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Friends today, we shall discuss ah the equal
area criteria for stability.
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We have studied that to analyze the stability
of a power system, we have to solve swing
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equations and solution of swing equations
is through a numerical technique and it is
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a time consuming process. The equal area criteria
of stability is very powerful tool to understand
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the basic concepts of stability. However,
as we will see that this criteria has its
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limited applications.
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Now, today in our study we will establish
this basic concepts pertaining to equal area
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criteria and we will analyze considering a
one machine swinging with respect to the infinite
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bus, then we will develop the equivalent of
one machine infinite bus system of a 2 machine
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system.
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Then we will study the applicability of equal
area criteria under what circumstances the
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provoke systems this criteria is applicable
and what are its limitations. Then we will
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illustrate the applications of this criteria
considering two examples, a sustained line
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fault and second a line fault with subsequent
clearing.
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The equal area criteria of stability is a
graphical method for determining whether the
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system is in stable or not, that is in any
stability studies our primary requirement
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is that for given operating condition and
for a given disturbance whether the system
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is stable or not, many times we are interested
in knowing if it is stable how much stability
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is and what is the stability margin.
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This information we can get by plotting the
swing curve but as I told you that the computation
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of swing curve is time consuming and for simple
system for a two machine system or a one machine
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connected to infinite bus, we can obtain this
information by applying a graphical method
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and that method is the equal area criteria.
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Now when this criteria is applicable, its
use wholly or partially eliminates the need
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of computing swing curve and thus saves considerable
amount of time. I am emphasizing here, that
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it eliminates the computation of swing curve
wholly or partially. Okay that we will see
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actually when we attempt some example.
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This criteria is applicable to any two machine
system for which commonly made assumptions
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are applicable right. We have studied the
commonly made assumptions for analyzing the
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transient stability problem and when these
assumptions are applicable, this can be applied
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to any two machine system.
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Now first we will mathematically establish
the equal area criteria of stability to establish
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we start with the swing equation of a machine,
here we are considering a machine connected
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to the infinite bus, infinite bus is one which
can be represented by a constant voltage source
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its internal impedance is 0 and its inertia
is infinite. Now you multiply this both sides
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of this equation by a term 2 delta d delta
by dt.
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When we multiply both the sides by this term
2 times d delta by dt, we get equation in
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this form 2 times d2 delta by dt2 into d delta
by dt equal to 2 times Pa by M, d delta by
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dt. Now this left hand side of this equation
can be identified as derivative of d delta
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by dt square, you look it very carefully that
is if you take this term and find out its
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derivative you will get the term 2 times d2
delta by dt2 d delta by dt, okay and right
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hand side we are writing as it is 2 times
Pa by M d delta by dt okay.
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The next step is you multiply both sides of
this equation by dt and we are getting differentials
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instead of derivatives that is when you multiply
both sides of this equation 5.3 by dt it becomes
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a differential that is d of d delta by dt
square equal to 2 times Pa by M, d delta.
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Okay, which is written here actually as d
of d delta by dt square 2 times Pa by M d
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delta okay. Now you integrate this equation
5.4, integrate it.
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When you integrate, you will get d delta by
dt square equal to 2 by M integral Pa d delta.
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This integration we are doing over certain
range of delta that is start with initial
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value of delta naught delta equal to delta
naught and go to some value of delta. I am
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just putting it general that is delta naught
to delta, I am not specifying what should
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be the value of delta here.
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Now from this equation 5.5, we can write d
delta by dt equal to square root of 2 by M
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integral Pa d delta, the limits are implied
okay. Now we look at this equation and we
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know that the condition which indicates the
stability of the system is that d delta by
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dt should be 0 that is a system starts where
it is perturbed right with the initial value
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delta naught and when it is, when delta increases
it will attain a maximum value then start
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decreasing that is the condition of stability,
it means that when it goes to maximum and
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start decreasing that maximum point, at that
maximum point d delta by dt is 0 therefore,
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the condition to indicate the stability is
that d delta dt should become 0.
