 Welcome to class 6 on power electronics topics in power electronics and distributed generation. In the last class, we were looking at distribution system example and the objective was to look at the sizing, the ratings of the protective elements that we would need if you are connecting a facility to a feeder. In this case, there is a fairly large facility being connected to the feeder and you want to look at what should be your rating of the fuse, the circuit breakers etcetera coming back in from the facility. In the last class, we had looked at what is the impedance looking back from the point of common coupling, taking the connection of the feeder as the point of common coupling to evaluate what should be the what would be the fault current that the fuse would see. Example, if you have a primary winding fault in the transformer, essentially the fuse would need to carry a fairly large current which would only be limited by what is the upstream impedance. So, we had looked at the values of the per unit values of the impendences seeing back from the point of common coupling and we were about to look at the distribution transformer and we will do that in today's class. .. So, if you look at the distribution transformer, it is a 2 MVA 11 kV slash 415 volt transformer. So, you know your nominal current level in your transformer, your primary current level I primary is 100 amps and your secondary current is 2 point almost 2.8 kilo amps. So, if you look at say taking your overall system base to be your distribution transformer base, then your Z base on your primary side would be 60.5 ohms and if you look at it on your and if you look at the low voltage side, your Z base would be 86.1 milli ohms. So, if you are looking at a transformer, you are given your transformer percentage reactance and winding resistance, you are given XL is 4 percent and your R of your winding is 1 percent. So, essentially this would then correspond to 2.42 ohms on the high voltage side for the reactance and 0.61 ohms on the low voltage side, on the high voltage side, on the low voltage side you would have 3.4 milli ohms and 0.86 milli ohms for the reactance and the winding resistance respectively. And the base quantities that we would consider for the analysis, the overall analysis would be your transformer quantities. So, your V base for your low voltage side is will take it as 415 by root 3, this is 240 volts and if your S base is 2 MVA and your corresponding I base is 27, 8, 2 amps and you get your Z base. So, 86 milli ohms is your Z base. So, if you look at it on the high voltage side your base voltage is 11 kV is line to line, your phase voltage is 11 by root 3 that would turn out to be 6.35 kV, your S base stays the same, your I base would be 100 amps and your Z base would be 60.5 ohms. So, depending on from what side you are normalizing you would have to use the appropriate base quantities. So, if you want to normalize the source side we calculated the source parameters in the last class, your voltage on a normalized basis is 11 kV voltage divided by the 6.35 on the high voltage side is 1 per unit. So, that is what you would have as a voltage base. And if you look at your source side impedance your XS is we calculated this to be 0.87 ohms in the last class and your base quantity is 60.5 ohms. So, this turns out to be 0.14 per unit or 1.4 percent. If you look at your RS, your RS turns out to be 0.5 ohms divided by 0.008 per unit. So, if you compare with the numbers that we had in the last class your XS and RS as seen by now your reduced power level distribution transformer compared to your substation power level the per unit count quantities it is actually lower now, because you are now changed over to a lower power base. So, this is what we had mentioned when you change over to a lower power base you would see this effect which is what you notice in terms of those numbers. If you look at the distribution transformer your XL and RL stays the same, because you are making use of the same the transformer values as the base quantities. So, your RW so it stays the same, because your base quantities are related to the transformer quantities itself. So, the next thing is your internal distribution. So, to look at your internal distribution you have to look at what it is quantities are related to you are told it was 1 percent for your reactants and 2 percent for your resistance, but now that is related to it is power rating which was 1 MVA power level. So, if you look at the base quantities for corresponding base quantities that would be 415 square divided by 1 MVA. So, this would be 0.172 ohms. So, your actual X line is so to bring it now to your common base you could you have to now normalize to your base value of your transformer. So, your X line is now so you can see that, because now you went from a 1 MVA base level where it was specified to a higher base your 1 percent reactants of the line now got scaled up to 2 percent, because you are going up to a larger base quantity. So, this is again consistent with so you could either go to the physical quantity and then derive by the actual base quantities or you could use the change of base equation that was mentioned in the previous class. So, the next part in this particular circuit is the load will assume that the load is a parallel RL load and the load is given to be 0.8 MVA 0.8 power factor lag 440 volts and will assume that it is a. So, if it was running at 440 volts your load current would be 0.8 at 440 volts if you are assuming it is just a RL load now you apply a reduced voltage of 415 volt. So, then maybe your current level will come down to 0.99 kilo amps say suppose it was a induction