Chapter two per unit analysis and the normalization process. per unit system analysis is a tool used in the simplification process of power system analysis. This involves expressing system quantities as fractions of a defined base unit quantity. More specifically the per unit system is a method of expressing quantities in an electrical system such as voltage current impedance MVA etc. as proportions of a predefined base quantity. Once these quantities are expressed as a ratio of their equivalent base values, this is called a per unit value.

Using per unit value simplifies the process in many ways, as you will see, and this is called normalization The per unit system has some advantages and the main ones are that it effectively removes the turns ratio of a transformer in circuit analysis. It is also used by manufacturers in specifying the apparatus in parameters in per unit, so you must know how to interpret these manufacturers quoted parameters. And lastly per unit values usually lie within a very narrow range so, you can see if anything is going wrong with your calculations almost immediately, whereas actual values can vary over a wide range. The main advantage though, and we're going to see how this works is the fact that it essentially removes the turns ratio of Transformers from circuit analysis. per unit analysis involves the Conversion of all system units to per unit value. So, we then perform a calculations which are much simpler in you in using per unit values.

And then in the end we are able to calculate or convert the per unit values to actual values. And this is called the normalization process. Now, the first step in the normalization process is to determine what our base values are for the system and we start by selecting a VA base as our step one. This is an arbitrary selection you can pick any VA that you want or s base if you want and using that apparent power base value, you can then start to calculate the per unit values for MVA. However, as I said, VA selection is purely a arbitrary now, as you will see, as we go through some examples, it's easier to select certain values that will make our calculations easier as we go along. But you don't have to select any particular one, you can pick 100 va, or 1000 va or 10 k VA or 10 m VA as your base value.

Once you've picked that, that is the base value you will use for all your system calculations throughout the entire system. Now, the other the other thing to keep in mind is that base values are magnitude only. That is, for example, a parent power is a phasor or a vector quantity, which has magnitude and direction, which in polar coordinate formation has a quantity at some angle, the base value however, involves only the quantity. In other words, it is a scalar quantity and it doesn't have direction. The next step that you're going to do is you're going to determine or pick a voltage base. Now, picking the voltage basis, like VA is purely arbitrary, you can pick one kV, or 10 volts or 100 volts.

It's entirely up to you. But there are some voltages in our system that will work out easier for calculations than others and you'll see that as we go along. However, again voltage bases are purely arbitrary and like the VA base, voltages are quantity plus angle. In other words, they are a vector quantity. However, the base quantity is just magnitude only, and that I is a scalar quantity and its its dimensions are in volts. Now the other thing to keep in mind that there is a voltage base for every voltage level in the system.

I'm going to repeat that because this is an important item and we'll spend some time developing it. There is a voltage base for every voltage level in the system. Now I've have a simple system drawn below here, and you can see there's two Transformers in that system. One Two Step the voltage up and the other two step the voltage down so there are three voltage levels in the system. Now, the other thing to keep in mind is transformer voltage level ratings are given in line to line and this is called transformer ratios. Now this doesn't mean Come to significant using single phase quantities which we're dealing with right now.

But it becomes very significant when you're into a three phase system. But we'll leave it for now and I'll return to it again once we start to just to discuss three phase systems. Now, the system diagram below has three voltage levels, it has the generation voltage level at eight kV, it has the transmission voltage level, which is listed as 80 kV and it has a load or distribution voltage level which is 16 kV. Now, we can pick as I said any voltage level that you want for the system. Now, it I could have picked 100 volts as my bass voltage and that is quite okay but you will see in The next thing that I have to say, it becomes very significant that you want to choose something that's easy to work with. For example, if we select the voltage rating of line one to be our base voltage, then v base one is 80 kV.

Now, as I said, there is a voltage base for each level of the system and is given and determined by the ratio of the transformers. So, if we were to look at the voltage base of the generator and its voltage basis, eight kV, and if we were to look at the distribution side where the voltage is stepped down, there's a voltage base for that voltage level. And because I've picked 80 kV for the line, voltage base, then I automatically select the voltage base for the two lower voltages because They are given by the ratios of the transformer. Let's take a closer look at our system here in regard to the voltage levels that we've been talking about our voltage base one, we chose to be a KV. So it's actually described as a or it can be described as a zone that involves everything that's at that 80 kV level.

