Chapter one per phase analysis. 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're dealing with a balanced system, the single phase system can easily be extrapolated back into a three phase system. The key to the process is to recognize a balanced three phase system. And a balanced three phase system consists of all sources produced equal magnitude and voltage or current. All sources produce three phase quantities 120 degrees apart.

And thirdly, all loads are equally measured face to face or face to neutral The procedure for a per phase analysis and this is what we want to return to every time we enter into the situation to be consistent. Number one, convert all Delta loads and sources to their equivalent y connections or configurations. then solve for a single phase usually the a phase or the read phase. Just pick a phase and solve it using only one phase disregarding the other two phases. If there is no neutral, then the return path has to be provided hypothetically as a neutral. Then when calculating the total system apparent power, it's three times the individual phases which goes for the power as well as the var load Then you can calculate the other two phases, the B's and the C's, or the twos and the threes by determining whether it's plus or minus 120 degrees and it has the equal magnitude of the phase that we just solve for probably the a phase.

And if necessary, If required, you can go back to the original circuit to determine the line to line values or internal Delta values. Let me demonstrate this procedure with this example case of a three phase system which consists consists of a generator, a three phase generator, that's why connected with a line to neutral load of one at zero degree volts. It has a Delta connected load minus j ohms that is connected back to the generator by a transmission line that has an impedance of J 0.1 ohms per phase and the load is also connected to back to a Delta connected generator Whose Line the line voltage is one at zero degrees through a second transmission line that has the same per phase impedance as the other transmission line of J 0.1 ohms per phase. We are going to give ourselves some tasks here and the tasks involve finding first of all the current that is flowing in each of the lines.

Secondly, we would like to find the voltage across our load And lastly, we want to find the total apparent power supplied by each generator. And those apparent powers are going to be designated as s subscript y at s subscript delta. So let's start the process. And we're going to start with the first step and that is to convert all the Delta loads and sources to their equivalent y connections. We have two Delta connected items in this circuit, one is the Delta connected load, and the other is the Delta connected generator on the right. Now in order to convert the Delta connected generator to a wide connected configuration, we're going to review the phase relationship voltages in a balanced circuit.

The phase to neutral voltages in a balanced system are equal in magnitude and 120 degrees apart. I've color coded them just to differentiate the terms as we go along. So the red phase of course will be red and the blue phase will be blue. I've color coded the white phase as green because if it was white, you wouldn't see it. So our face to neutral voltages as I said, you can see them here are equal in magnitude and 120 degrees apart. Therefore, the phase two phase voltage quantities actually form an equal lateral triangle whose contained angle inside the equal lateral triangle is 60 degrees.

Now the white to blue face to face both Agenda white phase voltage in the blue phase voltage form and a Sausalito triangle, you can see on the bottom of the, of the picture there. And that is actually made up of two triangles. And those triangles are a standard 60 3090 triangle for contained angles that is, and the ratios of a 60 3090 triangle is one, two and root three that is the magnitudes or the lengths of the size of a 6030 triangle are in the ratio of one, two and root three. What this means is the magnitude of the phase voltage to the phase two phase voltage, the ratio of the magnitude that is, in other words, white All over white to blue is in the ratio of two, all over two times the root of three. And that, of course, is the same as the ratio of the red phase voltage to the red, the white voltage and the blue phase voltage to the phase two phase blue red voltage.

Now we're talking magnitudes only here, which is the length of the phasers that you see in the pictures, which are the magnitudes of the voltages involved. That is to say then that the phase two phase voltage magnitude is equal to root three times the phase two neutral voltage, or the phase two neutral voltage is equal to the phase two phase voltage all over the root of three Next, let's have a look at the phase angle of the phase two phase voltage with respect to the phase voltage. And in order to do that, I'm going to look at the red to white voltage phase angle in comparison to the red to neutral voltage phase angle. And for that purpose, I'm going to move the red to white phasor over so that the two tails of the phasers red to white and the red to neutral are coinciding.

