 We proceed with our simulation of a two machine system. In the previous lecture we kind of I gave you an idea how you can formulate the equations. We will actually go through the all the equations once and then go ahead and actually simulate. I will show you the results of the simulation for various disturbances like load change as well as faults. Now, this is lecture number 37 and what we will be doing here is of course, continuing with our simulation of a two machine system. So, let us just have a relook at what the system we were trying to simulate. This is the simulation system simulation and this is the initial steady state operating condition. Now, we will take this as the beginning or the starting point of the simulation. In fact, you can say this is the load flow situation and of course, the first step of any simulation is to back calculate initial condition. So, if I know the voltages here then I can get the values of for example, all the states and the field voltages of both generators based on the voltage at the terminals and the currents in steady state which are coming out. And of course, from the conditions from all the you know specifications which are given for this scenario you should be able to find out the voltage and the angles as well as the current output of the machine which will enable you as I said to compute the initial conditions. You can refer to our simulation of a synchronous machine with an AVR connected to an infinite bus. We did that several lectures ago where actually showed you how you can from the terminal conditions calculate all the initial conditions. So, let us look at the synchronous machine equations for the two machine system. I mean this includes both the machines and also how you will interface them. So, of course, you know that the basic differential equations of a synchronous machine with the stator transients neglected we have just four differential equations per machine. So, when I say network transients neglected what I mean also is that stator transients also are neglected. When I say stator transients are neglected what I mean of course, is that d psi d by d t and d psi q by d t terms of the synchronous machine have been set to 0 and the corresponding differential equation has been converted to an algebraic equation. So, what we have essentially are the differential equations of the rotor windings g h f and k and we also have a differential equation corresponding to a simple static exciter plus AVR model. In the previous class I told you that the AVR plus static exciter is modeled just by a transfer function k a upon 1 plus s t a that is the form of the transfer function as a result of which it actually is embodying the differential equation in a state x c which is given here and of course, e f d the field voltage is equal to x c except when x c hits limits. If x c hits limits we kind of clip e f d to those limits. So, this is basically the static excitation system model this is the same for both the machines to keep things simple we have considered similar parameters for both the machines. The turbine governor model is also kind of a transfer function of course, you can write the state space equation corresponding to this transfer function that not that is something which I have not written here, but you can easily you know infer that this first order transfer function k 1 plus 2 s upon 1 plus 6 s is actually embodying a single differential equation. So, effectively you can write down this transfer function in state space form where the input is the error in the frequency or the deviation of the speed from the reference value and the mechanical power is of course, the set point the load set point P m 0 plus the output of this governor delta P m. So, P m of course, is limited and out here you see as in a static excitation plus AVR model the turbine governor model in fact, the governor model effectively has a gain k remember that the governor is a proportional type controller in this case. So, what we are considering is proportional type controllers in both synchronous machines. So, the turbine governor model the steady state gain is not infinity like an integral controller it has got a finite steady state gain. The mechanical equations of course, are the rate of change of the rotor position that is if theta is of course, the rotor position. So, if theta is equal to omega t plus delta then d delta by d t of course, is given by omega minus omega naught and the rate of change of speed is of course, the torque equation and in per unit of course, T m and P m are practically the same the assumption of course, is whatever transients we are going to study the speed deviation from the nominal will not be too much. So, mechanical torque in per unit is the same as is practically the same as the mechanical power in per unit. Now, if you look at the next equation rather a few more clarifications here remember that omega b in all what have written is the base angular speed omega naught is basically such that theta is equal to omega naught T plus delta is the rotor position. So, delta is equal to theta minus omega naught t in this study let us take omega naught to be equal to omega b. So, what basically we are saying is that the base frequency and omega naught are taken to be the same in fact, another the thing we will assume in this example to keep things simpler is basically assume that the steady state frequency is nothing but, the base frequency. So, this is some simplifications which we are going to consider. We have the algebraic equation for both machines in fact, along the d axis and the q axis. The first algebraic equation is something of course, you know we derive the synchronous machine