Chapter Seven, a symmetrical three phase fault analysis. With a symmetrical three phase fault analysis, there are some assumptions that are commonly made prior to doing the calculations. All sources are balanced and equal in magnitude and face. sources are represented by the Thevenin equivalent, just prior to the fault and at the point of the fault. This is important. You'll see in our calculations later how that works.

So I'm going to repeat it sources are represented by their Thevenin equivalent just prior to the fault and at the point of the fault. large systems may be represented by an infinite bus transport numbers are usually on nominal tap position resistances are negligible compared to react and says transmission lines are assumed fully transposed. So all three phases have the same impedance. Load currents are negligible compared to the fault currents. And lying charging currents can completely be neglected. So, let's take a look at an example system just so that we can see what kind of assumptions we're making and where we are going with the analysis.

Looking at this system, we have very typically in a system you have generating sources at both ends of the system on the left and the right You have transmission lines, one at 220 kV and the other 110 kV, and you have a series of Transformers that are hooking everything together. And that's how a system might look. We're going to assume just for the purposes of an example here, a fault might occur on line one. And just prior to that fault, all our assumptions have been made. So what we've done here, because firstly the system is balanced, is that we can do a per phase analysis on the system and replace the three phase three phases with a single phase equivalent circuit. The other thing that we have done too is we have converted all of our all of our system quantities to per unit values.

Which essentially gets rid of all the Transformers turns ratios. So we have a much simpler circuit here to deal with. And it's again a single phase circuit. If we again look at the fault occurring on the 220 kV line, what we can do is we can replace all of the system impedances and sources by their Thevenin equivalent, so that we would be left with one source one impedance. So, just prior to the fault, we know that we can replace the system by one source and one source impedance, which is the the Thevenin equivalent for the source and the Thevenin equivalent for the impedance. During the fault condition, we are going to short circuit, the system and we're going to Do it in various ways.

And things start to happen after the fault. And that's what we're going to analyze. But for the purposes of this analysis, we're going to start just prior to the fault to analyze this circuit. Just taking a small step backwards just prior to doing our Thevenin equivalent of the circuit. The circuit, just prior to the fault is made up of several impedances and several are a couple of sources here anywhere in most cases could be several sources. The fact of the matter is that all of these impedances we are given them by the various manufacturers or design people, and they're not necessarily just given as actual impedances themselves.

They are made up Each impedance is made up of a positive and negative and a zero sequence component. And those are usually specified by the manufacturer. As in the case of the impedance of transformer one, it's made up of a positive negative and a zero sequence. And every one of the positive negative and zero sequences that are in our system have positive negative and zero sequence components and they are usually specified by the manufacturer prior to us doing any studies. We're now going to look at the system post fault. In other words right after the fault occurs, and see how the system reacts up till this point.

Everything has been symmetrical. The system has been nicely balanced and compact Currently, we're able to analyze the circuits using balance conditions, which was a lot easier than where we're going now. However, we are now going to enter into a system that has generated voltages and generated currents that will be asymmetrical. Having said that, what we have to remember is that regardless of what happens or in and around the system with the voltages and occurrence and everything working towards an unbalanced condition, the system only generates positive sequence voltages and currents. In other words, it doesn't matter what happens to the system in the way of impedance changes due to a fault conditions in short circuits. We are only generating positive sequence flows.

Images and or currents, and for the time being, we're only going to be considering voltages. The same thing can be said for the currents. But you must remember that the system has only positive sequence generators in it. So as I said, we're going to break out the system into its positive, negative and zero sequence networks. And we will analyze each one of them individually. And then we can add them together at the end to find out exactly what's happened asymmetrically to the whole system.

Because we're only generating positive sequence generate or we only have positive sequence generators in the system. The positive sequence network will look like this. It will have a positive voltage generator, which we call here is the voltage generator that EDF and the system will have because it's a positive sequence network will help have only positive sequence impedance in that circuit. So this positive sequence network will consist of one voltage generator, which is the, by the way the Thevenin equivalent at the fault. So we're really looking at the fault voltage and it looking at only the positive sequence impedances and in fact, it's going to be the sum of all of the feminine equivalent positive sequence impedances in the circuit. And the circuit equation for positive sequence network is given by the voltage drop across the whole circuit is given by EF minus Zed one times I want or the positive sequence impedance times the positive sequence current.

