 in the last lecture we had tried to derive an expression for the self inductance of a coil of a single coil of n s turns placed on the stator of a machine having a salient pole rotor and that was of the form the self inductance of the stator that the expression that we derived had the form Ld plus Lq by 2 plus Ld minus Lq by 2 into cos of 2 times ?r. Now there are a few observations to be made in this machine we have considered the rotor to have 2 poles the salient pole structure had only 2 poles we could go back to that figure and take a look at that so this was the salient pole structure that we had considered and one can clearly see that this has 2 poles and this is the expression that arises for that machine. In the 2 pole machine the mechanical angle we have seen this earlier the mechanical angle is equal to the electrical angle if the number of poles increases then the electrical angle will be correspondingly more than the mechanical angle as compared to the mechanical angle and while deriving this expression we are trying to find out the flux linkage in the stator winding and in order to find out the flux linkage in the stator winding it is necessary to consider one span of the stator winding and one span of the stator winding is the distance the angular distance over which the flux changes by 180° of its electrical variation right and therefore if the number of machine poles is going to increase then we still need to consider only 180° of electrical variation which means that the ?R considered has to be basically the electrical angle of the rotor so if we all the time ensure that we take the electrical angle of the rotor to substitute in this expression this equation is valid for machine of any number of rotor poles this is the first observation that we make the second one we notice that the MMF waveform that we took was the fundamental component of the MMF waveform actual MMF waveform and in trying to determine the flux generated by the MMF waveform we are assuming that the flux crosses the air gap and connects with rotor as well the method by which we determined the mutual inductance is also to determine the flux density waveform around the air gap and by that flux density what we meant what we implicitly assume is that all the flux that for which this flux density is responsible all the flux enters into the stator from the rotor and similarly here all the flux that is generated due to the MMF crosses the air gap and goes into the rotor as well which therefore means that the expression we have derived is essentially the magnetizing inductance in addition to this kind of flux since the stator is essentially a cylindrical stator and we have seen in the animation that the stator winding goes through the stator along the axis in a particular slot and then it comes out of the stator and then travels along the circumference of the machine and then goes in through the slot 180 degrees away and then further goes along the circumference and may make another turn. So you have stator winding that lies outside the stator as well and the flux produced due to this part of the stator winding will not be linking or will not be going through the rotor and therefore such flux lines are not really considered in the expression that we have derived. So such flux lines will essentially be grouped under what is called as leakage flux and this leakage flux in machines can happen at a variety of locations this particular location is called as end winding leakage flux or the overhang leakage flux there may be other forms of leakage flux that exist inside the machine and since the leakage flux path is very difficult to determine it is difficult to define an analytically derived expression for the leakage flux and therefore it is usual to add as leakage inductance to this expression and therefore the self inductance is essentially a leakage inductance plus what we have derived as the magnetizing inductance and then we said that the machine for which we have derived this is essentially a single coil stator along with the salient pole rotor and we saw through the animation in the last lecture that coils on the stator are seldom single coils. The phase winding so to say if you consider a particular R phase in the three phase machine is almost usually a distributed winding which means that the phase occupies several slots more than one may be more two or may be more than two several more slots perhaps and the angular separation which is spanned by the coils of a particular phase is then called as a phase spread and it is for such a distributed winding that we really need to find out the self inductance which we shall attempt to do now. Now if you look at such a structure let us so here we have a finite element study conducted for a distributed winding on the stator with a cylindrical rotor. Now note that it is sufficient now to consider a cylindrical rotor because even for the case of a salient pole machine what we have achieved through the two reaction theory is that we are considering the salient pole the effect of the salient pole rotor as having been produced by two cylindrical rotors with different air gaps one air gap corresponding to LGD which we designate as LGD another air gap that we designate as LGQ and to find out the total flux we find the flux develop when a particular MMF component is applied on the cylindrical rotor with the gap LGD then we find the flux generated in the cylindrical rotor with the gap LGQ due to a different flux component and when we add the two we say that that is the total flux that is generated at a given angle a and therefore in order to do that it is sufficient first to consider the case of a cylindrical rotor and here we have a setup which is a cylindrical rotor. So here we see that the stator winding is distributed in three slots this is the return path of the stator winding and these are excited you find that flux lines go like this and the air gap flux density can now be determined but in our attempt to