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Friends, today we will study the transient
stability analysis of a multi machine system
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using a classical approach.
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We shall address to some of the important
issues related to the transient stability
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analysis of a multi machine system, I shall
discuss the node elimination by matrix algebra,
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classical model of the system using that model
we will develop the power output equations
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of generators in the multi machine system,
then we will write down the swing equations
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in a multi machine system and then for transient
stability analysis whatsoever preliminary
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computations are required to be made then
we will discuss those.
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One of the
requirements while pursuing the multi machine
stability problem is the elimination of some
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nodes in the system where there is no current
injection that is if you have a system right
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and at some nodes in the system if there is
no injection then these nodes can be eliminated
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okay.
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Now to understand how these nodes can be eliminated
we start with the general system and the mathematical
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model of the system is expressed as I is equal
to Y bus into V where I is the current vector
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and V is the voltage vector that is I are
the currents which are injected at the buses
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and V are the bus voltages okay, Y bus is
a matrix which is a square matrix okay and
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this is symmetric matrix you all know about
it.
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Now to understand how the nodes can be eliminated
what we will do is that we will partition
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this equation that is the current vector I
can be partitioned into two sub vectors similarly,
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the voltage vector will be appropriately partition
and then Y bus matrix is also partitioned,
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after partitioning this matrices we can write
down IA IX this is the I vector and IA is
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a vector, IX is also a vector.
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It can be partitioned as the sub matrix Y
matrix is partitioned as K, L, L transpose
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M, VA, VX that is we are when we are doing
the partitioning right of the system then
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when I partition this current vector then
the Y bus matrix is also appropriately partitioned
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and bus voltage that is also appropriately
partitioned. After this partitioning we can
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write down the equations, the two basic equation
in the form, IA equal to K times VA plus L
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times VX and IX can be written as L transpose
VA plus M times VX that is this is simply
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obtained by multiplying the uh this two matrices
on the right hand side of this equation okay
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therefore, I get IA is equal to K times VA
plus L times VX, IX is equal to L transpose
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VA plus M times VX.
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Now in case in case this IX is a null vector
in the sense all elements are 0, okay in that
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case what we can do is that we can write down
VX in terms of VA and other elements of this
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equation, what we do is that you subtract
both the sides of the equation by L transpose
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VA IX is anyway 0 and multiply by M inverse.
Now when you do this two operations that is
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you are subtracting both sides by this L transpose
VA right and then you multiply by M inverse,
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you will get an equation in the form minus
M inverse L transpose VA it is wrong it is
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L transpose VA.
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Okay this is equal to VX, now we have obtained
the expression or equation for VX in terms
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of other non-quantities okay. So that what
we do is that we substitute the expression
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for VX in equation 7.3 right. So that this
VX is eliminated and the moment we eliminate
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this VX you can write down the equation in
the form IA is equal to K times VA minus LM
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inverse L times VA. Okay and therefore this
equation can be put in the form IA equal to
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K minus LM inverse L transpose into VA right
that is if I write the equation in this form,
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IA equal to K minus LM inverse L transpose
this is L transpose into VA.
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Here, I and V have the same dimensions and
this is a matrix is going to be square matrix
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have the same dimensions as that of the K
and therefore now I can say that the my model
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becomes IA equal to Y bus reduced Y bus reduced
into VA. Therefore, this reduced Y bus this
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you can say this is Y bus reduced okay and
this reduced Y bus can be written in the form
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as K minus LM inverse L transpose. Therefore,
here the network which initially had certain
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number of nodes and those nodes where there
was no current injection okay. These nodes
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have been eliminated and therefore this is
now a reduced order model and there are no
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nodes where there is no injection. Actually
I will just illustrate this with the help
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of a example
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Now, let us consider that we have a simple
system is the third order system let us say
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okay and the Y bus matrix as original system
is minus 3.330, 3.330 minus 7.5, 2.5, 3.333
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and in, I can put 3 here also 2.5 and minus
10.833. This is let us say is Y bus of a system
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which has 3 nodes okay. Now let us assume
that the node number 3 is 1where there is
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no current injection and I want to eliminate
that node, what we do is that we partition
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this matrix because there are 3 nodes 1 and
2 and 3, okay therefore we partition it in
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the fashion.
