 In this class let us now continue with the equations that we started developing in the last lecture. We started developing equations for the three phase induction machine. We began with equations like VA equals VAS equals RS into IAS plus V times IAS. We had six such equations like this and then subsequently we expanded expressions for IAS in terms of IAS which is the phase current flowing through the phase A as well as the other currents IBS, ICS and then the three currents in the rotor and so on. And we saw how the first row of that expression will look like. Similarly one has to fill up all the other five rows of the system description. We can attempt to write these equations in a simplified notational form where we write V as the vector. Let me underline that to denote that it is a vector equals R matrix which is the resistance matrix multiplied by the current vector multiplied by P times that is d by dt of an inductance matrix multiplied by the I vector. This is a simplified notational form of the equation where the matrix R is a square matrix. Remember each row of this describes an equation like this and let us now write these different vectors and matrices. So you have RS and then 0, 0, 0, 0, 0 that means this corresponds to the AS phase and then BS, PS, AR, BR and PR. And being a balanced machine these first three rows would then look like this. The next three rows correspond to the resistance of the rotor equations of the rotor and therefore they would be RR and 0 and again 0, 0, 0, RR, 0, RR. So this would be the resistance matrix. This is multiplied by the vector I which means IAS, IBS, ICS, IAR, IBR, ICR. So you get RR into IAR for the first row RS, I mean RS into IAS, RS into IBS for the second row and so on. The inductance matrix which is L could be written as L11, M12, M12 transpose and L22 where each is a small sub matrix by itself where we can write L11 as LLS plus LMS. This is the leakage inductance per phase and this is the magnetizing inductance and then this is the mutual inductance between the stator phases MS again and here there is MS, LLS plus MS plus LMS and this is MS, MS here, MS here, LLS plus LMS. This is the self inductance matrix. So if you write the I vector here, this matrix is going to multiply the vector corresponding to IAS, IBS and ICS. So one can call this as the self inductance matrix of the stator. L22 is then similarly described, this pertains to the inductances of the rotor and therefore what you would have is LLR plus LMR and then MR, MR, MR here again LLR plus LMR, MR, LLR plus LMR. So this is the matrix that is going to multiply IAR, IBR and ICR and the form is very similar to that of this matrix. M12 on the other hand describes the mutual flux linkage between the rotor currents and the stator flux and at this point it is between the rotor flux caused due to stator current and therefore M12 is going to describe mutual inductance behavior and that we had seen the first row yesterday, I mean the last class. So that is MSR into cos of ?R and then you have MSR into cos of ?R minus 2p by 3 MSR cos of ?R plus 2p by 3 and then you have MSR cos of ?R plus 2p by 3 MSR cos of ?R that is MSR cos of ?R minus 2p by 3 this would be MSR cos of ?R plus minus 2p by 3 MSR cos ?R plus 2p by 3 and that is MSR cos of ?R. So this is the form that the M12 sub matrix is going to take and what occurs here is the transpose of this M12 matrix because whatever term comes here would come in the first column and so on. So one can write down this elaborate description of the induction machine where as what we had seen earlier the stator phase axis are AS, DS and CS this describes the spatial orientation of the stator axis and the rotor axis are displaced from the stator axis by an angle ?R that is the rotor axis and this angle is ?R and therefore you have BR here and CR here so all these angles are ?R. So this is how the machine description would look like this is called as the natural reference frame description of the induction machine natural reference frame that means the variables that are associated with it that is the variables I and V are the naturally occurring variables in the induction machine this is for example the actual voltage applied to the stator a phase and this is the actual flow of current in the stator a phase similarly in the rotor equation you would have the actual voltage applied to the rotor and current flowing in the rotor and so on. So since they are the natural variables it is then called as the natural reference frame description as before this is the electrical aspect of the induction machine that is the electrical subsystem of the induction machine which then says if you are going to apply a voltage VAS or VBS VCS and so on how much current would flow but in order to arrive at that you need to know what is the angle that is involved and the angle will in turn depend upon where the rotor is and therefore how the rotor is going to move. Therefore in order to complete the description you need to describe the mechanical subsystem as well and the link between the electrical subsystem and the mechanical subsystem is provided by the torque that is generated and therefore you need an expression for the torque as we have seen earlier can be described as half of I