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Friends, we should study today the modeling
of synchronous machine.
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Till now the synchronous machine was model
as a constant voltage behind direct axis transient
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reactance. The synchronous machine modeling
has been a challenge all through and lot of
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work has been done over the years to develop
more accurate models of the synchronous machine.
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Today in our study we will develop the basic
equations of synchronous machine and then
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we will go to dqO transformation which is
also commonly known as park’s transformation.
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Now this synchronous machine has two major
parts, stator and rotor. We shall represent
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stator has provided with 3 windings and we
assume that these windings are sinusoidally
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distributed. On the rotor, we have a field
winding on the direct axis and we have amortisseur
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or damper windings. In a synchronous generator
we provide dampers and these dampers can be
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represented by considering considering the
amortisseurs located on the d axis and on
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the quadrature axis.
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Now here in my presentation we will presume
or we will assume one amortisseur on the d
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axis and another amortisseur on the q axis.
The convention which we will follow here is
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that the q axis leads d axis by 90 degrees,
although there are some some you know cases
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where the q axis has been taken as lagging
the d axis but in IEEE standards consider
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the q axis leading the d axis by 90 degrees.
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Now here the d axis is along the axis of north
pole, it coincides with the axis of north
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pole then we will measure the angular position
of the, angular position of the direct axis
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with respect to the axis of phase A of the
stator that is here this straight lines shows
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the axis of phase A and the the angular position
of the rotor is measured with respect to the
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axis of phase A and we call this angle as
theta. Further we will be following the generator
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convention there is the stator currents are
leaving the terminals of the machine that
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is ia, ib and ic are leaving the machine terminals.
The rotor is rotating in the anticlockwise
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direction this is direction of rotation of
the rotor which we are presuming.
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Now the currents in the rotor circuits are
entering the rotor circuit, if you just see
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here this field winding the current is entering
the field winding and the applied voltage
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is efd the damper windings are closed circuits
amortisseurs are closed circuits. The current
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flowing is again into the amortisseur windings
closed circuit.
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Some of the important nomenclature are ah
will be use here, a, b, c stands for the stator
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phase windings, fd stands for field winding,
kd stands for d axis amortisseur circuit,
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kq stands for q axis amortisseur circuit,
this K stands for 1, 2, 3, n, the number of
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amortisseur circuits that is if I put one
amortisseur circuit on the d axis, k becomes
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1 I can say 1 d if there is one amortisseur
on the q axis it is 1q. Okay therefore in
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general the amortisseurs are represented by
putting substitute kd or kq, theta is the
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angle by which the d axis leads the magnetic
axis of the phase a winding in the electrical
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radians and omega r is rotor angular velocity
is electrical radians.
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The ea, eb and ec are the instantaneous stator
phase to neutral voltages that is the voltages
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which are shown here these are the instantaneous
values and they are with respect to phase
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to neutral there is a raise from neutral to
phase, instantaneous stator currents are shown
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as ia, ib and ic.
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The field voltage is efd, the field and amortisseur
circuit currents are denoted as ifd, ikd and
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ikq the rotor circuit resistances will be
denoted by Rfd, Rkd, Rkq with Rfd is the resistance
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of the field winding, Rkd is the resistance
of direct axis amortisseur circuit and Rkq
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is the resistance of the quadrature axis amortisseur
circuit.
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Now here, we will see that we have stator
windings, we have windings on the rotor and
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rotor is rotating and because because of this
we will find actually that the, we come across
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various types of inductances in the synchronous
machine, the inductances are the self- inductances
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of the stator windings, the mutual inductance
between the windings of the stator then mutual
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inductances between the stator winding in
the rotor circuits and self-inductances of
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the rotor circuits and mutual inductances
between the rotor circuits therefore we come
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across different types of inductances in the
stator in the synchronous machine.
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The, we represent this by double circuit same
circuit laa to denote that it is a self laa,
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lbb and lcc stand for self-inductances of
stator windings that is we will use double
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circuit notation to denote the self-inductances
or mutual inductances if there are self-inductances
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the two circuits will be same if they are
mutual inductances the two circuits will be
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different, like say lab lbc and lca stands
for mutual inductances between stator winding
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that is lab is the mutual inductance between
stator a phase and stator b phase, so on.
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Then lafd, lakd and lakq represents the mutual
inductances between the stator a phase and
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rotor windings that is lafd is the mutual
inductance between the stator a phase and
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field winding lakd is the mutual inductance
between the stator a phase and amortisseur
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on the d axis and similarly, lakd then lffd,
lkkd and lkkq represents the self-inductances
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of rotor circuit. Ra is armature resistance
per phase and we will represent this differential
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operator P which is your d by dt by the symbol
P,
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P is the differential operator.
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Now in the case of synchronous machine, the
self-inductances of the stator winding and
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the mutual inductances between the stator
windings and they they they are affected because
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of the the non-uniform air gap. As we know
that the magnetic field produced by the stator
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winding it passes through passes through the
stator core, through the air gap, through
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the rotor iron then air gap and again return
backs through the stator core right and therefore
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the flux produced by the stator winding will
be affected by the position of the rotor.
