 So welcome to 28 lecture, we will continue our discussion on time harmonic problem and the solution of diffusion equation. Now first application we are going to see 3, 4 applications of this time harmonic problem. So first we will see calculation of ready current losses which is very common in many you know applications and impact in electrical machines and equipment we want to minimize ready current, but in some other applications like heating applications you need ready current right. And then so calculation of ready current is important in electromagnetic devices. Now induced electric field intensity in absence of free charges is E is equal to minus daba A by daba T, the complete expression is minus del V minus daba A by daba T is it is not, but minus del V is equal to 0 because there is no charge accumulation. So charges are moving, so it is basically only currents are there either free current or induced current. You understand what is meant by you know there is no there are no accumulation of charges means no nothing like you know like capacitor for example, because capacitor like charges get accumulated, moment charges get accumulated you have del minus del V term right. But if you are considering a case in where only currents are flowing either source current or induced current, there is no accumulation anywhere charge accumulation. So del minus del V term is 0 right. So E is equal to minus daba A by daba T, so that is equal to minus j omega A d by daba T is converted to j omega in frequency domain. So eddy current due to induced electric field will be j A d is equal to sigma E that is minus sigma dA by dt this we have seen earlier in fact in the diffusion equation it appears is it not. So minus j omega sigma A is j A d right. Now eddy current loss is V into I is it not is the loss the corresponding j into E will be loss per unit volume, because E is volts per meter and j is ampere per meter square. So 1 upon meter and 1 upon meter square, so that is why j into A E will be loss per unit volume right. And why here j E complex conjugate, because these are phases so always when we take you know product in our electrical calculation we always take one of the quantities as complex conjugate. If you substitute E as j upon sigma j A d upon sigma we have j A d square by sigma d x d y d z right this is per unit volume we have seen that. Now we will take a very standard problem a bar plate problem which is a very representative problem for eddy current what we have is a current bar carrying suppose high current in the vicinity there is some structural plate mild steel this situation appears in many electrical equipment and machines is it not. Some current carrying conductor is close to a structural element conducting element and eddy currents are induced. So we are now interested in finding eddy current loss in this plate due to this current right. So this current is going to have magnetic field alternating magnetic field around it that alternating magnetic field is going to induce eddy current in this plate and that since the there is a conductivity associated with this we are interested in finding the losses. If you are interested further in calculation of temperature rise that also can be done we can actually these losses can be fed to a thermal FEM core or commercial software. There we have to give the loss per unit volume individual finite element as the input for the thermal model and then you will get the temperature rise. So that also is possible but in this course we are mostly restricting ourselves to electromagnetic field calculation and the corresponding performance parameter. So we will not get into the thermal analysis. Now here one more thing that we want to remember we are not interested in eddy current loss that is happening in this bar. So we will define sigma equal to 0 for this that is the you know while doing FE analysis one has to be clear what is the focus and correspondingly reduce the problem complexity right. Here we are interested in eddy current in this plate only. So we need not necessarily put sigma here otherwise it will you know you will get losses in this bar also. If you are interested in losses in bar as well then of course you have to give conductivity for this copper bar suppose this copper then you have to define conductivity there and then you have to worry about the skin effect and the corresponding size of element in this bar. So that actually is the complexity if you want to avoid that complexity you do not define conductivity for this bar you are then treating that bar as just the source which is producing or inducing eddy currents in this plate. But if you are interested in skin effect in the bar itself and effective AC resistance of bar this current bar then of course you have to define conductivity and then take care of skin effect and the corresponding mesh size in the bar. So that is the reason now say again we are not giving we are going to only study part of that code which is new right rest of the things will be similar. So what we are doing we are set in the code we will set up a loop and then one will go through each element for each element and if the material is 2 that is this plate mild steel plate then we will define mu as mu r times mu 0 relative permeability of mild steel into mu naught and sigma will be 1 upon this because sigma of mild steel is taken as 7 into 10 raised to 6 siemens per meter so 7 is nothing but 1 over this right and then if it is 1 and 3 that means either this is the this air part right all this air part or bar is 