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Now, let us see the lecture number 6 of module
2. In the previous lecture; that is lecture
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number five, we saw the various definitions
of the various type of stabilities, and there
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difference of classification, again with respect
to the disturbance means, magnitude of disturbance
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time frame, and again the quantities of interest.
In this lecture I will discuss about transient
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stability of the power system, and then we
will see what are the ways that we can analyze.
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First one is a classical approach for the
transient stability analysis, and the various
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assumptions are made in the classical approach.
First one is the mechanical input to the generator
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remains constant. Means governing system action
is neglected. Means if your system here this,
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whatever the generators here we are giving
the mechanical power, and that is coming through
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your turbine system, again turbine system
then governing system. So, this we are assuming
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for that analysis period, this pm is almost
constant, and that assumption is valid. Because
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here the transient stability our analysis
is concerned up to 5 to 10 seconds. So, by
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that here the governor action or prime over
action is almost negligible, because the time
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constant of this is very high. So, we can
ignore this, means your mechanical power input
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to the generators are remain constant. Second
assumption here is the mechanical damping;
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that is machine damping, and the AVR action
is also neglected means the damping of the
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machine that is D, and the automatic voltage
regulator action s also neglected.
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The synchronous machine model as constant
voltage source, behind a transient reactance.
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Means here it is your generator. So, we model
this generator with a some reactance here
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and this is your terminal voltage. So, this
reactance is nothing, but your x D prime.
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Again as in the previous lectures we have
seen the synchronous alternators may of two
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kind, two types; one is your cylindrical rotors,
another is the silent type of rotor of machine,
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where Xd and Xc are different, however in
the synchronous machine it is a Xs; that is
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constant and it is irrespective of the D and
Q Xs components. So, if it is a salient then
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it is Xd prime, and if it is your cylindrical
rotor machine then it will be Xs. So, it is
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your some, the constant voltage source Ef
behind this voltage behind this terminal voltage
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this t, and this is your sub transient reactance
of this alternator than we can model in this
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way. So, the machine can be represented by
the classical; that is also called the classical
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representation of the machine model. Also
we know in the power system, all that components
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they have some dynamics, and they have some
their equations and their dynamics are there.
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For your example your network dynamics, protection
system dynamics, your governor dynamics, your
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mechanical inputs dynamics.
So, all are having the dynamics, but in this
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classical approach, we also ignore that network
transients. Means those are neglected, thus
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the static model of network is consider. Only
we have to go for the static model in this
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classical approach. Load also may have some
dynamics; for example, your induction machine
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load, but also they are changing with the
change of voltage, as well as changing in
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your reactive power requirement. But in this
classical approach, we model load at the simple
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impedance or admittance form. The mechanical
angle of each machine rotor coincides with
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electrical phase of voltage behind transient
reactance. So, these are the five assumptions
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are made in the classical analysis of the
transient stability. So, analysis utilizes
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the static equation of network, and the dynamic
equation of machine. As I said, here the network,
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this we have to take static model, and others
we have to take here the dynamics of the machine,
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here that is included.
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Let us see a generator G here that is connected
with the transformer having impedance Xd,
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and here these blocks are basically the circuit
breakers; means we can trip this portion,
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we can trip this transmission lines as well,
and here we have the infinite system. Normally,
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most of the analysis people go for either
single machine infinite bus system; that is
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called SIMB single machine infinite bus machine
system. So, this is basically your Infinite
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bus, where normally this voltage angle is
zero. Means again what is the infinite bus,
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why it is called infinite, whether infinite
in terms of its location, infinite in terms
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of its, what is infinite term why. So, infinite
bus is a bus, which is if you are injecting
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power, or you are extracting power from that
bus. Normally they should not be any change
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in the voltage and the frequency of the system
of that bus, so this is called infinite. Means
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you can draw effectively infinite power, which
is basically not possible, but if you are
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drawing large power, and there is no change
in the voltage and the frequency of that system,
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then we can say that bus where we are drawing
the power, is known as your infinite bus.
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So, here basically this is another system,
and where this bus is located, so you can
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draw more power here at this bus. Means here
practically the voltage an angle is constant,
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what about the power you are taking.
So, we had the system; one generator, the
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transformer, and then we have the two lines;
line X is here that is impedance is Xl one,
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and another is here your Xl two. Now, this
machine as I said in the classical model,
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we can represent this machine as E and then
we can have a reactance here, and this is
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your voltage source; that is your Xd prime
. So, now, your this bus, this system we can
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represent here with this one. Now, this is
a transformer. So, this Pm we have assumed
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constant, but this electrical power P is changing
and it depends upon the system here the angle
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etcetera. Now, for the cylindrical rotor,
we can write the P. If you remembered already
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we have defined is P for any machines here
is V 1 V 2 upon X sine delta and this delta
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is the difference between the angle of the
two buses. And that bus is the element with
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Xs is considered. For this case we are just
considering here one bus here, and another
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we are talking here. So, the Impedance in
this bus, is let us suppose now here what
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will be this X.
Now this X component just we have to calculate.
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So, we have to write here now our V 1 in this
equation is nothing, but your E, because we
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have taken here E angle delta of this bus.
So, your this V 1 is E, and your V 2 is nothing,
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but your V, and X is the total reactance of
this whole system, including your this as
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well. So, this is your complete. So, we can
write now this can be simplified this P; that
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is Pe is equal to E multiplied by your voltage;
that is infinite bus voltage divided by the
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Xd prime sum. Means the total X here as I
said here is X and the sine delta basically
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angle here the zero and the internal here
angle is your delta. Now, here this Xd prime
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will be nothing, but the reactance of your
generator; that is Xd prime, the reactance
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of your transformer, here this is this. And
these two lines basically this is in series
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with this, and the impedance parallel effect
impedance is series with these three. So,
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these two are in parallel. So, it is a parallel
combination of the reactance of these two
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lines this Xl 1 and Xl 2 thus it is a parallel.
So, it is a total Xd prime will be this, and
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then we can write electrical power which is
flowing out from this system. Here Pe can
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be related here, and that is basically in
terms of you cylindrical rotor machine.
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To see the same system, if your machine is
a salient rotor type of synchronous machine,
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then we have also derived this. Here this
is your Eq prime, here now here Eq and the
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Vd both are the different one. Means here
this Eq prime multiplied by your V infinite
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plus some of the reactance s in that line,
and again here we have another component here.
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So, here this is delta. It is a 2 delta, and
already in the previous lecture I have shown
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that. So, if your round rotor case, if your
V infinitely. This X and E prime are constant.
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E prime is nothing, but in the previous case
it was E, if are constant, then we are write
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this Pe is nothing, but is a function of your
simply delta, or you can say sine delta. And
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if you draw then you will find here a curve
of your sinusoidal here. Means we are getting
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some constant k sine delta. Means if your
delta is changing, the power that will be
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keep on changing, this Pe. So, this is called
your P delta curve, and at the 90 degree pi
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by 2. This k term will maximum and it is called
Pe; that is Pe maximum. Means when it is your
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degree is 90 degree then we can have the P
maximum, means maximum power that we can flow.
