 We signed off in the previous lecture by discussing in a preliminary fashion, what happens if you neglect the stator transients that is neglect the d psi d by dt and d psi d by d q terms. In fact, set them to 0 and convert the differential equations corresponding to psi d and psi q into algebraic equations and then doing the short circuit analysis. Now, we saw of course, that the Eigen values which we obtained were very similar they were a subset practically a subset of the Eigen values obtained earlier that is because the transients associated with psi d and psi q are fast. In fact, we did not prove it we kind of intuitively guessed it something of course, which you can prove using the participation factors which we have discussed in the 8th or 9th lecture of this course. Now, in today's lecture we just continue our discussion we had in fact done a very preliminary analysis we did not complete our discussion about the short circuit analysis of a generator with the psi d and psi q transients neglected. We call that the number of states reduces from 6 to 4 we are of course, assuming the speed of the generator is constant. So, we are only looking at the flux transients and we will redo this example and we will try to understand what exactly information we are losing or what information we really do not lose much in case we make that approximation of neglecting the fast transients. Now, the reason of course, we are neglecting what are known as the fast transients is that the Eigen value corresponding to it is large. So, that is why the rates of change associated with that mode are fast. You may recall the Eigen value which is fast or has a high magnitude has a complex part which is practically equal to omega which is the speed of the rotating generator. Now, so today's lecture in fact, we will go on a bit ahead and also try to understand the behavior of a generator which is synchronized to a voltage source. So, we will go one step ahead and connect the generator to a voltage source. Now, so today's lecture in fact, we will first move on and do a study which we left half way last time. So, recall that when we do the short circuit analysis of a generator the response is given by this. So, this is what we got in the last lecture and recall that the current was looks like this. We will just look at the current we will just redo this again. The current looks like this there is a band there is a initials large jump in the current is I D of course, then there is a band corresponding to the state the oscillatory mode. In fact, that is the oscillatory mode of radian frequency 314 radians per second that is 2 pi into 50 there after the current settles down to E f t divided by x d. Now, if we do the same analysis with the stator transients or neglected or equivalently neglecting d psi d by d t and d psi q by d t in that case the current that you get looks a bit different. It is shown off as I said last time of the oscillatory band and in fact, it goes to a value of roughly between 4 and 5 initially then it drops down to around 3 and then gradually drops to the steady state value. In fact, what I mean will become clear here if you expand this you said that initially you see that there is a relatively sharp drop from roughly 4.25 to around 3.25 and then there is a gradual drop. So, this is the nature of the curve and then of course, it so there are in fact, two modes you see one quickly decaying mode it is sharply comes down here and then slowly it drops off. In fact, it is interesting to note that if you look at E f d which is 1 we have chosen it to be 1 divided by x t double dash in fact, it is 4.34 which is in fact, the value which we have seen here initially and after some time 3.33. So, you can see that initially the current takes the value E f d by x t double dash then it drops very quickly and then becomes roughly E f d by x d dash and after a long time in steady state in about 2 to 3 seconds in fact, it drops down to the steady state value which is 1 upon by x d. Now, this is the reason why a synchronous generator especially in short circuit studies is represented not sometimes not by its dynamical equations, but roughly by static equations algebraic equations will tell us that the current initially will be E f d by x d double dash and after some time after a few cycles it will be E f d by x d dash and in steady state it is E f d by x d. So, this is basically the reason in fact, why in short many short circuit studies we in fact, do that for what is known as a sub transient period we take the reactance of the generator to be very small that is in fact, the x d double dash value then after a few cycles we take what is known as a transient value of the reactance and then the steady state value. So, I hope from this dynamical study which is in fact, neglecting the stator transients neglecting the oscillatory band which comes if you consider stator transients shows that this approximation is I mean it gives a rough picture of what happens. So, you have got a initially large current then a smaller current then the steady state value. So, that is how the behavior of the machine is during short circuit in fact, neglecting the stator transients results in removal of the oscillatory band the decaying oscillatory band of 50 hertz in the current response. In fact, in the I d and I q in fact, I have not shown you I q. So, you can just have a look at what I how I q looks like I q with the stator transients neglected looks like this. In fact, these figures have got superimposed. So, I will just do this again. So, it is in fact, it