 We started off the past 10 lectures or so with a reasonably detailed model of synchronous machine. In fact, we considered several windings on the rotor. I mean in fact, one can imagine that the simplest model which could give most of the correct information, most correct information of how a synchronous machine behaves especially in steady state etcetera can be got without considering damper windings because eventually the current goes down to 0. Of course, during transient conditions as well, one it would be useful to work out some simple models of a synchronous machine. So, that well there are three reasons why one can look for simplified models. One of course, is that the data for detailed models may not be available. We may not get all the standard parameters sometimes in our studies. These this kind of data is not available. On the other hand, one would like to have all this data and use a detailed model. In some cases of course, you will find that you can model a generator by a simple more simple model than what we have considered. For example, hydroturbines it turns out can be represented by one less damper winding on the q axis. So, one can actually work with a simple simplified model which is suitable for a hydroturbine. Of course, if you one wants to do a theoretical analysis and try to get information about phenomena without worrying too much about you know getting a great deal of accuracy. In that case, one may wish to work with lower order models. They seem to give better insight. Right now of course, if we look at our synchronous machine model, it contains a large number of equations. So, you have got in fact, if you consider the zero sequence equations as well, you will have seven flux equations, seven differential equations of flux in addition to the electromechanical equations. And as we shall see later in this course, when you consider an integrated power systems, you have got lots of synchronous machines and so on. So, what you will find is that the number of equations becomes quite large and one may not be able to do you know any kind of theoretical or kind of insightful analysis into a synchronous machine without invoking fairly detailed and sophisticated numerical tools. So, I think it is a worthwhile to look at somewhat simplified synchronous machine models which ignore one or the other damper winding. In fact, actually we have already worked out one simplification which is okay when we study slow transients that is you know replacing d psi d by dt and d psi q by dt is equal to 0 and converting the corresponding equations into algebraic equation. That was one simplification which we did which was justifiable in case one used one was really interested in only slow transients. So, that is one thing which you have already done, what we will do now in this in today's lecture come up with more simple models in which we will be neglecting one or the other damper winding or even we may neglect all the damper windings and come up with a simple model. Now, one interesting thing which I want to do by the end of this lecture is using the you know the model which we have derived using two damper windings on the cube axis and one on the d axis along with the field winding. I want to come to a point where I can show you that you can get what is known as the classical model of a synchronous machine which we have used to derive or rather understand some important phenomena right in the beginning of the course. So, what are the approximation involved approximation involved in getting such a simplified model of a synchronous machine. So, that is something we will try to you know try to understand by the end of this by this lecture. Now, so today's lecture is titled simplified synchronous machine models just a one small point about the kind of terminology we will be using. We have derived what is known as or we shall call as the 2.2 model of a synchronous machine. A 2.2 model of a synchronous machine involves two damper windings on the cube axis this of course refers to the rotor windings and one damper winding and one field winding on the d axis. This is a fairly detailed model it is a quite a respectable model to use especially for steam turbine driven generators round rotor generators normally we would use such a full blown model. Now, what if we want to get a simplified model. In fact, before I go ahead with trying to reduce the number of rotor windings and getting simplified models with lesser number of rotor windings. Let us look at one of the approximations we have already made just for the sake of revision. So, let us just first look at the q axis per unit model. So, we will just look at what we have so far in the 2.2 model. We have got different two differential equations corresponding to the two damper windings in the q axis. We have an algebraic relationship relating psi q to the current and psi k and psi g and we have of course, a differential equation in psi q d psi q by d t. Now, in the d axis with an assumption that T d c double dash is equal to T d double dash we have a similar model. But of course, one of the windings here is you know is the field winding. The field winding is of course, affected by what voltage you apply to the field. So, that field the effect of the field winding in in this per unit equations is captured by E f d. So, if you look at what E f d is it is of course, related to the field voltage which is applied at the field winding. Of course, we have 0 sequence equations as well which you may require to use in case you are doing unbalance analysis. And we have the torque equation in per unit that is relating the rate of