 We have been looking at inductances in machines, we have been considering electrical machine with a cylindrical stator and a salient pole rotor with one coil each on them. We have derived expressions for the mutual inductance between the coil on the rotor and the coil on the stator. We have derived an expression for the self inductance of the rotor coil and we were beginning to look at an expression for the self inductance of the stator coil which we shall try to do in this lecture. The difficulties we encounter for this are one is that the flux density wave form flux density distribution around the air gap not have a constant wave shape and because of this it also further in addition to this it we could say that the flux levels in the machine also change with respect to rotor angle that means if the rotor is going to rotate then depending on where the rotor is you may have large flux levels in the machine or you may have low flux levels in the machine. For example let us look at this FEM result let us so here we have a finite element study of a simple machine geometry better to go to the next page so here we are. Now here we have the geometry that we were discussing a cylindrical stator and a salient pole rotor the rotor is horizontal in this position and has rotated by a certain angle rotates further and then becomes vertical rotates further and is again on the way to becoming horizontal again. Now here if you see the flux density levels in the rotor are quite high the flux density levels in this place in this place are quite high if you look at what that level might be this table says that the flux density in that area is likely to be greater than 1.5 Tesla in the middle and then goes down to about 1.27 Tesla or so and then that is distributed here in this position of the rotor also you find that there is considerable flux density here in these simulations the stator winding is excited there is no excitation on the rotor. So here also the flux density levels are high but as the rotor rotates further you now find that the flux density levels have come down here the flux densities in the middle of the rotor are likely to be in the range of about 1 Tesla and indeed when the rotor becomes completely vertical like this then the flux levels are quite low it is only in the range of about 0.6 Tesla in the middle and at the pole phase however it is little higher pole phase could probably have a flux density level of about 0.9 Tesla. Now as the orientation becomes this way now the flux density levels again increase and as the rotor aligns itself to the horizontal position the rotor flux density further increases. So you see that as the rotor moves from a horizontal position goes towards a vertical position and then rotates again to a horizontal position the rotor would have moved by 180 mechanical degrees if it becomes horizontal again and you see that flux density levels itself are undergoing a change and we know that inductance is nothing but flux linkage divided by how much current you are exciting it with this is the stator induct. And if this flux linkage is going to change with respect to rotor angle for a given excitation this then implies that inductance changes with respect to rotor angle indeed that is what is seen in this figure the inductances are also marked in this position we have a stator inductance at the level of 40.5 milli Henry it goes down to 40.31 then 34.09 at this position it is only 30 milli Henry and then it builds up again right. So we have therefore inductance changing with respect to the rotor angle how about the flux density distribution now let us look at how the flux density distribution around the air gap will look for a salient pole machine again we will see the results from a finite element study of a simple geometry now this plot shows how the flux density levels are going to vary the x axis indicates the circumferential distance traveled around the inner circumference of the stator the y axis denotes the modulus of the normal component of flux density. So let us just call it as b now you see this plot has been drawn for rotor angle at 0 degrees so you see that the flux density is low in this area becomes high here goes low here inverts in sign becomes high here again why is this high here it is in this region that the pole phase occurs. So if you notice here the pole phase occurs in this area this is the pole phase region of the rotor and you find that there is very high flux density levels in these areas and as you travel away the flux density becomes lower and that is what is shown in this plot flux density levels here are high and elsewhere it is quite low but what happens when the rotor moves this plot is drawn for a rotor angle equal to 30 degrees and you see here that flux density level here and here are high the rotor has moved such that the rotor pole phases are moving towards this end and wherever the pole phase is encountered those regions have high flux density wherever the other regions of the rotor are there the inter polar regions so to say they have low flux density is here as the rotor rotates further this plot is drawn for 60 degrees of the rotor angle here you see again