 In the last class we had been looking at deriving inductances for the case of an electrical machine, a simple machine with a cylindrical stator and a salient pole rotor. The cylindrical stator has a single coil and the salient pole rotor has another single coil. In such a system there are basically two coils one on the stator, one on the rotor, stator is cylindrical and rotor is salient pole. For this we were trying to determine the mutual inductance between the stator and the rotor coil which we can call as MSR. In order to determine the mutual inductance we have tried to understand how the flux density distribution inside the machine would look like. And after looking at the way the flux density distribution is we have seen that the flux density distribution is a wave form that travels along with the rotor that means as the rotor rotates the flux density wave form also rotates along with it and this flux density wave form can be approximated as a quasi square wave shown here in the orange line that is this wave form which is 0 over some interval it has a high value when it is under the phase of the pole and then 0 elsewhere it has a negative value under the phase of the other pole and 0 again. The interval over which it is 0 from 0 degree to the place where it first goes up that we denoted it by ?. So this is the kind of flux density distribution at this point it may be helpful to look at an animation to understand how the geometry is going to look like. So here we have an animation of the machine assembly this is not the way the actual machine may get assembled but it is essentially meant to understand what the geometry looks like. So here you see the machine shaft and on that the cylindrical the two rotor poles have now been inserted and on this the field winding is now getting wound these two are the slip rings which through which the excitation is then provided to the rotor field coils the rotor has now been wound and should be inserted into the stator here you see the stator of the machine the stator consists of two slots one here and one here through this the coil is getting wound you see that the coil is inserted into one slot goes around the circumference and then gets inserted into the next slot there are five turns in this so the coil gets wound and has now finished winding the rotor is inserted into the stator assembly now here you see that the rotor has now been inserted into the stator assembly and the rotor can now rotate because of a small air gap between the pole phase of the rotor and the stator here you see the rotor rotating and now here you can see the gap that is there between the rotor pole phase and that of the stator the normal drawings that we draw we depict this view of the machine and end on view of the machine where we show only the slot here and slot here some conductors in this slot and conductors in this slot this cylindrical otherwise cylindrical stator and then the rotor now once the field is excited this arrangement will produce a magnetic field that circulates through this rotor pole gets into the stator one half of those flux lines would go around this stator come back into this rotor pole the other half would travel around this through the stator again and come back into the rotor pole that field generated is shown as a blue glow that is going to happen in this animation I would like to reinforce here that this blue glow is only meant for animation it is not that in the actual machine when the field is generated there is a blue glow everywhere so now you see the field that is that has been generated in the machine and this field divides itself here and here in order to go around the machine so that was the animation to help us understand how the geometry of the machine looks like and this part of the flux density is the one that is under the pole phase other areas around the pole phase do not have much of flux density in the air gap so we approximate that flux density wave form to a quasi square like this and as the rotor rotates this flux density wave form moves or the flux density wave form rotates along with the rotor so this being the case the next question is how do we determine the mutual inductance of the machine of the two stator coil between the stator coil and the rotor coil now if you remember earlier we had done this exercise of determining the mutual inductance for these between the stator and the rotor for the case where stator was cylindrical and the rotor was also cylindrical and there we looked at the MMF generated by the rotor and as this wave form moves how we evaluated how the flux linkage is going to be and there we had looked at how much flux is going to be linked by the turns of the stator as this moves and we simply took the area finding approach now it would be B average multiplied by the area B here multiplied by the area which will be the flux linked and as this wave form moves you can then determine how much of flux still links this a similar approach could be followed here as well but this time we will take a different approach what we will do is we first consider as we have said before the rotor to be excited second we know the flux density wave form please note that this is the flux density wave shape around the air gap which means as you travel around the air gap of the machine for a particular position of the rotor you know what the flux density wave shape is and we also know that the flux density wave shape does not change wave shape does not change as the rotor rotates the wave shape bodily moves but it is not going to change in shape and therefore what we can do is to resolve this flux density wave form wave form into Fourier series and then use this Fourier series description to determine the flux linkage in the stator use this use to determine flux linkage in stator and hence the value of mutual induct this is the approach that we are going to take now therefore the end on view of the machine what we saw now looks like this we are now going to determine the flux the expansion for the flux