 last lecture we had started with the modeling of synchronous machine three phase synchronous machine and we derived the various expressions for inductances in the salient rotor synchronous machine and it is now time to assemble all these equations to form the voltage expression or the voltage equations for the induction machine which we will write down now. What we are having is the stator of the synchronous machine and symbolically we represent the salient pole rotor in this fashion. So what we have is the axis for the A phase axis for the B phase and the axis for the C phase. So this is the A phase axis B phase and C phase axis and then we have the rotor axis. So the rotor axis is here and this angle is ?r. So that was the system that we had considered. So if you take the A phase axis we symbolically represent the coil or the winding of the A phase in this manner winding is actually on the stator and we represent it like this to denote that the axis of the A phase winding is here and you have a coil there. What we are going to write is the expression relating to for example VAS is then RS into IAS plus P times ?AS and similarly you have all equations for the stator VBS and VCS that is RS into ICS plus P times ?CS that is on the stator and then we need to move over to the rotor windings on the rotor you have just one winding that is the field winding of the rotor and the field winding is there only along the D axis on the Q axis there is no excitation and there is no winding there. So the D axis or the field winding is present only along the D axis so that is then the field winding representation and therefore you would have VF equals RF since the field winding is on the rotor we will follow the convention that we have that the superscript variables denote the rotor and therefore IF plus P times ?F so these are fundamentally the equations which we now have to elaborate what is essential to note is that the way we have written the equations if you have an applied voltage VAS here this equation implicitly means that we are taking the flow of IAS into the winding here and therefore we are treating the machine as if it is a synchronous motor we are following motor conventions that means if you apply a voltage VAS the direction of IAS is assumed to be into the terminal which is marked as plus normally large power synchronous machines are operated in the generator mode you know that all the AC that you are getting all the supply that you are getting is from synchronous generators located at some place and all of them interconnected on a grid. So when you are operating them as a synchronous generator it is usual to take the direction of IAS to be flowing out of the terminal which is marked as plus and therefore if you want to convert these equations to the normal generator convention you will then have to say IAS is reversed and therefore it would be – RAS x IAS – RAS RCS whereas field winding is always it is an applied voltage on the field and field draws supply current from the energization that is given and therefore the field winding is always into the terminal mark plus so this expression normally follows this convention motor convention whereas IAS, IBS and ICS would follow the generator convention. But in order that we maintain the sense that we have used all through we will continue to use this convention with the understanding that if we want to study this as a generator and use the generator convention we will replace all variables of the stator that is IAS, IBS and ICS with the negative sign. So with that understanding we will proceed ahead again in these expressions you remember that in the last lecture we saw an image of a salient pole machine of the rotor in a salient pole machine and we saw that the rotor pole phase in a salient pole machine had a structure like this and on the pole shoe you had poles through which we said that there would be bars that are introduced and these bars are shorted by a ring on either side one ring on either side and we noted that these bars are called as damper bars and this is then called as a damper cage. Now if these are there then what can happen is whenever the rotor falls out of synchronism there would be induced voltage in these bars as long as the rotor rotates at synchronous speed the field in the air gap is also rotating at synchronous speed these bars are also rotating at synchronous speed therefore there is no relative movement between the bars and the field they are stationary with respect to each other and therefore there is no induced voltage in these bars. If however there is a change in the speed of the rotor it deviates from the synchronous speed then there will be induced voltages in this and therefore if there are voltages and you have a shorting ring on either side you can have some flow of current in these bars there would be currents that are flowing these currents would then distribute in some manner on the end rings both here and here and that in turn would cause its own mm. In order to model this what is done is sometimes you take the q axis of the rotor as well and then you introduce windings on the d axis of the rotor and on the q axis of the rotor. Since the damper bars are shorted by end rings on either side these windings are considered to be shorted and now you see that there can be an mmf along the q axis as well and along the d axis as well therefore there is at