Welcome to the next lecture in this course.
We have progressed a reasonable distance into the course, so let us stand back a little
and try to recapitulate what we have learnt so far and where we are with respect to the
objectives of this course. So our broad objective in this course is to
develop dynamic equation for electrical machines. For this purpose we started out by understanding
an electromagnetic circuit, and how equations can be written for an electromagnetic circuit,
we looked at a singly excited linear motion system, wrote down the electromagnetic equations
for a single excited linear motion system. We understood what an inductance was and how
it is related to flux linkage, how flux is generated, and we saw from the equation for
a single excited linear motion system that the inductance is an important aspect in describing
their behavior. We saw that the force generated, then is dependent on the rate of change of
inductance with respect to the displacement. And from there we went on to systems that
now rotate about an axis, rotary systems, and there we saw that the generated torque
is proportional to the rate of change of inductance with respect to angle. And these are basically
linear magnetic systems that we saw, and we made a small digression to see how we can
handle non-linear systems, where the BH curve is not linear, but a non-linear curve.
And then we came across ideas such as co-energy, and magnetic energy, and we understood how
one can develop an expression for the generated electromagnetic torque from either the co-energy
or the magnetic energy of a system. And then we said we are not really going to look at
non-linear magnetic systems, but we are going to confine ourselves, the linear magnetic
systems. And then in the linear magnetic systems case,
we saw that the inductance was an important aspect in describing their behavior. And therefore,
we then went around to deriving expressions for the inductance in case of rotational systems.
So we looked at several rotational systems, again we looked at systems where which had
a single coil on stator, and then we found out the expression for the inductance of this
single coil on the stator. We considered a salient pole rotor
and a cylindrical rotor; we looked at both these aspects. And then from the case of a
single coil on the stator we went to a situation where there was a single coil on the stator
plus a single coil on the rotor. And we derived expressions for the inductance of the stator
coil or the rotor coil, and then the mutual inductance between the two.
So having developed an experience with deriving inductance, expressions we went on to look
at machines or machine arrangements which are closer to real life. And they have not
a single coil, but they have a distributed winding on the stator. And then for this distributed
winding case, we then derived expressions for the self inductance, mutual inductance,
between the stator phases and between the stator and the rotor.
So now we have a good idea of how the expressions for the inductance in various situations look
like. So armed with all these things we are now ready to go back to developing the dynamic
equations for electrical machines. This information was required in order to write expressions
which we started out with. So let us now look at the case of a three
phase induction machine. A three phase induction machine, if you would remember from the first
course on electrical machines that you would have done. The three phase induction machine
has three windings on the stator, and three windings on the rotor.
Let us consider a round rotor induction machine, again you know from your machines from the
first course on electrical machines, that a squirrel cage induction machine can also
be considered as a three phase rotor. And therefore, if you know how to write down the
equations for a round rotor induction machine, you also know what to do for a squirrel cage
machine. So this is a more general case, and that is
what we will look at. So that has three windings on the stator, and three windings on the rotor.
So if you look at the geometry of the system, you have the stator that is there, and then
you have the rotor inside this, this is a uniform air gap. And on the stator and rotor
three phase windings are there. We have seen earlier animations on how this winding distribution
in a stator or the rotor is going to look like.
Rotor of course we saw for a alternator in the animations, but the stator which we saw
in the animations for the alternator is the same as the stator of an induction machine
both fall under the same AC machines group. And so, let us say that you have a uniformly
distributed winding, let us assume a uniformly distributed stator and rotor windings.
So if it is uniformly distributed, let us say that the R phase winding lies here, and
the return conductors of the R phase winding then lie here. And then the axis of this R
phase winding would lie along this axis. So let us call these windings as not RYB, but
ABC. So this is you’re a phase winding and the return conductors we will call them as
A’, that means the coil starts from here, goes in and then comes out this way, then
goes in here, comes out there so on so this is then the access of the stator as similarly
you then have the stator b phase who is access is going to be 1200 displaced from this so
let us take that access to lie along this direction so this is the access of the b phase
bs meaning b phase of the stator and so if this access as to be here then you need to
have the b phase winding phase here. And then you have the c phase winding whose
access is going to be another 120 0 displaced so that would lie here and if this access
as to lie here then this winding will have to occupy this region so this is a representation
of how the 3 phase winding be spread around the circumference of the stator and these
are the access of the 3 phase 3 phases a, b and c I have written b so this is now cs.
