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Welcome to lecture number seven of module
two. In previous lecture, just we have seen
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that equal area criteria can be used to analyze
the stability of the system, and that is only
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applicable for this single machine Infinite
bus system or two machine system. And the
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last example we saw, that if we are increasing,
the sudden increase in the power, how your
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system will be reacting, and what will be
the stability of the system. Now, let us determine
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what will be the maximum power that can be
increased, on any machine over on over what
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is loading, present loading, and that is basically
governed by the stability of the system.
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Let us take your machine is loaded again this
machine, now your ultra. This is your synchronous
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machine. It is a here, it is a reactance,
and this is the transformer, and then we have
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two lines, and then it is connected with the
Infinite bus here, that is your Infinite here.
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So, this is your voltage angle zero; that
is V infinite. Here it is Ef angle delta.
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So, in this case, here your Pm that is your
mechanical power; that is entering, and now
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I can say this is Pm not. For this case, your
Pm naught here where your machine is operating
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at angle delta naught, and this is your Pm
and this A point; means A point will be the
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operation when your feeding, this mechanical
power is Pm not. As again I explained that
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the operation point will be the Intersection
of this mechanical power characteristic; that
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is Pm line here, and then that is your Pe
line, this is your Pe curve, which is a function
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of sine delta. So, the Intersection here the
point A now the question I want to know, how
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much power that we can increase that is your
Pm you want to calculate, so that our machine
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should be stable without losing synchronous.
To know this we have seen that is equal area
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criteria, is used to determine the maximum
additional Pm, which can be applied for to
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stability to be maintained, with the sudden
change in power Input. The stability is maintained
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only If the area A 2. This area, this is your
A 2, at least equal to area A 1. Means this
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area here, it will at least equal this, means
this area may be larger, but it should not
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be less, but If this area is less than this
A 1, then your system is not stable. So, we
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want that maximum power that we can Inject,
suddenly we can change on the loaded power
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system, or on this generator, that can be
determined; that means, area A 1 will be equal
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to area A 2, and then we can what will be
the determine the Pm value. To determine we
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have to calculate this delta 1, or we can
have calculate this delta max, because this
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delta max is nothing, but here delta max will
be, or you can say is a pi minus delta max
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will be your delta 1. So, to determine delta
1, we want to know delta 1, then we can get
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the point Pe, because the Pe characteristic
we know. If we put this delta value in that
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one here; delta 1 in that Pe characteristic,
we will get the point Pe, and thus we can
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determine the Pm. We will see in the following
lines. So, here now this area one, here is
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escalating power. Means this area as I said,
so to maintain the stability area A 2 at least
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equal to area A one, can be located A above
the Pm, If area A 2 is less than area two.
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If this is less than area two the accelerating
momentum can never be overcome.
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The limit of stability occurs, when delta
max is at the Intersection of the line Pm
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here, means this Pm we want to determine,
and the Intersection of Pe and this Pm will
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give your delta max, and this will be always
here, will be delta will be more than ninety
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degree this side, means here in this zone,
here it is your pi by 2 that is 90 degree.
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Applying equal area criteria to above figure,
we can have this area A 1. This area that
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is acb here you can say area acb, or you can
say area one will be nothing, but integration
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from delta naught to delta 1 Pm minus this
Pe curve. So, means this is nothing, but here
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it is Integration of delta naught to delta
1. Here this Pm minus Pe d delta. So, this
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Pm is constant. So, Integration of this we
can take the Pm outside. So, it is a delta
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1 minus delta 2 first term, and another term
this Pe with the minus sign we are integrating
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from delta naught to delta 1 that is P max
sine delta d delta. Similarly, this area 2,
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it will be the Integration here from delta
1 to delta max. And now it is Pe minus Pm,
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because this curve is over, and then this
minus this Integration from here to here,
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will be your area 2. And then we can know
this Pe is a function of sine delta. So, it
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is integration from delta 1 to delta max P
max sine delta d delta minus this Pm is constant,
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so delta minus delta e delta 1.
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So, based on the equal area criteria to the
system is stable, at least the area A 2 should
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be equal to area A 1. So, this area one for
maximum Pm, the area A 1 should be equal to
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the area A two, and then from the previous
equation, If we will equate we will get this
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expression. Now, here the Pm we are getting
here, and that Pm is nothing, but again we
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can get the Pm is equal to Pmax sine delta
max. Means this value the Pm, will be again
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that will be your Pmax sine delta max we can
get this point, and then you can substitute
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this. So, we will get a expression of us a
delta max value, and this is a non-linear
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equation; means we have the sine delta max,
we have another delta max here, we have cosine
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of delta max, and cos delta naught is known
to you. Delta naught is original, where your
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machine was operating, that can be calculated,
because Pm is known to you, Pm naught here.
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So, this equation is a non-linear algebraic
equation, and can be solved by any Iterative
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techniques, including your Newton Raphson
method, Gauss Seidal methods, Gauss Iteration
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method so on and so forth. So, that we can
determine delta max. Once this delta max,
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means here our Intension is to find delta
max. Once you can determine the delta max,
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then the maximum power Pmax will be Pmax sine
delta here; this Pm, or you can get this one
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directly.
So, you even though where delta 1 is your
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pi max delta minus, or here you can write
this simply Pmax sine delta max. So, this
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will be the maximum power that you can load
above this Pm not, without losing their stability.
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So, this is one way to determine that how
much you can load suddenly. Again question
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here, if you are continuously increasing,
then your maximum loading here is up to Pmax.
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But the sudden increase here for the sudden,
means it will be accelerating and decelerating,
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because here is an acceleration power, here
machine is a decelerating, here this zero,
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and here is again escalating so on and so
forth. So, since why this area A 1 is called
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accelerating, because the Pm one, Pm is more
than Pe, so what will happen. Your machine
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will accelerate, and it will store the kinetic
energy; however this area A 2 is decelerating
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area, because your Pe is more means the power
electrical power that you are taking from
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the system is more, means here, then your
power that you are feeding to the turbine,
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so, this is a decelerating. So, accelerating
power must be equal to you decelerating power,
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then the Pm can be determined.
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Now, let us see another consideration; means
how this equal area criteria, can be useful
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for determining the other stability criteria.
Let us suppose there is some fault. So, if
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we consider the three phase fault; the three
phase fault can be a bus fault, it can be
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in the line fault, and it can be anywhere
in the line either in sending or in receiving
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end or in between line. So, let us first consider
the three phase fault at the generator terminal,
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here that is this terminal we are boarding.
Basically this is a generator, this is a GT
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generating transformer and this is a terminal,
where we can say this is bus one. So, again
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this system is similar to our previous system;
that here a generator is connected to infinite
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bus bar through two parallel lines; line one,
and line two. Assume that the input power
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again here is Pm is your constant, and the
machine is operating steadily delivering power
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to the system with an angle delta delta not.
Here means your delta naught is Ef means,
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if you are here I can say E delta naught during
their steady state, and that will be shown
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in the next term. Our temporary three phase
fault that is bolted fault. Bolted fault means
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the fault without any up fault resistors,
means directly three phase to ground fault,
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occurs at the sending end of one of the transmission
line at the bus one here; three phase fault
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let us suppose has occur here in line two.
