q d q
d d q q q d d q
S P jQ V e I e V jV e I jI e
V I V I j V I V I
(5.36)
10. 5.10
The electrical torque output is given as, by substituting equation (5.1) and (5.2) for
stator flux linkages,
2 2
*2
( )
e d q q d q s q q d s d d
d d q q s q d
j j
s t q t
T I I V R I I V R I I
V I V I R I I
P R I real E e I e
(5.37)
It can be understood from equation (5.37) that the per unit electric torque produced is
equal to the real power delivered by the dependent source j
qE e
.
Fig. 5.6: Phasor diagram of steady state synchronous generator connected to infinite
bus
5.2 Multi-Machine System Representation
In a power system there will be multiple synchronous machines with their
respective exciters, turbines, speed governors and load distributed throughout the
d-axisdV dI
tV
q-axis
tI
s tR I
q tjX I
qV
qI
qE
md fdX I
d q dX X I
R-axis
I-axis
V
11. 5.11
system. The synchronous machines and load are connected through power system
network that is through transformers and transmission lines. For a total system
representation the system network model also should be included along with the
synchronous machine, exciter, turbine, speed governor and static/dynamic load.
Let there be a system with n number of buses. Let m represent number of load
buses and gn number of generator buses and hence gn number of generators. The
dynamics of an th
i generator along with the exciter, steam turbine and speed governor
are given as
' ' "
' ' ' '
1' 2
'
( )
( ) ( )
( )
qi di di
doi di di di qi di lsi di di
di lsi
qi fdi
dE X X
T X X I E X X I
dt X X
E E
(5.38)
" ' '1
1( )di
doi qi di lsi di di
d
T E X X I
dt
(5.39)
' "'
' ' ' ' '
2' 2
( )
( ) ( )
( )
qi qidi
qoi di qi qi qi di qi lsi qi qi
qi lsi
X XdE
T E X X I E X X I
dt X X
(5.40)
2" ' '
2( )
qi
qoi di qi lsi qi qi
d
T E X X I
dt
(5.41)
i
i base
d
dt
(5.42)
2 i i
mi ei i i base
base
H d
T T D
dt
(5.43)
( )fdi
Ei Ri Ei Ei fdi fdi
dE
T V K S E E
dt
(5.44)
, Min MaxRi Ai Fi
Ai Ri Ai Fi fdi Ai refi ti Ri Ri Ri
Fi
dV K K
T V K R E K V V V V V
dt T
(5.45)
Fi Fi
Fi fdi
Fi
dR K
R E
dt T
(5.46)
1Mi HPi RHi HPi RHi
RHi Mi CHi SVi
CHi CHi
dT K T K T
T T P P
dt T T
(5.47)
CHi
CHi CHi SVi
dP
T P P
dt
(5.48)
12. 5.12
1
1 ,0 MaxSVi i
SVi SVi refi SVi SVi
Di base
dP
T P P P P
dt R
(5.49)
Stator algebraic equations are given as
" ' "
" '
1' '
( ) ( )
( ) ( )
di lsi di di
di di qi di qi si qi
di lsi di lsi
X X X X
X I E V R I
X X X X
(5.50)
" ' "
" '
2' '
( ) ( )
( ) ( )
qi lsi qi qi
qi qi di qi di si di
qi lsi qi lsi
X X X X
X I E V R I
X X X X
(5.51)
Here, 1,2......... gi n . These gn generators are connected to the network. In a
power system network with n number of buses, the injected current at all the n buses
can be written in terms of admittance matrix and the bus voltages. The current
injected at th
i bus is given as, where 1,2.........i n ,
1
ij j
n
j j
i ij j
j
I Y e V e
(5.52)
Where, ijj
ijY e
is the ( , )i j element of the admittance matrix. jj
jV e
is the bus voltage
at th
j bus. The complex power injected at th
i bus can be written as
*
1
i j iji
n
jj
i i i i i j ij
j
P jQ V e I VV Y e
(5.53)
The injected power at any bus is the difference between the power generated at
that bus minus the load power required at that bus. Hence, at th
i bus the injected real
and reactive power can be written as
1
cos
n
i Gi Di i j ij i j ij
j
P P P VV Y
(5.54)
1
sin
n
i Gi Di i j ij i j ij
j
Q Q Q VV Y
(5.55)
13. 5.13
Where, ,Gi GiP Q are the real and reactive power generated at th
i bus and ,Di DiP Q
are the real and reactive power loads. The real and reactive power generated at a
generator bus is given as
*
2
Real
sin( ) cos( )
i
i
j
j
Gi i di qi
di i i i qi i i i
P V e I jI e
I V I V
(5.56)
*
2
Imaginary
cos( ) sin( )
i
i
j
j
Gi i di qi
di i i i qi i i i
Q V e I jI e
I V I V
(5.57)
In case static loads they can be represented as a function of the bus voltage as was
explained in Chapter 4. Then equation (5.56) and (5.57) can be written as
1
sin( ) cos( ) ( ) cos
n
di i i i qi i i i Di i i j ij i j ij
j
I V I V P V VV Y
(5.58)
1
cos( ) sin( ) ( ) sin
n
di i i i qi i i i Di i i j ij i j ij
j
I V I V Q V VV Y
(5.59)
for, 1,2,......, gi n , generator buses
1
( ) cos
n
Di i i j ij i j ij
j
P V VV Y
(5.60)
1
( ) sin
n
Di i i j ij i j ij
j
Q V VV Y
(5.61)
For, 1, 2,......,g gi n n n , for load buses.
