2.1

Chapter 2 Analysis of Power System Stability by Classical Methods

2.1 Basic Model of a Synchronous Generator

As discussed in the previous Chapter, the first step in analyzing power stability

is to model the system components mathematically. The simplest yet very useful

representation of synchronous generator is the classical model advocated by E. W.

Kimbark and S. B. Crary [1], [2]. In this model the stability of a single generator

connected to an infinite bus is analyzed. Though this model is not appropriate for

current generators with fast acting exciters, governors, power system stabilizers, it is

nevertheless useful in understanding the basic phenomenon of stability.

Assumptions for representing the generator by classical model

Some assumptions are made to represent a synchronous generator

mathematically by a classical model. The assumptions are:

1. The exciter dynamics are not considered and the field current is assumed to be

constant so that the generator stator induced voltage is always constant.

2. The effect of damper windings, present on the rotor of the synchronous

generators, is neglected.

3. The input mechanical power to the generator is assumed to be constant during

the period of study.

4. The saliency of the generator is neglected, that is the generator is assumed to

be of cylindrical type rotor.

These assumptions make the representation of synchronous generator easy.

Assumptions 1 and 2 lead to neglecting the dynamics of exciter, damper windings and

rotor windings. Assumption 3, leads to neglecting the dynamics of turbine and

turbine speed governor. Assumption 3 is justified as the change in the mechanical

input power takes more time, in the order of 2 to 10 minutes, due to the involvement

of mechanical systems where as the electrical power output can change within in

milliseconds.

2.2

For the sake of understanding the classical model better the case of a generator

connected to an infinite bus through a transformer and transmission lines is

considered. This type of system is called as Single Machine Infinite Bus (SMIB)

system. The single line diagram of SMIB system is shown in Fig. 2.1. Since

resistance of synchronous generator stator, transformer and the transmission line are

relatively negligible as compared to the corresponding reactances, only reactances of

synchronous generator, transformer and transmission line are considered. Here,

infinite bus represents rest of the system or grid, where the voltage magnitude and

frequency are held constant. The infinite bus can act like infinite source or sink. It can

also be considered as a generator with infinite inertia and fixed voltage. In Fig. 2.1,

E represents the complex internal voltage of the synchronous generator behind

the transient reactance 'dX . The significance of '

dX will be explained in Chapter 3 but

for now it can be assumed to be a reactance of the generator during transients. T TV

is the terminal voltage of the synchronous generator. ,T LX X represent the

transformer and line reactance. The complex infinite bus voltage is represented as

0V . The infinite bus voltage is a taken as the reference because of which the angle

is taken as zero. The generator internal voltage angle is defined with respect to the

infinite bus voltage angle. The input mechanical power is represented as mP and the

output electrical power is defined by eP . H is the inertia constant of the generator.

H is the inertia constant of the grid equivalent generator connected at the infinite

bus and it’s value can be taken as . From Fig. 2.1, the real power output of the

generator, eP , can be computed as

0 0

'

'

( ) ( )Real Real

( ) ( )

sin

T

T

j j j jj jT

e TT L d T L

d T L

V e V e Ee V eP V e Ee

j X X j X X X

EV

X X X

(2.1)

The maximum real power output of the synchronous generator that can be

transferred to the infinite bus in this case is

2.3

max ', 90

d T L

EVP at

X X X

(2.2)

Hence, the synchronous generator real power output can be represented as

max sineP P (2.3)

Fig. 2.1: Single line diagram of the SMIB system

Till now the equivalent electrical representation of the synchronous machine is

discussed. The synchronous machine also has a mechanical system which has to be

modeled. The prime mover gives mechanical energy to the generator rotor and in turn

the generator converts the mechanical energy into electrical energy through magnetic

coupling. The dynamics of a rotational mechanical system can be represented as

2

2m

m e

dJ T T

dt

(2.4)

where, 2kg.mJ is the inertia constant of the rotating machine. The mechanical input

torque due to the prime mover is represented as N.mmT and the electrical torque

acting against the mechanical input torque is represented by N.meT . The angle m is

the mechanical angle of the rotor field axis with respect to the stator reference or fixed

reference frame. As the rotor is continuously rotating at synchronous speed in steady

'dX

TX

0V E

LX

T TV

eP

2.4

state m will also be continuously varying with respect to time. To make the angle

m constant in steady state we can measure this angle with respect to a synchronously

rotating reference instead of a stationary reference. Hence, we can write

m m mst (2.5)

