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Friends, during this semester we will be studying
a course on power system dynamics.
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This power system dynamics is also popularly
known as power system stability. The modern
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power systems are very widely interconnected
while this interconnections result in operating
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economy and reliability through mutual assistance.
They also contribute to the stability problem,
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what I am trying to emphasize here is that
the power systems are widely interconnected
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and the stability problem has become a very
important challenge to power system engineers
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because of this large scale interconnections.
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The powers for power system stability analysis
the mathematical model of the system is required
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to be developed. The mathematical model is
a set of nonlinear differential equations
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and a set of algebraic equations, for this
nonlinear differential equations and set of
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algebraic equations there is no formal solution
available. These equations are to be solved
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using non-linear using numerical techniques,
I am sorry using numerical techniques and
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the numerical techniques take lot of time
to solve the equations.
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Over the years the power system stability
has posed a problem to the power system engineers.
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The problem is posed in two respects; the
one is the modeling of the system. To get
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the correct assessment of power system stability
a detail model of the power system need to
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be developed. Once the mathematical model
is correctly developed, one has to obtain
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the solution through numerical techniques.
Because of the large size of the power system,
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the numbers of differential equations are
large in number and therefore solution through
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numerical techniques takes enormous time.
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Historically, historically the stability problem
has been attempted from 1920 onwards. Earlier
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we were not having digital computers and therefore
computations were mainly done using hand calculations
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or in those days slide rules were available
for calculating. Somewhere in 1950 or so the
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analog computers were developed and these
were used for simulating the power system
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stability problem.
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Then in late 1950’s digital computers came
in and the first digital computer program
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for power system stability was developed in
1956. At that time the program was mainly
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to analyze the transient stability of the
system. Over the years another development
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took place that is the implementation or application
of high response excitation systems. The high
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response excitation systems were capable of
improving transient stability of the system
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but with the application of the high response
excitation systems resulted into a problem
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of poor damping of the system oscillations.
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This problem of poor damping of system oscillations
has been overcome by implementing what is
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commonly known as the power system stabilizers.
During this semester we will try to understand,
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the development of mathematical model of the
power system, the mathematical model includes
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the mathematical model for synchronous machine,
excitation systems, voltage regulators, governors
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and loads. For having the correct assessment
of the stability the models need to be accurately
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developed.
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Earlier due to the computational difficulties,
many assumptions were made and therefore the,
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the assessment through computation and the
actual groundility there used to be lot of
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difference. Now, this course we will cover
in about 40 hours the text for, textbook for
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this course is power system stability and
control by Prabha Kundur. This is a very nicely
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written book. Over the years a number of books
have been develop, written a lot of research
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work or research papers are available, from
time to time i will refer to other books and
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research papers which are relevant to our
study.
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The first lesson on power system dynamics
is, introduction to the power system stability
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problem. Today, we will address to these aspects,
basic concepts and definitions classification
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to the power system stability rotor dynamics
and swing equation and swing curve.
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Now let us first define what is power system
stability? Power system stability may be broadly
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defined as that property of a power system
that enables it to remain in state of operating
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equilibrium under normal condition and to
regain an acceptable state of equilibrium
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after being subjected to a disturbance.
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If you carefully look into this definition,
you will find that one has to emphasize on
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ability to remain in operating equilibrium
and second point we have to emphasize is the
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equilibrium between opposing forces. A power
system is subjected to a variety of disturbances,
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it is never in steady state condition. Small
disturbances in the form of load changes continuously
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come, while large perturbations in the form
of faults, tripping of lines, change in large
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load and dropping of generators do come in
the system.
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A power system is designed and operated in
such a fashion. So that it can withstand certain
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probable contingencies. For the purpose of
understanding this problem, the power system
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stability problem is classified into 2 broad
categories, one is called voltage stability
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and second is called, I am sorry one is called
angle stability and second is called voltage
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stability. Earlier years the stability means
angle stability. During the last one decade
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the another type of stability that has come
into picture is the voltage stability.
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The angle stability is further classified
into small signal stability, transient stability,
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mid-term stability and long-term stability.
We will briefly define these different types
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of stability problems and try to understand
what we mean by the different terminologies.
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Voltage stability is also defined into two
broad categories, one is called larger disturbance
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voltage stability and second is small disturbance
voltage stability.
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Now, let us define what is rotor angle stability?
The rotor angle stability is the ability of
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the interconnected synchronous machines of
a power system to remain in synchronism. Here
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the emphasis on ability to maintain synchronism.
