 Welcome to the next lecture on modeling and analysis of machines. We had been looking at the equations for induction machine, induction machine model to be more specific what we are looking at is a three phase induction machine model. We started with the three phase machine and then we transformed the machine to stationary coordinates that is the a ? ? 0 coordinates and then we said that there are benefits to transforming the equations further to a set of rotating reference frames and therefore we went on to the rotating coordinates which is the dq axis and these rotating coordinates we said could be at arbitrary speed whereby you get a general machine equation which by appropriately substituting the value of speed the normal values of speed that are of interest in studying the induction machine are ? x equal to 0 in which case it goes to the stationary frame that means the rotating axis have a speed of 0 the speed of 0 that means that the frame does not rotate it is fixed and therefore it is a stationary frame we could fix ? x equal to ? s that is the synchronous speed in which case it becomes the synchronous reference frame or we could have ? x equal to ? r that is the rotor speed in which case it would be a rotor attached frame of reference. And then in the last lecture we had seen how the synchronous reference frame description can be made use of in order to orient the reference frame itself orientation of reference frame what ? x equal to ? s says is that the frame should rotate at synchronous speed but it does not really say where the d axis must be and we had seen in the last class that last lecture that the axis of the reference frame may be oriented along a particular space vector for example we had selected in the last lecture the reference frame where the d axis of the rotating reference frame is oriented along the rotor flux vector and that resulted in certain simplifications what we had seen and therefore from then on we had said that this is the basis of what is called as field oriented control of machines. So you have field oriented control it is more of reference frame orientation rather than field orientation you choose a reference frame such that it is oriented along the field either the rotor field or the stator field or maybe some other field and then that results in certain simplifications as far as motor control is concerned this is what we had seen in the last lecture but the machine equations themselves the way we had written it was let us write it down once again we had Vds and then Vqs we will neglect the zero sequence part so we leave that out and then you have Vdr and Vqr that is the stator vector matrix stator voltage vector this is equal to Rs plus Ls into P and then minus ?x into Ls and then here you have ?x into Ls this is Lsp then you have Lm into P Lm into ?x minus Lm into ?x Lm P and here you have Lm into P magnetizing inductance and this is magnetizing inductance into the difference of speeds the rotor impedance drop and then this Pdmf P here you have Lr into ?x minus ?r Rr plus Lr P you have Ids Iqs I this is then the set of machine equations that we were dealing with neglecting zero sequence that is there so this equation these set of equations neglects zero sequence terms and is written for arbitrary speed of the reference frame now one can see that the equation as is written is fairly big and there are a large number of individual terms and is fairly difficult to remember and to write now if you look back we had introduced the idea of space vector the space vector is if you look at the a beta axis so you have the beta axis here and the a axis here we said that if you are looking at for example the applied AC supply AC voltage to the machine you would have a V a and V beta that is applied and if you have the V a component here and the V a component here one can represent this as a space phasor which is the voltage phase phasor this is let us say the voltage phase phasor similarly you have a flow of I a and the in the a axis and you have I beta flowing on the beta axis coil those two together we can then form a space vector for the current and this space vector can then be written in a concise form as the space vector is nothing but V a plus j times V beta similarly the space vector of the current I can be written as I a plus j times I beta so instead of now representing four terms V a V beta I a and I beta individually one can sort of combine them into a more concise notation that is the space vector notation as space vector of V and space vector of I in the induction machine one can define similar space vectors you can define space vector for the stator voltage you can define space vector for the rotor voltage you can define space vector for stator currents or the rotor currents and in addition you can define space vectors for the stator flux space vector for the rotor flux all those can then be described in the last lecture we saw how one can substitute for these two terms idr and iqr in terms of this of the ?r terms idr and ?qr. Now what we can do instead is to combine all of them into a concise notation using space vector terms which is sometimes useful for understanding the dynamics of the machine and many times in literature you would come across the machine description written in the space vector form. So it is useful to have a look at how that can be done before proceeding ahead so that is what we are trying to do now. So for that even though we have drawn the drawing here in the aß axis one must remember that the space vector of voltage v or i is not dependent really on the frame of reference you are applying a set of three phase AC supply va, vb and vc and as a result of that the flux inside the machine is going to rotate at synchronous speed and this is that flux vector which is rotating at synchronous speed. Now if you have a flux vector you can describe it by the two terms va and vb if this is the axis that you want to choose or you may choose for example another set of axis let me draw the other set of axis as this may be for example your d axis and q axis this axis may be rotating. In which case now you had a v a term as from this vector if you then take the a axis part this is your v a but on this axis you will have to take a component here this then becomes your length vd and similarly this component becomes your length vq so depending on the axis you will then have vd