 In the last couple of lectures we have been looking at reactances of salient pole alternators we started with the equation of the salient pole alternator we derived an equivalent circuit out of that as can be seen one for the direct axis and one for the quadrature axis and that equivalent circuit we represented the stator voltage equations in a simplified form in terms of the flux linkages as can be seen in these two equations and in order to understand these two terms better that is d by dt of ?d and d by dt of ?q we looked at the circuit enclosed by this rectangle here and the same for the q axis and then we try to analyze the expression d ?d of s and d ?q s which then led us to various expressions and then combining of terms and simplifications and in that we defined various expressions t1, t2, t3 some of these and then we went on to study all the other expressions one can see all the derivations that we have done we then defined some more terms t4, t5 and t6 and looking at the q axis we defined t7 and t8 and we also discussed how these terms can then be derived by looking at the equivalent circuit on the d and q axis some of them by considering that the stator side is shorted and some by considering that the stator side is open so we have open circuit entries and then close circuit once and then based on all these we defined certain new expressions that is the direct axis reactance operational reactance is what we defined at the outset and then from that by looking at high speed operations that is responses varying which are going to vary very rapidly with time and then we came across another definition called the direct axis sub transient reactance which we called as xd double prime and then when the dampers are neglected what we have is the direct axis transient reactance xd prime and similarly on the q axis you then have xq double prime which is the direct which is the quadrature axis sub transient reactance which is xq multiplied by tq double dash by tq o double dash so these are all terms that we had derived in the course of analyzing the circuit enclosed within the green rectangle and now how do we use them or where is the significance of these things applied so all these terms we will find are very useful to describe the response of the alternator when subject to certain disturbances that may happen during the course of their operation so let us take a look at one of the specific and most important disturbances in the life of an alternator which is this so what we are going to look at is these short circuit of the three phase alternator that means we are having the alternator let us call this as the alternator and then it has the three phases of course the field excitation is also given that is another terminal and if this alternator is subjected to a sudden three phase fault how does the response look like it can be noted that this is also done as an experiment it is done as a regular test in order to examine the behavior of the alternator also to determine alternator parameters more specifically the D axis one all those terms that we have defined now that is XD double prime and XD prime and TD dash to all those things one can see that these numbers can be estimated by looking at the waveforms that arise during this experiment that we shall see as we go along so how do we analyze this the situation using all that we have learned so far so if we look at the alternator equations the equations can be written in this form this we have already seen this is the shortened form of the alternator voltage equations by making use of the flux linkage definitions so you have VDS is RA into IDS plus the derivative of the direct axis flux linkage minus speed into the flux linkage along the Q axis whereas for the Q axis the equation says RA into IQ plus the derivative of flux linkage along the Q axis plus speed multiplied by the derivative by the flux linkage on the direct axis now the experiment that we are going to consider is the application of a short circuit to an alternator with the assumption that the speed of the rotor that is you have the rotor rotating at some speed speed is constant this is a reasonable assumption because normally if you are going to subject an alternator to a fault the speed will take certain time to change whereas electrical variables are immediately affected and therefore during the interval when speed has not yet changed our analysis will still be holding good so if speed is going to be fixed ?R is going to be fixed we know that the alternator has to run or is operated at synchronous speed then ?R can be replaced by the actual synchronous speed we note that ?R here is the rotor speed which can be any number in general it can be a function of time as well but when we simplify it to this case we can replace it by ?S which is a constant equal to the synchronous speed in radian per second. So since we are replacing it by a fixed number we can transform these equations into the Laplace domain because by representing ?R as a fixed number ?S this system now becomes a linear system you do not longer have ?R which is a function of time multiplied by flux linkage is also a function of time and therefore the system becomes non-linear but now replacing it by ?S makes it a linear system and when we do that and change it to Laplace transforms what we have is VDS or as a function of S this is the Laplace transform of the direct axis voltage is RA x ID and the derivative now becomes S times ID assuming that the initial conditions are 0 – ?S x ?Q which is a function of S and VQ now this becomes Laplace transform RA as a function of S this is the Laplace transform the quadrature axis current then S times ?Q of S plus ?S x ?D of