 I welcome you to the next lecture on modeling and analysis of machines in the earlier lecture we had seen the derivation of expressions for the generated torque in a singly fed rotational system this expression we saw it to be half of I squared dl by d ? and we understood that this would exist therefore only when inductance can change with the angle of the rotor. If inductance does not change with the angle of the rotor then there is no this expression becomes equal to 0 if dl by d ? is 0 and therefore in a singly excited system it is essential that the rotor has saliency this we saw yesterday and we started looking at a system where the stator is cylindrical and the rotor is also cylindrical the stator has one coil that is placed there and the rotor also has one coil that is placed there. And we wanted an expression for the field energy of the system field energy Ws is where we stopped in the last lecture. So how to get field energy so let us look at field energy in a little more detail before we get to this system let us look at a single coil and understand how we derive the field energy expression for a single coil in a single coil if we picture is it in this manner you are applying a voltage V there is a current flowing I neglect resistance for the time being because we are not interested in the resistance losses which therefore means that the electrical equation becomes V equal to the induced EMF E across this because there is a flow of current ? there will be an induced EMF because of the way I is going to change d5 by dt and all that so this would then is also equal to P x ? right and power that is input is V x I and that is therefore I x P x ? in the system here let us also assume that there is no movement there is no element that physically moves that means mechanical output is not there from the system mechanical output equal to 0 and therefore all the electrical input that is there that has been given to the system goes entirely as field energy that is stored which means integral of Vi dt must be equal to field energy which is equal to I x integral of I P ? dt we had seen sometime in an earlier lecture that from this one can derive the energy in the field to be half Li square in the case where only one single coil is there now in a in an electrical electromechanical system like the one that we have there there may be something that can move for example in the in the singly excited linear motion system that we saw there is an iron bar that can move and if we are going to derive this expression here the implicit assumption that we are doing in deriving this now is that the location of the bar has been fixed for a given location of the bar there is a fixed value of inductance and you can then say that for that location any electrical input that you give if the bar is not going to move will have to be entirely stored as field energy now if you have more than one loops like this then for example let us say that you have a coil here and another coil place somewhere else you are applying a voltage V1 here and a voltage V2 here again assume that there is no movement no mechanical output under that case again the entire energy that is given through this terminal as well as this terminal has to go as field energy that is stored and the energy that is given to this can be written as V1 into I1 plus V2 into I2 integral with respect to time now how do we do this integration because unless you know if you want to integrate this then you must know V1 as dependent on T you need to know I1 as another dependency on T and then similarly V2 as a function of time I2 as a function of time if you know all these expressions then you can substitute into this expression and integrate you will get the energy stored in the field at any given instant but it so happens that in this kind of a system it is unnecessary to know all these things because the field energy system the magnetic field system can be assumed to be a conservative system which essentially means that there is no loss in the system in an actual case if the magnetic field is going to change and you have in the area here if you are going to have iron then as the field undergoes a change in the iron there will be loss there will be a hysteresis loss there will be eddy current loss and because of that the energy that you give to the system some of it will be expanded as heat if you assume that these losses do not exist because in actual systems all effort is made to minimize these losses you attempt to have very low hysteresis loss and you also want to reduce the amount of eddy current loss because all electromagnetic equipment the iron is going to be laminated the entire idea of having this is to reduce eddy current loss eddy current loss reduction and hysteresis loss is again avoided by having appropriate alloys of iron you choose material the material selection is appropriately done to reduce hysteresis loss you cannot eliminate it all together but definitely you can reduce the losses and because of that you can do the analysis by neglecting the loss that occurs in the iron that is there and if you neglect the loss then all the energy that you input goes towards the field or work that is done this also means that the work that is done because there is no losses the work that is done is independent of is independent of actual trajectory of the state this means that if you draw a graph of let us say with respect to time you draw a graph of Y instead of looking at this system where there is an I1 and I2 let us take the example of the first case where there is a single voltage and single current I you may do an experiment where this variable I starts from 0 and then reaches a value let us say I1 in this manner it does reach I1 this is one experiment you may do another experiment where this variable I starts increasing and reaches I1 like this a third experiment where I start and increases to I1 like this. Now the way in which I has reached I1 is different in all these three situations