transpose dl by d ? x I where L is the matrix L as is given here now if you evaluate this expression it will come to a fairly long lengthy expression one can do that if it is required but as of now we do not need that expression we will proceed further this expression for the torque is then for what is called as a two pole induction machine the expressions we have so far obtained all these expressions assume implicitly that the stator magnetic field and the rotor magnetic field are two pole fields if you remember we had described the induction machine stator like this assume it is a circular stator and then we had the a phase of the stator here and the return conductors here and then you had the next phase which was here and then the third phase that is sitting here if the machine is wound indeed like this then what would happen is around if you start from one point and then go around and complete the circle you would encounter exactly one cycle of variation of the MMF which means that this MMF is going to generate a magnetic field that has one north pole and one south pole and therefore this description whatever we have written is for a two pole machine how does this description change if one were to consider a machine with more number of poles if there are more number of poles then in the same 360 degrees of course more poles as we created which means that these would then become shorter and then you would have for example R phase and then a Y and then a B and then again this R Y and B so this completes one MMF from zero degrees it would come back to 360 degree this from the start of this to the end of this will be one cycle of the MMF waveform and you want to have one more cycle of the MMF waveform so you would have R again and then Y again then B and then R Y and B so the stator has to be rewound or redesigned if you want more number of stator poles but however we know that whatever is going to happen between the start of one cycle of the MMF waveform and the end of one cycle of the MMF waveform the same thing is going to happen this side also there is no difference between what is going what the machine is going to see here and what the machine is going to see here therefore it may be sufficient if you analyze what is going to happen here and then therefore it means that it is enough if you analyze one cycle of the MMF waveform it is not necessary to look at the other similar repeats of the MMF waveform here you have the first cycle and here you are having the second cycle if you are going to have more poles then you need to shorten this then you will have one two three and four pole structure and so on and therefore since it is sufficient to look at one cycle of the space enclosed by one cycle of the MMF waveform what we analyze is then a two pole equivalent the behavior will be the same the only difference now is that an angle of 180 degrees is equivalent to 360 degrees of the MMF waveform and therefore whatever angle we consider in these equations if we now let that to be the electrical angle instead of the mechanical angle the same description holds good for the induction machine having any number of separate poles so one can have a two pole machine or a four pole machine or any number of poles if it is two pole then ?r is equal to ?mechanical if it is four pole then ?r is equal to two times ?mechanical that can be seen from here and if it is six pole then the electrical angle is equal to three times ?mechanical in general the ?r electrical which is what we are going to substitute in these expressions can be written as the number of poles divided by two into the mechanical angle so the induction machine description can then be written like this what would happen to the expression for the generated electromagnetic torque the generated electromagnetic torque if it is going to be only a machine having two poles you would get an expression like this but if the number of poles increases then what is going to happen in the second set of poles is going to add the force is generated on this side is going to add to the force is generated by the first pole pair and therefore the torque generated would correspondingly double if you are going to have one more pole pair the torque generated would become three times and therefore in general what is written is P by two times where P is the total number of poles P by two into this would then give you the total generated electromagnetic torque so once this is known then you can write the expression for the mechanical subsystem then the moment of inertia multiplied by jd that is multiplied by d ? by dt is then the generated torque – whatever load torque is there load torque could again be dependent upon speed in some form this expression then describes the behavior of the mechanical subsystem note that what we put a speed here is the actual speed of the machine actual speed of the rotor which we can call it as ? mechanical if you take the derivative of this angle which this will then be the derivative will be the electrical speed that differs from the mechanical speed by the number of pole pairs so accordingly if you now want to compute the angle from this ?m is nothing but d ? mechanical by dt and in order to get in order to substitute this angle into that you