machine then instead of the current level dropping it might draw a constant power. So, the current might actually go up. So, you have to make the assumptions on about what your actual load is to determine what your current would be. So, your P load your actual real power of the load is 0.8 MVA into 0.8 power factor 0.64 megawatt and your reactive power of the load is your 8 MVA square minus 0.64 kilo watt square under square road. So, that turns out to be 0.48 MVA. So, if you look at your R and L of your parallel RL load your R load is 440 square divided by. So, you determine what you can determine what your R and L is and then if you look at it on your common base on your low voltage side your R load and per unit would be 0.3025 divided by 86.1 into 10 power of minus 3. So, this is 3.51 per unit and your X load in per unit is 0.403 divided by your base quantity on your low voltage side which is 86 milli ohms. So, this turns out to be 4.7 per unit. So, now that you have all the parameters of your circuit now normalized to a common basis you could actually then draw a final single line equivalent circuit of your system. So, if you look at your final single line circuit essentially you have the source which is now its reactance is J 0.014 the resistance is 0.008 your fuse f 1 is over here and then you have the distribution transformer impedance which was J 0.04 reactance and resistance of 0.01 then you had then you have circuit breaker 1 followed by your distribution panel which goes to a bank of multiple breakers. So, circuit breaker 2 is also essentially at the same point then you have your internal wiring your Z L which is J 0.02 reactance and resistance of 0.04 and your load is 3.5 per unit and your reactance is J 4.7. So, now you have the parameters of your circuit on a single line on a consistent base. So, now you could you make use of this to look at what happens if you have a fault at say the distribution transformer or just somewhere along the internal wiring sections or some somewhere closer to the load point. So, the next thing that we will look at is so we made use of if you have a fault over here you could evaluate what the rating of the fuse f 1 needs should be. So, you next need to evaluate what should be the rating for your circuit breaker 1 and circuit breaker 2. So, if you have a fault downstream of your transformer if your distribution transformer essentially this is for your C b 1 comma and C b 2 your I f is by Z of the source plus Z of the transformer. So, if you look at its magnitude this turns out to be 17.4 per unit which corresponds to 48 kilo amps RMS. So, you get your peak current that you would expect your circuit breakers to interrupt when you have a fault just downstream of your distribution transformer to be 48 kilo amps and you know what your ratings of the circuit breaker 1 is it has to carry 2 MVA of load current. So, for your C b 1 your rated current is 2800 amps that is the what it is rated to carry if your fault current is 48 kilo amps RMS your V rated is 415 volts and your isolation should be at least twice your rated voltage with some additional margin say you would you could take it as say 2 k V might be or 1800 volts might be the standard values that you might get at for this particular voltage. So, if you look at the circuit breaker 2 the circuit breaker 2 is now rated for feeding a internal distribution line which is rated to carry 1 MVA of power. So, if you look at your I rated for circuit breaker 2 it is actually now 1400 amps, but your I fault is still 48 kilo amps. So, you can see that even though the continuous rating of your circuit breaker 2 is lesser it is fault current that it has to interrupt is the same as the bigger the circuit breaker C b 1 your V rated and the isolation voltage stays the same. So, in the circuit that we were looking at. So, in fact all the devices all the breakers C b 2, C b 4, C b 5 plus any other breakers that might appear in parallel they might carry a potentially lower current levels on a continuous basis, but the sizing to interrupt a fault current is quite large depends on essentially the short circuit current level that can happen immediately after the transformer. So, the next point at which you would need to calculate a fault is if a fault occurs right at the load then you make use of the fault current level at the load to now decide on your circuit breaker 3. So, your fault current level is now your 1 per unit of your voltage divided by Z s plus Z of your distribution transformer plus Z of your internal wiring and your magnitude of I f turns out to be 10.6 per unit. So, this corresponds to 29.4 kilo amps RMS. So, if you look at your circuit breaker 3 it is I rated is now corresponding to your 0.8 m e a load. So, you get current level of roughly 1.1 kilo amp your fault current level is 29.4 kilo amps and your voltage rated voltage and the isolation voltage stays the same. In fact if you look at this particular circuit you will see that if you look at a breaker often the ratings of the breaker is related to how much fault current it needs to interrupt because the amount of energy that gets dissipated when the contacts open corresponds to the current that it is trying to interrupt. So, even though C b 4 and C b 5 might be carrying much lower current the cost can be quite substantial if your current level it needs to interrupt is quite high. So, sometimes people what they do is they could add say series inductance with just upstream of the breaker. So, the purpose of the inductance might be to reduce your fault current level. So, the cost of your inductance might be lesser than the cost savings that you incur in your circuit breakers