density wave form so let us redraw the flux density wave form around the air gap so the angle as you traverse around the air gap let us denote by a and the flux density wave form is given by is drawn like this 0 so you have a certain maximum value B hat and this value is – B hat it is 0 the flux density wave form has a value of 0 from 0 degrees to an angle equal to ? this is 180 degrees so this angle is 180 degree – ? this would be 180 plus ? and this is 360 degrees – ? so this describes the flux density wave form now if you proceed to the left of 0 degrees then what you would have is that the flux density wave form would then look like this so if you look at this wave form we at once notice that this flux density wave form is an odd function of a we also notice that this wave form is having half wave symmetry you may recollect from your first courses on network analysis if a wave form has this feature that it is an odd function of the independent variable then in the Fourier series expansion you will have only sinusoidal terms sinusoidal term if it has half wave symmetry then this wave form will have only odd harmonics therefore this wave form B of a can be written as ? n equal to 1 to 8n odd Bn sin n a is the expansion of this wave form you would have in a normal function other terms such as an cos n a this is not there because the function is an odd function of a there is no DC component because the average value of this function is 0 so in order to determine the Fourier components all that we need to do is determine this Bn so how do we go about doing that we know that Bn is given by 2 over 2 pi integral 0 to 2 pi B of a sin n a d a and because this wave form is an odd function of a this can be reduced to 4 over 2 pi integral 0 to pi B a sin n a d a and this function also has another aspect that it is symmetric about this line as well and therefore this can be further simplified as 8 by 2 pi integral 0 to pi by 2 B of a sin n a d a so this would then give us in the interval 0 to pi by 2 the function has a value equal to B hat over the interval ? to pi by 2 and 0 elsewhere and therefore this is sin n a d a which can therefore be written as 8 times B hat by 2n pi integral of sin n a is cos n a by n we have taken the n out and this is ? to pi by 2 and therefore that is 8 B hat by 2n pi let us take the minus sign also outside then it is cos n pi by 2 – cos n ? note that this is valid for odd n which means whenever n is odd cos n pi by 2 is 0 which therefore means that Bn is equal to 8 B hat by 2n pi into cos of n ? so that is the expression for Bn and therefore B of a can then be written as 8 B hat by 2 pi ? m equal to 1 to infinity cos of 2m – 1 ? into sin of 2m – 1 into a divided by 2m – 1 we use 2m – 1 instead of n because n has only odd values this is the expression for the flux density waveform that we have seen that is this quasi square wave flux density waveform has now been resolved into a Fourier series expression which looks like this now this is this expression is valid for this waveform in the sense this waveform is drawn for the case where this rotor as this is drawn the rotor that is shown here is in the horizontal angular location angular position in that case the maximum value of flux density would occur in this region the flux density will be 0 from here to somewhere here and then you have maximum flux density and then 0 flux density therefore this figure is drawn for the rotor horizontal. Now as the rotor begins to move what would happen as the rotor begins to move this flux density waveform would shift let us say the rotor from this horizontal has moved by a certain angle ?r in which case this rotor waveform is going to shift it will shift by an angle ?r and hence look like this so this shift being ?r what we have determined is the flux density waveform for ?r equal to 0 and we need to modify this flux density waveform expression the Fourier series expansion to take into account any angular position of the rotor and that is easily done because what we have determined is a function f of a and as the rotor rotates by a certain angle we have said that this waveform also bodily moves that means we are now looking at a function f of a – ?r and therefore we can write b of a, ?r the two variable function is nothing but b a – ?r which is the same as 8b hat by 2p sigma m equal to 1 to infinity cos 2m – 1 ? sin 2m – 1 now it is a – ?r this whole thing divided by 2m – 1 now this is your flux density waveform and this flux density waveform has to be integrated in order to get the flux linkage from this we need to get flux linkage so the goal is to find out flux linkage in the stator now how to do that let us look at our animation once again so here you see a snapshot of the geometry of the system as the system was rotating this is one particular frame in that what we see in this is that the flux in the machine travels through the rotor crosses the pole phase gets into the stator and comes back so essentially what we have said is in this region of the pole phase the flux density waveform is high and as you travel around here the flux density waveform becomes 0 then beneath this pole phase the flux density waveform acquires the negative value and then here again it is 0 and it is this waveform that we have resolved into Fourier series and therefore we now have a functional description of the flux density waveform as we travel around the circumference inner circumference of the machine as we travel around this and we now have to determine the flux linkage for the coil as shown here this is the coil that is the stator coil which goes into this slot comes out here goes in through this slot comes out on the other side and again around the circumference goes back and comes now how to find the flux linkage flux linkage is found by to find the flux linkage we have to find out flux crossing an area spanned by the coil and what we see here is that this area around the inner circumference what is shown here in green this area is the area through which flux is going to