1 and air is 3 then it is mu naught here anyway it is mu naught we have seen earlier copper also is mu naught right mu r is close to 1 and sigma is 0 air anyway it is 0 I just explained we are defining sigma 0 for bar also copper bar because we are not interested in eddy currents in this bar. So now here this is the element level coefficient matrix this is identical is it not I hope you remember this expression the element level coefficient matrix is form just 3 by 3 9 elements get form only by this one you know a statement in the code right and then this B is nothing but j delta by 3 ok what is mu is only this D so now this D we have seen here sigma delta by 12 and this matrix right sigma delta by 12 so sigma delta by 12 and this matrix so Syram had explained in one of his you know tutorials on FEM matrix can be written in Sylab a small matrix can be written just by this first row second row and third row separated by this semicolon so this is the only new line so when you are actually solving this diffusion problem right as compared to Poisson so you know you are this will be only this will be the only line which will be new so see how FEM coding is straightforward and simple if you understand simple Poisson's code which was given entire code was discussed and explained to you there you have to just add this right and with corresponding definition of sigma and all that now here we have found at the element level now global again same thing we first take using T matrix we take the global node numbers which will be second to fourth entries because T you remember is 4 by number of elements right 4 by number of elements first entry is the sub domain number and second third fourth entry are the global node numbers of that element right so by this command for one particular element under consideration we are basically in nodes we are taking three global node number of that element right and then we basically set up this again loop this we are known by using this command just one command you are forming global coefficient matrix C capital C and then by this command will form the corresponding global D matrix so this is the BJ that means source contribution to B and later after getting the final set of linear equation you have to impose the boundary condition if outermost here in this case is outermost box A will be defined as 0 so all those you know entries and then that also we have seen we have to find out those edges find out those edges and then the corresponding node numbers there start a node and end node of each edge and then all those nodes you define A equal to 0 so that procedure remains same then what we are doing we have got by doing all this forming this and imposing boundary condition then inverting the matrix finally we would have got the solution right so I am assuming now that we have written a code and we have got the solution now this is the post processing part so for again going from element to element if the element if the sub domain number of that element which will be the first entry is it not of T matrix if that is 2 that means that what is 2? 2 is the this plate where we are interested in eddy current so that means if that element is inside the plate right then you take its global node numbers take the corresponding nodes X and Y nodes coordinates X Y and Y coordinates of those 3 global node numbers of that element which is inside the plate in this part variable that means potential you get those magnetic vector potential solution that you obtained is it not 3 values and then take the average that means we are taking the average potential at the centroid right so and then the loss we are calculating as loss is equal to sigma delta so this is the formula sigma E square dx dy dz now dz here is 1 because the 2 dimensional approximation so we will be calculating per meter depth losses right that is why dz is 1 so basically it is E magnitude is omega times A right so omega so that means omega A square and we are calculating at the element level so omega A square and this A for that element we are taking the average value right so that is omega that means this is nothing but whole this thing is magnitude of E square into sigma into dx dy is delta the area of the element this is omega and it is 2 pi f right this command will add all the losses corresponding to the elements all the elements in that plate and then display the loss is it clear now this is the solution now for mild steel and aluminium now one case is the original case that we had considered this is a mild steel plate and is the current bar and then you have got a field now can you see clearly the skin effect and in case of aluminium the skin depth is more here it is something like 2.69 mm for mild steel here the you know it is more than 10 mm for aluminium so that is why you can see can you see one count in fact is diffusing out of the aluminium is it not slightly diffusing out so the diffusion is more because skin depth is more right and the corresponding losses after you know calculating fields and all that and by following the procedure that we just now saw the loss obtained for mild steel magnetic steel is 367 and for aluminium it is 76 right and why it is low because it is sufficiently thick aluminium is used if you use a very thin aluminium this loss will be quite high so whenever you use aluminium or copper the thickness should be sufficiently high it should be at least equal to you know depth of penetration of skin depth if not more to reduce the losses in the aluminium or copper to a reasonably low value