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In the previous two lectures, we have had
a brief overview of power systems and stability
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problems associated with it. What I really
told you in those lectures was a description
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of the phenomena, the basic phenomena. We
now go on to the nitigrities of trying to
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analyze these phenomenon, try to understand
how they occur and look at it in a more mathematical
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fashion.
Now, mathematicians will scoff at as in probably
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we horrified by the kind of the approximations
we will be doing in order to get a basic understanding
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of the phenomena. You will understand of course
that the nature of the models which we will
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use will be determined by that by which phenomenon,
we want to understand.
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So, today’s lecture we will learn about
some basic concepts of dynamical systems.
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We will also understand a few examples pertaining
to a power system, but the main issue which
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we will try to tackle in this particular lecture
is a general attitude, to get a general understanding
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of dynamical systems and how we can attack
the problem scientifically. In the past two
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lectures, I have not really given any mathematical
or rigorous explanation for the kind of phenomena
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we have seen, that is voltage instability
or loss of synchronism. I just gave you a
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kind of a hint that the reason why the systems
behave the way they do, they are due to the
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physical loss which govern there motion.
So, today what we will do is try to understand
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some kind of general ideas, understand general
ideas of dynamical systems. The first thing
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about dynamical systems is what is equilibrium?
Of course, we would like for example, a power
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system or any other system to be at an acceptable
equilibrium at all times. This of course will
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not be true because power system is always
subjected to some disturbance or the other.
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It could be a minor load change. For example,
you switch off a light in your house. It is
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a load change or they could be really large
faults involving tripping of components due
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to you know some short circuit and so on.
Those are large disturbances. So, what we
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need to do is understand what you mean by
equilibrium. Now, if you look at it using
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the very simple examples so what we will do
is just look at a very simple example of a
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ball on a hill. So, you can just concentrate
on what I am drawing here.
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Suppose, you have got a hill of this kind,
you got a valley here and a ball is somewhere
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on this hill. So, you got a ball on this hill.
The ball will be at equilibrium if
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its x coordinate is here. So, this is what
is known as equilibrium. So, I will call this
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x 1. Actually, what I have told you is not
strictly speaking true. What I should tell
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you is that if this ball is at this point
and its speed is equal to 0, then it is at
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equilibrium here.
We can always have ball which is rolling down
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and at this point, it has got some non zero
speed. In that case, it will continue moving.
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So, when we say something is at equilibrium,
we should really specify all the states corresponding
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to that equilibrium. So, what I should say
is that if x is equal to x 1 and x dot or
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which is nothing, but a notational simplification
of d x by d t. The rate of change of x is
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equal to 0. If these two things are satisfied,
you are at equilibrium. The ball will be at
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equilibrium. Is there another equilibrium?
Yeah, there is one more. For example, if I
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manage to place the ball here, I will call
this x 2. X is equal to x 2 and x dot is nothing,
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but d x by d t is equal to 0. This also defines
equilibrium.
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So, if I place a ball right there at the peak
of this hill very carefully, so that its velocity
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is 0 and and it is just exactly here, then
the ball will stay there. So, that is also
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equilibrium. There are two equilibria here
in this diagram. As far as I have shown you
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here, there are only two equilibria here and
here and there are two states of the system.
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What you mean by state? We will not go into
any rigorous definition, but in this context
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you see that these values of the states are
the minimum information I need to give you
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to know whether you are at equilibrium or
not. So, you should, you need to specify two
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things-the value of x and the value of x dot.
So, this particular system has two states.
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Now, you will notice that there is a qualitative
difference between the equilibrium here and
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here. In fact, it could have come to your
mind right away. This equilibrium 1 is different
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from this equilibrium 2. How is this equilibrium
different? Well, in case I place this ball
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at this equilibrium and I give it a small
disturbance, what would happen? It would kind
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of suppose, I gave it a kick while it was
here. Suppose, somebody came and gave it a
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kick, so it would go up, then come down and
then go up again and then, come down. There
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could be forces which you try to pull this
thing back to equilibrium, but of course,
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if there is no damping, we will set off an
oscillatory motion, but the fact remains that
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there are forces which you are pulling it
back to this equilibrium. So, this particular
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ball would eventually stabilize at this equilibrium.
The point is of course, there should be some
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friction here, so that it comes back to this
equilibrium. Otherwise, it will just go on
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continuing to oscillate. Contrast this with
a situation where the ball is here. If the
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ball is placed here at a 0 velocity, in case
I give a small disturbance, it will just roll
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off.
So, this particular equilibrium is such that
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if the ball is hit a small push can really
take you out of the equilibrium. So, in fact
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a kind of intuitive definition of stability
can be got right here is that if you have
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got equilibrium and if I give a small push
from that equilibrium, if the ball tends to
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come back to the equilibrium, eventually of
course, it should settle back at the equilibrium.
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Then, that equilibrium point is stable. To
put it more precisely, it is stable for small
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disturbances. So, that is one important thing
which you should keep in mind is that there
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could be many equilibria, but if those states
are subjected to a small disturbance like
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the ball being given a small push, are we
going to come back to the equilibria. So,
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that is the basic concept of stability. In
the example which I have shown you, this is
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unstable equilibria and this is stable equilibria.