change of the speed of the synchronous machine with the electromagnetic torque which is a function of course, of the fluxes and currents. So, this is where we are this is the kind of mountain average of synchronous machine modeling. Now, we look we slide down mountain average and look at approximate models. One of the models which you have already kind of beaten to death or rather one approximation which you have considered fairly in the past 2 or 3 lectures was replacing the d psi d by d t and d psi q by d t differential equation by algebraic equations. This was done simply by setting d psi d by d t equal to 0 and d psi q by d t equal to 0. So, in that equation we just just replace d psi d by d t equal to 0 and d psi q by d t equal to 0. So, this is an approximation which is valid in case we are talking of slow electromechanical transients slow electromechanical transients the swings which we were discussing in some of the lectures. Now, while operating near the nominal speed we could if we are all our electromechanical transients. In fact, R if R taking place if we are near the nominal speed we could in this particular equation replace omega by omega b. So, that becomes kind of a constant multiplication factor. Now, this kind of approximate model can be used if you are of course, near the nominal speed you cannot use this approximation of omega approximately being equal to omega b. If you are trying to simulate or you know you know understand the synchronous machine right from the process of synchronization that is starting rolling our rolling the generator and getting it near the synchronous speed. When this speed is not too close to the synchronous speed we cannot make that approximation of omega being approximately equal to omega b in this particular equation. But, as I mentioned if you are talking of transients in which you are not going to deviate too much from 50 hertz or 60 hertz whatever your nominal speed may be then of course, this equation is valid if one is starting slow transients. So, this is one of the approximations that we have made and we saw that it did not make much of a difference during our short circuit study. In fact, it made some difference alright, but it did not affect the modes associated with the slow transients. So, that was the basic effect of this approximation. Do not make this approximation if you are going to study fast transients say which occur in time scales of 1 or 2 cycles of course, that would not be correct it would not be right to set d psi d by d t and d psi q by d t equal to 0 in that case. So, this is one approximate model which we have done we move on to making approximations not in the stator coils psi d and psi q of course, are stator coils. What we will do is making makes a few approximations as far as the rotor coils are concerned. So, the simplest thing we can do is one of the damper windings we the effect of it is neglected. So, for doing that what we need to do is consider one less damper winding in our original you know derivation of the synchronous machine model, but rather than of course, redoing the whole derivation again a simple thing which can be done is of course, open the particular damper winding. So, suppose you have of course, got the d and q winding. So, in the q axis rotor windings are of course, 2 the g and the k winding and there is a field winding and a damper winding on the d axis. What I need to do is in this particular model in order to simplify it is get rid of this effect of this winding. So, what you can do of course, is set the resistance of this winding for example, this winding r k if the winding r k I set the resistance to be a very large value. In that case it is as good as opening the damper winding in which case of course, no current will flow in the damper winding that particular damper winding and its effect is in would get nullified. Now, important point which you should note is that when I said r k tending to infinity, what effectively happens to the time constants that is something I leave as an exercise to you, but you can show that in such a case t q double dash tends to 0. So, if r k tends to infinity t q double dash tends to 0, how do I know that? Well I know the equations relating the time constants, time constants in the standard parameter model, standard parameters with the original inductances and resistances of the winding. Recall this we have done it somewhere in the between the 10th and the 20th lecture, where we were modeling synchronous machines. Now, so setting r k tending to infinity would mean that you are setting t q double dash tending to 0. Now, if you do that what happens? So, if you look at the basic equation which is there, you can write this in fact as t q double dash is equal to minus i k plus psi d. So, this is the equation which is there of the flux of this particular winding and if t q double dash tending to 0, you can roughly say that this term here becomes 0 and as a result of it psi k becomes equal to psi d. So, from this you get this. Now, if psi k is equal to psi d, you can replace psi k by psi d in the algebraic relationship which relates psi d sorry psi q i q psi g and psi k. So, do you recall that equation? We will just have a look at it right away. So, what I am trying to say is of course that you get psi k is equal to I am sorry yeah it should be psi k equals to psi q this some small error which I made while writing this should have been psi q this should have been psi q. So, getting back to this slide which I have you get psi k is equal to psi q. So, you can substitute psi k by psi q here in this algebraic equation. The third equation