And you can see that the phase two phase voltage will lead the phase to neutral voltage by 30 degrees or the phase to neutral voltage is 30 degrees behind the face to face voltage Again, or the face to face is 30 degrees I had of the phase two neutral voltage. And what we normally say is that the phase two neutral lags the phase two phase voltage by 30 degrees or the phase two phase voltage leads to phase two neutral voltage by 30 degrees. Let's return to our three phase circuit that we have here. And before proceeding any further, I wanted to talk about the nomenclature that I'm going to use so that we will not get confused as to where we are and what voltages and currents that we're talking about. You will notice that I have placed point designations on the diagram for instance, where the white connected generator is connected to the line.

I've designated those points as a, b, and c. similarity for the load where the load is connected to the two lines, I've designated those points as a prime, B prime and C prime. And where the Delta generator is connected to the line, I've designated those points as a double prime to B double prime and C double prime. So when I'm talking about a voltage, for example, the voltage of the Y connected generator, the phase two neutral voltage of the generator is designated as the subscript a and similarily. For the Delta connected generator, the voltage across the lines or the phase two phase voltage of the Delta connected generator is designated as the subscript a double prime B double prime. And if I'm talking about current flowing, I'm talking about to current flowing, for example, in the first line as I subscript a dash a primed. So returning to our original task, I want to replace the Delta connected generator with a Wye connected configuration.

And we're going to use our phase relationships that we've already developed for phase two phase with relationship to phase two neutral values. And we know that the phase two neutral voltage will be equal to the phase two phase voltage divided by root three, and the phase to neutral is 30 degrees behind the face to face value. So if we replace the Delta connected generator with a lie connected generator, the voltage a double primed to N will now equal one over root three at minus 30 degrees. Next, we want to replace the Delta connected load with its equivalent y configuration. And for passive impedances connected as a delta, there is a transition formula that you see here. If the impedances of the Delta connected load are a, b and c, then we let the why connected impedances b, p, q and r, then passive conversions are such that P is going to be equal to A B all over the sum of the three impedances Q is going to be equal to a times c over the sum of the impedances.

And R is going to be equal to b times c all over the sum of the A, B C impedances. So we know that the impedances, the Delta connected impedances are all equal, and they are equal to minus J. And that makes our work a lot easier because the individual loads in the Delta circuit are all e equivalent. In fact, there are equal so that our formulas for p q and r now are equal to the same thing and that is minus j times minus j, all over the sum of minus j plus minus j plus minus J. If we want to reduce the figure of minus times minus j all over the sum of three minus J's. that's going to equal minus j times minus j, all over three J.

And of course, two of the J's can be canceled out. And that leaves us with the equivalent of minus j over three. So that the p q and r impedances, when we convert the delta to a y configuration, are all equal to the same thing. And they are equal to minus j all over three. So again, returning to our circuit, and this time, we're going to replace the Delta load with its equivalent y configuration. And the individual impedances in that y configuration are going to be equal and they are going to be equal to minus j all over three Having converted all the Delta loads and sources to their equivalent y configurations, we now can move on to the step two of our procedure, which is solve for a phase independent of the other two phases.

But before actually going there, I wanted to have a look at what happens to the currents. In a balanced three phase circuit here you see the phasers for each of the currents in a balanced three phase circuit. They are 120 degrees apart and equal in magnitude. If we were to add the three currents together, they would come to zero. So if we did this graphically, you can see that the red face added to the blue face added to the white phase returns us to zero. So, we can rewrite that equation actually and say that the red phase is equal to minus white minus blue.

So, if we looked at the minus white phase, and added the minus blue phase, it would result in the read phase. So in other words, we could say that all of the red phase currents returned by the white and the blue phases or each of the phases all of the phase currents returns by the other two phases. And if those add to zero, then there is no current flowing in the neutral. So if we were to return to our our circuit analysis here, we would See that if we removed the other two phases and remained with just the a phase, we would have no return path for the a phase current. So in analyzing a per phase circuit, we would have to temporarily replace the other two phases with a return path for the read phase or the a phase current that's flowing in the single phase diagram.