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One of the ways we can really mathematically
define in equilibrium is if we take the states
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for in this case x, which is the position
and x dot, which is the velocity, if the rate
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of change of position and the rate of change
of velocity which is nothing, but I will call
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velocity as dv in that case we will have.
So, equilibrium is the point at which the
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rate of change of the states is all equal
to 0. Now, as we saw in the previous slide
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in this particular example, there were 2 equilibria.
Now, this is the general you know definition
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of equilibrium and states. How do you really
analyze the small disturbance behavior of
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the states? So, that is something which we
will do slowly and understand it using a power
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system example, the simplified power system
example in this particular lecture, but before
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we move on to that, let us just look at some
more interesting things about stability. You
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take this example again.
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These are the two equilibria. This is unstable,
this is stable for small disturbances and
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this is unstable for small disturbances. However,
the small disturbance may be the point is
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if a ball is here and I give not a small disturbance,
but a big disturbance, somebody really kicks
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it hard, then you may find that this ball
if somebody kicks it, for example, this ball
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will tend to move upward. It is constantly
being decelerated because of the force of
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gravity, but that force is not adequate to
stop the balls motion before it reaches this
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point. So, what will happen is that if the
ball rolls over this hill and starts rolling
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down here and if that happens, this hill just
goes on like this or it goes on like this
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for example, you will find that the ball will
never come back to this particular equilibrium.
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So, although this equilibrium is stable for
small disturbances, it is not for large disturbances.
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A big enough kick to this ball can make it
roll over this hill and never come back to
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this equilibrium. So, this is a typical situation
where a system is small disturbance stable
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at this equilibrium, but if a large enough
disturbance is there, it will not return to
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its equilibrium. So, this is what is known
as large disturbance stability. So, I hope
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you got a kind of a feel for these two concepts.
Let us do one thing.
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Let us take a simple power system example
and go ahead in trying to analyze it. A very
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similar example will have two states. Consider
a single machine, a synchronous machine connected
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say, via a transformer to a transmission line
and a very large grid. The grid is very large
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and let say this practically a voltage source
of constant magnitude E, constant frequency,
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so that its phase angle never changes. So,
it is sinusoid, a three phase sinusoidal voltage
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source whose voltage magnitude frequency and
phase angle simply do not change whatever
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you do. So, it is a very large system. This
is a transmission line.
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Now, when we are trying to analyze this system
for small and large disturbances, what do
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we need to do? We need to really form a model.
Now, the whole idea of this course in the
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later lectures is to get good models of all
these elements like a transmission line, a
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transformer synchronous machine. One important
point which you should notice is that we will
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never be getting exact models. We will never
be using exact models, in the sense that every
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model will involve some approximation or the
other. For example, look at this example.
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This synchronous machine let me model as a
voltage source behind what is known as a transient
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reactants. How do we come to this model? We
will do that in the course as the course moves
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on.
Let me just tell you that this is not a very
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respectable model of synchronous machine.
I am just using it to highlight a certain
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phenomena. So, what I will do is a synchronous
machine is model as a voltage source E s with
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a constant magnitude behind a transient reactants.
So, although this E s is constant does not
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mean that the terminal voltage is a constant,
E s is constant, delta is a rotor position.
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So, what we have is if the rotor moves, it
directly reflects in the phase angle of the
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voltage which appears. So, this of course
is not very difficult at least intuitively
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to understand that this could be a kind of
a model of a synchronous machine, but there
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is absolutely no rigger. We are just I am
just putting it forth to you. The transformer
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is modeled just by its leakage and a transmission
line and this voltage source. A transmission
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line is just one lumped inductor x. This is
x transformer and I will call this x line.
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So, this is the electrical circuit model.
We have really made a lot of approximations
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to bring a transmission line model from something
very complicated. In fact, the transmission
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line is modeled by Maxwell equation, described
Maxwell equations. So, it is a big long story
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of how we come to a lumped reactance equivalent
of a transmission line. I hope I will get
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some time to tell you how it comes about,
but right now, let us just take this model
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of a transmission line as a simple lumped
reactance x l. So, I will further simplify
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this. I will just make it a bit. So, this
circuit, as far this circuit is concerned,
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this is how it looks. So, x is nothing, but
x dash plus x t plus x l.
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Now, what are the equations which describe
this system? Now, those who are purist would
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say, well this transmission line has to be
modeled by a partial differential equation,
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this transform model probably again one has
to use Maxwell equations. This synchronous
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machine well, I have just given you a model
and not told you how you know it. It really
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comes about. It involves of course, Maxwell
equations and Newton equations which described
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the motion of the rotor. So, just let me put
forth particular model.
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We will be doing the origin of this model
later on in the course, but right now you
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just take it from me. I will just explain
each element. This is omega B which is the
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base radian frequency, that is, 2 pi into
the frequency base. H is nothing, but the
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inertia constant of the machine which is actually
equal to half into J into the mechanical speed
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square divided by the volt ampere base. So,
its units are actually
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mega joules per MVA. So, these are the units.
In the expression for H, we should have omega
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m B square, that is, the mechanical speed
base square. Out here I have written it as
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omega m square, but what really needs to be
a, what is correct it is in fact omega m base
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square. So, please note that the minor error
here.
