 One of the most important loads category of loads are those formed by induction machines. In fact, we also discussed in the previous class that induction machines are used in another context as generators in many wind farms. Now, so it was in the previous class I told you how from a basic synchronous machine model for example, the 1.1 model the model in which you have got one rotor winding on the d axis and 1 on the q axis. You can modify it so as to get a induction machine model for that you would have to set the field voltage to 0, remove all kinds of salient behavior for example, x d and x u you would have to make to a to a value x and x d dash and x q dash also would be equal. So, if you did that kind of thing and even made a time constants on both axis the same you would come across induction machine model from the original synchronous machine model. So, let us just review what we are going to do in today's lecture. Today is the 32nd lecture and we will be continuing a bit of a discussion on induction machines and we will try to cover a bit of transmission line modeling also in this lecture. So, if you recall in the previous class we had done static and dynamic models of we had discussed the nature of static frequency and voltage dependent loads and we had begun on our induction machine model. Now in the previous class the point at which we left off was some of the issues when it came to induction machine models although I did mention that you can have in fact induction machine model obtained from the synchronous machine model there are few issues which naturally will come to your mind. For example, what do you mean by rotor angle in a in a situation where there is really no saliency in the machine. So, you cannot really align your axis to any particular in any particular direction with respect to the rotor because the rotor is exactly symmetrical. So, that was one issue which we need to tackle. So, if you look at the synchronous machine equations they seem to be having a rotor angle dependence. So, how do we actually show that induction machines in fact are not dependent on the rotor angle although you can get an induction machine model by you know just modifying a synchronous machine model. So, if you look at the basic equations which we had discussed last time that if the induction machine model is obtained from a synchronous machine model by neglecting E F D and setting all the transient time constants and transient reactances and of course, the steady state reactance is equal on the D and Q axis. So, this was the model for the 1.1 model on the D axis and for the Q axis this is what we have. So, what you notice here of course, is in both these equations both on the D and the Q axis the last equation for example, here is dependent on V D and the next one here the Q axis equations you have got V Q here. Now, if you assume a three phase sinusoidal source a sinusoidal balance source of a constant frequency and you use Parkes transformation that is depending on the rotor position. Then you will find that V D and V Q become delta dependent as I mentioned sometime back delta is only notional it is a kind of a abstraction in the context of induction machine it can be anywhere, but the point is does it affect eventually the observables are choice of the D the D axis and the Q axis. So, in other words thus the rotor angle matter eventually the answer is no and proving it is not is not just you know it is not cannot be shown for example, in a couple of steps here. What we will do is this something which will come in coming useful even later what we will do is reformulate the same equations using another transformation. Now, this seems to be a kind of a kind of a complication, but sooner or later as I just start working in power system dynamics you will kind of get used to this change of reference frames. So, our normal reference frame of course, before we go there let us just repeat what we did last time that is the x to be used in these equations is nothing, but in terms of the typical parameters of a induction machine x is equal to x s the leakage reactance of the stator plus the mutual reactance of the stator x dash in fact, is obtained from the second equation in this x r is the rotor leakage referred to the stator and t dash similarly is defined at the bottom with r r r r being the resistance of the rotor winding. So, this kind of these are the parameters which are used in the model 1.1 model, but for an induction machine they can be related to the cell the leakage as well as the mutual reactances as shown in this slide. Now, remember that torque equation is the same its i d i q minus i q i d and we have used in this equation the Parkes transformation. So, where theta is the position of the rotor it is omega t plus delta. Now, let me introduce you to you another transformation C k in fact, it is also called Kronz transformation the difference between the transformation before and now is that instead of theta which is omega naught t plus delta you have got in this transformation the argument of the cosine and sines contain just omega naught t they do not have delta. So, this is not the same as the previous transformation. So, if I use this transformation to transform to the new set of variables from a b c as shown in this slide that is instead of transforming to the small d q lower case d q 0 variables you transform a b c using C k into the variables f d f q f and f naught. And it is you can really show that eventually the lower case f d f q and f 0 are related to the f upper case d q and 0 by the relationship which is given below that is f q plus j f d upper case is equal to f q plus j f d lower case into e raise to j delta this is a compact way of writing this. In fact, you can really f a f b f c is