The negative signal network because it hasn't got any negative sequence generation and it consists only of negative sequence impedance and it looks like this, and the circuit will be given or the voltage drop on that impedance will be given by minus Zed two times I two or minus taught minus the negative sequence impedance times the negative sequence current. And similarly, the zero sequence network will look like this without any generation in it. And it has an equivalent circuit equation where the voltage drop is just the voltage drop across the zero sequence impedance times the zero sequence current. Now, at the risk of repeating myself, everything up to the point of the fault occurring, everything was nice and symmetrical and balanced and everything was humming along nicely how However, once a fault occurs, things start to happen to the circuit. And each fault type has a characteristic that we will have to look at.

And we'll have to introduce those parameters into our fault analysis. And we're going to do the fault analysis using each one of the sequential components. But we have to use the parameters that are generated by the various types of faults we're going to study. And we're going to look at for now, four of the basic fault conditions in doing our analysis. The first one is going to be a simple single phase to ground fault. The second one is going to be a single phase two phase fault.

The third one is going to be a phase two phase two ground fault. And the last one we're going to look at is a symmetrical actually three phase fault. condition where all three phases are shorted together. Looking at face to ground or a single line to ground fault, as I said, the type of fault that generates the parameters that have to be considered in, in doing our analysis. And in this case, we're going to ground the a phase, we could have just as easily grounded the B or the C, but the the analysis would be the same. So we're just going to pick the a phase for now.

And we're going to ground it and once we ground the a phase, we automatically set the parameters for this type of a fault. Which says because we've grounded the phase, the voltage at the at the fault, the a is going to be equal to zero. Now, what I'm talking about here is the aid of neutral voltage. I've left off the neutral part of the subscript Only for convenient because it gets cumbersome carrying it all the way through. But anytime you see a with a sub sorry, a letter with a subscript letter, that's considered if it's a voltage line to neutral voltage. So anyway, the V subscript a, the a voltage at the point of the fault is equal to zero because we ground in the system and it has to be zero.

We also considered the fact that the load currents don't enter into the calculation because they're so they're so small considered to the default values that we're going to set them equal to zero and that'll make our our analysis a lot easier. So, the B phase current is zero and the C phase current is zero. These are the parameters of face to ground fault. The current in the a phase all of the current is going to flow to the fault So I subscripts a is going to equal I subscript F for the fault current. We've already set or said that our a phase voltage at the fault is equal to zero, but the a phase voltage is also made up of the sum of the symmetrical components of the a phase. So the a zero which is the zero sequence voltage at the fault it plus the a phase, positive sequence voltage plus the a phase positive and negative sequence voltage all summed up is also equal to zero.

So, just keep that in mind as we go through our calculations. We also know from our synthesis equations that this is a fact we can calculate the positive negative and zero sequence current by virtue of knowing what our original or actual current values are. Are. And in this case, we can rewrite the synthesis equations in this manner, because all we've done here is substitute the current in the B phase and C phase which is zero. So all the terms that have zero in them don't add to the equation. So what is this, the three equations tell us is that the negative phase sequence a phase current is equal to the positive a phase sequence current is equal to the negative a phase sequence current, and all of them are equal to one third, the actual a face current.

We're now going to start to analyze each one of the sequence networks. And you remember the sequence networks from a previous slide. The positive sequence network is made up of The Thevenin equivalent of the voltage at the point of the fault, which is generating only positive phase sequence voltages in series with a positive sequence impedance, the negative sequence network is made up of only the negative sequence impedance and the zero sequence network is only made up of the zero sequence impedance. The equivalent circuit equation for the positive sequence network is the voltage drop across the two components of the generator and the voltage drop across the impedance is given by this equation here and the voltage drop across the negative sequence. Network is given by this equation as is the negative or zero sequence natural Work, voltage drop is given by this equation. Now because these individual networks are symmetrical and balanced in their own right, we're going to look at only the a phase of these symmetrical networks.

So we can rewrite the equations. And they have a subtle difference in that we're only talking about the a phase voltage and the a phase current in those networks. So we now are looking at the voltages and currents being replaced by the a phase voltages and currents. The impedances don't change, they're the same, no matter which phase you're talking about. The next thing we're going to do is we're going to do a series of just mathematical manipulation of the equations that we already have there. There's nothing, nothing magical about it other than this just using street equation, mathematics and moving terms around.