When the fault is at the sending end of the
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line, no power is transmitted to the infinite
bus. If the bus fault is here, what will happen.
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There will be power which will be flowing
to Infinite bus. So, your Pe at that case,
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it will be your zero. Since the resistance
is neglected of the line, the electrical power
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Pe is zero, and the power angle curve corresponding
to the horizontal axis. Means you can see
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what will happen.
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Your this Pm here A was your operating point,
because your Pe here is equal to your Pm not;
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that is Pm naught bus you are getting. So,
suddenly there is a fault, what is happening
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your Pe becomes zero. So, this is suddenly
coming to this point, and then till the fault
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is persisting here, it is going and at the
delta. Once fault is cleared, and then again
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it is going back to the Pe. Again the Pe is
going to be delivering to the Infinite bus.
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So, in this case the machine accelerates with
the total input power, as the accelerating.
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Means the total Pm, because we are not taking
Pe from that generator. So, this machine will
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get escalated, means your input power is more
than output power, so, whatever power you
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are feeding as you know the energy conservation
law, then this energy will be stored in terms
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of kinetic energy. And if the kinetic energy
of any system Increasing, means speed of that
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machine rotating mass is increasing. So, there
by it increasing its speed, storing added
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kinetic energy, and increasing the delta.
If machine is accelerating what is happening,
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now delta will be changing, the delta is angle
if you see what is now question is delta.
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Delta is, let us suppose this your reference
angle, any or rotary reference angle. Here
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this is your delta not, and your machine here
this is a rotating two fluxes here. This is
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your synchronously rotating here, and it is
actual rotor speed is omega r; this is omega
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not. So, this difference that angle; that
means, your delta is omega r minus omega not;
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that is t with delta not. So, if though machine
is accelerating this is increasing, this is
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your synchronous speed. So, what will happen
this term is increasing with the time. So,
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your delta is increasing. Here delta naught
is any represent, so this delta naught will
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be zero. So, once your machine is accelerating,
means your delta is accelerating, increasing,
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and once it is decelerating means your delta;
that is angle delta I have defined with this
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one, it is a decreasing. So, that is retaining
here the Increasing angle delta. When the
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fault is clear, and it was assumed that the
fault was temporary, and the fault it cleared
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without tripping the of the line, because
it was the bus fault, and the bus fault is
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cleared. So, both line assumed to be Intact.
Means your Impedance of the system is not
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going to be changed. So, again it will follow
the Pe characteristic. The fault is clear
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at delta one, which the operation to the original
power angle curve at point e; here this is
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your point d. Means your fault is cleared
here, means it will suddenly follow this delta
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curve, because this is Pe here Pe is zero.
So, this Pe it was fault running at this point,
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suddenly due to the fault this Pe becomes
zero; this is your zero Pe this is a Pe axis.
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And once fault is cleared, then it will again
just it will follow the Pe curve here; that
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is here it is your Pe curve it is a function
of sine delta. So, it is coming to your e
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point. The net power is now decelerating,
and the previously stored kinetic energy will
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be reduced to zero at point f, when the shaded
area is shown by the area two, equal to the
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area shaded in abcd shown by this. Now, you
can see this area A one, as I said this one
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is your accelerating power. Means rare in
this area A 1 your machine is accelerating,
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and you are storing kinetic energy. Now, once
this kinetic energy is stored in the machine,
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what is happening at the e point, you can
see this is now electrical power is more than
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your mechanical power. Means you are taking
more power. So, whatever the energy is stored
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in the machine that will be taken out, and
this area here it is called your decelerating.
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Means here it was accelerating, now from here
it will decelerate till f, when again this
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your acceleration will become zero.
So, again you can see the delta is keep on
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increasing here and it is delta max. Here
again your decelerating power is zero, then
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again it will find that this is more than
this it will try to retard, and again it will
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coming back. So, it is again there will be
oscillation around point A, and finally, due
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to the damping of the system, it will be damp
out, and your final steady state operate of
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the system it will be stable operation. Now,
the question again here arise, If this area
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is less than this area. If this is area is
more A 1. Means your area A 1 is greater than
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two, and till what point, till certain range
here, then machine will be unstable. So, we
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can the rotor angle would then oscillate back,
and force around here this not, means here
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it will be oscillating at its natural frequency,
because of Inherent damping in the system.
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The oscillation subside, and the operating
points return to the original power angle
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delta naught here; that is at point A.
So, by applying the equal area criteria, this
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area A 1 that is integration from delta naught
to delta naught to delta one, here that is
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Integration of here this completely Pm minus
this zero. So, this Pm delta, and area A 2
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will be delta 1 integration from delta 1 to
delta max. Here this Pe minus Pm d delta,
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means accelerating area one will be equal
to accelerating area two. Now you can see,
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how much you can move ahead this is f. Let
us suppose you have increased further, means
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it is not clear at this one. Let us suppose
your machine, fault is clear at this point
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what will happen. Now this area here it is
up to this area, and once fault is clear now
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it will follow here, and then it will go up
to here area now area two will this much completely,
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and it will be your area two is this much.
So, still your system is stable. Let us suppose
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further you have increased. Let us suppose
your delta 1 your fault is cleared after longer
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time. So, this area now is registered here,
and now you have come down here, and again
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you can see the area this is bounded here
A 2, is now less than your area A 1 complete
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this area, this area here is larger than this.
So, in this case your area is more than A
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2 and the system is unstable. So, what we
normally do. We try to determine what should
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be value of delta one. Here what I am going
to explain, that how much this delta 1 that
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can be maximum value means the fault can be
cleared, so that our system will be stable.
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Why we are very concern about this this delta
one; that is maximum delta 1 at which we can
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clear the fault at which our system will be
stable, this is known as the critical clearing
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time. Critical clearing angle delta If you
are calculating this delta If you are calculating
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in the time then it is called the critical
calculating time CCT, or here delta 1 is your
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critical clearing angle.
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In the next slide we will see; that is delta
C is critical clearing time, that is what
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will be this value you want to determine this,
so that you fault is cleared before that.
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So, in that case suppose your fault is cleared.
So, this condition will be always cleared
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here this area A 2 will be always you can
say more than your area A 1, means your system
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will be stable. So, from here this complete
area is always larger, and then A one. So,
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we can say our accelerating power will be
always less than your decelerating power.
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So, the system will be stable. To determine
this critical clearing time, that is very
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important. The critical clearing angle is
reached when any further Increase in delta
187
00:20:40,350 --> 00:20:48,040
one; that is delta 1 here, causes the area
A 2, representing decelerating energy to become
188
00:20:48,040 --> 00:20:54,900
less than the area representing the accelerating
energy; that is your A one. This occurs when
189
00:20:54,900 --> 00:21:02,230
the delta max here or the point f is at the
Intersection of the line point Pm, here this
190
00:21:02,230 --> 00:21:12,330
is line Pm, and this is curve Pe then it is
Intersection point f as shown in this figure.
