 So, welcome to this 29th lecture wherein we will see more about you know applications of non-homogeneous Neumann conditions which we actually analyzed in lecture 28. And then we will also then further see in this lecture how do we you know find out eddy current losses happening in a winding of a power transformer biasing FE analysis. So, now let us first you know see you know application of Neumann conditions. So, let us you know take a case of conducting material plate which is excited on its both surfaces by same magnetic field intensity H. Now this configuration is quite common for example any conducting plate in a rotating machine or a transformer for example a cold lamination or a winding conductor excited on its both sides by the same value of magnetic field intensity H. Now here B is equal to del cross A and then you know then it is H will be 1 over mu 0 del cross A and noting that A will be only in z direction because we are doing two-dimensional approximation right. So, then you get this and this in fact was derived in slide 7 of L28. Now in this case since H is only in x direction. So, only this component exists. So, you know you have H as this this component is not there on this surface. So, H is this. So, this is what we are imposing on this you know boundary and inside also H is going to be in x direction only and because of that what is going to happen H is going to reduce as you go from the surface to you know inside and that is the reason that daba Az by daba y also is going to change as you can see here. Now remember these are equi A or equi flux contours right. So, as I mentioned in one of the previous lectures lines and contours are used interchangeably in you know electromagnetic and particularly FE analysis. So, now here also you can see A is varying with respect to y because these colors are changing that you can easily understand from this. So, Hx is some Hx is imposed and then you have by the we are solving diffusion equation. So, the Hx is going to reduce as you go inside and then corresponding daba Az by daba y is also going to change as you go inside from both these surfaces when you go towards center daba Az by daba y is not going to be 0 or in other words Az is going to vary with respect to y as can be seen here. Now what is the application of what we just saw? Now let us take a conductor, a flat conductor which is excited by H on both sides and suppose let us take a very thick conductor. When we say thick conductor it is electrically thick I explained you in L10 then when you say electrically thick that means the thickness is much greater than the skin depth right. So, effectively suppose this thickness is much higher then what is going to happen? This H value is going to reduce to 0 before you go to the you know midpoint and similarly here the H value will go to 0 even before you go to the center line of this conductor. So, this in other words will represent two semi-infinite cases. We also know from the literature that for semi-infinite case the edulars per unit area is given by s square upon 2 sigma delta. Sigma is conductivity and delta is skin depth right. Since the considered plate can be analyzed in terms of two semi-infinite house spaces. So, that is the reason that you can see here as your thickness 2B this is the total thickness 2B as thickness 2B exceeds and becomes much greater than skin depth this normalized eddy current loss which is p upon s square upon 2 sigma delta which is for semi-infinite case it will approach the value of 2. Now, let us go further and have some more complication. How do we model transformer co-rejoins particularly for you know frequency response analysis when we want to actually you know want to find out transformer impedance as a function of frequency. Then we have to do time harmonic analysis with you know conductivity of the core considered and corresponding losses considered and hence the permeability will be complex permeability this also I have explained earlier that complex permeability will represent lossy magnetic material where in imaginary part of the complex permeability represents losses right. So, here again what we do we want to find out effective complex permeability of this core structure. Now, what is this cross section? This is a typical joint region of a leg or limb and yolk. So, vertical portion of the core and horizontal portion of the core of a transformer and if you take a cross section like this AA then you will get this kind of you know problem domain and these are the air gaps here. Here this is the air gap is in this region which basically in this cross section those air gaps will appear staggered like this. So, that mechanically hold this structure becomes stable air gap is never allowed to be along the same you know along the same line vertical line here. So, now here what we do is we again you know impose Hx here on both this you know horizontal boundaries top and bottom, but we do not actually you know have any you know information about HY because as you can see here because of this air gaps and all that you have these you know flux counters you know hitting this top and bottom you know surfaces or you know lines in 2D approximation at some angle. So, HY is varying all along these 2 you know lines. So, HY is unknown what you have imposed is only Hx, x