model. The second algebraic equation is actually got by assuming that d psi d by d t is equal to 0 and in that in this particular equation this term here should have been omega, but we have approximated it to be omega b. So, this particular algebraic equation has got constant coefficients into these variables psi q i d and v d. Similarly, on the q axis you have got this algebraic equation this of course, derived from the synchronous machine model and this algebraic equation is in fact, obtained by setting psi d psi d by d t and d psi rather d psi q by d t equal to 0. So, the same approximations apply here as well as here. So, we have got in fact, four differential equations per you know you have got four equations of the synchronous machine rotor windings and one differential equation for the static excitation plus AVR, one differential equation for the simplified governor model turbine governor model. So, that gives you basically six equations algebra differential equations per machine. So, there are two machines there will be 12 states in fact, but the algebraic equations per machine are four. So, four algebraic equations per machine and I will state down here. Now, remember one thing what I have shown you so far is the synchronous machines in the park's reference frame. So, when you derive it for each machine you will be using the park's reference frame native to that synchronous machine. But if I use a transformation C k instead of C p where instead of using theta as the argument of the cosine and sine terms here I use omega naught t. Remember delta is missing here you know if it was omega t plus delta it would be theta corresponding to a synchronous machine. So, instead of having two different transformations for two different machines what I will do is use a common transformation whenever I am going to interface the variables corresponding to the two synchronous machines. So, what this will become clear in the moment. So, the ABC variables of course, are the same all what we are doing is changing using the transformation C k to what is known as the d q variables. So, if you look at how these transformations and the variables are related you get the equation as shown here. Now, if you look at you know you just do C k inverse C p 1 or do C p 1 inverse C k you can get a relationship between f d 1 f q 1 and f d 2 f upper case d 1 and upper case q 1. So, this relationship is essentially a matrix relationship, but you can easily see that it essentially gives you can be compactly written by this complex relationship. Similarly, for machine 2 if I use you know C p 2 which uses the argument for the cosines and sin's as theta 2 is equal to omega naught t plus delta 2 will get basically the relationship between the capital or rather the upper case q and d variables as related to the local variables which use the local parks transformation. So, whenever I am going to what I will do is I will get the equations to the upper case q and d variables the lower case q and d variables are using a transformation which is native to that machine. The upper case d and q transformations for machine 1 as well as machine 2 both use C k which is independent of theta delta. So, whenever we are going to so f q 1 and f d 1 upper case are in fact variables which are obtained for common transformation. So, f q 1 f q 2 f d 1 and f d 2 upper case are in fact obtained from a the same transformation. So, the point is that if all the a b c variables are you transform using a common transformation C k then we can use k b a l and k c l that is Kirchhoff voltage and Kirchhoff current law while writing down when interfacing both machines. Now, the important thing is you do not have to transform all the differential equations to the upper case d q variables at all. What we need to do is whenever we are interfacing these variables whenever you are writing the algebraic constraints relating to these variables that time you should ensure that all the variables involved in that interface are obtained using a common transformation C k otherwise you will not be able to use the algebraic relationships which are obtained from Kirchhoff's voltage and current laws. So, that will become quite clear now if you look at the algebraic equations corresponding to both machines. What I have done is the algebraic the variables psi d lower case psi q lower case i d lower case and i q lower case have all been transformed the equations have been transformed the same relationship is there, but now I am using different variables I am writing the same relationship in terms of different variables. Now, the interesting thing is that if I write down these algebraic equations in terms of the upper case variables I will have to combine those equations in fact and you get this kind of form of the equations. Now, the interesting thing is that if you choose x d double dash and x q double dash to be equal what you find is that this capital D variables psi d is related to i d by simply x d double dash psi q is related to i q also by x d double dash. So, if you assume that this sub transient saliency is not there that is x d double dash and x q double dash are equal then an interesting thing which follows is that the algebraic equation corresponding to psi upper case d is related to i upper case d by a simple constant coefficient which is not dependent on delta. The other terms are all gathered out here and they are all states f 1 is a function of the states. So, this is a non-linear function of the states. So, the algebraic equations effectively have been rewritten in terms of the capital D q variables. Now, this of course, can be done for both the machines. So, when you are talking of machine 1 you will