POWER SYSTEM STABILITY STUDIES USING MATLAB
A Project
Report submitted in partial fulfillment of the
requirements for the award of the Degree of
BACHELOR OF TECHNOLOGY
IN
ELECTRICAL AND ELECTRONICS
BY
SHIWANI RAJ
(BTECH/15222/2019)
DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING
BIRLA INSTITUTE OF TECHNOLOGY
MESRA, OFF CAMPUS PATNA
2022
STUDENT DECLARATION
I, Shiwani Raj, certify that
a. The work contained in the project report titled “Power System Stability Studies using
MATLAB” is original and has been done by myself under the general supervision of my
supervisor/guide “Dr. Kumar Nikhil”.
b. The work has not been submitted to any other Institute for any degree or diploma.
c. I have followed the guidelines provided by the Institute in writing the project report.
d. I have conformed to the norms and guidelines given in the Ethical Code of Conduct of the
Institute.
e. Whenever I have used materials (data, theoretical analysis, and text) from other sources, I
have given due credit to them by citing them in the text of the project report and giving their
details in the references.
f.
Whenever I have quoted written materials from other sources, I have put them under
quotation marks and given due credit to the sources by citing them and giving required
details in the references.
Student Name
Student Roll No.
Shiwani Raj
BTECH/15222/19
Signature of the Student
DECLERATION CERTIFICATE
This is to certify that the work presented in the project report
entitled “Power System Stability Studies using MATLAB” in partial
fulfilment of the requirement for the award of Degree of Bachelor of
Engineering in Electrical and Electronics Engineering of Birla Institute of
Technology Mesra, Ranchi is an authentic work carried out under my
supervision and guidance.
To the best of my knowledge, the content of this project does not
form a basis for the award of any previous Degree to anyone else.
Date:
(Dr. Kumar Nikhil)
Department of Electrical & Electronics Engineering
Birla Institute of Technology
Mesra, off Campus Patna
Patna-800014
CERTIFICATE OF APPROVAL
The foregoing project entitled “Power System Stability Studies using
MATLAB”, is hereby approved as a creditable study of research topic and has
been presented in satisfactory manner to warrant its acceptance as prerequisite
to the degree for which it has been submitted.
It is understood that by this approval, the undersigned do not necessarily
endorse any conclusion drawn or opinion expressed therein, but approve the
project for the purpose for which it is submitted.
(Internal Examiner)
(External Examiner)
(Chairman)
Head of the Department
Department of Electrical & Electronics Engineering
Birla Institute of Technology
Mesra, off Campus Patna
Patna-800014
ACKNOWLEDGMENT
Before anything else, I want to thank God. I've put in a lot of work on this minor project. But this
wouldn't have been possible without the generous aid of many others. To everyone who made it
possible for me to complete this project, thank you from the bottom of my heart.
I would like to express my deepest appreciation to Dr. Kumar Nikhil, my advisor for my minor
project, for all of the help, motivation, and direction he has given me throughout the course of the
past year. For everything I've learned from him, both formally and informally, I will be eternally
thankful. It was thanks to his support that I was able to finish my assignment.
Date:
Shiwani Raj (BTECH/15222/19)
ABSTRACT
Modern power networks are highly interconnected, and this interconnectedness leads in increased
operating efficiency and dependability through mutual cooperation. They also contribute to the
issue of stability. The stability problem has become a fundamental challenge to power system
engineers because of these large-scale interconnections. The stability problem stems from the
synchronous machine's response to disturbances in the power grid. Stability in a power system
means it can go back to normal or stable after being disturbed. This paper discusses steady-state
and transient stability, as well as the swing's equation and numerical solutions using MATLAB/
Simulink. This research is divided into two parts. First, we looked at a single synchronous
generator coupled to an infinite bus as an example to demonstrate their features using Swing's
Equation. Secondly, we connected a Solar panel with the previous network and studied the features
of steady-state and transient stabilities. At last, we compared the results from both. In this paper,
we used standard numerical techniques to solve Swing Equation, and the Runge-Kutta method.
Using simulation, we obtained the swing curve.
Keywords—Steady State Stability, Transient Stability, Swing Equation, Runge-Kutta
method
TABLE OF CONTENTS
Contents
Page No.
I.
Introduction
II.
Power System Stability (PSS): What it means
III.
Stability Classification for Power Systems
IV.
Swing Equation
V.
Synchronous Machine Models for Stability Studies
VI.
Steady-State Stability- Small Disturbances
VII.
Description of the system for Steady-State Stability study
VIII. Transient Stability Studies- Three-Phase Fault
1-3
3
3-5
6
6-9
9-13
13-16
16
IX.
Equal Area Criterion
16-19
X.
Application to Three-Phase Fault
19-25
XI.
Description of the system for Transient Stability study
25-32
REFERENCES
APPENDIX
33
34-39
I.
INTRODUCTION
Since the 1920s, the importance of a reliable power grid has been acknowledged as a vital issue
for the safe operation of critical infrastructure. The significance of this phenomenon has been
demonstrated by numerous significant blackouts brought on by instability in the power grid. When
it comes to the control and operation of power systems, stability is among the most crucial
constraints.
By definition, a stable power system is one that has a tendency to reestablish equilibrium after a
physical disturbance. This study examines the power system's steady-state stability and transient