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So that K becomes now matrix minus 3.3300
minus 7.5 okay. L is a vector in this particular
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case 3.33, 2.5, j will be there in all the
cases okay and M is minus j times 10.833.
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Okay now you substitute the value of KL and
M in the expression, K minus L M inverse L
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transpose.
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Now we make the substitution here this is
a column column matrix this is simply a scalar
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quantity this is a row where when you multiply
this 2 it will be a 2 by 2 matrix and then
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subtract this form the elements of K matrix
and the resultant matrix which comes out to
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be j times 2.308, 0.769, 0.769 minus 6.923.
Now I have illustrated through a very a you
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know small example but this is applicable
to any system which may have ah any number
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of nodes to be eliminated and this simplifies
the process to a great extent okay. Now our
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next step will be in solving the multi machine
problem, we have to develop the classical
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model.
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When I say it is a they are using the classical
approach we will have to develop the model.
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Now in the classical approach for transient
stability analysis we make all those assumptions
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which are made earlier. These 5 assumptions
which we made earlier I will just reiterate
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them, first was that mechanical input power
remains constant, second was all asynchronous
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powers or dumping torques were assumed to
be 0, third assumption was asynchronous generator
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can be represented by a constant voltage behind
direct axis transient reactance, fourth was
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that synchronous power can be computed from
the steady state solution of network and the
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fifth was the phase angle of the voltage behind
transient reactance coincides with the total
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position with respect to the synchronously
rotating reference plane.
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These were the 5 assumptions which we had
made, the next assumption which we will make
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here is all loads because in any power system
we will have loads at various buses all loads
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may be considered as shunt impedances to ground
with values determined with values determined
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by conditions prevailing immediately prior
to the transient conditions
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that is in the system we have loads connected
and we will convert the loads loads by equivalent
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shunt impedances okay and while calculating
the shunt impedances, we will consider the
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condition prevailing prior to the transient.
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Okay whatsoever are the condition before the
occurrence of transient or steady state condition,
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okay using that condition we will find out
the equivalent load admittances or load impedances.
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Okay this is very important next assumption
which is made
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Now for analyzing the transient stability
or for any a stability problem we always require
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the initial operating condition or the steady
state operating condition prior to the occurrence
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of disturbance.
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Now for a multi machine system the steady
state pre fault conditions for the for the
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system are computed using load flow, there
is a mistake by mistake I did not write. The
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steady state pre fault conditions for the
system are computed using a load flow. Therefore,
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a load flow for the system becomes a pre condition
for starting the solution of any stability
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problem in a multi machine system. Okay now
the model for the system.
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Okay now the model of the system actually
with this all the assumptions which we have
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made it looks like this. The transmission
network is shown here, to this transmission
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network we have generators connected at certain
nodes. We can, so that these are the points
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at which the generators are connected. There
are certain points in the network where the
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loads are connected. Okay now these loads
are represented by constant impedances connected
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to the reference bus.
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Now this generators are replaced by ah voltage
behind transient reactance direct axis transient
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reactance. Now here, here we will not neglect
the armature resistance of the synchronous
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machine okay. Once we account for the armature
resistance then we are representing the generator
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by a impedance r1 plus j times xd1 prime r2
plus j times xd2 prime and we find out the
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voltage behind this impedance, if you neglect
this resistance then it becomes simply the
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voltage behind direct axis transient reactance
but here when we do the transient stability
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analysis there is no need to ignore the resistance
and we can account for it.
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Once you account for this then these are the
voltages behind the transient reactance.
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Therefore, now if you see here the network
which is enclosed in this rectangle is a passive
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network it does not have any source now, okay
all the sources have been taken out that is
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at this nodes we have connected the voltage
sources and all the loads have been replaced
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by constant impedances and therefore what
we exactly do when solving this problem is
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that initially you will perform a load flow
analysis okay and once you perform the load
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flow analysis you will find out the voltages
at all these nodes and all the nodes where
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the all the nodes where the loads are connected.
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Okay then using using this information the
loads will be converted into equivalent impedances
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and we will replace the generators by equivalent
voltage in series with impedance. Okay then
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then this becomes my model which is going
to be used for transient stability analysis.
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Now in this case what we have done is that
I have shown n generators okay and there may
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be ah r nodes where the loads are connected.