because potentially incoming feeder might go to a large number of branches. So, there are potentially a large number of breakers. So, you might see suddenly in the wiring inductances being added in some sections. So, this is intended to handle the fault current level and manage it to an appropriate level. So, at this particular point we did a per unit analysis of the system and we calculated the fault current levels. We did it for a balanced case and some of the assumptions that we had in when we do a per unit analysis this is a radial distribution system. So, we do not have loops. So, if you assume in general that there are no loops in with net transformer ratio gains when you are doing your per unit analysis and also if you are that is especially the case when you might have auto transformers tap changes etcetera you want to ensure that you are doing it at the nominal level. So, that you do not have circulation within loops also if you have phase shifters you have star delta transformers etcetera. Again you are assuming that the net phase shift in a loop is adds up to the same irrespective of which direction you are going. So, these are some of the assumptions that you are making and the even the fault current level is steady state fault current level. We are not looking at unbalanced effects at this particular point and to address the unbalanced effects we will have to look at sequence components. Also we are not looking at the dynamical effects. For example, depending on where the fault is occurring and depending on your x by r ratio of the line you can have peak fault currents which are much higher than your steady state fault current levels. So, the dynamic effects are also ignored and what we are calculating is the steady state fault current level. So, what we have looked at so far is the a three phase fault and if you want to look at unbalanced faults and a common fault is a single line to ground fault you will have to look at the sequence components. Again the sequence components along with on which the symmetrical components analysis is based one needs to keep in mind that it addresses the unbalanced issue, but it is still a steady state concept. So, the quantities that you are looking at are still phases RMS quantities etcetera rather than peak instantaneous values etcetera which you would need for dynamic analysis to look at what might be the worst case peak current that may be your breaker needs to interrupt. So, if you look at the symmetrical components essentially what you are saying is if you have unbalanced your phase quantities might be not equal and not phase shifted by 120 degrees you might have three arbitrary quantities say if you take the voltage your voltage V a V b and V c might be arbitrary. And what we are doing is we will take this three phases a V a V b V c and represent it in terms of nine quantities V a 0 V a plus V a minus V b 0 V b plus V b minus V c 0 V c plus V c minus. So, if you look at it you will think that this is actually a bad way to go you are taking three quantities and you are representing it as nine quantities which is making it more complex, but the main advantage that is there is that this V a 0 V a plus V b V a minus and the b and c corresponding quantities are all related the V a plus V b plus and V c plus form a positive sequence group in the sense that V b plus is V a plus with a phase lag of 120 degrees V c plus is V a plus with a phase lag of 240. If you look at the negative sequence component your V b plus is V a with a phase lead of 120 degrees and V c minus is V a with a phase lead of 240 degrees. And if you look at V b 0 and V c 0 they are actually equal to V a 0. So, if you look at the number of independent quantities over here you really have just three independent quantities. So, you are representing three arbitrary quantities in the phase basis in terms of three quantities on a sequence basis. So, based on what we just mentioned now if you represent 120 degree phase shift by the number a small a your V a is V a 0 plus V a plus plus V a minus. If you look at V b 0 is the same as V a 0 V b plus is 120 degree lagging V a plus which means that it is now a square times V a plus. And similarly you will get a times V a minus for the negative sequence value of the b phase quantity. And similarly you can write it now in a matrix form for your transformation now from a b c to 0 plus minus. And because you are now looking at a quantities for all the three phases you could also drop the subscript a and just look at call it V 0 V plus V minus the implicit assumption is that you are referring to phase a when you are looking at 0 plus minus. So, you could write now this in a matrix form as and essentially in a compact form you could call this as V a b c is the matrix a times V 0 plus minus. So, essentially what you are looking at is transformation going from a b c to 0 plus minus. If you look at the determinant of the matrix a you could calculate that fairly straight forward manner you would get 3 into a into 1 minus a which means it is non zero. And the matrix is invertible in fact what you have is a linear transformation from a standard basis to a new set of basis which leads to your symmetrical components. So, if you look at the matrix a if you this matrix a is actually symmetrical. So, you take a transpose would be the same as a in fact you can calculate what your a inverse is a inverse turns out to be this particular matrix over here. If you look at what it is it is actually one third a transpose conjugate sometimes people refer to the transpose conjugate as the Hermitian of a matrix. So, a h is essentially its transpose conjugate if you have a matrix which where