Dynamic equations (5.38) to (5.49), stator algebraic equations (5.50) to (5.51)
and network power balance equations (5.58) to (5.61) together are called as
differential algebraic equations (DAEs) and completely define the behaviour of multi-
14. 5.14
machine power system. In case the loads are dynamic loads then apart from equation
(5.38) to (5.49) and (5.58) to (5.61) the dynamic equations of the loads should also be
considered.
5.2.1 A special case of impedance loads
If loads are represented as constant impedance loads then DAEs given in
equations (5.38) to (5.49) and (5.58) to (5.61) can be simplified to a significant extent.
Suppose load at an th
i bus is represented by an impedance load then, the real and
reactive power drawn by the load is given as
2
i
Li Li
Li
V
P jQ
Z
(5.62)
* *
Li Li i
Li
i Li
P jQ V
I
V Z
(5.63)
Where, , ,Li Li LiP Q Z are the real power, reactive power and impedance of a load
at an th
i bus. Substituting equation (5.63) in (5.52), the current injected into th
i bus
can be expressed as
*
1
ij j
n
j ji
i Gi Li Gi ij j
jLi
V
I I I I Y e V e
Z
(5.64)
Let 1
Li LiY Z
, then
1
1,2,.......,ij jLi i
n
j jj j
Gi Li i ij j g
j
I Y e V e Y e V e i n
(5.65)
1 2
1
0 , ,.......,ij jLi i
n
j jj j
Li i ij j g g
j
Y e V e Y e V e i n n n
(5.66)
15. 5.15
Let a new admittance matrix be formed with the load impedances included as
shunt branches at each bus, represented as BUSY . The equation (5.65) and (5.66) can
be represented as
11
11
0
0
gg
gg
G
nGn
BUS
nn
n n
VI
VI
Y
V
V
(5.67)
The network voltages and the generator currents can be represented in terms of
the generator internal voltages by making modifications to equations (5.67). This will
eliminate the generator currents and terminal voltages in dynamic equations given in
equations (5.38) to (5.49) reducing the complexity. This also has another advantage
when transient stability analysis is done and this will be explained later.