pass from the rotor to the stator you can see here from the rotor to stator this flux is going to pass that has to cut across this area and what we have obtained is the flux density distribution as we travel around the circumference note that the axial length of the machine is not included in that expression which means essentially we are assuming that this flux density distribution is constant along the axis of the machine along this axis in the stator whatever value of flux density you encounter at one point it is the same thing as you travel along the axis but around the circumference we have determined what the distribution is and therefore what we can do in order to determine the flux density distribution is we will consider this is the mid point that is the center of the cylindrical arrangement and we will consider a small segment around the circumference of the machine that small segment let us say subtends an angle that small segment subtends an angle d a so let us call this angle as d a and that d a generates a small arc here and the length of that arc obviously from geometry you would know is r d a is the length of this arc and we will consider an elemental area around this inner circumference of the machine that area around the inner circumference of the machine is taken along this axial length this is the area around the inner circumference of the machine so if the stator length is l then the area along that axis is then given by r d a multiplied by l and we know the flux density at this particular angle a so let us say a equal to 0 is at this point where this coil starts so this is an angle a and at an angle a you consider an elemental movement of d a that subtends an arc equal to r d where r is the radius of the inner surface of the stator and along the stator then you get a length l so this area r d a now this occurs at an angle a and we know the flux density at a given angle a by our Fourier series expansion and therefore the flux passing through this elemental area is given by b of a multiplied by this elemental area due to this flux that is going to cross through this area the stator has a flux linkage which is the number of turns in the stator multiplied by this flux that is passing through so this would then be an elemental flux linkage in the stator so writing this expression down you have an elemental flux linkage in the stator is given by the number of turns multiplied by r into l into b of a d a that is the expression that we have written here if this is the flux linkage due to this elemental area the total flux linkage is then obtained as ? of s equal to integral of this note that we have done this expression for a starting from here at some point a if we want to get the flux linkage note that the coil one end of the coil is in this lot another end of the coil is in this lot and therefore the flux linkage that is going to be the flux that is going to be linking this is obtained as the flux crossing this inner circumferential area which is spanned by this coil and that area is obtained then by a sweeping a from 0 to 180 degrees and therefore this integration can be done as integral 0 to ? d ? s which is nothing but integral 0 to ? of this so which we will now expand further so this is ns r into l b of a we have determined in the earlier expression it is a ? ? ? r really and therefore that is 8b hat by 2 ? these are constants so we can take them out of the integration and ? m equal to 1 to infinity cos of 2m-1 ? by 2m-1 into integral 0 to ? sin of 2m-1 into ?- ? r that is your expression to get the total flux linkage and we need to perform this integration now to get the final expression and this is nothing but ns r into l 8b hat by 2 ? into ? m equal to 1 to infinity cos of 2m-1 into ? by 2m-1 integral of this remember the integration is with respect to a and hence ? r can be considered to be fixed and therefore integral of sin 2m-1 is – of this multiplied by cos of 2m-1 into a – ? r going from 0 to ? there is a denominator 2m-1 which we will take out and put it as square here so let us look at simplifying this expression alone the remaining terms are the same so this expression works out to then cos of 2m-1 into ? – ? r – cos of 2m-1 into – ? r that is the expression so let us simplify this further what you have is cos of 2m-1 p x cos ? r – sin of 2m-1 p x sin ? r that is the familiar cos a – b expansion and then you have – cos of 2m-1 into ? r cos ? r and therefore this is the expression now this part is cos of 2m-1 into ? so cos of an odd multiple of ? is like cos of ? cos of 3 ? cos of ? and so on so all of these are – 1 so this is – 1 sin of an odd multiple of ? is 0 so this term goes to 0 and therefore the term in the square brackets boils down to – 2 cos there is a 2m-1 here so – 2 cos of 2m-1 into ? r and therefore the expression for the flux linkage reduces to ns x r x l x cos of 2m-1 into ? cos of 2m-1 into ? r multiplied by or divided by 2m-1 whole square this whole thing multiplied by 2 that is the expression you get for the flux linkage. So let us simplify this a little by expanding it and we will write the terms individually so what you have is ns x r x l x 8 b hat by ? multiplied by for m equal to 1 what you have is cos ? cos ? r by 1 for m equal to 2 you have cos 3 ? cos 3 ? r by 9 plus cos 5 ? cos 5 ? r by 25 plus so on. So we see that the flux linkage that is arising due to exciting the stator has one term that is a fundamental component has a third harmonic and has a fifth harmonic and so on. We also notice that the third harmonic is attenuated by as 1 over n square that is the third harmonic is 1 over 9 fifth harmonic goes down as 1 over 25 since the nth harmonic goes down as 1 over n square it is normally approximated this expression is normally approximated using the fundamental component alone therefore what we write is ? s is nothing but ns r x l x 8 b hat by ? x cos ? cos ? r alone the others are neglected since they are generally smaller compared to the fundamental and this approximation gives us by enlarge sufficiently accurate answers for machine analysis and therefore this is all that is done b hat is nothing but ns x is the mmf that is there divided by the reluctance which is lg divided by ? 