Where, m is the angle between the rotor field axis and the reference axis rotating

synchronously at rpsms . If we differentiate (2.5) with respect to time we get

m mms

d d

dt dt

(2.6)

2 2

2 2m md d

dt dt

(2.7)

But, the rate of change of the rotor mechanical angle m with respect to time is

nothing but the speed of the rotor. Hence,

rpsmm

d

dt

(2.8)

Substituting, (2.8) in (2.6) we get

mm ms

d

dt

(2.9)

Similarly, substituting (2.7) in (2.4) we get

2

2m

m e

dJ T T

dt

(2.10)

if we multiply with m on both the side of (2.10) and noting that torque multiplied by

speed gives power, we can write (2.10) as

2.5

2

2m

m m e

dJ P P

dt

(2.11)

Now multiply on both the sides with the term 1

2 ms on both the sides of (2.11) and

divide the entire equation with the base MVA ( BS ), in order to express the equation in

per unit, lead to

2

2

1122

m msm m e

msB B B

J d P P

S dt S S

(2.12)

Let us define a new parameter named as machine inertia constant

2 21 1MW.S2 2

Base MVA MVA

ms ms

B

J JH

S

(2.13)

If we assume that on left hand side of (2.12) m ms as the variation of the speed,

even during transients, from synchronous speed is quite less. This assumption does

not mean that the speed of the rotor has reached the synchronous speed but

instead 21 1

2 2ms m msJ J . With this assumption if we substitute (2.13) in (2.12) we

get

2

2

1per unit

2m

ms m e

dH P P

dt

(2.14)

andm ms are expressed in mechanical radians and mechanical radians per second,

in order to convert them in to electrical radians and electrical radians per second

respectively we have to take the number of poles ( P ) of the synchronous machine

rotor into consideration. Hence, the electrical angle and electrical speed can be

represented as

2.6

elec.rad2

elec.rad/s2

m

s ms

P

P

(2.15)

Substituting (2.3) and (2.15) into (2.14) we get

2

max2

2

max2

sin per unit2

sin per unit ( 50 or 60 Hz)

sm

sm s

dP P

dt Hor

fdP P f

dt H

(2.16)

Equation (2.16) is called as swing equation. Equation (2.16), assuming all the

parameters expressed in per units, can also be written as

max

( )

sin

me s

m sm

d

dtd f

P Pdt H

(2.17)

Where, / 2me mP . In can be observed from (2.17) that, if max sinmP P then

there will be no speed change and there will be no angle change. But, if

max sinmP P due to disturbance in the system then either the speed increase or

decrease with respect to time. Let us take the case of max sinmP P , there is more

input mechanical power than the electrical power output. In this case, as the energy

has to be conserved difference between the input and output powers will lead to

increase in the kinetic energy of the rotor and speed increases. Similarly, if

max sinmP P then, the input power is less than the required electrical power output.

Again the balance power, to meet the load requirement, is drawn from the kinetic

energy stored in the rotor due to which the rotor speed decreases.

2.7

2.2 Small-Disturbance Stability Analysis of SMIB System

The classical model of synchronous generator in a SMIB system was derived in the

previous section. In this section we will try to understand how we can check the

stability of the generator in a SMIB system when subjected to small-disturbances. The

swing equation, given in (2.16), can be written as

2

m a x2s inm

s

H dP P

f d t

(2.18)

In the steady state, that is when the speed of the generator rotor is constant at

synchronous speed, the rate of change of rotor speed will become zero due to which

(2.18) can be written as

max sinmP P (2.19)

Since, the mechanical power input mP and the maximum power output of the

generator maxP are known for a given system topology and load, we can find the rotor

angle from (2.19) as

1 1

max max

sin sinm mP Por

P P

(2.20)

It can be observed from (2.19) that rotor angle has two solutions at which

max sinmP P . Now, let us label 1maxsin /o mP P and

1max maxsin /m oP P . There is a very important implication of the two

solutions ( o , max ) of equation (2.18) on the stability of the system. This can be

clearly understood from what is called as swing curve or P curve. The swing

curve is depicted in Fig. 2.2. The swing curve shows the plot of electrical power

output eP with respect to the rotor angle . As can be observed from Fig. 2.2 the

curve is a sine curve with varying from 0 to . The maximum power output of the

2.8

generator maxP occurs at an angle / 2 . On the same curve the input mechanical

power mP can also be represented. Since mP is assumed to be constant and does not

vary with respect to , a straight line is drawn which cuts the P curve at points A

and B. It can be seen from Fig. 2.2 that at points A and B the mechanical power input

mP is equal to eP . The rotor angles at point A and B are o and max , the solutions of

equation (2.19).