a inverse is equal to a h then it is called a Hermitian matrix here it is not exactly a Hermitian matrix. Because you have the factor 1 by 3, but it is nearly Hermitian because your a inverse is related to the inverse transpose of a. And if you take a times a inverse transpose you would get 3 times the identity matrix. So, if you look at it in compact form you can determine your 0 plus minus quantities as one third a h times v a b c. So, this is so the next thing that you could ask is what happens when you look at power in on a sequence basis rather than on a phase basis. So, your power is essentially s is p plus j q and on a per phase basis this is v a times i a conjugate plus v b i b conjugate plus v c i c conjugate. So, if you write it in a vector form this can be thought of as multiplication of a row vector times a column vector. And the row vector being v a b c transpose and the column vector being i a b c conjugate. So, the dot product of v a b c and i a b c conjugate gives your power. Now, you could substitute for v a b c and i a b c in terms of the change of basis transformation to go from your a b c reference to your sequence reference. So, you can write v a b c as a times v 0 plus minus and because you have the transpose over here the a moves over to the other side and a transpose is the same as a. And if you write i a b c conjugate that is a conjugate times i 0 plus minus conjugate. And we know that a conjugate is 3 times the identity matrix. So, your p plus j q is now 3 times your v 0 plus minus vector transpose into i 0 plus minus transpose. So, if you just expand this out this is 3 times v 0 i 0 conjugate plus. So, if you look at a typical circuit typical circuit you would have you might have unbalanced loads. But, under normal conditions your voltages should not be unbalanced if you want to maintain reasonable power quality to your loads. So, your v plus is the dominant term over here v 0 and v minus should be quite negligible. So, even with unbalanced loads your dominant part of your power comes from v plus i plus conjugate. So, your positive sequence would determine most of your power in under normal conditions. Of course, if you have a severe condition such as a fault then the value of v plus v minus etcetera can also become large. So, the next question that you could ask is then what happens when you have impedances and if you look at impedances in the normal per phase basis and then if you look at impedances on a sequence basis how would it transform. So, if you look at uncoupled impedance say v a is equal to z a i a v b is z b i b and v c is equal to z c i c. Essentially you can write it in a matrix form like this and then you can write your v a b c as a times your v 0 plus minus and your i a b c can also be written. So, this is actually i a b c can also be substituted as a times i 0 plus minus. So, you could then transfer over the a you get a inverse z a b c times a and if you carry out the multiplication for the matrix this particular matrix for that is given over here. Then essentially what you would get is a fairly full matrix of this particular form which is shown at the bottom of this particular section. So, if you look at this you will feel that this is may not be a good idea where you have a simple nice diagonal matrix and you do some transformation and you end up with a matrix which is full. You could then look at some simplifications where what happens when say the value z a z b z c are equal. So, in a situation such as that then you know that a property is of the quantity small a is 1 plus a plus a square is 0. So, when a b c are equal then the off diagonal terms would drop off again you would end up with a diagonal matrix, but again this looks fairly simplistic. The more interesting situation is when you have coupled system where you have mutual coupling between the phases. So, suppose you have v a is equal to i a times z s plus i b z m plus i c z m. So, now you have mutual coupling from phase b and c into your voltage and phase a and similarly for your b and c phase. So, if you look at your z a b c matrix you would get z s z m z m and if you transform it to your sequence components matrix you end up with something which is actually now substantially simpler you get z s plus 2 z m 0 0 0. So, you can see that you end up with a fair amount of simplification if you look at systems where you have mutual coupling. And for many practical power application where you are looking at machines, transformers etcetera you end up with coupling between the phases. And using a symmetrical components based analysis you can expect considerable simplification in terms of your circuit analysis where you have a diagonalized circuit in your sequence domain. Whereas, in your phase domain you end up with a full matrix for the impedances. So, if you look at specifically a power line. So, if you are looking at the sequence models of a power line your z plus is equal to z minus and this is z s minus z m. And from your courses on power systems analysis you would have looked at expressions for how to calculate the reactance of a line. So, you have expressions such as mu naught by 2 pi ln d by r prime d represents the distances between conductors and r prime is a effective radius. So, if you look at a situation where you are looking at the distance between conductors and overhead line the distance between the phase conductors might be small. But, then if you look at the height of the line above the ground what you might need to consider for a phase 2 single phase line to ground fault your z naught you will end up with a much larger value of distance compared to what you have for your positive and negative