0 x area under the pull phase. So this is flux and then divided by the area is b and therefore b hat is ns is ? 0 by lg and therefore the inductance the mutual inductance the mutual inductance is then given by ? s divided by is which is nothing but ns sorry this is not ns but nr because the excitation is on the rotor so you have nr so ns x nr x r x l x 8 by ? x b hat so that is is x 0 by lg and ? s by is would mean that this is is no longer there so this is then the mutual inductance multiplied by cos ? cos ? r and therefore this term one can call as the peak value of mutual inductance msr hat multiplied by cos ? r so what we have done now is we have found out mutual inductance between stator and rotor we may find out the self inductance of the rotor as well self inductance of rotor winding rotor coil how to find out self inductance again excite the rotor coil inductance is always the flux linkage divided by the excitation exciting amperes so self inductance of the rotor coil lr is then obtained as ? r which is the flux linkage in the rotor divided by the current in the rotor so in order to do this we need to get at the flux linkage that is going to be generated we know that the mmf of mmf generated by the rotor winding generated by rotor winding rotor coil is nr multiplied by is so at this point it would be helpful to go back to our image so what we have here is we are now saying that the rotor coil has been excited as before of course and then the flux is flowing through but what we want to determine is the flux linkage through the rotor itself flux linkage in this coil and this coil and the flux passing through this coil is therefore independent of the rotor angle we if you remember we said that this flux distribution rotates along with the rotor and therefore irrespective of where the rotor is the flux passing through this will always be the same because the entire air gap that is faced by this flux lies here and here that is on this pole phase to stator is one air gap this pole phase to stator is another air gap and irrespective of where the rotor is that air gap is going to remain the same which means that the rotor mmf that is so generated faces a constant air gap irrespective of rotation whether the rotor is in this position or in some other position the air gap is the same and therefore the flux generated does not change much and we also find that the flux is generated primarily in this region there is not much of flux in these regions and therefore the area that is now relevant to look at flux flow is the area that is covered by this pole phase and that pole phase which is lying there. So if we say that the pole phase area let the pole phase area be equal to some a then let the length of air gap equal to LG then the reluctance of air gap is equal to LG by mu 0 x a the total mmf that is acting is number of rotor turns multiplied by the rotor current this is mmf and therefore the flux that is generated is nothing but mmf divided by the reluctance and therefore it is nr x ir divided by LG x mu 0 x a this is your flux that is going to be generated and what is the flux linkage ?r ?r is nothing but number of rotor turns multiplied by the flux linkage and therefore that is nr2 multiplied by ir x mu 0 x a by LG and the inductance in this case it is the self inductance of the rotor winding is nothing but ?r divided by ir which is then given as nr2 mu 0 x a by LG. So this you see that this inductance is independent of rotor angle that is expected because irrespective of the rotor angle the flux linkage does not change why it does not change because the air gap is remaining the same irrespective of the angle at which the rotor is and why is the air gap same irrespective of the angle of the rotor that is because the stator is cylindrical if the stator were not cylindrical then as the rotor rotates the air gap could change from one angle to another so we have determined the self inductance of the rotor winding. So we have now a stator winding we have rotor winding we have determined the mutual inductance between these two we have determined the self inductance of this and we need to determine the self inductance of the stator winding in order to be able to write the full electrical system equations. Now determining this inductance of the stator winding is not very simple the reason is in the case of self inductance of the rotor winding we found that irrespective of the angle of the rotor the flux generated is always the same independent of angle therefore it is easy to determine in the case of MSR when we try to determine the mutual inductance between the stator and the rotor what we did was we excited the rotor and we try to find out the flux density distribution what we found was the wave form of flux density distribution around the air gap is the same irrespective of where the rotor is it is just that the wave form bodily moves but the wave shape remains the same. Now that enables us to describe this flux density wave form in terms of Fourier series and then we were able to arrive at the flux linkages but now in order to determine the self inductance of the stator what we need to do is to find out the flux linkage of the stator for unit excitation on the stator. Now if we excite the stator then depending on where the rotor is you may have less flux or more flux if the angle of the rotor is aligned along the axis of the stator you may get more flux if it is aligned away from that you will get less flux so the flux linkage is not going to be the same in fact the wave shape of the flux density distribution itself will undergo a change as the rotor rotates this makes it a little more difficult to determine the self inductance of the stator winding in the next lecture we will see how this can be handled in order to arrive at an expression for self inductance we will stop now for this lecture.