Fig. 2.2: Swing curve or P curve

Let us try to understand which of the solutions ( o , max ) leads to a stable operation.

Let us take first point A (corresponding rotor angle is o ). If we perturb the rotor

angle o by a small positive angle o , so that the new operating point is at C, then

electrical power output will also increase to max sin o oP . Since, in steady state

max sinm oP P , after perturbation max sinm o oP P , for a positive value of

o . Now the output electrical power will become more than the input mechanical

power and hence the rotor starts decelerating due to which the angle will be pulled

back to the point A. But since the rotor has certain inertia it cannot stop at the point A

and decelerate further due to which the angle moves to the point say D. At point D

m eP P that is input mechanical power will become more than the electrical power

output and hence the rotor starts accelerating. Again the rotor angle starts

increasing and will reach point A but due to inertia it cannot stop there and it will

eP

0 2

max

mP

maxP

0

A B

C´

´D

2.9

again move to point C. This phenomenon repeats itself indefinitely if there is no

damping to the rotor oscillations. This can be analogously understood from the

pendulum motion in vacuum. If a pendulum is perturbed from its steady state position

(the vertical position with the bob of the pendulum hung by a string attached to a

fixed point) then it swings to one extreme point reverses it direction pass through its

steady state position goes to the other extreme and reverse and pass through the

extreme and this happens indefinitely as it is oscillating in vacuum and there is no air

friction to stop the oscillations.

Now take the case of second operating point max . Again if we perturb max by a

positive angle max then the electrical power output will be max max maxsinP .

In this case however, unlike the earlier case, max max maxsinmP P and the rotor

starts accelerating due to which the angle will increase further and this will lead to

further decrease in the electrical power output. Hence, for a small positive

perturbation in the rotor angle at the operating point B leads to continuous increase in

the speed of the rotor there by leading to unstable operation of the generator. This

discussion leads us to an important conclusion that out of the two operating points A

and B, with rotor angles o and max , operating point A is stable and operating point

B is unstable for small disturbances. Hence, point A is called stable equilibrium

point and operating point B is called as unstable equilibrium point. Similarly, o is a

stable steady state rotor angle and max is an unstable rotor angle.

Though we have discussed about the implications of the two operating points A

and B through their physical effect when subjected to disturbance, we can also prove

that operating point A is stable as compared to operating point B mathematically.

2.2.1 Linearizing SMIB system

A linear autonomous system can be represented in the form of state space as

X AX BU (2.21)

Where, X is the vector of state variables, A is called as the state matrix, B is the

input matrix and U is the vector of control inputs. In an autonomous system, the state

2.10

vector and the input vector or not explicit functions of time. According to linear

control theory, for a system given in (2.21) the stability can be assessed by computing

the eigen values of the state matrix A . Eigen values for a given matrix A can be

computed as the solution of the equation given in (2.22)

0I A (2.22)

Where, is vector of eigen values and I is an identity matrix and is of the same

order of the state matrix A . If the order of the state matrix A is n n then there will

be n eigen values which could be real or complex. The stability of the system given

in (2.21) can be found out from the eigen values computed from (2.22). There are

three scenarios:

1. All the eigen values have negative real part

2. Some of the eigen values have positive real part

3. Some of the eigen values have zero real part

If all the eigen values have negative real part irrespective of their imaginary parts

then the system is said to be asymptotically stable. If at least one eigen value has

positive real part then the system is unstable. If at least one complex conjugate pair of

eigen values have zero real parts then the system is called as marginally stable with

sustained oscillations. Form the above discussion we may think that if we apply same

linear control theory to the SMIB system model given in (2.18) we can find the

stability of the system. The problem with SMIB model given in (2.18) is that it is

nonlinear and the eigen values can be found only for a linear system. Hence, we use

the concept of linearization.