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Now here in this diagram we saw the variation
of permeance with rotor position means you
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know that permeance is the reciprocal of reluctance.
Okay now here I am considering a salient pole
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machine and these are the pole location and
we are just showing the expanded version.
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Now the permeance is maximum when the, when
the permeance is maximum along the d axis
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or we can say the reluctance is minimum. This
graph shows the variation of permeance as
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with respect to the position that is angle
alpha which is measured with respect to the
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d axis which coincides with the North Pole
axis okay.
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We can easily see that this is maximum position,
when it coincide with the Q axis it is minimum
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and it again coincides with the d axis it
is maximum and this variation is of the form
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P equal to Po plus P2 cos 2 alpha that is
when alpha is 0, alpha is 0 its value is Po
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plus P2 and when alpha is ninety degrees its
value is Po minus P2 right that is cos2 alpha
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becomes minus 1 and it is this variation of
this permeance right is having a strong bearing
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on the variation of self-inductances mutual
inductances and so on.
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Now to understand the whole thing what we
start with this we first write down the stator
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circuit equations. The basic stator circuit
equations are ea is equal to p, psi a minus
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Ra ia, eb equal to p psi b minus Ra ib and
ec equal to p psi c minus Ra ic, ia ib and
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ic are the instantaneous value of the phase
currents and p psi a stands for d by dt of
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psi a, psi a, psi b and psi c are the flux
linking phase a phase b and phase c respectively.
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Okay that means straight forward that the
induced emf is d by dt of psi a and this will
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be equal to the terminal voltage plus the
resistance drop or now this equation is drawn
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considering the generator action okay.
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Now here, let us see actually that what determines
the flux linkage in the stator phase winding
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the flux linkage in the stator phase winding
can be written as psi a equal to minus laa
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ia, now here I will explain this minus terms
but laa is the self-inductance of phase a
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la into ia minus mutual inductance between
a and b and multiplied by ib minus iac ic
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plus lafd ifd where lafd is the mutual inductance
between a phase and field winding ifd is the
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field current similarly lakd, ikd, lakq, ikq.
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Now since we have assumed in the
basic model here that the flux linkages are
shown in the direction opposite to the current
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and that is why actually the negative signs
are appearing here that is in these terms
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you can just see these are the negative signs
while the currents are entering the other
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three rotor windings therefore they are the
positive signs. Now we will see that that
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these self-inductances mutual inductances
these are not constant these depend upon the
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position of the rotor with respect to the
windings, the stator windings and we will
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show that these depend upon the angular position
of the rotor and since the rotor is rotating
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the angular position of rotor keeps on changing
and therefore these inductances are going
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to be a function of angular position theta.
Okay now to understand this let us first start
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with the stator self-inductances, the stator
self-inductances.
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Now here the stator self-inductance is denoted
by the symbol laa okay and how when we define
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this stator self-inductance, the basic definition
is the flux linking the phase a winding divided
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by the current that is the self-inductance
of phase a winding with no currents on other
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windings that is when only current ia is flowing
and we find out what is the total flux linking
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the stator winding a that is the self-inductance
of stator winding a laa.
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Now when the current ia is flowing okay then
the MMF, MMF which is produced due to the
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flow of current is Na into ia and this MMF
is sinusoidally distributed along the surface
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of the stator or along the air gap okay because
the stator is suppose to produce a sinusoidally
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distributed MMF okay and this MMF has the
maximum value along the d axis right it is
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it is peak is along the d axis and when you
go away from the d axis both sides this is
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going to decrease.
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Now here this diagram shows, this diagram
shows the MMF produced by the stator phase
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a that is the I am just showing this is the
this is the axis of phase a okay and the MMF
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produced by the axis of phase a MMF produced
by the phase a or stator phase a is having
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its peak along the d axis, this is the d axis.
I am sorry, this is the not d axis, I am sorry
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this is the axis of the phase a, the axis
of the phase a. Okay it is a little correction
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this is the axis of phase a.
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Now what we do is we split this MMF into two
components both are having the sinusoidal
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distribution, one having its peak along d
axis another having its peak along q axis
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therefore this this graph red graph which
I have shown here, this shows the sinusoidal
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distribution having its peak coinciding the
d axis, this is the again sinusoidal distribution
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its peak is coinciding with q axis and the
q axis is leading d axis by 90 degrees okay.
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Now this can be seen here in this diagram
that the MMF produced by the stator right
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along its own axis that is the MMF produced
by a stator winding of phase a right is having
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its maximum value along its own axis that
is axis of phase a. Now we have assumed likely
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that the rotor is rotating in the anti clockwise
direction therefore axis of, now the d axis
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is shown here and q axis is leading points
and what we do is that this MMF is resolved
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into two components one along d axis another
along q axis. The d axis component is Na ia
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cos theta and q axis component is minus Na
ia sin theta, okay.