191
00:21:12,330 --> 00:21:20,350
Now for this critical value, here this point
cannot come below, because this is decelerating
192
00:21:20,350 --> 00:21:26,640
up to this point Pe is more than Pm. So, this
is decelerating. You cannot go beyond this
193
00:21:26,640 --> 00:21:31,280
because if you are coming here what will happen
your Pm will be more than Pe and again it
194
00:21:31,280 --> 00:21:35,860
will accelerate. So, your system will be unstable
if it is moving beyond this point. So, we
195
00:21:35,860 --> 00:21:42,620
have to this area A 2 should be less than
this delta max, and then we have to determine
196
00:21:42,620 --> 00:21:48,650
this delta c, so that we can know if your
fault is cleared less than that angle, then
197
00:21:48,650 --> 00:21:54,220
we can say our system is stable. To be very
critical for getting the maximum value of
198
00:21:54,220 --> 00:22:00,330
delta at which it is cleared for this stable
operation. This area A 1 should be equal to
199
00:22:00,330 --> 00:22:11,770
the area A 2 here. So, area A 1 that is your
ABCD will be equal to your DEF be, this is
200
00:22:11,770 --> 00:22:18,290
area two. So, this area for this case it is
you Pm Integration from delta naught to delta
201
00:22:18,290 --> 00:22:29,380
c, is equal to your Integration from delta
c to delta max. Here Pe minus Pm, so it is
202
00:22:29,380 --> 00:22:36,260
Pm max here this is nothing, but pe minus
Pm delta d delta.
203
00:22:36,260 --> 00:22:45,340
So, integrating both side, you can say left
hand side here, your Pm delta c minus delta
204
00:22:45,340 --> 00:22:51,740
naught will be there, your another side here
it is a Pm cos delta c minus cos delta max
205
00:22:51,740 --> 00:22:59,010
minus Pm delta max minus delta c. So, here
this you can say this delta c this delta c
206
00:22:59,010 --> 00:23:03,890
will be cancelled, and we are getting a 1
equation of delta c very easily, and that
207
00:23:03,890 --> 00:23:10,260
we can say cos delta will be this function
or we can say delta c will be cosine Inverse
208
00:23:10,260 --> 00:23:19,630
Pm upon Pe max, delta max minus delta naught
plus cos delta max, and delta max can be determine
209
00:23:19,630 --> 00:23:29,470
again here we know this is Pm, is your Pmax,
here sine delta max. So, if you have solve
210
00:23:29,470 --> 00:23:36,720
this knowing Pm Pm as well as Ps max, you
will get the two values of delta max. Means
211
00:23:36,720 --> 00:23:44,520
here in the previous case, If you will write
this equation this Pm will be equal to this
212
00:23:44,520 --> 00:23:53,320
is Pm at this f point, it is your Pmax here
sine delta If you will solve here, knowing
213
00:23:53,320 --> 00:23:59,430
this Pm, knowing this you will get the two
values of delta 1 you will get the delta one,
214
00:23:59,430 --> 00:24:04,690
another delta max and these value less than
equal to 180.
215
00:24:04,690 --> 00:24:09,570
So, since we are talking up to the pi degree,
so you will get the two values. So, first
216
00:24:09,570 --> 00:24:16,650
one the small value will be your delta naught
here in this case; that is here, and this
217
00:24:16,650 --> 00:24:22,120
value delta max you can get the another value
that will be again the pi minus delta naught
218
00:24:22,120 --> 00:24:27,590
will be the delta max, means here delta max
will be nothing, but your pi minus delta not.
219
00:24:27,590 --> 00:24:35,370
You can see from here it this angle will be
equal to this angle. So, now we can get this
220
00:24:35,370 --> 00:24:40,690
by, because you are knowing this delta max.
We know this delta naught and then Pm Pmax
221
00:24:40,690 --> 00:24:47,160
we know, and then we can determine the critical
clearing angle and this is known CCA, and
222
00:24:47,160 --> 00:24:52,880
that is one of the major system stability.
Means if your fault is cleared before this
223
00:24:52,880 --> 00:24:57,960
angle your system will be stable, and if it
is cleared after this your system will be
224
00:24:57,960 --> 00:25:04,340
unstable. In addition it is possible to find
the critical clearing time, because angle
225
00:25:04,340 --> 00:25:09,250
is very difficult to say what is angle and
alpha measure, and the time is very important
226
00:25:09,250 --> 00:25:13,490
means at what time your fault is cleared,
and that is called your critical clearing
227
00:25:13,490 --> 00:25:19,190
time and that can be calculated again by using
the swing equation.
228
00:25:19,190 --> 00:25:25,650
Your swing equation is nothing, but your H
upon pi is f naught, here the double differentiation
229
00:25:25,650 --> 00:25:31,020
of delta with respect to time that will be
equal to your Pm minus Pe, and this Pe is
230
00:25:31,020 --> 00:25:38,110
nothing, but your zero, because here in this
curve you can see this is your Pe is zero.
231
00:25:38,110 --> 00:25:47,780
So, you want to Integrate from here to here,
where the Pe is zero, or we can say this is
232
00:25:47,780 --> 00:25:52,920
Pe we can write the delta we can differentiate
this equation ,we can solve this equation.
233
00:25:52,920 --> 00:25:58,960
Here we will get the pi f not upon twice H
Pm t square plus delta not, because it is
234
00:25:58,960 --> 00:26:04,040
starting from delta not. So, delta naught
is added, and then this is with respect to
235
00:26:04,040 --> 00:26:11,140
time thus delta is increasing. So, delta is
equal to now delta c If you will put this
236
00:26:11,140 --> 00:26:16,370
value. So, at that time your t will become
tc, means at delta is equal to your delta
237
00:26:16,370 --> 00:26:23,180
c, your time will become your tc. So, the
tc will be nothing, but here, it is delta
238
00:26:23,180 --> 00:26:30,270
c minus delta naught here, twice H is multiplied,
divided by twice f Pm under root it will be
239
00:26:30,270 --> 00:26:37,220
your tc, and this is called your cct; that
is a critical clearing time. So, this time
240
00:26:37,220 --> 00:26:39,350
can also be determined.
241
00:26:39,350 --> 00:26:47,770
Now, let us see the effect of pre fault load,
means earlier your system was loaded at the
242
00:26:47,770 --> 00:26:55,670
Pm one. Now in this what we are going to see
let this loading. In this case again your
243
00:26:55,670 --> 00:27:01,250
fault same system; there is a three fault
at the bus say, the curve is similar to the
244
00:27:01,250 --> 00:27:08,340
previous one. So, it was operating at the
0.1, that is angle at delta naught the fault
245
00:27:08,340 --> 00:27:13,900
was cleared after the delta c and this regard
the A 1 it will equal to A 2 and system is
246
00:27:13,900 --> 00:27:20,930
stable. Now you want to your machine was loaded
more than that your Pm 1 value. Let us suppose
247
00:27:20,930 --> 00:27:27,870
your value is Pm 2 and that your value is
Pm 2 1.5 times more. Means this is more than
248
00:27:27,870 --> 00:27:33,680
what will happen. In this case also critical
clearing time let us suppose the same fault
249
00:27:33,680 --> 00:27:41,070
is clear for the same duration. So, what happens
you can say, here again the 0.2 it was delta
250
00:27:41,070 --> 00:27:49,150
not, and again this delta naught is different
from this delta not. But this duration from
251
00:27:49,150 --> 00:27:57,200
two to three is same from here as well as
here. So, this Pm 2 is 1.5 times more this
252
00:27:57,200 --> 00:28:03,790
I have already written Pm 2 is equal to 1.5
times of Pm 1 here.