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Omega is the electrical frequency, electrical
radian frequency of the generator. The electrical
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radian speed I should say. So, it is 2. If
there are four poles, of course the mechanical
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motion into P by 2 is equal to the electrical
speed. So, this is the mechanical speed, this
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is the electrical speed. Omega naught is 2
pi into f 0, where f 0 is the frequency of
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the infinite bus, that is the voltage source
whose frequency and voltage do not change
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at all. So, this is f 0.
Now, this is basically P m is the mechanical
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power input. So, this is the mechanical power
input, this is the electrical power output
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of the machine. Now, one important point is,
remember that P m and P e are in per unit.
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So, that is a very important thing. You should
note this P m and P e are in per unit, otherwise
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you will not find this dimensionally. So,
P m and P e are in per unit. In fact, the
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mechanical and electrical powers, this in
fact is in approximate equation. What should
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have been correctly written is T m minus T
e in per unit.
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So, sorry, in fact this is equal to T m minus
T e in per unit or both in per unit, but I
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have written it as mechanical power minus
electrical power. The approximation here is
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that mechanical power in per unit and mechanical
torque in per unit, they are practically the
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same. So, this is one approximation which
we have made right away. So, let me just rewrite
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these equations again.
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2H by omega B d omega minus omega naught is
approximately equal to the mechanical power
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minus the electrical power in per unit. So,
that is one important thing you should remember.
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In fact, the rotor position, the rate of change
of the rotor position is given by omega minus
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omega naught. Now, the rate of change of rotor
position as seen by a frame which is rotating
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at omega naught, so if the machine is rotating
exactly at 50 hertz and your omega naught
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also corresponds to 50 hertz. Then, this angle
delta will be a constant. However, if the
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frequency of this even transiently changes
the position of say, this mark will appear
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to move, so let me just put it you know in
a kind of physical way. If I am rotating at
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50 hertz and I am looking at this mark which
is also rotating at 50 hertz, this mark will
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appear stationery.
However, if there is a transient change in
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the speed, I will see this mark moving. So,
that movement of this mark is captured by
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this equation. Remember, the position of the
rotor, of the rotor of a synchronous machine
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also determines the phase angle of the voltage
which appears at the terminals. That is why
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what you see here is
this delta also has a bearing on the phase
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angle of the voltage of this voltage source
in this model of this synchronous machine.
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This again is not difficult to understand
intuitively because the position of the field
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which is on the rotor will really determine
the kind of the phase angle of the voltage
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which appears in the stator winding say, the
a stator winding.
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So, you know the point here I wish to make
here is that it is not difficult to see intuitively.
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We will do this in a bit more detail later
that the actual position of this will determine
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the phase at a given time. The position of
this will determine the phase of the voltages
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which I induced in the stator winding say
of the a phase. So, we will do this a bit
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later. We will keep this for later to show
this actually is true when we actually derive
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this synchronous machine model.
This is of course let me reiterate a very
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simplistic model of a synchronous machine.
This in fact is not even a respectable model,
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but I will use it to highlight the certain
phenomenon which is actually seen in practice.
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So, it is only used to show that a certain
phenomenon actually exists. Now, if we look
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at the system here, what is P e?
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00:24:11,880 --> 00:24:23,739
Now, P e is equal to E s. This is a power
transfer between two voltage sources which
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you are quite familiar with
x e. So, this is you can say the sinusoidal
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steady state power flow formula for a three
phase system, a three phase balance system.
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Now, of course again somebody may raise an
objection that well, you are studying system
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dynamics. Aren't you? So, why give a sinusoidal
steady state expression for power? Actually,
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you should actually write the differential
equations corresponding to this inductance
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here. After all this is inductance x e while,
so this we cannot treat it as a reactance.
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Then, uses sinusoidal steady state formula
and then, use it in our differential equations.
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So, unless I actually make this justification,
you will probably be unconvinced. So, I agree
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that I am not given you any such justification.
Let me just tell you right now
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that if the phenomena we are trying to study
using this model is much slower than the transients
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associated or the natural transients associated
with this electrical network here. This is
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just a simple transmission line. We can use
a kind of a quasi. This is a kind of a steady
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00:26:01,880 --> 00:26:08,769
state formula for analyzing these equations.
So, what appears actually a transmission line
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or transformer or a synchronous machine. It
should be made out of many many many differential
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equations and it in fact does. We have reduced
it to just two differential equations and
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this is you can call it a toy model. It is
not a very respectable model. It is a toy
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model which will help us to understand certain
phenomena. In fact, it highlights the phenomenon
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I wish to tell you quiet well.
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Now, how do I analyze this system? So, you
have got d delta by d t is equal to omega
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minus omega naught and 2H by omega B d omega
minus omega. When you first look at this equation
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and I tell you well, is this system stable
or not stable. Well, the correct way to approach
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00:27:06,619 --> 00:27:12,179
this problem is first of all take out what
are the equilibrium conditions for this system.