however at this instant of T after it has reached I1 the energy in all these three situations the energy stored in the field is still half Li2 that means the energy stored in the field is independent of how this current actually reaches the value of I1 it only depends on where it has reached if it has reached I1 it is half Li2 is the energy right so it is independent of how the actually I1 is going to change we make use of this idea in order to derive the energy for this system also how to do that now let us look at the expression for V1 V1 is R1 I1 plus P into ?1 V2 equals R2 I2 plus P ?2 this is nothing but R1 I1 plus P of ?1 we will expand and write it as L1 I1 plus M into I2 and here we will write R2 I2 plus P L2 I2 plus M into I1 the electrical power input is then given by R1 I1 square plus I1 into Lp L1 I1 plus Mi2 plus R2 I2 square plus I2 P into L2 I2 plus Mi1 now we started out by assuming that the resistance may be neglected so we are looking only at the energy stored in the field and therefore it does make sense to neglect the resistance if you have resistance you just add resistance loss to the whole thing. So this is the electrical input that is given the energy since there is no mechanical motion mechanical output is 0 so all the energy stored as field energy and that is therefore integral of this V1 I1 plus V2 I2 DT so how to do this integration since we know that the work done is the energy stored if you are giving an electrical input the work that you are doing is the electrical energy that you are giving to the system and that energy that is stored is independent of how I1 and I2 reach their end state and we will therefore assume that we do this experiment in stage 1 and stage 2. In stage 1 we allow I1 to go from 0 to the value that we are interested in order to find out the energy let me call that as I1 and in stage 2 and in this stage 1 I2 is held equal to 0. So stage 1 is you fix I2 at 0 and allow I1 to go from 0 to I1 and then in stage 2 you fix I1 at the value I1 and allow I2 to go from 0 to some I2 so in order to estimate the energy stored in the field when I1 has a value equal to I1 and I2 has a value equal to I2 in order to arrive at the expression you do you evaluate this integral in two stages the first stage is this and second stage is this. So if you do that how does this integration now get simplified you are essentially going to integrate this expression is nothing but I1 into P L1 I1 plus MI2 DT plus I2 P of L2 I2 plus MI1 DT this is the integral that we have to do now this is therefore split into two stages in the first stage I2 is held 0 so in the entire expression you put I2 equal to 0 and then this integral reduces to I1 times P L1 I1 DT and then the next term is MI2 but I2 equal to 0 so this term does not give anything in the second expression I2 multiplies this whole thing and therefore I2 being 0 this term does not give anything so in the first stage the integral is only this much plus in stage two what happens to this integral in stage two I1 has already reached a value equal to this I1 is equal to I and therefore P of L1 I1 is 0 so the first term does not give anything the second expression however is then I1 I1 has now reached a value I1 and then you have P MI2 DT plus in the second stage this term is definitely there so you have I2 P of L2 I2 DT plus is I2 P of MI1 I1 being fixed at this value equal to I1 P of MI1 is 0 and therefore there is nothing more to be done so this then gives us half L1 I1 square this then gives us I1 is I1 here this then gives us MI1 I2 and the third expression gives us half L2 I2 square so the field energy can then be written as half L1 I1 square plus half L2 I2 square this is the self inductance in both plus the mutual inductance term gives us MI1 I2 this expression note that it can also be written as follows WF in this case can be written as half of a row vector containing I1 I2 a square matrix containing L1 MM L2 and a column vector containing I1 I2 let us evaluate this and see what is the expression so this can be simplified as half of I1 I2 and this evaluates to L1 I1 plus MI2 and you have MI1 plus L2 I2 and this simplifies to half of I1 into L1 I1 plus MI2 that gives us L1 I1 square plus MI1 I2 plus MI1 I2 plus L2 I2 square which then essentially is half L1 I1 square plus MI1 I2 plus half L2 I2 square which is the same expression what we had earlier so this form however is very useful one can simplify the notation and then say that the field energy is equal to half of a vector I transpose square matrix L multiplied by I it would be useful to have the expression in this form in the next situations we will see in the later lectures for the present however this expression is of use. So having derived this expression let us put it back into the earlier system that we saw we are looking basically at the rotational system having one coil on the stator one coil on the rotor the rotor being cylindrical and this coil being placed in the rotor somewhere here coil on the stator is somewhere here and we had said that the power that is input V1 I1 plus V2 I2 is nothing but now we will include resistance also and therefore that is I1 square R1 plus I2 square R2 plus I1 into P P1 plus I2 into P of P2 and therefore we now substitute the expression for P1 and P2 what you have is I1 square R1 plus I2 square R2 plus I1 into P of L1 I1 plus MI2 plus I2 into P of L2 I2 plus MI1 and this expression as we have seen in the earlier systems this power that is input must be equal to resistive loss plus mechanical power plus rate of change of field energy and we know the field energy expression now so let us calculate the rate of change of field energy the rate of change of field energy is DWF by DT we are differentiating this expression the first term would give us which is okay let me write that again that is D by DT of half L1 I1 square plus half L2 I2 square plus M into I1 into I2 so this then gives us derivative of this in this term L1 is a fixed number because the