and vq where depending on where they are located what we had again seen in the last class was if you choose a reference frame such that it is oriented directly along the voltage vector then vq would be 0 there is no component along this axis whereas the axis on this if you call this as d then vd is equal to the space vector itself no component is there along the q axis. So in this manner depending on the axis that you choose the individual two terms will change but however what is important to remember is that the space vector itself is not something that is depending on the axis the length of the space vector and the location in with respect to the stationary frame. So one can then reduce this description to the space vector notation how does one do that to this vector though we have written it as v a plus j vb suppose you are now going to look at it in the dq reference frame can be written now as vd plus j times vq as well of course the d component is not the same as the a component the amplitude or the length of this vd part is not the same as this that is very clear this is your v a whereas this is vd obviously the lengths are different but however the space vector can still be written as vd plus j vq similarly this current can also be written as id plus j iq. So what one can do is from these equations you add this j times the second equation to the first and then you get the voltage space vector of the stator so the voltage space vector of the stator is equal to vds plus j times vq as which can then be written as rs plus ls into p into ids minus omega x ls into iqs plus lmp into idr minus lm omega x into iqr this is the first term vds and now we add j times the second equation which is here so that is j times rs plus lsp into iqs and then you have plus omega x ls into ids with j here and then you have plus j times lmp into iqr and then plus lm omega x j times into idr so that is what you will end up with now you see that we have already grouped similar looking terms together and therefore this can be simplified as rs plus ls into p this term is common multiplied by ids plus j times iqs and then similarly here you have plus j times omega x ls into ids plus j times iqs note that j multiplied by j is minus 1 so you have minus omega x ls into iqs that is what you have here the first term is accounting for this then plus lmp into idr plus j times iqr and then you have plus j lm omega x into idr plus j times iqr now we are in a much better position you have the voltage space vector here and this term is nothing but the space vector of i and therefore I can write this as rs plus lsp and here also you have the space vector of that current so plus j times omega x into ls into the space vector of the stator current and then you have the space vector of the rotor current that is lm into p plus j times lm into omega x multiplied by the space vector of the rotor current so this is a much more concise notation for representing the stator equation this is one part of it the next part belongs to the rotor so again what we do is take these two equations add j times the second equation to the first so this is stator so you have the rotor as rr plus lrp multiplied by idr and then minus lr omega x minus omega r multiplied by iqr and then you have lm p into ids and then minus ln omega x minus omega r into iqs so this is the first term vdr and then you add j times the second term that is rr plus lrp into iqr and plus lr into omega x minus omega r multiplied by idr this is j and then you have plus j times lm into p into iqs and then plus j times lm into omega x minus omega r into ids and so these two terms can then be combined to make use of the space vector notation so that is let us start from ids again so you have lm into omega x minus omega r into j times lm into omega x minus omega r into ids plus j times iqs and then plus lm p j times lm p into ids plus j times iqs no there is no j here just lm p and then you have rr plus lrp this term comes so you have rr plus lrp into idr plus j times iqr and then the last term here this is j times lr into omega x minus omega r into idr plus j times iqr so that is what we land up with and then let us write down the final result here so you have vr is then equal to j times lm into omega x minus omega r multiplied by the stator current vector plus lm p into the stator current vector and then you have rr plus lrp multiplied by the rotor current vector rotor space phasor and then you have j times lr into omega x minus omega r into the rotor space phasor rotor current space this can then be combined into two terms that is rr plus lr into p plus j times lr into omega x minus omega r multiplied by the rotor current space phasor and then you have lm p plus j times lm into omega x minus omega r multiplied by the stator current space so one can we have achieved a certain simplicity in the equations that are I think I had written the rotor current with a r in the superscript to denote the rotor so that is what we have now this is a form which is a little more convenient to handle because you do not have such an elaborate description you just have to deal with these two equations now frequently however one more simplification is done or maybe one can say it is a slightly different form where the this equation is written in terms of flux linkages so that also it is useful to have in mind and it results in a much more compact representation we will see how that happens we remember in the last lecture when we had discussed how idr and iqr were replaced by means of psi dr and psi qr we defined certain relationships for the flux vector so we had said that psi dr is given by lr into idr plus lm into ids similarly psi qr was represented as lr into iqr plus lm into iqs now one can just like we had vds plus j vqs represent the space phasor of the voltages one can define the space phasor of the rotor flux vector to be psi dr plus j times psi qr which means that is nothing but lr into idr plus j times iqr plus lm into ids plus j times iqs which is nothing but lr into the space phasor of the rotor current and plus lm into the space phasor of the stator current so these two together now form your space vector space phasor of psi dr similarly we can define psi ds and psi qs to be ls into ids plus lm into iqs lm into idr so maybe I should write it down so the space psi ds is written as ls into ids plus lm into idr similarly psi qs