So what I want to do is I want to take the last set of equations that we have there. And I want to take all the terms on the left side of those three equations, and I'm going to sum them up. So we'll have the VA zero plus VA one plus v two. And we're going to equate that to all the terms on the right hand side of those equations, what you have to do to maintain balance in the equation that we're developing here. So in taking the first set of terms, it comes from the the zero sequence network equation, and that means we're going to move minus at naught times one A naught down here. We're going to take the terms from the positive sequence network, right hands side of that equation, move it down here.

And we're going to take the right hand side of the negative sequence network equation and move it down here. So we have maintained our balance on the equation. The other thing I want to do is we have a series of three different currents on the right hand side of the equation we got one a zero we got one a one, sorry got ay ay ay ay ay ay zero, ie one, and ay ay ay ay two. We want to replace that with ay ay, and we have to carry the one third term from this equation up here so I'm going to substitute, one third ay ay ay ay ay ay ay ay ay ay into this equation now, and it's going to look like this. Essentially, I've gotten rid of the positive negative and zero sequence currents and replace them with the actual, a phase current.

And we're left with this equation which is still balanced and intact. still valid. Okay, continuing on with our mathematical magic here, we know that the terms on the left side of this equation are essentially just the addition of the three sequential voltage components, which add up to VA, which we know VA is equal to zero. So we can rewrite the equation so that everything on the right side of the equal sign is equal to zero. Now we're just going to again continuing on with mathematical manipulation. I'm going to bring the E f term over to the left side of the equation.

And I'm also going to drop the negative sign that it gets as it moves over there. But I'm also going to have to keep track of that because I'm going to change the sign on everything on the right side of the equation. But that's all going to be looked after. I'm going to collect the like terms. So I'm going to put all of the Zed or the sequential impedances. in brackets zeros at one to two.

And I'm going to collect them. The in terms of one third which is attributed each one of the components and ay ay ay ay. so we now can rewrite that equation so that f is equal to one third, ay ay ay ay ay. All times inside a set of brackets, zero plus said one plus two. We can rewrite that equation so that ay ay ay ay is on the left side of the equation, and everything else is on the right side of the equation, which gives us three times e subscript F, all over the sum of the sequential impedances And we know that the fault current is equal to I A. So, I A is equal to i f which is equal to three times e f all over the sum of the sequential impedances.

So, what is this all say? Well basically we have solved for the fault current that flows in a line to ground fault we are we know what the E f is which is the Thevenin equivalent of the positive sequence voltage generators and we are given the positive negative and zero sequence impedances for the circuits So, the fault current is going to be three times the F voltage at the fault all over The sum of the sequential component impedances. Also, the, the voltages, positive negative sequence voltages a phase, all in series add up to zero. And all of the sequential currents, the positive negative and zero sequence current are equal. So if we were to draw an equivalent circuit that we could use to model that, we would come up with something like this, which we can use Well, we're going to model a line to ground fault and analyze it for various various different components that might be given in the system.

Because VA zero plus v one plus v two is just purchase Voltage Law. All right. The series loop, so that satisfies the equation. The sequential components are all equal. So that's the same as a, putting the network components in series that gives us a series circuit. So, this circuit would represent how a system would react.

If a ground if a line was to be grounded. Also, the fault current is equal to the phase a current which is equal to the sum of the phase AC quencher components. We also know that each one of the components are equal to the other and they're all equal to one third ay ay ay ay. Or you can rewrite that so that the fault current is equal to i A is equal to three times the zero sequence current or three times the positive sequence current or three times the negative sequence current. So this would help in the analysis of a line to ground fault. Now let's look at a face to face or a single line to line fault.

And in this case, let's choose phases B and C. And basically they are connected together solidly that would produce a solid short circuit between phase B and phase C. And they would generate these parameters for this type of fault where the voltage at the point of the fault, the B is equal to VC because they're connected together. Because the phase A only has load current on it which is considered negligible compared to the fault current we're going to set ay ay ay at zero. And because of that, The return path for the current in IB is through ice or through the sea phase, we can safely say that the current flowing out of B as the current flowing into C or IB is equal to minus I see. And these are the parameters for a face to face fault. writing our synthesis equations which allow us to find the sequence component values from our actual values, we start with these equations which we've already devised in our symmetrical opponents study.