stability, along with its behavior in response to fault locations. Most systems' primary stability
issue is transient instability; hence most research and development effort has been directed toward
this type of instability. A system's stability can be described as its ability to maintain equilibrium
under normal conditions and recover to a satisfactory equilibrium level following a disturbance.
When talking about alternating current (a.c.) electrical power systems, the term "power system
stability" refers to a state in which all of the system's synchronous machines continue to follow
one another.
Different types of system instability have emerged as power systems have developed due to the
ever-increasing number of connections between them, the adoption of novel technologies and
controls, and the growing operation under extreme conditions. For instance, voltage and frequency
stability, as well as inter-area oscillations, are now considered far more severe problems than in
the past. This has prompted reevaluating how power system stability is defined and categorized.
Understanding the various types of instability and how they relate to one another is critical for the
effective layout and operation of power systems. Consistent language helps with developing
system design and operational requirements, utilizing standard analytical tools, and conducting
studies.
The steady-state stability of power systems was and remains a fundamental major challenge in
operation of the system. Contemporary electrical power infrastructures have significantly
developed due to growing interconnections, installing massive producing units, extra-high voltage
tie-lines, and other considerations. The power system's ability to reestablish synchronism
following tiny and slow perturbations, such as progressive power adjustments, often termed
steady-state stability [1].
Because of economic and operational constraints, power systems use the most significant
percentage of available transmission capacity and act near stability limitations with minimal
margins. In such a context, voltage instability has substantially threatened security measure of
power system. Several power grids now deploy stabilizing devices, fast-response excitation
systems, and faster relays to strengthen the security margin. Voltage stability and collapse issues
have grown in importance in power systems. [2].
Blackouts have occurred all around the world as a result of several major voltage breakdown
occurrences that utilities have experienced [3].
The fault can happen in any part of the power utility's network, including:
1. Close to the generator.
2. In the middle of the transmission line.
3. Near the infinite bus., etc.
A basic generator model comprises generators that behave as constant voltage sources for steadystate stability concerns.
The power system is intended to provide continuous power while maintaining voltage stability.
However, due to unfavorable events such as lightning, accidents, or other sudden events, a short
circuit between the transmission line's phase wires or between the phase and ground wires that
may occur is referred to as a fault. One or more generators may be severely disrupted due to an
error, resulting in an imbalance between generation and demand. If the problem persists and is not
resolved within the specified time frame, it can cause serious damage to the equipment, resulting
in power loss and a blackout. As a result, protective equipment is installed to detect faults and
clean / isolate the faulty portion of the power system as soon as possible before the disturbance
energy spreads throughout the system.
The core purpose of this stability investigation is to ascertain if the machine's rotors are reluctant
to resume constant speed operation. This causes a short but intense disconnect between the
synchronous speeds of the rotors. Since the last thirty years, the employment of power system
stabilizers (PSSs) has become increasingly common to boost the damping of a system's lowfrequency oscillations. PSSs are being employed by power companies worldwide as reliable
excitation controllers to improve system stability in the face of a variety of failure scenarios. This
paper aims to look into environmental factors in the stability analysis of power systems under
various fault scenarios.
Stability studies aid in determining circuit breaker critical clearing times, voltage levels, and
system transfer capabilities. The Transient Stability Study aims to improve the reliability of the
systems under study by minimizing the damaging effects of transients, which frequently occur in
power systems, using industry-standard mitigation methods.
II.
POWER SYSTEM STABILITY (PSS): WHAT IT MEANS
An electric power grid's capacity to recover to an operational state shortly after a physical
disturbance occurs while keeping the majority of system variables constrained to ensure that the
system is mostly unaffected is what is meant by "stability." Interruption criteria include load
changes, faults, line outages, generator outages, voltage collapse, and combinations. Power system
instability is classically defined as synchronous machines getting out of phase when a disturbance
hits.
III.
STABILITY CLASSIFICATION FOR POWER SYSTEM
The dynamic response of power systems in use today is a high-order multivariate procedure that
is affected by the large number of devices involved. Stability occurs when opposing forces are in
equilibrium with one another. Instability can manifest itself in a number of different ways, based
on the network's topology, the condition of the system's operation, and the nature of the
disturbance. Here, we propose a methodical framework for categorizing the stability of power
systems.
For categorizing the stability of power systems in accordance with the following factors, which
have been proposed below:

