 We continue our discussion on the modeling of a synchronous machine recall in the previous class that we had represented the electromagnetic effects of the rotor by 4 coils, 2 coils in the direct axis and 2 coils in the quadrature axis out of 1 coil in the direct axis. In fact, the direct axis is in fact, the axis of the field winding. So, of the 2 direct axis windings, one of the windings is actually the field winding. The other 3 coils on the rotor are in fact, representative of the damper bar effects and eddy current effects which may be existing on the rotor of a synchronous machine. Now, as I mentioned in the previous class, this is a model of the rotor circuits. There is no hard and fast rule that you can model it only by 2 coils, you can model by rather you cannot model it by 4 coils, you can model it by 4 coils, 6 coils, 8 coils and so on. You can make a more complicated model, but for most purposes it has been found that it is adequate to model the effects of the rotor electromagnetic circuits as these 4 coils, 2 on the d axis and 2 on the q axis. We further try to compute the mutual inductance. We try to get a circuit model of a synchronous machine. So, we need to calculate the self and mutual inductances between various coils which we have. Remember, there are 3 sets of coils on the stator, the phase windings that is and the 4 coils on the rotor. So, I request you to direct your attention to the sheet which we had shown in the previous lecture. You have got a synchronous machine, you have got 3 phase windings and you have got these 4 coils on the rotor. The d axis of the direct axis is the axis of the field coil and you have got these short circuited windings are in fact represented over the damper bar and eddy current effects on the rotor. Now, remember the axis of a winding, you will just recall what an axis of a winding means. The axis of a winding is in fact, if I show the axis is like this, what I really mean is you have a coil like this. For example, state the a winding is represented like this with an axis like this. So, what it really means is the a winding is wound like this and the direction of the flux is like this. So, this is what I mean by axis of the winding. Our basic effort in the previous class which we will continue in this class is towards trying to get a relationship between the fluxes and the currents. So, you have got the fluxes psi a psi b these are flux linked by the 3 stator windings and the 4 rotor windings. They are related to the currents the 3 currents on the stator winding and the 3 currents on the rotor winding by the inductance matrix. The inductance matrix is made out of LSS, LSR, LRS and LRR sub matrices and of course, the sizing is LSS is 3 by 3, LSR is 3 by 4 and so on. So, the first term in LSS that is LSS 1 1 is in fact, the self inductance of the a phase winding. So, that is basically what it means of course, the other mutual inductances and so on. Yesterday in fact, we tried to compute the mutual inductances between say the a winding and the field winding and we saw that it is related to the position of the rotor. In fact, it is related by the function cos theta. So, your mutual inductance between the phase a winding and the field winding is related by a relationship m a f cos theta where theta is the angle which the field winding makes with the a winding axis. You know this keeps on changing this theta is keeping on changing because the rotor is continuously moving. So, that was one term which we kind of derived in the previous class. Now, the basic idea of how to take out the mutual inductances is thus, what you need to do is you from the winding configuration of a particular winding, we compute the m m f if a current was flowing. So, if you have got a current i a in the a winding what we do is compute the m m f in the air gap due to the a winding current. Then what we do is compute the flux. So, if you look at this the basic way of doing things is first compute the m m f in space that is the air gap compute the flux in space or in the air gap. Then you compute the flux linked to a specific coil of a interest. So, suppose you are taking out the mutual inductance between the a winding and the b winding what you will do is first assume a current i a flows in the a winding compute the m m f in space compute the flux due to this m m f from the flux you compute the flux linked with the b coil. So, if you want to take out the mutual inductance between a and b coil this is the procedure which you will take out. So, for example, if you have got a current in the a winding you will have a flux like this. In fact, you will have a m m f which will cause the flux, but let us first assume it is cylindrical core. So, then you will get the flux of this nature. So, if you want to find out the self inductance of the a winding that is l a a what you will do is just find out the flux linked with the a winding due to current in the a winding. So, what you will have to find out is the flux you know the integration of flux density over this surface. So, that will give you how much flux is getting linked with this a winding is it clear. So, however, if you want to find out for example, the flux linked in the b winding due to current in the a winding. So, the flux of course, will remain the same due to the a winding current you assume only a winding current is there. So, you will get the flux as in the previous case you will get a flux, but now you have to find out the flux linked with the b winding. So, the b winding flux would be the integration of the flux density from this point over this surface. So, how much flux is getting into this like