We're going to rewrite the equations this time we're going to set it equal to zero. And we're going to replace the term or the current IC with the current minus IB and rewrite it The equations they would look like this. The first one would be one third inside the bracket zero plus IB plus minus IB which is equal to zero. The positive sequence current would be one third inside the bracket would be zero, the operator times B and the a squared operator times minus b, which if you rewrite it could look like one third. The a operator times a term I be inside the brackets one minus A outside the brackets. Looking at the negative sequence current and substituting the zero for ay ay ay ay ay, and the minus the term minus IB for ICS.

See, we're left with the term Finally is equal to minus one third a operator times I be, inside the bracket would be times inside the bracket would be one minus a. So the first equation says that ay ay ay ay ay ay ay not or the, the the a phase zero sequence current is zero. And that the because the second equation is just the minus the first equation, that the phase a positive sequence current is equal to minus the face a negative sequence current. So those two terms will just leave for now and they were just mathematically derived from our sentences equation and substituting our non values Let's leave the current equation For the time being, and let's look at the voltage equations and the voltage phasers. Synthesis equations are the same synthesis equations that we use for the current, and I've rewritten them here. I'm going to go through and I'm going to make a substitute because the B phase voltage is equal to the C phase voltage at the point in the fall.

I'm going to replace all the B phase voltages with the C phase voltages and rewrite the synthesis equations using just VA and, and VC. And you can see that the positive sequence synthesis equation works out to this. The negative one works out to that. And if you look closely at the last two terms of those two equations, they are identical but just changed places. So we could actually rewrite the last equation Which would equal the second equation if you can see they're the same terms there. So that this tells us that the a phase positive voltage sequence is equal to the a phase negative sequence voltage.

So we've got these three equations that we're going to continue to develop in the next slide. So just remember we have ay ay ay ay ay not is equal to zero. I a one is equal to minus ay ay ay ay ay two, and ba one is equal to VA two. Let's move these equations that we just developed over to the left so that we can have some room to look at them. And just for convenience, I'm going to rewrite the current equations for the positive and negative sequence. Once current components for the a phase like this, so we'll have 181 plus 182 is equal to zero.

So what we'd like to now do is devise or connect the network equations in such a way that they satisfy these equations. And what we come up with is this. We'll now have a look at how these equations are satisfied by these models that we've developed for the positive negative and zero sequence network. Starting with the easiest one, the zero sequence network, we know from the equation ay ay ay ay ay not as equal to zero tells us that there is no current flowing in the zero sequence network. So we can represent the zero sequence network as open circuited and not connected to anything and that satisfies The equation, the other current equation tells us that the current flowing in the positive sequence network is equal in the opposite direction to the negative sequence current. So that if we connect the positive and negative sequence networks in parallel, that satisfies the current equations that we have there.

Also, the voltage drop across the positive sequence network is given by the a one and it's also equal to the voltage drop across the negative sequence impedance, and that's in parallel. And so we satisfy the third equation which is VA one is equal to VA two. So far we've looked at two situations one is a face used to ground or single line to ground fault. And the second one is a face to face fault or single line to line fault. And just a little bit of a discussion here before going on. What we're what we're able to do with this and I call it modeling is we can now analyze what happens after a fault.

Prior to a fault. We are only generating positive sequence voltages and currents in the balance system. There is no negative or zero sequence. currents are things flowing, because everything is balanced. But after the fault occurs, we start to connect things in such a way that the circuits are no longer balanced. In other words, if we ground a phase, it's no longer a balanced system.

And if we Short out to the phases together then it's no longer a balanced system. And what we have done is devised a way of bringing in the positive negative and zero sequence networks and showing how they would be connected together to model that particular type of a fault. So, in the case of the the ground fault, you can see that the positive negative and zero sequence components are connected in series. Here positive negative and zero sequence networks are connected in series. And in the case of the face to face we got the positive and negative connected in parallel and the zero sequence phase open. So we can essentially find out just about anything we want to know about the circuits after the fault occurs by substituting in the known quantities that are given to us prior to the fault occurring It also will help us to understand what kind of things affect our circuit.