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Now for this d delta by dt2 become 0 which
is the condition to indicate the stability
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of the system, our requirement is that the
integral of Pa d delta from a initial value
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of delta equal to delta naught to some value
of delta m should be 0 that is if I, if I
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plot the area under the curve Pa that is accelerating
power starting from initial value of delta
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equal to delta naught to some maximum value
of delta m and in case the in the area under
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this curve is 0 right, then at the value of
delta equal to delta m the d delta by dt will
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become 0 right.
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Now this I will show here in this diagram,
this is the power angle characteristic of
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the system, this was the mechanical input
line. Now in this diagram I am simply showing
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that suppose initially the system was operating
at delta naught and the power angle characteristic
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applicable for the initial operating condition
was not this but something different. The
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moment fault occurs on the system the operating
point has shifted from this point to this
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point okay then delta will increase from initial
operating from the point a on the power angle
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curve Pe.
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The initial acceleration is given by this
or initial accelerating power is given by
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this line, this is the initial accelerating
power, okay therefore rotor accelerates when
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rotor accelerates the delta will increase,
speed increases then delta increases and now
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when it reaches the point b right, the accelerating
power becomes 0. Now at this point what is
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the rotor speed?
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The rotor speed is more than the synchronous
speed because, because from when it is travels
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from this point a to b, the rotor is accelerated
gain some kinetic energy it has gained some
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extra speed over and above the synchronous
speed therefore, at this point the rotor will
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not stop it will continue to, the angle will
continue to increase. Now the moment it crosses
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this point b we see that the accelerating
power becomes negative that electrical power
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output becomes more than the mechanical input.
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Therefore, now the rotor will be subjected
to retardation and when it reaches the point
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c point c, if suppose whatsoever the kinetic
energy it has gained if it is return back
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or is ah lost then at point c the rotor will
again attain a speed equal to synchronous
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speed and therefore for the system to be stable,
the rotor will swing from delta naught to
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delta m and the area under the accelerating
power.
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Now here the accelerating power is obtained
by separating therefore here I can say that
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accelerating power curve is given by this
difference, okay and therefore this area should
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be 0. Now for this area under the accelerating
power curve to become 0 requires that the
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this this positive area, we call this as a
positive area because accelerating power is
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positive this is called negative area, the
accelerating power is negative therefore these
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2 areas should be equal and this is what is
known as the equal area criteria of stability.
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Now this equal area criteria of stability
can also be interpreted in terms of the kinetic
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energy gain and kinetic energy lost right.
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Now we know actually that suppose a rotor
is subjected to a torque T, torque T and when
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it swings from delta naught to delta the,
the work done on the rotor is equal to or
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work done by the rotating body is equal to
T into d delta. This is actually when the
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delta is small and you integrate this expression
from delta naught to sum value delta then
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this becomes the work done when the rotating
body accelerates, okay or decelerates it depends
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upon the situation and hence we can say that
the area a1 represents kinetic energy gained
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by this machine or I can say this area a1
is directly proportional to kinetic energy
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gained there has to be some proportionality
constant I cannot say that a1 is in in joules
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or mega joules or so it will depend upon what
is the unit which we have attached to it.
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Now the derivation is which we have seen just
now is considering a machine connected to
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infinite bus. Now suppose you have 2 finite
machines, 2 finite machines a problem that
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of a synchronous generator supplying power
to a synchronous motor through a transmission
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line. Then this 2 finite machine system can
be replaced by an equivalent machine infinite
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bus system, this derivation is very straight
forward we take the swing equation of machine
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1, d2 delta 1 by dt2 equal to Pa1 by M1 that
is Pm1 minus P1 by M1. Then we take a swing
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equation of second machine ah d2 delta 2 by
dt equal to Pa2 by M2 that is Pa2 is Pm2 minus
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P21 okay.
