 Welcome to the next lecture in this course, we have progressed reasonable distance into the course, so let us stand back a little and try to recapitulate what we have learned so far and where we are with respect to the objectives of this course. So our broad objective in this course is to develop dynamic equations for electrical machine. For this purpose we started out by understanding an electromagnetic circuit and how equations can be written for an electromagnetic circuit, we looked at a singly excited linear motion system wrote down the electromagnetic equations for a singly excited linear motion system. We understood what an inductance was and how it is related to flux linkage, how flux is generated and we saw from the equation for the singly excited linear motion system that the inductance is an important aspect in describing their behavior. We saw that the force generated then is dependent on the rate of change of inductance with respect to the displacement and from there we went on to systems that now rotate about an axis rotary systems and there we saw that the generated torque is proportional to the rate of change of inductance with respect to angle and these are basically linear magnetic systems that we saw linear magnetic and we made a small digression to see how we can handle non-linear systems where the BH curve is not linear but a non-linear curve and then we came across ideas such as co-energy and magnetic energy and we understood how one can develop an expression for the generated electromagnetic torque from either the co-energy or the magnetic energy of a system and then we said we are not really going to look at non-linear magnetic systems but we are going to confine ourselves to linear magnetic systems and then in the linear magnetic systems case we saw that the inductance was an important aspect in describing their behavior and therefore we then went around to deriving expressions for the inductance in case of rotational systems. So we looked at several rotational systems again we looked at systems where which had a single coil on stator and then we found out the expression for the inductance of this single coil on the stator we considered a salient pole rotor and a cylindrical rotor we looked at both these aspects and then from the case of a single coil on the stator we went to a situation where there was a single coil on the stator plus a single coil on the rotor and we derived expressions for the inductance of the stator coil of the rotor coil and then the mutual inductance between the two. So having developed an experience with deriving inductance expressions we went on to look at machines or machine arrangements which are which are closer to real life and they have not a single coil but they have a distributed winding on the stator and then for this distributed winding case we then derived expressions for the self inductance mutual inductance between the stator phases and between the stator and the rotor. So now we have a good idea of how the expressions for the inductance in various situations look like. So armed with all these things we are now ready to go back to developing the dynamic equations for electrical machines this information was required in order to write expressions which we started out with. So let us now look at the case of a three phase induction machine a three phase induction machine if you would remember from the first course on electrical machines that you would have done the three phase induction machine has three windings on the stator and three windings on the rotor. Let us consider a wound rotor induction machine again you know from your machines from the first course on electrical machines that a squirrel cage induction machine can also be considered as a three phase rotor and therefore if you know how to write down the equations for a wound rotor induction machine you also know what to do for a squirrel cage machine. So this is a more general case and that is what we will look at. So that has three windings on the stator and three windings on the rotor. So if you look at the geometry of the system you have the stator that is there and then you have the rotor inside this this is a uniform air gap and on the stator and rotor three phase windings are there we have seen earlier animations on how this winding distribution in a stator or the rotor is going to look like rotor of course we saw for a alternator in the animations but the stator which we saw in the animations for the alternator is the same as the stator of an induction machine both fall under the same AC machine group and so let us say that you have a uniformly distributed winding let us assume a uniformly distributed stator and rotor windings. So if it is uniformly distributed let us say that the R phase winding lies here and the return conductors of the R phase winding then lie here and then the axis of this R phase winding would lie along this axis. So let us call these windings as not R y b but a b c. So this is your a phase winding and the return conductors we will call them as a dash that means the coil starts from here goes in and then comes out this way then goes in here comes out there and so on. So this is then the axis of the stator a phase similarly you then have the stator b phase whose axis is going to be 120 degree displaced from this. So let us take that axis to lie along this direction so this is the axis of the b phase bs meaning b phase of the stator and so if this axis has to be here then you need to have the b phase winding placed here and then you have the c phase winding whose axis is going to be another 120 degree displaced so that would lie here and if this axis has to lie here then this winding will have to occupy this region. So this is a representation of how the three phase winding would be spread around the circumference of the stator and these are the axis of the three phases a b and c I have written b so this is now cs. What happens in the induction machine is that these three phases will then be excited with three phase voltages which are displaced in time it is important to understand that the voltages which are used to excite these three phase windings are displaced with respect to time that means if I call va as vm cos ?t then vb is vm cos ?t – 2 ? 