 So, what we have been doing in the last class is to try and look at machines with a cylindrical rotor cylindrical stator and salient pole rotor. We have been trying to find out the self inductance of a single coil on stator. We found that this task is rendered difficult because the air gap variations are complex which therefore give rise to a changing flux density waveform as the rotor is going to rotate which therefore means that we will not be able to get an expression for the flux density waveform unlike the earlier cases and then do an integration because every instant the flux density waveform is different so which flux density waveform will you take. So this is rather difficult issue. In order to address this what has been done is that we take recourse to what is called as two reaction theory proposed by Blondel several years back. Essentially this attempts to solve this issue by this way. We have a fixed MMF distribution which is acting across air gaps that are varying as the rotor rotates. We say fixed MMF distribution because we are looking at a instantaneous value of Is that is flowing for a given value of Is we are now looking at what would happen if the rotor is going to rotate and how the flux density variation is going to be there. So for that fixed MMF distribution this MMF distribution is known and air gaps that are varying with respect to time varying with respect to rotor angle as the rotor rotates and therefore the situation is fairly involved you do not know how to get the flux density distribution. What this approach does is that it attempts to convert this to a constant air gap case is not a single air gap now but the machine then has two constant air gaps and across these air gaps there is a MMF applied so to say a component MMF of the original MMF a component MMF applied that varies with rotor position. Now since we are considering sinusoidal MMFs it is easy to determine what the MMF components are and since they are now applied across constant air gaps we can then arrive at flux density distribution and then do the integrations that we need to do. So this is an issue which can be addressed analytically and this is essentially what Blondel had then suggested which goes under the name of two reaction approach and we use this to arrive at the MMF distribution we said that the actual MMF is then resolved into two parts one acting along the pole axis and another acting along an axis which is at 90 degrees to the axis of the pole which is called as quadrature axis and so you have an MMF component acting along this axis and an MMF component acting along the pole axis the air gap along the pole axis is has a particular value which we call as LGD and the air gap on the quadrature axis is LGQ. So now the machine is visualized as a machine with two uniform air gaps it is as if the machine has two cylindrical rotors it is as if there are two cylindrical rotors one having an air gap LGD another having an air gap LGQ and the MMF that is acting along LGD is an MMF component which is called as the direct axis component and this is determined by the location of the rotor pole phase and this is called as a quadrature axis component is determined by an angle 90 degrees to the pole phase. So all this is what we have done and then the total flux at any given angle is the sum of the direct axis flux plus quadrature axis flux and the flux linkage is then the number of turns in the stator multiplied by this flux that is going through this is at a particular area at a given angle along the circumference of the stator and then if we want to find out the net flux linkage you then integrate over the curved surface area of the stator the inner circumference of the stator and we had derived an expression for that which we will take from here. So this was the expression derived in the last class which we will copy and carry forward. So in order to get the flux linkage then you integrate with respect to a over the range 0 to pi and this is the function which we have to integrate. So let us do this integration now this is equal to NS multiplied by f hat f hat is common so we take that out into cos of ?r integral of sin so that is a – cos cos a – ?r going from 0 to pi divided by lgd and then plus sin of ?r into sin of a – ?r going from 0 to pi divided by lgq and this is multiplied by ?r into l. So let us evaluate the first term within the bracket cos of a – ?r so that gives us cos of pi – ?r – cos of – ?r cos of pi – ?r is – cos ?r so this gives us – 2 times cos ?r and this term gives us sin of ? – ?r – sin of – ?r and that is 2 sin ?r. So substituting these expressions then what we get is this is equal to NS into f hat into µ0r into l multiplied by 2 cos ?r divided by lgd plus 2 sin ?r divided by lgq is the expression that you have this can further be simplified as NS into f hat into µ0r into l multiplied by cos ? can be expressed as 1 plus cos 2 ? by 2 so this is 1 plus cos of 2 times ?r divided by lgd plus 1 – cos 2 ?r divided by lgq that can now be written as f hat NS µ0r into l divided by I mean multiplied by 1 over lgd plus 1 over lgq plus cos 2 ?r means 1 over lgd – 1 over lgq f hat is given by let us take the expression from the earlier derivations okay we have taken a peak f hat which is the sinusoidal fundamental of the actual