Chapter Five be per unit and system circuit analysis. In the normalization process, step four is where you calculate your per unit values. And then Step five is a method of calculating the actual values from the per unit value. What isn't explained or it's implied. When you go from step four, before you go from step four to Step five, you're going to want to analyze your system and find various things that you're looking for such as current and voltages in different locations, using per unit values, because the first thing you did is you went to a per phase analysis because it was a balanced system, which reduced you to a single phase quantities and then go To per unit values, you've got rid of all the transformers. So now you have a very simple circuit to analyze.

That's when you do your analysis. Before you start to convert out of it again, you come up with all these new per unit values that you're looking for or the new values, and then they're all in per unit values, then you would calculate your actual values from these newly calculated per unit values. So I'm going to talk about step four a and the first step is to draw impedance diagram and solve for the per unit quantities that you're looking for. And in a balanced three phase circuit, you can solve in a per unit on a per phase basis. You're going to convert all your delta loads and sources to their equivalent y connections. And then you can Solve for your individual items current and voltages that you might be looking for by solving in one phase only using per unit values.

Let's look at a quick example of what I would mean. If you had this three phase system using two transformers and a generator and some load. The first thing you're going to do is convert that to a per phase equivalent circuit. And you can do that because it's a balanced system. So your per phase representation for this circuit would look like this much simpler. However, it still has two Transformers in it which you will I'd like to get rid of because it'll make the analysis simpler.

So you convert all of your impedances and voltages and currents and everything in that circuit that you have as a per phase representation to per unit values, and your Transformers will disappear. And your circuit will now look like this, which is the per unit representation. much simpler to analyze, you've got one circuit with basically four impedances and one generator that you can solve for. We are going to analyze this circuit by first converting it to a per unit representation. But before starting the process, let's make sure we understand exactly what we have in our three phase circuit diagram. We have a three phase generator connected to a boss and the generator Is 300 MVA and the bus is 13.8 kV and the internal impedance which is more of a reactance for the generator is J one ohm and that's high for a generator but we'll just leave that for now and and assume that is correct.

The bus is connected to a step up transformer which is 13.8 to 138 kV and the transformer is connected Why Why? But to keep our circuit simple we're going to assume the internal impedance of this transformer is zero. In other words, it doesn't have any internal impedance. The transformer is connected to a line and that line is 138 kV line of course and each per phase value of its impedance is J 24. ohms. Again reactance. The line is then connected to step down transformer hundred 38 to 27.6 kV again why why connected and again the transformer has zero internal impedance transformer sends out the power on a distribution circuit that has a per phase impedance or reactance of J, one Ohm feeding a load that's pure resistance, which is 10 ohms.

This is what our per phase diagram will look like for our three phase circuit. And in proceeding to the per unit equivalent circuit, I have to convert all my voltages and currents and impedances and everything in that circuit in that per phase circuit to a per unit value in order to do that. And in order to convert these units to per unit values, I'm going to have to establish my base values. That is taking the first step in the normalization process, I'm going to arbitrarily choose a VA base for this circuit, and I'm going to choose the VA or the MVA of the generator, and that is 300 MVA, and that's a three phase, MVA base, and I'm going to call it s bass. The voltages, I'm going to pick are going to be based on the transformer ratios, starting with the line voltages. I'm going to call that my v base one and I'm going to set that at 138 kV.

Now once I've picked 138 kV, the rest of the voltages will be determined by the ratios of the transformer. And because I chose 138 it's going to make my life a lot easier because I can just follow the ratios with a transformer give me a V base two of 13.8 kV and the base three of 27 six kV I have now selected all of my arbitrary choices. And I have now move on to my calculated values. And I'm going to calculate the impedance base for these circuits. And I'm going to use the formula kV base squared all over the MVA base, because I've got values in kV and MVA. So Zed base one for the impedance base of my line and my green zone, if you would, it's going to be using 138 kV and 300 MVA.

And that means I'm going to have to take 138 kV and square it and divide it by 300 MVA and that will give me 64 ohms. My next impedance value will be Zed base two That's going to be 13.8 k v squared over 300. MVA give me 2.64 ohms. And the last impedance base that I have to calculate is for the 27 six level, so that will be 27, six squared all over 300. And that is equal to 2.56 ohms. I will now proceed to calculate my current basis for this circuit, and I'm going to use the formula MVA base times 1000 all over root three times the KV base primarily because my MVA is given in MVA and my voltage is given in kV terms, so I can directly plug those into my formula.