equal to C p into f d f q and f 0 is equal to C k this is another transformation which gets you somewhere else it is another set of variables. So, it follows that f d f q and f 0 is nothing but C k inverse C p into f d f q and f 0. So, from this if you evaluate this it can be compactly written down in fact f 0 of course, in these both in these transformation f 0 are the same, but f d and f q upper case are related to f d and f q lower case by this transformation. Now, this is I will not derive it here, but you can compactly write it as shown in the slide that is f q plus j f d is equal to f q plus j f d into e raise to j delta. What I am trying to say here is since the rotor angle in the case of a induction machine or rotor position in case of an induction machine is a bit of an abstraction the rotor angle is a bit of an abstraction because there is no saliency in the rotor. It makes sense to reformulate our differential equations not in terms of the Parkes transformation, but in terms of the Kranz transformation which is a which contains omega naught t as the arguments of the sines and the cosines and omega naught of course, is a constant in this case there is no delta coming into the picture. So, we use such a transformation instead of Parkes transformation. So, this is what is very important. So, if I actually reformulate the equations you know in the capital D Q frame of reference I will call it the capital D Q frame of reference or the Kranz frame of reference. The equations look similar with this additional term. So, if I write down you know for example, d psi capital D by d t the equations for it look the same except here you get omega naught. So, this is one change you will see another important thing is I have also changed the variables psi capital G and psi capital F to the new frame. So, actually if you look at the new variables psi g k and psi f k in fact, they are related to these old variables. So, if I do that you know I just substitute for the old equations with the new variables your differential equations look like this in the new variables. So, you have got now new variables psi f k psi capital D on the d axis and similarly on the q axis psi g k and psi q. Now, what is the advantage of doing this? So, why are we writing down the equations in terms of different variables? Now, the important thing here is if your source is a three phase balance source of frequency omega naught you will find that v q v capital Q or upper case q and v capital D are in fact, independent of delta. In fact, you will find that there are constants which are independent of delta. So, although the earlier differential equations of the induction machine was certainly valid what we have done is come to a form in which the inputs v d and v q are in fact, constant in the new variables in the new transformed variables they are not dependent on delta. So, it is better to get rid of the concept of delta or the abstraction of delta in case you do not have a salient pole machine like in an induction machine. So, the important thing of course, is something which you can prove using this basic relationship which I have shown you here on the written slide you can show that of course, and i d plus j i q sorry is equal to i q sorry this should be q and this should be d and this should be q and this should be d. So, if I actually compute psi d i q minus psi q i d delta turnout that it is equal to psi d i q minus psi q i d. So, what I want to say here is of course, that the torque equation also the electrical torque is independent of delta it is not dependent on delta. So, I have formulated my equations in the Kranz variables or using a transformation which is you can say rotating at a constant frequency instead of using Parkes transformation I am able to formulate my equations of an induction machine. So, that it is independent of delta and of course, that also means that the rotor angle is not important in some sense. So, we know for example, that an induction machine under steady state conditions under load or no load in fact, under loaded conditions its speed is not equal to the frequency or the electrical speed of the machine is not equal to the frequency of the source to which it is connected to. In fact, if you load an induction machine you get a slip. Now, if you took the classical definition of delta that delta is nothing but theta is equal to omega t is equal to omega naught t plus delta. You see that in case the speed of the induction machine is different from the speed of the transformation and therefore, also as I have told you it is also equal to the frequency of the voltage it is connected to voltage source to which the induction machine is connected to. We will see that delta is constantly varying if omega naught equal to omega naught and you know that induction machine can operate stably even this is even if this is true in fact, there is always a steady state slip when you load the machine. So, what I have done here is come to a formulation using the Kranz transformation which is completely independent of theta and the induction machine in some sense can happily operate at a speed which is different from the omega naught. Now, this is not true of a synchronous machine in a synchronous machine in fact, you will find that you will find that the torque in fact, is a function of delta it is related in some way to delta. So, in case delta is not a constant in such a case you will find that the torque is also not a constant, but this is not true as far as an induction machine is concerned. So, it may be a good idea in an induction machine to formulate your equations in the capital D Q frame, but you could formulate your equations in the you know the Parkes reference frame as well, but the changing delta whenever there