 In the last lecture we had tried to derive an expression for the self inductance of a distributed winding on the stator of a salient rotor cylindrical stator machine. In order to do that what we had done was first look at the MMF distribution generated by exciting the stator winding as you travel around the circumference of the air gap. We plotted that distribution and we found out the fundamental component of that distribution and then that was resolved into two directions along the axis of the pole phase and along an axis 90 degrees to that of the pole phase and then we tried to find out the net flux linkage in the stator winding. In order to do that again we considered an elementary coil in the stator winding found out the flux linkage of the elementary coil and then using that expression we perform another integration over the span of the winding and when we did that what we have is an expression here this expression was the flux linkage of the elementary coil and then in order to get the flux linkage of the entire winding we integrate over the span of the winding being denoted by ? and we were integrating that and you got ? s equal to this much. Now let us try to simplify this relation and proceed further the integration that we have done when you substitute the limits the first term that is ? of ? – ? r going from ? – ? by 2 to ? plus ? by 2 can then be written as ? of ? – ? r plus ? by 2 – ? of ? – ? r – ? by 2 so this is of the ? a plus b – ? a – b form so this is 2 x cos of ? – ? r into ? ? by 2 which is nothing but – 2 cos ? r ? ? by 2 the second part of the expression can then be written as cos of ? – ? r going from ? – ? by 2 to ? plus ? by 2 which is cos of ? – ? r plus ? by 2 – cos of ? – ? r – ? by 2 this is the cos a plus b – cos a – b form so that is nothing but – 2 sin ? – ? r sin ? by 2 and therefore the expression for the flux linkage can then be written as ? s equals 2 and we will take this 2 also outside which means it is 4 f hat z ? r l into cos ? r by lgd plus sin ? r by lgq multiplied by sin ? by 2 because sin ? by 2 expression is common to both we can take it out. Now this can further be simplified as 4 f hat z ? r l into 1 plus cos 2 ? r by 2 times lgd plus 1 – cos 2 ? r by 2 times lgq into sin ? by 2 and we rewrite this as 4 f hat z ? r l 1 by 2 lgd plus 1 by 2 lgq plus 1 by 2 lgd – 1 by 2 lgq into cos 2 ? r into sin ? by 2. Now let us define 2 inductance terms one which we will call as ld there must be a term is here that is missing this is is there is also a term is that was missing there. Now we will define 2 inductance terms one of which we will call as ld which is 4 times f hat z ? r l by lgd and f hat being 8 times is is there inside that so that is right is would not come here explicitly is is already embedded in f hat so let us first expand this term f hat is nothing but 8 times z is by 2 ? sin ? by 2 this expression we get from our earlier derivation of earlier derivation as is seen here where we try to get the amplitude of the fundamental component of the mmf distribution that was our f hat and therefore we substitute that here and using this then what we have is ? s is nothing but 32 times z square sin square ? by 2 ? r into l divided by 2 ? into 1 by 2 lgd plus 1 by 2 lgq cos 2 ? r multiplied by is. Now we will define an inductance term ld as 32 z square sin square ? by 2 ? r l by 2 ? lgd and another inductance term lq as 32 z square sin square ? by 2 ? r l by 2 ? into lgq this inductance ld is representative of the assumed cylindrical rotor structure with an air gap of lgd which is a smaller air gap and another inductance lq which is representative of a cylindrical rotor structure which has an air gap of lgq which is a larger air gap and the larger air gap is the one that is present along the quadrature axis, axis 90 degrees to that of the pole phase so we will call this as quadrature axis inductance and we will call this since it represents the inductance of the smaller air gap and therefore the inductance along the axis of the pole phase we will call that as the direct axis inductance with this then the expression for ? s can then be rewritten as ? s is equal to ld by 2 plus lq by 2 plus ld by 2 minus lq by 2 into cos 2 ? r multiplied by is and therefore the ratio of ? s to is which is nothing but the self inductance of the stator is then given by this term therefore we can write ls is ld plus lq by 2 plus ld minus lq by 2 into cos 2 ? r note that the inductance in the inductance expression we have derived like this what we are essentially determining is the flux linkage and in determining the flux linkage we are assuming that the mmf generated by the stator produces flux which crosses the air gap and goes into the rotor and therefore all of this flux would link in some manner coils that might be on the rotor and therefore this flux linkage and hence the inductance that we have derived essentially represents the magnetizing inductance of the stator winding in order to determine the total self inductance therefore one has to add the leakage inductance component ll and leakage inductance which by definition closes itself in the air gap which might be in the air gap between the stator and the rotor or it might also be due to flux linkage which is there in the end windings so all those effects together add up and is lumped into one leakage inductance term which we add as ll so the total self inductance then is this leakage inductance component plus the magnetizing inductance term. Now note again that the form of the inductance then has one term which is constant which does not change and another term which varies as cos of 2 times ?r which means that over an angle 0 to 360 degrees if we now plot this inductance with respect to ?r let us say that this is 180 degrees and that is 360 degrees and 0 degrees