stator is going to be cylindrical rotor is also cylindrical therefore the self inductance of the stator is not going to vary with the rotor angle and the self inductance of the rotor will not vary with the rotor angle either irrespective of where the rotor is the stator will not know the difference because the shape is fully symmetrical and therefore L1 is fixed number which means the derivative of this will be 2 times I1 into L1 multiplied by half which is nothing but L1 I1 DI1 by DT plus L2 I2 DI2 by DT in the next expression you have MI1 and I2 I1 and I2 are obviously dependent on T they will change with respect to time mutual inductance will also change with respect to rotor angle and therefore with respect to time if the rotor is going to rotate one can see that the inductance will depend upon how much of flux that is generated by one coil is going to link the second coil obviously if the rotor were such that the rotor coil is placed here then all the flux that is generated by this coil almost all of it is going to flow in this manner then all the flux will link this coil therefore the mutual inductance will be highest all flux is linking this on the other hand if the rotor were oriented such that this coil instead of being there is placed here the flux lines generated by the exciting the first coil on the stator would be traveling like this will have a loop like this and one can see that there is no flux line that is trying to cut across the plane of this coil and therefore the flux linkage in the second coil is 0 which means the mutual inductance is 0 therefore in this system as the rotor rotates the mutual inductance which is nothing but a reflection of how much of flux generated by the first coil is going to link the second coil the mutual inductance therefore is going to change with respect to the orientation of the rotor and therefore in this expression mi1 i2 all of them can change with respect to time so when you differentiate it you will have different terms the derivative will have i1 i2 into dm by dt then plus mi2 into di1 by dt plus mi1 into di1 by dt all these terms will be there so this expression now describes the rate of change of field energy and what do we get from input power we can now we know that this component is the resistive loss so let us leave that aside we now need to look only at these two terms and identify what part of it is rate of change of field energy then the remaining part is the mechanical part so let us expand this expression as well if you call this let me in order to write it on the next section let me call this expression as e then e is i1 into l1 di1 by dt from the first term plus i1 into i2 into dm by dt plus i1 into m into di2 by dt plus i2 into l2 di2 by dt plus i2 into i1 into dm by dt plus i2 into m into di2 by dt so these are the terms that come if you differentiate this expression the first term is i1 l1 does not vary with time therefore the first term gives us i1 l1 di1 by dt in the second term both m and i2 are varying with respect to time so i1 into derivative of m multiplied by i2 plus i1 into i2 multiplied by derivative of m two expressions come two terms this one will have i2 l2 into di2 by dt because l2 does not change with respect to time and this term will again give two expressions which are listed there so now we need to compare these two expressions so essentially what we want to do is e-dwf by dt is expected to give us the mechanical output so if you subtract that this term goes away and then i2 l2 di by dt also goes away then mi2 di1 by dt mi2 this must be di1 by dt so mi2 di1 by dt gets cancelled and then you have mi1 there is a mistake here this should be i2 this term and this term gets cancelled and then you have one i1 i2 dm by dt getting cancelled with one i2 dm by dt and therefore the remaining expression is this which means that the mechanical output power is equal to i1 i2 dm by dt which can then be written as i1 i2 dm by dth multiplied by dth by dt and dth by dt is the angular speed omega and therefore this part must be the torque and therefore we can write torque that is generated the electromagnetic torque Te is nothing but i1 i2 into dm by dt in this case as we have already seen the mutual inductance will change with respect to the angle of the rotor because the flux linkages are going to change and therefore there will certainly be a generated torque for some given values of i1 and i2 let us also see how this mutual inductance is going to change with respect to time you know that in this location of the rotor as shown here the m is equal to 0 mutual inductance is 0 and if the rotor is going to be here mutual inductance is highest now in between the two mutual inductance will vary as mutual inductance will vary as some maximum value m hat multiplied by cos of the rotor angle that is if you consider the rotor angle to be 0 when it is here then if the rotor moves by some distance through some angle then the mutual inductance when the rotor is there is given by mutual inductance here multiplied by cos of this angle and therefore if you substitute this expression there you get generated torque equals i1 i2 dm by dth is nothing but – m hat sin ? r ? is nothing but ? r which is angle of the rotor so this is the expression for generated torque in the machine that we have just now seen with cylindrical stator and cylindrical rotor now it is interesting to look at this expression in little more detail if you look at this the torque the generated torque Te as a function of ? r this says that at ? r equal to 0 the generated torque is 0 that is extended this side ? r generated torque is 0 and the generated torque becomes highest at ? r equal to 90 degrees there is a negative sign here note that there is a negative sign and therefore the wave shape of torque if you draw this expression as a function of ? r then this is 90 degrees and then goes back to 180 and then 270 degrees and 360 degrees on this side that is how the wave shape is going to look this maximum