 Welcome to this 30th lecture. So, we will see here torque-speed characteristic of an induction motor. How do we estimate by using time harmonic formulation that we saw in the previous two lectures. So, now here governing partial differential equation is this. So, this is you are familiar with this except the change here is see actually it is sigma dA by dt is it not. So, dA by dt d by dt is replaced by j omega and again there is now one more S this is the slip of induction motor because depending upon the frequency of induced EMF you have to take either you know S as whatever slip frequency suppose it is rotor you have to take the corresponding slip if it is you know a stator then it will be S will be equal to 1. So, depending upon so we are generalizing this so S into omega rest is of course same. So, now we know from the induction motor theory that torque if we take the rotor losses then torque is given by p rotor losses upon S times omega. Now, here if you see this geometry there is this rotor which is steel and we are considering this as solid iron because here you know it is slip is very small. So, induced frequency is very small. So, the eddy current losses will be quite small. So, we can consider this rotor as solid is not laminated whereas this stator the outer most part this stator this steel is laminated steel. But here we are not interested in loss calculation. So, we will not be defining the conductivity for this stator also. And another thing there are two more simplifications in this model this you know stator does not have teeth. So, there are only these coils A plus A minus B plus B minus C plus C minus there are there are no teeth these are not the teeth. So, it is like you know the this outer part is just path for the containment of flux and so that the flux is contained within the geometry. And so basically you have this complete as sort of air gap and then you have this rotor is in the form of this aluminum ring. So, this entire this part is aluminum. So, instead of you know you have in squirrel cage induction motor for example, you have you know the cage having aluminum bars is it not which are short circuited at both ends. So, here we are simplifying that model as simply you know aluminum and then you have this all these as I explained A plus A minus and all those. Now, this source for these geometry is this website where you know these are what are called as team workshop problems. These are standard benchmark problems whenever you get on to any complex FEM you know best analysis. Here you have an option to verify your FEM methodology by solving some standard benchmark problems given in this website. So, I will just show you that website. So, here for example, you see here there are number of problems given at this website and then you know you can just open any of those problems and detail some theory is given problem definition is given results are given. So, you can model any of these problems which are of interest to you and verify your FEM analysis. So, these are very useful you know website with number of problems given static time harmonic transient couple problems couple circuit field problems. So, we have taken this geometry on this website and the problem is 30A. So, of course this is a simplified geometry. So, now this is a three phase two pole induction motor. Now, you can see here we are actually what we are going to do is we are going to define peak current for a phase. So, a plus is peak current. So, B plus and C plus coil will have negative half. So, that means B minus and C minus will have positive half current. So, this is positive peak and these two will be this will be positive half peak. So, together you know you will say this is plus plus plus this is say 1 per unit this is half half. So, together you know you will have flux contours which will be enclosing all these positive current. Of course, they are displaced in time this is positive peak this is half positive peak. But there will be you know you can expect to see flux pattern which will be you know like this. Similarly, here A minus will be negative peak and C plus and B plus will be also negative half peak. So, there will be another set of flux contours which will be enclosing these three coils. So, now, let us before going into this let us just compare the flux plot that we have got. So, as I was explaining you have got one set of contours like this and another set of contours like this. So, I hope you understood the geometry and the ampere turn definition. We basically solve this we discretize this. This will give you the global coefficient matrix which will have geometry and material information this will give you D matrix is it not we have seen it earlier D matrix this will give you B j matrix. So, now, here we already saw torque is given by rotor losses divided by S into omega. Now, here omega is the synchronous speed in radians per second. So, S is nothing but omega synchronous minus omega r that is the rotor speed upon omega. So, that is slip and f we are taking it 60 hertz because in this problem here frequency is 60 hertz. So, we are taking it as 60 hertz. So, slip is omega minus omega r upon omega where omega corresponds to synchronous speed. So, now, having got the solution A then what we can do is we can then find out E is nothing but minus daba A by daba T and that means it will be minus j omega