 we had seen in the earlier lecture how the idea of inductance is used in order to derive an expression for the force that is exerted. More specifically we saw a singly excited linear motion system which consists of a coil and an iron rod that is fixed to a substratum connected by a spring having a spring constant of k the iron bar could have a mass of m and you are exciting the coil with a voltage v in this case we wrote down the system expressions we started with the electrical system equation which said that v equals resistance multiplied by the flow of current current is assumed to be flowing inwards at this terminal and therefore the applied voltage v would equal resistance multiplied by i plus the derivative of the flux linkage as we saw this is an expression that would fit all systems and then from this expression onwards we went on to derive further equations we started with from this we derived an expression for power input and we said that this input power is equal to the resistance loss plus the mechanical power output plus the rate of change of field energy this equation is important and from this we wrote down if you get the input electrical power if you write vi and then this was equal to i multiplied by this expression and then we wrote down that the rate of change of field energy is nothing but the derivative of the field energy which we wrote down as half li square if it is a magnetically linear system this is valid for a magnetically linear system which means systems where the bh curve is a simple slope and then from this we identified the mechanical output power as let me call that as p mechanical that is half i square dl by dt from which we identified that the force that is exerted is equal to half of i square dl by dx and from this we see that the developed force it depends on the rate at which inductance is going to change with respect to x that means if you had locate if you had said that this was the origin of the bar beginning to move this is x equal to 0 and then the bar moves in the direction of x the expression for the developed force says that force is going to act along the direction of increasing x so if this is the force this is the direction in which it acts and this is greater than 0 would also mean that dl by dx is greater than 0 that means that as x increases the inductance increases which definitely is true if the bar moves inside the inductance as seen by this input will increase now all this can be obtained from this and then if you want to find out how this bar is going to move you write the mechanical equation if you call this force as fe then fe minus k times x because of the attached spring this must be equal to mass of this multiplied by the acceleration which is then d square x by dt square we had seen this in the last class so this is just to remind ourselves what we have done now this is not just an imaginary a hypothetical system this is really used in applications there are equipment known as solenoid valves solenoid switches which but which work based on this idea for example if you take a solenoid a solenoid switch then it would look like this you have a cylindrical iron structure in the center of which there would be a hole passing through the cylinder and around this around this central hole you then have coils that are wound and within this central hole there is an iron bar that would move so if you now energize this allow some flow of current into this this bar would be forced inward would be drawn inward and you can attach a spring here which will arrest the movement so it would move down against the spring force these are used for example in all your automobiles to start the engine in the starting sequence of the engine you have this with this will go down and close an electrical switch here which will then allow the DC motor to rotate and start the engine right so if you want to analyze this kind of a system and say how much force is going to be developed what will be the force which is developed when it hits the end you need to write down equations like this and then analyze the entire arrangement you need to see how the field is going to be developed in this how this bar will move what will be the input current that is drawn into this all this has to be analyzed now we have an FE analysis of this as you can see that this arrangement is entirely symmetrical about the axis in the center we call such systems as being axis symmetric systems it has an axis symmetric geometry and in order to do an FE analysis of this most of the FE software allow us to model only one one section of this which is assumed to be then swept around and therefore it may be enough to represent the section which is seen in the middle so you have half of the this iron rod and then there is a gap here this is going to hit somewhere at below below this and then these turns are distributed here all this is flow of current into this and then you have the cylinder as well so I am now drawing a view of the sectional sectional view this is how the system will be this section is as if you slice this slice this by a plane and then take one half of that alone this is how the system will look now we can see some of the outputs from this from the simulations that we have done in order to get an idea of how this input current is going to flow let us see the simulation results so here then we have some simulation results using FEM and as we saw on the board here we have a sectional view of the spring plunger system this is the outer shell and then here you have the iron rod that can move and the excitation is given here and you can see how the magnetic field is there you can see a large field value existing here which is about 0.5 Tesla as per the scale 0.4 Tesla as per the scale that is given here and the field decreases as you go to the extremities of this bar and here and due to this field then as we have seen through the equations this bar is drawn inwards and as it is drawn inwards as it goes close to the end stopper now you can see that a significantly larger field exists over a longer length of the