What happens in the induction machine is that these 3 phases will then be excited with 3
phase voltages which are displaced in time it is important to understand that the voltages
which are used to excide these 3 phase windings are displaced with respect to time that means
if I call Va as Vm cos ?t then vb is vm cos ?t – p/3 which means that the wave form
of phase v is shifted with respect to wave form of phase a by an angle 1200.
And then Vc will be = Vm cos ?t – 4p/3 which means that the wave form of phase c is further
phase shifted by 1200 from the wave form of phase b so these 3 wave forms are phase shifted
with respect to time and these are applied to phase windings that are displaced with
respective phase you have a phase here B phase winding as displaced 1200 with respect to
phase and c phase winding sorry that is a phase winding and then the b phase winding
is displaced 1200 with respect to phase. And c phase is further 1200 with respect to
phase so these are spatially displaced winding otherwise these windings are identical they
have the same we assume that these windings are identical otherwise they have the same
number of turns the phase spread is the same everything is identical about them except
that there axis is displacement state now one may wonder why we have put a phase here
and b phase appears to be leading that of a.
And c phase appears to be further leading that of b where as the excitation is lagging
you have a phase excitation b phase lags that by 2 p/3 c phase logs further by 2 p /3 this
is rather arbiter this is a physically available electrical machine and one can winde the axis
in whatever way we want it could be a here and then v here and c here or b here and c
here. In some manner it can be 1 and having wounded
it is a physically available machine somewhere and we can now connect any sort of voltage
that we want to it we can either have Va as Vm cos ?t and Vb as Vm cos ?t – 4 p/3 and
in which case this will become 2 p/3 so that is ac b sequence or ab c sequence this is
completely independent of how the machine is gone, this is separate this is separate
but why we have selected this would be evident if you now see what happens if you excite
it with this. Now having excited this with Va, Vd and Vc
let us now say that it is going to cos a current ia = Im cos ?t - ? then ib would be In cos
?t – ? - 2p/3 and ic would be Im cos ?t - ? - 4p /2 these are 3 currents that would
then flow in the machine having excited it with these voltages it means we are assuming
that the machine is a balanced machine balance voltages are applied to a machine it causes
balance flow of currents in the machine this is the nature of current.
Some phase angle would there depending up on what is the load the machine is operating
at that point, so if this is what is going to be applied let us understand what is the
meaning if this when we say that ia so much current is flowing this is the current flowing
in this phase a phase and we have said that the axis of the a phase winding is here that
means if a current is flowing in the a phase we seen how the MMF generated by a phase winding
is drawn with respect to the angle as we travel around the circumference that MMF plot as
a particular wave shape and we said that we consider the fundamental of that MMF wave
form. And therefore if this current that is flowing
we know that for any given current flowing in this phase winding the maximum value of
the fundamental component of the MMF occurs at this angle and that is sinusoidal distributed
that mean form this b it then drops down to 0 here and then reaches negative peak here
and then reaches 0 here and as this current is going to increases because this is im cos
?t this current may increase or decrease as t is going to change.
Whatever happens note that around the circumference if you travel at any given instantaneous value
of ia the maximum MMF will always be reached at this point and then it would go to 0 here
and negative maximum here what is that maximum value will depend up on the value of ia where
as the point that which the maximum occurs will always be here and that is why we call
it as a phase axis that means. The MMF generated by this will be oscillating
on this axis as the amplitude of ia is going to change this MMF will build up then it will
decrease and then revere build up in a negative direction decreased to 0 and then reveres
so it is an oscillating MMF that is generated along the a phase axis similarly whatever
MMF is developed along the by the ib current which is flowing in the b phase or rather
this phase would be oscillating along this axis always then would ne oscillating along
this axis this maximum would occur here and the MMF generated by the c phase would be
oscillating along this axis it is P could always occur here, but what we mean by saying
that Ia =Ivm Cos ?t – ? but Ib is phase hafted from this by 2p/3 it means that these
oscillations though they are taking place at specified access around the circumstances
of the machine. Or not oscillating
at the same phase this is displacement phase that means the maximum of this would occur
at a time instant different from that of this and the maximum of this would occur at a time
instant different from those these two let us what we mean so now let us having understood
