 our course so far is concentrated on the general analysis of dynamical systems. We now move on to an important part of this course that is the modeling of several power system components. In particular we will concentrate in the next few lectures on the modeling of a synchronous machine. Now, a synchronous machine is a very important you can say the most important component of a power system. The modeling of which requires some a little bit of effort will actually be taking out the transient model of a synchronous machine. We will of course, learn some interesting new concepts like the use of a time variant transformation in order to make the synchronous machine flux differential equations linear time invariant. But to put matters first in perspective just let us look at where we stand in relation to the content which we intend to cover in this course. So, let us just you know shift our attention to what point we are exactly in this course at the present time. So, we have actually done the analysis of dynamical systems in particular we studied the use of Eigen value analysis in linear time invariant systems and we also learned the techniques for numerical integration of ordinary differential equations. We now move to the modeling of a synchronous machine will be further of course, considering the modeling of several other components like excitation systems prime mover systems and in fact, transmission lines and so on. Of course, there is a lot more to come remember that the part b of this course you can say will really try to eventually try to get a few inferences about the nature of the behavior of interconnected power systems. We shall also look at power system stability analysis tools and how we can use all this analysis to improving system stability. So, now we really move on to modeling of a synchronous machine will be instructive to first look at how practical synchronous machine looks like. In fact, what I will show you right now is steam turbine driven generator it is of around 210 megawatts you can look at this photograph. What you see in white here is in fact, the synchronous machine and what you see in green are the various stages of the turbine. The steam turbine which drives this machine right on this side are the bearings and what you will see what is seen here right at the end is actually the slipping and brushes which convey the excitation voltage to the field winding which is of course, rotating in this synchronous generator. So, this is how a steam turbine driven synchronous machine looks like of course, our analysis will cover all means both steam turbine driven as well as hydraulic turbine driven synchronous generators. So, let us move on to the nuts and bolts of this just some basic review a synchronous machine. In fact, has got two kinds of windings you can say the armature winding usually present on the stator are in fact, three phase windings there three sets of windings and you have got a field winding which of course, is fed with d c voltage. The armature is laminated the armature is laminated because the armature winding which are housed on the armature in fact, see a time varying flux and important relationship which you should keep in mind is that the voltages which are generated the M f's the alternating voltages which are generated in the armature due to a rotating field the d c field created by the field winding as well as when the machine is loaded you will have a rotating field created by the three phase armature windings. The M f generated have the frequency which is given by omega e which is the electrical radian frequency is p by 2 times the mechanical rotational frequency where p is the number of poles. So, you have got omega e is equal to p by 2 omega m normally steam turbine driven machines are in fact, high speed generators they run at in our 50 hertz system they run at 3000 rpm and are two pole machines. On the other hand hydro machines are low speed and have salient pole construction which basically have many poles I mean many pole pairs. So, the you know you know steam turbine driven machine of course, we have got one pole pair in salient pole machines you can have many many pole pairs. Now, usually the salient pole what is the what are the other electromagnetic what is the other electromagnetic which environment which you see see of course, the field winding and the armature winding are housed on a ferromagnetic material, but we also have what are known as damper windings or basically squirrel squirrel cage kind of damper bars on the pole phase of salient pole synchronous machines. So, that is usually seen of course, in a steam gas turbine which is high speed you also have a round rotor construction they only two you will have only one pole pair that is two poles and they have got a cylindrical pole or a round rotor and the rotor is usually made out of solid steel forging. So, even the rotating you know the rotating field winding is housed on a solid steel body and because of that you can have eddy currents in the rotor during transients. So, that is one important point you can have eddy currents in the rotor during transients in a steam or gas turbine of course, in a low speed turbine salient pole machine the pole phases are laminated. Now, this is how usually your steam turbine or rather hydraulic turbine pole phase would look like. So, there is a pole phase you will find that the field winding will be housed here and this on this pole phase you have these damper bars which are short circuited at the at the end of the. So, your damper bars they are not. So, the damper bars are kind of embedded on the pole phase