 So from the last couple of lectures we have been looking at how to analyze the performance of the alternator when it is subjected to a sudden three phase shot and the problem that we have taken up in order to look at one of the applications of the equations that we have studied is the three phase short circuit of an alternator when it is operating on no load that means initially the alternator was not supplying any load that means the flow of current in the stator winding is 0 and at that point it is subjected to a sudden short circuit and in order to make the equations wheel D in the sense to be able to manipulate those equations by hand we can do it only under the assumption that speed is held fixed the assumption is not really a bad one considering that speed will not change immediately it will take some time to change and by that time lot of electrical phenomena happen in the machine and it is those phenomena that we are interested in analyzing by looking at the electrical circuit equations of the machine and therefore we started out assuming that speed is held fixed at the synchronous speed alternator was on no load and then is subjected to a sudden short circuit. So with that then we started looking at the equations that we have developed we started out with our familiar equations in the D and Q axis for the machine stator alone we had used this formulation which effectively avoids or illuminates the rotor variables and then we tried to since you have the derivative of the stator flux linkage which in turn involves the flow of current in the dampers and in the field winding we then saw the equivalent circuit along the D and Q axis and then from analysis of those equivalent circuits we derived expressions for these two terms that is P ? D of S and ? Q of S in terms of I D I Q and the field and we use those expressions in order to substitute into these equations which are for an alternator running at constant speed these are the expressions that we derived from circuit analysis I am sure you remember from the last few lectures so these when substituted into the first two equations give you a resulting set of equations where the variables are in terms of the Laplace transforms of stator current D and Q axis for the Q voltage also it is now D and Q axis along with the field equation I mean field voltage which we now put it in this form and in order to do the short circuit studies we said that this system can be considered to be a linear system and in a linear system one can arrive at a response of a sum of two inputs by considering the inputs to each individual input and then summing up the total response and we use that approach to determine what will be the short circuit current for this unloaded alternator and we came across the result that one can assess the behavior by looking at the system energized with VDS equal to 0 and VQS equal to minus of E by S so together when this response is added to the open circuit response the net VD and VQ becomes 0 and 0 and the open circuit response we know already there is no current flowing and therefore the response to this input as given by ID and IQ itself will then be the actual current that is flowing in the short circuit alternator and having written this equation we then got the expressions for ID and IQ by simple inversion of this matrix equation and this is the expression that we land up with and then we said that the stator resistance is really small and therefore the stator resistance which occurs in the denominator of these two equations in the term delta which is the determinant of the earlier expression. So resistance can be neglected and this is what we end up with and in these places we noted that you have 1 over XD and 1 over XQ, 1 over XD is present outside the bracket also so we then try to look at the expression 1 over XD, 1 over XD we know what XD of S is from our earlier discussions that is simply described by the steady state direct axis reactance multiplied by a function which is a second order function of S in the numerator and second order denominator and therefore reciprocal now looks like this we have defined the terms T dash DO, T double dash do all that in the earlier lectures where double dash represents the sub transient terms and single dash represents the transient terms and O represents open circuit so this is sub transient time constant in the open circuit condition this is the sub transient time constant in the short circuit condition. And we noted that one can derive these two terms T dash DO and T dash D by neglecting the RKD and LLKD neglecting the existence of that one can derive these two terms we will see as we go along some justifications for all that and then this expression we tried to simplify by converting the second order term into a number 1 plus a second order denominator by a first order numerator multiplied by S and first order numerator and then we derived expressions for A and B as well where A is given by this expression and B was this and making use of certain approximations that the sub transient terms are really much smaller than the transient terms whether in open circuit or short circuit condition the terms A and B can be simplified to these two expression A is then minus T D dash into 1 minus T dash DO by T dash D and B simplifies to minus T double dash D into T dash DO by T dash D minus T double dash DO by T double dash D and we note that A and B occurred basically here 1 over X D plus 1 into 1 plus AS by 1 plus ST dash D plus BS by 1 plus ST double dash so if we now substitute