So as we proceed through our per unit analysis, and we start to determine what the per unit values are, and including the step of choosing our base values, any calculations involving v base one which is a DK V, which is anything inside that green zone that's described with that box you see there has to use v base One or 80 kV as the voltage base. Similarly, there is a zone about the generation in which we will call it zone two or V base two and any calculations including per unit or base value calculations that involve the voltage base the base to we must use a KV for any calculations inside that red zone. And similarly, with anything involved with the distribution side of the voltage at 16 kV, anything inside that orange box calculations of say the impedances at per unit impedances or the basis that are using v base three, four our calculations, anything falling inside that zone must be Use 16 kV as the V base.

Now our VA base that we chose right at the beginning that is used for anything in our system. Once we've chose that VA base, it does not change for any calculations for the whole system involving the VA base that we chose the voltage basis change with the different voltage levels, but the VA base remains constant over the entire system that we're working with. And just for an example, we may have chosen that to be 100 va. Now, if we're working in the the base to zone and we have to use the VA base, it's still 100 if we're working in the V base three or 16 k v zone and we involve something uses the VA base, we are going to use 100. Va It is used for the entire system. Returning to the system here, I want to demonstrate something that I had said earlier.

And that is you can choose any voltage base that you want. It's entirely arbitrary. We chose 80 kV for our V base where the line is involved. And that worked out really nice because then we knew it automatically without doing any calculations, what the V bases were for the generation side and the distribution side. However, I could have chose another voltage base for the base one. Let's say I decided to choose 240 kV instead of 80 kV.

Well that's legitimate and you can use it, but you can see right away that is going to start to involve couple of calculations or extra steps in your calculation. Because one I want to find out what the voltage base for the generation side, which was automatically eight kV, if I had chosen 80 kV, I have to now use the 10 to one ratio for converting the 240 kV to the lower voltage V base. So, the V base two isn't just a simple matter of writing in a key V, I have to do a calculation dividing the 240 kV by the 10 to one ratio, and I come up with 24 kV and that's not a big deal but it is an extra step that you have to go through and it becomes more complex if you're using numbers other than 10 to one such as if we go over to the distribution side.

The voltage base three zone there is going to be given by again the voltage ratio at 216 which converts for a 240 kV 248 kV, which is still legitimate and you can use that. However, you can't it does involve an extra step. So it's much easier to start out using a D kV as your voltage base. Now, having said all that, our MVA or our VA base that we chose right at the beginning is still legitimate for the whole system. I never changed that all I changed was my voltage base. So the VA base that I chose right at the beginning of 100, va still applies to the entire system.

Once we have selected the VA base and the voltage basis for our system, then we proceed to find the basis of the other items in the system such as the impedance base and the current bases and possibly the power bases etc, etc. But once we have selected the VA base and the voltage basis, then the Other items, the other base values that is our set and are calculated from the VA and the voltage basis. And that brings us to the next step in our procedure we want to we want to determine what the impedance base and the current basis are for our system. And as I said, we have to use the VA base in the voltage basis in these calculations. And we just use the standard electrical type equations in calculating these bases and the electrical formulas that we would use are the equations are similar to these like if we're going to calculate the impedance space, we would use the voltage base over the current base which stems from the electrical equation, the impedance is equal to V over i.

And if we wanted to calculate the VA or the apparent power in The system we will just multiply the voltage times the current. And the current then is equal to from the second equation, they're equal to the s all over the voltage. In other words the VA all over the voltage. Now, we can use the current formula i is equal to s over V and substitute that in the first formula that we had for calculating the impedance and the impedance could be calculated using the voltage and the VA for the system, which is V all over the fraction s divided by V, which works out to v squared all over s. Now, in doing these calculations, you have to realize that we are using volts and volt amps. So if we are dealing with or we have been given kV Data voltage, we have to multiply the KV times 1000 and then divide that by the current.

