 So, welcome to the 23rd lecture, we will continue our discussion and or a finite element applications now. First we will see leakage inductance calculation of a transformer and then later on we will also see magnetizing inductance calculation of a rotating machine. So, now when we talk about a transformer and leakage inductance what we do is we in finite element simulation we take only one window this could be a 3 phase transformer we take only one window and then this is the first is the L V winding here which is mass by this flux line and the second this is the high voltage winding and then we actually you know o's the ampere turns to be equal and opposite in the FEM simulation. So, N 1 I 1 is equal to N 2 I 2 and one of the N 1 I 1 or N 2 I 2 we make it minus so that net ampere turns enclose in this if you take any closed flux contour in this core portion net ampere turns enclose will be 0. So, that is why you will see here there is not a single flux contour which is enclosing both the windings. So, many contours are enclosing the L V ampere turns rest are enclosing H V ampere turns right. But there is not a single flux line which is enclosing both windings. Now, in practice actually this is not the case because there are magnetizing ampere turns and that is why you will get some magnetizing flux no load flux in the core which will move around the core which will enclose both the winding and be responsible for induced voltage in both winding right. So, that is why it is very important in FEM to know what we want to model and accordingly define parameters which actually may not be the true case in practice because you will never have you know transformer working and you know currents being passed through both winding and there is no there are no magnetizing ampere turns. This condition would not be there because we have a practical core it will require some ampere turns to magnetize. But see since magnetizing ampere turns are negligible and here we are interested only in leakage field and the corresponding leakage inductance we are neglecting magnetizing ampere turns right. So, effectively in the transformer equivalent circuit we are assuming X m as is sort of infinity right because magnetizing current is 0 right. So, I hope you understood this leakage field plot. Then we go to this formula inductance we have seen in basics of electromagnetic L is given by mu n square A by L is also n psi by I flux linkage upon current and that is also given by n square upon reluctance because L by mu A is the reluctance. Now of course, here that time we had used instead of A we had used S I think that A and S both are same area right. So, if you are getting confused with this as magnetic vector potential symbol then better you know right here mu n square S by L ok, but remember this A is S or area. Now this area also this area is actually the area of the flux tube this L in this formula is really height of the corresponding flux tube. Now in this case flux psi in this formula flux psi is generally linked by all the turns when we say mu n square A by L and the corresponding n psi by I this some whatever that flux is there is linked by all n turns equally right. Now there could be a case as we will see later the number of turns in winding they are not linking the same amount of flux that is why then this area does not become the actual physical area, but then you know you have to have integration of flux linkages. Now if you want to calculate this analytically because we every time we do some FM analysis we should be clear about like the background principles so that we do not make any mistake in FM analysis. So, it is always better to understand you know analytical calculation which may be approximate right. So, for example, here this flux plot in this window can be approximated as shown here. Now this is the LV winding this HV winding and now you can see here because of end effect the flux size also is actually fringing at the end is it not. So, now with such fringing it will be difficult to find inductance by some simple analytical formula is it not. So, we have to make some approximations. So, what is the approximation we are making? We are making the flux as entirely axial. Effectively what we are doing this increase in the height of the flux tube at the ends because there is a this flux tube the height is effectively more than the physical height of the coils is it not. Physical of height of the coils is from here to here right, but actually flux is going beyond and there is a fringing. So, that actually is accounted by some extra height of the flux tube H equivalent which is more than the height of the winding HW. So, now going further in this plot now you can see here in the gap between the LV and HV ending the flux is uniformly spaced. Whereas as you go away from the gap into the winding initially the flux is closely spaced and then it becomes pass this space. Why that happens? Because as you go away from the gap the ampere turns enclosed by any flux line would go on reducing. And now we know this is we are actually neglecting magnetizing ampere turns effectively what we are doing we are taking permeability of core as sort of infinite. So, that means if permeability is infinite B by mu is tending to 0. So, H will tend to 0 in the core part is it not here in the core part H will tend to 0 because mu is taken as sort of infinite right. So, that means if you take now HL drop that means if you take integral H dot dl is equal to ni is it not that is the ampere circuit along. So, now since there is hardly any H in the core if you take any close flux contour entire HL drop will be only in this in this window region is it not H into L will be entirely only in this window region which is the air non magnetic because this is copper this is also LV and HU endings are made up of copper they are also non magnetic. So, they are also mu R is effectively 1. So, if you take any flux contour right integral H dot dl if you calculate along that that H dot dl will be entirely basically contributed by this the flux