In other words, how does the zero sequence impedance affect a fault and that would lead us to be able to understand the circuit protection, relay protection and how we're going to develop it. However, it's just a couple of comments at this point in time. There are many, many books out there on the detailed study of asymmetrical fault analysis. This is basically an introduction to it. Kind of to whet your appetite if you want to go further and and study it. Okay, we're going to move on to a another couple of parameters just to cement our knowledge of how the network secret the sequence networks react.

To various faults, we're now going to look at a face to face to ground fault, or this is where two of the lines are shorted together in the same time, I guess maybe fall to the ground or they're shorted to the ground. Anyway, you're connecting two of the phases together and connecting them to ground such as indicated in the diagram here. And we've chosen to use phase B and C as the ones that are shorted out to ground and phase A is essentially just connected to the load. The parameters that are generated that are characteristic of this fault are that voltage VB is going to be equal to voltage, the C at the fault because they're connected together and they're connected to ground so they're at a potential of zero. So VB equals c is equal to zero. And the only current flowing in a phase is a load current which we can neglect it or actually is Essentially equated to zero so i A is equal to zero.

We know that ay ay ay is made up of the sequential components of zero plus and negative or zero, positive and negative so that i A is equal to i a zero plus iy one plus i two, which is all equal to zero. We can now write the synthesis equations for this circuit. And we will use the voltages in this case so the zero positive and negative sequence voltages. Va zero VA one VA two are made up of the actual voltages or the asymmetrical voltages for phase A, B and C, as written here with the A operator. We can rewrite that equation or rewrite these equations by substituting zero in For the terms VB MVC, and we're left with this these three equations here, where if we get rid of the zero factors in the equation, they boil down to the fact that they're each one of the sequence voltage components are equal to one third VA.

So, we might say, are we not we can say safely that the zero sequence a phase voltage is equal to the positive phase sequence of voltage for the a phase is equal to the negative phase sequence for the a phase. So those three voltages are all equal. Once again, we have a set of equations that we would like to develop a model using our sequence networks now, that would satisfy these equations that we have here, which are the parameters or the characteristics of a face to face to ground fault. And if we connected our sequence networks, all in parallel, as indicated here, they would satisfy those equations on the left. You can see here that the currents for the positive negative and zero sequence, if they're complete connected in parallel, they have to add to zero which is purchase current law. So we've satisfied that middle equation.

The voltage drops across the three network components are equal. In other words, the zero sequence voltage drop is equal to the positive sequence of voltage drop is equal to the negative voltage drop as indicated here. So We can safely say that this model connecting all of the network network components in parallel satisfies the equations on the left. We're not going to look at a face to ground fault with impedance. And what that means is when a line is grounded, it doesn't necessarily form a zero impedance to ground sometimes your fault impedance included in that. Also, in the case of a generator that is, grid has an impedance to ground purposely wired in, it would have the same effect if we had a single face to ground fault, in that it limits the fault current.

But for now, let's just assume that the impedance to ground is Actual an impedance of fault impedance sets on saying that phase A to ground. So the phase a current would be made up of all of the fault current. And our parameters for this type of fault has to be rewritten. Previously we had zero impedance to ground so VA was zero, but in this case it is now the a phase times the fault impedance, which is just ohms law. We know that the sum of the sequential components for the voltage is equal to the voltage across the the impedance at the fault. It'd be previously it was zero but because we got impedance we have to include that in our equation.

The same synthesis equations that we use for our previous line to ground fault are going to be used. And in this case IB and IC are still equal to zero. So we still end up with the sequential components for positive, negative and zero sequence current are all equal, and they're all equal to one third, the actual phase current. Okay, just as we did with the line to ground fault, we're going to establish the positive sequence network and its equation which you see here, the negative sequence network and its equation, which is you, you see here, and zero sequence network equation, which you see here. I'm going to go through the same mathematical calculations. I'm going to sum up all the terms on the left side of those equations.