sequence components which flow within the lines themselves. So, you end up with z 0 which is typically greater than z plus. So, if you look at a typical number for a line you would see things like such as x 0 divided by x plus is around 3. So, it gives you higher 0 sequence impedance compared to your positive sequence impedance for your line. If you look at a transformer your z plus is equal to z minus your z 0 depends on the type of your construction and the transformer configuration. So, for example, if you look at a transformer you are familiar with core type or a shell type transformer. So, if you have a shell type transformer and if you have a 0 sequence flow that can flow in one of the windings of the primary or secondary winding of the transformer essentially you will end up with 0 sequence flux and the return path of the 0 sequence flux is essentially the same core the same path as for the positive or negative sequence. So, you end up with similar impedances for your 0 sequence as for positive or negative sequence whereas, if you look at a transformer which is of core type what is shown over here is what is in the white is a core and the hashed and the wavy lines correspond to the primary and secondary windings of the transformer. So, if you have again currents that can flow through the winding which can lead to a 0 sequence flux. Now, the 0 sequence flux are going up on all 3 limbs of the transformer essentially the path for it would now be through the air rather than through a magnetic path. So, in one case you would see a large air gap. So, you would end up with essentially much different value of your 0 sequence impedance also it depends on say for example, you take this core type and you put it in a in some sort of a cabinet or a enclosure. So, you might have walls of the enclosure which is sitting nearby and the flux that comes out of the transformer met now interact with the cabinet or the enclosure and cause heating or hot spots on the surface. So, depending on your application if you want to have 0 sequence flow in your windings it may be preferable to look at shell type or if you want you could look at a single phase transformers connected in the appropriate configuration. So, the type the value of the 0 sequence impedance depends on the type of transformer that you are using and whether you should be using such a transformer in case in the first place at all. What is shown over here is the core and shell type for 3 phase you can also have core and shell type for single phase also depending on whether the windings are enclosed by a core on all 3 sides or whether the windings are sitting on the 2 limbs of a single phase transformer. So, depending on the transformer configuration and also depending on the winding type you can have different impedances for the transformer. So, what is shown over here is say y y transformer where both the y points are grounded. So, if you look at the positive and the negative sequence model going from your primary to the secondary if you now do a per unitize analysis essentially the transformer becomes transparent. So, essentially you are modeling the leakage inductance and the winding resistance the leakage inductance term is a dominant term. So, you might call the impedance as Z L and you use the same Z L in your model for both positive and negative sequence. When it comes to 0 sequence you have a Z 0 and your Z 0 would roughly be equal to your Z L. So, for a grounded y y transformer your 0 sequence impedance turns out to be the same as what you had for your positive or negative sequence. If you look at say other configurations say if you have say a ungrounded y you can have a grounded y on the secondary or it can be even ungrounded essentially what you would have is the positive and negative sequence circuits stay the same as what we discussed previously, but the 0 sequence circuit is now open. So, essentially this would not allow 0 sequence current to flow through the circuit. Essentially in this model we are neglecting the magnetizing branch and the core loss branch assuming that its value is much larger than 1 per unit. So, it can be neglected. So, for 0 sequence practically you have now a open circuit even if you take 1 grounding of from 1 of the y points of the transformer. If you have now y delta type of transformer now if you ground the y point now from your primary side essentially you now have a path for your 0 sequence current to flow. So, you have now z naught connected back to your return on the primary side on your y side, but on the delta side there is no path for the 0 sequence to flow. So, it is open on the delta side. If you have a grounded y delta and if you are grounding through a impedance z n essentially now your 0 sequence currents coming through the individual phases would sum up you will have 3 i 0 now flowing through this particular impedance. So, essentially the model would be z 0 plus 3 times the impedance that you are connecting to the neutral point and the secondary side stays open. So, depending in this case say when you look at your positive sequence model the impedance wise it is still stays the same it is still the dominated by the leakage inductance. You have to keep in mind what the phase angle is going to be between your primary and secondary voltage. In the next class we will take a look at the phase angle depending on your winding configuration you can have different phase relationship between the angle between your primary and secondary. We look at it what are the possibilities with a standard y y or delta delta or delta y type of configuration in terms of phase relationship between your primary and secondary winding. Thank you.