derive the inductance we started out with the MMF distribution MMF distribution is nothing but the distribution of H multiplied by L a scaled version of the H field distribution and if we now look at the H field distribution for the if you start with angle equal to 0 from here and then go around the air gap and look at how H is going to vary that plot looks like this. So here is the H field variation the X axis is plotted in terms of angle that is it goes from 0 to an angle of 2p. Now you see that the H field variation has a step here and then is more or less constant over this region more or less constant over this region and then it displays three steps to reach this point again constant and then so on. So where is this H field distribution constant you will find that the H field distribution is constant after you cross this so from here to here the H field distribution is constant and when you reach this point as you traverse from here to here there is one step and then two steps and then the third step reaches the other level after which it is constant again and then one step down another step down back to this point and one more step down to reach the flat level here so that is how it goes. Now the question is why do these steps arise and why is this flat now if you want to evaluate H you would take an Ampere loop apply your Ampere rule which says that integral of H.dl around the closed loop is equal to the number of turns multiplied by current enclosed by that loop. So if we now let us say we take this flux line and let us form our Ampere loop around that flux line so which goes like that so as you travel around that you see that N x I is basically the number of turns in this slot plus this slot multiplied by the flow of current in each so that is H x L and we have already understood that H field within the iron regions is almost 0 negligible compared to what occurs in the air gap so it is essentially H in the air gap multiplied by the length of the air gap added these two you add in absolute terms that should equal the number of turns multiplied by current for these three slots. Now observe that this will be the same whether you take this flux line or that flux line or that or that or that because for all these cases the number of turns included is the same. On the other hand if you take a loop like this then you see that only one slot is included within this loop and therefore you would lose out these two and half of that is consumed here so that is the reason for the step that you see here. Now assume that the number of slots here is not three but more number of slots are present let us say four slots are present in which case what would happen is this would take four steps two three and then four steps if more number of steps are taken then correspondingly the number of steps will be more and therefore one can understand that in the limit where the number of slots are very large and conductors are uniformly distributed around that distribution angle then what you would have is that this H variation would be almost straight line here and then this area becomes flat in any case and then again a straight line variation another straight line variation here so the MMF or the H field wave form is trapezoidal. So that is the MMF wave form that we have to consider indeed what we can say is that in the limit let us say we have a circular stator we have a circular stator and we distribute the windings in a uniform manner all over this place and then the return conductors are distributed in a uniform manner here we can say that the there is a uniform distribution of conductors in the stator let it therefore be equal to Z conductors per radian. Now it is Z conductors per radian what we are essentially saying is that each conductor here which is there has a return conductor 180 degrees away that the coils are full pitched and then there are the next turn comes here next turn comes here and so on and this distribution is uniform so there are Z conductors per radian. Now it is easy to appreciate now here there is a rotor which is cylindrical a rotor that is cylindrical and at any angle let us take an angle here you have seen how the flux lines look like the flux lines are going around the distribution like this and therefore the flux lines if we plot would go this way that will be the nature of flux lines and if we consider any flux line at an angle of let us say with respect to the centre let us say that we want to find the mmf at this angle call that angle as some a with respect to a equal to 0 then if we now take an ampere in loop that goes around at that particular angle then we can see that the mmf or the h.dl would give us 2 times h into lg h at the air gap into lg that will essentially be equal to the number of conductors in this region that is 2 times Z into a into the flow of i here so multiplied by is and therefore hg into lg is equal to Z a is so that is the mmf this is then the mmf that is acting at a particular angle a so if you go on increasing a the mmf goes on increasing until you reach the end of the phase winding distribution after that irrespective of what loop you take it is always equal to Z multiplied by the entire phase spread from a equal to 0 to the phase spread note that a is the angle from 0 to this point which can alternatively be written as Z into ? by 2 into is where ? is the phase spread phase spread is this angle let me denote that by some other colour so this angle is then the phase spread ? and the maximum value of the mmf will therefore be Z ? by 2 into is this is not equal to this always this is the maximum value of the mmf and therefore the plot as we have seen looks like this it reaches a certain maximum let us assume that mmf is positive in this area so goes this way and then comes down like this so this interval over which the mmf changes from one level to another level is the phase spread and therefore this angle is ? this angle is ? by 2 so this is plotted with respect to a and this is a equal