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Now with this MMF’s then we can find out
what will be the flux produced at the air
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gap along this d and q axis. Okay, now here
we are showing that the MMFad that is MMF
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due to due to current flowing in the stator
a phase and it is component along d axis ad
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is equal to Na ia cos theta and these are
the peak values therefore when ia attains
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its peak value this will also become this
varying this is varying along as the ia is
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varying, then the along the q is minus ia
Na sin theta okay.
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Now the flux produced long these two axis
because of these MMF can be written as MMF
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into Pd that is phi gad, g stand for air gap
or gap flux okay the phi gad is equal to Na
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ia cos theta into Pd and phi gaq is equal
to minus Na ia sin theta into Pq. Now here
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this is the MMF and to relate this MMF to
the flux we are using this term Pd therefore,
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Pd is in general a permeance coefficient we
call it it is not only the absolute value
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of permeance but all other parameters which
relate flux to MMF because this is the MMF
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only this is the flux okay.
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Now what is done is we again make use of this
phasor diagram, the flux which is produced
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along d axis in the air gap, flux which is
produced along q axis in the air gap, we resolve
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them back along the axis of phase a. that
is when you resolve this right then this component
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will come out to be equal to phi gad cos theta
and the second component comes out to be phi
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gaq sin theta with negative sign because there
was negative sign already attached with it
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and therefore we can say that the air gap
flux due to current flowing in the stator
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winding a only comes out to be equal to Na
ia substituting these values on the previous
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equations in this form.
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The Na ia Pd cos square theta plus Pq sin
square theta and this expression when simplified
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it can be put in the form Na ia into Pd plus
Pq by 2 plus Pd minus Pq by 2 cos 2 theta.
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Now here this is very important point to understand
that the air gap flux produced, air gap flux
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produced by current flowing in the stator
winding of phase a is equal to is proportional
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to a term Pd plus Pq by 2 and another term
Pd minus Pq by 2, cos 2 theta that is this
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term does not depend upon angular position,
while this term depends upon the angular position.
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Now we define the inductance.
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The inductance, the self-inductance of the
stator phase a due to gap flux only the flux
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which is produced in the air gap, lgaa is
equal to Na affective number of terms into
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the gap air gap flux divided by ia and this
comes out to be we substitute the value of
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phi gaa it comes out to be Na square Pd plus
Pq by 2 plus Pd minus Pq by 2 cos 2 theta
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okay therefore this can be put in the form
that is this is your Lgaa is the self-inductance
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of phase a due to gap flux only which can
be put as a constant terms Lg0 plus another
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term Laa2 cos 2 theta right because as I have
seen the I have told you that the permanence
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of the air gap varies as a with the position
of the rotor and there we found actually that
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it has a second harmonic variation. Here,
also you find there is a constant term plus
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a quantity varying as a function of cosine
2 theta okay.
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Now to make the whole thing more complete
there is some a leakage flux which does not
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cross the air gap. Okay and this leakage flux
also contributes the self-inductance of the
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stator phase and therefore, when you account
for the leakage flux then we can say that
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the self-inductance Laa of the stator phase
is equal to self-inductance due to leakage
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flux plus lgaa which I have obtained in the
previous equation that is due to the gap flux
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and then when you combine these two terms.
We can see here that this mutual inductance
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or the this second term will not be affected
by the leakage and therefore this leakage
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term is combined and you find here that the
self-inductance can be written as Laao plus
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Laa2 cos 2 theta.
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Now this is the most important equation to
understand that how the self-inductance of
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stator phase varies as the position of the
rotor varies the angular position of the rotor.
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Now this angular position is measured with
respect to axis of phase a.
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Now this graph shows the plot for the variation
of the self-inductance of stator phase a as
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function of theta okay and you can identify
here that this is the term Laa2 which varies
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this Laa2 is constant and this is another
term which we call Laao and the total inductance
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of the stator phase is now written as Laa
equal to Laao plus Laa2. These two terms are
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00:27:19,289 --> 00:27:23,729
constant these constants these are constant
they do not depend upon the angular position
193
00:27:23,729 --> 00:27:30,019
that mean the total self-inductance depends
upon the angular position but these two coefficients
194
00:27:30,019 --> 00:27:37,019
are constant. Now when we perform the similar
exercise for phase b and phase c, since the
195
00:27:42,749 --> 00:27:48,049
the axis of the phase b and phase c are displaced
by 120 degrees with respect to axis of phase
196
00:27:48,049 --> 00:27:49,200
a right.
197
00:27:49,200 --> 00:27:56,200
Therefore, the expressions which we have written
here for self-inductance of phase b right
198
00:27:57,799 --> 00:28:04,799
will be of the same form except theta is replaced
by theta minus 2 phi by 3 and since these
199
00:28:08,489 --> 00:28:14,349
the everything remains same therefore these
terms are also same therefore it is not Lbbo
200
00:28:14,349 --> 00:28:21,239
but Lbbo is same as Laao okay similarly, we
write down lcc as Lao plus Laa2 cos 2 times
201
00:28:21,239 --> 00:28:28,070
theta plus 2 phi by 3 okay very straight forward.
