Download Chapter 2 Analysis of Power System Stability by Classical

Chapter 2
Analysis of Power System Stability by Classical Methods
2.1 Classical Model
As discussed in the previous Chapter, the first step in analyzing power stability
is to represent the power 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 whereas the electrical power output can change within in
milliseconds.
2.1
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 X d' . The significance of X d' will be explained in Chapter 3 but
for now it can be assumed to be a reactance of the generator during transients. VT T
is the terminal voltage of the synchronous generator. X T , X L represent the
transformer and line reactance. The complex infinite bus voltage is represented as
V 0 . The infinite bus voltage is 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 Pm and the
output electrical power is defined by Pe . 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, Pe , can be computed as
*
*


 ( Ee j  V e j 0 )  
 (VT e jT  V e j 0 )  
jT
j
Pe  Real  VT e  
 
   Real  Ee  

j( X T  X L )  
j ( X d'  X T  X L )  






EV
sin 
 '
X d  XT  X L
(2.1)
The maximum real power output of the synchronous generator that can be
transferred to the infinite bus in this case is
2.2
Pmax 
EV
, at   90
X  XT  X L
(2.2)
'
d
Hence, the synchronous generator real power output can be represented as
Pe  Pmax sin 
(2.3)
X d'
 Pe
XT
XL
VT T
E 
V 0
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
J
d 2 m
 Tm  Te
dt 2
(2.4)
where, J kg.m 2 is the inertia constant of the rotating machine. The mechanical input
torque due to the prime mover is represented as Tm N.m and the electrical torque,
acting against the mechanical input torque, is represented by Te N.m . 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
2.3
steady 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  ms t
(2.5)
Where,  m is the angle between the rotor field axis and the reference axis rotating
synchronously at ms rps . If we differentiate (2.5) with respect to time we get
d m d m

 ms
dt
dt
(2.6)
d 2 m d 2 m

dt 2
dt 2
(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,
m 
d m
rps
dt
(2.8)
Substituting, (2.8) in (2.6) we get
d m
 m  ms
dt
(2.9)
Similarly, substituting (2.7) in (2.4) we get
J
d 2 m
 Tm  Te
dt 2
(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
2.4
J m
d 2 m
 Pm  Pe
dt 2
(2.11)
Now multiply with the term
1
ms on both the sides of (2.11) and divide the entire
2
equation with the base MVA ( S B ), in order to express the equation in per unit, lead to
1
J mms 2
P
d m 1
P 
2
 ms  m  e 
2
2
SB
dt
 SB SB 
(2.12)
Let us define a new parameter named as machine inertia constant
1
1
J ms 2
J ms 2
MW.S
2
2

H
Base MVA
MVA
SB
(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
1
1
2
instead J ms
 J mms . With this assumption if we substitute (2.13) in (2.12) we
2
2
get
d 2 m 1
 ms  Pm  Pe 
H
dt 2
2
(2.14)
per unit
 m and 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
2.5
P
 m elec.rad
2
P
s  ms elec.rad/s
2

(2.15)
Substituting (2.3) and (2.15) into (2.14) we get
d 2 s

 Pm  Pmax sin  
dt 2 2 H
or
d 2  f s

 Pm  Pmax sin  
dt 2
H
per unit
(2.16)
per unit ( f s  50 or 60 Hz)
Equation (2.16) is called as swing equation. Equation (2.16), assuming all the
parameters expressed in per units, can also be written as
d
 (  s )
dt
d  f s

 Pm  Pmax sin  
dt
H
(2.17)
Where,    P / 2  m . In can be observed from (2.17) that, if Pm  Pmax sin  then there
will be no speed change and there will be no angle change. But, if Pm  Pmax sin  due
to disturbance in the system then either the speed increase or decrease with respect to
time. Let us take the case of Pm  Pmax sin  , 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 Pm  Pmax sin  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.
2.6
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
H d 2
 Pm  Pm a x s in 
 fs dt 2
(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
Pm  Pmax sin 
(2.19)
Since, the mechanical power input Pm and the maximum power output of the
generator Pmax are known for a given system topology and load, we can find the rotor
angle  from (2.19) as
 Pm
 Pmax
  sin 1 

