 We have derived the dynamic equations of electrical machines and we have started looking at the application of these equations and we started looking at the induction machine equations and what would be the kind of results that one can get if we use these equations for simulation of an induction machine. Towards that in the last lecture we started looking at the simulation results for a free running induction machine and here we have the data for the machine it is a 400 volt 4 pole machine with stator resistance rotor resistance etc as indicated and when we do simulate that for a free acceleration that means the load torque is 0 the applied load torque is 0 and we run the simulation we find that the machine accelerates in this manner what we have is here we have the x axis as time in seconds and the y axis as speed in rpm and we have seen that the speed rises in this manner with certain overshoots and undershoots etc and the speed settles down at approximately 1500 about 1499 rpm which is near about the synchronous speed because there is no load on the machine and then we saw that stator transients you see that initial high inrush current that flows and then settles down and I think we saw in the last class that the amplitude of the sinusoidal steady state stator current is about 8.8 ampere this of course one can verify using the equivalent circuit analysis if you consider that the phase voltage is applied across the equivalent circuit and all the equivalent circuit parameters are given here so one can do a simple circuit analysis assuming that the speed is going to be 1499 rpm one can find out by circuit analysis that indeed 8.8 ampere of steady state current flows in the machine from the equivalent circuit it would not be feasible to get how the speed is going to change that is if the speed is going to change increase in this manner and then oscillate and then settle down all this information is not available from the equivalent circuit you cannot solve for this situation what you can get from the equivalent circuit is a performance when the speed has settled down and you must know what is the speed at which it has settled down if you know what the speed is then using the equivalent circuit one can find out how much steady state current is going to flow in the stator and indeed how much steady state current is going to flow in the rotor and in this analysis since there is no load the rotor current settles down at approximately 0 ampere no current is flowing in the rotor because the rotor is running almost at synchronous speed that can be verified by zooming into the rotor current plot as well one can see really it settles down at about 0 ampere that means no rotor current that is flowing the generated electromagnetic torque this is torque in Newton meter now that shows an initial high peak so you see that the initial torque generated is very high that is the time during which acceleration occurs and then you see that there are oscillations in the generated torque as well which as it must be because it is in this region that the speed exceeds the synchronous speed so one can go back to the speed plot so one can see that the speed is beyond the synchronous speed in this region and you see speed is coming down and the rate at which speed comes down is a maximum here so you expect that maximum negative torque is available at this point which is somewhere after 0.05 second and one can see that indeed is the case so maximum negative torque is available when the rate of fall of speed is very high and negative torque is there in the synchronous machine because the speed now has gone beyond the synchronous speed and therefore since the flux is rotating at synchronous speed it now tries to retard the rotor and that is manifesting as a negative torque and then there are these torque oscillations and finally the torque settles down to 0 Newton meter it is 0 Newton meter because again there is no load applied on the machine then if we look at the loaded behavior of the machine what we have done in this case is added a load torque that is 0.002 times speed square obviously is in radian per second so if this is the load torque and remember this load torque is going to oppose the generated electromagnetic torque inside the machine and the resultant of these two is going to be responsible for the acceleration and with this then again if we look at the speed transient so the x axis is again time in seconds and the y axis is speed in rpm we see now that the speed rises in a manner which is very similar to what we saw in the last case so here again the speed rises goes beyond the point where it settles down and then there is a speed oscillation and finally the speed settles down and we see that the speed settles down at 1443 rpm that is a reasonable slip because of that slip it is now able to support the load torque which we now have given as 0.002 times omega square so here there are overshoots and undershoots about 1443 rpm if we look at the generated electromagnetic torque electromagnetic torque again shows initial high values which is when the rotor is accelerating so large torque for the rotor to accelerate and then finally the torque settles down at this level which is a value of 45.76 Nm so one can see that this is the steady state torque that is generated by the machine and this torque must be equivalent equal to the load torque and therefore the net torque is 0 so the speed is maintained at a fixed value. Now input you see the input flow of current drawn into the stator you see that there are again inrush currents very high inrush currents one must remember that the experiment that we are trying to do in this case is what is called as a direct on line start of the induction machine that means the machine which is may be connected to the load is there and you are simply trying to connect the supply voltage to the machine in one single step the entire voltage is simply switched on to the machine in that case what happens that is what we are trying to see here so you have initially very high inrush currents that are drawn again we see that the inrush current here it reaches a level of a steady state 17.8 ampere peak whereas initial inrush we see it is 150 ampere peak almost equal to 9 to 10 time the loaded