 lecture we had seen how one can make use of the inductance expressions that we have derived in order to write down the voltage equations for a three phase induction machine and then a three phase synchronous machine. We have seen that these expressions are fairly long and the system size, system description is fairly big. In the case of induction machines we have six variables on the electrical side and then one variable on the mechanical side which makes it a seven variable description. In the case of a synchronous machine we had considered two windings on the rotor and three windings on the stator which makes it a five electrical variable description and then you have the speed, so the six variables. So it is a six variable description and the description is also not very simple, it involves lot of rotor angles and the mutual inductances all those things are there. So the question is how do we simplify these machine description in order to make it more understandable and in order also to make it easily analyzable so that you can do as an easy simulation on any digital computer that you might have. So for this purpose let us look at these machines one by one let us take the induction machine first. In the case of the induction machine as I said we have a six variable that is electrical and then one mechanical variable as well and this induction machine is a three phase induction machine which means it has three phases on stator and three phases on the rotor that is what we have and in order that the induction machine works we have seen you would have definitely seen in our first course on electrical machines that the induction machine or any three phase AC machine a three phase winding produces what is called as a rotating magnetic field. And since we have now understood how the magnetic field and the magnetic circuit is to be analyzed we know that a magnetic field is or the magnetic flux is generated by an MMF that exists and therefore if there is a rotating magnetic field we can also talk in terms of rotating MMF waveform this rotating MMF waveform is the net of excitation on all the three phases for example in order to derive the expression for inductances. We have looked at the MMF distribution around the circumference of the induction machine of the electrical machine in the air gap and we had looked at one phase how the MMF generated by one phase is distributed around the air gap and this MMF waveform what we had drawn was a trapezoidal waveform for a distributed winding and this is for a specific flow of current IA some value for example let us say IA equal to 10 ampere this is the distribution having a certain amplitude and then a certain region where it decreases all this depends upon how many turns are wound in the phase that we are considering. We also said that in our analysis it is sufficient to consider only the sinusoidal fundamental component of this so you take the fundamental component this is the fundamental component of one particular phase if this current is at this level let us say this is MMF that is produced if the current now increases then a higher MMF would be produced higher amplitude MMF and if this current decreases a lower amplitude MMF would be produced if this current reverses then this MMF would then reverse and therefore what you have is that if this current is going to change in a sinusoidal manner you would have an oscillating MMF waveform around the circumference of the air gap this x axis refers to the angle around the circumference and this may be the MMF now if you consider the MMF of another phase that would also have a similar nature that means it would oscillate if you give a sinusoidal waveform to it but the difference being that MMF is displaced with respect to this MMF by 120 degrees and it would then oscillate here and if you take a third MMF that is further displaced by 120 degrees and therefore you would have something like this and so that again oscillates in its own plane the net result of all these things if you could do that is a sinusoidal waveform that then moves with respect to time so if you add up all these things at every given angle for sinusoidal excitations that are three phase that means 120 degrees displaced with respect to time then you would get a sinusoidally varying MMF waveform that moves with respect to time and has a constant amplitude so you get a moving waveform which in a circumferential air gap is nothing but a rotating MMF there are good animations of this in the NPTEL web based course on electrical machines one can take a look at this to understand how this rotating MMF is generated now this is a rotating MMF that means what you have is if you take if you mark an angle which represents 0 degrees as you travel around the circumference then the instantaneous amplitude that is this is a sinusoidal distribution around the circumference of the machine and this sinusoidal distribution of the net MMF will have its maximum amplitude occurring at some particular angle and as instance at different instance this is going to occur at different angles and hence you say that is rotating which therefore means that one can describe this net MMF as for example one can call it as F is equal to F hat into may be sin of a