 In the previous class, we could draw two inferences about the behavior of a synchronous machine from its constituent equations. In fact, the analysis which we did was a steady state analysis. What we did last time was, we obtained the open circuit voltage at the terminals of a synchronous machine when a voltage V f is applied at the field. The machine of course, is rotating at a speed say omega naught. In such a case, it was shown that the voltage open circuit voltage which appeared at the terminals of a machine was given by V a is equal to omega naught m a f i f sin omega naught t. This was with the assumption of course, that theta is equal to omega naught t. Of course, remember that i f in steady state is nothing but V f by R f. So, the expression which we got was of V voltage, the voltage which appeared across the A winding. Of course, if it is a star connected winding, the root 2 by 3 times this would be the line to line voltage. So, line to line voltage magnitude would be root 2 by 3 into omega naught m a f I am sorry, this should be root 3 by 2 m a f into I f. So, this would be the line to line voltage R m s magnitude whereas, the voltage across the A winding would be this. The second and more interesting inference about the steady state behavior was a machine was that if a machine was connected to a three phase balance voltage source and we studied a particular situation in which the rotor is aligned at an angle delta at the time when the voltage source to which the synchronous machine is connected to undergoes a negative to positive 0 crossing. So, if that situation exists t dash into omega naught is equal to V line to line R m s square into sin 2 delta by 2 into this term. Remember that this term is equal to 0 in case x t and x q are equal. So, this is a saliency dependent term. The other term for in fact, this is the power if you multiplied by omega naught it is the steady state power. The second term of this steady state power equation is V line to line R m s of the applied voltage into x d f into I f x d f into I f as we saw sometime back was nothing but the open circuit line to line R m s voltage. Please remember that we did an open circuit analysis of the machine and we got root 3 by 2 into omega naught into m a f into I f as a line to line R m s voltage magnitude. Now, root 3 by 2 into m a f is nothing but m d f omega naught into m d f is x d f. So, what we have here is the steady state torque torque of power expression looks like this with this should have a sin delta sin into sin delta term here which was missed out. So, this particular term has got V line to line R m s multiplied by the open circuit line to line R m s voltage into sin delta by x d. So, these were the 2 simple and important results of a synchronous machine. We now move on to an important part of our course in fact in the previous class we have previous lectures we have obtained the synchronous machine equation the differential equations both the flux equations and the torque equations. So, in fact we have come to a point where we can analyze the machine completely but there is one important engineering aspect which we should go through before we start actually applying these equations for the analysis of a synchronous machine. It may appear very trivial especially for a theoretically inclined person what we are going to do now. The basic point we are going to do is try to obtain the parameters which are required for the analysis of a synchronous machine. So, if I want to obtain certain parameters then I need to do a certain amount of testing and once I do a testing I will have to fit whatever I get as a test results into my model and obtain the parameters for it. So, this is what is going to be the aim of this particular lecture. In fact what we see in the literature on synchronous machine analysis is that often we will not be given the inductances of the machine. For example, LAA to LAB, LAB, LAA 0, MAF all these inductances are rarely given what instead are given are some of the parameters obtained from some tests which are applied to a synchronous machine. For example, you will be given things like the open circuit time constant of the synchronous machine or the short circuit time constants of a synchronous machine or the transient and sub transient reactances of a synchronous machine. Now, how do you correlate these parameters or these variables which are given I would not say variable the parameters which are given as a result of testing of a synchronous machine. How do you correlate them to the model that we have derived so far? So, that is basically the aim of this particular lecture. So, we will try to understand a synchronous machine in terms which is written the equations of which are written in terms of what are known as the standard parameters or the parameters obtained from standard tests. So, this is our lecture today the modeling of synchronous machine in terms of standard parameters. Now, suppose I have got of course the synchronous machine equations they are the d q 0 flux equations differential equations then the f g h and k flux equations the differential equations and there is a algebraic relationship between flux a linear algebraic relationship between the fluxes and the currents. So, my equations are constituting in fact, a combination of differential and algebraic equations. There is nothing sacrosanct about representing the algebraic equations you can always eliminate current by expressing it in terms of flux and ensuring that your flux equations are in pure state space form, but generally people like to