Chapter 4
Modelling of Exciter, Turbine and System Load
4.1 Exciter
Excitation system provides the current required for the field winding of a
synchronous generator to produce the rated terminal voltage at the generator
terminals. The basic blocks that are involved in the excitation system [1] are shown in
Fig. 4.1.
Limiting and
protecting
circuit
Voltage
transducers
Voltage
Regulator
Exciter
Generator
Power
System
Rate
Feedback
Power system
stabilizer
Fig. 4.1: Excitation system block diagram
Exciter
Exciter can be of three types DC exciter, AC exciter and Static exciter.
DC exciter: In this type of exciter a separately or self excited DC generator driven by
a motor or connected to the same shaft as that of the main generator rotor is used. In
case of separately excited DC generator the field winding of the DC generator is
energised through a permanent magnet AC generator, the three-phase out of which is
converted to DC through rectifiers. Example of DC excitation system is IEEE type
4.1
DC1A [1], [2]. This type of excitation was widely used up to 1960 but now-a-days
AC excitation or static excitation is being used. For older generating stations where
DC excitation is still used the voltage regulator alone is replaced with electronic
regulators.
AC exciter: In this type of exciter the AC generator whose armature is mounted on the
same shaft as that of the main generator, with its field stationary, is used for supplying
field current to the main generator field winding. The output of AC exciter generator
is converted to DC through rotating rectifiers, as the armature is now rotating at the
main generator rotor speed, and the output is directly connected to the main field
winding. The filed of the AC exciter generator itself is energised through a pilot
permanent magnet AC generator whose three-phase output is converted to DC
through rectification. Example IEEE type AC1A.
Static exciter: In this exciter the output of the main synchronous generator is
converted from AC to DC through static rectification and then the output is supplied
to the main generator field winding through slip rings. Example IEEE type ST1A.
Voltage transducers: Voltage transducers measure the three-phase terminal voltage
through potential transformers and convert them and filter them to DC which is
compared with the reference voltage and the error signal is used to control the main
exciter field winding current through the exciter.
Voltage Regulator: The voltage regulator can be of magnetic amplification type or
digital amplification type. At present digital amplification is widely used. The voltage
regulator amplifies the error signal generated by voltage transducers and the output of
the voltage regulator is used to control the pilot excitation there by controlling the
main generator field winding through the DC/AC excitation system.
Limiting and protection circuit: Limiting and protection circuits are used to limit main
generator field winding current, over excitation, under excitation, terminal voltage,
Volts-per-Hertz etc.
4.2
Rate feedback: Rate feedback is used to stabilize the excitation system and will be
explained shortly
Power system stabilizer: It is used to damp the power system oscillations through the
excitation system. Power system stabilizer will be explained in detail while discussing
small-signal stability.
4.1.1 IEEE Type DC1A
The modelling of IEEE Type DC1A [1], [2] will be discussed here. Though,
only DC excitation modelling is dealt here, the same approach can be extended to
IEEE Type AC1A and ST1A [1]. A separately excited DC generator excitation is
shown in Fig. 4. 2.
rf 1

La1
ra
iin1

ein1

Lf 1
K a11a1
eout1  v fd


Fig. 4.2: Separately excited DC machine circuit
The output of the DC generator, eout1 , can be directly applied to the main
generator field winding and hence eout1  v fd . DC generator field is excited through a
pilot generator as explained earlier. If ein1 is the voltage applied to the DC generator
field, with field current iin1 , then the following expression can be written
ein1  iin1rf 1  N f 1
d f 1
(4.1)
dt
4.3
Where, rf 1 , L f 1 are the resistance and inductance of the DC generator field
winding. N f 1 is the number of turns of the DC generator field winding. If the flux
produced due to field current iin1 is  f 1 and a fraction of it 1/  links the armature of
the DC generator then the armature flux a1  1/    f 1 . The induced emf in a DC
generator armature is directly proportional to the speed of the armature and the flux
linking it and can be written as
ea1  ka1a1
(4.2)
here, a1 is the speed of the armature and k is a proportionality constant
corresponding to specific machine. The leakage inductance of the DC generator
armature La1 and resistance ra1 are negligible. Hence, the following expression can be
written
ea1  eout1  v fd  ka11a1
(4.3)
From equation (4.3) we can express  f 1 in terms of the DC generator output voltage,
armature speed and proportionality constant as
 f 1  a1 

v fd
ka11
(4.4)
The DC generator field winding flux linkage is given as
 f 1  l f 1iin1  N f 1 f 1 
V fd
iin1

 N f1
v fd
ka11
(4.5)
ka11
l f 1  kg
N f 1 1
(4.6)
Hence, without considering the effect of saturation,
4.4
iin1 
V fd
(4.7)
kg
slope k g
v fd
iin1
v fd
iin1
kg
Fig. 4.3: Excitation saturation characteristics
As can be observed from Fig. 4.3, field current iin1 
v fd
kg
lead to a DC generator
output voltage v fd without saturation. If saturation is considered then an additional
current iin1 is required to produce the same output voltage v fd . Hence
iin1 
v fd
kg
 iin1 
v fd
kg
 f sat (v fd )v fd
(4.8)
iin1
is the function representing the effect of saturation.
v fd
Substituting equations (4.4) and (4.8) in (4.1) lead to
Where,
ein1 
rf 1
kg
f sat (v fd ) 
v fd  rf 1 f sat (v fd )v fd 
l f 1 dv fd
(4.9)
k g dt
Since, the DC generator output is directly connected to the generator field
winding the base used for the field circuits to express parameters in per units can be
4.5
used here as well. Hence, divide both sides of equation (4.9) by V fd _ base and multiply
by
X md
X
(note E fd  md V fd ) and this lead to
R fd
R fd
E
S ( E fd )
fd
K E 
VR