where a is this angle and we also know that as T increases this is going to be at different locations and because of that we call it as a rotating MMF if you now mark this angle at which this instantaneous peak occurs let us say if this angle may be 90 degrees so one can represent this amplitude by an arrow at another instance this occurs at a different angle so this arrow has moved at yet another instance it occurs at some other place so this arrow has moved and with respect to space then at any given angle if you want to find out what the amplitude is you can take the horizontal component of that and then that will then determine the instantaneous amplitude at any given angle. Now this interpretation looks very similar to that of your normal phasor representation which you use for an electrical let us say IA you would call it as some IM cos ? T and then you would represent this as a phasor with respect to a horizontal axis which is rotating with respect to time here also there is a very similar situation you have a sinusoidally distributed entity here which is MMF and that MMF we find is rotating in space and therefore if you represented by an arrow that arrow is really rotating in space and therefore we can describe this also by a phasor and this time this phasor is called as a space phasor as opposed to these variables which are then called as time phasor note that these phasors are rotating with respect to time these phasors are also rotating with respect to time here the variation of the phasor is sinusoidal with respect to time here the variation of the phasor is sinusoidal with respect to space and that is why this is then called as a space phasor. Therefore given this arrangement we can represent the net MMF generated by the three phases of the stator by an arrow which is inclined at some particular angle at a given instant of time at some other instant of time the maximum amplitude would occur somewhere else and therefore the phasor should be rotated right. So this is the effect that the three phases of the induction machine is going to give and it is because of this rotating phasor that the machine is able to start and run at some speed. So if this is going to be the net effect of excitation in the three phases we could arrive at this phasor amplitude and the angle of the phasor at any given instant by writing down a mathematical expression of the MMF generated in each phase remember that the axis of the a phase we took to be horizontal. So this is the as axis and axis of the b phase is 120 degree away from that of the a phase and axis of the c phase is 120 degree here. So this is bs and this is cs you might also remember a few lectures back we had also demonstrated that the net MMF generated by these three could be rotating in space by considering excitation along this axis that axis and the other axis and we saw that the net MMF is constant in amplitude but then rotated. So this is the net result a phasor that is rotating in space we can get this by knowing what is the excitation along this axis that axis and this axis. So you have three axis the result of which gives you one phasor. Now if this phasor is what we are going to get we could as well describe this phasor by considering a two axis system what we can do is take one axis along the same horizontal we will call this as the a axis we can take another axis which is 90 degrees to this and call this as beta axis then this phasor can also be described by the a axis projection of this vector and the beta axis projection of this vector. So you have an a component and a beta component which if you call this phasor as f I can call this as f a and call this as f beta. So if I know f a and f beta I know f right. So how to find out f a and f beta so instead of having to know a component, b component and c component we are now reducing it to two components alone and a component and a beta component. So these two components appear to be sufficient to describe the resultant mmf phasor f so how to get this a component a beta of the mmf waveform. So we are essentially saying that whether it is the a component beta component that you are considering or a b c component that you are considering the net mmf should always be the same which means that the number of turns in the phasor in the stator phase if you have some current ia flowing in the a phase and some current ib flowing in the b phase and some current ic flowing in the c phase at any given angle ? if you want to determine what the mmf is you would say ia x cos ? is the mmf component which is provided by a phase a axis excitation at the angle ? similarly the mmf component acting along the b axis excitation it is component at this angle and similarly c component along that angle multiplied by the number of turns is then the mmf component so ns multiplied by ia x cos ? plus ib x cos of 2 ? by 3 – ? plus ic x cos of 2 ? by 3 – ? is the mmf that is acting along this phase. So let us expand this expression so this is equal to ns x ia cos ? plus ib x cos 2 ? by 3 cos ? plus ic x cos 2 ? by 3 cos ? – ic x sin 2 ? by 3 so this is the mmf that is produced at this angle ? let us simplify this expression so this can be written as ns x ia cos ? cos 2 ? by 3 is – half and therefore this is – ib by 2 cos ? and then let us take this expression again cos 2 ? by 3 so – ic by 2 cos ? now let us consider this one sin 2 ? by 3 is sin 120 degrees that is ? 