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The equilibrium condition for this system
is, remember how do we get the equilibrium
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00:27:18,040 --> 00:27:23,320
conditions? By setting the rates of changes
of the states which are in this case the speed
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00:27:23,320 --> 00:27:35,679
deviation and the rotor angle delta equal
to 0 in which case you will get the equilibria
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00:27:35,679 --> 00:27:48,840
are omega is equal to omega naught and well,
P m is equal to E s ES sine delta by x e.
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00:27:48,840 --> 00:27:56,789
So, at equilibrium I will call this may be
omega e. Omega e and delta e are the equilibrium
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values of this particular system now.
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So, just look at it graphically. This is P
e. P e is nothing, but e S e sine delta e
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00:28:13,639 --> 00:28:20,919
by x e. So, this is electrical power verses
delta and this is P m say, a certain value
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00:28:20,919 --> 00:28:29,059
of P m. Then, the point at which they are
equal defines the equilibrium, in fact from
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00:28:29,059 --> 00:28:35,289
0 to 180, the two possible equilibria of delta.
Here this is one and this is another for this
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00:28:35,289 --> 00:28:47,080
particular value of P m. So, let me just,
so omega e is equal to omega 0 and delta is
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00:28:47,080 --> 00:29:05,869
equal to sine inverse E s E sine delta, sorry
rewrite this
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00:29:05,869 --> 00:29:17,370
P m c.
Now, the point what we are trying to do here
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is analyze the stability of this system. Now,
of course there are two equilibria. So, there
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00:29:22,289 --> 00:29:26,840
is one here and other here. So, the first
thing which I can try to understand let us
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not try to understand large disturbance stability
right away. Let us talk about the small disturbance
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stability of this system. How does this system
behave, in case you give small disturbances
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00:29:38,669 --> 00:29:44,669
if it is initially at this equilibrium? So,
the point here is suppose I am at this equilibrium.
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00:29:44,669 --> 00:29:51,460
If I am at this equilibrium, I will simply
not move. The reason is d delta by d t is
235
00:29:51,460 --> 00:30:00,570
equal to 0 and d. So, if these two things
are satisfied, I will just stay where I am.
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00:30:00,570 --> 00:30:08,989
So, the only way we can get a transient is
to give it a small disturbance.
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00:30:08,989 --> 00:30:18,600
If as a result of a disturbance, the system
slightly gets deviates from one of the equilibria,
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00:30:18,600 --> 00:30:23,049
so it is at one of the equilibrium. Let us
say delta 1 itself and it slightly deviates
239
00:30:23,049 --> 00:30:32,769
from there. So, let us call this deviation
to be small. So, this is a small deviation.
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00:30:32,769 --> 00:30:43,090
So, let us say
omega is a small deviation from omega e, the
241
00:30:43,090 --> 00:30:53,679
equilibrium value and delta is a small deviation
from its equilibrium value. So, these are
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00:30:53,679 --> 00:31:00,480
small deviations. Why are we talking about
small deviations? We shall see later that
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00:31:00,480 --> 00:31:09,620
if we talk about small deviations, the analysis
is practical. You look at this. The set of
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00:31:09,620 --> 00:31:13,989
differential equations is described the motion
of the system.
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00:31:13,989 --> 00:31:18,769
How do you solve this? Unfortunately, it is
not equally easy to solve this. You will have
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00:31:18,769 --> 00:31:24,230
to actually knew kind of integrate these set
of equations, but without actually doing the
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00:31:24,230 --> 00:31:30,269
integrations say, numerically or by some other
technique, can we try to understand the behavior
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00:31:30,269 --> 00:31:36,440
of this system. The answer is yes. For small
disturbances, yes quite easily. So, let us
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00:31:36,440 --> 00:31:40,470
talk about only very small disturbances around
the equilibrium. Why are we talking of small
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00:31:40,470 --> 00:31:47,009
disturbances? Because once we have small disturbances,
we can make certain approximations. So, you
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00:31:47,009 --> 00:31:48,559
take our original equations.
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00:31:48,559 --> 00:31:57,229
So, what I do is d delta by d t is equal to
omega minus omega naught which implies d delta
253
00:31:57,229 --> 00:32:08,980
e plus delta delta is equal to omega e plus
delta omega minus omega naught. So, actually
254
00:32:08,980 --> 00:32:14,510
omega e and omega naught are the same. Delta
e is a particular value. So, the derivative
255
00:32:14,510 --> 00:32:22,489
of that is equal to 0. So, you have got finally
for small disturbances, fine, for the small
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00:32:22,489 --> 00:32:26,860
disturbances, the other equation. This is
the first equation; this is the second state
257
00:32:26,860 --> 00:32:53,499
equation. You have 2H by omega B into d is
nothing, but 2H by omega B into is equal to
258
00:32:53,499 --> 00:33:00,559
P m minus P e. Now, P e is E s E. We will
assume these to be constant. All these things
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00:33:00,559 --> 00:33:12,710
are to be constant sine delta plus delta e
plus delta delta. So, that becomes equal to
260
00:33:12,710 --> 00:33:23,909
P m minus, well we can apply the formula for
sine delta plus delta delta e plus delta delta.