Okay therefore the system may have originally
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n plus r nodes. Now when you perform this
load flow analysis you will, you will find
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out that the Y bus matrix considering only
the line parameters. Okay therefore now when
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you perform the transient stability analysis
the Y bus matrix will be modified and it will
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be obtained considering the machine impedances
and this load impedances.
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Okay once you obtain this Y matrix accounting
for this machine impedances and the load impedances,
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the load buses can be eliminated okay and
therefore you will be left with only these
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n buses where the generators are connected
okay all the load buses you eliminated and
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the the buses where this generators are connected
they are also eliminated because there is
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no injection here also but ultimately you
will need or you will get a model which will
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have the n nodes, where the generators are
connected and these are the constant voltages
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that is the voltages of constant magnitude
their phase angles will very during the dynamic
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conditions okay.
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Now for this network we can write down the
equations in the from I equal to Y bus E,
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where I is the current vector are or current
injected at all the n buses and Y bus is the
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admittance matrix obtained after eliminating
all the nodes. Okay therefore using this we
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shall be in a position to obtain the equations
for for electrical power output from all the
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machines because we know actually that where
we have to obtain or we have to analyze the
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transient stability analysis we need the expressions
for electrical power output from the generators,
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okay. Therefore using this model which is
having only the n nodes where the generators
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or internal voltage of the generators are
connected and this exercise is a important
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exercise for which I have already the explained
the methodology.
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Now here, we will represent the uh diagonal
elements of the Y bus matrix which are called
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Yii will be equal to magnitude Yii angle theta
ii, which will have real and imaginary parts
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Gii plus j times Bii. Similarly, the diagonal
elements will be shown as Yij magnitude Yij
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angle theta ij, Gij plus j times Bij and these
voltages, Ei will be written as Ei angle delta
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i.
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Now here I am not writing this prime right
because once we understand that this is the
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voltage behind the direct axis transient reactance
and the state of resistance included. Okay
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then there because writing every time E prime
is little what inconvenient okay therefore
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we consider that these are the internal voltages
and the magnitudes of these voltages will
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remain constant. Now we can find out the electrical
power output Phi by this basic equation that
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is the real part of Ei Ii star real part of
Ei Ii star this is very important.
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Now here you look like this that this is the
bus at which the voltage source is connected
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say Ei this is i th bus, this Ii is the current
which is injected by this voltage source.
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Okay and we know actually the the complex
power injected is equal to Pi Si equal to
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Pi plus j times Qi and this can be obtained
by the formula formula always and in this
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particular system Ei into Ii conjugate, this
star stands for conjugate. Okay and therefore
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when I take the real part of this Ei Ii star
that becomes the electrical power output from
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the I th generator.
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Now you we make the we substitute the expression
for Ii, the expression for Ii is
we have the vector Ii can be obtained as Yi1
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Yi2 Yin multiplied by this vector E1 up to
that is you have this is this is the I th
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row of the Y bus matrix and when you multiply
this with the voltage vector, bus voltage
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vector you will get actually this Ii as Yi1
plus Yi2 E2 plus Yin En that is we can write
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down this Ii as summation of Yij Ej where
j varies from 1 to n, this is the way we will
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be writing the expression for Ii .
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Now when you make this substitution here for
the expressions for Ii that is you have to
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the conjugate of this current and then perform
simplifications you will get the expression
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Pei that is electrical power output of I th
machine as Ei square Gii plus j equal to 1
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to n Ei Ej cosine of theta ij minus delta
i plus delta j and j will not take the value
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equal to i and this equation is valid for
i varying from 1 to n for all the machines
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that is you put i equal to 1, 2, 3, 4. You
will find that you will be in a position to
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get the expression for electrical power.
This is a very important expression and for
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n machine system we will get n such equations.
Now you can if you just look this look at
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this expressions carefully, you will find
actually that this depends upon the magnitudes
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of all the voltages which are known to us,
they remain constant during the transient
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stability analysis that is during the transient
period. These are the Gii there is one term
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which is missing here actually just please
correct it. It is Yij it is there will be
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Ei Ej, Yij will also be there.
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Okay cosine of theta ij minus delta i plus
delta j therefore this, Yij is also known
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to us only thing which is not known will be
the quantity which vary the delta i delta
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ij therefore, what happens is that these angles
will be computed by using the numerical technique
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and at the end of each step of computation
these angles will be known. Therefore, you
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substitute the value of these angles which
are computed when you solve the swing equations
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right and you get the expression for Pei at
each step. Now, once we know the expressions
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for electrical power output of the machines
for all the n machines then we can write down
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the n swing equations okay.