3 by 2 so plus ib x ? 3 by 2 sin ? sin 2 ? by 3 is again so this is – ib x ? 3 by 2 sin ? and therefore we can now group these terms together ia – ib by 2 – ic by 2 this whole thing multiplied by cos ? plus 0 x ia plus ? 3 by 2 x ib – ? 3 by 2 x ic multiplied by sin ? is your resulting mmf. In this expression note that we have some current here multiplied by cos ? and some other current here multiplied by sin ? this therefore means that suppose I had an excitation that is acting along a axis and I have another excitation that is acting along the ? axis this current being ia – ib by 2 – ic by 2 that is here and this current being ? 3 by 2 ib – ? 3 by 2 ic then for that excitation being provided in a coil here and a coil here the mmf is the same and therefore we can say that this is equal to some ns x ia cos ? plus some ib sin 5 where ia is the current flowing here and ib is the current flowing there. So one can now regard the net mmf that is being produced as having been produced by 2 excitations alone something here and something here now the flow of current here is a combination of the current flowing in a b and c axis similarly the current flowing here is a similar combination of the 3 currents now we have looked at this total current being the total mmf being generated by the number of turns consisting in the a phase winding b phase winding and c phase winding. Now we can be a little more general and then say that you have i a as this current and i ? as this current we can now define a relationship saying i a i ? is equal to ia or rather we can write it as 1 – ½ – 0 root 3 by 2 and – root 3 by 2 multiplied by the vector ia ib ic this is what we have. So you can see that i a is nothing but ia – ib by 2 – ic by 2 that is what you have here and i ? is 0 times ia root 3 by 2 ib – root 3 by 2 ic and now you have this ns so multiply this by ns and we can in general have some other number of turns here which I will call as n a ? so whether you want to choose ns as equal to a ? or ns by n a ? as some other ratio could be decided on certain other issues which we will see as we go along. So the equation that is going to relate i a ib and ic has now been derived in this manner in order to ensure MMF invariance which is to say whether ia ib ic is going to flow or i a ib is going to flow in these two windings it does not make a difference as far as the net MMF is concerned that is the meaning of MMF invariance. So now we see that instead of considering these three variables you might as well consider these two variables however these two variables are not the real life variables they are fictitious variables we are only trying to imagine that a current i a and ib is flowing whereas actually the currents are that are flowing are ia ib and ic in the real three phase machine. However since this has only i a and ib number of variables is only two whereas here number of variables are three maybe we can try to recast the equation in terms of i a and ib solution might be simpler and then having found out i a and ib by solving the machine equations we might then try to get back i a ib and ic but you can at once see that it is you can go from ia ib ic to i a ib by these equations whereas you cannot go from i a ib to ia ib ic easily because this is not a square matrix if you have a square matrix then you can easily invert it and then you can get ia ib ic from i a ib but this is not square. So how does one solve the system solve this issue now you would know definitely you would have heard of something called symmetrical components the idea of symmetrical components is that any unbalanced three phase system of either voltages or flow of currents can then be resolved into a positive sequence component plus a negative sequence component and then a zero sequence component now you may remember what these three sequences are what one can show is if you look back on the subjects that you have learnt in order to find out what these things are you know that given va vb and vc you can determine v plus which is this and v minus which is a negative sequence component and v0 which is a zero sequence component or having been given v plus v minus and v0 you can also get va vb and vc these two are interchangeable if you know one you can find the other what can be shown is if you look at the equations deriving these expressions and if you look at the equations we have written here one can show that if you know v a and vb it is equivalent to knowing v plus and v minus and therefore if you want to determine a set which you can either do the transformation from here to here or you want to go back in order to make this description square the information that is missing is that of v0 so if you somehow include a zero sequence representation in these equations then we would hope to have a system where you can go from a bc to a beta or a beta to a bc and this representation if you remember how those are then derived in the symmetrical component expressions v0 is then written as one third of the phasor of va plus vb plus vc average so to say of va plus vb plus vc what we