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So, the area which is just this side it is
your stable zone, this and outside this is
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your unstable. Now, always what will be your
operating point. Let us suppose your mechanical
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power X Pm is this axis. So, your electrical
always we know for their steady operation,
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this Pm should be equal to your Pe. Then here
we can have your k; that is k term is here.
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This is nothing, but your k; that is Pmax
we can say Pmax sine delta. So, this is a
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curve of sine delta.
So, this will satisfy at a point, and that
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is basically the intersection of this curve,
this and this curve. So, this point A is your
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operation of point, and the angle that is
a torque angle or machine angle that will
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be your delta naught here. So, your machine
which was here this I wrote E angle here,
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now it will be delta naught, if your system
is without any disturbance and this is steady
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state it is operating with this, your previous
case, this your transmission like this. Now,
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and this is your delta angle where it is operating.
So, the weekend can go up to slowly if you
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are slowly and slowly we can increase, without
disturbing the system slowly power is increased,
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then we can go up to this point, and this
is normally called your steady state stability
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limit, but always we operate our power system,
must be low this steady state that is Pmax
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value, because always you know there is a
dynamics, always there is a change in the
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system. So, if you are operating here, there
may be possibility your system may be your
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unstable zone, because this area is unstable
and this side beyond this pi by 2, and then
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your system will be collapse, and it is not
possible to operate
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Let us see the power transfer on a transmission
line. So, far we had seen the power transfer
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including your machine, transformer, and the
transmission line. Here let us we have a bus
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i and the bus j, and we have the angle here
Vj delta i j, and here the voltage magnitude
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is Vi and angle is delta I, and this Z of
this line is R plus ZX. So, we can write here
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the Pij in this line with this formula. This
is nothing, here this Pij is nothing, but
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your Vi; that is complex into your i ij that
is your Vi minus Vj conjugate divided by your
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i Z prime and here basically the real of this,
and if you simply you will get this expression.
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So, basically it is a z square. If your R
is zero, means your line is loss less, means
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normally R is very small. So, we can ignore
compared to the X then we can get whole expression
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R is zero we will get this and this is zero.
So, we can get this X X will be cancelled
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and we will get the Pij is approximately VijVi
inot Vj divided by X reactance of this line
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sine delta, or we can say this Pm sine delta.
So, for the line as of full system this formula
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is equally valid. Only you have to remember
in this case, that this delta is the angle
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between the two buses, where these two voltages.
So, that is why here I write this P between
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two nodes. Node means two buses. Here this
voltage of one bus, voltage of another bus
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divided by X between these two buses. Means
there may be several for example, here one
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is this, there may be another line here there
may be some transformer here and then another
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bus here. So, if you are talking this is V
1 and V two. So, that this Whole will be the
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total of this line one line going transformer.
And the angle here is the difference between
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angles of these two buses. So, here X is the
total X and then it is your sine basically
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delta 2 minus delta one, or basically or basically
here delta 1 minus delta 2 what is the direction.
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Means if your power is your calculating the
power showing P in this one. So, it is delta
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1 minus delta 2 as I explained in the previous
lectures, the real power will flow from higher
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angle to lower angle. So, this expression
is equally valid, and this is the case when
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R is very small and that can be taken as zero,
then this full expression is valid. If R is
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very high, then we can apply this one, especially
for the distributed system you can write this.
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You have to take this expression which is
the original expression.
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Now, let us see another one. This steady state
stability limit that is Pm; that is your V
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1 V 2 upon X. It is sometimes called Vm sometimes
Pmax, so it is given in the different book
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here. Normally this Pm I will denote the mechanical
power. So, it is better to write the Pmax.
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So, it is the steady state stability limit.
If you differentiate the Pij in the previous
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case, if you differentiating this, that we
have if you are differentiating that the Pij
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with reference to the delta here that is again
here delta, I have just written it is nothing
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the delta between the bus 1 and bus 2 is your
delta. So, always you must be very much careful
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what is this delta. So, if you are differentiating
this, then the voltage here is a Vi Vj and
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here you X, it is your cos delta. And this
coefficient is called the stiffness of line
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or it is called the synchronizing coefficient
of the line. So, this term gets your idea
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that what is the stiffness, constant of the
line that is very much important, how much
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your system is stable.
Also in the classical approach, we may need
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more assumptions, and that is basically the
saliencies of synchronous machines are neglected,
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and resistance of lines are also neglected.
In the classical approach, as I said your
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two types of machines may be there, here it
is a stator, and here is your rotor which
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is rotating. So, if it is there the constant
air gap, then it is called in rotor, and if
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you are having some poles, and we have the
different air gap; that is rotating in the
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machine here. So, here we have the different
air gap, here we have different air gap and
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this called the saliency machine. So, we can
ignore means here we can write this Xd prime
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will equal Xq then we have to ignore saliency
we have to assume, but both quantities are
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very close, but normally it is not so easy,
it is very different. And also the resistances
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of the lines it is not only lines, even though
of the transformers we have to ignore, and
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the resistance of your synchronous machine
is also ignore.
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Now, let us see the single machine infinite
bus; this SMIB the single machine infinite
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bus system, where again the previous case
we use here a generator G. This is your machine
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is represented by its equivalent impedance;
that is nothing, but hour jxd or Xd prime
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here. You have to take the sub synchronous
transient reactance, and this machine is the
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internal here the prime that is equal to magnitude
this is complex quantity. So, the magnitude
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with the angle delta with reference to the
infinite bus. So, the Infinite bus basically
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denoted by, it has the infinite inertia, system
inertia; whole this system basically. This
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is a system where this bus is there, so it
has the infinite inertia, it has a constant
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frequency, and it has a constant voltage.
So, these three criteria basically gives you
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the criteria for the infinite bus, and the
voltage here as I said the voltage is V infinite,
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here I have written angle zero, and with the
reference to this angle we have taken this
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angle delta. This is your transformer this
is your terminal voltage of generator.
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This is another bus, where this is a line
there and it has some impedance. Now, this
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machine here is nothing, but your spring mass
system. Here there is some inductance it was
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A the total this we can denote whole this
machine. This is your infinite bus where this
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V infinite angle zero, and this is another
mass. So, this is just like a masses spring
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system, and then it will have some, and then
we can analyze dynamics of the system, as
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you know the dynamics of the system can be
retained and solve the differential equation.
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Now, there is a two type of concept, if there
is a three phase. Let us suppose at this bus
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as we shown here. If there is three phase
fault, there will be this machine delta will
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be changing, why. This electrical power due
to this fault, more power will be flown as
189
00:20:47,600 --> 00:20:55,299
i said the power here is your nothing, but
your E prime v that is infinite the total
190
00:20:55,299 --> 00:21:03,429
X completely X sine delta. What will happen
now this is in steady state, and this Pm is
191
00:21:03,429 --> 00:21:08,730
almost constant. We have assumed this now
this three phase fault is there, what will
192
00:21:08,730 --> 00:21:12,820
happen. This X will be change, and this X
will be reducing.
