 Welcome to the next lecture on the modeling and analysis of electrical machine. What we had started doing in the last lecture was try and develop the steady state models of electrical machine. The DC machine of course was a simple model which we did at the time of deriving the dynamic equations of the DC machine. The induction machine and the synchronous machine we have developed equations based on a totally different idea or a totally different approach as compared to the machine representations which one would have derived in a first course on electrical machine. And these equations apparently do not look like anything which is like what one would have done in the first course on electrical machine. And therefore we wanted to find out how these equations or these models look like under steady state operations. Therefore we tried to reduce these equations to more simpler forms under steady state and then we wanted to see whether the conventional known equivalent circuits of machines that one has studied for example the induction machine equivalent circuit whether it can be derived from these equations. And if so is it exactly the same as what one would have done in the earlier course on electrical machines or is there any difference. So those are the things that we wanted to investigate and we started looking at the induction machine equations. So let us look at that once again. We just started that in the last lecture so let us see those equations again. So steady state analysis and we are looking at induction machine. So in the steady state representation what we are looking at is a per phase representation that is what one would look at as an equivalent electrical circuit and to arrive at that we have these two equations Vds is equal to Rs plus Lsp into Ids minus ?s Ls Iqs minus Lmp Idr minus Lm ?s Iqr and we are looking at steady state and therefore d by dt terms can be made 0. Why is it that d by dt terms can be made 0. These equations are written in the synchronous reference frame and in the synchronous reference frame we have already seen that all the variables reduce to dc terms. An excitation an AC voltage that is applied to the stator at 50 hertz say in the synchronous frame looks like a dc term. And therefore if one were to do a differentiation operation on a dc term the result is 0 and therefore this term goes to 0 and similarly this term goes to 0 and therefore under steady state what you get is Rs Ids minus ?s Ls Iqs minus Lm ?s Iqr this is Vds and similarly Vqs would give us Rs Iqs plus ?s Ls Ids plus ?s Lm into Idr. So these are the two expressions for the stator voltages similarly you will have two expressions for the rotor voltages that is Vdr equals Rr into Idr again d by dt terms can be neglected because we are in the synchronous reference frame so Rr into Idr plus or rather minus Lm ?s – ?r Idiqs – Lr ?s – ?r into Iqr and similarly Vqr is Rr into Iqr plus Lm into ?s – ?r into Ids plus Lr into ?s – ?r into Idr. So these are the equations for the rotor voltages under steady state and what we want to do since we are looking at the per phase equivalent circuit in the natural reference frame in the three phase reference frame we need to transform these equations to the normal phase voltage of a phase stator and the phase voltage of the rotor and we know that we have gone through the dq equations Vd, Vq and V0 this was obtained as cos ? – sin ? cos ? and root 2 by 3 times 1 – 1.5 – 1.5 V0 root 3 by 2 – root 3 by 2 1 by root 2 1 by 2 and 1 by 3 times Va, Vd and Vc. So this is the transformation that takes us from the ABC reference frame all the way to dq and what we want to do is exactly the inverse now we want to go from the dq equations that are there back to ABC so that we can find out what Va is in terms of all these terms that we have represented. Similarly what ia is in terms of all these terms that have been represented. So if you want to do the inverse what you do is Vd, Vq and V0 we multiply by the inverse of this matrix that is cos ? – sin ? 0 sin ? cos ? 0 0 0 1 so this matrix and this matrix are nothing but the transpose of each other similarly you multiply again by the transpose of this matrix which is 1 0 1 over root 2 0 – 1.5 root 3 by 2 1 over root 2 and – 1.5 – root 3 by 2 1 over root 2 the scalar remains the same so that is the root 2 by 3 so you get Va, Vb and Vc and what we are interested in is only Va and therefore these two terms we are not interested in because we need to develop a per phase equivalent circuit and therefore we only look for the expression for Va and Va is nothing but root 2 by 3 times 1 0 1 over root 2 multiplied by this one and therefore we can reduce that Va is root of 2 by 3 times the first term alone is going to be of importance the second term in this is multiplied by 0 and third term is going to come here and this term is going to be 0 because V0 is 0 so this reduces to Vd cos ? – Vq x sin ? and by the same token what you would have is Ia equals root of 2 by 3 times Id cos ? – Iq sin ? and further we are looking at the synchronous reference frame and therefore ? equals ? synchronous x time so what we need to do is now substitute these expressions for Vd and Vq into what we have got when we say Va we have not yet not really specified whether it is Va of the stator or Va of the rotor it could be either one depending upon what you substitute here if you make this as Vds that is pick up the expression from here and then Vq then what you get is a phase voltage of the stator if you substitute these two equations what you get is a phase voltage of the rotor so let us substitute that and what we get is Va equals root of 2 by 3 x this expression multiplied by cos ? so that is Rs x Ids cos ? – ?s Ls x