This is the primary requirement for the operation
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of a power system where, where all the machines
of the system remain in synchronism.
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Now for the system to remain in synchronism
we have to study the torque balance of synchronous
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machines, that is synchronous machine is the
primary component here, where we have to maintain
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equilibrium between the torque supplied by
the prime over and the electromagnetic torque
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developed by the synchronous generator. To
analyze this power system stability, we have
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to first understand the dynamics of the rotor
and develop a mathematical equation to describe
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the dynamics of the rotor.
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For developing the basic equation, we make
use of the principle of dynamics, elementary
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principle of dynamics as for the rotational
dynamics we all know that the accelerating
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torque is the product of moment of inertia
and angular acceleration that is for any rotating
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body the accelerating torque is equal to the
moment of inertia multiplied by angular acceleration
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and this is the fundamental law on which actually
the swing equation is best. Now as you know
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that synchronous machine may operate as a
synchronous generator or a synchronous motor.
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When we look at the rotor of the synchronous
generator there are two torques which act
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on the synchronous generator rotor are, one
is the mechanical torque which acts on the
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system and another is the electrical torque
or we can call it electromagnetic torque which
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acts on the, these two torques operate or
act on the rotor in opposite direction.
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The mechanical torque is provided by the prime
over and electrical torque is developed due
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to interaction of magnetic field and a stator
currents. The rotor rotates in the direction
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of mechanical torque that is if you just look
here, if I show the direction of rotation,
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it is in the same direction as the mechanical
torque is applied to the system.
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Under steady operating condition, these two
torques are equal and the rotor of the synchronous
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machine rotates at synchronous speed. However,
however when disturbance is occur, there exist
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a unequilibrium between the two torques, these
two torques are not equal and this this difference
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is called accelerating torque.
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When I look at the synchronous motor thus
the driving torque is developed by the flow
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of electric power from the supply and it meets
the load torque, therefore load torque becomes
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the breaking torque while the driving torque
is the electrical torque and the rotor rotates
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in the direction of in which the electrical
torque is developed.
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The swing equation or actually I will the
swing equation is to be derived, a differential
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equation can be written relating the accelerating
torque moment of inertia and acceleration.
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This can be written as J into d 2 theta m
by dt2 equal to Ta, where Ta is difference
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of Tm minus of Te. Tm is positive for generator
operation and T is also positive for generator
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operation, while for motor operation they
take negative signs. When they use mks system
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of units, J the total moment of inertia expressed
is in kilogram meter square, theta m the angular
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displacement of rotor with respect to the
stationary axis in mechanical radians. Now
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I am here emphasizing that in this equation
theta m is measured with respect to a stationary
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axis
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Time is measured in seconds. Tm is the mechanical
or shaft torque supplied by the prime over
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less retarding torque due to rotational losses
in Newton meters, that the unit for the torque
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is Newton meters, this Tm is torque supplied
with the prime over to less rotational losses.
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It is the torque which is available for rotating
the rotor. Similarly, Te is the net electrical
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torque or electromagnetic torque in Newton
meters. Ta is the net accelerating torque
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in Newton meters.
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Now, if you see this equation then this theta
m increases continuously even under steady
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state conditions because theta m is measured
with respect to stationary axis, okay. Now
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instead of measuring the angle with respect
to a stationary axis, the angle can be measured
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with respect to a synchronously rotating axis,
therefore theta m can be defined as omega
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sm into t plus delta m, where omega sm is
the synchronous speed of the machine. This
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is measured in radians per second.
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Delta m is angular displacement of the rotor
in mechanical radians from the synchronously
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rotating reference axis. Now this equation
two, when you when you take the derivatives
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of this equation with respect to time, the
first derivative is written as d theta m by
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dt2 equal to omega sm plus d delta m by dt,
that is we can see the rotor speed is equal
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to synchronous speed plus this additional
torque.
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Now we take the second derivative, d2 theta
m divided by dt2 is equal to d2 delta m by
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dt2, okay. Omega sm being constant its derivative
is 0. Now we substitute the value of rotor
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acceleration in the equation 4, we get expression
as J times d2 delta m by dt2 equal to Ta Tm
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minus Te. Now at this point I want to emphasize
that all the terms in this equation are torque
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terms in Newton meters. In power system studies,
we are more comfortable with the terms in
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power, power may be watts, kilowatts, megawatts
and therefore, what we do is we multiply this
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equation 5 by omega m that is d theta m by
dt.