S. So this is the Laplace transform representation or form of the D and Q axis equation now you see that what we have here we have S ?D of S and S ?Q of S and this is exactly the expressions that we had analyzed over the last couple of lectures we had tried to represent S ?D of S and S ?Q of S in terms of the other inputs to the smaller sub-circuit that we saw the circuit enclosed in the green rectangle and therefore those expressions are now reproduced here S times ID of S is nothing but S x XDS XD as a function of S is the direct axis operational reactance divided by ?S is the synchronous speed x IDS plus the other factor B of S multiplied by the Laplace transform of the field voltage and similarly S ?Q was then derived as S x XQS by ? x IQS. Now since we have derived expressions for these what we can do is substitute these expressions into the Laplace transform a form of VD and VQ and therefore VDS as a function of S can now be written as Ra x IDS as a function of S plus S times ID is nothing but the expression that we have here and therefore that is S XD of S by ?S x IDS as a function of S plus V S x VF of S so this is a representation or the expression for S times ID and then you have – ?S x ?Q of S which is nothing but S x XQS by ?S x IQS as a function of S. So we have now eliminated the flux linkage variables in terms of the flow of currents the plus transform of the currents that is ID of S and the field voltage. Now this speed of course can be removed and these the first two terms being multiplied by IDS what we can do is combine them as Ra plus S x XDS multiplied by ?S x IDS as a function of S and then – S x XQ of S multiplied by IQS function of S and then plus V of S x VF of S so this is what the direct axis voltage equation simplifies to and then you can write for the quadrature axis as Ra IQS as a function of S. Now S times ?Q which is nothing but S times XQS by ?S x IQS as a function of S and then plus ?S x ?D which is S x XD of S by ?S x IDS of S plus V x VF. So as before we can combine the IQ terms together so that is Ra plus S times XQ by ?S multiplied by IQS of S and then this ?S cancels due to the presence of ?S here and therefore that is nothing but S times XD of S into IDS of S plus ?S V as a function of S multiplied by VF as a function of S. So these two expressions now have got rid removed the dependency on ?D and ?Q and therefore we can now write this in a matrix form as VDS as a function of S and VQ as a function of S is then equal to Ra plus S times XD of S by ?S and then for IQ you have S times XQ of S and this is S times XD of S and then here you have Ra plus S XQ as a function of S by ?S this is multiplied by IDS and IQS and then you have the other vector which is the field input so B of S and ?S into B of S multiplied by VF of S. So we can now write we have therefore written the voltage equations of the alternator in a slightly different form which is more amenable for the operations that are to follow. So having written it in this form let us now see what is it that we are going to do we started out by saying that we want to analyze the performance of the machine under short circuit conditions right. So we start out with the unloaded alternator under steady state unloaded alternator in the sense we have the alternator operating on essentially open circuit means no flow of currents are there and under steady state at constant speed. So here is the equations of the alternator which is the general large signal equations I have not represented the expressions for the dampers because we are going to consider simplifying this for steady state under the steady state situation and under steady state the dampers are essentially open circuited because there is no induced EMF in the damper bars or in the rotor ion and therefore no currents flow in them and therefore in this equation I have not considered to represent the dampers and when we are looking at steady state please note that what we have written these equations are in the synchronous reference frame and in the synchronous reference frame under steady state all the terms IDIQIFVDVF they are all DC terms and since they are DC under steady state the derivative of DC term goes to 0 so this term is 0 this term goes to 0 all the terms which involve P which is d by dt will have to go to 0. So this is also 0 this is 0 this is 0 and therefore under steady state this dynamic equation reduces to a much simpler form which is as represented here VF is then simply RF into IF that is what has been put here I have used uppercase letters to denote the steady state and then if you look at VD you have RS into ID-Omega LQ into IQ or XQ into IQ so that is what you have here and VQ is nothing but Omega LD into ID that is XD into ID and then you have RS IQ so that is here and then Omega LMD into IF that is XMD into IF. Now we are considering an unloaded alternator that means that ID and IQ initially therefore and the unloaded condition they must be 0 because there is no current flowing because there is no loading and if ID and IQ are 0 one can see what happens to these expression there is some field current flowing and therefore VF is RF into IF so this is fine it remains like that as far as VDS is concerned ID is 0 IQ is also 0 and therefore VDS now becomes 0 in that expression for VQS IQS is 0 IDS is 0 IF alone is there so this VQ now is simply equal to XMD into IF and indeed if the alternator is on open circuit what the alternator will produce is an induced EMF as is seen on the stator and when you transform this to the synchronous reference frame we find that it manifests only on the Q axis there is no voltage on the DX so this term XMD into IF we may call this as the induced EMF