At an th
i generator bus an additional node is created marked as i with a
voltage " " 2
ij
i di qiE E jE e
, representing the generator internal voltage. The
impedance between i i nodes is given as Si diR jX . Hence, gn numbers of
additional nodes are added to the existing n nodes. The currents injected at these
nodes are given as
" " 2
" " 2
1
1 1
1,2 ......,
1,2,.......,
i
i
i
i
j
j
i Gi di qi i
Si di
j
j
di qi i g
Si di Si di
g
I I E jE e V e
R jX
E jE e V e i n
R jX R jX
i n
(5.68)
Note that since i is an additional node connected to an th
i generator node, it is
simply represented with a number same as that of the generator node to which it is
connected but with a prime to differentiate it. Also, now the additional nodes are the
generator nodes injecting a current into the network and at all other network buses the
16. 5.16
current injection is zero. Equation (5.68) can be included in the equation (5.67) which
leads to
1
(1, 1) (1, )1 1 1 1
1 ( , 1) ( , )
11
1 1
0 0
0 0
1 1 0 0
0 0
1
0
0
0
0
g
g g g
g
g
g
n nS d S d
G n n n n
Sng dng Sng dng
Gn
S
n
n
n
R jX R jX
I
R jX R jX
I
R j
1
( 1,1) ( 1, )
( ,1) ( , )
0
(1,1) (1, ) (1, 1) (1, )
1 ( ,1) ( , ) ( , 1) ( , )
0
0 0
0 0
g g g
g
BUS BUS g BUS g BUSd
BUS g BUS g g BUS g g BUS g
Sng dng
n n n
n n n
Y Y n Y n Y nX
Y n Y n n Y n n Y n n
R jX
1
1
1
( 1,1) ( 1, ) ( 1, 1) ( 1, )
( ,1) ( , ) ( , 1) ( , )
g
g
g
n
n
n
BUS g BUS g g BUS g g BUS g
n
BUS BUS g BUS g BUS
E
E
V
V
V
Y n Y n n Y n n Y n n
V
Y n Y n n Y n n Y n n
Let,
1 1
1
0
1
0
g g
S d
II
Sng dng n n
R jX
Y
R jX
,
(1, 1) (1, )1 1
( , 1) ( , )
1
0
0 0
1 0 0
0
g
g g g
g
n nS d
IN
n n n n
Sng dng
n n
R jX
Y
R jX
17. 5.17
(1,1) (1, ) (1, 1) (1, )
( ,1) ( , ) ( , 1) ( , )
( 1,1) ( 1, ) ( 1, 1
( ,1) ( , )
BUS BUS g BUS g BUS
BUS g BUS g g BUS g g BUS g
BUS
BUS g BUS g g BUS g g
BUS BUS g
Y Y n Y n Y n
Y n Y n n Y n n Y n n
Y
Y n Y n n Y n n
Y n Y n n
) ( 1, )
( , 1) ( , )
BUS g
BUS g BUS
Y n n
Y n n Y n n
1
1
1 1
1 1
1
, ,
g
g
g g
g g
G
n
G N
n
Gn n
n n
n n
V
I E
V
V
I E
V
I E V
Then,
'
1 1
0
g gg g
II ING
T
IN BUSn n n nn n n n
Y Y
Y Y
N
EI
V
(5.69)
From equation (5.69) it can be seen that,
1' T
BUS INY Y
NV E (5.70)
1' T
G II IN BUS INY Y Y Y
I E (5.71)
From equation (5.70) and (5.71) it can be observed that the network voltages
and the generator currents, represented by the vector NV and GI , can be obtained
directly from the internal voltages vector E without the need to solve stator
algebraic and network power balance equations.
18. 5.18
5.2.2 Initial Conditions
Before doing any stability analysis the system along with its generators, loads
and network should be initialized. These are the steps which should be followed for
initializing the system
Step 1
The network real and reactive power balance equations given in equations (5.58) to
(5.61) should be solved. Solving equations (5.58) to (5.61) is nothing but doing load
flow analysis. After load flow analysis all the network bus complex voltages, real and
reactive powers injected at a bus are know.
Step 2
Find the internal load angle of each generator. This can be done as following
1,2,.....i i Gij j j
i i si qi Gi gE e V e R jX I e i n
(5.72)
After load flow analysis the network bus voltages are known. The currents generated
by each generator are also known. With this information an internal voltage can be
defined for all the generators as given in equation (5.72). The internal load angle i of
an th
i generator is the angle of the complex internal voltage ij
iE e
.