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The swing equations for each machine will
be 2 times Hi upon omega s, d omega i by dt
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equal to Pmi, that is the mechanical input
to i th machine minus the electrical output
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of the i th machine again here. Yij term is
missing. Okay that is what we have done is
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that this expression is same as I have computed
this is the expression for electrical power
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output from i th machine okay then this is
the main equation and we write down this d
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delta i by dt equal to omega I, which is equal
to the actual speed of the rotor minus synchronous
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speed because d delta i by dt is the access
speed of the rotor right and this it is access
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00:29:14,260 --> 00:29:18,870
over synchronous speed. Therefore, I am writing
d delta i by dt equal to omega i which is
201
00:29:18,870 --> 00:29:22,940
equal to d theta i by dt minus omega s.
202
00:29:22,940 --> 00:29:29,940
Okay therefore what we see here is that we
have written 2 equations, 2 first order equations
203
00:29:31,880 --> 00:29:38,620
for one machine and therefore, if you have
n machines in the system then you will get
204
00:29:38,620 --> 00:29:45,620
total 2 n such first order equations. Now
we need the information of what is the mechanical
205
00:29:58,420 --> 00:30:05,420
power input to the machines Pmi and we assume
that this Pmi is constant over the transient
206
00:30:09,930 --> 00:30:14,610
period that is mechanical power input to all
the machines are assumed to be constant right.
207
00:30:14,610 --> 00:30:21,420
Therefore, there is a necessity to find out
what will be the mechanical power input initially
208
00:30:21,420 --> 00:30:22,320
okay.
209
00:30:22,320 --> 00:30:28,960
Now how to do it because under steady state
conditions the electrical power output is
210
00:30:28,960 --> 00:30:34,990
equal to the mechanical power input and the
system is in steady state condition that is
211
00:30:34,990 --> 00:30:40,530
the derivative terms on the right left hand
side of the equations are all 0 okay.
212
00:30:40,530 --> 00:30:46,830
Therefore, with this condition we can write
down the expression for mechanical power input
213
00:30:46,830 --> 00:30:53,830
as Pmio equal to Ei square Giio, I am putting
o here means this is the this is the element
214
00:30:56,750 --> 00:31:03,750
of the Y bus matrix before the occurrence
of transient again here you have Yij okay,
215
00:31:04,180 --> 00:31:11,180
that is the complete expression of wherever
actually ah angles were delta i, I put delta
216
00:31:14,550 --> 00:31:21,550
io, delta j, delta jo, now theta ijo here
you will find actually that when the system
217
00:31:26,150 --> 00:31:30,550
is subjected to disturbance.
218
00:31:30,550 --> 00:31:35,810
Okay therefore during fault conditions the
Y bus matrix of the system will be different,
219
00:31:35,810 --> 00:31:42,810
during post fault condition it is again going
to be different right and therefore we will
220
00:31:43,500 --> 00:31:50,500
have 3 different ah Y bus matrices one pre
fault condition, one during fault condition,
221
00:31:50,790 --> 00:31:55,660
third post fault condition. Now here to compute
the mechanical power input initial mechanical
222
00:31:55,660 --> 00:32:02,660
power input Pmio right. We use the conditions
which are prevailing the in the system before
223
00:32:03,870 --> 00:32:08,880
the occurrence of disturbance and this is,
I am denoting by putting a further subscript
224
00:32:08,880 --> 00:32:15,880
o, okay.
225
00:32:18,450 --> 00:32:25,450
Now the equations which we have written were
2n equations, these 2n equations can be written
226
00:32:27,130 --> 00:32:34,130
in the general form as X dot equal to a function
a non-linear function X, Xo, t where, X is
227
00:32:42,270 --> 00:32:49,270
a vector of dimensions 2n into 1X is a vector
of dimension 2n into 1 where, n is the number
228
00:32:52,500 --> 00:32:59,360
of machines and these are all first order
nonlinear differential equations or we can
229
00:32:59,360 --> 00:33:06,360
say a set of 2n first order couple non-linear
differential equations they are all couple.
