 So, welcome to lecture 35. In the previous lecture, we saw coupled voltage fed coil as an example of coupled circuit field computation. So, we will now see the corresponding example here and code. And as we have been doing earlier also, we will now describe only the new part of the code, what is new here and rest coding will be similar to what we have seen earlier in so many lectures. So, now for voltage fed coil, let us take one example of a transformer. This is a transformer core window with two coils LV and HV. And now we want to, now only the window is modeled. You have to remember that there will be LV on this, LV and HV on this side also. The corresponding other half will be there. So, now what are the new things in this code? First of all, earlier whenever we did for example, leakage reactance calculation in transformer, we were knowing current in both the winding. And in fact, we made ensure that n 1 i 1 is equal to minus n 2 i 2. So, that we have perfect balance of ampere turn. But here in this case, current is unknown. Current is unknown here because here in general, the current that is fed, it is the suppose primary winding of transformer is fed by this voltage source. Voltage is known, current is not known. So, that is why primary winding current is not known. As well as secondary winding which is short circuited, that current also is not known. So, here that means there are two current variables which are unknown. So, now that is why here what we are doing is, we are actually seeing this now 3 and 2 are the corresponding, 3 is the primary winding, 2 is the say secondary winding or to be more general, 3 is low voltage winding because we do not know which is primary and which is secondary, secondary. You can always either excite LV or HV. So, let us you know go by winding wise. So, 3 is LV, 2 is HV. So, if it is LV winding because 3 is LV, then the number of turns TDLV, turn density, D stands for density. So, turn density, earlier we were calling this as ampere turn density because current was known. So, ampere turns we were dividing by area. Here current is not known. So, we are only taking turn density. That is why number of turns of LV divided by the cross sectional area of LV. So, it is just turn density. So, turns divided by area because current is unknown. Similarly, when it is HV winding because for all HV winding elements the corresponding first entry in the T matrix for all those elements the number will be 2 because HV is 2. So, turn density for HV will be number of HV turns divided by cross sectional area of HV winding. And then for all the elements in LV and HV winding we will have that density defined when we execute this for loop. Then what we do? This element level coefficient matrix exactly it is identical. This statement we have seen number of times earlier. So, there is no change here. What is the change? This B matrices at the element level B1 and B2. B1 is for LV, B2 is for HV. Those get changed. And again here what is the change? Instead of ampere turn density we have turn density of LV and turn density of HV and into delta by 3 because that we have seen earlier. Why this delta by 3 come? Because of this number of turns this is this is turn density turn density into delta by 3. So, that is what this B matrix is. So, turn density N by S into delta by 3. So, turn density into area of the element by 3. Similarly for the HV winding. So, here question asked by one student is we are taking this as minus. So, effectively because we know ampere turns are equal and opposite. So, here this turn density is taken as minus. So, the corresponding current will be positive, induced current. So, then this minus with that current of the HV will balance the LV ampere turn. So, now let us form global level matrices. So, now this B will be simply augmentation of B1 and B2 together. So, this is global B matrix and what are these capital B1 and B2? They will be formed by basically combine B1 and B2 at the element level as we have seen earlier so many times. The K matrix capital K matrix is this. So, it consists of capital C which is basically the corresponding matrix here which is this C matrix because of these terms C minus B and then on the in the second row you have square root of minus 1 is this omega is this B dash because G and B are same. So, B dash is nothing but G dash and then R external and J omega again J omega L external and then you have the right hand side matrix will be now remember this has dimensions of nodes plus 2 because nodes is the, nodes is the dimension for this. Suppose there are say 1000 nodes. So, N nodes will be 1000 and plus there will be additional two entries one corresponding to the LV terminal voltage and second corresponding to the HV terminal voltage. So, that is why this the dimension of this US will be number of nodes plus 2 by 1. So, nodes will be this whole matrix 0 what is shown here will be the corresponding matrix nodes and then this U matrix is this U1 and U2. So, U basically has two entries U1 and U2. Now, here what we are doing LV inner winding is short circuited LV is short circuited. So, we are actually short circuiting LV winding in by using this command the voltage across LV winding is made 0 and voltage across HV winding