mmf distribution so f hat is nothing but 4 by ? into NS into Is by 2 so we substitute that here so you have 4 by ? into NS Is by 2 multiplied by NS µ0r into l multiplied by 1 over lgd plus 1 over lgq plus cos of 2 ?r 1 by lgd – 1 by lgq now what we can do is define one inductance ld which then corresponds to this air gap lgd as 4 by ? NS µ0r into l and another inductance lq as 4 by ? divided by lgd 4 by ? NS µ0r into l divided by lgq so having defined these two symbols then we can write this flux linkage as Is into ld by 2 plus lq by 2 plus cos of 2 ?r into ld by 2 – lq by 2 and therefore the total inductance ls can then be written as Is over Is flux linkage per unit flow of current which is nothing but ld plus lq by 2 plus ld – lq by 2 into cos of 2 ?r this is the expression for the inductance that we get by using what is called as two reaction theory. Is this meeting our requirements we have earlier seen that the actual variation of inductance let us now look back at the inductance variation that we saw in the last lecture for this kind of a geometry for the geometry that we have here that is a cylindrical stator with a salient rotor one coil on the stator we want to find out the stator inductance and we found that the stator inductance looks like this let us copy this figure so we found that the stator inductance looks like this note that the x axis is marked in angles degrees and represents the rotor angle as the rotor is going to rotate the angle that it moves through is what is marked in the x axis in the y axis the self inductance of the stator coil is marked we find that it is at the rotor angle equal to 0 the inductance is at its highest the blue line here represents the actual value of inductance that we get through an FEM simulation the red line indicates a sinusoidal approximation to this blue line we find that the approximation is pretty good we find that the maximum value of inductance occurs at a rotor angle 0 and this goes through two cycles of variation when the rotor angle moves over 0 to 360 degrees and that is precisely what this term is going to give us this term if you notice is cos of 2 times ?r that means as ?r goes from 0 to 180 degrees cos 2 ? will go through one cycle and 180 to 360 it goes through another cycle so this term gives us 2 times over 0 to 360 degrees and it has an average value which is Ld plus Lq by 2 that means the maximum value here would be Ld the minimum value here if you call it as Lq then this way form can be described as an average value plus a cos sinusoidal variation of an amplitude equal to Ld minus Lq by 2 that is the amplitude of the variation so Ld minus Lq by 2 into cos 2 ?r floating over an average of Ld plus Lq by 2 so we see that by using this approach we are indeed able to predict the exact variation or a very good approximation to the actual variation of inductance. So we have now arrived at an expression for the inductance of the stator so now let us stand back and look at what we have done so far we have been trying to get expressions for the inductances that are there in a simple elementary electrical machine the elementary electrical machine that we considered had one coil on the stator and one coil on rotor if you remember the first geometry that we considered was a cylindrical stator and a cylindrical rotor as well how did we try to find out the inductances involved in all these cases please remember that we will have self inductances of stator coil of rotor coil and then we would have a mutual inductance between the stator and the rotor coil now in the case of cylindrical stator and cylindrical rotor if you remember what we have done the self inductance of the stator and of the rotor can be very easily found because the air gap is uniform there is no involved variation of the air gap as the rotor rotates as the rotor rotates the air gap remains the same and therefore it is very easy to derive the expressions for inductance in this case which is what we have done and we found that the self inductance of the stator is some constant value and the self inductance of the rotor is another constant value we derived an expression for the mutual inductance if you remember how we did this we took the actual flux density variation around the air gap in this case which would be a square wave let us say we excite the stator and try to find out the flux linkage with the rotor a stator coil is placed here and then a rotor coil let us say is placed there we excite the stator coil find the flux linkage with the rotor we took the actual flux density variation which would be a square wave in this manner and then as the rotor coil moves the rotor coil is initially positioned here and as the rotor coil moves to some other location say here and here as the rotor rotates it is equivalent to this coil moving on the x axis as this moves we then try to find out the net flux by taking this area of flux density we try to find out the flux linkage we got an expression for the mutual inductance that was linear with respect to the rotor angle and one could have drawn the expression the drawn the sketch of variation of mutual inductance with respect to angle you would have got a linear variation next we looked at the case of cylindrical stator and salient pole rotor the rotor inductance in