A and then you can then substitute then find out the amplitude of E and then substitute it here sigma is the conductivity. Now, in terms of you can directly calculate this in terms of j also then in that case the formula becomes rho j square dv where rho is the resistivity because E is nothing but j upon sigma. So, this will be j square upon sigma square. So, j square upon sigma because this sigma 1 will go. So, it will become j square by sigma 1 upon sigma is rho. So, sometimes you will find in the text and in some commercial software they will code this formula rho j square dv. So, rho j square and sigma square they are the same and as I previously mentioned this rho j square or this sigma E square that has the unit of watts per meter cube. The dimensions of this geometry are given here. So, various radias are given here 233.25.25.7. So, remember this is a very simplified geometry just for the verification purpose. So, as we have been earlier also discussing the advantage with finite element method is it is like sort of geometry independent the procedure. So, if your finite element solution works for this geometry which is very sort of trivial or simplified geometry it will definitely work for a complicated geometry real geometry of a motor with all the details modeled correctly. That is the advantage of finite element analysis. So, this we have already seen this plot. Now, this is for a slip of 0.05 which is the actually practical slip value. Whereas slip of 0.47 is not practical because the torque split characteristic if you see the operating region or the stable region is this region is it not this is unstable region. So, s is equal to 0.5 will come somewhere here which is unstable. But just for comparison purpose we have plotted for s is equal to this value also which corresponds to speed of 200 radias per second for 60 hertz case because this for this value that problem 30a has the corresponding value for which you can actually verify. So, then you get losses in aluminum and losses in rotor steel as this. This is calculated from this formula. This is just for verification purpose we are taking this. Similarly, this slip is not possible and you can also see that because of slip being high the induced currents and the corresponding distortion will be high. That is why you can see that the flux pattern here is considerably distorted as compared to s is equal to 0.05. So, then our purpose is to plot torque split characteristics. So, what we need to do is for we vary slip we vary slip in this formulation for every slip we basically solve this PDE. For every slip we can calculate eventually this PR from PR we calculate a torque and then we plot it here. So, this is rotor speed is nothing but we can actually this is nothing but slip is it not this is representing slip. So, then for every slip or the rotor speed we can get one torque value and this is that is how you can. So, that as many FEM simulations will be required as many points you want to plot on this torque split characteristics. But it is fairly easy because this is a quasi static calculation is it not this is not really a transient. Once you have your code running you have to just you know in fact you can set a outer to loop with s varying from some 0 to 1 and in one go you can actually get all the torque value is it not. So, this is fairly straight forward here this just shows now here for example, this corresponds to the synchronous speed. So, that is equal to s is equal to you know 0 here. So, this is just explained here omega is 120 F by P in RPM divided by 60 you get RPS revolutions per second. So, that comes to twice F by P and then this is revolutions per second. So, if you want to find out radiance per second then you have to multiply this by 2 pi because one revolution has 2 pi radians. So, 2 pi into 2 F by P. So, this gives you you know 4 pi 60 by 2 because F is 60 in our case and that gives you 377 radians per second that is what you know this point is 377 radians per second. So, this is how we can actually use FE methodology to plot torque split characteristic of induction motor. Now, we go to the next topic axi-symmetric problems. Now, the axi-symmetric problems are quite often encountered in our electrical engineering when we deal with coils mostly circular coils are there and if we can simplify some geometrical details and make it sort of you know symmetrical about axis then this axi-symmetric formulation is really good because it gives by doing 2D simulation you are effectively getting 3D fields because fields are independent in phi direction is it not. So, that is why axi-symmetric problems can be quite formulation can be quite handy in analysis of electrical machines and equipment. Now, here you know we are assuming symmetry about z axis which is quite logical then let us see some theory B is equal to del cross A and now here you know J is now this is axis symmetry J current is in phi direction is it not axis is in z direction. So, the current will be in phi direction it is symmetry about z axis current will be in phi direction. So, J also will be in phi direction current density also will be in phi direction if current density is in phi direction A also will be in phi direction magnetic vector potential we have seen in basics of electromagnetic A is the direction of A is direction of current or J is it not. So, that is why in this