to 0 a equal to ? to ? and so on. So if you look back on how we have computed the self inductance for the single coil case we determined how the mmf distribution is going to look like and then we found out the fundamental component of the mmf distribution so that is exactly what we need to do here we need to find out the fundamental component of this level as we have said is Z ? by 2 into is what we are drawing is the mmf plot and this angle is ? by 2 so how to find out the mmf distribution for this one can write down the expression equations describing this way form and then do the Fourier series analysis for this that is one way of doing it a simpler way of doing it would then be to first consider the derivative of this way form the derivative of this way form would then look like from 0 to ? by 2 the slope is constant so it is a constant value and then from ? by 2 to this value the slope is 0 so derivative is 0 and then here to here the derivative is again constant and then goes to 0 at this point and then goes high again until this and so on this is your way form and what is the value of the derivative is the slope so Z ? by 2 is the amplitude here into is divided by ? by 2 so it is Z times is the amplitude so this way form looks much simpler to handle in fact we can do a little better what we can do is take the derivative of this also the derivative of this is an impulse function so the impulse function here it is a step going down so an impulse function pointing down this is also an impulse function pointing down this is an impulse function pointing up so here again is an impulse function pointing up note that the step levels are all the same it goes from Z is to 0 and this value is – of Z is so this is – of Z is and here it is Z is again so the impulse functions will then have a value the impulse function this function would be described as Z – of Z is into ? and the impulse function occurs at an angle ? by 2 and therefore it is a – ? by 2 so the strength of the impulse is – Z is and it occurs at an angle ? by 2 so it is Z is ? of a – ? by 2 and 90 degrees of the wave form occurs here this is ? by 2 so here you have ? by 2 so what we could do is we could find out the Fourier series of the impulse function train which is an easier exercise to do mathematically and to get to this quasi square wave form which is the first derivative of the MMF wave form what we can do is integrate that Fourier series of the impulse function distribution and then to get to the Fourier series of the actual MMF wave form integrate once again because we have done two derivations now we go back and do two integrations. So how do we find out the Fourier series of this impulse function wave form we note again that the impulse function wave form if you had drawn for the – a axis also so look at the – a axis then this function would have continued here and then gone up 0 over some distance and so on so if you now take the derivative again there would be an impulse function here of the same amplitude and then an impulse function when it goes down again so one can see that the impulse function wave form is an odd function of a whatever occurs here is a negative of what is happening here and therefore if we are going to determine the Fourier series of the impulse function in that Fourier series we will have only sinusoidal terms the cosinus sinusoidal terms will not be there and therefore this can be described as sigma of n equal to 1 to infinity bn sin n a so we need to find out what bn is and to do that what we do is bn is 2 by 2 pi integral 0 to 2 pi if the derivative function we call as f double dashed of a sin n a d a and since the function also has half wave symmetry these two are negatives since it has half wave symmetry one can write it as 4 by 2 pi integral 0 to pi f double dashed a sin n a d a and indeed this integration is double the value from 0 to pi by 2 and further we know that we are going to be looking only at the fundamental component so it is sufficient if we determine b1 so let us write b1 as then 8 by 2 pi integral 0 to pi by 2 f double dashed of a sin a d a and this f double dashed of a a is an impulse function which has which occurs at beta by 2 and therefore one can write this as 0 to 2 pi 0 to pi by 2 delta of a minus beta by 2 into sin a d a multiplied by minus z is and therefore we get 8 minus 8 z is by 2 pi into sin of beta by 2 is the b1 term now how to get this is b1 of the second derivative function so let us indeed call this as bn double dash to make to remember that we are talking about the second derivative function and from this we need to go to the first derivative function b1 dash so which means essentially f double dashed of a looking at the fundamental component alone can be written as minus 8 z is by 2 pi sin beta by 2 into sin a and we want to find out then f dashed of a that is the fundamental component of the first derivative function for that all we need to do is to integrate this and that would give us 8 z is by 2 pi sin beta by 2 cos a and then we need to integrate once again to get f of a which is 8 z is by 2 pi sin beta by 2 into sin a so this is the function that we were looking at that represents the fundamental of the mmf distribution now we can therefore write this as a fundamental the amplitude of the function f hat multiplied by sin a and this is the actual fundamental distribution we need to resolve this into two components to be acting along the air gap representative of the direct axis and another component acting along an air gap representative of the quadrature axis therefore what you have this is similar to what we have done earlier fd of a is then f hat multiplied by cos ?r into cos of a – pi by 2 – ?r and fq of a which is a variation with respect to a is f hat into sin ?r multiplied by cos of a – ?r similar expressions we have derived in the earlier lectures you may look back at those expressions this d expression can be simplified as f hat cos ?r into sin of a – ?r what do we do with this now