Iqs cos ? – Lm ?s Iqr cos ? and then Rs – Vq sin ? so this is – Rs Iqs sin ? – ?s Ls Ids sin ? – ?s Lm Idr cos ? – ? so that is the expression we have which can then be simplified as root 2 by 3 x Rs x we combine these two terms so that is Ids cos ? – Iqs sin ? and then we combine these two terms ?s Ls so that is – ?s Ls x Ids sin ? – Iqs cos ? and then – ?s Lm x Iqr cos ? this is what one gets if we multiply these two terms now what you see here is Ids cos ? – Iqs sin ? and that is exactly what we have here Ia is Id cos ? – Iq sin ? and that root 2 by 3 factor is here and this refers to ds and qs and therefore we can write this as Ias so this term becomes Rs x Ias when you come here what you have is Ids sin ? – Iqs cos ? so imagine you have to differentiate this term with respect to angle considering or with respect to T considering the fact that ? is given by ?s x T so if you differentiate this what you would get is root of 2 by 3 x – Ids sin ? differential of cos is – sin and then – Iq cos ? and ? is ?s x time so d ? by dt which is multiplied by ?s so that is exactly what you have here – ?s x Ids sin ? – Iq cos ? so this term is nothing but – Ls x d by dt of Ias so that includes the root 2 by 3 factor as well and then we look at this term so again you have Iqr cos ? – Idr sin ? – sin taken out but now that this is Idr and Iqr this term will give us Iar and diar by dt so this is nothing but ?s x Lm diar by dt so this whole long expression can be compressed into these 3 terms Va is Rs Ias plus Ls x d by dt of Ias plus this so therefore let us write that down Rs x Ias plus Ls x d by dt of Ias plus this ?s comes out as a result of multiplying this so we do not write that explicitly Lm x diar by dt so that is what you get now let us look at the rotor voltage Var is again nothing but root of 2 by 3 times Vdr x cos ? – Vqr x sin ? which can be taken up from these 2 expressions if we look at these 2 expressions what we find is that instead of ?s you have ?s – ?r here instead of ?s again you have ?s – ?r therefore fundamentally these 2 equations look pretty much the same except for the fact that instead of ?s you now have ?s – ?r and therefore one could anyway let us write that down so Vdr is Rr x Idr cos ? – Lm x ?s – ?r x Iqs cos ? – Lr x ?s – ?r x Iqr cos ? plus – Rr x Iqr sin ? and then plus Lm x ?s – ?r x Ids plus here it is minus because you are subtracting and then one more – term Lr x ?s – ?r Idr x sin ? so this is what you will end up with and that can be written therefore as Rr x Iar these 2 terms together would give us Iar and then this term give Lm x ?s – ?r x Iqs cos ? plus Ids sin ? and then – Lr x ?s – ?r x Iqr cos ? plus Idr sin ? so what we have is this expression which one can write now as Rr x Iar – Lm now this term ?s – ?r ?s is the synchronous frequency ?r is the rotor speed and in the case of induction machines the ratio ?s – ?r x ?s is written as a slip S and therefore ?s – ?r can be written as S ?s and therefore this is S ?s multiplied by this expression and then you have – Lr x S ?s multiplied by the second expression and we know what is occurring inside these 2 is nothing but the derivative of I and therefore we can write this as Rr x Iar – Lm multiplied by S what we get here is Iqs and Ids and therefore it is the derivative of the stator current that is a phase stator current ?s comes as a result of differentiation that we have seen already and therefore this is nothing but plus d by dt of Ias and here what you get is the derivative of the rotor current and therefore this is nothing but plus Lr x S x d by dt of Iar so these are the 2 expressions that we land up with so what we will now do is convert to the phasor notation to make it easier to handle so the first equation is the equation for the stator voltage Vas so Vas as a phasor is then equal to Rs x Ias where I is a phasor and now you want to represent this one Ls d by dt of Ias so in phasor notation that then becomes plus j times ? Ls x Ias and j times ? I mean ? times Ls is nothing but x S that is the stator impedance inductive impedance and then you have plus j times xm x Iar so this equation in phasor form has now been written similarly you have Var is equal to Rr x Iar plus S times Lm now dias by dt into Lm can now be represented as j times xm Ias phasor plus j times xr multiplied by the slip into Iar phase now you may have a doubt that comes up at this instant now we are looking at doing a differentiation of Iar and the rotor current in reality is not at the frequency of the supply but it is at the frequency of slip so one may ask why slip frequency is not coming so that in fact is indicated by S here but equation wise what you are differentiating is this expression Iqr cos ? and idr sin ? and ? is nothing but ? S x t and therefore when you differentiate what you will get is only the supply frequency this S comes as a result of this ? S – ? R and therefore when you look at it here what you get is xr and that is the stator current phase now the rotor in the induction machine is invariably shorted and therefore Var is 0 and therefore that is what it is so now what we can do is manipulate these two equations what we will do is we will add to this j xm x Ias and subtract jxm x Ias so that makes no difference as far as the equation is concerned and similarly here what we will do is divide this expression by S and therefore this S goes away this S goes away and what you get is Rr by S into all those terms and now what we will do to this expression is add and subtract xm x Iar and – xm x Iar so with the result the first equation Vas now becomes Rs x Ias plus these two terms can be combined and the self inductance or the self reactance – the xm value will give the leakage inductance