is a slip is of no consequence eventually. So, that is what really I wish to tell you here. So, you can get the induction machines as equations as I mentioned to you a few more points. First thing is what is the torque equation of a of the machine? I told you we are working with the torque equation in fact, is exactly the same this is the per unit torque of a machine in fact, this is correct where t is nothing, but what I had written sometime before it is this. So, this is the differential equation which you have to use along with the differential equations of the various fluxes that is psi d psi q psi g k and psi f k alternatively you can also use psi d psi q psi g and psi k, but remember delta will keep on varying in this equations. So, in steady state you will not find psi d psi q psi g and psi k as constants. Since, this is not a function of delta t is not a function of delta you do not really actually have to write the separate equation for delta itself. Nothing if you formulate your equations this way you do not really have to write the delta equation at all the differential equation corresponding to delta at all because delta never appears in any of these equations. Now, one of the important things of course, is this is the equation of an induction machine in fact, it is derived from the equation of the synchronous machine, but if you want to operate you want to really study the operation of a motor instead of a generator it is important to remember that we when we first formulated this swing equation or this in case of a synchronous machine the directions of t m and t were as follows. We assume that your machine is moving in this direction t is like this t m is like this. So, it is correct to say that the speed of the induction machine is equal to t m minus t in a motor in a motor remember that t m is in fact, in the opposite direction of the speed of rotation. So, if you are in a motoring mode t m is in fact, minus t l dash where t l dash is the mechanical load on the machine. So, you do not have t m in a synchronous machine a synchronous generator t m would have been the prime over torque, but now the load torque on a motor would be minus t l. So, if I am writing down the equation this is the speed of the machine rate of change of the speed of the machine this is not slip remember this speed of the machine is equal to minus of t l in case it is a motor with t l in this direction then you will have minus t l minus. Now, just remember of course, that whenever we are writing this is a correct equation there is nothing wrong in this equation. So, it is minus t l minus of this psi d i q minus this. So, this is correct there is nothing wrong in this equation you see the direction of speed t is this expression for t is in this direction t in this direction is given by this expression t l is in this direction. So, this is the correct equation, but one more small or other somewhat minor point is in case you are studying a motor another change you would probably like to do is assume that the currents i d and i q are going into the motor rather than coming out of the motor in a generator convention we had assumed that the currents are going out. So, these currents are in fact the currents coming out of the machine. So, all the in all the equations wherever current appears it is referring to the current going out of the machine. So, of course, if you change the direction of the current this will have a positive sign here and similarly in other places you will have to change the sign in case you change the direction of the current. So, this is something which you should remember that in case you are taking the direction of current inward then you have to change the sign of this then our equations are absolutely self consistent there is no problem with them. So, this is regarding the motor convention. The second point which I would like to talk is about T L itself. T L itself is the mechanical load on the machine a particular load has a you know normally we would like to characterize each mechanical load also by some torque speed characteristic. Now, for example, a fan you know you can have a if you have a fan you will find that its torque speed characteristic is something like this the torque versus the speed. So, this is the torque speed this is the load this is the load torque versus speed characteristic and of course, if you have got an induction machine which is driving this fan then the operating speed is given by the point at which both these things intersect. So, this is the operating speed of the machine. So, or rather the steady state speed of the machine. So, this is the steady state speed of the machine. Now, one of the points which you should appreciate at this juncture is that in case. So, if you look at a torque speed characteristics of an induction machine the electrical torque versus speed and this is the load characteristics. This is the load versus speed for a fan type load for a maybe you may can also I think of constant power torque loads for example, if you are lifting up something you know like a through a lift then the torque is a constant it is not a function of the speed torque is nothing, but the mass into gravity means its proportional to the mass and the gravity which is being lifted the mass which is being lifted against the gravitational force. Now, the operating speed as I mentioned was this. Now, suppose an interesting point here is that suppose this of course, frequency here is omega naught at when the slip is 0 or the speed of the mechanical speed of the machine becomes equal to omega naught the electrical torque becomes 0. Now, an interesting thing is if a frequency changes what happens to the torque? If frequency changes you will find the torque speed characteristic in fact, the