is an algebraic equation and then if you do that you can write psi q as a function of psi q. Is a function of i q something into i q plus something into psi g. So, the effect of psi k gets subsumed in psi q. So, what are these coefficients here and here? You can work it out I will just write down what you get eventually what you will get eventually is if you look at the slide your equations of the 2.1 model 2.1 because now we have got just 1 damper winding on the q axis will be given by one differential equation corresponding to psi g. One of course is corresponding to psi q and the algebraic equation relating psi q i q and psi g becomes much simpler and you notice that x q double dash no longer appears in this equation. So, for this particular model 2.1 model you do not require one set of time constraints and one set of one basically you not require one time constant and one reactance. So, it does not figure in the equations any longer. If you look at the slides again the last equation of course is still the differential equation in the flux psi q remember that you can make an approximation that you can replace d psi q by d t equal to 0 and get back to the approximation which we mentioned right at the beginning of this lecture that would be if you are studying slow transient. So, the last equation also could be simplified in case you are talking about slow transients. Now, one of the things which I should mention at this point is that the model 2.1 model which I have just discussed so far is found to be adequate in the sense that it can represent the transients which occur in hydro driven hydro turbine driven synchronous generators. So, hydro turbine synchronous generators are often represented by 2.1 model and do not be surprised if the data sheet of a hydro turbine generator has one less time constant and one less reactance. So, that particular data will not be given because it is adequate to model hydro turbine just by 2.1 model. So, this is something you ought to keep in keep in mind. Now, an interesting observation here which you may have noticed is that if you look at the algebraic equation relating psi q and psi g and i q of course, it does not contains x it does not contain x q double dash there is no need of having x q double dash, but if you look at the algebraic equation so I will just write it down here psi q is equal to x q dash i q plus x q minus x q dash upon x q into psi g if you look at this particular equation and compare it with the algebraic equation which we had for 2.2 model. So, if you look at the screen here you will see that the equation which was there before I will just write it a bit quickly out here well it looks pretty complicated, but something you will probably able to suggest is that if I had directly set x q double dash is equal to x q dash in this equation in that case you would have directly got this equation. Note that this whole this term would have disappeared this is in fact that. So, this would be 1 so we will be really getting back to this equation. So, an interesting observation here is that if I want to go from 2.2 model that is the original 2 dampers on the q axis model to 1 damper winding on the q axis then what you can do actually is simply set simply set simply set x q double dash is equal to x q dash in the 2.2 model equations. Now, why am I telling you to do this well you may have already written a program or you know kind of a tool which does Eigen value analysis or synchronization the simulation numerical integration of the differential equations in the 2.2 model and somebody tells you suddenly well a particular generator has to model not by 2.2 model, but it has to model by 2.1 model in that case you do not have to rewrite your program with lesser number of equations you would need to tinker with your program instead of doing that you just set x q double dash is equal to x q dash in the data. So, just look at these equations again you will find that because of doing this you will get rid of the effect of psi k. Of course, this equation would still be there, but it would be kind of decoupled from every means the effect of psi k would not be effectively seen by a synchronous generator state of windings at all. So, the there is no harm in just setting x q dash is x q double dash is equal to x q dash and removing the coupling which psi k has with the rest of the equations. So, this is a pragmatic way of using a program or a tool which is program for 2.2 model directly make it suitable for use to directly make it suitable for use with 2.1 model just set x q double dash is equal to x q dash. Of course, you need not do it you can actually even do another thing that is reprogram everything for a lower order model that is also possible. So, in case you do that you will have to use these equations the lesser number of equations in your programming. So, this is what basically is you do when you neglect the effect of one damper on the q axis. Now, before we go ahead let us just see what happens in case we try to simulate a system with a lower order model of a synchronous machine. So, what we will do is take the data which we have been using for studying the synchronization transient. In fact, the data is for a round rotor machine, but all the same will still use it for you know just comparing what happens in case you take a simplified model. So, what I will do is I have programmed the program for 2.2 model, but I will simply make the simplification x q double dash is equal to x q dash and effectively get 2.1 model. So, I will just do that. So, we will get back to a silap program and what I will do is of course first of