and then they are short circuited at the end. So, they basically are like a cage on a induction machine. The basic function of these damper bars is basically to you know damp out transients you know normally both the field wind the flux created by the field winding as well as the three phase rotating flux as well as the rotary motion of the rotor are all of the same speed. So, you do not have actually any may have generated on the on the bars of this damper bars, but during transients of course, you have got currents in these and they are intended that intended to damp out any transients which are occurring. So, in fact in a cylindrical rotor machine the solid steel body of the rotor itself creates eddy currents which perform the same function. On a cylindrical rotor the field winding is housed in slots of this kind. So, of course, the slots have a wedge which is not very clear in this there were wedge to keep the windings in place. So, you have got the windings are basically placed in the slots in a cylindrical pole machine. Now, just to give you a good feel consider 210 megawatt machine generator that is the typical parameters could be for example, in fact this 210 megawatt machine is typical generator which was used in the electrical power grid the many many many of them at present in the Indian power grid. The MVA is 247. So, you can actually calculate the rated power factor from this and this the rated voltage is 15.75 kilo volts the rated current is almost 1 kilo well 10 kilo amperes roughly that is 9000 more than 9000 amperes and it is of course, a 50 hertz machine. Now, when I say rated current it means it is the maximum continuous rating of the armature current. So, when I say rated current is the armature current this of course, does not mean that you cannot have currents higher than this for a short while every electrical equipment can tolerate about you know 10 to 15 percent over current, but only for about 10 to 15 minutes. So, the maximum continuous rating is around 10 kilo amperes of course, there is a lot of heating due to this. So, a machine like this is cooled by hydrogen similarly, the field has got a rated current of 2600 amperes when I say rated current what it means is that the current which flows in the field winding in order to get the rated voltage under loaded condition that is full load conditions or the rated load conditions. The rated voltage similarly, is 310 volts and the efficiency can be very high it is in this case around 98.5 this is only of the generator I am not talking of the turbine etcetera. So, this is the efficiency only of the generator one important point which you should note which we shall again discuss later is that the field current rated current is 2600, but under no load conditions the amount of current required in the field to generate rated voltage again under let me repeat under no load conditions that is open circuit conditions the amount of current needed to generate this rated voltage at the armature terminals is in fact much lower than this it may be half or slightly lower than half of this. Remember that once we load the machine the flux picture in the machine gets changed because of the current in the armature winding. So, you may actually have to have more current in the field to get this rated voltage. So, this current in fact is the current required under rated conditions. So, under no load conditions to get 15.75 you may require less than half of this current and corresponding less than half of this voltage. So, this is an important point which you should remember now there are three kinds of limits which you may encounter I will just write this on a separate sheet of paper. There are three kinds of limits which you may encounter one is the inner generator is the armature current limit or rather I should say the armature heating limit which is of course, determined by the armature current. The second limit which you will encounter is the field current limit. So, you cannot have current more than a certain value others it will cause heating simple I square r heating which may be more for which there is of course, a limit. So, you should ensure of course, that the current will never go above the rated current that is in this case in the example which I said 9050. For of course, for a short while as I said you can have a slight over rating similarly a similar thing applies for the field also the field current should not exceed its rating. But of course, for a short while you can have a higher than the you can have higher than the rated current for say about 10 to 15 minutes that is that is permissible. Now, there is so these are the two things which prevent you from you know there they kind of limit the amount of you know power and reactive power you can have. In in in short you can never exceed the MVA rating of the machine. So, your root of p square plus q square p is the real power and q is the reactive power should not exceed 247. Otherwise you will find that since the voltage will always be maintained here about this value the current will get exceed current rate current will exceed its rating. So, your MVA is typically will always be of course, you have to maintain root of p square plus q square less than 247. Now, the field current limit also prevents you from increasing the q beyond a certain point remember in a synchronous machine by changing the field current you can change the reactive power. But you cannot go on increasing the reactive power after a point you will find that the field current hits its limit. So, even though you have very low real power you cannot have q very high because