3 which means that the wave form of phase b is shifted with respect to wave form of phase a by an angle 120° and then vc will be equal to vm cos ?t – 4 ? 3 which means that the wave form of phase c is further phase shifted by 120° from the wave form of phase b. So these three wave forms are phase shifted with respect to time and these are applied to phase windings that are displaced with respect to space you have a phase here b phase winding is displaced 120° with respect to space and c phase winding sorry that is a phase winding and then the b phase winding is displaced 120° with respect to space and c phase is further 120° with respect to space. So these are spatially displaced winding otherwise these windings are identical they have the same we assume that these windings are identical otherwise they have the same number of turns the phase spread is the same everything is identical about them except that their axis is displaced in space. Now one may wonder why we have put a phase here and b phase appears to be leading that of a and c phase appears to be further leading that of b whereas the excitation is lagging you have a phase excitation b phase lags that by 2 ? 3 c phase lags further by 2 ? 3 this is rather arbitrary this is a physically available electrical machine and one can wind the axis in whatever way we want it could be a here and then b here and c here or b here and c here in some manner it can be want and having wanted it is a physically available machine somewhere and we can now connect any sort of voltage that we want to it we can either have va as vm cos ? t and vb as vm cos ? t – 4 ? 3 and in which case this will become 2 ? 3 so that is a cb sequence or a bc sequence this is completely independent of how the machine is want this is separate this is separate but why we have selected this would be evident if you now see what happens if you are excited with this. Now having excited this with va vb and vc let us now say that it is going to cos a current I a equals I m cos ? t – 5 then I b would be I m cos ? t – 5 – 2 ? 3 and I c would be I m cos ? t – 5 – 4 ? 3 these are the three currents that would then flow in the machine having excited it with these three voltages that means we are assuming that the machine is a balanced machine balanced machine balanced voltages are applied to the machine it causes the balance flow of currents in the machine this is the nature of current some phase angle would be there depending upon what is the load the machine is operating at that point. So if this is what is going to be applied let us understand what is the meaning of this when we say that I a so much current is flowing this is the current flowing in this phase a phase and we have said that the axis of the a phase winding is here that means if a current is flowing in the a phase we had seen how the mmf generated by a phase winding is drawn with respect to the angle as you travel around the circumference that mmf plot has a particular wave shape and we said that we consider the fundamental of that mmf wave form and therefore if this is the current that is flowing we know that for any given current flowing in this phase winding the maximum value of the fundamental component of the mmf occurs at this angle and that is sinusoidally distributed that means from this peak it then drops down to 0 here and then reaches negative peak here and then reaches 0 here and as this current is going to increase because this is I m cos omega t this current may increase or decrease as t is going to change whatever happens note that around the circumference if you travel at any given instantaneous value of I a the maximum mmf will always be reached at this point and then it would go to 0 here and negative maximum here what is that maximum value will depend upon the value of I a whereas the point at which the maximum occurs will always be here and that is why we call it as a phase axis that means the mmf generated by this will be oscillating on this axis as the amplitude of I a is going to change this mmf will build up then it will decrease and then reverse build up in a negative direction decrease to 0 and then reverse so it is an oscillating mmf that is generated along the a phase axis similarly whatever mmf is developed along the by the I b current which is flowing in the b phase or rather this phase would be oscillating along this axis always I mean would be oscillating along this axis this maximum would occur here and the mmf generated by the c phase would be oscillating along this axis it is peak would always occur here but what we mean by saying that I a is I m cos omega t-5 but I b is phase shifted from this by 2 ? by 3 it means that these oscillations though they are taking place at specified axis around the circumference of the machine are not oscillating at the same phase this is displaced in phase that means the maximum of this would occur at a time instant different from that of this and the maximum of this would occur at a time instant different from those of these two it is what we mean. So now let us having understood that it is