E


 
 1

r X
 R
 X
X md
1
 rf 1 f sat  fd E fdVbase _ fd   md V fd 
ein1  f 1  md V fd 
 Vbase _ fd


R fd Vbase _ fd
k g  R fd
 X md
  R fd


(4.10)
TE


l d X
....  f 1  md V fd 

k g dt  R fd

TE
dE fd
dt
 VR   K E  S E ( E fd )  E fd
(4.11)
In equation (4.11) K E , TE are the gain and time constant of the DC exciter. The
saturation function S E ( E fd ) can be mathematically approximated as
S E ( E fd )  Ae
BE fd
(4.12)
Where, A, B are constants which can be computed. In the steady state a
maximum scaled DC field winding input voltage VR max will lead to a maximum scaled
DC generator output armature voltage E fd hence
0  VR max   K E  S E ( E fd max )  E fd max
(4.13)
Usually the value of saturation function S E ( E fd ) at two different values,
S E _ max ( E fd max ) and S E _ 0.75max (0.75E fd max ) for two different E fd is specified
S E _ max ( E fd )  Ae
BE fd max
S0.75 E _ max ( E fd )  Ae
(4.14)
B 0.75 E fd max
4.6
The constants A, B can be computed by solving equations (4.13) and (4.14).
The voltage regulator can be simply represented by a first order system with a
gain K A , by which it amplifies the error signal, and a time constant TA . If the input to
the voltage regulator is an error signal Vin and the output of the voltage regulator is
the input to the DC generator field winding VR then the dynamic equation of the
voltage regulator can be expressed as
TA
dVR
 VR  K AVin
dt
VR _ min  VR  VR _ max
(4.15)
So far a separately excited DC generator was considered. A self excited DC
generator can also be modelled in the same way. In case of self excited DC generator
since armatures it self supplies the field current as well assuming there is residual
field voltage to build up the armature voltage gradually. The voltage regulator output
comes in series with the field winding and hence
v fd  iin1rf 1  N f 1
d f 1
dt
 ein1
(4.16)
with this modification the dynamic equation of the exciter in per units can be written
as
TE
dE fd
dt
 VR   K E  S E ( E fd )  E fd
(4.17)
where, K E  K E  1 , and is a small negative number. From equation (4.17) it can
be observed that since K E is small negative number and with small positive S E ( E fd )
the closed loop transfer function, with input VR and output E fd , has a positive eigen
value making the system unstable. Even without this problem to make the entire open
loop excitation system with the voltage regulator and exciter stable a feedback system
is used to stabilize. A transformer as shown in Fig. 4.4 is used to feeds back the output
4.7
voltage E fd to the input. The transformer secondary side produces a voltage VF
which is then subtracted from the input voltage VR .
It 2

Rt 2
Lt1
Lt 2
N 2 : N1
VF
rt1
I t1

Ltm
E fd


Fig. 4.4: Stabilizing transformer circuit
From Fig. 4.4, the following voltage equation can be written
E fd  Rt1 I t1   Lt1  Ltm 
dI t1
dt
(4.18)
Where, Rt1 , Lt1 , Ltm are the resistance, leakage inductance and magnetising
inductance of the primary side of the feedback transformer. I t1 is the primary side
current. If the secondary side the leakage inductance Lt 2 is very large then the
secondary side current cannot change from its initial conditions and in this case it is
zero then with turns ratio N 2 : N1
VF 
dI
N2
Ltm t1
N1
dt
(4.19)
or
d 2 I t1
dVF N 2

Ltm
dt
N1
dt 2
(4.20)
Differentiating equation (4.18) and rearranging we get
4.8
dE fd 
d 2 I t1  Rt1 N1
1
 
VF 

2
dt
dt   Lt1  Ltm 
 Ltm N 2
(4.21)
Substituting equation (4.20) in (4.21) we get
dE fd 
 R N
dVF N 2
1

Ltm   t1 1 VF 

dt
N1
dt   Lt1  Ltm 
 Ltm N 2
(4.22)
let,
TF 
Lt1  Ltm
N L
, K F  2 tm
Rt1
N1 Rt1
(4.23)
Then, equation (4.22) can be written as
E fd 
1
dVF
K V
  VF  F  R   K E  S E ( E fd ) 

dt
TF
TF  TE
TE 
(4.24)
Defining,
RF 
KF
E fd  VF
TF
(4.25)
RF is called as the rate feedback. Differentiating equation (4.25) and rearranging we
get
TF
dRF
K
  RF  F E fd
dt
TF
(4.26)
Finally the IEEE type DC1A exciter dynamics can be expressed by the following
three equations
4.9
TE
TA
dE fd
dt
 VR   K E  S E ( E fd )  E fd
(4.27)
dVR
K K
 VR  K A RF  A F E fd  K A Vref  Vt  , VR _ min  VR  VR _ max
dt
TF
(4.28)
dRF
K
  RF  F E fd
dt
TF
(4.29)
The control block diagram corresponding to the dynamics equations (4.27) to (4.29) is
given in Fig. 4.5.
VR max
Vt
Vref
 Vin