253
00:28:03,790 --> 00:28:11,760
Now, you can see this now area A 1; that this
time is same, this area is increased by 1.5
254
00:28:11,760 --> 00:28:20,870
times means this area was A 1. Now this area
one A prime mean A 1 prime, will be 1.5 times
255
00:28:20,870 --> 00:28:26,070
of A 1, because this height was Increased
by “one point five” times where this axis
256
00:28:26,070 --> 00:28:33,690
same. So, this area is increased. Now you
can see, and at the same time your area 2
257
00:28:33,690 --> 00:28:40,710
is in decrease. Means here what is happening,
this area is now shorter, smaller. So, we
258
00:28:40,710 --> 00:28:48,490
can see this again, that the Increase in the
pre fault load by 50 percent, the Pm 2 Increases
259
00:28:48,490 --> 00:28:55,250
the acceleration power, one and half times
here. It is increased by one and half times.
260
00:28:55,250 --> 00:29:01,740
So, based on the swing curve equation during
the fault, here Integration of d delta upon
261
00:29:01,740 --> 00:29:09,210
dt square, Pm upon m, then we can write this
delta this delta naught Pm t square upon 2
262
00:29:09,210 --> 00:29:15,360
m. Again it is written in terms of m your
momentum. And therefore, we can see this is
263
00:29:15,360 --> 00:29:25,580
your angle. The change in the power angle
delta here, this delta also increases by a
264
00:29:25,580 --> 00:29:30,890
factor 1.5 times consequently, at each side
of accelerating area rectangle one to three
265
00:29:30,890 --> 00:29:37,390
has increased by 1.5 times this is Increased
by 1.5 times the accelerating of one two three
266
00:29:37,390 --> 00:29:42,530
of here means this area is Increased than
this one, is now much larger than allowable
267
00:29:42,530 --> 00:29:49,010
decelerating area 4 5 8, means this area is
much less compared to this area.
268
00:29:49,010 --> 00:29:56,700
So, what happens if your pre fault load is
more, your system is again critical clearing
269
00:29:56,700 --> 00:30:01,910
time is lesser. Means you must clear your
fault well before here so that we can have
270
00:30:01,910 --> 00:30:08,860
area A 1 better more than your area A 2, means
we should have some margin we will see. Now
271
00:30:08,860 --> 00:30:16,550
question what is this area, because you A
1 is equal to A 2, your machine will here
272
00:30:16,550 --> 00:30:21,050
decelerating is equal to your accelerating
power, and finally, your machine will oscillate
273
00:30:21,050 --> 00:30:28,010
around the point one, and finally, it will
be stable, it will be stable. So, area 6 7
274
00:30:28,010 --> 00:30:39,000
8 is called the margin, and that is known
as stability margin, stability margin of the
275
00:30:39,000 --> 00:30:45,870
system. So, if this margin is larger, than
we say our system is more stable. So, this
276
00:30:45,870 --> 00:30:51,820
gives your relative stability. Means this
margin is an Indication of stability of the
277
00:30:51,820 --> 00:30:57,560
system. in this case you can say the stability
margin, If area is equal to this stability
278
00:30:57,560 --> 00:31:04,450
margin is zero, means in the case at the certain
critical clearing time, as I said here in
279
00:31:04,450 --> 00:31:09,850
this critical clearing time, this area is
equal to area two there is no margin here
280
00:31:09,850 --> 00:31:17,710
left out. So, if your system is cleared after
this angle your system will become unstable,
281
00:31:17,710 --> 00:31:22,420
but If it is cleared here, then we will have
some margin here, and we can say our system
282
00:31:22,420 --> 00:31:24,340
is stable.
283
00:31:27,270 --> 00:31:32,820
Let us say the three phase fault in the different
scenarios. So, far just we have discussed
284
00:31:32,820 --> 00:31:38,210
the three phase fault; that is occurring at
the bus one, but let us it is a fault is occurring
285
00:31:38,210 --> 00:31:43,680
in the line, and followed by the tripping
of the line. So, during the pre fault again
286
00:31:43,680 --> 00:31:51,240
your system as the previous, just we have
consider all the slides, same here your generator,
287
00:31:51,240 --> 00:31:56,670
generating transformer two lines and connected
to the Infinite bus system. So, during fault
288
00:31:56,670 --> 00:32:01,850
what is happening, now fault earlier it was
at bus one, now I am taking fault as line
289
00:32:01,850 --> 00:32:08,620
two, and that is your three phase fault. And
this fault is cleared only after tripping
290
00:32:08,620 --> 00:32:14,470
of this line. So, this is your post fault
scenario, means your line here is tripped
291
00:32:14,470 --> 00:32:22,110
fault is cleared. Now what is happening, the
system impedance here, is a different your
292
00:32:22,110 --> 00:32:28,330
due to the fault Impedance is different, and
now after tripping this Impedance is a different.
293
00:32:28,330 --> 00:32:36,520
Now in this case what will be your X, and
using this X here it your Xd prime; that is
294
00:32:36,520 --> 00:32:44,720
of your generator, your XT of your transformer,
and plus here Xl 1 parallel to Xl two, If
295
00:32:44,720 --> 00:32:52,900
you remember this value with I was using in
lecture number six. So, this is your X in
296
00:32:52,900 --> 00:33:00,400
this case. Now in this case what will be happening,
your X will be your Xd prime plus your Xt
297
00:33:00,400 --> 00:33:06,260
plus your Xl 1, only one line is there. So,
now you can say this Impedance is lesser than
298
00:33:06,260 --> 00:33:10,390
this X one, because the parallel Impedance
of the two systems is always less than Impedance
299
00:33:10,390 --> 00:33:18,620
of small. So, here this value is now more
compared to this. Now during the fault again
300
00:33:18,620 --> 00:33:22,650
the Impedance of the system is reduced, whether
the reducity is less than this because this
301
00:33:22,650 --> 00:33:26,430
is a faulty, and less power is flowing into
the system.
302
00:33:26,430 --> 00:33:32,809
To understand this now you can see these curves,
we have now three curves; one is your pre
303
00:33:32,809 --> 00:33:38,830
fault, another is your post fault, and then
another is during the fault. So, let your
304
00:33:38,830 --> 00:33:46,680
curve is A, outer curve from 0 to pi, and
that is denoted by your a and it is called
305
00:33:46,680 --> 00:33:52,590
the pre fault curve. Means we are all the
Impedances are intact, where X is your Xd
306
00:33:52,590 --> 00:33:59,120
prime plus Xd plus the parallel Impedance
of the two lines. Since Impedance is less
307
00:33:59,120 --> 00:34:04,910
in that case, so what is happening this, this
curve will be higher why. Again you remember
308
00:34:04,910 --> 00:34:12,230
this is your VE upon Xd prime, means X prime
simply sine delta, If this value is small
309
00:34:12,230 --> 00:34:19,829
then Pe will be larger value. So, it is outer
where it is follow. Now after the fault one
310
00:34:19,829 --> 00:34:25,020
line is tripped. So, what is happening your
Impedance is Increased, that let us suppose
311
00:34:25,020 --> 00:34:29,810
your point your c curve that is curve here,
that is following this, means you can see
312
00:34:29,810 --> 00:34:34,260
this curve.