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Now suppose we define here the relative angle
between the two machines as delta equal to
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delta 1 minus delta 2. Now whether these 2
machines are going to remain in stable condition
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or become unstable is going to determine not
by delta 1 and delta 2 individually but the
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difference between this 2 angles delta 1 and
delta 2, in case this difference difference
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right remains within certain limits system
will continue to be stable and therefore the
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this 2 differential equations which we have
written right, if we can be written in this
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form that is you can subtract these 2 equations
you will find that d2 delta 1 delta d2 minus
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d2 delta 2 by dt2. Okay this difference can
be written as d2 delta by dt2 because delta
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1 minus delta 2 is delta and on this right
hand side you have Pa1 by M1 minus Pa2 by
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M2.
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We can multiply both sides of the equation
by this product M1 M2 or M1 plus M2 and we
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may get the expression in this form. M1 M2
or M1 plus M2 d2 delta by dt 2 equal to M2
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Pa1 minus M1 Pa2 or M1plus M2 .Okay now if
we replace Pa1 and Pa2 by Pm1 minus P1 and
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Pm2 minus P2, we get this expression that
is I have put here in this expression wherever
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you had Pa1 Pm1 minus P1, Pa2, Pm2minus P2
and then simplify it in this we find here
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that we can write this as M1 M2 or M1 plus
M2 d2 delta by dt2 equal to M2 Pm1 minus M1
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Pm2 divided by M1plus M 2,that is in this
term we have the inertia constants and mechanical
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powers while here in this term we have inertia
constants and electrical powers P1 minus P1
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and P2 okay.
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Now this equation may be considered to be
the swing equation of a machine connected
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to infinite bus, where where the equivalent
inertia constant is M1M 2 or M 1 plus M2,
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the equivalent mechanical input is given by
this expression and equivalent electrical
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output is given by this expression.
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Therefore, we can say that this expression
written as M times d2 delta by dt2 equal to
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Pa Pm minus Pe, where Pm is given by this
expression, Pe is given by this expression,
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we can say that Pm is the equivalent mechanical
input Pe is the equivalent electrical output
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and the equivalent inertia constant is M1
M2 over M1 plus M2. Now at this point you
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can understand that the equivalent inertia
constant is as if we are connecting 2 resistances
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in parallel to find out the equivalent resistance.
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Suppose you have 2 resistances r1 and r2 and
you put them in parallel the equivalent resistance
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is r1, r2 or r 1 plus r2 therefore this is
the equivalent inertia constant is primarily
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determined by or is going to be closer to
the smaller one that is suppose I put M1 M2
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M plus M2 values, if M2 is large as compared
to M1. Okay then we will find out the value
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let us say I will just take example take just
values actually let us say M1 is 5 and M2
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is say 50 what will be the resultant? It is
going to be less than 5 but it is closer to
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5 not closer to a 50 therefore, the equivalent
machine will have the inertia which is less
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than, less than the the inertia of a smaller
machine or inertia of a machine which has
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a smaller inertia constant.
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Now here at this stage, to illustrate the
application of equal area criteria for analyzing
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stability of a system, I will consider an
example, the example we consider is a simple
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example. Let us consider this example, we
have a synchronous generator connected to
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a double circuit transmission line to an infinite
bus, we put infinity here to show that it
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is infinite bus its inertia constant is infinite.
To illustrate the application of equal area
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criteria of stability what we will consider
is that yes there is a machine infinite bus
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system. We will consider a fault at the middle
of the transmission line, one of the transmission
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lines at the point P.
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We will consider a three phase fault, a balance
3 phase fault okay. I will consider the unsymmetrical
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faults in my next lecture because in any system
we do come across symmetrical, unsymmetrical
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faults and unsymmetrical faults are more frequent
in occurrence okay. Now let us assume some
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parameters okay, let us say Xd prime for this
system is 0.2, reactance of this transformer
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is say 0.1 that is impedance is j times 0.1.
