 In the last lecture we had seen how to derive the expressions for the generated electromagnetic forces and we found that in order to derive those expressions we need to know how to express the field energy in terms of the excitation systems that are there whether it is a single coil or whether there are more than one coil we need to know how to write down the expression for field energy and once we know the expression for field energy then we get an expression for the rate of change of field energy with respect to time and we use that we separate that out from the applied electrical input V into I and that then enables us to get an expression for the generated forces in the electromechanical system. We saw that the field energy WF can be written in a vector form using vectors for the various excitations that are given into the system. So if I is a vector then half I transpose inductance matrix multiplied by I where I is a vector of flow of input currents I1 I2 etc may be up to IN. This is for the case of a linear system and we then said that before we go ahead with the derivation of various expressions for electrical machines in the linear magnetic domain let us briefly pause and look at what the expressions will be if the magnetic system is not linear and we started looking at the non-linear magnetic systems magnetic circuit non-linear magnetic let us say and they are then characterized by a relation between current flux linkage. Let us consider a single excitation I and a flux linkage side then the relationship between them is not linear and is a non-linear curve had it been a linear system the curve would have been may be something like this straight line whereas now if we are going to look at non-linear magnetic this curve is not a straight line and it bends over that means that as you go to values of higher and higher flux linkages you require a flow of current I which is more than what you would have required to increase the flux in the lower levels of flux linkage so that is what is meant by this and while looking at the expression for field energy if we say that you have a coil with an applied voltage V and a current that is flowing in I with no mechanical output and no resistance which means resistance loss dissipation is 0 mechanical output power is 0 then all the electrical energy that is supplied goes towards increasing or decreasing the magnetic field and therefore if we are going to write V equals d Psi by dt then input power V into I is I into d Psi by dt and which is nothing but the rate of change of magnetic field energy with respect to time and we can write therefore dWf in the differential form this expression can be converted to the differential form which says that dWf is equal to I d Psi so if you then want to get the expression for the field energy you then need to integrate this expression dWf is a small change in the field energy is given by I into d Psi and if you want to find out the total field energy for a given current flowing which means that you enable the linkage to go from maybe 0 to some value then the field energy is given by the integral of this expression I d Psi going from 0 where Psi goes from 0 to some capital value Psi so if this is your final flux linkage Psi this integration would then give you this area so that area represents the field energy let us remove this line no longer necessary this is the field energy and we saw yesterday in the last lecture we can define another term which is then called as the co-energy which we can call as Wc the co-energy is given by the expression Psi di or rather a small change in the co-energy dWc is given by the expression Psi di and therefore co-energy is then up sorry dWc is then Psi di and then the integral Wc is given by integral of Psi di going from 0 to some final current I and by the very form of this expression one can understand that this integral Psi di would result in this area a small differential di integrated from 0 to this value of I would then give this area and therefore that area is field energy Wf and this area represents the co-energy Wc we assume in drawing this figure that the flux linkage Psi results due to flow of current I that means at this current you get this flux linkage so obviously from this graph from the picture that we have drawn one can see that Wc plus Wf is equal to Psi multiplied by I that is the total area of this rectangle that we have is Psi multiplied by I that is divided into two parts by this non-linear curve which is field energy and co-energy these two terms are useful to us in deriving expressions for the mechanical work done in mechanical forces let us see how that can be done initially when we wrote down the expression we said that no resistance and no mechanical output is there let us retain this no resistance but now if mechanical output is allowed in the system then what we would have is dwf by dt which is the rate at which field energy is increasing is then given by V into I which is the electrical input that we are giving minus mechanical power because from the total electrical input that you give something goes away as mechanical output then it is a remaining that goes towards the field energy so this expression does indeed make sense and we can then write this as this is V into I minus now let us look at a system which has a linear displacement which is what we started out with in the beginning and you can write that therefore as the force that is developed f multiplied by force into the velocity force multiplied by the velocity gives you the mechanical output and therefore one can write that as f into dx by dt we also know that voltage you are applying a voltage here and this voltage because you are applying directly across this loop this voltage has to be