steady state current that is flowing so as we noted last class these heavy inrush currents that might flow can cause other detrimental effects on your supply system like voltage drops and then because of voltage drop some other equipments may trip or lamps may flicker and all kinds of effects like that will happen so here again there would be a DC inrush initial you see that this is 150 ampere whereas here this peak is only 100 ampere so it indicates not really symmetric and therefore there is likely to be a DC component which then starts at some value and then decays down to 0 as it reaches steady state and this behavior what you get here this can of course be analyzed by means of the equivalent circuit again you must know what is the speed at which it is going to settle down then once that is known one can find out the steady state performance using your circuit analysis so this is the stator current and here one can see the rotor current that reaches 15.6 ampere peak and one can see in 0 to 1 second there are about two cycles of oscillation so that clearly indicates that we have a very low frequency wave form here about 2 hertz or so whereas the supply side on the stator this is clearly a 50 hertz wave form so one can see that because of the small slip that is there the flow the rotor current has slip frequency and now one can look at what happens if you transform these variables to some other reference frame this first plot gives us the stator current wave form stator current is what we have seen here the 50 hertz wave form whatever is shown here if you were to transform the stator current that is three phase stator current into the a beta reference axis two phase axis what we can see here is that the two phase wave forms apply like I mean appear like a pair of sine and cosine waves if you look at it closely you can see that the wave forms are basically displaced by 90 degrees one can see that if you take one wave form let us take this one so if this is the case you can see that the blue color wave form looks like this whereas the green color wave form looks something like this right so the phase difference is plus or minus 90 degrees and these are sine cosine pairs that are available which is what we expected if you look back or if you remember back to the lectures where we discussed the nature of the wave form due to this two three phase to two phase conversion we remember that these were of sine cosine pair and that exists right from the beginning right you can see that there are only two this red line that you see here the red line that you see in the middle that then corresponds to the zero sequence term the red line is the zero sequence that is I0 and because we are supplying a voltage system that is a three phase voltage balance three phase voltage zero sequence response from the machine is again zero and no zero sequence current flow this is again true also because even if you were to supply an unbalanced excitation as long as the machine is star connected as long as the machine is star connected even if zero sequence component where to exist in the applied voltage zero sequence flow of current cannot be there star connected plus isolated neutral in the case of induction machines neutral is invariably isolated so that is your response or the flow of currents drawn in the a beta reference if you look at the rotor current in the a beta reference frame please note that what we are looking at is the rotor attached a beta zero reference frame in that again you see that the stator I mean the a beta currents form a sine cosine pair but they are of low frequency this is the same as the frequency of the actual stator current which means that if you go from the natural three phase reference frame to the two phase a beta reference frame attached to themselves rotor attached to rotor stator attached to stator frequency does not change frequency remains the same but the three phase variable get converted to variables get converted to phase variables and here again you see that the zero sequence component I0 is 0 this again is the result of balanced system and considering a star connected winding on the rotor the x axis is of course time in second and the y axis is rotor current rotor current in ampere now we may do the other thing we may consider then all the machine responses as looked at from the stator referred frame for the induction machine then for the rotor variables it would be a pseudo stationary reference frame which is then attached to stator and here we see that even though these are rotor currents this is rotor current in ampere and you have time in seconds. So this is the rotor current referred as seen from the stator a beta zero reference frame and you can see that this interval is 0.1 second and in 0.1 second you basically have 1, 2, 3, 4, 5 cycles which means what you have here is 50 hertz waveform that is indeed what happens if you look at the rotor phenomena as seen from the stator you only see a flux rotating at synchronous speed and which would lead you to infer that whatever is generating the flux could be something that is stationary but excited at 50 hertz and that is the physical equivalence of the a beta zero reference frame for the rotor by the same token one can also look at stator current in the rotor a beta zero reference frame that means you are looking at the dynamics of the machine or what is happening inside the machine sitting on the rotor and when you sit on the rotor the frequency of the flux the rotating air gap flux will now be at slip frequency and therefore you see that the rotor current or rather the stator current as seen from the rotor side is at slip frequency the amplitude would be determined by the manner in which you convert or transform the stator current to the rotor reference free and that is what it is so here again the x axis is time in seconds and the y axis is then stator current in rotor attached frame in ampere so one can by this means transform the machine equations and the responses of machine to look at the dynamics of the system either from completely the stator side or from the rotor side whichever is advantageous to the particular problem at hand one would probably use an appropriate description. Now one can of course go to the synchronous