Tip:
Highlight text to annotate it
X
Friends, we should study today the modeling of synchronous machine.
Till now the synchronous machine was model as a constant voltage behind direct axis transient
reactance. The synchronous machine modeling has been a challenge all through and lot of
work has been done over the years to develop more accurate models of the synchronous machine.
Today in our study we will develop the basic equations of synchronous machine and then
we will go to dqO transformation which is also commonly known as park’s transformation.
Now this synchronous machine has two major parts, stator and rotor. We shall represent
stator has provided with 3 windings and we assume that these windings are sinusoidally
distributed. On the rotor, we have a field winding on the direct axis and we have amortisseur
or damper windings. In a synchronous generator we provide dampers and these dampers can be
represented by considering considering the amortisseurs located on the d axis and on
the quadrature axis.
Now here in my presentation we will presume or we will assume one amortisseur on the d
axis and another amortisseur on the q axis. The convention which we will follow here is
that the q axis leads d axis by 90 degrees, although there are some some you know cases
where the q axis has been taken as lagging the d axis but in IEEE standards consider
the q axis leading the d axis by 90 degrees.
Now here the d axis is along the axis of north pole, it coincides with the axis of north
pole then we will measure the angular position of the, angular position of the direct axis
with respect to the axis of phase A of the stator that is here this straight lines shows
the axis of phase A and the the angular position of the rotor is measured with respect to the
axis of phase A and we call this angle as theta. Further we will be following the generator
convention there is the stator currents are leaving the terminals of the machine that
is ia, ib and ic are leaving the machine terminals. The rotor is rotating in the anticlockwise
direction this is direction of rotation of the rotor which we are presuming.
Now the currents in the rotor circuits are entering the rotor circuit, if you just see
here this field winding the current is entering the field winding and the applied voltage
is efd the damper windings are closed circuits amortisseurs are closed circuits. The current
flowing is again into the amortisseur windings closed circuit.
Some of the important nomenclature are ah will be use here, a, b, c stands for the stator
phase windings, fd stands for field winding, kd stands for d axis amortisseur circuit,
kq stands for q axis amortisseur circuit, this K stands for 1, 2, 3, n, the number of
amortisseur circuits that is if I put one amortisseur circuit on the d axis, k becomes
1 I can say 1 d if there is one amortisseur on the q axis it is 1q. Okay therefore in
general the amortisseurs are represented by putting substitute kd or kq, theta is the
angle by which the d axis leads the magnetic axis of the phase a winding in the electrical
radians and omega r is rotor angular velocity is electrical radians.
The ea, eb and ec are the instantaneous stator phase to neutral voltages that is the voltages
which are shown here these are the instantaneous values and they are with respect to phase
to neutral there is a raise from neutral to phase, instantaneous stator currents are shown
as ia, ib and ic.
The field voltage is efd, the field and amortisseur circuit currents are denoted as ifd, ikd and
ikq the rotor circuit resistances will be denoted by Rfd, Rkd, Rkq with Rfd is the resistance
of the field winding, Rkd is the resistance of direct axis amortisseur circuit and Rkq
is the resistance of the quadrature axis amortisseur circuit.
Now here, we will see that we have stator windings, we have windings on the rotor and
rotor is rotating and because because of this we will find actually that the, we come across
various types of inductances in the synchronous machine, the inductances are the self- inductances
of the stator windings, the mutual inductance between the windings of the stator then mutual
inductances between the stator winding in the rotor circuits and self-inductances of
the rotor circuits and mutual inductances between the rotor circuits therefore we come
across different types of inductances in the stator in the synchronous machine.
The, we represent this by double circuit same circuit laa to denote that it is a self laa,
lbb and lcc stand for self-inductances of stator windings that is we will use double
circuit notation to denote the self-inductances or mutual inductances if there are self-inductances
the two circuits will be same if they are mutual inductances the two circuits will be
different, like say lab lbc and lca stands for mutual inductances between stator winding
that is lab is the mutual inductance between stator a phase and stator b phase, so on.