From P curve is can be observed that if we perturb the rotor angle form

point A by a small angle then the rotor angle starts oscillating between points C

and D. If observed closely it can be seen that P curve falling between the points C

and D is almost linear if is very small. Hence, if the rotor angle perturbation

is very small then we can assume that the system behaves in a linear fashion around

the operating point A between points C and D. Mathematically this can be written by

supposing the angle o is perturbed by an angle then we can express the swing

equation as

2.11

20

max2sinm o

s

dHP P

f dt

(2.23)

Expanding the sin o in (2.23) we get,

2 20

2 2

max (sin cos cos sin )s

m o o

dH H d

f dt f dt

P P

(2.24)

Since, is a very small perturbation angle, we can approximately write

cos 1 andsin . With these approximations substituted in (2.24), we can get

2 20

max max2 2sin cosm

s s

dH H dP P P

f dt f dt

(2.25)

Since at initial operating point A,

20

m a x2s in 0m o

s

dHP P

f d t

2

max2cos o

s

H dP

f dt

(2.26)

We can label, max 0 cossP P d= . Ps

is called as the synchronizing torque or power, as

in per units torque and power will be same. Ps

is the torque required to pull in the

rotor in to synchronization. Substituting Ps

in (2.26) and rewriting it we can get

2

2. 0s

s

H dP

f dt

(2.27)

2.12

Equation (2.27) is a linear differential equation of order two. We can solve equation

(2.27) through Laplace transformation. Applying Laplace transformation to equation

(2.27) and converting it from time domain to frequency domain we get ( s j )

2 ( ) ( ) 0ss

Hs s P s

f

(2.28)

Solving (2.28) we get

ss

fs P

H

(2.29)

Since, the swing equation is of second order it has two roots. Now if sP is positive

then there will be two roots on the imaginary axis with value

ss

fs j P

H

(2.30)

and if sP is negative i.e. 0 90 then the root are

,s ss s

f fs P P

H H

(2.31)

So in this case one root has positive real part and other root has negative real part

which means the system is unstable. But even if sP >0 the system has roots on the

imaginary axis due to which the system has sustained oscillation or the system is

marginally stable. In case of synchronous generator, the rotor has damper windings

due to which it acts as a induction motor during transients and this effect has to be

considered while evaluating the stability of the system. This effect of damper winding

is called as damping torque which depends on the rate of change of rotor angle.

2.13

d

dP D

dt

(2.32)

where, D is the damping coefficient. After including damping torque, the swing

equation (2.27) can be written as

2

20s

s

H d dD P

f dt dt

(2.33)

Applying Laplace transformation to (2.33) we get

2 0s ss

f fs Ds P

H H

(2.34)

Compare (2.34) with the characteristic equation of a standard second order system, as

given in (2.35)

2 22 0n ns s (2.35)

from (2.34) and (2.35) we can obtain

.sn s

fP

H

(2.36)

2 .s

s

fD

H P

(2.37)

Where, n is the natural frequency and is damping ratio of the oscillations of the

system given in (2.34). The eigen values of the system given in (2.33) can be

obtained from the characteristic equation given in (2.34) as,

21n ns j (2.38)

2.14

If 0sP then , 0n so the system has two complex conjugate roots with

negative real parts and hence the system will be stable. To observe the time response

of the synchronous generator in a SMIB system when subjected to a small disturbance

, we can apply inverse Laplace transform to (2.34) with an initial condition of

1

max

sin mo

P

P

then we obtain

2 10 2

20 2

sin ( 1 ) cos1

sin ( 1 )1

n

n

tn

tnn

e t

e t

(2.39)

We can observe from (2.39) that if 0 then the time response is a damped sinusoid.

That is from an initial condition of o and 0w , when subjected to a disturbance ,

there will be oscillations in the angle and speed w around the initial angle o , 0w

and these oscillations will be damped out after certain time and finally the angle

and speed w will settle down to the initial angle o and initial speed 0w . Here, the

initial speed of the generator rotor is nothing but the synchronous speed that is

0 sw w= .

2.3 Large-Disturbance Stability or Transient Stability

We have analyzed the stability of SMIB system when subjected to small

disturbances. In this section we will analyze the stability of SMIB system when

subjected to large disturbance like short circuit faults. The main difference between

small-disturbance and large-disturbance stability analysis is that in small-disturbance

we can assume that over small range around an operating condition the power system

can be considered linear and thereby can be linearized around that operating

condition. Whereas in case of large disturbances the change in the angle or speed of

the generator will be high and we can no longer assume that the system is linear.

2.15

Hence, analysis of large-disturbance stability will be different from that of small-

disturbance stability.

Transient stability analysis is the study of the system stability when subjected to

server faults like three-phase to ground short circuit or major generator outage etc.