want is only a representation of the symmetrical component and therefore it need not necessarily be one third it could be any number k which is then used to multiply va, vb and vc and therefore if you want to have a system that is a square matrix one can therefore do it in this way so you have i a i beta we said that the missing information was that of i0 so you add an i0 component now the number of turns is still there to be found out that is then equal to the number of turns on the stator here you have one minus half minus half zero root 3 by 2 and minus root 3 by 2 if you want the exact i0 then you should have put 1 by 3, 1 by 3, 1 by 3 but that is not really necessary what we want is a representation of that so I could as well put kk and then this would be i a, i b and i c so with this description added with the description of representation of i0 added you now have a square this become square and therefore one can invert this so if you want if you have i a, i beta, i0 you can multiply by the inverse and then get i a, i b, i c but then we now have to determine what should be the value of k as I said we could fix k to be any value which is which gives us some ease of operation so what ease of operation will you look at now let us say you do want to find out i a, i b, i c from i a, i beta, then you would need to multiply this by the inverse of the matrix here if you find the inverse of this what you would find this for example let us say this matrix is m and you want to find the inverse of this matrix then what you would see is that m inverse is given by 2 by 3 times 1 minus half and then 0 root 3 by 2 minus root 3 by 2 and then what happens is you get 1 by 2k this is the inverse of the matrix that we have written down here you can try to evaluate the inverse and verify for yourself. Now one can see that this inverse except for these 3 numbers here this matrix and this matrix appear to be related as transpose of each other appear to be so except for this row and that if you really wanted to be so then what we have to do is if you can set k equal to 1 over 2k then this matrix and this matrix will really be m and m transpose not exactly except for this number 2 by 3 and therefore if k has to be equal to 1 over 2k what we are saying is k square is half and k equals therefore 1 by root 2 so if you put k equal to 1 by root 2 then what you have is these 3 numbers become 1 by root 2, 1 by root 2 and 1 by root 2 and here also this becomes 1 by root 2 and therefore what we can say is m inverse is equal to 2 by 3 times m transpose so this is a rather good advantage for us it is enough if we remember 1m and then the inverse is simply two thirds of m transpose. So the choice of k has enabled us to get this simplification this is fine next what do we need to do we now have this description n aß and ns completed right how to determine what is n aß and what is ns that is the next issue that we have to address so let us do that there are two approaches that can be taken in order to resolve that issue we are looking at some sort of equivalence between a winding that is placed on the aß axis bs axis and cs axis that means three windings which are there originally and we want to transform this to two windings which is aß and aß how do you establish the equivalence we have said that the first basis of establishing equivalence is that the mmf generated by the two should be the same only then these two are equivalent. Now other than that one can also look at the electrical active electrical power consumed by these two by the entire set here on the one hand you have these three which are going to consume electrical input power here you have only these two which are going to consume electrical input power so what one can say is that say case one is when the net power or the total power is unchanged which means the net power taken by aß, bß and cß together must now be equal to the net power drawn by aß and aß which we have and now of course there is one more variable 0 phase also or one can look at another case where we can say that per phase power is unchanged. So this situation is called power invariance this is non invariance so if we take these two how does the system of equations look like so let us consider that you have let us take the first case one if the net power which is drawn by the three phases must be equal to these two what we have is I transpose a, b, c that means I is a vector consisting of Ia, Ib, Ic transpose a, b, c multiplied by v, a, b, c this is a vector consisting of va, v, b, b, c remember that active power is given by va x Ia x vb x Ib x vc x Ic which is the same as I transpose a, b, c multiplied by va, b, c what we are saying is this is equal to I aß now the 0 is also added transpose multiplied by v aß 0 this is what we are saying and here we have already derived the relationship between I aß 0 and Ia, b, c what we have is Iaß 0 equals ns by n aß multiplied by this matrix m multiplied by the vector Ia, b, c so let us substitute for I aß 0 there so what you have is Ia, b, c transpose va, b, c equals ns by n aß multiplied by m into Ia, b, c transpose v aß 0 and this is nothing but ns by n aß into Ia, b, c transpose m transpose v aß 0 so what we find is that if the left hand side