that it is always an MMF acting along this access one can now draw this by a set of phases. Those phases would then be the phasor of Ia
if we take as the reference then the phasor of Ib large that by 1200 we can see that Ib
large by 1200 for an assumed direction of rotation that is anticlockwise of the phases
Ib would then come here Ic would come here, now note that this is a phasor diagram these
are time phasor this phasor diagram as nothing to do where the special orientation of the
windings in the machine this is a physical arrangement how the windings are placed whereas
this is an interpretation. This is a representation of what happens to
the excitation with respect to time, so the phasor representation the interpretation of
the phasor representation is that at any given instant, if you want to find out the value
of Ia what you would do is then you draw this horizontal axis and find out the horizontal
component of this phasor and that then represents the instantaneous value of that particular
NTT, so at this instant normally the phasor diagram is drawn for the case where P equal
to 0 or let us called ?t = 0. So at ?t = 0 what this diagrams says then
is that the AC has current Ia will have its maximum amplitude whereas the d phase and
c currents would have each an amplitude equal to this much and instantaneous value equal
to this much not the amplitude instantaneous value equal to this, this is negative and
since this angle is 1200 what it means if that the maximum value of Ia is Im amplitude
of ya is Im then at t = 0 the value of Ia = Im the value Ib = Im into Cos of 1200 that
is – ½ of Im and c is – ½ of Im again. As can be seen so what you have is Ia = Im
Ib = - Im / 2 and Ic = - Im / 2 these are instantaneous values of currents, now these
are going to flow in windings that are phase displace like this and each of them would
produce an MMF along it shown axis, so Ia would produce an MMF along this a axis Ib
will produce an MMF along this axis Ic would produce an MMF along this axis so let me draw
these axis again here so you have the F phase axis and then the b phase axis
and then the c phase axis. And what this phase is that there is Im acting
along this axis on the b phase it is – Im / 2 acting along this axis would mean that
Im / 2 acting here and – Im / 2 acting along this axis would men Im / 2 acting here, and
the net MMF that is generated is then the some of these three space phasors and this
angle is 600 so Cos of 60 is ½ so what you get is the resultant phasor of this acts along
the same axis and has a magnitude three times Im / 2 this is at ?t = 0, now let us consider
what happens. When ?t = say 300 ?t = 30 so let us see this
these at ?t = 300 this diagram which is a time phasor if you recollect what this phasor
representation is the phasors are arrows that are rotating with respect to time in the anti
clock wise direction and the diagram is drawn as a free is obtained at ?t = 0 so at ?t = 30
it then means that this diagram are rotated by 300 so Ia would have gone here Ic would
have come here and Ib would have come here, so this phasors are now rotated they are now
long at here. But they are also sitting at this position
now again to find out what is the value of instantaneous Ia Ib and Ic what we have to
do is. Take the horizontal component projection so
this angle is now 300 and this angle is 900 and this angle is 300 and therefore Ia would
then be Im into Cos of 300 Cos 300 is v 3 / 2 so you have v3 / 2 Im the horizontal component
of this phasor being at 900 is 0, so this is 0 and Ic this angle is again 300 so it
is – v3/ 2 I now these current are acting along the same axis the axis has not rotated
this is an axis of coils that are bound one the stator which is stationary and this axis
remains at the same position now therefore on a phase you have rv3/2 IM here. The net
MMF therefore the sum of these two phases so you can add this by the parallelogram law
you would get a phasor at looks like and since this angle is 60° the resultant what you
would find is at 30°. And what is the magnitude of this that magnitude
is again 3Im/3, so what we find is that at ?t=0 the net MMF was lying along the A phase
axis at ?t=30° the net MMF lies at 30° to the A phase axis the angle made is 30° the
magnitude remains the same and therefore we can extrapolate one can verify for various
positions of this phasor diagram as time increases this then traces a circle as we go around
and the angle made is always equal to ?t. Which is what one says as a rotating MMF wave
form in an induction machine, now is this MMF rotating in this direction in the anti
clockwise direction, it happens to rotate in anti clockwise direction because of the
displacement of the axis. Had you taken the B phase axis here and C phase axis there the
net MMF would have rotated in this direction. Since we want the rotation since we want the
rotation to be anti clock direction we have chosen A here B here and C here. If we are
chosen at other way the rotation would have been the clockwise direction and what happens
if the MMF rotates in anti clockwise direction in the case of an induction machine we know
that the MMF that is rotating in the air gap is the one that is inducing the voltage in
the rotor and therefore some currents flow in the rotor.
