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Okay, friends today we will continue with
the study of modeling of synchronous machine.
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Today we will address some of these issues
per unit system for rotor circuits, per unit
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power and torque and alternative transformations.
We have discussed dqO transformation and we
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will also talk about the alternative transformation
which has also been proposed in the literature
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and then we then, we will talk about the steady
state analysis of the synchronous machine.
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Now we have developed the per unit equations
for
the stator circuits. Okay we have also developed
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that to the per unit equations for rotor circuits.
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Now these terms were defined and the definitions
were given like this Lafd bar equal to Lafd
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divided by Lsbase, ifdbase divided by s base
that is these three terms were defined that
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they are basically in the stator circuit equations
you see this that the stator circuit flux
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linkage equations psi d, psi q and psi o.
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Therefore in these three equations we had
defined these per unit inductances now these
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per unit inductances are the mutual inductances
okay.
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Similarly, when we talked about the per unit
rotor flux linkage equations, we had define
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these terms these mutual terms Lfkd, Lfda
these are all per unit quantities which were
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defined.
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Now, now we want to establish the per unit
system for the rotor. Now when I am trying
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to say per unit system for the rotor in the
sense that we have chosen the base quantities
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in the stator circuits. Now how should we
choose the base quantities for the rotor circuits,
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the rotor circuits are rotor field winding
and rotor amortisseurs. Okay now the primary
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consideration for choosing the base quantities
in the rotor are to simplify the the flux
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linkage equations to make this flux linkage
equation as simple as possible.
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Now one requirement which we have put is or
one way of simplifying is we make mutual inductances
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between different windings reciprocal. We
make mutual inductances between different
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windings reciprocal. Now when I say that we
make these mutual inductances reciprocal in
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the sense that you there is a mutual inductance
between the stator winding and the field winding
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okay.
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Now the if I write down the mutual inductance
between the stator winding and field winding
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I can write down Lafd. When I write down between
field winding and stator winding it can written
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as Lfda we want to make these two inductances
equal. So that that is what is the meaning
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of reciprocal okay another thing which we
would do to simplify the equations will be
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that the mutual inductances between stator
and rotor circuits on each axis are to be
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equal that is when I say each axis means,
we have two axis d axis and q axis. The on
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the d axis we have the stator direct axis
winding, field winding and amortisseurs on
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the d axis therefore, there exists mutual
inductances between these windings we would
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like to make them equal.
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Similarly, there exists mutual inductance
between the q axis winding on the stator and
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q axis amortisseur we want to make these mutual
inductance also equal therefore we are trying
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to do two things, one is that we will try
to make or we will make the mutual inductances
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reciprocal and the mutual inductances on each
axis equal okay.
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Now this will simplify the equations to a
great extent. Now how do how do we make the
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mutual inductances reciprocal, now to make
the mutual inductance reciprocal you just
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look at the suppose you look actually this
equation I just put this, this is slightly
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I have to make that I think Lakd. Okay, now
this Lakd and another is Lkda we want to make
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them equal okay. Now these quantities were
defined now when to make them equal reciprocal,
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okay so that what is to be done is that we
equate the expression for this with the expression
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for Lkda, okay.
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Now we will start first
that is let us say that we want to make the
per unit mutual inductance Lfkd equal to per
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unit mutual inductance Lkdf right that is
the amortisseur and field winding, okay that
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is these are the two, you know mutual inductances
one between between field winding and amortisseur
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another is between amortisseur and field winding
and want to make these reciprocal.
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Now these quantities were defined now as per
the definition this quantity L bar fkd that
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is the per unit mutual inductance between
field and amortisseur windings were defined
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like this that is Lfkd upon Lfdbase, ikdbase
upon ifdbase that is in the denominator here
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is Lfdbase, ifdbase here you have Lkdbase,
ikdbase. Now this is by definition which we
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have defined actually when we wrote the equations
for flux linkage equations in the rotor circuit.
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Now when we equate these two, okay we will
get this expression that is you just cross
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multiply
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You will find that it becomes Lkdbase into
ikdbase square equal to Lfdbase, ifdbase square
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you can just check it it comes out to be like
this that is ikdbase comes from this side,
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so okay and this these two cancel out they
are equal okay and we get this equation. Now
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here what we do is we multiply both sides
of this equation by omega base, once you multiply
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by omega base it becomes L omega base into
Lkdbase, ikdbase square similarly, omega base
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equal to Lfdbase into ifdbase square.
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Now you can identify this quantity omega L
into i this quantity this is actually the
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voltage this is a voltage therefore I take
this base frequency in radiance per second
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the base inductance and multiply this by current,
base current this will come out to be base
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voltage okay and now we are taking these quantities
for amortisseur the quantities are for amortisseur
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Lkd ikd therefore the base voltage will become
the amortisseur circuit base voltage this
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d axis amortisseur circuit base voltage similarly,
here.