need to multiply it by the number of pole pair so if one takes care to do that then the description that we have now holds good for a machine with any number of pole appropriately it could be taken care so this is the natural reference frame description for the induction machine now let us look at how the description will look like for a synchronous machine so a synchronous machine description and in this let us look at the salient pole synchronous machine this is a more general case because here we are as we have seen in the earlier derivations of the mmf inductances and all that the salient pole synchronous machine has a non uniform air gap and hence is described by two inductances one along the direct axis and quadrature axis called Ld and Lq if it is going to be a cylindrical rotor induction machine then the air gap is uniform and we can simply arrive at that description from this by saying that Ld is equal to Lq and therefore let us look at the salient pole machine the salient pole machine again has a stator and the rotor is salient pole so let us assume that the rotor is at this position and as before we assume that the a phase of the stator is here the return conductors are here and then the next phase of the stator comes here and the return conductors are here and then the C phase of the stator that is the blue phase comes here and then it is return conductors are here so this is the same description that we followed earlier now the rotor has an angle the rotor axis is situated there and that makes an angle ?r with respect to the stator a phase axis so this is ?r now on the rotor let us assume that two sources of excitation are there one which will generate an excitation along this axis that is axis of the rotor and another coil which generates an excitation along this axis which is 90 degrees to that of the rotor so with this system we need to now start writing the system equations and we start as before we did as in a similar manner to what we did for the induction machine the starting point is always VAS is equal to RS x IAS plus P times ?AS this is for the stator a phase and similarly you get an expression for VBS RS x IBS plus P times ?BS then VCS is RS x ICS plus P times ICS these are the three windings on the stator and now you have two more windings on the rotor which one generating an MMF along the pole phase axis another generating an MMF perpendicular to the pole phase axis we will call them as windings F1 and F2 so VF1 is then RF1 multiplied by IF1 plus P times ?F1 VF2 is RF2 times IF2 plus P times ?F2 so this is an equation set similar to what we wrote for the induction machine now there are one or two points to note in the equations that we have written down these equations assume that if you have a coil on the stator which is going to generate an MMF along the AS axis so let us say this is a coil on the stator that is your AS axis and VS and CS this means that there is a coil on the stator generating an MMF along the AS axis the actual location of the coil would be as we have seen earlier displaced around the circumference so this expression VS is valid for a voltage that is applied to this coil plus minus having shown like this and the flow of current considered into the terminal mark plus and similarly if you have a BS coil the voltage is plus minus here and the current is flowing into the terminal mark plus and the dot points of the coils are such that it is dot at both places where current is flowing in similarly you have the C phase coil current flowing in and plus minus this way this is this convention holds good for the induction machine as well now what this convention means is that we have assumed that the reference direction of current is one that flows into the terminal mark plus and therefore the equations we have written down follow the motor convention so what we are developing essentially are equations for a synchronous motor in this case and the equation that we have derived earlier is for an induction motor so both are for motors let us have that in mind and if it is going to be a generator what we are going to do is only say that current is flowing out therefore the currents in the stator become negative if it is a generator let us not discuss more about that now for the present let us go ahead with developing these equations so the next thing as we have done for the induction machine is to write down expressions for the flux linkages let us start with ?as is again going to have a self flux linkage self flux part and a mutual flux part the self flux part arises from the current flowing in itself that is is and the mutual flux part is due to the currents flowing everywhere else that means IBS ICS and that the two currents in the rotor IF1 and IF2 so how do we write down the expression for self flux that is fairly easy this is as we have seen the flux linkage is inductance multiplied by the flow of current and for a geometry like this which has a cylindrical stator and a salient pole rotor we have already derived the expressions for the self inductance and we know that the self inductance LAS for example is given by a leakage inductance plus Ld plus Lq by 2 plus Ld minus Lq by 2 into cos of 2 times ?r this is an expression which we have derived earlier so the flux