 the last lecture in this course on modeling and analysis of machines we had looked at inductances we had looked at how inductances can be divided we had seen that inductance can be divided into different forms for convenience one is the leakage inductance the leakage inductance is then the effect on the electrical electrical network due to flux which is not really flowing through the iron and linking other circuits that may be there but predominantly for a flowing through air and then we have magnetizing inductance this indicates the flux that is linking everything else and majorly flowing through iron which is then which is then brought into existence by your excitation so that represents the magnetizing inductance and then we have also mutual inductance when you have two or more electrical circuits which are linked by the magnetic flux then how flux that is linking a different electrical network due to excitation and another network that is described by the mutual inductance and then we had also seen how the idea of magnetic circuit is born this arises because of your excitation and the flow of the magnetic flux you have a core let us say and a source of excitation we have seen in the last class that if some current flows into the score you have a magnetic field set up and in order to understand how much of the field is there how many how many flux lines are there how much is 5 how much is the flux density that is B in order to arrive at these numbers one can redraw this arrangement in the form of equivalent electrical circuit where one can talk of the MMF source and then you have a resistance which is then called as the reluctance this is then your MMF source and equivalently the source divided by the value of reluctance gives you ? and ? divided by the area of the circuit gives you B so all these things can be done to analyze how this the system here is going to behave now in order to see the real effect of these things let us do a numerical example which will help to understand all these ideas so in the numerical example let us say we have a system like this a core let me draw a three dimensional extension of the core that is your core and then coils are wound around this okay so let us say that this coil has 10 turns that is the number N and then if you excited flux flows around this path and the mean length of this flux path which is called as a mean magnetic path length let us say is 375 mm and I is 6 ampere we are interested to find out what is B in iron within the iron region for this let us consider that this material out of which this is made has a relative permeability relative permeability equal to 5000 so under the circumstances we want to know what is the value of B that is how much flux density is there within this material so how do we do that we know from the earlier analysis we did in the last lecture that the mmf which is N x I is given by flux multiplied by the reluctance R and the reluctance R is given by ? the reluctance is given by length in the core divided by ? of the core into area of the core which can be written again as since ? divided by area is nothing but B we can call that as flux density in the core multiplied by length of the core divided by ? of the core which is BC x length of the core divided by ?R x ?0 we know the value of ?R and therefore this is BC multiplied by length of the core is 375 mm that is a mean magnetic path length you convert it to SI units that is in meters so that is 0.375 meters divided by ?R is given to be 5000 so 5 x 103 and ?0 is 4p x 10 to the power of –7 that is equal to N x I Ni is 10 x 6 so that is 60 so if you evaluate this you will get this is BC x let me look at that number so that is 59.6831 which then means that BC is equal to 1.005 Tesla so you get a value of B in this core which is equal to 1 Tesla now let us see this is good if you are going to introduce an air gap air gaps are very important in electrical machines and we need to understand what is the effect of having an air gap. So let us say that you are now introducing an air gap in the ion path so that you now have an air gap and the length of this air gap that length is let us call it Lg, Lg is 1mm so 1mm is a very small air gap that is introduced and let us see with that small air gap what happens to these numbers okay so if there is an air gap the expression then becomes Ni equals ? multiplied by the reluctance of the core plus the reluctance of the gap that is a net reluctance look at the equivalent now you have not just the reluctance of the core you have reluctance of the core and the reluctance of the air gap coming in series and this is N x y and therefore one can write the expression Ni equals ? into reluctance of the core plus the reluctance of the air gap that is equal to ? multiplied by the reluctance of the core is as before Lc by µr x µ0 x AC now note that since we are introducing an air gap in the ion path Lc which was 375mm earlier now you have to subtract that 1mm so it becomes 374mm and the remaining 1mm is passed on to the air gap so that is Lg divided by µ0 x A of the air gap and the flux is going to pass through the air gap as it passes through the ion so let us assume that the area through which flux passes in the air gap is the same as the area of the core and therefore AC and what is AG these are same and therefore one can further simplify this expression since AC and AG are the same you can take that out call it A then you have Lc by µr x µ0 plus Lg by µ0 now you know the values of Lc Lc is 374 Lg is 1mm µr is known µ0 is known so let us put the numbers into this expression by A is nothing but B and therefore that is the flux density multiplied by let us get the numbers and that works out to 59.5239 plus 795.7747 n x y remains the same you have the same number of turns the same current flowing and therefore if you divide that you get B equals 0.07 s one can see that the value of B has dramatically come down and that has been brought about because earlier this number is the same 59 now that the air gap has reduced I mean Lc has reduced a small difference between this number and this number but otherwise they are almost the same but the major difference has been brought about by the addition of this term which is due to the air gap and so if you are going to introduce