write the flux equation in terms of differential equations in flux which contain current and separately write current as algebraically related to the fluxes. Now, let us say I want to obtain these parameters of a synchronous machine they have several ways you can do it one you can of course, you know do an electromagnetic analysis of a machine and try to using say some computational methods for electromagnetic fields accurately calculate the inductances and other parameters of the machine, but another simpler and better way of doing things is to actually physically test the machine and obtain its parameters. So, for example, one of the ways you can do it is what is known as the stand still frequency response test of a synchronous machine. So, let us see what we can do suppose you have got a synchronous machine which is at stand still that is its speed is 0 and what we do is also for example, keep the field voltage 0 that is short circuit the field binding in such a case the stator flux equations can be written down like d psi d by d t there is no speed emf term because speed is 0. So, suppose you have a situation where speed is equal to 0 v f is equal to 0 this one the equation of a synchronous machine the stator fluxes are going to look like this of course, if you assume that the resistances of a synchronous generator are very small for large synchronous machines they are indeed very small resistances then you get d psi d by d t is equal to minus v d and d psi q by d t is equal to minus v q. We shall arrange our tests in such a way that the fluxes the 0 sequence fluxes and currents are 0. So, we shall arrange everything in such a manner. So, we do not have to bother about the 0 sequence variables here. Now, suppose I excite the stator winding in a certain fashion for example, I take my stator winding and I excite it in this fashion I connect the v a b c windings in star keep the star point open and of course, this is I a this is I b and this is I c and I apply a voltage here which is equal to v and I short this winding here. So, I apply a voltage here v short this winding. So, this is what I will do the other thing I will do is I will align my field winding axis to the a winding axis. So, what I will do is the field winding is aligned to the a axis. So, what is theta? So, theta will be 0. See remember it is it is also at stances. So, theta will always remain 0. Now, what I will do is apply voltages of various frequencies here and try to take out the frequency responses between the currents and the voltages. In this case, since v b is equal to v c, and v s, s is the Laplace variable is equal to v a of s minus v b of s. This is condition number 1, this is condition number 2 and there is a third condition I a I will call as I and I b which will be equal to I c will be equal to half of I of s. So, if you look at this figure here this is I. So, the current will split equally between these two windings and you will have these conditions. So, there are 3 or 4 conditions let me just write them down here v b is equal to v c and this. Now, since theta is equal to 0, we have I of d is equal to root 2 by 3. This current will take in this direction, current will take in this direction into the machine that case you will have minus I cos of 0 theta 0 remember plus I by 2 cos of minus 120 degrees plus I by 2 cos of plus 120 degrees. So, this will be equal to minus root 3 by 2 rather I should say minus of root of 3 by 2 into I. So, we have got let us just write it again v d I d of s is minus root 3 by 2 into I and I q you can take this out very quickly I will just write it down minus of I into sin of 0 plus I by 2 into sin of 120 minus plus of I by 2 into sin of minus 120. So, this turns out this turns out to be equal to 0. Similarly, v d is equal to root 2 by 3 into v a cos of 0 plus v b into cos of minus 120 plus v c into cos of plus 120 and that comes out to be equal to root 2 by 3 times v. So, what we have is this and v d of s where s is a Laplace variable is this. So, if you apply Laplace terms forms here on the basic time dependent quantities these are the time dependent quantities we will have this this is of course of s. So, what we have here is v d of s upon I d of s is nothing but minus 2 by 3 times v of s by I of s. So, what we will do is in this setup this test setup I will apply a voltage v this is voltage which I will apply v and obtain the current I what I will do is I will do this for various frequencies of voltage which are applied. So, what we will get is some kind of the frequency response v s by I s. So, this is what we will get if we do this test since v f is shorted that is the field winding is shorted here you do not have any term corresponding to v f. So, this test is done with the field winding shorted. So, what we can do is obtain v of s by I of s and recall from one of the equations which we wrote previously we have minus d psi d by d t is equal to v d and minus d psi q by d t is equal to v q this is if of course resistance is neglected and the speed is 0. So, these two things are obtained from those equations. So, the point is once I get this transfer function can I correlate it with what we get from actually what I get from the model what is the transfer function I get from the model we shall see shortly. But, basically what I wanted to say is since psi d and v d are related in this fashion we can get the transfer function from this because we have this relationship and we also have this relationship I will just write it in Laplace domain minus of psi d of s is equal to v d of s upon s. So, actually