always an mmf acting along this axis one can now draw this by a set of phases those phases would then be the phasor of I a if we take as the reference then the phasor of I b lags that by 120 degrees we can see that I b lags by 120 degrees for an assumed direction of rotation that is anticlockwise of the phases I b would then come here I c would come here. Now note that this is a phasor diagram these are time phases this phasor diagram has nothing to do with the spatial orientation of the windings in the machine this is a physical arrangement how the windings are placed whereas this is an interpretation this is a representation of what happens to the excitation with respect to time. So the phasor representation the interpretation of the phasor representation is that at any given instant if you want to find out the value of I a what you would do is then you draw this horizontal axis and find out the horizontal component of this phasor and that then represents the instantaneous value of that particular entity so at this instant normally the phasor diagram is drawn for the case where t equal to 0 or let us call ? t equal to 0. So at ? t equal to 0 what this diagram says then is that the a phase current I a will have its maximum amplitude whereas the b phase and c phase currents would have each an amplitude equal to this much and instantaneous value equal to this much not the amplitude instantaneous value equal to this much this is negative and since this angle is 120 degrees what it means is that the maximum value of I a is I m amplitude of I a is I m then at t equal to 0 the value of I a is equal to I m the value of I b is equal to I m into cos of 120 degrees that is minus half of I m and c is minus half of I m again as can be seen. So what you have is I a equal to I m I b equal to minus I m by 2 and I c equal to minus I m by 2 these are instantaneous values of current. Now these are going to flow in windings that are phase displaced like this and each of them would produce an mmf along its own axis. So I a would produce an mmf along this a axis I b will produce an mmf along this axis I c would produce an mmf along this axis. So let me draw these axis again here so you have the a phase axis and then the b phase axis and then the c phase axis and what this says is that there is I m acting along this axis on the b phase it is minus I m by 2 minus I m by 2 acting along this axis would mean that I m by 2 acting here and minus I m by 2 acting along this axis would mean I m by 2 acting here and the net mmf that is generated is then the sum of these three phase phases and this angle is 60 degrees this angle is 60 degrees so cos of 60 is half. So what you get is the resultant phase of this acts along the same axis and has a magnitude 3 times I m by 2 this is at omega t equal to 0. Now let us consider what happens when omega t equal to say 30 degree so let us erase these at omega t equal to 30 degrees this diagram which is a time phasor if you recollect what this phasor representation is the phasors are arrows that are rotating with respect to time in the anticlockwise direction and the diagram is drawn as freeze obtained at omega t equal to 0. So at omega t equal to 30 it then means that this diagram would have rotated by 30 degrees so I a would have gone here, I c would have come here and I b would have come here so this these phasors are now rotated they are no longer here but they are all sitting at this position. Now again to find out what is the value of instantaneous I a I b and I c what we have to do is take the horizontal component projection so this angle is now 30 degrees and this angle is 90 degrees and this angle is 30 degrees and therefore I a would then be I m into cos of 30 degrees cos 30 degrees is root 3 by 2 so you have root 3 by 2 I m the horizontal component of this phasor being at 90 degrees is 0 so this is 0 and I c this angle is again 30 degrees so it is minus root 3 by 2 I m. Now these currents are acting along the same axis the axis has not rotated this is an axis of coils that are wound on the stator which is stationary and this axis remains at the same position. Now therefore on a phase you have root 3 by 2 I m on the b phase it is 0 so no current due to the b phase c phase is minus root 3 by 2 I m therefore root 3 by 2 I m along the c axis but minus that means it is minus root 3 by 2 I m here. The net mmf therefore is the sum of these two phases so you can add this by the parallel law you would get a phasor that looks like this and since this angle is 60 degrees the resultant what you would find is at 30 degrees and what is the magnitude of this that magnitude is again 3 times I m by 2 so what we find is that at omega t equal to 0 the net mmf was lying along the a phase axis at omega t equal to 30 degrees the net mmf lies at 30 degrees to the a phase axis the angle made is 30 degrees the magnitude remains the same and therefore we can extrapolate one can verify for various positions of this phase a diagram as time increases this then traces a circle as you go around and the angle made is always equal to omega t which is what one says as a rotating mmf waveform in an induction machine. Now why is this mmf rotating in this direction in the anticlockwise direction it happens to rotate in the anticlockwise direction because of the displacement of the axis had you taken the b phase axis here and c phase axis there the net mmf would have rotated in this direction since we want the rotation to be anticlock direction we have chosen a here b here and c here if we had chosen it the other way the rotation would have been the clockwise direction and what happens if the mmf rotates in the anticlockwise