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Now when you do this you will find this is
very important relationship that is the base
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voltage in the direct axis amortisseur circuit
multiplied by the base current in the direct
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axis amortisseur circuit this product must
be equal to the the base voltage in the field
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circuit into base current in the field circuit,
it is very simple that is the the base vi
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in the amortisseur circuit should be equal
to base vi in the field circuit
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Now if you do this you will find actually
that the mutual mutual inductances right between
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the field winding and the amortisseur winding
will become reciprocal. Okay therefore this
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is one important requirement that base VA
in the field circuit should be same as the
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base VA in the amortisseur circuit now we
do one similar exercise for mutual inductance
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between the stator winding that is the stator
direct axis winding and the field winding
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that is stator and field that is Lafd, we
want to make this as Lfda that we want to
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make these two per unit mutual inductances
equal right now these quantities were again
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defined when we wrote the equations for flux
linkages okay.
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Now when we equate this two terms and follow
the same procedure as we have done for the
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for the mutual inductance between field and
amortisseur. Okay therefore here again we
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come to very important result and that result
is that vibase, v is that is the efd base
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into ifd base that is the vibase in the field
circuit comes out to be equal to 3 phase VAbase
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for stator, 3 phase VAbase for the stator
that is in the stator circuit when you assume
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that the the base voltage as peak value of
the phase voltage, the the base current in
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the stator circuit is the peak value of the
phase current and we have established that
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this quantity is equal to 3 times VA, 3 times
VAbase for the stator that is 3 phase VAbase
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for the stator the total volt ampere rating
right.
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Therefore, this is a very important relationship
because in the we have also established the
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relationship that the base VA in the kd circuit
direct axis amortisseur should be same as
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that in the field circuit it means now we
can conclude actually that these 3 windings
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one in the field winding, another stator winding,
third amortisseur winding, direct axis amortisseur
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winding.
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Now if you want to make this mutual inductances
reciprocal then the volt ampere rating or
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volt ampere base in the stator circuit must
be equal to the volt ampere base in the field
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circuit and volt ampere base in the amortisseur
circuit because in the stator with the the
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volt ampere base is equal to the 3 phase VAbase
3 phase VA total MVA rating.
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Okay this is exactly similar to what you normally
do in any system we take total 3 phase MVA
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as the base quantity and then we talk about
the base for the rotor circuits we find actually
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that but rotor circuit now if the field circuit
if you consider it it has a applied voltage
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Vefd and the current flowing is ifd therefore
efdbase into base ifdbase becomes the this
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product becomes the VAbase for the field circuit
therefore this one condition will be established
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that if we want to make, we want to make the
mutual inductances between the circuits reciprocal
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then this is the criteria required for choosing
the VAbase for these circuits, okay.
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Now this the similar exercises have to be
done for for quadrature axis, stator winding
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and the quadrature axis amortisseur because
so for we have talked about the d axis and
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here also we established the relationship
and the relationship is that ekqbase into
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ikqbase is equal to 3 by 2 esbase into isbase
not now this quantity is nothing but the 3
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phase VAbase for the stator therefore ultimately
what is our conclusion that all the circuits
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state all the all the circuits that is all
which are in the rotor field circuit, amortisseur
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circuits right these would have their VAbase
equal to the 3 times or 3 phase VAbase of
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the stator circuit this is the and this will
make all the mutual inductances reciprocal.
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Therefore, now I conclude here that in in
order to satisfy the requirement that per
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unit mutual inductances between again the
emphasis is that we are trying to make the
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per until mutual inductances reciprocal actual
mutual inductances which exist, we have seen
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actually that they are not reciprocal when
you see the actual equations which we have
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written right they are they appear in the
equations in a different fashion between different
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windings the reciprocal okay and the requirement
is that VAbase must be same and equal to the
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stator 3 phase VAbase, this is the main conclusion
okay.
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Now the next requirement which we have posed
here is posed here is that the mutual inductances
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on the same axis we want to make them equal
first we make them reciprocal and then requirement
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was there want to make them equal.
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Now to make them equal we we define the per
unit self inductance of the d axis winding,
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d axis stator winding as leakage inductance
per unit leakage inductance plus per unit
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mutual inductance that is when you look at
the d axis winding on the stator. Okay the
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total per unit inductance is defined as Ld
bar you know this bar stands for rotor super
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bar stands for per unit.
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Now the total flux which is produced right
which links the d axis winding does not pass
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through the field circuit or the amortisseur
circuits d axis amortisseur circuit there
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is some flux which is the local leakage flux
therefore this Ll is accounts for the inductance
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of the d axis stator winding due to leakage.
Okay therefore this is this becomes our mutual
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inductance Lad bar similarly, for q axis.
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Now here our requirement is that in order
to make all per unit mutual inductances between
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stator and rotor circuits in the d axis equal
that is we have in the stator circuit a fictitious
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d axis winding okay and the rotor circuit
we have on the d axis two windings okay and
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we want to make them equal for making them
equal or this by definition this is actual
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value of mutual inductance Lad divided by
the Lsbase okay and this this is equal to
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Lafd which has which was earlier defined by
this expression when we wrote the wrote the
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mutual or we can say when we write the expressions
for the psi d psi q and psi naught right,
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the Lafd was defined like this per unit value
of Lafd was defined by this expression that
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you can refer to the pervious equations okay.
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Similarly, now we want to make Lad to equal
to the per until Lakd okay while Lakd was
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defined by this expression. Okay now what
we want that we equate these 2 as we did in
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the previous case and when we equates these
2 we come to one very important relationship
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that the ifdbase should be equal to Lad upon
Lafd, isbase. Similarly, ikdbase comes out
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to be Lad upon Lakd divided by Lsbase while
these quantities are the actual inductances,
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actual mutual inductances right while what
we trying to make them equal are the per unit
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mutual inductances. Okay, now this this gives
a relationship that suppose I have chosen
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the stator base current then the base current
in the field circuit should be given by this
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equation.
