 we are looking at the steady state modeling of alternator that is synchronous machine. So let us continue with what we were discussing, we had written down a few equations, so we will start with those equations again, we had said that the stator voltage VaS is given by root of 2 by 3 into RS into IDS – XQS into IQS cos ? – RS into IQS – XDS IDS – XMD into IF – into sin ?, so this was one equation and then the second equation that we had was IAS is root of 2 by 3 times IDS cos ? – IQS sin ? and then we have the expression for the generator torque, electromagnetic torque TE is given by XMD into IF – into IQS plus LDS – LQS into IDS into IQS, so these are the three expressions that we landed up with when we tried to convert the DQ axis differential equations to the steady state formulation. In this steady state formulation we must understand that IDS IQS since they are the variables in the synchronous reference frame these quantities are essentially constants at steady state they are DC variables and IF – he is also a constant because the field current is excited with DC in the synchronous reference frame because the rotor is already rotating at synchronous field and in that reference frame we are exciting it with DC and therefore IF – he is constant in the synchronous reference frame. So with this what we had tried to do was to develop a phasor diagram because the synchronous machine is analyzed in steady state using phasor diagram, so what do we do with respect to the phasor diagrams in the last class we saw a brief review of phasors and therefore this expression ID cos ? is basically since ? is equal to ? synchronous multiplied by time what you have here is IAS is root of 2 by 3 times IDS x cos ? T – IQS x sin of ?S x T, so this itself represents a sinusoidally varying sinusoidal with respect to time right and since it is something that is varying sinusoidally with respect to time one can look at its RMS value and to get the RMS value you have to divide this by root 2 and therefore the RMS of this is then given by IDS by root 3 times cos ? ST or rather IDS by root 3 is the RMS value of the first term and IQS by root 3 is the RMS value of the second term. Since each of them are varying sinusoidally with respect to time we can represent each one as a phasor for example if we take let us draw the DQ axis and let this be the D axis and that is the Q axis it is of course rotating with synchronous speed but at any given instant let there be DQ axis position this way. Now this term is root 2 by 3 times multiplied by IDS cos ? T since it is a sinusoidally varying term we can represent this as a phasor and since it is just cos ? T this can be represented as a phasor lying along the D axis the length of the arrow would then be equal to the RMS value of this sinusoidally varying term and the RMS value is given by IDS by root 3 so let us call this as capital ID so this length is nothing but ID. Now if you look at the second term this is root 2 by 3 times multiplied by IQS sin ? T this is again sinusoidally varying but it is sin ? T and in the last lecture we saw how you can represent a term varying as sin ? T that too with a negative sign and therefore that term would then be represented by a phasor that can be drawn here so this length of this is then represented by the RMS value which is IQS by root 3 so let us call this as IQ if you draw the phasor of this variable IQ that then lies along the Q axis as it is here because it is – sin ? T and what this expression says that IAS is the sum of these two which means that we can write IAS phasor as ID plus J times IQ so ID is along the real axis because that term corresponds to cos ? T IQ is along the imaginary axis or the DX Q axis because it is a term that is – sin ? T which can alternatively be written as an ID phasor plus an IQ phasor where IQ is the phasor representation of the Q axis along the Q axis so this itself is IQ so the length is given by IQ the length is given by ID as a phasor one can represent these two terms as a phasor ID and a phasor IQ the phasor IQ lies along the Q axis the phasor ID lies along the DX now one must remember we have talked about space phasor earlier these are not space phasor these are since we are dealing with cos ? T sin ? T and so on these are time phasor we are drawing time phasor diagram and it is not really a space phasor diagram one can however give it a space phasor orientation because the DQ axis itself is rotating synchronously in along with the rotor and therefore one can interpret this but then nevertheless what we are looking at is since there is a variation with respect to time we are looking at it as a time phasor. So having written this that IA is the phasor sum of ID and IQ we will now go back to look at this expression itself now this expression also is some term multiplied by cos ? T and some other term multiplied by sin ? T and therefore VAS is again a sinusoidally varying term which therefore if you want the RMS value of the sinusoidally varying term you have to divide this by root 2 and once you do that if you club the IDS and IQS terms together there is a bracket missing here so let us now write VAS so the okay so VAS can be written as root of 2 by 3 