lines in the window portion is this clear. So, that is the reason any of these flux lines which are enclosing either LV full LV turns or full HV turns they are enclosing the same turns. So, integral H dot dl is equal to ni is it not. So, since number of turns enclosed are same by any of these lines H value will be same in this entire gap is this clear see integral H dot dl is equal to i is it not and here since it is a number of turns are more actually more than 1. So, it will be actually integral H dot dl is equal to N into i right. Now, as you go inside the winding either this way or this way into LV winding or HV winding progressively any flux all these flux lines will go on linking lower and lower ampere turns because here the ampere turns in this line will enclose these many ampere turns this line will enclose only these many ampere turns is it not remember the current is actually into or out of the paper is it not because this is a 2D simulation. So, always the currents are either dot or cross. So, any flux line is enclosing current that means current is either coming in or going out is it not. So, this flux line for example encloses or links only these many ampere turns right. So, that means I see these ampere turns are lesser than the full winding the corresponding H into L again L is same L is same is it not. So, if ampere turns are less Ni is less H will be less and correspondingly B will be less. So, that is the reason that you get this MMF diagram like this MMF because H into L is constant here as you go into the windings on both sides H would reduce and it will go to 0 at this point because at this point the ampere turns enclose by this flux line will be this flux line if you have here ampere turns enclose are 0. So, I hope you understood this MMF magnitude motive force diagram. So, now here now see this we have to calculate the total inductance because of all this leakage field that is our objective is it not. In this part since everywhere the flux is uniform right. So, the corresponding contribution by that gap is also you know simply E gap is this gap into the corresponding mean diameter D gap D is the mean diameter right into pi. So, pi into that mean diameter will give you the circumferential length is it not pi into mean diameter of the gap will give you the circumferential length into this gap will give you the corresponding cross sectional area of that gap is it not. The cross sectional area through which the flux is crossing will be given by pi into mean diameter of the gap into the this radial depth. So, in this slide let us understand the areas through which leakage flux is passing through. Let us take for example, the LVHV gap. Now, LVHV gap from the top will be seen like this D gap is this gap D gap is the mean diameter of that gap and pi into D gap will be the circumferential distance. So, now if you develop this area along a horizontal line then that area would be like this and at an instant of time flux is going up is directed upwards. So, in this developed area you will have flux represented by dots coming out. So, now if we have understood this then the area of the LVHV gap through which the leakage flux crosses here would be simply pi into D gap is this length and into T gap. So, this is the cross sectional area corresponding to this LVHV gap through which this leakage flux is crossing. So, that is why T gap into D gap which is the mean diameter into pi is the corresponding area cross sectional area through which this flux is passing. But this is the T LV is the corresponding radial depth here, THV is the corresponding radial depth of HE winding, D LV and DHV are the corresponding mean diameter of LV will be somewhere here, HE will be somewhere here. But here one-third appears why because flux is not uniform. If you actually integrate the flux linkages over this LV area or HV area and after some simple equations you will see that it is one-third which is obvious because you know the flux density is flux lines are the density of flux lines is reducing. So, it is quite logical that it will definitely be less than one. Actually it comes close to one-third the derivation is there you know if you want the derivation is in this book in chapter number 3. So, you can verify by referring the book. So, if you understood that the area that appears in the inductance formula is given by this and this one-third appears here because of this non-uniformly spaced flux lines in LV and HV radial depth and here no one-third no term appears multiplying term appears constant appears because flux is uniform in the gap because any of these flux lines is linking full LV or HV ampere turns. Going further now we need to worry about this height we have already you know agreed that we should have only axial field and then we you know increase the height of that flux tube more than the physical height of winding to take care of the fringing fields at the end. So, now again there is one sort of empirical formula here given by Rogoski factor the height of the winding increases by this hw by kr, hw is the physical height of the winding from here to here kr by this formula will come less than one and that is why h equivalent will be more than hw right. So, that is why it is h equivalent is more than hw. So, now here again it is a function of you know so many constant again the same radial depth of various things height of the windings and this formula is used to find out the effective height of the flux tube if you want to neglect the fringing effect and account them by only the axial field ok. Now, we can actually before we see the FEM calculation we can for the transformer this is the dimensions of the transformer right. So, let us not go into details of all these dimensions and coordinates there are basically two windings lv and hw winding with corresponding radial depth heights of winding these are window only remember we have modeled only one phase there is another phase here right and we are modeling only half that means the other lv and hw winding of this phase is on this side right, but