for that you require very high field current. So, our p and q in some sense are limited by both these limits. So, you have to find out the you know safe operating region for a synchronous machine and ensure that these limits are not exceeded. In some machines since in some steam turbine generator driven generators you will find that you have another limit is called the core end heating limit. This is caused by higher axial flux in the end region of a synchronous machine. So, if you got a synchronous machine this is called a end region the end region of a synchronous machine where actually the windings come out and go in again. So, that is called a end region of a synchronous machine. Now, it is seen that under if you under excite the machine that is reduce the field current such that the synchronous machine starts absorbing reactive power you have a situation of higher axial flux. Of course, the explanation of this is beyond the scope of this course, but I refer you to a paper which appeared quite some time ago by Fanham and Swarthout just a moment. So, I will just write down this reference for you. If you are interested in knowing the effect of all these limits I refer you to this field excitation in relation to machine and system operation. So, this is the title of the paper by Fanham and Swarthout and it appeared in IEEE transactions A IEEE because there was no IEEE at that time in 1953 the page number 1 2 1 5 2 1 2 2 3. So, this is a copy of this paper. So, I request you to those who are interested in finding out more about why we have this heating limit you can of course, refer to this paper. Now, of course, this core end heating is not a limit seen by all machines it is only for specific usually steam turbine driven machines. Now, one the main objective of this lecture really is to discuss the modeling of a synchronous machine. So, before we go into modeling we should of course, look at what are the basic assumptions that we have. Now, remember that no modeling can be perfect in the sense that we for example, in fact, if we try to do some kind of perfect modeling you will find that the problem becomes altogether intractable. So, if somebody says that well I will use Newton's law as well as Maxwell's equations and derive all the equations of a synchronous machine right from scratch with no assumptions whatsoever you will find that it is almost impossible to get anywhere. So, what we will do is do a simplified modeling with the understanding that the modeling which we are going to do will be valid for certain kind of trends. In particular we will be actually taking out what is known as a lumped circuit model of a synchronous machine. So, obviously this kind of model will not be adequate to you know study phenomena which are which are affected by parasitic and so on. So, what we will have is a lumped kind of circuit model and roughly you can say that phenomena or phenomena which have rates of change which are or which kind of last more than say 1 millisecond can be studied reasonably well by this kind of model. Now, I will just read out the main assumptions which we are going to make. One of the important assumptions we are going to make is that there is going to be a sinusoidal distribution of MMF in an air gap. So, the air gap in the synchronous machine will see a sinusoidal magnetic motive force. So, this is an assumption actually we do have almost sinusoidal distribution of MMF because the winding arrangements which are used in a synchronous machine. Another assumption we will make is that saliency is restricted to the rotor that is we will not be considering the saliency due to the slots on the stator winding on the stator. So, the as you know the armature winding is housed on the stator and you have got slots in it to house the windings. Now, we will neglect the saliency which occurs or the changes in reluctance which occur due to the slots on the stator. So, only saliency which will be considered is due to the rotor structure. Another very important assumption which we will make is that saturation and hysteresis are neglected. So, we can consider the machine as more as a linear magnetic circuit and because of this assumption we will be actually be able to simplify our analysis to a very great extent. Of course, at some point or the other you may have to relax this assumption of saturation to get realistic answers, but to get a basic model we will make this assumption. What we will do is we will create this model based on this assumption and try to have some kind of correction to that model in order to take into account saturation. So, this is a kind of pragmatic approach we can follow and as I said sometime we will be getting a lump circuit model will not be modeling things like parasitic capacitances or winding capacitances and so on. So, our model is effectively restricted to the study of relatively slow transients not fast transients. One more thing of course, which I mentioned sometime back is that if you look at the angular position or the rotor position and if you look at the phase of the EMS which are developed or for example, speed of or rather the frequency of EMS which are developed on the armature winding they have this relationship. This is something which we did sometime back the electrical speed or the frequency of EMS generated on this armature windings is equal to p by 2 times omega m where omega m is the mechanical frequency in p is the number of poles. So, just remember this before we go on for the actual analysis what we will do initially is kind of do a simplified analysis of a synchronous machine of a two pole machine and we will