 Welcome to this next lecture on the course on modeling and analysis of electrical machines. We have looked at many of the elementary ideas in modeling machines and it would be helpful at this stage to review what we have done before we go ahead with further analysis. So if we remember we started out looking at electrical machines, we said that electrical machines are electromagnetic equipment which have an electrical input or an electrical side and the mechanical side and it is important that there is a magnetic field established inside the machine through which the electrical port interacts with the mechanical port and it is therefore extremely essential that we understand how this magnetic field is established and impacts the interaction between the electrical and the mechanical sides of the machine. And we said that in this interaction if we ultimately want to reduce this behavior to an equivalent set of equations which establish the relationship between what is happening here and what is happening here and to do that we said that if you want to establish a set of equations we need to look at it as an electrical network from this side which then has an impact on the mechanical side and therefore we need to derive an equivalent electrical network and in deriving the equivalent electrical network we said that inductances are very important. You have a resistance as well but it is the inductance that is going to determine the nature of interaction between the electrical and mechanical sides. We said that the inductance is a representation from the electrical network of what is happening at the magnetic field. So it is this element that then provides a link between the magnetic field and the electrical aspect. In order to establish this the importance of the inductance we looked at certain simple systems. One was the linear motion system where we found that the force is equal to half of I2 into dl by dx and then we looked at a rotational system where we said that torque is equal to half of I2 dl by dθ in one case where it was a cylindrical stator plus coil and then a salient pole rotor. Then we also said that if it is going to be a cylindrical stator and a cylindrical rotor then in order to generate some useful mechanical output the rotor must also have an excitation system then in which case we also derived another expression which is Is into Ir into d by dθ of the mutual inductance between the stator and rotor. So if you look at all these expressions we can understand that it is the variation of inductance with respect to displacement as the rotor moves or as this iron bar moves with respect to motion if the inductance is going to change that is when you get a useful mechanical output and therefore it is necessary to establish in electrical machines how this inductance is going to change with respect to rotor angle. After we determine this of course then we need to use our equations which say that applied voltage is equal to the resistance drop plus the rate of change of flux linkages this was the equation that we wrote for the linear motion case and this was the equation with that we wrote for these two cases as well. So once we know how the inductance is going to change with respect to angle then it becomes easy to substitute that in this equation and finally derive an overall system of equations that links the electrical side and the mechanical side. So this is your electrical equation and the mechanical equation for example for the linear motion case would read something like F- the load I will call that as F load is equal to mass into d square x by dt square and we have also seen some examples of how to solve this and the nature of the response that one gets that is force with respect to time and motion of the system displacement with respect to time all this can be obtained by solving these equations. If the system is non-linear if it is a magnetically non-linear system then one cannot write equations in this manner one has to take records to energy behavior energy based approach and for this we then saw how one can describe the field energy and then the co energy that is WF and WC and if you know WF and WC we have seen how one can find out the torque from WF or WC. So if the system is non-linear then you would go about using these kind of approaches whereas if the system is linear then you can write down equations in terms of an inductance and this earlier analysis shows us that it is important to know how inductance is going to change with respect to displacement either linear or angle in order that we understand the machine understand estimate how the machine is going to behave. So with this understanding then let us go ahead and try to see how the electrical machine is going to behave. Electrical machines as we know are mainly rotational so we will be looking at rotational electrical machines in this course not the linear machines and rotational machines as you would have learnt already consist of a stator and a rotor right on the stator then you have windings that are placed on the stator the rotor also may have windings this is stator and that is the rotor the rotor may also have windings and there is going to be a magnetic field produced by the stator as well as the rotor and they would interact. Now let us look at before we go to the actual machine let us develop it stage by stage and let us first look at a machine with a cylindrical stator and a cylindrical rotor you can see that it is a perfect approximately shape here and we assume that this air gap is uniform throughout this is your air gap that air gap is assumed to be uniform in the case of cylindrical stator and cylindrical rotor that is the system and let us say that this system has a slot here and a slot here and there is a coil which goes around these two slots which means we are taking a section of the cylindrical machine the cylinder is long like this and what we are drawing is a sectional view on the section what you see is one slot that has been running axially another slot below that runs axially and there is a coil that goes like that this coil let us say is stator now has a coil of NS turns this is a simple system let us not look at a coil on the rotor now coil is there on the stator we want to determine inductances so if you have a coil like this the only thing that is now there is we need to determine the self inductance of this coil since there is only one coil there is nothing else to say about it only self inductance parameter is sufficient so how to find out self inductance we know that flux linkage flux linkage is given by inductance multiplied by I and