During the fault what is happening, again
313
00:34:34,260 --> 00:34:40,480
there is a bolted fault, Impedance is again
increased tremendously and the power which
314
00:34:40,480 --> 00:34:46,810
is showing Inside the Infinite bus is reduced,
and let us suppose this one curve. There is
315
00:34:46,810 --> 00:34:52,889
a curve, and where it is denoted at the during
fault and it is denoted at the Pe 2, that
316
00:34:52,889 --> 00:35:01,320
is during post and one is pre fault. Now,
you can see the loading of the system was
317
00:35:01,320 --> 00:35:09,360
Pm not, before fault, your operating point
was A, because it is Intersection of pre fault
318
00:35:09,360 --> 00:35:16,720
curve A and your Pm naught curve, and this
is your operating, that is pre fault operating
319
00:35:16,720 --> 00:35:27,040
point, and the angle of the excitation voltage
your delta naught is this one. Now, let us
320
00:35:27,040 --> 00:35:32,930
fault we have to apply. Once fault has come
what will happen, during the fault that which
321
00:35:32,930 --> 00:35:39,500
is occurring here, this is suddenly it is
shifting to another curve, because it will
322
00:35:39,500 --> 00:35:44,980
always follow the Pe curve here. Now the Pe
curve it is now during the fault we have this
323
00:35:44,980 --> 00:35:49,350
curve, that is curve B, and it will be following
this curve.
324
00:35:49,350 --> 00:35:58,119
Now, what will happen, now in this case your
Pe which is fed to the Infinite bus is reduced,
325
00:35:58,119 --> 00:36:03,900
because here it is Pm is more and Pe is less,
so machine will accelerate. Accelerate means
326
00:36:03,900 --> 00:36:12,250
here your delta will Increase. Let us suppose
at delta one, your fault is strictly cleared
327
00:36:12,250 --> 00:36:18,840
and this fault clearing is associated with
this tripping of the line, means suddenly
328
00:36:18,840 --> 00:36:24,970
we have to reach the post fault condition;
that is point e of this curve that is the
329
00:36:24,970 --> 00:36:33,860
post fault curve Pe. And then what will happen,
this is your accelerating energy, this curve,
330
00:36:33,860 --> 00:36:40,369
then here it is going to decelerate till f,
till the point when the A 1 will be equal
331
00:36:40,369 --> 00:36:47,010
to A 2, and then again it will go back, and
then it will be again oscillating, and your
332
00:36:47,010 --> 00:36:53,300
final point of operation will be now here.
It will be another, it is your k point will
333
00:36:53,300 --> 00:36:58,130
be here the final point, because it will be
oscillating around this point, and your system
334
00:36:58,130 --> 00:37:03,350
is stable, if damping is there and the system
will subside.
335
00:37:03,350 --> 00:37:11,050
So, this is the condition when it is the fault,
means during fault, curve means this during
336
00:37:11,050 --> 00:37:15,960
fault, this is post fault, this is a pre fault.
So, this is retaining in this with the fault
337
00:37:15,960 --> 00:37:21,570
location away from the sending end, the equivalent
transfer reactance between bus bare is Increased,
338
00:37:21,570 --> 00:37:26,320
lowering the power transfer capability. And
the power angle curve is represented by the
339
00:37:26,320 --> 00:37:31,560
curve b during the fault, because less power
is fed. Finally, the curve c represent the
340
00:37:31,560 --> 00:37:37,010
post fault power angle curve, assuming the
faulted line is removed. When the three fault
341
00:37:37,010 --> 00:37:44,000
phase occurs, the operating point shift Immediately
to point B on curve B. here that is coming
342
00:37:44,000 --> 00:37:48,380
during the fault, and the axis of mechanical
Input power of this machine accelerate the
343
00:37:48,380 --> 00:37:54,590
rotor, there by storing kinetic energy, and
the angle this is Increased here up to this
344
00:37:54,590 --> 00:37:59,560
point. So, this is your point of movement,
here it is coming back, and then finally,
345
00:37:59,560 --> 00:38:03,910
attained it would be attaining now, later
that it will come follow this one. It will
346
00:38:03,910 --> 00:38:12,150
come here and then go back. So, in this curve
we will see, that this area again is here
347
00:38:12,150 --> 00:38:19,250
again, I can say in this vertical hashed line.
This is basically your margin that the system
348
00:38:19,250 --> 00:38:24,130
had. Means your system is stable no doubt,
and this area if this area is larger, we can
349
00:38:24,130 --> 00:38:28,080
say our system is more stable. If this area
is more stable if this area is very small,
350
00:38:28,080 --> 00:38:33,180
then we can say our system is less stable.
So, what happens there is a possibility, that
351
00:38:33,180 --> 00:38:38,310
to fault even though if it is not cleared
then here delta 1 less than that r at delta.
352
00:38:38,310 --> 00:38:42,520
So, this value will be kept on Increasing.
So, there is a possibility, let us suppose
353
00:38:42,520 --> 00:38:48,050
our delta is increased here. So, this machine
will go up to accelerating here, and this
354
00:38:48,050 --> 00:38:55,900
complete area, will be equal to If area A
2. Then this angle again theta delta c is
355
00:38:55,900 --> 00:39:01,710
called as a critical clearing. Means the critical
clearing angle is the angle that beyond that
356
00:39:01,710 --> 00:39:06,660
fault cleared that system will be unstable,
and if it is cleared before that angle, or
357
00:39:06,660 --> 00:39:11,570
you can say that time then your system will
be stable, and there will no margin means
358
00:39:11,570 --> 00:39:13,350
that is a critical clearing time.
359
00:39:13,350 --> 00:39:20,950
So, to calculate this, the critical clearing
angle, means I want to calculate this delta
360
00:39:20,950 --> 00:39:28,610
c at which this area up to this point; means
that is your Pm line. This area A 2 will be
361
00:39:28,610 --> 00:39:34,560
equal to your area A one, then we can say
our system is margin less stable, means angles
362
00:39:34,560 --> 00:39:40,430
beyond this If fault is cleared system is
unstable. To know this critical clearing time,
363
00:39:40,430 --> 00:39:46,670
because we have to plan our system; means
your protective device, Including your relays,
364
00:39:46,670 --> 00:39:50,830
as well as circuit breakers that must clear
the fault, if any fault occurs before this
365
00:39:50,830 --> 00:39:57,220
angle, and we can also calculate the time.
So, this delta c is the dead line, less than
366
00:39:57,220 --> 00:40:00,610
that fault must be removed otherwise your
system, this generator will be out of the
367
00:40:00,610 --> 00:40:08,080
unstable. So, this area A 1 will be equal
to your area A 2 and then we can determine
368
00:40:08,080 --> 00:40:15,360
again your delta c as the previous example.
Again this area is nothing, but now Pm minus
369
00:40:15,360 --> 00:40:25,170
Pe again the Pe here of the during fault that
is Pe 2; means I can say this area A 1 will
370
00:40:25,170 --> 00:40:35,410
be your Integration, from here delta naught
to delta c, and this one is your Pm minus
371
00:40:35,410 --> 00:40:41,520
Pe 2 d delta delta. So, this is your area
A 1.
