 So, welcome to the last lecture of this course L40. In this lecture what we will do is we will study theory related to virtual work method and also we will compare the results of virtual work based force calculation with MST based force calculation as well as you know that from with that from an analytical formula. So, to start with let us understand this energy balance in electromechanical devices. So, now electrical energy is equal to field energy plus mechanical energy in any electromechanical device like a rotating machine for example. So, now electrical input is basically now this WE is the electrical input to the motor and then that will be equal to the field energy plus mechanical energy. These also will include losses one can in a more accurate model losses in the corresponding three components can also be accounted for in this equation. So, now DW ME change in mechanical energy will be equal to the change in electrical energy minus change in field energy. Now DW E is we know it is EIDT right E is the voltage I is the current. So, EI gives you power power into time is energy and we also know the and E also is basically EMF. So, induced voltage EMF is nothing but rate of change of flux linkages lambda is flux linkage and then that is why DW E will be given by I d lambda by dt into dt. So, that gives you I d lambda as change in electrical energy. So, but one may argue that E is equal to d by dt of Li and then that can be written as Ldi by dt right if we assume L as constant and you know here lambda is substituted by Li because L is flux linkage upon current is L. So, that is substituted here and then we get this if we assume L as constant. So, then DW will be EI dt and then we substitute here E substitute expression of E here. So, Ldi by dt into I dt and then you get lambda di. So, we are getting that means two expressions one is I d lambda and again we are getting another expression as lambda di. Now, let us understand what is this lambda di. However, we should keep in mind that in moving systems L is function of time and thus representing E as Ldi by dt is physically incorrect because E should be then written as this because L also is changing with time in a typical rotating machine. Also co energy this actually later on we will see that this actually is representing the co energy. So, this I d lambda is energy lambda di is co energy. Now, in basics of electromagnetic we had seen you know I is corresponding to H lambda is corresponding to B. So, this is nothing but HDB this is representing HDB. So, HDB is change in corresponding energy density. Here it is this is representing BdH. So, this is co energy density and the corresponding it is showing the change in co energy density. If it is H if it is you know taken as BdH whereas lambda di is has the unit of energy. So, this is co energy. So, also it should be remembered that co energy which is integral lambda di is a non physical quantity which will be demonstrated later in this lecture. Now, consider a relay structure as shown here and this is a hinged armature with this as pure and now once you pass the current in this coil here this armature will get attracted and suppose that it moves along this x direction by a very small delta x distance then the operating point would shift from A to B. So, initially if the you know characteristic curve was this for a particular gap after it has moved by a delta x the operating point shifts on this curve from A to B along this line. Now, here why it becomes like this because of the fact that reluctance is going down because this air gap is reducing. So, then your flux in the circuit will increase you know when you go from A to B flux is going up and current is going down. So, basically then you know we have two options to simplify our force calculations either you know we can assume flux as constant that means we approximate this A B instead of A B we can approximate it as A E. So, that will be constant flux condition when the operating point shifts from this curve to this curve or the second case is we can approximate constant current condition when the operating point shifts from this blue curve to black curve then it will be a moving the operating point would be moving from A to B dash. So, in both these cases there is approximation in the first case when we are when we are assuming constant flux condition. So, we are basically neglecting this E A B area and the corresponding energy whereas in the second case when we are assuming constant current case we are in fact considering additional area and the force calculated would be higher. In the first case the force calculated would be lower than actual. So, now we will see the advantage and disadvantage of using both these approaches and we will eventually you know understand that keeping constant current that means considering this shift as A to B dash this condition is you know easier to implement while doing calculations and then we use we end up using co-energy based force computation method. So, now let us see in detail the two approaches that I just now briefly discussed. So, approach one is constant flux case this is the characteristics for the initial position of the armature and A is the operating point of the magnetic circuit in this position and then it goes to the other curve and the new operating point is this E and it is this AE they are on the same horizontal line. So, field energy at the initial position corresponds to area of OACO this highlighted area. Field energy after the moment of delta X the field energy would be represented by OECO area. So, change in field energy will be simply DWF as difference in the two you know areas that I just now discussed and remember this is why these are all field energies because we are calculating integral you know I d lambda. So, integral I d lambda is the energy field energy that is also equivalent to integral HDB because I corresponds to H lambda corresponds to B. So, this is nothing but also integral HDB but then that would represent energy density because unit of HDB will be you know energy you know the joules per meter cube whereas lambda into I the unit is energy that is joules. So, going further so change in electrical energy fed by the source here in this case is 0 because lambda is not changing because we are considering constant flux case. So, d lambda is 0 so, I d lambda