And the rotor tends to get dragged along the direction of the MMF that is rotating in the
air gap, so if the MMF rotates in the anti clockwise since the rotor also would move
in the anti clockwise since and therefore we want to rotation of the rotor to be in
the anti clockwise since so this particular convention has been chosen in order to meet
that requirement but if one does not care for that one can choose it in the other direction
as well. So in the induction machine coming back to
this again this is the distribution of windings on the stator and the axis that are there
in the stator. Similarly we have windings on the rotor as well and those windings would
also be distributed on the rotor they would have their own axis about which their MMF
act. And since the rotor can now rotate with respect
to the stator in general the axis of the rotor windings may not be aligned to that of the
stator and hence let us assume that the stator I mean the MMF axis of the rotor are displacement
with that of the stator we call this as ar and then therefore that of the B phase br
and that of the C phase cr. And the rotor has now displaced with respect to the stator
by an angle ?r. So in the rotation of the rotor what we have
done is we have taken one snap shot of the machine at a particular instant and in that
particular instant we have drawn this arrangement where the axis of ar is displaced with that
of as by an angle ?r and under this condition we now want to write down we are looking to
write down the equations of the induction machine.
Since 6 coils are there three on the stator and three on the rotor we are going to have
six equations the describe the induction machine, so how to write those equations. We know that
from our earlier experience with writing down equations for voltages applied to a coil that
is connected somewhere we know that the general form of the equation. Is if you take for the A phase va is then
ra ia plus the rate of change of flux linkages ?a, this is called the A phase of the stator
ras, ias and p?as. Similarly you would write an expression for the B phase rbs, ibs+p?bs
note that we are using lower case voltages for a, b phases and lower case currents for
this it means that we are writing down expressions for instantaneous voltages that are applied
instantaneous currents that are flowing these are instantaneous value.
Similarly vcs is rcs ics+p?cs, further let us consider that the machine is a balanced
machine that means the phase windings are identical expect that they are displaced in
space of course and since they are identical the three resistances are expected to be the
same, so we will remove the subscript a, b and c from this and call this as simply rs
so it is simply rs ias, ibs and ics. Similarly now there are three more windings on the rotor
so which will be var=rriar+p?ar and vbs is rr ibr+p?br and vcr equal.
So these are six voltage equations that we have written one for each phase of the induction
machine on the stator and one for each on the rotor. Having written this of course this
is a you then you need to determine the expression for f a s f bs and so on. So now let us consider
the expression for f a s is the flux linkage of the phase a, now flux linkage on the phase
a arises due to self flux + mutual flux. Self flux arises due to current flowing in
itself mutual flux arises due to current flowing in all the other systems that are there, all
other phases that are there in the stator as well as well as in the rotor. So what is
an expression for the self flux? If you remember we have derived inductance self flux linkage
is nothing but in general inductance multiplied by current which is the same as the number
of turns multiplied by that is flux that is flowing through it.
So we take this form so we need to know what is the inductance now we have derive expressions
for the inductance of a winding that is distributed on the stator and the rotor being salient
and when you see that case we saw that the inductance of a particular phase stator is
a leakage inductance + if the rotor is salient then we had an expression like ld + lq / 2
+ ld – lq / 2 x cos 2 ?, in the case of an induction machine the rotor is not salient
the rotor is cylindrical and the air gap is uniform that means whether you take the d
access representing air gap or the q access representing air gap they both are the same
which means that ld and lq are identical and therefore this c is to exist ld and lq are
identical and therefore this is ld + lq which is 2 x ld.
We will simply call this as the magnetizing inductance LMS of the stator where LMS would
be ld as for our derivations earlier. Maybe will put that down LMS what we derived as
LD we call it as LMS. So the self inductance of a stator winding is nothing but the leakage
inductance plus a magnetizing inductance for which we have derived expressions earlier.
Now looking at the mutual flux let us consider the mutual flux linkage between one stator
phase and another stator phase again we have derived an expression fort this earlier for
the case of course of a cylindrical stator and a salient rotor machine.
The expression was the mutual flux linkage is ld cos ?r cos d – ß r – lq f ? r sin
of d – ?r so this was the expression for the mutual inductance between two stator fields.
Again for the induction machine we have seen that ld = lq therefore this expression can
be written as ld = lq which is equal to LMS so LMS x cos ? cos d – ? – sin ? sin d
– ? that is nothing but cos of ? r + d – ? r which is LMS x cos of d, where d is the separation
of the access of the two phase windings two windings that are under considering.
