 we have been looking at the utilization of the machine equations that we have derived so far. We have seen how the equations for induction machine could be used to solve the dynamics of induction machine namely speed increases, speed falls, wave forms during non sinusoidal supply voltage conditions most specifically inverted driven induction machines and so on. Using the synchronous reference frame approach one can also derive small signal model for the induction machine which can be used for control system design and applications. Therein one can look at the vector control formulation where by aligning the reference frame along a particular space vector one can derive decoupled and simpler models with which one can derive control system representation. Now we will go on to look at the alternator machine model and see what are the uses that it is applied for some specific applications examples of use of the alternator equations. Now when we look at alternators in the case of alternators we have a field winding on the rotor and a three phase winding on stator. So since there is a field winding on the rotor it is natural and the most used approach it is very natural to align the D axis of the synchronous frame. So the D axis of the synchronously rotating reference frame is aligned with the field winding on the rotor and then you have the Q axis is aligned on the inter polar region. So this is a simplistic way of representing the alternator since there is no winding on the Q axis actually existing on the machine similar to the field winding. So the alternator may be represented at that is the rotor of the alternator may be represented as having a D axis and then a Q axis with the field winding being represented here and no winding on the Q axis because there is no explicit source of excitation or impressed source of excitation. But however we have seen that in the alternators you have arrangements called damper windings or damper bars which basically form a closed circuit arrangement it looks like we have seen it looks like the squirrel cage of the induction machine and since the rotor rotates at synchronous speed and the field of the alternator energised by the stator also rotates at synchronous speed there is no relative movement between the air gap field and the rotor and therefore the squirrel cage so to say does not have any induced EMS and therefore this arrangement is normally equivalent to an open circuit under steady state that is when the alternator is really rotating at synchronous speed the field is also rotating at synchronous speed there is no induced EMS and there is no flow of AC current and hence one may say that this is really an open circuit which is there but however when there is a disturbance in the alternator when whenever there is a relative movement between the rotor and the air gap field which may happen due to faults that may be there on the stator side due to increase in mechanical input there may be a momentary acceleration of the rotor until it settles down again to synchronous speed during such times there is always a relative movement between that of the stator field and the rotor and therefore there is induced EMS in these bars and therefore since they are closed there is a flow of current in the dampers which serve to decelerate the machine which serve to bring the rotor back into synchronism and therefore any model of the alternator which aims to simulate the dynamics of the rotor must include a representation for this arrangement of dampers as well without that it is incomplete and now if we say that we need to include a representation for dampers then the issue is how to represent it. Now we have seen that in the case of the rotor if we draw the rotor as follows you have the rotor structure which looks like this we are drawing one pole of the rotor and then we said that there are slots that may be provided here through which the damper bars run and then that is then shorted by an end ring on either side therefore it forms a closed circuit. Now this will in effect cause some field along the polar axis and some field along the inter polar axis that is the field that is produced by this flow in these bars can be thought of as having been produced by a winding on the direct axis and a winding on the q axis as well but however if we look at other effects that may be there for example if the rotor is going to run at a slightly different speed as compared to the stator flux we may have induced flow in the rotor itself in the rotor iron itself the rotor iron especially if you have solid rotor alternators under ordinary circumstances when the rotor is rotating at synchronism since there is anyway no induced emf you need not have an arrangement on the rotor where the main field pole is being made of is being divided into rotor lamination right because in any case since the rotor is going to rotate at synchronist speed there will be no eddy currents induced no emf is there no eddy currents are there and therefore it is not really necessary to have a laminated arrangement in the main physical rotor as such but however the rotor shoes this area where the rotor pole shoes are there that alone may be laminated in order to account for variations in the air gap and therefore induced emf that may arise due to that because of effects due to stator