Now that might seem pretty mundane and obvious, but it has to be stated because you can get mixed up with the all the zeros that are involved in the various items that we're going to look at that right now. If we are going to calculate the current from say the KV, a and the KV, then we multiply those items by 1000. But if you put them in the equation that the 1000 cancels out, so we can just take the caveat and the KV directly into our equation, and we don't have to multiply by 1000. Of course, this is a little bit more complex. And as you move into some of these more complex formulas, you have to be careful and in determining the impedance. We said that we could take the voltage squared all over the apparent power.

But if we were given a kV and M VA for our system, we have to multiply the KV values times 1000. And we'd have to multiply the M VA by a million in order to come up with just the VA and the voltage. Now, if you put those in the equation, you can see that the thousands and the millions cancel out. So then you could actually use the KV and the MVA directly into the formula, because the magnitudes of zeros kind of look after themselves, but you have to remember these formulas, the impedance is calculated by the KV squared all over the MVA. So, now, we can go back to what we had said before at the beginning of the slide, that the impedance space is going to be used or calculated sorry, by using the voltage basis and the S basis for Our system and as impedance base is given by the voltage base squared, all over the S base.

It's also equal to the KV base squared all over the MVA. The current base is given by the space divided by the V base. And it can also be given by the KV a base divided by the K v base. Because the thousands multiplier looks after this looks after themselves, and they get cancelled out of the equation, of course, and that is also equal to the MVA base times 1000 all over the KV base. These are the formulas or equations that you must use to determine the impedance base in the current basis. If you don't memorize these formulas, at least, keep handy if you're working in the per unit normalization process, because these are the equations used in calculating the impedance base and the current base for a single phase system.

Now, remember that if you look at these formulas, each one of these formulas involves the use of the voltage base. So there is a an impedance base and occurring base for every voltage zone in the system. So if we were working in the zone v base one or where the line is or in the 80 kV v base area, then the voltage basis that we must use our 80 kV and the formulas then would look like this. Zed base is 80 squared all over the MVA base And the current base is equal to the KV a base all over ad. Similarly, if we're working in the eight kV zone or the bases eight kV, we have to use eight kV in the position of the formula that uses the V base. And lastly, if we're working in the base zone of 16 kV, we have to use 16 kV for our voltage base in our formulas.

We now have all that is required in order to calculate the per unit values for our system. And in calculating the per unit values per unit values, or quantities are calculated by dividing the actual value by its equivalent base value. So in other words, if I had a voltage somewhere in the system I wanted to find that voltage, whether it's given in kV or just volts, I want a per unit value I divide that voltage by the voltage base and the resultant is a per unit value. So, what that means in the way of an equation is that the quantity in per unit is equal to the actual value of the quantity divided by the base value of the quantity. For example, if I had a VA or an apparent power listed somewhere in our system, and I wanted to calculate the per unit value, I would just take the actual value divided by the base value now, the actual value of VA is a phaser.

So it will be it will have a quantity or a magnitude and it will have a double reaction or an angle, if it's in polar coordinates, and the tag on on the end or the descriptor is VA, so it would be 100 va or 250 va, but it is a phaser. And you divide that by the base value, and the base value is just a magnitude or a quantity, also listed as VA for a parent power, but it's an apparent power base. So the the the equation has a phaser in the numerator and a scalar. In the denominator. The two VA quantities of course cancel out and you're left with something that is most likely less than one. So it per unit values are usually zero point something in this case I've just said xx if x s they Have a direction or an angle and that direction or angle is the same angle as the actual quantity.

However, the magnitude is the is the quotient of the actual quantity divided by the base quantity. And that is in per unit values. So if I had 100 va, and I wanted to find out what the per unit value is i divided by the base value I'd get something that was a decimal fraction, and that would have the same direction as the actual quantity and that is per unit. The other thing to keep in mind is that the base value depends on the location in the system. Components connected to the eight kV bus, such as the generator must use VB To etc, which in our cases, eight kV components connected to the 16 kV bus such as the motor must use v base three and components connected to the 80 kV bus, such as the line must use the base one. The final step of the normalization process is that we want to convert from the per unit values back to actual values.