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Similarly, if I choose in the base current
in the stator as is base then the base current
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in the amortisseur circuit, direct axis amortisseur
circuits will be given by this expression.
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Okay similarly you can do for q axis also
that is q axis will give you that ikq base
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should be equal to Laq divided by Lakq, isbase
I hope you understand this one is this is
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this is a straight forward okay, what we are
trying to do is that what we want to make
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them equal. We want to make the per unit per
unit the mutual inductances on the same axis
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d axis equal that is right, we want to make
Lad per unit equal to Lafd which is the definition
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of Lad same as Lfd we want to make this Lakd
and Lafd equal to make them equal you equate
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these quantities which was the basic definitions
for this per unit Lafd and per unit Lakd.
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Okay, and we finally get the relationship
that the field currents base value of field
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current base value of the amortisseur circuit
current and the base value of the quadrature
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axis amortisseur therefore current are expressed
in terms of the stator, the stator circuit
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base current okay Similarly, we had also established
that VA base for all the rotor circuits in
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terms of the 3phase VA base of the stator
and therefore since actually when I talk about
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the VA that the field voltage base into ah
field voltage current base okay.
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Now this product is equal to the three phase
VA of the stator current right but the individually
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currents will be defined by this expression
now once the current is current base is given
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for these circuits the , we compute the voltage
base for the rotor circuits using the previous
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relationship I hope it is clear to all of
you now once we have done this and define
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the the model for the synchronous machine
is complete, the synchronous machine model
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is complete we for all for all further discussions
what we will do is that this super bar which
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we had put to identify these quantities as
the per unit quantities we will drop it but
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we will always keep in our mind that these
are per unit quantities once these are per
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unit quantity because every time writing super
bar is not very convenient okay.
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Now further we can express the per unit power
and torque per unit power and torque the in
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our previous derivations the terminal power,
total terminal real power was expressed as
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3 by 2 ed, id, eq, iq plus 2 times eo, io.
Okay that you can recollect now here what
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you do is that you replace all these quantities
by that is you divide by base 3 phase base
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VA of the stator here that is you divide this
by base volt ampere, you divide all these
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quantities by the base voltage and currents
when you do this exercise the power in per
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unit can be written as ed bar id bar plus
eq bar iq bar plus 2 times eo bar io bar.
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Okay that is only thing which happens is that
3 by 2 term which appears in this expression
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disappears in the expression for per unit
generator power, okay other expressions are
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same except that the per unit quantities this
exercise I will just suggest you to verify,
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you substitute the values and then you verify
that you get these 3 by 2 is then we obtained
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the per unit electrical torque or air gap
torque which will come out to be equal to
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per unit direct axis flux linkage psi d into
iq minus psi q into id. In fact this this
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is the most important expression
when we talk about the synchronous machine
model why is this expression is, so important
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can you tell me.
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Other, other stator circuit rotor circuit
equations are also ah equally important right
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because that gives the model complete model
we have to write down the stator circuit equations,
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rotor circuit equations, stator circuit flux
linkages, rotor circuit flux linkages. Okay
197
00:26:53,700 --> 00:26:58,320
and all these things have to be written in
per unit by establishing proper per unit base
198
00:26:58,320 --> 00:27:03,210
but when I say that the per unit air gap torque
is given by this expression and this is quite
199
00:27:03,210 --> 00:27:10,210
important, can you tell me why? yes instead
of calculating the torque in terms of three
200
00:27:14,280 --> 00:27:21,280
phase we are getting the torque just by 2
values. No, because the torque is computed
201
00:27:24,180 --> 00:27:31,180
in per unit using the per unit flux linkages
and per unit currents this expression is going
202
00:27:32,770 --> 00:27:36,090
to be used in our swing equation.
203
00:27:36,090 --> 00:27:40,420
When you have written the swing equation in
the swing equation, we have to write down
204
00:27:40,420 --> 00:27:45,470
mechanical torque minus electrical torque
and this this is the electrical torque that
205
00:27:45,470 --> 00:27:49,790
will come in the swing equation okay.
206
00:27:49,790 --> 00:27:56,790
I will just mention briefly about alternative
per unit systems and transformation that is
207
00:28:03,400 --> 00:28:09,830
the dqO transformation which we discussed
till now and another form of dqo transformation
208
00:28:09,830 --> 00:28:16,830
which has been discussed in the literature
and also used by some people. Here, in this
209
00:28:21,770 --> 00:28:28,770
the the dqo quantities are expressed in terms
of this transformation matrix into the phase
210
00:28:32,680 --> 00:28:36,960
currents that is the phase variables ia, ib
and ic.
211
00:28:36,960 --> 00:28:43,960
Now if you see this transformation matrix
the main difference here is that the dqO transformation
212
00:28:45,270 --> 00:28:52,270
which we used right this quantity or this
term kd this is a kd was a kd or kq they have
213
00:28:53,630 --> 00:29:00,630
both taken equal to 2 by 3. Here they are
taken as square root of 2 by 3 and correspondingly
214
00:29:01,830 --> 00:29:08,090
the the last row that is the third row the
terms become instead of 1 by 2, it becomes
215
00:29:08,090 --> 00:29:13,210
square root of 1 by 2 and so on.
