 We are now in a position to write down the equations of a synchronous machine together and try to draw some inferences from the equations which are derived in the d q variables. Remember that we have done a time variant or theta variant transformation of variables from the A B C of the A B C variables or the phase variables to the d q 0 variables. The transformation which we have used is defined using the rotor position theta that theta is of course, the electrical angle and it would be instructive to just look at the transformation once. So, we can express f d f q and f 0 in terms of f A f B and f C and vice versa. In this case of course, this transformation is what we refer to as C p and this of course, is cos theta cos theta minus 2 pi by 3 cos theta plus 2 pi by 3 and similar things for sign. Now, the inverse transformation of course, is given by this. So, the basic idea which we have tried to get is that we can do the analysis in the d q 0 variables and after obtaining the results in the d q 0 variables, we try to move on and see what are the values in the A B C variables. So, why do of course, we go to the d q 0 variables. The reason is that the differential equations which we get in the d q 0 variables are in fact, time not theta dependent that is one major advantage of using the theta dependent transformation of variables. So, what are those equations? The basic equations of a synchronous machine in d q 0 variables are as follows. So, the stator flux equations are given by this d psi d by minus of d psi d by d t minus omega psi q minus r A i d is equal to v d and similar equations for psi d and psi 0. Omega here is of course, d theta by d t theta being the electrical angle and omega being the electrical angular frequency. Now, remember that these equations are obtained using the transformation as discussed before with k d and k q equal to root 2 by 3 and k 0 is equal to 1 by root 3. In fact, by doing this we make this transformation C p and this inverse transformation C p inverse transpose of each other. That is C p inverse becomes equal to C p transpose for this choice of k d k q and k 0 of course, but of course, it is not necessary for k d and k q and k 0 to be these values in the sense that even if you choose any non zero k d k q k 0 value absolutely arbitrarily still we will achieve one thing that is time variance of time invariance of the resulting differential equations. So, of course, what I will write here now what will do in this course will be for the specific choice of k d k q and k 0, but you notice that all the equations are time invariant or theta invariant. Now, the rotor equations recall that we have represented the rotor as two coils on the d axis and two coils on the quadrature axis h g and k are representing the effect of damper bars as well as eddy current effects in the machine. Now, V f is the applied voltage on the field binding if you recall an earlier diagram. So, your this is I f and this is V f. So, this is the applied voltage on the field binding. Now, of course, the equations which relate if you recall if you look at these equations they are not in pure state space form for example, the differential equations are in psi, but you also have these terms i d i q i 0 i f i g and i k i f i h i g and i k. These are in fact related to the flux variables by this relationship which is of course, also theta invariant. So, this is l d l q l 0 this is a diagonal matrix I have just shown the non zero terms in bold. This is of course, a symmetric matrix because you have chosen our k d and k q values such that we get this symmetry. In fact, the special choice of k d and k q was made in order to get this symmetry and of course, the interesting property for C p and C p inverse and C p. Now, remember that m d f here is related to m a f the mutual inductance coefficient by this formula. So, our m d f is m a f divided by k d m a f in fact, is the mutual inductance between the a winding and the field a winding and the field winding in fact, is the maximum value of the inductance of the a winding and the field winding divided by k d k d of course, we have taken to be root 2 by 3. Similarly, you can define m d h m q g and m q k and of course, remember that l d itself is can be written down as l a a 0 minus l a b 0 plus 3 by 2 times l a a 2 and similarly, l d a l q and l 0 are defined in this fashion. So, these are of course, the original self and mutual inductance terms when written down using the a b c variables, but of course, when you go to the d q 0 frame of reference or the d q 0 variables this relationship is like this and it of course, does not depend on theta. In fact, you can aggregate all the windings on the d axis that is the d f and h winding are on the d axis the direct axis and the q g and k are on the quadrature axis and psi 0 of course, is the 0 sequence flux. So, if you relate the d axis fluxes they are only dependent on the d axis currents. Similarly, the q axis fluxes are dependent on the q axis currents alone and psi 0 is