202
00:28:28,070 --> 00:28:35,070
Now next very important point we have to understand
is the stator mutual inductances, the stator
203
00:28:36,299 --> 00:28:41,609
winding mutual inductances again we will see
that the stator winding mutual inductances
204
00:28:41,609 --> 00:28:48,609
also are function of rotor position that will
be function of theta. Now here, here when
205
00:28:51,070 --> 00:28:58,070
the when the axis of the rotor is in the middle
of the axis of stator phase a and stator phase
206
00:29:04,349 --> 00:29:11,349
b then at that position the mutual inductance
between a and b will be maximum for example
207
00:29:15,619 --> 00:29:22,619
the mutual inductance between phase b and
c when you try to see it will be maximum when
208
00:29:22,989 --> 00:29:29,989
theta is theta is 30 degree minus 30 degrees
and 150 degrees these they are the positions
209
00:29:31,219 --> 00:29:38,219
which we have to see. Okay, using this information
the flux linkage is, flux linkage is of phase
210
00:29:48,710 --> 00:29:49,940
b.
211
00:29:49,940 --> 00:29:55,229
When current is flowing in phase a is are
obtained that is we want to to find out the
212
00:29:55,229 --> 00:30:02,229
flux mutual flux right that the flux linking
flux, linking phase b due to current flowing
213
00:30:03,809 --> 00:30:10,809
in phase a okay and then once you find out
this flux. Okay, we can find out the mutual
214
00:30:13,099 --> 00:30:17,359
inductance because the the inductance is the
flux linkage by the current mutual inductance
215
00:30:17,359 --> 00:30:23,919
will be the flux linking phase b due to current
in phase a and then you divide by the current
216
00:30:23,919 --> 00:30:30,919
you will get the mutual inductance. Here here
following the same approach as we have done
217
00:30:33,700 --> 00:30:40,700
for done for obtaining the self-inductance
the, the air gap flux flux again the gap flux
218
00:30:48,779 --> 00:30:55,779
linking phase b with when current is flowing
in phase a is obtained in this form that is
219
00:30:56,659 --> 00:31:03,659
this is obtained in terms of these 2 components
phi gad and phi gaq that is this is the air
220
00:31:06,619 --> 00:31:13,619
gap flux along d axis this is the air gap
flux along q axis and after making the substitutions
221
00:31:18,359 --> 00:31:24,690
we find actually that mutual flux comes out
to be equal to Na ia minus Pd plus Pq by 4
222
00:31:24,690 --> 00:31:31,690
plus Pd minus Pq by 2 cos 2 theta minus 2
phi by 3, okay.
223
00:31:32,690 --> 00:31:38,070
Now you can easily see here actually that
if you substitute here to make this quantity
224
00:31:38,070 --> 00:31:45,070
one to make this quantity one. You can find
it out actually ah what should be the value
225
00:31:45,229 --> 00:31:52,229
of theta right and since this term is minus
here Pd is always greater than Pq permanence
226
00:31:54,029 --> 00:31:59,830
along the d axis is more then the permanence
along q axis and therefore this is minus to
227
00:31:59,830 --> 00:32:06,830
have this also minus so that the total quantity
is added up. Okay you can find out the value
228
00:32:07,509 --> 00:32:14,450
of this angle theta and you will find actually
that when theta occupies either 30 degrees
229
00:32:14,450 --> 00:32:21,450
or 150 degrees it will be maximum. Now this,
this mutual inductance can be obtained as
230
00:32:25,279 --> 00:32:32,279
lgba divided by after dividing the the expression
for phi gba by ia okay.
231
00:32:34,119 --> 00:32:41,119
Therefore, the expression for Lgba comes out
to be in this form. Okay, now again it can
232
00:32:43,239 --> 00:32:50,239
be written as minus 1 by 2 Lgo, Lg0 plus Lab2,
now if you very carefully examine then this
233
00:32:54,609 --> 00:33:01,609
Lab2, Lab2 will be of the same amplitude as
Laa2, Laa2 right.
234
00:33:07,849 --> 00:33:14,849
Similarly, you can find out the mutual inductance
between b and a and we this ba mutual inductance
235
00:33:18,599 --> 00:33:25,599
between the phase a and b that is equal ba
or ab, they are always equal okay and the
236
00:33:27,450 --> 00:33:34,019
expressions are written in the form minus
Labo minus, now here actually when you have
237
00:33:34,019 --> 00:33:41,019
written in this form what we have done is
that we have accounted for some some leakage
238
00:33:42,529 --> 00:33:49,529
flux which also leaks to windings right because
there are, there is a air gap flux and there
239
00:33:50,809 --> 00:33:55,489
is some flux which does not cross the air
gap and once you account that we can write
240
00:33:55,489 --> 00:34:02,009
down these mutual inductances in this form,
okay again you can see that this depends upon
241
00:34:02,009 --> 00:34:09,009
theta. Similarly, you can write down for bc
and cb it comes out to be in the similar form
242
00:34:11,700 --> 00:34:18,700
and similarly lac and lca can be written like
this okay.