And for each component, its actual value may be found by multiplying its per unit quantity by the base value for that quantity at the at the connection location because remember, we have to decide which voltage zone that it's in to use that in the calculation process. So, the actual cap calculation of the actual quantity is just the reverse process. So, the actual value of the quantity is given by the quantity in per unit times the base value of that quantity. So, that completes the normalization process, it is a five step process. The first three steps are determining the base values that we need to use to calculate the per unit quantities. The Step four is a per unit calculation.

In other words, calculating the per unit values which is given by the actual value divided by the base value, then the final step in the normalization process is converting back to actual values and the actual values are given by the per unit values times the base values. Now that we have developed the normalization process, SAS and procedures, we want to take a closer look at Transformers and what happens to Transformers when we use per unit values. In order to proceed with a discussion about a transformer in this case, we're going to look at an ideal transformer. And we want to review a couple of identities that are characteristic of an ideal transformer. And that is the turns ratio is given by the ratio of the number of turns in the high side over the number of turns in the low side, or the number of turns in the primary over the number of turns in the secondary.

It's also given by a ratio of the primary voltage over the secondary voltage or in this case, the high side voltage over the low side voltage. The current if a load is connected to the secondary Have a transformer or the low side of the transformer is going to be given by the inverse of the turns ratio. And that is indicated with a equation that you see in front of you. And if a an impedance is connected to the secondary of the transformer or the low side of the transformer and that secondary impedance is given by Zed subscript two, then the impedance will be reflected to the primary side and it will be reflected by a factor of the square of the turns ratio. Let us consider, for the moment a single phase transformer and we're going to put a low side impedance on that transformer we're going to call it Zed L and I'm going to start doing some Per Unit analysis of this transformer circuit.

And in doing so I'm going to set a VA base and I'm going to call that s base. And I'm going to choose my voltage base to be the low side bass of the transformer voltage. And I'm going to call that the L base. Now we know that the ratio of the transformer is given by the high side voltage to the low side voltage. So once I've set the low side base voltage, I've automatically set the high side voltage base and I'm going to call that V h base. Now I'm going to redraw this transformer but I'm going to move the impedance from the low side to the high side everything else is remaining the same.

So the voltage bases are still going to remain the same. The only thing is I've moved zet L over to the high side, I'm going to call it Zed H, because they reflect an impedance is going to be different than a low voltage impedance, and it is going to be given by the voltage ratio of the transformer or the transformer ratio times Zed L. Now I'm going to look for the low voltage impedance in terms of the high voltage impedance. So it's just basically almost the same using the same formula actually, and just cross multiplying and putting Zed l on the left side and Zed h on the right side, I'll have to invert the fraction, of course, however, all I've done is rearrange the terms so that I'm calculating the low voltage impedance in terms of the high voltage impedance. Now I'm going to get Calculate the impedance space on the high side of the transformer.

And that's given by the high side voltage base squared all over s base. And I also want to calculate the low voltage impedance base, and that's going to be given by the low voltage base squared all over s base. So these formulas are these equations are established just using the identities of an ideal transformer or calculating voltage basis which we've already established in the first part of this chapter. I now want to calculate what the low voltage impedance is, in terms of per unit quantities. So I'm going to call Zed l per unit and that's going to be equal to the actual value of Over the base value, or Zed L over Zed l base. I'm going to rewrite that equation I'm going to replace v Zed lbs with a one of my pre established equations, which is the lb squared all over SPS, and I'm going to rearrange the terms so that I only have one fraction line, and that will then equal Zed L times s base all over v l base squared.

I'm now going to rewrite the equation I'm going to replace Zed l with one of my pre established equations. And that is going to give me the L base squared all over the H base squared, times Zed h times Essbase all over the L base squared. And clearly you can See that there is a common term in the numerator in the denominator. So I can cancel those two common terms out, and I'm left with Zed h times s base all over V h base squared. Now I want to find h ZH per unit or the high side impedance in terms of per unit value, and that's going to be equal to Zed H, all over h base. And I'm going to replace Zed h base with one of my pre established formulas.