1  Pm 
 or   sin 


 Pmax 
(2.20)
It can be observed from (2.19) that rotor angle  has two solutions at which
Pm  Pmax sin  .
Now,
let
us
label
 o  sin 1  Pm / Pmax 
and
 max    sin 1  Pm / Pmax      o . There is an 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 Pe 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
2.7
generator Pmax occurs at an angle    / 2 . On the same curve the input mechanical
power Pm can also be represented. Since Pm 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
Pm is equal to Pe . The rotor angles at point A and B are  o and  max , the solutions of
equation (2.19).
Pe
Pmax
C´
A
Pm
B
´D
0
0
 max 


2
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 Pmax sin  o   o  . Since, in steady state
Pm  Pmax sin  o  , after perturbation Pm  Pmax sin  o   o  , 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
Pm  Pe 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
2.8
again move to point C. This phenomenon repeats itself indefinitely if there is no
damping to the rotor oscillations. This can be understood analogously from the motion
of the pendulum 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 , that is point B in Fig. 2.2.
Again if we perturb  max by a positive angle  max then the electrical power output
will be Pmax sin  max   max  . In this case however, unlike the earlier case,
Pm  Pmax sin  max   max  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)
2.9
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
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)
I  A  0
(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 it 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
2.10
supposing the angle  o is perturbed by an angle  then we can express the swing
equation as
2
H d  0    
 Pm  Pmax sin  o    
 fs
dt 2
(2.23)
Expanding the sin  o    in (2.23) we get,
H d 2 0
H d 2

 f s dt 2
 f dt 2
(2.24)
 Pm  Pmax (sin  o cos    cos  o sin   )
Since,

is
a
very
small
perturbation
angle,
we
can
approximate
cos  1 and sin   . With these approximations substituted in (2.24), we can get
H d 20 H d 2

 Pm  Pmax sin   Pmax cos 
 fs dt 2  fs dt 2
(2.25)
Since at initial operating point A,
H d 2 0
 P m  Pm a x s i n  o  0
 fs dt2
H d 2
 Pmax cos o
 fs dt 2
(2.26)
We can label, Ps = Pmax cos d0 . P is called as the synchronizing torque or power, as
s
in per units torque and power will be same. P is the torque required to pull in the
s
rotor in to synchronization. Substituting P in (2.26) and rewriting it we can get
s
2.11
H d 2 
 Ps .  0
 f s dt 2
(2.27)
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 )
H 2
s  (s)  Ps  (s)  0
 fs
(2.28)
Solving (2.28) we get
s 
 fs
H
Ps
(2.29)
Since, the swing equation is of second order it has two roots. Now if Ps is positive
then there will be two roots on the imaginary axis with value
sj
 fs
H
Ps
(2.30)
and if Ps is negative i.e.  0  90 then the root are
s 
 fs
H
Ps , 
 fs
H
Ps
(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 Ps >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
2.12
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.
d
dt
Pd  D
(2.32)
where, D is the damping coefficient. After including damping torque, the swing
equation (2.27) can be written as
H d 2
d 
D
 Ps   0
2
 f s dt
dt
(2.33)
Applying Laplace transformation to (2.33) we get
 fs
s2 
H
Ds 
 fs
H
Ps  0
(2.34)
Compare (2.34) with the characteristic equation of a standard second order system, as
given in (2.35)
s2  2n s  n2  0
(2.35)
from (2.34) and (2.35) we can obtain
n 

D
2
 fs
H
(2.36)
. Ps
 fs
(2.37)
H .Ps
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,
2.13
s n  jn 12
(2.38)
If Ps  0 then n ,   0 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
 Pm  then we obtain

 Pmax 
   o  sin 1 
  0 
  0 

1  2
n
1  2

ent sin (n 1 2 )t  cos1  

ent sin (n 1 2 )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 w0 , when subjected to a disturbance  ,
there will be oscillations in the angle  and speed w around the initial angle  o , w0
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 w0 . Here, the
initial speed of the generator rotor is nothing but the synchronous speed that is
w0 = w s .
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
2.14
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.
Hence, analysis of large-disturbance stability will be different from that of smalldisturbance 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.
d 2  fs

 Pm  Pmax sin  
dt 2
H
Multiply both sides of (2.40) by 2
2
(2.40)
d
dt
d d 2  f s
d

 Pm  Pmax sin    2 
2
dt dt
H
 dt 
(2.41)
d  d 
d d 2
Equation (2.41) can be further simplified by noting that
2
, as
.
dt  dt 
dt dt 2
2
d  d  2 fs
d