linkage arising due to that is this self inductance multiplied by the current IAS we need to then look at the mutual flux and let us consider the mutual flux arising between AS winding mutual flux arising in AS winding due to the current flowing in BS winding again we have derived the expression for the mutual inductance between two windings on the stator with their axis displaced by an angle ? with the arrangement of salient pole machine and we have found that that expression is Ld into cos of ?r cos of ? minus ?r minus Lq into sin ?r into sin of ? minus ?r this is again an expression that we have derived so let us try to simplify this expression this can be written as Ld into cos ?r multiplied by cos of ? minus ?r can be written as cos ? cos ?r plus sin ? sin ?r minus Lq into sin ?r multiplied by sin ? cos ?r minus cos ? sin ?r so expand this further so this gives us Ld cos ? cos ? and here you get Lq sin ?r cos ? so this is one part and then you have Ld cos ?r sin ?r sin ? we can take this out and then you have minus Lq sin ?r cos ?r multiplied by sin ? so this can be written as Ld cos ?r is 1 plus cos 2 ?r by 2 plus Lq into 1 minus cos 2 ?r by 2 this whole thing multiplied by cos ? in both places so that is cos ? and then you have here it is cos ?r into sin ?r so that sin 2 ?r so Ld sin 2 ?r by 2 minus Lq sin 2 ?r by 2 multiplied by sin ? so let us look at this expression further so that expression can be further written as Ld plus Lq by 2 plus Ld minus Lq by 2 into cos 2 ?r multiplied by cos ? here you have cos ? and then you have Ld minus Lq by 2 into sin 2 ?r multiplied by sin ? so that becomes Ld plus Lq by 2 cos ? plus Ld minus Lq by 2 into cos 2 ?r cos ? plus sin 2 ?r sin ? that is nothing but cos of ? minus 2 ?r so this is the expression that we get for the mutual induct and in the case of the synchronous machine of a balance synchronous machine three phase synchronous machine we know that ? is equal to 120 degrees between the A and B phases so if we call LAB as the mutual inductance between the A phase and the B phase of the stator that can then be written as Ld plus Lq by 2 into cos of 120 degrees is minus half so minus half plus Ld minus Lq by 2 into cos of 2 ?r minus 2 ? by 3 ? is 120 degrees and minus half so this can be written as minus 1 by 4 times this expression so this is between A and B. If we consider A and C so let us write LAC then what happens is ? which is the angular separation between the axis of the A phase and C phase now becomes 240 degrees and therefore you have Ld plus Lq by 2 into cos of 240 degrees plus Ld minus Lq by 2 into cos of 2 ?r minus 240 we can write it as 2 ?r minus 4 ? by 3 LAB we know must be the same as Lba as well that means mutual inductance between B and A phase that is A excited flux linkage in B and similarly LAC must be the same as Lca as well what remains is Lbc to find out Lbc we can use the same expressions but we must understand how this what is the geometry for which this expression has been derived we have we can see that by looking at this expression let us remove the rotor for the time being we are looking at the stator flux linkages. So if you remember what we have done to derive this expression for the mutual flux linkage is that we have considered that a winding is there along the axis 1 and then the rotor is situated at an angle ?r with respect to the first winding and a second winding is located whose axis is located at an angle of ? this was the geometry for which we have derived the expression for mutual inductance. Now this figure as such can be used to determine the expression for the mutual inductance between this A phase and that of the B phase the rotor makes an angle ?r with respect to that of the A phase the C phase on the other hand is here this is BS and this is CS we now want to derive an expression for mutual inductance between B and C. Now that would mean that if you want to use the same expressions what we have derived those expressions stipulated that the rotor angle that is considered is the angle made between the axis of the first coil and that of the rotor. Now the first coil is BS and the rotor is sitting here and therefore the angle we need to consider is now this angle and the separation between the two axis is still the same ? 120 degrees and therefore the expression for mutual inductance would arise from whatever we have derived if we substitute instead of ?r this angle and that angle is nothing but 2 ? by 3 – ?r but the rotor is now on the other side of the axis and therefore this is a negative angle. So if we substitute instead of ?r this expression we should get the expression for mutual inductance between the B and C phases and therefore this can be written as Ld plus Lq by 2 x cos of 120 degrees ? is still 120 degrees plus Ld – Lq by 2 x cos of ? instead of ?r we are going to substitute that therefore – 2 x 2 ? by 3 – ?r and because that is negative this now becomes plus and therefore the second part of the term becomes Ld – Lq by 2 x cos of 2 ? by 3 – 4 ? by 3 – 2 ?r. So this is simply Ld plus Ld – Lq by 2 x cos of 2 ?r this term remains as it is and cos of 120 is again – ½ therefore you get – of Ld plus Lq by 4 plus Ld – Lq by 2 x cos of 2 ?r. So with this then we have derived expressions for the mutual flux