dependent on i 0 alone. So, I just rearrange the equations and they look like this of course, remember that the d q variables are coupled, but why are this differential equation remember the differential equation in psi d has got this psi q term or this what we kind of inferred to be the speedy m f terms which comes because of taking the derivative of the transformation. We have done of course, this in a previous class. Now, our discussion would not be complete unless we talk about how omega or theta vary. In fact, omega itself is d theta by d t this is the electrical angle theta, omega is the electrical angular frequency. So, we got this equation sometime back 2 by p p is the number of poles j d omega by d t j is the moment of inertia is equal to t m minus p by 2 into psi d i q minus psi q i d it is a very neat term for the t e dash. So, p by 2 times t e dash is actually the electromagnetic torque which we get. Now, we can write these in per unit by dividing the both the sides by the torque base torque base is nothing but m v a base divided by the mechanical rotational angular frequency. So, if you do that you can actually get this in this form where t m is the per unit torque mechanical torque the prime mover torque and h is nothing but half of j omega mechanical base p square divided by m v a base. So, this is basically if you divide both sides by the torque mechanical torque base. So, torque base is the torque base. Note that we have not written down the synchronous machine equations in pure state space form that is x dot is equal to a x we actually written it in a kind of composite form we have written d psi by d t is equal to in terms of psi q if we for example, takes the fluxes as the states then we also use the auxiliary variables i d i q and i 0 and similarly i f i g i h and i k, but then we relate the fluxes and the currents through an inductance matrix. So, we are not written this down in a pure state space form we are now in a position we are now in a position to actually draw inferences about the behavior of a synchronous machine that is how what we have you know that is why we are doing this modeling after all we are trying to get how the synchronous machine behaves. So, what we will do we will keep our ambition a bit low at present we will use this flux equations all the equations which we have written down to actually at least first infer the steady state behavior of a synchronous machine. So, how do I infer the steady state behavior of a synchronous machine now the thing is that when do you say that a system x dot is equal to f of x is in steady state where x is d x dot is equal to d x by d t we would say of course, the system is in steady state when this equals to 0. Now the interesting thing about this whole set of equations in the d q frame d q 0 frame of references that although in the a b c frame when we say we are in steady state psi a psi b psi c are sin a solids it turns out that if we are having a balance system then psi d psi q and psi 0 in steady state become constants. Let me just give you an example suppose you have got a set of voltages v a v b and v c and these are a balance set of voltages. So, what I can write down v a for example, as v a is equal to v max the phase to neutral max as suppose I have got a star connected machine this the stator is connected say in star then this is v sin and it is say the voltages applied to the stator have a frequency omega t and v b is equal to v m suppose is a balance set of voltages. In that case what is v d and v q and v 0 v 0 is easy to get remember the transformation is this and of course, we are taking special values of k d k q and k 0 k d and k q are root 2 by 3 and k 0 is equal to 1 by root 3. So, I will just write down what the expressions for v d. So, v d will be root 2 by 3 v a of t into I am sorry into cos of theta plus v b of t into cos theta minus 2 pi by 3 plus v c into cos theta plus 2 pi by 3. So, you have got v d is equal to this. So, of course, if v a of t and v b of t and v c of t are like this the question is what is v d. So, what is v d and what similarly we have got v q is equal to root 2 by 3 v a v 0 is equal to. So, if the speed of rotation of the machine theta is equal to the speed of rotation of the machine theta is equal to omega 0 t itself. Suppose, the speed of the machine is omega 0 t. So, if theta is equal to sorry the speed of the machine is omega 0 and theta is equal to omega 0 t in such a case we will have v 0 equal to 0 that is almost obvious it is a balance set of sinusoid. So, we will have v a plus v b plus v c is equal to 0 and we will have v q will be equal to root 2 by 3 into sin square theta plus. So, it will be v m sin square theta plus sin square theta minus 2 pi by 3 plus sin square theta plus 2 pi by 3 and that will give you in fact this is remember there is an identity which we discussed sometime back this sum of these 3 terms is actually root 3 by 2. So, this is a trigonometric identity there is a of course, a multiplication factor of v m here. So, we