least one coil on the q axis and one coil on the d axis. Now if one wants more accurate results of simulation perhaps one need to consider more number of shorted representations on the d axis and on the q axis but in general people have found that sufficient accuracy of simulation is obtained with one coil on the d and q axis. Of course these bars have no effect when the machine is operating in steady state since at steady state the rotor speed is the synchronous speed and therefore these will have no induced mmf in them and therefore it is as good as being open circuited for the present however we will not consider the presence of these coils we will neglect this representation and then consider the mmf that is the electrical equations of the synchronous machine without looking at the damper. So that is the environment in which we are going to look at the synchronous machine equations. So let us look at the set of equations you have the voltage vector therefore VAS, VBS, and VF this is then equal to the resistance drop so RS and then 0, 0, 0, 0 RS, 0, 0, 0, 0, 0 RS these are for the three stator windings and then you have the field winding so RF is multiplied by IAS, IBS, ICS, and IF plus you now have P times ?AS so that is P times ?AS is nothing but as we had seen yesterday self inductance of the A phase multiplied by the flow of current in the A phase mutual inductance and so on so those expressions are going to be pretty long so we will write that in the adjacent portion of the board. So here you are going to have IAS, IBS, ICS, and IF so what is going to come here if we call this as the matrix L then L is described by the first term is the self inductance which comprises of a leakage inductance LL plus LD plus LQ by 2 plus LD minus LQ by 2 into cos of 2 ?R so this is one term and then you have the mutual inductance here and therefore the mutual inductance term is minus LD plus LQ by 4 minus LD minus LQ by 2 into cos of 2 ?R minus 120 degrees so this is one term this is the mutual inductance between the A phase and the B phase which we have derived yesterday and then we are going to have the mutual inductance between the next two phases that is minus LD plus LQ by 4 plus LD minus LQ by 2 into cos of 2 ?R minus 240 degrees so this is one term this is the mutual inductance between the A and C phases so let us try to simplify the notation otherwise the equation will look too big so we will simplified by saying LD plus LQ by 2 is written as LA and LD minus LQ by 2 is written as LB then what we have here is LA plus LB cos 2 ?R and therefore this equation can be written as minus LA by 2 plus LD cos 2 ?R minus 120 degrees we will follow another notation in order to make it more concise we will write ?1 equal to ?R minus 120 degrees and ?2 ?R minus 120 and ?2 as 2 ?R minus 240 degrees so that this then becomes cos of ?1 and then you have minus LA by 2 plus LB cos ?2 now we get little more space so this is the inductance self inductance of the A phase mutual between A and B mutual between A and C and then you need to write down the mutual between A and the field winding so the A and the field winding are separated by the rotor angle and therefore that is nothing but MSR x cos ?R so for the B phase the mutual between A and B is the same expression you have LA by 2 plus LB cos ?1 and then you have the self inductance of the B phase which is the leakage inductance plus the inductance magnetizing which is nothing but LA plus LB cos so you get ?2 here and then this one is minus LA by 2 plus LB cos 2 ?R and here you will have MSR x cos of 120 degrees minus ?R and then the third expression here is between C and A so minus LA by 2 plus LB cos ?2 and then between B and C this is LA by 2 plus LB cos 2 ?R C phase is LL plus LA by 2 plus LB cos ?1 and then you have MSR cos 240 degrees minus ?R and then the last row corresponds to the field and therefore you have between field and A it is MSR cos ?R and then MSR cos 120 minus ?R and then MSR cos 240 degrees minus ?R and now you have the leakage inductance of the field plus the magnetizing inductance of the field that is going to come here now note that the magnetizing inductance of the field is not something that is dependent on the rotor angle for reasons that we had discussed yesterday irrespective of where the rotor is the field always sees a constant air gap and therefore there is no dependency on the rotor angle. So this matrix is then the inductance matrix which is needs to be substituted in this place in order to form the full set of equations. So these equations are then in the natural reference frame that means the stator variables are in the stator reference rotor variables are in the rotor reference and the actual numbers have to be substituted. Now synchronous machine is rarely analyzed in this reference frame and the synchronous machine rotates at only one particular speed that is the synchronous speed and therefore the only frame of reference other than the stator attached natural reference frame may be which is of use is the synchronous reference frame there is no use in having the arbitrary speed reference frame is not of much use or any other frame of reference is also not used by enlarge it is only the synchronous reference frame that is used and therefore we need to in order to look at those equations we need to transform these equations into the synchronous reference and that if you