slotting because of that there will be local flux variations and that may induce an emf and to not allow that to damage or to cause increase in heating one may have lamination in the pole shoes but however the main central core of the rotor which is this area that may not be laminated and if it is not then if the rotor undergoes any disturbance in speed then there could be substantial eddy currents induced in the rotor and therefore fields produced due to that which act in a very similar way as compared to fields or induced emf and hence the flow of current in the damper bar so the issue is then how to represent this we may go for an arrangement where if you have the dq axis of the rotor this is your d and q fixed to the rotor in order to represent the fields produced due to a disturbance in the rotor speed or in the air gap field you may have a coil represented on the d axis and a coil represented on the q axis this is the barest minimum in addition however if there is going to be substantial induced emf in the rotor itself may be one may go in for additional coils that are represented on the d and q axis of course being solid rotor and therefore flow of current can be there even on the damper bars they are shorted and therefore these coils are invariably represented as shorted one so all these are then representations of damper circuits to be more to be more general this may be formed within the damper bars or it may be formed in the solid rotor part of the alternator that is somewhere in these regions all together then produce a field which then tends to retard against the oscillation of the rotor so depending on the accuracy that one wants one may go in either for single coil representation one on each axis or you may go for two coil representations on each axis and so on right so for our discussion now we will look at only one coil representation we will assume that the effects of the dampers can be sufficiently represented by having one coil each on the d and q axis so if that is the case how does the alternator equation look like we have of course derived this earlier but let us look at the expression once again so here we have the alternator equation equations considering one damper circuit on in each axis we have here the d axis equation v d of s that is when we represented this way it means that the voltage along the d axis attached to the rotor but of the stator variable so similarly you then have voltage of the stator along the q axis so you have the d axis expression on the rotor that is equal to RS plus LDSP LDS is then the direct axis synchronous inductance and then you have minus LQS ? where ? is the rotor speed LMD x P these two represent the mutual effect between the field and the d axis of the stator between the direct axis damper and that of the d axis of the stator and then the q axis you have a rotational emf term now note that both for the field and for the direct axis of the damper you have the same LMD term now this arises as we have discussed earlier this arises because the equations that we are we have written down they are equations referred to stator turns and as we have said before once you refer it to the stator turns then all the mutual inductances become equal to the magnetizing inductances mutual inductance becomes equivalent to the magnetizing inductance along the respective axis so that is why you find these term then if you take the q axis you have RS plus LQS P LQS then is the quadrature axis synchronous inductance and then you have LMD x ? referring to the field pdmf and then pdmf due to the damper and then the di by dt term due to the damper in the same axis and then one can represent the field equations and so on field equations equations for kd kd and kq represent the these two represent damper winding so kd represents the d axis damper winding and kq represents q axis damper winding that is what is then shown here if these are not there we will have only this of the alternator equation where you have only d and q axis of the stator and then the field now in addition we have one coil representing damper on the d axis one coil representing damper on the q axis so LDS is then equal to the leakage inductance plus LMD which is equal to the magnetizing inductance on the d axis similarly field these are inductances which are referred to stator terms in the earlier notation we use to write this as LF dash meaning it refers to the stator means the inductance referred to stator terms and similarly this would have been called as LKD dash this would have been LQS and LKQ dash but however in the normal notation of the alternators that we are going to see now whatever simplification of the equations that we are going to see this dash is used to refer to something else and therefore we will omit the use of dash in this region and we hope that there is no disturbance arising due to that so LF really represents the stator terms referred field inductance LKD represents the stator terms referred inductance of the damper along the d axis and so on. Now having seen this it is useful to reduce this to an equivalent electrical circuit we can see that this entire representation can be reduced to two equivalent circuits one along the d axis and one along the q axis so how do we do that we take a look at the first equation VD of S that is VDS we see