this. So, that is the flux linked with the b winding. Obviously, the flux linked in the a winding due to the a winding current is different from the flux linked in the b winding due to the b winding due to the a winding current. So, mutual inductance between a and b the a and b winding would be different from the self inductance of the a winding itself. So, that is an obvious thing. Now, just before we things start getting a bit complicated when you have got saliency. Now, once you have got saliency the step from getting the flux rather getting the flux from the mmf is non trivial. I mean you have to just think over it a bit, because in this particular case which I have considered cylindrical rotor. Once you have got the mmf waveform you know the reluctance of the air gap is uniform in a cylindrical rotor machine almost uniform in a cylindrical rotor machine. In that case getting the flux from the mmf is very straight forward, but if you have got a salient pole machine things can be a bit complicated. So, even getting the mutual inductance between the a winding and a b winding would depend on the rotor position, because the salient the salient the rotor would define saliency in a particular direction. So, just look at before we go ahead with trying to compute this just a minor point about the convention I am going to use. Suppose you have got mmf of you know of this nature this is the angular position the x axis is the angular position measured from this vertical axis. So, if my mmf is like this and reaches the peak value at alpha at an angle alpha in that case I will represent this is a represented I will represent the mmf by a vector which is at an angle alpha. So, just remember this convention which I even use last time. So, if my mmf direction is shown like this what it means is of course, that the air gap distribution of the mmf is like this with the peak at the angle alpha with respect to the vertical. So, now let us go on to this problem of trying to compute the mmf rather the mutual inductance between the a winding and a b winding when there is saliency. So, now things become a bit tricky, because suppose you have got a situation like this this is your rotor. So, you have got the a winding it say let carries a current let say it carries a current of I a. So, where does it create an mmf well it creates an mmf in this direction. So, I will represent the mmf as a vector. So, it is a peak value occurs here and then it drops of sinusoidally as with the angular position. So, mmf is like this, but now the second step of getting the flux is non trivial, because if the if the rotor position was like this that is if theta is equal to 0 you would get a different value of flux if theta is something else you will get some different some other value of flux. So, in fact how do you then compute the flux in the core for any arbitrary position. So, that is one worry which you have now one you know reasonable way of doing this is actually compute the flux for two extreme positions and then compute the equivalent flux. So, I will just this this appears a bit a bit vague. So, I will just tell you what I mean suppose you have got current in the a winding and the rotor has got this position the salient pole is here then the mmf is like this and the flux also is of course, like this. So, the flux vector the same convention I am using the flux vector also will be like this mmf is also like this, but if I if the rotor position is like this the mmf is still here the same flux which will be caused will be much much see I am not shown it on this side is a mirror image the flux caused will be smaller than this. So, if this was so a rotor in this position would would cause of flux like this a rotor in this position would cause a much smaller flux. So, if you are if you have got a salient pole machine and theta is 90 degrees you will get this. Now, what we do is instead of computing the flux for every position what we do is we compute the flux for two extreme positions what I what I mean to say is this. Suppose you have got an mmf like this the mmf is maximum in this direction what I do is compute the component of mmf in this direction what do I mean by that it is easy to see. So, if this mmf is mmf a I call this mmf a and this is theta the component of mmf in this direction is mmf a cos theta and the component of mmf in this direction here is mmf a sin theta. Now, the thing what so what does this actually mean this means that if a mmf I will just draw this here if a mmf is like this it is maximum at the 0 position. So, if a mmf is like this I have actually written down this mmf this is the 0 position I am writing down this mmf this is mmf a as if it is made out of two components. So, it is made out of two components this is 0 this is theta. So, it is made out of two components this is mmf a cos theta and of course, you have got mmf a sin theta of course, the peak value of this and this will be different. So, the if this is mmf a this will be mmf a sin theta cos theta and this is mmf a sin theta the magnitude of this will be different what I want to say is that the sum of these two waves will give you this wave. So, what I have done is what I am trying to do is I am splitting up the mmf into two components I will what I will do is one component is along this d axis the field winding axis one is along the quadrature axis. Now, what I do is I know that if if this rotor salient pole is aligned with the mmf I will get a certain amount of flux I know if the rotor is aligned at this position I will get a certain amount of flux. So, what I do is to compute the flux I know this how much flux I will get if the mmf is in