you can get the relationship psi d of upon I d of s by actually doing a measurement of this transfer function. So, what I can do is take out this frequency response of this transfer function and try to fit it into the transfer function I get from the model which we have. So, that is what we can do in fact it is an easy it is easy to see that in case I align my field winding at theta is equal to 90 degrees. If I align my theta at 90 degrees in such a case I can get this transfer function the reason of course, is that under these situations I d will turn out to be 0. So, what we will get effectively is the transfer function which is of importance is actually going to be this. Now, it is in fact one important point which I missed out was in case the field winding is aligned axis is aligned to the a winding v q is also 0. So, this is something I did not show. So, in fact that is something I did not explicitly show, but it is easy to see. So, now I have got these transfer functions from these tests the question is from these transfer functions can I obtain the parameters of a synchronous machine. Now, the obvious thing which we ought to do whenever we try to equate the transfer functions which are obtained by testing. In fact you will get what is known as the frequency response of this transfer function by testing. So, suppose you get some frequencies once of any transfer function like this. Suppose there is a transfer function y s upon u s which is of this kind for various frequencies you evaluate this transfer function by putting s is equal to j omega and you get the magnitude of this transfer function and you get the phase of this transfer function. What you need to do is you can fit say a rational function like 1 plus s t 1 upon 1 plus s t 2 and try to see by choosing appropriate values of t 1 and t 2 can I get a response which is identical to the one obtained from measurements. The answer is that in general in physical systems whenever you take out these transfer functions actually the frequency response of transfer functions by testing the physical system you will find that it usually will not match exactly with the transfer function obtained analytically via some model because there are always some kind of approximations implicit whenever we obtain a model of a synchronous machine. So, for example if we take the model of a synchronous machine that is we have got what are known as the flux and the current relationships we have got the differential equations of the flux fluxes that is we have got equations in d psi by d t we also have got a flux current relationship. The question is can I get the same transfer function analytically the answer is yes you can in fact if you look at the rotor equations of a synchronous machine in the direct axis the rotor equations are given by these two differential equations and the quadrature axis by these two differential equations. Moreover as I mentioned sometime back the currents which appear in these differential equations are related to the fluxes by this relationship. So, this is the relationship you have got for the fluxes. So, if for example I wanted to find out the transfer function between psi d and this what would I do what I need to do is take out the apply Laplace transformation to the direct axis differential equations as a result of which I will get s into psi f of s plus r f into i f of s is equal to v f of s. Now of course, if voltage applied to the field winding is 0 in that case we said this equal to 0 in such a case you will have psi of f s is equal to minus of r f by s into i f of s. Similarly, the other equation that in psi h of s differential equation in psi f of s if you take the Laplace transform you will get minus of r h upon s into i h of s. Now you have got these equations and you wish to now obtain this transfer function. So, what you need to do is use these equations in conjunction with the algebraic relationship which is again sorry. So, you have got this relationship too. So, you can substitute what we have got sometime back this into this. So, just to cut a long story short what you will have is if you substitute for psi f and psi h here you will finally, get psi d of s is equal to l d i d of s plus m d f into i f of s plus m d h into i h of s. And as far as these two equations are concerned in psi f and psi h it is easy to see that you will have this is by simply transferring the or expressing psi f of s in terms of i f and i h. So, this is basically what you get of course, this is multiplied by just use the word capital. So, as to denote that these are Laplace variables Laplace transformed currents. Now what you can do is express i f and i h of s in terms of i d of s and then what you do is effectively eliminate them from the first equation. So, you can write get psi d of s wholly in terms of i d of s. So, what you effectively have to do if you focus on this equation which I have to get i f and i h in terms of i d I will have to take this this term on to this side then invert this matrix and write i f and i h in terms of i d of s and substitute here. So, I have to substitute for i d and i h. So, I will cut a very long story short and directly tell you that psi d of s will have the following form. In fact, it will be l d into a transfer function it will have this form this is of course, obtained provided you keep the field winding shorted otherwise you will have another term in v of s. So, you got a what is known as a second order numerator and a second order denominator polynomials which relate psi d of s to i d of s. Now, just we will do it