KA
1  sTA
VR


VR min
1
sTE
E fd
KE

S E ( E fd )
VF


Rf
KA
1  sTA
KF
TF
Fig. 4.5: Control block diagram of IEEE type DC1A exciter
4.2 Model of Turbine
So far the modelling of synchronous machine along with the exciter was
explained. Now, modelling of prime mover will be explained. Prime mover can be a
diesel generator, gas turbine, hydro turbine or steam turbine. Prime mover provides
the mechanical power required as input to the synchronous machine to rotate the rotor
4.10
of the synchronous machine at synchronous speed. Here, we will discuss about hydro
and steam turbine.
4.2.1 Hydro Turbine
Hydro generating stations consists of a reservoir, gate to release water to
generating station, penstock (a conduit) to carry water from the reservoir up to the
hydro turbine. The turbine can be impulse-type (Pelton wheel) or reaction turbine
(Francis or propeller).
The water which flows from the reservoir through the
penstock is directed on to the turbine blades, which are spoon-shaped buckets, and
this force makes the turbine rotate. The hydro turbine and the rotor of the synchronous
machine are mounted on the same shaft.
Hydro turbine modelling
Let H be the height from the gate to the water level in the reservoir also called
as head, U the water velocity and G the gate position (full open means full rated
mechanical power input to the hydro turbine and full close means nil power to the
hydro turbine). The velocity of the water and be expressed as
U  Ku G H
(4.30)
Where K u is a constant of proportionality. If there is a small change in the velocity
of water it can be expressed as a sum of linear changes in the gate position and head
as
U 
U
U
H 
G
H
G
(4.31)
The partial derivatives in equations (4.31) should be evaluated at an initial operating
condition from which speed of the water changed by a small amount. The initial speed
of water, for a given initial gate position Go and head H o , is given as
4.11
U o  K u Go H o
(4.32)
Hence, equation (4.31) can be written as
U 
1
K u Go H  K u H o G
2 Ho
(4.33)
Equation (4.33) can be normalized by dividing it with equation (4.32)
U H G


U o 2 H o Go
(4.34)
Let U / U o be represented as a normalized value as U . Similarly, H and G
represents the normalized values of small changes in head and gate positions. So,
equation (4.34) can be written as
U 
1
H  G
2
(4.35)
The turbine mechanical power is proportional to the product of pressure and flow and
hence, with proportionality constant K P
Pm  K P HU
(4.36)
Again a small change in Pm can be expressed in a normalized form with the initial
condition Pmo  K P H oU o , as
Pm  H  U  1.5H  G
(4.37)
From Newton’s second law ( Force=Mass×acceleration ), the acceleration in water
column due to change in water head can be written as
4.12
 LA
d U
  A ag H
dt
(4.38)
Where, L, A,  are the length of the conduit, area of the pipe and the mass
density of the water. ag is the acceleration due to gravity.  LA represents the mass of
the water in conduit and  ag H represents the incremental force. To normalize
equation (4.38) divide both sides by A ag H oU o . After rearranging the normalized
equation (4.38) is written as
TW

LU o d U
d U
 H
 H or TW
dt
ag H o dt
(4.39)
TW has seconds as its units which can be verified from equation (4.39). It is called as
water starting time and it the time required for a head H o to accelerate the water to a
velocity U o . If equation (4.39) is converted in to Laplace domain then
TW sU ( s )  H ( s )
(4.40)
Substituting, H  2  U  G  , from equation (4.35) in equation (4.39) and
rearranging leads to
U ( s ) 
1
G ( s )
1
1  TW s
2
(4.41)
From equation (4.35) and (4.37) it can be observed that
Pm  3U  2G
(4.42)
Substituting equation (4.42) in (4.41) and eliminating U lead to
4.13
Pm ( s ) 
1  TW s
G ( s )
1
1  TW s
2
(4.43)
Now this model is a linear model and hence is only valid for slow gradual small
changes and not for sudden large changes. Equation (4.43) can be represented in time
domain as
TW
d Pm
d G 

 2 Pm  G  TW

dt
dt 

(4.44)
The gate position G is the corresponding power output of the turbine that is a per unit
change in gate position leads to per unit change in the mechanical power output Pm .
Also, since the model is valid for perturbations around an operating point we can
write the actual parameters as
Pm  Pmo  Pm ,
(4.45)
G  Go  G,
With the assumption given in (4.45) we can write equation (4.44) as
TW
dPm
dG 

 2   Pm  G  TW

dt
dt 

(4.46)
Since the synchronous machine model is represented in per units the turbine model
should also be presented in per units. To do this we can divide both sides of equation
(4.46) by base MVA or the rating of the synchronous machine. Also it can be noted
that in per units power and torque are same hence
TW
dG pu 