Step 3
Calculate the ,d q axis components of voltages, currents and fluxes
sindi i i iV V (5.73)
cosqi i i iV V (5.74)
sindi Gi i GiI I (5.75)
19. 5.19
cosqi i i GiI I (5.76)
' '
di di si di qi qiE V R I X I (5.77)
' '
qi qi si qi di diE V R I X I (5.78)
' '
( )fdi qi di di diE E X X I (5.79)
' '
1 ( )di qi di lsi diE X X I (5.80)
' '
2 ( )qi di qi lsi qiE X X I (5.81)
Fi
Fi fdi
Fi
K
R E
T
(5.82)
( )Ri Ei Ei fdi fdiV K S E E (5.83)
Ri
refi i
Ai
V
V V
K
(5.84)
i base (5.85)
SVi refiP P (5.86)
CHi SViP P (5.87)
Mi CHiT P (5.88)
ei MiT T (5.89)
The synchronous machine model along with the exciter, turbine and speed
governor model given in (5.38) to (5.49) can be simplified significantly my making
some assumptions. This simplification can lead to significant reduction in
computational complexity [1]-[2]. The various models that can be derived by
simplifying equations (5.38) to (5.49) is given below
5.3 Sub-transient Model with Stator and Network Transients
Neglected
The rate of changes of the stator flux linkages ,d q (here dot means rate of
change with respect to time) in equation (3.149) and (3.150) in Chapter 3, can be
neglected. The reason behind this is that the stator, which is connected to the rest of
20. 5.20
the network electrically, has very fast transients as compared to the rotor. Compared
to the slow dynamics of the rotor the stator dynamics can be considered to be very fast
and hence any change or disturbance in the stator or the network can be considered as
an instantaneous change without any dynamics. This leads to reduction in
computational complexity drastically as the network transient also need not be
considered along with stator transients. With these assumptions the synchronous
machine dynamics along with exciter, turbine and governor are given as
q d s dV R I (5.90)
d q s qV R I (5.91)
0o s oV R I (5.92)
' ' "
' ' ' ' '
1' 2
( )
( ) ( )
( )
q d d
do q d d d q d ls d d fd
d ls
dE X X
T E X X I E X X I E
dt X X
(5.93)
" ' '1
1( )d
do q d ls d d
d
T E X X I
dt
(5.94)
' "'
' ' ' ' '
2' 2
( )
( ) ( )
( )
q qd
qo d q q q d q ls q q
q ls
X XdE
T E X X I E X X I
dt X X
(5.95)
2" ' '
2( )
q
qo d q ls q q
d
T E X X I
dt
(5.96)
base
d
dt
(5.97)
2
m e base
base
H d
T T D
dt
(5.98)
( )fd
E R E E fd fd
dE
T V K S E E
dt
(5.99)
R A F
A R A in R A F fd A ref t
F
dV K K
T V K V V K R E K V V
dt T
(5.100)
F F
F fd
F
dR K
R E
dt T
(5.101)
1M HP RH HP RH
RH M CH SV
CH CH
dT K T K T
T T P P
dt T T
(5.102)
21. 5.21
CH
CH CH SV
dP
T P P
dt
(5.103)
1
1SV
SV SV ref
D s
dP
T P P
dt R
(5.104)
" ' "
" '
1' '
( ) ( )
( ) ( )
d ls d d
d d d q d
d ls d ls
X X X X
X I E
X X X X
(5.105)
" ' "
" '
2' '
( ) ( )
( ) ( )
q ls q q
q q q d q
q ls q ls
X X X X
X I E
X X X X
(5.106)
Equations (5.90), (5.91) can be substituted in (5.105) and (5.106) which lead to
" ' "
" '
1' '
( ) ( )
( ) ( )
d ls d d
d d q d q s q
d ls d ls
X X X X
X I E V R I
X X X X
(5.107)
" ' "
" '
2' '
( ) ( )
( ) ( )
q ls q q
q q d q d s d
q ls q ls
X X X X
X I E V R I
X X X X
(5.108)
Now by multiplying equation (5.107) by complex number 1j and adding both
equation (5.107) and (5.108) lead to
" ' " " ' "
" ' " '
2 1' ' ' '
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
q ls q q d ls d d
q q d q d d q d
q ls q ls d ls d ls
d s d q s q
X X X X X X X X
X I E j X I E
X X X X X X X X
V R I j V R I
(5.109)
Rearranging the terms, we can get
"
" ' " " ' "
" " ' '
2 1' ' ' '
"
" "
( ) ( ) ( ) ( )
( )
( ) ( ) ( ) ( )
E
q ls q q d ls d d
q d q d q q d
q ls q ls d ls d ls
d q s d d q
d q s d d q
X X X X X X X X
X X I E j E
X X X X X X X X
V jV R jX I jI
E V jV R jX I jI
…………………………………………………………………………… (5.110)
22. 5.22
Equation (5.110) can be further simplified by assuming that the effect of
saliency during sub-transient is negligible on the voltage "
E that is it can be assumed
that " "
q dX X . This assumption is very useful in multi-machine system stability
analysis. With this assumption equation (5.110) can be written as
" " " "
d q d q s d d qE E jE V jV R jX I jI (5.111)
Where,
" ' " " ' "
" ' " '
2 1' ' ' '
( ) ( ) ( ) ( )
,
( ) ( ) ( ) ( )
q ls q q d ls d d
d d q q q d
q ls q ls d ls d ls
X X X X X X X X
E E E E
X X X X X X X X
Equation (5.111) when expressed in RI-axis can be represented as
" " "2 2 2
j j j
d q d q s d d qE jE e V jV e R jX I jI e
(5.112)
Equation (5.112) can be represented as a dynamical electrical circuit, as shown
in Fig. 5.7. Where, " " 2
j
d qE jE e
is a dependent voltage source behind the sub-
transient reactance "
dX producing a current 2
j
d qI jI e
with a terminal voltage
of 2
j
d qV jV e
.