243
00:34:20,269 --> 00:34:27,269
This diagram shows the variation of mutual
inductance as a function of theta between
244
00:34:28,059 --> 00:34:34,859
the 2 stator phases that is here we have shown
the Lab and you can easily see that first
245
00:34:34,859 --> 00:34:41,859
thing which we see here is that the the mutual
inductances all through negative. Okay and
246
00:34:42,750 --> 00:34:49,750
its variation is shown in this form therefore,
this this quantity a constant quantity is
247
00:34:50,279 --> 00:34:56,099
Labo and over this is you superimpose this
sinusoidally varying quantity and variation
248
00:34:56,099 --> 00:35:03,099
is as a function of 2theta. Therefore what
we have seen till now that the self-inductances
249
00:35:06,710 --> 00:35:13,119
of the stator phases or stator winding and
mutual inductances between the stator winding.
250
00:35:13,119 --> 00:35:18,579
Now we will consider the mutual inductances
between stator and rotor windings, stator
251
00:35:18,579 --> 00:35:25,539
and rotor winding. Now so far the mutual inductances
between stator rotor windings are concerned
252
00:35:25,539 --> 00:35:32,539
that they are function of angular position
but they are not because of the variation
253
00:35:32,619 --> 00:35:38,890
in permanence here because so far the rotor
is concerned, rotor will always see the same
254
00:35:38,890 --> 00:35:45,890
permanence because the stator stator is having
the ah a uniform shape right and therefore,
255
00:35:49,519 --> 00:35:56,099
so far the so far actually the rotor is concerned,
rotor windings are concerned right the the
256
00:35:56,099 --> 00:35:59,390
there will be no variation in permanence.
257
00:35:59,390 --> 00:36:04,609
Now here the mutual inductances between stator
and rotor windings vary because of angular
258
00:36:04,609 --> 00:36:11,569
position. Now for example, if you take take
the stator phase a and field winding in case
259
00:36:11,569 --> 00:36:17,299
the axis of these windings coincide they will
have maximum mutual inductance in case the
260
00:36:17,299 --> 00:36:23,420
axis of stator winding of phase a and the
field winding they are in quadrature, the
261
00:36:23,420 --> 00:36:30,420
mutual inductance will be 0 right and since
the rotor is having rotating it occupies different
262
00:36:31,349 --> 00:36:37,500
positions therefore, when it coincides where
the direct axis of the rotor coincide with
263
00:36:37,500 --> 00:36:44,500
the stator phase a axis or b axis or c axis
they will have maximum mutual inductance and
264
00:36:46,210 --> 00:36:53,210
when the quadrature axis of the rotor coincides
with the stator phase axis, okay phase a axis
265
00:36:53,519 --> 00:36:59,329
or phase b axis or phase c axis then the mutual
inductance will be 0. Okay therefore, we can
266
00:36:59,329 --> 00:37:06,329
write down this mutual inductance Lafd equal
to L, Lafd into cos theta.
267
00:37:08,720 --> 00:37:15,720
When suppose as we know that the theta is
measured right considering the axis of phase
268
00:37:15,849 --> 00:37:22,849
a as reference and theta is the angle between
the d axis and axis of phase a. Okay therefore
269
00:37:24,359 --> 00:37:31,359
when theta is 0, the mutual inductance between
stator stator winding and the field winding
270
00:37:31,859 --> 00:37:38,859
is maximum. Okay and 90 degrees it is now,
so the the amortisseur on the direct axis
271
00:37:44,089 --> 00:37:51,089
is is also going to have the inductance, mutual
inductance in the form Lakd cos theta right
272
00:37:54,960 --> 00:38:01,960
because this the the direct axis amortisseur
is having axis coinciding with the field winding
273
00:38:02,200 --> 00:38:06,279
right and therefore the variation is going
to be similar.
274
00:38:06,279 --> 00:38:13,279
Now the mutual inductance in the quadrature
axis amortisseur and the stator winding will
275
00:38:14,609 --> 00:38:20,930
be written by the formula Lakq cos of theta
plus phi by 2, why this theta is replaced
276
00:38:20,930 --> 00:38:26,359
by theta plus phi 2 by 2 because q axis is
leading d axis by 90 degrees therefore, this
277
00:38:26,359 --> 00:38:33,359
can be written as minus Lakq sin theta.