E and therefore what we have is VF equals RF into IF and VDS equal to 0 please note that VDS this S stands for the synchronous reference for the stator not in the synchronous reference frame it is not a function of S and you have VQS as equal to E and ID for the stator is equal to IQ of the stator equal to 0 this then is the initial condition with which we start out doing our experiment now note that under the assumption of fixed speed of operation we have transformed a non-linear system or we have converted a non-linear system into a linear system so what we now have is a linear system and we now need to analyze the behavior of the short circuit of a linear system now we know that for a linear system if you have a system here a system let us say some S well let us not use S because it could be confusing a linear system let us call this as a system A let us say you apply an input I1 and it produces a response O1 and then you have the same system A you apply an input I2 and it produces a response O2 then if you have the linear system A to which you supply an input I1 plus I2 we know from our earlier studies of linear system that the response must be O1 plus O2 this is usually called as a principle of superposition so that is what we are now going to use for our study of short circuit of the alternator how do we use that now if you have the alternator now operating on open circuit let us say that you have the de-axis of the alternator we have seen an equivalent circuit for the de-axis so let us just represent that equivalent circuit by a box and you have two terminals of the equivalent circuit for the stator representing the stator this is the de-axis equivalent circuit from the initial conditions that we have seen under unloaded conditions we find that there is an induced in the alternator that will manifest at the terminals of the stator and that magnitude is E suppose now I connect a voltage source here whose magnitude exactly is equal to E which is the induced generated by the alternator also we will find that no flow of current is there into the machine why because the induced EMF is exactly equal to the voltage that is applied since the two voltages are equal no current will flow in the machine which is the situation we have in the initial condition right so this is your input 1 what we want to do is look at the alternator and we want to consider the response to a short circuit at the terminal so the terminal is going to go to a dead short now this means that the EMF or the voltage that is applied at the terminals of the alternator is equal to 0 and 0 volt can be represented as E-E and therefore the response that we are getting can now be considered to be the superposition of this is not a to be the superposition of the responses due to the two circuits here on the one case you are applying E and in the second case you are applying – E and therefore when you add the two E-E the resultant is 0 and therefore the response produced in let us say we will call this as case 1 the response produced in case 1 plus the response produced in case 2 must equal the response produced in case 3 but we already know the response produced in case 1 we know that the currents are 0 and therefore 0 plus whatever is produced here must be the response that is there in this which means that this response can be obtained by simply applying a voltage of – E to the D axis of the circuit and looking at the response of this circuit alone because this is 0 and how about the other axis you are applying – E on the D axis and now when you are considering a short circuit of the three phases of the stator the D axis goes to 0 D axis voltage that is applied goes to 0 the Q axis voltage is also 0 but however we see that I am sorry this is the Q axis equivalent circuit not the D axis I have been saying it is D axis that is wrong this is Q axis equivalent circuit the Q axis equivalent circuit has an initial voltage let us say E now you may apply – E to get the net voltage to 0 on the D axis we find that the voltage initially itself is 0 and where we want to go to is also 0. So there is no change in the voltage that is applied to the D axis so we do not have to consider anything special for that as far as the field is concerned the field voltage is VF is finally also VF so therefore there is no change in the field voltage as well and therefore the response to short circuit can be obtained as considering the response to Q axis excitation as – E so initially it was 0 and when you apply the voltage – E the sum total which is 0 plus whatever is now going to come is the short circuit current and therefore this response itself is the short circuit current however in the alternator the normally in the alternator is analyzed such that the flow of alternator stator currents is out of the stator but all along right from the beginning we have been following the convention that the direction of current is taken to be flowing into the machine and therefore what we will get by this analysis is really current that is flowing into the machine and therefore if we want current flowing out of the machine we simply have to take the negative of whatever current that we are going to end up with. So this is what we are going to do which means that if we look at the earlier equation we had derived an expression for the D and Q axis here what we are now saying is that for the system that we are now going to consider there is no voltage applied along the direct axis this is 0 VQ of S is applied a voltage – E which means in the Laplace domain it would then be – of E over S where S is the Laplace variable and we are not applying any change to VF therefore there is no change to VF so this is 0 so essentially