multiplied by let us bring the similar terms together so you have RS x IDS cos ? – IQS ? so that is these two terms we are now we have taken and then you have – XQS IQS cos ? and then you have – XDS IDS sin ? and then you have – XMD x IF- x sin ? this is what we have now if you look at this expression this term RS multiplied by IDS cos ? – IQS sin ? we have already seen that IDS root of 2 by 3 multiplied by IDS cos ? – IQS sin ? is nothing but IAS and therefore we can write this as RS x IAS and then you have this next term which is XQS x IQS x cos ? multiplied by root of 2 by 3. Now what we had said earlier is IQS x sin ? is a phasor that is lying along this axis whose length is IQS by root 3 if we take the RMS you need to divide by this root 2 therefore what you are left with is IQS by root 3 and the phasor of IQ lies along the Q axis if that is so the description is – QS sin ? if we have the same length but lying along the – D axis that would then be written as IQS x – cos ? and therefore this term can be written as since it represents the same length but lying along – D axis which represents then a rotation by 90 degrees rotation of the Q phasor by 90 degrees can then be written as J x IQ phasor J x IQ because the Q axis phasor is rotated by 90 degrees and this is then XQS where the first term will then if you can write this as RS x IAS where this is the phasor VAS right. So VAS being RS x IAS in the phasor notation will simply be RS x phasor of IAS and here this is IQS x cos ? because that lies 90 degrees ahead – cos ? one can write it as J x IQS now let us come to this one you have IQS that is what we have written so here we go this is IDS x sin ? now IDS this representation as a phasor for IDS represents IDS cos ? and if this phasor is rotated by 90 degrees you would then get a phasor representation that has to be given as IDS sin ? with a – sin that is just the same as what you had here – IQS cos IQS x sin ? but now instead of IQS it is IDS and therefore this term can be represented as plus J x IDS J x IDS would be the phasor which represents – IDS sin ? x IDS and then you have this term this is also sinusoidally varying and that is the – sin ? term and this – sin ? term just like what we had in the case of IQS this is also – sin ? and therefore this was represented as a phasor along the Q axis similarly this term is also a – sin ? term and therefore that would also be represented along the Q axis and it is RMS value will be ? 2 x 3 x that RMS value is this amplitude divided by ? 2 and therefore if we represent an induced EMFE which is sinusoidally varying with respect to time given by the amplitude of 2 x 3 x XMD x IF dash the RMS value of that would then be given by XMD x IF dash x 3 since RMS value you have to divide by root 2 and therefore this term in phasor notation can simply be written as J x this number E because it J x the number E it is understood that it lies along the Q axis so this is then the phasor representation of the phase voltage that is applied which can then be used to construct phasor diagram we will do so in the moment but before we get to the phasor diagram itself we need to look at one more aspect the synchronous machine the synchronous machine as the case of any other machine can be made to work as a motor or it can be made to work in the generating mode if we remember we have all along been saying that we are basically making use of motor convention in developing these equations that means we are considering that currents into the windings are positive that means if I am going to draw a phasor like this and I consider this to be greater than 0 then it means that what I am really saying is this much current is flowing into the machine that is not the only thing because we are considering motor conventions and we have derived the expressions in that form torque that is developed is positive if it is a motor that means if the machine is really supplying some load torque making the load rotate by taking electrical input then this number Te if you evaluate from this expression it will turn out to be greater than 0 right so now let us look at the synchronous machine equations in a slightly different format we know that Vds we have Vds and Vqs Vds is basically given by Rs x Ids – xqs x Iqs okay and then Vqs is given by Rs x Iqs xds x Ids xmd x If this is what we have written earlier normally in synchronous machines the stator resistance drop is quite small and the machine when we are operating it connected to the infinite grid the phasor representation of the machine is normally drawn for the case where the machine is connected to the infinite grid infinite grid being infinite grid the supply voltage is always fixed frequency is always fixed and therefore Vds and Vqs would really be a representation of the phase the stator phase voltage that is applied and since it is applied from the grid the supply voltage remains fixed it does not change since we are looking at an infinite grid and because the resistance drop is small normally one can neglect the resistance drop so if you neglect the resistance drop this is the equation that we land up with Vds is simply proportional to Iqs and Vqs is proportional to or is equal to