261
00:33:23,909 --> 00:33:42,100
So, we will go to P m minus E s E by x e sine
delta e cos delta delta minus, sorry plus
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00:33:42,100 --> 00:33:59,979
cos delta e sine delta delta. So, what we
have is P m minus E S E by x e sine delta
263
00:33:59,979 --> 00:34:14,050
e plus, sorry minus E S E by x e cos delta
e. Is this a variable cos delta e and sine
264
00:34:14,050 --> 00:34:21,010
delta e? No, these are cos delta e. E is a
particular value, so it is evaluated. It is
265
00:34:21,010 --> 00:34:26,350
not a variable into delta delta.
So, this is using the formula that cos delta
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00:34:26,350 --> 00:34:35,419
delta is approximately 1. If delta delta is
very small and sine delta delta approximately
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00:34:35,419 --> 00:34:52,750
delta, so what we have here is the second
differential equation becomes
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00:34:52,750 --> 00:34:58,950
if P m is equal to con is a constant. Then,
these since at equilibrium E S E sine delta
269
00:34:58,950 --> 00:35:04,760
e by x e should be equal to P m. These two
can get cancelled. So, what we have here is
270
00:35:04,760 --> 00:35:17,599
basically E S E by x e cos delta e delta delta.
This is a constant value evaluated at the
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00:35:17,599 --> 00:35:22,530
equilibrium point. So, what we have? I will
call this capital K.
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00:35:22,530 --> 00:35:27,770
So, what we have here is the differential
equations for small disturbances around an
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00:35:27,770 --> 00:35:42,390
equilibrium are d delta by d t is equal to
delta omega and d 2H by omega b b delta omega
274
00:35:42,390 --> 00:35:53,380
by d t is equal to minus K delta delta.
Now, the reason why I got you up to this point
275
00:35:53,380 --> 00:35:57,549
is this particular differential equation is
actually easy to solve. For small disturbances,
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00:35:57,549 --> 00:36:04,320
one can get a kind of an exact solution for
this differential equation. So, what is the
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00:36:04,320 --> 00:36:09,680
solution of this? Now, how to get the solution
of this in this particular case or in a general
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00:36:09,680 --> 00:36:15,720
case is something I will try to tell you later
on in this course. Right now, let me just
279
00:36:15,720 --> 00:36:27,260
suggest a solution. This looks very much like
the equation of a spring mass system M and
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00:36:27,260 --> 00:36:33,600
K, where M is 2H by omega B and K is the spring
constant of the spring. The differential equations
281
00:36:33,600 --> 00:36:38,790
which you get for this are very similar to
this differential equations which are update.
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00:36:38,790 --> 00:36:45,390
So, let me not derive any solution for this.
Let us assume that the solution for this particular
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00:36:45,390 --> 00:36:56,539
system of equations let us guess it. So, let
us say delta delta is equal to A sine omega
284
00:36:56,539 --> 00:37:01,780
t. I will call omega n t. This is the different
omega n from the omegas we have been talking
285
00:37:01,780 --> 00:37:10,869
of plus 5. So, what will be delta omega if
this is delta? You will have since this is
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00:37:10,869 --> 00:37:21,750
the derivative 5, so these are well, we do
not know. What we do not even know whether
287
00:37:21,750 --> 00:37:25,180
this is the correct solution or not, but if
you plug it in it actually satisfies this
288
00:37:25,180 --> 00:37:31,290
equation. The first one is obviously satisfied
if this is true, but you just plug it in this
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00:37:31,290 --> 00:37:40,760
equation. So, what you have is 2H by omega
B. You will have omega n square A sine. This
290
00:37:40,760 --> 00:37:53,579
will be minus is equal to minus K A sine omega
n plus 5.
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00:37:53,579 --> 00:38:04,720
So, obviously this is satisfied if omega n
square is equal to yeah. So, omega n square
292
00:38:04,720 --> 00:38:16,589
is equal to K k by 2H. This is omega B. So,
this is true. Of course, these equations get
293
00:38:16,589 --> 00:38:24,260
satisfied and indeed our guess solution is
correct if omega n is taken to be this. So,
294
00:38:24,260 --> 00:38:32,220
let us just move on. So, have we got the solution
yet? No, we have just said that.
295
00:38:32,220 --> 00:38:44,539
Out of this, we have got what omega n should
be, that is it should be square root of omega
296
00:38:44,539 --> 00:38:53,780
B K by 2H can be positive or the negative
square root. The things we do not have yet
297
00:38:53,780 --> 00:38:59,760
are A and 5. We only know that this particular
solution, this solution is in fact correct.
298
00:38:59,760 --> 00:39:11,060
So, how do we get this a and 5? Well, let
us say that at time t is equal to 0, we are
299
00:39:11,060 --> 00:39:14,549
not at equilibrium. We have been slightly
displaced from the equilibrium. So, at time
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00:39:14,549 --> 00:39:19,869
t is equal to 0 suppose, remember that the
whole idea of doing this analysis is to see
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00:39:19,869 --> 00:39:24,480
how we displaced from the equilibrium. If
we are at a equilibrium, of course there are
302
00:39:24,480 --> 00:39:28,789
no transients at all because the rates of
change of all the states are equal to 0. So,
303
00:39:28,789 --> 00:39:51,230
suppose t is equal to 0, we have
and in that case, just substituting it in
304
00:39:51,230 --> 00:40:18,609
this particular equation delta is equal to
A sine 5, right. So, actually if you look
305
00:40:18,609 --> 00:40:26,210
at the equations of this, this is delta delta,
this is delta omega. So, what we have here
306
00:40:26,210 --> 00:40:31,960
is time is equal to 0. This is how things
will look like.