have of course, an additional subscript of 1 everywhere. So, of course, I have not done that, but remember that this has to be done for both machines. So, what you will have effectively is that these 4 algebraic equations for each machine for both machines will be 4 algebraic equations, but remember that the variables in these 4 algebraic equations will be having a common transformation from the ABC variables. So, you can actually use these upper case d and q variables directly whenever you are using any algebraic constraints like K v l or K c l. If I neglect of course, if I neglected the stator transient that is d psi d by d t are set to equal to 0 and d psi q by d t are set equal to 0 then there is no real logic in retaining the first transients associated with the network. Remember that the network is was already simplified you know we may got an lumped R L model from what was essentially a partial differential equation model. So, that is of course, a big jump I am a big models in simplification. Now, we go ahead and do another modeling simplification that is we neglect the network transients. So, if you neglect the network transients then the equations are essentially like this v d 1 and v q 1 and v d 2 and v q 2 are the terminal voltages of the synchronous 2 synchronous machines and both are of course, obtained using a common d q transformation. So, you can get them and subtract them they are kind of compatible. So, we can use the voltage on either side of course, is obtained from a common transformation. So, this equation is actually valid we could not have for example, written down here small or lower case d 1 and q 1 and lower case d 2 and q 2 that would be incorrect. So, remember that this change is there. So, what we have effectively done is obtained these algebraic equations of the network from the essentially from the differential equation of the lumped R L model of the interconnection. There are some more algebraic equations what we have done essentially is what is v d 1 and v q 1 since there is a resistive load. If you focus on this diagram which is here if this is a resistive load it is a unity power factor load, but I also additionally say that it is a you know kind of a resistive load. So, the voltage here is equal to the current into the resistance. If the resistance is R L 1 then you know the voltage and the current are simply related by an algebraic relationship. Now, so the what is the current flowing through this it is the current of the generator minus the current through the line. If I call this current through the line as I L in that case you will have v q 1 is equal to R L 1 into the difference of the generator current minus the line current. Now, this 1 and 2 actually are superfluous actually this should be just I L q this 1 is not necessary. So, this is a small error which you should correct v d 1 is equal to R L 1 into I d 1 minus I L d please remember to remove this 1 here it is not required there is only one line current. So, there is no need of this additional subscript 1 out here. Similarly, v q 2 is equal to R L 2 into the current through the resistance that is the generator current plus the interconnectional line current again please neglect this superscript 2 here which appears in this term I L q. So, just remove this superscript I am sorry. So, you can remove this subscript. So, you have got this these algebraic equation please remember to make these corrections. So, eventually you have got 16 states what are the 16 states. In fact, these are delta 1 omega 1 then the 4 rotor windings are the first machine then the state corresponding to the simplified AVR in exciter model and x g 1 is the state corresponding to a simplified turbine governor model. Similarly, you have got the same thing for the other machine. So, there will be a total of 16 states and the algebraic equations R 14 you could have counted all the algebraic equations 4 for each generator that will become 8 plus 6 equations which I just described some time back 2 of course, for the line and the 4 equations which I just mentioned some time back. Remember that the number of variables algebraic variables also are 14. So, it should be possible whenever you are writing a program to actually use you know you can solve the algebraic equations and write them in terms of the states. So, this is something you can do remember that at every point at every instant of your simulation on numerical integration remember that psi d 1, i d 1, psi q 2, psi q 1, i q 1 and so on are dependent on the states. So, if you recall that if you look at these equations here you see that eventually psi d this equation is dependent on the states. So, there is an overall dependence on the states just remember that there is an overall dependence on the states. Now, we begin our simulation remember all the assumptions we have been which we have made we have made lots and lots of them, but some of the important ones are that we are considering we are going to consider relatively slower phenomena we are not going to consider very fast. Basically we are going to study electromechanical transients and what we have of course, done is also assumed in some cases replaced omega by omega b. So, what the assumption is that you are not deviating too much from omega b. So, all our studies are for frequencies which are near the nominal frequency is it now let us get back to our case studies. So, I will just get you to that slide remember we have made the simple models of the AVR and turbine governor. So, this is something you can watch on your screen. So, once you have made the simplified models and obtain the differential equations and of course, made a few more assumptions will you