component. So, again you know as we have seen this earlier these Hx and HY expressions we have seen in L28. So, those were this. So, Hx is this and we are imposing this HY is this and it is not 0 and it is unknown. So, now this corresponding to this boundary condition Hx the matrix the right hand side matrix Bb of the final linear system of equation. So, this will contribute to the B matrix overall B matrix on the right hand side of our final linear system of equations. So, this Bbe of I this I stands for you know node number B is the right hand side matrix small b is the contribution of boundary condition to this right hand side B matrix and this is related to element E under consideration and the corresponding node I. So, there are 3 nodes. So, one of the nodes is being considered and now this comes out of you know when we write weighted residual statement in Galerkin's method for you know each node then for ith node we will get Bbe as this again for this you refer slides 8 and 9 of this L28. And now here An hat will be simply Ay hat because for this line and the corresponding surface in 3D the unit normal will be in y direction we are considering these as xy plane. So, this is x and this is y direction vertical is y direction. So, now actually if you substitute this here instead of this we substitute Ay hat dx then this dot product will be 0 only this dot product will be non-zero and now in place of this we substitute hx 1 over mu 0 into this is hx which is the imposed boundary condition. So, that is hx and integral n i will remain because here it is 1 over mu 0 dabaiz by daba y. So, 1 over mu 0 dabaiz by daba y is equal to hx. So, that hx comes here and then n i e dx and over the edge. Now again as we saw in this L28 this the contribution to this for any node will come from the corresponding you know contour integral over the considered element. So, remember for each node of the element under consideration we will get this kind of statement corresponding to the you know boundary condition. And now here the contributions of this inner segments will get cancelled as explained in L28 and contribution will remain only for this edge 1, 2 and this edge is this edge 1, 2 for element 1 refer L28 for more details. So, now this and we also saw in this lecture L28 how to evaluate this. So, then you know we can calculate contributions of all such boundary edges by using this procedure for the imposed boundary condition. Because here there will be number of elements here triangular elements is it not and for each of those you know nodes of individual elements the corresponding edges will contribute to this b capital B small b matrix. And all those contributions can be calculated by this procedure and then the overall finally global matrix can be calculated accordingly. And then further calculations can be made you can you know get the value of you know b you can first you will get the value of a from this analysis from a magnetic vector potential you can calculate b. And then from b you can find out the average value of b for this whole you know domain under consideration. And then that average value of b divided by this imposed Hx will give you complex permeability for more details you can see this reference. Now let us go to the next topic which is very important how do we calculate eddy current losses in a coil. Again here as an example we are taking a high voltage winding of a power transformer. Now here just a three-dimensional view is shown here not everything is shown only the core leg is shown that is the vertical part of the core. And surrounding it we have LB and HV windings. Now the leakage field at an instant of time is vertically will be vertically upwards and here we are not showing the fringing field at the end we are just showing the axial field but of course there will be radial flux at the ends. So now our objective is to calculate eddy current loss of high voltage winding which is subjected to this alternating magnetic field B which is leakage field. And then we have already you know seen in lecture 10 the eddy loss formulae we will use those here. So before that let us understand this geometry little bit more. So now this high voltage winding is you know given here now of course this is not to the scale. So this point is this and this point is this. So this is the total radial depth of this HV winding. Now high voltage winding will have number of conductors placed like this this is called as one disk this is one disk this is second disk and this is say nth disk. Now individual conductor as shown here is a rectangular conductor and dimension chosen are this. Now this is the actual leakage field plot for a transformer and now you can see here there is a axial flux in the middle portion of the winding and at the ends you have fringing and predominant radial flux. And then we have seen in lecture 10 eddy loss formula for a thin conductor of thickness T which is less than or equal to skin depth and excited by B0 on both sides the thickness T and B0 is there like this. Then P eddy is given by this formula rho is the resistivity B0 is the peak value of flux density here T is the thickness of the conductor which is perpendicular to the direction of flux. Now here the flux is in this vertical direction. So T will be this perpendicular direction to this. So that is