the expressions for A and B here what do we land up with you get 1 over X D of S as 1 by X D multiplied by 1 plus now AS by 1 plus ST dash D A is nothing but minus T dash D into S by 1 plus ST dash D this is the part that is already there in this expression this is the part that is already there here now A one part of A we have considered the other part of A is then into 1 minus T dash DO by T dash D and then you have the expression for B which is minus T double dash D into S then divided by 1 plus ST double dash D so that is again borrowed from this expression multiplied by T dash DO by T dash D minus T double dash DO by T double dash D now what we could do is take this 1 over X D within this in which case what would happen is that this term can now be taken inside this becomes 1 by X D here and this is multiplied by 1 by X D and this is also multiplied by 1 by X D and now what we notice if you look at this term that becomes 1 by X D minus 1 by X D into T dash D T dash DO and we know that X D into T dash D by T dash DO is nothing but X D dash which is the direct axis transient reactance and similarly if we look at this term that can be written as 1 by X D into T dash DO by T dash D minus 1 by X D into T double dash DO by T double dash D this term we have already seen what it is it is X D dash or rather 1 over X D dash and this term we know that X D multiplied by T double dash D divided by T double dash DO is the same as X D double dash what we really have is if you take this inside there is a small error this term when it is taken inside it is T double dash DO by T double dash D multiplied by T dash DO by T dash D that is what we are going to have here this is T dash DO by T dash D and this is T dash DO by T double dash D so this into T dash D by T dash DO this is really X D double dash which is the direct axis sub transient reactant and therefore this expression 1 by X D of S simplifies to as we see here it is 1 by X D plus 1 by X D dash minus 1 by X D into T dash DS by 1 plus ST dash plus this term and similarly IQ of S by using the same kind of elimination methods one can write it in this form both these expressions really come from the expression here what we have here as ID and IQ so we have then expressions for ID and IQ as is mentioned here E by S into this term and E by S multiplied by this term where a is then given by 1 over TA which is omega S into RA by 2 into 1 by X D dash plus X Q dash now how does this come about if we look at this expression the expression for ID of S here this is S square plus omega S square multiplied by this term now RS is a very small number and therefore this term is going to give an effect only when because you are having something multiplied by a small term this will have an effect only when the X D and X Q change very fast when there are very fast changes which happens in the sub transient interval and therefore this can be replaced in an approximate sense by 1 over X D dash plus 1 over X Q dash so we end up with the result ID of S and IQ of S expressed in this form after all the simplifications are done now being Laplace transform expressions these can then be transformed into the respective equations as a function of time inverse through the inverse Laplace transform route and in the inverse after applying the inverse Laplace transform which is fairly easy to see we end up with an expression for ID and IQ in this manner that is ID of T is E into 1 over X D plus 1 over X D dash minus 1 over X D into E power minus T by TD dash and then this difference into E power minus T by TD double dash minus E by X D double prime into E power minus alpha T cos omega ST and similarly IQ of S can then be written in this fashion now ID and IQ can be combined using the normal route that is what you would have is if you want to get IA we know the transform that we have to apply IA, IB and IC is then given by root of 2 by 3 times 1, 0 and 1 over root 2 and minus half root of 3 by 2 and so on you have other numbers here and then you have cos ? minus sin ? sin ? and cos ? 0, 0 and 1 and then you have here ID, IQ and I0 now the I0 term again is 0 under this condition and therefore if we look at the expression for IA of T, IA is nothing but root of 2 by 3 into ID cos ? which is cos ? T and IQ sin ? which is sin ? T now in order to look at the situation where the alternate we are saying essentially that the alternator is subjected to a short circuit and one can look at the short circuit behavior at various instance in the sense if you look at the induced EMF wave form which is going to be a sin wave one may consider the short as having occurring at the 0 crossing of the sin wave or at any other instance along the sin wave. So in order to have some flexibility on that this in turn will depend upon where the rotor is with respect to the angle equal to 0 reference that we have taken that is the horizontal a axis. So in order to bring in that effect that the rotor might have been really somewhere else we substitute ? as ?S into T plus some angle ? which is then the angle that the rotor makes with respect to the horizontal a axis at T equal to 0 if we say that ? equal to 0 then the rotor was aligned along the a axis at T equal to 0 whereas one can put any other number also for a any other suitable angle. So we then have to add up ID and IQ and we get an expression for IA as is in this form. So what we see is that IA consists of a cosinsoidally varying entity which is then whose amplitude is determined by all these expressions the amplitude is determined by these expressions