will have v m into root 3 by 2 and it is easy to show that v d is equal to 0. So, in case v a v b v c are like this v 0 v q and v d are equal to 0 and in more importantly sorry v 0 is equal to 0 and v d is equal to 0, but v q comes out to be v m into root 3 by 2. In fact, this v m is nothing but the maximum of the phase 2 neutral voltage for a star connected system here. So, maximum of the phase 2 neutral voltage. So, it is easy to see that v q in this case is equal to the line to line r m s voltage of the system. So, actually what I wanted to show you here that if you have got a balance set of sinusoids then it turns out that v 0 v q and v d are constants since constants. So, if you have got a if you are steady state in your system is a balance three phase sinusoidal steady state it turns out that the d q 0 variables are in fact constant. So, this is a just a simple example to prove that or show that actually. Now, let us actually try to get some inferences from of the steady state behavior from our differential equation. So, one thing I showed you is that if you are in a balance three phase sinusoidal steady state then the d q variables become constants and if they become constants of course the d by d t's of all the d q variables are going to be 0. Moreover, the field winding of a synchronous machine as well as the damper winding currents also become constants in steady state. In fact, the damper winding currents are 0. So, i g i h and i k are 0 in steady state. So, if you have all these conditions it should be possible to infer the steady state behavior of a machine. So, what the first thing we will try to do is what if we want to see this what happens in case you have got a field voltage v f applied to the machine at the field windings in steady state what are the voltages induced on the stator. So, that is a very basic study that we can do. So, what we do is take the differential equations of the synchronous machine set all the d by d t is equal to 0. So, if you do that I will just draw a few d by d t's. So, if for example, if you take the stator flux equations since d by d t is equal to 0 I will have omega psi q minus r a i d is equal to v d and you will have omega psi d minus r a i q is equal to v q and of course, minus r a i 0 is equal to 0 omega is of course, the angular frequency electrical angular frequency please remember that. Now, similarly we have got I just mentioned sometime ago. So, this of course, implies that I 0 is equal to 0 you also have I g these are the damper winding currents I h is equal to I k is equal to 0 this is of course, simply obtained by setting the d by d t is corresponding to the field winding equal to 0. So, in fact, what you have is d psi h by d t. So, if you said this this to be 0 it automatically means I h is equal to 0. So, similarly you can show that I g and I k are also equal to 0 I f is not equal to 0 it is going to be a constant in steady state. So, what we have is these two equations here these two equations here and then we will have. So, we said this equal to 0 steady state conditions. So, that means that v f by r f is equal to I f. So, this is a second equation of importance. Now, we also have from the flux current relationship if you look at this flux current relationship which we have we have got now this equal to 0 this equal to 0 this equal to 0 and this equal to 0 and only non zero currents are I d I q and I f, but if you are also we are actually taking out the open circuit voltages induced on the stator of a synchronous machine. Then we also have I d is equal to 0 and I q is equal to 0. So, what we have under open circuit conditions the differential the algebraic equations which are there are very simple. So, we have got omega psi q is equal to v d omega psi d is equal to v q v f by r f is equal to I f. Then we have psi d is equal to l d into into I d, but of course, I d is equal to 0. So, the only term which comes is m d f into I f all the other currents are 0 psi q turns out to be 0 because it is l q psi q is nothing, but l q into I q is equal to 0. So, l q plus m q g into I g plus m q k into I k, but all these are in fact 0. In fact, this is also 0 and this is also 0 under open circuit conditions I d and I q are also 0 under open circuit conditions. So, what you have here is this is also 0. So, under open circuit conditions you will have from all this you can infer that you will have v d equal to 0 and v q is equal to m d f by r f into v f. There is a small correction here I have forgotten to multiply the term v q is equal to m d f r m d f by r f into v f I forgotten to multiply it by omega naught which is the speed of the rotational speed of the synchronous machine. So, v q the expression here m d f by r f into v f has to be multiplied by omega naught. So, this is v q and of course, v 0 turns out to be equal to 0 is it. So, what we have here eventually is a situation where we can find out v a from the values of v d and v q. So, what is v a from the transformation which we have defined v a is