The large-disturbance (here after will be called as transient stability) of a SMIB

system can be analyzed through a method called as “equal area criterion”. Equal area

criterion can be obtained from the swing equation as explained below.

Consider the swing equation.

2

max2sins

m

fdP P

dt H

(2.40)

Multiply both sides of (2.40) by 2d

dt

2

max22 sin 2s

m

fd d dP P

dt dt H dt

(2.41)

Equation (2.41) can be further simplified by noting that 2 2

22 .

d d d d

dt dt dt dt

, as

2

max sinsm

fd d dP P

dt dt H dt

(2.42)

or

2

max sinsm

fdd P P d

dt H

Integrating on both sides and taking square root we get

0

max sinsm

fdP P d

dt H

(2.43)

2.16

For the system to be stable the rotor angle should settle down to its steady state

value and hence (2.43) should be zero (rate of change of angle with respect to time

should become zero in steady state)

0

max0 sin 0m

dP P d

dt

(2.44)

2.3.1 Equal area criteria

Let us try to understand the implication of (2.44) on system stability through a three-

phase to ground fault in the SMIB system as shown in Fig. 2.3. Let us now draw the

( P ) curve, for the system shown in Fig. 2.3, as shown in Fig. 2.4. Here, mP is the

mechanical input and is assumed constant throughout the analysis. The stable

operating point is 1max( sin / )o mP P and the unstable operating point is

1max max( sin / )mP P .

Fig. 2.3: SMIB system with a three-phase to ground fault

In steady state before the fault is applied the rotor angle is at the stable

equilibrium point A. When a three-phase to ground fault is applied on one of the

transmission lines, towards the generator, then the electrical power output of the

LXSX

TX

0V E

ePLX

mP

L L L G

fault

2.17

generator will be zero and hence the operating point moves to point B on P

curve. Since, electrical power output max sinP has become zero and the mechanical

input power mP is constant the rotor angle starts to increase as can be observed

from swing equation. The imbalance in input and output power will lead to rotor

gaining kinetic energy and the rotor starts accelerating. Hence, the operating point

starts moving from point B on the P curve along the -axis towards infinity. If the

fault is cleared say at point C, which corresponds to an angle c , then since electrical

power output is not zero but it is max sin cP the operating point moves to point E on the

swing curve.

It can be observed that at point E the electrical power output is greater than the

mechanical power input and hence the rotor starts decelerating. Since, rotor has finite

inertia it cannot change its speed suddenly and hence the rotor acceleration will

decrease slows due to which angle will swing from point E to point F. The rotor

angle will increase until the rotor losses all the kinetic energy it gained during the

fault period is exhausted. Once, the rotor exhausts the kinetic energy gained during

fault period it will swing back from point F and will move towards the point A. But, if

the rotor swing causes the rotor angle to cross the unstable operating point max from

point E then the system will become unstable. Hence, the rotor can swing maximum

up to max before becoming unstable. Now if we apply (2.44) for this system, so that

the system is stable, then we can express (2.44) as

max

0

max sin 0mP P d

(2.45)

we can dived the intervals of integration of (2.5) in to o c and

maxc which lead

to

max

0

max maxsin sin 0c

c

m mP P d P P d

(2.46)

2.18

max

0

max maxsin sinc

c

m mP P d P P d

(2.47)

Fig. 2.4: Swing curve to demonstrate equal area criterion

From Fig. 2.4 we can immediately observe that

0

max sin Areaof ABCDc

mP P d

(2.48)

max

max sin Areaof DEFc

mP P d

(2.49)

From (2.48) and (2.49) it can be understood that the energy gained during acceleration

(Area ABCD) should be exactly equal to the energy lost during deceleration (Area

DEF) for the system to be stable. Hence, we can write

Area of P curve during acceleration = Area during deceleration (2.50)

Because of (2.50) we call this method of assessing the stability as equal area criterion.