T and that is what is written here P eddy axial and remember this is loss per unit volume. So P eddy axial will be omega square omega is 2 pi f square then thickness square by vertical component vertical component of B upon 24 times resistivity. For the radial flux now at the ends particularly you have the radial flux component. So for the radial flux component the perpendicular dimension because flux is now like this the perpendicular dimension will be this W the width. So that is why here it is omega square W square Bx square upon 24 times rho. So we can individually calculate losses due to axial and radial fields and then basically sum them up for a winding under consideration. So if we have understood this let us go further. Now let us go again go back to the problem domain this is the whole HEA winding and now what is the title written here eddy current loss using magnetostatic FEM simulation. Now this is you know something it looks odd that we are finding eddy current losses. So that means it is a time varying case but we are you know using magnetostatic FEM solution for that. Now this is allowed because seeing the thickness of the conductor and dimensions overall dimensions of the conductors are such that no those dimensions this dimension and this dimension they are comparable or less than the skin depth you can consider them as you know electrically thin conductors and then the eddy currents induced in this individual conductors are not significant and it can be assumed that those eddy currents do not influence the leakage field much. So then in that case if eddy current reaction effect can be you know neglected then we can effectively use magnetostatic FEM simulation and then calculate you know values of Bx and By at the conductor centers and then you use this to formulate for calculating eddy current loss due to axial and radial fields for each conductor. So effectively what we are doing we are actually calculating field values using static simulation and then we are using eddy current theory because this is a classical eddy current theory this you know these expressions are derived starting from Maxwell's equations and solving diffusion equation. So we are using static FEM simulation and then using classical eddy current theory to calculate the eddy current losses of each conductor analytically and then we will sum of sum up all the losses for the entire winding. So now how do we do that? So we need to find out each this conductor and the corresponding representative point so what we will assume for the entire conductor we will assume that the flux density is that at the centroid. So then we have to first find out in which element this center point lies and then using that information of element we get the corresponding Bx and By of that element and then use it further in the eddy current loss calculation. How do we find out the you know element corresponding to this you know point or where in which element this point lies we have seen it earlier in one of the lectures. Now going further how do we find out you know how do we know that our FEM obtained eddy current losses are you know okay and they are not you know way out because of some you know mistake done in you know defining some quantities. See it is always you know in any kind of FE analysis always you know you should have some verification either you should verify it with approximated analytical solution or if you have some you know verification problem in the literature and all the geometrical and electrical parameters are given then you can first solve that problem get confidence and then you know you do analysis for your problem. So here in this case I am using analytical formula to verify my FE analysis. So now in analytical formulation obviously we have to do some approximation. So what is the approximation we are doing? We are actually you know analytically integrating you know the effect what this entire you know radial depth of the high voltage winding. Now we have seen this in one of the earlier lecture. This is the ampere turn diagram associated with the leakage field in a transformer with this LV and HUE windings and this is the gap between the LV and HUE windings. So in the gap region the flux is uniform and in LV and HUE windings as you go away from the you know gap towards you know the other ends then the leakage field reduces. So if we analytically integrate this flux then what we will get? We will get integral b0 square mean as b gap gp stands for gap b gap square by 3. Now this you know derivation is given here in this reference. So basically what we are doing? We are analytically integrating because why we are doing this? Because we need to integrate over the entire winding analytically. So in our formula for ideal loss we have omega square t square b square. So when we want to integrate over the whole winding you basically find out the mean value of you know b0 square and use that mean value which is representative to you for whole this winding. So then the integration becomes you know straightforward. In fact the integration does not really remain integration because we have already taken into account you know that integration by using you know the mean value of b0 square. And remember bgp b gap is given by root 2 mu 0 ni by hw. hw is the height of the winding and this