216
00:29:13,210 --> 00:29:20,210
Now with this transformation if you write
the inverse transformation that is you write
217
00:29:20,640 --> 00:29:27,640
down the phase currents ia, ib and ic in terms
of dqo quantities right. Then this this this
218
00:29:30,900 --> 00:29:37,900
becomes the inverse of our transformation
matrix, this is the transformation matrix
219
00:29:38,300 --> 00:29:44,880
we use therefore if you invert this matrix
right then we can write down using this inverted
220
00:29:44,880 --> 00:29:51,880
matrix the relationship between the phase
currents ia, ib and ic in terms of dqO components.
221
00:29:54,740 --> 00:30:01,740
This inverse if you just see here then this
inverse is nothing but the transpose transpose
222
00:30:02,710 --> 00:30:05,110
of the transformation matrix.
223
00:30:05,110 --> 00:30:12,110
You just see here actually that this row is
the column here right while this row appears
224
00:30:14,480 --> 00:30:21,480
as column in the inverse okay and therefore
this is this inverse this is the inverse of
225
00:30:21,720 --> 00:30:27,840
the transformation matrix but is also the
the transpose of the transformation matrix.
226
00:30:27,840 --> 00:30:34,730
Now whenever you have this type of situation
we call this transformation as orthogonal
227
00:30:34,730 --> 00:30:41,730
transformation now when you resort to this
transformation one very important result which
228
00:30:43,260 --> 00:30:45,650
we get is like this.
229
00:30:45,650 --> 00:30:52,650
The terminal power Pt right which when we
write down as ea into ia plus eb into ib plus
230
00:30:53,840 --> 00:31:00,840
ec into ic this is a terminal power okay.
This Pt is average or instantaneous, instantaneous
231
00:31:02,020 --> 00:31:07,020
but what is average?
232
00:31:07,020 --> 00:31:14,020
The average and instantaneous become equal
in a 3 phase system, 3 phase system the total
233
00:31:15,790 --> 00:31:22,790
power is total or the average power is constant,
instant we say instant is power we calculate
234
00:31:24,940 --> 00:31:31,860
this sum will come out to be equal to average
power right therefore now if you substitute
235
00:31:31,860 --> 00:31:38,860
here that is you write this equation, that
is this equation can be written as Pt can
236
00:31:44,880 --> 00:31:51,880
be written as ea, eb, ec multiplied by ia,
ib, ic is it not this expression is written
237
00:32:02,450 --> 00:32:06,620
you can put in the matrix form this is row
vector multiplied by this column vector this
238
00:32:06,620 --> 00:32:13,620
will give you ea, ia plus eb, ib plus ec,
ic. Now you replace this row vector by dqO
239
00:32:18,220 --> 00:32:25,220
components you replace this also by dqO components
and multiply these 2 matrices when you do
240
00:32:27,870 --> 00:32:33,630
this multiplication the end result comes out
to be of this form that is the Pt will come
241
00:32:33,630 --> 00:32:40,630
out to be equal to ed, id plus eq, iq plus
eo, io it is something like this that when
242
00:32:41,700 --> 00:32:48,700
I had 3 voltages and 3 currents okay I got
the terminal power by multiplying this quantity.
243
00:32:49,110 --> 00:32:54,900
Similarly, I have now the direct axis voltage
quadrature axis voltage and zero sequence
244
00:32:54,900 --> 00:33:00,250
voltage similarly the direct axis current
quadrature axis current and zero sequence
245
00:33:00,250 --> 00:33:07,250
current I multiply and add this comes out
to be equal to right now for this this type
246
00:33:07,510 --> 00:33:13,520
of therefore this transformation is called
power invariant transformation, power invariant
247
00:33:13,520 --> 00:33:20,520
transformation and it has some merits which
has been discussed in literature but it has
248
00:33:21,860 --> 00:33:28,860
some demerits also and that is why in most
of the models which have been developed the
249
00:33:33,740 --> 00:33:40,740
dqo transformation which was discussed earlier
is used where we use kd and kq equal to 2
250
00:33:40,750 --> 00:33:47,750
by 3. Okay having developed the synchronous
machine model you have developed is synchronous
251
00:33:55,950 --> 00:34:02,950
machine model completely in using per unit
quantities and the model is developed in dqo
252
00:34:08,629 --> 00:34:12,649
frame of reference okay.
253
00:34:12,649 --> 00:34:19,649
Now this model is the complete model and now
what we will study from this point onwards
254
00:34:21,530 --> 00:34:28,530
will be first we will study the steady state
aspect okay and then we will go how this model
255
00:34:29,350 --> 00:34:36,350
can be simplified for our stability studies.
Okay that is this is the complete model actually
256
00:34:37,600 --> 00:34:44,600
right but sometimes we can make some assumptions
and simplify the model okay and how do we
257
00:34:46,250 --> 00:34:51,810
simplify this model which will be suitable
for stability studies that is what we will
258
00:34:51,810 --> 00:34:55,720
study in the subsequent lectures.