is a very small current in fact, this 0.04 it may not be visible very clearly on your scene this 0.04. So, it is practically 0. So, I d I q is practically 0, but here to you are not seeing any oscillatory band because they have neglected the stator transients. Interestingly I a looks like this. So, if you look at I a it looks like this. So, you have got in fact, I a remember is in steady state a sinusoid I a is the current in the a phase it is in fact, thus. So, in fact, under short circuits conditions and neglecting the stator transients you get what is a looks like a symmetrical 50 hertz waveform. So, you see symmetrical in the sense that there is no dc offset in this. So, you do not see any dc offset. On the other hand, if you did consider stator transients like the original program did let us just do it once more. And if you plot I a what you see is a totally different picture you have got a dc offset in the phase current. So, it may be said that by neglecting the stator transients. So, you see the dc offset as well. So, by neglecting the stator transients we are neglecting an oscillatory 50 hertz component in the d q currents I d I q. And but the phase current equivalently when we neglect the stator transients we are in fact neglecting the dc offset. So, dc offset in the phase currents manifest as an oscillatory component in the d q variables. So, this is an important interesting interpretation of the results. Now, of course, it is not difficult to prove this it is fairly straight forward to prove that this is true. I will now show you an experimental verification of a short circuit applied on a synchronous machine. In fact, we will be doing the short circuit on a medium sized machine of course, this is still a small synchronous machine as compared to the machines which are used in power systems which can go up to hundreds of megawatts. Remember of course, the parameters of a small synchronous machine can be significantly different from a large synchronous machine very notably all the lossy elements that is resistance of windings the friction of bearings etcetera tend to be higher in smaller machines. Nonetheless, we still can show you some interesting features or rather show to you that the short circuit current of a open circuited synchronous machine does have the signatures which we have discussed in the simulation in theory we have done sometime ago. So, in this particular experiment you have a DC prime mover which I am pointing out at connected to a synchronous machine. The synchronous machine will be run up to roughly the rated speed of the machine by the prime mover and we will excite the machine not of course, to its full voltage will give a low voltage short circuit. So, that the currents are safe for you know for the purpose of this lab demonstration. So, I will start this demonstration now remember that of course, we will the machine is already been started and it is running and it is also been excited to give a very low voltage not of around 5 to 10 volts at the above sorry about 20 volts line to line RMS at the terminals of the machine. So, we will apply a dead short circuit under these conditions. So, let us observe the current waveforms for the phase A, B and C under these circumstances. So, we will start our video now. So, this is the DC machine prime mover that is the synchronous machine and of course, this synchronous machine has a self excited DC generator as its excitation it is not a static excitation system it is a rotating excitation system. So, we use a DC self excited DC generator for self excitation the machine is already on running and the voltage is not a very high voltage just around 20 volts line to line which appears across the 3 phases 3 phase output of the 3 windings of the synchronous machine. This is an MCB which is connecting all the 3 phases really connecting all the 3 phases together to create a short 3 phase short and these are the current probes which are going to the oscilloscope at which we will observe the short circuit currents. So, now I will actually perform the short circuit by actually connecting all the 3 phases together. So, the MCB will have to be switched on to the on position and we will simultaneously view the short circuit. So, it will be anytime now yeah he has given the short circuit and you see that large envelope in the beginning which decays you will also see a DC offset it is not very clear here. So, I will show you the actual figures. So, if you this is what we actually captured we switched gave a short then remove the short by opening the MCB again and gave another short. So, we we actually did it twice one of the things you will notice in the waveform this is expanded the expanded part of this waveform is shown below you will see that there is a DC offset as well as the envelope sinusoidal envelope is quite large in the beginning DC offset and a large envelope. The DC offset in all the 3 phases are not the same it depends on the instant at which you apply the short circuit this is something you can think over why it is. So, in fact you will notice that when you applied the short circuit the first time and the second time the waveforms look a bit different that is because the DC component when we applied the short circuit here is different from the DC offset component which decays out here that really depends on the instant of short circuit what point on the sinusoidal waveform the input voltage did you apply the short circuit it is different for when we did the short circuit at this instant and at distance. So, you actually see different DC