we need to look at the system that is comprising of this as the input vector the matrix given here and this current there is no contribution due to this because we are not going to touch the field and so what we have is then this expression VD of S is 0 VQ of S is – E by S in this since it is a simple system only two variables are there we can write this by inversion where IDS and IQS this is equal to now RA plus S times XQ of S by Omega and then you have – XQ XD of S and then RA plus S times XD of S divided by Omega S multiplied by 1 over delta where delta is the determinant of this matrix this multiplied by 0 and – E over S where delta is RA plus S times XD divided by Omega S this is Laplace transform multiplied by RA plus S XQ by Omega plus XD of S into XQ of S so this is a first simple step where we write the expression for ID and IQ in terms of the input variables by simply considering the matrix inverse from which you can write an expression for ID ID is nothing but this RA plus XQ is getting multiplied by 0 so that is not there is simply – XQ into – E by S divided by the determinant so this is your expression for ID as far as IQ is concerned you have this term RA plus S times XD by Omega S into E is a – sign divided by delta so this is then the solution of ID and IQ of S but however this is a fairly involved expression and it is in the Laplace transform domain so what we nearly need to do is try and simplify these expressions to get them into a more manageable form and represent the variation with respect to time after all what we can measure on the stator side is the stator currents IA, IB and IC with respect to time therefore we need to do an inverse Laplace transform and before that we need to simplify these expressions so that they are in a more manageable form so that involves lot of approximations based on what real life alternators are going to look like. So let us expand this expression first look at ID the denominator delta is given here which we expanded you have RA or what is written here as RS is the same as RA so RS squared and then S XD into RA by Omega S plus S XQ so that is what you have here and then you have the product of these two terms and with an XD XQ so if you take that out you have S squared by Omega squared plus 1 for this XD XQ multiplied by this term and then you have E by S that is here XQ is present in the numerator now normally in alternators at least the large alternators the stator resistance is a very small number and therefore in the denominator expression this stator resistance term which comes as a square of the stator resistance which is even more smaller this is negligible in compared to all the terms that are there so RS squared is simply neglected therefore you get a simplified expression neglecting RS squared and then we also take XD and XQ out of this since you have these terms here XD XQ if you take out of this then you have S times RA by Omega S divided by XD XQ so this becomes 1 over XQ and 1 over XD that is what you have here S into R into Omega and then this becomes S squared plus Omega S squared divided by Omega S squared and so that S squared plus Omega S squared is what is here and when you multiply throughout by Omega S squared the numerator here becomes S RA into Omega S so RA into Omega S RS squared is of course neglected so that you get this Omega S squared what you have been what you have seen in the denominator here now comes over to the numerator so E Omega S squared and since we have taken XD XQ outside XQ is removed so what you are left with is XD of S now XD of S itself is a fairly complicated expression if you remember what we have done this is the operation this is a direct axis operational reactance of the machine so it is of the form 1 plus S T dash D into 1 plus ST double dash D by 1 plus ST dash DO into 1 plus ST double dash DO into XD. So let us look at that expression that now comes as 1 over XD so if you take a look at this 1 over XD term it is 1 plus T dash DO and divided by all this so we really need to simplify this part of the expression is 1 over XD let us keep it outside and this term is a second-order numerator by a second-order denominator so what we can do is convert this into partial fractions but however it is a second-order numerator by a second-order denominator and therefore what we do is take a one term outside and then this term can then be expressed as in this form that is so this essentially can be represented as if you take 1 and then it is 1 plus ST dash DO into 1 plus ST double dash DO minus 1 plus ST dash D into 1 plus ST double dash D divided by 1 plus ST dash D into 1 plus ST double dash D. So we are doing this so that we can get a numerator order lesser than the denominator order so that partial fraction expansion can be applied obviously if you add these two you see that if you are going to multiply this then this term goes away and what we are left with is the original term so now this one is what we have represented here and so if you look at whatever is there within this expression so here if you are going to multiply the first two terms this product of 1 is cancelled with the product of this one so all we will be left with is a square term and the S term in the numerator and S square term is T dash DO multiplied by T double dash DO minus T dash D T dash D into T double dash D so that is what you have here and then you have the S terms so T dash DO will come due to a product of this T double dash DO will come due to a product of this so those are the two terms here and this comes with a minus sign that is there here. So the numerator when written in this form you see that it has S squared and S and therefore one S can be taken out