the sum of these two and we know that this is representative of the emf that is induced just now we have written some expressions where this represents the emf that is induced and let us therefore represented by the symbol E though really it has to be root 3 times V well we could write it as root 3 times V as well to maintain equality of symbols. Now imagine a situation where the synchronous machine is has just been synchronized to the grid so if it has you know that when you want to synchronize a machine to the grid it can be done only if the voltages induced in the synchronous machine are equal to the grid voltage is equal to the grid voltage only when these two are equal in phase sequence amplitude and frequency you can close the switch and synchronize it to the grid and therefore under that condition when all these three are equal and you close the switch the synchronous machine basically floats on the grid there is no flow of armature that is Ias in the machine Ias is 0 that means here Ids is 0 and Iqs is 0. So what we can see from this is under this condition when Ias is 0 and Ids is 0 and Iqs is 0 Vds becomes 0 and Vqs is directly equal to the induced Tms so that is the situation that exists when a machine is just synchronized to the grid and if you look at the expression for Te in the expression for Te what you see is Iqs is 0 when the machine is just synchronized as we have been discussing Ids and Iqs are 0 therefore Te is equal to 0 when the machine is just synchronized obviously because no flow of current is there. But what can be done in the synchronous machine you know that when you synchronize the machine you would have adjusted the field excitation so that you get the voltage that you want at the speed that is required. Now from that level of induced Tmf or from that level of field excitation that you have given after synchronizing one can either increase the field excitation or decrease the field excitation Ias dependent on is equal to Xmd x If by R3 at the instant of synchronization you would have fixed the field current at some value such that it generate rated voltage. Now you can try to increase the field excitation beyond that because the machine is synchronized to the grid the voltage at the phase at the output of the machine cannot change it is already connected to the grid that means Vqs cannot change why Vqs alone cannot change there is Vds as well and square root of Vds2 vqs2 is then nothing but the representation of Va therefore Va does not change one can argue maybe if you change the excitation Vds may change as well in addition to Vqs. But if you are just changing the field excitation one can look at the expression for Te now this torque is normally not so significant as the torque component due to this therefore let us say for a moment we neglect this portion then if you increase if field excitation alone Qs is not going to flow because of that or if you are not increasing the mechanical input Te still has to be 0 because Te has to be equal and opposite to the mechanical input shaft torque that is going to drive the generator under steady state and therefore if these two are equal this must be equal to 0 if we know is not equal to 0 because we are supplying some field current and therefore Iqs will have to be 0 if Iqs is 0 then Vds is obviously 0 and therefore this cannot change and Vds is 0 so if the stator voltage that is once it is connected to the grid if the stator voltage therefore has to be fixed what it really means is that Vqs remains fixed even though you attempt to change E by changing this excitation and how will that remain fixed that will therefore mean that some current Ids will have to start flowing if you change this value of E. So if you try to increase E then Ids which was 0 at the time of synchronization Ids must now start becoming negative if you try to decrease E Ids must become greater than 0 so that the sum is always equal to Vqs in normal operation you then have the resistive term as well but if the resistive term is going to be small then this behavior will still hold good. So what we can conclude is that we have derived the expressions for motor operation so may be let us try to divide this into 4 parts this represents motor operation motor mode in this mode if you try to over excite this machine that means field current is being increased from the level that was required for synchronization Ids will become negative so Ids is negative and we are looking at it as a synchronous motor that means the torque that is output from the machine must be greater than 0 TE must be greater than 0 if TE is greater than 0 you want Iqs to be greater than 0 if this is going to be a small component so if Ids is less than 0 and Iqs is greater than 0 you get the motoring mode of operation over excited motoring mode of operation over excited motoring mode of operation. Now on the other hand what one can do is under excite the machine how can you under excite after you synchronize the machine to the grid you would have done it with a certain field current now try to reduce the field current so if you reduce the field current E goes down and since Vqs has to be maintained at the same level Ids has to become