is written as ls into iqs plus lm into iqr and therefore the space vector of psi s is written as psi ds plus j times psi qs which if you look at these two equations since you are adding j times this to the first equation you have ls into the space vector of is plus lm into the space phasor of ir. So if you now look at these two equations space phasor of the rotor voltage space phasor of the stator voltage so let us take the space phasor of the stator voltage first so we have vs and if you look at this here ls p into is is occurring here and similarly lm p into ir is occurring here and that is nothing but p of ls into is plus lm into ir which is nothing but the derivative of this and therefore we can write this as p times psi s so we have sort of accounted for this term and this term and then you have the stator resistance into is that must come so the stator resistance into the stator space phasor and then you have here ls into is and lm into ir both multiplied by j omega x so you have plus j omega x into is suppose you are looking at the description of the machine in the stator attached reference frame that means reference frame is not rotating it is a 0 speed then omega x is 0 then this term is not there and this equation now looks like a very familiar equation whenever you apply a voltage to an inductance and a resistance the voltage is nothing but a resistive drop plus the change in the flux linkage remember our first equations that we wrote for a single coil and linear excited machines and all that this was a form of the expression this form is very intuitive to write also and is a very the notation is very concise. Similarly let us look at the second expression space vector space phasor of the rotor we are you now have similarly rr into the space phasor of ir that is there the term analogous to this expression and then you have lrp into ir and then lmp into is so lr into ir lm into is is nothing but the rotor flux vector and therefore you have p times ?r I think it to be consistent with our notation we were using this on top so we have accounted for this term we have accounted for this term we have accounted for this term and then this term is nothing but lr into ir lm into is so that is again the space phasor of the rotor flux and so this is nothing but j times ?x – ?r into the rotor space phasor. So we had a huge equation earlier and all that has been compressed into neat forms that are shown so this says that the applied vr consists of a resistive drop plus a differentiation due with respect to time of the flux linkage in the rotor and also a speed emf which arises due to rotation so this is again what we had seen earlier. Now this kind of notation as we have seen is very concise and can be used to analyze or represent induction machine much more simpler fashion of course if you want to do the control as we saw in the last lecture one has to orient the reference frame of appropriately and there one has to deal with the dq axis and therefore this has to be resolved into its individual equations. Now these dq representations whatever we have seen now the utility of this does not rest with the machine descriptions alone. Let us look at some other aspects of the machine representation of the dq representation. Now we know that in 3 phase systems you have when you connect a load there is a certain active power that is being given what we have done as a part of this analysis is we have converted this 3 phase system to a 2 phase system and therefore we have aß axis we also have the 0 axis and the 0 axis term reduces to 0 if is or maybe I should say 0 axis term is equal to 0 if the system is balanced then you deal only with aß and bt and many times 3 phase systems are balanced or we tend to analyze balance system and therefore one can look only at these two term and what do you have an a and bt so you have one coil here in the a axis another coil here on the b axis and therefore since you are supplying a voltage here and supplying a voltage here the power delivered is simply equal to v a into i a plus vß into iß now we have actually converted your 3 phase system to the 2 phase system if you go back use the inverse relationships and try to derive what would be this expression in terms of 3 phase variables you will find that this is equal to v a into i a plus v b into i b plus vc into ic this is the familiar expression for the active electrical input to a 3 phase system and this is active power delivered and similarly people have defined the reactive power delivered as being equal to vß into i a minus v a into iß this is the definition people have defined it like this and this definition has found usage in many applications notably reactive power compensation applications reactive power compensation active filters and so on one can of course try to substitute the inverse going from here to a b c and try to find out what is the resulting expression in terms of the a b c variable but normally this computation is done after transforming a b c variables to the aß reference in which case these kind of expression would be sufficient to determine the behavior of the 3 phase system going further what can also be done is you can transform these variables to the synchronous reference frame which is the dq reference frame and there again instead of the aß axis you have d and q axis and therefore the active power delivered in the dq reference frame will then be equal to so the active power delivered in the dq frame is then equal to vd into id plus vq into ic and similarly if you want to define the reactive power then you have vq into id minus vd into ic this is your reactive power now because the synchronous reference frame is a rotating reference frame and you can position your d axis wherever you need you can choose to have your d axis aligned with either the space vector of voltage or the space vector of y or you can choose to have the q axis aligned with them in which case suppose let us say for example you are aligning the d axis of the reference frame along the space phasor of v then vq becomes 0 in which case active power is simply vd into id and if this is the case vq is 0 that means the reactive power would then be minus vd into ic so one can now deal with much simpler expressions and if you are going to control id then it directly results in controlling how much active power you want to deliver and if you control ic it amounts