In the last lecture, we had looked at the
induction machine equation being re-written in a more compact form using the space vector
notation. We saw how large system of equations having four independent variables neglecting
the zero sequence components of course, where the input voltages are Vds, Vqs.
And then the rotor voltage is Vdr and Vqr with the responses Ids, Iqs, Idr, and Iqr.
All of them can be combined into one single need space vector based equation. That is
the usual form of representation in order to derive and understand control structures,
which one would learn if you were to study a course on motor drives.
Having done that, we started looking at the modeling of alternators and in the alternators,
we said that primarily, the stator of the alternator looks very similar to that of the
induction machine, both are three phase wound status distributed binding and the rotor is
quite different from that of the induction machine. The rotor contains a source of excitation,
which is absent in the case of induction machine. The rotor now has a field winding and then
we said that the synchronous machine rotor predominantly for large machine is of two
varieties, one is the cylindrical rotor stator and then cylindrical rotor of a synchronous
machine and the other is the salient pole variety. In the last lecture, we were looking
at how the cylindrical rotor of the induction machine would look like and we a representation
of the cylindrical rotor machine, which we have here now just to briefly look back. . The rotor basically has a cylindrical surface,
which can be seen here. You see that the outer surface of the rotor is by enlarge a smooth
cylinder, except for the slots that are there in which the rotor coils have to housed, but
for those slots, it is a very smooth cylindrical surface and this arrangement shown is for
a four pole cylindrical rotor and we have one pole that is here, another pole phase
here, a third pole phase here, and a fourth pole phase here.
They all have to generate a magnetic field, so around this pole you have rotor field coils
that are wound through these two slots and then through the next two slots. For generating
the rotor field here, you have a field winding and these two slots connected in series with
the winding here and then the third pole phase and then the fourth field winding. They may
all be connected in series to ensure that the same current flows through them.
And then the direction in which field current flows is so adjusted that if you get a south
pole field here due to flow of current in the orientation that is shown, the flow of
current in the adjacent one is such so arranged, that it generates a north pole field and then
here it is a south pole and then here it is a north pole field. And these green lines
represent show how the field lines are going to flow in the rotor structure.
So you have north pole field flux lines that are coming in to the south pole and then flowing
along the rotor out through the north pole phase and similarly here as well. So, this
gives rise to a four pole field structure. Here is the centre shaft, which will then
be driven by a suitable mechanism. This whole cylindrical structure is then inserted into
the stator bore and the whole arrangement then forms here synchronous machine. If this
were not the case, the alternative arrangement is then a salient pole rotor. Which would then look like this, so here we
see a salient pole rotor structure the centre hollow represents the place where the shaft
will be there is the shaft is not shown here where as in the earlier rotor case, you had
the shaft coming out and now here what we see is explicit poles. If we call this as
the North Pole and then here you have a south pole this one would then be a north pole and
this would be a south pole and you find that this pole surface is smooth.
Here again the pole surface is smooth, where as between this pole and the next pole, there
is only a big air gap that is seen this region, there is a big air gap. Similarly, between
this pole and this pole there is a big air gap here. So, between adjacent two poles,
there is large amount of air gap that is available in this kind of arrangement. Field is then
wound around each pole, you see that coils have been wound around this and this winding
is then connected in series with coils that are wound around this pole in series with
coils that are wound here and then wound here. So, this then provides the way to excite this
field system, so some flow of current is then given through these winding, because of which
this may generate a north pole, south pole, and so on and this structure is then rotated,
so that the field that is generated bodily rotates along with the rotor. The field is
again a DC field excitation, just like in the case of a cylindrical rotor arrangement.
Now apart from this we also see that there are some round holes that are provided in
the pole shoe. This region is called as the pole shoe and this is then the pole phase.
So, in this, there are round holes that are provided. These holes run through the pole
along the length of the machine, so along these dotted lines inside this hole runs through
the pole. Similarly, on each pole, these holes have been provided.
It might actually be a slot for example, the pole phase, if we are to draw the pole phase
here, it may be a slot like this in this manner. So, by and large the surface is smooth, but
for the small opening. This slot is actually used to drive a bar through it, so there is
a bar that is driven through the slot and which comes out here and similarly a bar here,
which comes out of this, a bar that comes out of this, and so in each pole through this
holes that are provided, you then have bars are coming out.
These bars are known as damper bars, what is done with these bars is? If these bars,
once you provide all these bars here, here, and so on. Then there is a ring that is usually
used in order to short all these bars together. There is a ring that shots all the bars here
and there is another ring that shorts all the bars on the other side as well. So, these
damper bars actually form a short circuited cage like structure.