278
00:38:35,289 --> 00:38:42,289
Now what we have seen here is till now, we
have obtained the expression for the self-inductances
279
00:38:46,960 --> 00:38:53,960
of the stator windings, mutual inductances
between stator windings and we have also obtained
280
00:38:54,099 --> 00:39:00,849
the mutual inductances between the rotor windings
and stator winding and we have seen that all
281
00:39:00,849 --> 00:39:07,849
these are function of angular position. Okay
282
00:39:15,920 --> 00:39:22,920
now we again come to our fundamental equations
that is the stator voltage equations stator
283
00:39:23,039 --> 00:39:30,039
circuit equation ea equal to p psi a minus
Ra ia and we have seen that the flux linkage
284
00:39:32,029 --> 00:39:39,029
of phase a is now written as minus laa ia
minus lab ib minus lac ic plus lafd ifd plus
285
00:39:41,339 --> 00:39:48,339
lakd ikd plus lakd ikq therefore, now in this
equation we substitute the expression for
286
00:39:50,230 --> 00:39:55,510
laa, lab, lac, lafd, lakd, lakq.
287
00:39:55,510 --> 00:40:02,510
Okay, then we get the expression for flux
linkage of phase a as minus ia into Laa plus
288
00:40:12,019 --> 00:40:19,019
ib into Labo Lab2 cos 2 theta plus phi by
3 this plus sign has come because there was
289
00:40:20,359 --> 00:40:25,119
negative sign here earlier. Okay when you
see this mutual inductance there was a negative
290
00:40:25,119 --> 00:40:32,119
sign therefore it becomes plus here. Similarly,
you have ic into Labo plus Lao cos 2 theta
291
00:40:32,839 --> 00:40:39,839
minus phi by 3 and so on, that is what we
have done is that in this in this basic equation,
292
00:40:41,029 --> 00:40:48,029
we have substituted value of all the inductances
okay, which were all found to be function
293
00:40:51,240 --> 00:40:58,240
of theta. Similarly we can write down the
flux linkage of phase b and flux linkage of
294
00:40:58,259 --> 00:41:04,140
phase c, they are exactly similar except you
will find that theta is replaced by theta
295
00:41:04,140 --> 00:41:10,940
plus 2 phi by 3 or theta minus 2 phi by okay.
296
00:41:10,940 --> 00:41:17,940
Now after having written the equations for
the stator windings, voltage equation for
297
00:41:20,289 --> 00:41:27,289
the stator windings we can write down the
rotor circuit voltage equations. The, in the
298
00:41:33,250 --> 00:41:39,539
rotors on the rotor we have considered 3windings,
1 filed winding and 2 amortisseurs. Okay therefore
299
00:41:39,539 --> 00:41:46,539
efd the voltage applied to the field winding
is equal to P psi fd plus Rfd ifd. Now here,
300
00:41:48,250 --> 00:41:55,250
since we have assumed that the current is
entering the field winding and therefore the
301
00:41:55,339 --> 00:42:00,579
term here is P psi fd plus this is a simple
Rl circuit.
302
00:42:00,579 --> 00:42:05,920
Suppose you have Rl circuit, then the applied
voltage is equal to the rate of change of
303
00:42:05,920 --> 00:42:12,920
flux linkages plus voltage drop in the resistance.
Okay then the other 2 equations these relate
304
00:42:14,210 --> 00:42:20,130
to the direct axis amortisseur, winding amortisseur
circuit and quadrature at the amortisseur
305
00:42:20,130 --> 00:42:26,779
circuit here since there is no external applied
voltage therefore we have 0 term here. Okay
306
00:42:26,779 --> 00:42:33,779
therefore, there are 3 basic rotor circuit
voltage equations we have 3 basic stator circuit
307
00:42:35,700 --> 00:42:42,700
voltages equations. Now let us write down
the expression for these flux linkages psi
308
00:42:43,849 --> 00:42:50,849
fd psi kd and
psi kq.
309
00:42:52,230 --> 00:42:59,230
Now psi fd can be written as Lffd that is
the self-inductance of field winding into
310
00:42:59,230 --> 00:43:06,230
ifd plus mutual inductance of the mutual inductance
between between field winding and amortisseur
311
00:43:14,180 --> 00:43:21,180
that is Lakd into ikd okay and there will
be no there will be no flux linking the field
312
00:43:24,039 --> 00:43:31,039
winding due to the quadrature axis amortisseur
because the the 2 axis are that that the the
313
00:43:33,509 --> 00:43:40,089
displacement of 90 degrees between the 2 axis
okay and therefore there is no flux linkage
314
00:43:40,089 --> 00:43:47,089
contributed by by amortisseur on the quadrature
axis to field axis flux linkage.
315
00:43:48,519 --> 00:43:55,519
Then these 3 terms are here Lafd into ia cos
theta Lafd into ib cos theta minus 2 phi by
316
00:43:59,220 --> 00:44:05,950
3and Lafd plus ic into cos theta plus 2 phi
by 3 that is when the 3 stator currents are
317
00:44:05,950 --> 00:44:12,950
carrying the values ia, ib and ic and depending
upon their mutual inductances this will be
318
00:44:14,470 --> 00:44:20,799
the flux linkage in the stator winding. Now
one point which I wanted to mention here is
319
00:44:20,799 --> 00:44:27,799
the so far the self-inductances of the rotor
circuits are concerned that is self-inductance
320
00:44:31,730 --> 00:44:38,730
of field winding, self inductance of amortisseurs
they do not depend upon the angular position
321
00:44:44,410 --> 00:44:51,410
because because so far actually the the the
magnetic circuit is concerned for computing
322
00:44:54,079 --> 00:45:00,539
the self-inductances of rotor circuits are
concerned, these the self-inductances are
323
00:45:00,539 --> 00:45:06,269
constant and the since that we have assumed
like with uniform internal surface of the
324
00:45:06,269 --> 00:45:11,869
stator okay and therefore no variation of
reluctance so far actually the rotor circuits
325
00:45:11,869 --> 00:45:14,019
are concerned.