this curve would then look like this cos 2 ?r variation happens over and above a dc value which is leakage inductance plus this ld plus lq by 2 so let us denote that by this average line now over and above this you now have ld minus lq by 2 x cos 2 ?r and therefore at an angle ? equal to 0 this inductance is really maximum and then at an angle ? equal to 180 degrees you find that this becomes cos of 360 degrees and therefore this completes one cycle of variation in a span of 360 degrees so if we start here and then it goes up to here so and then in the next 180 degrees it completes one more cycle so you have two cycles of variation within 0 to 360 degrees of the mechanical rotor angle if we neglect the leakage inductance if we are plotting only the magnetizing inductance part then this level would denote ld this would denote lq then the mean is given by ld plus lq by 2 and this amplitude is then given by ld minus lq divided by 2 so this would then be the form of the expression now if we look back at what was the plot obtained through finite element study for a particular winding we saw again a similar distribution from 0 to 360 degrees you see that there are two cycles of variation therefore there must be a cos 2 ? term and a mean term which you have and the expression is as what we have direct therefore whether the winding is going to be concentrated or distributed the form of the expression remains the same we had earlier looked at a single coil on the stator and a salient pole rotor as it rotates what will be the nature of inductance variation that was the plot that we saw earlier now we find that the distributed winding also behaves in the same way in fact if we now look at those two expressions if we make an equivalent number of turns by looking at these expressions then one can arrive at an equivalent number of turns between which the distributed winding must have in order to be considered as a concentrated winding so the nature of the expression still remains the same now AC machines do not just have a single winding on the stator they have many windings on the stator normally a three phase AC machine has three windings on the stator consisting of R, Y and B phases therefore we have now looked at the self inductance of the rotor we have looked at self inductance of the stator now we need to also see what will happen if you have one more winding on the stator which will then mean there could be a mutual inductance effect between these two winding so how does the expression for mutual inductance look like now let us understand how the machine geometry looks like before we get into the mutual inductance part of it so let us now see how the machine is going to look like let us see a picture now here you have an image which we have seen previously in animation now it is just a snapshot of that animation in this case we have a four pole stator winding and what is shown is just one pole pair portion of it so we see here that the coil enters here one side of the first coil so to say and then the coil closes with the other coil side there may be many more turns in a coil which is multi turn coil but here we are only considering a single turn to simplify the view and therefore it goes around comes here and then around the circumference on the other side it goes that completes one turn and then the coil has a second turn which goes around here and then comes up now this would be the way the winding would be represented or would look like in the actual machine now in this case what happens is that you would have within this region within this circumference of the stator you would have a large mmf and the mmf decreases as you go outside the axis of the mmf would then be situated there and from the center if you draw a line to this point and you take it out so that would be the direction of the axis of the mmf produced by this winding. Now in the electrical machine there are likely to be more than one machine windings and let us look at the developed view of this that is the way normally machine windings are represented this is just to see how the spatial axis are then separated now that particular one winding is then shown here you can see that this coil comes in this turn comes in and then travels around goes to slot number 7 comes back and goes here. So what we have shown in the earlier figure in this figure is the R phase shown only up to this point beyond this is not shown now in this case what we have been seeing so far is that if we plot the mmf distribution of this stator winding since there is going to be flow of current in this the flow of current in this bar let us say is going in to this and then this current goes along this bar here also you have current flowing in and here you have current flowing out here you have current flowing out. The nature of the mmf then since currents are flowing in here you have a variation assuming that the slotting is not there and whatever is spread uniformly this we have discussed earlier the mmf distribution may be drawn as a straight line till this point and then it remains constant until you reach here and now the currents are flowing in the opposite direction so it decreases linearly and then it remains constant over this region and so on. So now you have this is the trapezoidal nature of the mmf which we have seen when we considered the distributed winding self inductance same thing that happens here and if you plot the fundamental component of this that would reach a peak right at the middle and therefore the axis of the R phase winding is considered to be along the place where the maximum of the fundamental component occurs and that is at this point. Now if you consider the coil that is marked in green so you have current flowing in here current flowing in again current flowing in and at