all show you the original speed transient under synchronization. We have been doing this in the past few lectures. So, I need not explain everything right from scratch. It is basically a transient which shows the synchronization of a synchronous machine to an infinite bus followed by torque and field voltage increases. So, this is increase in torque and field voltage. So, these are the transient which you see. Now, in case you take this program and in that set x q double dash which is here equal to x q dash. So, if you do that how does it affect. So, let us just try if we can do it of course we have to hope and pray that. So, the program runs without any problem it is just we have just change the data we are not really change anything in the program. So, if I plot this now and look at what we get well you notice something well the original program the damping was much higher. Whereas, now in all the transient which are considered one thing you can notice is that the transient takes a longer time to die down. In fact, in a synchronous machine with one damper winding less it appears that the damping has reduced this is not very surprising. In fact, you know one may say that the damper winding derives its name in some sense from the damping effect that it has. So, if you actually reduce the amount of damping in that case that is by removing one damper it is not surprising that we get this larger time for decay of the transient. So, this is something which is not at all surprising this is effectively the response will get the top one which you see here the one which is taking a long time to die is effectively the response without a damper winding one of the damper winding. So, let us go back and also do the Eigen analysis you do exactly the same thing and do an Eigen analysis Eigen value analysis using a linearized model which we discussed in the previous class. So, now I will just run the program. So, I just take out the Eigen values and the Eigen values are these well not much can be understood from this unless we compare it with the Eigen values with the 2.2 model. So, what I just now showed you were the Eigen values with 2.1 model I will rerun the program to see the Eigen values with 2.1 model. So, what you notice here of course, is that there is significant change in the real part of the Eigen value corresponding to the electromechanical oscillation which is seen in the speed transient. So, this is consistent with what we have seen before in fact of course, we need to just check out the operating point which we are talking of here. So, what we see here is of course, that if I take 2.2 model the real part of the Eigen value is 1.4 roughly whereas, the Eigen value here is 1. So, you see that with 2.1 model the damping of the swing mode is slightly lower. So, this is something we see it seems to be consistent with what we see here we see a slower rate of decay when we consider a lower order model. Now remember here that we are not doing any closed loop feedback control of the field voltage the field voltage is practically a constant we are either giving step changes or we are keeping it a constant. We shall see later in the later part of this course that the feedback control of the field voltage in a synchronous machine can again affect the damping of the electromechanical or what is known as the swing mode that oscillatory mode which is has a frequency of around 10 around 10 radian that is 12 radians per second here. Now, let us go one step ahead let us go to a model now what I will do is go a bit faster and talk about what next can be done you can in fact go to what is known as the 1.1 model. In 1.1 model we have done the corresponding thing with the damper winding H. So, what I have done is set R H tending to infinity and therefore, got rid of the differential equation corresponding to psi capital K and as a result of which we get a simplified model with just one differential equation and as I mentioned last time the same thing can be done here you can get this lower order model directly by setting x d double dash is equal to x d dash. So, if you have already programmed 2.1 model or 2.2 model to get to 1.1 model you all you need to do in that program is to set x d double dash is equal to x d dash alternatively you can program a lower order model. So, both of the thing both these things can be actually done we can go one step further we can actually you know get rid of the last remaining winding damper winding on the q axis that is the g winding, but at this point let us just take a small diversion not a very big one if from this 1.1 model 1.1 model means you will be using these equations on the q axis and these equations on the d axis. So, that is 1.1 model I will show it to you again to be using these equations in the q axis and these equations in the d axis. Now, one question which I would like to ask you is can we can we get the equations of an induction machine from 1.1 model of a synchronous machine. The answer is yes from the 1.1 model equate x d dash and x q dash and induction machine can be modeled a simple induction machine model could be by considering one damper winding on the d axis and one damper winding on the q axis remember induction machine does not have any field winding. So, what you can do is of course, these windings are absolutely symmetric if you look at a normal synchronous induction machine both the d and q axis appear similar there is nothing to distinguish at d and q winding. So, you need to set this I will call it x dash and what you need to do is of course, do this equate effectively what I have done is got from 1.1 model simply by