Fig. 5.7: Dynamic electrical equivalent circuit of synchronous generator sub-transient
model
2
j
d qI jI e
"
dX sR
2
j
d qV jV e
" " 2
j
d qE jE e
23. 5.23
The per unit torque is equal to the real power delivered by the dependent source
" " 2
j
d qE jE e
, which is
*
" " 2 2
" "
j j
e d q d q
d d q q
T real E jE e I jI e
E I E I
(5.113)
5.4 Transient or Two-Axis Model
The sub-transient generator model can be further simplified. The sub-transient
time constants " "
,do qoT T corresponding to the damper windings 1 ,2d q are very small as
compare to the transient time constants ' '
,do qoT T . Hence, the sub-transient time constant
" "
,do qoT T can be set zero in equation (5.94) and (5.96). This leads to
" ' '1
1
' '
1
( ) 0
( )
d
do q d ls d d
d q d ls d
d
T E X X I
dt
E X X I
(5.114)
2" ' '
2
' '
2
( ) 0
( )
q
qo d q ls d q
q d q ls q
d
T E X X I
dt
E X X I
(5.115)
Substituting equation (5.114) and (5.115) in equation (5.93) to (5.106) the following
expression can be obtained
'
' ' '
( )
q
do q d d d fd
dE
T E X X I E
dt
(5.116)
'
' ' '
( )d
qo d q q q
dE
T E X X I
dt
(5.117)
base
d
dt
(5.118)
24. 5.24
2
m e base
base
H d
T T D
dt
(5.119)
( )fd
E R E E fd fd
dE
T V K S E E
dt
(5.120)
R A F
A R A in R A F fd A ref t
F
dV K K
T V K V V K R E K V V
dt T
(5.121)
F F
F fd
F
dR K
R E
dt T
(5.122)
1M HP RH HP RH
RH M CH SV
CH CH
dT K T K T
T T P P
dt T T
(5.123)
CH
CH CH SV
dP
T P P
dt
(5.124)
1
1SV
SV SV ref
D s
dP
T P P
dt R
(5.125)
' '
d d d qX I E (5.126)
' '
q q q dX I E (5.127)
Substituting equation (5.90) and (5.91) in equation (5.126) and (127) lead to
' '
d s d q q dV R I X I E (5.128)
' '
q s q d d qV R I X I E (5.129)
Multiplying equation (5.129) with 1j and adding it to equation (5.128) gives
' ' ' ' '
d q s d d q q d q d qV jV R jX I jI X X I E jE
(5.130)
Expressing equation (5.130) in RI-axis lead to, with the additional assumption that
' '
d qX X .