Then lafd, lakd and lakq represents the mutual inductances between the stator a phase and
rotor windings that is lafd is the mutual inductance between the stator a phase and
field winding lakd is the mutual inductance between the stator a phase and amortisseur
on the d axis and similarly, lakd then lffd, lkkd and lkkq represents the self-inductances
of rotor circuit. Ra is armature resistance per phase and we will represent this differential
operator P which is your d by dt by the symbol P,
P is the differential operator.
Now in the case of synchronous machine, the self-inductances of the stator winding and
the mutual inductances between the stator windings and they they they are affected because
of the the non-uniform air gap. As we know that the magnetic field produced by the stator
winding it passes through passes through the stator core, through the air gap, through
the rotor iron then air gap and again return backs through the stator core right and therefore
the flux produced by the stator winding will be affected by the position of the rotor.
Now here in this diagram we saw the variation of permeance with rotor position means you
know that permeance is the reciprocal of reluctance. Okay now here I am considering a salient pole
machine and these are the pole location and we are just showing the expanded version.
Now the permeance is maximum when the, when the permeance is maximum along the d axis
or we can say the reluctance is minimum. This graph shows the variation of permeance as
with respect to the position that is angle alpha which is measured with respect to the
d axis which coincides with the North Pole axis okay.
We can easily see that this is maximum position, when it coincide with the Q axis it is minimum
and it again coincides with the d axis it is maximum and this variation is of the form
P equal to Po plus P2 cos 2 alpha that is when alpha is 0, alpha is 0 its value is Po
plus P2 and when alpha is ninety degrees its value is Po minus P2 right that is cos2 alpha
becomes minus 1 and it is this variation of this permeance right is having a strong bearing
on the variation of self-inductances mutual inductances and so on.
Now to understand the whole thing what we start with this we first write down the stator
circuit equations. The basic stator circuit equations are ea is equal to p, psi a minus
Ra ia, eb equal to p psi b minus Ra ib and ec equal to p psi c minus Ra ic, ia ib and
ic are the instantaneous value of the phase currents and p psi a stands for d by dt of
psi a, psi a, psi b and psi c are the flux linking phase a phase b and phase c respectively.
Okay that means straight forward that the induced emf is d by dt of psi a and this will
be equal to the terminal voltage plus the resistance drop or now this equation is drawn
considering the generator action okay.
Now here, let us see actually that what determines the flux linkage in the stator phase winding
the flux linkage in the stator phase winding can be written as psi a equal to minus laa
ia, now here I will explain this minus terms but laa is the self-inductance of phase a
la into ia minus mutual inductance between a and b and multiplied by ib minus iac ic
plus lafd ifd where lafd is the mutual inductance between a phase and field winding ifd is the
field current similarly lakd, ikd, lakq, ikq.
Now since we have assumed in the basic model here that the flux linkages are shown in the direction opposite to the current
and that is why actually the negative signs are appearing here that is in these terms
you can just see these are the negative signs while the currents are entering the other
three rotor windings therefore they are the positive signs. Now we will see that that
these self-inductances mutual inductances these are not constant these depend upon the
position of the rotor with respect to the windings, the stator windings and we will
show that these depend upon the angular position of the rotor and since the rotor is rotating
the angular position of rotor keeps on changing and therefore these inductances are going
to be a function of angular position theta. Okay now to understand this let us first start
with the stator self-inductances, the stator self-inductances.
Now here the stator self-inductance is denoted by the symbol laa okay and how when we define
this stator self-inductance, the basic definition is the flux linking the phase a winding divided
by the current that is the self-inductance of phase a winding with no currents on other
windings that is when only current ia is flowing and we find out what is the total flux linking
the stator winding a that is the self-inductance of stator winding a laa.