torque speed characteristic in fact changes and it becomes 0 at some other mechanical speed because of this you will find. So, if there is a small change in the torque you will find this point slightly shifts and operating speed also changes in fact, the amount of the even the torque changes. So, what you find is if you are driving a fan type load you will find that if the electrical frequency changes from omega naught to omega naught dash. So, the source frequency changes from omega naught to omega naught dash the power output of such a motor would reduce and therefore, the input power also would reduce to a certain extent. So, in fact, if I got a fan type mechanical load it also implies that the steady state load of your machine the electrical load of your machine is frequency dependent. So, this is something which in fact, this is one of the mechanisms by which load becomes frequency dependent. We will conclude this our discussion of induction machine with a simple interesting dynamical example this is not something to do with our classical power systems analysis, but nonetheless we are at a point where we can actually analyze this system that is the behavior of a induction machine which is connected to just a set of capacitors. So, if you have got an induction machine and you connect to say a star connected bank of capacitors like this the stator winding is a connected to a star connected bank of capacitors and suppose the induction machine is being driven is not a motor suppose it is a it is basically like a generator it is being driven by some prime mover at a constant speed let us assume this. So, there is a prime mover which is rotating the induction machine at a constant speed and the induction machine stator is not connected to a voltage source, but it is connected to a bank of capacitors in such a case a very interesting thing is which is observed in practice is that the induction machine self excites that is you will find that some voltage appears here. So, if there is some residual magnetism in the machine or there is some residual charge on the capacitor you will find that automatically if you of course, connect an appropriate value of these C's this is a balanced star connected C you will find that the machine suddenly self excites. This is by the way a very from physical perspective a very very interesting phenomena it practically says that if you have got a bank of capacitors which is basically just you know metal and dielectric and you have got a machine which is just a ferromagnetic material in copper and you rotate the machine low and behold you have got some voltage appearing at the terminals of the machine you know without having any other magnet or battery available with you. So, this is something very interesting about self excitation as a phenomena and what I wanted to tell you here today is that with the tools which you have right now at your disposal you should be able to show that self excitation occurs. What exactly is self excitation you rotate a machine with you know practically you know with no self excitation source except the fact that the machine is being rotated by a prime mover at a certain speed say the rated speed and what you want to show is that if there is some you know small however, small and initial condition you will find that voltage grows and the machine self excites eventually of course, what you normally find in practice is that the voltage grows up to a point and settles down because of saturation of the ferromagnetic material which constitutes the induction machine. So, if you look at it from a mathematical perspective an analytical perspective what you have to do is write down the equations of the induction machine. So, we have for example, written down the equations of induction machine you do not worry about the torque will assume that the prime mover is somehow maintaining the speed of the machine a constant. So, the equations of the induction machine are as shown in these slides. So, the first three equations two differential equations and one algebraic equation and the q axis equations. So, these are the equations of the induction machine you will have to just interface them with the equations of the capacitor bank in the d q frame of reference. Now, remember we are not written down the zero sequence equation will assume absolutely a balance setup and therefore, the zero sequence equations are completely decoupled. So, we do not have to include them in this analysis the decoupled completely from this our set of d q equations. So, what do you need to do well what you need to do here is write down the equations of the capacitance capacitors. So, if you look at the equations as they given here you will find that the equations of the capacitor are c c and c d b n by d t is equal to the current I a I b and I c. So, what you get here is if you transform these using the transformation c k what you will find is this something I leave as an exercise to you d v d by d t is equal to. So, in fact this will be omega s omega naught. So, if actually if I write these equations again. So, we will have d v d by d t is equal to minus omega naught v q plus I d by c and you will have of course. Now, we can write this in per unit form in which case you will have d v d by d t is equal to minus omega naught v q. Now, these are in per unit plus I d per unit divided by c per unit, but in case you are in per unit you can also write it the c in per unit is the same as the susceptance in per unit. So, this one. So, normally you will be given the susceptance in per unit. So, that is why I am writing it in this form. Now, the differential equations corresponding to this plus the differential equations corresponding to psi d psi q psi g k and psi f k can be combined together