bar and this gap is now very very small and you can see that the force that has been developed is 3.37 Newton meter in the earlier case when the gap was large and the inductance is only 0.0037 Henry 3.7 milli Henry whereas now the inductance has increased when the bar has reached the end nearly the end the inductance has gone up to 0.01 Henry and the linkage has also gone up 0.06 Weber as against 0.022 Weber one can also see that the field in this area the field in the region here is bending out of the air gap it is not really flowing straight along this but you find all kinds of field lines that are going out whereas here you do not see that effect field lines are going straight from the bar on to the other level right this kind of effect where the field lines move out of the ordinary route these are called this behavior is called as fringing this happens if the air gap is large you can see for example in this case also the air gap here is pretty large and you can see many of the field lines going through other routes and not really through this iron bar directly on to this plane below here again you can see field lines going outwards. The analytical expressions that we wrote for inductance for example in the earlier lectures we wrote that inductance is then some mu into n squared into area divided by the length of the iron path or length of the air gap path all these are valid when the air gap if there is an air gap the air gap is small if the air gap is small then all the field lines that go through this iron bar cross the air gap and get into the other area and not many lines flow through in the other way but if the air gap is large then the field is more distributed and it is very difficult to find out what the inductance exactly is your hand calculation will not match the actual inductance that you get. So this shows how the force is there and we have seen from the expressions that the force is given by half I squared into dl by dx and as this bar approaches the as the air gap becomes very small dl by dx also becomes very large and therefore the force developed is significantly higher than in the earlier case of 3.37 Newton meter. Now all this is an analysis of how the field plot is going to look when this iron bar is at different locations this is not going to say for example how much flow of current is going to be there how fast the bar will move right and how long it will take for the iron bar from beginning this position all the way to travel to this position how long is it going to take all these are not shown by this figure for that we have to do a simulation we have to solve all these equations that we have written down earlier by some means these are normally difficult to solve by hand so you have to take recourse to some kind of simulation software in order to solve these equations and if you do that here you can see the results of how this works this line that is in green shows the displacement x beginning from 0. So if you let x equal to 0 be somewhere there in fact that distance is about 1.5 inches that is considered here and then energize the system we have looked at initial or to put it the other way the initial distance between the bar and the stopper stopper is when the bar goes and hits in the end this distance is 1.5 inches and we are supplying a DC voltage equal to 12 volt right and the resistance of the coil is 2 ohms now with this one can see how the simulation has happened the energization happens at this instant at t equal to 1 second until then the DC supply is not switched on and you can see that the initial input current drawn is equal to 0 no current flows obviously before you switch it on as soon as you switch it on this bar is fairly far away from the rest of the system and therefore the inductance is very small since the inductance is small flow of current I attempts to increase fast and as this increases there is also an exerted force on this bar and the bar begins to move you can see this plot this plot shows the displacement versus time until then the displacement is equal to 0 that means bar is not moving and from this point onwards bar has begun to move in response to this ?2 and along with movement of the bar the inductance also increases if the inductance is going to increase then ? which is the linkage of flux which is equal to inductance multiplied by I that also increases and after some time the rate at which ? increases is so high that the flow I begins to fall right and in response to all these this bar then moves all the way from 0 goes up to 1.5 inches hits the stopper and there it stops you can see that the input I drawn here this axis gives you I this drawn increases up to 5 ampere or so and then drops down at this instant the bar has hit the limit and if it has hit the limit it is not going to move anymore therefore the inductive effect is no longer there and from this instant moving of the inductance does not become different it is a fixed value and therefore I increases at the normal increase of increase having L by R as the ratio so here it increases and then settles down to a level of 6 ampere which is the same as 12 divided by 2 ohm that is 6 ampere so at the end it has to settle down at 6 ampere because no more inductance is going to change everything has settled down you are giving DC current and therefore it has to settle down at 6 ampere but until then you can see that there is significant events that are happening this bar has moved from 0 all the way to 1.5 inches and stop and how that happens is displayed by this graph. Now in the analysis of electromechanical systems these are the kind of results that we are looking for we want to know how the situation how this electromechanical system behaves with the increase in T if you did not know this you would have said that applied voltage is DC and this has a resistance equal to 2 ohms which implies that it should draw 6 ampere and you see that it is not drawing 6 ampere immediately there are lot of things happening here ultimately if everything settles down no more inductance changes there then this I settles