The fundamental system parameter that can be used to observe the instability, and thus the
resulting physical properties of the instability.

The degree of the perturbation is analyzed, which influences the calculating methods and
stability prediction.

In order to determine stability, the devices, procedures, and period must be considered.
Fig 1- Stability Classification for Electric Power Systems
The stability of system is primarily determined by synchronous machine behavior following a
fault. The power system stability can be broadly categorized into two subtypes, distinguished by
the severity of the disturbance they experience.
1. Steady-State Stability: It refers to a synchronous machine's capacity to continue operating
while subjected to a steadily rising load. Despite minor disruptions, the system's engine
and the external tie line can remain synchronized (load fluctuation, turbine regulator,
voltage regulator).
The objective is to determine the maximum load that may be applied to a machine without
causing a disruption in synchronization, considering that the load has been steadily raised.
The maximum power that can be delivered to the system without compromising the
system’s stability is defined as the steady-state stability limit.
2. Transient
Stability: Transient stability necessitates responding to substantial
disturbances, which can significantly change power angles, rotor speeds, and power
transfers. In most cases, a transient stability phenomenon will occur in less than a few
seconds. The maximum amount of power a system can transfer before it becomes unstable
in response to a sudden or big change is called its "transient stability limit.
Typically, the mechanical and electromagnetic qualities of synchronous machines and the
impedance of the circuits linking them are the only determinants of a system's transient
stability. No thought is given to how the excitation or control system will react to problems
that change the generator's speed or output of electricity.
Transient stability is usually thought of as the ability of a system to stay in sync during an
interference event. Studies of transient stability are typically conducted assuming that the
time constants of the excitations and the prime mover are significantly larger than the
duration of the disturbance that triggers off the instability.
Studies of transient stability focus on the seconds right after a disturbance, when the system
is most likely to be damaged by things like the loss of generation, line-switching activities,
faults, and sudden changes in load. Following the interruption, the synchronous machine's
frequency will temporarily diverge from the synchronous frequency. The purpose of
transient stability studies is to determine how quickly the machine will reestablish its
synchronous frequency after an interruption.
IV.
SWING EQUATION
During normal functioning, the rotor's axis is always perpendicular to the magnetic field's axis.
Power angle, often called torque angle, is the angle between them. Because the air-gap mmf is
rotating at a steady pace, any disturbance will cause the rotor to either slow down or accelerate up.
The swing equation represents the motion of two points in relation to one another. If, after this
period of oscillation, the rotor locks back into synchronous speed, the generator will remain stable.
If the disturbance causes no net change in power, the rotor will return to its previous position. A
new working power angle between the rotor and the synchronously rotating field is created if the
disturbance is caused by a change in generation, load, or network conditions.
A mathematical expression for the swing equation is written as
 d 2
 Pm  Pe
180 f 0 dt 2
V.
(1)
STABILITY STUDIES USING SYNCHRONOUS MACHINE MODELS
The classical model is the simplest for stability analysis because it ignores saliency and represents
the machine as a constant voltage E' behind the direct axis transient reactance X’ d.
Fig 2- Single machine coupled with the infinite bus
As demonstrated in Figure 2, a transmission line runs from a generator to a huge substation in a
vast power grid.
The voltage as well as frequency of the substation bus are considered to be constant. This is
referred to as an endless bus because its properties stay constant independent of the power supplied
or used by any device connected to it. The generator is represented by a constant voltage behind
the direct axis transient reactance X. The node indicating the generator terminal voltage V can be
deleted by changing the Y-connected impedances to an equivalent delta. Admittances are given by
y10 
ZL
jX Z s  jX d' Z s  Z L Z s
y20 
jX d'
jX d' Z s  jX d' Z L  Z L Z s
y12 
Zs
jX Z s  jX d' Z L  Z L Z s
'
d
(2)
'
d
Fig. 3 is an equivalent circuit showing the internal voltage at node 1 and the infinite bus at node 2.
The nodal equations are
I1  ( y10  y12 ) E '  y12V
I 2   y12 E '  ( y20  y12 )V
Fig 3: Single machine's equivalent circuit coupled with the infinite bus
In terms of the bus admittance matrix, the above equations are represented as follows:
(3)
 I1   Y11 Y12   E ' 
 
 
 I 2   Y21 Y22   V 
Where the bus admittance matrix has diagonal entries Y11  y10  y12 and Y22  y20  y12 .
Y12  Y21   y12 is the off-diagonal elements.
Z L and Z S are considerably inductive in most systems. When all resistances are ignored,
11  12  90 , Y12  B12  1/ X12 , yielding a simplified mathematical equation for power.
Pe 
E' V
X 12
sin 
(4)
All stability issues can be reduced to this most elementary form of the power flow equation. This
relationship reveals that the transfer reactance, as well as the angle formed by the two voltages are
critical components in determining the amount of power transferred. Figure 4 depicts the power
angle curve, which relates to Pe &  .
Fig 4: Power-angle curve
The power output of the generator can be raised incrementally until the maximum amount of
electricity is sent. At an angular displacement of 90 , the steady-state stability limit is attained,
and maximum power is produced.
Pmax 
E' V
X 12
(5)
Increasing the shaft input by advancing  beyond the Pmax point will result in a drop in electrical
power output. Acceleration causes the machine to become out of phase with the infinite bus bar.
An Expression for Electric Power in Terms of Pmax
Pe  Pmax sin 
(6)
The transient reactance X d' of a generator controls the maximum current that can flow during a
short circuit. Therefore, the machine is represented by the voltage E behind the reactance X d' for
transient stability concerns where saliency is disregarded. Assuming that Vg is the terminal voltage
of the generator and I a is the steady-state current of the generator before the fault, we may
calculate E prime as
E '  Vg  jX d' I a
(7)
Due to the low resistance of the field winding, it is assumed that the voltage E' remains stable
during the onset of the disturbance. The steady-state and transient power-angle curves look similar,
but the transient peak is higher.
VI.
STEADY-STATE STABILITY STUDY WITH SMALL DISTURBANCES
The ability of the power system to maintain synchronism in the face of minor disturbances is
referred to as "steady-state stability." It is convenient to assume that the disturbances that cause
the changes vanish. The system's motion is free, and stability is guaranteed if the system returns
to its original state. In a linear system, such behavior can be determined by examining the system's
characteristic equation. The automatic controls, such as the voltage regulator and governor, are
assumed to be inactive.
To demonstrate the problem of steady-state stability, we will investigate the dynamic behavior of
a one-machine system connected to an infinite bus bar, as illustrated in Figure 2.
Electrical power, Pe  Pmax sin 
(8)
Substituting the electrical power from eq (8) into the swing equation eq (1) given in results in
 d 2
 Pm  Pmax sin 
 f 0 dt 2
(9)
The swing equation is a nonlinear power-angle-dependent function. The swing equation can be
linearized for minor disturbances with only a slight loss of accuracy, as shown below.
Consider a minor power angle deviation  from the initial operating point 0 , i.e.
   0  
(10)
Substituting in eq (9), we get:
 d 2 ( 0   )
 Pm  Pmax sin( 0   )
 f0
dt 2

 d 2 ( 0   )
 Pm  Pmax (sin  0 cos   cos  0 sin  )
 f0
dt 2
Since  is acute, cos   1 and sin    , we have
 d 2 0
 d 2 