or the leakage inductive impedance of that so this is nothing but jxl x Ias plus jxm x Iar plus Ias and in the second term 0 is equal to Rr by S into Iar plus similarly here if you look at these two terms xr – xm would be the leakage inductance and therefore jxl x Iar this is xls this is xr leakage inductance of the rotor again remember we are at the outside when we started with the synchronous reference frame machine equations we had already referred everything to the stator and therefore all these terms are stator turns referred entity Rr for example is Rr that is the rotor resistance referred by number of turns to the stator turns therefore it is really Rr dash in the conventional sense and then you have these two terms so that is jxm x Ias plus Iar so this is what you get and this can be represented as a simple equivalent circuit in this manner so let us say you have you are applying Vas here that Vas is consumed as a stator resistance term the current flowing in is Ias so that is consumed as a stator resistance term and then you have a leakage inductance this is Rs xls and then you have an xm through which both Ias and Iar are going to flow so if I now represent this by an inductance here and this current is Ias this current is Iar then these two together flow in this inductance and if you now apply KVL around this loop so this is Iar so if you now apply KVL around this loop you precisely get this equation. Now to focus on the second side you have the rotor voltage Var which is 0 alright but the rotor voltage is then consumed as Rr by s multiplied by Iar and then a leakage inductance xlr and now if you write KVL around this loop Iar is already flowing here therefore this voltage is equal to resistive drop plus the inductive drop plus this voltage which is nothing but xm into Ias plus Iar and the rotor is usually short circuited and therefore Var is 0. So that is the equivalent circuit which one gets on a per phase basis for the induction machine which definitely looks very similar to what we would have we have studied in the first course on machines except for this additional resistance which comes in the normal exact equivalent circuit of the induction machine where is this resistance why is this resistance not there in the formulation that we have done this resistance is not there because all this analysis neglects core losses since we have not considered the effect of core losses in this development it is not represented in the formulation that we have done it does not come out in the equivalent circuit also. So if we neglect core losses whatever we have done we have derived as the induction machine equivalent circuit is still applicable in comparison with what we had studied earlier in machines first course on machine. So from this then one can go about to predict all the machine behavior under steady state one can derive the speed torque relationships and then the various approaches to speed control one might have and then solve I mean if you are going to connect this induction machine to a big electrical network one can then solve it as an equivalent electrical network to find out how the machine is going to behave at a particular slip. Now the derivation must have also made it very explicit that this equivalent circuit is valid only under sinusoidal steady state this is important that is what we started out with right at the beginning of the course that the conventional equivalent circuits that one have studied are valid only under steady state and that to sinusoidal steady state this derivation makes it very evident that it is indeed so you cannot use this equivalent circuit for example to find out how speed is going to change if you give a supply voltage here if all of a sudden you switch on an AC voltage here how a speed going to change this equivalent circuit does not predict. You have to make use of all the elaborate dq formulation equations that we have derived and combine it with the mechanical equations of the machine in order to find out how the system is going to behave. Sometime later on in one of the lectures we will see a response a solution of these equations done through a computer to see what kind of responses one gets when machine undergoes an acceleration or a deceleration but that is for a later date and for now we have achieved our job of deducing the steady state equivalent circuit from this representation this is for an induction machine a three phase induction machine. So how does this work for the synchronous machine let us now look at steady state representation or the model for a synchronous machine let us get rid of this also in the case of synchronous machine let us write down the equations first on the d and q axis. So now we move on to synchronous machines steady state analysis and we are going to look at synchronous machines and in case in that we are going to look at salient pole machine as we saw earlier the cylindrical rotor variety is easily deduced from the salient pole equations by setting the direct axis synchronous inductance equal to that of the q axis synchronous inductance so it is enough if we solve this the other case is easily done. So we have again the equations on the d and q axis vds is nothing but rs x ids and then plus lds x p ids plus lmd x if – lqs x omegas x iqs so this is as far as the stator t axis is concerned and vqs is rs x iqs plus lqs x p iqs plus lds x omegas ids plus lmd x omegas x if these are the two equations we get for the stator of the synchronous machine. Now in the case of synchronous machines we have apart from these two we have the field equation as well which is the rotor expression and the rotor expression is not going to give us major information because the rotor