offsets in the different phases, but one thing is clear of course that there is a DC offset which you see very clearly here this is a DC offset and also the overall envelope of this waveform decays with time. So, both the overall envelope as well as the there is a DC component which decays with time. So, this is a as as shown in the simulation results and as discussed in the theory. So, in some sense it validates our theory the only of course difference in the simulated result and this is that this is a relatively smaller machine. So, its parameters are not actually the similar to the parameters of a very large machine, but none the less you see these signatures in the current waveforms. Now, there is another figure I would like you to see is that of the field current under these conditions. So, when I gave you a short circuit if there is a short circuit applied at the three phase stator terminals your field current is also affected and what you notice here this is something I did not show in the live experimental demonstration, but we captured this waveform by performing the experiment again this is the field current which you see initially there is some current value in the field field winding as soon as you apply the short circuit there is a oscillatory response normally of course the dc the field current is the dc current, but you see an oscillatory response which dies out with time as well as an exponential decay. So, eventually of course the field current goes to its original value that does not change, because the field voltage being applied is constant we do not have any AVR or anything of that kind it is a constant field voltage applied by a self excited dc generator. So, you see this 50 hertz oscillatory component in I f remember that the dq currents the field current the dq fluxes and the field fluxes as well as a torque see an oscillatory decaying component which is near about 50 hertz which has got a frequency near about 50 hertz this manifests as a decaying dc offset in the phase currents. So, this is something which you should keep in mind. So, in the field current you do see this 50 hertz oscillation decaying 50 hertz oscillation as well as an overall decay. So, this envelope here decays as well as there is a this you know you can say the there is also one more you know mode which is very clearly seen which is decaying this way. So, this is basically a summary of the experimental results obtained for a short circuited generator. So, this is as far as our analysis of a short circuited generator is concerned one of the important ideas which we learnt in this lecture was that we could under certain circumstances neglect the fast transients, but that would be equivalent to neglecting the dc offsets in the phase currents or the oscillatory band which you see in the in the dq currents. Now, we move on to another interesting simulation. In fact, this time it will be a simulation that would be corresponding to the connection of this generator a generator under open circuit conditions which is rotating at rotating to a voltage source. So, what we will do now in this subsequent simulation which I will show you is using the same equations which we have derived. A machine initially will assume to be open circuited it is running slightly at a frequency which is slightly higher than that of a 3 phase balanced voltage source to which it is going to be connected. We will assume that it is in steady state and under open circuit conditions the machine has been excited it has been excited in such a way that you get one per unit line to line voltage across its terminals the windings of course, we assume to be star connected. So, in some sense we have we prepare the generation generator for connecting it to a synchronous to a voltage source the voltage source has a frequency say omega naught whereas, let us assume that the synchronous machine is rotating at a slightly higher frequency than omega naught. So, in fact, you can you can look at it this way if you have got a voltage source a star connected voltage source. So, this is a star connected voltage source it is a balanced voltage source. So, we will not consider 0 sequence quantities. We will assume that the voltage source is such that V a to neutral is equal to root 2 by 3 sin omega naught t. We will assume it is a source which has got a frequency and of course, V b n is sin 2 pi by 3 or 120 degrees and V c n is root 2 by 3. Now, what we have if V a and V b and V c n are like this let us assume that the synchronous machine is rotating and the theta of the machine the angular position of a machine is omega naught t plus delta. So, if this is omega naught t this is omega naught t plus delta. Now, of course, in case the speed of the machine the speed of the machine is nothing but omega which is nothing but d theta by d t if it is not equal to omega naught in that case delta will be time varying. So, what we consider initially is that you have got a synchronous machine which is to be connected to this voltage source. So, we will be connecting it to this voltage source and it is rotating at a slightly higher frequency than omega naught which is actually the frequency electrical frequency of this voltage 3 phase voltage source. Now, in case V a n V b n V c n are like this we have done in an earlier lecture that V d by applying the transformation V d is nothing but minus 1.0 sin delta see remember theta is omega naught t plus delta. So, this source has got V d n V q like this. So, recall that in our analysis of a short circuited generator where it is simply put V d