230
00:33:06,600 --> 00:33:13,600
Now X transpose is actually the X is the state
vector the state variables are omega 1, delta
231
00:33:16,640 --> 00:33:23,640
1, omega 2, delta 2, omega n, delta n right
and let me just summarize here, how we obtain
232
00:33:31,570 --> 00:33:38,290
the mathematical model of the system. The
first step is therefore a given system you
233
00:33:38,290 --> 00:33:45,290
perform a load flow study and find out the
initial steady state operating condition okay,
234
00:33:46,740 --> 00:33:53,740
once you have done this the next step will
be to replace all the loads by constant impedances.
235
00:33:56,230 --> 00:34:02,520
Then next step will be to knowing the information
which you have obtained in the steady state
236
00:34:02,520 --> 00:34:09,520
ah condition you replace all the generators
by constant voltage behind a impedance r plus
237
00:34:12,380 --> 00:34:19,169
j times Xd prime okay. Once you have done
this thing, you can illuminate all those nodes
238
00:34:19,169 --> 00:34:26,169
where they are no current injections or no
source is connected okay using the matrix
239
00:34:26,860 --> 00:34:28,960
algebra.
240
00:34:28,960 --> 00:34:35,960
Once you have done this thing you can find
out the set of non-linear coupled differential
241
00:34:36,770 --> 00:34:43,770
equations for the machine okay. Once these
equations are there our task is to solve these
242
00:34:43,880 --> 00:34:50,880
equations using numerical technique and plot
the graph relating delta versus time that
243
00:34:52,860 --> 00:34:57,280
is the we have to plot swing curves for all
the machines and then we examine the swing
244
00:34:57,280 --> 00:35:04,280
curves to see whether the system is stable
or not okay this is what is step.
245
00:35:04,550 --> 00:35:11,550
Now here before we do all the transient steady
analysis one has to perform some preliminary
246
00:35:14,300 --> 00:35:21,240
computations. The preliminary preliminary
computations to be computed done are, all
247
00:35:21,240 --> 00:35:27,380
system data are converted to a common base
this is a pre important requirement that when
248
00:35:27,380 --> 00:35:32,670
you assemble the system data you will find
actually that data are given on the basis
249
00:35:32,670 --> 00:35:39,670
of there name plate ratings. Suppose you have
a generator of 5hundred MVA, okay then the
250
00:35:40,050 --> 00:35:45,500
generator parameters will be given on the
basis of name plate rating of the machine.
251
00:35:45,500 --> 00:35:52,500
Okay but once you are consoling a system you
have to convert all the data on a common base
252
00:35:52,790 --> 00:35:59,790
and the common base generally chosen is hundred
MVA, okay 100 MVA is common because it looks
253
00:36:02,110 --> 00:36:07,140
to be little convenient for computations okay.
Otherwise there is no such hard and fast rule
254
00:36:07,140 --> 00:36:13,100
you choose the common base and when you say
the one is MVA base another is the voltage
255
00:36:13,100 --> 00:36:19,700
voltage in a particular circuit you will choose
and then you will compute the per unit values
256
00:36:19,700 --> 00:36:23,660
of all the system parameters on common base.
257
00:36:23,660 --> 00:36:28,390
The second step is the loads are converted
into equivalent impedances or admittances
258
00:36:28,390 --> 00:36:34,240
the needed for this is obtained from the load
flow study, okay this is what I have already
259
00:36:34,240 --> 00:36:41,240
told you. Now how do we convert actually the
loads into constant admittances these steps
260
00:36:42,250 --> 00:36:49,250
are also important this we can write down
here, that at suppose the load is represented
261
00:36:49,810 --> 00:36:56,810
as PL plus j times QL at any bus as load is
a PL plus j times QL.
262
00:36:58,450 --> 00:37:05,450
This load can be written as VL into IL star
because this is the complex complex load right
263
00:37:09,360 --> 00:37:16,360
and with this can be written as the bus voltage
, of the load bus, the voltage of the load
264
00:37:18,740 --> 00:37:25,740
bus into the conjugate of the current drawn
by the load IL star. Okay using this expression
265
00:37:25,790 --> 00:37:32,790
we will compute the value of elements of the
admittance which is going to replace the load
266
00:37:41,360 --> 00:37:47,240
by a constant admittance. Now the admittance
will be written as YL plus j times YL equal
267
00:37:47,240 --> 00:37:52,280
to GL plus j times BL, okay.