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Now when we multiply this by this term omega
m, our equation becomes J into omega m d2
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delta m by d2 equal to omega m Tm minus omega
m Te. Now here, this speed into torque, this
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is the power, the speed is measured in radians
per second, power or torque is measured in
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Newton meters this product is power in watts.
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Now we will represent this term omega m Tm
as Pm, omega m Te as Pe, where Pm is shaft
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power input to the machine less rotational
losses. We are emphasizing all the time, the
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term rotational losses here. The meaning here
is that this is not the torque supplied by
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the prime over but it is the prime over torque
minus rotational losses.
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Pe is
the electrical power crossing the air gap
also called air gap power. I mean in this
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equation 7, we substitute the expression for
power and the equation become J into omega
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m d2 delta m by dt2 equal to Pm minus Pe equal
to Pa. Now this term J into omega m, J is
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the moment of inertia in kilogram meter square,
omega m is the speed in radians per second.
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Now this product is called angular momentum.
Now in practical situations, the rotor speed
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omega m is very nearly equal to the synchronous
speed.
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The difference is very small, the difference
may become large only when machine loses synchronism
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right and therefore, what we do is for the
purpose of simplicity, if it represents omega
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m in this equation by this omega sm, then
this product of omega sm and moment of inertia
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J, it will be denoted as the inertia constant
M
and therefore my equation take the shape M
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d2 delta m by dt2 equal to Pm minus Pe, where
m is known as inertia constant.
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Strictly speaking this coefficient should
be J into omega m and since omega m is not
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constant therefore this coefficient term,
strictly is not constant but by assuming making
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this assumption omega m is equal to omega
sm, this coefficient become constant and it
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simplifies our analysis, further the error
incurred by making this simplification is
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negligible. Okay therefore we are justified
in assuming this coefficient of the acceleration
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term as constant and this is called inertia
constant M.
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Now at this stage, I would like to emphasize
that this term M, varies over a wide range
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depending upon size of the machine, type of
machine because we have two main types of
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synchronous generators hydro machines, hydro
generators and turbo generators and these
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two machines have widely different value of
inertia constant M and this inertia constant
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M will be different depending upon the size
of the machine.
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Now to overcome this problem, we define another
term or another inertia constant which is
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denoted by the symbol H. You have to we have
to very clearly understand the definition
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of H, the inertia constant H or the constant
H is defined as the stored kinetic energy
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in mega joules at synchronous speed divided
by machine rating in MVA. For any synchronous
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machine, when is when the rotor is rotating
at synchronous speed, we can find out what
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is the kinetic energy stored in the rotor
and we divide this kinetic energy stored in
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the rotor by machine rating. If we denote
the kinetic energy stored in mega joules and
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machine rating in MVA, then the unit of this
term becomes mega joules per MVA. We can also
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represent in kilo joules per kilo watt, per
KVA not kilo watt, per KVA but still the machine
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ratings are normally in MVA, we prefer to
use the rating in MVA and energy stored in
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mega joules, okay. Therefore, this can be
written as 1 by 2J omega sm square upon the
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machine rating in MVA as mach.
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Now we will develop a relationship between
the inertia constant M and constant H. So
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that the M which is there in our differential
equation will be replaced by constant H, from
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this equation 10, we can write the inertia
constant M as 2H upon omega sm into ah machine
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rating S mach.
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Now we substitute for inertia constant M in
our equation, we find here that this becomes
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2times H omega sm d2 delta m upon dt2 equal
to Pm minus Pe upon of S mach. Now if we assume,
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MVA based for the system as the machine rating
in MVA, then Pm upon S mach becomes the per
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unit mechanical power, similarly Pe divided
by S mach becomes the per unit electrical
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power okay. Therefore, this equation can now
be written in terms of power expressed in
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per unit, further as you all know that in
power systems the per unit system of calculations
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are very convenient and therefore the equation
is manipulated and written in terms of per
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unit power.
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Therefore, the equation 13 here is written
as 2H upon omega sm d2 delta m divided dt2
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equal to Pm in per unit minus Pe in per unit.
Now every time writing per unit is not very
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convenient, therefore what we do with that,
we drop this nomenclature here but we keep
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in mind that power is expressed in per unit.
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Therefore with this, we can write down equations
describing the rotor dynamics of the synchronous
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machine as 2H upon omega sm d2 delta m by
dt2 equal Pm minus Pe. At this stage we have
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to very careful and understand very clearly
that this Pm and Pe are expressed in per unit
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while expressing Pm and Pe in per unit, the
MVA base is the machine rating in MVA. For
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any machine which I am writing this differential
equation, if we assume the MVA base okay then
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the MVA base is the machine rating in MVA
and therefore Pm and Pe are expressed in per
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unit in terms of MVA base of the machine.