n V q equal to 0 and in the open circuited conditions we assume that a very large generator very large resistance R L is connected in star at the terminals of the machine. But now you are connecting it to a voltage source. So, V d n V q are specified. So, what we are going to do is try to simulate a machine which is connected to a voltage source whose V d n V q values are given here. This is unlike what we did for an open circuit generator and a short circuit generator where we did not have a voltage source connected at the terminals. We just connected to a resistance whose value we flipped from say 1000 per unit to 0 you know to simulate open circuit and a short circuit. Now, once this is for obtaining V d V q from V a V b n V n V b n and V c n you of course have to use this use the transformation C p. I will not do it in this class we have done it sometime earlier. Now, what we will do of course, here is once we have got this how do you actually simulate this machine connected to a voltage source well it is not very difficult. So, let for a short we had done the equations of a machine where we had written psi d dot where psi d dot is psi d psi d by d t psi h psi f psi g plus a 2 into i d i q plus b 1 into V d V q and plus b 2 into e f d. So, these are the 2 inputs which we have now remember that in the previous discussion when we are considering an open open circuit generator we had subsumed this into this because V d a and V q were related to i d a and i q because we considered that a star connected resistance is connected at the generator, but now we have independently going to specify V d a and V q. Now, these are the equations the flux equations, but very importantly whenever you are considering the synchronization of a machine it is important to consider the transients associate the electromechanical transients. So, what we will do here is of course, that we have to write down the equations of the motion of the machine and recall that we had obtained in that the equations in per unit where omega b is the electrical base frequency omega is the electrical radian frequency is equal and T m in per unit the per unit mechanical torque minus psi d i q minus psi q i d this of course, a per unit equation. Another equation which we have is since theta is equal to omega naught plus delta it follows that d theta by d t is equal to d theta by d t is nothing but omega naught plus d delta by d t. So, another equation which we have is d delta by d t is equal to omega minus omega naught where omega is nothing but d theta by d t. So, this equation along with this equation and this equation determine the behavior of a synchronous machine. In the short circuited and open circuit analysis we had assumed that the speed is constant will not be making that assumption here. Now, these equations are of course, coupled why are they coupled because v d is in fact, a function of delta v d and v q are functions of delta that is what we just did sometime back and in a 1 if you recall the equations of psi d and psi q omega appears. So, there is coupling here also. So, a 1 in fact is a function of is a function of omega and in the electromechanical the mechanical equations psi d and psi q of course, are coupled to those equations of course, we have to give the initial conditions. The initial conditions are such e f d is 1 speed is almost equal to omega base of the rated speed and let us assume that this frequency of this let this is an assumption we will make that omega naught which is the infinite bus or the voltage source. In fact, the voltage source since it is a perfect voltage source it can be called as an infinite bus it is frequency is nothing, but omega. So, we will just assume this. So, what we are effectively assuming is omega is approximately omega naught it may be slight what we really are going to do in this simulation is have omega slightly higher than omega naught. Now, e f d is equal to 1 if speed were equal to omega the base speed then the line to line voltage which would appear across the star connected synchronous generator would be in fact, 1 per unit. So, the voltage source which we have the voltage source or the infinite bus which we have has got line to line voltage 1 per unit. The voltage source the generator itself has a open circuit voltage line to line voltage also of 1 per unit slightly higher in fact, because we are going to have omega slightly higher than omega naught. So, what we will be doing is connecting the synchronous machine under these circumstances. The fluxes all the fluxes we assume are in steady state under open circuited conditions. So, we pre calculate the steady state open circuit conditions and then apply or connect the machine to the infinite bus. Of course, since the speed of the machine is slightly greater than the infinite bus if we delay in fact, if we assume that initially delta is equal to 0 if we delay a bit the connection of the machine to the infinite bus delta will change. Remember that d delta by d t is equal to omega minus omega naught. So, delta is in fact, changing linearly. So, in fact, if we delay this or connect the synchronous machine when delta is large you will get a correspondingly large transient. So, we will assume that we are at a steady state. So, I will pre calculate the steady state values of a synchronous machine under open circuit conditions and then what we will do is connect the machine under various values of time. So, remember that we assume that at t is equal to 0 delta is equal to 0. So, as time increases delta will change because delta is equal to omega