greater than 0. So Ids greater than 0 is the case for an under excited synchronous motor or under excite synchronous machine and you want to look at it as a synchronous motor so Iqs is greater than 0 so using the conventions that we have used if we have a case where Ids is less than 0 and Iqs is greater than 0 one can say that we are looking at an over excited synchronous motor. If Ids is greater than 0 and Iqs is also greater than 0 we can say we are looking at it looking at the machine operating as a under excited synchronous motor we can alternatively look at a generator mode of operation there also one can have over excitation and under excitation an over excited generator synchronous generator in the convention that we have used would then mean Ids is still less than 0 because we are talking about an over excited case but it is a synchronous generator and the torque must therefore be negative in our convention and therefore Iqs has to be less than 0 and an under excited synchronous generator will then mean Ids is greater than 0 and Iqs is less than 0 so with these broad ranges that have been defined for Ids and Iqs one can now try to draw the phasor diagrams for the various modes of operation how does one do that so let us take the first case over excited synchronous motor so and let us write that over excited synchronous motor let us take the d and q axis so that is your dq axis as before this is d and that is q so we know that over excited synchronous motor Ids is less than 0 therefore d axis current would then be drawn here this is Ids is less than 0 so it has to be in the negative side and similarly Iqs is greater than 0 so we draw the q axis current here Iq and the sum of this Id plus jIq is then the armature current phasor that is Ia. Now the next thing is to look at this expression this expression says that Vas is Rs into Ias plus this plus this and j times e now j times e always has to lie along the q axis because we are taking the dq axis as reference j times e is a pure imaginary number so we will have j times e here I can represent it as a phasor itself with respect to time j time e and then to that you have to add j times Ids xds Ids is here now j times Ids will then be a phasor that leads Ids by 90 degrees and therefore we that would then be j times Ids xxds and then you have j times Iq xqs Iq is a phasor that is here j times Iq will lead it further by 90 degrees so j times Iqs xqs we are adding right you have e plus jId plus jIq and then you have to add plus R into I so R into I and the resultant this quantity is then your armature volt Vas so Vas is here and this angle is now called as the load angle which is normally given the symbol ? now this procedure gives you how this phasor diagram can be drawn where the location of the exact phasor will lie and the length of all these things will depend upon the actual magnitude of current that are flowing okay so this is an over excited synchronous machine similarly one can draw the phasor diagram for the synchronous motor similarly one can draw the phasor diagram for a synchronous generator so that is your d axis and q axis so as before we draw the phasors of Id and Iq first so here you have Ids is less than 0 and Iqs is also less than 0 so you have Ids here and Iqs here so the phasor of Ia will then lie here this is the phasor of Ia and then you have the induced emf is in all cases along the q axis and therefore you have the induced emf here that is E and now you need to add J times Ids x Xds so we had used yellow for that so let us use yellow and you have Ids here J times Ids will be a phasor that leads Ids by 90 degrees so J times Ids and then you have J times Iq x Xqs Iq is here and J times Iq will lead this by 90 degrees and therefore Iqs and then you need to add R x I is here so R x I will be in this direction in the same direction as this so this resultant is now your voltage Vas so in this way the phasor diagram can be drawn normally when you talk of the generator mode of operation as we have been saying many times earlier you consider that the flow of current out of the machine is greater than 0 that is the reference direction for current whereas what we have used is current into the winding being greater than 0 so if you consider the reference phasor in the other direction you will have to think of the phasor for armature that is Ia as this is your Ias generator convention whereas this is Ias motor convention right so in this manner one can draw the phasor diagram now obviously one can see that in the generator mode you have a larger emf and a smaller grid voltage because of the drop that is coming in from the induced emf to the grid of course that will depend on the excitation condition so let us now look at the next phasor diagram which is the over under excited synchronous motor so these diagrams correspond to the motoring mode and what we are going to draw here they correspond to the generator mode so the under excited synchronous motor the under excited synchronous motor has ids greater than 0 so ids is greater than 0 and then you have Iqs also greater than 0 so you have Iqs phasor here and the resultant of these two is the armature current so that is Iqs Ias and as before