 We have been looking at steady state analysis, we have seen the induction machine how the equivalent circuit can be obtained from the dynamic equations that we have derived. Once the equivalent circuit is obtained of course one can go ahead and use the conventional steady state analysis method be it phasor diagrams or an equivalent circuit study in order to understand the behavior of the induction machine in a larger electrical network. And then we went on to study the synchronous machine and in the synchronous machine we have tried to derive the phasor diagram of the synchronous machine from the steady state equations which we have got from the dynamic equations. By neglecting D by DT terms we arrived at the steady state expressions for the phase voltage and then we converted that to the phasor representation and we derived the phasor diagram for various modes of operation. We saw how one can represent the synchronous motor operation that in that the over excited and under excited modes of synchronous motor operation and then from there we went on to look at the synchronous generator operation where we said that if IQS is negative it results in a synchronous generator mode. So these are the phasor diagrams and then one more expression which is of importance in the synchronous machine analysis is the representation of power flow. So we are looking at steady state analysis of synchronous machine. We are specifically looking at the salient pull machine with the understanding that the cylindrical rotor machine performance can be easily obtained by setting XD equal to XQ. So the equation for the stator voltage phasor what we had derived was VS is equal to R I think we called it as VAS so VAS is equal to that is stator A phase is RA into IA phasor plus J times XQ into IQ phasor plus J times XD into ID phasor plus J times E and ID is a phasor that lies along the D axis and IQ is a phasor of the A phase current that lies along the Q axis. So let us draw the phasor diagram for a particular case let us assume that we are looking at an under excited motor. So if it is an under excited motor let us see what the phasor diagram will be and from that how to derive an expression for the input electrical power that is drawn. So we draw the phasor diagram first therefore we do the D and Q axis so this is the quadrant and then you have the D axis along here and the Q axis along here. We have also seen in the last lecture how one can derive the phasor diagram when the D and Q axis locations are not known only if you know the applied voltage to the machine stator voltage and then the input flow of current drawn to the machine and you know the phase difference between the two based on that how do you draw the phasor diagram that we have already seen in the last lecture. So for this development let us assume that the D and Q axis are known. So if it is known let us further make another assumption that we are going to neglect armature resistance. So this is again an assumption that is normally done in the case of synchronous machines so it is the same thing here and under excited would then mean that ID is greater than 0 and the motor would mean that IQ is greater than 0. So we start with the representation of ID and IQ so let us say that you have ID that is drawn here and IQ that is here so which therefore means that the phasor of IA would lie here. So this is the current phasor IA yeah so let us do this diagram on a bigger page so here you have let us the D axis and then the Q axis and we draw the phasor diagram so the ID phasor is here and then the IQ phasor is here. So these two together would then give us the phasor of the armature current so there you have IA and then what we have is let us keep a copy of this expression so that we do not have to move back and forth and here it is. We have neglected the armature resistance and therefore this term is not there the induced voltage or the stator voltage consists of these three terms now and we have already drawn the IA phasor so J times XQIQ would then be well let us draw the phasor of E first so you have the induced EMFE that is E we said that E must lie along the Q axis because it is J times V and then you have JXD ID that is added to this ID lies along the D axis so JXD ID will be a phasor that is 90 degree leading this one and therefore you have JXD ID here and then you need JXQIQ which is a phasor that leads IQ by 90 degrees so you have XQIQ here and what this equation says is that the sum of these three phasors XQIQ, XD ID and E is nothing but the voltage at the armature and therefore the voltage at the armature is then given by this phasor this is now your VAS phasor so having drawn this we have identified that this angle that angle is called as the load angle which is given the symbol ? that is normally the symbol given for the load angle and this angle between V and I is then the power factor angle power factor angle which is normally given the symbol ? so if we now want to write an expression for power that is drawn from the AC source. So power drawn from source is VA multiplied by IA these are the magnitude x cos ? for one phase this is per phase so all three phases concerned all the three phases combined it would then be three times VA x IA x cos ? so in order to get this we take the magnitude of VA and IA and cos ? is this angle cos ? is the angle contained between V and I so this can be interpreted in the following way that is you take the line along VA take this line and then VA x IA cos ? will mean the projection of IA on this line so this length is what we are concerned about so now instead of looking at this length I mean another way of arriving at this length will be to consider ID and IQ these