You see that there are bars here, which are shorted on either side. So it forms a cage
like structure, much like the squirrel cage of an induction machine. You know what a squirrel
cage rotor looks like? The squirrel cage rotor looks essentially like this. It is a set of
bars running through the rotor with a shorting ring on either side they are called end rings
and that is the rotor, which is used in the induction machine.
The induction machine does not have a field winding whereas here, the main source of rotor
field is the field winding. These bars are provided as the name says to damp out the
rotor movements. Now, what happens is, if you want to, for example, you that the synchronous
machine rotor runs at synchronous speed, and if you want to supply a higher load on the
electrical side, you have to increase the input on the mechanical side as well, which
is what is responsible for higher electrical output.
Now when you supply an increased mechanical input, what will happen to the rotor is that
the rotor because of a higher input now will accelerate a little and as it accelerates,
the rotor tends to move away from its synchronous locked position. As it begins to accelerate,
then the angle between the voltage vector are supplied on the stator side and that of
the field it increases and therefore more mechanical input is able to absorb and deliver
higher electrical output. Ultimately, the rotor still has to run at
synchronous speed if it is connected to the grid, so the rotor will momentarily accelerate
and then decelerate again, but settles down at a slightly different position. During this
acceleration and deceleration the rotor may actually oscillate and these oscillations
are to be damped out and it for that reason that this arrangement is given. So, whenever,
the speed goes away from synchronous speed, this damper bars act very much likes the squirrel
cage of an induction machine. In the squirrel cage of an induction machine,
EMF can be induced only when the speed is different from that of a synchronous speed
and that is exactly what is happening here. If the rotor accelerates, the speed is different
from that of the synchronous speed and now an induction machine action comes into play,
which now tries to bring the rotor, which tries to decelerate the rotor and the rotor
locks back into synchronism. So, this arrangement is also symbolically represented in the figures
that I can show here, so you see here. In this picture that in these slots bars have
been inserted in all the slots and then in the next picture here, you see that rings
have been put on either side, which short all these bars. If this ring is the end ring
and these bars the ones that are mend to provide the damping action. So, this then describes
how the salient pole rotor will be. Of course, these bars are not restricted to the salient
pole arrangement alone, even if the rotor is cylindrical one may desire a damping action
in which case this arrangement would need to be there, even for the cylindrical rotor
case. Now, the question is, how do we model this
sector that is this alternator. So, let us start looking at the equations that one can
use in order to describe this. So, in the alternator then, the stator that is available
is cylindrical and you have windings that are placed distributed on the stator and A
phase winding here, and then you will have the B phase winding here and C phase, so B
phase and C phase, what we do as we have being doing before, we draw spacial axis in order
to describe the axis of each phase and so the A phase winding would then have an axis
here, so this is the A phase axis and then you have the B phase axis, and then you have
the C phase axis. The rotor, because it is associated with the
field structure, it will generate a north pole along some direction and south pole along
some other direction, so we associate and axis with the rotor to denote where the rotor
angle is, where the rotor is angularly located and that axis we will associate with the north
pole of the field that is there in the rotor. For analysis, we are going to consider a two
pole alternator. So there is only one north pole and one south
pole and that north pole is associated with an axis, which we will then call as the rotor
axis, so the rotor axis is, let us say the rotor axis can be drawn here, so this is the
axis of the rotor and this angle, which the axis makes with respect to the axis of the
A phase is then called as the rotor angle. Now, we need to write down equations that
describe the behavior of this machine. So as before what need to start out with is
the voltage equation we need to write down expressions for the voltage that is applied
to the machine and there are three phases, so you have Vas, Vbs, and Vcs and there are
three phases. So if you take for example, Vas the expression can very easily be written
as the stator resistance, we will assume that the three phases are identical except that
they are displaced in space of course. The resistance of Vs is the same, so you have
RS multiplied by Ias + T times f here, which is f as is the flux linkage of the stator
A phase and derivative of that gives you the induced voltage, so P times f A and now therefore,
we will have to write down and expression for f As. f as is now going to comprise of
flux that is produced due to current flowing in A phase plus due to current flowing in
B and C and due to current flowing in the rotor as well.
So how do you get these expressions? So we know that flux linkage is equal to inductance
multiplied by the flow of current, so self inductance multiplied by current flowing in
the A phase, let us call it Las and then mutual inductance between A and B multiplied by the
current flowing in B phase + mutual inductance between A and C multiplied by the current
flowing in B phase + mutual inductance between A and C multiplied by the current flowing
in C phase + mutual inductance between A phase and the rotor.