307
00:40:31,960 --> 00:40:38,549
So, if I am given, if I give you this delta
and omega 0 that is time t is 0 to 0, can
308
00:40:38,549 --> 00:40:47,980
you take figure out what is 5? Yes, you can
find out what A is and what is 5. Therefore,
309
00:40:47,980 --> 00:40:54,150
we get the complete solution because if I
know what a is in that case and if I know
310
00:40:54,150 --> 00:40:58,829
what 5 is, you know what omega n is. So, I
have got a complete solution for how this
311
00:40:58,829 --> 00:41:07,410
system behaves. So, this is what we get as
the final solution of this particular system.
312
00:41:07,410 --> 00:41:13,789
Now, what I have shown you is that this particular
system is stable. Well, not really stable,
313
00:41:13,789 --> 00:41:18,490
it has got an oscillatory motion. Is it really
surprising? The answer is no.
314
00:41:18,490 --> 00:41:31,069
Suppose, I am at this, I am at this equilibrium.
This is P m, this is P e, this is delta and
315
00:41:31,069 --> 00:41:37,170
I am at this equilibrium. If I give a small
push, if I give a small push to this system,
316
00:41:37,170 --> 00:41:42,520
that is del. If this gets slightly away from
this equilibrium because of a disturbance
317
00:41:42,520 --> 00:41:48,150
and I am suppose, at this point let us say
delta is slightly away from the equilibrium
318
00:41:48,150 --> 00:41:54,900
and say, speed deviation is 0. In that case,
P e becomes greater than P m, the machine
319
00:41:54,900 --> 00:42:00,050
will decelerate and try to come back to this
equilibrium. Similar thing happens here if
320
00:42:00,050 --> 00:42:06,690
you are here, P e is less than P m. So, the
machine will accelerate and come back to this
321
00:42:06,690 --> 00:42:13,289
equilibrium.
So, however the motion is kind of oscillatory,
322
00:42:13,289 --> 00:42:17,990
the restoring torques which are there are
in fact proportionally. For the small disturbances,
323
00:42:17,990 --> 00:42:22,790
they are practically proportional to the angular
deviations and therefore, you get a kind of
324
00:42:22,790 --> 00:42:29,790
spring mass behavior of this system. Of course,
what about this equilibrium? In this equilibrium,
325
00:42:29,790 --> 00:42:37,420
it is not oscillatory. For example, my delta
is here, speed deviation let us say is 0,
326
00:42:37,420 --> 00:42:45,900
but my delta deviation is there and slightly
deviated from the equilibrium. In that case,
327
00:42:45,900 --> 00:42:50,900
P e is less than P m. So, the machine will
accelerate. If it accelerated, it actually
328
00:42:50,900 --> 00:42:55,000
goes away from the equilibrium point. So,
one would expect that this particular equilibrium
329
00:42:55,000 --> 00:43:05,480
point is not stable, it is unstable. So, how
come I got this solution? It seemed all fine,
330
00:43:05,480 --> 00:43:15,680
right. This is my solution where A and 5 are
obtained
331
00:43:15,680 --> 00:43:29,400
from the initial conditions of that is from
this, you get A and 5. So, everything is fine.
332
00:43:29,400 --> 00:43:36,910
So, this seems to say it is oscillatory, where
omega n is. So, where am I going wrong? I
333
00:43:36,910 --> 00:43:42,910
mean this particular equilibrium point is
unstable. Remember that this K if we recall
334
00:43:42,910 --> 00:43:54,970
our derivation, this K is nothing, but E E
s cos delta equilibrium upon x e. This delta
335
00:43:54,970 --> 00:43:59,940
e here is dependent on the equilibrium point.
In fact, this particular equilibrium point
336
00:43:59,940 --> 00:44:10,950
is greater than 90. So, this k is negative.
If K is negative, omega n is a complex number.
337
00:44:10,950 --> 00:44:16,990
So, what we have is omega n is some complex
number.
338
00:44:16,990 --> 00:44:26,900
So, in that case, things get a bit complicated
because now our solution is delta delta is
339
00:44:26,900 --> 00:44:39,250
equal to A sine, a complex number omega n.
I will call it let omega n be, so at the other
340
00:44:39,250 --> 00:45:04,539
equilibrium. At the other equilibrium, delta
delta is A sine J. Remember omega n is complex
341
00:45:04,539 --> 00:45:23,240
plus 5 which is nothing, but A. We can expand
this. This is nothing, but E, (Refer slide
342
00:45:23,240 --> 00:45:35,289
time: 45:13-45:29) right.
So, this is nothing, but small values, sorry
343
00:45:35,289 --> 00:45:49,220
it should be j into j omega t here and it
should be again, j into j omega t. Remember
344
00:45:49,220 --> 00:46:00,369
that this sine x is nothing, but equal to
E raise to j x minus e raise to minus j x
345
00:46:00,369 --> 00:46:11,849
upon 2 j. So, this is what I view. So, it
is nothing, but (Refer slide time: 46:03-46:14).