see the field coil it is already in the synchronous reference frame because it is attached to the rotor which rotates only at the synchronous speed and therefore we really do not need to touch the rotor part the stator variables however have to be transformed from the stator attached free phase reference to the synchronous reference frame and therefore what we can do to transform the stator alone we already know how to do it. So if you are going to have a stator variable let us say IAS, IBS and ICS if this is going to be multiplied by the transformation matrix which is root of 2 x 3 x 1 – 0.5 – 0.5 and then 0 root 3 x 2 – root 3 x 2 1 over root 2 1 over root 2 and 1 over root 2 this operation transforms the 3 phase variables into the a beta or the 2 phase variable reference frame and then from there we will have to move to the synchronous reference frame and that transformation also we have done earlier that transformation to the synchronous reference frame having the d axis aligned with that of the rotor field q axis this way is then cos ? r sin ? r 0 – sin ? r cos ? r 0 0 0 and 1 so this is now going to give us IDS, IQS and I0S. So in order to convert ABC to DQ and 0 we need to derive the transformation matrix let us say C as this matrix multiplied by this matrix that will give us the total matrix C which one can use to transform the stator variables to the DQ reference frame but that is only the stator variables ABC but whereas this description is a 4 x 4 element here this has 4 columns and 4 rows so what do we do with the remaining one we do not need to do anything to the field and therefore the overall description that is used to transform the 4 x 4 synchronous machine description into the synchronously rotating reference frame will then be the matrix C here and then 0 1 and 0 so this row is then going to correspond to the field variable which we are not doing anything with and therefore that entry becomes 1. So if we do this this can easily be written as like you have V ABC is going to be transformed to the synchronous reference frame and the synchronous reference frame we let us say this matrix is then called as D matrix then what we need to do is the ABC variables is nothing but D transpose or D inverse x V DQ 0 and then DQ 0 here that is nothing but R x D transpose I DQ 0 DQ 0 plus P times this L matrix into D transpose x I DQ 0 and DQ 0 so this is what we are going to get by substituting the transformation matrix in this equation and then you have to multiply this by the D matrix in order to get an expression for V DQ 0 these are steps that we have already done and we do not need to repeat the entire algebra we will just write down the results the results of this operation will then be given by these equations. So you have V DS V QS and V 0S and VF this is RS 0 0 0 this matrix of course does not change being a diagonal matrix here you have I DS I QS I 0S and I F plus you have an inductance matrix which is L leakage inductance of the stator what we have written there as LL I am now putting it explicitly as LLS plus 3 by 2 times LD and then 0 here you have LLS plus 3 by 2 times LQ and then 0 here you have root 3 by 2 times MSR 0 and 0 have 0 0 LLS 0 here again root 3 by 2 times MSR and then this is the leakage inductance of the field plus the magnetizing inductance of the field into P of I A I DS I QS I 0S and I F so this then gives the inductance expression and then you have another part which is the speed EMF terms so that comes as 0 minus LLS minus 3 by 2 LQ here you have LLS plus 3 by 2 LD and 0 0 0 here root 3 by 2 MSR and then these 2 rows are 0 this is multiplied by the current vector and by the speed Omega S Omega S is the synchronous speed now if we look at this expression what we see is an arrangement very similar to what we had in the case of induction machines what we see is that the DI by DT terms for example if you take the D axis DI by DT terms comes from those coils which are also on the D axis you know that the field is on the D axis so you have a term here D by DT of the field current and then some inductance into D by DT of I DS on the Q axis you have only one term which relates to D by DT of the Q axis current on the stator there is not anything else on the rotor because Q we have not considered a Q axis excitation source on the rotor right there is not anything on the Q axis for the rotor so nothing is there similarly for V0S there is no term because it is at a completely different axis it is considered and therefore only the leakage inductance figures here if you look at the field winding field winding is also along the D axis and you have a stator winding fictitious winding on the D axis so you have a DI by DT term for the D axis from the stator and then you have of course the field inductance we noted earlier that the speed MF terms arise on a particular axis due to the excitation on the 90 degree axis and therefore if you look at the D axis equation you get a speed MF term due to current flowing on the Q axis of the stator on the Q axis you would have speed MF terms due to currents flowing on the D axis so you have one stator D axis fictitious winding and the field also is on the D axis on the rotor there are no speed MF terms because the rotor coil is not a fictitious one it is already there on the D axis only it is not fictitious it is not being transformed to something else and therefore there are no