Now when the current ia is flowing okay then the MMF, MMF which is produced due to the
flow of current is Na into ia and this MMF is sinusoidally distributed along the surface
of the stator or along the air gap okay because the stator is suppose to produce a sinusoidally
distributed MMF okay and this MMF has the maximum value along the d axis right it is
it is peak is along the d axis and when you go away from the d axis both sides this is
going to decrease.
Now here this diagram shows, this diagram shows the MMF produced by the stator phase
a that is the I am just showing this is the this is the axis of phase a okay and the MMF
produced by the axis of phase a MMF produced by the phase a or stator phase a is having
its peak along the d axis, this is the d axis. I am sorry, this is the not d axis, I am sorry
this is the axis of the phase a, the axis of the phase a. Okay it is a little correction
this is the axis of phase a.
Now what we do is we split this MMF into two components both are having the sinusoidal
distribution, one having its peak along d axis another having its peak along q axis
therefore this this graph red graph which I have shown here, this shows the sinusoidal
distribution having its peak coinciding the d axis, this is the again sinusoidal distribution
its peak is coinciding with q axis and the q axis is leading d axis by 90 degrees okay.
Now this can be seen here in this diagram that the MMF produced by the stator right
along its own axis that is the MMF produced by a stator winding of phase a right is having
its maximum value along its own axis that is axis of phase a. Now we have assumed likely
that the rotor is rotating in the anti clockwise direction therefore axis of, now the d axis
is shown here and q axis is leading points and what we do is that this MMF is resolved
into two components one along d axis another along q axis. The d axis component is Na ia
cos theta and q axis component is minus Na ia sin theta, okay.
Now with this MMF’s then we can find out what will be the flux produced at the air
gap along this d and q axis. Okay, now here we are showing that the MMFad that is MMF
due to due to current flowing in the stator a phase and it is component along d axis ad
is equal to Na ia cos theta and these are the peak values therefore when ia attains
its peak value this will also become this varying this is varying along as the ia is
varying, then the along the q is minus ia Na sin theta okay.
Now the flux produced long these two axis because of these MMF can be written as MMF
into Pd that is phi gad, g stand for air gap or gap flux okay the phi gad is equal to Na
ia cos theta into Pd and phi gaq is equal to minus Na ia sin theta into Pq. Now here
this is the MMF and to relate this MMF to the flux we are using this term Pd therefore,
Pd is in general a permeance coefficient we call it it is not only the absolute value
of permeance but all other parameters which relate flux to MMF because this is the MMF
only this is the flux okay.
Now what is done is we again make use of this phasor diagram, the flux which is produced
along d axis in the air gap, flux which is produced along q axis in the air gap, we resolve
them back along the axis of phase a. that is when you resolve this right then this component
will come out to be equal to phi gad cos theta and the second component comes out to be phi
gaq sin theta with negative sign because there was negative sign already attached with it
and therefore we can say that the air gap flux due to current flowing in the stator
winding a only comes out to be equal to Na ia substituting these values on the previous
equations in this form.
The Na ia Pd cos square theta plus Pq sin square theta and this expression when simplified
it can be put in the form Na ia into Pd plus Pq by 2 plus Pd minus Pq by 2 cos 2 theta.
Now here this is very important point to understand that the air gap flux produced, air gap flux
produced by current flowing in the stator winding of phase a is equal to is proportional
to a term Pd plus Pq by 2 and another term Pd minus Pq by 2, cos 2 theta that is this
term does not depend upon angular position, while this term depends upon the angular position.
Now we define the inductance.
The inductance, the self-inductance of the stator phase a due to gap flux only the flux
which is produced in the air gap, lgaa is equal to Na affective number of terms into
the gap air gap flux divided by ia and this comes out to be we substitute the value of
phi gaa it comes out to be Na square Pd plus Pq by 2 plus Pd minus Pq by 2 cos 2 theta
okay therefore this can be put in the form that is this is your Lgaa is the self-inductance
of phase a due to gap flux only which can be put as a constant terms Lg0 plus another
term Laa2 cos 2 theta right because as I have seen the I have told you that the permanence
of the air gap varies as a with the position of the rotor and there we found actually that
it has a second harmonic variation. Here, also you find there is a constant term plus
a quantity varying as a function of cosine 2 theta okay.