once. So, just take a deep breath I will just actually write down the equations which you get if you use actually evaluate this term which I have been mentioning some time back that is you will get if you actually evaluate this you will get this t d dash t d double dash t d zero double dash and t d zero dash in terms of the original variables that is m d s m d f m d h l d l f f l f h and l h h and of course, r f and r h. So, this particular equation. So, in fact, you will get this as l d into 1 plus s t d dash plus t d double dash plus t d dash t d double dash. So, the general form into s square this is square. So, this is the general form, but the point is I have defined this new time constants we shall see the significance of this time constants a bit later, but the fact is that transfer function which you will get when you actually evaluate this you know when you actually substitute for i f and i h here you by using this particular equation will look like this where I will call this as a d of s square. So, what is a d and b d a n and b n well hold your breath this is what it will look like. So, a n is nothing, but this it is a very complicated looks a very complicated equation in terms of the basic parameters of the machine and b n. In fact, looks like this it is again looks very complicated a d equals l f f l h h upon r f r h and b d is equal to l f f r h plus r f into l h h upon r f r h. The expression for a d is wrongly written the numerator should be l f f into l h h minus l f h square. So, please note that this error. So, although these expressions look very complicated they can actually be found out by applying by in fact substituting for the field and h damper winding currents in terms of i d f s. So, let me just tell you that. So, the point is that this second order transfer function we have got the coefficients b n a n b d n a d can be written in terms of these, but the general you can factorize them in this form and I define this time constants which will come out as a result of this factorization as t d dash t d double dash for the numerator polynomial and t d 0 dash and t d 0 double dash in the denominator polynomial. So, let me just summarize what I have done. I have got the transfer function in terms of the basic variables of the machine r f r h l f f l f h l h h. So, these are the basic parameters of the machine it turns out that psi d by i d from the model which I have used will give you a second order it will have the form in which you have got a second order polynomial in s divided by a second order polynomial in the denominator both in the numerator and the denominator you will find it that a second there are second order polynomials. So, I can actually what I can do is I can do an experiment do a test on the machine obtain the frequency response of the machine that is obtain the frequency response of psi d of s upon i d of s for various omegas that is I do an experiment obtain psi d. In fact, I should say j omega upon i d this is basically a transfer function. So, if I apply various the voltages voltage as I showed sometime back or various frequencies and a compute this transfer function or other I obtain this transfer function from the measurements which I take. So, what I need to do is actually measure v the magnitude and phase of v and measure the magnitude and phase of i for various frequencies for the test I described sometime back. So, once I do that I can actually try to correlate the frequency response with the transfer function which I have got. In particular I could get the parameters of the transfer function which I theoretically obtained by comparing it with the transfer function I obtained by measurement. So, I have to in some sense fit the model to the experimental data. Now, one important point which I should mention here that I took when I model a synchronous machine I model in the q axis for example, two damper windings it may. So, happen especially for if you look at hydro turbine driven generators that you will be able to fit the data obtained from experiment to the model very easily by just one damper winding on the q axis. So, that that can happen. So, what we have assumed is a kind of a model in which there are two damper windings on the q axis one damper winding on the d axis and a field winding and obtain the transfer function and the transfer function show obtained when we correlate it with experiment it may so turn out that we get a lot of error or other we are not able to fit the experimental data to the model which we have got. So, this may indicate that our is assumption which we made right in the beginning of having two damper windings on the q axis and one field winding and one damper on the d axis may require revision you may require more windings or in some cases you can even model a synchronous machine adequately with just one winding on a certain axis. So, this does happen, but what we will take right now what has been found in the literature in or reported in the literature is that this two damper windings on the q axis and one damper winding on the d axis with along with the field winding is adequate or rather models a steam turbine driven generator quite well. For a hydro turbine driven generator you can in fact show that you can in fact observe that one damper winding on the q axis is adequate to model the hydro turbine driven generator in fact even if you have modeled a higher order machine you know machine with more number of windings one can always reduce the order of the or rather reduce the order of the model by open circuiting one of the damper