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Now here, this delta m is expressed in radians
per second, I am sure radians not radians
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per second, this is the correction omega sm
is expressed in radians per second. The units
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of these two terms should be consistent. We
know a term known as angle in electrical radians,
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speed in electrical radians per second.
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Therefore, if we use instead of the angle
in mechanical radians and speed in mechanical
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00:32:29,570 --> 00:32:36,570
radians per second, we can use delta in electrical
radians and omega sm in electrical radians
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per second and therefore what we do is that
keeping in mind that these 2 terms would have
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the consistent units, we drop the subscript
m and we write the equation in the form 2
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times H upon omega s equal to d2 delta by
dt2 equal to Pm minus Pe. In this is, this
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00:33:01,899 --> 00:33:07,879
equation is known as the swing equation of
the synchronous machine.
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This is applicable to generator as well as
to motor, only difference is when we write
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this equation for motor Pm becomes negative
and Pe also become negative. So, that this
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00:33:24,279 --> 00:33:31,279
term becomes Pe minus Pm. Now here, we can
express delta in radians, omega s in radians
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00:33:40,460 --> 00:33:47,460
per second. If we do in this manner, we can
write the equation in the form H upon 2, H
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00:33:48,519 --> 00:33:54,700
upon phi f that is omega s can be replaced
by 2 phi f and you will have the equation
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00:33:54,700 --> 00:34:01,700
in the form H upon phi f d2 delta by dt2 equal
to Pm minus Pe okay, many time we express
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00:34:06,759 --> 00:34:13,759
delta in electrical degrees and omega in electrical
degrees per second.
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00:34:20,740 --> 00:34:27,740
In that case omega s will be replaced by 360
into f. So that the resulting equation will
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00:34:32,850 --> 00:34:39,850
be in the form H upon 180 f d2 delta by dt2
equal to Pm minus Pe. We do use swing equation
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00:34:51,660 --> 00:34:58,660
either in this form or in the previous form
but while using this equation, we have to
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00:34:59,880 --> 00:35:06,880
very careful in expressing the delta in proper
units. You can in this case delta is in the
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00:35:12,150 --> 00:35:19,150
electrical degrees while in the equation 16
coefficient is H upon phi f delta is in electrical
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00:35:27,540 --> 00:35:34,540
radians. As we will see here actually that,
this is the basic swing equation on which
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00:35:39,680 --> 00:35:46,680
or around which the stability analysis of
the power system depends
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00:35:50,910 --> 00:35:57,910
Now we have 2 terms here Pm minus Pe, this
Pe which is the electrical power output of
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00:36:00,840 --> 00:36:07,840
the machine is, is non linearly related to
or it is non-linear function of delta and
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00:36:14,100 --> 00:36:21,100
therefore, this differential equation becomes
a non-linear differential equation or we can
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say that the swing equation is a second order
non-linear differential equation. Suppose,
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00:36:37,430 --> 00:36:44,430
you have any system a number of machines number
of synchronous machines then we have to write
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00:36:46,010 --> 00:36:53,010
down swing equation of each machine while
in this, we have to write down the correct
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00:37:01,200 --> 00:37:05,320
expression for electrical power output.
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00:37:05,320 --> 00:37:12,320
Now when this swing equation is solved, the
solution of the swing equation is known as
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00:37:21,280 --> 00:37:27,810
swing curve that is what we will get actually
when we solve this equation, if you solve
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00:37:27,810 --> 00:37:34,810
this equation you will get the delta as a
function of time delta as function of time
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00:37:35,670 --> 00:37:42,670
and
therefore we can define this swing curve.
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00:37:45,420 --> 00:37:52,420
When the swing equation is solved we obtain
the expression for delta as a function of
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00:37:53,470 --> 00:38:00,470
time. A graph of the solution is called swing
curve of the machine and inspection of the
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00:38:02,620 --> 00:38:09,620
swing curves of all the machines of the system
will show whether, whether the machines this
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is the wrong mistake, I have whether the machines
remains in synchronism after a disturbance.
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00:38:26,560 --> 00:38:33,560
I am writing here whether the machines remain
in synchronism after a disturbance again this
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00:38:38,040 --> 00:38:45,040
is a big mistake here after a disturbance.