nothing, but root 2 by 3 into f v d. So, v d is equal to 0. So, what we have here eventually is a situation into cos theta plus root 2 by 3 into m d f by r f into v f into sin theta plus 0. So, if you have this it turns out that v a is m d f root 2 by 3 m d f by r f into v f into v f sin theta. So, actually sin theta is theta if the machine is rotating at a angular frequency of omega naught and theta suppose is equal to omega 0 t what you have is under open circuit conditions you will have a voltage induced which is nothing, but a sinusoidal voltage and this is the coefficient corresponding to it. Now, if you look at this expression v a is nothing, but root 2 by 3 m d f m d f is nothing, but m a f by k d this is root 2 by 3 itself into v f by r f v f by r f is the current of course, into sin omega naught t. So, if you look at this particular equation this is nothing, but m a f into i f into sin omega t the expression for v a which is m a f into i f into sin omega naught t has to be multiplied by omega naught. So, there is minor correction here v a the expression for v a which is given as m a f into i f into sin omega naught t has to be multiplied by omega naught. So, actually it is understandable why this is true. So, if you have got current i f in your field winding m a f i f is the flux linked with the a winding because of the field winding and because of that you have got v a having this form. So, this is the open circuit voltage on the a winding of course, b is equal to nothing, but root 2 by 3 in fact, it is root 2 by 3 m d f by r f into sin into v f sin theta minus 2 pi by 3 and similarly v c v c will be phase shifted by 2 pi by 3. So, this is what we get when we apply the equations under steady state under open circuit conditions when v f is applied to the stator. So, let me just revise what are the steps we just put d by d t is equal to 0 why do we put them to be equal to 0 because under balanced 3 phase sinusoidal conditions the d q variables become constants in steady state. So, if they are constants the d by d t is become equal to 0 remember d psi a by d t is not constant in steady state. So, in fact, the a b c variables are in sinusoidal steady state corresponds to the d q variables being in you can say d c steady state d c steady state means the values become constants. So, by d by d t is become equal to 0. So, that is one important thing now we go on a step further let us see what happens when we connect the machine to a voltage source we are not having an open circuit machine remember we are connecting it to a voltage source. So, what let us assume we have got a star connected machine. So, the a winding b winding and the c winding are connected to voltage sources. So, this is there is a voltage source say a star connected voltage source. So, I have got a voltage source of this kind and let us assume that this voltage source is V a V max this is a constant voltage source three phase voltage source and a balance 1 that 2 and V c is equal to V m sin omega naught t plus 2 pi by 3. So, if you have got this voltage source balance voltage source connected there is a three phase balance voltage source connected here and let us assume that theta is equal to omega naught t plus delta. Now, what I am trying to say here is that the let us suppose we are operating at a situation suppose we have we have got a steady state situation in which whenever this voltage V a n has a zero crossing and I just take a quick snapshot of the machine I find that the field winding which is here is at an angle delta. Let me repeat what I say what I said suppose I am operating at such a steady state such that if I take a snapshot of the machine when the phase voltage V a is crossing its zero has a zero crossing here I take a snapshot at that point I see that the field winding the field winding is aligned at an angle delta with the a axis winding. So, if this is your field winding remember this is your field winding. So, this is something I am telling you it is not something which is obvious I am just telling you that a situation the situation I am studying is such that when the V a is having a zero crossing this is at an angle delta what basically the speed of the machine is also omega naught t omega naught. So, the voltages which are applied and the rotational speed of the machine are in fact the same only thing is that at time t is equal to zero or whenever V a hits the zero if I take a snapshot this is at an angle of delta. So, such a situation suppose exists in such a case what will be V d V q and V naught. So, we can apply of course, this transformation this transformation which we have studied with of course, k d and k q as root 2 by 3 and k naught is equal to 1 by root 3, but of course, I will not go through the complete manipulations let me try to just show by inspection what we will get. Now, you know that V a is equal to from the C p transformation is nothing, but root 2 by 3 times V d cos theta plus V q sin theta plus V naught. V naught of course, is zero because V a