an air gap in this loop then what you have is a substantially lower flux density as compared to what it was without the air gap how will you now increase this level this B here to the same levels as it was earlier the ways to do it would be to increase n x y you cannot do much with this even if you are going to increase the value of µ0 µr it is not going to help because this number is going to stamp out whatever you do here right so the only way to increase B is then to increase the value of ni or of course the other way would be to decrease this number which means you have to decrease the value of Lg which means a smaller air gap will give you a higher flux density or you need to increase the number of turns and or increase I which means effectively increasing the excitation to the loop here this is important right but you can also notice another thing in this this number is fairly small as compared to this number which means even if µr is going to change in some reasonable range normally depending on the level of B µr may change your BH loop in physical materials is not really linear and therefore µr is subject to change even if µr is going to change you see that over the addition of these two the effect is not likely to be very high because this number is substantially higher than this number and therefore if you have an air gap change in the iron material due to levels of B being higher or lower is not going to substantially affect the value of B which is an effect which is good because it helps you fix the level of B irrespective of changes in the material. So using this one can now estimate inductances one can find out what is the value of the inductance of this wire loop as you know inductance is given by n2 divided by the reluctance and the reluctance is that is n2 divided by length of the core divided by µ into area of the core this is for the case where there is no air gap if there is an air gap we will have to add that. So first let us look at the case without the air gap and let us say that the area of the core is 625 mm2 so if you substitute the numbers then the answer that we have is that this inductance is 1.04 mH fairly good induct on the other hand if you introduce an air gap then the inductance is given by n2 divided by the net value of r which is rc plus rg and if you do that sorry rc plus rg one can compute these values you find that the reluctance is 0.07 mH you can see that the inductance has come down substantially because of the addition of air gap so this is another important effect of having an air gap the inductance comes down substantially. Now as I mentioned before inductances and air gaps are very important to the behavior of electrical machines and we will see now why inductance is very important air gap is of course very important because air gap is the one that allows the rotor to rotate as I mentioned earlier if there was no air gap you would not be able to move anything right. So to understand this inductance effect of inductance in the behavior of electrical machines let us take a very simple system okay we can call that as a singly excited linear system linear in movement okay that is let us describe the system I have a coil wound around something a hollow length that is wound and within that hollow length I have an iron bar that can move inside this hollow and this iron bar is fixed to the wall fixed to a fixed level through a spring okay and this iron bar let us say has a mass of m the spring has a spring constant of k. Now in this we want to excite the system with a voltage V plus minus and in response to which it will draw some current I now intuitively let us see how this is going to behave we know that if a flow I is going to happen through a loop you would have seen this example you would have seen this image that if you have a wire loop and you have some flow of current I in this we have already seen that a wire loop in which some current I is flowing is going to generate magnetic field and this field looks very similar to what would you what you would have had you put in a normal magnet right. So you are going to generate a magnetic field if this current is going to flow this way then you effectively have flux lines that would be going into the board which means that this is going to look like a south pole from this side on the other hand if you have a wire loop through which current was flowing in the other direction then flux lines would come out of the plane of the board and this would look like a north pole okay. Now this loop is going to behave in a similar manner you are having some flow of current here and that is going around the loop which means that this will generate a flux it will generate a magnetic field that is going to go around it and you know that if you are going to have a magnetic field and iron nearby it would get attracted right. So you would have this iron bar let us call I will write this here that is an iron bar this iron bar would get attracted to this wire loop it would tend to since it is a hollow loop here so this will tend to get attracted and go within this hollow but at the same time it is now restrained by the spring which will then as it moves this will exert a force withdrawing it back intuitively this is what we see will happen but our interest is to describe the behavior of this by means of equations so that we can now estimate how much applied voltage should be there in order to generate how much force how much flow of current as this bar moves how will the system respond how much current will flow how much force is experienced by the bar all this we want to find out by means of having equations. So how does one start looking at this obviously this system is an electromechanical system you have an electrical side to it and you have a mechanical aspect to it also there is something going to move so you need to have systems describing what is happening here and what is happening here will in turn depend upon what is going to happen here. So let us see how to write down the system equations the first part is the electrical circuit where you write down equations for the applied voltage V you know that if there is a voltage V that is applied similar to the equations that we have written in the earlier lectures you have V should obviously