372
00:40:41,520 --> 00:40:51,359
Similarly your area A 2 will be your Integration
from delta c to delta max, and this is your
373
00:40:51,359 --> 00:41:01,320
Pe now curve, it is your c curve that is Pe
three minus Pm d delta. So, if we equating
374
00:41:01,320 --> 00:41:07,000
this I am integrating for this case you can
say integrating this Pm is constant. So, we
375
00:41:07,000 --> 00:41:13,850
will get this Pm delta c minus delta not;
this term. Another term here is with the negative
376
00:41:13,850 --> 00:41:21,590
sign, integrating from delta naught to delta
c of this Pe 2 and Pe 2 is your P 2 max, Pe
377
00:41:21,590 --> 00:41:27,930
2 max we are integrating here. Similarly,
for another area two we can write in this
378
00:41:27,930 --> 00:41:35,480
fashion and again you can see this Pm delta
c here Pm delta c it is this side it is plus
379
00:41:35,480 --> 00:41:42,710
plus it will cancelled out. So, we can determine
this delta c that is a critical clearing angle,
380
00:41:42,710 --> 00:41:49,580
it is a cosine Inverse this Pm delta max minus
delta naught plus Pm 3 plus max cos delta
381
00:41:49,580 --> 00:41:58,420
max minus P 2 max cos delta naught divided
by P three max minus P 2 max. So, we can determine
382
00:41:58,420 --> 00:42:01,220
the critical clearing angle for the different
decision.
383
00:42:01,220 --> 00:42:06,980
Similarly, we can go for the different one
let us see in that case just fault is cleared,
384
00:42:06,980 --> 00:42:11,630
by the tripping of the line, but there may
be the possibility that the circuit breaker
385
00:42:11,630 --> 00:42:17,190
those are in the line, they may have some
re closing facility. Why we go for the re
386
00:42:17,190 --> 00:42:23,260
closing, because whenever there is temporary
fault occurs. There may be possibility, that
387
00:42:23,260 --> 00:42:28,990
this circuit breaker is opening and fault
is automatically cleared, and after the closing,
388
00:42:28,990 --> 00:42:33,970
it was found that there is no fault. For example,
let us suppose your there is a flux over on
389
00:42:33,970 --> 00:42:38,210
the Insulator, due to the fogging, due to
the motion on Insulator, there may be possibility
390
00:42:38,210 --> 00:42:44,720
that is line, due to the higher line voltage,
and there is a flash over on the line Insulator,
391
00:42:44,720 --> 00:42:50,630
and there will be the fire. So, your circuit
breaker will be clear, simply that it will
392
00:42:50,630 --> 00:42:56,740
be opening, and then it will try after few
cycles. Again depends upon the circuit breaker
393
00:42:56,740 --> 00:43:00,730
consideration type and make etcetera. So,
it will try to re close the fault to see,
394
00:43:00,730 --> 00:43:05,470
whether fault is cleared or not for example,
let us suppose some word, has come to the
395
00:43:05,470 --> 00:43:11,540
wire solitary wire. So, there is a depth three
phase wire, three phase circuit. So, once
396
00:43:11,540 --> 00:43:17,930
this any live animal or bird there is on that
phase here. Once that will die automatically
397
00:43:17,930 --> 00:43:23,330
that is again all these phrases are operating
in Isolation. So, the fault is automatically
398
00:43:23,330 --> 00:43:28,390
cleared after one saw. So, the circuit breaker
normally for ESB transmission line, they will
399
00:43:28,390 --> 00:43:30,320
have re closing facility.
400
00:43:30,320 --> 00:43:37,800
So, with that we can see the line re closing
as well. Means here our pre fault system as
401
00:43:37,800 --> 00:43:44,430
usual in the previous case, during fault we
have the three phase fault. Now we hear the
402
00:43:44,430 --> 00:43:50,350
fault is cleared by that tripping, and again
it is the line is re closed with the help
403
00:43:50,350 --> 00:43:56,100
of re closing facility of the circuit breaker
after this cycle. And again is your system
404
00:43:56,100 --> 00:44:03,130
configuration here is similar to the pre fault
condition. So, now, what will be the scenario
405
00:44:03,130 --> 00:44:11,270
to see this. Again we have now three you can
see three categories; here this is equal to
406
00:44:11,270 --> 00:44:17,680
this, means your pre fault, and your post
fault with the line re closing is your same
407
00:44:17,680 --> 00:44:25,000
Impedance. So, same curve, and that curve
is here, it is written that Pe of Pre fault.
408
00:44:25,000 --> 00:44:32,280
Now, during the fault, again it is similar
to this. And the post fault that the fault
409
00:44:32,280 --> 00:44:44,920
which has occurred here sorry. So, this fault
here, you can see this Impedance is same.
410
00:44:44,920 --> 00:44:51,040
Here this line is cleared. So, this is fault
clear. So, this Impedance is just like in
411
00:44:51,040 --> 00:44:52,730
the previous case post fault.
412
00:44:52,730 --> 00:44:59,650
So, just we have here curve this curve here,
is a basically the fault clear at that time
413
00:44:59,650 --> 00:45:04,619
fault is clear, but it is not a post fault.
Post fault is again here If this condition
414
00:45:04,619 --> 00:45:12,180
where is again it follows the same Pe. So,
in this case here your operation is at point
415
00:45:12,180 --> 00:45:19,380
A. Again it is a Intersection of Pm not, and
you are the Pe angle; that is pre fault, and
416
00:45:19,380 --> 00:45:25,070
this, after this there is some fault your
system is coming here, that is your during
417
00:45:25,070 --> 00:45:30,700
the fault condition, it will be try to follow
here. And then fault is clear, there is a
418
00:45:30,700 --> 00:45:36,390
clearing angle here, it will try to go at
this point, because the Pe during the once
419
00:45:36,390 --> 00:45:43,940
fault is cleared it is following this, now
till it is accelerating this line was opened
420
00:45:43,940 --> 00:45:49,690
it follows this and at the same time circuit
breaker try to close It. Means it is suddenly
421
00:45:49,690 --> 00:45:56,640
here it is Increase here before that here
it is a decelerating completely, it is your
422
00:45:56,640 --> 00:46:04,690
after few cycle it Is. What is happening now,
If this is coming and now again following
423
00:46:04,690 --> 00:46:09,950
this, and then it is coming back, because
this area is now more. So, it is more accelerating,
424
00:46:09,950 --> 00:46:15,980
because we are taking more power here. So,
now, area again for stable, here this area
425
00:46:15,980 --> 00:46:21,440
one will be equal to area two. And in this
case you can see area of margin here it is
426
00:46:21,440 --> 00:46:25,690
very high, because line is not true. So, you
can feed more power. So, your system is more
427
00:46:25,690 --> 00:46:31,320
stable. So, the outage of line, again create
some criticality of the line. If you saw in
428
00:46:31,320 --> 00:46:37,119
the previous case it was this area only, now
we have this area. So, your system is more
429
00:46:37,119 --> 00:46:44,109
stable. So, outage of lines sometimes creates
some Instability and the system is very near
430
00:46:44,109 --> 00:46:52,100
to the stability margin. So, we want this
Rc is called angle of re closure. We want
431
00:46:52,100 --> 00:47:02,010
to know, and this your delta d angle between
clearing and the re closure this angle. We
432
00:47:02,010 --> 00:47:06,910
can determine, and we can know that it is
cleared at this point, and then we can see
433
00:47:06,910 --> 00:47:10,190
what should be the critical clearing time
in this case.