plus V b plus V c is equal to zero. So, this is zero also since V a is equal to V m sin omega naught t is equal to root 2 by 3. So, we have V d cos theta I have told you is omega naught t plus delta plus V q sin omega naught t plus. So, can you guess what V d and V q will be? So, the question I am asking you is if I give you the right hand side and I tell you with the left hand side is this can you find out what V d and V q are please recall the trigonometric identity cos a plus b is equal to cos a cos b minus sin a sin b and sin a plus b is equal to cos a sin b plus cos b sin a. If you know this that is very easy to see that V d is if the left hand side and right hand side are to be equal then V d will be root 3 by 2 V m into minus sin delta and V q is equal to root 3 by 2 times V m cos delta. So, V d and V q become this in fact V m remember is the maximum voltage of the phase 2 neutral. So, if it is a star connected winding in fact, this is the line to line RMS voltage this taken together is the line to line RMS voltage anyway. So, now that we have got V d V q and V naught can we take out again the steady state situation steady state values of current voltage and so on well yes. In fact, again we set the d by d t is equal to 0 set the d by d t is equal to 0 for all fluxes because they become constants in steady state. So, remember of course, by doing this we will again get the same condition that the damper winding and the fact that V 0 is equal to 0 V 0 is also equal to 0 and d psi 0 by d t is equal to 0 would also mean that I 0 is equal to 0. We will also have from the flux differential flux field flux differential equations V f by R f is equal to I f. So, V f by R f V f by R f is equal to I f this is another equation we have. So, this is an important equation and if we neglect resistance of the stator then we also have minus omega psi q is equal to V d and omega psi d is equal to V q. Now, one of the things which I mentioned sometime back is that theta is equal to omega naught t plus delta. So, it follows of course, this should be omega naught because d theta by d t is equal to omega naught is that I just discovered a small error in what we have done slightly earlier I will just write that down here to theta we have taken as omega naught t this is the earlier example. So, in fact this should be into omega omega naught. So, this is a small error please note the small error in what we did previously. We continue our example of a synchronous machine connected to a voltage source. So, if we neglect the resistance of the stator we in fact get this particular set of equations and because of this now we have got we can get the value of psi d and psi q. How? So, remember that psi d equal to l d into i d plus m d f into i f and psi q is equal to l q i q and of course, it is not related to the since the other currents on the q axis are all 0 you do not have any other terms. So, what we can do from all this is actually obtain the torque. So, torque is equal to if you recall I am sorry this electrical torque t dash actually t dash is equal to psi d q minus psi q i d and this is in fact the electrical torque for a two pole machine for a or in general the torque electromagnetic torque is p by 2 times t dash this is something you should remember. So, if you are working with a two pole machine then of course, we finally have. So, if you want to take out t dash it will turn out to be psi d psi d is equal to nothing but v q divided by omega naught v q divided by omega naught into i q i q is nothing but psi q by l q and psi q minus psi q psi q is nothing but minus of v d by omega naught you can actually have a look at these equations which are just done in the previous slide into i d i d is nothing but I just rewrite this. So, t dash is equal to v q by omega naught into psi q by l q minus or you can say call it plus v d by omega naught into i d i d is nothing but psi d minus m d f i f by l d. So, if we just continue with these manipulations we will get t e dash is equal to we can substitute for psi q again here. So, we will have v q into psi d minus I am sorry v q into psi q psi q is nothing but minus v d by omega naught l q plus v d by omega naught psi d is nothing but. So, if we regather all these terms again this is turning out to be quite an avalanche of equations, but we just be steady. So, minus v q v d by omega naught I think this is in omega naught square l q plus v d p q by omega naught square l d minus v d by omega naught l d into m d f into i f. So, this is our these are the expressions we will get finally, from this actually v d and v q have already computed from a previous manipulations. In fact, they are nothing but v d is equal to root 3 by 2 v m sin delta minus of it and v q is equal to root 3 by 2 v m cos delta cos delta. So, we can substitute this on to this and finally, obtain the electrical torque. So, what we get is essentially t dash is equal to it is it will be 3 by 2 1 upon x d x q omega naught into l q is x q. So, you will have into v m square sin