this case the rotor self inductance of the coil on the rotor is easy to determine because as the rotor rotates because the stator has a smooth surface the air gap faced by the flux passing from the rotor to the stator will always remain the same irrespective of where the rotor is and therefore the self inductance of the rotor is going to remain constant this again we have seen then we looked at expressions for the mutual inductance between the stator and rotor we could have derived it in the manner similar to what we have done here but we took a different approach we knew the flux density wave form because as the rotor rotates the flux density wave form did not change and therefore we could write an analytical expression for the Fourier series of the flux density wave form and having known the expression for the Fourier series you go about determining the flux linkage on the stator and thereby we could get an expression for the mutual inductance which consists of a fundamental term and several harmonics and we said that by enlarge in most machines we have a sinusoidal variation of the mutual inductance and therefore the fundamental component alone would be sufficient in this case also the mutual inductance variation would look something like this if you think of the linear variation of the linear expression that you had one could have looked at the fundamental component alone which is what we have done here we have explicitly determine the Fourier series and looked at the fundamental component. In order to determine the self inductance we found that the problem is not so simple because of the very involved variations in the air gap that the stator would see as the rotor is going to rotate and therefore we cannot do it in a simple way we applied to reaction theory so all this we have done why have we done all this because we wanted to write an expression we wanted to describe these electrical machines by suitable equations and those equations are of the form applied voltage is equal to rotor resist I mean the resistance multiplied by the flow of current plus the rate of change of flux linkages and these flux linkages are dependent on inductances and therefore unless we know the inductances you cannot write this expression so in order to write this expression we have gone about trying to determine the inductances of various types of situations we looked at simple situations first. Now in all these situations what we have done is taken a single coil on stator so that it is easy for us to understand at first. In actual machines this is not really the case actual machines have a distributed stator winding most of them so in order to understand how this stator is going to look let us look at a simple animation where the stator winding is distributed over two slots per pole per phase so we are going to look at an animation where one phase alone is going to be wound and that phase per pole occupies two slots this is a four pole salient rotor structure so let us look at the animation for that so here we see a four pole rotor one pole two three and four field wound are getting wound they are all connected in series field winding have been wound and you see a view of the rotor please remember that what is shown here is intended for the purpose of understanding how the machine structure looks and is not indicative of how the machine will actually be assembled from various different rotor parts and stator parts. So let us continue with the animation this is how this is going to look and here you have the shaft and the stator coming here this stator has several slots as you can see it is in spite of these slots we have said at the outset that we are neglecting the slotting effects in the stator so what we are assuming is that really the stator surface the inner surface is really smooth all along and we have certain armature that is wound in these slots so let us continue with the animation now here you see the stator being wound the stator slots are shown here also for ease of understanding because normally in books it is this that can be easily represented the stator slots are also numbered here and we have seen that the stator has four poles and this machine the system that we have here has number of slots equal to 24 so four poles that means the number of slots per pole is equal to 24 divided by four that is six slots per pole and in that interval three phases have to be accommodated so six divided by three is then two slots per pole per phase and therefore you will find that if you take this to be the R phase R phase will occupy two slots under each pole so let us continue with the animation you see the actual arrangement here whereas what you see here is known as a developed view that means if you assume that this is going to be slit along the axis and then rolled out this is what you would see you see that the stator bar in this case it is a single turn so goes here enters this slot and then travels along the circumference and enters this slot and then comes out travels along the circumference there that is what is being shown here as entering into this travelling along the circumference into the next slot and then coming out so now you see that the stator winding is distributed in two slots it is not a single slot winding and then two