P
P
sin



m
max
dt  dt 
H
dt
2
(2.42)
or
 d  2 fs
d  
 Pm  Pmax sin   d
H
 dt 
2
Integrating on both sides and taking square root we get

2 f s
d

 Pm  Pmax sin   d
dt
H 0
(2.43)
2.15
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)

d
 0    Pm  Pmax sin   d  0
dt
0
(2.44)
2.3.1 Equal area criteria
Let us try to understand the implication of (2.44) on system stability through a threephase 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,
Pm is the
mechanical input and is assumed to be constant throughout the analysis. The stable
operating point is  o (  sin 1  Pm / Pmax ) and the unstable operating point is
 max (   sin 1  Pm / Pmax ) .
XS
XT
 Pe
L  L  L G
 Pm
E 
fault
XL
XL
V 0
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
2.16
generator will be zero and hence the operating point moves to point B on P  
curve. Since, electrical power output  Pmax sin   has become zero and the mechanical
 Pm 
input power
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 Pmax sin  c 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 kinetic energy 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
  P  P
m
max
sin   d  0
(2.45)
0
we can dived the intervals of integration of (2.5) in to  o   c and  c   max which lead
to
c
  P  P
m
0
max
sin   d 
max
  P  P
m
max
sin   d  0
c
2.17
(2.46)
c
  P  P
m
max
sin   d 
0
max
  P
max
sin   Pm  d
(2.47)
c
Pe
Pmax
E
A
Pm
B
0
0
D
F
C
 max 
C 

2
Fig. 2.4: Swing curve to demonstrate equal area criterion
From Fig. 2.4 we can immediately observe that
c
  P  P
m
max
sin  d Areaof ABCD
(2.48)
0
max
  P
max
sin  Pm  d  Areaof DEF
(2.49)
c
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
2.18
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
Pe  Pmax sin   0 . Now substituting this in equation (2.46) leads to
c
 max
 P d    P
m
max
o
(2.51)
c
solving (2.51) for
cos  c 
or
sin   P m  d
c
will give
Pm
( max   o )  cos  max
Pmax
 Pm

 max   o   cos  max 
 Pmax

 c  cos1 
(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  Pe  0 
d 2  f s

Pm
dt 2
H
(2.53)
Integrating (2.53) twice, on both left hand and right hand side of the expression, gives
the expression
 
 f0
2H
Pm t 2   0
(2.54)
2.19
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
t  tc . Substituting  c and tc in (2.54) gives
tc 
c 0  2H
 fs Pm
s
(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 Pmax 2
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 Pmax 3 . 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 Pmax 2 . As the mechanical input power Pm is greater than Pmax 2 sin  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 transfer Pmax 3 . 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,  o  sin 1  Pm / Pmax 
and  max
    sin 1  Pm / Pmax 3  . It has to be noted that  max in (2.6) corresponds to the
 corresponds to the unstable
pre-fault system unstable equilibrium point whereas  max
2.20
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
Pe
Pmax
Pmax 3
E
D
A
Pm
Pmax 2
0
F
B
C
0
C 

 max


2
Fig. 2.5: Swing curve during pre, post and during the fault
c
 (P
m
 Pmax2 sin  ) d 
o

 max
  P
max3
sin   P m  d
(2.56)
c
Solving (2.156) for  c will give
cos  c 
   o )  Pmax 3 cos  max
  Pmax 2 cos  o
Pm ( max
Pmax 3  Pmax 2
 P (    o )  Pmax 3 cos  max
  Pmax 2 cos  o 
or  c  cos 1  m max

Pmax 3  Pmax 2


(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
2.21
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 ( x)
dt
(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 (to , t f ) with an initial value of x  x 0 ,at t  t0 with
a small time step t . This can be expressed as
 dx
x1  x 0  
 dt

 t
0
x x 
(2.59)
Here, x1 is the variable value at time instant t  t0  t . This process has to be
repeated iteratively until the final time t f is reached. Equation (2.59) is nothing but
Taylor series expansion of the variable x around the operating point (to , x o ) 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:
 dx
x p k 1  x k  
 dt
xc
k 1
x x
1  dx
x  
2  dt

k

 t

k
x x p
k 1
dx

dt
(2.60)
x xk

 t


(2.61)
2.22
Where,
x kp1 is the predicted and xck 1 is the corrected value of the variable x
at the
time instant tk 1  tk  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 (to , t f ) 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 (to , t f ) . The second order differential equations of the SMIB
system are given below:
d
   s  
dt
(2.62)
d   f s