setting x d dash is equal to x q dash and t d dash is equal to t d dash is equal to t dash is equal to t q dash you can in fact, from the 1.1 model get a induction machine model. So, 1.1 model of a synchronous machine get directly yield to you the induction machine model you will of course, have to set e f d is equal to 0 this is something you need to do and effectively equate the d and q axis parameters set e f d also equal to 0. So, these are the things you would need to do in case you wanted to derive an induction machine model directly from 1.1 model of a synchronous machine. So, this is a interesting diversion which this is apt at this point. Now, we can go ahead and get rid of even the remaining damper winding on the q axis. So, what we are going to do now is talk about a simplified synchronous machine model which does not have any damper winding at all. So, this is the what you will get effectively what you have done is set t q dash tending to 0 basically if you have r g tending to infinity that is opening the last winding remaining winding of the q axis in that case t q dash tends to infinity and in that case we can kind of convert the differential equation corresponding to the g winding that is d psi g by d t t q dash is equal to psi g plus psi q and if you do that if you set d q t q dash tending to 0 in that case you will get psi g is equal to psi q and if you use this fact in the algebraic equations which you have seen earlier say if you look at it to 2.1 model if you look at this algebraic equation here if you set psi g is equal to psi q a few manipulations will get you to psi q is equal to x q i q. So, what you have in the q axis is just one algebraic equation and one differential equation corresponding to the stator winding that is psi q. So, one differential equation and one algebraic equation in fact the differential equation also can be removed by setting d psi q by d t is equal to 0 in case you are studying slow transients. So, this is what you get as 1.0 model in fact 1.0 model equations will have the d the d axis equations looking like this and the q axis equations looking like this. So, that effectively tells you what is the 1.0 model of a synchronous machine in fact this is you can say the last or the most basic of synchronous machine model where here now you have just the field winding and the stator winding and all the damper windings have been got rid of. Now, if you look at the behavior of a 1.0 model synchronous machine. So, what I do is in this generators program I set to get 1.0 model what will I what I need to do is set first x d double dash is equal to x d dash that is what I have done first. The second thing you can do directly something which I have not mentioned here is that in the 2.1 q axis equations if I directly set x q dash is equal to x q you can get rid of you can really come directly to this algebraic equation and 1.0 model. So, that is what I will do right now what we will have here is I will set x q dash equal to x q and of course, I also need to set x q double dash is equal to x q. So, that we get 1.0 model. So, this is 1.0 model x q double dash is equal to x q dash is equal to x q. So, by doing this we are directly modifying a program we are not without modifying the program of 2.2 model we can effectively get the response of the simplified generator model and I have also set x q double x d double dash is equal to x d dash. So, let us try to simulate this in psi lab and if I plot this now you get a totally different response. In fact, what you are seeing is that the system is losing synchronism. So, in case you and this the response is completely different. So, of course, one of the reasons why this could be happening let me just expand this view is a numerical reason. One of the reasons which why this could be happening is a numerical reason that is see if you look at the first curve here it is a 2.2 this one is with 2.1. Now, we have gone to 1.0 and you see growing oscillations. Now, is this correct or not that we can say by actually finding out the Eigen values of the system. For example, we try to try to find out the Eigen values of the system for 1.0 model. So, I modified the data. So, that we will take out the Eigen values of the system. Now, what you notice is the real part of the Eigen value has become quite small it is minus 0.4. Now, one of the things you should remember here is that we are using Euler method to simulate the system. And if the damping that the real part of an Eigen value is low Euler method may actually show it to be not a damped oscillation, but an increasing oscillation and this is what exactly is happening here. So, although Eigen analysis shows that the system is in fact stable what you get here is a growing oscillation. So, growing oscillation of course, will eventually cause a loss of synchronism. So, this is kind of spurious, but remember one thing that if we neglect the effect of all damper windings in 1.0 model, the damping of the swing mode may come down to a low enough value not necessarily to make the system unstable as is seen in this Eigen analysis, but which may cause misbehavior or other wrong information to be displayed in case we do a simulation using Euler method. Recall in the first 10 lectures of this course we have discussed numerical integration techniques and there we did find, we did discuss this that in case you use Euler method on a poorly damped system with a poorly damped oscillation in that case you can in fact get wrong information. Now, the solution of this of course is to keep on reducing your time step of the simulation till Euler method starts giving