Chapter5

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)

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 qfd fd

X XX 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)

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 2t 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)

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-ax

is

tV

5.5

Noting that mdfd fd md fd

fd

XE 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

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 21, cost t t

tt t t

P Q PI

V V I

(5.23)

0q t s q tE V R jX I (5.24)

1 cos sintan

cos sinq t s t

qt s t q t

X I R IE

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-ax

is

tI

s tR I

q tjX I

qV

qI

qE

md fdX I

d q dX X I

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

fdmd

IX

.

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-ax

is

tV

d-axis

RV

IV

qV dV

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 ( )

jt 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 cosj

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 2j

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 0jV 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 jtV 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

0jV e

qX sR

2( )j

jd q tV jV e V e

2

j

qE e

2( )j

d qI jI e

TX TR

P jQ

5.9

jt j

t

P jQI e

V e

(5.31)

2 2 2( ) ( )j j j

q d q s q d q

j jt 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

2j

j j jq 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 jj

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

cosd 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 jt 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)

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 js 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 jqE 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-ax

is

tI

s tR I

q tjX I

qV

qI

qE

md fdX I

d q dX X I

R-axis

I-axis

V

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 thi 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 didi lsi

qi fdi

dE X XT X X I E X X I

dt X X

E E

(5.38)

" ' '11( )di

doi qi di lsi di did

T E X X Idt

(5.39)

' "'

' ' ' ' '2' 2

( )( ) ( )

( )qi qidi

qoi di qi qi qi di qi lsi qi qiqi lsi

X XdET E X X I E X X I

dt X X

(5.40)

2" ' '2( )qi

qoi di qi lsi qi qi

dT E X X I

dt

(5.41)

ii base

d

dt

(5.42)

2 i imi ei i i base

base

H dT T D

dt

(5.43)

( )fdiEi Ri Ei Ei fdi fdi

dET V K S E E

dt (5.44)

, Min MaxRi Ai FiAi Ri Ai Fi fdi Ai refi ti Ri Ri Ri

Fi

dV K KT V K R E K V V V V V

dt T (5.45)

Fi FiFi fdi

Fi

dR KR E

dt T (5.46)

1Mi HPi RHi HPi RHiRHi Mi CHi SVi

CHi CHi

dT K T K TT T P P

dt T T

(5.47)

CHiCHi CHi SVi

dPT P P

dt (5.48)

5.12

11 ,0 MaxSVi i

SVi SVi refi SVi SViDi base

dPT 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 qidi lsi di lsi

X X X XX I E V R I

X X X X

(5.50)

" ' "" '

2' '

( ) ( )

( ) ( )qi lsi qi qi

qi qi di qi di si diqi lsi qi lsi

X X X XX 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 thi bus is given as, where 1,2.........i n ,

1

ij j

nj j

i ij jj

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 thj bus. The complex power injected at thi bus can be written as

*

1

i j iji

njj

i i i i i j ijj

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 thi bus the injected real

and reactive power can be written as

1

cosn

i Gi Di i j ij i j ijj

P P P VV Y

(5.54)

1

sinn

i Gi Di i j ij i j ijj

Q Q Q VV Y

(5.55)

5.13

Where, ,Gi GiP Q are the real and reactive power generated at thi 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

*

2Real

sin( ) cos( )

ii

jj

Gi i di qi

di i i i qi i i i

P V e I jI e

I V I V

(5.56)

*

2Imaginary

cos( ) sin( )

ii

jj

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( ) ( ) cosn

di i i i qi i i i Di i i j ij i j ijj

I V I V P V VV Y

(5.58)

1

cos( ) sin( ) ( ) sinn

di i i i qi i i i Di i i j ij i j ijj

I V I V Q V VV Y

(5.59)

for, 1, 2,......, gi n , generator buses

1

( ) cosn

Di i i j ij i j ijj

P V VV Y

(5.60)

1

( ) sinn

Di i i j ij i j ijj

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-

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 thi bus is represented by an impedance load then, the real and

reactive power drawn by the load is given as

2i

Li LiLi

VP jQ

Z (5.62)

* *Li Li i

Lii Li

P jQ VI

V Z

(5.63)

Where, , ,Li Li LiP Q Z are the real power, reactive power and impedance of a load

at an thi bus. Substituting equation (5.63) in (5.52), the current injected into thi bus

can be expressed as

*1

ij j

nj ji

i Gi Li Gi ij jjLi

VI I I I Y e V e

Z

(5.64)

Let 1Li LiY Z , then

1

1, 2,.......,ij jLi i

nj jj j

Gi Li i ij j gj

I Y e V e Y e V e i n

(5.65)

1 21

0 , ,.......,ij jLi i

nj jj j

Li i ij j g gj

Y e V e Y e V e i n n n

(5.66)

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

VIY

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 thi generator bus an additional node is created marked as i with a

voltage " " 2ij

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 11 ,2 ......,

1, 2,.......,

ii

ii

jj

i Gi di qi iSi di

jj

di qi i gSi di Si di

g

I I E jE e V eR jX

E jE e V e i nR jX R jX

i n

(5.68)

Note that since i is an additional node connected to an thi 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

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 10 0 0 0

1 1 0 00 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

VY n Y n n Y n n Y n n

VY n Y n n Y n n Y n n

Let,

1 1

10

10

g g

S d

II

Sng dng n n

R jX

Y

R jX

,

(1, 1) (1, )1 1

( , 1) ( , )

10 0 0

1 0 00

g

g g g

g

n nS d

IN

n n n n

Sng dngn n

R jX

Y

R jX

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 nY

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 gg g

Gn

G N

nGn n

n n

n n

V

I E V

VI E

V

I E V

Then,

'

1 10

g gg g

II ING

TIN 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' TBUS INY Y

NV E (5.70)

1' TG 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.