minus omega naught and omega is slightly greater than omega naught. So, I will first what we will do is synchronize the machine to the voltage source the 3 phase balance voltage source. In fact, since the initial speed of the machine is slightly greater than the infinite or the voltage source once you connect it you will notice that if the machine does synchronize properly then you will find that the speed of the machine is equal to that of the in steady state it becomes equal to that of the voltage source. So, good if you have if the machine synchronizes. So, this is what is known as synchronization it machine kind of even if initial speed is slightly higher than the frequency of the voltage source it is to be synchronized to once you connect it it locks on to the frequency of the voltage source. In fact, if you recall the experiment demonstration experimental demonstration which I showed to you in the first lecture of this course we in fact did just this. So, let us just see this particular simulation before we go ahead let us see the program which actually implements this one problem in trying to write a program for this particular set of equations set of differential equations which we have seen is the nonlinearity of the differential equations because the product terms. So, we cannot directly apply Eigen value or Eigen vector analysis and write down the response you know they are product terms which appear here in fact V d and V q are sin of delta and have sin of delta and cosine of delta terms. So, you have got what is known as a non-linear set of coupled differential equation ordinary differential equations we will have to apply numerical techniques to solve this equations. Now, one of the problems in applying numerical techniques in this particular equation set of equations is that it is stiff and of course, we spent quite a bit of time in the first in around the 5th to 10th lecture trying to understand how we can analyze stiff systems without a significant loss of accuracy. Now, one of the things we have already done for short circuit analysis is removed a stiff part of the system what was the stiff part of the system the differential equations corresponding to psi d and psi q. So, what we did was you know assumed in these equations that psi d dot and just replace this by 0 made these algebraic equations the first two equations become algebraic equations simply by setting these 2 to 0 d psi d by d t and d psi q by d t then we can express psi d and psi q in terms of these fluxes and thereafter we can do the simulation of the reduced system wherein we have removed the stiffness because we have in fact neglected the fast transient. So, that is what we will do the reason why we do it is somewhat opportunistic as far as this particular lecture is concerned. If I want to solve a set of non-linear differential equations which are stiff then it is a good idea to use an implicit method like trapezoidal rule, but we have seen in the lectures in which I described to you non simulation of non-linear systems. If you want to apply implicit methods like trapezoidal rule in each step you will have to solve non-linear algebraic equations in order to obtain the value of the states at that step. So, that becomes a fairly complicated programming exercise. So, what I have done is I have taken the simpler path that is I have removed the fast transients of the system, removed the stiffness to some extent by neglecting d psi d by d t and d psi q by d t and then used a simple explicit method like Euler method with a small enough time constant small enough time step. The reason of course is that Euler method is not very accurate it is not a very it is a first order method it is not very accurate. So, I have to keep the time step a bit low, but at least by doing this removing this stiffness we have achieved we have avoided trying to have a complicated program in trapezoidal rule which will require us to solve non-linear algebraic equations at every step. So, let us try to simulate the system with Euler method of course with d psi d by d t and d psi q by d t neglected. So, let us now look at this program. So, please pay attention to this program. So, what we will do is simulate a synchronized generator now we are doing actually simulation that is now numerical integration in this program. The base frequency we take as 2 pi into 50 the speed of the generator initially is omega which is at present I will of course change this value when we will try to change this value later probably is slightly higher it is 10 percent it is in fact 1 percent higher than omega b these are the values of the parameters which I have chosen in fact they are the same as what we use for a short circuit study. The important thing of course is T m is equal to 0 see if a machine is running under open circuited conditions the mechanical power equals 0 of course in real life since we are you will require little bit of mechanical power to overcome the friction even though you are running at no load. So, under open circuited conditions you are in fact at no load. So, in real life you have to have some little bit of mechanical torque to overcome the friction, but of course we have not modeled friction here. So, T m in fact has to be 0 otherwise of course the machine would keep on accelerating. So, mechanical power is equal to 0 initially the machine is under open circuited conditions. So, for equilibrium we have to have mechanical power equal to 0 since electrical power is also 0 now we do the same things as we had done before we are going to calculate the time