Now to make the whole thing more complete there is some a leakage flux which does not
cross the air gap. Okay and this leakage flux also contributes the self-inductance of the
stator phase and therefore, when you account for the leakage flux then we can say that
the self-inductance Laa of the stator phase is equal to self-inductance due to leakage
flux plus lgaa which I have obtained in the previous equation that is due to the gap flux
and then when you combine these two terms. We can see here that this mutual inductance
or the this second term will not be affected by the leakage and therefore this leakage
term is combined and you find here that the self-inductance can be written as Laao plus
Laa2 cos 2 theta.
Now this is the most important equation to understand that how the self-inductance of
stator phase varies as the position of the rotor varies the angular position of the rotor.
Now this angular position is measured with respect to axis of phase a.
Now this graph shows the plot for the variation of the self-inductance of stator phase a as
function of theta okay and you can identify here that this is the term Laa2 which varies
this Laa2 is constant and this is another term which we call Laao and the total inductance
of the stator phase is now written as Laa equal to Laao plus Laa2. These two terms are
constant these constants these are constant they do not depend upon the angular position
that mean the total self-inductance depends upon the angular position but these two coefficients
are constant. Now when we perform the similar exercise for phase b and phase c, since the
the axis of the phase b and phase c are displaced by 120 degrees with respect to axis of phase
a right.
Therefore, the expressions which we have written here for self-inductance of phase b right
will be of the same form except theta is replaced by theta minus 2 phi by 3 and since these
the everything remains same therefore these terms are also same therefore it is not Lbbo
but Lbbo is same as Laao okay similarly, we write down lcc as Lao plus Laa2 cos 2 times
theta plus 2 phi by 3 okay very straight forward.
Now next very important point we have to understand is the stator mutual inductances, the stator
winding mutual inductances again we will see that the stator winding mutual inductances
also are function of rotor position that will be function of theta. Now here, here when
the when the axis of the rotor is in the middle of the axis of stator phase a and stator phase
b then at that position the mutual inductance between a and b will be maximum for example
the mutual inductance between phase b and c when you try to see it will be maximum when
theta is theta is 30 degree minus 30 degrees and 150 degrees these they are the positions
which we have to see. Okay, using this information the flux linkage is, flux linkage is of phase
b.
When current is flowing in phase a is are obtained that is we want to to find out the
flux mutual flux right that the flux linking flux, linking phase b due to current flowing
in phase a okay and then once you find out this flux. Okay, we can find out the mutual
inductance because the the inductance is the flux linkage by the current mutual inductance
will be the flux linking phase b due to current in phase a and then you divide by the current
you will get the mutual inductance. Here here following the same approach as we have done
for done for obtaining the self-inductance the, the air gap flux flux again the gap flux
linking phase b with when current is flowing in phase a is obtained in this form that is
this is obtained in terms of these 2 components phi gad and phi gaq that is this is the air
gap flux along d axis this is the air gap flux along q axis and after making the substitutions
we find actually that mutual flux comes out to be equal to Na ia minus Pd plus Pq by 4
plus Pd minus Pq by 2 cos 2 theta minus 2 phi by 3, okay.
Now you can easily see here actually that if you substitute here to make this quantity
one to make this quantity one. You can find it out actually ah what should be the value
of theta right and since this term is minus here Pd is always greater than Pq permanence
along the d axis is more then the permanence along q axis and therefore this is minus to
have this also minus so that the total quantity is added up. Okay you can find out the value
of this angle theta and you will find actually that when theta occupies either 30 degrees
or 150 degrees it will be maximum. Now this, this mutual inductance can be obtained as
lgba divided by after dividing the the expression for phi gba by ia okay.