slots again here and then again here there is going to be two slots and two slots here so now you have two slots per pole there are four poles in the machine so one pole if it is going to influence the section from here to here then R phase occupies two slots out of this the next pole influence starts here and therefore R phase occupies two slots again here again the third pole and the fourth pole so two slots per pole and this is how the winding in one phase is going to look this is the winding in the next phase so the B phase now that is occupying two more slots in one pole area and this is wound so you can see how this happens here the actual situation looks like this whereas the developed view this is how it happens and then the third phase we have shown this by green one can call it RBY perhaps because yellow would interfere with yellow this has been made as green so the G phase now each of these you can see occupies two slots to R to B to G so the winding is now distributed over two slots and this figure here shows how the finished armature stator is going to look like so now let us continue with the animation so one can now see how the stator will look like from various views and now it is time to assemble the machine so the stator is there and the rotor along with the shaft and the rotor structure is assembled into the machine and the rotor now begins to rotate so this is how a general machine would look if it is a large machine the number of slots over which a particular phase is distributed may be many more here we have number of slots for a particular phase the R phase is only two slots in a particular pole area if it is a large machine that number of slots may be much higher it may be four slots per pole per phase or may be even more so this is how the system is going to look so what we have is a distributed stator winding in the animation that we saw we had two slots per pole per phase and if you would recollect there are four poles that means the first pole if you designate as north pole then the subsequent pole would be a south pole and then a north pole here and a south pole here so which means if you look at the flux density distributions the flux density would go from north to south and then from south to north that means one cycle of variation has been completed and then again north to south to north is another cycle of variation that has been completed. So over one mechanical 360 degrees you have two cycles of flux density variation this is what is referred to as mechanical degrees and therefore electrical degrees since two cycles complete cycles of flux density variation is there one can say that two electrical 360 degrees are there in one mechanical 360 degrees and therefore one says that ? electrical is equal to two times ? mechanical. Now if we say this then it means that over this part the flux density goes from north to south that means you are going through 180 degrees of flux density variation and in that 180 degree of flux density variation there are six slots which means that the slot pitch which means that the slot pitch is equal to 180 divided by 6 which is 30 degrees that means each slot occupies an angle equal to 30 degrees electrical or it would be 15 degrees mechanical one can see 360 divided by 24 360 mechanical degrees in one circle there are 24 slots so if you do this 12, 2 is 24, 30 so 15 degrees is the angle between two slots and that 15 degrees now becomes 30 electrical degrees. So we say that the angular separation between two slots which is called as a slot pitch is equal to 30 degrees and if you take the R phase occupies two slots which means it is distributed over two slots and the number of slots are the electrical angle equivalent to the number of slots over which a particular phase winding is distributed is called as phase spread I am sure you would have looked at all this in your earlier courses on electrical machine so this is just to review and recollect the terms which are involved in electrical machine that you would have seen so phase spread so in this animation what we have seen is a 60 degree phase spread winding the winding occupies 60 degrees of phase spread therefore what we have found so far is that if you take a particular cylindrical stator let us make this a little more circular so if you take a cylindrical stator and you are now going to have the R phase distributed over a certain angle which is corresponding to the phase spread so let us say that the R phase is distributed over this angle what do we mean by these two what we are saying is that the animation had a four pole arrangement we are now looking at a simplified case this is two pole structure so which means that electrical angle is equal to mechanical angle and therefore if you have one coil starting here one coil side starting here the other side of that coil would lie 180 degrees away assuming it is a full pitched coil so all this belt the belt that is shown here will be one set of coil sides the return conductors of all these coils are now going to lie at this point so one side goes in and then comes out here and then is connected to the next conductor going in coming out next conductor going in coming out and so on so this is then your phase spread so if you take the center this angle is called as