And in calculating my I base one, it's 300 times 1000, all over root three times 138 given me 1250 amps as a base value for the green zone, calculating the base value for current in the red zone is 300 times 1000 for 300,000 all over root three times 13.8 give me 12,500 amps. And lastly, my I base three will be 300,000 all over root three times 27 six, which gives us 6250 amps. The per unit representation for this circuit would look like this, which is pretty much identical to the per phase circuit minus the transformers. And since the Transformers have no internal impedance they don't add any other impedances to the system here if they did, they would they would be added in series with The generator impedance and the distribution impedance that you see there, but we have zero impedance for the Transformers so we don't have to worry about them.

The line per unit impedance is given by taking the actual value divided by the base value which is J 24 divided by 64 which is J point 375 per unit the generator per unit impedance is given by the same calculation and it'll come out to J 1.56 per unit and the distribution align impedance is J one Ohm divided by the the base impedance and add that voltage which is 2.56 ohms, which is equal to j point three nine per unit and the load is 10 ohms. divided by 2.56, which is 3.91 per unit. And I've maintained the color code for the zones just to show you that we have to change the base values depending on where they land in the circuit. The generator is going to be given by 13.8 at zero degrees divided by 13.8 kV, which is equal to one at zero degrees per unit.

Our task will be to solve for current voltages and the apparent impedance delay that's delivered to the to the circuit here. And that's what we call step four a and I've already gone through the explanation for that. First, let's calculate what the current will be in per unit values. And that's simply ohms law which is going to be The generator voltage divided by the total impedance of the circuit. And we know that the generator impedance is one at zero degrees per unit and we know the total load is just the sum of the per unit values. And that will work out to be 3.91 plus j 2.3 to seven.

And that would give us one and zero degrees all over the total impedance, which would result in point two two at minus 30.8 degrees per unit. Next, we're going to find out what the load voltage is. And that's given by the subscript L. And again, we're going to use ohms law and that tells us the voltage drop Cross the load is given by the current times the load. And we know what those two values are. And that would be the current being 22 at minus 30.8 degrees times the per unit load for the system which is straight 3.91. And that would result in point eight six at minus 30.8 degrees.

The generator, apparent power that's delivered but to the circuit, it's just the voltage times the current and that is equal 2.22 at minus 30.8 degrees. We have successfully solved for all the required values or the tasks that we needed. In other words, we found the current and the voltage and the apparent power that was delivered to the system. And we did it using all per unit values. And all our answers are in per unit. Now, you may have noticed something familiar by about this circuit, because you have seen it before, back when we first started looking at normalization, and actually we analyze this circuit as a single phase circuit before we contemplated the three phase circuit that we just analyzed here.

And I'm going to walk back and have a look at the comparison on this. Because the single phase analysis is just the analysis of this three phase circuit if we had use single phase or per phase values, which you could have done, but you would have ended up with single phase or lying to neutral values. I want to take a step backwards for a moment. And we'll look at the per phase circuit for this three phase circuit that we just analyzed. And we chose to use a three phase value for the MVA base which was 300. And we chose to use line to line values for the analysis of this circuit.

Now, we could have used single phase values and we could have used line to neutral values for this circuit. And this is what we would have come up with, we would have used 100 MVA and we would have used the line to neutral voltages for this circuit. Now as I said, this may have looked familiar to you because This is the same circuit that we use when we first started looking at normalization. And we were looking at normalization from a single phase perspective. And at that time, we chose 100 MVA for the VA base for this circuit. And we chose the line to neutral voltages at eight and 16, which I knew were related to the three phase values.

So we analyze both circuits using both these type of voltages and MVA. And if you can go back and look at your circuits or if you have a really good memory, one of the things that you will see is that the per unit values for each of these circuits didn't matter whether we use the line to line voltages, or three phase voltages, or line to neutral or single phase values, the per unit values were identical So our per unit circuit looked like this. And it didn't matter which voltages or which MVA that you used. Where the difference appears, is when we convert back to our original values, and this is very important. If we were going to convert back using our convert back to actual values, starting with the current we would take the per unit current and multiply by its base value for that circuit.