this direction and I know how much flux I will get if I am in this direction. So, the thing is I have split up the mmf into two components I know the ratio of the mmf and the flux when the mmf is in the direct axis I know the ratio of flux and the mmf when the mmf is in the q axis. So, I just remember these two ratios. So, what I do is I got the mmf split it up into two parts I know what mmf what flux I will get in this direction I know what flux I will get in this direction I will superpose the solutions. So, this saves us the trouble of computing the flux for each position. So, what I have done is if I know what the flux the flux I will get for a certain mmf in the direct axis and I know what the flux I will get for an mmf in the quadrature axis I will compute the flux individually in these two axis and then superimpose the solution. So, what I will do is. So, what I have effectively done is got the flux which will result in this direction got what flux will result in this direction. So, obviously, going to be smaller and then get out get the resultant flux in the core. So, this is the basic idea. So, from an mmf due to a coil due to a coil I can get the flux how do I get the flux not by remembering the flux for every position, but for every position I compute the component of mmf in the direct and the quadrature axis and I just remember these two ratios the or two constants which really relate the mmf and the flux in these two directions. So, I just remain to remember the relationship between the flux and an mmf in this direction and this direction rather than remembering it for every rotor position. So, after I compute this please look at this carefully this is the flux component in this direction this is the flux component in this direction I can compute the resultant flux the resultant flux direction and magnitude is like this. So, once I compute that the resultant flux direction and magnitude is like this I can really draw the flux in the core not really draw it, but I know that well it is going to have a peak value in this direction and going to drop of sinusoidally as we change the angular position. So, now I know that the flux is going to be like this in this direction in the core now the question is the second point which you you should do in order to compute the mutual inductance is you have computed the flux the flux in the core you have computed the direction and magnitude. Now, you compute the flux linked with a specific coil. So, what you do is now for example, if you want to find out how much of this flux is linked with the b coil. So, you want to find out how much of this flux is linked in the b coil what you need to do is find out the component of this in this direction. So, what I am asking you is suppose I know that the flux is like this compute the component of flux in this direction in fact, it will be like this. So, in this direction the b axis b winding axis the amount of flux linked by the b winding would be proportional to the component this something you can actually work out is going to be given by the component of this vector along the b axis. So, this is how you would compute the mutual inductance between the a winding and the b winding when there is saliency. So, now what we will do is simply start writing down I will not derive each and every inductance, but tell you the basic form of the equation. So, if you recall what I am trying to do is first of all I will tell you how L s s looks like, L s r looks like L r s and L r r now please remember I am not deriving every inductance out here that will take a very long time, but basic way of doing things please remember compute the mmf due to a current where the mmf is then compute the flux in the core in the air gap. The flux in the air gap is actually found out by in case you have got saliency in that case you take out the mmf component in the d axis and the q axis you know the ratio between the mmf and the flux in the d axis the mmf and the flux in the q axis what I mean is if an mmf existed along the d axis what would be the flux. If an mmf existed along the q axis what would be the flux and then superimpose the solutions two solutions. So, this saves us the trouble of trying to remember all rather the flux for every rotor position. So, you remember the flux and mmf ratio only for two positions and try to derive what is going to be the flux for a given mmf for an arbitrary position of the rotor. So, now I will just write down this LSS which is the LSS remember is the self inductance or rather the matrix which relates psi s, i b, psi c to i a, i b and i c. So, LSS will have this form I will not read out each component, but you will notice that there is a self inductance term LAA 0 plus LAA 2 cos 2 theta. Now of course, if there is no saliency LAA 2 is 0, then there is a mutual inductance between the A winding and the B winding plus LAA 2 into cos 2 theta minus 2 by by 3 this is something you ought to derive I have not of course, derived every component of this inductance matrix. So, if saliency is not there at all you just have a constant matrix you do not have things like LAC and LCC the reason is that it is assumed to be a symmetric machine all it is a balanced machine. So, this will be equal to this, this will be equal to this, this will be equal to this and this this and this will be the same. So, a self inductance of the A winding in case there is no saliency the same as self inductance of the B winding and a self inductance of the C winding in case there is saliency you will notice a cyclic nature with respect to theta. So, whatever happens in this row the row gets kind of shifted in this