is 0. So, now we know from the very first slide that DWM change in mechanical energy is change in electrical energy minus change in field energy and since DW is 0 you have DWM as equal to minus DWF. Now DWM is nothing but FDX force into the distance. So, FDX you get it as minus DWF and then eventually then you can calculate F as minus DWF by DX with lambda the total flux linkage being held as constant. So, now you know this effectively what we are doing the difference in two energies is basically this area Ae Oae O is the change in the field energy and corresponding energy density would be integral HDB in joules per meter cube. Now this is you know absolutely there is no problem here from the point of here you know basic principles. The force should be you know positive number now DWF by DX is negative because WF is reducing. So, minus of this minus gives you plus F. So, that also is you know verified. So, in approach to now what we will do here we are keeping current as constant. So, we are going from A to B dash. So, now originally we were at operating point A then we went to operating point B dash and the field energy at the initial position is represented by OACO area field energy after the moment is area OB dash DO. So, change in field energy is the difference of the two areas. So, OB dash DO minus OACO. So, if we actually you know represent the areas as given here. So, C E B dash DC C E B dash DC as 1 then O A E O this blue one as 2 and this triangle more or less this is a triangular area that triangular area is E A B dash E is 3 then you know DWF is given by 1 minus 2 areas the difference in areas 1 and 2 and 3 does not contribute. Now, this DWE we have seen earlier it is I D lambda. So, now once we have moved from A to B dash this much electrical input has increased in the system. So, this is representing I D lambda. So, that is nothing but area 1, area 1 is this and area 3 which is E A B dash E. So, that is 1 plus 3. So, now DWE minus DWF which is DWM can be represented as 1 plus 3 just now we saw this is 1 plus 3 DWE DWF is 1 minus 2 from the previous slide. In the previous slide we just saw this DWF is 1 minus 2 will be equal to the change in mechanical energy. So, 1 plus 3 minus of 1 minus 2 will be simply 2 plus 3 and that is what we are getting this area was O A E O is area 2 and this E A B dash E was designated as area 3. So, you get 2 plus 3 and now you can see here this area is basically representing the change in co-energy because now if you actually you know find out this area by you know difference of 2 areas which are those areas. Now, what you do you calculate integral B D H or integral lambda Di then the original operating point was A. So, the corresponding area will be the area under this curve under O A curve with H or I as the reference axis. So, then we get integral B D H or integral lambda Di. So, this was the initial co-energy when the operating point is approximated to shift to point B dash then the area co-energy is represented by area under this curve which is under the curve O E B dash. So, effectively now you can see there is increase in co-energy there is increase in co-energy and you know that is something you know that means effectively the potential energy of the system has increased which should not be the case and in that sense co-energy is non-physical. So, you cannot get both co-energy also increasing as well as kinetic energy in terms of moment also increasing. So, both you know are increasing this case which is non-physical but then you know this approach has some advantage we will see. So, DWM then is given by DW minus DWM that is area O A B dash O A B dash O which is nothing but change in co-energy and this is nothing but DWM is FDX DWM is FDX and this difference is DWCO change in co-energy. So, finally you get F as DWCO upon DX with current held constant. It is basically dual of energy a non-physical quantity which I explained and co-energy is used with virtual work principle to calculate the forces and torques in rotating machines. So, in terms of FEM coding what is the advantage of using co-energy we will see now. In simulations of moving systems it is easy to keep current as constant since it is a circuit quantity and therefore co-energy based calculations are easy with current being held constant. In contrast magnetic circuit change leads to corresponding change in flux linkage lambda and this lambda is not easy to control it is not easy to enforce constant flux condition during flux during the FEM simulations. That is why we use co-energy based force computation method and that is being demonstrated now. Now, in order to verify our virtual work based you know force computation method first let us calculate analytically the force value and in L39 we have already calculated force using MST using the MST method. So, then we will compare all the three values. So, analytically we can calculate because this is a fairly a simple you know problem and corresponding geometry. So, this geometry was already described in L39. So, this is a core or is also armature but it is this armature is not hinged it is free and it will give a moment you pass the current this core will get attracted to this you know to this C shaped core. So, now the air gap here is 0.04 meters reluctance is L by mu 0 times area S. So, now there are you know two air gaps. So, the total reluctance in this magnetic circuit because since there are two air gaps there are two lengths two air gaps. So, 2 into 0.04 upon mu 0 into area what is the area? Area through which the flux is crossing. So, one of four dimension of the area is this 0.01 and other dimension is 1 meter depth in Z direction because we are going to calculate as usual the performance parameter that is force per unit length in Z direction this is in XY plane. That is why it is 0.01 into 1 then this gets calculated as this number MMF magnitude motive force is ampere turns is equal to 100 or the number of turns into I 362.04 this was actually calculated in L39. So, Ni is this. So, flux is MMF upon reluctance. So, then that gets gets calculated as so many Weber's and B is flux upon area. So, flux is this upon area 0.1 into 1 I just now explain why area is this. So, the flux density is 0.569 Tesla in each gap in each of these two gaps. So, energy in two gaps together is twice B square upon twice mu 0 is the