 Welcome to lecture 34. We will continue our discussion where we stop in the last lecture. In this lecture, we will briefly review the application of FE modeling for analysis of rotating machines. As an example, we will take permanent magnet machines for the purpose. Next, we will see how in a typical electrical equipment or machine, various engineering fields like thermal, structural fluid, etc. are coupled to electromagnetic fields. Circuits are also coupled to electric and magnetic fields. And in this course, we will study FE formulation for voltage and current-fed devices. To start with, we will study FE formulation for voltage-fed devices. And for current-fed device in the corresponding formulation, we will see in a later lecture. Any rotating machine in general can be modeled by this equivalent circuit. You have terminal voltage and induced voltage and the current flowing. Depending upon whether it is motor or generator, the direction of current will be decided. Now, the way it is shown, current is shown, this is like a motoring operation. So, now you have here, now we are discussing here surface mounted permanent magnet machines. That is on the rotor surface, you have permanent magnets mounted. So, what I assume is those who are listening to this course, they have some background in these machines and they know some basic operational and design or construction details of these machines and then they will appreciate what is being said here. So, here, now this is a motoring operation and then you have the current and induced voltage waveforms as shown here. So, to produce constant torque, current should be fed into machine as shown in this figure because now here induced voltage is of trapezoidal form. So, to have constant torque, which is proportional to product of induced EMF and the current, then you have to have current which is rectangular pulse and that can be easily controlled by using power electronic converter. So, as long as you maintain this constancy of induced EMF and constancy of current in this figure, then you will get constant torque. So, as induced EMF opposes the flow of current, terminal voltage V t is varied to maintain the required phase current. So, how actually the power basically gets applied is moment the rotating machine starts rotating, you have induced EMF, is it not? That induced EMF is going to oppose this current that is flowing in and if to maintain this current constant V t has to be adjusted by the source. So, that even current is maintained constant. Effectively, this source is going to supply additional e into i power, is it not? Because when the speed was 0, e is 0. When there is some speed, the corresponding induced voltage is there and e into i is the opposition to the source effectively. So, the source has to basically supply power e into i, i being constant here, is it not? And that e into i power is basically converted into mechanical torque at the shaft. So, the power transferred from electrical to the mechanical system is e into i. Now, this was the motoring operation. Same thing is true for generating mode with some changes. Now, here again the same equivalent circuit is there. So, current direction is kept same, but actually you know when it is acting at generator, the current will actually reverse. The current actually will go from e to V terminal. That is the reason that you know this in the current direction is actually shown opposite. That is why the. The sign of current is changed when compared to the previous slide. But if you reverse the current here, representing the actual generating operation, this current again will come like this as in the previous slide and then you know same logic applies. Now, here again you know to produce constant torque, phase current should be as shown in the figure and in phase with the induced EMF. The induced EMF feeds the current to the terminal voltage and to vary the terminal voltage, mechanical system must vary the speed to and then the corresponding power transferred from mechanical to the electrical system is again e into i. So, the power flow is just reversed here. So, the induced EMF can be is generally sinusoidal in case of permanent magnet synchronous machines and that can be achieved by generally using distributed winding in space and but it can be also achieved by using concentrated winding with means such as you know pole shaping. Whereas, in case of brushless DC machines, you generally have trapezoidally distributed EMF in the space and that is generally achieved by using concentrated winding. So, sinusoidal induced EMF machines when fed with sinusoidal current excitation produce constant torque and similarly trapezoidal induced EMF machines with square wave excitation as we have seen in the previous slide produce constant torque. And machine torque ripple can be predicted from induced voltage waveform similarly from the clogging torque of the machine. So, this the importance of this analysis is that is why it is clear now that basically if we can know the induced EMF waveforms, we can predict the machine performance parameters accordingly. And in such FEM analysis, generally what is done is even though we may have motor, it is generally simulated as generator to extract induced EMF waveform. This is for convenience only. So, now we will continue our discussion further and now we will see a new topic coupled field systems. Now, in the second lecture, I had explained you this particular diagram also wherein coupled fields in case of transformers are explained. So, now in any electromagnetic device be it transformer, motor, generator you have so many physical field for example, in case of transformer you have magnetic field coupled with electric field, thermal field, fluid field, acoustic and then circuits. So, and then there are coupling parameters between any of these two systems and so any of these systems and then for example, between magnetic field and thermal field, the coupling parameters are AD and hysteresis losses basically they are produced due to alternating magnetic field that produces temperature rise and temperature rise in turn may change permeability and conductivity and that would change the magnetic field. So, these are the coupling parameters. Similarly, you can see the coupling parameters written for the other sets of coupled fields here. So, in this course now what we are going to do is we are only going to see coupled circuit field computation. We are not going to see thermal and mechanical fields coupled to you know electromagnetic fields. But for more discussion on this how to couple for example, thermal and mechanical fields you can see this reference here and many other good books and papers on the subject some of which are mentioned in this reference. Right. So, now we will start our discussion on coupled circuit field computations. First we will see voltage field coupled circuit field analysis which is quite common most of the devices. In fact, all our devices and equipments are typically voltage fed because there is some voltage source feeding our transformers, motors, generators. Right. So, the governing field equation is this which we are now quite familiar with is del square i equal to minus mu j. Right. Now, this j is now written as Ni by s, s being the area and s typically is in this case coil area. Right. To which the current is flowing. So, Ni ampere turns divided by the area. So, the functional for the above equation is this also we have seen and we are quite familiar with now. So, this is the functional. Instead of earlier we had