 Pm  Pmax sin  0  Pmax cos  0 
 f 0 dt 2  f 0 dt 2
During the initial state of operation
 d 2 0
 Pm  Pmax sin  0
 f 0 dt 2
In terms of small changes in the power angle, the above equation simplifies to the linearized
equation, which is:
 d 2 
 Pmax cos  0   0
 f 0 dt 2
(11)
The parameter Pmax cos  0 in the eq (11) represents the slope of the power-angle curve at 0 . It is
referred to as the synchronizing coefficient, and its symbol is  s . This coefficient is critical in
determining the stability of the system and is demonstrated by the expression
s 
d
|  Pmax cos  0
d 0
(12)
Substituting eq (12) in eq (11)
 d 2 
  s   0
 f 0 dt 2
(13)
The solution of the differential equation of second order presented above is dependent on the roots
of the characteristic equation given by
s2  
 f0

s
(14)
For negative  s values, the system is no longer stable, the response increases exponentially, and
only one root is located in the right half of the s-plane. For positive  s values, two roots are located
on the j-w axis, and the motion is undamped and oscillatory. The system is marginally stable, and
its natural frequency of oscillation is determined by
n 
 f0

s
Figure 4 reveals that the range of values for  s (i.e., the slope
(15)
d
) that are positive lies between
d
0 and 90 degrees, with the maximum value occurring when there is no load (  0  0 ).
Induction motor activity occurs between the rotor and the resulting rotating air gap field whenever
there is a difference between their angular velocities. A torque is generated to the rotor to reduce
that difference. This is known as the damping torque. Damping power is roughly proportional to
speed deviation.
d  D
d
dt
(16)
The damping coefficient D can be calculated from either design information or experimental
results. We also ignore the speed and torque characteristics of the prime mover and the load
dynamics, which contribute to additional damping torques. The oscillations will die down, and the
operation will return to the equilibrium angle due to the damping power when the synchronizing
power coefficient P is positive. There is no synchronization loss, and the system remains stable.
Once dampening is taken into consideration, the linearized swing equation is simplified to

 d 2 
d 
D
  s   0
2
 f 0 dt
dt
(17)
d 2   f 0 d   f 0

D

 s   0
dt 2

dt

(18)
In terms of the standard second-order differential equation, we have
d 2 
d 
 2n
 s2   0
2
dt
dt
(19)
where n is the natural frequency of oscillation, which is given by eq (15), and  is dimensionless
damping ratio, calculated by
 
D  f0
2  s
(20)
Characteristic equation is given by
s 2  2n s  s2  0
(21)
Under normal operating conditions,  
D  f0
 1 , the roots of the characteristic equation are
2  s
complex
s1 , s2  n  jn 1  
2
(22)
 n  jd
where d is the damped frequency of oscillation given by
d  n 1  
2
(23)
It is evident that if the synchronizing power coefficient,  s , is positive, the roots of the
characteristic equation will have a negative real part for positive damping. The system is stable,
and the response is constrained.
VII.
DESCRIPTION OF THE SYSTEM FOR STEADY-STATE STABILITY STUDY
As shown in Fig. 1, the system consists of a 60Hz synchronous generator with inertia constant H
= 9.94 MJ/MVA and a transient reactance X' d =0.3 pu connected to an infinite bus through a
purely reactive circuit. The generator delivers real power of 0.6 pu, 0.8 PF lagging to the infinite
bus at a V = 1 pu. The pu damping power coefficient is assumed D = 0.138. Considering a minor
disturbance of Δ∂ =10° =0.1745 radian, the equations describing the motion of the rotor angle and
the generator frequency are obtained.
Fig. 5- One line diagram of the system
Transfer reactance between the generated voltage and the infinite bus is
X  0.3  0.2 
0.3
 0.65
0.2
The per unit apparent power is, S 
Current,  
0.6
 cos 1 0.8  0.75  36.87 
0.8
S * 0.75   36.87 

 0.75   36.87 
V*
1.0  0 
Excitation Voltage,  '  V  jX   1.0  0   ( j 0.65)(0.75   36.87 )  1.35 16.79 
Initial operating power angle  16.79  0.2931 radians.
Synchronizing power coefficient   s   max cos  0 
(1.35)(1)
 1.9884
0.65
Undamped angular frequency of oscillation and damping ration
 fo
n 
 

s 
 fo
D
2
s

 (60)
(9.94)
1.9884  6.1405 rad/sec
0.138
 (60)
 0.2131
2
(9.94)(1.9884)
Linearized force -free equation which determines the mode of oscillation is
d 2 
d 
 2.62
 37.7  0 , where  is in radians
2
dt
dt
Damped angular frequency of oscillation is
d  
n
1   2  6.1405 1  (0.2131) 2  6.0 rad/sec
Corresponding to a damped oscillation of frequency of
fd 
60
 0.9549 Hz
2
The equations describing the rotor's motion relative to the synchronously rotating field in electrical
degrees and the angular frequency deviation in Hz are given by
  16.79  10.234e1.3t sin(6.0t  77.6966)
f  60  0.1746e 1.3t sin 6.0t
MATLAB SIMULATION
Fig. 6- Natural machine responses to rotor angle and frequency of Fig 5
RESULT
The above response demonstrates that the rotor would oscillate, or swing, relatively slowly due to
damping after a little perturbation, before returning to steady state functioning at synchronous
speed. In a matter of seconds, the response will settle. The frequency of the oscillations is on the
order of 0.955 Hz, which is relatively low. A second-order differential equation or a twodimensional state equation resulted from modeling a single-machine system with passive control
devices.
VIII. TRANSIENT STABILITY STUDIES
The capacity of the electrical grid to recover from a major disturbance, for example, a short circuit
on a transmission line, while maintaining synchronization, is known as transient stability. Disasters
like cascade failures and widespread blackouts can result from a breakdown in transient stability.
Therefore, transient stability preservation is fundamental to the proper functioning of power
systems. Transient stability studies examine whether or not synchronism is maintained after a
major disturbance to the machine. Possible causes include unexpectedly high demand, a drop in
generation, the removal of a substantial load, or a breakdown in the system. Most disturbances
have oscillations that are too large to be linearized, necessitating the solution of the nonlinear
swing equation.
IX.
EQUAL AREA CRITERION
For a rapid assessment of stability, the equal-area criterion can be utilized. This method uses a
graphical visualization of the energy stored in the rotating mass to determine whether the machine
retains its stability after a disruption. The process only applies to a one-machine or two-machine
system connected to an infinite bus. Because it provides detailed insight into the dynamic behavior
of the machine, the approach is considered here for the analysis of a single machine connected to
an extensive system.
Consider a synchronous machine coupled to an infinite bus. The swing equation without damping
typically stated in eq (1)
 d 2
  m  e   a
 f 0 dt 2
where a represents the acceleration power. We can deduce from the preceding equation that
d 2  f 0