try to infer later what would happen if you have got many poles. So, what we will do essentially is consider first of all a two pole machine actually whatever we will be doing does not we do not lose generality because later on we will see that it really does not matter whether it is a two pole machine or four pole machine as long as we take into account the correct equations, but the basic derivations look practically similar. We represent a synchronous machine this is a schematic representation of a synchronous machine you have got three windings on the stator this is the A winding B winding C winding we assume that the rotor moves in this direction. So, whatever happens in the A winding happens to the B winding after sometime. So, there is a phase lag of 120 degrees between the B and the A winding and the C and the B winding respectively. Now, one important point now this is an important thing what are the electromagnetic circuits on the rotor of course, there is a field winding. So, you have got this what you see here is the field winding. So, that field winding is of course, fed from a DC voltage source. Now, one more interesting point which you should notice that you also have other electromagnetic circuits though they are not necessarily in the form of windings what we have are for example, in a synchronous machine a steam turbine driven generator may have eddy currents on the rotor in case of hydraulic driven turbines you have got actually a squirrel cage winding squirrel cage like winding on the pole phase. So, what you really see is that the effect of these the eddy currents or what are known as the squirrel cage currents or the damper currents as these all alternative names. The effect of this is captured using these coils which are short circuited on them. So, after all a damper winding or eddy currents are a kind of a circulating current. So, you have got what basic damper winding is represented by a coil of this kind of course, these this is an approximate representation. For example, in steam turbine driven generators you do not actually have you may not have any specific damper bars the eddy currents are there, but their effect is captured by this winding as well as windings on this axis. So, you can have eddy currents on this axis you can have circulating currents on this axis as well as this axis. This let me just clarify one point the axis on which of this field winding is also called a direct axis the axis quadrature to this direct axis is called a quadrature axis. So, D axis direct axis quadrature axis. So, what we do is model three windings for the phase windings which are on the stator and you have got a field winding and these three other windings are are cumulatively kind of you know representing the damper winding and eddy current effects. Now, you may ask why two why not three or why two why not one the answer to this question is that this is a model and we will try to fit our observations into the model. So, what we will do is we will assume that there are two windings which completely capture the response of a two windings of the quadrature axis and two windings on the direct axis which completely capture the effects of the rotor circuits. So, this is an assumption in the sense that somebody may say you can have more the answer is yes you can have more, but often it is seen as as far as steam turbine driven generators are concerned you require these these many windings and their parameters to represent the kind of dynamics which you will see from actual measurements observations in actual tests which you will do. In fact, for hydraulic turbine machines it may be even possible to represent the whole system by a lower order model that is using lesser number of circuits. In fact, in some cases you may neglect the k winding when we are modeling a hydraulic turbine because a model consisting of these number of windings with their parameters can accurately represent a represent the kind of observations which are seen in the tests. So, right now let us derive the model for you know these three state of windings a field winding and these other three circuits which represent eddy ends cage or damper winding effects. And then if required later we can simplify things by opening one of these windings and get a lower order model, but let us initially get a higher order model and try to represent our system. Remember of course, that when I say that this is a winding this is schematic representation I have talked of the axis of the field winding this is the axis of the a winding this is the axis of the b winding this is the axis of the c winding what do I mean. For example, when I say that this is the a winding and this is its axis what I mean is of course, that your a winding is wound like this is the overhang portion of the end region of the machine this is just one winding. So, this winding means a schematically represented like this and its axis is like this. So, this is what I mean by axis of a winding and a schematic representation like this is actually implying a set of winding which are in this plane and the flux they cause if current flows through them will be like this. So, this is what I mean by axis of a winding. Now, what we need to do of course, when we are representing this as a circuit is compute the various mutual and self inductances between all these windings. So, you may want to for example, have the mutual inductance between this winding and this winding or this winding and this winding and of course, these things will be dependent upon the angle theta. Right now we are talking of a two pole machine. So, there is no distinction