And then I'm going to rearrange the, the term here or the fraction here so that I have only one fraction line and that is going to be equals Ed h times s base all over the H base squared. Now, you might notice something here and that is the equations are both equal to the same thing. What does that mean? That means that Zed h per unit is equal to Zed l per unit Zed h per unit is equal to Zed l per unit, this is very significant. The reason that it is significant is the fact that the high side per unit value is equal to the low side per unit value, which essentially tells us that we no longer have to worry about the ratios of the transformer. And indeed if we are going to draw an equivalent circuit using per unit values, we don't even put in the transformer coils because the do not come into the calculation.

The impedances are have to be in per unit values in order for this to happen. So, the actual equivalent circuit for a transformer when using per unit values is two wires, and we do not have any transformer coils to worry about. If we place the impedance, it could be the impedance of the transformer or a load impedance or any impedance in the circuit in regard to the transformer, we can place it anywhere on the circuit. And it doesn't matter which zone the per unit value is in. It could be on the V base one side, or it could be on the V base two side. It doesn't matter because the Zed h per unit is equal to the Zed Per Unit, this may seem odd, but in when you're analyzing a per unit circuit, this is indeed the case where it becomes significant is when you want to find the actual values, because when you want to find the actual values of the per unit impedance say, then we have to multiply by Zed base and then it will depend on where you want to place or where that load is placed in the circuit, because the actual value is equal to the base impedance times the per unit impedance and the base impedances of a per unit circuit depend on which zone you are in.

But as far as analyzing a circuit is concerned, in per unit values, You do not have to worry about transformer ratios or where the impedance is with respect to a transformer connection. Let's look at a very quick example. A transformer is rated at 2000 va and it has a ratio of 200 volts to 400 volts and has an internal impedance of J 4.0 ohms as seen on the low voltage side, the internal impedance of the transformer as seeing from the high voltage side is equal to the low voltage impedance times the square of the turns ratio which is J 4.0 times 400 over 200 squared, which is equal to j 16.0 ohms. Now, the rating values of apparent power and voltage are used as the basis for our per unit calculations. In other words s base is going to be 2000 va, the low voltage base is going to be 200 volts and the high voltage base is going to be 400 volts.

Let's compare the basis and the per unit values on both sides of those transformers. The s base on the low voltage side is 200 va and it's 200 va on the high voltage side as well. The voltage basis on the low voltage side is 200 volts and 400 volts on the high voltage side. The I base is calculated from these values and I base is equal to s base over V base which is 10 amps on the low voltage side and five amps on the high voltage side. The Zed bass is calculated using the formula v squared base, all over s base, which gives us a 200 or sorry, the 220 Ohm value on the low voltage side and an at home value on the high voltage side, calculating the per unit impedance on both sides, we take the actual value divided by the base value.

And on the low voltage side, it's j 4.0 all over 20 which gives us j 0.2 per unit. And on the high voltage side, it's j 16. All over ad which equals j 0.2 per unit. These two values are the same as it should be. You notice in the table with a transformer per unit impedance is the same regardless of which side of the transformer is removed. To gain the conversion complications are absorbed into the base relationships.

So the equivalent for this transformer would be what you see here in the diagram. We have gone through the normalization process in theory. So now let's apply some numbers to our diagram or system that we have here. And let's say that if the line impedance was j 24 ohms. The generator internal impedance is J, one ohm. The distribution line is J one ohm and the load at the end of the distribution line is 10 ohms.

We're going to let the generator output voltage be eight kV at zero degrees. So the first step in our process is just specify a VA or an S base and in this case, we're going to choose s base to be 100 MVA. The next step is to determine the voltage basis and we've already gone through that process. And again, we're going to make life easy for us by choosing a V base one to be 80 kV which automatically says that the V base two is eight kV and V base three is 16 kV. We now want to determine the impedance bases and the current basis from our VA base and each of the voltage basis. So the impedance base for zone one in green is going to be 80 k v squared all over 100 MVA which is equal to 64 ohms and that is a scalar quantity the impedance base in the generator zone the red zone or V base two is given by eight kV squared over 100 MVA and that's equal to 0.64 ohms.