( m   e )
dt 2

Multiplying both sides of the above equation by 2
2
d
, we get
dt
d  d 2 2 f 0
d

( m   e )
2
dt dt

dt
This can also be expressed as
2
d  d    2 f 0
d
( m   e )

 
dt  dt  

dt
 d   2  2 f 0
 d 
( m   e ) d 
 
dt





Integrating both sides, we get
2
2 f 0
 d 

 

 dt 

 (
m
  e )d
0

2 f 0
d


( m   e ) d 
dt
 0
(24)
Equation (24) calculates the machine's relative speed in relation to the synchronously revolving
reference frame. This speed must become zero at some point following the disruption for stability.
As a result of (24), we have for the stability requirement,

 (
m
 e )d  0
(25)
0
Figure 7 depicts the machine working at the equilibrium point 0 , which corresponds to the
mechanical power input m 0  e0 . Let's say the horizontal line represents an input power that
suddenly increases by m1 . Because m1  e0 , the rotor's accelerating power is increasing, and so
is the power angle 5. The initial acceleration surplus energy stored in the rotor is

 (
0
m
  e )d   area abc = area 1
(26)
Increases in  result in a corresponding rise in electrical power, and when   1 equals the new
input power m1 , no further power gain is seen. In spite of the fact that the accelerating power is
currently zero, the rotor is operating at a speed greater than synchronous speed, so both  &
electrical power e , are increasing. Now that m  e , the rotor will slow down until it reaches
  1 , which is the synchronous speed. According to (11.79), the rotor has to make a full swing
through point b before the rotating masses give up an equal amount of energy. The power lost by
the rotor when it slows to a synchronous velocity
 max

( m1  e )d  area bde = area  2
(27)
1
As a result, the rotor will swing to position b at an angle of &max.
area 1  area 2
(28)
Is therefore called as the equal-area criterion. If allowed to return to its normal frequency, the rotor
angle will oscillate between 0 and  max . The machine's dampening will eventually cause the
oscillations to die down, and a new steady state would be created at b .
Fig 7- Equal-area criterion with a sudden change in load
X.
APPLICATION TO THREE-PHASE FAULT
In Figure 8, a generator is linked to an infinite bus bar by means of two parallel lines. Consider the
case when the input power Pm is held constant and the machine is running smoothly, supplying
power to the system at an angle of  0 as depicted in Figure 9. Bus 1 experiences a temporary threephase bolted fault at the transmitting end of one of the lines.
Fig 8 - Three-phase fault at F in a single machine system connected with an infinite bus
Power to the infinite bus is stopped when the fault happens at the sending end of the line, at point
F. When resistances are ignored, the electrical power Pe is zero, as is the horizontal axis of the
power-angle plot. With the full input power serving as the accelerating power, the machine speeds
up, accumulating more kinetic energy and raising the angle by a factor of  . Both lines are
believed to be functional once the issue has been fixed. Once the error is fixed at 1 , the process
returns to the initial power-angle curve at position e. When the shaded area (defg), represented by
A2, equals the shaded area (abcd), given by A1, the net power is now reducing, and the stored
kinetic energy will be null at point f. The rotor continues to slow down as it follows the powerangle curve back via points e and a, as P is still bigger than Pm . The rotor's angle would then bob
up and down at its fundamental frequency, or delta 0. The inherent damping causes oscillation to
die down and the operating point to realign with the power angle  0 .
Fig 9- Three-phase fault at the sending end fulfils the equal-area criterion
With any additional increase in 1 , the area A2 representing decelerating energy becomes smaller
than the area representing accelerating energy, and the critical clearing angle is reached. This is
depicted in Figure 10, where the maximum value of delta occurs at the intersection of the line Pm
and the curve Pe . In Figure 10, we use the equal-area criterion to get
c
 max
 P d  
m
0
( Pm sin   Pm ) d 
c
We have integration on both sides
Pm ( c   0 )  Pmax (cos  c  cos  max )  Pm ( max   c )
When we solve for  c , we get
cos  c 
Pm
( max   c )  cos  max
Pmax
(29)
Fig 10- Equal-area criterion for critical clearing angle
By using the equal-area criterion, we determined the critical cleaning angle for the machine to
remain stable. An analytical solution for critical clearing time can be derived in the special scenario
where the electrical power Pe is zero at the time of the fault. During a fault with Pe  0 , the swing
equation given by eq (1) becomes
H d 2
 Pm
 f 0 dt 2