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The transmission line is loss free and its
reactance is j times 0.4 and further we assume
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that the infinite bus voltage V is 1.0 and
we consider this infinite bus voltage as reference,
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so that V 1.0 angle 0this is the terminal
of the synchronous generator. We denote the
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terminal voltage by the symbol Vt and let
us assume that Vt magnitude is given to be
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equal to 1.0 per unit and also it is assumed
actually that this machine delivers electrical
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power Pe equal to 1.0 per unit okay.
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Now for this system, for this system we will
obtain the power angle characteristic for
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3 conditions, one is pre-fault operating condition,
second is during fault or fault on operating
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condition. Now once the fault is there in
the system the fault will be cleared by operating
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these circuit breakers at the two ends of
the transmission line and therefore the third
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operate, third power angle characteristic
which we will obtain is post-fault power angle
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characteristic, although we are considering
a simple system but this exercise is required
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to be performed even for a multi machine system.
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We have to obtain the expressions for power
outputs of machines under pre-fault condition,
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during fault condition and post-fault condition
and the approach we are following here will
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also be applicable for a classical multi machine
stability problem.
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I am using about classical means here, we
will be making those basic assumptions that
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is synchronous generator can be represented
by a constant voltage behind direct axis transient
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reactance.
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Now with this information given how to find
out the pre-fault power angle characteristic
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Therefore, here our primary requirement is
that to get the pre-fault power angle characteristic
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what do we need is the voltage behind transient
reactance which is not given, what is given
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is the voltage at the terminal of the machine.
Now we can use this information to compute,
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to compute ah pre-fault power angle what is
first step is you draw from the one line diagram
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a reactance diagram.
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The reactance diagram can be drawn as a voltage
source, we will denote this voltage source
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as E prime, reactance Xd prime the value of
this impedance is given to be is given as
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j times 0.1, the transformer reactance is
.1 per unit that its impedance is I am sorry,
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the generator reactance is 0.2 not point .2
transformer is 0.1. These 2 transmission lines
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and infinite bus voltage, we denote this as
V at this point which is the terminal of the
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synchronous generator the voltage is Vt and
this magnitude of this voltage Vt is 1.0 while
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this magnitude of V is 1.0 and delta 1 is
taken as our delta is 0 the delta 1 is 0,
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okay. I can call it delta, now the with this
information or with this equivalent circuit
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what do we do is we find out what is the phase
angle of this Vt with respect to infinite
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00:28:29,110 --> 00:28:30,399
bus voltage.
202
00:28:30,399 --> 00:28:37,399
This can be obtained simply by using the relationship
that the power output P can be written as
203
00:28:40,489 --> 00:28:47,489
Vt magnitude into V divided by the reactance
connecting the terminal voltage or terminal
204
00:28:48,460 --> 00:28:55,460
of the generator to the infinite bus and let
us say that the, the phase angle between Vt
205
00:28:57,950 --> 00:29:04,950
and infinite bus voltage is alpha. Now we
substitute the values of power because it
206
00:29:07,639 --> 00:29:14,639
is we are supplying 1 per unit power Vt magnitude
is given as 1 this is one and the reactance
207
00:29:15,090 --> 00:29:22,090
between the infinite bus voltage and the generator
terminal is how much
208
00:29:23,279 --> 00:29:30,279
0.3 sin delta because here we are substituting
only the reactance value there is it is not
209
00:29:32,989 --> 00:29:35,230
impedance it is the magnitude okay.
210
00:29:35,230 --> 00:29:39,470
Therefore with this, you can calculate the
value of I am sorry there is a mistake here
211
00:29:39,470 --> 00:29:46,470
this we will call alpha comes out to be equal
to you calculate it and alpha in this case
212
00:29:49,119 --> 00:29:56,119
comes out to be equal to 17. 458 degrees okay.