to controlling how much of hours you want to give so this is an idea that is used in the case of active filters that is widely seen in the literature today. So the use of dq references goes beyond the normal machine analysis alone and it has influence other fields of study as well now we have seen a large amount of material on the induction machines and we should try to move on to the synchronous machines to see how those can be represented by suitable equations now the synchronous machine differs from the induction in one main aspect that is the rotor of the machine now the synchronous machine the stator of the synchronous machine is very much similar to that of an induction machine where it has a three phase distributed winding on the stator so there is not much of difference between an induction machine stator and the synchronous machine stator in the idea but as far as the rotor is concerned it is quite different in the induction machine the rotor has no independent source of excitation which is why it is called an induction machine in the first place the field that is generated by the stator because it is revolving induces a voltage in the rotor and because the rotor is short circuited the voltage induced allows flow of the rotor currents and thereby generates a torque whereas in the synchronous machine the rotor contains excitation that means it produces its own magnetic field and in large synchronous machines this is normally done by energizing the rotor with a DC excitation so the rotor produces a direct field it is not an alternating field and the rotor is of two types the rotor can either be of cylindrical type or it could be of salient pole type we need to see how these two look like in order to develop the idea a little further which I will show you by means of some images so here we have some images of the rotor of a synchronous machine so this rotor this type of rotor it is a not a real synchronous machine rotor but this is a model to show how the salient features the important aspects of a cylindrical rotor of a synchronous machine will look like so the idea is that if you see the surface of the synchronous machine rotor see the surface of the synchronous machine rotor is about cylindrical there are no major differences no major ups and downs in the cylindrical machine in the rotor surface and therefore this rotor is called as a cylindrical rotor note that the surface of the rotor is not perfectly smooth there are small insertions here those are the slots on the surface of the cylindrical rotor machine slots are needed in order to hold field coils you need to put the field coils somewhere the field coils are the ones that are going to generate the magnetic field and they have to be housed somewhere so you need proper housing mechanisms housing means and therefore you have slots on the rotor so this is a four pole cylindrical rotor and how is the field winding done so let us see how the field winding is going to be done you have a slot here and a slot here so you have a field turn that is going to go like this into the turn slot here and you may have multiple turns that make up a single field coil and that in turn will be connected in series with the next field coil which will occupy the two slots here so that makes up one field coil and let us say that there is this will obviously flow along the that is this will go along the length of the rotor and then make its loop on the other side of the rotor and come back that is the idea of the coil and let us say you are going to have flow of current in the coil in this manner so you are having currents that are flowing in and that means the field generated by the coil will be downwards and therefore this part of the rotor surface will then generate a south pole field and the next side this has to generate a north pole field so here you have coils that are going to be laid out like this of course they will come all the way from here and then go like this along the length of the rotor and complete the turns there and then you have the next coil coming from here and then going this way and you want this to generate a north pole field that means the lines of field must be going out of the surface of the rotor that means you need to have currents that are flowing like this in the rotor similarly you will have a coil here several turns and similarly one more set of coils here and coils that are placed here so the direction in which excitation is arranged would be such that you generate a north pole here a south pole here and a north pole here so you have a four pole system all these may be connected in series so that you ensure same current flows here or it may also be connected in different ways but the idea is the ampere turns in each pole is to be ensured to be the same that is how the machine would then work so you have a four pole structure whose surface is more or less cylindrical more or less cylindrical surface the only reason for this being non cylindrical in a sense is to have these slots which you cannot avoid now the rotor as is seen here is inserted into the stator bore of the machine and then it may rotate so in south the field that is generated will then be south pole field would come like this and then emanate out of the north pole and similarly it would go this way and emanate out of the north pole here similarly it would go like this and then you would have the field coming out this way out of the surface of the machine now when this rotor is going to rotate the field structure generated by this rotor bodily rotates along with the rotor so while in the induction machine you have a rotating field due to the stator that is excited by three phase in the synchronous machine you have the rotor which generates a DC field all these flow of currents in the rotor coils are direct currents there is no alternating current that is given here and therefore the field structure generated with respect to the rotor is stationary always this pole is the south pole always this pole is the north pole and so on and this bodily rotates at synchronous speed that is how the system is arranged we will stop here now and then in the next lecture we will look at the other variety of cylinder of the synchronous machine rotor that is a salient pole rotor and how that is geometrically designed and how it looks like before we go on to modeling the machine at this we will stop here for now.