reasonably correct results, but of course that will take a long time to simulate if I really go on reducing my time step. So, one thing which you should remember now it is important as a student to distinguish between two issues we are talking here. One is that with a lower order synchronous machine model 1.0 model the amount of damping or the amount of rate of decay of the swing mode has come down substantially that is one aspect that is the physical aspect. One more aspect is about the analysis the damping has really come down to such an extent that if you try to simulate Euler using Euler method with a time step of we will just I just forgot what the name of the variable is for time step I will just get that with the time step of 0.005 a system with poorly damped oscillation like this one is likely to show. So, let us just see this again t step a system which has poor damping is likely to show is likely to cause incorrect information being shown in a simulation in which Euler method is used with a time step of this kind. So, really what you need to do as I mentioned some time back is go on reducing the time step or use a better numerical integration method, but of course we have been using Euler method simply because it is an easy method to use it is sacrilegious to use Euler method for any realistic or you know practical or industry grade program I mean because of this particular problem that is not very accurate it can give wrong information. So, this is another illustration of this. So, although the system is stable remember all the Eigen values for all operating points is of course the Eigen values for a particular operating point the Eigen values have negative real parts. So, it is a stable system, but simulating it may result in problem if you are using methods like Euler method. So, that is the summary of what we have been discussing so far. Now, 1.0 model in some sense is the limit of what simplification we can have in a synchronous machine if you want to you know get even you know a realistic picture in steady state you have to use at worst you will use this simplified model. Of course, for theoretical studies involving the swing mode we can use a simplified model which includes no differential equation no electrical differential equation no equation differential equation corresponding to the flux how do we get to that model well this is not a very respectable model because. So, I will just tell you how you can get to what is known as 0.0 model which is what is known as the classical model a classical model is not a respectable model for getting realistic and quantitatively correct answers as far as synchronous machine behavior is concerned for example, you cannot use classical model to understand the nuances of what happens during a short circuit in a synchronous generator. So, there are very great limitations in applying this classical model, but all the same I will just tell you what it is and how it is obtained and what kind of assumptions are inherent when you get to classical model. Remember that right in the beginning of the course when I was talking about introducing you to things like loss of synchronism and the you know the origin of swings power swings or rotor angle swings in a power system I had used in fact the classical model it is not a respectable model I called it toy model then I will call it toy model even now, but let us just see what really was involved in getting to this classical model what you need to do is from 1.0 model you can look at the screen again from 1.0 model 1.0 modeling requires these equations on the d axis and these equations in the q axis what you do is you assume that the field winding resistance is very small as a result of which you can show using the basic equations which relate the standard parameters and the resistance and inductance parameters of a synchronous machine you can show that t d dash becomes a very large value it becomes tending to infinity. If t d dash tends to infinity you are in effect saying that the field winding flux does not change. So, psi f here is 0 d psi f by d t is equal to 0. So, psi f becomes a constant so what you have is you have got rid of the differential equation the last remaining differential equation of the rotor field winding you have effectively said that psi f is simply a constant. Now, a further simplification of course which we have been which I talked to right in the beginning was you set d psi d by d t equal to 0 as well if you are going to study slow transients this is an approximation to make. So, what you have eventually is for the classical model setting d psi d by d t and d psi q by d t we have got this model. So, this is a classical model of a synchronous machine which does not have any differential equation no electrical differential equation. We have got the mechanical differential equations, but there is no electrical differential equation and a further simplification which you can make is r a is equal to 0 and x q dash x q is equal to x t x t dash. Now, this is a absolutely an ad hoc assumption this is no you know kind of justification which I can give you. So, classical model is obtained by a large number of approximation. So, if you want to get from 2.2 model to classical model you have really made a huge a large number of approximations. So, what you will get of course, if you just go ahead with what you have got psi d is equal to x t dash i d plus e dash e dash is that is actually proportional to psi f which is assumed to be a constant. So, e dash is also a constant and you also have psi q is equal to x t dash i q this is got by approximating x q is equal