 So, welcome to lecture L24. In this lecture what we will do is we will see 2 case studies one on inductance calculation for an induction motor and second would be for inductance calculation for gap coefficient reactor. So, in the first case study let us understand how to calculate no load inductance of a 3 phase induction motor. So, what we will assume here is we will assume that rotor speed is synchronous. Effectively what we are assuming is slip is 0 and hence there are no induced currents in the rotor structure and flux in the motor structure is completely due to the stator winding currents. So, in such case wherein we do these approximations then magnetizing inductance can be calculated by using a simple magnetostatic simulation. So, what is the governing partial differential equation? It will be say del square a equal to minus mu j and I have as I have already told you in some other lecture that 1 over mu this mu is taken as 1 over mu on this left hand side to make it more general. So, that changes in material properties can be easily accounted for in the FEM formulation. So, remember j is in z direction because this is a 2 dimensional approximation and hence a also will be in z direction. So, and of course, then that is why it becomes a very simple formulation in which unknowns are only magnetic vector potential values at various nodes in the domain direction of a is already fixed which is in z direction. Now, let us go further. Now, this slide shows mesh of the chosen induction motor geometry. Geometry details are taken from this reference and we are modeled only one quarter of the full geometry. So, now let us revisit calculation of flux from magnetic vector potential which we had seen in basics of electromagnetic. Now, this we will study with reference to a coil in a rotating machine. Now, this is a coil in a rotating machine. So, this is one coil side, this is the other coil side and these two are under two opposite poles. So, now we know psi flux is close counter integral a dot dl and in this case it will be ac1 minus ac2 multiplied by l. Why? Because these two horizontal segments of this contour will not contribute to this line integral because the length vector and a vectors they are orthogonal to each other. So, dot product will be 0. So, this we have seen in earlier. So, now here as I mentioned to you these two coil sides of a coil are under two opposite poles in a two pole machine and then the flux is like this for current direction shown. So, now this let us go further. So, now remember this is the axis of symmetry and this is coil side 1, this is coil side 2. So, this coil side 1 corresponds to this coil side here and this coil side 2 corresponds to this. So, now consider that each coil side in the figure is as discretized into two finite elements as shown here. So, coil side 1 is divided into two finite elements similarly coil side 2. So, the return path for this the current in this element would be this and for current in this finite element the return path would be to this element of coil side 2. Now, let ac1 and ac2 be the average values of magnetic vector potentials in this coil side 1 and coil side 2 respectively. So, now ac1 then can be given as integral acd as integrated over the coil area and they divided by ac because we are calculating the average value. So, this is the average value of magnetic vector potential in coil side 1. Now, further you know simplifying this now since this coil side is made up of two elements. So, now instead of integrating writing it like this we can write summation over two elements and then instead of the integral over the whole coil area it will be integration over element area and e goes from 1 to 2. So, now we can further simplify this as we can split this summation and then we can write summation over individual elements as shown here. So, remember this a bracket 1 and a bracket 2 these are the magnetic vector potentials over element 1 and element 2 respectively whereas ac1 is the overall average magnetic vector potential over this coil side 1. So, further you know expanding it what we have done this a1, a bracket 1 and a bracket 2 they are written in terms of shape functions like this n1 a1 plus n2 a2 plus n3 a3 for element 1. Similarly, the other magnetic vector potential is n1 a1 plus n2 a2 plus n3 a3 for element 2. And now we know that due to symmetry ac1 will be equal to minus ac2 because by symmetry the magnetic vector potential here and the corresponding point mirror point here they will be exactly opposite in science. So, then the flux is counter integral close counter integral a dot dl. So, that will be ac1 minus ac2 into L and putting this relation you will have twice ac1 into L. Now in place of ac1 this now substitute this expression. So, this twice L into ac1 so ac1 is 1 by ac into this whole bracket further simplifying it now what we are doing here we are bringing out magnetic vector potentials out of the integrals because only n i's are the functions of x and y. So, similarly for the element 2 here. So, now and we know you can refer lecture 18 slide number 4 we have seen this formula there integral over the element area n i ds is simply delta by 3 area of the element by 3. So, then substituting that then you simply get this a1 of element 1 into area of element 1 divided by 3 plus this plus this and similarly for the element 2. So, delta 1 and delta 2 are the areas of the triangular elements 1 and 2. So, now same expression which was derived on the last slide is written here and now what we do this term this term we can write as a product of 2 matrices and multiplied by delta by 3. So, a1 a2 a3 rho vector into column vector 1 1 1 into delta by 3 is nothing but this. Similarly, this expression we can write the way it is written here. So, now this whole thing can be written elegantly using 2 summation signs as here and remember what we are calculating this is psi. So, psi is twice L by SC summation over 2 elements and then for individual element summation over for its 3 nodes delta e by 3 a i e. So, a1 e a2 e a3 e you will get when i runs from 1 to 3 and then you will get this entire this expression when you actually expand this expression. So, now this is for you know 1 ton we considered only 1 ton coin if there are n tons and that too if there were you know more than 1 pole pair then that has to be you know taken into account. So, here for simplicity to start with we are taking number of pole pairs as 1. So, then total flux linkages which is nothing but number of turns into flux will be given by n capital N Np psi and Np here is 1. So, that