So you have VA zero plus v one plus v two is going to be equal to everything on the right side of those equations. And because I can now make a substitute for the positive negative and zero sequence currents in terms of ay ay ay ay ay ay, I'll do that. And that will introduce a one third into the term from our current equations up above. So, this is no different than we did before. And we ended up with essentially the same equation. Okay, all of the terms on the right hand side of the equation r equal to the sum of the voltage symmetrical components VA zero plus v one plus v two is equal to i a Zed F. So I will equate that to everything that was on the right hand side of that equation now.

So we have this equation here. Which, if we multiply both sides of the equation by three, we'll get rid of the fraction of one third on the right hand side of the equation. I'm going to bring the term three to the left side of the equation and I'm going to change the sign so I'm going to, same as I did before, bring everything. In terms of ay ay ay ay ay ay ay on the right hand side of the equation. I'm going to collect like terms. Everything that's connected with or multiplied by ay ay ay ay ay ay, which happens to be Zed not plus Zed one plus two, plus three Zed F. So we that can be rewritten as ay ay ay ay ay, or the fault current, or the current in phase A is equal to three times the positive sequence Thevenin equivalent voltage all over does that one Plus said two plus said not plus three said F. What does this mean in terms of our modeling?

Not a whole lot of difference, but there is a difference. And we know that the current in the a phase is equal to the fault current And we say or we are going to have a look at this particular model that will model a line to ground fault through some type of impedance whether that's grounding impedance of a generator or the actual ground impedance itself. We're going to call it Zed F. And the model is going to look like this. And let's see if it satisfies our equations. We know that all of the sequential components for the fit positive, negative and zero, sequence phase a current are all equal. So they would be satisfied by this type of a circuit putting the three circuit components in series.

Because if you have ay ay ay ay ay not there is a one there and a two there. They are definitely equal because they are in series. So we've put our three network components in series with each other. We know that the fault current is equal to i a, is equal to some of these sequential currents, or each one of the sequential occurence are equal to each other and they're equal to one third ay ay ay ay ay ay ay. Or you can say the fault current or the i a current is three times any one of the sequential component currents. So we can say now, if we put this impedance Z three times that F in series with this circuit, we have still, we're still satisfying the equation that the sequential component currents are equal, but now we have these voltage currents or sorry, voltage values here.

That have to be considered. And you can see that V zero is the voltage drop in the zero sequence network. v one is the voltage drop in the positive sequence network, and the two is the voltage drop in the negative sequence currents. And that has to satisfy our equation that is equal to i a Zed F, which is the voltage drop across our impedance three times our impedance. That's what comes out of the equation. So if we take the fault impedance, whether that's the impedance of grounding of a generator, or the actual grounding, impedance through the line that's grounded.

We multiply it by three and put it in series with these networks. Then it satisfies all the equations. So, if we are modeling a after a fault condition in a circuit that is grounded through an impedance, this is the circuit that would model and we can then proceed to analyze this given the quantities of the voltages and the impedances that usually come from the manufacturer. Last fault that I want to look at I'm going to look at it very quickly. It is a three phase fault. And as I said in in previous lessons, this type of fault seldom occurs naturally because as natural occurrences happen in a system, one or two of the phases might be shorted together or shorted to ground, and the line would essentially trip off before the third one was faulted.

Where you do see this happen is when, for some reason, during maintenance procedures, ground clamps are left on the line or the generator and the system is closed into it, then you'll get a three phase fault condition. So it's worth having a look at. What we have here is a phases A, B and C are actually connected together instantaneously to give a short circuit, so the VA is going to be equal to VB is going to be equal to the C is going to be equal to zero. The currents flowing in the three phases all add to zero because they're essentially connected together and current shops current law states that the currents will add on to zero. So these are two of the characteristic equations for that type of fault. Now, there you can go through a fairly long series of equations and mathematical manipulations, which I'm not going to do now.

Suffice it to say this is good, what the outcome looks like is that the zero and the negative sequence currents are zero, they always come out to zero given these parameters. So, these are the equations that have to be satisfied by our or what we like to model them. And we can model them in this fashion, such that the negative and zero sequence networks have no current flowing in them. They're essentially zero and zero voltage drop across the impedances. So they don't become significant. In any calculations for a three phase fault situation, all you have to do is deal with your positive sequence voltages and currents and your Thevenin equivalent for the voltage at the point of the fault.

So, f is going to be equal to the positive sequence, a phase current times the positive sequence impedance for that circuit. This ends chapter seven