Chapter two per phase analysis. It is simple and I'm sure that every power system person at one time or another has performed this, but because it is ultimately the foundation of power system analysis, I wanted to firstly review it, but also give it some formal process and structure. So, it can be used consistently and accurately. Simply put, per phase analysis is the reduction of a balanced three phase system to a single phase circuit for ease of computation. Then because we are dealing with balanced systems, the single phase solution can easily be extrapolated back to all three phases. The point to remember here is that per phase analysis isn't only useful For balanced loads and balance generators, what happens with an asymmetrical fault analysis that you get in in the real world is that you can use symmetrical components to break that asymmetrical system down into three symmetrical components a positive negative and a zero sequence, then each one of those can be treated as a balanced system and a per phase analysis applied to it.

And then we can work backwards and get the total solution after we've solved each one of the symmetrical components. So per phase analysis is very, very useful. In fact, it's almost an absolute when doing fault analysis. I'm going to go through a an overview of how the per phase system analysis works. And then I will go back and Do it in a little bit more detail and add a little bit more meat to the, to the process so that you can return and use it the same way every time. However, in a per phase analysis, the main thing we have to remember is we're working with a balanced system, that is all the loads are equal in magnitude, all the voltages generated are equal in magnitude and 120 degrees part.

The internal impedance in this case is designated by Zed s, and the load is designated by Zed. Now, the load is connected y configuration as well as the generators are connected y in configuration. If they aren't, in reality, they can be converted from a delta to a why there's a one to one conversion factor and we'll go through that as we go. Look at the per phase analysis process in more detail. However, what we want to end up with is basically a circuit that looks like what you have in front of you here in this slide. So, the line current, l is equal to the phase current IP is equal to the current flowing from a through to ass.

And it's equal to i l or the line current in three phase situation. Now, this current returns to the source via the other two phases because we have a balanced system. That is how the current returns to the source. There is no current flowing through the neutral or even if an was connected to ass and or grounded. There is no neutral or ground current flowing because the system is balanced. So we don't have a connection there, we don't have current flowing there.

Also, we know that the per phase voltage is equal to the line the line voltage divided by the root of three. We can now remove the other two phases, and we can see that just the we have put a connection between asset and because we do need a return path, and if we've removed the other two phases, we've removed that return path, so we have to replace it at least for our calculations. And it doesn't change the calculations in any way shape or form. It's just returned to the three phase system, we'll remove that connection. They have the impedance voltage is equal to the phase voltage again is equal to the line the line voltage divided by the root of three. And the phase voltage is given by v now we don't have to call it VP Because there is only one generated voltage, it's a single phase circuit.

And the power in the per phase calculation or the power per phase is equal to one third of the total power consumed by all three phases because remember, we do have three loads that are connected, and they're each drawing the same amount of power. So, if we want the total load for the whole system or the all three phases, we have to multiply the per phase power consumption by three or if we want to know the per phase power consumption, we have to take a total load and divide that by three. As I said before, per phase analysis allows the analysis of a balanced three phase system with the same effort as a single phase system. key here is the fact that we're dealing with a balanced three phase system. So, let's define what a balanced three phase system looks like. All sources produce equal magnitude voltage or current.

All sources produce three phase quantities that are 120 degrees apart. All loads are equally of equal value, measured face to face or measured face to neutral. There is no mutual inductance between the phases all neutrals are at the same potential. And all phases are completely decoupled. per phase analysis involves a five step procedure. First, we will convert all Delta loads and sources to the equivalent y connections.

We then solve for one of the single phases in this case we'll pick the a phase, and we'll solve it independently of the other phases. The total system power can then be determined by multiplying the phase voltage and phase current by three phases B and C can be determined by the hundred and 20 degrees phase shift of a balanced system. Then, if necessary, we can go back to the original circuit to determine line to line values or internal Delta values. Let's put our procedure into practice by looking at an example. Here we have a Delta connected load being fed by two sources through two transmission lines. AY connected three phase generator with line to neutral voltage of one volt at zero degrees.

On the left hand side of the circuit. A delta connected load was impedances minus j ohms which is a capacitive load essentially. And it's connected through a transmission line with the with the impedance of J 0.1 ohms per face. The load is also connected to a Delta connected generator whose voltage line to line is one volt at zero degrees, and it's connected to the load through a second transmission line of J 0.1 ohms per face. So our task will be to find number one, the line currents, number two, the load voltages. And the total power supplied by each generator, the Y connected generator and the Delta connected generator.

Okay, starting the analysis process, and we remember that the per phase analysis is a five step operation. And we'll just bring it up here so that we can remember what it is. First of all, we want to convert all of the Delta loads and sources to their equivalent y configurations. In this circuit, we have two items that are Delta connected. One is the load and the other is the generator on the right hand side. The load I'll deal with in a couple more slides.