And finally, the impedance base in the distribution zone which is a 16 kV zone is going to be 16 k v squared all over 100 MVA, which is 2.56 ohms. We now can move on to the calculation of the current basis and the current base for sole one, which is the green zone is equal to 100,000 all over 80, which is equal to 1250 camps. The current base for the red zone or the generators only as 100,000 over eight, which is equal to 12,500 amps. And the eye base for the orange zone is equal to 100,000 over 16 which is equal to 6250 amps. We now can move on to Step Four of the normalization process and that is to find the per unit quantities for all of the system components. And what is very significant here is the fact that once we convert all the units in our circuit to per unit values, we no longer have to use the coils of the transformer.

In other words, the transformer ratios no longer have to be considered. So our circuit now is a single phase with one loop that goes through the generator internal impedance through the line impedance and then through the distribution impedance to our load. So, we can then calculate the load quite easily in per unit values. So, the line per unit impedance is given by J 24 divided by the base impedance of 64 ohms which is equal to j 0.375 per unit the generator internal impedance is in the red zone or the generator zone. So, it's impedance of j one home is going to be divided by the base impedance of point six four ohms which gives us j 1.56 per unit. The distribution per unit impedance is given by J One om divided by 2.56 ohms which is equal to j 0.39 per unit our load again is in the orange zone.

So, it is 10 ohms which is the actual value divided by the orange zone is bass impedance which is 2.56 and that is equal to 3.91 per unit the generator output voltages, eight volts at zero kV and our base voltage is eight kV. So, that means that eight kV divided by eight kV is one and the per unit value absorbs or adopts the same, the same direction as the phasor of the actual unit and that is zero degrees. So, our generator output voltage is one and zero degrees per unit. Now, we'd like to solve for the required values of current voltages and apparent power that's delivered into the system. The current is simply the current flowing in one circuit, which is there. And that's given by ohms law, which is the all over Zed v being the generator voltage in per unit, and the total per unit in the sum of the total per unit impedances because they are all in series, the generator per unit impedance is one at zero degrees.

The total load is made up of the generator impedance plus the line impedance plus the distribution impedance plus the load impedance, which equals three point nine one plus j 2.327 per unit. And if we put those into our equation, we get one at zero degrees all over 3.91 plus j 2.327, which is equal to 0.22 at minus three 0.8 per unit. And that is the current that's flowing in our circuit in terms of per unit values. The voltage, in this case, we're going to calculate the voltage drop across the load. It's given by the current flowing through the load times the load impedance. And that is equal to zero Point two to add minus 30.8 degrees times 3.91 which equals 0.86 at minus 30.8 degrees per unit the generator of apparent power delivered is just the voltage times the current, which is one at zero degrees times point two, two at minus 30.8 degrees, and that's equal to 0.22 at minus 30.8 degrees per unit.

So now we've calculated all of the required components or, or values that we were looking for in this circuit, but they are all in per unit values. Next slide. We're going to convert them back to actual values. We can now go to Step five of the normalization process and this is where we convert the per unit values back to their actual values. And it's a very simple process all you got to do is take the per unit values and multiply by their equivalent base values depending on which zone that they are in in our circuit. So starting with the line current, the line current and per unit value is point two two and minus 30.8 degrees and we can multiply that by the base value for the current which is 1250.

And that gives us 275 amps at minus 30.8 degrees. The current in the generator, section or zone is given by the same current the same per unit current times 12,500 amps which gives us two Thousand 750 amps at minus three 0.8 degrees. similarity the current in the load side or the distribution side is the same current in per unit value times 6250, which gives us 1375 amps at minus 30.8 degrees. The voltage in the load is just the per unit voltage that we developed or we found in the previous slide times the base value in that area, which is 16 kV that gives us 13.7 at minus three 0.8 degrees kV. The generator, apparent power that's delivered to the circuit is just the per unit apparent power that's delivered times 100 MTA, which gives us 22 MVA at minus 30.8 degrees. So there you have it.

It's a pretty simple process. You don't have to worry about the ratios of the transformer as long as you follow the step by step procedure in normalization process. This ends chapter two the normalization process