is 100 volt that is why here HV winding is supplied HV winding is supplied and we are just taking 100 volts. So, this basically completes our couple circuit field formulation with definition of voltage that is applied to one of the winding and one winding is short circuited and you have two currents in currents in LV and HV winding I1 and I2 they are unknown. So, here I also will be I1 and I2 and A will be again our normal A suppose if there are 1000 nodes in the field model this will be 1000 by 1. If there are 1000 nodes in the field model and there are two winding. So, this again will be 1002 by 1. Then you know we what we do? So, we are calling this as AI matrix combination of this matrix and this matrix we are calling this as AI this whole matrix is K which contains this our normal global coefficient matrix of the field model. And then and you are this matrix right hand side matrix you are calling it as us where it has two terms corresponding to the two windings considered and this is usual matrix corresponding to all the nodes in the field model. So, now this AI matrix which is the unknown all these variables are unknown. So, you have AI will be inverse of whole K into us right and then you get then A as the first node N nodes 1 to N nodes and underscore nodes will give you A and then I1 and I2 will be the corresponding entries at the position N nodes plus 1 N underscore nodes plus 2 right. So, then by after having got the solution by this command by these three commands we will get in A I1 and A2 the field solution and the circuit unknown quantities I1 and I2. So, having seen this theory now you will see examples of implementation of this coupled circuit field computation with binding applied with a voltage source. So, now let us see a case where you know you have inner winding is short circuited that is what we had in the formulation. So, inner winding is LV which is short circuited right. So, this inner winding is LV which is short circuited and HV winding is supplied with the voltage. Now, the inner winding if you apply KVL to the inner winding then it will be the corresponding IR plus induced voltage will be equal to 0 right. So, this is the induced voltage in the LV winding it is D by DT of N psi right N psi is the total flux linkage. So, now we are neglecting IR draw because anyway we had neglected conductivity we did not define conductivity in the FE analysis also right. So, we are neglecting the conductivity. So, we are considering the winding as lossless right. So, that is why corresponding IR draw is 0. So, what it means is the total induced voltage will be equated to will be equated to 0 when we short circuited the inner winding. So, what it means is the total flux linkage associated with the winding is 0. So, in this representative figure and this is for the three phase case of course, we are modeling here single phase, but this is for three phase case. Here what we will get is the flux will actually flow the leakage flux will complete the path like this and in this case it will complete the path like this in case of three phase excitation three phase transformer. So, what will happen is now you can see this LV winding the total flux linkage is 0 you can see that there is not a single flux come to which is you know being enclosed by this inner winding. So, the total flux link by the inner winding is 0 which should be the case because total flux linkage is 0 rate of change of flux linkage is 0. So, flux linkage is 0 same thing you are observing here in the FE plot also. Can you see here that this is the and of course, here you have to remember we had modeled only this much part it was a single phase case and we had modeled only this, but then we basically got this solution also for the other half by just sort of taking mirror image of the solution and how do you obtain the mirror image? Take any point here suppose the magnetic vector potential is plus a here the corresponding mirror point here the value will be minus a because you know currents are opposite if here the current is dot corresponding current will be cross in the corresponding winding right. So, just by you know taking negative of all the magnetic potential in this half we can plot field solution for the second half and that is how we just plotted this total field distribution and this line is just the symmetry line is not the flux line that you should note. So, now as we have seen here can you in this figure can you see that all the flux which is there in the winding the L V and H V and in the gap that is all returning to the core like this just neglect this single contour because the value is very close to 0. So, that that is the reason that you have you see here that all the flux is returning through the core all the flux is returning to the core. Now, it is a three phase case here representative when this is enclosed in a tank for this one and this is the flux will return to the air outside air. Now, we will see the other case outer winding short circuited right. So, for outer winding short circuited what we have to do we have to make this 0 and this corresponding voltage we have to excite the L V winding with corresponding voltage which will typically go in the turns ratio the turns ratio