it is this region where the pole phase occurs is having high flux density again the inter polar regions have low flux density that are there now here this plot is drawn for 90 degree of the rotor position that means rotor is vertical. So here again we see that this region corresponds to the pole phase area so flux density levels are high this region here again is the other pole phase flux densities are high this is the inter polar region flux densities are low now you see that as the rotor rotates the flux density distribution around the inner circumference of the stator the wave shape itself undergoes a change wave shape undergoes a dramatic change therefore we will not be able to resolve this into a Fourier series expansion a fixed Fourier series expansion the wave shape is going to depend on where the rotor is and if we are to go and expand this by Fourier series we will have to do a separate Fourier series expansion for every rotor angle and it is not going to lead us anywhere and the wave shape is also not going to be fixed how this shape is going to look like will depend on the particular machine that you are looking at depending on how brought the pole is how much the inter polar air gap is how the rotor is shaped how the stator is shaped all those will have an effect on how this distribution is going to look like therefore it is impractical to try and determine the flux density wave form and hence the flux linkage of this machine though you find that this is appears to be a difficult issue to handle all hope is not lost if we however look at the variation of inductance now let us look at another plot which shows the variation of inductance so here you have a plot which gives which shows how the inductance is going to vary for the stator as the rotor is going to rotate now the x axis gives the rotor angle and the y axis describes the inductance now you see that the inductance wave form is not all that distorted even though the flux density wave forms here wave shapes are going to change as the rotor rotates the inductance wave shape appears like a rather smooth curve and in fact it may be feasible to approximate this by a sinusoidal wave form and indeed if you look at the next plot where a sinusoid has been superimposed on this wave form this variation of inductance you see that this inductance variation could probably be nicely approximated by a sinusoidal wave shape the red line here shows a sinusoidal approximation to the blue line which is the inductance wave form if we look at this therefore since this wave form is going to be a sinusoidal wave form we could probably handle this situation as follows instead of looking at the flux density wave form we start with the mmf wave form generated by the stator winding for the inductance expression we derived earlier we looked at the mutual inductance between the rotor and the stator we derived an expression that consisted of a fundamental component and then a third harmonic and then a fifth harmonic and so on and we said that it is good enough to approximate this by the fundamental alone and here we have a sinusoidal variation it should be enough to get a sinusoidal variation by looking at the fundamental component of the mmf wave form that is generated so we take the fundamental component of the stator mmf wave form the stator mmf for a given angle of the rotor the stator would generate an mmf that is that has different values as you travel around the circumference inner circumference of the stator and what we are saying therefore is that let us start by assuming that this variation of mmf that happens as you travel around the stator is sinusoidal in nature that is what we mean by the fundamental component of the stator mmf so we are assuming a sinusoidal variation around the air gap this assumption is not bad because as you look at the actual mmf in machines do not just have a single coil on the stator electrical machines have a distributed stator which is there and as you go for the distributed stator arrangement distributed stator winding arrangement you find that the mmf that is generated by the stator winding becomes more and more closer to a sinusoidal distribution and it is enough therefore if we consider a sinusoidal variation that is fine so we consider a sinusoidal variation now what where does the sinusoidal mmf act as the rotor is going to rotate we find that the air gap that the mmf is going to face will change so let us look at this picture now let us say we look at a given angle on the circumference of the stator let us say we are looking at this point here the same angle as you travel around the circumference now at this point you see that the air gap that this region is going to face is pretty large whereas when the rotor rotates to this position the air gap around this region is very small which is that of the pole phase again here the air gap is pretty small but here the pole phase is just leaving this region here again the air gap would have increased in comparison with this here again the air gap is fairly large and here is a still larger air gap so as the rotor is going to rotate at any given angular location on the inner circumference of