And in the lines, the base value for current was 1250 amps given us 275 at minus 30.8 degree amps. Now what you will notice if you look at both circuits now this one and the one we did when we were doing normalization, the line current looks the same and that should not be a surprise to you, because when you were doing just per phase analysis before we even got into per unit analysis, the per unit line current is the same whether you're looking at a three phase quantity or a single phase quantity. So, incoming out from per unit to single phase are two actual values your delay the currents are going to be the same. So the line current is definitely the same as the single face current. The generator current is the same as the single phase current as well as the current flowing in the distribution system to the load is the same as a single phase current.

The voltage is however, the load voltage, we are going to multiply the per unit value by the line the line base value so we will get a line to line voltage value for the actual value, and if we wanted the line to neutral value, we would just have to divide by root three. So didn't matter which way we did our calculations, you just have to be aware of what you're going to get once you convert from a per unit value back to your actual value. And it's the same thing with the generator. apparent power that's delivered to the system. If we use the three phase base value, we will come up with a total three phase VA that's delivered to the system. However, if we wanted to find a single phase value, we just have to divide by three.

The point here is you're going to if you're working in a three phase system and you're working with line to line values, then it it's better to work with line to line values in three phase values to start with and then you can use them directly without having to make further conversions once you get the actual values. So the bottom line here is that this is a three phase example. And this is the single phase example, they are pretty much the same answer, you just have to remember what that answer means. And if you're in a single phase example, you're going to have to multiply by a factor of three or root three to come up with the answers that you may want. And the reverse. If you're in a three phase example, you're either going to, you're going to have to divide by a factor of three or root three or multiply by root three.

Bottom line is, it makes no difference how you do your analysis as long as you know which values base value here using what you're using. And those are the facts. The the measurements that you're going to use coming out of it i'd like to return to step four a again, because it is at this point that the system should be sufficiently simplified in order to carry out systems circuit analysis. But I want to look at one more common element that takes place in in various three phase circuits in the representation. interconnections among system components in a power system may be shown in a so called one line or single line diagram. The single line diagrams represent all three phases of a balanced system, but they do it with a single line rather than using three lines.

And this is a typical single line schematic For a circuit and we've looked at this circuit before, but once the system components are replaced by their per unit values, the single line diagram of a particular power system or network can be replaced by its equivalent per phase diagram using per unit values. The analysis then is greatly simplifies as you can see, there is only one phase to worry about and no transformer ratios to worry about. Replacing the single line diagram by its equivalent per phase diagram may seem obvious and it is but there are some step by step rules that you should follow. And these become second nature to you once you start dealing with a system but it helps to go back to it especially if you get into a very complex system. There are only three steps involved In a single line diagram conversion to a per unit diagram and these becomes second nature to you as you start to deal with them.

Firstly, components are represented by their single phase per unit impedances you cannot have a per unit diagram unless all the components in the system are converted to per unit values. Once you do that the turns ratio of the Transformers disappear and you don't have to worry about them anymore. So, you have to convert all your values to per unit values. The two ended connections of a transformer or lines and breakers. That's how they're represented in a single line diagram. These two ended connections look the same in a per unit diagram and open ended components that are showing such as generators and and loads etc and motors.

They have to have a return path. If you're going to have a per unit diagram, that is a complete circuit. So you replace or you put in the neutral connection so that it'll look like a, a single line, or sorry if it'll look like a single phase circuit that you can analyze using per unit values. And I'm going to demonstrate that now. This is what the starting point is and that is your single line schematic diagram. And it gets converted to a per unit diagram and look like this.

Following the general or three rules that we just looked at. components are represented by their single phase per unit impedances. In other words, all impedances and all components are converted From their actual values to per unit values, to ended components, such as Transformers in lines and breakers, they're showing the same way in the in the per unit diagram as they are in the single line diagram. generators, loads, motors are usually open ended in a single line diagram, we just provide a return path for them. That makes it into a circuit rather than just a single line diagram. And we provide that return path so that we can do our analysis and it looks more like a circuit in this system circuit analysis, and we have to note this, and we've discussed this before this analysis is for a balanced system only training Former voltage ratios are given as line to line voltages regardless of the connection configuration.

There is no neutral or ground current flowing because it's a balanced system. So grounds are emitted from the diagram in our analysis. You are going to convert all your delta loads and sources to their equivalent wide connections prior to doing your calculation. You're going to solve for a single circuit. And quite likely it'll be a your circuit one or whatever terminology you're going to use. And then the other two phases can be determined by our phase relationship equations.

And the fact that these phasers are 120 degrees apart. We then can proceed to Step five and convert everything back to their actual quantities depending on which base values you are using. This ends chapter five B