dTM
 2 TM  G pu  TW

dt
dt 

(4.47)
4.14
It has to be noted that TM is the per unit mechanical torque output of the hydro
turbine. Equation (4.47) represents the dynamic behaviour of hydro-turbine [3].
4.2.2 Steam Turbine
The steam generating system consists of a boiler to generate steam, steam chest
with a valve to store the steam and release it in to the turbine as shown in Fig. 4.6 [3].
The steam from the steam chest initially enters a high pressure turbine through the
steam valve nozzle at high temperature and pressure with a high velocity and this gets
converted in to rotational kinetic energy through blades connected to the turbine shaft.
Now as in the case of hydro turbine the position of the steam valve is proportional to
the mechanical power output of the turbine. Only part of the input power gets
converted into mechanical power at the high pressure turbine shaft and the steam then
enters reheater were the temperature and pressure of the steam are increased and then
again this reheated steam is forced in to a low pressure or a series of intermediate and
low pressure turbines. Each of these turbines convert the steam into rotational energy
through their blades connected on the shaft. The connection can be tandem connected
that is all the high pressure, intermediate pressure, low pressure turbine and the
generator are on the same shaft as is shown in Fig. 4.6 or cross connected that is there
may be two shafts cross connected with same steam input valve and two generators
connected to each shaft respectively.
Fig. 4.6: Steam generation system schematic diagram
The steam chest with a valve and the high pressure turbine can be modelled as a
linear first order system. Let PSV be the change in the steam valve position and
PCH is the change in the output power of the steam chest which gets converted in to
4.15
the mechanical power at the turbine shaft. If TCH is the time delay of the steam chest
then we can represent the dynamics of the steam chest along with steam valve as
TCH
d PCH
 PCH  PSV
dt
(4.48)
A fraction of the power output of the steam chest, K HP PCH , gets converted in to
mechanical power at the high pressure turbine shaft and the remaining power that is
1  K HP  PCH
enters the reheat chamber. The reheat chamber itself has a time delay
TRH . After the time delay the steam output from the rehear chamber is forced into the
low pressure turbine (or a series of intermediate turbines followed by a low pressure
turbine. But here only low pressure turbine is taken and the explanation can extended
to any number of turbines in tandem or cross connected) where it gets converted into
incremental mechanical power PRH . The reheat chamber can also be represented as
a linear first order system as
TRH
d PRH
 PRH  1  K HP  PCH
dt
(4.49)
The total shaft mechanical power or torque (as torque and power in per units on the
generator MVA basis are same) is the summation of high pressure and low pressure
turbine mechanical power and can be written as
TM  PRH  K HP PCH
(4.50)
Substituting equation (4.50) in (4.49) by eliminating
PRH we can represent the
dynamics of steam turbine system as
TRH
 K T
d TM
 TM  1  HP RH
dt
TCH


K HPTRH
PSV
 PCH 
TCH

4.16
(4.51)
TCH
d PCH
 PCH  PSV
dt
(4.52)
In case reheater dynamics are not considered then TRH is zero and the dynamic
equations can be written as
TCH
d TM
 TM  PSV
dt
(4.53)
Where, TM  PCH . Equations (4.51) and (4.52) can be converted to equations (4.54)
and
(4.55);
by
defining
actual
variables
as
TM  TM 0  TM , PCH  PCH 0  PCH , PSV  PSV 0  PSV , where TM 0 , PCH 0 , PSV 0 are the
initial operating conditions,
TRH
TCH
 K T
dTM
 TM  1  HP RH
dt
TCH

dPCH
  PCH  PSV
dt

K HPTRH
PSV
 PCH 
T
CH

(4.54)
(4.55)
4.2.3 Turbine governor
The gate position G or the steam valve position PSV can be automatically
controlled to change the power input according to the output power required through
governor mechanism. The speed governor mechanism [4] is shown in Fig. 4.7. The
speed governor mechanism consists of a fly ball mechanism which detects the change
in the shaft speed above the reference speed. Levers with connecting rods and a main
piston operated through high pressure oil to either open or close the value or gate. If
the speed of the shaft increases above the reference speed the centrifugal force on the
fly balls will increase and they move away from each other there by moving the point
b downwards and this lead to lowering of point c and hence point d. As point d move
down wards the high pressure oil enters the main piston chamber through the lower
opening and hence pushes the point e and piston upwards and there by reducing the
steam input. Increases in shaft speed above the reference speed indicates that the input
power is more than the output power required and hence speed increasing and to bring
back to the reference speed the input mechanical power should be reduced i.e. the gate
4.17
or valve should position should change to reduced the input power and this is what
happen in the governor mechanism.
Fig. 4.7: Speed governor mechanism
Similarly if speed is decreasing then more input power is required and it can be
observed that with decrease in speed the fly balls will move towards each other and
the point c moves upwards and leads to movement of point d upwards now the high
pressure oil will enter the main piston chamber through upper opening and there by
forcing the piston downwards and opening the valve. The actual mechanical power
input reference can also be changed at reference speed by changing the set point
through speed changer i.e. moving point a downwards or upwards and thereby
decreasing or increasing the power input.
The governor can be mathematically modelled by assuming that the change in
the positions of the points a, b, c, d and e can be related to each other linearly
depending on the length of the rods separating them. If the point b moves a distance
yb then it is related to the change in point a ( ya ) and c ( yc ), as it is directly
connected to them, as
yb  K ba ya  K bc yc
(4.56)
4.18
where, K ba , K bc are constants. Similarly we can write
yd  K dc yc  K de ye
(4.57)
The point a is related to the change in the reference power and if the reference power
is Pref then
Pref  K a ya
(4.58)
The point b changes according to the change in the speed of the shaft and is
proportional to it and hence
  base K b yb
(4.59)
Point d effects point e through the high pressure oil servo mechanism. The point e
moves according to the rate of change of oil flow in the main piston chamber and the
flow in turn is effected by point e. Also it can be noted that point d and e move in
opposite directions and hence the following expression can be written
d ye
  K e yd
dt
(4.60)
Substituting, equation (4.57) in (4.60) lead to
d ye
  K e yd   K dc K e yc  K e K de ye
dt
(4.61)
From equation (4.56), yc can be expressed in terms of ya , yb and this can be
substituted in equation (4.61) which lead to
4.19
 y  K ba ya
d ye
  K dc K e  b
dt
K bc