between mechanical angle and physical angle electrical angle or the mechanical speed and the frequency of the EMF which are generated. So, they are exactly equivalent. So, right now let us restrict our attention to a four pole machine or two pole machine I am sorry. Now, so what we need to do if you want to describe the machine is basically get a relationship between flux psi a and the rest of the currents I a, I b, I c, I f, I h, I g, I k are in fact the three windings which capture the effect of a d currents as well as damper bar currents in damper bars. So, you have got psi a is equal to a matrix or rather you can say psi a, psi b, psi c, psi f, psi g, psi k are equal to a matrix L into the currents. So, the structure of this L matrix is in fact, L s s, L s r, L r s and L r r remember I a, I b, I c are what are known as the stator currents. So, I will call them as I s, together they are what are known as I s, I r is consisting of I f, I h, I g and I k these are the different windings which we have seen in the previous slide. Similarly, you have psi s and psi r. So, L s s is the matrix relating psi s and I s and it is a 3 by 3 matrix. Similarly, L s r is 3 by 4 matrix L r s is 4 by 3 and L r r is a 4 by 4 matrix because it is the relationship between psi r which is 4 components and I r also which is 4 components. So, now the important thing we need to derive are these components which are there here. So, that is the key challenge now. Now, why is it a challenge? One of the reasons why it is a challenge is the mutual inductances in some cases or even sometimes the self-inductances are dependent on the rotor position. Why is that so? Well, you can easily imagine that the amount of flux linked in for example, this these coils here due to a current here will change according to the position of the rotor. These are of course, on the rotor. So, they will change as per the rotor position that is one thing. Another problem which may occur is because of saliency you will find that the mutual inductance between or the even the self-inductances may change. For example, if you look at a coil, suppose you have got the A winding housed in say just for the timing let us assume it is housed in one slot. So, if you got the A winding housed in this slot it will try to assign if there is a current flow in this A winding. So, the A winding the current say enters here goes in the machine comes out at the end region at the other end and then again enters the machine slot and comes out here. So, if you look at this kind of suppose this is the A winding it will create a flux in this in this direction. So, you will have a flux like this flux of this kind. So, this A winding tends to cause a flux in this direction. Of course, the amount of flux you will you can well imagine that the amount of flux which you will get under in this condition will be much more than the flux you will have under this situation. Why is it so? Because the reluctance of path is going to be more. So, if you have got again current in the A winding suppose it is just one winding you will find that the current the flux which will be caused tends to stick to the ferromagnetic path you know this is the air gap. This air gap is more on this axis and less here. So, you will find that there is flux which is which goes like this, but the flux here will try to tend to stick to this you know of course, trying to find out the flux path requires you to do a fully electromagnetic analysis. But you will notice that the reluctance is likely to be more and the flux created is likely to be less if the rotor is in this position for a salient pole machine. This is not true of course, for a cylindrical pole machine because the you will find that practically the reluctance is the same you know for a cylindrical machine. So, the self inductance of this A winding will also depend on the rotor position. So, that is one important point which you should remember. Now, let us first let us just stick to a cylindrical pole machine. Now, one of the important things which we wish to do is try to see how the flux will be in case for under several circumstances. See what we need to do is if you want to find out the inductance of self inductance or mutual inductance what we need to do is first of all for example, if I want to find out the mutual inductance the mutual inductance between the A winding the phase A winding and say the field winding and what you need to do is assume that a current flows in the A winding that will of course, create a flux an MMF in the air gap. So, let us assume that is only current in the A winding. So, you will have current in the A winding that will create an MMF that will be NA into IA, but of course, this I should call this is not the this is a equivalent number of terms. So, you will have some constant into IA is the MMF which is caused in the air gap. Of course, when I say MMF is caused in the core it is a function of the angular position I will just try to elaborate on this in a few seconds from now. Now, this is the MMF which is caused which is proportional to by some constant NA into IA the MMF will cause some flux, but this is all this is a function of the angular position let us say I call this angular position phi for the time being I call this angular position will call this beta may be this angular position is beta. This will cause a flux this also will have a angular distribution it is not a constant value everywhere in the air gap, then you will have to find out what is the component of flux which is linked with the field winding. So, once you calculate the component of flux which is linked with