otherwise there will be excessive losses in aluminium if it is very thin aluminium plate is used. So before going further you know we will see what happens when aluminium plate thickness is very small I was mentioning in the previous you know discussion that aluminium sheet thickness should be sufficiently you know they are comparable to at least skin depth right now here you can see this graph when the aluminium this aluminium plate thickness is like less than 5 mm then you know the loss is quite high right so that that is the reason that you need to have sufficient thickness of the electromagnetic shield now this aluminium and copper they are called as electromagnetic shield because they work as shield by virtue of the eddy current induced in them and the repelling action there are other shields which are called as magnetic shield right for example if suppose this is a structural component and you want to shield it you could just put a some magnetic shield that is high permeable magnetic material just you know on this plate on this side so it will take all the flux through it and will not allow to go into the structural plate which is to be shielded so that is type of magnetic shielding right so here it is when aluminium and copper are used they are called as electromagnetic shield right because they work on the principle of eddy current and the corresponding repelling effect of the shield right then let us go further now we will talk of non-homogeneous Neumann boundary condition so this little advanced topic but still I think it is important to understand because we have always been talking of homogeneous Neumann condition is it not? Dirichlet conditions are two types one is homogeneous non-homogeneous when voltage is 0 it is homogeneous Dirichlet when voltage is non-zero it is non-homogeneous that is obvious and homogeneous Neumann condition also we have seen in case of parallel plate capacitor when we neglected the fringing effectively on the vertical plate vertical sides of that parallel plate configuration we discussed earlier that we have homogeneous Neumann condition then one may be wondering when would this non-homogeneous boundary conditions would be helpful right so let us see that application now so now let us see you know let us start with first B is equal to del cross A and that is why H is 1 upon mu naught del cross A in free space right now if you expand this del cross A with A only having z component then we will have only two components when we expand the curl is it not there will be only two components this we have seen earlier so now this is the geometry and suppose you want to impose non-homogeneous Neumann boundary conditions on the top and bottom why do you want to do that we will see it later how that is useful that is equivalent to you know imposing h t magnetic field intensity on the top and bottom plate right h t constant h t on this you know line and constant h B on this line or surface in 3D right always remember this is a 2D approximation actually it is 3D so this when we are saying this top and bottom they are actually plate in 3D right or surfaces so now if we say h is equal to h t A x on the top plate that is the boundary condition that we are imposing then effectively you know h t A x that means in this case this has to be 0 that means daba A z by daba x we are we have to impose it at 0 is it not effectively then only h will become h t A x at so this has to be 0 then only this this A y component will be 0 and then h you can impose as just h t A x in x direction so then that effectively means that you are imposing this equal to 0 on the top boundary right and effectively you are imposing daba A z by daba A x as mu 0 times h t on the on the top surface so effectively this is non-homogeneous boundary condition you are imposing because daba A z by daba y is equal to mu 0 times h t right and this is this is a non-homogeneous and this is daba A z by daba y so that is a normal component here right okay so similarly h is equal to h B A x hat and then again you know it is the same thing only thing here this daba A z by daba y will be different than this because if h t and h B are different these derivatives will be different as value right so on the bottom boundary so now the element elemental entries in this matrix capital B small b for a nodar now we are discussing Galerkin method here because you remember this equation this expression that we had you know got when we saw Galerkin equation and then we know we said that this is the we in fact invoke first divergence definition and then the divergence theorem and then we got in fact this in that derivation for you know Galerkin method under weighted residual technique as a as a measured weighted residual technique so these are also told here slide 3 of lecture 22 okay so we had got it here so in the process of this Galerkin method we get this now we have to all other things remain same because now we are imposing the boundary conditions is it not so these boundary conditions how they reflect in our final set of matrix equation that only we have to see is it not so now let us see what happens to this now we are we are taking so this for top edge for example so 1 upon mu 0 daba A z by daba y is h t and daba A z by daba x is equal to 0 and a n will be a n hat will be a y hat normal to this will be a y and d tau will be d x because this counter will be d x so that gives so this expression becomes this whole thing becomes now this you know this since a a n hat is a y so only this dot product will be non-zero this dot