the stator as the rotor rotates because the rotor is a salient pole rotor this the air gap is going to vary and therefore even if we know what is the mmf at this place and we assume that the mmf variation around the circumference is going to be sinusoidal we still do not know what is the air gap on which this mmf is going to act again if we look at the flux density distributions which determine which are which is the result of flux flowing in the air gap what we see here is that wherever the phase of the rotor pole occurs those regions seem to have fairly high flux densities wherever the inter polar region occurs those regions seem to have low flux densities this is something that is valid across all these wave forms inter polar region low flux density wherever there is the rotor pole phase you have high flux density which means and the inter polar area or the phase of the rotor pole where it occurs will depend upon the rotor angle so as the rotor rotates now the inter polar region is here in this area but if you look at here this area will now slowly contain the pole phase here it has the pole phase fully and as the rotor rotates this area becomes inter polar region again. So if we look at the way the machine behaves it appears as if inter polar regions are associated with large air gaps low flux density whereas pole phase regions are associated small air gaps high flux density and where the inter polar region will occur and where the pole phase will occur will depend upon where the rotor is at any given angle that may be an inter polar region or it may be a pole phase region so that depends on the rotor angle. So even though we know the mmf that is there at a particular angle because we have assumed the sinusoidal variation of the mmf around the air gap the gap the actual gap across which this mmf is going to act will depend upon where the rotor is at some instance at a given angle it may be a pole phase at some instance it may be an inter polar air gap. So in order to address this issue the idea that is used is that we assume that there are two air gaps in the machine assume that there are two air gaps in the machine and these two air gaps are uniform air gaps it sort of amounts to saying that you assume that instead of the salient pole rotor that is there you have instead of the salient pole rotor structure you now have two cylindrical rotors one of them giving rise to a small air gap and another giving rise to a large air gap. We further note that the inter polar regions are always associated with low air gap with large air gap and low flux density pole phase regions are associated with small air gaps and high flux density therefore the small air gap equivalent is representative of the pole phase regions the large air gap rotor is representative of the inter polar region and the region of the pole phase is going to be rotating as the rotor rotates the inter polar region also changes as the rotor rotates and therefore in order to find out the MMF that is acting along the small air gap cylindrical rotor or the large air gap cylindrical rotor how do we find out what is the MMF that is acting what we try to do is resolve the sinusoidal MMF sinusoidal stator MMF we have assume that the stator MMF has a sinusoidal distribution around the inner circumference of the stator. We now resolve this into two parts one of which lies or not may be a better word would be one of which is oriented along the pole phase another which is oriented along the inter polar region and then we say that that component of the MMF which is oriented along the pole phase is the one that acts along the cylindrical rotor with small air gap the stator MMF part that is oriented along the inter polar region acts along the air gap which is larger the actual flux. Now the MMF acting along the small air gap will produce its own flux and the MMF acting along the inter polar region for the large air gap will produce its own flux then the net flux that is there at a given angle is then the sum of these two fluxes flux acting along the small air gap rotor plus flux acting along large air gap rotor large gap these two together is then flux at any given angle. The idea is that the flux generated at any given angle it is not possible to determine just from the MMF distribution alone because the salient pole rotor is we see that the pole phase regions are the ones that are associated with higher flux density inter polar regions are the ones associated with low flux density and as the rotor rotates where the pole phase occurs where the inter polar region occurs both are going to change. So we now consider the rotor to be composed of two cylindrical rotor one having a small air gap one having a large air gap and resolve the MMF along these two axis and then you find out the net flux this approach has been called as the two reaction theory and is attributed to Blondel so it is called as Blondel's two reaction approach. So how do we then do this now let us say that we are now drawing let us take a previous figure that we have used so this shows now the geometry of the system that we are having and our as you travel around the inner circumference of the stator you