Chapter Six problems and solutions using symmetrical components. In this problem, we are given three asymmetrical currents as you can see plotted there. The eighth phase we've have at 10 a quantity of 10. at zero degrees, the B phase is 10 at 230 degrees and the C phase is 10 at 130 degrees, it's important to make sure you note the rotation is ABC. So the first step is we select our formulas for calculating the symmetrical components. So we're going to calculate the a phase symmetrical components and then the B and the C phases are just Hundred and 20 degrees removed from that, the zero sequence current is given by the formula one third. The a phase plus the B play phase plus the C phase.

The, a phase positive sequence component is given by one third, the ay ay ay ay phase plus the A operator times the IB phase, plus the A operator squared times the C phase. And the negative sequence component is given by one third. The a phase plus the A operator squared b phase plus the A operator, C phase. These are standard formulas that we've already developed so we just we know what the ay ay ay ay. and the is In the ICR, because we were just given them and we know what the operator does, the operator shifts vectors or phasers through 120 degrees. So starting with the zero sequence component, we know that we zero sequence component is one third 10 at zero degrees, plus 10 at 230 degrees plus 10 at 130 degrees.

And we can just for the purposes of adding those components together, we've converted the, the vectors to their rectangular coordinates rather than polar coordinates and we get one third plus minus 6.428 minus j 766 minus 6.428 plus j 7.2. Six, six, and it's very evident that the J component cancels out. So we're really only going to be left with something that lies along the zero axis either plus or minus. So we add up the, the 10 and the minus 6.4 to eight, and then minus 6.4 to eight, and we come out with one third minus 2.856, which also equals to minus 0.952. And if we want to convert that back to polar coordinates, we have point nine five to 180 degrees. So in calculating the positive phase sequence for the a phase, it's just those vectors but in this case, the IB the current and the B phase is going to be shifted through 120 degrees.

By the AR operator and the IC is going to be shifted at 240 degrees by the operator. So we're left with those polar coordinates. Again, I'm going to convert them to rectangular coordinates. And you can see, just as in the case, we just finished that the J component cancels out. So we're again left with something that's going to be along the zero axis either plus or minus, and it works out in this case to 9.899 at zero degrees. Carrying on similarly with the negative sequence component, we have IE a two is equal to one third times 10 is zero degrees times 10 at 470 degrees because we have the operator squared times IV and the A operator times I see is equal to 10 at 250 degrees, we then convert the last two components, we can convert all the components to rectangular coordinates.

And we come up with this long number which we can see very quickly, the J factors cancel out and you're only left with something that's going to be along me the positive or negative axis, the real axis, so you're left with actually 1.053 at zero degrees. before going any further and doing our calculations, there's a quick check that we are on the right track and that is, we know if we add the zero sequence and the positive sequence and the negative sequences together, we should come out with the actual thing. So we can add a zero which is a zero component, zero sequence component of the a phase worked out to minus 0.952. And a positive sequence component for the phase worked out to 9.899. And the negative sequence component, I two worked out to 1.053. If you add those together, they come out to 10.

And lo and behold, at zero degrees, that's what our a phase is. So we must be doing something right because it checks up. Now we had to come up with the, the other phases of our zero sequence components, because we're not quite finished yet. We didn't get the a phase. But we know in the fact of the zero sequence component that B and C phase are indeed equal to the a phase. So we already have the answer that we were looking for as far as the zero sequence compete components are concern.

The zero sequence components for the A, B and C phase are equal and they're 0.9 sorry, 0.952 at 180 degrees. Now the positive phase sequence we already calculated out was zero degrees, it was 9.899. So the B phase has to be 120 degrees lighting from that and equal to it because they are symmetrical components. So the magnitudes have to be equal. We know that the rotation is counterclockwise. So the B phase is 120 degrees lighting the a phase, which puts it at 240 degrees, because A is zero, then B is at 240 degrees.

That leaves us with just the C phase of the positive phase sequence and again, there's symmetrical components. So the magnetic tubes are equal, and it has to be leading the a phase by 120 degrees. The negative sequence component we've already calculated the a phase not 1.053. at zero degrees, the B phase is 1.0853. Because the magnitudes are the same, again the rotation is in that direction, the sequencing is reversed on the negative face sequence. So the B phase leaves the a phase by 120 degrees. And similarly the magnitudes of the sea phase of the negative sequence component is the same as the other two 1.053 and that is going to be 240 degrees because they are all 120 degrees apart.

Now, I've taken the liberty of expanding the magnitude as far as the the diagram is concerned just so that you could see it a little bit better, but it is actually expanded by 10 times what the negative sequence component vectors are with respect to the positive sequence vectors. In this problem, we are given the positive negative and zero sequence a phase voltages, which are 50 at zero degrees for the positive sequence voltage a phase 20 at 90 degrees for the negative sequence a face and 10 at 180 degrees for the negative sorry for the zero sequence a phase voltage, the component superscripts are written as one, two and 04 We're positive, negative and zero sequences, just so we can keep track of things as we go along. The subscript A n indicates that we're dealing with the a phase and its face to neutral voltage that we're using.