Therefore, the expression for Lgba comes out to be in this form. Okay, now again it can
be written as minus 1 by 2 Lgo, Lg0 plus Lab2, now if you very carefully examine then this
Lab2, Lab2 will be of the same amplitude as Laa2, Laa2 right.
Similarly, you can find out the mutual inductance between b and a and we this ba mutual inductance
between the phase a and b that is equal ba or ab, they are always equal okay and the
expressions are written in the form minus Labo minus, now here actually when you have
written in this form what we have done is that we have accounted for some some leakage
flux which also leaks to windings right because there are, there is a air gap flux and there
is some flux which does not cross the air gap and once you account that we can write
down these mutual inductances in this form, okay again you can see that this depends upon
theta. Similarly, you can write down for bc and cb it comes out to be in the similar form
and similarly lac and lca can be written like this okay.
This diagram shows the variation of mutual inductance as a function of theta between
the 2 stator phases that is here we have shown the Lab and you can easily see that first
thing which we see here is that the the mutual inductances all through negative. Okay and
its variation is shown in this form therefore, this this quantity a constant quantity is
Labo and over this is you superimpose this sinusoidally varying quantity and variation
is as a function of 2theta. Therefore what we have seen till now that the self-inductances
of the stator phases or stator winding and mutual inductances between the stator winding.
Now we will consider the mutual inductances between stator and rotor windings, stator
and rotor winding. Now so far the mutual inductances between stator rotor windings are concerned
that they are function of angular position but they are not because of the variation
in permanence here because so far the rotor is concerned, rotor will always see the same
permanence because the stator stator is having the ah a uniform shape right and therefore,
so far the so far actually the rotor is concerned, rotor windings are concerned right the the
there will be no variation in permanence.
Now here the mutual inductances between stator and rotor windings vary because of angular
position. Now for example, if you take take the stator phase a and field winding in case
the axis of these windings coincide they will have maximum mutual inductance in case the
axis of stator winding of phase a and the field winding they are in quadrature, the
mutual inductance will be 0 right and since the rotor is having rotating it occupies different
positions therefore, when it coincides where the direct axis of the rotor coincide with
the stator phase a axis or b axis or c axis they will have maximum mutual inductance and
when the quadrature axis of the rotor coincides with the stator phase axis, okay phase a axis
or phase b axis or phase c axis then the mutual inductance will be 0. Okay therefore, we can
write down this mutual inductance Lafd equal to L, Lafd into cos theta.
When suppose as we know that the theta is measured right considering the axis of phase
a as reference and theta is the angle between the d axis and axis of phase a. Okay therefore
when theta is 0, the mutual inductance between stator stator winding and the field winding
is maximum. Okay and 90 degrees it is now, so the the amortisseur on the direct axis
is is also going to have the inductance, mutual inductance in the form Lakd cos theta right
because this the the direct axis amortisseur is having axis coinciding with the field winding
right and therefore the variation is going to be similar.
Now the mutual inductance in the quadrature axis amortisseur and the stator winding will
be written by the formula Lakq cos of theta plus phi by 2, why this theta is replaced
by theta plus phi 2 by 2 because q axis is leading d axis by 90 degrees therefore, this
can be written as minus Lakq sin theta.
Now what we have seen here is till now, we have obtained the expression for the self-inductances
of the stator windings, mutual inductances between stator windings and we have also obtained
the mutual inductances between the rotor windings and stator winding and we have seen that all
these are function of angular position. Okay
now we again come to our fundamental equations that is the stator voltage equations stator
circuit equation ea equal to p psi a minus Ra ia and we have seen that the flux linkage
of phase a is now written as minus laa ia minus lab ib minus lac ic plus lafd ifd plus
lakd ikd plus lakd ikq therefore, now in this equation we substitute the expression for
laa, lab, lac, lafd, lakd, lakq.