which consists of a simple number e by xd and then it also consists of something that decays exponentially with time with the time constant Td dash another term of the amplitude decays again exponentially with time with the time constant Td dashd and then you have two more terms which decays exponentially with respect to time with the time constant of a let us try to simplify this expression further what we can say is 1 by 2 x 1 by xd double dash minus 1 by xq double dash as equal to a and then 1 by 2 x 1 by xd double dash plus 1 by xq double dash if you call it as b then we find that 1 by xd double dash can be written as a plus b and then 1 by xq double dash can be written as b minus a so one can now write this as minus e e to the power of minus a T into a plus b cos omega st into cos of omega st plus lambda and this can be written as e into e power minus a T into a minus b into sin omega st sin omega st plus lambda which then can be simplified each of these terms can be simplified as e e to the power of minus a T into let us take a outside and then you have cos omega st cos omega st plus lambda plus sin omega st sin omega st plus lambda so this is one term and then you have b minus a so what one can do is this is minus b times e e to the power of minus a T into cos omega st cos omega st plus lambda and then because there is a minus sign here as well so this becomes plus take the minus sign so this is really plus so this is minus here sin omega st sin omega st plus lambda and this can then be simplified you have in the form cos a cos b plus sin a sin b so that is nothing but cos of a minus b and this is cos a cos b minus sin a sin b which is cos of a plus b so this will give us a term cos of 2 times omega st plus lambda and this will give us a term cos of omega st plus lambda minus omega st which is cos of lambda so if you look at these two terms this term and this term together they then can be separated into two terms one involving just cos lambda this is not a sinusoidally varying term whereas this term is sinusoidally varying but at a frequency two times omega therefore we get an expression for Ia of T in this manner what we note from this is that Ia of T consists of one part which varies at the fundamental frequency that is omega st. If you look at the fundamental frequency part fundamental frequency component this then consists of a steady amplitude component which is nothing but e by xd multiplied by this cos omega t plus lambda and then it consists of another component which is exponentially decaying with T dash d as the time constant and then another term which is also exponentially decaying with T double dash d as the time constant remember in our derivation we said that T double dash d is much smaller than e dash d therefore this component is expected to decay down much faster than the second term that is this term this is decaying much faster whereas this does not decay at all and reaches a steady value after the expression for Ia of T has a double frequency component which depends upon the difference of xd double dash and xq double dash this part obviously can be written as xq double dash minus xd double dash divided by xd double dash into xq double dash. So the amplitude of the double frequency component second harmonic will really depend upon the difference of the sub transient reactances along the direct and quadrature axis normally this value is pretty small the sub transient reactances themselves are small and this further is the difference of these two which is very very small and therefore the double frequency component is normally expected to be really small in amplitude and will not may not show up in the actual wave form of Ia of T. The next term that you have is a simple dc decaying term there is no oscillatory component in this but we see that the existence of this term depends upon what value of ? you assume if ? equal to 0 then this term attains the highest amplitude at T equal to 0 cos ? then becomes 1 and as ? goes to 90 degrees then this term really becomes 0. So this term is something that is going to depend upon the instant of switching that we are going to have. So these are the three terms that are there in the expression therefore for Ia. Now if we draw the graph of this wave form then you notice that these exponential amplitudes exponentially decaying terms obviously they will decay down to 0 as T becomes much larger than Td dash and Td double dash finally what we will be left with is only the steady state term which is there here and therefore if we find out what is the steady state amplitude of the resulting wave form of Ia subjected to a three phase shot at the alternator side alternator stator then from the steady state amplitude it should be feasible to get what is the value of xd and if we look at the decaying part of the wave form there should be some region where from the decay rate of the amplitude of the fundamental component of this Ia it should be possible to estimate what is Td dash. And similarly if we look at again the exponentially decaying form we noticed again that Td double dash is going to be much smaller than Td dash and therefore we expect that this part which decays with the time constant Td double dash is going to persist only in the initial small interval and looking at the decay rates in that initial small interval it should be feasible to get a value for Td double dash and once you know Td dash and Td double dash we can then find out all these of course in order to find out x double dash d and xd dash we also need