these two points currents are flowing out and therefore for the green phase the mmf distribution would look like this it is again linearly varying assuming that is uniformly distributed and then constant here and then it linearly varies here and that would have its axis situated exactly here along this line and now you see that there is a separation between the axis of these two winding. These two windings do not will generate maximum mmf along two different directions if excited what we are now interested in is assuming that the R phase winding is excited we know what is the variation of mmf around the air gap we want to find out what is the flux linkage on this green stator phase. If we find out the flux linkage then the ratio of flux linkage to the exciting current in the R phase will give us the mutual inductance. One can also look at a picture of the machine as it would appear in three dimensions so as to get an understanding of how the spatial axis are split so here you have again we have seen this as an animation before so now you see that the R phase winding occupies one phase spread here and another phase spread somewhere here and then the green winding occupies a phase spread here and then a phase spread here now the blue winding occupies here and here. Now if we look at the axis of the R phase winding that would come as we have seen in this case right in the middle of the last coil side of the R phase and here the first coil side of the R phase as we progress from left to right so that means in this diagram that would come you have seen an R phase here which enters here and you have the other set of R phase coil side so that would come right at somewhere here as far as the green phase is concerned that axis would come somewhere here assume that you have a line connecting the center of the stator to these two dots and going out so those two will then represent the axis the angle between the axis is 30° I mean is 120° remember that this is 120° electrical now if you look at this arrangement you can see that there are 24 slots in 360 mechanical degrees and therefore the slot angle will be 360° by 24 which is 15° so this is 15° mechanical being a 4 pole machine this is also equivalent to 30° electrical and now one can see if the slot angle is 30° electrical you now have here between the axis of the R phase and the green phase you have 1, 2, 3 and 4 slot angles which means that the phase displacement of the axis of these two coils is 120° that is what we have written here this being a 4 pole machine this is the arrangement so let us instead consider only a 2 pole machine to simplify our drawings so if it is a 2 pole machine then what would happen is you have the stator which is there and now in this 2 pole machine you are going to have two phases let us consider the R phase let us say the R phase is distributed here and the return conductors are placed here the axis of the R phase winding would then be as we have seen in between the last conductor of one phase spread and the first conductor of the other phase spread right in the middle of that so you have the axis of the R phase sitting here. Now the green phase axis is 120° displaced away from this so if we are to have that then you need to have a coil spread which has its axis along this direction this being 120° if that is the case then you must have the coil here this figure means that you are having conductors that are uniformly spread in this region uniformly spread here uniformly spread and uniformly spread let us assume that this uniform spread is as before Z conductors per radio we have seen all this before. Now we are going to find the mutual inductance between these two winding mutual inductance between the two winding and in order to do that we are going to consider R winding to be excited and find the flux linkage with G winding this is our approach. So how to do that as before we try to draw the MMF distribution of the R phase we know that the MMF distribution the wave shape is known and then we resolve that MMF distribution into Fourier series and consider only the fundamental and we know that the fundamental of this MMF distribution will be f hat the amplitude of the fundamental of the MMF distribution will be f hat is equal to 8 times Z by 2p x is x sign of beta by 2 where beta is the phase spread this we know. Now what we want to do is to find out flux linkage with the G phase if you remember how we did the X derivation for self inductance we took an elementary coil in the winding in which we want to find out the self inductance on the flux linkage there and then we did another integration so we will follow the same approach now we need to consider an elementary coil and as before we will consider this axis for a that is a equal to 0. We will consider an elementary coil at some place here which is at an angle gamma from the alpha reference axis so a small angular displacement of d gamma here and then the elementary coil at gamma having number of turns equal to Z times d gamma so what we have is a coil here the return conductors of that coil will be at this point we want to find out the flux linkage how to find out the flux linkage we will have to integrate over an area spanned by this coil that is located at gamma so in order to derive the expression for mutual inductance we consider the flux linkage in this elementary coil at gamma and that can be written as d psi s which is elementary flux linkage that is then given by the number of turns in this elementary coil that is Z times d gamma multiplied by the flux that is going to pass through at any given angle alpha this expression is similar to the ones that we have written earlier that is fd at alpha divided by the reluctance let us call it rd and then fq at alpha divided by the reluctance rq we know that rd is the