326
00:45:14,019 --> 00:45:21,019
Similarly, similarly the mutual inductance
between the rotor circuits there is mutual
327
00:45:21,119 --> 00:45:28,119
inductance between the field winding and amortisseur
on the d axis they will be fixed, they do
328
00:45:28,579 --> 00:45:33,789
not depend upon the rotor position right.
Therefore for example, Lfkd is a mutual inductance
329
00:45:33,789 --> 00:45:38,200
between field winding and direct axis amortisseur
this is a constant quantity they will not
330
00:45:38,200 --> 00:45:41,160
depend upon the rotor position.
331
00:45:41,160 --> 00:45:48,160
Now this similarly you can write down the
flux linkage of amortisseur on d axis and
332
00:45:49,430 --> 00:45:55,059
flux linkage of amortisseur in the q axis
okay.
333
00:45:55,059 --> 00:46:02,059
Now with this with this we have developed
the complete mathematical model that we have
334
00:46:07,079 --> 00:46:14,079
written three stator circuit equations, we
have written the rotor circuit equations we
335
00:46:14,289 --> 00:46:21,289
have expressed all the inductances as function
of currents and I am sorry ,not all the flux
336
00:46:23,990 --> 00:46:29,849
linkages as function of currents and the self
and mutual inductances.
337
00:46:29,849 --> 00:46:36,849
Now this that is ah this is what is called
complete model of the system however the basic
338
00:46:39,039 --> 00:46:46,039
problems which arise are due to due to the
variation of these inductances with the variation
339
00:46:48,180 --> 00:46:55,180
of rotor angular position and to overcome
this problem and seeing very carefully the
340
00:46:58,180 --> 00:47:05,180
expression for, you see the expression for
psi fd, we find here a term Lafd that is the
341
00:47:08,160 --> 00:47:14,039
along with this term where ia cos theta plus
ib cos theta minus 2 phi by 3 therefore, this
342
00:47:14,039 --> 00:47:21,039
has prompted us to obtain a transformation
and once we go we use this transformation,
343
00:47:21,599 --> 00:47:28,599
we will find that the equations can be simplified
and we can make these equations equations
344
00:47:30,200 --> 00:47:34,960
where they do not exclusively depend upon
or the inductances do not depend upon the
345
00:47:34,960 --> 00:47:37,400
angular position.
346
00:47:37,400 --> 00:47:44,180
Okay the the transformation is of this form
that is we define, we define this term ia
347
00:47:44,180 --> 00:47:51,089
cos theta plus ib cos theta minus 2 phi 3
plus ic cos theta 2 phi by 3 multiplied by
348
00:47:51,089 --> 00:47:58,089
some constant kd as id that is these 3 terms
ia cos theta ib cos theta minus 2 phi by 3,
349
00:48:00,509 --> 00:48:07,440
ic cos theta plus 2 by 3 this complete term
multiplied with some constant kd is denoted
350
00:48:07,440 --> 00:48:11,920
by a term id.
351
00:48:11,920 --> 00:48:18,920
Similarly we denote another term iq as minus
ikq multiplied by ia sin theta plus ib sin
352
00:48:19,289 --> 00:48:26,289
theta minus 2 phi by 3, now this with this
with this assumption or this transformation
353
00:48:32,190 --> 00:48:39,190
if we consider balance three phase currents
that is ia equal to Im sin omega s t, ib equal
354
00:48:40,240 --> 00:48:47,240
to Im sin omega s t minus 2 phi by 3 ic equal
to Im sin omega st plus 2 phi by 3. Okay that
355
00:48:48,210 --> 00:48:55,210
is, we are assuming that the 3 stator currents
are balanced with this 3 stator currents to
356
00:48:56,619 --> 00:49:03,619
be balanced, okay what we do is if you substitute
and find out the expression for id, the id
357
00:49:05,990 --> 00:49:12,990
will come out to be as kd into 3 by 2 Im sin
omega st minus theta. This is very important
358
00:49:15,299 --> 00:49:22,299
term that is with this transformation, this
id current id is equal to kd into 3 by 2 Im
359
00:49:24,440 --> 00:49:27,289
sin omega st minus theta.