346
00:46:11,849 --> 00:46:21,630
So, now your solution of delta delta if you
are at the other equilibrium in which omega
347
00:46:21,630 --> 00:46:27,510
n turns out to be complex, the solution comes
out to be like this, where omega of course
348
00:46:27,510 --> 00:46:36,930
is a real number, but omega n of course is
a complex number. In fact is the purely imaginary
349
00:46:36,930 --> 00:46:41,170
number. Not only a complex number, but it
is purely imaginary too. You look at this
350
00:46:41,170 --> 00:46:50,040
solution. If this is a real number, either
this or this is going to be positive. So,
351
00:46:50,040 --> 00:47:00,510
minus omega will be positive or this or this
will be positive. Now, remember that e raise
352
00:47:00,510 --> 00:47:19,309
to 5 t for example, grows with t. So, of course,
e raise to minus 5 t decays. So, your solution
353
00:47:19,309 --> 00:47:24,180
for delta delta at the other equilibrium for
a small disturbance is going to be a super
354
00:47:24,180 --> 00:47:32,900
imposition of these two terms. If the initial
conditions are such that k 1 and k 2 are non
355
00:47:32,900 --> 00:47:42,150
zero, remember k 1 and k 2 are obtained from
A and 5 and if this k 1 and k 2 are non zero,
356
00:47:42,150 --> 00:47:48,700
you will find that there will be one component
of this solution which grows with time. It
357
00:47:48,700 --> 00:47:53,720
increases with time.
So, if you are at the other equilibrium point,
358
00:47:53,720 --> 00:47:59,780
what we have really shown is that you will
be unstable; you will really grow with time.
359
00:47:59,780 --> 00:48:06,160
So, to summarize at least the small signal
part, remember what we have done really is
360
00:48:06,160 --> 00:48:11,329
analyze the small disturbance stability of
a single machine connected to an infinite
361
00:48:11,329 --> 00:48:20,820
bus. Using a very simplified model, we have
seen that for a particular small disturbance,
362
00:48:20,820 --> 00:48:26,950
the system not for a particular small disturbance,
for small disturbances, the system gives an
363
00:48:26,950 --> 00:48:32,799
oscillatory behavior for the equilibrium which
is lower than 90 degrees and if it is of course
364
00:48:32,799 --> 00:48:37,150
the equilibrium point which is greater than
90 degrees which we have shown between 90
365
00:48:37,150 --> 00:48:46,580
degrees and 180 degrees, the system is unstable,
that is if I give a small push, the deviation
366
00:48:46,580 --> 00:48:50,730
grows with time. If the deviation grows with
the time means, it will not come back to that
367
00:48:50,730 --> 00:48:56,559
particular equilibrium. So, after two equilibriums
which we have got between 0 and 180 for delta
368
00:48:56,559 --> 00:49:02,779
for a given mechanical power, you see that
one of this system is one of the equilibria
369
00:49:02,779 --> 00:49:04,640
is actually unstable.
370
00:49:04,640 --> 00:49:13,500
So, to summarize a single machine infinite
bus behaves like more or less like a spring
371
00:49:13,500 --> 00:49:19,670
mass system at the equilibrium point delta
1.
372
00:49:19,670 --> 00:49:26,700
Now, last disturbances something we will learn
in the next lecture, but let us do one practical.
373
00:49:26,700 --> 00:49:31,329
In this particular lecture, we have been looking
at more of the mathematical treatment of a
374
00:49:31,329 --> 00:49:34,859
particular set of a differential equations.
I said at the beginning that although, we
375
00:49:34,859 --> 00:49:41,230
are going to use a very toy model of the synchronous
machine, it is actually going to replicate
376
00:49:41,230 --> 00:49:45,339
what we actually see in practice, a particular
phenomena which we see, that is the oscillatory
377
00:49:45,339 --> 00:49:50,200
behavior etcetera is actually seen in practice.
So, if you recall in my previous lecture,
378
00:49:50,200 --> 00:49:59,020
I had shown you a small particular graphic
of a disturbance did occurred in Tata power
379
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system and you could see that there were oscillations.
So, these oscillation actually occur in practice.
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00:50:03,660 --> 00:50:09,309
One good check we can do right away you know
with this toy model is what is the frequency
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of oscillation, what are the frequency of
oscillation if you are likely to face.
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So, for example, you take this particular
system, you call this E s angle delta. This
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00:50:30,820 --> 00:50:39,099
is e e, e angle 0 and this is the infinite
bus or constant voltage source. We saw that
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00:50:39,099 --> 00:50:53,240
omega n is equal to root of yeah omega n is
equal to root of omega b capital K upon 2H.
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This is nothing, but omega B. This K is nothing,
but
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00:51:10,510 --> 00:51:20,380
So, if I take omega B for 50 hertz system,
the base frequency is nothing, but roughly
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00:51:20,380 --> 00:51:30,990
2 pi into 50 which is nothing, but roughly
314, roughly. H the typical value of the inertia
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00:51:30,990 --> 00:51:42,680
constant of a machine let say, it is 4 mega
joules per MVA. Let us say E s and e b are
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00:51:42,680 --> 00:51:54,549
1, cos delta e, let us say delta e is 50 degrees.