constants T d dash T d double dash T q dash T q double dash and recall that our state space equation would be like this the equilibrium conditions of course are given by you know by as we have discussed in the previous lecture we effectively have to set x dot is equal to 0 that is all the rates of change of the states have to be set to 0 and the corresponding algebraic equations have to be solved in order to get the steady state values of the states. In fact, this x s s here in fact is the steady state variable steady state values of the fluxes under open circuit conditions. So, initially of course remember that we are under open circuit conditions. So, what I have assumed is that v d and v q are related to i d and i q by this relationship v is equal to R L i R L being a very large value 1000 to simulate an open circuited generator. Thereafter what we do is we give the initial condition corresponding to the steady state values under open circuited conditions of the synchronous machine the initial speed is omega which is slightly greater than omega naught the infinite bus or rather the voltage source electrical frequency is omega naught which is equal to the base frequency the initial value of delta is 0. Now, we do the numerical integration now remember that this is the a 1 matrix we reduce a 1 matrix by neglecting the fast transient. So, this is the step which does this. So, we apply Euler's method here. So, this line effectively tells you that I am using effectively Euler method here to simulate with the time step of 0.005 that is 5 milliseconds. So, we move on down here electrical torque is calculated. So, I am using a reduced order model remember with this d psi d by d t equal to 0. Now, once the short once at time t is equal to t 1 if you look at this initial simulation under open circuited conditions is carried out to time capital t 1 at time t is equal to t 1 we connect the machine. So, our Euler simulation includes includes b 1 b 1 equivalent in fact where b 1 is in fact related to I will just come to b 1 here once again. So, if you look at where b 1 is it will be here here is b 1. So, b 1 is nothing but minus v sin delta and so on. This is basically v d and v q. So, as soon as you at time t is equal to t 1 b 1 effectively we switch we kind of have a switch here in which we remove this R L which appears in a 2 we remove this R L which appears in a 2 and we apply the voltage. So, recall that we had simulated a open circuit by subsuming the effect of v d and v q under open circuited conditions in a 2 by you know representing v d and v q as R into i d and i q R L into i d and i q. Now, we remove that R L from a 2 and we connect the voltage source. So, we have got this second input b 1. So, you can of course, look at this program it is a bit involved I will not you know explain every step, but you can just have a look at it nonetheless. So, what I will do is after t is equal to 5 I will actually start increasing the mechanical torque of the machine at t is equal to 10 I will increase it even further, but before we do that let us simulate the system only to the point of synchronization. So, I will limit my simulation to 5 seconds. So, let us try to do that. So, there I run it it is already done. So, what I have done effectively is that one thing which you probably I forgot to tell you that t 2 has been chosen to be 5 and t 1 has been chosen to be 0.2. So, t 1 is the point at which I remove R L the. So, you know R L of course, is very large value of the star connected resistance at the generator and connect the generator to a voltage source. So, we are from open circuited condition we are going to a short circuit open circuited condition we are going to a condition in which we have connected the machine to the voltage source. So, t 1 is the time at which we do that t 2 is the n time of the simulation. So, of course, you have plot this for example, omega this is how what it looks like. So, please pay attention to this initially the speed is 10 percent higher than 314. 314 is roughly the speed at 50 hertz. So, at time t is equal to 0.2 we connect the machine and what you notice is that the frequency locks on to the frequency of the infinite bus. In some sense as I mentioned sometime back this is in fact, the phenomena of synchronization with the synchronous machine locks on to a voltage source and what you see of course, is a relatively low frequency swing it is a low frequency swing which you see or one may say a low frequency oscillation which precedes the steady state. If you look at the frequency of this this is roughly slightly more than it is more than 1 hertz because this is roughly 0.75 seconds and this is 1. So, roughly this is you know the frequency turns out to be 1 upon 0.75 which is a period the answer you multiply by 2 star by. So, the frequency is roughly I am sorry pi is 3.14. So, roughly the frequency is around 8 actually this is a very rough kind of calculation, but this is approximate order slightly greater than 1 hertz you see a swing this in fact, called a swing. So, it is associated with the electromechanical transients in the machine. Now, in fact, if you look at delta we will close this window of omega and plot delta this is how it looks. So, till point in fact, till 0.2 seconds you see that delta is increasing linearly this is increasing linearly because the speed of the synchronous machine is 10 percent more than the infinite bus it is really increasing linearly I have zoomed this and after sometime at this point we of course, connect the machine. So, it kind of synchronizes now if