if we are somehow able to establish how much flux linkage is there then you can divide it by I and get the value L that is what we need to do so how to find out how much flux linkage is going to be there so let us say now that the stator is excited with stator excited by a current IS ampere the flow of current in the stator is IS let us say that the flow of current is such that it is coming out of this slot so this is a dot here and it is going into this therefore this is cross here and we know that with this kind of an excitation this coil will produce a field which is oriented in the horizontal direction towards this you can see that most of this is iron all this stator is composed of iron the rotor is also composed of iron and this is the air gap and we know that the permeability of iron is much larger compared to that of air and therefore for all purposes of analysis we assume that mu of iron is infinite very large mu which therefore means as we have seen in the earlier analysis whenever there is an iron element the MMF that is generated by this is dropped primarily across the air gaps and iron consumes negligible MMF and therefore if we now look at what is the nature of the field that is generated by this excitation you would find that the field lines go this way cross the air gap radially and then continue into the rotor cross the air gap radially again and then come out like this and so you get field lines that go this way and we are saying that this MMF is dropped primarily across the air gap here and this line if you see crosses the air gap here and then crosses the air gap here as well both lengths are the same and therefore what we can do is apply Ampere's law around this loop when we get NS x IS is the amount of Ampere's contained in that slot this should be equal to H at the air gap multiplied by length of the air gap this is at on the left side and similarly H at the air gap multiplied by the length of the air gap on this side and since these two lengths are equal H values at these two sides would also be equal and therefore it is two times Hg x Lg and therefore you get Hg is equal to NS x IS divided by two times Lg G stands for air gap now we know H and since the material is air we know that B is given by µ0 x H and therefore we can write Bg as equal to µ0 NS x IS divided by two times Lg this is the air gap field how much flux density is present in the air gap and we see that flux density is independent of where we are measuring it because H independent of which part you take whether you take this radius or this radius or another line here or another line here at all these places H values are numerically it is always equal to NS IS by two times Lg therefore the field in this region is the same it is uniform flux density in this area uniform flux density here so from this then we have found out what the flux density is if you draw this flux density waveform as a function of angle that means let us say that you are looking at a reference axis here which is a equal to 0 and then you start moving along the circumference of the stator go around like this and come back you would have covered 360 degrees this is a is then the circumferential angle that you encounter as you travel along the circumference of the stator or the rotor now because there is a field density you have a density of flux Bg that exists along the air gap one can now try to find out as you walk along the circumference how this air gap flux density is going to change and that is the expression that we have derived here this expression says that the air gap flux density is independent of angle and therefore if you plot this flux density as a function of a if you plot Bg this says that Bg has a constant value whose value is equal to µ0 NS IS by two times Lg from a equal to 0 to a equal to p that is a equal to p is at this point as you move across on to the other side now you see that if flux lines are entering here flux lines must be leaving here if flux lines are entering here they should be leaving here in this case this current is flowing out so you have flux lines that are entering this the flux lines would be oriented this way and therefore flux lines enter the rotor here flux lines leave the rotor here and therefore this flux line if you take this flux line as something greater than 0 then this will be less than 0 so you would have a flux density waveform that looks like this which is a square wave of amplitude µ0 NS IS by two times Lg so we know how the flux density is going to vary now we need to find out what is the flux linkage that is there in this coil now flux to find out the flux linkage in that what we need to do is to find out the flux that is crossing an area spanned by these two sides for which we will take an area which lies along the circumference we consider this circumference half the circumference of the stator and remember this machine is a cylindrical machine and therefore this into the length of the machine will give you an area through which this flux is going to go so that is what we need to find out and since the flux density is uniform along this in order to arrive at the total flux that is passing through this area it is enough to take the flux density and multiply by the circumferential area through which it is going if the flux density is not uniform if it is going to be some way in which it is going to change along a then you cannot do it this way you will have to do an integration which we will see later but for now in this particular case the flux density is uniform and therefore the total flux that is crossing the air gap is nothing but bg multiplied by that area the area is pi into r multiplied by the length of the machine and therefore the flux linkage side is nothing but the stator number of turns multiplied by this flux which is bg into pi into r into l the inductance is therefore psi divided by i and i is there in the expression for bg therefore you get µ0 ns square into pi rl divided by 2 times lg µ0 ns multiplied by ns into pi rl divided by this 2 times lg comes as it is so this is then the inductance of the winding that you have put here note that this inductance is independent of the rotor angle which is fairly evident if you look at the way it is we have already observed in an earlier lecture that inductance in a linear system that is where the electromagnetic circuit the material is linear not linear motion but the material is linear in such a system inductance is a function of geometry that means of the physical arrangement and the layout of the various elements which are there now if you look at this case you have a