For the purpose of solving the second order
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00:39:03,160 --> 00:39:10,160
differential equation, using numerical techniques
or standard practice is to represent the second
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00:39:15,370 --> 00:39:22,370
order differential equation in terms of 2
first-order differential equations because
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00:39:23,870 --> 00:39:30,870
when we apply the numerical techniques for
solving the differential equations using digital
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00:39:32,040 --> 00:39:38,750
computers, it handles first-order differential
equations.
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00:39:38,750 --> 00:39:45,020
We can write down the second order differential
equation as 2 first-order differential equations
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00:39:45,020 --> 00:39:52,020
by defining by defining a term omega but the
omega is the actual speed of the machine and
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00:39:58,280 --> 00:40:04,130
therefore we can define these two differential
equations in the form, 2H upon omega s, d
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00:40:04,130 --> 00:40:11,060
omega by dt that is you are having the d2
delta by dt2 therefore omega is defined as
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00:40:11,060 --> 00:40:18,060
d delta by dt okay. So that now here, we have
one variable omega, then the second equation
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00:40:20,450 --> 00:40:27,450
becomes d delta by dt whose expression comes
out to be omega minus omega s okay.
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00:40:41,680 --> 00:40:46,860
Now suppose you have a practical system where
you have more than one machine that is called
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a multi-machine system. In a multi-machine
system, the output and hence the accelerating
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00:40:57,410 --> 00:41:04,410
power of each machine depend upon the angular
position and, to the more rigorous also upon
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00:41:06,070 --> 00:41:13,070
the angular speeds of all the machines of
the system. Thus for a 3- phase system, this
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00:41:17,420 --> 00:41:24,420
is written for mistake here again. Thus for
a 3 machine system, for a 3 machine system,
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00:41:25,390 --> 00:41:32,390
there are 3 simultaneous differential equations.
The equations will look like this.
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00:41:35,730 --> 00:41:42,730
M1 d2 delta 1 by dt2 equal to Pm1 minus Pe1
which is function of delta 1, delta 2, delta
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00:41:50,400 --> 00:41:57,400
3and it is also function of d delta 1 by dt,
d delta 2 by dt, d delta 3 by dt. I will tell
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00:42:00,340 --> 00:42:07,340
actually why this Pe1 is function of this
derivative terms also the, for the second
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00:42:10,100 --> 00:42:17,100
machine we have to put inertia constant of
that machine and the electrical output of
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00:42:18,610 --> 00:42:23,030
this machine is function of all these angles
and these derivatives.
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00:42:23,030 --> 00:42:30,030
Similarly, for the third machine, we have
Pe3 function of all these derivatives. Now
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00:42:34,620 --> 00:42:41,620
since the system, we are considering here
is in dynamic condition. Okay when system
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00:42:42,030 --> 00:42:49,030
rotor is in dynamic condition, it develops
some damping torque and this damping torque
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00:42:49,780 --> 00:42:56,780
is proportional to the speed deviation. Then,
let us correct it is proportional to the speed
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00:42:59,670 --> 00:43:06,670
or speed deviation with respect to the synchronous
speed. Now this d delta 1 by dt, d delta 2
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00:43:08,040 --> 00:43:14,030
by dt, d delta 3 by dt, these are the rotor
speeds with respect to the synchronously rotating
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00:43:14,030 --> 00:43:17,990
reference speed.
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00:43:17,990 --> 00:43:24,990
Now to simplify our stability analysis, we
ignore this damping terms. Once we ignore
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00:43:27,930 --> 00:43:33,950
this damping terms, the electrical output
will exclusively be function of these angles
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00:43:33,950 --> 00:43:40,950
only okay. Now I would like to tell something
more about the electrical power output.
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00:44:00,990 --> 00:44:07,990
The important characteristic that has a strong
bearing on power system stability is the relationship
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00:44:09,760 --> 00:44:15,310
between interchanges power and the position
of rotors of the synchronous machines,
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this relationship is highly non-linear. I
have told you here the electrical power Pe
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00:44:20,430 --> 00:44:27,430
right is function of rotor angels and it is
non-linear. Now if I take a simple example
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00:44:28,190 --> 00:44:35,190
to illustrate this, let us consider this case,
we have 2 machines connected by a transmission
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00:44:45,980 --> 00:44:48,210
line.
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00:44:48,210 --> 00:44:55,210
We can represent this machine 1 by a voltage
source in series with a reactance. Similarly,
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the machine 2 can be represented by a voltage
source in series with a reactance. The machine
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00:45:03,070 --> 00:45:10,070
1 is supplying power to machine 2. This can
be understood as machine 1 is a synchronous
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00:45:13,940 --> 00:45:20,940
generator machine 2 is asynchronous motor.