I mean you can write the whole thing down as in the form x dot is equal to A x. There is no other source of excitation there is no other source of excitation will assume speed is constant. So, we do not write down the prime over mechanical equation we just assume that the speed is constant. Now, what you need to do is of course, what are the equilibrium conditions x is equal to 0 I mean all the states are equal to 0 is an equilibrium condition. Then what you need to do is find the Eigen values of A for different values of B c and what you will find very surprisingly is that such a system will have Eigen values with real part less than 0 if B c is greater than a certain value. In fact, B c is greater than a certain value which is related to the reactances you can actually get the condition explicitly. So, very very interesting thing can be analyzed mathematically that self excitation will occur in an induction machine if connected to a bank of capacitors and driven by a constant speed prime over you will find that at a certain value of those capacitors the system is unstable at the equilibrium point as a result of which you will find that you know you will find that some voltage appears you know for any non-zero initial condition which exist. So, it may be some residual flux in the machine or some residual charge on the capacitors and the machine will just simply self excite. So, it is a very you know interesting and exciting phenomena. Of course, one important point which in this case is very critical is that the equations which I have derived for the synchronous machine as well as the induction machine are shown to be linear. Now, remember that if you have got a linear system and it is stable it you will find that for any non-zero initial condition of the states it will simply the system will simply blow up it will just go on increasing to infinity. This does not happen in practice and the reason is that the ferromagnetic parts of the machine tend to saturate. Now, if that happens it what it also means is that our model here linear model here is inadequate to capture that phenomena because it is a linear model. So, in fact sometime before I had mentioned to you that there are you know ways which are not very theoretically you know provable kind of ways to account for saturation of a synchronous of a synchronous or any induction machine. If you actually did manage to account for saturation and make this model non-linear in that case you ought to have been able to show that there would be another equilibrium point and the system would go to that new equilibrium point that equilibrium point is not a zero equilibrium point. So, what I mean to say is that with saturation the system becomes x dot is equal to f of x this is a linear system which we are handling right now. What we have we what the thing which you can prove is that if b c is greater than a certain value you will find that at the equilibrium point x is equal to zero this system is unstable and therefore, the system tends to self-excite. If you take a non-linear system first of all you may be having more than one equilibrium point if you take into account saturation and one of the equilibria is stable and the other is unstable. So, if you look at self-excitation you will find that the system kind of you know settles down to a new you know let us say an equilibrium periodic equilibrium. So, that is one interesting thing which you can further chew upon. I will now show you a an videographed experiment which really shows the self-excitation phenomena what we will be seeing in this demonstration is in fact, self-excited induction generator which is driven by a DC generator which maintains the speed practically a constant rather induction self-excited we will just do this again once more. What I will be showing you now is a experimental demonstration of self-excitation phenomena in this experiment we have an induction generator which is driven by a DC motor which keeps the speed of keeps the speed of rotation constant and we will connect a bank of capacitors across the induction machine. What we will see is of course, that if the amount of capacitance is adequate you can actually have a voltage build up at the induction generator at the output of the induction machine terminals. So, that really shows you what is known as self-excited induction machine phenomena. So, now let us see the demonstration. So, what you are seeing here is a DC machine DC motor which is cup and this of course is the bank three banks three banks of delta connected capacitors each with a switch which we shall connect across the induction machine terminals. So, the DC motor will drive this induction machine which you are seeing and we shall of course, monitor the output voltage through one of these probes which is fed to this oscilloscope. So, what we will do now is start this machine DC motor there is one more motor one more machine right in the middle also, but that is playing no role as it is not energized there is one in the middle also here. So, the DC motor has been started the machine on the left. So, we build up the speed and you see that the output voltage of the induction machine there is probably some residual magnetism because which you get little bit of voltage, but it is not really much. So, what is happening is now I will switch on one of the banks of the capacitors, but you see that really there is no voltage really building up at the induction machine terminals there is just a little bit probably due to the residual magnetism. One of my students is removing an additional probe which is not required for this experiment it is no bearing on the outcome of the experiment of course. Now, we will switch on the second bank and what you see is the voltage builds up spontaneously