now when you expand this curl del cross A then you have A as only with A phi component. So, indeed only there is a phi component and this rho comes here you know when you this is the basic definition of curl in cylindrical coordinates here 1 upon rho comes here 1 rho comes and here also 1 rho comes. So, this is basic definition of curl in cylindrical coordinate system. Now, if you actually expand this determinant then you will get this 2 components A rho and a z components. So, which is obvious because J is in phi. So, that will lead to B and B having 2 components in rho and z direction like when J earlier case whenever we use to approximate in Cartesian system we use to take J in z direction is it not and then B used to be having x and y components. Similarly, here you have J in phi we have rho and z component rho and z component. So, now here this rho this d by dz of rho. So, this rho can be taken out is it not rho is constant when you are taking derivative with respect to z. So, this rho and this rho will get cancelled. So, that is why finally you get this expression. So, now further going for the now we calculate J, J is del cross H and H is B upon mu right. So, then you know you again use curl expansion and then you get now here B is having 2 components rho and z right. So, that is what is written here 1 upon mu this is this one more by mu is coming because of this B rho, B rho is this and B z B z is this. So, this component right and B phi is 0 because there is no B phi component right. Now, again you expand now here J will be in only phi direction and that can be verified here that you know if you take a rho then d d by d by phi of this will be you know because with respect to phi there will not be any variation is it not because it is axis symmetric. So, d d by d by phi of this will be 0 and this d by d by z of 0 is anyway 0. So, that is why a rho component is 0. Similarly, you can see a z component this anyway is 0 d phi d by d by d by phi of B rho again since it is symmetrical about z axis again d by by d by phi of this will be 0. So, only component that remains is this a phi which is d by d by phi of this and minus d by d by z of this right that is what is written here and then we actually substitute this components here B rho and B z and then we are just getting that as J phi going further now what we do. So, if we have to use the same formulation that we develop for Cartesian systems in x and y then we have to make this symmetrical. Now, here for example you know here it is derivative a partial derivative with respect to O A phi whereas here it is a partial derivative with respect to rho A phi. So, it is not you know symmetrical is it not. So, we can make this symmetrical. So, that we can apply the same formulation that we developed in x y system for this rho z system how we can do that we multiply this numerator and denominator by rho because rho is constant is it not rho is just constant and that rho can be taken inside because d by d z for d by d z rho is constant is it not. That is why we take this rho inside and divide by rho. Now, you can see this is symmetrical and then here with respect to our earlier Cartesian systems what are the changes here rho is of course replacing x z is replacing y is it not right. So, I hope you get this. Cartesian systems of course these axis this is some coil if this was x y system this would have been x this would have been y in cylindrical system. Now, this is rho and this is z is it not. So, this is x is nothing, but to go and y is nothing, but z in cylindrical systems. So, that is why rho is replacing s x z is replacing y and rho a is replacing a earlier we had only a here and there a was what a z is it not because there is a two dimensional approximation in x y. So, we always took current in z direction. So, a z was in z direction. So, this was actually a z there. So, that a z is now replaced by rho a phi. So, now then if we do this you know sort of manipulation then we can exactly use the earlier code for this with no change only thing that we have to remember is the solution that we get is not a phi but it is rho a phi. And then if you want to find out a at any point then you have to divide that solution value by rho at that point is it not because the solution that we are going to get is rho times a phi. So, now at any point suppose after having got solution suppose you want to find out actual a phi at this point what you have to do you have to actually take that corresponding rho at that point. So, that rho a phi divided by rho will give you the a phi at that point clear. So, if this is then it is fairly straight forward. So, now what another approximation we do for simplifying or our calculation is we consider rho as constant for elements. That means suppose you have some element here triangular element actually speaking the rho is changing at every point within the element is it not. But if the element is fairly small we can assume for you know that rho is constant for that element and that and equal to the centroid rho value. So, that at the centroid then if we do all this simplifications and then manipulation as explained earlier. Then our element coefficient matrix and the corresponding entries will remain exactly the formula remains exactly same except we have in the denominator one rho term the radius that is the point at which calculations are being