to know the open circuit open circuit numbers which are not really available from this stage. So let us now look at how a numerical example will look like so we will go to an example where we take an alternator this is really a 2mv alternator which is then fairly large 400 volts 50 hertz the 4 pole machine now all the data that is given here are referred to the stator turns and what we see is Rs is 0.76m ohms now somewhere along the derivation we made an assumption that Rs is small and hence can be neglected in comparison with the other terms we see that Rs for large machines is really small and then the stator leakage inductance is 12.7 micro Henry which is quite small compared to LMD and LMQ that we have here 520 micro Henry and 384 micro Henry the field resistance is again much smaller than even the stator resistance and it is only 0.157m whereas this is 0.76m field inductance is given here now if you look at the resistance RKD and RKQ we see that they are of the order of 16m ohms and 2m ohms and again if you look back at the derivations that we have done we had said that RKD and RKQ are much larger than RF and therefore certain simplifications were made in analyzing the D and Q axis is equivalent circuit and that is really seen here RKD and RKQ are much larger than RF which is the field resistance which is only 0.157m ohms these are 16 and 2m ohms respectively and here again we have the leakage inductances of the damper that are given with these then one can evaluate what the various values of the different expressions for the TD-TDDD etc and the direct axis short circuit transient TD- then is obtained as 0.6311 seconds and here we made an assumption that the sub transient time constants are much smaller than the transient values transient time constants which we see here this is 0.63 seconds whereas TD- is 0.03 and similarly the open circuit numbers TD-DO is 3.8 seconds whereas TD-DO is 0.03 seconds so that again is to show that this assumption is not a bad assumption and then along the Q axis you have TDQO and TDQ which are derived from the earlier numbers so these are the two values for that with all that you then see that the direct axis synchronous reactance of course is given from the earlier numbers 0.1688 ohms XD- is given as 0.0275 and this is 0.0246 and you see that the difference between these two terms is not really large and therefore the double frequency term in the expression for IA is expected to be fairly small so with all these numbers then if we plug these numbers into the expression for IA as we have derived earlier what exactly happens so here we have the flow of current IA of T calculated from the expressions that we have derived we see here that there is a steady state part so after all the exponents have decayed down to 0 what results is this is then called as the final steady state short circuit current this region is the exponentially decaying part the envelope as you can see seems to show a nice exponential decay as you have here and in this case the short has been done assuming that ? equal to 90 degrees and therefore there is no DC component that can be seen if ? were something else then there would be an exponentially decaying DC component either decaying from this side or from this side and then the entire way form will be bent over from one side finally reaching the steady state value if however we now since this looks like an exponential decay we cannot superimpose on this the expression which has the exponentially decaying term that is this part 1 over xd ? – 1 over xd into e power – t by t double dash d if we draw this alone this will form an exponentially decaying line which should form the envelope of this curve and here that is also shown by the green line so the green envelope that is drawn here is the envelope as determined by the t dash d decay component that we have seen we notice however that there is a small difference that we see here if we zoom into that region we see that this exponential envelope while it sort of grazes mean all the amplitudes of the sine wave the maximum amplitudes of the sine wave as it reaches near t equal to 0 in this region you find that the actual amplitude of the sine wave is larger than what the exponential envelope itself says it appears that the actual envelope joining all the amplitudes maximum amplitudes of the sine wave should really bend over a little more and go higher than that and indeed if we plot add to this the other exponentially decaying term that is with respect to t double dash d then you see that that is really what happens here this red line now shows the envelope after adding the t double dash d decaying term so this envelope which is given by the red line as you can see it decays down and meets this green line somewhere there which means the exponential decay of the t double dash d term that is the term corresponding to the sub transient that decays down and becomes 0 somewhere at this instant of time after that this does not contribute anything and therefore there is no effect of this term so this region can then be thought of as being called as the sub transient region where the effect of the t double dash d term is then felt so that is then the initial portion of this graph somewhere there the rest of this region until the exponent part exponent with t dash d decays down that is then the transient interval or the transient region and finally you have the steady state region that is there now you can see that the initial value of the Ia in the worst case is then determined by this