Therefore I can say here that the terminal
213
00:30:02,149 --> 00:30:09,149
voltage Vt is equal 1.0 angle 17 .458 per
unit. Now the using this information we find
214
00:30:18,889 --> 00:30:25,889
out the current supplied by the generator
or I can be written as Vt minus V they are
215
00:30:31,989 --> 00:30:38,989
all phases divided by the total reactance
between the 2 machine or impedance, the impedance
216
00:30:39,299 --> 00:30:45,179
here is how much j times 0.3.
217
00:30:45,179 --> 00:30:52,179
Okay now if you substitute the value of Vt
and V, the current comes out to be equal to
218
00:30:55,090 --> 00:31:02,090
1.012 angle 8.729 degrees, okay. Now once
you get the current
219
00:31:17,960 --> 00:31:24,960
I supplied by the generator we can find out
now the internal voltage E prime that is E
220
00:31:26,179 --> 00:31:33,179
prime can be written as Vt plus j times Xd
into I
okay and you substitute the value of Vt Xd
221
00:31:43,269 --> 00:31:50,269
and I the calculated value of E prime comes
out to be equal to
that is magnitude of E prime because at this
222
00:32:00,220 --> 00:32:02,179
stage I am not interested in phase angle right.
223
00:32:02,179 --> 00:32:09,179
Now E prime comes out to be equal to 1.05
per unit and the pre-fault power angle characteristic
224
00:32:11,619 --> 00:32:18,619
Pe1 becomes now 1.05 into 1divided by total
reactance between infinite bus voltage and
225
00:32:24,109 --> 00:32:31,109
internal voltage that is coming out to be
how much .5 sin delta here now because when
226
00:32:35,460 --> 00:32:42,059
you are talking in terms of the relations
with the power considering the terminal voltage
227
00:32:42,059 --> 00:32:44,269
of the machine, we have written alpha.
228
00:32:44,269 --> 00:32:51,269
Now delta is the angle and therefore the pre-fault
power angle characteristic is now 2.10 sin,
229
00:32:57,419 --> 00:33:03,559
okay this these steps are extremely important.
We will follow we may have to follow the similar
230
00:33:03,559 --> 00:33:10,559
steps for a multi machine system also. The
next step is we want to find out the power
231
00:33:17,840 --> 00:33:23,119
angle curve or power angle characteristic
when fault is on.
232
00:33:23,119 --> 00:33:30,119
When the fault is on, on the system we can
write down or we can again draw the reactance
233
00:33:32,379 --> 00:33:39,379
diagram, since the I have consider the fault
at the middle of the line therefore I will
234
00:33:54,070 --> 00:34:01,070
divide this line reactance into 2 parts and
show as .2 .2 on both the sides of the faulted
235
00:34:01,830 --> 00:34:08,830
point or voltage at the point P.
236
00:34:17,350 --> 00:34:23,320
Now since, we have considered a balanced three
phase fault this P will be connected to the
237
00:34:23,320 --> 00:34:30,320
reference bus directly connected there is
no impedance involved however when we consider
238
00:34:33,359 --> 00:34:40,359
the unsymmetrical fault we will see that to
analyze the or to obtain the power angle characteristic
239
00:34:43,919 --> 00:34:48,010
during fault conditions there will be some
impedance connected between the faulted point
240
00:34:48,010 --> 00:34:55,010
P and reference bus. This impedance will depend
upon the type of fault but for a 3 phase fault
241
00:34:57,450 --> 00:35:00,100
the impedance to be connected is 0.
242
00:35:00,100 --> 00:35:05,020
Now here we are considering a three phase
metallic fault okay there is no fault impedance
243
00:35:05,020 --> 00:35:12,020
. Now in this the total reactance of the this
these two components that is the Xd prime
244
00:35:14,540 --> 00:35:21,150
and the transformer can be combined and this
can be written as impedance is 0.3, this is
245
00:35:21,150 --> 00:35:28,150
j times 0.4, this is j times 0.2, this is
j times 0.2 and this voltage is this voltage
246
00:35:39,300 --> 00:35:46,300
magnitude is this is V voltage, this is E
prime okay. Now what we do is we will try
247
00:35:49,910 --> 00:35:55,450
to simply this network, so that we can find
out the power angle characteristic during
248
00:35:55,450 --> 00:36:01,070
fault on period.