193
00:21:12,820 --> 00:21:23,090
So, means more power is flowing. So, your
Pm is not equal to Pe during this three phases
194
00:21:23,090 --> 00:21:29,690
fault. So, this to make it what will happen
this Pe this delta will keep on increasing,
195
00:21:29,690 --> 00:21:34,630
because this value is larger, if it maintain
the Pe constant this delta, you can say this
196
00:21:34,630 --> 00:21:41,730
delta will increase, and now you can say delta
is increasing. So, over time period during
197
00:21:41,730 --> 00:21:47,600
the fault, this delta again can follow the
different trajectory. One condition you can
198
00:21:47,600 --> 00:21:56,580
see if delta is falling like this, and finally,
it is stable. No doubt, there is a possibility
199
00:21:56,580 --> 00:22:05,030
at this point, this fault is clear. If the
fault is persisting, and this fault is not
200
00:22:05,030 --> 00:22:11,429
severe, so delta will increase and then it
will be stable, but if this fault is clear.
201
00:22:11,429 --> 00:22:17,290
So, then delta again may come to its initial
value, may be, again depends upon the, whether
202
00:22:17,290 --> 00:22:23,179
fault is removed or not. If fault is removed,
then it may come to its original value, and
203
00:22:23,179 --> 00:22:28,960
the system is called to be stable, but this
delta can increase like this, till this fault
204
00:22:28,960 --> 00:22:34,230
is again cleared, the system is this delta
is keep on increasing and it losing their
205
00:22:34,230 --> 00:22:39,679
stability, because here if you see, if this
is your delta and this delta is keep on increasing,
206
00:22:39,679 --> 00:22:46,130
then system is unstable and system is unstable
then it is called there is a loss of synchronism.
207
00:22:46,130 --> 00:22:53,270
So, it is losing synchronism operation, and
then we can say system is unstable.
208
00:22:53,270 --> 00:23:00,160
To see the system modeling, again this whole
the rotating mass, here it has some inertia
209
00:23:00,160 --> 00:23:06,330
constant. This rotating mass we have the synchronous
generators. The systems also we are have some
210
00:23:06,330 --> 00:23:10,820
damping. Damping means we have some resistances,
means there is some damping factor as well,
211
00:23:10,820 --> 00:23:19,510
and let us say it is D, and here this your
Tm this mechanical torque that is mechanical
212
00:23:19,510 --> 00:23:25,530
power divided by this normal frequency that
is omega not. And this omega naught is nothing,
213
00:23:25,530 --> 00:23:31,730
but it is equal to your 2 pi f not, and f
naught is your synchronous frequency. This
214
00:23:31,730 --> 00:23:40,049
synchronous frequency is decided as I said;
it is 120 f upon p; that is how much you are
215
00:23:40,049 --> 00:23:44,320
rotating. This rotating is p this is giving
your frequency based on the rotation. So,
216
00:23:44,320 --> 00:23:52,710
number of poles of the machine. So, this is
a rotating mass, and again the two forces;
217
00:23:52,710 --> 00:23:59,190
one is Tm, another is your Te. So, your electrical
power which is coming through this one going
218
00:23:59,190 --> 00:24:03,840
that side, it is an electric power and the
input we are giving in terms of mechanical
219
00:24:03,840 --> 00:24:11,690
power. So, at the balance as I said here again
this Te must be equal to your Tm, or we can
220
00:24:11,690 --> 00:24:16,340
say this Pe will be equal to your Pm. Means
your electrical power will be equal to your
221
00:24:16,340 --> 00:24:23,010
mechanical power, and then it is called steady
state. But during the steady state it may
222
00:24:23,010 --> 00:24:27,770
not be equal it will try to equalize this
Te.
223
00:24:27,770 --> 00:24:33,020
As I said this Pm we have achieving the constant
in this transient stability analysis, so only
224
00:24:33,020 --> 00:24:39,660
parameter which is changing your Pe. So, various
in this system modeling that will be used,
225
00:24:39,660 --> 00:24:45,650
this Pe in per unit electrical power output
that is power which is coming out, Pm is mechanical
226
00:24:45,650 --> 00:24:52,530
power input that Te is electrical torque,
Tm is mechanical torque input again, delta
227
00:24:52,530 --> 00:24:57,690
is the rotor angle. And this angle is the
angle between the two here your reference
228
00:24:57,690 --> 00:25:04,049
axis, and here your rotor axis. So, this is
the angle between the reference where you
229
00:25:04,049 --> 00:25:08,340
are measuring all these angles, and this is
the angle between your, and this is called
230
00:25:08,340 --> 00:25:16,940
your torque angle, all load, angle between
the two axis here. Your h is the inertia constant
231
00:25:16,940 --> 00:25:21,790
of machine; whole system basically here we
are talking. If all machines are coming into
232
00:25:21,790 --> 00:25:26,870
1 1, one combined machine then it is inertia
constant of complete system, or if you are
233
00:25:26,870 --> 00:25:30,960
analyzing one machine with one machine, then
it is inertia constant of one machine.
234
00:25:30,960 --> 00:25:36,940
So, D is your damping coefficient as I said.
Here this is your V infinite is here infinite
235
00:25:36,940 --> 00:25:43,179
voltage. This is terminal of this generator.
This is the Xprime; that is a transient reactance
236
00:25:43,179 --> 00:25:50,720
of synchronous machine, and here is E is behind
the transient reactance. So, what about the
237
00:25:50,720 --> 00:25:58,690
power that is here coming out, again this
will be equal to your Pm in the steady state.
238
00:25:58,690 --> 00:26:06,160
This Xl this include your the transformer
and the line reactance s, as I said we had
239
00:26:06,160 --> 00:26:11,370
one transformer and then we had the two lines,
again the same example and here it was Ef.
240
00:26:11,370 --> 00:26:19,150
So, this is your Xt plus and Xl 1 plus parallel
to Xt 2 parallel. So, it is X is the combine
241
00:26:19,150 --> 00:26:25,250
of the transformer as well as the two transmission
lines, those are in parallel.
242
00:26:25,250 --> 00:26:31,810
So, for the transient analysis, normally we
use the swing equation. Let us what is the
243
00:26:31,980 --> 00:26:36,799
swing equation. Under the normal operating
condition, the relative position of rotor
244
00:26:36,799 --> 00:26:45,330
axis, and the resultant magnetic field axis
is checks. So, the rotating axis of the rotor,
245
00:26:45,330 --> 00:26:50,929
and the magnetic axis is field, is fixed.
The angle between these two is known as the
246
00:26:50,929 --> 00:26:59,400
power angle or the torque angle. During any
disturbance, rotor will accelerate or decelerate.