equal to the induced emf and this induced emf by definition we know should be equal to dpsi by dt and therefore voltage is equal to dpsi by dt and when you substitute this what you get is dwf by dt is equal to i into dpsi by dt minus f into dx by dt now this expression since all are differentiations with respect to time one can write it in the differential form as dwf equals i dpsi minus f into dx so let us look at this graph once again or may be we will draw it once again here this is your axis for i and here you have psi now if we represent the relationship between psi and i as this line this line is obviously valid for some value of x say x0 right if the value of x is going to change then you would get some other curve at another value x1 so obviously here we have drawn the curve x1 such that for the same current i this location x1 gives rise to a flux linkage that is lower than the at the position x0 it could have been also the other way around it is just arbitrarily we have chosen x0 and x1 essentially we are saying then that the energy in the field depends not only on how much flux linkage is there because it represents the area above this curve so for a given flux linkage some let us call that as psi1 for a given flux linkage the area above the curve obviously now depends upon where the curve is if it is this curve then you are talking about this area if it is this curve then it is that area plus what is there here therefore the field energy WF then depends upon we can write this as a function of psi as well as x if x is going to change then this curve itself changes if flux linkage is going to change then we are shifting this value at which we are computing the field energy and therefore this area obviously is going to change so if this is the relationship if this is the dependency of WF then we know from our elementary understanding of differentiation that a differential change in WF can then be written as the partial derivative of WF with respect to psi multiplied by d psi this partial derivative being evaluated from this expression by keeping x constant plus the partial derivative of WF with respect to x multiplied by dx this partial derivative being evaluated keeping psi constant this is a general expression if a function is going to depend on a variable 1 and a variable 2 then a small change in this WF can then be written as the derivative with respect to one variable keeping the other fixed and the derivative with respect to the other variable keeping the first one fixed but then the other expression also gives us the same thing and therefore by looking at the expression for DWF from here as well as here we can conclude that the this variable I must be given by doh WF by doh psi keeping x constant and then the force is given by – of doh WF by doh x keeping psi constant so if we know what is the expression for the field energy and that expression is given in terms of flux linkage and the displacement then what this says is that we can find out the mechanical force exerted by differentiating this expression with respect to x while keeping the flux linkage fixed during the differentiation if this is going to be a system which is going to rotate then obviously instead of f dx you would have torque into d ? and therefore here you would have doh WF by doh ? keeping psi constant and therefore you would land up with the expression torque is equal to – doh WF by doh ? keeping psi constant so in this manner then if you have an expression for field energy in terms of the two variables psi and x one can find out how much force is going to be developed or how much torque is going to be developed similar expressions can be derived based on co energy we have seen that co energy is given by psi into I – field energy that is the expression we have seen here WC plus WF is equal to ? into I indeed we can recall again that the sum of area above the curve and sum of area below the curve together gave this rectangular area which obviously has to be equal to ? into I and therefore DWC a small change in WC can be written as ? into DI plus I into d ? – DW and from our earlier expressions we already know what DWF by dt is or DWF is so we substitute this expression there and which results in ? DI plus I d ? – I d ? plus f dx so these two terms obviously cancel out so you get ? DI plus f dx is your DWC now again if you look at WC which is this area now if you choose the curve x one it is this area if you choose the curve x not determined by x not then you have to add this additional area in order to determine WC so obviously now WC depends upon at what value of I you are going to evaluate WC and at the same time which curve you are going to use so in a similar manner as before we can write WC to be a function of I as well as x and since we know that it is a function of two variables I and x we can also write a change in this function can be written as DWC by ? I keeping x constant multiplied by DI plus DWC by ? x keeping I constant multiplied by DX and these two expressions have to be the same because both represent differential changes in the co energy and therefore by comparing these two what we can say is that ? is equal to DWC by ? I keeping x constant and force is equal to DWC by ? x keeping I constant which means that if you go through the field energy route one can evaluate force as minus of change of field energy with respect to x keeping I constant or if you go through the co energy route then force can be evaluated as the rate of the change of the differential of co energy with respect to x keeping I constant. Now both these must yield the same answer the same expression which one you will use depends upon which is more easier for you to handle and these two should give the same expressions. Now it