259
00:34:55,720 --> 00:35:01,710
Now let us quickly do the steady state analysis
because steady state analysis is one which
260
00:35:01,710 --> 00:35:08,710
is known to everybody all of you know this
analysis but now we will develop the steady
261
00:35:09,940 --> 00:35:16,940
state equations for the machine starting from
our dynamic model of the synchronous machine.
262
00:35:21,020 --> 00:35:27,720
Okay now when we talk about steady state conditions
we are presuming that the synchronous machine
263
00:35:27,720 --> 00:35:34,720
is operating under steady state condition
and the stator currents and voltages are balanced
264
00:35:35,900 --> 00:35:41,570
3phase currents and voltages that is this
is the assumption that these steady state
265
00:35:41,570 --> 00:35:48,570
operating condition and the stator and stator
quantity stator currents as well as the stator
266
00:35:50,250 --> 00:35:52,170
voltages are balanced.
267
00:35:52,170 --> 00:35:59,170
With this assumption once it is steady state
this derivative terms will be absent, in the
268
00:35:59,530 --> 00:36:05,170
steady state condition we do not have anything
like d by dt of psi suppose, I take this psi
269
00:36:05,170 --> 00:36:10,040
will be constant therefore this derivative
term derivative term will be absent, second
270
00:36:10,040 --> 00:36:17,040
is that all all zero sequence terms will also
be absent because it is a balanced 3 phase
271
00:36:19,330 --> 00:36:20,890
system okay.
272
00:36:20,890 --> 00:36:27,890
With this assumption the stator circuit equations
can be written now in per unit again remember
273
00:36:30,010 --> 00:36:35,450
that these are all per unit stator circuit
equations the bar is not written here while
274
00:36:35,450 --> 00:36:40,010
the stator circuit equations are written in
the form ed equal to minus omega r psi q minus
275
00:36:40,010 --> 00:36:47,010
Ra id eq equal to omega r psi d minus Raid
and efd equal to Rfd, ifd it is very simple
276
00:36:50,580 --> 00:36:57,580
actually, you can just in all the 3 equations
the p psi d terms have been p psi terms that
277
00:36:58,700 --> 00:37:04,380
is the derivative terms have been set equal
to 0 okay.
278
00:37:04,380 --> 00:37:09,930
Now under a steady state operating condition
the rotor speed is equal to synchronous speed
279
00:37:09,930 --> 00:37:16,690
therefore, we substitute here omega r equal
to omega s and in a per unit system we will
280
00:37:16,690 --> 00:37:23,690
see that omega r is equal to omega s equal
to 1. Okay now these are the 3 voltage equations
281
00:37:28,390 --> 00:37:35,390
these 2 voltage equations for the stator circuit
this voltage equation for the field winding
282
00:37:35,780 --> 00:37:40,970
and these are the flux linkage equations that
is psi d and psi q in per unit quantities
283
00:37:40,970 --> 00:37:47,970
they are all per unit terms we have not done
anything except that except that in these
284
00:37:52,190 --> 00:37:59,190
equations the the derivative terms have been
set equal to 0 right. Another thing while
285
00:38:01,860 --> 00:38:08,860
writing here we have also set the mutual inductances
per unit mutual inductances on the d axis
286
00:38:11,630 --> 00:38:15,940
windings are equal right therefore, you will
find here actually when you write down this
287
00:38:15,940 --> 00:38:18,110
psi d is Lad ifd.
288
00:38:18,110 --> 00:38:25,110
We will write down psi fd again I have put
Lad because the mutual inductance is mutual
289
00:38:30,220 --> 00:38:37,020
inductance between the stator d axis winding
and the field winding have been made equal
290
00:38:37,020 --> 00:38:43,770
in per unit in per unit system okay. Therefore,
this simply appears as Lad okay. Further actually
291
00:38:43,770 --> 00:38:50,770
when we have written these equations suppose
there is one amortisseur on the d axis k becomes
292
00:38:52,080 --> 00:38:59,080
one here, suppose you take 1 or 2 amortisseurs
on the q axis one will, one equation will
293
00:39:01,100 --> 00:39:08,060
be for psi 1q there is a second amortisseur
will be psi 2q right and they will be written
294
00:39:08,060 --> 00:39:15,060
in the by the same equation minus Laq iq why
the same Laq will come because mutual inductances
295
00:39:19,480 --> 00:39:26,480
between the 2 amortisseur on the q axis, when
expression per unit are equal.
296
00:39:28,370 --> 00:39:35,370
Therefore, this is okay now here we can just
derive one simple expression for field current
297
00:39:40,990 --> 00:39:47,990
what we do is that in these equations just
start with this. Suppose you take my interest
298
00:39:52,890 --> 00:39:59,890
is to write down what will be the value of
field current required or field current in
299
00:40:01,100 --> 00:40:07,120
per unit required to produce the required
steady state operating condition. So that
300
00:40:07,120 --> 00:40:13,870
using this expression you can write down the
expression for ifd that is ifd is written
301
00:40:13,870 --> 00:40:20,870
first here as psi d plus Ld id divided by
Lad and then this psi d that is the per unit
302
00:40:23,290 --> 00:40:30,000
flux linkages in the d axis will be replaced
by the this equation psi d equal to minus
303
00:40:30,000 --> 00:40:37,000
Ld id not not here actually this psi d because
we will resort back to our stator winding
304
00:40:40,730 --> 00:40:47,730
equations that is psi d can be written using
this equation. Okay that is eq equal to omega
305
00:40:48,320 --> 00:40:55,320
r psi d minus Raiq okay and when we make these
substitutions here we get very important relationship
306
00:41:00,280 --> 00:41:06,610
is this relationship that is the field current
which is required to produce certain value
307
00:41:06,610 --> 00:41:13,610
of eq the iq, id with these quantities known
they are all the circuit machine parameters.