you look at the electrical torque as well and have a look at the electrical torque well it is got superimposed on this. So, we will just do it again. So, at of course, initially the electrical torque is 0, but as soon as you connect the machine it kind of oscillates there is a torque transient. In fact, you see that the torque becomes positive and then negative and so on and in fact, it is slightly positive to begin with and then it goes negative that is one of the reasons of course, is that the speed of the machine is slightly greater than that of the infinite bus. So, in fact, the machine gives out some of some energy eventually to the infinite bus because it is overall kinetic energy reduces that is eventually the speed settles down to the speed of the infinite bus a little bit. So, you see this torque transient. So, actually this is the electrical torque in fact, we will hold this figure and we will just do one more thing we will suppose our time of connection was when delta was almost 0 that is the voltage sources the open circuited voltage source of a synchronous machine is almost I mean if delta is smaller it really means that the open circuit voltage of the machine is practically equal to that of the infinite bus if delta is 0 it if I in fact, it is equal to. So, in fact, if we you know at do the synchronization a bit early delta movement from 0 is reduced and you are connecting the synchronous machine to the voltage source when delta is smaller. So, one can expect that there will be a smaller transient of course, the ideal situation for a bumpless transfer would be that you know your delta is practically equal to 0 that is the phase angle of the open circuited voltage of the synchronous machine just before it is synchronized is equal to the voltage of the infinite bus the phase of the voltage of the infinite bus. So, that would be an ideal situation in fact, those who are aware of the dark lamp method of synchronization or the synchroscope method of synchronization would recall that we synchronize as close to 0 delta equal to 0 as we can. Now, suppose I make this delay in synchronization smaller I mean in the sense that I make T 1 as 0.1. So, delta will deviate less from 0 in the time period 0.1. So, in that case if I rerun this gain what you will see is of course, that the electrical torque would be smaller in fact, the bump should be smaller that is what you should see. So, we have done the synchronization a bit earlier here and the overall torque transient certainly has reduced. So, you can reduce you can make it a bumpless transfer if we get the speeds almost equal of the machines equal and we also synchronize around delta is equal to 0. So, absolute bumpless transfer will occur for example, if in this program I make if I make the machine synchronized when the speed is exactly equal to the synchronous speed speed of the infinite bus and delta also is 0. So, delta is 0 if I do the synchronization right at time t is equal to 0 because I have said that my initial value of delta is 0 at t is equal to 0. So, if I do the synchronization under these circumstances of course, you should have a nice really nice transfer in fact, you will not get any transient. You see a transient, but if you look at the scales they are effectively indicative only of numerical error. In fact, this is 0.001 the electrical torque 0.0015 here. So, practically it is a bumpless transfer this is a very very small value which you are seeing here. So, this is basically what we get to see here we will now move on to try to seeing what try to see what happens when we synchronize the generator, but we will do something more we will synchronize the generator and now increase the power. So, I increase the electric mechanical input power to the turbine and therefore, the generator gets more mechanical power if the mechanical power increases of course, the electrical power also will increase because eventually in steady state mechanical electrical power is the same. So, what you will find is that the mechanical power will increase then the electrical power also increases. So, let us do that. So, what I will do now is I will do the simulation. So, I will not worry too much about the synchronization transient maybe I will just get back to what we were sometime ago. What I will do is simulate for a longer time say I will do it for 10 seconds I will simulate this for 10 seconds the n time is 10. What I will do is now at t is equal to 5 seconds after t is equal to 5 seconds I will give a step change in the mechanical power of course, it is not easy to give a step change in the mechanical power, but we will just for the sake of analysis we will assume that I am able to give a step change normally you will ramp it up with a certain rate of rise. So, real turbine you can only ramp up the power you cannot give a step change, but that is we will just see what happens we will keep the e f d still at 1 per unit. So, what we will do is we will simulate only for 5 seconds. So, saved it and now I rerun it. So, if I plot now if I plot the electrical torque what you notice here is that I simulate it for of course, 10 seconds this is the initial synchronization transient the electrical torque becomes equal to the mechanical torque which is 0, but at time t is equal to 5 I give a step change in T m and the machine oscillates and go as it settles down to this new value of T m the new value of T m is 0.25 per unit. So, it of course, mechanical power becomes equal to the electrical power. So, electrical power also becomes 0.25. So, if I look at instead of looking