  K e K de ye

(4.62)
K K
K KK
  dc e yb  dc e ba ya  K e K de ye
K bc
K bc
ya , yb can be represented in terms of Pref ,  from equations (4.58) and (4.59)
this when substituted in (4.62) lead to
d ye
K dc K e
K KK

  dc e ba Pref  K e K de ye
base K b K bc
dt
K bc K a
or
PSV
RD
TSV
1/





K a 
K bc K a K de
1 K bc K a K de d ye

 Pref 
ye
K e K de K dc K ba
dt
K b K ba base
K dc K ba
TSV
d PSV
1 
 PSV  Pref 
dt
RD base
(4.63)
(4.64)
If we define the variables as PSV  PSV 0  PSV , Pref  Pref 0  Pref ,   base   then
we can express equation (4.64) as, with limits on the valve position included,
TSV

dPSV
1 
Max
  PSV  Pref 
  1 , 0  PSV  PSV
dt
RD  s 
(4.65)
RD is the speed regulation and it depends on the droop, the amount by which the
frequency falls from no load to full load without change in the input power. RD can
be defined in terms of droop as
RD 
2 droop
(4.66)
base
4.20
Where, droop has units of Hz/per megawatts. If droop is defined in terms of
percentage i.e. the percentage by which the frequency falls from no load to full load
then
RD 
%droop
100
(4.67)
So far we have discussed about the modelling of the synchronous generator along
with exciter, hydro/steam turbine and turbine governor. Next we will discuss about
modelling of loads.
4.3 Load Representation
The loads can be represented as either static or dynamic loads. Lighting and
heating loads can be represented by static models. Whereas, to represent motor loads
like synchronous motors and induction motors the motor dynamics have to be taken
into consideration.
4.3.1 Static load representation
Static load representation is a simple way of representing the effect of load on
the system stability. The effect of change in voltage and system frequency can be
incorporated in the static load model. Usually instead of representing each static load
separately an aggregate load is represented like all loads at particular distribution
station can be clubbed together and represented as a single load. The typical way of
representing static loads is either as a constant impedance load or constant current
load or constant power load. If Po , Qo ,Vo are the initial real power, reactive power
and voltage then the effect of change in the voltage on the static load can be
represented as [5]
V 
P  Po  
 Vo 
np
(4.68)
4.21
V 
Q  Qo  
 Vo 
nq
(4.69)
Where, P, Q are the real and reactive powers at changed voltage V . In case
equation (4.68) and (4.69) are representing single loads then np, nq when taken as 0
represents constant power load, taken as 1 represents constant current and taken as 2
represents constant impedance loads. For aggregate loads np, nq may be chosen to
approximate the actual behaviour of the load and may be in the range of 0.5 to 1.8 for
np and in the range of 1.5 to 6 for nq . A polynomial mode of load representation [5]
can also be used which includes all the types i.e. constant impedance (Z), current (I)
and power (P) called as ZIP load as shown in equation (4.70) and (4.71)
  V 2

V 
P  Po  PZ    PI    PP 
  Vo 

 Vo 


(4.70)
  V 2
V
Q  Qo  QZ    QI 
  Vo 
 Vo

(4.71)




Q

P



Where, PZ , I , P , QZ , I , P are constants representing the approximated fraction of the
aggregated load taken as constant impedance, current and power loads. Though
constant impedance are practical loads like lighting and heating, constant current and
power loads are not real loads but are used to represent dynamic loads like motors as
static loads assuming that after a disturbance the motor loads will reach their steady
state rapidly. Hence it is necessary to take the effect of frequency variation also on the
loads to improve the accuracy and can be represented as
  V 2
V
P  Po  PZ    PI 
  Vo 
 Vo




P
 P  1  K Pf f 


  V 2

V 
Q  Qo  QZ    QI    QP  1  K qf f 
  Vo 

 Vo 


4.22
(4.72)
(4.73)
In equations (4.72) and (4.73), the effect of change in the bus frequency from
the steady state frequency, f  f  f 0 , is included in the static load model with
constant K pf taking a value between 0 to 3.0 and K qf a value between -2 to 0 [6].
The bus frequency information is not available but can be obtained by the rate of
change of the bus angle which can be computed. For a very wide variation in voltage
the aggregate static load is converted into constant impedance load for computational
simplicity. In fact constant impedance load representation has several computational
benefits which will be discussed later. Though IP part of ZIP load represents the static
model of motor loads this representation is not accurate and can lead to erroneous
conclusions. Hence, for large motor loads it is better to take the dynamics of the
motors into consideration for stability studies. In this context we will discuss the
modelling of synchronous motor and induction motor.
4.3.2 Model of synchronous motor
The dynamic equations (3.149)-(3.157) and (3.162) given in Chapter 3
representing the synchronous generator can also be used for representing the
synchronous motor as well with a modification in equation (3.162) given in Chapter 3
as shown in equation (4.74)
2 H d
 Te  D   base   Tm
base dt
(4.74)
This change is required because in a motor electrical energy is converted to
mechanical energy and hence electrical torque is the input whereas mechanical torque
is the output. The converse is true of a synchronous generator. It has to be also noted
that in case of a synchronous motor current into the stator should be taken as positive
unlike the case of synchronous generator. The IEEE type DC1A exciter model can
also be used for the synchronous motor.
4.23
4.3.4 Model of induction motor
The stator of the induction machine is similar to a synchronous machine with a
three-phase winding distributed symmetrically with 120 separation. The rotor of an
induction machine also has three phase windings distributed symmetrically with 120
separation with same number of poles as that of stator (a set of three phase windings
distributed symmetrically with 120 separation produces a pair of poles in the air gap
rotating at synchronous speed). The three-phase rotor windings terminals are brought
out to be connected to external resistance in case of a wound rotor induction machine
and short circuited in case of a squirrel cage induction machine. For modelling
purpose both can be considered same with some assumptions. Some points to be
noted which can help us in modelling induction machine are [7]

The dq 0 transformation used for modelling synchronous machine can also be
used for modelling induction machine as well.