two components so IA x cos ? must be the same as the sum of these two projections as well on this line. So we can write this in turn as VA multiplied by now look at ID so this component and then you have another component here so IQ if you put it on the VAS line would result in a length of this much and ID when you projected on to the VA line is this much and the sum of these two will then be the same as the length of IA on this that we want so that can be obtained as IQ this length IQ and it is if you take it on the VA axis IQ x cos ? will give us this length so IQ cos ? and then – ID is here and this part is negative and therefore that is ID x cos of this angle. So this angle is what we need to find and that angle is 90 – ? so ID x cos of 90 – ? that is ? so VA x IA cos ? can alternatively be written as VA x IQ cos ? – ID sin ? now the question is what you do with ID and IQ now if we go back to our original equations we would see from the earlier equations that we wrote on synchronous machines that what you have is VDS is equal to RS IDS – XQS IQS and VQS is nothing but XDS IDS plus this is neglecting armature resistance otherwise those terms will also come in sorry RS IDS is therefore not there you neglect armature resistance this is also 0. So from this what we can do is derive this expression IQ is nothing but VDS by XQS I think I need to break that here VDS is nothing but VA so you have the armature vector V armature AC supply voltage phasor VDS is nothing but this along the D axis which is nothing but VA x sin of ? so VA sin ? divided by XQ is nothing but IQ and similarly ID is VQ – E divided by XD from the earlier equation and VQ is nothing but VA cos ? – E divided by XD and what we do is substitute these two expressions in the earlier one and therefore the power drawn is nothing but VA multiplied by IQ cos ? instead of IQ we substitute this so VA sin ? cos ? by XQ – ID sin ? ID is given here VA cos ? – E divided by XD into sin ? so what we what this can be written as VA sin ? cos ? is nothing but sin 2 ? divided by 2 XQ and then you have – VA cos ? sin ? by XD plus E times sin ? by XD which can therefore be written as VA multiplied by VA sin 2 ? by 2 into 1 over XQ – 1 over XD plus E sin ? by XD so this therefore consists of two terms one is E times VA sin ? divided by XD and plus E times VA by 2 sin 2 ? 1 by XQ – 1 by XD now this is the expression for the power drawn from AC source by a synchronous machine now this is an expression which you might have already come across in your first course on electrical machines if you consider a cylindrical rotor machine then XD is equal to XQ and therefore this term goes out if it is a cylindrical rotor machine only this term remains so EV by X into sin ? this D is the synchronous reactance in the case of cylindrical rotor now in the case of salient pole machines XD is the direct access synchronous reactance XQ is the quadrature access synchronous reactance so in the case of salient rotor machine we find that an additional term is also there as compared to what is there in the synchronous machine now this expression we have derived for the case of under excited synchronous motor I will leave it to you as an exercise to derive this expression from the other phasor diagrams as well that is the phasor diagram for an over excited synchronous motor or an under excited or over excited synchronous generator as well you will find that the expressions look the same the method of derivation is also along the same line so in this manner one can then do the steady state analysis of the synchronous machine we have already seen what is there for an induction machine now this is an attempt or the derivations that we have done where primarily to show that the equations that we have derived reduced to the same forms with which we are already familiar from our first course on electrical machines so it is not something that is dramatically different as far as steady state analysis is concerned but what these formulations whatever we have derived do give us is information about the transient behavior of machines which our study in the first course of electrical machines did not give us now these equations can give us transient behavior of electrical machine so how to evaluate or how to estimate what this behavior is going to be now all these equations that we have seen are not something that one can solve by hand because they are fourth order non-linear equations so invariably you would need to take the course to some software in order to compute the solution of these differential equations. So let us now look at the equations for induction machine before we go to the synchronous machine again so if you remember the equations for the induction machine one has to first take stock of what reference frame are you going to do the solution in now we have seen that for the induction machine we have used many reference frame we have used we have looked at stator attached reference frame this can be the natural frame itself or the a ? 