is important to understand that keeping ? constant or keeping I constant here does not have anything to do with the physical system attempting to keep ? or I fixed it is just a requirement for differentiation of WC or WI and has nothing to do with the physical system. So let us take an example to see how this can be done for example let us look at our linear system which we started in the beginning you had a plunger and a coil this was going to move inside if you excited and we know that in this case because there is only one excitation and this bar is going to move therefore inductance will change with respect to the position x we know that the field energy is given by half L as a function of x multiplied by I square that is the expression for field energy and our expression that we have derived for the mechanical force says that field energy is the rate of change of WF with respect to x while keeping ? constant and we know that ? is equal to inductance multiplied by the flow of I and therefore we can substitute in this expression in terms of ? or I is equal to ? divided by L of x so we substitute this there which means WF is equal to half of Lx multiplied by I square I square is ? square divided by L square which means this is half of ? square divided by Lx and now we need to take the derivative of this with respect to x keeping ? constant and the force is the negative of that expression so let us do it here so force is equal to – do WF by ?x keeping ? constant which is – do by ?x of ? square by 2 into 1 over L of x from the earlier expression and ? is kept constant during the differentiation and therefore this part can be taken out so you have – ? square by 2 into ? by ?x of 1 over Lx which is nothing but – ? square by 2 and the derivative ? by ?x of 1 over Lx is – of 1 over L square x multiplied by dL by dx we write it as derivative with respect to x because L is a function only of x there is only a single variable function and therefore we do not write ?L by ?x it is dL by dx which therefore is ? square by L square of x into dL by dx by 2 and we know that ? equals L of x into I and therefore ? square is L square of x into I square so we substitute that here and you get this is half of I square into dL by dx and if you remember look back on our earlier lectures the force that we derived for the singly excited linear motion system is the same expression right so this shows that one can get the expression for force from the field energy in this manner now how about the other expression we have also said you can derive the force from the co-energy expression and the co-energy expression says that force is obtained as ?wc by ?x keeping I constant and in this case wc the system we are looking at now is a linear system linear in I and ? I mean a linear relationship between I and ? we are expressing it in terms of an inductance coefficient and saying that the inductance coefficient is a function only of displacement and therefore if it is only a function of displacement this curve will be a straight line for a given position x if the position x is going to change then you might get another straight line for some other displacement. So if you now consider this curve for a linear magnetic system and you look at a value of ? here and a corresponding value of I now you see that this the rectangle ? I is obviously divided into two halves by the diagonal of this rectangle which is this curve and therefore because it is a diagonal of this rectangle the area above the curve must be equal to the area below the curve which means that in this case wf must be equal to wc and therefore wc is also equal to half of lx into I2 in this case therefore it is simple and therefore if you want to now evaluate the force is nothing but ? wc by ?x keeping I constant I is constant so keep it out of the differentiation process and therefore force is straight away half of I2 dl by dx so we get this in a single step if you use the co energy function through this route therefore one can use energy or co energy in order to find out the expression for force. Let us do one more example to illustrate the process let us say that we have a system which is a rotational system where the relationship between I and ? is described in this manner a0-a1 cos of 2 times ? multiplied by ? to the power of 1.6 now the expressions the system that we considered earlier is a magnetically linear system whereas what we are looking at now is a magnetically non-linear system and in this case we want to find out the expression for force as before we can use either the field energy or the field co energy so let us do using both of them now the field energy wf if you would recall is given by the integral id ? now this expression only gives us the relationship between I and ? it does not give you the field energy directly in the earlier example we started out with the expression for the field energy now you have to determine the field energy given this expression and therefore first do an integration id ? so let us say this goes from 0 to some value ? which is integral from 0 to ? substitute for I from this expression so this is a0-a1 cos 2 ? multiplied by ? to the power of 1.6 d ? which is now in this integration ? is the only thing that is going to change so the angle is fixed so we can take this out of the integration so this then becomes a0-a1 cos 2 ? multiplied by ? to the power of integral of ? to the power of 1.6 d ? is ? to the power of 1.6-1 that is 2.6 divided by 2.6 going from 0 to ? and obviously at the lower limit this function reduces to 0 because you are multiplying by 0 and therefore one can write down this expression as ? to the power of 2.6 by 2.6. In this case what we are interested in determining is the torque T and the torque T is