The rotor is cylindrical and hence has a uniform air-gap due to which there
will be no difference between d -axis or q -axis inductances and time
constants.

In case of synchronous machine the d -axis was aligned with the rotor of the
machine as it rotates at synchronous speed in steady state for varying loads. In
case of induction machine the rotor speed is not synchronous speed and its
speed depends on the load torque required. Even though the rotor of induction
machine rotates at slip speed the rotor winding produces a synchronously
rotating field with the same number of poles as that of the stator winding field.
Hence, a synchronously rotating dq -axis can be taken which is neither fixed
to rotor or fixed to stator and the three-phase rotor and stator quantities can be
referred to this axis.

The current in the rotor of an induction machine does not depend on the
excitation system like in the case of synchronous machine but instead depends
on the relative speed between the synchronously rotating stator field and the
rotor speed that is the slip speed.
4.24

The same model can be used for induction motor and generator and the main
difference lies in the slip.
With these assumptions let us look in to the modelling of an induction machine.
The basic diagram of induction machine with all the relevant windings represented is
shown in Fig. 4.8. The stator three-phase voltages and currents are represented by
va , vb , vc and ia , ib , ic . Similarly, rotor three-phase voltage and currents are represented
by v A , vB , vC and iA , iB , iC . The rotor along with its windings will rotate at a speed r .
Let the angle with which the field axis of rotor A-phase leads the field axis of stator aphase be  then with continuous time we can write
s
d  axis
ib
r
iA
r
iB
vB
v A vb

va
ia
vC vc
ic
iC
Fig. 4.8: Induction machine with all the windings represented
  r t
(4.75)
The slip is defined as the ratio of the difference between the synchronous speed and
the rotor speed to the synchronous speed. Hence, slip can be expressed as
s
 s  r
s
(4.76)
4.25
From equation (4.76), the angle  in equation (4.75) can be represented in terms of
slip and synchronous speed as
  r t  1  s  s t
(4.77)
If now we consider dq -axis rotating in the air-gap at synchronous speed and is
separated by an angle  r with the field axis of rotor A-phase. Then the following
expression can be written
 r  s t    ss t
(4.78)
d r
 ss
dt
(4.79)
In the same way we have written voltage equations for synchronous machine stator
windings we can also write voltage equations for the windings shown in Fig. 4.8 as
 va   rs 0 0  ia 
a 
 v    0 r 0  i   d   
 b   s   b  dt  b 
 vc  0 0 rs  ic 
c 
(4.80)
 v A   rr 0 0  iA 
A 
 v    0 r 0  i   d   
 A   r   B  dt  B 
 vC  0 0 rr  iC 
C 
(4.81)
Where, rs , rr are the stator and rotor winding resistances. a ,b ,c , A, B ,C are the
flux linkage of the three-phase stator and rotor windings, respectively. Let the self
inductance of the stator a-phase winding be laa and this can be assumed to the same
for all the three-phases. The mutual inductance between any two phases, let it be lab ,
of the stator will be same and independent of the rotor position as the air-gap is
uniform. Let the maximum value of the mutual inductance between any rotor and
4.26
stator winding be laA then the flux linkage equation of stator and rotor can be written
as
laA cos   120  laA cos   120   iA 
 a  laa lab lab  ia  laA cos 
 
    l l l  i   l cos   120 l cos 
l

cos

120







b
ab
aa
s
ab
b
aA
aA
aA
  
 
iB 


 c  lab lab laa  ic  l cos   120  l cos   120 
laA cos cos   iC 
aA
 aA
(4.82)
laA cos   120  laA cos   120   ia  l AA l AB l Ab  iA 
 A  laA cos 

  
 
    l cos   120 l cos 
l
cos


120







B
aA
aA
aA
ib   l AB l AA l AB  iB 
 
 C  l cos   120  l cos   120 
laA cos cos   ic  l AB l AB l AA  iC 
aA
 aA
(4.83)
The Park’s transformation used in synchronous machine modelling can also be used
here but a separate transformation has to be used for stator and rotor windings. Let
rotor Park’s transformation be Tdqr 0 and that of stator be Tdqs 0 and can be expressed as
Tdqs 0


 cos(s t ) cos(s t  120) cos(s t  120) 

2
   sin(s t )  sin(s t  120)  sin(s t  120) 
3

1
1
1


 2
2
2

(4.84)
Tdqr 0


 cos  s  r  t  cos  s  r  t  120  cos  s  r  t  120  

2
   sin  s  r  t   sin  s  r  t  120   sin  s  r  t  120  
3

1
1
 1

2
2
 2

(4.85)
Applying transformation given in (4.84) and (4.85) to flux linkage equations (4.82)
3
and (4.83) and also assuming that lss  laa  lab , lrr  l AA  l AB and lm  laA we get
2
flux linkage equations in dq0 reference frame as
4.27
ds  lss 0 0  ids  lm 0 0  idr 
  
  
 
qs   0 lss 0  iqs   0 lm 0  iqr 
  0 0 0  i  0 0 0  i 
 0s 
 0s 
 0r 
(4.86)
dr  lm 0 0  ids  lrr 0 0  idr 
  