product will be 0 so only one term will remain this term and that is what one term a y hat dot a y hat right now this close integral now we have to evaluate this now let us take this this geometry now this top surface or top boundary there may be number of you know elements here like this right so I am just drawing on this top counter suppose there are such 1 2 3 3 elements now we have to evaluate for each of this you know elements and the corresponding nodes what would this boundary condition mean now we are we are also seen in case of wetted residual technique that we form we write wetted residual statement for each node you remember we in wetted residual method we go for each node if if an in linear formulation if an element has 3 if the element has 3 nodes then there will be 3 wetted residual statement for each node is it not so this when we do that for each node we will get this one expression corresponding to the boundary and then here it will be suppose it is node 1 then this will be n 1 and n 1 right so now in that case we are evaluating for node 1 this expression in the process of Galerkin's technique and the geometry is this with 3 elements so node 1 is here right this is the node 1 so now here this daba az by daba y into mu 1 upon mu 0 is ht is it not so this when you substitute it there for this you know for node 1 so node 1 is part of now this element element number 1 right and we had already seen in that wetted residual technique this common edges inside common edges they do not contribute to the you know that matrix boundary matrix is it not because this a n's are oppositely directed and this h is common so the corresponding function is common and this a n's being opposite they cancel so that is why the line integral over edges 1 4 and 4 2 1 4 4 2 or 2 5 or 3 5 so that is why the line integral over all these contributions will get cancelled only the contributions on the boundary will remain right so if we have understood this then we are just substituting so what we have to evaluate we have to evaluate only on the outermost boundary and there we are just taking now these 2 segments 1 2 and 2 3 right so now h the close integral for node 1 in element 1 will be ht node 2 to node 1 we are going in you know this direction anticlockwise direction so node 2 to node 1 n 1 1 n 1 of 1 right and 1 upon mu 0 daba az by daba y is nothing but ht and ht is constant boundary the value of ht on the boundary is you are specifying it is constant so it comes out of the integral right now this n 1 1 these are our standard expression for n 1 shape function is it not and then we are just integrating with respect to x so it will it will reduce to this into x plus y 2 minus y 3 x square by 2 into x 3 minus x 2 y x integration with respect to x right and then we you know put the limits x 2 because this is from x 2 to x 1 because node 2 to node 1 so x 2 to x 1 so we just in place of x we substitute x 1 minus x 2 here and here y in place of y you can either substitute y 1 or y 2 it they are same because y is constant is it not so since we are talking of node 1 we have put here y 1 so now this is what this whole integral when evaluated will reduce to this right and then that will effectively go into that which matrix corresponding to the entry of b corresponding to node 1 in the b matrix will basically be resulting from this expression similarly now for node 2 now see node 2 is common for elements 1 and elementary node 2 is common for element 1 and elementary so now this expression will get evaluated for node 2 twice 1 as part of element 1 as part of element 3 element 2 there is a it is a part but element 2 there will not be any contribution to the boundary because none of these edges 3 edges of element 2 are on the boundary clear it is a good question but you understood the answer because this element 2 all these 3 sides of this element 2 they are basically common to some you know 2 elements is it not and again by this logic the contributions will get cancelled so for node 2 only the contributions will be from these 2 so contributions from element 1 with respect to element 1 and element 3 so that is what it is element 1 element 3 into 1 into 2 right and h t remains same similarly for node 3 it will be node 3 will be you know again from 3 to 2 we will evaluate for corresponding to element 3 element 3 integral node 3 to node 2 h t integral n 3 3 dx and similar to node 2 this element this node 1 and node 3 they may be common to some other element is it not there will be some element like this here or some element like this here so again there will be additional contributions for node 3 from this segment here and for node 1 from this segment which will be part of some other element clear so that is how you basically evaluate and then this we have done it only for the top edge and that too only for only for you know I have shown it for 3 nodes here there may be more number of nodes although for all those nodes these expressions will have to be evaluated and they will go into the corresponding b capital B small v matrix which is the right hand side matrix corresponding to the boundary conditions right and remember this b is only for the boundary conditions there will be some b if there is a j existing somewhere in the domain then there will be b j right and if it is a linear formulation if that j exist it will go as j delta by 3 at all the respective those 3 nodes and that will get upended to the b matrix right so I think we will stop here and then we will continue the rest discussion in the subsequent lecture.