that can be represented as here is your VDS that is being consumed in a stator resistance and then an inductance of the stator so one can see that VDS is RS into IDS that is the stator resistance part and then there is also a speed EMF term which arises due to flow of currents in the q axis this term is being multiplied by IQS this term is multiplied by IKQ both are due to q axis flow of currents that will not be there in these flow of currents is not there on the d axis and therefore if we are to draw an equivalent circuit on the d axis it is appropriate to represent it by means of a control voltage source so this control voltage source will have a value –LQS into ? into IQS –LNQ into ? into IKQ after that the circuit has LDS that is seen here LDS P into IDS LDS in turn consist of a leakage inductance plus magnetizing inductance and therefore we can write this as a leakage inductance and a magnetizing inductance so this is LLS and this is LMD this is RS and the EMF source as we have said it is the control voltage source controlled by flow of currents in the q axis so if the current that is flowing into this is IDS then we can see that the first equation is already met VDS is nothing but IDS multiplied by RS that accounts for this term and then LLS plus LMD multiplied by this d by dt of IDS so LDS plus LMD is then this term so this equivalent circuit accounts for this term and this term and these two PDMF term now what is left is these two terms which arise due to field current and KD damper now let us look at the equation for KD that is this equation normally VKD which is the voltage applied to the damper winding this is equal to 0 because the damper form a short circuited set of winding whether it represents the rotor body or the damper bars both are anyway short circuited. So this means that 0 equals LMD into P IDS now LMD into P IDS would already be here and then RKD into IKD plus LKD PIQ LKD again is given by LLKD plus LMD and therefore this equation can be adequately represented in the equivalent circuit by having one more branch like this so this is RKD and this is LLKD if we call this as IKD then this is the current flowing in and if you apply KVL around this loop one can see RKD into IKD plus LLKD plus LMD into IKD P of that and then you have the field winding also so if we now represent the field winding here this is LLF and RF and then this is now your field voltage VF this is the field current that is flowing in IF so you have field current coming here IDS coming here IKD coming here which means the flow of current through this branch LMD is nothing but the sum of all those. So now if you look at this equivalent circuit one can see you can take any equation and see that it is represented 0 equals LMDP into IDS plus LMDP into IF plus LMDP into IKD now you see that all that is accounted for by this branch IDS into LMDP IKD flowing into LMDP and then IF flowing into LMDP plus RKD into this so this loop accounts for this equation. Similarly now if you take this loop it accounts for the first equation and if you take this loop it accounts for the field equation so this equivalent circuit is able to represent the first equation second the third and this fourth now similarly we can draw an equivalent circuit in the Q axis the Q axis equivalent circuit will then have VQS that means the Q axis stator winding put in the Q axis and then you have the resistance of the stator and then the leakage inductance of the stator well you also have an induced EMF so let us represent the induced EMF first and then the leakage inductance this is the magnetizing inductance along the Q axis unlike the D axis there is no field winding there is only a damper and that can then be represented in this manner so you have RKQ and then LLKQ this is LMQ this is LLSRS and then your induced EMF so this current that is flowing in is now IQS this current that is flowing out of the damper bar is IKQ that is the rotor current and together these two currents flow into LMKQ so what the flow of current in this branch is IQS plus IKQ that would flow here so if you now apply KVL around this loop one can see that you get this equation with the source representing the speed EMF LDS ? x IDS plus LMD ? x IKD that is the value of this source with that if you apply KVL around this loop you get the second equation that is here and if you apply KVL around this loop you get the KQ equation here. So this is these two equivalent circuits then represent the alternator model together in the D and Q axis now what we can do is similar to the definition of flux linkages that we have been writing down for the induction machine along the D and Q axis one can write now ? DS this is the stator flux linkage along the D axis this can be written as LDS x IDS plus LMD x IF plus LMD x IKD similarly one can write an expression for ? QS as LQS x IQS plus LMQ x IKQ now with this then if you look at the first equation if you look at this equation you can see that VDS consists of RS IDS and then this PDMF terms if you look at all the D by DT terms that are occurring that is the first one third and the fourth one P of LDS P IDS can be written as P LDS IDS similarly this term can be written as PLMD IF and this is PLMD IKD and that is what you see exactly here P of LDS IDS PLMDIF LMD IKD all that would then simply mean that it is