0 reference frame we have seen the rotor attached frame I mean the arbitrary frame could then be converted to a rotor attach frame by setting the speed of the reference frame to that of the rotor and then we have seen the synchronous reference frame which where both stator and rotor variables are referred to this frame where the frame rotates at synchronous speed so now the question is in each of these reference frame you have equations that look like v equals r into i plus some l into di by dt plus a g matrix into i multiplied by ? and then you have an expression for the developed electromagnetic torque as well now vi they are all vectors and l and g are square matrix so how does one solve this set of equations so since Te and then you have j d ? by dt is Te minus the load torque and this is the equation that one has to solve and you have ? that appears in the voltage equations also so voltage equations multiplying this vector I therefore this renders the equations nonlinear and even if it were linear it would be difficult to solve it by hand so what one would do is to take recourse to a numerical integration method so in the numerical integration method just to give a brief overview of what it would be what we are looking for is a solution of the state variables of the system state variables of the system are those that define the state they are in this way of where we have written down the equations the state variables will be the variables i both referring to may be I will write it as a vector i which has all the stator current and the rotor current variables along with the speed so you may typically have if you neglect the 0 sequence terms you typically have about 5 variables to solve for 5 state variables so let us say we are looking at one of the state variables which is may be say ids and if we plot this as a function of time you may find that the vector the ids undertake some root and finally settles down somewhere so what we want to do is to compute this entire variation of ids right from t equal to 0 to some t equal to n time and the way that is done is you break this entire range of interval over which you are interested for t starting from t equal to 0 you divide this into small steps of time and what we do is estimate what will be the solution at each of these instance by approximating the differential equation to a difference equation so it goes like this in an approximate way of saying that you find out the slope at this point so find out the slope here this slope multiplied by the step length will give you what is the increment will give you this increment slope multiplied by this length so that when added to the initial condition at t equal to 0 will give us this value so we say that at the first step the value of ids is here but obviously that is not right ids the actual value of ids is lower down and what we are getting is somewhere higher so that is obviously wrong but however you notice from the figure that had we taken the time step to be here then doing the same operation that is multiplying the time step with the slope will give us an id value which is here and that appears to be almost right so the accuracy with which one can solve these differential equations using a numerical integration method depends very much on what time step you choose how many discrete instance into which you discretize the x axis so we choose a small enough step time and having selected this arrived at the value of ids here what you do now is you know this value this value is known so now find out what is the slope of ids at this point and from the slope of ids at the next time instant estimate what will be the second value of ids now find the slope again and estimate what will be ids at the third instance and so on like this one discretizes the x axis and tries to find out the solution step by step so how that is done you basically have to write a code which looks something like this you say that for time equal to 0 to end time 0 to end time is where you want to find out the solution what you do is find slope this is nothing but di by dt and d omega by dt in order to do this let us look at the differential equation so this equation can be recast in this following form p i is equal to v – r x i – g i omega therefore the slopes di by dt can be obtained if you know the voltages at that instant of time and we know the vector i at this instant of time and we know omega at this instant of time because it is l x p i now you do not know what i and omega are at a given instant of time because that is where you want to obtain the solution you may know values of i and omega at the earlier instance of time so there are various algorithms which help you address this issue one may first make an estimate of what i and omega are and then using that estimate try and find out what the slope is once you know the slope you can find out what the actual value of i and omega will be at this instant of time if they do not match revise your estimate and redo until you convert so whichever way you use what one has to do is find out the slope and once you find out the slope then this may require using information about previous values once that is done then find state variable values at next instance and then repeat this entire loop so that is the way one would go about doing so you can write a code there are various software available to do this job one may write a code and may be C or C++ or one may make use of specialized mathematical analysis software to do this job some of the well known software that are available are the software MATLAB and simulink this can be done in MATLAB as well one need not go to simulink alternatively one can use software like Psylab which is another software that provides an environment very much similar to that of MATLAB alternatively one can use another software that belong to the same set of analysis software that is called Octave these are open source software that are available so one of these can be used or if that is not there one can do the entire programming on C or C++ and save your results and use so whichever way it is done this is the approximate way in which you would go about solving the set of equation. So let us see how a typical induction machine analysis would look like let us take an example where we have an induction machine the machine that we are looking at is a 400 volt induction machine 50 hertz supply