the field winding while we will be able to compute the mutual inductance between A and F. So, that is the basic you know procedure we need to apply before we get actually what is the mutual inductance between the winding A and the winding F. Now, so how do I actually proceed for example, let us just start off with trying to find out the mutual inductance between the winding A and the winding F. So, if I want to find out this mutual inductance of the winding. So, actually I am trying to take out the first component of this rather yeah the first component of this LSR matrix. So, I am trying to get the relationship in flux linked I am sorry I am trying to get the value of the flux linked with the field winding due to a current I A. So, I am sorry I have what I should be saying is I am trying to find out this component first. So, just let us see the steps which we need to apply. So, if you have got a current in a in the A winding let us say the A winding is present in this slot we will just assume for the timing it is concentrated though a concentrated winding strictly speaking will never cause a sinusoidal M M F in the core, but let us say you have got this A winding in this slot. Now, we need to find out what is the M M F due to this A winding first step. So, to do that of course, it is going to be proportional to the current and a number of turns and how these windings of the A how these A winding is distributed. So, let me just redo this in the slightly smaller sheet. So, that smaller diagram. So, if you have got a current flowing in the A winding let us say is distributed actually, but let me just represent it right now by a concentrated winding. So, this is your A winding and a current flow. So, this now what I have told you before is that the M M F will assume cause by any winding is going to be sinusoidally distributed in the core in the air gap. So, what will assume is that the M M F which is created by the A winding is going to be sinusoidally distributed, but of course, the B winding also causes the sinusoidally distributed M M F in the air gap, but one difference is there. The M M F which is caused by the A winding will be maximum or it tends to cause if you have got a cylindrical rotor of course, it will cause maximum flux in this direction. So, we expect the M M F to be maximum here. So, then it tries to cause a flux like this. So, your flux caused is like this. So, what we see is that the M M F caused is maximum here. Let us call this position as the beta position is maximum here and it tends to drop as beta increases. It becomes 0 here because there is very little flux which is cause in these directions by the current A. In fact, it is practically 0. So, there will be no flux entering the core in this direction. So, what we will assume is that what we really see is that the M M F is tends to cause a flux maximum flux. So, M M F will be maximum in this direction. In fact, it tends to cause maximum flux in a cylindrical machine also in this direction. So, M M F is maximum when beta is equal to 0. So, I will just take this beta is equal to 0 point here. So, this is beta is equal to 0. M M F is maximum here. It becomes 0 when beta is equal to 90 degrees and it tends to cause flux in the opposite direction because it causes a flux which comes out of the air gap here and goes in here. So, the M M F is negative after 90 degrees in this region reaches a peak here. So, this is basically how your M M F looks like 0, 90, 180 and back here. So, just to make our analysis a bit simplified whenever I say that the M M F whenever I will kind of represent this kind of M M F by a vector in this direction. So, if I show a M M F by a vector in this direction, what does it mean? It is maximum M M F is in this direction and drops of sinusoidally like this. So, the M M F of the B winding will be can be represented like this in this direction. So, when I show a vector like this, it means this. If I show a vector like this, what does it mean? It reaches its peak value at 120 degrees here. This is 120 degrees. So, an M M F caused by the B winding current in the B winding will be in this will be represented by this arrow. So, when I should represent a M M F by this arrow, I really mean that the flux distribution in the core is something like this. It is speaking here at 120 degrees. So, this is due to B winding, this is due to the A winding. So, this will simplify our analysis a bit by if we represent a M M F by a single arrow, but a single arrow means this. So, that is something which you should remember and the distribution we assume is sinusoidal. So, suppose I want to take out the mutual inductance between the A winding and the field winding, what do I need to do is first take out the M M F. Now, once I take out the M M F, I need to compute the flux. So, the next step is compute the flux. Now, the flux if it is a cylindrical rotor machine is quite straight forward because the reluctance of a cylindrical rotor, cylindrical rotor machine does not change with this beta. So, a flux will be practically for all practical purposes, if the if there is a current flow only in the A winding, then the flux will be simply a replica or scaled you know scaled waveform which is simply going to be similar to this blue waveform which I have drawn here. So, a flux also will be something like this in the air gap. So, this is what a cylindrical rotor machine, the kind of flux which will be created in the air gap due to the current in winding A. Now, whenever I am going to try to take out the flux linked to the field winding, see after all what am I going to do, now I have got the flux in the