can be replaced by just 1. So, it will not appear in the expression further. So, n and this psi this expression we are just putting it here. Now, this expression can be further written elegantly as done here where now this g e element level matrix g e is given by this twice n l by sc delta e by 3 column vector 1 1 1. So, now this you know size of a e t you know it is 1 by 3 because a e is a column vector a e is a column vector and when it is transpose will be 1 by 3. So, a e is 3 by 1 a transpose will be 1 by 3 and g e will be 3 by 1. So, product of these 2 will give you 1 by 1 matrix. So, now this is over this product is over 1 element and then we are summing it over both the elements. So, now if we actually go to the global level then we can actually from this local element level matrices we can get global matrices corresponding to both elements together and then this size of global matrices will be 1 by 4 will be for a transpose and g will be 4 by 1. Now, why 4 by 1 and 1 by 4 because here there are 2 elements but the total number of nodes are not 6 total number of nodes are 4 only. So, 1 2 3 and 4 because 2 nodes are common. So, the size of a matrix is 1 by 4 a e transpose is 1 by 4 and similarly g matrix would be 4 by 1. So, for these 2 element example we have quickly understood how we can go from element level to global level. Now, we actually you know go to the induction motor example that we saw in the first 2 slides. So, we are talking of jth phase coil. So, there are 3 phases. So, we are talking of jth phase. So, j can be either 1 2 or 3. So, now just reproducing the expression that we have already derived earlier with now NP taken explicitly here because NP now we are not assuming as 1 there can be more than 1 pole pair. So, NP is number of pole pairs and now this capital N which is number of turns is given by for an induction motor NCS by NPL. NCS is total number of conductors in a slot and divided by NPL is the number of parallel parts. So, NCS upon NPL will give you the electrical number of turns and here SC we should remember that it is coil sides total cross sectional area and now this we have already seen here you know from local. Now, what is the difference here as compared to the last slide earlier it was e was going from 1 to 2. Now, here e is going from 1 to N e, N e is the total number of elements in the domain. Now, here you see this we are doing it for jth phase coil. So, gj e is written here and when you go to the global matrix a this becomes a transpose. So, a is at here this a will be N into 1 and this gj so this you know a transpose will be 1 by N and gj will be N by 1. So, again this will give you 1 into 1. So, lambda flux linkages would be just a scalar number 1 into 1. So, now this lambda j is a transpose gj. So, the size of a transpose is 1 by N we just I explained and gj is N by 1 where N is the number of nodes in the whole domain and gj matrix will have non-zero entries only for nodes within the jth phase coil. So, corresponding to the nodes of the jth phase coil gj entries will be non-zero rest entries will be 0. Similarly, for some other coil some other nodes the entries corresponding to some other nodes corresponding to that coil will be non-zero. So, 3 global matrices can be computed for 3 phases as I just explained briefly and then the magnetizing inductance of jth phase coil can be simply calculated by flux linkages upon current. Now, you can see here using the formulation just described we have you know got this flux plot and this the flux count tools are you know as expected we are getting. So, then the inductance calculated would also be then you know as per expectation. Now, we will see the second case study of this lecture which is inductance calculation for a gap core shunt reactor. Now, what is this shunt reactor? This shunt reactor has a central core with air gaps. So, these are the air gaps. So, these are although they are called as air gaps there is some you know non-magnetic and non-metallic material in all these gaps. So, that this whole thing is mechanically stable and these you know gaps basically are adjusted the length of or the height of the gaps is adjusted to get a particular value of inductance. Now, what is the inductance? Inductance is mu 0 n square s by L because we have seen in basics of electromagnetics that inductance is given by n square upon the reluctance and here L upon mu 0 s is the reluctance is it not and reluctance in this whole magnetic circuit is mainly and predominantly calculate contributed by these air gaps which also have explained during the basics that reluctance of air gap is very high as compared to the magnetic material because all this rectangular bridge that you see in cross section that is the magnetic material high permeability magnetic material. So, if the relative permeability of this magnetic material is 1000 in a very simplified calculation one can say that 1 mm of air gap is equal to 1000 mm of this high permeable ion material. So, most of the reluctance is offered to the flux by these air gaps and hence most of the energy also is stored in all these air gaps. Now, remember this each of these packets rectangular packets that you see in this cross section is actually a circular packet right like this. So, this is having actually a circular cross section when you take a cross section and look on this from the top. So, you will get a cross section which is circular when you take a cross section here and look from the top you will get a circular cross section right. So, now the field solution can be calculated by using Poisson's equation in magnetostatics. Now, let us you know see the understand the dimensions. So, first of all this you know core packet the diameter is this is D and diameter is phi 80 mm coil depth is 175. So, this distance is 175, core to coil distance is 135, core to coil this distance is 135, coil to yoke this distance is 150 and coil to end limit. So, this distance also is 150 right. So, now ampere turns for this shunt reactor are number of turns is 2800 and current is 68.73 and area of cross section is pi D square by 4 and if you substitute 0.58 meters here then you will get 0.6242 meter square as area and total length of the air all 8 air gaps together because there are 8 air gaps here. So, 1, 2 and this is 8 air gap. So, total length of all air gaps together is 320 mm. So, if you just put it in this formula mu 0 mu 0 n square S by L and put in all the values then you will get inductance as 8.13. But we will see later that this is far from the actual value because of you know these air gaps flux