is not really connected to the AC circuit the rotor is fed from the DC side and under steady state when we are looking at it it is going to carry a steady field excitation so which is not normally considered therefore in the steady state analysis. In fact if you look back at your earlier electrical machines exposure you would not have discussed anything with respect to the field other than saying the field induces an emf in the stator so here also therefore in steady state analysis there is nothing to consider in the field so we will leave that out. Now again we are looking at steady state and the synchronous machine equations are already in the synchronous reference frame and in the synchronous reference frame as we had noted earlier in the case of induction machines all terms are DC terms and therefore derivatives goes to 0 and therefore this term goes to 0 this term goes to 0 this goes to 0 so these are all the terms that go to 0 so VDS is nothing but RSIDS-LQS?QS and VQS is given by these three terms. Now what we want we want to get the a phase voltage and how to get the a phase voltage we proceed with the same route if you want a phase voltage it is 2 x 3 times all this and therefore VAS is nothing but route of 2 x 3 times this x cos ? so RSIDS cos ?-LQS?SIQS? that is VDS cos ? and then you have to do – VQ sin ? so – RSIQS sin ? and then – LDS?SIDS?S-LMD?SIF? and now again note that ? is equal to ?S x T since the reference frame is synchronous reference frame and now you have an RS term here and an RS term here and this is IDS cos ? – IQS sin ? x 2 x 3 so you have RS x IAS you take that out plus route of 2 x 3 times this term – LQS?S ? – LDS?S?IDS sin ? – LMD?S?IF?S this is what we have now we know that IAS is IAS is route of 2 x 3 x ID cos that is already written there well okay so ID cos ? – IQ sin ? so having written it in this form what we can see is ? is nothing but ?S x time so you are obtaining IAS as an addition of some term depending on cos ? cos ?T and another term depending on sin ?T which means the resultant if you add cos ?T term and sin ?T term in some ratio resultant will be another sin wave which is phase shifted with some phase shift but at the same ?S as well and we know that if you are going to write a sin wave form for example some flow of current I and it is some let us say cos ?T the term that you put here will represent the amplitude of the sin wave and if you want to get at the RMS since it is a sinusoidal wave form you take the amplitude and divided by root 2 and therefore if you want the RMS of this wave form what you have to do is you have to take the amplitude and divide by root 2 and root 2 is already there so if you divide by root 2 this goes away and then you will just have 1 over root 3 okay. Now it is necessary to before we go ahead remember or try to recollect what is the phasor notation right so let us briefly look at the phasor notation before we move ahead now if you have some variable V let us say is equal to Vm cos ?T and you want to represent this as a phasor what you do this is considered to be a phasor is nothing but an arrow which is drawn in the complex plane. So this is your real axis and this is the imaginary axis so that is the plane that we are talking about the phasor is nothing but an arrow represented in the complex plane and that arrow is assumed to rotate at the angular frequency that is mentioned here ? so we can draw let us in general it is given as ?T ? so now you can draw an arrow which is displaced from this axis by an angle of ? that is mentioned here the length of the arrow is normally taken to be the RMS value of the sinusoidal wave form so the length of the arrow is Vm by ?2 that is this length right so if this is the representation of Vm cos ?T ? let us say now you want to represent a term V1 which has which is Vm cos ?T the phasor for this will then because ? equal to 0 the phasor length will be Vm by ?2 and it will lie entirely on this axis so this is V1 phasor okay suppose you have another V2 and V2 is given by Vm into sin ?T where will this phasor be so this can be written as Vm into cos of 90° plus ?T but cos of 90 plus ?T will give you – of sin ? but what we want is plus sin ? so you put another – sin here to get ? sin ? and if one does that what you see here is you have Vm cos 90 plus ?T so cos of ?T itself would be located along the horizontal here and therefore cos of 90 plus ?T would have been located here but then you have – of cos 90 plus ?T and therefore V2 phasor would really lie here this is your V2 phasor so if you now have – Vm sin ?T if we now have – Vm sin ?T then that phasor would be represented as since Vm sin ?T is below – Vm sin ?T will then be here and therefore – Vm cos ?T will then be a phasor lying along the other direction so it is important to understand or recollect this because what we are going to be dealing with is these kind of phasor. So this term Ia is equal to ?2 by 3 sin ? I mean into id cos ? – iq sin ? can now be converted into a representation as phasors because ? is nothing but ?T and this is also ?T and that is going to come in useful in simplifying this equation further into the phasor notation because as far as synchronous machines are concerned one does not really make use of an equivalent circuit in the steady state analysis rather what one looks at is the phasor representations in the steady state analysis so that one can solve whatever electrical network that is going to be connected one can understand how it is. So to get to that phasor we briefly recollected how the phasor representation is we will make use of that and develop the phasor notation for the salient pole synchronous machine in the next lecture we will stop here for today.