now has to be equal to the right hand side we need to define va, b, c as ns by n aß x m transpose v aß this is what we get now what we have already for I is that we have Ia, b, c is equal to n aß by ns into m inverse into I aß 0 I have just rewritten the expression that we already have in terms of Ia, b, c so it is n aß by n aß multiplied by m inverse so what we see from these two equations is the transformation of the equation which is going to relate va, b, c to v aß 0 appears to be a little different from the relation between Ia, b, c and Ia, b, c now we could retain it like this but it would be simpler if the relationships between v and I are identical which means that what we want is ns by n aß must be equal to n aß by ns multiplied by I am sorry this into m transpose is equal to this multiplied by m inverse so this must be equal to this if the two equations are to look identical that is the manner in which aß has to be transformed to a, b, c is the same for v and I then this part should be equal to this part. But we already know from our earlier expressions that we have derived m inverse is nothing but 2 by 3 times m transpose so this can be written as n aß by ns into 2 by 3 times m transpose and therefore what we get is that if this has to be true then ns square by n aß square should be equal to 2 by 3 or ns by n aß should be equal to root of 2 by 3. So if you choose ns by n aß that is the number of turns on the aß winding here and of course the number of turns in the aß winding is the same as the number of turns on this winding these two are identical similarly the turns on a, b, c are identical the ratio of those two turns is what we have derived if it is equal to the square root of 2 by 3 then what we are saying is that the net input power is invariant it is the same in the a, b, c system as well as in the aß system. So this is one approach another approach is as we have seen the second case that is what we are saying is case 2 is i a, b, c transpose multiplied by v a, b, c this is the three phase system and per phase active power is this divided by 3. If we look at the aß system i aß 0 transpose into v aß 0 by 2. Now why are we saying this is 2 in spite of having written aß and 0 reason is we know that if we have a system of ac voltages which are balanced then the 0 sequence part will go to 0 and by enlarge we deal with electrical systems which are of this nature and in most cases therefore 0 sequence will go to 0. And therefore the active electrical input power is then consumed only by the aß and b and not by this part which is this and therefore we consider this to be having only two phases. So if we have this and then we substitute the expression for i aß 0 again as before what we have is i aß 0 is ns by n aß. So ns by n aß multiplied by m into i a, b, c transpose multiplied by v aß 0 into 1 by 2 which is the same as ns by n aß multiplied by i a, b, c transpose m transpose v aß 0 multiplied by half and this has to be equal to i a, b, c transpose v a, b, c by 3 and looking at this equation we can therefore conclude that these two expressions will be identical if v a, b, c is equal to 3 by 2 times ns by n aß multiplied by m transpose into v aß 0 and we already have a relationship for i a, b, c, i a, b, c is n aß by ns into m inverse into i aß 0. So again looking at the same argument that we would like the relationship to transform v aß 0 to a, b, c to be the same as the relationship that transforms i aß 0 to i a, b, c we would need to have this part of the expression and this part of the expression to be the same and therefore we can derive another condition. So what we want is 3 by 2 times ns by n aß multiplied by m transpose should be equal to n aß by ns into m inverse and we know that m inverse is two thirds of m transpose so 2 by 3 times m transpose therefore what you have is these two terms will be equal if ns square by n aß square is equal to 4 by 9 which means that ns by n aß is equal to 2 by 3. So on the one hand if you want to have invariance of the electrical input power you need to choose ns by n aß to be root 2 by 3 if you want to have input power per phase to be equal then you need to choose ns by n aß to be just 2 by 3 and what is the implication of these two so with this then we can now try to find out what is the relationship between v and i. So let us look at v a, b, c v a, b, c in the second case is 3 by 2 times ns by n aß is 2 by 3 times m transpose into v aß 0 which is nothing but m transpose into v aß 0 similarly obviously i a, b, c will also be equal to m transpose into i aß 0 because we have intentionally said that the equations transforming v and i should be the same one can now look at the inverse also then v aß 0 is equal to m transpose inverse into v a, b, c which is then m transpose inverse is m inverse transpose into v a, b, c and m inverse is nothing but 2 by 3 m transpose into v a, b, c so this is 2 by 3 times m into v a, b, c. So you get this extra factor 2 by 3 as compared to what you had here similarly you will have an expression for i aß 0 also. So in the next lecture we will see how the equations relating v a, b, c and v aß 0 for the invariant case are going to simplify with the choice of ns and n aß 0. We will stop here for this lecture.