energy density into volume because this energy per unit volume into volume s into area into length into 2 because there are two gaps. So, that gives after substituting all the values you get the energy as this and then force is energy upon length force into distance is energy. So, it is you know energy by this distance because this entire core moves in this direction by 0.04 meters. So, we are here we are not considering in small incremental distance we are considering the total moment through this you know 0.04 meters and we are calculating the force for that complete moment of 0.04 meters and then you get the force as 2.57 into 10 raise to 4 Newton per meter because again I explained you we are calculating per unit length in z direction because of 2D approximation. So, you get Newton's per meter we have we had already calculated using the MST approach in L39. So, this slide in fact is from L39 only. So, we had seen it there and we in fact calculated their force as 2.7 into 10 raise to 4 Newton per meter. Now, we will calculate using co-energy using the co-energy approach or the virtual work method approach seen in this lecture L40. So, here again these are the 2 you know FEM simulations one with you know higher distance and one with smaller distance after virtually displacing this you know core or armature or even it is called as in literature plunger. So, by some distance small distance along this direction and then find out the energies in both these positions take the difference which will give you DWCO divided by the distance by which this core plunger or armature is moved which is in fact 5 mm in this case and co-energy at position 1 is this co-energy in position 2 is this. Now, you can see here co-energy has increased. So, as was explained here in this case when operating point is approximated to to have shifted from A to B dash there is increase in this co-energy which is represented by this shaded area. That is what we are also getting by simulation there is increase in co-energy and then force is calculated as this. So, it is coming as 2.28 into 10 raise to 4 whereas the MST based force calculation gave the force as 2.71 into 10 raise to 4 and our analytical formula based value of force was 2.57 into 10 raise to 4. So, all these 3 values 2.28, 2.71 and 2.57 in one of the previous slides all are in the same range. So, that gives us confidence in all our calculation approaches. Now, these are very simple problem to analyze as I have been mentioning in this course you start with very simple problems verify with analytical formulae or some other methods get confidence in the procedure that you are adopting and then go on to solve more complex problems. Finally, we will see how do we calculate reluctance torque on a simple rotating machine. Now, here this is like a rotating structure which moves in this gap. Now, this is again a core which is excited by this coil. So, flux is established and this rotating structure tends to rotate toward minimum reluctance position. So, that means when it is in this position torque would be acting to move this so that it gets aligned and becomes vertical. So, what we will do again the same thing we can develop our own code or use any commercial FAA software to model this. So, using a code developed based on the theory that was discussed we get here co-energy for position 1 as this for position 2 as this and then the torque is dwco by d theta. So, the difference between these two numbers divided by the angular displacement which is you know 0.18 radian or 10.31 degrees you get torque as 0.2895 Newton meter. So, this is how we can calculate torque using co-energy based approach. With this you know last case study our course is completed. Now, I just want to you know summarize using this slide what are all things that we have covered. So, in first 10 lectures we covered basics of electromagnetics relevant to electrical machines and equipment finite element method and finite element analysis. And then we you know in these 10 lectures some of the concepts were explained through practical examples and virtual electromagnetic laboratory developed in E department of IIT Bombay. In lectures 11 to 16 we understood fundamental principles of finite element method in the next 5 lectures we studied two dimensional finite element formulation and the corresponding coding procedure and basically we studied that with reference to Poisson's equation. Then we basically in these lectures we demonstrated few course using freeware platforms like Gmesh and Sylab. Then we went on to you know Sol Laplace and Poisson's equations in these three lectures and basically we calculated leakage inductance of a power transformer, we calculated magnetizing inductance of an induction motor and also we understood how we can use finite element method effectively for high voltage insulation design. Then in further lectures from 26 to 40 we analyzed many case studies and some extensions of standard and you know simple FEM formulations for Laplace and Poisson's equation to you know more complex cases involving axisymmetric problems, permanent magnets, eddy currents. Then we also studied how do we couple circuits and fields to solve practical problems which may involve transients and nonlinearities. Then we also got exposure to some advanced topics in which we studied how do we you know do FE coding for simulating rotating parts where that is basically rotor structure in a typical rotating machine and we also studied how to impose periodic boundary conditions which basically you know reduces computational burden. Then we went on to see how to compute torque speed characteristics of rotating machines. How do we impose tangential magnetic field excitation for some specific problems. For example, we studied how do we calculate effective complex permeability for analysis of core joints in transomers. And finally, we studied how do we calculate forces using two approaches, Maxwell stress tensor best approach and virtual work best approach. So in all we basically you know had about 20 case studies throughout this course and I hope this course will be useful to not only undergraduate and postgraduate students but practicing professionals for solving you know real life problems either by developing their own course or by effectively using available commercial FEM software. Thank you very much.