f
d 2
 Pm 0
2
dt
H
We have integration on both sides
f
d  f0


Pm  dt  0 Pmt
dt
H
H
0
t
Again Integrating, we have

 f0
H
Pm t 2   0
As a result, if  c is the critical clearing angle, then the critical clearing time is
tc 
2 H ( c   0 )
 f0 Pm
(20)
Fig 11- Three-phase fault at F in a single machine system connected to an infinite bus
Figure 11 depicts a fault location F that is mainly found some distance from the sending end. Power is
supplied to the system at an angle of  0 , as illustrated in Figure 12, assuming that the input power Pm is
constant and the machine is functioning efficiently. Curve A represents the power-angle curve for the
state before the fault occurred.
Fig 12 – Equal-area criterion for the three-phase fault far from the sending end.
Because of the increased equivalent transfer reactance between bus bars when the fault is located
at F, far from the sending end, the power transfer capability is reduced, and the power-angle curve
is represented by curve B. Finally, if the faulty line is eliminated, the power-angle curve depicted
by curve C is what remains. The operational point moves to point b on curve B as soon as the
three-phase fault occurs. When the mechanical input exceeds the electrical output, the rotor spins
faster than usual, storing the extra kinetic energy as angular momentum. At 6, it is assumed that
the issue has been repaired by removing the line that was damaged. As a result, the operating point
on curve C is shifted abruptly to e. Now, the net power is going down, and the kinetic energy that
had been stored will be gone at point f when the shaded area (defg) is the same as the shaded area
(abcd). Since Pe is still bigger than Pm , the rotor keeps slowing down, and the power-angle curve
is followed back to point e. The angular position of the rotor will then cycle back and forth around
point e at its intrinsic frequency. The dampening of the machine will reduce these oscillations, and
a new steady state operation will be established at the point where curves C and Pm intersect.
The critical clearing angle is attained when the area representing decelerating energy becomes
smaller than the area representing accelerating energy. As seen in Figure 13, this occurs when max,
or point f, connects line and curve C.
Fig 13– Equal-area criterion for critical clearing angle
When we apply the equal-area condition to Figure 13, we get
c
 max
0
c
Pm ( c   0 )   P2 max sin  d  

P3max sin  d   Pm ( max   c )
By integrating both sides, and then solving for  c , we get
cos  c 
XI.
Pm ( max   0 )  P3max cos  max  P2max cos  0
P3max  P2 max
DESCRIPTION OF THE SYSTEM FOR TRANSIENT STABILITY STUDY
As shown in Fig. 3, the system consists of a 60Hz synchronous generator with inertia constant H
= 5 MJ/MVA and a transient reactance X' d =0.3 pu connected to an infinite bus through a purely
reactive circuit. The generator delivers real power of P e=0.8 pu and Q=0.074 PF per unit to the
infinite bus at a V = 1 per unit voltage.
Fig. 14- One Line Diagram of the system
One of the lines experiences a three-phase fault in the middle; the fault is repaired, and the
concerned line is cut off. Both the critical fault clearing time and the clearing angle were evaluated.
The current entering the infinite bus is

S* 0.8  j 0.074

 0.8  j 0.074 pu
V*
1.00
In the absence of a fault, the reactance of the internal voltage to the infinite bus is
X1  0.3  0.2 
0.3
 0.65
2
The transient internal voltage is given by
E '  V  jX1I  1.0  ( j 0.65)(0.8  j 0.074)  1.1726.387 pu
The power-angle curve before the fault occurs is the same as previously, as demonstrated by
P1max  1.8sin 
The initial operating angle for the generator is calculated as
1.8sin  0  0.8
  0  26.388  0.46055 rad
The circuit in Figure 15 is the result of a fault on one line, located at point F. If the Y-circuit ABF
can be transformed into a junction C, then the transfer reactance during fault may be calculated
with relative ease. For the final circuit, see Figure 16.
Fig 15- Equivalent circuit of a three-phase fault in the center of the line
Fig 16- After a Star-Delta transformation, the equivalent circuit
Equivalent reactance between generator and infinite bus is given by
X2 
(0.5)(0.3)  (0.5)(0.15)  (0.3)(0.15)
 1.8 pu
0.15
As a result, the power-angle curve during a fault is given by
P2max sin  
(1.17)(1.0)
sin   0.65sin 
1.8
After the fault has been repaired, the affected line can be cut off. Therefore, the transfer reactance
after this fault is
X 3  0.3  0.2  0.3  0.8 pu
The Power-angle curve is given by
P3max sin  
(1.17)(1.0)
sin   1.4625sin 
0.8
In reference to Figure 13
 max  180  sin 1 (
0.8
)  146.838  2.5628 rad
1.4625
The critical clearing angle can be calculated as
cos  c 
0.8(2.5628  0.46055)  1.4625 cos146.838  0.65cos 26.388
14625  0.65
 0.15356
As a result, the essential clearing angle is
 c  cos 1 (0.15356)  98.834  1.725 rad
The critical clearing time is given by
2 ( c   0 )
 f0 m
tc 
2(5)(1.7250  0.4605)
 0.289  0.30 s
 (60)(0.8)
 tc 
(a) There is a three-phase fault in the middle of one of the lines, and it is cleared by isolating
the faulted circuit at both ends at once. In 0.3 seconds, the fault is cleared. By employing
the modified Euler technique (swingmeu) with a step size of 0.01 seconds, the numerical
solution to the swing equation with a period of 1.0 seconds is found. The swing curve is a
measure of how stable the system is.
From the above calculation, we can get the power-angle curve before to the fault occurrence
P1max  1.8sin 
While the generator has been operating at the given initial power angle
 0  26.388  0.46055 rad
0  0
The circuit in Figure 15 is the result of a fault on one line, located at point F. The accelerating
power equation can be derived as
Pa  0.8  0.65sin 
The initial derivatives when using the modified Euler's method are
d
dt
0
 0
d 
dt