speed MF terms here. So one can sort of extrapolate what we had the arrangement of the equation what we had studied in the induction machine case to this case also in the case of induction machines we did not have a situation where the surface of the rotor was non uniform induction motor always has a rotor which is cylindrical and therefore the D axis inductance is equal to the Q axis inductance whereas here we are looking at a case where the D and Q axis are different and therefore these two inductances are different D axis has a separate inductance and this term what we see here LLS plus 3 by 2 times LD is then called as the direct axis synchronous inductance and ?s times this term is the direct axis synchronous it will be reactive so that is the normal and then you have LLS plus 3 by 2 times LQ which is then called as the quadrature axis synchronous inductance these are terms which you would have already come across in relation to the salient pole synchronous machine analysis which you would have studied the steady state analysis which you would have seen in your earlier electrical machine course. So this can then be we will give it another symbol as LDS which is the direct axis synchronous inductance and LQS which is the quadrature axis synchronous inductance then this term is replaced by LDS and this term is replaced by LQS we must note that this direct axis synchronous inductance consists of a leakage part and this part which is the magnetizing part but that is 3 by 2 times whatever we determined as the magnetizing inductance of the D axis alone by our analysis that is that is the magnetizing inductance of a particular phase winding 3 by 2 times that is what is then the magnetizing inductance along the D axis with the leakage inductance is then called as the direct axis synchronous inductance. Now if you look at the expression for the generated electromagnetic torque TE we know that TE from our earlier study of induction machines you know that TE can be written as I transpose into G times I where G is the description that is given here speed matrix and therefore if you now do I transpose GI what you would get is IDS into IQS multiplied by LDS – LQS plus IF into root 3 by 2 MSR into IDS so this is the expression for the generated torque note that this consists of two parts one is what is coming due to MSR another that arises due to the difference LD – LQ if LD where to be equal to LQ and when will that be equal LD will be equal to LQ whenever the rotor is fully cylindrical and we noted at the beginning of the discussion on alternators that the rotor of the alternator is of two types there are two varieties that are available one is the cylindrical rotor alternator the other is the salient pole alternator. So if you take a cylindrical rotor alternator then LD may be considered equal to LQ in which case this term becomes 0 it is only in the salient rotor alternator where D and Q axis are different that this term now comes into existence and therefore this part which is now coming in is then called as the saliency torque because it arises due to a salient arrangement where D and Q axis are different this is then the regular electromagnetic torque which arises because of the interaction of the field winding and the stator current excitation this is sorry that is not IDS but IQS. Now note that this torque does not have the field current in it so this torque can be produced or is produced even without field excitation but then it occurs at synchronous speed whereas this torque is produced because of field excitation and normally this component is far higher than this component however where you have a machine with this type of arrangement of rotor you can get a higher generated torque as compared to a cylindrical rotor machine which is again another aspect why they say that this variety of machine is more stable it can take more load than the cylindrical rotor variety. Now this is the machine equations in the synchronous reference frame with ID, IQ and IF as the variable now just like in the case of an induction machine where we saw that different variables may be introduced in order to describe the equations one may change the set of variables to the variables based on flux linkage and for that purpose we may define flux linkage variables as ?DS is equal to LDS x IDS x LMD x IF and similarly ?QS is LQS x IQS x LMQ x IQS x IQS so this is all that would be there but I think may be before we go to this stage we will have to take one more step before defining the flux linkages. Now the equations that we have arrived at we have arrived at it from the natural reference frame description where the stator variables are described in their own reference frame and the rotor variable is described in its own reference frame and in this stage we went from the three phase variable to the two phase variable of the stator the rotor was kept untouched and if we look at the way we have transformed ABC to a ? we have used this root 2 by 3 here and therefore what we have done here is made use of the power invariant transform if we use this what happens as if you recollect back at the equations which we had developed for this purpose in this case if the number of turns on the ABC phases is let us say N3 phase this is transformed into the two phase representation a ? 0 and the number of turns on the a ? 