cylindrical stator and a cylindrical rotor now imagine that you are rotating the rotor by a certain angle if you rotate the rotor by a certain angle still electromagnetic field that is there would not differ in anyway because the field would not be able to distinguish the angle of the rotor as it is placed here or as it is placed here because in any case it is always fully symmetric arrangement and therefore we expect that since the field is not going to change depending on what exactly the angle of the rotor is the inductance should also remain the same it would not be affected along similar lines then we can derive an expression suppose you have a coil on the rotor you have a coil on the rotor what would be the self inductance of this if you excite this coil this will generate a flux which would look very similar you assume that there is dot here and across here and this would also generate a field which goes in this manner which crosses the air gap along the radius and then completes the circle around the stator crosses the air gap here and goes around we can analyze it in the same way all these equations instead of ns is it would now have nr into ir because the excitation is from the rotor side assume that number of turns in the rotor is nr and therefore you will have nr ir equal to times hg into lg and then hg will be nr ir by 2 times lg the graph of the flux density will remain the same similar instead of ns it would be nr is it will be ir and so on and finally you will land up with an inductance equation which says that lr is equal to µ0 into nr2 multiplied by ? into r into l divided by 2 times the length of the air gap. So here again we see that the inductance is independent of the rotor angle and this is also obvious looking at the way the machine is here also whether the rotor is at one position or another angle the field distribution is unchanged because the system will not be able to distinguish the difference between having the rotor here or having the rotor here they are all the same and therefore ls is given by the simple expression lr is also given by a similar expression independent of the angle of the rotor. However if you now look at mutual inductance there has to be a difference if you take a cylindrical stator and a cylindrical rotor let us say you have a coil that is placed here in you have a slot here and then you have a slot here on the stator you have a slot here and you have a slot here how to find out the mutual inductance there is a coil that we will assume is here and here for the self inductance what we did was you excite if you want to find out the self inductance of the stator what we did is excite the stator first stator is excited by some current is and once you excite the stator it will have generate a certain magnetic field and it will have some flux linkage and then the flux linkage divided by the stator current gives you inductance that is the self inductance if you want to find out mutual inductance between the stator and the rotor you excite the stator excite stator find flux linkage of the rotor this current is is then the mutual inductance between stator and rotor is the flux linkage on the rotor divided by the current that you have given on the stator this will be the mutual inductance one can do in the other way also you could excite the rotor with IR find out the flux linkage on the stator ?S then the mutual inductance is ?S divided by IR this way also one can do which method you would like to depend upon which one is easier to analytically direct both will give the same expressions so in this case if one is going to do this what you would need to do is now let us say for example we are going to excite the rotor and if you excite the rotor this is a dot here and across here therefore field lines would be going this way so you have field lines that are going here and these field lines would then cross the air gap like that and then close the circuit around this way now if this is what is going to happen now if you plot this field variation let us call as we have done earlier this angle is a equal to 0 and then you start walking around on the circumference of the stator you go through various angles a and then you try to find out how much flux how much field is there on these area now we have just now seen that independent of where the rotor is the magnetic field that is generated that is BG values around the rotor is not going to change because the stator is fully cylindrical the rotor is also cylindrical air gap is the same everywhere so irrespective of where you put the rotor the flux density around the air gap is not going to change however as you rotate the rotor the flux density distribution moves along with the rotor obviously if you look at the rotor as is drawn here at these locations the flux density is going to undergo a change until this point the flux density was such the flux lines were moving out but at this point flux lines are moving in that represents this graph flux density is greater than 0 until pi and beyond pi flux density becomes negative now if you rotate the rotor by some angle then if this slot moves to here then at this point the flux density would undergo a change similarly the other side of the slot would come here other slot would come here then at this point the flux density would undergo a change as we rotate the rotor the place where the flux density undergoes a change from greater than 0 to less than 0 that undergoes a change which means that if you draw as a function of a BG now this position is here let us say that we determine the rotor angle by a line that is drawn at 90 degrees to this axis that is then the rotor angle ?r taken with reference to this horizontal right. Now if this angle is ?r then it is obvious that this angle is also ?r so that means now the flux density distribution would go from negative to greater than 0 at an angle ?r so the flux density distribution would then look like this this angle will be ?r and it would be negative and it would go again to positive at 2 ? plus ?r the amplitude of course remains the same µ0 nr2 Ir divided by 2 times lg this is ? ?r and this is negative of ?m2 so as the rotor moves this flux density distribution bodily moves along the a axis whereas you have the stator 2 sides which are fixed it is only the rotor flux density distribution that moves the stator is always one coil side is at a equal to 0 another coil side is at a equal to ? this is one coil side of the stator one another coil side of the stator and now we need to find out what is the flux linkage that now links that links these 2 sides and one can see that