247
00:26:59,400 --> 00:27:05,400
There by what will happen, there will be change
in the rotor angle. So, with acceleration
248
00:27:05,400 --> 00:27:11,520
and the deceleration, with respect to the
synchronizing rotor air gap mmf and a relative
249
00:27:11,520 --> 00:27:17,910
motion became basically. The equation describing
this relative motion is known as swing equation.
250
00:27:17,910 --> 00:27:26,130
Means during any disturbance, rotor will decelerate
or accelerate. Again depending upon the Pm
251
00:27:26,130 --> 00:27:33,210
and Tm balance; means whether the Tm is more.
Means input is more than the electrical output,
252
00:27:33,210 --> 00:27:40,530
then machine will accelerate .I f your input
is less than your output then it will be decelerate.
253
00:27:40,530 --> 00:27:49,020
Then accelerate with respect to the synchronizing
rotating air gap mf, and our relative motion
254
00:27:49,020 --> 00:27:55,210
begins, and the equation describing this relative
motion is known as swing equation.
255
00:27:55,210 --> 00:28:01,900
If after this oscillatory period the rotor
locks back into the synchronous speed, the
256
00:28:01,900 --> 00:28:06,559
generator will maintain its stability. If
the disturbance. There will not involve any
257
00:28:06,559 --> 00:28:12,250
net change in power. The rotor returns to
its original position. If the disturbance
258
00:28:12,250 --> 00:28:21,230
is created by a change in generation, load,
or network condition, or network topology.
259
00:28:21,230 --> 00:28:27,130
The rotor comes to a new operating angle,
relatively synchronously revolving field.
260
00:28:27,130 --> 00:28:32,919
Swing equation can be expressed in terms of
the second ordered non-linear differential
261
00:28:32,919 --> 00:28:43,590
equation as here H over pi f naught second
derivative of delta with respect to time;
262
00:28:43,590 --> 00:28:50,910
that will be equal to Pm minus Pe delta and
that is a function of your delta minus this
263
00:28:50,910 --> 00:28:58,049
damping constant and then we have D delta
upon dt. Basically this D delta upon dt is
264
00:28:58,049 --> 00:29:08,620
nothing, but your changing speed. So, if we
neglect this term this speed E; the damping
265
00:29:08,620 --> 00:29:14,309
effect as I said in the classical analysis
we always neglect this. So, this can be neglected.
266
00:29:14,309 --> 00:29:25,900
So, we will have the equation here like this.
Now you can see earlier I was using this H,
267
00:29:25,900 --> 00:29:32,110
and I now I am using you know. So, this M
and H they are defined they are related, as
268
00:29:32,110 --> 00:29:40,020
this M angular momentum is equal to H upon
pi f not, and this omega is nothing, but this
269
00:29:40,020 --> 00:29:46,100
del D delta upon dt. So, whole this equation
can be replaced and can be retained in terms
270
00:29:46,100 --> 00:29:47,460
of omega.
271
00:29:47,460 --> 00:29:55,260
So, this M is angular momentum, and this is
defined as inertia i of whole system multiplied
272
00:29:55,260 --> 00:30:03,780
by angular speed, and its unit is your mega
joule second per electrical radian or degree.
273
00:30:03,780 --> 00:30:11,340
i is your moment of inertia, here you delta
is theta minus omega naught t, and theta is
274
00:30:11,340 --> 00:30:17,440
angular displacement of rotor, and omega naught
is the speed of the synchronous rotating frame.
275
00:30:17,440 --> 00:30:25,330
Now, let us again I use this H and how we
have to get the information of this H. Now
276
00:30:25,330 --> 00:30:32,059
let us define the per unit inertia constant
H; that is H is defined at the stored kinetic
277
00:30:32,059 --> 00:30:38,929
energy in mega joule divided by the MVA rating
of the machine. If you will see the unit of
278
00:30:38,929 --> 00:30:44,730
this here the unit is nothing doubt here the
unit of the kinetic energy is mega joule.
279
00:30:44,730 --> 00:30:52,440
So, I can write here mega joule and over your
MVA. So, this your power this is energy, and
280
00:30:52,440 --> 00:30:58,850
you know the energy here it is nothing, but
I can say MWh. So, here the total, this is
281
00:30:58,850 --> 00:31:06,169
nothing, but your second. So, the unit here
mega joule per MVA, or it is second, and H
282
00:31:06,169 --> 00:31:13,309
is very widely used, because H gives what
is the inertia constant, and this kinetic
283
00:31:13,309 --> 00:31:22,210
energy is stored, it is nothing, but I can
write here half i omega square; this KE. And
284
00:31:22,210 --> 00:31:28,950
this will be equal to i can write half i omega
into omega, and this is nothing, but your
285
00:31:28,950 --> 00:31:36,020
arm, and here I can say half M omega, and
this is nothing, but your G, and G is your
286
00:31:36,020 --> 00:31:39,750
rating or installed capacity of your synchronous
machine.
287
00:31:39,750 --> 00:31:48,460
So, knowing this M, knowing your G and omega;
that is a synchronous speed we can get H,
288
00:31:48,460 --> 00:31:54,850
or knowing H we can get M vice versa. So,
the swing equation can be written as in terms
289
00:31:54,850 --> 00:32:04,020
of H. Again if you are using delta in electrical
degrees, then it is one Atf here double differentiation
290
00:32:04,020 --> 00:32:12,740
of delta with respect to time, and here you
power minus, this mechanical power minus your
291
00:32:12,740 --> 00:32:17,270
electrical power in per unit. So, here we
are talking just about the per unit, it is
292
00:32:17,270 --> 00:32:22,169
divided by the base unit. So, it is per unit
always, because this is H in per unit, this
293
00:32:22,169 --> 00:32:25,799
unit at all in per unit. So, it is not in
megawatt. So, you have to write the base,
294
00:32:25,799 --> 00:32:32,210
but you have to used, and base is of course,
is a MVA rating of the machine. So, simply
295
00:32:32,210 --> 00:32:36,980
we have divided here. So, if you are using
the delta in electrical radians, then here
296
00:32:36,980 --> 00:32:40,460
it is a pi f and then we can write this one.
297
00:32:40,460 --> 00:32:47,169
Now, the various types of machines are used;
synchronous machines I am talking, and then
298
00:32:47,169 --> 00:32:51,799
the inertia constant is different for the
different type of machine, and this inertia
299
00:32:51,799 --> 00:32:59,429
constant H here it is basically 3 to 10 second
for steam turbine generator system. It is
300
00:32:59,429 --> 00:33:06,970
from two to four for a hydro generator system.
Normally if H is large then we say this machine
301
00:33:06,970 --> 00:33:14,030
is large, if it is a small, then machine is
small. You can see the Induction motor, this
302
00:33:14,030 --> 00:33:19,730
inertia constant is 0.5 second means its size
is very small. So, this inertia constant gives
303
00:33:19,730 --> 00:33:25,900
information about the size of the machine.