value for a given current is i1 i2 x m hat this is – i1 i2 x m hat now if you see this at this point and this point the generated torque is 0 so it is here and here as well which means one would think that if the rotor is at this angle there is no tendency to move if the rotor is at this angle there is no tendency to move but let us look at it a little more closely suppose the rotor is here and let us say that due to some disturbance the rotor is shaken a little and let us say the angle of the rotor from 0 becomes this value if it becomes this value the torque is greater than 0 and therefore the rotor will try to rotate in the same direction of increasing angle please remember that we are having a situation like in the earlier case of the linear motion force acts to increase displacement this is the notation we are following that is positive force acts to increase the displacement similarly positive torque acts to increase rotor angle. So if you if the rotor happens to be displaced by a small angle and at this angle the generated torque is greater than 0 it will now try to move the rotor in increasing in the direction of increasing angle and therefore it will come back to 0 whereas due to some disturbance if it happens to be displaced on the other side then the torque generated is negative and therefore it will try to bring the rotor back in decreasing angle so it will come back here but whereas on at this point if you see if the rotor is here and then due to some disturbance it happens to move a little and if the rotor happens to move to this angle then the torque generated is greater than 0 and it acts in the direction of increasing angle therefore the rotor will try to move further away as it increases the generated torque increases therefore it will move further and further away and finally come to this point. If the disturbance is such that it makes the rotor to come to this angle some small disturbance then the generated torque is negative and it will make the rotor move this way in decreasing in the direction of decreasing angle because torque is negative and therefore the rotor will move further and come to this point. So this angle is what is called as an unstable equilibrium point and this one is then a stable equilibrium so the system that we have looked at which consists of a single coil on the stator and a single coil on the rotor if you excited it will at best move and then settle down at either 0 degrees or at 360 degrees it will come and settle down somewhere if there will not be any tendency to go on moving so it will settle down this system will not go on moving so how to get a system that is going to move all the while in order to do that you have to have a means of generating a rotating magnetic field and if you want to generate a rotating magnetic field a singly excited stator is insufficient we will have to go for more excitation systems on the stator and then probably more excitation systems on the rotor as well before we get to that it would be instructive to look at energy we have been looking at field energy and it is instructive then to look at systems that have non-linear magnetic though in most of the analysis that we are going to do we are going to assume that the system is linear it is instructive and it is useful to look for a short while at what will happen if the system is non-linear because in actual cases all the systems are of this variety and one needs to know how to handle those systems it is not easy to handle these systems by analytical means because the expressions are not amenable to and analytical handling and therefore one has to take recourse to specialized software like the FEM software that we have seen during the various lectures earlier but one needs to know how to go about doing in order to do this we will find that the idea of energy place an important part energy stored in magnetic field work done is going to come from energy that is stored. So let us just briefly look at this before we end the session today we will look at this in little more detail in the next lecture but this is just to introduce you to the idea if you write the expression V equal to PSI which is what we started out with in the lecture today and power that is input is V x I that is I x PSI then the energy that is input is integral of VI DT which is integral of I x PSI now in order to be a little more exact what we could say is this integral is VI DT we will take a change in the variable of integration and then designate the limits as from 0 to T this is I PSI DT again integral from 0 to T and since this is integral of a differential one can write this in the differential form itself as integral from 0 to PSI I times D PSI again make a change in the variable it is I times some D x may be where x takes the value from 0 to PSI. Now it is interesting to note you can just look over this until we get to the next lecture if you plot PSI versus I when the system is not linear this curve is going to look something like this and what is the implication of this integration I times D x from 0 to PSI this means that we are taking a small differential D PSI or DX along this path and integrating from 0 to some value PSI that is there this integration you can appreciate will then give us this area associated with the curve right. So one can therefore infer that energy is given by this integral and therefore this integral is nothing but your field energy this area represents field energy then this area which can be represented as integral 0 to I PSI DI is then given a name co-energy WC and we will see in the next lectures that the energy WF and the co-energy WC this is called as co-energy this WF and WC play a very important role in determining the generated torque if it is a rotational system or the force if it is a linear movement system in a non-linear arrangement in a non-linear system how we use this in order to in order to estimate the generated torque we will see in the next lecture we will close for this lecture.