It's just Straight mathematical conversion, pretty simple. And we already dealt with the relationship between face to face and face to neutral voltages. And we know that the face to face values are related to the face to neutral values by square root of three, and we know that the one will lag or lead the other by 30 degrees. Okay, let's return to our example problem and start the analysis process. before going any further, I wanted to describe the labeling standard that I've adopted here. As you can see, in the diagram, I've put several points at various locations.

For example, I've got point A, B, and C. As indicated on the diagram. I've got points a prime, B prime, C prime On the diagram, and I have a double prime B double prime C double prime as points. So when I'm describing the voltages and currents, for example, on the left hand side of the of the diagram, I have the face to neutral voltage of that generator described as the subscript A n. And on the right hand side, the voltage generator is a Delta generator. So the phase two phase voltage of that delta generator is described as the subscript a double prime B double prime. And when we get into the currents, as a current flows from A to A prime, I'll describe it as I subscript a dash a primed. So the first step that we wanted to take was to replace the delta j generator with a Wye configuration.

Now, we have to determine what the face to neutral voltage is for that configuration that we just established given that the face to face value is one at zero degrees. from our previous discussion, we know that the magnitude of the face to neutral voltage is equal to the face the magnitude of the face to face voltage divided by root three and the face to neutral voltage will lag the face to face by 30 degrees. So we can describe if we go through the mathematical process that the face to neutral voltage is given by one over root three at an angle of minus 30 degrees. Next You want to replace the Delta load with its equivalent y connected configuration. If you are going to do that the replacement values are as indicated in this diagram, it's fairly straightforward. All you got to do is replace the A B's and C's and the pq and RS with the values that are that are in our sample problem.

Now, we don't want to be confused and and go through a whole whole conversion process for anything except just one phase because we're going to ultimately end up with a one phase and the balance condition. So really, all we need to do is we want to find out what P is which is the phase two neutral impedance and the phase two neutral impedance. P is given by the A impedance times the B impedance over the a plus b plus c impedance. Now let's return to our example diagram. And we will carry forward the conversion formula from our previous slide. And we know that from the given values that a, which is the impedance from a prime to B prime is minus j or one at minus 90 degrees ohms.

B is equal to A, which was designated by the impedance a prime C prime and it's also equal to minus j and C is the impedance from C prime to be primed and it is designated as minus j We can now replace the Delta load with a Y connected load. And we want to determine what a prime and is in a way of impedance, which is our P and our formula, p is equal to A primed and primed. And if we substitute the A, B and C on the right hand side of our conversion formula, we have this. So the impedance a primed and is equal to minus j times minus j all over minus j plus minus j plus minus j. So, we can further reduce that to a simple factor. But before doing that, I want to just very quickly go over what the J operator is in a way of a quick review and then return and and reduce this equation to something a little bit simpler.

The j operator is used in conjunction with phasers. Remember a phaser has a magnitude and a displacement angle. Now the J operator operates on the phaser. But it only changes really the angle of that phaser. So let's have a look at it. If we have a phaser that is called plus one or one at zero degrees, it would look like this on our diagram.

If we were to multiply this vector or this phasor by the J operator, it would swing it through 90 degrees. And we would end up with this vector or this phaser which is described as plus j Or one at 90 degrees. Now if we operate on that phaser twice, in other words, multiply it by J and then multiply it by J again, we would end up with a phaser in this direction, which takes our original phaser swings it through 90 degrees, and swings it through 90 degrees again. So it ended up with a minus one configuration, or one at 180 degrees. If we were to operate on the phasor, three times j times j times j, the original phaser would then be swung through three sets of 90 degrees, or it would be described as minus j or one at 270 degrees. Now, let's return To our sample problem and apply some of these j operators in order to reduce the formula that we have derived before.

Getting back to our PR phase example, if we look at the fraction that we generated in the bottom left hand corner of the slide, that can be reduced by collecting terms in the denominator. So, you are left with minus j times minus j all over three times minus j and you can treat the operators as numbers and the numerator and denominator cancel up one minus j each and you're left with minus j over three which is the face to neutral impedance of the load. And we know that the B leg and the C leg are equivalent to the a leg, and they're all equal to minus j over three. We can now move on to the second step of our per phase analysis and that is to solve for phase A independent of the other phases. This is the circuit that we have after conversion of our delta loads and sources to y configurations.

We now are going to consider only the a phase and the return path the courses to neutral we have an array we know what the face to neutral voltage of the source on The left is we know what the voltage of the source on the right is. And we want to calculate the current through the lines. So we revert to current shops. Current law, which states that all core currents in a circuit, flowing into, a point or a node must add to zero. It's this equation. The current through line one which is ay ay dash, a prime, plus the current through the load, which is I, and a prime end and the current through the second line, which is ay, ay double prime minus A prime must add to zero.

Next, fall back to ohms law, which states that the current through an impedance is equal to the voltage drop across that impedance over that impedance. So we take each one of the currents and derive them from the voltages or the voltage drops across them. And we're left with this equation. The current through line one is given by the term on the left, which is the A prime and minus V AN all over j 0.1 plus the current through the impedance, load impedance, which is V A prime to n minus j all over three and the current through the second line, which is the A prime and minus the a double prime and all over the impedance of that line, those have to add to zero. And looking at this equation, you can see that we already know what the A n is. And we know what a V a double prime Dan is.