P times ? DS and therefore this expression VDS VDS can be written as it can be written as RS x IDS plus P times ? DS and then you have the PDMF terms which are here now this PDMF terms LQS ? x IQS and LMQ ? x IKQ now that you can see here you have LQS x IQ that is this term LMQ x IKQ that you have here only thing is in the first equation there is a multiplying factor ? as well in both the term therefore this first equation can be completed as this – ? QS. Now similarly one can write an expression for VQS as RS x IQS plus if you look at the derivative terms you have LQS you have LQS P IQS and then LMQ P IKQ LQS IQS and LMQ IKQ are already here derivative of that would mean derivative of this flux linkage and therefore this expression can be written as P times ? QS and then you have the PDMF terms one can see boils down to LDS IDS LMD IF LMD IKD that is exactly what is there in the flux linkage term all those are multiplied by speed and therefore what we can write is plus ? DS now having written these two forms now let us compare this with the equivalent circuit again you have VDS that is applied here RS IDS is available here PDMF is available here which means that the voltage available between these two nodes is nothing but PS IDS RS IDS is already available ? IQ is available here and VDS is nothing but the sum of these three therefore this expression this voltage between these two nodes must represent PS IDS similarly this voltage between these two nodes must represent PS QS by the similar argument now having seen this let us look at that part of the equivalent circuit which is only this region that is let me do it in a different color so this region of the equivalent circuit similarly let us look at this part of the equivalent circuit alone so we will redraw that equivalent circuit here what we have is the leakage inductance and then the magnetizing inductance and then the damper representation and then the field representation so this is the circuit that we have this is LLS this is LMD this is LLKD this is RKD this is LLF and RS remember that these are all terms that are referred to stator turn so we are not really putting dashes everywhere but that is understood this voltage is nothing but PS IDS so if we look at this equivalent circuit we see that there is current flowing here this is IDS and there is current flowing here this is IF VF is the source of excitation that is given here so this circuit being a simple circuit which has one input here we may consider IDS as another input so if we consider IDS as one input and VF as another input then PS IDS can be written as a function of these two input variables that is in this form now we have switched to the Laplace representation where P is nothing but the differential operator so D by DT of PS IDS can be represented in the Laplace domain as S times PS IDS where this is the Laplace transform of the direct axis flux linkage now this voltage can then be written as some transfer function multiplied by ID of S plus another transfer function multiplied by VF of S which one can obtain by superposition so in order to obtain this A of S what do we do we short one of the input make this since this is source of voltage that is to say in other words what we are doing is we are assuming that there is a current source here that is ID of S there is a voltage source here that is VD of S may remove this there is a voltage source here that is V of S and what we want to write is an expression for the voltage here that is S side ID of S so how does one do that this being a voltage source you apply superposition because there are two sources in order to determine the response due to this source we deactivate this source that means short this source and if you short this source what we have is this impedance comes in parallel with this branch that is coming in parallel with this branch and therefore this voltage is nothing but ID of S multiplied by this way of S which is nothing but S times LL of S that is the impedance of this plus the impedance of all these three branches put together which is the reciprocal of one over the sum of all the admittances that is one by RKD plus S times LLKD is the impedance of or rather the admittance of this branch one by RF plus SLF is the admittance of this branch remember this is shorted one by SLMD is the admittance of this branch so add all the three admittances and then take the reciprocal again you get the impedance this impedance plus this impedance is the total impedance here multiplied by ID is the voltage available due to this source alone then we need to find out the contribution to this voltage from the other source so deactivate this source deactivating this meaning this flow of current goes to 0 that means this is an open circuit only this is there and therefore the voltage across this is the same as the voltage across these two branches because this is an open circuit and that is then given by this voltage multiplied by this impedance divided by the sum of these two impedance that is the total this plus this which is nothing but therefore this B of S is the numerator is then this total impedance which is a parallel combination of these two so S times LMD which is this multiplied by this RKD plus SLKD divided by the sum of these two impedances RKD plus SLKD plus LMD this is this impedance