308
00:41:16,710 --> 00:41:23,710
Now stator per unit stator resistance per
unit direct axis inductance, per unit mutual
309
00:41:28,260 --> 00:41:33,510
inductance between direct axis and the field
winding these are all now circuit parameters
310
00:41:33,510 --> 00:41:39,340
or synchronous machine parameters they are
all known. Okay and therefore when system
311
00:41:39,340 --> 00:41:46,340
is operating with some load the that load
will determine iq and id, okay and to have
312
00:41:48,260 --> 00:41:54,680
then for a particular value of eq this is
the value of ifd required therefore for all
313
00:41:54,680 --> 00:41:59,290
our studies actually as we will see further
when we talk about the models of the synchronous
314
00:41:59,290 --> 00:42:05,270
machine for our stability studies this information
is required but this is simply obtained using
315
00:42:05,270 --> 00:42:12,270
those equations that is the stator circuit
equations flux linkage equations and making
316
00:42:14,930 --> 00:42:19,500
some appropriate substitutions.
317
00:42:19,500 --> 00:42:26,500
Now we use this omega r is equal to omega
s and this omega s into Ld we will represent
318
00:42:30,350 --> 00:42:37,350
by xd omega s into Ld will be represented
by xd right. Similarly, you can see omega
319
00:42:38,190 --> 00:42:45,190
s into Lad will be represented by xad reactance
omega L is equal to reactance always okay.
320
00:42:46,080 --> 00:42:51,270
Therefore if this equation is written again
exactly the same but instead of writing in
321
00:42:51,270 --> 00:42:58,270
terms of the per unit per unit d axis inductance
and mutual inductance which is written in
322
00:43:02,530 --> 00:43:09,530
terms of Xd and Xad Xd is the direct axis
direct axis reactance while Xad is the mutual
323
00:43:12,500 --> 00:43:19,500
reactance between between the field and the
d axis winding.
324
00:43:24,760 --> 00:43:31,760
Now we will quickly see, now the phasor representation
before we talk about the phasor representation
325
00:43:33,070 --> 00:43:40,070
let us reinstate the some of the facts. Under
steady state conditions the id, iq these two
326
00:43:45,540 --> 00:43:50,710
currents these two currents are constant in
magnitude therefore basically the current
327
00:43:50,710 --> 00:43:57,710
flowing in the fictitious d axis winding and
the fictitious q axis winding are dc current
328
00:43:58,210 --> 00:44:00,710
direct currents.
329
00:44:00,710 --> 00:44:06,730
Similarly, the ed and eq they came out to
be constant voltages that we have already
330
00:44:06,730 --> 00:44:13,730
established under steady state operating conditions.
Now the question arises is that the when we,
331
00:44:14,030 --> 00:44:21,030
when we draw a phasor diagram right all the
phasor are corresponding to a sinusoidally
332
00:44:21,370 --> 00:44:26,620
varying quantity that is all the quantities
which vary sinusoidally and having the same
333
00:44:26,620 --> 00:44:33,620
frequency right can be represented in the
phasor diagram. Okay but here also although
334
00:44:36,170 --> 00:44:40,710
these quantities are constant in magnitude
because of certain trigonometrical relations
335
00:44:40,710 --> 00:44:47,710
which exist we will be in a position to represent
this d and q axis voltage and currents as
336
00:44:47,900 --> 00:44:54,900
phasor but with the with the clear clear understanding
in our mind that d and q axis quantities are
337
00:44:55,710 --> 00:44:56,330
scalars.
338
00:44:56,330 --> 00:45:03,330
\The ed, eq, id, iq they are scalars right
but because there exists trigonometrical relationships
339
00:45:06,040 --> 00:45:13,040
between the certain phasors and the d and
q axis components therefore we can represent
340
00:45:15,120 --> 00:45:22,120
these quantities in the phasor diagram. Now
to start with let us say that we have a balanced
341
00:45:22,280 --> 00:45:29,280
steady state operation the phase voltages
are given by these equations. Okay that is
342
00:45:31,030 --> 00:45:36,400
ea is equal to Em cos omega s t plus alpha
alpha is some arbitrary angle a reference
343
00:45:36,400 --> 00:45:43,400
angle or may be to start with and this eb
is lagging by 2 by 3, 2 phi by 3,120 degrees
344
00:45:44,200 --> 00:45:51,200
ec by leading by 2 phi by 3. Now you apply
dqO transformation to this equation.
345
00:45:52,490 --> 00:45:59,490
When you apply the dqo transformation you
will get the d axis component ed and q axis
346
00:46:02,300 --> 00:46:09,300
component eq as Em cos omega st plus alpha
minus theta Em sin omega st plus that is in
347
00:46:10,420 --> 00:46:17,420
fact when you apply this transformation right
to arrive at this result we have to make use
348
00:46:18,870 --> 00:46:25,870
of many trigonometrical identities that is
when you when you when you put these equations
349
00:46:27,730 --> 00:46:34,730
right ea eb and ec this this column vector
in that transformation matrix.