delta cos delta plus in fact, we will have here minus 3 by 2 times 3 by 2 times 1 upon x d into v m square sin delta cos delta divided by omega naught minus root 3 by 2. So, actually it is plus we are evaluating this expression please remember that. So, I am evaluating right now this expression here. So, it will be root 2 by 3 root 3 by 2 v m x d into x d f by omega naught into i f into sin delta. Now, does this expression look familiar to you? This expression in fact, is something we probably done in our undergraduate years. What we are really having here I will just rewrite this. This is nothing but t dash is nothing but or t dash I will call it t dash into omega naught this is the power you can say. So, if you look at the power it is nothing but 3 by 2 times v m square. So, if you look at this 3 by 2 times v m square. So, actually v m is the phase 2 neutral peak value. So, 3 by 2 times the peak value of the phase 2 neutral in fact, is rather root 3 by 2 times the peak of the phase 2 neutral is in fact, the line to line RMS voltage. So, actually we can write this is v line to line RMS square into sin 2 delta by 2 cos delta sin delta the product is sin 2 delta by 2 into 1 upon x q minus 1 upon x d plus v line to line RMS into x d into x d f into i f. So, in fact x d f into i f is the open circuit voltage I should say open circuit line to line RMS voltage. We have just done this a few minutes ago. In fact, when we had obtained our open circuit voltages formula in fact, v a is nothing but omega naught m a f into i f sin omega t which is which also means that open circuit line to line voltage is this. This should have a sin into sin delta term here which was missed out. So, what we have obtained is the steady state let us say it is called a power angle characteristic. So, the power angle characteristic of us in steady state is given by these equations. But of course, if you are going to do only a steady state analysis to some extent that is keeping our ambitions too low as far as this course is concerned. What we will do in the next lecture is try to build a ground for doing transient analysis of a synchronous machine where we will be not setting the derivatives of the d q 0 as well as the field flux and the damper winding flux is equal to there. We will not be doing that, but one step which we need to take before we actually start getting the transient analysis of a synchronous machine is try to you know kind of rephrase or rewrite your synchronous machines in a more user friendly or engineering friendly format. What I mean is that the synchronous machine equations as they have been written right now are in terms of inductances and resistances. These often are not easy to obtain the values of the inductances and the resistances themselves are not easy to obtain by simple tests. What we instead can do is to prescribe a few tests from which we can back calculate these inductances and resistances. So, let me repeat although we have obtained the synchronous machine equations we need to obtain synchronous machine parameters from a real machine. So, if I want to do the analysis somebody has to give me the parameters. So, I should define some tests and methods to back calculate the inductances. In fact, you may argue of course, that these inductances can be analytically derived. Well, while that is to some extent true from electromagnetic analysis you can actually compute fields and compute the mutual inductances and self inductances and so on. It is in general quite a difficult process. Moreover, remember that we have actually represented a synchronous machine by windings and sometimes the windings are actually representing the effect of eddy currents. So, they are not actually there actually no windings in some cases. So, in such a case to obtain the mutual or the self inductances analytically is going to be a bit tough. So, as I mentioned sometime back we have to prepare a ground for inferring these values from measured tests. So, whatever tests we do we will try to infer these values and we will find out a way to fit the measured you know responses into this model. Remember that whatever model we make this is obvious something which is very important in engineering whatever model we have made is based on of some assumptions. So, we for example, as I mentioned right in the beginning of synchronous machine modeling that our assumption that we can represent the synchronous machine effects by two windings on the rotor and two windings on the rotor in the d axis and two windings on the q axis. In addition of course, to the stator windings was an assumption we assume that we will be able to fit our results or rather in we will be able to fit the results in obtain in tests to our model. So, that is the basic point which is implicit in all our model derivation. So, now in the next class we will try to actually take this a bit forward try to see how we can measure parameters and fit them into our model or rather obtain the parameters in our model from the measured tests.