direction. So, you will notice that this is the basic nature of the self inductance if you find out the mutual inductances between the or the relationship between the rotor axis currents and the rotor windings the mutual inductances and the self inductance of the rotor winding this is the basic structure you will come across. LF is the self inductance of the field winding LFH is the mutual inductance between the field winding and the H winding there is no mutual inductance between the field winding and the G winding. So, that is one thing which you should remember this the mutual inductance between the F winding and G winding is in fact, 0 why is that so just think over it with the assumptions we have made it can be easily shown that there will be no flux linked, but maybe I will just try to tell you why it is so suppose you have got the field winding it creates an MMF in this direction the MMF it creates will also cause a flux in this direction. So, if you got a flux in this direction how much of that flux gets linked in this 90 degrees to it where the other windings rotor windings exist it will be 0 because the component of the flux along this will be equal to 0, but of course, the correct way of actually saying this is the MMF there is a MMF distribution with a peak in this direction it will cause a flux distribution which again has a peak here. Now, if it has a peak here the amount of flux linked by a coil which is has an axis in this direction if you integrate the flux density along this surface you will find that it is 0. So, the mutual inductance between this winding and a q axis winding is actually 0. So, that is why this off diagonal elements are in fact, 0. So, this is the nature of the self inductance matrix the mutual inductance between the various stator windings and the rotor windings have this form in fact, it is an interesting thing is L S R is equal to L R S transpose and L S R itself can be partitioned into two components. So, you have got the mutual inductance between the A winding and the D axis windings what are the D axis windings F and H. So, those for those who have just forgotten or just missed out. So, the what we are doing is now finding out the mutual inductance between the A winding B winding and C winding with the D axis coils and similarly, we have to find out the mutual inductance between A B and C and these coils. So, what I am saying is that so, we can sub partition this matrix into the mutual inductance between the stator and the D axis rotor windings and the stator and the q axis rotor winding. The q axis partition matrix here has got this form again I am not deriving it, but you have to use the same procedure as I mentioned some time back of computing the mutual inductances. Similarly, the D axis L S R D that is the mutual inductance between the stator phase windings and the rotor windings on the D axis that two of them. So, this is a basically three by two matrix it looks like this. There is no difference between the mutual inductance between the A winding and the field winding and the B winding in the field winding except that this dependence on theta is different. So, that is the only difference otherwise the constant of here is the same this coefficient here is the same only the theta dependence is different that is not surprising because the A B and C windings are absolutely identical other than the fact that they are they are not in the same spatial position. Now, the next thing I need to take out of course, just getting the relationship between the fluxes and the currents is not enough I need to apply the physical laws to obtain the relationship the rate of change of flux and applied voltage of course, that is really obtained from Faraday's law. So, just of course, we will just pay an attention since we are developing what is known as circuit model of the machine we need to pay little bit of attention to the sign conventions and the dot conventions. So, please assuming that a dot convention like this is used the dot is on the top dot dot here is the dot what is the dot mean actually the thing is that in case two windings have the same dot and current is entering both windings in the dot they will create fluxes which tend to enhance or rather in the same direction that is what it means. So, if you have got two coils with the dot on top if I align the two coils and the dots are on the top then a current entering the dots will cause a flux in the same direction. So, both the fluxes tend to add up that is what the dot convention means after all again let me repeat if I have got two coils with dots on the top and I align them and current enters the dot at the dot into one coil and it creates flux in a particular direction. Then the second coil also if it if the dot and the same at the top of the coil if the current enters the dot in the second at the dot of the second coil also it will cause the flux in the same direction that is what the dot convention effectively means. So, if you look at the dot convention here the dot is here the dot is here the dot is here dot is here dot is here and dot is here and dot is here. So, just look at how the dots are placed. One important point which you should note that this current convention I am using is that the current is going out of the dot. So, whenever I am going to I will use this convention derive the equations and whenever I get the answer for the current remember that it pertains to the current flowing out of the dot of the winding. So, this is what I mean. So, the if theta were to be 0 if theta were to be 0 and currents were entering the dot they would cause fluxes in the same direction that is what