at the terminals of the induction machine. So, this is an example of self excitation if I switch off the capacitor bank one of the one of this capacitor banks then you see that the machine gets the excited and you will voltage will no longer be sustained. So, we will do this experiment again we will switch on this capacitor bank again, but remember that the capacitor bank may have some residual charge on it. So, what we need to do is discharge these capacitors the de-agent energize capacitors. So, we will just discharge it through a resistance. So, we see what that is what we are doing right now we will discharge it. There is no need of doing that because it is not energized it was not energized anyway. So, now we will just redo that. So, you see that the induction machine self excites again. So, you can have an induction machine self excitation by connecting capacitors across it. So, this is so much about induction machine models remember that when you have a large induction machines you may have to model them in power system studies. So, for example, large power plant auxiliaries or induction machines in large industries you may have to model it by a dynamical model. Alternatively you can try to even model smaller induction machines or in some cases even larger machines depending on the nature of the study you are interested in by a static polynomial kind of model. I told you that an induction machine would say with a fan load if you look at its static characteristics it shows a frequency dependence. So, let us now move on to another you know equipment or another component of a power system which is absolutely important is a transmission line. You know transmission line is an element it is a distributed parameter element in the sense that it is defined by you know if you look at the if you look at it from first principle in fact all electrical equipment from first principles would satisfy Maxwell's equations. But if you look at some the low frequency behavior of transmission lines you can model them by distributed parameter equivalent where you have got shunt capacitances and series inductances, but these are distributed parameter devices. So, the basic mathematical equations which describe a transmission line are given as follows. So, if I just write down this the equations. So, if you have got a point k this analysis practically follows what is given in Sauer and Pies book. So, if this is I call x is equal to 0 point of the transmission line this is x is equal to d it is a transmission line element with of length d in that case and x is measured in this fashion. So, you have got this is v of x and we will assume current is in this direction I of x then the equations of the transmission line are this is something of course, I would not derive it is a big exercise in itself to from the basic Maxwell's equations to show that these are the roughly the equations which describe the transmission line for 50 hertz kind of behavior. So, these are the equations of the transmission line they are partial differential equations and the current directions and voltages at a distance x are described by these equations. Now, R L G and C are per unit length parameters. So, just one small note of caution because you may think L is in Henry no it is Henry per unit length and C is capacitance per unit length. So, just this is an important thing these are the partial differential equations which describe transmission line you know especially for power you know power frequency behavior. So, most of our these equations are valid for that kind of thing. Now, the solution of the general solution of these equations is something which you have done in your undergraduate years in a course on electromagnetic or perhaps even in maths or power systems itself is the current at any time t and distance x from from one end is given by just a moment we just have this in view. So, v of x at a distance x from this side is given by so, we are just following the notation of software and pi where in this when you are this of course is true only for a lossless line that is R is equal to G is equal to 0 the resistance and the conductance shunt conductance per unit length is actually 0. So, in that case we get what is known as the wave solution of these partial differential equations and where C is equal to 1 by root L C and Z C is equal to root of L of C this is the dimensions of ohms of a resistance and this is in fact the speed propagation of velocity. So, it is got the dimension of distance per unit time. So, of course it is important to note that this equation this solution is valid only for a lossless line. EHV lines in fact you will find that they come to this close to this lossless behavior. The resistance is low compared to the reactive components that is the resistance per unit length as compared to the x per unit length is much smaller. Remember that this functions f 1 and f 2 really dependent are dependent on the what are the conditions on the boundary. So, if I tell you about the conditions at the boundary for example, what is connected at one end and what is connected at the other end and I tell you its behavior with respect to time. In that case I should be able to tell you what the behavior of i at any other i or v at any point on the line. So, this is a solution which is valid for a lossless line. Now, the point of course which I want to make here is that this is the solution of a transmission line under transient conditions of a lossless line under transient conditions. If you are talking about the sinusoidal steady state, if you are talking of sinusoidal steady state behavior in such a case you must be recalling that if I take a transmission line this is k and this is m. I will be able to mimic or at least capture the