direction in the case of an induction machine we know that the mmf that is rotating in the air gap is the one that is inducing a voltage in the rotor and therefore some current flow in the rotor and the rotor tends to get dragged along the direction of the mmf that is rotating in the air gap. So if the mmf rotates in the anticlockwise sense the rotor also would move in the anticlockwise sense and therefore we want the rotation of the rotor to be in the anticlockwise sense so this particular convention has been chosen in order to meet that requirement but if one does not care for that one can choose it in the other direction as well. So in the induction machine coming back to this again this is the distribution of windings on the stator and the axis that are there in the stator similarly we have windings on the rotor as well and those windings would also be distributed on the rotor they would have their own axis about which their mmf act and since the rotor can now rotate with respect to the stator in general the axis of the rotor windings may not be aligned to that of the stator and hence let us assume that the stator the mmf axis of the rotor are displaced with that of the stator we call this as AR and then therefore that of the B phase BR and that of the C phase PR and the rotor has now displaced with respect to the stator by an angle ?r. So in the rotation of the rotor what we have done is we have taken one snapshot of the machine at a particular instant and in that particular instant we have drawn this arrangement where the axis of AR is displaced with that of AS by an angle ?r and under this condition we now want to write down we are looking to write down the equations of the induction machine since six coils are there three on the stator and three on the rotor we are going to have six equations that describe the induction machine. So how to write those equations we know that from our earlier experience with writing down equations for voltages applied to a coil that is connected somewhere we know that the general form of the equation is if you take for the A phase VA is then rA x IA plus the rate of change of flux linkages IA this is for the A phase of the stator RAS IAS and P times IAS similarly you would write an expression for the B phase RBS IBS plus P times IBS note that we are using lower case voltages for AB phases and lower case currents for this it means that we are writing down expressions for instantaneous voltages that are applied instantaneous currents that are flowing these are instantaneous value similarly Vcs is Rcs x Ics plus P times Ips further let us consider that the machine is a balanced machine that means the phase windings are identical except that they are displaced in space of course and since they are identical the three resistances are expected to be the same so we will remove the subscript AB and C from this and call this as simply RS so it is simply RS x IAS IBS and Ics similarly now there are three more windings on the rotor so which will be VAR equal to RRIAR plus P times IAR and VBR is RRIBR plus P times IBR and VCR equals so these are six voltage equations that we have written one for each phase of the induction machine on the stator and one for each on the rotor having written this of course this is a straight forward expression but then you need to determine the expression for IAS IBS and so on so now let us consider the expression for IAS IAS is the flux linkage of the phase A now flux linkage in the phase A arises due to self flux plus mutual flux self flux arises due to current flowing in itself mutual flux arises due to current flowing in all the other systems that are there all other phases that are there in the stator as well as in the rotor now what is an expression for the self flux now if you remember we have derived induct self flux linkage is nothing but in general inductance multiplied by current which is the same as the number of turns multiplied by flux that is flowing through it so we take this form so we need to know what is the inductance now we have derived expressions for the inductance of winding that is distributed on the stator and the rotor being salient and when you see that case we saw that the inductance of a particular phase stator is a leakage inductance plus if the rotor is salient then we had an expression like LD plus LQ by 2 plus LD minus LQ by 2 into cos of 2 theta in the case of an induction machine the rotor is not salient the rotor is cylindrical and the air gap is uniform that means whether you take the D axis representing air gap or the Q axis representing air gap they both are the same which means that LD and LQ are identical and therefore this ceases to exist LD and LQ are identical and therefore this is LD plus LQ which is 2 times LD divided by 2 simply some LD we will simply call this as the magnetizing induct LMS of the stator where LMS would be LD as per our derivations earlier may be we will put this down LMS what we derived as LD we call it as LMS so the self inductance of a stator winding is nothing but the leakage inductance plus a magnetizing inductance for which we have derived expression earlier. Now looking at the mutual flux let us consider the mutual flux linkage between one stator phase and another stator phase again we have derived an expression for this earlier for the case of course for a cylindrical stator and a salient rotor machine the expression was the mutual flux linkage is LD cos theta r cos of ? – ? r – LQ sin ? r sin of ? – ? r so this was the expression for the mutual inductance between two stator phases again for the induction machine we have seen that LD is equal to LQ therefore this expression can be written