268
00:37:52,280 --> 00:37:59,280
This can be simply obtained that in this expression
in this expression you can substitute the
269
00:37:59,460 --> 00:38:06,460
value of IL, we know that the current drawn
by an admittance is equal to admittance into
270
00:38:07,680 --> 00:38:13,910
the voltage and admittance load admittance
multiplied by the voltage that will be the
271
00:38:13,910 --> 00:38:19,360
current and admittance is given by this formula,
load voltage is VL therefore IL star is going
272
00:38:19,360 --> 00:38:26,360
to be VL star into YL star IL star is VL star
into therefore VL into VL star becomes VL
273
00:38:29,930 --> 00:38:36,740
square and IL then YL will become GL minus
times j times BL, YL star.
274
00:38:36,740 --> 00:38:43,740
So that we can write down this load as VL
square into GL minus j times BL right and
275
00:38:46,080 --> 00:38:53,080
therefore, using this expression equation
7.8 right, we can say that the load admittance
276
00:38:54,880 --> 00:39:01,880
is equal to PL minus VL square minus j times
QL by VL square that is you can say that the
277
00:39:03,000 --> 00:39:09,770
ah real part is PL divided by VL square and
reactive part is BL divided QL minus VL square
278
00:39:09,770 --> 00:39:13,230
okay and the proper sign is to be consider.
279
00:39:13,230 --> 00:39:18,640
Therefore, this is one important step and
this voltage is VL which are substituting
280
00:39:18,640 --> 00:39:25,640
in this expression are the obtained by load
flow analysis. Okay another computation which
281
00:39:26,340 --> 00:39:33,340
is required to perform is the voltage behind
the transient reactance. Okay now here you
282
00:39:34,490 --> 00:39:41,060
can use this expression Ei prime equal to
Vti plus j times Xdi prime into Ii.
283
00:39:41,060 --> 00:39:48,060
Now here I have not considered the armature
resistance right but you can modify this by
284
00:39:48,530 --> 00:39:55,530
putting here instead of j times Xdi prime,
we can put here Ri plus j times Xdi prime
285
00:39:56,790 --> 00:40:01,590
okay and since the all information is are
known, current is known, terminal voltage
286
00:40:01,590 --> 00:40:08,590
is known, you can find out the while finding
out this internal voltage we will use this
287
00:40:16,080 --> 00:40:22,020
expression, the current injected by the generator
is Pi minus j times Qi divided by Vti star.
288
00:40:22,020 --> 00:40:28,810
This is very standard equation which can be
used to compute the value of current and this
289
00:40:28,810 --> 00:40:35,810
current is required in the equation 7.10 to
compute the internal voltage. The internal
290
00:40:39,230 --> 00:40:46,230
voltage which you compute, the internal voltage
which you compute right will have its magnitude
291
00:40:47,300 --> 00:40:54,070
and phase angle right and this phase angle
which you get become the initial value of
292
00:40:54,070 --> 00:41:01,070
deltas, okay and the magnitudes will remain
constant during the transient.
293
00:41:11,100 --> 00:41:18,100
Now once you have done these preliminary calculations
okay we obtain using this information, the
294
00:41:18,770 --> 00:41:25,770
swing equations. Now you know that we require
swing equations under 3 different conditions,
295
00:41:28,940 --> 00:41:35,940
one is the pre fault operating condition,
second is the during fault and third is the
296
00:41:36,620 --> 00:41:42,620
post fault and therefore when you solve the
swing equations suppose I start solving the
297
00:41:42,620 --> 00:41:48,540
equations at time t equal to 0, okay now if
the perturbation or disturbances occur at
298
00:41:48,540 --> 00:41:55,540
time t equal to 0 right then immediately from
that initial steady state condition I will
299
00:41:56,450 --> 00:42:00,830
be using the swing equations, I will be solving
the swing equations which are applicable during
300
00:42:00,830 --> 00:42:05,960
the during the fault on period.