so if you take an S term outside out of this part you can write it as S into this S square goes away and this S goes away so now what you have here what you have within this bracket the numerator has this is now S multiplied by S the numerator has is one degree lesser than the denominator so we apply the normal methods of simplifying it this is really one plus S T dash D and T double dash D so we now have to determine these constants A and B how do we do that in order to determine A we multiply this expression by one plus S T dash D and substitute S equal to minus one over T dash D and what happens if we do that so let us substitute S equal to minus one by T dash D so you have minus one by T dash D multiplied by T dash D O into T double dash D O minus T D dash T double dash D plus T dash D O plus T double dash D O minus T dash D minus T double dash D so what we have done is substituted S equal to minus one by T dash D so this whole expression in order to determine A this whole expression divided by minus one by divided by one minus T double dash D by T dash this comes because of the denominator term that is there here so we can simplify this further so let me make use of some slightly bigger space available above so that is one by one minus T double dash D by T dash D into T dash D O T double dash D O by T dash D plus with the minus sign and then T double dash D plus T dash D O plus T double dash D O minus T dash D minus T double dash D and therefore we can cancel of this T double dash D and this T double dash D and from this if we take minus T D outside so we write this as minus T D divided by one minus T double dash D by T dash D this is then T dash D O multiplied by T double dash D O by T dash D whole square plus T dash D O divided by T dash D since we are taking minus sign so this becomes minus and then there is one more minus sign T double dash D O by T dash D and then you have T dash D there so this becomes a plus one this is what you get and one can see that this numerator can be expressed as the product of two terms this term can this entire term can be written as one plus not one plus one minus one minus T dash D O by T dash D into one minus T double dash D O by T dash D and therefore this term A can indeed be written as minus T dash D minus T dash D into one minus T dash D O by T dash D into one minus T double dash D O by T dash D and then you have the denominator term. Now in a similar way one can try to simplify or one can try to derive an expression for this B what you need to do is multiply this expression by one by one plus ST double dash D that means this will go away and substitute S equals minus one over T double dash D and then simplify that expression so that will come as B equals minus T double dash D into one minus T dash D O by T double dash D into one minus T double dash D O by T double dash D divided by one minus T dash D by T double dash D so having determined this expressions for A and B terms we now apply some simplifications normally again for alternators the terms involving double dash T double dash D and T double dash D O are normally considerably smaller than T dash D and T dash D O these are sub transient time constants sub transient time constants these are much smaller than transient time constants later on we will see a numerical example of data pertaining to an alternator where we will really show or see how these numbers are different so if we use these approximations let us look at the term for A in the expression for A what we are saying is a sub transient time constants are much smaller than the transient time constants and therefore if T double dash D O is going to be much smaller than T dash D and it is one minus a very small number and therefore this term becomes approximately one and the denominator also you have T double dash D by T dash D so this also is therefore very small so these two terms can simply be assumed to be equal to one itself and therefore A simplifies as minus T dash D into one minus T dash D O by T dash D now as far as the expression for B is concerned to apply these simplifications now let us look at this we can expand this term so that here the denominator becomes T double dash D minus T dash D divided by T double dash D and T double dash D is much smaller than T dash D and therefore what you get here is T double dash D minus T dash D and here also one can write it as T D double dash minus T double dash D so those two are removed and you have in the numerator T double dash D minus T dash D O denominator you have T double dash D minus T dash D and then you have this expression now here this term is much smaller than this this term is much smaller than this and therefore what we can do is this term simplifies to minus T double dash D because you neglect this term T dash D O and the denominator simplifies to minus T dash D and therefore this part can be written as minus T double dash D into T dash D O by T dash D that is what you have here into one minus this term which remains as it is and again we take this fraction into here which means T dash D O by T dash D and when this is multiplied with this term since this is a much smaller fraction than this this is approximated to this to be to remain as it is and this is actually T dash D O so we have now simplified the expression for essentially 1 over X D of S into a much simpler form splitting it into two different parts and we now have to take this to the expression for ID of S that we have derived earlier and see how further that can be simplified this we will look at it in the next class and then go further on to look at a numerical example of how the waveforms will then turn out to be we will stop with this for today's lecture.