Why this steam turbine generator system is
304
00:33:25,900 --> 00:33:32,570
having high inertia, because you know here
we have this generator, that generator is
305
00:33:32,570 --> 00:33:38,210
coupled with the various types of turbines,
different stages of turbines. So, they also
306
00:33:38,210 --> 00:33:43,039
high speed and rotating mass are there. So,
the total H of the system is more compared
307
00:33:43,039 --> 00:33:47,940
to the hydro, where one the hydro turbines
are there, and then it is coupled with your
308
00:33:47,940 --> 00:33:53,070
turbine system here. So, it has a lesser time
constant H, means inertia constant.
309
00:33:53,070 --> 00:33:57,890
So, it is 2 to 4 second however the steam
turbine depending upon the size, if it is
310
00:33:57,890 --> 00:34:03,590
very large then H very high, and if it is
small then it is again small. Again, in terms
311
00:34:03,590 --> 00:34:07,840
of synchronous motor, because we can use the
synchronous motors as well. So, the average
312
00:34:07,840 --> 00:34:15,599
value of H, it may two again 2 second. Synchronous
condenses, if it is large then it may be average
313
00:34:15,599 --> 00:34:21,929
value is 1.25, and if it small then it is
one. What is the synchronous condenser. To
314
00:34:21,929 --> 00:34:30,239
understand here the synchronous condenser
are nothing, but they are some sought of alternators,
315
00:34:30,239 --> 00:34:35,659
but mechanical power output, this electrical
power output is zero. Means if you are using
316
00:34:35,659 --> 00:34:43,919
here this generator, and there is no here
Pe is zero. You are rotating to provide the
317
00:34:43,919 --> 00:34:50,299
reactive power only Q, then it is known as
synchronous condenser. So, you can either
318
00:34:50,299 --> 00:34:54,960
generate, or absorb reactive power by this
synchronous alternator. We know this by changing
319
00:34:54,960 --> 00:35:00,719
the excitation system we can generate or we
can absorb the reactive power. So, it is acting
320
00:35:00,719 --> 00:35:05,809
as some variable capacitor along with the
inductor. See one zone it is giving, another
321
00:35:05,809 --> 00:35:11,849
it is also absorbing. So, it can do, and we
are not taking any real power load then it
322
00:35:11,849 --> 00:35:15,710
is called synchronous condenser.
323
00:35:15,710 --> 00:35:23,150
Now, let us see the physical interpretation
of swing equation; that is here. Means we
324
00:35:23,150 --> 00:35:28,749
can derive this swing equation in terms of
change in the speed with respect to time,
325
00:35:28,749 --> 00:35:35,739
the derivative of speed with respect to time
this M value this Pm minus Pe, and that Pe
326
00:35:35,739 --> 00:35:42,609
is a function of delta. Again this Pe is nothing,
but Pe; that is approximate the V 1 V 2 upon
327
00:35:42,609 --> 00:35:49,430
X sine delta. Normally these voltages are
constant, we assume that constant, so it is
328
00:35:49,430 --> 00:35:53,819
a function of sine delta, and it is a non-linear
function. So, this differential equation is
329
00:35:53,819 --> 00:36:00,450
a non-linear differential equation. So, here
it is your rotor, and rotor is rotated by
330
00:36:00,450 --> 00:36:05,869
the mechanical input, and then we are taking
the electrical output, and this machine is
331
00:36:05,869 --> 00:36:14,529
rotating with a speed of omega. Now, if this
Pm is equal to Pe what will happen. This component
332
00:36:14,529 --> 00:36:21,630
will be equal to zero, means if Pm is equal
to Pe, then this component here M d omega
333
00:36:21,630 --> 00:36:28,369
upon dt will be zero and this. So, that d
omega upon dt is zero; means your synchronous
334
00:36:28,369 --> 00:36:33,819
machine is running at the synchronous speed;
neither it is accelerating nor it is decelerating,
335
00:36:33,819 --> 00:36:40,339
so, it is a steady state operation.
Let us take if mechanical power is more, means
336
00:36:40,339 --> 00:36:47,239
here this more than this output, what will
happen. This machine will rotate, and accelerate,
337
00:36:47,239 --> 00:36:53,920
why it is. So, because always we know this
energy is conserved, when this power is coming
338
00:36:53,920 --> 00:37:00,640
here, it is going less. So, that power will
be store in this machine itself. And how an
339
00:37:00,640 --> 00:37:05,989
alternating, or you can say rotating mass
will store the energy, and that is no doubt
340
00:37:05,989 --> 00:37:11,089
in terms of kinetic energy means that energy
will be stored in the kinetic energy, so the
341
00:37:11,089 --> 00:37:15,660
kinetic energy will increase. Therefore, the
speed of the machine will increase, so this
342
00:37:15,660 --> 00:37:21,049
is same concept. So, if the Pm is more than
Pe, than this component will be more than
343
00:37:21,049 --> 00:37:27,849
zero. What does it mean. Your rate of change
of speed is more, means your this omega will
344
00:37:27,849 --> 00:37:33,499
be now more than your synchronous speed, and
then this rotor will accelerate, or you can
345
00:37:33,499 --> 00:37:39,569
say it will be speeding up. The reverse is
also true. Means your mechanical power is
346
00:37:39,569 --> 00:37:44,420
less than electrical power, means you are
taking more power, than your input, what will
347
00:37:44,420 --> 00:37:49,749
happen. The energy which is here in this mass
will be taken out, and then means kinetic
348
00:37:49,749 --> 00:37:54,460
energy will be reducing; therefore, the speed
of system will be reducing, and then we can
349
00:37:54,460 --> 00:37:59,289
say this change in omega with respect to t
will be less than zero, or the rotating here
350
00:37:59,289 --> 00:38:05,390
speed will be less than its normal or can
say frequent synchronous speed of the system,
351
00:38:05,390 --> 00:38:10,630
and the rotor will decelerate, or it will
be slowing down.
352
00:38:10,630 --> 00:38:19,249
So, the swing equation of the two coherent
machines. So, far we have used H. here it
353
00:38:19,249 --> 00:38:25,109
is H or M whatever you are talking about.
So, if it is one machine here, and then it
354
00:38:25,109 --> 00:38:34,239
is connected with your Infinite bus here.
Now, if there are two machines; means here
355
00:38:34,239 --> 00:38:41,079
you having two machines, and then you want
to include and combine together, then we can
356
00:38:41,079 --> 00:38:48,170
say this two coherent machine can be combined,
and their inertia constant can be also added
357
00:38:48,170 --> 00:38:59,969
together. So, here it is your Pm one, another
machine here it is your Pm two, and this another
358
00:38:59,969 --> 00:39:06,210
here, this is synchronous machine. So, if
we are having two coherent machines; like
359
00:39:06,210 --> 00:39:12,630
here machine one, I can say G one. Here it
is your G two, and it is connected by a, its
360
00:39:12,630 --> 00:39:20,759
equivalent here internal impedance. Here again
internal of this machine, and this is your
361
00:39:20,759 --> 00:39:27,170
infinite bus; means voltage angle is constant.