' ' '2 2 2
j j j
d q d q s d d qE jE e V jV e R jX I jI e
(5.131)
25. 5.25
Fig. 5.8: Dynamic electrical equivalent circuit of synchronous generator transient
model
The dynamic electrical equivalent circuit of the synchronous generator transient
model is shown in Fig. 5.8. The electrical torque is given as
*
' ' 2 2
' '
j j
e d q d q
d d q q
T real E jE e I jI e
E I E I
(5.132)
5.5 Flux Decay or One-Axis Model
The transient model can be further simplified by neglecting the damper
windings 1q and considering only the dynamics of field winding alone. This is also
justified as the time constant '
qoT is less compared to the time constant '
doT and hence
can be set to zero in equation (5.117) and with this assumption equation (5.116) to
(5.127) can be written as
'
' ' '
( )
q
do q d d d fd
dE
T E X X I E
dt
(5.133)
base
d
dt
(5.134)
2
m e base
base
H d
T T D
dt
(5.135)
'
dX sR
2
j
d qV jV e
' ' 2
j
d qE jE e
2
j
d qI jI e
26. 5.26
( )fd
E R E E fd fd
dE
T V K S E E
dt
(5.136)
R A F
A R A in R A F fd A ref t
F
dV K K
T V K V V K R E K V V
dt T
(5.137)
F F
F fd
F
dR K
R E
dt T
(5.138)
1M HP RH HP RH
RH M CH SV
CH CH
dT K T K T
T T P P
dt T T
(5.139)
CH
CH CH SV
dP
T P P
dt
(5.140)
1
1SV
SV SV ref
D s
dP
T P P
dt R
(5.141)
' '
d d d qX I E (5.142)
q q qX I (5.143)
Substituting equation (5.90) and (5.91) in equation (5.142) and (5.143) lead to
d s d q qV R I X I (5.144)
' '
q s q d d qV R I X I E (5.145)
Multiplying equation (5.145) with 1j and adding it to equation (5.144) and
expressing in RI-axis gives
' ' '2 2 2
j j j
q d q q d q s d d qX X I jE e V jV e R jX I jI e
(5.146)
The dynamic electrical equivalent circuit of flux decay model is given in Fig. 5.9.
27. 5.27
Fig. 5.9: Dynamic electrical equivalent circuit of synchronous generator flux decay
model
5.6 Classical or Constant Flux Linkage Model
This is the simplest model but not very accurate. This model is useful in case of
very large power systems. In case of dynamic studies with a time period less than that
of '
doT are considered then the field winding dynamics can also be neglected. In the
flux decay model in equation (5.133), if we assume that during the period of study the
voltage behind the transient reactance '
dX , ' '
d qE jE , is constant and only the
electro-mechanical equations (5.134) and (5.135) need to be considered. The transient
voltage is computed at steady state, ' '
_ _d ss q ssE jE , and kept constant throughout the
study. The input mechanical torque mT is assumed to be constant and hence the
turbine and turbine governor dynamics are eliminated. Hence, the dynamic equations
of the model can be written as
base
d
dt
(5.147)
2
m e base
base
H d
T T D
dt
(5.148)
The dynamic equivalent circuit of the classical model is shown in Fig. 5.10
'
dX sR
2
j
d qV jV e
' ' 2
j
q d q qX X I jE e
2
j
d qI jI e
28. 5.28
Fig. 5.10: Dynamic electrical equivalent circuit of synchronous generator classical
model
'
dX sR
2
j
d qV jV e
' ' 2
_ _
j
d ss q ssE jE e
2
j
d qI jI e
29. 5.29
Example Problems
E1. A three-phase, 50 Hz, synchronous generator is connected to an infinite bus
through a transformer and two parallel transmission lines. The generator is
transferring a complex power of 1 0.25j to the infinite bus.
The generator parameters are given below:
' '
0.8, 0.7, 0.2, 0.3, 0.55, 0.0025, 20, 1.0,
0.36, 0.125, 1.8
d q ls d q s A E
E F F
X X X X X R K K
T K T
= = = = = = = =
= = =
Initialize the synchronous generator
Sol:
The current drawn by the infinite bus can be computed as following
1 0.25 1.038 14.036
0
t t
P jQ
I j
V
f
¥
-
= = - = -
(E1.1)
The terminal voltage of the synchronous generator is given as
0.2LX
0.15TX 1 0V
t tV
1 0.25P jQ j
30. 5.30
( )0.5 1.0915 13.24t t t L t tV V j X X Iq f¥ = + + ´ = (E1.2)
The internal voltage of the synchronous generator is given as
( ) 2.4536 52.607q t t s q t tE V R jX Id q f = + + = (E1.3)
The rest of the synchronous generator variables can be initialized as following
sin 0.6923d t tV V (E1.4)
cos 0.8438q t tV V (E1.4)
sin 0.9463d t tI I (E1.6)
cos 0.4086q t tI I (E1.7)
' '
0.4699d d s d q qE V R I X I (E1.8)
' '
1.1288q q s q d dE V R I X I (E1.9)
' '
( ) 2.5482fd q d d dE E X X I (E1.10)
' '
1 ( ) 1.0341d q d ls dE X X I (E1.11)
' '
2 ( ) 0.3269q d q ls qE X X I (E1.12)
0.1769F
F fd
F
K
R E
T
(E1.13)
( ) 2.5482R E E fd fdV K S E E (E1.14)
1.2189R
ref t
A
V
V V
K
(E1.15)
314.159 /i base rad s (E1.16)
1.0SV refP P (E1.17)
1.0CH SVP P (E1.18)
1.0M CHT P (E1.19)
1.0e MT T (E1.20)
31. 5.31
E2. A 5 bus power system has two synchronous generators and one load as shown in
the figure below. The generators are connected at bus-1 and bus-4. The load is
connected at bus-5. The per unit reactances of the connecting branches are mentioned
in the figure. The load is consuming a complex power of 1.8 0.65j at a voltage
0.98 11.53 . Assuming the load to be constant impedance type, convert the system
admittance matrix to reduced admittance matrix with only internal nodes.