and it is a 4 pole machine that means we expect that the synchronous speed will be 1500 rpm the stator resistance of the machine is known the rotor resistance of the machine is also known as 0.74 ohms it is referred to the stator turns the stator leakage inductance is given rotor leakage inductance referred to stator turns is given the value of LM that is magnetizing inductance is given and moment of inertia is given note that from the data that are given you can straight away draw the equivalent circuit so if one were to draw the equivalent circuit what it would look like is a stator resistance and then a leakage inductance the magnetizing inductance and then the rotor resistance the rotor leakage inductance and then the rotor side is short circuited because it is an induction machine so this is your RS this is XLS leakage reactance this is XM and this one would then be RR dash divided by the slip this value that is given here is RR dash so you divided by the slip when you put it into the equivalent circuit and then this is XLR dash so essentially what we need in order to solve these equations is the same data what is required or what you obtain from the equivalent circuit as well only thing that is missing as we noted when we derived this equivalent circuit is the resistance representing core loss that would have come here but now that is not there because we are neglecting core loss so having got this now you see that an additional piece of information that is required that is moment of inertia because we are now trying to look at the dynamics of the machine so speed change becomes important the equivalent circuit does not require the moment of inertia information because it does not attempt this kind of a study so the experiment that we are now trying to do is you take an induction machine and you have three phase three terminals of the induction machine the rotor is of course shorted and you have a three phase source are connected sinusoidal in nature and then you have a switch which in which you connect the machine like this so what we are trying to do is if you close the switch all the three switches at T equal to 0 seconds the shaft of the machine is assumed to be not connected anywhere so it is a free running machine so what happens to the induction machine how does it accelerate how much input current is drawn and what happens to the current in steady state these are the questions that we are looking at for example we are asking how long does it take for the machine to run up from 0 speed to the rated speed of operation there is no load torque that is applied that means load torque TL is 0 so it is a free running machine and what we see here if we write code to solve the equations as we have just outlined in the earlier case and obtain the results for speed after solving the differential equations here we have chosen to select the value of time from 0 seconds to 0.5 seconds so half a second of study we have done this is please note that the intervals that are marked here are only for the purpose of drawing the graph it does not mean that we have discretized the entire 0 to 0.5 seconds in steps of 0.05 seconds you can see that the first mark here is 0.05 and then it is 0.1 0.15 and so on this is only a mark provided for the convenience of drawing and this is not the discretization step in this case for example what I have used to solve this is at a resolution of ?t which is a time step is 0.1 ms maximum may be less than that otherwise now what we see from this graph is that the speed starts at 0 and then accelerates very rapidly and reaches a value which is 1500 rpm this level is about 1499 rpm because there is no load torque that is applied the system or the machine speeds up to reach almost synchronous speed and it then overshoots the synchronous speed as well so here you have the synchronous speed or the speed at which the system settles down finally so that is the speed at which it settles down and that value is well to be very exact it is 1000 as I said 499 rpm and the speed overshoots and then there is an undershoot and then there is an overshoot again it sort of settles down in this manner why does this happen so we know that speed is increasing very fast here and as the speed reaches 1500 rpm you know in an induction machine the speed can never go beyond 1500 rpm because it is the revolving magnetic field that tries to draw the rotor along drag the rotor along with itself if the rotor tries to rotate faster than the synchronous speed then induced emf will be present in the rotor to retard the rotor and therefore it cannot run beyond synchronous speed it may be able to run below synchronous speed provided there is a load which retards the rotor sufficiently that the developed torque is equal to the load torque it cannot run exactly at synchronous speed because at synchronous speed the induced emf would be 0 and no current would flow and therefore no torque would be developed but if the rotor is an ideal rotor that means no frictional last terms are there then if it has gone to synchronous speed it can simply continue running at synchronous speed that is approximately the situation that we have but the rotor having reached synchronous speed here there it has been accelerating all along and there is a momentum that has been gathered so that momentum takes it beyond synchronous speed and once it is going beyond synchronous speed the retardation force sets in and therefore the rotor now decelerates and reaches synchronous speed here and then it undershoots as well because it is now retarding and it undershoots then it accelerates again and finally because of these oscillations that result these damp out and finally it settles down