0 axis is represented as N2 phase then this ratio if you recollect the older lectures is equal to root of 3 by 2. So when we transform and when we go from a ? 0 to dq the number of turns remains the same there is no change so after having come from ABC to the dq reference frame for the stator variables the number of turns have undergone a change that is here you have the actual three phase winding whereas what you have here is a fictitious winding the number of turns in this fictitious winding is root of 3 by 2 times the number of turns of the actual three phase system whereas on the rotor we have maintained the same number of turns. Now since we are we normally do the analysis or study of the alternator behavior by looking at it from the stator it makes sense to refer all these terms to the stator number of turns all these terms in the sense stator is already in the stator number of turns for the rotor it makes sense to refer it to the rotor turns I mean refer it to the stator turns and the stator turns to which we have to refer it to is the number of turns that are available in the fictitious coil because we are now looking at it in the synchronous reference frame and therefore how does one refer this what we do is suppose now in the equivalent rotor term that is referred to the fictitious winding let us say I f is then called as I f dash that is the referred variable to the stator winding then I f dash into root of the number of turns there is root of 3 by 2 times n 3 phase this should then be equal to I f multiplied by the actual number of turns in the rotor this preserves the mmf equality and therefore what we can write is I f is then equal to I f dash into root of 3 by 2 times n 3 phase by n rotor similarly we need to transform the applied voltage as well here you have vf which is the actual voltage applied to the rotor turns equivalently when you look at it in a turns referred manner to the stator turns what you would have is vf dash divided by root of 3 by 2 times n 3 phase should be equal to the actual vf divided by nr which then means vf can be written as vf dash into root of 2 by 3 into nr by n 3 phase so if we now substitute instead of I f and vf these expressions that means we are looking at the phenomena of that is the excitations that are given to the rotor in terms of the stator number of turns so what we do is substitute instead of vf dash so this term now becomes vf dash multiplied by root of 2 by 3 into nr by n 3 phase and I f now becomes I f dash multiplied by root of 3 by 2 into n 3 phase by nr similarly this also becomes I f dash multiplied by root of 3 by 2 into n 3 phase by nr and similarly here also is I f dash multiplied by root of 3 by 2 into n 3 phase by nr and what happens as a result of this we can get rid of this term by multiplying throughout this equation by the inverse of this and if we do that what happens here is this then becomes 3 by 2 and this becomes n 3 phase by nr whole square this again this becomes 3 by 2 and this multiplied by n 3 phase by nr if we retain this I f alone here that is equivalent to multiplying that term here so this becomes 3 by 2 into n 3 phase by nr and this one becomes multiplied by 3 by 2 into n 3 phase whole square by nr square and this can be absorbed into the term here so this now becomes 3 by 2 into msr n 3 phase by nr so that is the simplification or the change that you get by replacing the original variables of the field by stator turns referred excitation variables. Now what we see here is that therefore you have 3 by 2 times msr into n 3 phase by nr msr is nothing but the mutual inductance between the stator phase and the rotor 3 by 2 occurs because it is a 3 phase system now and this mutual inductance is multiplied by the turns ratio and here also the mutual inductance is multiplied by the turns ratio. Now what happens when you multiply mutual inductance by the ratio of number of turns that can be understood by looking at a simple system here let us say you have a core you have a coil here and then you have another coil here this may contain n1 turns and that may contain n2 turns note that the way we have drawn the system whatever magnetizing flux is produced by this coil will have to link this coil also that means these two coils are there in one common flux part all the flux that is generated by this all the magnetizing flux that is generated by this links this coil also under that situation if you were to write the self inductance of this coil as l1 that is the magnetizing inductance of this coil as lm1 then lm1 is given by n1 square divided by the reluctance of this flux part the mutual inductance on the other hand between this coil and this coil is given by n1 into n2 by the reluctance of the flux part these expressions we derived right at the beginning of these lectures and therefore if this is the mutual inductance multiplying the mutual inductance by the ratio of number of turns m divided by n2 multiplied by n1 directly gives us lm1 and therefore the same understanding is there here also if you multiply the mutual inductance by the ratio of the number of turns then what you get is the magnetizing inductance on the d axis so similarly here and here come the magnetizing inductances of the d axis. Now how that can be used in order to simplify the expressions and get a more concise form of the synchronous machine equations and using the linkages to describe the machine equations that we will see in the next lecture.