249
00:36:01,070 --> 00:36:08,070
This diagram can be redrawn as, now we will
denote this node as 1 denote this node as
250
00:37:05,060 --> 00:37:12,060
2, okay this node as 0 and this as now here,
if you examine this network then these 3reactances
251
00:37:20,440 --> 00:37:27,440
which are connected in star, j times between
1 and 3, 3 and 2 and 3 and 0.
252
00:37:29,960 --> 00:37:36,960
Okay this star can be replaced by an equivalent
delta, the equivalent using the standard star
253
00:37:50,240 --> 00:37:57,240
delta transformation technique this exercise
I have obtained and I find the equivalent
254
00:37:59,660 --> 00:38:06,660
network after this star delta transformation
as, E prime this reactance comes out to be
255
00:38:54,010 --> 00:38:59,500
when you calculate actually comes out to be
or this impedance comes out to be j times
256
00:38:59,500 --> 00:39:06,500
0.65, this comes out to be j times 0. I am
sorry it is j times 1.3, this is j times 0.266
257
00:39:15,760 --> 00:39:22,760
and what is this value? j times 0.2 and this
is your V and this is E prime.
258
00:39:24,290 --> 00:39:31,290
Now these terminals are this is your 1, this
terminal is 2 and the node 3 has been eliminated
259
00:39:31,480 --> 00:39:38,480
by star delta transformation this is our reference.
Now the power angle characteristic will depend
260
00:39:39,460 --> 00:39:45,530
upon the reactance connecting these 2nodes
1 and 2, one is the node of at which E prime
261
00:39:45,530 --> 00:39:51,410
is connected two is the node at which the
infinite bus voltage is connected this characteristic
262
00:39:51,410 --> 00:39:58,410
is not going to be affected by the shunt branches
because whatsoever is the, you now current
263
00:40:04,140 --> 00:40:09,160
which is flowing through the certain branches,
okay is not going to affect what is the power
264
00:40:09,160 --> 00:40:16,160
which is going to be transferred and therefore
the the power angle characteristic, the power
265
00:40:23,240 --> 00:40:30,240
angle characteristic during fault condition
can be obtained as now I will call it as Pe2
266
00:40:31,770 --> 00:40:38,770
equal to 1.05 which is the E prime into 1
divided by the reactance connecting node 1
267
00:40:42,150 --> 00:40:49,150
and 2 and that is 1.3 sin delta and this comes
out to be there.
268
00:40:52,720 --> 00:40:59,720
We computed and its value comes out to be
0.808 sin delta or what you observe here is
269
00:41:08,220 --> 00:41:15,220
that this Pe2 which is the equal to Pe max
into sin delta right, the the maximum value
270
00:41:18,870 --> 00:41:25,450
of this power angle characteristic will depend
primarily upon what is the reactance connecting
271
00:41:25,450 --> 00:41:32,450
the internal voltage of the synchronous machine
to the infinite bus voltage more is this reactance
272
00:41:35,270 --> 00:41:39,340
less will be the voltage okay.
273
00:41:39,340 --> 00:41:46,340
Now we can very easily obtain the post fault
power angle characteristic under the post
274
00:41:51,940 --> 00:41:58,940
fault condition one of the transmission line
is cleared or it is removed and therefore
275
00:42:00,540 --> 00:42:05,930
the reactance connecting the two nodes, internal
voltage of the synchronous generator and infiltrate
276
00:42:05,930 --> 00:42:12,930
bus voltage that comes out to be how much
.7 therefore our characteristic is now this
277
00:42:13,800 --> 00:42:20,800
divided by .7 sin delta. Okay now these are
the 3 important power angle characteristics
278
00:42:24,590 --> 00:42:31,590
which one has to compute or obtain for analyzing
the stability of a machine infinite bus system
279
00:42:33,100 --> 00:42:40,100
either we apply criteria or you directly solve
the some equation, it is the a material. The
280
00:42:45,570 --> 00:42:52,570
these 3 characteristics if you draw it can
be shown like this.