  
 
qr   0 lm 0  iqs   0 lrr 0  iqr 
  0 0 0  i  0 0 0  i 
 0r 
 0s 
 0r 
(4.87)
Similarly, applying transformations to voltage equations given in (4.80) and (4.81)
leads to
 vds   rs 0 0  ids  0  s
  
  
 vqs   0 rs 0  iqs   s 0
 v  0 0 rs  i  0 0
 0s 
 0s 
ds 
0  ds 
  d  

0  qs   qs 
dt
 
0  0 s 
 0s 
 s  r  0  dr 
 vdr   rr 0 0  idr  0
dr 

  d  
  



0  qr   qr 
 vqr   0 rr 0  iqr   s  r  0
dt
 v  0 0 rr  i  0
 
0
0  0 r 
 0r 
 0r  
 0r 
(4.88)
(4.89)
The rotor power input can be expressed in dq -axis by using equation (4.89) as
S rotor
vA 
  vB 
 vC 

T
iA  vdr 
i    v 
 B   qr 
iC  v0 r 
T
 T  
r
dq 0
1 T
idr 
 
Tdqr 0 iqr 
i 
 0r 
(4.90)
d qr 
 d
3
3
rr  idr2  iqr2    dr iqr  qr idr  s  r    idr dr  iqr

2
2
dt
dt 



In the steady state
d dr d qr
,
are zero, then from equation (4.90) it can
dt
dt
observed that there are two parts of which the first part corresponds to the rotor
copper losses and the second part corresponds to the product of torque produced and
the difference of the rotor speed with respect to the synchronous speed. If the torque is
defined at the relative speed of the rotor speed with respect to the speed of dq -axis,
4.28
r  s  , then from equation (4.90) the torque can be expressed for a
P number of
poles as
te 
3P
 qr idr  dr iqr 
22
(4.91)
Per unit representation
The same base quantities can be taken for both stator and rotor.
Let,
Vbase = peak value of rated phase voltage, V
I base = peak value of rated current, A
fbase = rated frequency, Hz
Z base 
Vbase
Z
V
 , base  s  2 fbase elec.rad/s , Lbase  base H , base  base ,
base
base
I base
3
3P
sbase  Vbase I base , Tbase 
base I base N.m
2
22
With the base voltage and current defined as above and given the transformations in
(4.85) and (4.86) it can be observed that a per unit phase voltage and current in abcreference frame leads to per unit voltage and current in dq 0 -reference frame as well
hence same base quantities can be used in both abc and dq 0 -reference frames.




i
i
i
i
I ds  ds , I qs  qs , I dr  dr , I qr  qr ,

I base
I base
I base
I base






 ds  ds , qs  qs , dr  dr , qs  qr ,

base
base
base
base

baselss
baselrr
baselm
rs 
Xs 
, Rs 
,
, Xr 
, Xm 
Z base
Z base
Z base
Z base 

t
r

Rr  r Te  e , r _ pu  r  1  s 

Z base
Tbase
base

Vds 
v
v
vds
v
, Vqs  qs , Vdr  dr , Vqr  qr
Vbase
Vbase
I base
I base
4.29
(4.92)
Then equations (4.86) to (4.89) can be written in per units as
Vds  Rs I ds  qs 
1 d ds
base dt
(4.93)
Vqs  Rs I qs  ds 
1 d qs
base dt
(4.94)
Vdr  Rr I dr  s qr 
1 d dr
base dt
(4.95)
Vqr  Rr I qr  s dr 
1 d qr
base dt
(4.96)
 ds  X s I ds  X m I dr
(4.97)
 qs  X s I qs  X m I qr
(4.98)
 dr  X m I ds  X r I dr
(4.99)
 qr  X m I qs  X r I qr
(4.100)
The stator transients of the induction machine can be neglected as was done for the
synchronous machine stator transients. Hence, equations (4.93) and (4.94) are
simplified to
Vds  Rs I ds  qs
(4.101)
Vqs  Rs I qs   ds
(4.102)
Eliminating I dr , I qr from equations (4.99) and (4.100), substituting them in (4.97) and
(4.98) and in turn substituting them in equations (4.95) and (4.96), noting that for a
squirrel cage rotor Vdr  Vqr  0 , leads to


(4.103)


(4.104)
dVd'
T
  Vd'   X s  X s'  I qs  base sT0'Vq'
dt
'
0
'
0
T
dVq'
dt
  Vq'   X s  X s'  I ds  base sT0'Vd'
4.30
Where,
Vd'  

Xm
X
X2 
Xr
 qr , Vq'  m  dr , X s'   X s  m  , To' 
base Rr
Xr
Xr
Xr 

The per unit torque is give as
Te 
te
Tbase
3P
 qr idr  dr iqr 
2
2

  qr I dr  dr I qr
3P
base I sbase
22
(4.105)
The equation of motion of the induction machine is given as
J
d m
 te  t m
dt
(4.106)
Where, m is the speed of the rotor in mechanical radians per second. Dividing both
side by Sbase 
2
Tbasebase we get
P
2