of course, considering three phase balanced disturbances there would not be any unbalanced disturbances synchronous machine data. And I have done the programming for analyzing the system the first result which I want to put forth to you are the Eigen values of the system for obtained after a small signal analysis around the equilibrium point corresponding to this situation. So, if you look at the situation again this is the equilibrium situation I have back calculated the condition linearized the differential equations we have done this before. So, I am not doing it again and I have actually tried to obtain the Eigen values of the system for this equilibrium point. So, if I give small disturbances the behavior will be captured by the linearized kind of response of the system for small disturbances. So, if you look at the Eigen values of the system there is some interesting things you will see if you have no governor if you have got no governor there are two zero Eigen values then you have got you will have two zero Eigen values and then you have got a low frequency oscillatory mode which is not so well damped. So, this is a low frequency oscillatory mode which is not so well damped I am sorry you also have other Eigen values some with relatively large real part and some very large Eigen values on the right hand side. So, without governor if I consider network and stator transients that is I do not neglect d psi d by d t d psi q by d t as well as d i d and d i q by d t of the network in that case you will have some very high frequency or high magnitude Eigen values corresponding to fast transients. So, this is what you get what are these two zero Eigen values remember I had mentioned to you that whenever you have got a system like this this is an analogy a two mass spring system you the thing is that if you have got essentially a swing mode which is an oscillatory mode as well as two zero Eigen values why do you get these two zero Eigen values you have got essentially if there is no friction of the surface then your response has two components corresponding to the zero Eigen values they are repeated Eigen values and in this case in fact you can show that you are going to get for the spring mass system response terms e raise to zero t and t e raise to zero t we have discussed this long time back when we are considering linearized analysis when you have got repeated Eigen values you can in some cases get this kind of response. In fact if you do not have friction we do get two zero Eigen values two for a two mass spring system this something you can verify by writing down the equations. So, what is this mean is that if I give if for example these two masses have equal initial velocities say v zero is equal to one and v v two zero is equal to one both these masses will move if there is no friction they will keep on moving getting what I am trying to say. So, they will keep on moving in case there is no friction. So, what you will find is that the displacement from an arbitrary reference will have a variation of t because if they given initial condition in these states which are equal you will find that this just keeps on moving. So, it is not surprising that you have got these two zero Eigen values and therefore, you have got this kind of response. So, remember that when you do not have a governor what does it mean? It means that you do not have any control over the load generation balance you do not have any control. So, it is equivalent to having the situation where you have got constant external forces acting on this mass spring system, but no friction there is no frequency dependence of these external forces. So, if there was friction of course things would be different in fact by putting a governor what you are doing is essentially making some external force on this system which is proportional to the change in velocity. So, in fact a governor changes the mechanical power with in response to omega minus omega omega f minus omega what it means essentially is that you are introducing a kind of a viscous damping coefficient into your torque equation and as a result of which when you have a governor you do not have these two zero Eigen values. So, that is an interesting point. So, just remember this remember that in case you have got a load which is frequency dependent or you have got a turbine governor enabled you will get only one zero Eigen value because in that case you do not expect this t e raise to zero t term to be there. In fact if in case you have friction and you have got an initial velocity you will find that eventually this mass will grind to some halt grind to a halt after some time. So, you cannot have this kind of term in case there is this viscous damping introduced by either load dependence on frequency or mechanical power dependence on frequency which is introduced by a governor. So, in this case of course our load is a constant resistance. So, of course you do not expect that the load to be frequency dependent. So, if you do not have a governor you will have two zero Eigen values and t e zero t kind of response. Remember of course that even if you put a governor there is still be a zero Eigen value or in effect a response term which is equal to e raise to zero t why is that. So, remember that if in this mass spring system if my response is x 1 t if your response is x 1 t and x 2 t you can add an arbitrary constant term to both of them and still your differential equations will be satisfied. So, an interesting point is that if your solution to your delta 1 and delta 2 or in this case for a spring mass system the displacements x 1 and x 2 is x 1 t then x 1 t plus x 0 also will satisfy your differential equation. This