We will call the rotor as the field winding that is normal nomenclature that is used for
synchronous machine. So, we will put that as Maf multiplied by I. Now, we need to derive
expressions for each one of these, L, M, Maf, and so on. So, how to derive expressions,
so we need to consider particular rotor geometry first and what we will do, we will consider
a salient pole rotor. Why should we consider the salient pole? If you remember that when
we started discussing about distributed field and determination of inductances and so on,
we wrote down expressions for inductance for the salient pole rotor and cylindrical stator
machine arrangements. And if you remember, we have in order to describe
the salient pole rotor; we had inductances, which were given symbols as Ld and Lq. Ld
represents the direct axis inductance and Lq represents quadrature axis inductance.
The direct axis inductances then is indicative of the smaller air gap that is present along
the field pole and the quadrature axis inductances indicative are an inductance that would arise
due to the larger air gap that is present 900 to the field pole.
These two are distinct and different because the rotor is salient pole, because in a salient
pole machine, the air gap along the direct axis is smaller than the air gap, which at
900 to it, because you have large air gap in between two poles. If you take a cylindrical
rotor structure, the air gap wherever you look at whether it is at a pole that is at
the pole phase or at 90 0 to the pole, the air gap is always the same and therefore you
expect Ld to be equal to Lq. So, in the resulting expressions of inductances
that we have determined, if you simply say Ld = Lq, you get equations for the cylindrical
rotor machine. If you keep Ld and Lq distinct, you get equations for the salient rotor machine,
and therefore it is good to derive equations for the salient pole machine and the cylindrical
rotor machine equations can be obtained very simply by making Ld = Lq. So, we will start
by looking at expressions for the salient rotor machine.
In the case of the salient rotor machine, again we have derived all the expressions
earlier, so we just have to remember and recollect those equations back. So, if you have an arrangement
like this, where there is a particular phase axis, a salient pole rotor, so let me symbolically
draw the salient pole rotor this way, salient pole rotor having its rotor axis oriented
here. Then, we have already determined an expression for the self inductance of the
stator A phase. So, let me write that down, what it would look like. So, the self inductance Las is then given
from the expressions that we have derived earlier, we had Ld +Lq/2 + Ld ñ Lq /2 x Cos
of 2 x ? R. If you remember that, this is expression that we wrote, where Ld and Lq
are inductances that taken to account equivalent air gap along the direct and quadrature axis.
Now this expression talks about flux that is generated due to excitation of the A phase,
flux that is generated and crosses into the rotor before coming out, so the inductance
that is described here essentially is a magnetizing inductance.
So, this is the magnetizing inductance component of the self inductance. In addition, there
would be leakage flux generated by exciting the rotor and this leakage flux is something
that does not cross over into the rotor and link the other coil as well. It links only
the A phase and there is no analytical expression for it. What is normally done is, you simply
add a term leakage inductance to it Ll is a leakage inductance of the A phase.
So, this is the expression that we had already derived. So, going forward, we need to get
the expression for the inductance of the phase B. Phase B has an axis here. So, this expression
can be derived very easily from the earlier expression, because it is the same phase,
which is displaced at a different angle. The inductance of the A phase when the rotor is
located here is that where ? R is the angle between the axis of the phase and the rotator
axis. Now you have the B phase axis here and the
rotor makes an angle with the B phase axis, which is this angle
and it is behind the B phase. In the earlier case when we determined the expression, the
rotor axis was ahead of the A phase, but now it is behind the B phase and this angle is
what you knee to determine for the angle ? R and this angle is nothing but 120 0 ñ ? R.
This angle is 120 because in a three phase machine, the phase axes are displaced by 120
0. Here it is already ?R and therefore this must be 120 ñ ? R, but however, with respect
to the B axis, the rotor axis is behind and therefore this angle is minus.
So, minus 120 ñ ? R, therefore in the same expression, if we substitute instead of ? R
-120 ñ ? R, you should get the equation for the self inductance bs and that is therefore
Ld + Lq x 2 + Ld ñ Lq x 2 x Cos of two times ñ of 120 ñ ? R is nothing but ? R ñ 120,
then +Lq and this is therefore, this can be simplified as 2 ? R ñ 240 0 so, 2 ? R ñ
2400. Similarly, if we want to determine the inductance
Lcs, now if you take a look at this figure again, you have the C phase axis here and
the rotor is behind that of the C phase axis by this much angle and that angle is nothing
but 240 ñ ? R and by the same argument, the rotor is now behind the C phase axis and therefore
what you are looking at is -240 ñ? R. So, let us put that down. You have Ld + Lq x 2
+ Ld ñ Lq x 2 x Cos of two times ? R ñ 240 0 + leakage inductance and 2 ? R ñ 240 is
nothing but 2 times ? R ñ 480 you can add another 360 0 to it and therefore or subtract
360, so this becomes ñ 1200. So, these are the expressions for the self
inductance of the three phases. Apart from this, then there is a field winding and so
the field winding will have a self inductance. Now is that inductance going to vary with
rotor position or not? Now the self inductances of the stator vary with respect to the rotor
angle because of the saliency of the rotor. Obviously, if this rotor were aligned along
the A phase axis, the inductance of the stator A phase would be maximum.