the induced emf is always along the q axis and therefore you draw the induced emf phasor here and induced after that your Ij times ids ids is here j times ids would be 90 degrees and j times ids therefore comes here and then you have j times Iqs so that comes here and then you have Rs x Ias Ias so Rs x Ias there and this resultant is then your voltage stator voltage Vas and this angle now becomes your load angle load angle is the angle that is between the stator voltage phasor and the induced emf phasor so in this case that would be the load angle so let us go to the next one the synchronous generator the under excited synchronous generator in the case of under excited synchronous generator what you have is ids greater than 0 and Iqs lesser than 0 so ids is greater than 0 Iqs is lesser than 0 so some of this is the armature current phasor that is here and as before the induced emf phasor is along the q axis and then you have j ids j times ids would be in this direction j times Iqs will be in this direction and then you have R x I that is here and therefore this will then be the stator phase voltage so this is e and this is j times id multiplied by xd this is j times Iq multiplied by xq and this is Ia times Ra this is all phasor so this would then be the phasor diagram and you have the load angle that is here so what we see is if the mode is that of a synchronous motor the load angle is positive this way and if the operation is that of a generator the load angle is negative this again is based on the convention that we are going to fall so in this manner one can try to draw the phasor diagrams of the machine we can I mean as in this case what we have done if you want to draw Ias in the generator convention that is just the negative of this and therefore the generator convention current would then be here this is Ias generator convention this is Ias in the motor convention now this is fine but normally when you want to draw the phasor diagram of a synchronous machine you do not know where the d and q axis are what we will know is the relationship between the phase voltage and the phase flow of current Va and Ia is what we would know we know for a we would know for example that the phase voltage if we take that as a reference we would know that the phase current may be it is acting in a lagging mode so we know that Vaas is here and we know that Ias is here so if this is the case how do you find out what is the magnitude of the induced Tms normally in the phasor diagram analysis one is interested to know what would be the induced Tms and what would be the load angle corresponding to a particular state of operation so if this is all that is known how do you construct the phasor diagram in order to find out where the induced Tms will be and how much it is so that can be done in a simple way in this manner what we do is we take this expression Vaas that is equal to Rs x Ias plus j xq x Iqs plus j e and then you have this term j times id xxd what we do is add a term j times id xxq and subtract this term j times id xxq so the net equation does not change and we can simplify this as Rs multiplied by Ias remember these are phasor so id plus j times e so we keep this here and now you have if you take this term these two terms you have j times xq into id plus Iq and then you have j times id xxd – xq so this equation simplifies like this now id plus Iq is nothing but Ia and therefore we can write this as Ia and so we write this as Rs x Ias or Ias plus j times xq into Ias plus j times e plus j times id xxd – x now what is interesting in this form of the expression is that you know that id lies along the d axis the phasor id lies along the d axis and therefore j times id must lie along the q axis and this j times e also lies along the q axis so both these term together lie along the q axis we know where Vas is and where if we take Vas as the reference phasor we know where Ias is because you can measure you know how much phas lag or phaselit that is going to be there so this term is nothing but Vas – Rs x Ias – j times xq x Ias is equal to this term so you know Vas and therefore you will know where Ias is so let us draw the phasor diagram in this manner so Vas is here and then let us say Ias is here – Rs Ias will be a phasor that is in the same sense as same angle as this but – and then jxq Ias will be a phasor 90 degree to Ias and because it is negative it would be in this direction so – jxq Ias this would then be at 90 degrees to this so once we reach this point we know that that is equal to this term and this term lies along the q axis and therefore this point must lie along the q axis so this axis is then the q axis and therefore 90 degrees to this must be the d axis and once you know d and q axis you can resolve this into id and iq and then construct the phasor diagram as we have done earlier so in this manner one can draw the phasor diagram from a set of measurements that are made even when we do not know where the d and q axis is in this case we started with the d and q axis drawn first whereas now we are able to arrive at where the d and q axis are and thereby finish the phasor diagram we will stop with this for today and continue in the next lecture.