350
00:46:35,280 --> 00:46:41,510
You will find actually that this trigonometrical
terms will have to be multiplied with the
351
00:46:41,510 --> 00:46:48,510
elements of that matrix transformation matrix
and finally to come to this result you may
352
00:46:48,560 --> 00:46:55,560
have to use number of trigonometrical identities.
Okay and the final result is going to be in
353
00:46:55,690 --> 00:46:59,520
this form okay.
354
00:46:59,520 --> 00:47:06,520
Now we can represent the angular position
theta as omega r into t plus theta o and omega
355
00:47:10,050 --> 00:47:17,050
r is same as omega s because it is a steady
state operating condition. Now when you make
356
00:47:18,510 --> 00:47:24,170
this substitution ed comes out to be equal
to em cos alpha minus theta o eq comes out
357
00:47:24,170 --> 00:47:30,060
to be equal to em sin alpha minus theta. Okay
and therefore in this equation when we find
358
00:47:30,060 --> 00:47:37,060
here in this equation the there is no term
like omega st right and therefore ed is constant
359
00:47:42,460 --> 00:47:49,460
it depends upon the peak value of the stator
phase voltage and the alpha and theta naught,
360
00:47:52,720 --> 00:47:59,050
okay that is why I said that this ed and eq
come out to be scalar quantities under steady
361
00:47:59,050 --> 00:48:02,000
state operating conditions.
362
00:48:02,000 --> 00:48:09,000
Now ed and em the ed is d axis voltage and
is in per unit. Okay now we are more more
363
00:48:18,230 --> 00:48:25,230
comfortable when we use the RMS value of the
voltage rather than using the peak value of
364
00:48:25,670 --> 00:48:32,670
the voltage. Now suppose, I assume the peak
value of the stator voltage as my stator base
365
00:48:32,980 --> 00:48:39,980
voltage then RMS value RMS value of the stator
base voltage when it is expressed per unit
366
00:48:41,420 --> 00:48:46,180
will come out be same as the peak value. It
is something like this that I have chosen
367
00:48:46,180 --> 00:48:53,180
the base voltage in the stator circuit and
that base voltage is peak value of the per
368
00:48:54,910 --> 00:49:01,470
phase per unit of the phase to neutral voltage.
369
00:49:01,470 --> 00:49:07,810
Okay now if I want to express ah any quantity
in terms of this then they are expressed in
370
00:49:07,810 --> 00:49:14,810
terms of the peak value. Now if suppose I
have chosen the stator base voltage as peak
371
00:49:16,150 --> 00:49:23,150
voltage I want to chose the the base voltage
in terms of RMS value, then it is automatically
372
00:49:26,490 --> 00:49:33,490
going to be the base voltage which is the
peak value divided by root two and therefore
373
00:49:37,640 --> 00:49:44,640
if you express these quantities that is ed
and eq not in terms of peak values but in
374
00:49:44,770 --> 00:49:51,770
terms of RMS value the equation will remain
same because the base values of the stator
375
00:49:52,510 --> 00:49:57,430
voltage, when it is expressed as a in terms
of peak or where it is expressed in terms
376
00:49:57,430 --> 00:50:03,990
of RMS value they will have a relationship
of one by root 2. Suppose I take the stator
377
00:50:03,990 --> 00:50:10,990
base as 100 volts then this is a peak value
then RMS value, RMS base value will be equal
378
00:50:12,900 --> 00:50:19,770
to 100 by root 2 and therefore this equation
can be written in terms of the RMS value now
379
00:50:19,770 --> 00:50:25,440
I put et, et as the RMS value and it remains
same there is absolutely no change this is
380
00:50:25,440 --> 00:50:30,070
also per unit, this is also per unit okay.
381
00:50:30,070 --> 00:50:35,720
Now when we represent these quantities in
the phasor diagram it will look like this.
382
00:50:35,720 --> 00:50:42,720
Let us say this is the terminal voltage phasor
I represent this Et tilde putting this as
383
00:50:44,590 --> 00:50:51,290
a phasor quantity because we use bar earlier
super bar earlier for representing the bar
384
00:50:51,290 --> 00:50:56,250
per unit quantities because that is why we
are putting Et tilde and let us show that
385
00:50:56,250 --> 00:51:03,250
this is my d axis, this is the q axis, q axis
lead d axis by 90 degrees and here the angle
386
00:51:05,210 --> 00:51:09,720
is alpha minus theta because you have seen
here actually that the angles comes out to
387
00:51:09,720 --> 00:51:16,720
be alpha minus theta o right. Then, then ed
is represented by Et cosine alpha minus theta
388
00:51:19,930 --> 00:51:26,930
o and eq is equal to et sin alpha minus theta
o therefore, this phasor diagram, phasor phasor
389
00:51:30,200 --> 00:51:37,200
diagram is expressing these quantities ed,
eq in terms of Et.