be equal to R into some resistive drop where R mean R into I which is the resistive drop R is the resistance of that wire loop so you have R into I plus you have the term that now occurs due to change in the magnetic field due to the expression we saw earlier that is P into ? this term comes about because of Faraday's law which says that whenever there is a change in flux linkage there is an induced TMS flux linkages will change because in the circuit here there is some current going to flow and we already have understood that because of this current there will be a field and this bar will get attracted now as the bar gets attracted the inductances will change and therefore that will result in change in I and hence change in the magnetic field and because there is a change in the magnetic field you will end up with this term P ? right now this is the electrical equation ? on the other hand we have seen can be written in terms of the inductance and the flow of current I remember P is a notation which stands for the operator D by dt the differentiation operator with respect to time ? is given by inductance multiplied by I inductance is going to depend now on where this bar is if you place the iron bar far away from the wire loop it would have display some inductance but as the bar approaches closer and closer to the wire loop then the inductance will change inductance is likely to increase because the bar is now in the vicinity and the reluctance is experienced by the loop will be smaller. So you have L x I so if you substitute this expression back there you have V ? R x I ? P of L I L is some function of x where x denotes the location of the iron bar let us say that this position of the iron bar now corresponds to x equal to 0 and x increases along this direction okay the iron bar is expected to move so this is what it is P is the operator D by dt so if you now expand this expression you have R x I plus this is differentiation of L and I which both depend on T and therefore you have to apply the differentiation rule and this therefore becomes I x D by dt of L plus L x D by dt of I normally in electrical in the network analysis what you would have seen V is equal to RI plus LDI by dt now that is true only as long as L is fixed if L is a fixed number then this term goes to 0 you have the familiar RI plus LDI by dt now that L is not fixed you have to account for this term as well so this is the expression for input voltage now let us look at what is the input power that is consumed input power is from the electrical side there are no other inputs to the system and the input power is therefore V multiplied by I this is input power instantaneous power instantaneous electrical power that is V x I which is nothing but I square x R plus I square x DL by dt plus LI DI by dt this is input electrical power now in this system that we have here you are giving electrical input and that has to be consumed by the rest of the system so how is this input power going to be consumed obviously this wire loop has certain amount of resistance so if there is going to be flow of certain I through the resistance then there will be a resistance drop and therefore a loss in the resistance that loss as you all know would be given by I square x R apart from that there is going to be mechanical output this bar is going to move as soon as you apply this V and there is therefore a mechanical output so this input that is given V x I has to be consumed as resistive loss plus mechanical power mechanical power and then what else now when this rod is going to move we expect the variable I to change input current is going to change and we know that if input current is going to change the magnetic field associated with that is also going to change and field is nothing but stored energy and therefore there is as I is going to change there is a rate at which the stored energy is going to change there is a change in the stored energy and therefore that is then reflected here that is rate of change of field energy if the iron is also associated if it is a non ideal iron where iron normally has hysteresis loss as the field is going to change increase and decrease it would not retrace this path back and there is therefore a hysteresis loss in the iron apart from that if the field is going to change in the iron there are going to be eddy currents that are set up and therefore eddy current loss will be there so all that will also come into this but for this analysis let us neglect hysteresis and eddy current losses let us neglect iron losses on the whole and therefore if that is neglected input V x I will have to be consumed as resistive loss mechanical power and rate of change of field energy in this expression it is easy to identify what is resistive loss it is very obvious that I2 x R is the resistive loss therefore this part has been identified but we do not know what are these two yet right so we need to find out what this is how to find out we know something about magnetic field energy let us see how to find out the stored energy in the magnetic field let us call that as W max if you have a wire loop some coil that is there for the moment let us assume that resistance is not there resistance is 0 negligible we have already identified the resistance part we are not interested in that anymore so let us say this coil has no resistance and you apply a voltage to it V and this voltage is going to change it is a changing voltage that you are applying if that is so then in response to this voltage V there should be some current I that is flowing which will then set up a magnetic field we have already seen those things and because this voltage is changing this I will change magnetic field will also change and therefore there is an induced EMF here and by lenses law induced EMF will have the orientation like this is E and therefore V has to be equal to E there is nothing else in between and therefore we can write V equals E which is equal to P of ? and now input power is equal to V into I which is equal to I into P of ? and therefore one can find out input energy how to find out energy energy is nothing but integral of input power and therefore I can write this as integral of Vi dt if you start this whole system at some T equal to 0 and you want to find out the