Now the basic idea is that we have to clear the fault below a critical clearing angle c

eP

0 2

max

mP

maxP

0

A

B

CC

E

D F

2.19

such that area of acceleration is equal to area of deceleration. If the fault is cleared

later than critical clearing angle then area of acceleration will become more than area

of deceleration due to which the rotor angle will swing beyond the point F and will

become unstable. In order to find the critical clearing angle c we can use equal area

criterion i.e. solve (2.46) for c . During fault the electrical output power is zero hence

max sin 0eP P . Now substituting this in equation (2.46) leads to

max

max sinc

o c

m mP d P P d

(2.51)

solving (2.51) for c will give

max maxmax

cos ( ) cosmc o

P

P

or 1max max

max

cos cosmc o

P

P

(2.52)

c is the critical clearing angle at which if the fault is cleared the system becomes

stable. We need to find the critical clearing time corresponding to the critical clearing

angle c . For this let us take swing equation during the fault 0eP

2

2s

m

fdP

dt H

(2.53)

Integrating (2.53) twice, on both left hand and right hand side of the expression, gives

the expression

2002 m

fP t

H

(2.54)

2.20

Since, we want to find the critical clearing time corresponding to the critical clearing

angle c , we can substitute c and let the time at this critical clearing angle be

ct t . Substituting c and ct in (2.54) gives

0 2sc

cs m

Ht

f P

(2.55)

So far a three-phase to ground fault near the generator bus was considered due to

which the electrical power output during the fault was zero. In case the three-phase to

ground fault is at some distance from the generator terminal then the electrical power

output will not become zero but will transmit power along the healthy transmission

line at a reduced level which depends on the distance of the fault from the generator

terminal.

Now consider SMIB system shown in Fig. 2.3. Instead of fault being at the

generator terminals let it be at the center of the second transmission line due to which

the maximum power that can be transferred to the infinite bus reduces to say max 2P

and after the fault is cleared the faulted line is disconnected from the system hence the

maximum power that can be transferred to the infinite bus post fault also reduced, due

to increased reactance, say to max3P . Hence, we will have three P curves that is

pre-fault, during-fault and post-fault as shown in Fig. 2.5.

In pre-fault case the system is stable at operating point A. At the time of the

fault the operating point moves to point B on the P curve with maximum power

transfer max 2P . As the mechanical input power mP is greater than

max 2 sinP the rotor

will accelerate and the operating point movies from point B to point C where the fault

is cleared followed by tripping of the faulted line. At operating point C as the fault is

cleared the rotor angle will move to point E on post fault P curve with maximum

power transfermax 3P . Due to inertia of the rotor the rotor angle will swing from point E

and will move up to the point F before becoming unstable. Here, 1maxsin /o mP P

and 1max max 3sin /mP P . It has to noted that max in (2.6) corresponds to the pre-

fault system unstable equilibrium point whereas max corresponds to the unstable

2.21

equilibrium point of post-fault system. Applying equal area criterion to this system

and equating the area of acceleration and area of deceleration we get

Fig. 2.5: Swing curve during pre, post and during the fault

max

max2 max3( sin ) sinc

o c

m mP P d P P d

(2.56)

Solving (2.156) for c will give

max max3 max max 2

max3 max 2

1 max max3 max max 2

max3 max 2

( ) cos coscos

( ) cos coscos

m o oc

m o oc

P P P

P P

P P Por

P P

(2.57)

2.3.2 Numerical integration of swing equation

The equal area criterion gives an analytic way of assessing the stability of the

system. The transient stability of the system can also be found out by numerically

integrating the swing equation. Since, swing equation is nonlinear we cannot solve for

the solution of swing equation through analytical methods. In order to solve the swing

eP

0 2

max

mP

maxP

0

A

B

C

C

E D F

max 2P

max3P

2.22

equation numerical methods [3] have to be used. Let us look at a well known

numerical method called as Euler’s method.

Euler’s method

Let a nonlinear differential equation of the form given in (2.58) exist, where ( )f x is a

nonlinear function of x

( )dx

f xdt

(2.58)

we can find the solution of (2.58) through Euler’s method. The idea is to integrate

(2.58) between the time instants ( , )o ft t with an initial value of 00,atx x t t with

a small time step t . This can be expressed as

0

1 0

x x

dxx x t

dt

(2.59)

Here, 1x is the variable value at time instant 0t t t . This process has to be

repeated iteratively until the final time ft is reached. Equation (2.59) is nothing but

Taylor series expansion of the variable x around the operating point ( , )oot x with

higher order terms discarded. Since, the higher terms are discarded (2.59) may

introduce error. Hence, to reduce the error modified Euler’s method can be used.

Modified Euler’s method has two steps: predictor step and corrector step. These steps

in a generalized for are as following:

1

k

k kp

x x

dxx x t

dt

(2.60)

1

1 1

2 kkp

k kc

x xx x

dx dxx x t

dt dt

(2.61)

2.23

Where, 1px is the predicted and

1cx is the corrected value of the variable x at the time

instant 0t t t . The time step t has to be carefully chosen, the smaller the

values better the accuracy. Ideally t should approach zero but then the number of

steps required to integrate (2.58) between the time limits ( , )o ft t will become infinite.