For this simple system, if we develop an expression
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00:45:24,450 --> 00:45:31,450
for electrical power output from machine 1,
then this can be derived by writing or by
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00:45:34,200 --> 00:45:40,970
first drawing a phasor diagram.
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00:45:40,970 --> 00:45:47,970
The phasor diagram can be simply drawn in
this fashion. We can start with the terminal
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00:45:52,650 --> 00:45:59,650
voltage of the machine, synchronous machine
2. Let I is the current, the internal voltage
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00:46:05,480 --> 00:46:12,480
of the synchronous motor is represented by
this phasor Em, to this Em we add this voltage
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00:46:13,220 --> 00:46:20,220
drops, where the voltage drops which take
place in the internal reactance of the motor
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00:46:21,720 --> 00:46:28,720
9 reactance and generator reactance to get
the internal voltage.
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00:46:28,860 --> 00:46:35,860
Therefore, this phasogram diagram shows that
Em to this Em, if we add this voltage drop
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00:46:42,290 --> 00:46:49,290
I into XM, we get the voltage ET 2, to this
we add the voltage drop I into XL, we get
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00:46:50,420 --> 00:46:56,150
the voltage ET 1 and to this we add the voltage
drop I into XG, we get the internal voltage
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00:46:56,150 --> 00:47:03,150
of the synchronous generator. Delta which
is the phase difference between EM and EG
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00:47:07,360 --> 00:47:14,360
is called power angle and this delta is sum
of all these three angles delta G delta L
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00:47:14,530 --> 00:47:21,530
and delta M.
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00:47:22,180 --> 00:47:29,180
If we plot, a graph relating power angle delta
and power P or electrical power Pe, we call
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00:47:40,900 --> 00:47:47,900
this as Pe, then this graph comes out to be
a sine curve, the expression for Pe is the
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00:47:58,550 --> 00:48:05,550
internal voltage of the generator, the magnitude
of this voltage, internal voltage of the motor
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00:48:07,940 --> 00:48:14,940
EM divided by the total reactance X into sine.
Now, if we see here as the delta increases,
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00:48:26,900 --> 00:48:33,900
the power output increases and becomes maximum
at delta equal to 90 degrees, on the same
300
00:48:43,820 --> 00:48:50,820
diagram if I draw the mechanical input line,
the mechanical input is not function of delta
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00:48:58,640 --> 00:49:05,640
and therefore it comes out to be a line parallel
to delta axis okay.
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00:49:31,080 --> 00:49:38,080
In this diagram, we will write this as Pm
mechanical power, now this mechanical power
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00:49:40,570 --> 00:49:47,570
line and this power angle characteristic intersect
at 2 points, this is 1 point I call it “a”,
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00:49:47,830 --> 00:49:54,830
this is another point we call it “b”,
“a” is stable equilibrium point while
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00:49:59,480 --> 00:50:06,480
“b” is unstable equilibrium point that
is we will denote this “a” as stable equilibrium
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00:50:10,530 --> 00:50:17,530
point normally called SEP and “b” is unstable
equilibrium point.
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00:50:25,530 --> 00:50:32,530
The system, if it is operating at this point
“a” and if it is perturb then it develops
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00:50:35,210 --> 00:50:40,950
the forces so as so that it returns back to
this operating point “a”. However if the
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00:50:40,950 --> 00:50:45,670
system is made to operate which is at the
point b which is also the equilibrium point
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00:50:45,670 --> 00:50:52,670
but if suppose some perturbation is given
then the system will lose stability. It cannot
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00:50:52,890 --> 00:50:57,460
come back it cannot rest develop restoring
forces to come back to the positing point
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00:50:57,460 --> 00:51:04,460
“b” and therefore, our stable operating
point is a and we shall represent this operating
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00:51:06,890 --> 00:51:13,890
angle as delta that is when the system is
under steady conditions the mechanical power
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00:51:16,300 --> 00:51:21,050
is equal to electrical power and operating
angle delta naught.
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00:51:21,050 --> 00:51:28,050
Now I conclude here, what we have learn in
this lecture. I have tried to give you the
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00:51:28,780 --> 00:51:35,780
basic classification of the power system stability.
We have defined the stability involved terms,
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00:51:38,600 --> 00:51:45,600
we have developed swing equation of the machine
and we have also defined a very important
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00:51:47,490 --> 00:51:54,490
term inertia constant H. We will continue
further in the next lecture.