is because whenever x 1 and x 2 appear in your differential equations or even in this synchronous machine example you will find that everything depends on the difference of the angles the absolute value of delta 1 and delta 2 does not determine the electrical powers etcetera. None of the components in this system are dependent solely on the absolute or the actual angle at phase angle at a given point. For example, the power flow the steady state power flow here depends on the difference of the angles at the two ends it does not depend for example, on this alone. So, what it follows is that if your delta 1 and delta 2 are solutions are the responses following a disturbance or even the steady state responses. If I add a constant term a constant term is x 0 e raise to zero t which is nothing but x 0 itself then also your differential equations are satisfied. Therefore, do not be surprised when you do an Eigen value analysis and find out that you are getting a zero Eigen value whenever you are studying this kind of system with the kind of differential equation and differential algebraic formulation which we have written. So, of course, whenever you do an Eigen analysis also another thing you try to remember is that you may not get this is exactly zero whenever you do a numerical analysis. You may get this slightly non zero but that will be because of numerical errors precision errors. So, of course, when you have a governor now coming back to this problem having a governor introduces changes this zero Eigen value to a negative real Eigen value which implies that they cannot be any t e raise to zero t kind of terms in your response. Another thing is that two additional Eigen values come about because there are two extra states which we have considered with when you model a turbine governor. If you assume of course, assume that the mechanical power is constant that is what this without governor means you will not have these additional two Eigen values but when you have a governor you will have these additional Eigen values. Again remember that when we consider network transients we have got these high frequency terms. If you neglect network transients in that case you will find of course, that the high frequency terms are missing but other than that you have got again these zero Eigen values. You also have the low frequency oscillation terms these are also called a swing modes and you have also got additional modes because of their many many more states. But the important modes which you need to remember are these zero Eigen values and these swing modes these are associated with electromechanical transients. Remember of course, then how do I know that I cannot I have just stated this fact I am not proved that these Eigen values are associated with the electromechanical variables deltas and omegas. But in fact, you can actually prove this using participation analysis which we discussed sometime back. You can associate the modes with electromechanical transients primarily with the electromechanical transients but of course, this does not mean that these modes are not dependent on the other states. But when I say electromechanical modes these are essentially primarily associated with deltas and omegas. The other states also influence these modes but they are primarily associated with delta and omega that is what I want to say. So, with governor and without governor you have similar behavior with and without network transients considered you still have a similar behavior. So, we can to some extent justify this approximation of neglecting stator and network transients. So, we go to case 1 now I will quickly go through a few simple disturbances. Suppose, I have got this I give a step change in this load I change this resistance from 0.52 per unit to 0.7 per unit. So, actually the load reduces because the resistance has reduced sorry resistance has increased. So, if there is a load throw off that some part of this load has been removed. In that case if I do not have governors if I look at various quantities first thing you will find is the rotor angular difference increases or rather decreases why does it decrease from 45 degrees it comes to roughly 40 degrees. So, the angular difference delta 1 minus delta 2 both are you know you know rotating and the angular difference between them is around 40 degrees after the disturbance while it was 45 degrees before the disturbance why does that happen well one way of looking at it is that when you throw off some load here the load here reduces on this line reduces. So, the angular difference between these 2 buses reduces and the corresponding rotor angle also reduces. Now, remember when you say no governor you are not changing the mechanical power. So, if I do not change the mechanical power remember this is like a friction less system. In fact, there is no dependence of the external powers mechanical or load powers on the frequency and if there is a load generation mismatch what you of course find is omega 1 omega 2 as well as the center of inertia frequency which is defined as h 1 omega 1 plus h 2 omega 2 divided by h 1 plus h 2 both linearly increase. So, you see this linear increase and of course, if you allow this to continue then after a point your generators will be tripped out the reason is that if your frequency goes greatly away from the nominal frequency there is a chance that you will damage the turbine and generator system why is that. So, actually turns out that the turbine is made out of blades they are made out of blades and you have got a steam flow through this through these blades along this periphery the steam flow is not uniform it is not perfectly uniform there are variation natural