J base
t
t
d P m
 e  m
base
Sbase
dt base Tbase Tbase
(4.107)
1
2
J base
2
let, H  2
and also noting that r  m we get
Sbase
P
2H
dr _ pu
dt
 2 H
ds
 Te  Tm
dt
(4.108)
Algebraic equations relating the stator voltage Vds , Vqs to the dynamic state Vd' , Vq' can
be obtained by eliminating I dr , I qr from equations (4.99) and (4.100), substituting
4.31
them in (4.97) and (4.98) and in turn substituting them in equations (4.101) and
(4.102) leads to
Vds  Rs I ds  X s' I qs  Vd'
(4.109)
Vqs  Rs I qs  X s' I ds  Vq'
(4.110)
If the three-phase stator voltage are given as
va  2Vt cos s t   
vb  2Vt cos s t    120 
(4.111)
vc  2Vt cos s t    120 
Where, Vt is the rms phase voltage. Applying parks transformation defined in
equation (4.84) to equation (4.111) and dividing both sides by base voltage lead to
Vds  Vt _ pu cos   
Vt 
 ,  where, Vt _ pu 
V   

Vsbase 
 qs  Vt _ pu sin   
(4.112)
Hence, the dq components of the voltages can be used to represent a phasor with real
and imaginary parts in per units as
Vds  jVqs  Vt _ pu  cos   j sin    Vt _ pu e j
(4.113)
With the same argument the stator currents in dq -axis can be represented in per units
as
I ds  jI qs  I t _ pu  cos   j sin    I t _ pu e j
4.32
(4.114)
Where, I t _ pu is the stator current in per units with a phase angle  . Since stator
voltage in dq -axis can be expressed as phasors we can add equations (4.109) and
(4.110) by multiplying equation (4.110) with complex number j1 , then we get
Vt _ pu e j  Vds  jVqs   Rs  jX s'   I ds  jI qs   Vd'  jVq' 
(4.115)
Hence, a simple electrical equivalent circuit can be formed for the induction machine
with an internal transient voltage behind the transient reactance X s' as shown in Fig.
4. 9.
X d'
V
ds
Rs
I
d
 jI q   I t _ pu e j
V   jV  
 jVqs   Vt _ pu e j
d
q
Fig. 4.9: Electrical equivalent circuit of an induction motor
Hence, the complete induction machine dynamic model in per units is defined by
1 d r base  r

s
base dt
base
(4.116)
ds
  Te  Tm 
dt
(4.117)
2H
T0'
T0'


(4.118)


(4.119)
dVd'
  Vd'   X s  X s'  I qs  base sT0'Vq'
dt
dVq'
dt
  Vq'   X s  X s'  I ds  base sT0'Vd'
Te   qr I dr  dr I qr  Vd' I ds  Vq' I qs 
(4.120)
Vt _ pu e j  Vds  jVqs   Rs  jX s'   I ds  jI qs   Vd'  jVq' 
(4.121)
4.33
Initial Conditions
To find the initial conditions of a induction motor, the dynamics given by equations
(4.118) and (4.119) can be equated to zero, which lead to
V   X
V   X
'
d
s
'
q
s

 X  I  
 X s'  I qs  base sT0'Vq'  0
(4.122)
sT0'Vd'  0
(4.123)
'
s
ds
base
Now multiply equation (4.123) by complex number j1 on both sides and add this to
equation (4.122).
Vd'  jVq'  j  X s  X s'   I ds  jI qs   jbase sT0' Vd'  jVq'   0
Now, V   Vd'  jVq' , I t _ pu e j  I ds  jI qs
(4.124)
and Vt _ pu e j  Vd  jVq , then equation
(4.124) leads to
Vd'  jVq'   j  X s  X s' 
V

I t _ pu e j  I ds  jI qs 
1  jbase sT0' 
(4.125)
Substituting equation (4.125) in equation (4.115) lead to

 X s  X s'   I e j
Vt _ pu e j    Rs  jX s'   j

1  jbase sT0'   t _ pu

(4.126)
Equation (4.126) can be further simplified and expressed as

I t _ pu e  


j
 R  
s
base
1  j sT 
sT X   j  sT R
base
'
0
'
s
'
0
base
'
0
s  Xs 


 Vt _ pu e j


The real power input to the induction machine is given as
4.34
(4.127)

Pt  Real Vt _ pu e j  I t _ pu e j 
*



'
'
'
'
  Rs  base sT0 X s   base sT0 base sT0 Rs  X s   2

 Vt _ pu
2
'
' 2
'



R
sT
X
sT
R
X



 s base 0 s   base 0 s s  

(4.128)


If from load flow analysis the Pt , Qt ,Vt _ pu are defined then from equation
(4.128) the slip can be found out. Equation (4.128) lead to a quadratic expression of
slip and hence there will be two solutions of which the feasible solution should be
considered if it exists for the given conditions. The other way is if we define a slip
with terminal voltage given then real power can be computed from (4.128) and
similarly reactive power can also be computed.
References
1. IEEE Standards 421.4, IEEE Recommended Practice for excitation system
models for power system stability studies, 1992.
2. IEEE Committee report, “Excitation system models for power system stability
studies,” IEEE Trans. Power Appar. Syst., PAS-100, pp. 494-509, February
1981.
3. IEEE Committee Report, “Dynamic models for steam and hydro turbines in
power system studies,” IEEE Trans., Vol. PAS-92, pp. 1904-1915, December
1973.
4. O. I. Elgerd, Electric Energy System Theory, an Introduction, McGraw-Hill
Book Co., New York, 1982.
5. IEEE Task Force Report, “Load representation for dynamic performance
analysis,” IEEE PES winter meeting, New York, January 26-30, 1992.
6. C. Concordia and S. Ihara, “Load representation in power system stability
studies,” IEEE Trans., Vol. PAS-101, pp. 969-977, April 1982.
7. D. S. Brereton, D. G. Lewis and C. C. Young, “Representation of induction
motor loads during power system stability studies,” AIEE Tans., Vol. 76, Part
III, pp. 451-460, August 1957.
4.35