does not remain entirely within this you know central this you know magnetic path flux actually comes out of this and it fringes between these you know air gaps as well as there is some flux in the coil and that flux outside the core also contributes to the inductance which is not accounted here in this formula. I also again mentioned during the basics of electromagnetics that inductance wherever there is a flux at the corresponding flux tube you can associate some inductance to that. So, this inductance is only controlled is actually this inductance calculated this way is representing inductance of flux only contained within this central this magnetic path and not you know the flux which is outside. So, finite element method will help you to you know account all the flux and get accurate value of inductance as we will see now. So, this is what the flux solution obtained by the flux plot is a flux plot obtained using finite element you know core. Now, here we have modelled only half the geometry right. So, here there is another half identical but we have because there is a symmetry we have taken only half the you know geometry. So, remember we are actually solving this in x y coordinate system. So, what we will be getting energy is per meter depth. So, we have to multiply by pi into mean diameter which is a circumferential length to get the entire energy you know stored in respective you know parts of this geometry. So, now the inductance calculation we will see how to do. So, now we have divided the whole geometry into 5 parts core air gaps in the core, core winding gap winding and rest window portion. So, I will explain you these 5 parts again. So, when I say core it is only this core magnetic part high permeable magnetic part here. Second is air gaps. So, basically it is all these air gaps. Third is core to coil. So, this region here and this region here. Next is the actual coil and then rest of the portion means this portion between coil and end limit. So, these are the 5 parts and remember we have modelled only one half of this whole geometry and then the other half we will take care by multiplying by you know pi into mean diameter. So, that will give us the total circumferential length because we have to remember that this core entire core with gaps as well as this coil they are circular. So, they are symmetrical about this central axis. Now, you can see here maximum energy is you know stored in the air gaps 1.92 into 10 raised to 4. Then it is you know 6.15 into 10 raised to 3 between core to winding because the flux is fringing between the gaps. So, there are also considerable energies there. But in the winding it is further less whereas in the core and in the rest of the winding portion it is very very less. In core obviously it is very less because the permeability is very high and the reluctance is very low. So, the energy stored is very low. For the same flux density the energy density will be given by half mu h square is it not. So, since mu is high for the same b h value will be very small and if h value is small half mu h square will be very small which is the energy density that energy density multiplied by volume will give the energy. So, now total stored energy is addition of all these pi numbers and now remember this is the total energy j because why we have already this the values given by the FEM code where joules per meter in each individual area to that to those you know values we have multiplied by corresponding pi into mean diameter. So, you can easily verify these are the corresponding mean diameters of those you know phi you know sections here, phi parts of the geometry. So, total stored energy is addition of all these phi. So, the i is 68.73 and now if you equate half L as square as this then L you will get as 11.64 and now actually by FEM if we just calculate if the inductance by considering only energy in air gaps which is 1.92 into 10 raise to 4 then we get you know inductance as 8.13 which matches with the one calculated by simple analytical formula. But as I said earlier other 4 energy values are not accounted here that is why we get this value much smaller as compared to the actual value which is 11.64 the same actually you know solution we can get by further reducing the model. So, earlier we made only you know we took the we basically obtained a solution by considering only half the model. Now in this case we will consider only quarter model because then we can because we can exploit a symmetry around this horizontal axis in the middle of the coil. Earlier we exploited symmetry along this axis now we are exploiting symmetry along this horizontal axis. So, number of ampere turns then we will have to define just 50 percent here in this coil block right and the total energy will be the energy obtained from this one fourth model you have to multiplied by 2 to basically get the energy for the both the parts this part as well as the part down right and then the total energy that was obtained was 2.78 into 10s to 4 which is quite close to that obtained considering the half model. And then the you know inductance we can calculate as usual as done in the previous slide and we get it as 11.77 Henry. So, the last point that I want to discuss here is you know why to have so many air gaps in the middle core structure why not have one single air gap that question may arise in minds of those who are actually not knowing the design aspects of shunt reactor typical shunt gap core shunt reactor. What happens is there are two problems associated with this if you have a single air gap of course it is much easier from the point of your design as well as from the point of your manufacturing. But then there are two issues one is now you can see the fringing is very high and the effective you know length of this flux counters also is more right. So, the inductance that you would get by only by having only this one air gap would be quite different than that obtained by using multiple gaps. So, that is one thing that one has to remember. Secondly, because of this heavy fringing flux that flux hits this winding winding area particularly the middle of the winding can you see here there is a heavy flux which is incident on the this winding area that can cause excessive eddy current losses and hot spot and that can even damage this winding within no time. So, that is why we should have multiple air gas which are of smaller heights so that you can avoid all these all such quality problems and you can estimate the inductance also quite accurately you know and get the value within the expected range. Thank you.