reference frame the synchronous reference frame is the reference frame that rotates at synchronous speed and if you rotate at synchronous speed all the fluxes that you will see around you are stationary with respect to yourself and therefore you see now that all these excitations are then DC flow of current you have IQS here and you have IDS here these are then DC currents and the zero sequence is zero in this case also because system is a balance system so you have again time and current in ampere on the y axis now in this case both the currents IDS and IQS have settled down at a negative value this will depend really upon where your reference frame is with respect to the stator current phasor. So if you now draw the phasor diagram which we will now see before that the rotor current is also DC quantities because whether it stator or rotor the moment you are in synchronous reference frame all variables look stationary and therefore you expect both IDR and IQR also to be DC quantity the phasor diagram which is the space phasor diagram I have drawn it normalized with respect to themselves the stator current normalized with respect to the magnitude of the stator space phasor and therefore all of them would have an amplitude of one maximum. So here you see that VS is the stator voltage phasor IS is the stator current phasor and this phasor diagram can be obtained from your ID IQ plot the stator current phasor I in the synchronous reference frame is nothing but ID plus J times IQ and that is what is shown here both are negative and therefore you get please note that 0 of the imaginary axis is here 0 of the real axis is here so this point indeed is the origin and with respect to that you can then have the stator voltage phasor and the stator flow of current the rotor current phasor and these two are phasors of the stator flux and the rotor flux as determined by the earlier equations that is the ?DR for example is then LR into IDR plus LM into IDS and similarly you define all of them ?QR then ?DR plus J times ?QR is then the flux phasor for the rotor and similarly one can define a flux linkage for the stator as well and these are the two flux vector. Now what we have seen is basically the induction machine with the DOL start that is a direct online start first we saw free running acceleration and then we have seen the loaded case with some particular load speed or characteristic this is a sinusoidal analysis that means you are supplying sinusoidal voltage to the machine and you want to see what how it behaves the nice thing about these equations is these equations are not restricted to sinusoidal analysis we have never ever in the development of equations assume that V and I are sinusoidal we have always spoken about instantaneous values of voltage applied instantaneous flow of currents and therefore these equations are general and can be applied to non sinusoidal conditions as well and therefore one can look at inverter fed operation just to briefly recollect what this is we are looking at an inverter which consists of 6 switches each switch having its own anti parallel diode so you have the diode here and this rail then forms the DC bus and you connect a DC voltage here these points then act as the output points of the inverter this is a three phase inverter and this is then given to the induction machine which is connected here and in this simulation that we are seeing these inverter devices are switched on in this manner this device is one switched on first and then two then three four five and six and then it goes back to one so that means one switch is turned on or off every 60 degrees and each switch having been turned on remains on for 180 degrees and that is then generally referred to as the square wave mode of operation which is what we have mentioned here so this is the inverter fed operation of the induction machine the machine is connected to the same load that we had seen earlier that is load torque is defined as 0.002 times omega square and then here you have time versus speed in rpm and one can see how the machine accelerates and settles down so the settling down value is more or less the same as what we have seen with the sinusoidal case so if it is the same with as with the sinusoidal case what is the difference so here you see the speed variation the y axis shows speed in rpm the x axis shows time in second in the sinusoidal excited case you found that the speed the machine accelerates and then settles down the speed is really flat fixed speed whereas here you see that the speed is oscillating so it is not a constant speed it is an oscillating speed that means there is a speed ripple the speed ripple in this case is about 1437.5 or so here and then goes up to 1440 almost 40.5 so it is about 3 rpm ripple that you have in speed now whether this ripple is okay or not will depend upon what application this machine is going to run maybe in some applications it is not acceptable to have this kind of ripple maybe in many applications it may be allowable to have this kind of ripple but whatever it is one must know what the machine is going to do to decide whether it is acceptable or not and if you look at this ripple you see that this is 0.34 seconds and here you have 0.35 seconds so in this time then you have one cycle two cycles and then three cycles of the ripple wave form three cycles in 0.1 seconds 0.01 seconds which means there are going to be six cycles in 0.02 seconds that means in one cycle of the main wave form there are six cycles of the speed ripple which means what we have is a sixth harmonic ripple in speed so why is there a sixth harmonic ripple one can then look at how the machine is going to behave in order to determine that one can see that the input voltage wave form has will have a fifth harmonic voltage and the seventh harmonic voltage and they react with the fundamental to get a sixth harmonic torque so all that can then be seen so if there is going to be a ripple in speed that can only come if there is a ripple in the generated electromagnetic as well so if you look at the torque transient you see that the wave form broadly looks something similar to what we have in the sinusoidal case but all along you see that there are oscillations these small oscillations are there and