So, these two machines can be combined together,
362
00:39:27,170 --> 00:39:32,950
and then again we can basically this, we can
write a single machine here, with the some
363
00:39:32,950 --> 00:39:39,319
here equivalent and this machine is like this.
So, here this is your, now this Pm equivalent,
364
00:39:39,319 --> 00:39:46,069
and again the electrical power which is going
the P equivalent here. So, I can say P 1 two,
365
00:39:46,069 --> 00:39:50,469
here Pm 1 2 I can write.
So, we can represent these two machines here
366
00:39:50,469 --> 00:39:55,369
like this one here, if both are in coherent
means they are oscillating and rotating in
367
00:39:55,369 --> 00:40:05,670
the same speed. So, if inertia constant this
here H one, here is S two, and the electrical
368
00:40:05,670 --> 00:40:14,369
power which is feeding is P 1 and here is
Pe two, then we can write this H 1 two, will
369
00:40:14,369 --> 00:40:22,049
be nothing, but H 1 S 2 over H 1 plus S 2
we can have this one expression for here.
370
00:40:22,049 --> 00:40:31,739
And this your Pm 1 2 can be related at the
Pm 1 S 2 minus Pm two, means Pm 2 here multiplied
371
00:40:31,739 --> 00:40:40,729
by this H one. So, this Pm 1 multiplied by
H 2 minus Pm 2 multiplied by H 1 divided by
372
00:40:40,729 --> 00:40:49,239
H 1 plus S two. Similarly the Pe 1 2 just
combine here, will be the P 1 here multiplied
373
00:40:49,239 --> 00:40:57,329
by this S 2 minus here Pe 2 multiplied by
H 1 and then divided by here and the angle
374
00:40:57,329 --> 00:41:07,229
here. If this is the angle E 1 angle delta
one, here this is E angle delta two. So, the
375
00:41:07,229 --> 00:41:12,749
angle here the difference will be your delta
1 minus delta 2 with reference to zero, and
376
00:41:12,749 --> 00:41:17,109
then we can go for equivalent. So, this is
again why we are going for if we are having
377
00:41:17,109 --> 00:41:24,809
two machines, then you can again go for, this
is nothing, but your single machine infinite
378
00:41:24,809 --> 00:41:33,109
bus system SMIB, and then you can go for classical
approach and then you can analyze it. So,
379
00:41:33,109 --> 00:41:39,819
for analysis for the transient stability for
this single machine infinite bus system, we
380
00:41:39,819 --> 00:41:44,789
can use the equal area criteria, and that
is very popularly used.
381
00:41:44,789 --> 00:41:51,359
So, the equal area criteria, is a quick prediction
of stability. It is a very quickly you can
382
00:41:51,359 --> 00:41:56,779
analyze, whether your system is table or not,
following the fault. This method is based
383
00:41:56,779 --> 00:42:03,339
on the graphical interpretation of the energy
stored in the rotating mass, as an aid to
384
00:42:03,339 --> 00:42:12,469
determine if the system machine maintains
its stability after a disturbance. The method
385
00:42:12,469 --> 00:42:17,890
is only applicable to one machine system connected
to an infinite bus or two machines system.
386
00:42:17,890 --> 00:42:24,369
Means if the two machines basically what is
one machine infinite bus system, you can see
387
00:42:24,369 --> 00:42:29,099
again the same example which I explained earlier
one. Here, one generating system Ef. Here
388
00:42:29,099 --> 00:42:36,420
we have the infinite bus, means this is also
a machine, which have the high inertia. Means
389
00:42:36,420 --> 00:42:42,170
large inertia, infinite inertia, and where
the voltage and the frequency are the constant.
390
00:42:42,170 --> 00:42:47,229
So, it is only applicable for the two machines
systems means this is two machines, or other
391
00:42:47,229 --> 00:42:53,029
words you can say it is the single machine
infinite bus system. So, it is only limited
392
00:42:53,029 --> 00:42:57,479
to that, it is not possible to go for the
multiple machine system. If the machines are
393
00:42:57,479 --> 00:43:02,170
several and connected by the different transmission
lines, then we have to go for another analysis.
394
00:43:02,170 --> 00:43:07,930
We have to solve the equations the non-linear
equations by help of several differential
395
00:43:07,930 --> 00:43:14,950
equation solution methods. So, the swing equation
here as again I can write; that is written
396
00:43:14,950 --> 00:43:29,839
by this one, and this Pm minus Pe is nothing, but it is a it is called the accelerating power. So, whether Pa is
397
00:43:29,839 --> 00:43:35,660
zero or it is positive or negative, it decides
whether your machine accelerating, your machine
398
00:43:35,660 --> 00:43:43,109
is decelerating, or it is in steady state
operation. So, this equation, now you can
399
00:43:43,109 --> 00:43:51,759
see, this equation can be written as in this
form. Means here from this equation you can
400
00:43:51,759 --> 00:44:00,650
see I can write this d omega over dt, is some
constant integration of Pm minus Pe angle
401
00:44:00,650 --> 00:44:05,180
delta, we can difference. We can derive this
equation no doubt very easily. We have to
402
00:44:05,180 --> 00:44:11,579
write the two differential equation and finally,
or you can differentiate this equation. Here
403
00:44:11,579 --> 00:44:17,589
you will find this equation without any problem.
So, this equation gives the relative speed,
404
00:44:17,589 --> 00:44:23,089
what is this. This is your relative speed
of the machine, with respect to synchronously
405
00:44:23,089 --> 00:44:31,420
revolving frame reference, for stability this
speed must be zero, at some time after this
406
00:44:31,420 --> 00:44:38,769
disturbance, therefore, we have for the stability
criteria this must be zero, means we can have
407
00:44:38,769 --> 00:44:44,859
this delta from delta naught where it was
operating to the delta, if we are integrating
408
00:44:44,859 --> 00:44:53,239
for that period then it will be zero. To understand
this it is very easy.
409
00:44:53,239 --> 00:44:59,119
Let us consider the sudden increase in power
input, means here this is your system, this
410
00:44:59,119 --> 00:45:05,640
is infinite bus, where we have suddenly. Earlier
it was operating at the Pm not, and then we
411
00:45:05,640 --> 00:45:13,509
have increase this Pm one. Suddenly we have
increase in power input. Consider the machine
412
00:45:13,509 --> 00:45:16,839
operating at the equilibrium point delta not,
what is the delta naught
413
00:45:16,839 --> 00:45:22,819
You can see here, this is your delta not.