exponential envelope leading and hitting this axis and that value if you look at the expression for Ia this value will then arise out of the amplitude of the omega s term that is here this expression the value of this expression at t equal to 0 will then sort of determine what can be the maximum amplitude of the omega s term that is the sinusoidal term varying at fundamental frequency so if we look at this term at t equal to 0 one should get what will be the amplitude of this so let us evaluate this term at t equal to 0 so what you have is root of 2 by 3 multiplied by e into 1 by xd plus 1 by xd dash minus 1 by xd plus 1 by xd double dash minus 1 by xd dash so evaluating it at t equal to 0 this is what one lines up with and here we see 1 by xd cancels with this 1 by xd 1 by xd dash cancels with this 1 by xd dash so what you are left with is root of 2 by 3 into e by xd dash so we find that the initial value the highest amplitude of the fundamental frequency decaying component can be very high as compared to the final steady state value you can see here that xd has a value of 0.1688 ohms whereas xd double dash has a value of 0.0246 ohms so this is quite small almost 5 to 6 times smaller even more than that smaller than xd and therefore initial value of the flow of current when subjected to short circuit can be substantially higher 7 to 8 times higher than this so which is another interesting result that in order to determine how much flow of current the system various elements along the line like may be circuit breakers and so on that they have to withstand one has to really look at the sub transient reactants the sub transient reactants is the one that is going to determine how much flow of current will be there when it is subjected to a short circuit. Of course after all the components decay down what you have is only the steady state component which lasts for a long time so we have seen that the equations that we have derived describing the alternator has enabled us to define these further terms like xd dash xd double dash and then xq double dash t dash do t double dash do and so on all these t dash d double dash d all these are useful in defining the behavior of alternators when subjected to sudden disturbances we have looked at a specific example of a case where the alternator is subjected to a very severe disturbance that is a full short circuit at the terminals of the machine. Now we have done this by analysis by equations and therefore we have only derived an expression for ia of t if we do a simulation of the alternator equations when subjected to a short circuit one can solve the differential equations on using a suitable numerical integration algorithm which using which we can then evaluate all the variables id iq ikd ikq if also we can estimate how the field flow of current is going to behave with respect to these things so all these will give us useful information about the behavior of the circuit. It is also interesting to see how ikd and ikq behave when the alternator is going to be subjected to a dead short like this so for that let us look at the plot of ikd and ikq so here we have a plot of ikd and ikq this blue graph represents ikq and the red graph represents ikd. Now we have we can see from this graph that these currents ikd and ikq which are essentially the flow of currents in the damper circuit they decay down to 0 pretty fast in this experiment the alternator was run in open circuit until this time and it was then subjected to a short at this instant of time that is 0.2 seconds it was subjected to a short which is why you see that no current flows until this time and then there is some large current flow in ikd and ikq but however they decay down to 0 pretty fast and after that there is no damper current which means that in this interval one can neglect the existence of the damper bars if you remember in our analysis of the equivalent circuits of the d and q axis when we derived the expressions for t-d we said that the expressions for t-d terms t-d and t-do basically came by neglecting the existence of rkd and rkq that is those elements and then analyzing the equivalent circuit after that now that can be done because after sometime as we see from the simulation results no flow of current is there in this part of the circuit and therefore it is as good as being open circuited and you can see that this effect is therefore felt in the other earlier graphs of ia that we saw initially there is a fairly high decay rate and then there is a fairly slower decay rate and finally there is no decay at all. So the initial decay rate is then arising because of this flow of current ikd and ikq which decay down very fast and go down to 0 so this essentially again this is to say that this region is then the sub transient region where there is flow of ikd and ikq in the circuits of the damper after this there is no flow of current in damper then this is then the transient time period and after that is then the steady state part. So these equations that we have derived help us to analyze the alternator under various disturbance conditions one can then work on these equations to represent it in various forms as is done normally in analysis of large systems one can simplify these forms to arrive at equations describing the movement of the load angle and so on which is again one of the forms that is used in analyzing large systems. Now those simplifications I will leave it to you as an exercise to see how that can be done we will stop with this for today and continue in the next lecture.