as LD equal to LQ which is equal to LMS so LMS x cos ? cos ? – ? – sin ? sin ? – ? that is nothing but cos of ? r – ? – ? r which is LMS x cos of ? where ? is the separation of the axis of the two phase winding two winding that are under consideration in the case of a balanced induction machine we can see that the axis are separated by 120 degrees and therefore this is nothing but LMS x cos of 120 degrees which is – LMS x 2 so the mutual inductance MS is nothing but – LMS x 2 this is let us say because we are considering a phase displacement of 120 degrees this is the mutual inductance between a and b if you consider mutual inductance between a and c then instead of ? being 120 ? would be 240 degrees and still cos of 240 is the same – LMS x 2 and therefore mac is also equal to mab and I will leave it to you to show that mbc is also equal to mab therefore all the mutual inductances between the three phases either between a phase of the stator and b phase of stator or b and c or a and c they are all same and is equal to – LMS x 2 and therefore we can write this as the self inductance of the stator multiplied by current flowing through the stator plus mutual inductance less since all the three mutual inductances are between stator coils are the same we will simply call it as MS for the time being into IBS plus again MS x ICS so this is self flux these two are mutual fluxes mutual fluxes arising due to currents flowing in the stator winding now you need to talk about mutual fluxes due to currents flowing in the rotor winding so if we look at this figure we see that the axis of the stator a phase and the rotor a phase are displaced by an angle ?r and therefore if the mutual inductance between these two coils if the two axis are aligned such that they are in the same line then the mutual inductance will obviously be the highest now if they are displaced then the mutual inductance would obviously change we have seen for a winding on the stator and rotor the mutual inductance can be described as MSr x cos ? and therefore the mutual inductance between the stator winding a phase and the rotor a phase can then be described as MSr x cos ?r where MSr is the peak value of the mutual inductance and then between the a phase of the stator and b phase of the rotor the angle difference is 120 plus ?r and therefore multiplied by Iar of course this is MSr x cos ?r plus 2px3 multiplied by Ibr and then between a phase of the stator and c phase of the rotor so that would be MSr x cos ?r plus 4px3 multiplied by Ibr so this then is the expression for the flux linkage of the stator a phase now similarly one can write an expression for the flux linkage of the stator b phase that would also have a self flux component so Ls x Ibs and this Ls is the same as that Ls because we are considering again balance 3 phase winding so the winding will have the same self inductance plus mutual inductance of the stator x Ias plus mutual inductance in the stator x Ics plus now the b phase of the rotor has of the stator has an axis here and the a phase of the rotor has an axis here the angular separation between the two is 2px3 – ?r so I can write it as MSr cos ?r – 2px3 x Iar between b phase of the stator and b phase of the rotor the angle is still ?r and therefore this can be written as MSr cos ?r x Ibr and then you have MSr cos of between b phase of the stator and c phase of the rotor the angle is 120° plus ?r and therefore this is ?r plus 2px3 x Icr so this is then the expression for the self flux linkage of the b phase in the stator similarly one can write the expression for b phase c phase in the stator Cs is Ls x Ics plus MS yas plus MS ybs plus MSr x cos of between Cs and ar the angle is 120 plus ?r so ?r plus 2px3 x Iar then MSr cos of between Cs and br it is 2px3 – ?r so ?r – 2px3 x Ibr plus MSr x cos of ?r x Icr so this would be the expression for the flux linkage in the c phase now these expressions are fairly lengthy and one looks for ways where you can represent them in a more compact form and therefore we take recourse to vector notations we define the vector V as a vector of voltages so it is defined as Vas, Vbs, Vcs, Var, Vbr and Vcr. So this is a vector having 6 elements and then we define the vector of currents as Ias, Ibs, Ics, Iar, Ibr and Icr the vector of currents and what we are looking for is a relationship between this voltage and current note that p times Ias is nothing but p times some inductance multiplied by some currents that is what we have p of Ias is p of L x some currents and therefore a relationship between these two can be expressed by looking at all these equations. So how to write that equation let us look at the first row of this we can write Vas, Vbs, Vcs and so on that is the vector the first term if you are going to write V equal to something into I you can see that here it is R into Ias plus p of Ias which is p of Ls into Ias so we write this as Rs plus p of Ls multiplied by Ias we will write it as another vector there and then you have p times Ms into Ibs, p times Ms into Ics and then you have p times Msr cos ?r and then you have p times Msr cos ?r minus 2p by 3 and then you have p of Msr cos ?r plus 2p by 3. Now the matrix ends here and is multiplied by a vector of I so what you have is Rs plus p Ls multiplied by Ias which is the first element of the vector Pms multiplied by Ibs and so on so that is how the first row can be written and the other rows can be written in a similar manner. So in this way one can write down a description of the induction machine and in the next lecture let us see how to proceed further what we do with this kind of a description of an induction machine we will stop at this point.