to x t dash and absolutely ad hoc assumption. We also from we also have 0 is equal to this is by neglecting d psi d and d psi q by d t terms this is by neglecting d psi d by d t we also make r a equal to 0. So, we will have minus omega b v d and psi q we will also have is equal to omega b psi d minus omega b psi q and what you can do is effectively substitute for psi d and psi q in these equations. So, what you will have eventually is if you do that you will have omega b v. So, what you will have 0 is equal to minus of x t dash i q minus of b d and what you will have here is x t dash i d minus of b d and what plus e q dash minus v q. Now, we also have d omega by d t 2 h into d omega by d t by omega b into d omega by d t is equal to t m minus psi d i q minus psi q i d. Now, if you look at. So, you have got these two algebraic equations and this differential equation. Now, if you are talking of a synchronous machine connected to an infinite bus or a stiff voltage source as the one which you have used for all our studies. So, far in that case v d is equal to v line to line r m s of that source into sin delta minus of it. So, this is the source whose characteristics I have discussed in the previous lectures. So, if this is true then it is easy to show it is quite easy to show that psi d i q minus psi q i d from psi d i q minus psi q i d from all the equations which we have got from these equations and these equations. What you will get eventually is this is the electromagnetic torque is nothing but v line to line r m s by x t dash e dash sin delta. So, from this and this this is what we have. So, the classical model effectively the torque the torque equation or the electromechanical equations are d delta by d t is equal to omega minus omega naught omega naught is the frequency of the infinite bus and you have got 2 h by omega b and d omega by d t is equal to t m minus v line to line r m s e dash sin delta upon x dash. This is the transient reactance of the generator we can call it just x dash. Now, another interesting thing is of course from these algebraic equations we look at these algebraic equations you can write this very compactly we will have e dash minus v q plus j v d this is just multiplying the second equation by the complex number j e dash minus this is equal to j x dash x dash is nothing but x d dash we need not apply the subscript any longer because we have equated the d a and q axis completely this is what we get. And you know from what we have here we have v q plus j v d can be written compactly as v line to line r m s into e raise to it is simply that and just have a look at it. So, what we have here is eventually e dash into a minus v line to line r m s into e raise to minus j delta is equal to j into x i q plus j i d and what we have from here effectively is e dash e raise to j delta minus v line to line r m s is equal to j x into i q plus j i d where i q plus j i d where i q plus j i d is nothing but i q plus j i d into e raise to j delta. So, what we have here is effectively a synchronous machine model in which the electrical equations are simply given by this or effectively an electrical circuit. So, what we have effectively is an electrical circuit. So, if you want to represent the synchronous machine electrical equations it is simply a phasor that is an algebraic equation this is v line to line r m s of the infinite bus this is e dash angle delta and the differential equations are d delta by d t is equal to omega minus omega naught omega naught is the frequency of the infinite bus and d omega by d t itself is nothing but t m minus e dash v line to line r m s sin delta by x dash. So, this is exactly the toy model which we used in this in fact, right in the second lecture with the kind of model which we have used here. In fact, this model predicts an oscillatory response for delta and omega whenever there is a disturbance. So, what we have with 2.2 model is not only this electromechanical swing in fact, if you recall whenever you use a higher model with damper windings the Eigen values the Eigen values are these. So, there are many many modes here amongst them we have got what is known as the swing mode. If you take the classical model directly what you will have is simply just because there are 2 differential equations just one mode and that will really represent as we have seen right in the beginning where we have we did analyze this toy model we do get the electromechanical oscillation. So, classical model is just to theoretically highlight the fact that you do have such a mode which is mainly associated with the electromechanical variables delta and omega. So, just have a look at this classical model again this should have been x dash here these are the differential equations. So, this is the classical model which is just for theoretical studies, but please do not it probably will give you fairly it will probably give you wrong quantitative answers wrong in the sense highly imprecise answers in case you try to use it for practical studies, but nonetheless just using the classical model we can in fact show that a phenomena called swings occurs which is also evident in the 2.2 model, but the 2.2 model will also bring into picture many other modes which are present. So, this brings to an end practically an end our discussion of synchronous machines we will just revisit a few minor points tomorrow in the next lecture and thereafter we will move on to another physical subsystem which is of importance in a power system that is the excitation system of a synchronous machine. We will discuss its modeling and also discuss how it looks physically.