Sol:
The system admittance matrix can be written as
6.67 6.67 0 0 0
6.67 13.16 4 0 2.50
0 4 13.55 5 5.55
0 0 5 5 0
0 2.5 5.55 0 8.05
BUS
j j
j j j j
Y j j j j
j j
j j j
é ù-
ê ú
ê ú-
ê ú
ê ú= -ê ú
ê ú-ê ú
ê ú
ê úë û
(E2.1)
Since the load is considered as impedance type, the load impedance can be included
in the admittance matrix. The load admittance should be added to the 5th
row-5th
column of the admittance matrix. The load admittance can be found as following:
*
5
1.8 0.65
1.932 0.2828
0.98 11.53
L
P jQ j
I j
V
- -
= = = -
-
(E2.2)
0.2j0.25j0.15j
1.8 0.65j
2 3 4
5
1
0.4j 0.18j
5 0.98 11.53V
'
0.20dX j
'
0.35dX j
32. 5.32
5 0.4720 0.1704L
L
V
Z j
I
= = + (E2.3)
1
2.0444 0.8295L
L
Y j
Z
= = - (E2.4)
The system admittance matrix can now be changed as
6.67 6.67 0 0 0
6.67 13.16 4 0 2.50
0 4 13.55 5 5.55
0 0 5 5 0
0 2.5 5.55 0 2.0444 7.2205
BUS
j j
j j j j
Y j j j j
j j
j j j
é ù-
ê ú
ê ú-
ê ú
ê ú= -ê ú
ê ú-ê ú
ê ú+ê úë û
(E2.5)
In order to reduce the system to internal nodes first add additional buses representing
generator nodes. The generator internal nodes are connected to the respective
generator terminal buses through the direct axis transient reactance. With the
generator internal nodes added the system admittance matrix changes to
'
GG GN
NG BUS
Y Y
Y
Y Y
é ù
ê ú= ê ú
ê úë û
(E2.6)
Where,
1
0
0.2
1
0
0.35
GG
j
Y
j
é ù
ê ú
ê ú
ê ú=
ê ú
ê ú
ê úë û
,
1
0 0 0 0
0.2
1
0 0 0 0
0.35
T
GN NG
j
Y Y
j
é ù
ê ú-
ê ú
ê ú=
ê ú
ê ú-
ê úë û
'
11.67 6.67 0 0 0
6.67 13.16 4 0 2.50
0 4 13.55 5 5.55
0 0 5 7.8571 0
0 2.5 5.55 0 2.0444 7.2205
BUS
j j
j j j j
Y j j j j
j j
j j j
é ù-
ê ú
ê ú-
ê ú
ê ú= -ê ú
ê ú-ê ú
ê ú+ê úë û
33. 5.33
The reduced admittance matrix can be computed as
( )
1' 0.0343 2.0289 0.0281 0.0788
0.0281 0.0788 0.0230 1.5819RED GG GN BUS NG
j j
Y Y Y Y Y
j j
- é ù- +
ê ú= - =
ê ú+ -ë û
(E2.7)
In the reduced system only two buses are present, which are the internal buses of the
respective generators, hence the size of the admittance matrix is 2 2´ .
34. 5.34
References
1. M. Stubbe, A. Bihain, J. Deuse and J. C. Baader, “STAG-A new unified
software program for the study of the dynamic behavior of electrical power
systems,” IEEE Trans., Vol. PWRS-4, No. 1, pp. 129-138, 1989.
2. P. Kundur and P. L. Dandeno, “Implementation of advanced generator
models into power system stability programs,” IEEE Trans., Vol. PAS-102,
pp. 2047-2052, July 1983.