air gap. Now, what I want to do is how much of this flux links with the field winding. So, what do I do is I where is the field winding suppose I want to take out the mutual inductance when the field winding is aligned at an angle of theta. Suppose the field winding is aligned at an angle theta, what do I mean by it is aligned at an angle of theta, the field winding is somewhere here is wound here. So, what the field winding is here at an angle theta. So, how much of this flux is actually linking this field winding. So, I will just redraw this again. So, you have got the A winding, it is causing a flux which is as I mentioned I am representing it by an arrow like this. What it means of course is the peak value is in this direction and drops of sinusoidal as beta increases. Suppose the field winding is here is wound here I am showing it again as just one winding one turn, but normally it is wound over the periphery and has an axis here. Similarly, the A winding though I have shown it as just two conductors actually it is wound over the periphery of the machine and it has its axis like this. So, the field winding axis say at a given point of time the field is aligned, remember the field is rotating this rotor is rotating. So, this field is aligned at an angle theta and the flux and this is the angular position theta from this. So, this is beta is equal to 0, beta is equal to theta. Now, the thing is how much of this flux is actually linking with this winding. So, your flux is going in actually if you look at the flux it is in this direction I will just redraw it here. So, that it becomes easy to understand the flux due to the A winding is going in is of this kind like this and maximum of course here. Now, if you have got a coil which is aligned at theta how much of the flux will actually get linked with it. The answer is that the flux total flux you will have to compute this basically the flux density and integrate it over this area this whole area. You can imagine that it is going to be dependent upon this theta. So, what is it going to be? For example, if theta is equal to 0 all this almost all of this flux will get linked with the field winding. In fact, there will be some leakage, but it will be maximum the linkage will be maximum here and reduce reduce reduce and in fact, become 0. So, if your theta is equal to 90 degrees you will find that the total flux linking this winding. So, you know to link what you need to have is the flux goes suppose this is the winding the flux needs to go in. Now, if you look if the flux is like this and the winding is like this nothing is going in it just the flux is in this direction. So, no flux is actually linking this winding if theta is equal to 90 degrees. So, you can imagine that your flux in fact the linkage of flux is dependent on theta. So, it is maximum here minimum here 0 here in fact, the linkage becomes negative because the flux direction is reversed as far as the field winding is concerned when the field becomes 180 degrees. So, one can expect that the mutual inductance between that is the mutual inductance between a winding and the field winding will be in fact, have the form m a f cos of theta. So, that is the most this is an important point that the variation of the mutual inductance between the a winding and the field winding is likely to be of this nature. Let us now look at another term in our inductance matrix that is the mutual induct the self inductance of the a winding this is again one simple relatively simple thing to you know imagine. For example, you got the a winding again I have shown it as a packed or concentrated winding actually it is distributed along the periphery. Now, if you have got the a winding like this it causes an m m f whose vector representation I have just discussed the vector representation sometime ago is this. Now, if it is a salient pole machine again the flux cos will depend on this position because as I said sometime back your the flux length will be position dependent as I said something here like this. The flux will be maximum here will be minimum here and then again maximum here. So, if you have got a current flow and you have got a salient pole machine you will you are again your flux link with the a winding due to a current in the a winding will be dependent on the rotor in case of a salient pole machine. It is not true of course, for a cylindrical pole machine you are likely to have practically the same flux even in this even when the field winding is rotated by 90 degrees. So, it is a cylindrical pole machine that will not be much of an issue, but if it is a salient pole machine you will find I will just use this itself you will find that the flux link is dependent on position. So, you expect the self inductance of the a winding due to of the a winding in a salient pole machine will be maximum at theta is that is when the rotor position theta is equal to 0 will be minimum in this position that is when theta is equal to 90 degrees. So, this is theta and then maximum again when it is 180 degrees. So, one can expect that the self inductance of the a winding in general will be of this form it will be L a a self inductance of the a winding is equal to L a a 0 plus some other term L a a 2 let us say in this particular into cos of 2 theta. So, our variation will be something like this in fact of course, I am not proved it should be a sinusoidal or cos sinusoidal variation, but this follows if you assume that the mmf is actually distributed sinusoidally mmf fluxes etcetera all distributed sinusoidally you will find that in fact the flux linked and the mmfs are distributed