this dot convention would mean. Now, so I just let me repeat again in case you have not got the point suppose a current is entering the dot it creates a flux say in this direction upwards it tends to create a flux like this. I take this field winding if I align it here and a current enters the dot of the field winding it again will cause a flux in this direction that is what the dot convention means. Now, since I have taken the current flowing out of this winding I need to write down my equations a bit carefully. You must be familiar with the typical way of representing mutually coupled circuit if you have got these dots currents are entering the dot then the equations defining the voltages v 1 and v 2 v 1 is in this direction that is the v 1 is the potential of this point with respect to this point v 2 is the potential of this point with respect to this point and the dots are here the currents are entering the dots then the relationship between v and i is like this v 1 is equal to l 1 1 into d i 1 by d t v 2 is equal to l 2 1 into d i 1 by d t plus l 2 2 into d i 2 by d t or just to write it in a in a more familiar notation v 1 is d psi 1 by d t and v 2 is d psi 2 by d t where psi 1 and psi 2 the other flux linked to the first and second coil respectively and the flux is written down like this. So, I am just writing down these equations there is no change I am just writing down these equations in two sets. So, v 1 is equal to d psi 1 by d t v 2 is equal to d psi 2 by d t where psi 1 is l 1 1 l 1 l 1 1 i 1 plus l 1 2 i 2 and psi 2 is equal to l 2 1 i 1 plus l 2 2 i 2 normally under such circumstances l 1 1 and l 2 2 both will be greater than 0 normally the self inductance is in such a circuit will be greater than 0. Now, if I have got currents if my current direction I reverse, but my inductance inductance matrix is same as before then I will have to write down my equations like this. So, my current has got reverse. So, my definition of flux in some sense has got reverse. So, your v 1 and v 2 are now given by this. So, this is Faraday's law would be correct like this if you use this convention that the currents are flowing out of the dot. So, this is also known as generator current. So, remember if you are using the generator convention the correct representation of the equations would be like this is it. Now, what are the equations if you if these are the we have already found out that we have found out the inductance matrices that is psi s psi r this in fact L s s and L r s are transpose of each other i s of course, is i a i b i c i r is i f h g and k similarly psi s and psi r. Now, the thing is that what are the equations if this is the convention if this is the dot convention and you are assuming the currents are coming out of the dots all the currents are flowing out of the dots that is you are using generator convention what are the equations which describe this Faraday's law. So, Faraday's law says that minus d psi s by d t minus of course, if there is a resistance in the winding r s i s is equal to v s of course, this is something I have introduced this is the resistance of the winding and minus d psi r by d t minus r r i r is equal to v r. So, this is basically your Faraday's law as applied to this system. So, basically it is a reiteration of this is it ok. Of course, if there was see the only difference here is I have put a resistance. So, if actually there was a resistance here of r r 1 and r 2 in that case you would have to modify these equations and just write them as v 1 is equal to. So, this is basically how things will look like when you have got this. So, if your current is flowing like this your drop is going to be like this and your drop here is going to be that. So, we will have v 1 is equal to minus r 1 i 1 v 1 is equal to minus i 1 r 1 and of course, this is minus d psi 1 by d t and minus d psi 2 by d t because your current is going in this direction, but we have kept a inductance definition same as before. So, the basic equations are these now what is r s actually r s is a matrix r s is the is actually r a r b and r c the phase winding resistances. So, what we have here is r s is nothing but you know r a r b and r c of course, are the same. So, I will not represent them separately. So, r s i s and r r of course, is r f 0 0 r h 0 0. Now, one interesting point is suppose I want to represent a synchronous machine not by 2 windings on the d axis and 2 windings on the q axis, but by only 1 winding on the q axis. What would I do? Well, I would go ahead with my formulation as we have been doing so far and then put r k tending to infinity or r k very large. So, it is a good idea to actually build a model using these rather large number of coils and in case you want a simpler model you just set some of the resistances tending to be a very large value that will kind of open the winding. Once you open a winding there is no current flow through it and there is no current flow through it will of course, not affect the M M F s or the flux. So, this is a interesting trick to do later on when we want to get a lower model lower order model from this relatively higher order model we can actually just you know change the parameters and get it. So, we do not have to re derive the whole equations with just one coil less you know you can just set one of the resistance to be very large. Now, the next step of course, is what is V s that is if you recall what I am doing here. So, r s we have discussed in r r we have discussed what is V s? V s is in fact, the applied voltages. So, what are the applied voltages? V a, V b and V c these are the applied voltages to the stator winding and of