We're going to find the current that is flowing in both lines. And in doing so, we have to start off by assuming that the current is flowing in some direction at an instant in time, and it doesn't matter what that assumption he has because if we have picked the Wrong direction, the mathematics will look after itself, as long as we adhere to the laws of physics. So I'm going to assume that the current flowing in the line on the left is going to be flowing from left to right. And I'm going to designate that current is I subscript a dash a primed and it's going to create a voltage drop across the impedance of that line as indicated in the diagram. Next, I'm going to assume the current that is flowing through the load from the neutral to point a primed and I'm going to call that I subscript in dash a primed and it's going to produce a voltage drop across the load as indicated in the diagram.

Lastly, I'm going to assume the current flowing in the right hand line as flowing from right to left and I'm going to designate that current is ay ay ay double prime dash a primed and it's going to produce a voltage drop across the line impedance as indicated in the diagram. Having assumed those directions of current I'm going to apply purchase current law at the node A prime or at the point A prime and purchase current loss as the sum of currents flowing into a node have to add to zero. So the currents i a dash a primed plus i N dash a prime plus ay ay double prime dash a prime has to equal zero. Next I'm going to apply ohms law which is which says the current is equal to the voltage over the impedance or the voltage drop across the ammeter gives us the current if we divide the voltage drop by the impedance.

So the first term or the first current in that equation can be replaced by the voltage drop across the line impedance over the line impedance. And the Nord find out I'm going to say the voltage at point A prime is going to be given by the A prime and so the first term is going to be simply the difference in voltage, that is v n minus V A primed and all over j 0.1, which is the impedance of the line. The next voltage drop is going to be described by the voltage drop across the load in our single phase diagram, which is going from neutral which is zero volts and I'm going Subtract the A prime to n. And I'm going to put that all over the impedance of the load, which is minus j, all over three. And lastly, we're going to do the same thing with the other line, and the voltage drop across that line impedance is given by the a double prime and minus V A prime and, and that's all over the impedance of the line j is 0.1.

And if we add up all those terms, kerchief current law says it has to equal zero. Now, this equation, as it turns out, will be a bit cumbersome with that value of zero in the center term. So I'm going to multiply both sides of the equation by minus one and if I multiply it by minus one, it's the same thing as reversing the numerator terms. So the first term is going to be reversed, we get rid of the zero in the center term. And the last term, we reverse the numerator value, we now have this equation, which is also equal to zero, and we can replace the original equation with these values. I'm going to begin the process of simplification now of this equation.

And ultimately, we want to find out what the voltages are in this equation. Well, we already know what the A n is, it's the voltage of the generator on the left hand side, and that's one at zero degrees. Similarly, we know what the a double prime and e is, which is the voltage of the generator on the right hand side, one over root three at my Is 30 degrees. And the other thing I'm going to do is I'm going to extract the fraction out of the term. And I'm going to pull it over to the right hand side, which now will mean that the first term will be v A prime n minus one at zero degrees times the fraction one over j 0.1. The center term, I'm going to do the same thing, I'm going to pull a fraction over to the right hand side.

So that's the A prime n times the fraction, three, all over a minus J. And the last term, again, I'm going to do a similar thing. I'm going to pull the fraction out which happens to have the same fraction as the first term one over j 0.1. And that's going to be times V A primed in Minus one all over root three at 30 degree or volts. And that whole equation, of course, is equal to zero. Now, in looking at the fractions contained by the first term in the last term, which is one over j 0.1, that is equal to j 10 all over j times j, which is equal to j 10 all over minus one because j times j is minus one, and that's equal to the value of minus 10 j.

So, in the first and the last term of my equation, which I have one over j 0.1, I'm going to replace that with minus 10 J. Now, the center term still has a fraction in it and that is three all over minus two I would like to reduce that so that it's going to be easier to work with. And that is equal to j three all over minus j times j, which is equal to j three all over one, which is equal to three J. So I can replace three over minus j with three J. Now, the equation is much easier to work with. The equation is much easier to work with and simpler because we now have one equation with one unknown.