This is your delta variation, this is your
414
00:45:22,819 --> 00:45:29,410
Pe, this is nothing, but your Pe delta curve
of the combined component system, including
415
00:45:29,410 --> 00:45:35,729
you are here this infinite bus, here another
terminal, here your generating terminal, here
416
00:45:35,729 --> 00:45:42,869
your generator, and this is your transformer
included everything. So, it is operating here
417
00:45:42,869 --> 00:45:48,709
as I said, here it is operating, because mechanical
power Pm naught is this axis. So, the operating
418
00:45:48,709 --> 00:45:54,499
point will be your A, where it is intersection
of this P delta curve with this your Pm naught
419
00:45:54,499 --> 00:46:00,789
curve that is constant. So, it is a constant
we have assumed. So, this delta naught is
420
00:46:00,789 --> 00:46:08,059
your initial operating angle, and corresponding
to this mechanical power Pm naught that will
421
00:46:08,059 --> 00:46:13,700
be not equal to your Pe naught; means your
electrical power is equal to your mechanical
422
00:46:13,700 --> 00:46:20,719
power in the steady state, as shown in the
figure, consider a sudden step. Sudden step
423
00:46:20,719 --> 00:46:28,630
here, you can see suddenly it is increased
here, to Pm 1. Now, what will happen, now
424
00:46:28,630 --> 00:46:38,279
its stable, this is Pm, so your B point is
the post increased stable point, this B point,
425
00:46:38,279 --> 00:46:46,579
because at that point here this Pm 1 will
be equal to Pe, now it is I can say delta
426
00:46:46,579 --> 00:46:52,289
one, where it is equal.
So, at the Initial it was Pm not, it was equal
427
00:46:52,289 --> 00:47:05,150
to your Pe delta not. So, this B point is
your another post increased the power stable.
428
00:47:05,150 --> 00:47:13,940
So, just we have seen the sudden change means;
since Pm 1 is greater than Pe 1. The accelerating
429
00:47:13,940 --> 00:47:20,390
power on the rotor is positive, because we
have increased the Pm, means at that point
430
00:47:20,390 --> 00:47:25,630
your Pe was Pe not, so once you are increasing
your machine will accelerate, because if Pm
431
00:47:25,630 --> 00:47:31,529
is more than Pe as in previous slide I showed
you that it will be increasing, and it will
432
00:47:31,529 --> 00:47:37,299
be accelerating, so your power delta angle
delta will increasing. Means here your this
433
00:47:37,299 --> 00:47:46,289
delta, here it will be increasing like this.
So, it is increasing. The axis energy is stored
434
00:47:46,289 --> 00:47:54,069
in the rotor during the initial accelerating
is here area A one; that is that is area ABCD;
435
00:47:54,069 --> 00:48:08,170
means here this area, this energy is that
is ABC, that area you can say area A 1 is
436
00:48:08,170 --> 00:48:18,299
the stored energy of this machine; that is
ABC and it is basically integration of delta
437
00:48:18,299 --> 00:48:26,699
naught to delta 1, means here is a delta not,
means this curve minus this curve if you are
438
00:48:26,699 --> 00:48:32,349
integrating from here to here, means your
area A 1 will be nothing, but integrating
439
00:48:32,349 --> 00:48:38,880
from zero naught here, to delta 1. Here this
is a constant curve. So, this is your Pm 1
440
00:48:38,880 --> 00:48:45,209
minus this curve; that is Pe, and again it
is a delta.
441
00:48:45,209 --> 00:48:53,699
So, this area is just this area here it is
written as this one. With the increase in
442
00:48:53,699 --> 00:49:01,279
the delta the electrical power will increase,
and when this delta is equal to delta one,
443
00:49:01,279 --> 00:49:07,229
the electrical power of machine matches the
new power that is Pm one, even though the
444
00:49:07,229 --> 00:49:12,690
accelerating power is zero at this point,
the rotor is running above the synchronous
445
00:49:12,690 --> 00:49:21,589
speed. Hence the delta and the electrical
power E E will continue to increase, and now
446
00:49:21,589 --> 00:49:27,709
this Pm 1 will be less than and causing the
rotor to decelerate the synchronous speed
447
00:49:27,709 --> 00:49:35,549
until again this delta is delta max. As a
result the rotor must swing fast point B,
448
00:49:35,549 --> 00:49:42,299
until an equal amount of energy is given by
the rotating mass. The energy given by the
449
00:49:42,299 --> 00:49:52,650
rotor as it is. Now what happens, now thus
from here it has to go here, and your rotor
450
00:49:52,650 --> 00:49:58,940
is now accelerating. Once it is accelerating
here, it has reached, but your rotor is accelerating
451
00:49:58,940 --> 00:50:05,289
at this point B, it is satisfied this, but
it is your machine is accelerating. So, what
452
00:50:05,289 --> 00:50:13,829
will happen it will try to accelerate and
it may go up to this point d. And here, again
453
00:50:13,829 --> 00:50:17,739
now speed of the machine will be synchronous
speed.
454
00:50:17,739 --> 00:50:23,269
Here it was synchronous speed it started accelerating,
due to this mechanical power is more, and
455
00:50:23,269 --> 00:50:29,219
here it is, it saw that Pm is Pe, then it
will be retarding, but the delta will be keep
456
00:50:29,219 --> 00:50:33,900
on increasing and it will be increasing up
to d, and again it will be, at this point
457
00:50:33,900 --> 00:50:41,579
it will see that your Pe is more than Pm one.
So, it will again retard it back. So, this
458
00:50:41,579 --> 00:50:47,089
is basically it will be going back and the
system damping will try to stabilize at certain
459
00:50:47,089 --> 00:50:52,900
point b. So, the deceleration here back to
the at the synchronous here area A two; that
460
00:50:52,900 --> 00:51:01,119
is the area BDE will be nothing, but the integration
from the delta 1 to delta max. Here at this
461
00:51:01,119 --> 00:51:07,329
delta max what is happening, the machine is
not accelerating, machine speed has. It is
462
00:51:07,329 --> 00:51:12,329
rotating at the synchronous speed, but their
electrical power is more. So, it is integration
463
00:51:12,329 --> 00:51:19,519
between this and this. So, it is Pm minus
Pm one. It is this curve minus this curve,
464
00:51:19,519 --> 00:51:25,019
if you are integrating from here to here.
This delta 1 to delta max; that is Pe minus
465
00:51:25,019 --> 00:51:33,400
Pm one. If you are integrating with delta
then you can see the total.
466
00:51:33,400 --> 00:51:41,519
The resultant is the total swing is to point
b; the angle delta max at which the point,
467
00:51:41,519 --> 00:51:47,739
you can add these two areas; that is delta
2 to delta 1; that is Pm 1 minus Pe delta,
468
00:51:47,739 --> 00:51:58,410
the pervious area plus here from this area,
if you are, means from the previous equation
469
00:51:58,410 --> 00:52:05,969
here basically. From this equation if you
can write this equation in here. Means, it
470
00:52:05,969 --> 00:52:16,680
0 naught to 0 1 Pm 1 minus e. Here again delta
1 to delta max, here the same equation, what
471
00:52:16,680 --> 00:52:23,410
is happening. now you can see this is nothing,
but your minus area one. So, I can write this
472
00:52:23,410 --> 00:52:31,150
is nothing, but area A 1 minus area 2 is equal
to 0, this is the area which we have taken.