course, what is V r? Let us just be a bit careful about this what is V r? Well, if you have got if you look at these rotor windings which represent damper and eddy current effects they are short circuited. So, for these three windings it is 0 and now if I have got a voltage say I will just magnify this I am just magnifying this coil here the dot is here current is I f this is the f field winding you need a voltage source like this to drive current into the field winding the d c voltage source which is connected this I will call V f. So, this voltage V f the potential of this point with respect to this is V f, but of course, normally when we are using this you know dot convention when I say V 1 it is usually the voltage at this dot with respect to the other point. So, when I when I show a dot here it is a voltage dot at this point with respect to this, but of course, if I have indicated this small V f to be connected like this obviously in V r I have to put minus V f what what I am saying. So, because the voltage is connected like this the dot is here, but I am connecting the voltage source like this V f whatever the magnitude of this voltage source has to be put as minus with a minus sign here. So, now we have got our complete equations. So, I know this so if I give you this this and the position of the rotor I in some sense I have got told you these equations will tell you how these how the rotor fluxes with will behave. Now, one small point regarding the number of poles remember that the flux the inductance matrices which you have taken out we in fact we were discussing only with a two pole machine that is mechanical angle and the electrical angle were equivalent. However, you consider the case of a four pole machine rather first we will see of course, a two pole machine in a two pole machine if this turns around by 180 degrees the current of course, in the coil remaining the same suppose this is the north pole and this is the south pole of this rotor the fluxes coming out of this north pole like this because of a current here. Suppose I turn it around by 180 degrees there will be a complete flux reversal in the in the air gap. So, a 180 degree turn results in a complete flux reversal in this particular case any winding if you look at if you have this is a four pole structure there is a north this is south this is north and there is a south. In this case even a 90 degree change in the mechanical angle will change the flux picture completely reverse the flux picture you can just chew over it a 180 degree change in the rotor position will cause a flux reversal here I mean whatever the flux values were at various positions on the rotor they will become negative of that. In this case when you have got four poles even a 90 degree change will cause a complete flux reversal. So, remember whenever we are calculating the inductance matrices we have calculated for a four pole machine if I have calculated for a two pole machine and I want to see what is the inductance matrices for a four pole machine we do not have to change anything, but remember that the theta which we are using has to be the electrical angle. So, the inductance matrices which I have derived are still valid the structure of and the nature of the each element is still the same, but for example if you have got cos 2 theta somewhere in the inductance that theta should be the electrical angle not the mechanical angle. So, you can be absolutely blind to you know you do not have to change the inductance matrices at all, but they are going to be a function of theta not theta m the mechanical angle that is one thing you should remember. So, inductance matrices which I which I have derived have the same structure, but they are going to be dependent on theta not theta m. So, as far as the machine is concerned synchronous machine is concerned the electrical equations are concerned what you are going to get out of the machine eventually I mean what you are going to connect to the grid the machine which is going to be connected to the grid. You will get actually you have got what are known as you have got the phase a terminal pair of phase a terminal the pair of phase b terminal a pair of phase c terminal. So, you have got this is what is coming out. Similarly, you have got the field winding. So, this has to be connected to a DC voltage source this is to be connected to the grid. Inside it may be a 2 pole machine or a 4 pole machine, but the point is the mutual inductances between a and f b and f a and b all are going to be functions of theta and not theta m. So, the inductances matrix can be used as they are. So, it does not matter how the windings are the you know what is how things are connected inside for example, when you take a 4 pole machine you there will be 2 sets of a b c winding. So, you may connect them in series of parallel or you know that actually in principle it can be anything, but what comes out the relationship between the current in the a or the flux in the f when there is a current in the a winding the inductance matrices which will use will be the same as before, but remember you have to use wherever theta appears it is the electrical angle. So, that is one important thing which you should remember. So, none of what we have what we have done is actually invalid just remember that theta has to be the electrical angle. What about the you know there of course, we have talked of the flux equations of the machine they are derived from Faraday's law the rate of change of flux is proportional to the voltage. So, what about the mechanical equations now it is very important to study the mechanical equation we will see that