windings. For example, if I set R h equals to infinity or a very very large value it is equivalent to opening that damper winding. So, depending on what experimental data we get we will be able to in most cases or most practical generators fit you know the experimental data to the theoretical transfer functions. So, what we will get in fact after doing all these measurements is the parameter values l d t d t d double dash by fitting it in fact this fit may not be exact, but you can always tune these values of t d dash t d double dash t d zero dash and t d zero double dash tune it in such a way that they match with what is obtained in by doing the experiment. So, we have got the transfer function psi d s upon i d s from experiment and we are tuning in the parameters of the model. So, that the model and the experimental data matches. So, that is done by what is known as the standstill frequency response test which I mentioned sometime back. Now, that is only one issue I will get l d t d dash t d double dash t d zero double t d zero dash and t d zero double dash by fitting the model to the experimental frequency response. Now, once you have got these transfer these time constants can you for example, back calculate all the parameters of the original model for example, what were the parameters in the d axis of the original model you have got l d m d f m d h l f f l h h l h f and yeah that is it l f h and l this is of course, equals. So, I should not call l f h is equal to and then you have got r h and r f of course, there is a parameter r a the resistance of a synchronous machine and that we assume also has been measured can be measured separately the resistance of the state of winding is something you can measure separately. So, we will not worry about resistance of the state of winding now. So, we have got 1 2 3 4 5 6 these are equal so 6 7 8 8 parameters in our original synchronous machine model by doing the tests we can fit the experimental data and obtain these or rather tune the time constant. So, that they fit exact more or less they fit very well with the experimental data. So, we will try to tune these parameters. So, we here you have got 1 2 3 4 5. So, actually you have got from the tests you will get the parameters of these parameters these parameters of the transfer function. So, they are only 5. So, if you just you know obtain these parameters you will of course, manage to replicate in the model using the model or transfer function which gives almost the same responses that obtained experimentally, but you will not be able to get all the parameters of the original model. Now, this is not very surprising it is in fact, a transfer function model in some sense is collapsing the whole state space model into a input output relationship and some of the nuances which are there present you know the nuanced information which is present in the state space model is in some sense destroyed because of this. So, just by doing this frequency response test just this is one test you know of the frequency response we are not going to get all the parameters required of the original state space equation or let me put it this way we will not get all the parameters specified in the original state space in which the states are the damper winding field winding fluxes. So, before we go into a more deeper discussion into this particular aspect let me just tell you one thing that if I obtain these parameters L D T D dash T D double dash T D 0 double dash and T D 0 dash am I going to get a workable state space model of a synchronous machine. The point is I cannot get the original model of a synchronous machine because all these parameters are not obtainable from these, but it would be useful to understand what exactly can I obtain from this these measurements we have got this time constants from this time constants I cannot back calculate 8 values from these 5 values. Remember of course, that whenever we have a transfer function can we obtain the state space representation. So, if I have got a transfer function representation of a dynamical system in fact, that is what I have got right now if I actually do this frequency response test I can infer the transfer function of a synchronous machine a particular transfer function of a synchronous machine on the d axis with the field winding shorted. So, the question is from the transfer function can I get back the state space with reference to a previous discussion I just one can say that obviously, one cannot get the original state space because the number of parameters obtained from this test are not adequate they are not adequate to obtain in fact, all the parameters of the original state space model. So, what we have done in fact, let me just tell you from the original state space which had in fact, 8 parameters which I just listed down I got a transfer function in fact, I took a I got this transfer function by shorting the field winding I got a transfer function in 5 parameters. Now, I cannot get back of course, to this state space of 8 parameters using these 5 parameters of the transfer function, but the question is can I get to a state space in which we are just 5 parameters the answer is yes it is possible. So, the key to this is to remember that if you have got a state space representation of a system like this the transfer function y s upon a u of s is nothing, but c into s i minus a inverse b. Now, one thing one interesting thing is if I use a transformation of variables x is equal to r into z in such a situation you will have the same system when written in terms of the z