If the rotor were aligned 90 0 to the A phase axis inductance of the stator A phase would
be minimum, but as far as the rotor is concerned, irrespective of where the rotor is located
it always sees the same air gap. So the flux that is generated by a unit excitation given
to the rotor will always be the same or the flux linkage will always be the same, irrespective
of where the rotor is and therefore the inductance of the rotor coil is independent of the rotor
angle. Therefore, we can say that the field inductance, self inductance, is Lf is simply
= a leakage inductance of the field plus we will call this as a magnetizing inductance
of the field. So, this is a simple expression for the self
inductance of the field. So, the self inductances are over. Now we need to determine expressions
for the mutual inductance. Again, let us look back on what we have done. We had already
studied the case where you have two stator windings. Let me redraw this figure again. You have the stator and then you have a winding
here, which is let us say phase A and then you have another phase, which is phase B and
phase A what have its axis here. This is the A phase axis and this is then the B phase
axis. The rotor is situated at an angle that is then your rotor angle. So, under this condition
assuming that it is a salient pole rotor again, where we had looked at the case where these
axes are displaced by an angle ?. We have determined the expression for mutual inductance
for these two coils and that mutual inductance is then given by Ld x Cos ? Cos of ? ñ ?. ? is
nothing but ? R ñ Lq x sign ? R sign of ? ñ ? R. This was the expression for the mutual
inductance between these phases A and what well one and the other.
Now in the actual machine what we are having is three such phases, so you have A phase,
B phase, and your C phase. So A, B, and C and we need to determine therefore the mutual
inductances between A and B, A and C, and then B and C. Of course, the mutual inductance
between B and A will be same as that of A and B, so that is prevail there is nothing
to derive. So, now if we want to get this for Mab, what do we do with this expression?
The figure is as it is. There is no change in the figure A is here and B is here, that
is the situation for which we have derived this expression.
All that we need to do is substitute ? = 1200 because in a three phase machine, the two
axis are displaced by 120 0. So, Mab is nothing but Ld x Cos of ? R x Cos of 120 ñ ? R ñ
Lq x sign ? R sign of 120 ñ ? R. So, this can be written as Ld x this of the form Cos
A, Cos B, which is nothing but Cos of A + B + Cos A ñB divided by 2. So, Cos of A +
B would then give you Cos 120 and then Cos A ñ B + Cos of A ñ B will be Cos ? R ñ
this, so 2 ? R ñ 120 0. And then ñ Lq. This is sign A, sign B, and
therefore that is nothing but Cos A ñ B ñ Cos A + B/2 of course if there is a by 2 here
and here also by 2 Cos of A ñ B is ? R ñ this, so 2 ? R -1200 Cos A ñB ñ Cos A +
B, so minus R Cos this plus this is 120 0. So, that is what you have, which is nothing
by Mab Cos of 120 0, Cos of 120 0 is -1/2, so you have -1/2 here. So, that simply gives
us ñLd /2 ñLq/2, so this minus and this minus goes away, but you still have this minus
here, So you have - Ld/4 ñ Lq /4 and then you have
Ld/2 ñ Lq / 2 x Cos 2 ? ñ 120, so you have +Ld ñLq/2 x Cos of 2 ? ñ 120 0. So, this
expression can then be combined. It is ñof Ld + Lq/4. So, that is the expression for
the mutual inductance between the phases A and B. What about mutual inductance between
phases A and C. In order to determine mutual inductance between A and C, the rotor position
is still the same, and it is with respect to A phase, so that is also not a problem.
The only thing that we have to now do is d is between A phase and C phase. That is the
angle we need to consider and d for considering A and C is 240 0. So, what we need to do is
substitute 240 here. So, what you have is Ld x Cos ? R, Cos of 240 ñ ? R ñ Lq sign
? R sign 240 ñ ? R. So, this is again the same kind of relations
where you have Ld x 2 multiplied by Cos of 240 0 Cos A + B + Cos A ñ B and therefore
+Cos of 2 ? R ñ 240 and ñ Lq/2 x Cos of 240 0 ñ Cos 2 ? ñ 240 0. So, Cos of 240
is again ñ1/2. This is again -1/2, so you have -Ld +Lq/4. This is Cos of A ñ B. There
is some problem in the Cos of A ñ B ñ Cos A + B, so it should be the other way around.