down at 6 ampere it is important to understand what we are attempting to get at by the analysis that we are doing in all these lectures we will now switch back to our the other analysis let us go back to the board so we have seen so far how linear motion systems can be analyzed they have application value by themselves but we will now move on to rotational systems most of the electrical machines are not linear but they are going to rotate about certain axis so in this again let us consider singly excited systems as we did for the linear k so in the case of a rotational system what we really have is a stator which is the fixed part of the machine which does not rotate and then you have a rotor inside this you have a rotor with a shaft there is an air gap here so let us say that all this region is iron outside all these is iron region here you have air I have of course exaggerated the air gap nobody would make a machine with such a large air gap so here is iron here is iron and then you have a shaft to which you can connect some mechanical loads this is the kind of system that we would have and we want to know how to analyze or represent this behavior of the system as an equation and we say that there is an excitation mechanism there is a wire loop that goes here so one must understand what this figure represents this is generally called as a slot in machines of course you would have seen all that at the UG machine course and this is really a structure that is cylindrical and we are taking a section of the cylinder right and this is how it looks let us say it has one coil which starts as one coil side here goes along the length of the machine completes on the other side and then comes back here may be it has several turns as well finally you get one terminal here and another terminal here what we want to see is how this system is going to behave when you give an electrical input at across these two terminals so again we write the same equations the applied voltage V across these two terminals can be written as R into I plus P times ? and then input power is Vi equal to I2 R plus I into P of ? and we know that input power has to be equal to the same ways in which it is then consumed you have a resistive loss component you have a mechanical power output hopefully it is there and then you can have a rate of change of field energy and in this system also because we are considering a linear magnetic system so far we can write field energy let us denote that by WF WF is then simply equal to ½ Li2 and we are seeing that the expression development is the same as what we had done earlier and therefore we expect to land up with a similar result that mechanical power is equal to ½ I2 DL by DT which can now be written as because this is a system that is going to rotate then you can write this as ½ I2 DL by DT into DT by DT now DT by DT is nothing but angular velocity so this is angular velocity and if you say mechanical output is something multiplied by angular velocity then this something has to be torque therefore we can write that in the rotational system torque is equal to ½ I2 DL by DT this is an analogous expression to what we had earlier in the linear motion system you have force equal ½ I2 DL by DX now since you have I2 here it means that whether you excite the system with plus here and minus here or you reverse it and say plus here and minus here it is immaterial as far as this expression goes because you have I2 so whether this current is greater than 0 or less than 0 where it is coming out the developed torque will always be greater than 0 but looking at this expression and looking at the geometry that we have drawn here it would be interesting to note that the inductance in a linear magnetic system the inductance is a function of geometry alone it means that depending on how the different elements of the system are arranged you would have a fixed inductance in the earlier case we saw that if you have let us look at the system here we saw in the simulation also if the bar was at a certain height you have some inductance and if the bar moves down the inductance increases and it is therefore only dependent on where the bar is once you fix the position of the bar the inductance is fixed in the system here the inductance if at all you say that it is ½ I2 DL by DT how can angle undergo a change angle can undergo a change if the rotor is now going to rotate if the rotor undergoes a rotation then the angle of the rotor undergoes a change and we are going to look at how the inductance of this is going to change if the rotor is going to rotate but here if you look at it irrespective of how the rotor is going to be turned what is the angular location of the rotor the system geometry does not change as far as this loop is concerned because the rotor is completely fully cylindrical it will not be able to make out the difference between the rotor at some angle and the rotor having rotated to another angle and therefore in this case DL by DT is really 0 and therefore you do not expect this system to rotate at all on the other hand if you have a rotor that looks slightly different let us say that the rotor is not fully cylindrical like this but it looks like this let me also erase these lines can create some confusion and perception so if the rotor is like this then it does make a difference whether the rotor is here or whether the rotor is here the air gap that is there at a given angle is completely different and therefore the inductance that this coil would experience is different for different angular locations of the rotor and therefore there is some value for DL by DT and in this case therefore the rotor will definitely move it is therefore important that inductance some inductance changes with respect to rotor position if inductance does not change with respect to rotor position then the machine will not rotate you will not able to develop a generated torque however one can now look at a system where the rotor incidentally this kind of rotor is called as a salient pole rotor so in the salient pole rotor inductance will depend