434
00:47:10,190 --> 00:47:16,000
We can again similarly determine. Now again
you want to get the what should be the critical
435
00:47:16,000 --> 00:47:22,670
clearing time here delta cr, having the re
closing facility this, this angle is given
436
00:47:22,670 --> 00:47:27,210
to you that is delta dt it is to note, because
it depends upon the circuit breaker configuration
437
00:47:27,210 --> 00:47:31,600
automatically, it may re close may be one
cycle re closing may be one and half or two
438
00:47:31,600 --> 00:47:35,390
cycle re closing. Means one cycle it will
wait and then again it will re-close. And
439
00:47:35,390 --> 00:47:41,410
again the fault is persisting what will happen,
this will open, circuit breaker will open
440
00:47:41,410 --> 00:47:47,490
and it will not re close again and again,
it is only once. So, in this case we want
441
00:47:47,490 --> 00:47:53,010
to determine what will be the critical clearing
angle this one. So, that knowing this specialty
442
00:47:53,010 --> 00:47:59,130
in your circuit breaker. You want to determine
that your circuit breaker fault must be clear
443
00:47:59,130 --> 00:48:05,440
before this angle, otherwise your system become
unstable. So, in this case your area again
444
00:48:05,440 --> 00:48:14,570
using equal area criteria this area, should
be equal to area this now. Your area defghd.
445
00:48:14,570 --> 00:48:19,840
So, this is the complete area now you want
to know.
446
00:48:19,840 --> 00:48:28,690
So, area one or you can say A 1 should be
equal to area A two. Now this area A 2 will
447
00:48:28,690 --> 00:48:37,010
be nothing, but the Pm minus this Pe 2 of
integration, that will be equal to from here
448
00:48:37,010 --> 00:48:42,540
to here. Now we have the different curves
and different area. Now we can divide this
449
00:48:42,540 --> 00:48:48,920
area in two parts; one is here one, here is
another is two. So, one is your nothing, but
450
00:48:48,920 --> 00:48:56,260
your either Pe 3, because it is a Pe 3 here,
Pe 3 max minus Pe, and it is from delta cr
451
00:48:56,260 --> 00:49:06,980
to delta rc re closing angle plus this area
this area is from rc to delta max Pm 1 here
452
00:49:06,980 --> 00:49:13,990
minus Pm delta. And now you have to solve,
because this Angle you know you can determine
453
00:49:13,990 --> 00:49:21,150
this knowing the Pm and Pm 1 max, you can
calculate delta cr knowing all this value.
454
00:49:21,150 --> 00:49:28,260
So, here we can get the value, we can substitute
rc cr here and we can get this value delta
455
00:49:28,260 --> 00:49:33,530
cr; that is your critical clearing angle,
and then we have to design our system based
456
00:49:33,530 --> 00:49:34,550
on that one.
457
00:49:34,550 --> 00:49:42,730
Now, same theory can be extended, can be applied
to the two machine phase, in which here this
458
00:49:42,730 --> 00:49:48,590
two machine having the different Inertia constant.
Here it has a different Inertia constant,
459
00:49:48,590 --> 00:49:56,590
and then we can add, and we can apply together.
So, in this case we want to apply the equal
460
00:49:56,590 --> 00:50:02,030
area criteria for the two machine case, is
a special case, and this two machine case
461
00:50:02,030 --> 00:50:07,690
can be represented by equivalent a single
machine connected to Infinite bus. Here we
462
00:50:07,690 --> 00:50:13,940
have the generator one, and another is generator
two. This generator two is having all the
463
00:50:13,940 --> 00:50:21,349
parameter; that is denoted by the two here
suffix. So, here it is having M one, it is
464
00:50:21,349 --> 00:50:27,810
having D 1 damping constant, it is having
delta on side it two, so it is two. This case
465
00:50:27,810 --> 00:50:34,940
it is your M 1 it is your D 1 and delta 1.
So, we can write the dynamic equation; that
466
00:50:34,940 --> 00:50:39,820
is a swing equation, even then including damping.
So, here the damping term is Included. Normally
467
00:50:39,820 --> 00:50:45,520
till now we have ignored the damping, because
it is related with the speed of the system.
468
00:50:45,520 --> 00:50:51,910
So, this is M 1 delta 1 double dot that is
double differentiation of delta 1, delta 1
469
00:50:51,910 --> 00:50:58,640
of here it is delta 1 written; that is feeding
power Pm one. Here the mechanical power that
470
00:50:58,640 --> 00:51:04,090
is coming to this generator is Pm 1 minus
Pe. Similarly for generator two we can write
471
00:51:04,090 --> 00:51:11,170
this M 2 delta 2 double dot plus d 2 delta
2 dot plus that will be equal to your Pm 2
472
00:51:11,170 --> 00:51:20,869
minus Pe two; that is coming from this side
If losses are neglected what will happen.
473
00:51:20,869 --> 00:51:27,450
This Pe 1 Pe 2 in this system the total Pe
here that is inducting that must balance that
474
00:51:27,450 --> 00:51:33,420
must conserve, energy cannot be going anywhere.
So, losses are zero means the energy which
475
00:51:33,420 --> 00:51:37,440
is feeding that will be the balance crowd,
means whatever you are feeding it must be
476
00:51:37,440 --> 00:51:44,400
taken by some machine. So, this Pe 1 plus
Pe 2 equal to zero. Similarly the energy,
477
00:51:44,400 --> 00:51:49,690
mechanical energy that is coming into the
system the Pm 1 plus Pm 2 is also equal to
478
00:51:49,690 --> 00:51:56,380
zero. Now, substituting equation three into
two here, even two substitute the equation
479
00:51:56,380 --> 00:51:59,720
three here.
So, substitute the equation three into two,
480
00:51:59,720 --> 00:52:07,000
we will get here your M 2 delta 2 double dot
plus D 2 here this side, and this we are just
481
00:52:07,000 --> 00:52:14,720
replacing, complete this 2 equations; that
is Pm 2 and Pe 1 in terms of P 1 we can write
482
00:52:14,720 --> 00:52:22,920
this minus Pm 1 this, and it will be plus
Pe one. Assuming the uniform damping, means
483
00:52:22,920 --> 00:52:29,230
damping here is a D 1 upon D 1 that will be
equal to D 2 upon M two, or you can Ignore
484
00:52:29,230 --> 00:52:33,710
the damping. If you are also Ignoring that
is no problem, then it is very simple, but
485
00:52:33,710 --> 00:52:42,320
if your damping is uniform. So, D 1 upon M
1 and D 2 upon M 2 is equal. Also we define
486
00:52:42,320 --> 00:52:50,119
the delta 1 minus delta 2 that is a delta
1 2. So, we can write a equation M naught
487
00:52:50,119 --> 00:52:59,310
delta 1 2 double dot plus delta 2 delta naught
delta 1 single dot is equal to Pm 1 minus
488
00:52:59,310 --> 00:53:08,890
Pe 1, and this Pm naught will be the basically
the parallel combination of this angular momentum
489
00:53:08,890 --> 00:53:14,930
that is M 1 and M two. So, M naught will be
equal to your M 1 into M 2 divided by M 1
490
00:53:14,930 --> 00:53:21,040
plus M 2. Similarly D naught will be D 1 into
M 2 divided by M 1 plus M 2.