Cos of A ñ B is Cos of 2 ? R ñ 240 0 ñ Cos A + B is Cos of 240 0. So, this is -1/2
then, so you have ñLd + Lq/4 and then you have +Ld ñ Lq/2 x Cos of 2 ? R ñ 240 0.
So, this is the expression for mutual inductance between the A and C phase. The third expression
that we need is the mutual inductance between the B and C phases. So, how does one determine
that? So, if you now look at this figure what we need is the inductance between B and C.
So, the angle between B and C is 120 0, just like it was between A and B. So, d is 120
0, but however, the rotor is now lagging behind are lying the B axis by certain angle and
therefore the angle that we will have to consider is ñof 120 ñ ? R.
Therefore what we will have to write is Mbc is Ld x Cos of, instead of ? R here, what
you will need is ñ of 120 ñ ? R or ? R ñ 120. Therefore Cos of ? R ñ 120 0 x Cos of
d ñ ? R, d is 120. So 120 ñ ? R. ? R is now ? R - 120 . So, 120 ñ ? R + 120 0 and
then ñ Lq sign ? R ñ 120 0 x sign of this expression and that is nothing but 240 ñ
? R. So, that is your expression for the mutual
inductance between B and C phases. So, this should then simplify as Ld/2. Again you have
Cos A and Cos B. So that is nothing but Cos of A + B, the ? R for this, let me rewrite
this as 240 ñ ? R. So, you add these two and that would then be ? R and minus ? R cancel
out 240 ñ 120 is 120 0 + Cos A ñ B, so this minus that, that is Cos R ? R ñ ? R is 2
times ? R ñ 120 ñ 240 ñ 350 0 and then -s Lq /2 x Cos of A ñ B. This - that is 2
times ? R ñ 360 0 ñ Cos A + B, so minus Cos 120 0.
That is the expression you get, which then simplifies to 2 ? R ñ 360 is nothing but
2 ? R itself, so that is 2 ? R and this is as usual ñR. So, you have ñR Ld + Lq /4
+Ld ñLq /2 x Cos of 2 ? R. That is then the mutual expression for the mutual inductance
between the B and C phases. So, we now have self inductance of the A phase. We have mutual
inductance of A and B. We have mutual inductance of A and C. When we write the flux linkage
for the B phase, then you will get the mutual inductance for B and C as well and now we
need to have an expression for the mutual inductance between the field and the A phase.
Now as far as the field is concerned, you are going to have a salient pole rotor that
is rotating and we have said that the field produced by the salient pole rotor is primarily
a DC field and the field bodily rotates with the rotor as the rotor moves. If that is the
case, then the field generated by the field structure is not going to change with respect
to time. Looked at from the rotor, it always generates a constant field provided the excitation
is also fixed and therefore the total flux generated by the field structure is the same,
but however, if the field structure were aligned on this axis, all of its flux will link phase
A and as the structure moves, rotates, then less and less of its flux is going to link
phase A. Therefore the mutual inductance will be maximum
when the field structure is aligned along the axis and that value of maximum mutual
inductance, let us call it as Maf itself. But however, that value of mutual inductance
is not going to remain as the rotor rotates the flux that links the stator A phase will
be whatever is generated by the rotor field. One part of it that lies along the A phase
axis is what you are going to link that field and therefore the total flux multiplied by
Cos of this angle is what is going to be responsible for linking phase A and therefore the mutual
inductance will then be Maf multiplied by Cos of ? R between the rotor field and A phase
of the stator. By the same token, if you then have the field
somewhere else then the mutual inductance between the field and another phase of the
stator will depend upon the angle between the two and therefore if the rotor is aligned
here, at ? R, mutual inductance between rotor and the B phase will then be the same Maf,
because had the field being aligned along the B phase axis, the flux linkage of the
B phase would have been the same as what was there for A phase.
Therefore the maximum value of mutual inductance for all the three phases with the rotor will
remain the same. Since that mutual inductance is a common valve between that of the stator
and the rotor structure, we will simply call it as Msr. That is a better notation I think. So, this is between mutual inductance between
A and C. Between B and C, will then be Msr x Cos of this angle, which is 120 ñ ? R will
be Mbf and then with the C phase, it would be Msr Cos 240 0 ñ ? R, will then be Mcf.
So, with this then we have derived all the expressions for inductances between the stator
phases and between the rotor and the stator as well. So, we now have to put all of them
together and formulate the voltage equations of the machine, which we will do it in the
next lecture.