sinusoidally you will find that this is the kind of thing you can expect. So, your self inductance term will look like this now more difficult perhaps to imagine is what is the self inductance or rather the mutual inductance between the a winding and the b winding. So, if what is the mutual inductance between the winding the b winding. So, if you have got if you are looking at current flow like this what is the flux linked with the b winding do you remember the b winding b or the c winding sorry I will just this is the b winding I am sorry this is the c winding the axis of the b winding is in this direction. So, when I am talking of flux linked in this b winding which is housed here again I am showing it as a concentrated winding actually it is distributed this is the b winding in green the b winding what is the flux linked with the b winding when current is I a flows through the a winding. So, how do you solve this problem now which is a cylindrical rotor machine there is no major problem as far as deriving this answer. So, if you take a cylindrical rotor machine you have got the a winding and you have got the b winding the a winding causes an M M F in this direction represented by this vector it means that of course, the maximum is in this direction it drops of sinusoidally and the flux also since actually the reluctance is the same all throughout the air gap in a cylindrical pole machine the flux also is can be represented by this vector. But, as I mentioned in case of the mutual inductance between the a winding in the field winding the flux which is flowing in this direction not the amount it links with the b winding is not the same I mean for example, we had discussed this before if I if a flux like this is caused due to the a current in the a winding how much of it links with the b winding now what does the b winding see basically the b winding when I say linked means what is the flux which is passing through the b winding in this direction in this direction. So, only a you can say a component of this is actually in this direction to get this what we do is compute the this flux vector which are drawn here it represents actually sinusoidal variation with a maximum here can be resolved into a vector in this direction that is you take this direction is the axis of the b winding it can be resolved in this fashion. So, I do not know whether it is very clearly visible or not basically what I am saying is that the flux in a cylindrical pole machine caused by the a winding is in this direction I will just draw it by a slightly thicker line so that you can see it there. So, the flux vector the more vector means that the flux is maximum here and drops of sinusoidal here. So, the flux vector is this I can resolve this flux in the b axis direction and the direction 90 degrees to it this is the b axis. So, what really you see is that if there is a current in the a winding and it causes flux like this only a component of this will actually be there which will link with the b winding remember if I am trying to look at the fluxes link to the b winding I should not look at the fluxes which are like this I should which are entering b after all eventually I will be using Faraday's law and Faraday's law relates the flux which is going like this into the winding. So, even if you have got a flux which is maximum here and it is dropping down like this the amount of flux which is linked to the b winding in fact is only a component of this flux. So, how much is that component well you know that the angle between this and this is 120 degrees. So, the flux linking the b winding due to a current in the a winding for a cylindrical pole machine will be simply yeah is it going to be a constant the answer is yes because even if this rotor position moves if this rotor moves this will not change. So, for a cylindrical rotor machine L a b that is the inductance between the a and the b winding will be lesser than the self inductance that is for sure for a cylindrical pole machine this L a b will be constant. Now, you may ask the question what about what happens when you have got a salient pole machine is then the mutual inductance between the a and the b winding a function of the position the answer is yes. The a winding and the b the inductance between the a winding and the b winding is in fact a function of the position because even of the a winding produces a flux in mmf in this direction the flux need not be having a vector representation in this direction because even though mmf is like this the flux is depends on the reluctance of the path and the reluctance of the path is not uniform. So, what we will do in the next lecture is try to get the mutual inductance between the a winding and the b winding in case you have got salient pole machine. So, just to summarize everything quickly the mutual inductance between the a winding and the field winding is in fact the function of cos theta the self inductance of the a winding or in fact the b winding or the c winding for a salient pole machine it is actually a function of cos 2 theta for a cylindrical pole machine it is in fact a constant it is not dependent on theta at all. So, we will actually take a few more representative cases in fact what we are doing now is finding out the mutual inductance between the a winding and the b winding when there is salience in the rotor. So, that is something we will resolve in the next class and there on we will then calculate the we will get the differential equations of the machine and then I will introduce you to a very interesting and important transformation of variables called the d q transformation.