I'm going to rewrite the, the positive negative and zero sequence voltages in right now they're in right there in polar coordinates, I'm going to write them in rectangular coordinates just so we'll have both and we can move back and forth to them. 58 zero is just 5020 at 90 s plus j 20. And 10 at 180 is just minus 10. We know that the asymmetrical face to neutral voltage of the a phase is made up of the sum of the positive negative and zero sequence Voltage phasers, and that is 50 plus j 20 minus 10, which is equal to 40 plus j 20. And you can convert that back to polar notation, which is 44.72 at 26.6 degrees and that's in bolts. The Bluetooth neutral asymmetrical voltage is given by the quantity a operator squared times the positive sequence voltage plus the operator times the negative phase sequence.

And that's plus the zero sequence voltage. Making this substitution we have 50 at 240 degrees plus 20 at 210 degrees, minus 10. convert that all over to rectangular coordinates and I'm not going to repeat the big long thing it's right there. If you collect like terms you end up with 52.32 minus j at 53.33 converted to polar notation is 74.7 at minus 134.4 degrees and that is also in volts. The asymmetrical seat a neutral voltage is made up of the A operator times the positive sequence voltage a voltage the plus the A operator squared times the negative a to neutral voltage plus the zero sequence voltage and that's just a straight substitution with moving the angles accordingly to the operators. The operators leaves us with 50 and 120 degrees plus 20 at 300 30 degrees minus 10 convert that over to rectangular coordinates and you get that great big long number and collect like terms you end up with minus 17.68 plus j 33.3, which is equal to 37.7 at 117 degrees and that is in faults.

So analytically, we have determined what the asymmetrical face to neutral voltages were, or are, we now have the task of showing graphically the sum of the given symmetrical components, which determine the line to neutral voltages. So I'm going to colorize the positive negative and zero sequence voltages just so that we can plot them and see them a little bit better than just having them all in the same color. Then I don't Don't have to keep writing the subscripts and superscripts. And and everything else noted here that we're dealing with the a phase of the positive sequence voltages, the a phase of the negative sequence voltages and the a phase of the zero sequence voltages. And they have they were given at the beginning and they are symmetrical components. So we can very easily determine what the B and C phases of the symmetrical components are.

Again, these are the a phase that were given to us at the beginning of the problem. We know that the C phase of the positive sequence phase sequence is 120 degrees removed from the a phase and the B phase is 120 degrees removed the other direction. And that is our positive sequence voltages. They're they're equal in magnitude and 120. degrees apart. And the sequence is because of the rotation is counterclockwise. So you have an ABC sequence rotation counterclockwise, the negative sequence in blue, the B is 120 degrees ahead of the A, because it's negative sequence, the beat the C is 120 degrees lagging because again it is negative sequence and that those three vectors are equal in magnitude and 120 degrees apart.

However, the sequence sequencing is a CB, which is the opposite to what the positive sequence was. And we now go to the zero sequence. And the zero sequence is the easiest of all of them, because they are all equal and they are in phase. So these are positive, negative and zero sequence. vectors are phasers. Let's see how we add them together.

Graphically, we know that the asymmetrical voltage for the positive phase sequence is made up of the sum of the A phases of the positive negative and zero sequences. So we can drop the a phase vector down or phaser. Take the negative phase sequence B, put it down, and we'll move the a phase of the zero sequence over. And we'll see that the asymmetrical voltage is made up of the sum of those three. And it is equal if you measured it and measured the angle is 44.72 at 26.6 degrees, which is what we came up with in our analytical calculations. Now let's take a look at the B phase.

Two B phase is again, made up of the positive negative and zero sequence Some. So we'll move over the B phase of the positive sequence, move over the B phase of the negative sequence components and move the a phase or sorry, the B phase of the zero sequence component. And that gives us our asymmetrical blue to neutral voltage as you see here. And if you measured it, it would come out to exactly what we calculated before 74.7 and minus 134.4 degrees. That leaves us with just the C phase, which again is made up of the some of the symmetrical component phases. We will move over the red or the positive sequence c phase and over the blue negative sequence of the C phase and the green or the zero sequence.

Phase that over, and our sum looks like that, which you have measured works out to 37.7 degrees, so 37.7 and 117 degrees. And there we have it, we've gone through the solution we've calculated analytically. The final result and we've plotted the final result. And they both agree with each other. In this problem, we have a generator. That has terminal a open, and the other two terminals are connected to each other with a short circuit.

From this connection, to also ground. Typical values for the symmetrical components. of current. In phase A are, the positive sequence current, ay ay ay ay ay ay one is equal to six. Hundred amps at minus 90 degrees. The negative phase sequence current for the a phase is equal to 250 amps at 90 degrees and the zero sequence current for the a phase is equal to 350 amps at 90 degrees.