what we really have is a distributed stator winding and we want to find out the flux linkage of this distributed stator winding the net flux linkage of this stator winding the approach that we will adopt for this purpose is let me copy this figure it is some of the things that are not required complete this and then we need to draw the spread so now we are ready so this is the machine that we have we want to find out the flux linkage of this distributed winding what we will do is we will consider a small coil elementary coil out of this distributed winding now let us say that we consider this is your angle a equal to 0 let us take some angle this is some a and at this angle a let us consider a small differential angle d a that differential angle is d a now if that differential angle is d a in this small angle there are certain turns that have been included how many turns will be included we know that the uniform that the stator conductors are distributed uniformly at a rate of z conductors per radian and this is an interval of d a radians therefore the number of conductors and indeed the number of turns because each of these conductors would have a return conductor on the other side and therefore number of turns is equal to z times d a so on second thoughts let us call this angle not as a we will call it as beta or we will call it as gamma beta means something else for us so let me call this as gamma okay beta is the phase spread this angular separation is beta so the number of turns would then be z d gamma now these number of turns form an elementary coil one can imagine that there is a single coil here with the return conductors here with the number of turns being z d gamma so let us now find out what is the flux linkage of this elementary coil and let us call that as d psi s is the flux linkage of this elementary coil that is obtained by the number of turns z d gamma multiplied by the direct axis mmf divided by the reluctance of the direct axis mmf at a particular angle a plus the quadrature axis mmf divided by the reluctance of the quadrature axis at a particular angle a is your net flux linkage this gives you the total flux and this gives you the number of turns and therefore it is flux linkage and rd of a is nothing but the length of the air gap along the direct axis divided by mmf into r into l into d a where r is the radius of the inner circumference of the stator and then rq at a is lgq divided by mu 0 rl d a and therefore this expression reduces to d psi s equals z into mu 0 rl d gamma multiplied by f at cos ?r sin a- ?r divided by lgd plus f at sin ?r multiplied by cos a- ?r divided by lgq d a this is our expression. Now in order to find out the flux linkage of this elementary coil you need to integrate this so the flux linkage of this elementary coil let me call it as ? psi s that ? psi s is then integral of d psi s and in doing this integration we have to consider the coil as is there here this is one side and that is the other side and we need to integrate over this range in order to get the flux linkage and that range is given by a varying from we know that from a equal to 0 where this coil elementary coil is located is an angle ? and therefore this varies from ? – ? to ? and that is therefore z mu 0 rl d ? into integral ? – ? to ? f at cos ?r sin a- ?r divided by lgd plus f at sin ?r cos a- ?r divided by lgq d a. And now let us do this integration the first term will have f of cos ?r f at cos ?r integral of sin is – cos a- ?r you have lgd going from ? – ? to ? this is the first term and this can be written as – f at cos ?r by lgd multiplied by cos of ? – ?r – cos of ? – ? and cos of ? – ?r is the same as – cos ? – ?r so this becomes plus cos ? – ?r and therefore this first term reduces to f at cos ?r – 2 times cos of ? – ?r by lgd the second term is f at sin ?r multiplied by sin of a – ?r divided by lgq going from ? – ? to ? which is f at sin ?r by lgq into sin of ? – ?r – sin of ? – ?r and sin of ? – ?r is – sin ? – ?r so this gives us 2 times f at sin ?r sin of ? – ?r divided by lgq and therefore the total expression then is ? ?s is equal to this z times µ0 rld ? µ0 rld ? into 2 times – you can take f at also outside – cos ?r cos ? – ?r by lgd plus sin ?r sin of ? – ?r divided by lgq so this is the flux linkage of this elementary coil which we have determined and now in order to determine the flux linkage of the entire phase spread we need to sweep this over the entire phase spread this angle ? can be anywhere from this level to that level and how do we do that therefore the flux linkage ?s of the stator winding can be determined as integral of this ? ?s the angle going from what we need to do is to integrate from here to here and that this angle that is the angle at this point is nothing but ? – ? by 2 from here to the centre point here is ? – ? by 2 to ? plus ? by 2 and therefore ?s going from ? – ? by 2 to ? plus ? by 2 this time the integration is with respect to ? it is ? that goes from ? – ? by 2 to ? plus ? by 2 and therefore that would give us ?s is equal to f hat z ?r into l multiplied by 2 into cos ?r by lgd multiplied by sin ? – ?r the – sin remains going from ? – ? by 2 to ? plus ? by 2 plus sin ?r by lgq into cos of ? – ?r and therefore this becomes – going from ? – ? by 2 to ? plus ? by 2 and this term gives us sin of ? plus ? by 2 – ?r – sin of ? – ? by 2 – ?r so this is ? – ?r – ? by 2 and this can therefore be written as – ? plus ? – ? so 2 times sin a cos b that is sin of ? – ?r cos ? by 2 is what you would get to be more specific what another way in which this could be written ? plus ? by 2 – ?r is – sin of ? by 2 – ?r and this expression is ? – ? by 2 plus ?r that is sin of ? by 2 – ?r which gives us – 2 sin ? by 2 – ?r and this can be taken forward to further simplify the expressions let us look at further simplification in the next class and see how the inductance term is really going to look like for this case of distributed stator wind you will stop here and continue in the next class.