variation in the steam flow. So, your steam flow may be 0 to 360 may be like this. So, it may be of course, this is an exaggerated electrical engineering or electrical engineers description of what happens. Now, what happens is that in case your rotational frequency deviates from nominal if it deviates from nominal you will find that this turbine blades start getting periodic forces across means are applied across the blades. So, if your steam flow is changing there is a variation of steam flow and you are running at the base speed you will of course, see that the force on the blades is actually changing. Now, the thing is if you deviate from this omega b the frequency of the force which these blades see will change. Now, what happens is in case the frequency with of the forces which these blades see starts coinciding with the natural vibration of these blades then there will be a huge displacements see you are giving a periodic force which is equal to the natural vibration frequency of the blades if that happens at a certain frequency if the rotational frequency deviates from omega b then you will find that the frequency of the forces on the blades also changes and if it coincides with the natural frequency of vibration the blades will get damaged. So, nobody allows especially a steam turbine in a steam turbine generator system nobody allows you to deviate the frequency to deviate much away from the nominal speed because typically you will find that after around a 1 hertz deviation from the nominal 1 or 1 and half hertz from the nominal you may start coinciding with the resonant frequency the forces which are incident on these blades start coincide the frequency of that starts coinciding with the natural frequency of the blades. So, that may damage the blades. So, no after a certain speed deviation either on the positive direction or the negative direction you may have to trip out the generator you either have an over speeding control which suddenly changes the mechanical power like a governor or you have to trip out the generator. So, this is not acceptable of course, remember that because the angular speeds are changing does not mean that the machines are losing synchronism. So, if you look at the difference frequencies it is constant in fact omega 1 minus omega 2 is constant only thing is that center of inertia speed is increasing because unbalanced external force because you have changed the load generation balance also the line power which is line power comes to a constant. So, actually from a point of view of synchronism synchronism is if you look at it we are not gone out of we have not gone out of step the mechanical powers are constant the center of inertia frequency is changing, but the relative speeds are in fact constant. So, we are not lost synchronism we have what we are saying is that the overall frequency has changed. So, that is something you have to keep in mind we are not talking of a relative motion instability we are talking of continuous change in the center of inertia speed. So, both machines are almost together decelerating or in this case they accelerating together they are not they are moving together, but they are both accelerating. If you look at a case two with the governors are enabled what you notice is of course, the mechanical the transient looks a bit different and in fact it settles down to a different value you look at the angular difference it settles down to 32 degrees instead of 40 degrees in the previous case. The angular speeds reach an equilibrium there is a decrease in this there is an increase in the center of inertia speed as well as the speed of each individual mass rotor mass, but eventually the speed reaches an equilibrium. The equilibrium speed is not equal to the previous speed. In fact, if you note something that the governors are proportional type governors in the sense that the mechanical power is changed using an algorithm like this the governor has got this kind of you know transfer function and of course, the turbine has got a transfer function. So, the turbine governor transfer function is k 1 plus 2 s upon 1 plus 6 s. Now, this k is a proportional controller effectively it is also the inverse of the droop characteristic of the machine. Now, the point is that you can change the mechanical output of the machine only if omega f and omega are not equal. So, in case this proportional kind of governor will always give you a steady state error. So, your frequency does not come back to the original frequency there is a frequency change that is something which you should remember. The mechanical speeds the difference speeds a relative motion is stable it is not unstable. The speeds the difference speeds become equal to 0, but the overall speed of both the machines settles down to a higher value. Here the mechanical power are changing why are they changing because we are having a governor. So, if you have got a governor your mechanical power will change and of course, the line powers also will change because governor is present here as well as here. So, the point is that if governors are present on both machines each of them will change the mechanical power. If they change the mechanical power the overall power flow situation changes. So, load has changed mechanical power here is change and the mechanical power here is change. So, if all the three things change you will find the frequency goes to a new equilibrium and you know your line power also settles to a different value the angular difference between the two buses also will settle down to different values. Now, of course, the point which you should note here is that governors on both machines are present. Now, what if we had just