Okay, then we get the expression for flux linkage of phase a as minus ia into Laa plus
ib into Labo Lab2 cos 2 theta plus phi by 3 this plus sign has come because there was
negative sign here earlier. Okay when you see this mutual inductance there was a negative
sign therefore it becomes plus here. Similarly, you have ic into Labo plus Lao cos 2 theta
minus phi by 3 and so on, that is what we have done is that in this in this basic equation,
we have substituted value of all the inductances okay, which were all found to be function
of theta. Similarly we can write down the flux linkage of phase b and flux linkage of
phase c, they are exactly similar except you will find that theta is replaced by theta
plus 2 phi by 3 or theta minus 2 phi by okay.
Now after having written the equations for the stator windings, voltage equation for
the stator windings we can write down the rotor circuit voltage equations. The, in the
rotors on the rotor we have considered 3windings, 1 filed winding and 2 amortisseurs. Okay therefore
efd the voltage applied to the field winding is equal to P psi fd plus Rfd ifd. Now here,
since we have assumed that the current is entering the field winding and therefore the
term here is P psi fd plus this is a simple Rl circuit.
Suppose you have Rl circuit, then the applied voltage is equal to the rate of change of
flux linkages plus voltage drop in the resistance. Okay then the other 2 equations these relate
to the direct axis amortisseur, winding amortisseur circuit and quadrature at the amortisseur
circuit here since there is no external applied voltage therefore we have 0 term here. Okay
therefore, there are 3 basic rotor circuit voltage equations we have 3 basic stator circuit
voltages equations. Now let us write down the expression for these flux linkages psi
fd psi kd and psi kq.
Now psi fd can be written as Lffd that is the self-inductance of field winding into
ifd plus mutual inductance of the mutual inductance between between field winding and amortisseur
that is Lakd into ikd okay and there will be no there will be no flux linking the field
winding due to the quadrature axis amortisseur because the the 2 axis are that that the the
displacement of 90 degrees between the 2 axis okay and therefore there is no flux linkage
contributed by by amortisseur on the quadrature axis to field axis flux linkage.
Then these 3 terms are here Lafd into ia cos theta Lafd into ib cos theta minus 2 phi by
3and Lafd plus ic into cos theta plus 2 phi by 3 that is when the 3 stator currents are
carrying the values ia, ib and ic and depending upon their mutual inductances this will be
the flux linkage in the stator winding. Now one point which I wanted to mention here is
the so far the self-inductances of the rotor circuits are concerned that is self-inductance
of field winding, self inductance of amortisseurs they do not depend upon the angular position
because because so far actually the the the magnetic circuit is concerned for computing
the self-inductances of rotor circuits are concerned, these the self-inductances are
constant and the since that we have assumed like with uniform internal surface of the
stator okay and therefore no variation of reluctance so far actually the rotor circuits
are concerned.
Similarly, similarly the mutual inductance between the rotor circuits there is mutual
inductance between the field winding and amortisseur on the d axis they will be fixed, they do
not depend upon the rotor position right. Therefore for example, Lfkd is a mutual inductance
between field winding and direct axis amortisseur this is a constant quantity they will not
depend upon the rotor position.
Now this similarly you can write down the flux linkage of amortisseur on d axis and
flux linkage of amortisseur in the q axis okay.
Now with this with this we have developed the complete mathematical model that we have
written three stator circuit equations, we have written the rotor circuit equations we
have expressed all the inductances as function of currents and I am sorry ,not all the flux
linkages as function of currents and the self and mutual inductances.
Now this that is ah this is what is called complete model of the system however the basic
problems which arise are due to due to the variation of these inductances with the variation
of rotor angular position and to overcome this problem and seeing very carefully the
expression for, you see the expression for psi fd, we find here a term Lafd that is the
along with this term where ia cos theta plus ib cos theta minus 2 phi by 3 therefore, this
has prompted us to obtain a transformation and once we go we use this transformation,
we will find that the equations can be simplified and we can make these equations equations
where they do not exclusively depend upon or the inductances do not depend upon the
angular position.