 Pm  Pmax sin  
dt
H
(2.63)
Here,     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:
 p k 1   k 
d
dt
 
  k
 p k 1   k 
k
t
d me
dt
(2.64)
  k
  k
t
(2.65)
The corrector step is given as


1  d
k 1
k
c   
2  dt


k 1
  p
  p k 1

d
dt
  k
  k
2.23


 t



(2.66)


1  d 
d 
k 1
k
 c   

k 1
2  dt   p k 1
dt   k
  p

  k



 t



(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 a steady
state value 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 dynamics, 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 smalldisturbance 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
2.24
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 Dd = 10 , find the change in internal angle and
speed of the generator with respect to time.
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
Pmax = 1 pu
(E1.1)
Pm = 0.8 pu
(E1.2)
2.25
The initial internal angle, d0 , can be computed as
æ P ö
d0 = sin-1 ççç m ÷÷÷ = sin-1 (0.8) = 53.13
çè Pmax ø÷
(E1.3)
The synchronizing torque or power is given as
Ps = Pmax cos (d0 ) = 0.6
(E1.4)
The linearized swing equations, as given in (2.27), can be written as, with
H = 5 s, f = 50 Hz
0.0318
d 2 
 0.6  0
dt 2
(E1.5)
The solution of (E.5) in Laplace domain gives two solutions, which are
s 
f
H
Ps   j 4.3146
(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
n  4.3146 rad / s or 0.691 Hz
(E1.7)
(b)
If the damping coefficient is considered as 0.1 then the swing equation given in (E1.5)
changes to
d 2 
d 
 3.14
 18.85  0
2
dt
dt
(E.8)
2.26
Equation (E.8) can be written in Laplace domain as
s
2
 3.14s  18.85   ( s )  0
Compare
(E.9)
with
the
(E1.9)
standard
second
order
characteristic
equation
s 2 + 2xwn s + wn2 = 0 , then
wn = 4.3146 rad / s or 0.6913 Hz
(E1.10)
x = 0.3616
(E1.11)
(c)
The solution of the swing equation given in (E1.8) in time domain for a given
perturbation in angle of Dd = 10 can be given be written as, as mentioned in (2.39),
  0 

1  2

ent sin (n 1 2 )t  cos1  

(E1.12)
 53.1310.7249e1.57t sin(4.05t  68.80)
  0 
n 
1 
2

e nt sin (n 1   2 )t

(E1.13)
 314.159  4.658e 1.57 t sin  4.05t 
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  j 0.6 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.
2.27
Sol: The current drawn by the grid is given as
I =
P - jQ
= 0.8 - j 0.6
V¥0
(E2.1)
X d'  0.25
E 
Pm  0.8
X T  0.5
L  L  L G
fault
X L  0.5
 P  jQ  0.8  j 0.6
X L  0.5
V  10
Fig. E2.1: A single machine connected to infinite bus system
The generator internal voltage and angle can be computed as
Ed = V¥ + j ( X d' + X T + 0.5 X L ) I
=1.6 + j 0.8
=1.788926.56
(E2.2)
Hence,
d0 = 26.56 = 0.4636 rad
(E2.3)
dmax = p - d0 = 153.44 = 2.6780 rad
The critical clearing angle is given as
 Pm

 max   o   cos  max 
 Pmax

 c  cos1 
 0.8
 cos1 
 2.678  0.4636  cos  2.678 
 1.7889

1.4748 rad or 84.5 
2.28
(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
X tr = 2.7503
(E2.5)
Hence, the maximum power that can be transferred during the fault is given as
Pmax 2 =
EV¥
= 0.6504
X tr
(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
Pmax 3 =
EV¥
= 1.4311
'
X d + XT + X L
(E2.7)
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
æ P ö
d0' = sin-1 ççç m ÷÷÷ = 0.5932 rad = 34
çè Pmax 3 ÷ø
(E2.8)
'
dmax
= p - d0' = 2.5484 rad = 146
Hence, the critical clearing angle can be computed as
 Pm ( max
   o )  Pmax 3 cos  max
  Pmax 2 cos  o 

Pmax 3  Pmax 2


 1.6998 rad or 97.39
 c  cos 1 
2.29
(E2.9)
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. 215220.
2.30