281
00:42:56,470 --> 00:43:03,470
I will call this is a Pe1 this is pre-fault
power angle characteristic in this particular
282
00:43:14,800 --> 00:43:21,800
case this is 2.1 P maximum value
under faulted condition or the power angle
curve under fault condition was found to have
283
00:43:30,480 --> 00:43:37,480
the maximum value equal to around .808 and
therefore the characteristic can be shown
284
00:43:38,210 --> 00:43:45,210
to be like this and this is 0.808 after the
fault is cleared, the amplitude of this power
285
00:43:56,470 --> 00:44:03,470
angle characteristic how much what is the
value of this1.5, 1.5 that is very good and
286
00:44:08,720 --> 00:44:15,720
therefore the third characteristic can be
plotted here, then this is 1.5 this is your
287
00:44:28,810 --> 00:44:35,810
Pe3, this is your Pe Pe2.
288
00:44:42,170 --> 00:44:49,170
In some of our representations what we will
do is that the Pe1 is written as Pmax sin
289
00:44:57,740 --> 00:45:04,740
delta Pe2 will be written as r1 times Pmax
sin delta, okay and Pe3 will be written as
290
00:45:19,060 --> 00:45:26,060
r2 times Pmax sin delta. Okay now this r1
and r2 these are the multiplying factors constants
291
00:45:33,780 --> 00:45:40,780
they will always be less than 1 and now I
shall take these 2 examples to illustrate
292
00:45:43,300 --> 00:45:50,300
how we apply the equal area criterion of stability.
293
00:45:53,790 --> 00:46:00,790
Now let us consider the first case where,
we
assume that there is sustain fault, fault
294
00:46:03,850 --> 00:46:10,850
is not cleared. Okay now in this particular
case the diagram shows that this is the output
295
00:46:13,440 --> 00:46:18,990
on normal conditions or can say pre-fault
power angle curve this is the output when
296
00:46:18,990 --> 00:46:24,890
fault is on this is the during the fault power
angle characteristic, the mechanical input
297
00:46:24,890 --> 00:46:31,890
line is shown here Pm the initial operating
angle is operating angle is delta naught where
298
00:46:35,670 --> 00:46:40,330
the pre-fault power angle characteristic intersect.
299
00:46:40,330 --> 00:46:47,330
Now the movement fault occurs the operating
points shifts from A to B then it moves from
300
00:46:50,160 --> 00:46:57,160
B on this during the fault power angle curve
and as you have seen that when its comes to
301
00:47:00,100 --> 00:47:07,100
this point C right the accelerating power
becomes 0 and it swings beyond this point
302
00:47:10,310 --> 00:47:17,310
and its swings to the point up to say D, this
is corresponding to angle delta M if these
303
00:47:17,980 --> 00:47:24,980
two areas become equal the rotor will swing
up to an angle equal to delta M and then from
304
00:47:26,500 --> 00:47:33,500
this point it will start returning back why
it starts returning back because it is now
305
00:47:34,060 --> 00:47:41,040
subjected to retardation electrical power
is more than the and when it comes back to
306
00:47:41,040 --> 00:47:48,040
this A what will happen will it stop here,
no it will continue to move and ultimately
307
00:47:48,540 --> 00:47:54,840
it is going to settle to this point C because
because system has some damping which we have
308
00:47:54,840 --> 00:48:01,840
not considered while writing the swing equation.