301
00:42:05,960 --> 00:42:12,830
Then once I know that okay after this much
time the fault is cleared right it means the
302
00:42:12,830 --> 00:42:19,830
movement I reach that particular point i will
change over from the fault on period swing
303
00:42:20,440 --> 00:42:27,440
equations to post fault swing equations and
continue to solve it right. Now one should
304
00:42:29,570 --> 00:42:36,570
know actually that suppose the fault occur
in the system, how to obtain the ah the swing
305
00:42:37,390 --> 00:42:43,700
equations which will be applicable during
fault conditions, pre fault conditions we
306
00:42:43,700 --> 00:42:45,290
have already obtained.
307
00:42:45,290 --> 00:42:52,290
Now the faults can be ah symmetrical faults
or unsymmetrical faults. Now in case it is
308
00:42:54,730 --> 00:43:01,030
a symmetrical fault then the problem is bit
simple, if it is unsymmetrical then also it
309
00:43:01,030 --> 00:43:06,140
is not difficult because we have already seen
how to account for the unsymmetrical fault.
310
00:43:06,140 --> 00:43:12,910
Now if it is suppose I take fault at a particular
bus, let us say that fault occurs at a particular
311
00:43:12,910 --> 00:43:19,910
bus then voltage of that bus will collapse
it will become 0. Therefore, the occurrence
312
00:43:20,150 --> 00:43:27,010
of fault at the particular bus is accounted
by considering or by is by setting that voltage
313
00:43:27,010 --> 00:43:30,140
of that bus to 0.
314
00:43:30,140 --> 00:43:37,140
Okay then once the voltage of that bus becomes
0 we have to modify the Y bus matrix which
315
00:43:38,680 --> 00:43:44,480
will be applicable to this period, this particular
fault on condition. This modification are
316
00:43:44,480 --> 00:43:50,550
not difficult one can, once you see that a
particular bus is grounded okay looking into
317
00:43:50,550 --> 00:43:55,940
that condition you can modify the Y bus matrix.
Now simplest way is that if you have Y bus
318
00:43:55,940 --> 00:44:02,560
matrix for the pre fault condition then if
suppose at a particular bus let us say ah
319
00:44:02,560 --> 00:44:08,770
bus number five bus number five if suppose
there is a 3-phase short circuit then the
320
00:44:08,770 --> 00:44:15,770
row and column corresponding to the bus number
5 can be deleted from the pre fault Y bus
321
00:44:17,980 --> 00:44:24,980
matrix and the remaining, the remaining Y
bus matrix right will become the Y bus matrix
322
00:44:25,700 --> 00:44:28,150
applicable to post fault condition.
323
00:44:28,150 --> 00:44:34,910
In fact actually suppose one particular bus
is grounded or since it is sorted it gets
324
00:44:34,910 --> 00:44:41,910
get grounded it they the number of buses which
remain now will become less by one right therefore
325
00:44:42,530 --> 00:44:49,530
Y bus matrix also reduce in dimension. Similarly,
once you consider the post fault condition
326
00:44:49,710 --> 00:44:56,350
right then, you can obtain the Y bus matrix
during the post fault condition because under
327
00:44:56,350 --> 00:45:01,640
post fault condition what happens one line
may be out therefore for the remaining system
328
00:45:01,640 --> 00:45:03,580
you have to find out the Y bus matrix.
329
00:45:03,580 --> 00:45:10,580
Now this Y bus matrix will not be assembled
fresh one can be modify the existing Y bus
330
00:45:11,090 --> 00:45:17,970
matrix by modifying the elements of Y bus
matrix some elements will be affected, for
331
00:45:17,970 --> 00:45:23,860
example actually if suppose there is a Y bus
matrix and if the line connecting the node
332
00:45:23,860 --> 00:45:30,320
one and two that is tripped.
Okay then what is that is going to affect
333
00:45:30,320 --> 00:45:37,170
the or which elements of the Y bus matrix
will be affected that is Y11 will be affected
334
00:45:37,170 --> 00:45:44,020
because that line is out, Y22 will be affected
and Y12 will be affected others will not be
335
00:45:44,020 --> 00:45:51,020
affected, right. Therefore once we know that
particular type of line element has been removed
336
00:45:55,100 --> 00:46:02,100
right you can easily find out the Y bus matrix
applicable during the post fault period and
337
00:46:07,710 --> 00:46:13,490
once we know the Y bus matrix which are applicable
during post fault, during fault and during
338
00:46:13,490 --> 00:46:19,350
pre-fault conditions that you already known
because our equations that is the equations
339
00:46:19,350 --> 00:46:25,510
for electrical power output are written using
the reduced Y bus matrix therefore this reduction
340
00:46:25,510 --> 00:46:32,510
process has to be done for all the 3 conditions
pre-fault fault on condition and post fault
341
00:46:40,990 --> 00:46:43,110
condition.