energy at some T equal to T1 at T equal to T1 this value of I had reached some value I1 so I at T equal to T1 is equal to I1 now that is the situation at which you want to evaluate what is the field energy that is there now if you look at this system this is just one loop of wire we have said already said there is no resistance in this that means the energy that is input into the system has no loss is not lost anywhere there is no mechanical part to this so nothing moves and therefore all the energy that is input from this electrical side has to get stored in the system and the only place of storage is the field that is generated so all this energy is stored as magnetic field energy and therefore this integral Vi dt is the magnetic field energy that is stored in the system and we will further assume that the magnetic system is linear which means that the relationship between if you plot flux linkage versus the flow of current I it means that this relationship is linear straight line and if it is so one can therefore describe this as ? equals L into I where L is the inductance of this loop okay so we can write this as integral 0 to T1 Vi dt is equal to integral 0 to T of I into L into I dt where ? HP ? so D by dt P of L into I dt where ? has been replaced by L into I and because the system is linear L is a fixed number and therefore can be taken out of the differentiation so this is nothing but 0 to Ti of L into Di by dt and now how to write this term let us call this as I so integral 0 to T of I dt since it now contains two functions within the integral we integrate by parts integration by parts so that will give us integral of this L into I square that is I take let me put it it is I into L into integral of this function which is nothing but I – again integral 0 to T integral of this function is I you have L as it is into derivative of this function that is Di by dt into dt which is the same as I and therefore let us call I as this integral this is I L Di by dt and this integral is the same as I so what you have is two times I is equal to I square into L or this integral is Li square divided by 2 so this integral is nothing but Li square divided by 2 which is nothing but your stored energy so this is of course a result that you might have known already I am sure you would have seen or heard or derived that the inductance is half Li square and but this is a more effective way of deriving that expression inductance at the stored energy is given by half Li square in this expression what we are having is the rate of change of field energy that means as the field energy is going to change you need to differentiate this term corresponds to the derivative of the field energy change so let us see what that is W mag is half Li square what we are looking at is DW mag by dt which is half of remember now L is not a fixed number L can change with respect to time therefore this is I square DL by dt plus 2 I L Di by dt right so this is I square by 2 DL by dt plus Li Di by dt this is rate of change of magnetic field energy now let us compare that with this expression in this expression we have seen that the input Vi that is the total electrical power that is input is equal to the resistive loss and we have identified what the resistive loss is that is your resistive loss this term has to be then the sum of mechanical power plus the rate of change of field energy and mechanical power plus rate of change of field energy is equal to I square DL by dt plus Li Di by dt and we know that DW mag by dt is given by this expression which means mechanical power is equal to this term minus that term this minus that and therefore these two cancel out you have I square DL by dt here and half I square DL by dt here therefore mechanical power is this minus that which is half of I square into DL by dt so you see that mechanical output depends upon the rate of change of inductance and that is one of the reasons why inductance is very important in electrical machine. Now this is mechanical power can be written as half of I square DL by dx into dx by dt we know further that power in mechanics is equal to force multiplied by velocity and here you have velocity which is dx by dt and therefore this term which is half I square DL by dx must be equal to force therefore we can write force is equal to half I square DL by dx. So we have now derived an expression for force so using this how do you now solve or how do you estimate the behavior of the entire system that is then done by taking this as one equation applied voltage is equal to resistive drop plus P into ? and then this is the second equation that you need to describe ? in terms of Y and L the next equation is the equation for force which is force is half I square DL by dx and you need to connect this up with the mechanical system for the mechanical system this developed force please remember that what we have is a bar which has been connected to some substratum through a spring of constant k and therefore f – k x is the net force acting on this bar this must therefore be equal to mass of the bar m multiplied by the acceleration d square x by dt square. So these equations together now if you solve one can find out what is happening to the total system you have this coil and the applied voltage V applied voltage V is related to I by this expression so if you solve it you will know what is the behavior of I with respect to time so you can plot now by solving this I as a function of time you can plot force as a function of time this is what is required in order to find out the behavior of the mechanical system and this is what is required to find out the behavior of the electrical system. So to solve the system as a whole you need an electrical equation you need a mechanical equation and you need something that is going to link the electrical and mechanical parts in short this is the sort of approach that we will be adopting for analyzing all the electrical machines later on so it is important to understand this very well we will see various varieties of systems in the lectures to come and we will also see what to do if the material is not linear here we have assumed that the system is linear if it is not how do you address the analysis all this we will see in the lectures to come we will close for today.