one governor or a governor present only on one machine basically your mechanical power of only one machine would change only one machine would take on the load change and of course, if you just had governor on one machine only one person would contribute to the you know to trying to get the frequency to equilibrium. So, if you keep the same gain and just disable one governor your frequency deviation will be more only one generator will respond to the load changes. It can also be inferred that since the load the amount of power change of each turbine is proportional to k into omega r f minus omega by changing k or omega r f you can in fact, change the amount of contribution of each generator. So, why is that so because remember that the speed eventually if you are remaining in synchronism speed of both machines is going to be the same. Now, if speed is the same and omega r f of both machines is the same then the amount of power change each generator takes up is proportional to omega r f minus omega into k. So, if k 1 and k 2 are different I mean you have got two different gains or two different groups on the generators of each machine in that case you will find that the each generator takes on different load change a kind of extreme condition is where the gain of one generator of one governor is much higher like one of the governors for example, has a steady state gain of 300 and the other has got a gain of 10. In that case the governor assuming omega r f's are the same the governor with a larger gain takes on more of the load it takes it increases its mechanical power more whereas, the other generator does not increase its mechanical power as much. Now, the extreme condition is where you have got an integral controller on one of the governors and integral controller essentially has got infinite gain in steady state you know 1 upon s remember. So, you will get infinite gain in steady state. So, if you got one governor which has got an integral gain it will take on the complete load change and the other governor will be left high and dry in the sense it will not contribute anything to trying to maintain frequency of course, this is assuming omega r f on both machines is the same. Now, just think of think about it can you have integral controllers on both governors this is something you need to think about this is a question which not obvious you think over it a bit you cannot have in fact both governors having integral gain why it is. So, is something you can think about now the third case is not a load change what we will do is we will have a three phase fault it is a balance fault at bus 1 at 1 second and we will clear the fault and after 80 milliseconds that is four cycles. Now, of course, remember that post fault conditions assume to be same as pre fault conditions. So, it is a kind of temporary fault it is we just come back. So, when the fault is cleared. So, you can consider this as a situation where in bus 1 suppose you have got an open line this is your oops bus this is a generator 2 is a generator 1 the same system this is a load this is a load. So, this is your same system and you have got a transmission line out here which is open circuited here at this end. So, short transmission line and you have got of three phase fault on this line near about this bus. In that case you will have a short circuited this bus a three phase short circuited this bus which has to be cleared by opening the line. The post fault and the pre fault situation is going to be the same when before the fault you had connected an open line here it is a short open line. So, it is like having nothing here you had a fault for some time which is cleared by opening the line. So, what you have is pre fault and pre post fault situation is the same assuming of course this is a short line and open circuited short line is effectively doing nothing to the system initially. So, under these circumstances what you find is if you are clearing time is this much you will find that delta 1 minus delta 2 in fact is stable. Since pre fault and post fault systems are equal you will find that there is no change in the load generation balance situation nor are the flows change. So, losses also do not change. So, angular speed simply comes back to the original value they are transients, but overall angular speed settles down. So, it is a stable large disturbance it is stable for this large disturbance omega 1 minus omega 2 is a relative speed it is also stable. The powers again are stable actually this simulation has been done considering network transients we have model network transients. So, what you see essentially in the beginning of this transient here for example, are some high frequency components. So, if you consider network transients you will get these high frequency components there is a high frequency component here as well. So, this is how your delta 1 minus delta 2 if you neglect the network transients you get almost similar response, but remember there is some there are some differences for example, the maximum deviation here is around 70 degrees whereas, when network transients were considered it was almost 60 degrees. So, there is a 10 degree difference. So, when network transients are neglected you do have some difference in the result. In fact, network transients neglected gives you a somewhat larger angular deviation and, but otherwise the response looks almost the same you just see that there is no high frequency you know transient you know because you have neglected network transients. Now, we will consider when we increase the fault clearing time to 0.28 seconds, but the network transients of course, are neglected the interesting thing is that the angular