360
00:49:27,289 --> 00:49:34,289
Now if you assume kd equal to 2 by 3, if you
assume kd equal to 2 by 3 then the the peak
361
00:49:37,470 --> 00:49:44,470
value of id will be same as Im okay and therefore
in the Park’s transformation, Park’s transformation
362
00:49:46,890 --> 00:49:53,890
kd and kq are taken equal to 2 by 3, that
is iq will also be taking the same minus iq
363
00:50:01,170 --> 00:50:04,789
into 3 by 2 Im cos omega st minus theta.
364
00:50:04,789 --> 00:50:11,710
Okay and therefore, if I take kq equal to
2 by 3 then the peak value of iq will be same
365
00:50:11,710 --> 00:50:18,710
as Im. Okay now to make this model complete
complete and assuming that suppose the 3 currents
366
00:50:24,140 --> 00:50:31,140
are not symmetrical right then we can define
one, 0 sequence current io as 1 by 3 ia plus
367
00:50:31,420 --> 00:50:38,420
ib plus ic and with this definition the transformation
looks like this it is very interesting thing
368
00:50:41,069 --> 00:50:47,049
this transformation looks likes this that
we can say id, iq, io, a vector consisting
369
00:50:47,049 --> 00:50:51,999
of d axis current, q axis current and i0.
370
00:50:51,999 --> 00:50:58,999
These 3 currents can be written in terms of
the phase currents ia, ib and ic in terms
371
00:50:59,380 --> 00:51:06,380
of this matrix and this is called transformation
matrix that is transformation matrix is 2
372
00:51:06,660 --> 00:51:11,980
by 3 the first row is cos theta cos theta
minus 2 phi by 3, cos theta plus 2 phi by
373
00:51:11,980 --> 00:51:16,119
3. Okay similarly similarly the second term
and third term.
374
00:51:16,119 --> 00:51:23,119
With this now inverse transformation that
is if you write down the expression for phase
375
00:51:23,259 --> 00:51:29,829
currents in terms of the dqO currents then
this can be written in this form cos theta
376
00:51:29,829 --> 00:51:34,630
minus sin theta 1like this this is called
inverse that is sometimes if you know the
377
00:51:34,630 --> 00:51:39,940
value of id, iq, io we can find out the phase
currents.
378
00:51:39,940 --> 00:51:46,940
Now interesting thing which happens is that
if I substitute the values of the phase currents
379
00:51:47,910 --> 00:51:54,910
in terms of dqO components right then I get
the expression for flux linkage in the d axis
380
00:51:56,249 --> 00:52:02,589
called psi d that is if the the all these
fluxes flux linking phase a phase b and phase
381
00:52:02,589 --> 00:52:08,130
c right they are also transformed currents
are also transformed that is by applying this
382
00:52:08,130 --> 00:52:15,130
transformation. We find that the flux linkage
is can be written in terms of ah the constant
383
00:52:16,999 --> 00:52:23,999
coefficients that is psi d is equal to minus
Lao plus Labo plus 3 by 2 Lao Laa2 okay then
384
00:52:25,400 --> 00:52:32,400
Lafd ifd plus Lakd ikd that is here the coefficient
of id is a constant term it does not depend
385
00:52:34,029 --> 00:52:41,029
upon angular position theta similarly, for
psi q and psi o right we define we define
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00:52:47,619 --> 00:52:54,619
these terms Ld equal to Lao plus Labo plus
3 by 2 Lao similarly Lq and L0 are defined.
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00:52:56,170 --> 00:53:03,170
When you make this substitution we can write
down the flux linkage psi d as minus Ld id
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00:53:03,910 --> 00:53:10,910
plus Lafd ifd plus like this. Similarly, when
you apply the dqO transformation the flux
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00:53:19,990 --> 00:53:26,990
linking the rotor circuits are also expressed
in terms of the rotor currents rotor currents
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00:53:29,089 --> 00:53:36,089
rotor circuit currents and the dqO components
of currents and again you find actually that
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00:53:37,049 --> 00:53:42,940
these flux linkages as well as these 3flux
linkages with the transform quantities are
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00:53:42,940 --> 00:53:49,940
are independent of rotor angular position
and this is what it helps the whole thing
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00:53:53,579 --> 00:53:59,920
and once you substitute these expressions
in our stator circuit equations.
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00:53:59,920 --> 00:54:06,609
We get these equation in the form ed equal
to P psi d minus psi q p theta minus Ra ideq
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00:54:06,609 --> 00:54:13,609
is equal to P psi q minus psi d p theta minus
Ra iq and eo equal to P psi o psi 0 minus
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00:54:15,200 --> 00:54:22,200
Ra io. Now these are the 3 basic equations
which are written in terms of transform quantities
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00:54:22,890 --> 00:54:29,890
or dqO terms or sometimes in case in dq of
frame of presentation. Okay today I have discussed
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00:54:36,029 --> 00:54:43,029
the basic circuit equations of the synchronous
machine and discuss the dqO transformation
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00:54:46,140 --> 00:54:53,140
and its importance. Ultimately, we have obtained
the stator circuit equations in terms of the
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00:54:54,819 --> 00:55:01,819
transform quantities. Thank you!