the very important phenomena in fact, involve the mechanical equations. So, we have to use of course, Newton's law in that case the basic equation for rotation of a electrical machine is J which is the moment of inertia into the rate of change of the mechanical speed in radians per second J into d omega by d t is equal to t mechanical minus t electrical what is t mechanical the torque provided by the prime mover t e is the electromagnetic torque. So, please remember if this is the direction of motion of the machine this one the electrical torque electromagnetic torque. So, we are considering generator convention electromagnetic torque is in this direction and the mechanical torque is in this direction. So, for a generator which is rotating at in this direction the rate of change of mechanical speed follows this law. Now, at this point just remember that the J am now somebody may of course, just require a clarification what is this J. So, actually if you look at any synchronous machine you have got the rotor of the machine and you also have got the rotor of the prime mover which consists of the turbine blades and so on. So, you have got two blocks the rotating masses at they are two actually two rotating masses connected by a shaft. So, if you assume that the shaft is rigid then the mechanical speed of the turbine and the generator are the same. So, omega m is the same for these two and this particular equation in this particular equation J actually denotes a total moment of inertia of this generator rotor as well as the turbine rotor. Later on we shall see that there are phenomena in which you cannot treat this shaft as rigid. So, there will be actually the motion of this and this will have to be described by two equations with the individual moments of inertia. But for the timing we will assume that the turbine and rotor generator rotor are in fact, one rigid mass and J denotes in fact, the inertia of the masses taken together omega m is of course, the speed of both the masses they assume to be rigid. So, they are rotating exactly the same speed if the shaft is elastic this assumption will not be true. So, one equation is adequate if the shaft is rigid we assume that the shaft is rigid. Now, something from basic electromagnetics you recall this formula electrical torque is equal to electrical torque in the direction opposite to what theta m is. So, theta m is of course, measured in this direction in this direction T e is in this direction. So, according to the you know the basics of electromagnetic energy conversion T e is equal to minus of the partial derivative of w is w dash the co energy with respect to the mechanical angle and of course, the co energy has to be expressed in terms of the current. So, this is the correct law the electrical torque in this direction is the partial derivative of the co energy negative of the partial derivative of the co energy with respect to the angle where the angle is measured in this direction this is consistent with what we have been doing before. The co energy of course, is expressed in terms of the currents if this itself can be written as minus of p by 2 p is the number of poles of the rate of change of co energy with respect to theta where theta is the electrical angle. So, which so this term I will call as T e dash. So, T e is equal to p by 2 into T e dash co energy expressed in terms of current is a linear machine is half of the current vector into this inductance matrix into this current vector again. So, this is this is a generalization of half L i square you know. So, it is actually a it is this and this of course, is the L's are a function of theta L r r of course, is not a function of theta we have seen that before at this point let me just recap what we have done we have computed the inductance matrices we did not compute all the terms, but I indicated how you can actually compute the mutual inductances and self inductances not only for a cylindrical pole machine where there is no saliency or very little saliency, but also for a salient pole machine. Salient pole machine mutual inductance computation is a bit tricky, but by you know simplifying the problem of taking out the components of M M F in the direct axis in the quadrature axis we can actually quickly come across the come to the correct mutual inductance terms. So, once we get the mutual inductance terms we have derived rather we have used Faraday's law to relate the rate of change of flux to the voltage we of course, use the current convention that currents are flowing out of the windings. So, we have computed the equations that way we have kind of you know also kind of got a relationship between the electrical circuit and the mechanical circuit the link of course, being that the electromagnetic torque is dependent on the current. So, the flux equations and the mechanical equations get coupled because electrical torque is dependent on the fluxes or the currents and of course, the fluxes are going to be dependent on theta. Now, the way we formulate our equations I have not just completed it fully you will find that the equations are functions of theta and theta is continuously changing with time. That makes actually our life a bit painful because the relation between flux and the current will keep on changing with theta. So, I will introduce you to a very important concept or a very important time variant transformation called it DQ transformation which will effectively make our equations a bit easier to understand. So, that is what we will do in the next class and using this transformation we will try to infer some of the you know the behavior of the synchronous machine.