indeed in steady state one can see all these ripples that are there this ripple frequency if you zoom in and see will be sixth harmonic frequency which therefore causes a sixth harmonic ripple in speed one therefore looks at methods by which the speed ripple can be reduced torque ripple may be reduced and so on there are various then schemes where one can look at these things state or current so if there is going to be a ripple in the generated electromagnetic torque we have seen that the expression for the generated TE depends upon is going to be something like Iqs x idr – ids x iqr so Iqs and ids must in turn have some ripples which means that the actual input current drawn should also have some ripple or harmonics so here we see that the input current drawn is not really a sinusoidal current but has lot of harmonic that are there here again there is an inrush current definitely inrush current will always be there because you are supplying full voltage in one stretch to the induction machine it will draw an inrush current to support the huge torque and here we see a zoom wave form of the stator current so this is current in ampere and time in seconds so here we see that this wave form is clearly non sinusoidal is rich in harmonics what happens because of these harmonics invariably they cause lot of heating and I2r losses iron losses eddy current losses in the machine and therefore may be detrimental and these harmonics also result in harmonics or oscillations in the generated torque TE and therefore in the speed wave form as well but if you are going to supply an induction machine through an inverter and not specially controlled inverter in any particular way simple square wave operation you just cannot avoid giving significant harmonics at low frequencies and that therefore results in this kind of distorted stator current wave form. Again after studying all these things by means of all the simulations one will then need to answer this that for a given application whether this kind of harmonics are acceptable or not if they are not one has to go around finding ways to minimize this harmonic and make the current more sinusoidal one can see the flow of current in the rotor so this is rotor current in ampere time in seconds even though you still have the same low frequency which obviously you must have because speed settles down at speed with a small slip 100440 rpm so you expect that the frequency will still be slow mean low and frequency is indeed low but it again has oscillations which are at high frequency thereby contributing to ripples in the generated TE wave form and the speed wave form. So this in short is an example of what kind of results one can get by using the equations that we have developed so far for the induction machine and what kind of studies it can be put to use you can use it to determine the transient behaviour of either sinusoidally excited systems or non sinusoidally excited systems but as far as the machine is concerned what we have modelled is a balance three phase machine the voltages that are applied may be balanced or unbalanced in whatever manner we want. So one can then use this to study other aspects of machine behaviour as well for example one can say that you have an induction machine being connected to some load and may be you know it is a direct mains connected machine it has been supplied with AC voltage from some substation let us say and all of a sudden let us say that there is a fault on the lines and all the three lines short to ground. What will happen to the fault current that is going to flow in this on this side there is a source three phase source and that will send some current into the fault the induction machine one may say that there is no source of excited it is not the alternator therefore may be it would not but then there is a magnetic field inside the machine which is rotating at synchronous speed and magnetic field represents stored energy so if there is a fault what you would find is that the induction machine also will contribute to the fault current though it would likely to be a short duration even it may be worthwhile to study what kind of fault performance this kind of a system gives so if you want to study that you need these kind of dynamic equations to see what is going to happen but then that is not all if you have a system like this because of the fault what is likely to happen is that at this point you might have a circuit breaker so if you have a breaker here which is there in the line as soon as fault current is detected the breaker may open and if the breaker opens still fault current if there is no breaker on the machine side fault current will still be supplied by the machine but then what would happen is the breaker may reclose and then you are now having a fairly severe situation you are trying to close this breaker when the machine is rotating and there is flux inside the machine now this kind of a situation can cause a more severe interest current than what was happening at direct online start because it may so happen at the time at which you decide to close the breaker the supply voltage and the induced voltages inside the machine because of the flux being there they might add up and cause a larger current that flows into the machine so these kind of studies will have to be done a priori in order to determine what kind of behavior to expect from a given system so that you can probably decide to choose an appropriate breaker or fuse ratings and so on so this is another aspect of the machine dynamic equations that we have studied of course as we have seen in the earlier when we develop the synchronous in the induction machine model in the synchronous reference frame one can take that along further and then see how the machine has to be controlled if you are going to have an inverter fed induction machine and you are interested in closed loop control you might want to look at strategies like field oriented control or vector control or may be direct or control or some other control strategy that you might invent yourself you need to understand and manipulate these machine equations so machine control is another big area where these machine equations are employed to a large extent so let us not I mean let us stop the induction