In the case of a balanced induction machine we can see that the access are separated by
1200 and therefore this is nothing but LMS x cos of 120 0 which is –LMS / 2, so the
mutual inductance ms is nothing but – LMS / 2 this is let us say because we are considering
a phase displacement of 1200 this is the mutual inductance between a and b if you consider
mutual inductance between a and c then instead of d being 120 d would 2400 and still cos
of 240 is the same – LMS / 2 and therefore mac is also equal to MAB.
And I leave it to you toi show that mbc is also equal to mab therefore all the mutual
inductances between the three phases either between a phase of the stator and the b phase
of the stator or b and c or a and c they are all same and is equal to – LMS / 2 and therefore
we can write this as the self inductance of the stator multiplied by current flowing through
the stator plus mutual inductance lessons all the three mutual inductance are between
stator coils are the same we will simply call it as ms for the time being in to ibs + again
ms x ics. So this is self flux these two are mutual
flux, mutual fluxes arising due to currents flowing in the stator wind now you need to
talk about mutual fluxes due to current flowing in the rotor winding, so if we look at this
figure we see that the access of the stator a phase and the rotor a phase are displace
by an angle ? r and therefore if the mutual inductance between these two coil if the two
access are aligned such that they are in the same line then the mutual inductance will
obviously be the high. Now if they are displace then the mutual inductance
would obviously change we have seen for a winding on the stator and the rotor the mutual
inductance can be described as msr x cos ? and therefore the mutual inductance between the
stator winding a phase and the rotor a phase can then described as a msr x cos of ? r where
msr is the peak value of the mutual inductance. And then between the a phase of the stator
and b phase of the rotor the angel difference is 120 + ?r and therefore x Iar of course
this is Msr x cos ?r +2p/3 x Idr and then A phase of the stator and c phase of the rotor
so that would be Msr x cos ?r +4p/3 x Icr. So this is then the expression for the flux
linkage of the stator A phase. Now similarly one can write an expression for the flux linkage
of the stator D phase it could also have a flux components.
So Ls x Ibs and this Ls is same as the that Ls because we are considering again balance
3 phase windings, so the windings will have the same self inductance + mutual inductance
stator x Ias + mutual inductance stator x Ics + now the B phase of the rotor as stator
as an access here and A phase of the rotor as an access here. The angle of separation
between the two is 2p/3 – ?r. So I can write it as Msr x ?r - 2p/3 x Iar between B phase
of the stator and the B phase of the rotor the angle is still ?r.
And therefore this can be written as Msr cos ?r x Ibr and then you have Msr cos of between
B phase of the stator and C phase of the rotor the angle is 1200 + ?r and therefore this
is ?r + 2p/3 x Icr. So this is then the expression for the self flux linkage of the B phase similarly
one can write expression for the C phase in the stator Ias is Ls x Ips + Ms x Ias +
Ms x Ibs + Msr x cos between Cs and Ar the angle is 120 + ?r, so ?r +2p/3 x Iar.
Then Msr cos between Cs and dr 2p/3 – ?r, so ?r - 2p/3 x Ibr + Msr x cos ?r x I, so
this would be the expression for the flux linkage in the C phase now these expression
are fairly lengthy and looks for ways where you can represent them in a more compact form
and therefore we take request to vector notations. We define the vector v as the vector of voltages
so it is defined as Vas, Vbs, Vcs, Var Vbr and Vcr. So this is a vector having 6 elements
and then we define the vector of currents. As Ias, Ibs, Ics, Iar, Icr vector currents
and what we are looking for is the relationship between this voltage and current. Note that
3 times Ias is nothing but 3 some inductance x some currents, so that is what we have fs
l x some currents and therefore a relationship between these two can be expressed by looking
at all these equations. So how to write that equations let us look at the 1 row of this. We can write Vas, Vbs and so on that is the
vector the 1st term if you are going to v = something x I you can see that here it is
r x Ias + p fas which is p Ls x Ias,s o we write this as Rs + p (Ls) x Ias we write it
as an another vector there and then you have P Ms x Ibs P Ms x Ics and then you P Msr cos
?r and then you have P Msr cos ?r - 2p/3 and then you have P Msr cos ?r + 2p/3. Now the
matrix ends here and this is x I so what you have is Rs + p Ls x Ias P Ms x Ibs and so
on that is how the 1st row can be written and other rows can be written in the similar
manner. So in this way one can write down the description
of the induction machine and then the next lecture let us see how to proceed further
what we do with this kind of a description of a induction machine we will stop at this
point.