are going to traverse a particular angle that angle a equal to 0 starts here let us assume that it starts here and as you travel around as you travel around you go from a equal to 0 to 180 degrees and then come back as 360 degrees and if we have the stator coil being excited. We are going to plot the MMF distribution we have assume that it is we are looking at the fundamental component of the MMF distribution so that variation can be plotted in this manner so you have the angle traveled around the circumference and you have the MMF let us call it F this is 90 degrees and then 180, 270 and 360 degrees so the MMF distribution since we have considered to be sinusoidal will then look like this it reaches a peak at 90 and then 180 it goes to 0 which is negative peak at minus 270 and then goes to 360 degrees this is the fundamental component of the stator MMF this is the variation of the stator MMF as you travel around the inner circumference note that this reaches a peak at an angle equal to 90 degrees that means at a equal to this angle this is a equal to 90 degrees it reaches a peak so this is then the axis of the stator winding stator coil that is there this is the axis of the stator coil that is this line that is the stator coil axis and let us further consider that the angle of the rotor when it is horizontal is 0 that means this axis is also coinciding with the rotor angle axis this refers to ?R equal to 0 and the rotor at present is aligned the midpoint of the pole phase defines let us say we define the rotor angle so this is now your rotor angle ?R what we have said is that we take this fundamental component of the MMF distribution of the stator and we find out a resolve this along the pole phase of the rotor and another component which is perpendicular to the pole phase of the rotor and you see that this component then lies always on the interpolar axis interpolar area this component as the rotor is going to rotate you always resolve this MMF one along the pole phase axis if the rotor were rotated to some other axis it would still be resolved along that pole phase axis and 90 to that would always lie along the interpolar area so this is what we do how to do that now let us say that the peak of this MMF distribution is F hat then F hat lies along this axis F hat lies along the axis of the stator winding and a component of F hat that lies along this ?R can then be given as F hat cos ?R and that part lying along this axis the interpolar axis will then be F hat sin ?R we will call this axis which lies along the pole phase as the direct axis lying along the pole phase we will call the axis lying in the interpolar region as the quadrature axis since it is 90 degrees to the direct axis therefore we call this as Fd we call this as Fq now Fd has a maximum now the original MMF distribution is a sinusoidal distribution around the air gap and therefore the resolved MMF should also be a distribution around the air gap and the MMF resolved along the direct axis has its maximum value at the direct axis and is subsequently sinusoidally distributed elsewhere and therefore we can write this as the total distribution as cos of a- ?R this would then be cos of a- ?R so this expression describes the original stator MMF resolved into a component along the direct axis and distributed sinusoidally around the inner circumference of the stator this expression describes the stator MMF resolved along the interpolar axis and distributed sinusoidally around the inner circumference if these are two parts if this these two represent the stator original stator MMF resolved into two components then it also stands to reason that if you add these two you get the original distribution back do we get that let us try to add these two terms Fd plus Fq is then equal to F had cos ?R cos of a- ?2- ?R plus F had sin ?R cos of a- ?R cos of a- ?2- ?R can be written as cos of a- ?R cos of a- ?R cos of a- ?R cos of a- ?R cos of a- ?R cos of a- ?R is nothing but – sin of ?R – a which is the same as sin of a- ?R and therefore Fd plus Fq can be written as F had multiplied by cos ?R sin a- ?R plus sin ?R cos a- ?R so this is of the form sin a cos b plus cos a sin b and therefore this is F had into sin of a- ?R which is nothing but F had sin a which is the way form that we have drawn here and therefore Fd plus Fq gives you the original MMF back to represent the whole thing in this graph essentially what we have done is that may be take a green. Now we see that the rotor has moved to some particular angle ?R and therefore the pole phase axis would lie somewhere here at this angle and F had into cos of ?R would then give you this magnitude and therefore what we are assuming is that this MMF on the direct axis is distributed as a sinusoidal function like this in some manner and then another term F had into sin of this angle would come somewhere here and then you have another sinusoid which is something like this and then we are saying that the sum of these two sin is the original sin. So that is a pictorial representation then of this expression. So if we then look back what we have said is that the salient pole rotor can now be thought of the machine with the salient