upon where the rotor is the angular position of the rotor and this the generated electromagnetic torque which depends purely on the variation of inductance right is then called as a reluctance torque. Re reluctance torque exists in systems where inductance is going to change the inductance in the sense self inductance self inductance changes with the angle of the rotor on the other hand if you have systems which are cylindrical where the rotor is also a cylinder please note that what we are drawing is a section of the machine the machine is a cylinder and we are taking a section and we are viewing that section end on so this is the axis of the machine. So if you have a system like this and you intend to have it rotating then you need to put one more coil at least here you need to therefore have some slots it is difficult to have a coil when there is no slot it has to be housed somewhere so there would be a slot and then you have a coil that is running through this now we can look at whether this system will be able to move earlier we said that when this is not there it is purely a cylindrical arrangement no torque is developed but now will it be able to move let us see whether this will move so in order to derive an expression in this case again we have to write down equations and we start with the familiar equations that we had written earlier now there is an excitation source on the stator there is one on the rotor as well so there will be two equations in the electrical part so we write V stator that is the voltage applied to the stator is equal to stator resistance multiplied by the current that is flowing through it obviously if you are going to have a wire look here it will have some resistance there is some current flowing in and therefore there is a resistive drop plus P times the rate of that is P times the linkage of flux in the stator that is rate of change of stator flux linkage and similarly you have another expression VR which is the voltage applied to the rotor that is RS RIR plus P times IR this form of the expression as I said is fairly you it will be repeating this is a general expression voltage is always equal to a resistive drop plus rate of change of flux linkages that are there but this alone is not enough you then need to have an expression for ?S and ?R ?S which is the flux linkage of the stator can be written as self inductance of the stator multiplied by the current flowing through it again we are assuming a magnetically linear system but the linkages will not be just the result of flow of current in the stator alone there can be flux linkages here when there is current flowing in the rotor as well therefore there is a mutual flux linkage between the two and therefore it is a mutual inductance between the stator and rotor multiplied by the current IR and similarly you write an expression for ?R which is the flux linkage for the coil in the rotor that is then LR multiplied by IR plus MSR multiplied by IS so these are the expressions that need to be substituted for ?S and ?R here so far the procedure is the same as what we had done in the linear motion case and in the singly excited case we wrote the expression for the applied voltage as VRI plus PSI but there are two voltages now so you have two equations and ? is not just one there are two size flux linkage for the stator and the rotor and therefore you have both of them the next stage is to find out how much input we are giving V into I here again there is one stator coil one rotor coil therefore total power input the total power input will then be given by VS into IS plus VR into IR that is the net power you are giving some power to the rotor you are giving some power to the stator as well and this power has to be then equal to the resistance loss in the machine plus the rate of change of magnetic field energy plus the mechanical output power if we expand this is then equal to IS square into RS plus IS into P times IS plus IR square into RR plus IR into P times IR this is what we said must be equal to resistance loss plus P mechanical plus rate of change of field energy so DWF by DT the resistive loss as is usual is easy to identify it is I2R here and I2R here that is the resistive loss component and therefore this term plus that term must equal mechanical power plus DSI by DT now the question is how do you find out what is the expression for field energy in this case we said that field energy is simply half LI square but then there was only one excitation system a single coil that was there on the stator and therefore it is simply half LI square but now you have a coil on the rotor and you have a coil on the stator so we need to derive an expression for the field energy in a multiple coil system and more than one coils are there how do you get the field energy once we get that we will be able to put it in and solve that expression we will do that in the next lecture so in this lecture what we have seen is that we started with the singly excited system and reviewed how the force expression is obtained and we have seen physical examples of how this expression could be of use it can be used to design electromagnetic equipment and from the linear system we went into the linear system where you have a rotational movement the rotational systems are by far the most frequently used and in the case of rotational system we saw why a singly excited system with a cylindrical rotor is not likely to move at all you need some kind of saliency in order that you get it moving and it is important to remember that inductance is a function of the geometry of the system if the geometry is going to change then the inductance will change and again it is important to remember it is a function of the geometry alone if it is a magnetically linear system and that is what we are looking at and inductance changes with the rotor position means that there is going to be a generated electromagnetic torque. In the next case we need to derive the expression for the field energy and substituted here we will see that in the next lecture and for this lecture we will close now.