And if we have one equation with one unknown, we should be able to solve for that unknown quantity. And I'm going to do that right now. The first step is to move the unknown quantity over to the left hand side of the equation and all of the other values which are known values to the right hand side of the equation. I am then going to cross them Multiply, which will leave the unknown quantity all by itself on the left hand side. And all of the quantities on the right hand side of the equation are no one values. So you should be able to reduce that to one simple value.

And I'm going to do that. And you'll have to trust me that the final solution is V at a prime and is equal to 0.9 at minus 10.9 degree volts. So we now know what the voltage of the A primed and is, we also know what the voltage of the A n is, and we know what the voltage of the a double prime is. So we know all of the voltages of our circuit. So we should be able to and we will be able to solve for the currents. We now know what the voltages are.

In our per phase diagram, and having solved for all of the voltages, we then can now solve for all of the currents, because each one of the currents that we identified at the beginning of the analysis is given by the voltage drops across the impedance all over the impedance. And we know the voltage drops, cuz they're identified in the diagram. And if we reduce that fraction, we come out with the line voltage I subscript a dash a prime is equal to 0.21 at 214.3 degree amps. We didn't have to solve for the current flowing through the load. However, we can do it because we know what the voltage drop across the impedance is here and that is given by that fraction which reduces to 2.7 at 79.1 degree amps and the current flowing in the other line is given by this equation which when reduced gives us 5.95 at 0.35 degree amps.

Once we have found all the per phase voltages in phase currents, we then can calculate the apparent power flow in the circuit and that is given by s is equal to three times the per phase voltage times the phase current. So, looking at our system here, the first thing we want to do is find the apparent power delivered by the why connected generator which is given by three times the voltage a times the current that's flowing a dash a prime. And we already have solved for those values. So that the final analysis is just taking the known values that we have and plugging them into the formula, and we find that the apparent power delivered by the Y generator is 5.1 at j 3.5, the A, then we're going to find the apparent power that's delivered by the Delta generator, and that's given by three times the a double prime to n times the phase current.

And we know what the phase current is, because we are we have already solved for it and the formula then would reduce to three times one over root three at minus 30 times 5.95 and 0.35 degrees which gives us 15.8 at minus j 2.94 v A. We now proceed to Step four of our analysis, we've already discovered what the values for phase A is. And all we have to do is look at the phase relationship of BNC with respect to A, and we will be able to calculate the values for the B in the C phase. So, we know that the a phase a phase to neutral voltage is point nine at minus 10.9 degree volts. And if we plot that on a phasor diagram, if you would, it would look like this and the rotation is counterclockwise. So, in coming up with the B phase, phase to neutral voltage, the magnitude is the same which is point nine but it is 204 degrees I had of the a phase or it's 120 degrees behind whichever way you want to look at it.

But graphically, it looks like the green arrow there in our diagram. And finding out what the C phase is, again, its magnitude is the same, and it's at 109.9 degree volts, because it is displaced by 120 degrees from the a phase, we can also find the phase two phase value. In other words, V at A prime B prime, which is equal to 1.56 in magnitude and its angle is 19.1 degree volts. And where do these terms come from? Well, we know that the phase two phase value is equal to root three times the phase two neutral value. So root three times point nine works out to 1.56.

And we know That the phase two phase value in phase relationship leads the phase two neutral value by 30 degrees. So we just add 30 degrees onto minus 10.9. And we come up with 19.1 degree volts. The final step in our procedure is and again, if it's necessary to go back to the original circuit to determine the line to line values, or the internal Delta values of, say our load or while we know what our voltages in our delta voltage generator. So basically, that's pretty easy to do once the per phase values are calculated than any of the above voltages and currents can be determined by the phase relationship. So we now have a method for doing perf analysis, which essentially takes this circuit and reduces it to something a little bit simpler.

However, we still have the problem of Transformers in our circuit. And there is there is a tool for working with transformers, and that is called per unit analysis, and we're going to move on that into the next chapter. This ends the chapter on per phase analysis.