342
00:46:43,110 --> 00:46:50,110
Now I will suggest you to refer to these problems
or these two examples which are given in here,
343
00:46:53,850 --> 00:47:00,850
one example is given in the book written by
P. M. Anderson and Fouad, power system control
344
00:47:03,370 --> 00:47:10,370
and stability volume one, example 2.6 and
2.7, these two examples this is example which
345
00:47:15,230 --> 00:47:22,230
is very very commonly used for illustrating
the transient stability analysis. It is a
346
00:47:25,500 --> 00:47:32,500
9 bus system, it has total 3 generators, 9
buses okay and in these two examples all the
347
00:47:35,360 --> 00:47:42,360
step by step computations have been carried
out to illustrate the steps involved in solving
348
00:47:44,580 --> 00:47:47,480
the transient stability problem.
349
00:47:47,480 --> 00:47:54,480
One more simple example which is given in
the ah basic book on power system analysis
350
00:47:56,450 --> 00:48:03,450
that is elements of power system analysis
by Stevenson, everybody is aware of this book
351
00:48:03,450 --> 00:48:10,450
Stevenson here he has solved 5 bus system.
Okay and in this 5bus system he has illustrated
352
00:48:15,090 --> 00:48:22,090
all the steps which are required to obtain
the complete transient stability solution.
353
00:48:25,440 --> 00:48:32,440
Now once the model is obtained and you solve
the swing equations, couple a set of the coupled
354
00:48:38,060 --> 00:48:43,050
swing equations, you will get the swing curves.
355
00:48:43,050 --> 00:48:49,430
Okay the swing curves will plot actually as
delta one delta two delta three as function
356
00:48:49,430 --> 00:48:56,430
of time. Now to examine whether the system
remains stable or not we have to see the relative
357
00:48:58,150 --> 00:49:05,150
variation of the swing curves that is not
the absolute values of delta 1, delta 2 delta
358
00:49:05,950 --> 00:49:10,090
n that is important what is the relative value
that is more important.
359
00:49:10,090 --> 00:49:17,090
Suppose the two machines may accelerate simultaneously
okay but the angles may remain very close
360
00:49:18,630 --> 00:49:23,420
to each other or angle difference may remain
very close then system is not losing synchronism
361
00:49:23,420 --> 00:49:29,240
therefore, when you examine the stability
of the multi machine system we have to examine
362
00:49:29,240 --> 00:49:36,240
the ah the plots of relative ah angles that
is we plot actually delta 12 that is delta
363
00:49:40,260 --> 00:49:41,520
1 minus delta 2.
364
00:49:41,520 --> 00:49:44,740
You plot delta 1 minus delta three suppose
there is a three machine system they are only
365
00:49:44,740 --> 00:49:51,670
a 3angles involved and once this curves are
plotted in case actually the curves or we
366
00:49:51,670 --> 00:49:57,840
can say delta 12 reaches the maximum value
and decreases it shows the stability condition,
367
00:49:57,840 --> 00:50:04,840
in case delta 12 increases continuously it
shows unstable condition. Okay with this let
368
00:50:06,940 --> 00:50:09,869
me conclude what we have done today.
369
00:50:09,869 --> 00:50:16,869
We have an we have given or we have studied
the basic steps involved in analyzing the
370
00:50:21,960 --> 00:50:28,960
transient stability of a multi machine system.
Here we have used a classical model for analyzing
371
00:50:31,940 --> 00:50:38,940
the stability I have mentioned that how do
we obtain the required required models for
372
00:50:41,670 --> 00:50:48,670
pre-fault, fault on and post fault operating
conditions. Okay and the swing equations which
373
00:50:49,880 --> 00:50:56,880
we obtained for the system are solved by using
numerical techniques and the stability of
374
00:50:59,130 --> 00:51:06,130
the system is understood by examining the
swing curves which we plot and particularly
375
00:51:08,680 --> 00:51:15,680
the is the swing curves are relating the relative
relative angles of the machines, okay thank
376
00:51:19,230 --> 00:51:26,230
you very much.