Our challenge will be to find the current in each phase of the generator and the current into the ground. In starting this solution we're going to look at the Firstly, the a phase and the a phase we know is made up of the positive, negative and zero sequences for the a face. And we've been given that at the beginning of the problem and that is equal to 600 and minus 90 or minus j 600. Plus the negative phase sequence for the a Phase II Is 250 amps at plus 90 degrees and plus the zero sequence current, which is 350 amps at 90 degrees or j 350. And if you add up those three figures they add not surprisingly to zero because the a phase is open circuited. So we've solved part of the problem already in in finding out what the phases but really that was given to us at the beginning all we've done is verify it with the with the symmetrical components.

Now we'd like to find out what the current in the B phase is. Now the current in the B phase is going to be the symmetrical or sorry the arithmetic sum of the symmetrical components for the B phase. It will be the positive sequence Phase plus the negative sequence B phase plus the zero sequence B phase. And in order to do that, I worked up a little bit of a, what I would call it a cheat sheet here. And what it is showing is a positive sequence set of phasers, a negative set of phasers, negative sequence set of phasers, and a zero sequence set of phasers. It also lists the quantities for the positive negative and zero sequences, phase currents that were given at the beginning of the problem, because I just covered them up with this bit of a diagram.

But really, these aren't necessarily the phasers related to the problem. All I'm showing is how in the positive sequence the A B and C phases are related to each other so that I can solve this problem and if you look The positive sequence because the phases are 120 degrees apart, and the sequences A B C with a rotation in the counterclockwise direction, the phasor be both magnitude and angle is equal to b a phase magnitude and angle plus 240 degrees of the B phase is equal to the a phase angle, but we have to add 240 degrees to it. Similarly, the C phase of the positive phase sequence because it is bounce is equal in magnitude to the a phase and the angle is equal to the angle of the a phase. But we have to add 120 degrees. So those formulas work out for the positive sequence.

The negative sequence has a reverse sequence to the counterclockwise rotation in that it is a CB so that the B phase will be equal to the a phase plus 120 degrees, the C phase will equally a phase plus 240 degrees. Okay, so that's pretty straightforward. It's a little bit difficult to follow when you when you start to substitute numbers into it. That's why I put the, what I call a cheat sheet up here. Now the zero sequence components are, are a no brainer, they're all equal because they're all in phase and they're all equal in magnitude. So a, b and c are always equal.

And I rewritten as I said, the, the positive negative is zero sequence currents that were given in the problem at the beginning. So going back to our formulas, if we want to find out with wood The IB one is IV, one is equal to 600. And it's minus 90 degrees the a phase is 90 degrees, but we have to shift it by 240 degrees to symmetrical components and we know what the a phase is the a phase is minus j 600 or 600 and minus 90. So, I b one will be 600 at 150 degrees, which is equal to minus 519.6 plus j 300 which is just a conversion to rectangular coordinates so that we can make an addition when we get to the end. So, we found out what the current in the positive or the IB one is in terms of the positive sequence current Now I b two is equal to i b one, but we have to add 120 degrees to it and it is 90 sorry is it yes it is 90 is 250 degrees 250 at 90 degrees, so we have to add 120 to 122 90 degrees gives us 250 at 210 degrees, which equates in rectangular coordinates to minus 216.5 minus j 125.

Now, the zero sequence component is simple, it's identical to what the a phase is it's 350 at 90 degrees, which is equal to zero plus j 350. So now if we want to find out what the B phase Is the B phase current is simply the addition of the real terms of those three equations. In other words, the IB one IP to IP zero real quantities add up to minus 736.1. The j terms will then arithmetically add to 525. And if we convert that back to polar coordinates is 904.1 and 144.5 degree amps, and it's indicated on the diagram there. Carrying on in similar fashion and solving for the positive phase sequence c phase, the positive sequence c phase is equal to the positive sequence a phase Plus 120 degrees, and that is minus 90 plus 120 degrees would equal 30 degrees.

So I see one is 630 degree amps, or 519.6 plus j 300. The negative sequence, C phase is given by taking the negative phase sequence, a phase 250 and the angle is going to be the angle of the a phase plus 240 degrees. So that's 90 plus 240 is equal to 330 degree amps or 216.5 minus j hi Hundred and 25. And the easiest one, of course is the zero sequence, C phase which is identical to the a phase, which is 350 at 90 degrees or zero plus j 350. Now the C phase is going to be the arithmetic sum of the IC one IC to IC zero. So again we add up the real terms and the J terms and we come up with 736.1 plus j 525, which equals 904.1 at an angle of 35.5 degree, tamps and that is what our C face is.

The neutral current is the earth. Just the phasers, some of IB plus IC, which is equal to j, one 1050 amps. Also, you can add up all the zero sequence current, or take three times any one of them and we can take three times see a phase zero sequence current and that will give us three times j 350 which equals the same answer which it should j 1050 amps. So we solved the questions or the requirements of the problem. This ends chapter six