length of the air gap lgd divided by mu0 rld alpha and rq is lgq divided by mu0 rld alpha and therefore this expression becomes Z times d gamma multiplied by fd at alpha is nothing but f had cos theta r multiplied by sin of alpha – theta r divided by lgd plus f had sin theta r multiplied by cos of alpha – theta r divided by lgq and multiplied by mu0 rld alpha so from this is the flux linkage in the elementary coil due to the flux passing at a given angle alpha along the span of this elementary coil and in order to get the flux linkage of this elementary coil let us call that by the term delta psi s that is then integral of this d psi s the integration performed with respect to alpha as we move along the circumference of the stator the inner circumference of the stator in this case alpha goes from ? – ? to ? no it is not ? – ? to ? it goes from ? to ? plus ? remember that this coil ultimately has to have an axis 120 degree displaced with respect to the first coil that we are considering when finding out the self inductance of this coil we had considered integrating along this area and now when this axis is shifted we need to do the integration there and therefore this is ? to ? plus ? and that can then be done as z times d ? remember alpha is the variable of integration not ? at this point of time therefore this is mu0 r l multiplied by f had into cos theta r and then integral of sin ? r is – cos of alpha – ? r going from ? to ? plus ? and then plus sin ? r sin of ? – ? r divided by lgq this again goes from ? to ? plus ? so if we look at the first part of the term what we get is that is cos of ? – ? r plus ? – cos of ? – ? r that is equal to – 2 times cos ? – ? r and then we look at the second term that is sin of ? – ? r plus ? – sin of ? – ? r that is – 2 times sin of ? – ? r and therefore this expression ? sin can be written as 2 times f at z d ? mu0 r l multiplied by cos ? r cos of ? – ? r divided by lgd – sin ? r sin of ? – ? r divided by lgq this is the flux linkage that is cos that arises in this elementary coil at an angle ? and therefore if we want to find out the flux linkage of the entire phase spread this again has to be integrated now the variable of integration is ? and therefore ? s is nothing but integral of this ? ? s now the integration variable is ? that varies from ? – ? by 2 to ? plus ? by 2 note that we go back to the figure ? varies from this the angle between a equal to 0 axis and the mid point the median line of this phase spread is equal to ? and from ? you are now looking at ? by 2 this side to ? by 2 that side and therefore ? varies from ? – ? by 2 to ? plus ? by 2 and that is what we have written here and that then becomes 2 times f at z mu0 r l into cos ? r sin of ? – ? r by lgd this going from ? – ? by 2 ? plus ? by 2 – sin ? r by lgq into cos of ? – ? r it is integral of sin therefore that becomes plus ? – ? by 2 ? plus ? by 2 and again let us take a look at the first term that is sin of ? – ? r plus ? by 2 – sin of ? – ? r – ? by 2 that is a plus b – ? a – b form so that is 2 times cos of ? – ? r into sin ? by 2 and then if we look at the second term that becomes cos of ? – ? r plus ? by 2 – cos of ? – ? r – ? by 2 that is cos a plus b – cos a – b form that is – 2 times cos of sin of ? – ? r into sin ? by 2 and therefore this expression ?s can now be written as 2 times we take this 2 also outside which means this becomes 4 times f at z mu0 r into l divided by cos ? r cos of ? – ? r – divided by lgd – sin ? r sin of ? – ? r divided by lgq multiplied by sin of ? by 2 and now let us expand this f at we know that f at is equal to 8 times z sin ? by 2 x is divided by 2 ? so if we substitute this expression for f at then what you get is ?s is equal to 32 times z square sin square ? by 2 divided by 2 ? into mu0 r into l multiplied by cos of ? r cos ? – ? r by lgd – sin ? r sin of ? – ? r divided by lgq multiplied by is now when we did the derivation for the self inductance of a particular face winding we define two inductance terms one was ld which was equal to 32 times z square sin square ? by 2 x mu0 rl divided by 2 ? into lgd and we define the term lq which was equal to 32 times z square sin square ? by 2 x mu0 rl divided by 2 ? into lgq so if we look at these definitions and our expression that we have derived now what we can see is that ?s can be written as ld multiplied by cos ? r cos of ? – ? r – lq sin ? r sin of ? – ? r this whole thing multiplied by is remember this expression is flux linkage in one face by exciting another and therefore the expression in this is nothing but the mutual inductance between the two phases and therefore we now have an expression for mutual inductance which is the ratio of ? s by is which is ld cos ? r cos of ? – ? r – lq into sin ? r sin of ? – ? r where remember ? is the separation in the axis of the two phases we have this figure here and this angle from here to here is what we call as ? which is the spatial separation between the two axis now if we call this as r phase and we call this as the g phase then this is the mutual inductance between r and g where the rotor axis is situated at an angle ? r somewhere in between so using this expression ? is the spatial separation using this expression we can now derive specific values for the case of a three phase machine where normally ? is 120 degrees in the case of three phase machine and the same expression can be used to derive the mutual inductance between r phase and g phase r phase and b phase and g and b also and that will then help us to write the expressions for the electrical equations for the machines having done this our next task will be to write the expressions for the electrical equations and see how we can compute the behavior of the machines using that we shall see that in the next class we will stop here for today.