So the only thing we do not know is the A prime n or the voltage drop across the load. So if we substitute in this equation, for the known factors, we can rewrite this equation. Now, what I've done is I've substituted in the voltage sources in the spots in the equation where they're located. And you can see we're left with basically one unknown And all we have to do is solve for that one unknown, then we can back calculate for the currents of the individual legs of the circuit. Now, you're looking at this and you're wondering, okay, how did I come up with minus 10? j in the first term?

And how did I come up with three j in the middle term? And how did I come up with again minus 10 in the third term? Well, I've manipulated the J factor. And I'm going to run through that now for you. Okay, so in the first term, we have the voltage over the impedance of the line, and then the one over the impedance of the line is one over j, one over j 0.11 over j 0.1 is equal to j 10 over j times j, which is equal to j 10. minus one, or over minus one, which is also equal to minus 10 J. So basically, I've just substituted one over j 0.1 in those two terms of the equation and replace it with the term minus 10 J.

It's just manipulating the J operator. Similarly, the middle term of that equation, I started out with one over minus j over three, which is equal to j three, all over minus j times j, which is equal to j three all over one, which is three j, and I've substituted in that equation. So far, we have an equation with one unknown and that unknown is The voltage across our load which is the A prime end. If we collect like terms collect the, the A primed and on the left hand side of the equation and all the other terms on the right. It's just a straight algebraic manipulation where we can transfer everything that's known to the right hand side of the equation, and no one unknown on the left, all we have to do is solve the complex numbers on the right hand side of the equation, which amounts to that the load voltage is point nine volts at an angle of minus 10.9 degrees.

We already know what the A n is one MC at zero angle. We've already calculated the solution. voltage on the right and converted it to a light connected source. So, it is one over root three at a phase angle of minus 30 degrees. So, we now know all of the voltages at the various points in our single phase circuit. So, you can now go to the individual currents and solve for them.

The current through line one is given by the voltage drop across the impedance of line one over the impedance of line one which is now just a series of complex numbers which we can solve and it comes out 2.21 amps at 214.3 degrees the current through the load is now the voltage across the load which we know over point three three at minus 90 degrees, which is the impedance of the load. And that can be solved for 2.7 at an angle of 79.1 degrees. Lastly, the current through the line from the Delta generator is given by the voltage drop across the impedance of that line over the impedance of the line, which again is we have all of the the various unknowns, we now know them, we've calculated them and we can reduce the equation to 5.95 amps at point three five degrees. We now want to calculate the power delivered to the load and The power comes from two sources in our case, one is the light connected voltage source on the left hand side of the circuit.

And the other source is the Delta connected voltage source which is on the right hand side of our circuit diagram. We know that the power delivered by the wind generator will be three times the face to neutral voltage times the phase current. In this case, it's three times VAN times I, a dash a primed. We already know what the current is, and we already know what the voltages are. So it's just a matter of substituting the values in our equation and we come up with With the power delivered by the why connected generator is 5.1 plus j 3.5. Va the Delta generator is the power delivered by the Delta generator is calculated by the equivalent face to neutral voltage of that generator which we have figured out and the current flowing through the a phase.

So it's three times v a double prime to n times i a double prime to ash a prime. And we know what the answer to the current is. We've already figured it out. And we've already figured out the voltages, so it's just a matter of substituting them into the equation and solving the complex arithmetic and we get the point delivered by the DEP Delta generator is 15.8 minus j 2.94 va. The fourth step is if we want to go backwards now from a single phase circuit into the three phase circuit and calculate out the values for the B phase and C phase. And that's simply done by knowing the fact that we're we have a balanced system and the phasers are 120 degrees apart.

So we know what the phase two neutral voltage of the load is. It's point nine and minus 10.9 volts. So we can say that the voltage the blue phase voltage to the load is V. be primed and point Nine times 229.1 volts, which is just the voltage of va va primed and plus 240 degrees and the the the C phase voltage is given by a phasor that is 120 degrees from the phase to neutral voltage for a. And you can see it on our diagram there as the phasers rotate the phase two phase voltage can be calculated and on our diagram which I haven't got on here, but if you remember from our diagram, the phase two phase voltage is a prime B prime which is 1.56 at 19.1 degrees how do we figure that while we go Back to our relationship of our face to neutral and face to face quantities.

The phase, the magnitude of the face to face quantity is root three times the face to neutral quantity for the magnitude only. And that comes out to 1.56. And if we want to calculate or we want to figure out what the angle is, we know that the the angle leads the Face Face angle leads the face to neutral voltage by 30 degrees. Now, if necessary, we can go back to the original circuit and determine the line to line values or the internal Delta values. And I'm not going to go through that exercise but just remember the face, face to face and face to neutral relationship and all of the values can be figured out from there. This ends chapter two