Hence, there is a trade off between the accuracy and the number of steps required.

Now modified Euler’s method can be applied to integrate swing equation between the

time limits ( , )o ft t . The second order differential equations of the SMIB system are

given below:

me s me

d

dt

(2.62)

max sinme sm

d fP P

dt H

(2.63)

Here, me me s that is the change in the rotor speed from the synchronous

speed. Now the predictor step of Euler’s method when applied to (2.62) and (2.63)

will give:

1

k

kme me

k kp

dt

dt

(2.64)

1

k

kme me

k k meme p me

dt

dt

(2.65)

The corrector step is given as

2.24

1

1

1 1

2 kp k

kkme mep

me me

k kc

d dt

dt dt

(2.66)

1

1

1 1

2 kp k

kkme mep

me me

k kmec me

d dt

dt dt

(2.67)

By numerically integrating the swing equation for a specified period like for 5

seconds, if we observe that the rotor angle and the rotor speed are settling to the

steady state values when subjected to a disturbance then we can say that the system is

stable. But if we observe from numerical integration of swing equation that the rotor

angle and speed are either continuously increasing or decreasing as time tends

towards infinity then that means that the system has become unstable.

2.4 Disadvantages of Classical Method of Stability Analysis

So, far we analyzed the small-disturbance and transient stability of a SMIB system

where the generator is represented by a classical second order electro-mechanical

model. There are several disadvantages of using this simplistic model of synchronous

generators. The disadvantages are:

1. The rotor flux decay, the damper windings on the generator rotor and the

effect of rotor core on the stability are totally neglected in the classical model.

But they can effect the stability of a system significantly.

2. The internal voltage of the generator behind the synchronous reactance is

assumed to be constant on the basis that the rotor field current is held constant.

This assumption is not true and the rotor field current is controlled through an

excitation system and automatic voltage regulator (AVR). Also, it is well

observed phenomenon that high gain fast acting exciters can cause small-

2.25

disturbance instability [4]. Hence, the exciter and AVR dynamics have to be

taken into account.

3. It is assumed throughout the analysis that the mechanical power input is held

constant. Again the input mechanical power depends on turbine speed

governor and turbine dynamics. By controlling the mechanical power input

according to the variation in the electrical power input we can improve the

transient stability of the system. Hence, the dynamics of speed governor and

turbine have to be considered.

4. Important generator external controllers like power system stabilizer (PSS)

improve the system stability significantly and need to be modeled.

5. Dynamic loads like induction motors, synchronous motors, power electronic

devices etc can affect the stability drastically. These are not taken into

consideration in classical model.

Due, to these limitations the classical method cannot give an accurate idea about the

stability of the system. In this regard, we have to go for detailed modeling of

synchronous generator, excitation system, speed governor, turbine, external

controllers and loads. In next few Chapters we will concentrate on detailed modeling

aspects of power system components.

Example Problems

E1. A three-phase, 50 Hz, synchronous generator is connected to an infinite bus. The

maximum real power that can be transferred to the infinite bus is 1 per unit. The

mechanical input to the generator is 0.8 per unit. The inertia constant of the generator

is 5 seconds.

(a) Find the natural frequency of oscillations of the system and comment on

stability.

(b) If the damping coefficient is taken as 0.1 then find the natural frequency,

damped frequency and damping ratio of the system oscillations.

(c) For a small disturbance of 10dD = , find the change in internal angle and

speed of the generator with respect to time.

2.26

Sol:

The maximum power that can be transferred by the generator to the infinite bus and

the input mechanical power to the generator are given as

max 1P pu= (E1.1)

0.8mP pu= (E1.2)

The initial internal angle, 0d , can be computed as

( )1 10

maxsin sin 0.8 53.13mP

Pd - -æ ö÷ç ÷= = = ç ÷ç ÷çè ø

(E1.3)

The synchronizing torque or power is given as

( )max 0cos 0.6sP P d= = (E1.4)

The linearized swing equations, as given in (2.27), can be written as, with

5 , 50H s f Hz= =

2

20.0318 0.6 0

d

dt

(E1.5)

The solution of (E.5) in Laplace domain gives two solutions, which are

4.3146s

fs P j

H

(E1.6)