made or the element at which element in which calculations are being made. So, here since we are assuming rho for the that element as constant and at that centroid that is why you know you can just write here one rho value. Earlier we had only one over mu is it not in x y system we had only one over mu here. Now, here we have one over rho mu. So, that is why here also mu gets accompanied by rho and here you know actually it is mu as a function of b is written this is actually not valid yet because we are not seen really non-linear analysis. So, do not worry about this b just consider this as just mu this is this has to be considered in this form mu as a function of b when we actually see non-linearity and we want to model non-linearity which we will see in a later lecture. So, right now this you consider this as just as mu because mu is constant over each element and it is a linear system. So, mu is not a function of b source matrix remains the same no change the way we calculated earlier that means this J phi will result into J delta by 3 and the same procedure applies. So, now for the going for the calculation of b from a you get you know now here earlier we had a as n 1 a 1 n 2 a 2 plus n 3 a 3 now we will have rho times a as this and then you know you have this is the expressions for n 1 n 2 n 3 with x and y now replaced by rho and z. So, this is n 1 this is n 1 this is n 2 this is n 3 with x and y replaced by rho and z. Now, we have already calculated this b b is this. So, we are taking that here right and then we are actually multiplying and dividing by rho to you know fit this thing fit into this and then now we actually take this derivatives with respect to z and rho. Now, this leads to the exactly same expression as earlier for b except you have gone 1 over rho which is logical is it not and this a 1 q 1 a 2 q 2 a 3 q 3 and this q 1 q 2 q 3 are the standard and p 1 p 2 p 3 are the standard once as earlier in terms of coordinates. But, there x and y will be replaced by rho and corresponding rho and z for the vertices 3 vertices clear. Once you have done that then of course, calculation of energy and inductance it is straight forward. Now, in Cartesian system we had b square upon twice mu as energy density energy per unit volume. So, to that we are multiplying with volume d x d y into d z, but we are taking per meter depth that is why one appears here is it not. So, this is the total energy b square upon twice mu is energy density energy per unit volume. In cylindrical system it is the same thing b square upon twice mu d rho d z. So, then how do we you know in terms of coding what changes we will have as compared to our 2 d Cartesian code. We will have here this mu 0 gets multiplied by accompanied by rho as we have seen earlier it denominator rho comes that is the only change because of this rho here in the denominator and b, b remains same j delta by 3. Similarly, here now b z and b rho instead of b y and b x you are now b z and b rho and this expressions we have already seen this expression we derived. So, corresponding you know expression is this again here additional this rho comes here which is variable c x here this is the only thing mu. Similarly, here for b rho you get this there is another there is a c x term here in the denominator and then b net is square root of b z square and b rho square plus b rho square and then you calculate the energy for the each element b square upon twice mu multiplied by 2 pi rho b square upon twice mu 0 into 2 pi rho for each element and then we have to integrate. So, this for one element then you integrate over sum over all elements. So, we will get the total energy. So, in this slide we analyze inductance of our cap core reactor which was discussed in L24. There is this central core with non magnetic caps whose total length is designed to have a particular value of inductance. Most of the energy stored in these gaps because of their high inductance and hence they did predominantly decide the inductance value. The central core and the coil are circular and therefore symmetric about the axis. A top and bottom yokes and two side limbs are flat structures with rectangular cross section. Hence truly speaking the entire geometry is not axis symmetric. Earlier we analyzed the same problem music 2D Cartesian system in which the computation is done per beta depth in z direction. This has essentially assumed that the entire geometry is infinite in extent in z direction which is not true. However, this approximation is not bad given the large diameters of the coil and the core. Thus we understood that both Cartesian and axis symmetric 2D approaches are approximate as summarized here. But they are good enough for inductance calculation. For more accurate computation one has to do 3D FEM best analysis. The values of inductance is given by both methods are close to each other. In the method based on Cartesian coordinates we had obtained energies per meter depth in various parts and multiplied them by corresponding beam diameters later to obtain the total energy. In the axis symmetric approach the multiplied elemental energy is by corresponding beam diameters at the element level inside the FEM core. So this is another difference between the two models. So in the next lecture we will see permanent bandwidths and corresponding FE formulation for permanent bandwidths. Thank you.