machine discussion at this point we have seen applications of machine equations from this side and the next thing to look at will be the alternator equations now the alternator equations also we had developed and let us now look at applications of the dynamic equations of the alternator that we have studied so alternator dynamic study we had already looked at the steady state analysis of the alternator and then we have seen how the steady state equations could result in the phasor diagram and using the phasor diagram one can then do all your steady state circuit analysis system analysis where the alternator is connected to the grid and so on. So when we talk about dynamic study of the alternator it is very important to consider dampers especially in the case of grid connected operation most of the alternators are operated in the grid connected manner they are the ones that are supplying all the electricity that are of use to you except that nowadays you have solar power and wind power where the generation is not through the alternator but most of the power in the world comes from alternator. So they are connected to the grid and what one may need to look at is at the time of synchronizing the alternator to the grid what happens to the machine normally if you synchronize it in a as it is expected to be done manner then you would not have much state of current flowing during synchronization but sometimes what happens is the synchronization due to some error or the other happens in the wrong way maybe the phases are interchanged or the phase sequence is not right or the phase they are not at the same phase at the time of synchronization then huge currents can flow and it can result in severe alternator damage. So how does the current flow what kind of currents and what kind of shaft torque may result one will need to make use of the equations that we have seen now in all these cases then you will have the rotor rotating at speeds that due to the extra excess torque that is generated or the fluctuation in the fields that are there it would be very much essential to look at how the dampers are going to behave. So let us then put down the equations of the synchronous machine with respect to dampers with respect to these kind of analysis there is another important aspect of alternator study because they are generally grid connected one of the important aspects of alternator performance is if you take the alternator three phase machine connected to a bus or may be a transformer and then connected to a bus there may be several lines that are then going on may go somewhere this may go somewhere may be another bus and so on. Now what is a very severe duty for the alternator I mean whenever you design a machine and you are going to design the switch gear to protect the machine and make it operate in a safe manner on the grid one has to answer the question what is the worst case situation that we are likely to encounter and in that analysis what one does normally is to look at therefore fault performance of the machine. So if there is going to be a ground fault on the lines what happens to the induction machine so we are looking at what is then called as a three phase short circuit of the machine and this test three phase short circuit sometimes called as sudden three phase short circuit is a very important experiment that is done on the alternator to determine what will be its behavior when subjected to this kind of a disturbance and there is lot of analysis that has been done invariably in such cases when there is a fault that occurs the speed of the machine does not undergo a change immediately because such alternators are invariably quite large associated with a huge moment of inertia and therefore when there is a fault at the stator the entire system which is rotating at some speed you do not expected to suddenly change speed. So in this kind of analysis what we are looking at is then to make use of the dynamic equations and predict what kind of flow of stator currents are likely to happen in the system when the speed is constant speed may be considered constant and then there is a sudden three phase fault at the induction machine that is very severe duty and it so turns out in trying to analyze this behavior by means of analytical expression there are several important features of the alternator behavior that can be defined which and several definitions emerge out of this analysis which then go towards defining a particular machine one then speaks of transient impedances sub transient impedances that are associated with a particular machine. So far in the machine models that we have derived we have not come across or we have not defined any of these terms all that we had for an alternator for example if it is a salient pole machine we talk about the direct axis impedance and then the q axis impedance whereas now in order to quantify the behavior of the alternator when connected to the grid and subject to disturbance you will now have to come across different terms new terms which will which one designates as transient or sub transient impedances now these definitions and the behavior also has something intimately to intimately related to the behavior of damper bars behavior of damper inside the alternator during the time that is when subjected to disturbance these bars that are there will be in action for a short period of time until the speed becomes again close to synchronous speed and after that the dampers are not active but still because of the disturbance that is caused some stator current disturbance and field current disturbances are likely to be there so how do you identify what is the sub transient impedance what is the transient impedance why they arise in order to understand all this what we will now do is proceed by studying the three phase short circuit behavior of the alternator starting from the dynamic equations of the machine including the representation of damper so we will look at one damper circuit on the axis and one on the q axis make the equations for this that is developed by inspection the equation for this system and from that equation we will then study how the machine responds to a sudden three phase short circuit this we will see in the next class we will stop here for now.