because there is a root 2 factor here i is the rms current. So always remember in this formula always these are the peak values. So that is why root 2 into the rms value. So we are considered peak value and s is the cross section area of each term. So let us understand this formula now. Again what we are doing? In analytical formulation we are considering only axial field. So axial field when we consider in this way and this is the corresponding thickness which is perpendicular dimension to this field and that is why this t appears here. So t square this we had you know seen it earlier omega square t square b square upon 24 rho. But here since we are taking b0 square mean as this. So that is why b square gets replaced by this bgp square by 3 here. And now this is you know for you know if there are n turns then we have to multiply it by n obviously because this we are calculating for each turn multiply it with number of turns multiply it with 3 for 3 phases. So then this will give you ad law square unit volume for all turns for the 3 phases. So now we have to multiply then by the volume. So volume is nothing but cross section area of each turn into pi into mean diameter. Now again we are taking mean diameter as the diameter as mean diameter. So that again simplifies the calculation. So this entire expression from here to here gives you ad loss for you know 1 turn. That ad loss for 1 turn is multiplied by n turns to give you for n turns and then it is multiplied by 3 to give you you know losses for all 3 phases of the high voltage winding. So now the final verification. Now see in the axial formulation we have considered only axial field right in the analytical formulation. In the analytical formulation we have considered only axial field. So in the FEM also we should have axial field only so that then the comparison is you know valid. So then what we do in case of FEM simulation this is we basically you know take the horizontal boundaries at top and bottom very close to these low voltage and high voltage windings and impose homogeneous Neumann condition right. So the derivative of you know potential with respect to the normal is 0. So effectively you know you will have you know all the flux lines you know almost going straight and impinging on both the horizontal boundaries normally. So now the since we are we have constrained axial the flux to remain axial throughout the this geometry in high voltage winding also the flux is axial. So then the ad loss calculated by FE analysis we can compare with the analytical formulation value. So before doing that let us understand this formula for EAD in FE analysis. Now here we are not analytically integrating or you know calculating the average value and you know doing what we did for approximate analytical formulation. We are going each conductor wise at every conductor as I explained earlier we are finding BX and BY and then correspondingly here we are you know taking that. Now here you know we need to worry about only BY because BX B horizontal or X direction is almost 0. So B is entirely in Y direction vertical direction. So only BY will come here and corresponding thickness because thickness of the conductor will be the perpendicular dimension to BY. So omega square T square B square upon 24 times rho is the ad loss per unit volume of that conductor into the conductor area because S is the cross sectional area of each conductor. Some these over all conductors and then so this is ad loss per unit length because area is now already considered. So now this becomes ad loss per unit length and now that you multiply here by point to mean diameter of the entire winding. So now this becomes ad loss for ad loss for the entire winding and that is for one phase of the HV winding and then multiply by 3 for all the three phases of the high voltage winding. So this is how you calculate the PAD. For more details of all these you know theory and corresponding you know explanation you can refer this reference and get more insight into the whole thing. So then now here and of course here I already explained you B is the peak magnetic flux density at the center of each conductor and then we calculate ad loss per unit volume for each conductor, sum it then multiply by area, sum it over all conductors, multiply by pi into mean diameter. So this gives you total ad loss for the entire one phase of the HV winding and then for three phases. Now here you can now see analytical and FEM results are compared and now you can see they are quite close. So this gives us confidence that finite element analysis procedure that we have adopted is okay and now for any complication either in geometry or ampere turn distribution you know we can use this procedure because in one of the very first lectures I told you one of the major advantages of FEM is it is more or less independent of you know geometrical complications. The procedure is same and that gives you know the distinct advantage to this numerical technique finite element method. So this is how you know you can validate your finite element analysis procedure by taking a simplified you know flux distribution by doing some approximations and comparing the values and get confidence and then go on to solve more complex problems. So we will stop here and start a new topic in the next lecture. Thank you.