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 ji 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 thi generator is the angle of the complex internal voltage ijiE 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)

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)

FiFi fdi

Fi

KR E

T (5.82)

( )Ri Ei Ei fdi fdiV K S E E (5.83)

Rirefi i

Ai

VV 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

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 fdd ls

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

dt X X

(5.93)

" ' '11( )d

do q d ls d dd

T E X X Idt

(5.94)

' "'

' ' ' ' '2' 2

( )( ) ( )

( )q qd

qo d q q q d q ls q qq ls

X XdET E X X I E X X I

dt X X

(5.95)

2" ' '2( )q

qo d q ls q q

dT E X X I

dt

(5.96)

base

d

dt

(5.97)

2m e base

base

H dT T D

dt

(5.98)

( )fdE R E E fd fd

dET V K S E E

dt (5.99)

R A FA R A in R A F fd A ref t

F

dV K KT V K V V K R E K V V

dt T (5.100)

F FF fd

F

dR KR E

dt T (5.101)

1M HP RH HP RHRH M CH SV

CH CH

dT K T K TT T P P

dt T T

(5.102)

5.21

CHCH CH SV

dPT P P

dt (5.103)

11SV

SV SV refD s

dPT P P

dt R

(5.104)

" ' "" '

1' '

( ) ( )

( ) ( )d ls d d

d d d q dd ls d ls

X X X XX I E

X X X X

(5.105)

" ' "" '

2' '

( ) ( )

( ) ( )q ls q q

q q q d qq ls q ls

X X X XX 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 qd ls d ls

X X X XX I E V R I

X X X X

(5.107)

" ' "" '

2' '

( ) ( )

( ) ( )q ls q q

q q d q d s dq ls q ls

X X X XX 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 dq ls q ls d ls d ls

d s d q s q

X X X X X X X XX 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 dq 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 XX 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)

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 dq ls q ls d ls d ls

X X X X X X X XE 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 2j 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, " " 2j

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 2j

d qV jV e

.

Fig. 5.7: Dynamic electrical equivalent circuit of synchronous generator sub-transient

model

2j

d qI jI e

"dX sR

2j

d qV jV e

" " 2j

d qE jE e

5.23

The per unit torque is equal to the real power delivered by the dependent source

" " 2j

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

" ' '11

' '1

( ) 0

( )

ddo q d ls d d

d q d ls d

dT E X X I

dt

E X X I

(5.114)

2" ' '2

' '2

( ) 0

( )

qqo d q ls d q

q d q ls q

dT 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

dET E X X I E

dt (5.116)

'' ' '( )d

qo d q q qdE

T E X X Idt

(5.117)

base

d

dt

(5.118)

5.24

2m e base

base

H dT T D

dt

(5.119)

( )fdE R E E fd fd

dET V K S E E

dt (5.120)

R A FA R A in R A F fd A ref t

F

dV K KT V K V V K R E K V V

dt T (5.121)

F FF fd

F

dR KR E

dt T (5.122)

1M HP RH HP RHRH M CH SV

CH CH

dT K T K TT T P P

dt T T

(5.123)

CHCH CH SV

dPT P P

dt (5.124)

11SV

SV SV refD s

dPT 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 2j j j

d q d q s d d qE jE e V jV e R jX I jI e

(5.131)

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

dET E X X I E

dt (5.133)

base

d

dt

(5.134)

2m e base

base

H dT T D

dt

(5.135)

'dX sR

2j

d qV jV e

' ' 2j

d qE jE e

2j

d qI jI e

5.26

( )fdE R E E fd fd

dET V K S E E

dt (5.136)

R A FA R A in R A F fd A ref t

F

dV K KT V K V V K R E K V V

dt T (5.137)

F FF fd

F

dR KR E

dt T (5.138)

1M HP RH HP RHRH M CH SV

CH CH

dT K T K TT T P P

dt T T

(5.139)

CHCH CH SV

dPT P P

dt (5.140)

11SV

SV SV refD s

dPT 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 2j 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.

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)

2m e base

base

H dT T D

dt

(5.148)

The dynamic equivalent circuit of the classical model is shown in Fig. 5.10

'dX sR

2j

d qV jV e

' ' 2j

q d q qX X I jE e

2j

d qI jI e

5.28

Fig. 5.10: Dynamic electrical equivalent circuit of synchronous generator classical

model

'dX sR

2j

d qV jV e

' ' 2_ _

j

d ss q ssE jE e

2j

d qI jI e

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 14.0360t t

P jQI j

Vf

¥

- = = - = -

(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

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)

0.1769FF fd

F

KR E

T (E1.13)

( ) 2.5482R E E fd fdV K S E E (E1.14)

1.2189Rref t

A

VV 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)

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 úê ú-ê úê ú= -ê úê ú-ê úê úê 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 0.651.932 0.2828

0.98 11.53LP jQ j

I jV

- -= = = -

- (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

5.32

5 0.4720 0.1704LL

VZ j

I= = + (E2.3)

12.0444 0.8295L

LY 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 úê ú-ê úê ú= -ê úê ú-ê úê 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 YY

Y Y

é ùê ú= ê úê úë û

(E2.6)

Where,

10

0.2

10

0.35

GGj

Y

j

é ùê úê úê ú=ê úê úê úë û

,

10 0 0 0

0.2

10 0 0 0

0.35

TGN NG

jY Y

j

é ùê ú-ê úê ú=ê úê 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 úê ú-ê úê ú= -ê úê ú-ê úê ú+ê úë û

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 NGj 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´ .

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.