at this value now how many oscillations are there and how much overshoot is there would depend upon how much moment of inertia the system has and how it overshoot so that will come out from the solution of the machine so this is the speed versus time plot now here you see how the input flow of current to the machine is these three plots show the phase currents of the machine that is if one were to see what is the current that is flowing here Ia, Ib and Ic that is now shown in these three plots now you see here that the steady current drawn is quite small in fact if you had zoomed in to this I have not shown you that plot but if you look at this figure this steady current which is a small number is about 8.1 ampere peak and the waveform is sinusoidal now comparing to this 8.1 ampere peak look at the first peak that is reached here the first peak that is reached is quite high in the range of about 140 ampere which is quite high compared to what it has under steady state so the initial flow that is happening this is called as inrush current the inrush current is quite high which is one of the reasons why if you have an induction machine directly started from the supply you may find that everywhere else the supply voltage dips because it draws such a huge inrush current and that inrush current has to come through your lines and it causes a voltage drop along the lines so supply at your installation is likely to dip when this is there. And how long does this inrush current last it lasts till about let us say almost 0.04 seconds or so after which it becomes small and settles down so why does this inrush current last inrush current arise because initially the rotor has not yet started moving and the rotor has not started moving the induced emf has not developed in the machine and therefore there is this huge input current that is drawn and therefore you expect this inrush phenomena to settle down only after the rotor speed has settled down and that we can verify from the earlier plot you see that the rotor speed reaches near about the synchronous value only at about 0.05 seconds and then there are some oscillations so these you can see are faithfully reflected here and only when the speed has settled down say at about this time which is about 0.225 seconds you see that there is no further increase in the stator current it is a nice sinusoid after that one can also see that if you look at this graph there is a DC component in this that attempts to decay down and go to 0 this waveform is not really symmetric about the x axis you have 0 here the first peak is probably about 150 whereas this peak is only about 100 so there is an offset in this waveform and this DC component decays down to 0 there is no offset here similarly on this side you can again see that there is another offset which decays down to 0 and here perhaps there is no offset these are two identical almost 100 ampere here and 100 ampere here the offset if at all is there is quite small. So initial offset will be there because you are switching on to an RL circuit with initial conditions as 0 ampere everywhere so to accommodate the initial condition the system introduces an initial offset which then decays down to 0. Now this draws 8.1 ampere peak so after the simulation study where the system goes through the initial acceleration phase and then settles down we find that it settles down at 8.1 ampere peak this result could have been obtained by analyzing your equivalent circuit itself what one could do is now you know that the speed is 1499 rpm which means the slip is approximately 0 if the slip is 0 rr dash by s goes to infinity so this branch becomes an open circuit that is the rotor branch becomes an open circuit so what we are left with is only the stator resistance leakage impedance and the magnetizing impedance so this boils down to essentially an RL circuit and you are applying the phase voltage here you know all the numbers they are all given so try to solve for how much input current will be drawn and you will find that the input current drawn is about 8.1 ampere sinusoidal 8.1 ampere peak that is the current that will be drawn so again we see that the set of equations that we have one solve completely and run up to steady state gives us the results that are exactly estimated by what you get from the equivalent circuit itself so this is the stator input current how does the rotor current look like now here you see that as far as the rotor is concerned we see that there is no load on the machine if there is no load on the machine the rotor rotates almost at synchronous speed and almost at synchronous speed would mean that there is no induced emf inside the machine inside the rotor if there is no induced emf in the rotor in steady state we expect that no current flows in the rotor or negligible current flows in the rotor and that indeed is shown in these three plots this is the 0 level for this plot 0 for this plot is here and 0 for this plot is here so we see that the rotor currents settle down to near about 0 value under steady state initially of course there is a large rotor current that is flowing obviously because there is an electromagnetic field that rotates at synchronous speed rotor is not yet at synchronous speed and therefore there is considerable induced emf and large currents that flow which in turn give rise to the generated torque which causes the acceleration of the machine itself so apart from this initial high values we find that the rotor current settles down to 0 what happens to the other variables how do they behave and what information we can get from the other expressions and other simulation studies we will see in the next lecture we will stop here for today.