j A dv. Now, we have Ni by s A dv and this dv will become ds when we actually do 2D approximation. Right. Then this will become dx dy. Now, after applying the FEM procedure this first term will become this. So, del square, del A square will basically will get reduced to this. Right. And this as we have seen over each element we will have 9 terms and then this over all the elements then this energy will get summed up. Right. So, this is one energy term. The second energy term is on account of source j which is in fact Ni by s. So, here it is again summation i goes from 1 to 3 Ni A i dx dy and here this is this leads to 3 terms corresponding to 3 nodes of a triangular element. Right. We have seen that earlier j gets a portion to j delta by 3 to the 3 nodes. Now, also remember that this is the functional when it gets minimized with respect to variable A i's. This one A i from here and one A i from here will go in the process of minimization. Right. And that is the reason that you will be left with only Ni s delta by 3 corresponding to this term. Right. Because integral Ni dx dy is nothing but delta by 3 into this constant N by s. Right. And this i is the variable which needs to be determined. So, that comes in the matrix equation as i. So, B E capital B at the element level is N s by N by s delta by 3. Right. That and then when you form global matrix equation then that will become B. So, basically then this becomes C A minus B i equal to 0 and these are these are global level matrices. Right. And this is local. Now, this C into A C basically we have seen del Ni dot del nj will give the corresponding C the coefficient matrix for element and I already explained you 9 terms will get out of these two summations and that will form the element coefficient matrix. And when you actually sum up all the elements that will lead to global coefficient matrix capital C. And A of course is the unknown magnetic vector potentials at all the nodes in the domain. And of course, this formula will have to use it whenever we are left with determining integral Ni dx dy. So, using the same thing we in fact can derive this as delta by 3. So, now what is new here is the circuit equation. Now, this device here is some device which is you know meshed and this is the corresponding finite element mesh that is represented here that is fed by this external circuit with this voltage of U. Right. So, these are external and L external and this device is represented by here induced voltage right at this terminal. So, the this voltage is the corresponding field parameter in this circuit equation. So, that is that is nothing but this that induced voltage at the terminals of the device is nothing but this. So, this is basically n d psi by dt. And n d psi by dt the d by dt basically is replaced by j omega. So, j omega n psi is the corresponding terminal voltage induced voltage at this device right. And then we are basically writing a KVL for this loop. So, external source voltage is equal to the corresponding terminal induced voltage at the device plus R external into i j omega L external into i j omega L into i. Now, here remember this U and U i psi they are in the curly brackets meaning that they are the column vector right. And then we what do we do we basically replace this term by this term. And again we have invoked here the theory that we saw in lecture 24. And in fact in the previous topic also we have invoked this. Basically what we are doing is we are calculating we know the flux passing through any two point is given by in case of two dimensional case it is given by a 2 minus a 1 into L right. So, now here so that is what you see here a into L right. Now, here what we are doing we are assuming that we are modeling the both sides of the coil. So, you know it is like suppose this is a coil current going in here and current coming out here. So, dot and cos and dot. So, when we are taking you know a this is what is what are we doing here we are taking average of a integral a ds upon s. So, average of a we are calculating over say this coil there are so many there could be so many elements here. So, we are actually averaging a was element wise and then corresponding coil wise and dividing it by total area s of this coil and suppose this is this is plus a because this current is say cross. So, say let us say this is minus a at that point because direction of a is nothing but direction of current right. So, suppose this we take as current as negative. So, this will be some minus a this is dot. So, this will be some plus a. So, the difference of these two a's will basically represent a flux passing to this area and corresponding voltage induced in this coil that is what is being done here. So, when you actually do this integral a ds and then this integration will be over this entire now coil area. Both coil sides. So, there will be plus a here minus a here. So, plus a minus of minus a will give you the total flux linked by this coil right. So, that is when we are doing this integral a ds over s we are considering this whole area of the coil both parts being considered. So, that will give you total magnetic potential difference between these two points and that into l will give you the flux and into n will give you the flux linkage right and then into j omega. And then here of course, remembering that l will be eventually taken as 1 because we are going to do per meter depth calculation. So, this l will be 1 going further. Now, we actually this a we substitute by our well known expression integral ni ai right as a and into ds is dx dy. So, that is what we have done and then this replaces to this integral ni dx dy will replace will be get replaced by the matrix g dash where g is nothing but n n by s delta by 3 1 1 1 again by the same logic that integral ni dx dy means ni raise to it is really ni raise to 1 l is 1 and by this formula again as we have got it here n by s delta e by 3. So, you will get n by s delta by 3 corresponding to this n by s and delta by 3 will be this integral right this on left hand side you have 1 by 1 because only one voltage. So, this is 1 by 1 right and then this these two terms as it is. So, now, we are ready to combine both field and circuit equations. So, now, first the equation that is seen here first we will see that the circuit equation j omega g dash into a plus r external j omega l external into i is equal to u. So, these are the circuit equation and then you have this field equation c into a minus b i equal to 0 c into a minus b i equal to 0. So, now, we have got combined set of equations you do field and circuit right and now you can see here b b and g incidentally here they are equal because l is equal to 1. So, you have to remember that in the expression for this b there was no l. So, it was simply n by s delta by 3 but in the expression of g you have because there is a l here and since l is 1 this gets reduced to because l is 1 here. So, that is why g becomes equal to e but if you take for example 5 meter depth if you are doing calculations for 5 meter depth then of course, g and b will not be equal because then g will get multiplied by 5 right and then since it is l is equal to 1 in this case g and b matrices they become equal right. So, now, we will when we use this equation for coding for a particular problem then we can get various analysis done particularly in case of transformers and rotating machines excited by external circuits. Here we will see how do we do short circuit analysis of a single phase transformers. So, we will see this in the next lecture. Thank you.