pole rotor can now be thought of as having a machine with two cylindrical rotors one cylindrical rotor being associated with a small air gap and being representative of the inter polar regions and one cylindrical rotor being associated with the large air gap being I am sorry small air gap being representative of the pole phase regions and another rotor being associated with the large air gap representative of inter polar regions and we need to add the two fluxes that are going to be generated. Now having come this far let us now see what we would do to arrive at the fluxes that are generated. Now let us say that the air gap along the direct axis air gap representative of the direct axis is LGD, LG was the air gap that we had used here we use the suffix D to represent direct axis and then the air gap representative of the quadrature axis let us call it as LGQ. So if this is the case now we have considered we are considering two cylindrical rotors having a uniform air gap of LGD and LGQ along LGD the MMF FD is acting along LGQ the MMF FQ is acting. So in order to find out the flux that is generated we need to divide the MMF by the reluctance and how do we find out the reluctance of the air gap LGD that is nothing but let us call it as RGD is nothing but LGD divided by Mu0 into area. Now how do we get the area we will take the approach that we have used earlier if you remember in order to find out the flux passing through a particular area let us go back to our figure of the last lecture here we have it in order to find out the flux that is passing through a particular area what we have done is we considered an elemental small area that lies along the axial length of the machine and the arc here is subtended by a small angle d a which is at a particular angle a from your reference a axis. So we use the same approach here also now let us say we want to consider the flux at a particular angle a then we consider an elemental angle d a and along the length axial length of the machine the area associated with that will then be Rd a that is the segment length multiplied by L that is the axial length of the machine. So this gives you the area of that small elemental segment LGD by Mu0 into that area gives you the reluctance offered by a small portion of air gap having this area similarly you would have Rgq which is then given by Lgq divided by Mu0 into RLd a note that if you see the difference between Rgd and Rgq it arises due to the air gap only otherwise the rotors are of the same dimension the two equivalent cylindrical rotors only the air gap is different. So this then helps us to write down the flux that is generated the flux generated at this elemental area due to the d axis is then given by Fd, Fd is a function of a that means the direct axis MMF distribution is again sinusoidally varying around the air gap at a particular angle a it would have a certain value and that value divided by the reluctance Rgd gives you the flux that is generated due to the direct axis MMF add that to the flux that is generated by the quadrature axis MMF to get that you have to divide by Rgq this is the flux that is generated at this angle a passing through this elemental area now if you want to find out the flux linkage due to this flux then d ?s is nothing but the number of turns in the stator multiplied by this d ?d ?q is what we have written so this is just d ? ns multiplied by d ? and therefore this is ns multiplied by Fd a by Rgd plus Fq a by Rgq. Now this is the flux linkage cause due to the elemental flux that is crossing that particular area now the MMF is distributed all along the circumference again let us go back to the figure of our earlier lecture in order to find out the flux linkage of the stator coil we need to integrate over the area spanned by the stator coil we did that in the earlier case also to find out the flux linkage but there the flux was generated by the rotor and the flux density waveforms are constant and so on and so forth but now we have determined how this MMF variation is there and due to that MMF with the two cylindrical rotors we now know what is the flux that is going to pass through a given elemental area and in order to find out the total flux passing through this area and therefore linking this coil here this coil we have to integrate from a equal to 0 to a equal to ? and therefore that is exactly what we need to do next in order to find out the total flux linkage ?s that is nothing but integral from 0 to ? of d ?s which is ns multiplied by integral 0 to ? fd of a we can get from our earlier expression so fd of a is f at cos ?r multiplied by this we have reduced that to sin of a – ?r so this is f at cos ?r multiplied by sin a – ?r divided by lgd – f at sin ?r multiplied by cos of a – ?r divided by lgq into ?rld a so this expression we have to integrate in order to get the total flux linkage and then flux linkage divided by the excitation current is would then give you the inductance so let us do this integration in the next class and see what is going to be the nature of the inductance and whether it has any approximation to the inductance variation that we have already seen we will stop here for today.