at the torque I look at the speed what you notice of course, is if you there is a initial initial synchronization transient after which the speed settles down to the speed omega naught even after increasing the electrical torque the speed of the machine does not change. I mean this there is of course, a transient here, but eventually it again settles down to omega naught that is because the machine in some sense is synchronized and locked on to the voltage source which has a constant frequency. Remember that if you are in synchronism the machine tends to stay synchronized unless of course, I give a very large disturbance. If you look at delta on the other hand it settles down to a value you see the oscillation here and also you know you see an exponential mode also in addition to the oscillation. So, I have got a damped oscillation and exponential growth that is of course, because the many modes associated here. This is of course, a non-linear system, but you do sometimes of course, if the system is not too non-linear see near the equilibrium point the appearance of all these typical transients like exponent exponential growth and decay or oscillatory growth and decay. So, this is what you are really seeing and the delta settles down to a value of approximately 0.4. So, this is the settling value of delta. Now, what we will do next is do something more at time t is equal to 15 seconds or time t is equal to 10 seconds I will increase mechanical power to 1 or let me do one thing I will increase it to 1, but I do not I keep E f d at 1 also. So, I do not change the field voltage I still keep E f d at 1 and I increase the mechanical power to 1 at time t is equal to 10 seconds. Now, if I do that of course, I have to simulate for a time longer than 10 seconds I will just simulate for 15 seconds. So, if I increase the power beyond a certain point. So, what I do I will do is plot time versus delta and what I see here is oops what you see is really this angle delta see this is basically the transient which we observed last time I have simulated for a slightly longer time after 10 seconds I applied the torque t is equal to 1 per unit and what you see is of course that delta goes on increasing. So, that is an interesting point that delta keeps on increasing this is because. So, the machine will eventually lose synchronism. So, in fact if you look at that is of course, because we will have to plot it again you will find that the speed of the machine in fact goes on is in fact equal to 3 1 4 once you synchronize it then you if you increase the torque again the speed remains the same, but well the machine seems to be slipping out of synchronism, but of course this is not apparent in this figure probably if we simulate for a longer time this will become apparent. So, let us do one thing we will simulate this for 25 seconds. So, if I do this I just rerun this again so hopefully you will be able to see something. So, what we see is eventually the speed you know. So, if you recall what we saw last time was this something like this we simulated it up to certain point thereafter the speed just goes on increasing and the system loses synchronism. So, this is what we get in case we will just plot it again if we apply a torque of t m is equal to 1 per unit without changing the field voltage. So, in fact is this surprising for example, I will do also I will show you also the plot of delta it goes on increasing. So, the machine has lost synchronism. So, this is not actually very surprising that is because the reason why this is happening is the mechanical torque which you are giving is in fact greater than the maximum possible torque which we can develop in this machine if E f d is kept constant at 1 per unit. So, I leave in this is an exercise for you to just check out that in case E f d is 1 and the values of x t x q which we have here and E f d also is 1 what is the maximum electrical power that we can push through this generator. In case we try to push something more without a corresponding increase in E f d then what we notice is that the machine loses synchronism. So, in fact if you recall something which we did quite some time ago that is in the first lecture I showed you a demonstration of a synchronous machine which loses synchronism if you go on increasing the power output beyond a point mechanical power that is the prime over power if you go on increasing then beyond a point it loses synchronism. So, today after lot of model development and understanding how to analyze a dynamical system we have come to a point in which we can simulate this phenomena. So, we will spend a little bit more time on this again in the next class I would like to show you what happens in case we increase T m, but also correspondingly increase E f d in that case do we remain in synchronism or not. So, that is something we will do in the next lecture. We will also thereafter look at how we can obtain lower order models of synchronous machines for certain theoretical studies. So, when we do some kind of theoretical study or when we are explaining a concept sometimes it is better to use a model which is much much more a toy model rather than the full blown model consisting of six flux states. So, that is something we will try to do in the next lecture I hope now with this couple of lectures that is the one last time and this time I hope you are getting a feel and some fun out of understanding some of the phenomena associated with the synchronous machine by a rigorous analysis of the simulated as well as the analytical treatment which we are doing here.