the flux linkage is if this level is given by Bm then the total flux that is passing through this area is then Bm multiplied by in the earlier case it was simply Bm multiplied by ?r into l but now since some part of this is going to have negative flux density you need to subtract that so in this case then it would be length multiplied by ?r into this span is ? – ?r which will give you this area and you have to subtract this area which is – ?r so the area that is enclosed by we need to look at the effective area as this as the rotor moves as this ?r moves to some position then rotor the flux density distribution would be here and the effective flux density is B multiplied by ? – ?r is the area on top negative value of that Bm multiplied by ?r is this value then this – that is then going to give you your net flux linkage this is this multiplied by ns is the flux linking the stator so this is then your stator flux linkage and Bm can then be written as ?0 nr2 ir by 2lg so that can be substituted there in order to get ?s so this would give us not nr2 it is nr so ns into µ0 nr into ir divided by 2lg into ?r – 2r ?r this is ?s and therefore msr is equal to ?s msr is equal to ?s divided by ir which is equal to this so one can then derive an expression like this for the mutual inductance as a function of angle so mutual inductance thus therefore depend on the rotor angle at ?r equal to ? by 2 at ?r equal to ? by 2 you see that ?r – ?r would be there so mutual inductance is equal to 0 msr equal to 0 at ?r equal to ? by 2 and that other angles it then varies in this manner right so one can determine mutual inductance like this in this particular case it is fairly straight forward because the air gap is uniform all around. Now there are occasions where air gap is then not uniform these kind of machines are called as cylindrical rotor machines the stator is always assumed to be cylindrical in our analysis cylindrical stator one can of course have a non cylindrical stator and a cylindrical rotor or a non cylindrical rotor and a cylindrical stator that also we would see as we go along right because these two are this is cylindrical this is also cylindrical air gap is uniform and because air gap is uniform it is rather easy to derive values or expressions for self inductances and reasonably easy to derive the relationship for the mutual inductance as well. In reality while some machines are of this variety cylindrical stator cylindrical rotor type mostly I mean the induction machines are of the cylindrical stator plus cylindrical rotor variety whereas if you look at synchronous machines there are both varieties available. You have cylindrical stator plus cylindrical rotor these are simply known as cylindrical rotor machines it is implicitly assumed or understood that stator is cylindrical you also have cylindrical stator plus salient pole rotor so these are called as salient pole machine and these are more difficult to analyze. You have DC machines which are then salient pole stator and cylindrical rotor so here the saliency is on the stator rotor is cylindrical in the case of alternators also you could have an arrangement where salient pole stator plus cylindrical rotor this kind of an arrangement is called an inverted synchronous machine though it would work predominantly like the synchronous machine it is normally not the arrangement that is used that is why it is called as an inverted synchronous machine. So these are the broadly different varieties of machines you see that many of them have cylindrical stator and cylindrical rotor while some cases you have one of them are is having a non cylindrical structure there are other machines which fall under the group of variable reluctance machines where you have salient pole stator and salient pole rotor saliency on the stator as well as saliency on the rotor a typical example of this is a stepper motor there are also other machines called switched reluctance machines. So here you have saliency on the stator as well as the rotor now in our analysis what we are going to do we will not be looking at variable reluctance machines but we will be looking at modeling the other three varieties and of these varieties we have seen our goal was to find out expressions for the inductance depending on the rotor angle and we have seen that it is rather easy to do that for the case where stator is cylindrical and rotor is cylindrical that means for the case of induction machines it is rather simple to derive expressions for the self inductance for the synchronous machine where cylindrical stator and cylindrical rotor are there also it is fairly straight forward but not in this case and this case. So here we have one member salient another member which is non salient so in the next lecture then we will try to look at how one can derive inductances if one of them is not a cylindrical member. We need expressions for self inductance mutual inductance and in self inductance we note that there is a leakage inductance and magnetizing inductance leakage inductance by definition is refers or is an effect of that part of the flux which does not cross over the air gap and link the other system that is there does not link these NR turns in the derivation that we have done even for the self inductance we have assumed that the flux that is there crosses the air gap links the rotor and then goes back into the stator and therefore the expressions that we have derived for the self inductance really refer to the magnetizing component of the self inductance. We have not done anything about the leakage inductance part it is very difficult to establish leakage inductance analytically because we really do not know how this flux is going to go and how much is the length of the flux part and therefore leakage inductance is not normally derived by analytical expressions in this manner empirical relations are used in order to determine leakage inductances and therefore we will simply in our analysis we will simply refer to this leakage inductance as LL and what we have derived as the inductance refers to this part and therefore the total self inductance will then be the leakage inductance LL plus this expression that we have derived that is µ0 ns2 x RL x P divided by Lg this is the stator inductance the rotor inductance would then have some other LL plus nR2 divided by Lg mutual inductance of course is this expression. So with this then we will end the lecture for today and in the next session we will see how to derive expressions for the inductances in the case where you have salient pole arrangement in one of the members we will stop here for today.