 (60)
5
0
(0.8  0.65sin 26.388)  19.2684 rad/sec 2
Predicted values after the first stage (t1  0.01) are
1p  0.46055  (0)(0.01)  0.46055 rad = 26.388
1p  0  (19.2684)(0.01)  0.1927 rad/sec
The derivatives at the end of the interval are calculated by using the expected value of 1p , and
1p :
d
dt
 1p  0.1927 rad/sec
1p
d  
 (60)

(0.8  0.65sin 26.388)  19.2684 rad/sec 2
dt 1p
5
The corrected value is then determined by taking the average of the two derivatives.
1c  0.46055 
1c  0.0 
0  0.1927
(0.01)  0.4615 rad
2
19.2684  19.2684
(0.01)  0.1927 rad/sec
2
The operation is repeated for each succeeding step until the fault is cleared at t  0.3 s . The
postfault accelerating power equation is given by
Pa  0.8  1.4625sin 
The operation is repeated with the new accelerating equation until the final time t f  1.0 s is
reached.
MATLAB SIMULATION
Fig. 17- Swing Curve for machine of Fig. 14. At 0.3s, the fault was cleared.
RESULT: The obtained swing curve demonstrates that the power angle returns after a maximum
swing, suggesting that the oscillations will decrease and a new operating angle will be attained
upon integrating system damping. As a result, the system is stable during the fault-clearing time.
The critical clearing angle is 98.83 degrees, and the critical clearing time is 0.4 seconds.
(b) By running the simulation over and over, swing plots can be made for the critical clearing
time and when the fault is fixed in 0.5 seconds.
MATLAB SIMULATIONS
Fig. 18- Swing Curve for machine of Fig. 14. At 0.4s, the fault was cleared.
RESULT: In this case, tc = 0.4 seconds, as shown by the swing curve, is the critical clearing time.
Fig. 19- Swing Curve for machine of Fig. 14. At 0.5s, the fault was cleared.
RESULT: As seen in the swing curve at tc = 0.5 seconds, the power angle the power angle 
rises exponentially. As a result, the system is unstable during this clearing time.
REFERENCES
1. H. Saadat, Power System Analysis. McGraw-Hill, 2002.
2. J. G. Calderon-Guizar, G. A. Inda-Ruiz, and G. E. Tovar,” Improving the steady-state
loading margin to voltage collapse in the north-west control area of the Mexican power
system,” Electrical Power and Energy Systems, vol. 25 (8), pp. 643–649, Oct. 2003.
3. N. Amjady and M. Esmaili, “Improving voltage security assessment and ranking
vulnerable buses with consideration of power system limits,” Electrical Power and Energy
Systems, vol. 25, pp. 705–715, Nov. 2003.
4. Steady State Stability Analysis of Power System under Various Fault Conditions by Md
Multan Biswas, Kamol Kanto Das Bangladesh University of Engineering and Technology
(BUET), Dhaka, Bangladesh
APPENDIX
MATLAB CODES
% STEADY-STATE STABILITY- SMALL DISTURBANCES
clear all
clc
E=1.35; V=1.0; H=9.94; X=0.65; Pm=0.6; D=0.138; f0=60;
Pmax= E*V/X; d0=asin(Pm/Pmax); % Max. power
Ps = Pmax*cos(d0); % Synchronizing power coefficient
wn = sqrt(pi*60/(H*Ps));% Undamped frequency of oscillation
z = D/2*sqrt(pi*60/(H*Ps));% Damping ratio
wd=wn*sqrt(1-z^2); fd=wd/(2*pi); %Damped frequency oscillation
tau=1/(z*wn); % Time constant
th= acos(z); % Phase angle theta
Dd0=10*pi/180; % Initial angle in radian
t = 0:0.01:3;
Dd= Dd0/sqrt(1-z^2)*exp(-z*wn*t).*sin(wd*t + th);
d = (d0+Dd)*180/pi; % Power angle in degree
Dw=-wn*Dd0/sqrt(1-z^2)*exp(-z*wn*t).*sin(wd*t);
f = f0+ Dw/(2*pi); % Frequency in Hz
subplot(2,1,1), plot(t, d), grid
title('Natural Response of the rotor angle for machine')
xlabel('t (sec)'), ylabel('Delta (degree)')
subplot(2,1,2), plot (t, f), grid
title('Natural Response of the frequency for machine')
xlabel('t (sec)'), ylabel('Frequency (Hz)')
subplot(111)
MATLAB CODE FOR TRANSIENT STABILITY
% TRANSIENT STABILITY- THREE-PHASE FAULT
clc
global Pm f H E V X1 X2 X3 tc tf Dt
Pm = 0.80; E = 1.17; V = 1.0;
X1 = 0.65; X2 = 1.80; X3 = 0.8;
H = 5.0; f = 60; tf = 1; Dt = 0.01;
%(a) Fault is cleared in 0.3 sec.
tc = 0.3;
swingmeu(Pm, E, V, X1, X2, X3, H, f, tc, tf, Dt)
%(b) Fault is cleared in 0.4 sec.
tc = 0.4;
swingmeu(Pm, E, V, X1, X2, X3, H, f, tc, tf, Dt)
%(c) Fault is cleared in 0.5 sec.
tc = 0.5;
swingmeu(Pm, E, V, X1, X2, X3, H, f, tc, tf, Dt)
disp('Parts (a) & (b) are repeated using swingrk4')