so the voltage available across these two is the sum numerator divided by denominator consists of this plus this which is what is here RF plus SLF plus this impedance itself now let us take a look at the first of these two terms that is the term AS now AS can be written as SLS plus we further reduce this expression so the denominator will now consist of product of two of these terms taken at a time so what would come for example here is this term multiplied by this term that is what you get here and then similarly as a numerator for this you get SLMD multiplied by RF plus SLF and if you look at the term here it is RKD plus SLKD multiplied by SMD this denominator which consists of the product of these three things because divided again goes to the numerator so this is the term you have that plus SLS now this entire term can be imagined to be consisting of a numerator and denominator the denominator of the entire expression will be the same as this denominator the numerator will consist of this denominator multiplied by SL of SLLS plus this term so let us now look at the denominator alone the denominator which is this term if we expand all the terms together what you have is obviously one can see it is a second order expression so you have a squared term coming from all these three terms so you have a squared into LLKD into LLF this a squared term and then you have an S term which then comes out of RF LLKD and LLF RKD that is occurring here contribute S term this and RKD RF term comes here from this term you get a squared LMD into LLF here also you have a square LMD so LMD is taken common and LLF LLKD now is seen here and you have SLMD RF plus RKD so that is the expansion of this what we can do is take this now you have RKD RF here this can be taken out so we can rewrite this expression as RKD RF multiplied by this term becomes a one and then take the S terms together S times LKD RF LLF RKD and then you have LMD RF and LMD RKD so that is now divided by RKD RF and then the S term as well divided by RKD RF now this term can be written as LMD into LLF plus LMD into LLKD plus LLKD into LLF that is what this expands into and what we can do is simplify this as you can combine these two terms together that is LMD into LLF plus LLKD into LMD plus LLF so what we have done is taken these two terms together and then taken LLKD common LMD plus LLF is here this is retained as it is so with that what we can do is RKD RF remains as it is you have one now these terms can be simplified to a form that looks by an L by R ratio so if you take this RF term and this RF term together divide by RKD RF what you get is LMD plus LLKD RF cancels and you get RKD here similarly take these two terms together then you have RKD being same in the numerator and denominator that is removed so you have LLF plus LMD by RF so these look like an L by R ratio similarly here we take this LMD plus LLF outside so S square multiplied by LLF plus LMD and then you have this RF from the denominator is retained here then the remaining term here would be this LLKD plus LMD LLF divided by this since we have taken this common outside so 1 over RKD now this expression now note that they are all in the form some L by R here L by R here again you have an L by R here this LMD plus LLF by RF is the same term that occurs here also similarly you have another inductance here divided by a resistance and therefore the expression the denominator itself can be written in the form as shown here where T1 is a time constant having the form of inductance ratio of inductance to resistance that is nothing but LLF plus LMD divided by RF so if we look at what this is this is nothing but the LMD is the direct inductance plus the leakage inductance of the field together it constitutes the self inductance of the field the ratio of that to the field resistance and therefore this is nothing but the field time constant similarly we can define a time constant T2 as LMD plus LLKD divided by RKD the numerator is nothing but the inductance of the D axis damper circuit denominator is the resistance of the damper and therefore this is nothing but D axis damper time constant there is yet another term T3 which is LLKD plus LMD LLF by LMD plus LLF together the whole thing denotes an inductance divided by this resistance of the damper so this is another L by R ratio. Now let us look at how one can write this or how this expression is related to the equivalent circuit itself now if you look at the equivalent circuit here what we have is LLKD plus LMD LF by LMD plus LF that is what we have here so if one were to imagine another equivalent circuit another circuit that looks like this you have your LMD here and then RKD LLKD LLF shorted upon each other and then we find out what is the inductance as seen by the two terminals of the resistance you can see that LLF occurs in parallel with LMD and those two together occur in series with LLKD so this is nothing but the inductance as seen by the resistance of the damper so that is what is shown here. Now we need to move further ahead and try to reduce these expressions to a more manageable form for the numerator also and for the other terms and then we will be able to derive certain interesting expressions relating to this equivalent circuit so that we will continue in the next lecture we will stop here for now.