Since, complex pair of poles or roots are on the imaginary axis the system will have

sustained oscillations. The natural frequency of oscillations is given as

4.3146 / 0.691n rad s or Hz (E1.7)

2.27

(b)

If the damping coefficient is considered as 0.1 then the swing equation given in (E1.5)

changes to

2

23.14 18.85 0

d d

dt dt

(E.8)

Equation (E.8) can be written in Laplace domain as

2 3.14 18.85 ( ) 0s s s (E1.9)

Compare (E.9) with the standard second order characteristic equation

2 22 0n ns sxw w+ + = , then

4.3437 / 0.6913n rad s or Hzw = (E1.10)

0.3616x= (E1.11)

(c)

The solution of the swing equation given in (E1.8) in time domain for a given

perturbation in angle of 10dD = can be given be written as, as mentioned in (2.39),

2 10 2

1.57

sin ( 1 ) cos1

53.13 10.7249 sin(4.05 68.80 )

ntn

t

e t

e t

(E1.12)

20 2

1.57

sin ( 1 )1

314.159 46.58 sin 4.05

ntnn

t

e t

e t

(E1.13)

The plots of angle and speed with respect to time for the given perturbation in angle

of 10dD = are given in Fig. E1.1. It can be observed from Fig. E1.1 that, both the

angle and speed settle down to the initial conditions when subjected to a perturbation.

2.28

(a)

(b)

Fig. E1.1: Generator (a) internal angle (b) rotor speed

2.29

E2. A three-phase, 50 Hz, synchronous generator is connected to an infinite bus

through a transformer and two parallel transmission lines. The input mechanical

power to the synchronous generator is given as 0.8 pu. The grid is consuming a

complex power of 0.8 0.6j pu. Find the critical clearing angle for

(a) A three phase-to-ground fault at the generator terminal, where system returns

to its pre-fault condition after fault clearing.

(b) A three phase-to-ground fault in the middle on the second transmission line

and after fault the second transmission line is disconnected from the system.

Sol: The current drawn by the grid is given as

0.8 0.60

P jQI j

V¥

-= = -

(E2.1)

Fig. E2.1: A single machine connected to infinite bus system

The generator internal voltage and angle can be computed as

'( 0.5 )

1.6 0.8

1.7889 26.56

d T LE V j X X X I

j

d ¥ = + + +

= +

=

(E2.2)

0.5LX ' 0.25dX 0.5TX

1 0V E

0.8 0.6P jQ j

0.5LX L L L G

fault

0.8mP

2.30

Hence,

0

max 0

26.56 0.4636

153.44 2.6780

rad

rad

d

d p d

= =

= - = = (E2.3)

The critical clearing angle is given as

1max max

max

1

cos cos

0.8cos 2.678 0.4636 cos 0.4636

1.7889

1.4748 84.5

mc o

P

P

rad or

(E2.4)

(b)

In case of a three phase-to-ground fault in the middle of the second transmission line

the transfer impedance between the generator internal bus and the infinite bus is given

as

2.7503trX = (E2.5)

Hence, the maximum power that can be transferred during the fault is given as

max 2 0.6504tr

EVP

X¥= = (E2.6)

After the clearing of fault the second transmission is disconnected from the system

hence the maximum power that can be transferred after the fault cleared is given as

max 3 '1.4311

d T L

EVP

X X X¥= =

+ + (E2.7)

2.31

After the fault is cleared the system will settle at a new operating point. The internal

angle at the new operating point is given as

' 10

max 3

' 'max 0

sin 0.5932 34

2.5484 146

mPrad

P

rad

d

d p d

- æ ö÷ç ÷= = = ç ÷ç ÷çè ø

= - = =

(E2.8)

Hence, the critical clearing angle can be computed as

1 max max3 max max 2

max3 max 2

( ) cos coscos

1.6998 97.39

m o oc

P P P

P P

rad or

(E2.9)

2.32

References

1. E.W. Kimbark, Power System Stability, Volume I: Elements of Stability

Calculations, John Wiley (New York), 1948.

2. S. B. Crary, Power System Stability Volume I: Steady State stability, John

Wiley, New York, 1945.

3. Ram Babu, Numerical Methods, Pearson Education, India, 2010.

4. C. Concordia, “Steady-state stability of synchronous machines as affected by

voltage regulator characteristics,” AIEE Trans., Vol. 63, May 1944, pp. 215-

220.