variables will be z dot is equal to r inverse a r z the same system plus r inverse b u and y is equal to c into x which is nothing, but c into r into x into z. The transfer function of this system is c into r into s i i is of course, an identity matrix into r inverse a r inverse r inverse b which is nothing, but c into s into r inverse this will be r r inverse into r into r inverse a r into b inverse. So, this is of course, using simple rules of matrix algebra. So, we will get c into s i minus a inverse b. So, what you have here is the same transfer function this transfer function and this transfer function in fact match exactly. So, the same system rather if you have got a transfer function you can either get to a state space of this kind or you can get a state space of this kind and both are in fact valid representations of your system, but the point we should remember is that the original variables and the new variables are related by this relationship here. So, let me get back to our original issue from the state space of 8 variables we got into a transfer function of using 5 parameters. So, using these 5 parameters of course, it will not be able to back calculate all the 8 variables of the original state space equations, but you can if you so wish write down the state space equation using only 5 parameters, but the state variables of this state space system using 5 parameters is not the same as the state variables here they will in fact be some kind of linear combination of the states in this. So, your state space equations here and the state space equation here are written in terms of different variables. So, that is one important thing you should keep in mind. So, using these transfer I can get a state space representation using just these 5 parameters. In fact, it is a nice thing if you can write down the state space with lesser number of parameters, but it is important that the state space equations which you will get will be in terms of states which are related in some way by some linear transformation to the original state space variables that is psi f psi h psi d, but of course, which is require lesser number of parameters. Now, in power system analysis it is often required to at least you know have a nice neat interpretation of the states. So, often people would insist that well I do not want a state space which use only 5 parameters, but the states are not easy to interpret you know the states for example, psi f and psi f psi h for example, are variables which are easy to interpret they are in fact the fluxes linked with the f and g by f and h winding, but of course, if I write the state space in terms of only 5 parameters the states in this state space representation will be a linear combination of these states and it may be difficult it is not really very nice to have for example, state space equations in which one of the states is say 5 times the field flux plus 3.5 times the you know the damper winding flux. So, there is a in fact problem here and power system engineers have tried to solve this problem in a bit round about fashion. In fact, the solution to this problem is to in fact try to do try to in fact obtain more transfer functions by various other tests may be similar setups some more tests and get get more parameters which can be correlated to the original 8 parameters which I have already said. So, if I through more tests obtain more parameters then I would should be able to obtain the original 8 parameters which I mentioned by back calculating. Remember that all the transfer functions which we have for example, if you recall this is the coefficient of the numerator polynomial in s. So, if we have got in fact, you know just t d dash and t d double dash one will not be able to get all these parameters. If you have got for example, sorry t d dash t d double dash t d 0 dash and l d and t d 0 double dash you are not going to get all these 8 parameters which are which are a part of this. So, it is not possible to back calculate 8 parameters, but one can take out other transfer functions you can you know you know try to think of other tests. For example, you do this test with the field winding open in that case you will get another transfer function you can actually obtain the same transfer function using the analytical model which we have then correlate both of them. So, you can actually have many more tests and actually get all the parameters, but unfortunately power system engineers have with a limited number of tests and limited number of parameters attempted to get an approximate synchronous machine model. And therefore, because of that we will in fact have to make certain approximations in the kind of state space equations we are finally, going to get which are in terms of what are known as the standard parameters. So, there is only one way of getting meaningful state space representations with a larger number of parameters than what are obtained by measurement and that is by making certain approximations. So, we will redo or rather recap what you have done today in the next lecture and try to obtain a state space representation of a synchronous machine a meaningful state space representation of a synchronous machine may be with a few approximations in fact, with a few approximations which will be in terms of states which we can directly interpret. So, there will be not some transform states which use lesser number of parameters, but we will try to get approximate state space equations with a few approximate with lesser number of parameters. So, this is what we will do in the next lecture.