disp'Press Enter to continue')
pause
tc = 0.3;
swingrk4(Pm, E, V, X1, X2, X3, H, f, tc, tf)
tc = .5;
swingrk4(Pm, E, V, X1, X2, X3, H, f, tc, tf)
tc = .41;
swingrk4(Pm, E, V, X1, X2, X3, H, f, tc, tf)
% This program solves the swing equation of a one-machine system
% when subjected to a three-phase fault with subsequent clearance
% of the fault. Modified Euler method
function swingmeu(Pm, E, V, X1, X2, X3, H, f, tc, tf, Dt)
%global Pm f H E V X1 X2 X3
clear t
if exist('Pm')~=1
Pm = input('Generator output power in p.u. Pm = '); else, end
if exist('E')~=1
E = input('Generator e.m.f. in p.u. E = '); else, end
if exist('V')~=1
V = input('Infinite bus-bar voltage in p.u. V = '); else, end
if exist('X1')~=1
X1 = input('Reactance before Fault in p.u. X1 = '); else, end
if exist('X2')~=1
X2 = input('Reactance during Fault X2 = '); else, end
if exist('X3')~=1
X3 = input('Reactance after Fault X3 = '); else, end
if exist('H')~=1
H = input('Generator Inertia constant in sec. H = '); else, end
if exist('f')~=1
f = input('System frequency in Hz f = '); else, end
if exist('Dt')~=1
Dt = input('Time interval Dt = '); else, end
if exist('tc')~=1
tc = input('Clearing time of fault in sec tc = '); else, end
if exist('tf')~=1
tf = input('Final time for swing equation in sec tf = '); else, end
Pe1max = E*V/X1; Pe2max=E*V/X2; Pe3max=E*V/X3;
clear t x1 x2 delta
d0 =asin(Pm/Pe1max);
t(1) = 0;
x1(1)= d0;
x2(1)=0;
np=tf /Dt;
Pemax=Pe2max;
ck=pi*f/H;
for k = 1:np
if t(k) >= tc
Pemax=Pe3max;
else, end
t(k+1)=t(k)+Dt;
Dx1b=x2(k);
Dx2b=ck*(Pm-Pemax*sin(x1(k)));
x1(k+1)=x1(k)+Dx1b*Dt;
x2(k+1)=x2(k)+Dx2b*Dt;
Dx1e=x2(k+1);
Dx2e=ck*(Pm-Pemax*sin(x1(k+1)));
Dx1=(Dx1b+Dx1e)/2;
Dx2=(Dx2b+Dx2e)/2;
x1(k+1)=x1(k)+Dx1*Dt;
x2(k+1)=x2(k)+Dx2*Dt;
end
delta=180*x1/pi;
clc
fprintf('\nFault is cleared at %4.3f Sec. \n', tc)
head=['
'
'
time
'
s
delta
Dw '
degrees
rad/s'
'
'];
disp(head)
disp([t', delta' x2'])
h=figure; figure(h)
plot(t, delta), grid
title(['One-machine system swing curve. Fault cleared at ', num2str(tc),'s'])
xlabel('t, sec'), ylabel('Delta, degree')
cctime(Pm, E, V, X1, X2, X3, H, f)
% Obtains the critical clearing time
% This program solves the swing equation of a one-machine system
% when subjected to a three-phase fault with subsequent clearance
% of the fault.
end
function swingrk4(Pm, E, V, X1, X2, X3, H, f, tc, tf, Dt)
%global Pm f H E V X1 X2 X3
if exist('Pm') ~= 1
Pm = input('Generator output power in p.u. Pm = '); else, end
if exist('E') ~= 1
E = input('Generator e.m.f. in p.u. E = '); else, end
if exist('V') ~= 1
V = input('Infinite bus-bar voltage in p.u. V = '); else, end
if exist('X1') ~= 1
X1 = input('Reactance before Fault in p.u. X1 = '); else, end
if exist('X2') ~= 1
X2 = input('Reactance during Fault X2 = '); else, end
if exist('X3') ~= 1
X3 = input('Reactance after Fault X3 = '); else, end
if exist('H') ~= 1
H = input('Generator Inertia constant in sec. H = '); else, end
if exist('f') ~= 1
f = input('System frequency in Hz f = '); else, end
if exist('tc') ~= 1
tc = input('Clearing time of fault in sec tc = '); else, end
if exist('tf') ~= 1
tf = input('Final time for swing equation in sec tf = '); else, end
Pe1max = E*V/X1; Pe2max=E*V/X2; Pe3max=E*V/X3;
clear t x delta
d0 =asin(Pm/Pe1max);
t0 = 0;
x0 = [d0; 0];
tol=0.001;
%[t1,xf]=ode45('pfpower', t0, tc, x0, tol); % During fault solution (use with MATLAB 4)
tspan = [t0; tc];
% use with MATLAB 5
[t1,xf]=ode45('pfpower', tspan, x0); % During fault solution (use with MATLAB 5)
x0c =xf(length(xf), :);
%[t2,xc] =ode45('afpower', tc, tf, x0c, tol); % After fault solution (use with MATLAB 4)
tspan = [tc, tf];
% use with MATLAB 5
[t2,xc] =ode45('afpower', tspan, x0c); % After fault solution (use with MATLAB 5)
t =[t1; t2]; x = [xf; xc];
delta = 180/pi*x(:,1);
clc
fprintf('\nFault is cleared at %4.3f Sec. \n', tc)
head=['
'
'
time
'
s
delta
Dw '
degrees
rad/s'
'
'];
disp(head)
disp([t, delta, x(:, 2)])
h=figure; figure(h)
plot(t, delta), grid
title(['One-machine system swing curve. Fault cleared at ', num2str(tc),'s'])
xlabel('t, sec'), ylabel('Delta, degree')
cctime(Pm, E, V, X1, X2, X3, H, f)
end
% Obtains the critical clearing time