390
00:51:37,280 --> 00:51:44,280
Similarly, you can draw a phasor diagram showing
the currents. Let us say this is my terminal
391
00:51:44,619 --> 00:51:51,619
voltage phasor so this is the current phasor
and let us say that phase difference between
392
00:51:53,320 --> 00:51:58,680
the voltages and current is phi right.
393
00:51:58,680 --> 00:52:05,680
Then we can express express the d and q axis
components of current as Et, I am sorry current
394
00:52:10,369 --> 00:52:17,369
can be expressed as id equal to It sin delta
i plus phi iq is equal to It cos delta i plus
395
00:52:21,770 --> 00:52:26,950
phi. This exercise is done similarly to what
we did for voltages right therefore in the
396
00:52:26,950 --> 00:52:33,950
phasor diagram I can simply show It equal
to id plus j times iq okay and the diagram
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00:52:34,610 --> 00:52:39,220
I have just now shown to you, okay.
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00:52:39,220 --> 00:52:46,220
Now here the angle between the q axis and
the terminal voltage we will denote by delta
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00:52:47,550 --> 00:52:54,550
i. Okay now one small exercise which is important
exercise to be done is done is to establish
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00:53:00,170 --> 00:53:07,170
the position of d and q axis with respect
to the terminal voltage that is I know the
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00:53:07,750 --> 00:53:12,500
terminal voltage and with this terminal voltage
as phasor I have to establish what is the
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00:53:12,500 --> 00:53:16,670
position of q axis what is the position that
is if I establish the position of q axis I
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00:53:16,670 --> 00:53:23,670
can automatically get the position of d axis,
now to establish this, now the some of these
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00:53:27,170 --> 00:53:32,970
are equations are all repeated that you know
the ed is given by this expression, we have
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00:53:32,970 --> 00:53:34,360
already established.
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00:53:34,360 --> 00:53:41,360
To establish the relationship what we do here
is we will define a voltage eq, a phasor eq
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00:53:43,580 --> 00:53:50,580
as terminal voltage plus Ra plus j times Xq
into It. This is an this relationship is a
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00:53:51,900 --> 00:53:58,800
very well known to all of us right that is
just start like this that my my intension
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00:53:58,800 --> 00:54:05,800
is that to establish relationship between
the terminal voltage and the d and q axis,
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00:54:06,090 --> 00:54:13,090
we define a voltage eq as Et plus this impedance
Ra plus j times Xq into It.
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00:54:16,540 --> 00:54:23,119
With this definition and we substitute the
expression for ed Et as ed plus j times eq
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00:54:23,119 --> 00:54:30,119
and It as id plus j times iq and we make simplifications
that is you multiply and simplify this expression.
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00:54:32,100 --> 00:54:39,100
You will find that eq comes out to be j times
Xad, ifd minus Xd minus Xq id what this what
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00:54:41,520 --> 00:54:48,520
does we get from this equation, what this
equation conveys to us. This equation conveys
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00:54:55,500 --> 00:55:02,500
that this voltage eq which I have assumed
right is in quadrature with d axis that is
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00:55:04,359 --> 00:55:07,940
the axis this is along q axis not part of
it it is of course quadrature of d axis but
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00:55:07,940 --> 00:55:14,600
it is along q axis because our reference with
reference to the reference this quantity is
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00:55:14,600 --> 00:55:20,500
I had by 90 degrees that is j term is existing
because the other terms are all scalars here
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00:55:20,500 --> 00:55:23,080
actually the terms are all scalar.
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00:55:23,080 --> 00:55:30,080
Okay and therefore the phasor diagram which
I draw here showing this Et plus Ra It plus
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00:55:31,380 --> 00:55:38,380
Xq It this gives a voltage Eq that is with
respect to this terminal voltage Et the position
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00:55:39,369 --> 00:55:46,369
of Eq comes out to be along q axis right and
therefore we have now established a relationship
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00:55:49,090 --> 00:55:56,090
between between the terminal voltage and q
axis and once we establish relationship d
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00:55:59,130 --> 00:56:06,130
axis position is known and all further computations
can be done the moment we know the dq and
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00:56:07,859 --> 00:56:14,859
other quantities. Friends I will conclude
my presentation here by by summarizing what
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00:56:23,030 --> 00:56:30,030
we have done in this lecture. First thing
that we have established the base volt amperes
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00:56:31,930 --> 00:56:33,510
for the rotor circuit.
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00:56:33,510 --> 00:56:40,510
We have also established the how to choose
the base current for the rotor circuits in
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00:56:43,270 --> 00:56:50,270
order to make the mutual inductances reciprocal
and also to make the mutual inductances on
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00:56:56,990 --> 00:57:03,990
d axis and mutual inductances on q axis equal.
Okay then we have also established the steady
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00:57:06,510 --> 00:57:13,510
state relations and developed a simple phasor
diagram to show that, to show that the the
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00:57:19,700 --> 00:57:26,700
q axis can be obtained by obtaining the position
of a phasor which is equal to terminal voltage
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00:57:28,869 --> 00:57:35,869
plus plus an impedance equal to Ra plus j
times Xq multiplied by the current that is
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00:57:38,160 --> 00:57:45,160
if you compute this quantity eq equal to terminal
voltage plus impedance that is Ra plus j times
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00:57:47,990 --> 00:57:54,990
Xq multiplied by the current this gives me
the position of q axis, okay. Thank you!