is say 10 then if this is the you know short circuit voltage for H V winding the short circuit voltage for the L V winding will be 10 volts and this will be 0 right. So, we have to only make that these two changes in the core and then we will get this result outer winding short circuited. So, in case of outer winding again the same same equation will come with resistance drop neglected the total induced voltage will be equal to 0 for the outer H V winding short circuited case and hence the total flux linkage associated with H V winding is 0 as you can see here. So, this is going up this is going down this is going down this is going up. So, the total flux linkage between these two halves of H V winding if you add up all these arrows going up and down that will lead to 0. So, the total flux linkage of H V is 0 and that is what we are seeing here also that all the flux that is there in H V gap and L V winding that is returning to the core again neglect this one single contour anywhere the value is close to 0. So, then we have seen with these two examples how do we actually simulate coils excited by voltages and find unknown current because by doing this what we would have got we not only got the flux plot, but we also got the currents in the winding and then we will easily calculate the impedance actual impedance because V by I the excited winding for you this is outer winding is short circuited. So, inner winding is supplied in this case. So, the supplied voltage by the corresponding inner winding current will give you the total leakage impedance of that winding set of winding and for the transformer because leakage impedance is for the transformer. So, the inner winding supplied voltage divided by the corresponding current which is I 1 will give the leakage impedance of that transformer. Now, we will see another example wherein we have not only you know currents unknown, but then there are you know eddy currents also in the system. In this case in the same case wherein you are exciting one of the winding L V and H V of one of the L V and H V winding and is the tank right. So, this tank there are eddy currents induced in this tank given due to this leakage field. So, these alternating leakage field will basically be incident on this tank which is made up of mild steel material it is conductive. So, under short circuit condition we want to estimate what is the loss in this tank. So, moment you want to find loss in the tank then there are induced current. So, then we have to solve diffusion equation. Similarly, we solved only Poisson's equation is it not we solved only this was Poisson's equation only this was Poisson's equation there was no eddy current because in fact conductive conductivity will all together be neglected is it not. But then if we have to now take into account eddy currents induced in any part of your geometry in the field model then we have to have the corresponding diffusion term which is you know mu sigma dA by dt is it not that term which represent the induced eddy current is also we have seen diffusion equation. So, this that diffusion equation will be this term plus the diffusion term minus mu sigma dA by dA by t term right. So, here then what you have to do you have to basically couple the diffusion equation with circuit equation. Previously we only coupled Poisson's equation with circuit equation. So, now we have to couple the corresponding set of equations obtained to FEM formulation after considering the diffusion term that whole set of equation have to be coupled with the corresponding circuit equation right. So, if you do that then we will also get the eddy currents induced in this tank part right. And remember now we have modeled this external part also because if your thickness of this tank is small is thin the flux will go out and that is the reason that you have to model this. If the thickness of your tank material is quite high much more than the skin depth then you can actually terminate your boundary at the outer thickness of the tank itself right. So, that then you need not model this, but if you are if you are interested for example, in this analysis to estimate the effects of regression in this tank thickness or material type on the corresponding eddy current losses and also the impedance of the transformer. Then you have to actually model the outside part also and then impose a equal to 0 on this outside outermost boundary and then you can simulate any flux that is coming out of this tank and then you know find out the effect. So, remember this eddy currents induced in this tank also have effect on the leakage impedance of the transformer right. Because this basically eddy current losses they represent not only the they will affect this leakage field pattern, but also the also that is representing RAC the part of effective AC resistance of the transformer. Because tank losses will get reflected in effective AC resistance. So, basically that is the reason that all these all such eddy current losses on account of leakage field will influence to some extent the leakage impedance of the transformer. So, all such analysis can be done by using this formulation in which we are coupling or solving diffusion equation along with the circuit equation.