491
00:53:21,040 --> 00:53:29,660
So, this is your equal length system, and
we can again apply our equal area criteria.
492
00:53:29,660 --> 00:53:36,260
Now, for the single machine, it is very easy,
no doubt to know, whether system is stable
493
00:53:36,260 --> 00:53:41,730
or not. We can analyze and we can calculate
your critical clearing time, we can see the
494
00:53:41,730 --> 00:53:46,760
stability of the system, and we can only know
the relatively stable system. We want to get
495
00:53:46,760 --> 00:53:53,670
what is the delta variation, it is not a time
domain simulation at all. It is only giving
496
00:53:53,670 --> 00:53:59,310
Information about the stability of the system,
means your system is stable or it is not a
497
00:53:59,310 --> 00:54:04,790
stable. But that it is not only our concept.
We want to know what is the system behavior
498
00:54:04,790 --> 00:54:11,270
during the period, means what is the speed
deviation, how speed is changing, what is
499
00:54:11,270 --> 00:54:17,800
your delta is changing that is also require,
and for that we have to go for solving differential
500
00:54:17,800 --> 00:54:22,240
equation for different Intervals. So, this
is the numerical solutions to the swing equations.
501
00:54:22,240 --> 00:54:28,540
We have written the swing equation with this
during the pre fault condition.
502
00:54:28,540 --> 00:54:35,609
During the faulted condition this equation,
and post fault here If know line re closing
503
00:54:35,609 --> 00:54:39,380
we are not considering means fault is cleared,
by the outage of the line means this line
504
00:54:39,380 --> 00:54:46,040
is tripped, after certain time only post fault
is this only one line is existing along with
505
00:54:46,040 --> 00:54:53,619
this generator transformer, and this is your
base equation. So, the equation one two three
506
00:54:53,619 --> 00:55:00,780
here, can be sold by the numerical methods,
for solving non-linear differential equation,
507
00:55:00,780 --> 00:55:05,470
why this delta here this is also delta this
non-linear function here, this non-linear
508
00:55:05,470 --> 00:55:10,839
non-linear. So, this differential equation
is a non-linear differential equation of second
509
00:55:10,839 --> 00:55:16,580
order, because of D two. And that can be solved
by the various methods. Methods may be your
510
00:55:16,580 --> 00:55:21,520
Eulers method, Trapezoidal methods that is
called kutta methods. We can show numerical
511
00:55:21,520 --> 00:55:26,410
results of the power angles with respect to
time; that is a swing curve. Our critical
512
00:55:26,410 --> 00:55:31,970
clearing time can be determined by this method.
In addition this method can be applied to
513
00:55:31,970 --> 00:55:35,830
the analyzer transient stability in a multi
machine system.
514
00:55:35,830 --> 00:55:42,470
For the single machine system, they as said
the equal area can be easily determined, and
515
00:55:42,470 --> 00:55:47,869
we can say our system is stable or not stable.
We can also expect what is the margin available
516
00:55:47,869 --> 00:55:53,070
with you, if fault is cleared at particular
time so we can know the relative stability,
517
00:55:53,070 --> 00:55:57,020
but we want to know the performance that is
a time domain simulation is required. So,
518
00:55:57,020 --> 00:56:04,450
we have to solve these three differential
equations. More over whole this swing equation
519
00:56:04,450 --> 00:56:11,580
here, we have modeled a simple classical approach,
and that is not valid. Means we have to go
520
00:56:11,580 --> 00:56:15,839
for the detailed modeling of the machine.
Here we have to Ignore we have made seven
521
00:56:15,839 --> 00:56:21,810
six seven assumptions, that can be Incorporated
and then there will be several hundreds of
522
00:56:21,810 --> 00:56:27,780
machines in a big power system. So, all these
will be solved for the system, then we can
523
00:56:27,780 --> 00:56:32,790
see the performance that the delta etc variation
of each Individual generators, and then we
524
00:56:32,790 --> 00:56:40,290
can say, whether the system is stable or not.
So, in this lecture we have seen the equal
525
00:56:40,290 --> 00:56:45,830
area criteria for the application for the
various, configuration of various fault with
526
00:56:45,830 --> 00:56:52,070
re closing, without re closing, fault at bus,
fault at line, and then we can saw this equal
527
00:56:52,070 --> 00:56:57,220
area criteria can give you the relative stability.
It will give Information about without, whether
528
00:56:57,220 --> 00:57:01,930
it means whether your system is stable or
not at the same time. It can also with the
529
00:57:01,930 --> 00:57:07,390
margin that is stability margin available
with the system, and it is very useful and
530
00:57:07,390 --> 00:57:14,849
very fast no doubt, but only the limitation
of this is that, we have made several assumptions;
531
00:57:14,849 --> 00:57:22,290
those may not be valid for the power system.
And also it is not possible to formulate to
532
00:57:22,290 --> 00:57:28,400
form the system into single machine Infinite
bus, equivalent and then we have to go for
533
00:57:28,400 --> 00:57:34,540
the complete and the real time petrol system
for the simulation. And for that we have to
534
00:57:34,540 --> 00:57:39,660
solve these differential equations for all
the machines, not only one machine, this is
535
00:57:39,660 --> 00:57:44,190
the case of one machine. So, we have to go
in this Infinite bus system there will be
536
00:57:44,190 --> 00:57:50,380
several generators, and we have to include.
Again this generator is not a classical generator.
537
00:57:50,380 --> 00:57:53,710
Classical generator means is a generator which
we have model in the classical form, means
538
00:57:53,710 --> 00:57:58,740
we have taken only this generator with the
some Inertia constant, then we can go for
539
00:57:58,740 --> 00:58:03,920
several order of modeling of synchronous machine.
It will equip with your excitation as well,
540
00:58:03,920 --> 00:58:07,440
it will be equipped with the governing system.
So, all dynamics will be, dynamical equation
541
00:58:07,440 --> 00:58:13,990
will be clubbed together along with algebraic
equation, because here the power flow of the
542
00:58:13,990 --> 00:58:18,190
different transmission lines, they are not
connected with the transmission lines generating
543
00:58:18,190 --> 00:58:22,310
stations, the different transmission lines
that they will be there. So, we have to form
544
00:58:22,310 --> 00:58:28,640
this algebraic equation, and then we can solve.
Some times what we do, we this all the buses,
545
00:58:28,640 --> 00:58:34,410
those are not connected we try to eliminate.
We will see in the later sections. Here this
546
00:58:34,410 --> 00:58:39,500
they will be eliminated only the generator
terminator buses are kept Intact, where there
547
00:58:39,500 --> 00:58:43,839
is a fault is occurring, and then we can analyze
the system behavior, means we can see the
548
00:58:43,839 --> 00:58:47,760
relative motion of the Individual generator
with reference to any particular generation,
549
00:58:47,760 --> 00:58:53,660
and then we can say whether your system is
stable or not stable. So, with this, now this
550
00:58:53,660 --> 00:58:57,600
chapter seven, lecture seven is closed.
Thank you.