deviation under this circumstance grows in an unbounded fashion. So, if you have a larger clearing time the fault is cleared, but after a long clearing time. So, a longer clearing time means that because of this sudden disturbance the deviation of the states from the post fault equilibria are going to be more. So, if your deviation is too much you may not you know pull back into synchronism as a result of which the relative angular motion is becomes unstable. So, please remember this is relative angular motion becoming unstable if you look at the angular speeds they too you know deviate from each other although omega c o i is relatively you know within bounds omega 1 and omega 2 the relative angles increase. This is contrast this with the earlier situation in the earlier situation when we had a load through off without governors omega c o i went on increasing, but the relative motion between them was stable this is exactly the converse situation. So, what we have seen is that if you give a disturbance like a fault which is not cleared for a long time you may go out of synchronism and you will see the relative motion is unstable. So, relative out of synchronism refers to relative motion. If you look at the mechanical the electrical power output of each machine you will find that it is oscillating there is a kind of a wild oscillations. In fact, it even reaches negative values positive and negative values. So, if you lose synchronism this is what your mechanical power is going to look like. So, if your mechanical power is of course going to look like this you cannot continue operating in this fashion. So, although I was kind of shown you a simulation which in which the loss of synchronism kind of remains for several seconds no action is taken you are just allowing the angles to you know increase. Another word for it is the poles of both machines are slipping with respect to each other. So, you just allowing that to continue if you allow it to continue there will be wild oscillations in power and you cannot operate in this fashion you will damage the shaft for example of the machine. So, what usually you do is that in case you detect an out of synchronism condition you will have to separate out the two machines. Now, if you separate out the two machines you will have to essentially trip the interconnection between them. But remember if I trip the interconnection between them I will have basically another problem to contend with what is that the local load generation balance in these two islands which have got which have kind of disconnected is not you know there is a fairly large generation imbalance in both these islands and you know then you have to really exert the governors as well as any emergency load shedding schemes may be required to quickly get an equilibrium in the center of inertia of each individual system or you know now they are disconnected systems. So, I have to ensure that the frequency of each system reaches an equilibrium. So, you have to really control the prime over and load power on both systems once you disconnect. So, let us what I was was I telling you you are interconnected you gave a fault the fault got cleared you lost synchronism the wild variations in power. So, you cannot operate that way you have to trip out the interconnection once you trip out the interconnection you are left with another problem that is of load generation imbalance in the two disconnected islands which has to be really sorted out. So, loss of synchronism is not a nice scenario and even if you manage to island or trip out or disconnect the two systems when they are lost synchronism why do you do you need to disconnect because there is going to be wild oscillations in power voltage and so on. So, you have to disconnect once you disconnect you have got the problem of maintaining the frequency in the two individual systems which boils down to trying to maintain the load generation imbalance balance in these two systems. Now of course, an interesting point is that if your power system experiences a loss of synchronism situation usually there are out of step relays which detect this and trip also an interesting point is that the wild deviations in voltage and current cause distance relays in a transmission system to mistake it for a fault and it causes transmission line tripping. So, whenever there is a loss of synchronism scenario you may have you know what you call an uncontrolled splitting of the system due to distance relay operations. So, you may what I said that you need to disconnect the two machines they may disconnect on their own in the sense that relays may actually trigger and trip out in case such a situation occurs. So, there will kind of summarize this very interesting study which we did on electromechanical transients in a two machine system. Just remember of course, that we have considered the electromechanical both the relative frequency relative motion as well as the center of inertia motion. There are other modes and patterns in this motion they are not very well observable in this response. Those patterns also can go unstable for example, if you do not design your control system properly for some inappropriate gain of AVR and govern you may find some other modes which are not necessarily electromechanical modes also go unstable. So, these kind of things also can occur will summarize at least this electromechanical transients part in the beginning of the next lecture and then move on to some understanding what happens when you have more than two machines like our grid which has more than 500 or so large machines synchronous machines all interconnected to each other. So, this something we will discuss in the next lecture.