linkage what you would have in the ?as expression then is ?as is then Las x Ias plus Lbs x Ibs sorry Lab x Ibs plus Lac x Ics. Las is the expression for mutual inductance that we have derived self-inductance that we have derived here Lbs Lblab is the expression for mutual inductance we have derived here Lac we have derived here and then there are 2 more terms one that would describe the mutual inductance between that of the a phase coil and the coil acting along the axis of the rotor that can be described as a maximum value of mutual inductance we will call it as MAF1 if these 2 axis are aligned then the mutual inductance of that those 2 coils would become the highest and as it goes out the mutual inductance will change as cos of ?r multiplied by the current flowing here which is that of If1 and then the mutual inductance between this winding and the winding which lies at an axis 90 degrees to that of ?r on the rotor. So if we call that as MAF2 then that is cos of ?r plus ? by 2 x If2 cos of ?r plus ? by 2 is minus sin ?r so we can write it as minus sin ?r into If2 so this would then be the expression for the flux linkage of the a phase in the stator the flux linkage for the b phase can be written as LAB into IAS plus LBS into IBS LBS being the self inductance of the b phase winding which will have the same form as that of LAS except that ?r will have to be replaced by a suitable angle and then you will have LBC times ICS the expression for LBC again we have derived which is of this nature and then you have a mutual inductance between the b phase of the stator and that of the rotor that would then again be MAF1 had the rotor been aligned with that of the b phase axis the inductance would have been maximum and since the a phase coil stator and the b phase winding of the stator are not really different in the number of turns the maximum value of mutual inductance would still remain the same and therefore this is still the same MAF1 multiplied by cos of ?r – 2p by 3 times If1 and then you would have MAF2 multiplied by MAF2 would be the mutual inductance between the b phase and the coil here this angle being 90 degree and therefore that would be cos of ?r – 2p by 3 that is this angle – 90 degrees so –p by 2 times If2 that can further be simplified and similarly one can write the expression for ?cs as well I will leave it to you as an exercise to derive the expression for ?cs for the rotor ?f1 if we write the flux linkage in the rotor would again have a self flux component and a mutual flux component the self flux component can be written as a leakage inductance of the F1 coil plus the magnetizing inductance of the F1 coil multiplied by If1 and then you have mutual flux components now if you consider mutual fluxes there are 3 coils on the stator and one more winding on the rotor as well but the rotor winding it has an axis 90 degrees with respect to that displaced by 90 degrees with respect to that of F1 and therefore the mutual flux arising due to that on the F1 coil will be 0 the stator being cylindrical the mutual flux linkage between these 2 is 0 and therefore 0 multiplied by If2 and then you have the mutual flux linkages between the stator windings and F1 and those are described by the same mutual inductances that we have written here whether it is between 8 flux linkage cos due to current in the rotor or rotor flux linkage cos due to current in the stator the mutual inductance term remains the same therefore similar mutual inductance term would then arise. If we look at the flux linkage in the coil the second coil of the rotor you would have no flux linkage due to current in the first coil on the rotor but there would be a self flux linkage component I mean there would be a self flux linkage component which is If2 and then you would have mutual flux linkage component similar terms arising due to Ias, Ibs and Ics. So having written all this one can now complete the expression for the electrical subsystem description of the synchronous motor once we write this then the expression for the generated electromagnetic torque again can be written as ½ of I transpose dl by dθr multiplied by I for a 2 pole machine and if you are going to have more number of poles that can be multiplied by the number of poles there so it is p by 2 times this expression. This equation again these set of equations as for the induction machine they are said to be in the so called natural reference frame because the variables are as they would be in a real life induction machine naturally found variable so it is called as natural reference frame. What we see from these two descriptions is that the description of the induction machine and the synchronous machine is fairly elaborate you have a large number of system variables the description is quite huge and the expression for the generated electromagnetic torque would also be an unwieldy expression consisting of many variables that are in that expression. Therefore we look at some ways to simplify this expression in order to make handling these expressions easier in order to make solving these expressions easier one looks at a different way a different approach to solving these set of equations. What is that we shall see in the next lectures to come for this lecture we will stop here.