behavior under sinusoidal steady state conditions what I mean by sinusoidal steady state conditions. For example, I apply a voltage source here and a weight for all the transients to settle down and then I measure for example, the terminal behavior here. I find out for example, what is the voltage which appears here say this line is open circuited. Then when I say sinusoidal steady state means that after all the transients have died down what is the behavior. So, you have already done this sometime previously in your undergraduate years. So, if I call this as you know Z S and Y S S H Z series and Y shunt you will find that Z series is equal to Z bar into sin of gamma bar D by gamma bar D sin H hyperbolic cinch function and this Y S H is equal to Y bar by 2 hyperbolic tan function. So, in fact, so if you take the sinusoidal steady state response you can use this pi circuit you can use this pi circuit with an impedance Z series impedance and the shunt admittance given by these values where Z bar is equal to R dash D. D is of course, the distance and Y bar is equal to G dash D plus J omega S and omega S is of course, the frequency of the sinusoidal sources. So, when you are talking of sinusoidal steady state the frequency which corresponds to those sinusoid sinusoids are omega S and of course, root of Z Y Z bar Y bar divided by D square. So, this is the sinusoidal steady state representation of a transmission line as a two port network. Remember that this is this solution is valid only for sinusoidal steady state conditions. So, as is mentioned in the book by sovereign pi it we should resist the temptation of trying to use this lumped pi equivalent two port equivalent of a transmission line under sin which is valid under sinusoidal steady state conditions. We should resist the temptation of using it for all kinds of transient analysis. So, often you will find you know for example, very implicitly people for example, represent a transmission line even under transmission conditions by a series inductance a short transmission line by a series inductance. And then write down the transient equations L di by dt lumped transient equations like this. Now, is this correct or no that is the basic doubt you may have. So, let me just tell you what the situation is you have got the differential partial differential equation corresponding to the behavior of a transmission line which to some extent can be said to be physically valid for the studies normally encountered at for power system in power system analysis. You can use the distributed parameter differential partial differential equation model it has got a wave like solution in transients it has got traveling wave kind of solution. Under sinusoidal steady state conditions you will find that the behavior of a transmission line can be represented by a two port network consisting of lumped a lumped pi network. Now, can we the question is can we utilize this lumped pi equivalent which comes the R and L corresponding to the lumped pi you know you have got the this impedance z series and y shunt which we just wrote down. Can we use them for example, if you look at this y series and shunt if you look at the equations which come about you will get basically a plus j b then you divided this b divided by omega s that will give you the equivalent inductance to be used in this sinusoidal steady state model and then used use the lumped parameter differential equation once we get the L from this and then use it in our analysis for transient behavior we should avoid this temptation because of rather we in fact we do it sometimes. So, the question is is it valid the answer is strictly speaking you know it is not valid, but under certain circumstances you will get close behavior to what you observe rather you can use this kind of approximation of representing a transmission line by lumped parameters obtained from a sinusoidal steady state model. But of course, you will be committing some errors because this is not really how the transmission line behaves during transient it is a traveling wave behavior. So, there is a example given in sovereign pi's book which is very apt and in which he is asked us to find out yeah it is basically an example here from sovereign pi you have got a voltage source V s and it is connected to a resistance of 10 ohms this voltage source is switched on this voltage source is a sinusoidal voltage source it is 230 kilo volts line to line. So, we are just talking of a single phase model of a transmission line it is a 230 kilo volt line to line system. So, the phase to neutral of this will be 230 divided by root 3 and the peak value is 230 into root 2 by 3. So, this is the voltage source here it is a sinusoidal voltage source of 60 hertz. So, your 2 pi into 60 into t this is switched on through a resistance of 10 ohms to a 100 mile transmission line distributed parameter transmission line with parameters L and C given as 1.5 milli Henry per mile and capacitance per unit length is 0.02 micro farad per mile. Now, the thing we need to check is when we do the simulation of this how to do the simulation of this is something we will discuss in the next class when you do the simulation of this and compare it with what we get if we simulate a lumped equivalent same same system except that now we are going to use a lumped equivalent I am sorry you will have an induction series induction lumped equivalent of the lossless line. So, we have to compare the behavior of this model of a distributed parameter model of a transmission line with what happens if you take a lumped equivalent of a transmission line based on the sinusoidal steady state representation. So, that is an interesting thing and we will compare the results and discuss this problem in more detail in the next class.