given by – of ? wf by ? keeping ? constant so this is – of ? by ? of this expression keeping ? constant so ? is constant it may be taken out of the derivative and therefore this is nothing but ? to the power of 2.6 by 2.6 multiplied by the derivative of this term the derivative of this term is – of a1 the derivative of cos is ? to ? multiplied by 2 and cos is ? so this then becomes plus but then you have a negative sign outside here so that negative sign continues to remain so what you have is – of 2 a1 sin 2 ? multiplied by ? to the power of 2.6 divided by 2.6 is the expression for the generated electromagnetic torque T one can do this using the co energy route so let us do that also the co energy just like what we have done here we have first determined the expression for the field energy so just like that we need to determine what is co energy first co energy is given by integral of ? di and therefore we need to get an expression for ? here this is integral going from 0 to I ? is given by I to the power of 1 by 1.6 divided by a0 – a1 cos 2 ? to the power of 1 by 1.6 that is we are converting this expression writing ? in terms of I multiplied by di in this integration again angle is kept fixed so that can be taken out of the differentiation so you get 1 by a0 – a1 cos 2 ? to the power of 1 by 1.6 multiplied by I to the power of 1 by 1.6 integral of that so 1 by 1.6 plus 1 that becomes 2.6 by 1.6 divided by the same exponent so 2.6 by 1.6 this is your expression going from 0 to I and since at the lower limit it is 0 the expression obviously is 0 so we replace the upper limit here so this is your expression for Wc. And in order to derive the expression for the torque we see that the torque is given by ? Wc by ? ? this is an expression for the force for a linear system for a rotational system it would be ? Wc by ? ? and so you need to differentiate this expression with respect to the angle keeping I constant which means this term is kept constant so the torque is then I to the power of 2.6 by 1.6 divided by 2.6 by 1.6 multiplied by d by d ? of a0 – a1 cos 2 ? to the power of – 1 by 1.6 so we can write this as I to the power of 2.6 by 1.6 divided by 2.6 by 1.6 into a0 – a1 cos 2 ? to the power of – 2.6 by 1.6 multiplied by – 1 by 1.6 so let us now simplify that expression let us simplify this expression what we have is torque we have obtained as – 1 by 1.6 multiplied by I to the power of 2.6 by 1.6 divided by 2.6 by 1.6 multiplied by 1 by a0 – a1 cos 2 ? to the power of 2.6 by 1.6 when you take the differentiation you need to differentiate this term as well so this term gives you the differential of cos is sin – sin so a1 sin 2 ? multiplied by 2 and the differential of cos is – sin and therefore that also goes away so you have this multiplied by 2 times a1 into sin 2 ? this is the expression that you have and we look at our original expression I is given as a0 – a1 cos 2 ? into the power of 1.6 and therefore if you look at this term the torque can be written as – 1 by 1.6 multiplied by 1.6 by 2.6 multiplied by I divided by a0 – a1 cos 2 ? this whole thing raised to the power of 2.6 by 1.6 multiplied by 2 a1 sin 2 ? so if you now compare these two expressions one can see that what is written inside here is nothing but ? to the power of 2.6 and this cancels out so what you have is – 2 times a1 sin 2 ? multiplied by ? to the power of 2.6 divided by 2.6 let us hope it was the same expression now we compare the earlier expression that we got here through the field energy route it also is the same as what we have got here now in this case it has so happened that the algebra involving the co-energy route is more involved than the algebra involving the field energy route one can choose either of these two routes but the answer is ultimately the same one can we now have an expression for the co-energy I would like to see one more thing the field energy is given by a0 – a1 cos 2 ? multiplied by ? to the power of 2.6 by 2.6 and the co-energy is given by this expression I to the power of 2.6 by 1.6 divided by a0 – a1 cos 2 ? 1 by 1.6 x 1.6 by 2.6 what we want to find out is what is this expression Wf plus Wc obviously you would know Wf plus Wc must equal ? into I we just want to verify that fact so that we are sure of what we are doing and let us try to add those two up Wf plus Wc is a0 – a1 cos 2 ? to the power of ? into ? to the power of 2.6 divided by 2.6 plus I by a0 – a1 cos 2 ? to the power of 1 by 1.6 to the power of 1 by 1.6 x I to the power of 1 multiplied by 1.6 by 2.6 is what you have here and one can add these two expressions I will leave it to you to verify that this indeed will give you an expression ? multiplied by I which is the total rectangular area that is contained within this curve. So in this class in this lecture then we have seen how to make use of the relationship between field energy and field co-energy and use those to derive an expression for the mechanical torque or the mechanical force that is there in the system. These expressions may not be useful to us in the magnetically linear case but in the non-linear case we do not have any other way to do it this would probably be the best way to do it. If you know the analytical expressions one can undertake the differentiation as such but if one does not have analytical expressions to express WF or WC one has to take recourse to some kind of numerical method which can enable us to do all these differentiation. So we will stop here for this energy looking at how one can get the forces from the energy. From the next lecture onwards we will not be looking at magnetic non-linear systems we will still be looking at linear systems and we will go ahead with our analysis of electrical machines which are magnetically linear. We will stop here.