the phase spread for this R phase similarly now you will have a Y phase and a B phase to cover the entire circumference of the stator so this being the case what we now have to see is we have understood now how to determine the inductance for a single coil that is present on the stator with the rotor being salient now that the actual machine is going to have a phase spread coils distributed over certain angle how to determine the inductance of this system in order to do that the way we proceed we could proceed is that we will make use of our understanding of how to determine the inductance for a single coil after all a spread like this is made up of several single coils and we assume that is we neglect slotting as usual which therefore we further assume that the conductors are uniformly distributed over the entire phase spread over the phase spread so uniform distribution of conductors that implies that we have neglected slotting effects other assumptions which we have made earlier continue to hold that we neglect saturation we neglect iron losses and under these assumptions we want to determine the inductance now how to do this how to determine the inductance of this case we have seen that since the rotor is salient rotor is salient we have seen that the air gap variation is not going to be simple and the flux density waveform will be highly difficult to describe by a single analytical equation as the rotor is going to change the angle the flux density waveform will change and therefore it is not analytically it is not feasible to describe it by a single expression and we have seen how we have handled this in the earlier case we resorted to two reaction theory so we will use the same thing here also and this implies that you transform or look at the salient pole rotor structure as now having been decomposed into two cylindrical rotor structures one having an air gap LGD another having an air gap LGQ the important thing to note is that now instead of the salient rotor which is the actual machine we are considering a description in terms of two equivalent cylindrical rotors the net effect of which is going to produce the behavior of the salient pole machine or net effect is going to describe the inductance of the actual salient pole machine and since we are now looking at two cylindrical rotors in order to describe what is going to happen here it will be sufficient to look at the cylindrical rotor alone and try to find out how the flux density waveform or the MMF waveforms are going to look like so the next task is to find out how the MMF waveform looks like because now that the problem has been converted to a cylindrical rotor structure we go back to the case when we first looked at cylindrical stator cylindrical rotor and then we wrote down the MMF variation from that the flux density variation and then try to determine the flux so the problem is now simplified a little now the question is how do we find out what is the MMF variation of a uniformly distributed stator phase winding and to do that how are we going to do it we will write down the actual expressions in the next class we will just look at an understand try to get an understanding of how that will be done since we have a cylindrical rotor structure so let me draw a cylindrical rotor here assume that the air gap is uniform and what we want to do first is to derive or to arrive at the variation of MMF around this air gap and how to do that in order to do that we apply our amperes law so we intuitively understand how the flux lines in this kind of system is going to look like this distribution here and here means that let us say current are coming out of the plane on this belt and going into the plane on this belt and therefore the whole arrangement looks like an inductance with coil with where I is coming out here and then going into the slot so you would have the net flux lines going horizontally on this axis so you would expect the flux lines to flow horizontally from left to right and indeed that is how it would look so if we now try to apply amperes law it would be convenient to take an amperian loop along the flux lines so which goes this way and then if you want something here then you take a flux line that goes like this and we see that irrespective of what flux line you take may be even something that goes here as long as the flux lines cross the air gap in this region that is in this region the net amperes enclosed by this loop always remains the same it encloses all of these conductors and hence the total current enclosed by any of these loops is the same and therefore the MMF acting along this gap and that gap together would always add up to the total current here and since these two air gaps are equal we can say that that MMF is spent half here and half here but then flux here is going in and there it is going out so the MMF acting would be reverse and therefore all along this region it would be the same MMF that is acting that is equal to half of the total MMF but as we enter the regions in this area we need to see how the MMF variation is going to occur and for that we will try to write down the expressions in the next class draw the waveform and then try to determine the inductors so with this we will stop for this class and continue in the next session.