473
00:52:31,150 --> 00:52:36,130
So, now what is happening from here we can
say, the area 1 is equal to the area two,
474
00:52:36,130 --> 00:52:40,599
and this is the case of the stability, if
the system is stable that must be satisfied.
475
00:52:40,599 --> 00:52:48,099
So, if the area here that is accelerating
energy, kinetic energy, here and is equal
476
00:52:48,099 --> 00:52:54,880
to area 2 then it is known as system is stable,
if area 1 is equal to area two, and this is
477
00:52:54,880 --> 00:53:01,690
known as equal area criteria. The rotor angle
would done oscillate back and forth between
478
00:53:01,690 --> 00:53:04,390
angle here.
So, basically your rotor will be oscillating
479
00:53:04,390 --> 00:53:09,809
between these two points. It is not going
beyond that range it is here again less than
480
00:53:09,809 --> 00:53:16,019
this, and its natural frequency. The damping
frequency in the machine will cause these
481
00:53:16,019 --> 00:53:21,130
oscillations to subside, and the new steady
state oscillations would be established at
482
00:53:21,130 --> 00:53:26,890
the point b after certain time. So, the machine
is here basically if you see it is oscillating
483
00:53:26,890 --> 00:53:33,219
your delta here, from here it is oscillating
and finally, it is new delta 1 it will be
484
00:53:33,219 --> 00:53:39,650
stabilized. So, if you are plotting with respect
to time what will happen. Here this is let
485
00:53:39,650 --> 00:53:44,890
us suppose your time, this is delta. Means
you are machine here operating initially at
486
00:53:44,890 --> 00:53:50,229
delta not, when the disturbance has taken
up. Means just you have to increase of the
487
00:53:50,229 --> 00:53:56,309
sudden input of the generating system, what
will happen, here it will be oscillating,
488
00:53:56,309 --> 00:54:02,839
and the finally, it will be steady state at
the lambda naught delta naught. So, this is
489
00:54:02,839 --> 00:54:07,229
basically, and this damping is system damping,
because there is some damping in the system,
490
00:54:07,229 --> 00:54:11,569
and it will be stabilized and the system will
be stable. But there is possibility if there
491
00:54:11,569 --> 00:54:17,859
is again excessive change.
This point d has gone beyond this then machine
492
00:54:17,859 --> 00:54:25,400
will not come back, and then it is set the
machine is system is unstable. So, in this
493
00:54:25,400 --> 00:54:31,229
lecture we saw that transient stability analysis,
using the classical approach we made certain
494
00:54:31,229 --> 00:54:37,749
assumptions, but it is very advantageous if
you can form the single machine infinite bus
495
00:54:37,749 --> 00:54:42,009
system, and then you can see the behavior
of that machine with reference to certain
496
00:54:42,009 --> 00:54:49,420
fault in the system. It is quickly giving
the idea with your system is stable or not
497
00:54:49,420 --> 00:54:54,150
without integrating the system. Means without
solving the differential equation here, with
498
00:54:54,150 --> 00:54:59,709
the help of the two areas we can know, whether
system is stable or not.
499
00:54:59,709 --> 00:55:09,430
In this case we saw that area one, here is
equal to your area 2 for the stable system,
500
00:55:09,430 --> 00:55:18,289
but if this area is more than your area two,
the system is again this area kinetic, this
501
00:55:18,289 --> 00:55:23,880
area basically always gives information that
whether you are at the verge of the stability,
502
00:55:23,880 --> 00:55:28,309
or whether your system is very strong in the
sense of the stability. So, this equal area
503
00:55:28,309 --> 00:55:34,920
criteria is used to know that whether how
much you can load your system; for example,
504
00:55:34,920 --> 00:55:42,819
in this case you can see your system is operating
here. Now, it is always advantageous to know
505
00:55:42,819 --> 00:55:48,439
that how much you can load in the system.
Means in this case it is a Pm one, there is
506
00:55:48,439 --> 00:55:56,059
a possibility you can go up to Pm two. Here
it is you Pm 2 this point, and then there
507
00:55:56,059 --> 00:56:01,430
is a sudden change here your what happen,
your system may go somewhere here, and then
508
00:56:01,430 --> 00:56:05,569
you are somewhere angle here. So, what now
what will happen now this area is the different
509
00:56:05,569 --> 00:56:12,069
area, then what you have area here. So, with
the help of we can know what is the maximum
510
00:56:12,069 --> 00:56:20,640
loading, that suddenly we can put on the generator,
without losing the stability of the system.
511
00:56:20,640 --> 00:56:27,219
And this classical approach as I said it is
limited to the single machine infinite bus,
512
00:56:27,219 --> 00:56:32,630
or the two machines here, this is another
machine, and this is another machine then
513
00:56:32,630 --> 00:56:39,209
we can analyze, and then we can know using
your equal area; means your area one; that
514
00:56:39,209 --> 00:56:44,499
is your energy storage here; that is accelerating.
So, energy that you are giving that is increase,
515
00:56:44,499 --> 00:56:51,609
that energy in terms of accelerating and here
is a decelerating. So, in this lecture we
516
00:56:51,609 --> 00:56:57,569
have seen, that only just I have taken only
one case, when there is a sudden change in
517
00:56:57,569 --> 00:57:05,430
the input to the power. In the next lecture,
we will see that how much we can load, and
518
00:57:05,430 --> 00:57:10,979
at the same time we can also see if there
will be any fault in that system here. There
519
00:57:10,979 --> 00:57:17,619
may be some fault here at this line at the
end, and if it is tripped out, then your whole
520
00:57:17,619 --> 00:57:20,920
your reactance the topology of the system
will change.
521
00:57:20,920 --> 00:57:25,559
There may possibility to do this fault this
circuit breaker is tripped, and then we will
522
00:57:25,559 --> 00:57:31,160
see what will happen during the fault, what
will happen the after the fault, and what
523
00:57:31,160 --> 00:57:36,969
will be the system stability criteria. We
can also determine what will be the critical
524
00:57:36,969 --> 00:57:42,119
clearing angle. Means at what time you will
clear your fault; otherwise you system will
525
00:57:42,119 --> 00:57:48,839
unstable, and then we will see with the help
of equal area criteria, and then you can determine
526
00:57:48,839 --> 00:57:55,140
your critical clearing time; normally it is
called CCT. So, to know their stability of
527
00:57:55,140 --> 00:58:02,170
the system; one approach here is the CCT,
means what is your critical clearing time.
528
00:58:02,170 --> 00:58:07,099
This also gives the stability majors, at the
same time some another approached is used;
529
00:58:07,099 --> 00:58:12,869
that is called energy module functions, how
energy margin you have, and that is also used
530
00:58:12,869 --> 00:58:18,789
to access system stability in event of the
transient faults; transient disturbances,
531
00:58:18,789 --> 00:58:25,729
or I can say large disturbances and for that
you have to analyze your transient stability analysis.
532
00:58:25,729 --> 00:58:26,889
Thank you.