Okay the the transformation is of this form that is we define, we define this term ia
cos theta plus ib cos theta minus 2 phi 3 plus ic cos theta 2 phi by 3 multiplied by
some constant kd as id that is these 3 terms ia cos theta ib cos theta minus 2 phi by 3,
ic cos theta plus 2 by 3 this complete term multiplied with some constant kd is denoted
by a term id.
Similarly we denote another term iq as minus ikq multiplied by ia sin theta plus ib sin
theta minus 2 phi by 3, now this with this with this assumption or this transformation
if we consider balance three phase currents that is ia equal to Im sin omega s t, ib equal
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
is, we are assuming that the 3 stator currents are balanced with this 3 stator currents to
be balanced, okay what we do is if you substitute and find out the expression for id, the id
will come out to be as kd into 3 by 2 Im sin omega st minus theta. This is very important
term that is with this transformation, this id current id is equal to kd into 3 by 2 Im
sin omega st minus theta.
Now if you assume kd equal to 2 by 3, if you assume kd equal to 2 by 3 then the the peak
value of id will be same as Im okay and therefore in the Park’s transformation, Park’s transformation
kd and kq are taken equal to 2 by 3, that is iq will also be taking the same minus iq
into 3 by 2 Im cos omega st minus theta.
Okay and therefore, if I take kq equal to 2 by 3 then the peak value of iq will be same
as Im. Okay now to make this model complete complete and assuming that suppose the 3 currents
are not symmetrical right then we can define one, 0 sequence current io as 1 by 3 ia plus
ib plus ic and with this definition the transformation looks like this it is very interesting thing
this transformation looks likes this that we can say id, iq, io, a vector consisting
of d axis current, q axis current and i0.
These 3 currents can be written in terms of the phase currents ia, ib and ic in terms
of this matrix and this is called transformation matrix that is transformation matrix is 2
by 3 the first row is cos theta cos theta minus 2 phi by 3, cos theta plus 2 phi by
3. Okay similarly similarly the second term and third term.
With this now inverse transformation that is if you write down the expression for phase
currents in terms of the dqO currents then this can be written in this form cos theta
minus sin theta 1like this this is called inverse that is sometimes if you know the
value of id, iq, io we can find out the phase currents.
Now interesting thing which happens is that if I substitute the values of the phase currents
in terms of dqO components right then I get the expression for flux linkage in the d axis
called psi d that is if the the all these fluxes flux linking phase a phase b and phase
c right they are also transformed currents are also transformed that is by applying this
transformation. We find that the flux linkage is can be written in terms of ah the constant
coefficients that is psi d is equal to minus Lao plus Labo plus 3 by 2 Lao Laa2 okay then
Lafd ifd plus Lakd ikd that is here the coefficient of id is a constant term it does not depend
upon angular position theta similarly, for psi q and psi o right we define we define
these terms Ld equal to Lao plus Labo plus 3 by 2 Lao similarly Lq and L0 are defined.
When you make this substitution we can write down the flux linkage psi d as minus Ld id
plus Lafd ifd plus like this. Similarly, when you apply the dqO transformation the flux
linking the rotor circuits are also expressed in terms of the rotor currents rotor currents
rotor circuit currents and the dqO components of currents and again you find actually that
these flux linkages as well as these 3flux linkages with the transform quantities are
are independent of rotor angular position and this is what it helps the whole thing
and once you substitute these expressions in our stator circuit equations.
We get these equation in the form ed equal to P psi d minus psi q p theta minus Ra ideq
is equal to P psi q minus psi d p theta minus Ra iq and eo equal to P psi o psi 0 minus
Ra io. Now these are the 3 basic equations which are written in terms of transform quantities
or dqO terms or sometimes in case in dq of frame of presentation. Okay today I have discussed
the basic circuit equations of the synchronous machine and discuss the dqO transformation
and its importance. Ultimately, we have obtained the stator circuit equations in terms of the
transform quantities. Thank you!