309
00:48:05,430 --> 00:48:12,430
Now here
depending upon actually the power angle curve,
in case actually the height of this power
310
00:48:17,010 --> 00:48:22,850
angle curve is less you will find actually
that you may have to swing to to larger angle
311
00:48:22,850 --> 00:48:29,390
to make this 2 areas equal. Now maximum swing
up to which rotor is going to be subject to
312
00:48:29,390 --> 00:48:36,030
retardation is where where the power angle
characteristic intersects with the mechanical
313
00:48:36,030 --> 00:48:43,030
input line at this point. In case suppose
the energy or kinetic energy gained is not
314
00:48:46,390 --> 00:48:51,500
returned back when it comes to this point
then suppose the movement it crosses this
315
00:48:51,500 --> 00:48:58,250
point, you will find that the machine is again
subjected to acceleration and therefore rotor
316
00:48:58,250 --> 00:49:02,000
is going to lose synchronism.
317
00:49:02,000 --> 00:49:09,000
Now this is shown in this diagram, where the
situation is slightly shown to be different
318
00:49:11,930 --> 00:49:17,820
where even when the rotor has come to the
point “e” these 2areas are not equal and
319
00:49:17,820 --> 00:49:24,820
system may lose synchronism. Now the third
case and the last case which I am discussing
320
00:49:24,880 --> 00:49:31,880
here is that we have a situation where we,
we switch off the faulted line or fault is
321
00:49:36,440 --> 00:49:40,410
cleared.
Now you start looking at this there are 3power
322
00:49:40,410 --> 00:49:46,250
angle curves this is the mechanical input
line initially we are operating at this point
323
00:49:46,250 --> 00:49:52,000
A, the movement fault occurs you shift to
the point B on the during the fault power
324
00:49:52,000 --> 00:49:59,000
angle curve. Now when you are moving on this
curve at this point C at on the angle equal
325
00:49:59,730 --> 00:50:06,350
to delta C the fault is cleared. We call this
as a fault clearing angle, the angle at which
326
00:50:06,350 --> 00:50:13,350
the fault is cleared then you will shift from
this point now to the post fault power angle
327
00:50:14,380 --> 00:50:20,210
curve then that is this is the, this curve
is the post fault output again as you know
328
00:50:20,210 --> 00:50:27,210
that this is this this is the area a1 that
is this is bound by these 2 angles and this
329
00:50:28,250 --> 00:50:32,670
power angle characteristic and mechanical
input line this becomes the accelerating area
330
00:50:32,670 --> 00:50:38,600
from “e” it will continue to swing it
come to the point f and the maximum angle
331
00:50:38,600 --> 00:50:44,180
becomes delta M at this point we find actually
that these 2 areas equal and therefore the
332
00:50:44,180 --> 00:50:50,550
maximum swing is up to delta M and the system
will return back.
333
00:50:50,550 --> 00:50:57,550
Now a new new stable operating point is now
where the post fault power angle characteristic
334
00:51:00,390 --> 00:51:07,390
intersects the mechanical input line that
is in this diagram this is the new stable
335
00:51:07,440 --> 00:51:13,960
operating point therefore, the rotor is going
to swing around this point okay oscillate
336
00:51:13,960 --> 00:51:20,960
around this point and because it has some
damping, it will settle to this new condition.
337
00:51:24,990 --> 00:51:29,360
Now with this I will just summarize what we
have discussed today.
338
00:51:29,360 --> 00:51:36,360
We have established the equal area criterion
of stability. We have also obtained for a
339
00:51:39,480 --> 00:51:46,480
two finite machine system an equivalent a
machine infinite bus system. We have also
340
00:51:48,490 --> 00:51:55,490
obtained for a given particular system the
pre-fault, during fault and post fault power
341
00:52:01,910 --> 00:52:08,910
angle characteristic a simple method is you
can say discussed here and at the end I have
342
00:52:10,780 --> 00:52:16,920
considered the 2 cases, one is considering
a sustained fault and another is the fault
343
00:52:16,920 --> 00:52:23,920
cleared after small amount of time, small
time. Okay I conclude my ah presentation here
344
00:52:26,080 --> 00:52:33,080
and thank you very much.