we do not have to model that because the we can get the answer by considering only this window model of winding right. So, now, we if we substitute all the values of you know radial depth of the winding now this is in centimeter here that is why it is multiplied afterwards by 10 raised to minus 4 because this is in centimeter this is also in centimeter so this becomes centimeter square. So, that centimeter square is converted into meter square by multiplying by 10 raised to minus 4. So, basically it is the radial depth into mean diameter of lv gap radial depth into the corresponding mean diameter of the gap and then radial depth of the hw winding and mean diameter of hw winding in centimeter and then this is one-third and one-third factors come as we discussed in the last slide into pi because pi into mean diameter we saw that in the previous slide pi into mean diameter is the circumferential distance that into the radial depth will give the area and in case of lv and hw winding you have to multiply additional by one-third term right. So, then you get this area as this and then l equivalent of that completely axial flux tube is 1.52 now this 1.52 this is in meters so 1.52 divided by 0.965. So, the height of the winding or that flux tube is 1.575 which is more than the physical height of the winding which is 1.52 meter right and then you substitute in the formula l is equal to mu 0 n square a by l all the quantities are known now number of turns. Now, this number of turns is the hv number of turns. So, if you take here lv number of turns you will get inductance referred to the lv side here I am taking hv number of turns here. So, this inductance comes referred to the hv side. So, 4.2 10 raised to minus 7 mu 0 right into number of turns square into this area. Now, what is this area remember this area is the effective area of this whole flux tube from here to here is it not is the effective area of this entire flux tube right. So, into a divided by height and then you get this. So, same thing we can now calculate by using finite element method right. So, again just for the sake of completeness all the parameters are given of this transformer it is 3 phase 31.5 mv 8 132 by 132 kv transformer vector group is yd 1 hv star winding hv is hv is star winding lv is delta winding and vector group is 1 and frequency is 50 hertz hv current is 137.78 amperes you can calculate based on this rating number of turns in hv winding is 980 and that in lv winding is 424. And then the radial depth and mean diameters are given here right which in fact, we use these things we use for you know see 5 lv is 52 mm and 638 mm is the mean diameter that is why we what we use here in analytical formula. Radial depth is 52 mm in centimeter is 5.2 and mean diameter is 638 mm which is 663.8 centimeter right and see why we are not taking this this gap here and this gap between lv and hv because there is no flux here and here there is no flux. So, that physical area does not contribute to the area in the inductance formula is it not even in this case you see now this all these radial field engine flux we have made it vertical is it not. So, effectively in this area there is no flux that is an approximation in analytical formula right that you do not have to do in case of in FEM that will calculate on its own okay. So, then you have this radial depth and mean diameters defined here and then you have this n1 i1 is equal to n2 i2 right. So, i lv you calculate n hv by n into i for hv divided by n lv will give the lv current right. So, these we are exactly matching. So, that even if there was some difference if you make here suppose instead of 318.45 you make it 318.2 for example, what will happen is this entire in this core there will be lot of magnetizing flux you will see because small difference in ampere turns will make those ampere turns as magnetizing ampere turns and since the core permeability is defined as very high even a very small difference will set up reasonably high flux density in the core that is why we need to exactly match n1 i2 is equal to n1 i1 is equal to n2 i2 is it clear. So, now you know we will see how do we do the sila coding now here we are only explaining what is new in this code rest of the code remains same the procedure is same. So, here basically what is new is you have to define ampere turn density in the winding right that is the main new thing. So, for example, here H V turns is 980 current is 137.78 amps similarly for L V turns is 424 current is this which we calculated on the previous slide area is 52 mm in radial direction into the height height of L V and H U winding is same. Now, we define the ampere turn density. So, far we go element wise we set up a far loop go across all the elements see which of those elements lie in this sub region number 4 which corresponds to L V this we would have taken care in G mesh while doing G mesh is it not. So, there we would define L V as sub region number 4. So, automatically all the elements will get in the L V will get you know sub region number 4. So, all those you know elements where L V is there that we will define ampere turns of L V divided by this area. So, ampere turns divided by area is the density always remember when it is more than one turn you have to take ampere turns upon area as a density current density not just current upon the area. So, ampere turns divided by area is J. Similarly, you know sub region number is 3 for H V corresponding formula here only observe here minus sign that means ampere turns of H V are taken as negative of ampere turns of L V. That means effectively currents current directions of L V and H V are in opposite direction if current of L V is going in current of H V will come out that is it. So, net ampere turns will be 0 that is why there will not be any magnetizing ampere turns which is justified when we are calculating the leakage inductance of transfer. If we are analyzing the core then that magnetizing and ampere turns are most important under no load condition for example, after getting the solution by the