So, cos delta e would be cos 50 is approximately
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00:51:54,549 --> 00:52:05,609
equal to cos 45 is approximately equal to
yeah 1 by root 2, yeah. X e let say, it is
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00:52:05,609 --> 00:52:12,240
a cumulative impedance of the transient reactance
of the generator, the transformer as well
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00:52:12,240 --> 00:52:15,539
as the transmission line. Let us say, it is
0.5 per unit.
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00:52:15,539 --> 00:52:35,760
So, in that case, what we have is 314 by 2
into 4 1 into 1 into cos 50 by 0.5 is nothing,
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00:52:35,760 --> 00:52:49,859
but 314 by 8. This is 1 by root 2 that is
0.7. Very roughly this is nothing, but 314
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00:52:49,859 --> 00:53:02,309
into 0.7 divided by 4 nothing, but half of
root of 4 into 0.7 which is nothing, but roughly
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00:53:02,309 --> 00:53:16,089
very very very very roughly. So, this is nothing,
but let us say 15 divided by 2 around 7.5
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00:53:16,089 --> 00:53:20,401
radians per second.
So, the frequency of these oscillations is
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00:53:20,401 --> 00:53:29,140
going to be roughly around 1, say 1.2 hertz.
So, this is basically the natural frequency
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00:53:29,140 --> 00:53:35,050
of oscillation which you will see if the system
is subjected to small disturbances. So, if
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00:53:35,050 --> 00:53:39,920
you are at the stable equilibrium point and
if I give a push, in that case you will get
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00:53:39,920 --> 00:53:46,339
a oscillation. So, the machines are rotating
at 3000 rpm for a two pole machine or 1500
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00:53:46,339 --> 00:53:51,470
rpm for a four pole machine, but over and
above that you will find that the speed of
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00:53:51,470 --> 00:53:57,799
the machines over and above the 3000 rpm or
1500 rpm depending on the pole number, you
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00:53:57,799 --> 00:54:04,170
will find that there is an oscillation taking
place of 1, around 1 to 2 hertz. So, actually
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00:54:04,170 --> 00:54:10,700
these oscillations have been observed in practice.
So, give a disturbance to a machine, it slightly
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00:54:10,700 --> 00:54:16,569
oscillates. According to the solution which
we have got for this particular system, if
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00:54:16,569 --> 00:54:21,020
the system oscillates, it keeps on oscillating
because you have got A sine omega t omega
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00:54:21,020 --> 00:54:30,860
n t plus 5. So, it just keeps on oscillating,
but actually there are reasons because of
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00:54:30,860 --> 00:54:36,200
which usually these oscillation dies unusually.
So, these oscillations die down. How do they
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00:54:36,200 --> 00:54:39,799
die down? Unfortunately to the equations which
I have given you in the toy model do not tell
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00:54:39,799 --> 00:54:45,890
you how they will die down because that particular
component of the model has not been included.
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00:54:45,890 --> 00:54:51,119
So, the point which I wanted to say here is
we have taken a very simple system and shown
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00:54:51,119 --> 00:54:58,730
that we give a small disturbance to it and
it oscillates. Now, the real intention of
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00:54:58,730 --> 00:55:05,160
doing this particular example was not so much
to trying to tell you about the behavior of
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00:55:05,160 --> 00:55:10,750
the electro mechanical behavior of a single
machine connected to a voltage source, but
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00:55:10,750 --> 00:55:16,490
the real intention was to tell you about how
we can actually systematically analyze this
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00:55:16,490 --> 00:55:22,400
system. In fact, we wrote down the differential
equation and got the equilibrium linearised
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00:55:22,400 --> 00:55:29,630
system or what we say is we saw how we can
analyze the system for small disturbances.
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00:55:29,630 --> 00:55:35,040
We guessed the solution. Now, this is one
thing which we will try to do later again
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00:55:35,040 --> 00:55:38,589
where we will not guess the solution, but
actually derive it. We just guess the solution
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00:55:38,589 --> 00:55:45,150
is A sine or omega t n or omega n t plus 5.
In the next several lectures, we will actually
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00:55:45,150 --> 00:55:52,720
see how we can derive the solution for such
system. The interesting thing which came out
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00:55:52,720 --> 00:55:58,509
of this without having to go into very detailed
models, we could infer that if such a phenomena
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00:55:58,509 --> 00:56:04,089
does occur for typical values of system parameters,
you will get oscillations of around 1 hertz
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00:56:04,089 --> 00:56:08,759
and these are actually seen in practice.
Now, in the next lecture, we will just look
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00:56:08,759 --> 00:56:12,550
at the equations for last disturbances. I
told you last disturbance behavior can be
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00:56:12,550 --> 00:56:18,259
quite different from the small disturbance
behavior and we will try to see whether we
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00:56:18,259 --> 00:56:26,250
can infer certain phenomena for last disturbances.
So, to sum up in this particular lecture,
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00:56:26,250 --> 00:56:33,200
we have really tried to see the concepts of
equilibrium and small disturbances stability
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00:56:33,200 --> 00:56:34,230
using an example.