normal procedure that we have seen then we can calculate this energy in each part of the problem domain. So, here first energy is calculated in the core B square associated with each element square B square upon twice mu is the energy density is it not B square upon twice because this half twice mu is the energy density into the area now energy density is energy per unit volume. So, we have to multiply by volume. So, here we are multiplying by area. So, into a length we have to multiply later by mean diameter pi into mean diameter which we will do here energy between L V and H V winding same procedure corresponding elements in sub region number 1 2 3 and 4 are taken and the corresponding energies are calculated for each element total energy associated with core and that we are multiplying by pi into mean diameter because here it was only multiplied by area. So, we have to multiply by circumferential length to get the volume. So, that volume into the corresponding energy density B square upon twice mu would give you the energy. Similarly, the other three energies in the air, air means in the gap between the L V and H U winding, air means gap between L V and H U winding which is the main contributor of the you know energy is it not and then three is H U winding, fourth is L V winding and then we are adding all the energies and then we are displaying all the energies here which are here. The L V H V gap has maximum contribution and the reason is obvious again you can see here the contribution of all these flux lines which are uniformly spaced and density being high will be maximum that is what we are seeing here the radial depth of the H U winding is more 65 versus 52 mm also mean diameter of H V is more than mean diameter of L V is it not. So, again the area will be more. So, that is why it is 383 versus 238 then you know this total energy when you sum up and then you equate it to half Li square you will get L then the total leakage of inductance of the transformer and since it is H V current here this is referred to H V side. So, now in this slide let us compare the values of the leakage inductance calculated by the analytical formula and by the Sylab code. So, the two values as you can see here they are quite close to each other which verifies both approaches of calculation analytical and F A M. In analytical formulation we did number of approximations and the main approximation being that all the leakage fluid is axial whereas in finite element method we did not do that approximation then is axial as well as a radial field. However in F A M also we did one approximation it is a two dimensional approximation actual field is three dimensional we have taken only one cross section. So, to that extent the value calculated by 2D F A M will have an error as compared to more accurate 3D field based computation. You can also use you know freeware F A M software as listed here and corresponding website links are given. So, for magnetostatic problems, electrostatic problems and one time harmonic eddy current problems some of these softwares can be used very easily for computing performance parameters. So, those who are interested they can you know go to these websites download them and try such simple you know F A M simulations and then go on to use them for more complicated problems. So, before you know going further into other problems like motor analysis and all that, let us quickly see the transmission line conductor analysis. Now in power system books typical books they will calculate internal inductance of a conductor. So, any conductor you take there is a internal flux and there is external flux is it not. So, if you take some conductor like this. So, there is going to be internal flux here and then there is going to be external flux is it not. So, the internal flux will contribute to what is known as internal inductor and the flux which is outside the conductor that flux will give you the external inductance of the conductor and the total inductance of the conductor will be L internal plus L external. So, now here in this slide we are just calculating the internal inductance. Now how do we calculate again you know we use the same concept of stored energy we calculate the stored energy inside this conductor because of its internal flux. How do you do that? Now H at any point inside the conductor at radius for example here this radius is rho in cylindrical coordinates. So, H at that rho is given by integral H dL is equal to I is it not. So, H into 2 pi rho will be the current enclosed. So, then that gives you know by rearranging H you will get it this then stored energy density will be half mu naught H square why again mu naught is copper mu r is very close to 1 is it not. Then substitute the value of H you will get half mu naught H square as this then the stored energy will be these energy density energy per unit volume multiply by or do volume integration. So, d rho rho d phi d z in cylindrical system this is the volume incremental volume is it not. So, if you you know simplify this and integrate and remember d z it is we are doing it per meter depth that means for a 1 meter length of the transition line in z direction. So, because this x y this is like a rho phi plane is it not this rho phi plane and then what you do not see is the z z direction z into the you know this plane. So, that is why z is taken as 1 0 to 1 and then if you simplify you will get stored energy as this and then that you equate it to half L internal I square you will get L internal as mu 0 upon 8 phi 8 phi right. So, now you can see here the internal inductance is constant unless it is not a function of geometrical details it is always it is constant. If you increase the diameter of the you know this conductor right the total stored energy E will remain same mu 0 upon 16 phi and that into I square because current is constant and that is why the L internal is always constant is independent of the dimension. So, this ends the lecture number 23 we will see in lecture 24 rotating machine inductance calculation. Thank you.