Chapter Eight, a per unit example problem. This is the system that we're going to analyze. It's a three phase system consisting of a generator, a 90 MVA generator connected to the system at 22 kV with an internal impedance of 18% on its own base, which we're going to consider as purely reactive and it is point one eight per unit. It's connected to the system through two Transformers both transformers are y to delta configuration, T one is a 50 MVA transformer with a low voltage of 22 kV and a high voltage of 220 kV. It has an internal impedance of 10% on its own base, which converts to a reactance of point one per unit. The other transformer t three is a 40 amp VA transformer same ratio and configuration as T one and it has an internal impedance of 6.4% on its own base which is again reactive and it is point 064 per unit.

These transformers are connected to the rest of the system through two transmission lines one is a 220 kV transmission line that has an impedance of 48.4 ohms on each phase and the other is 110 kV transmission line which has 65.43 ohms on each of the phases. These lines are connected to two transformers, which are delta y and configuration T two is a 40 MVA transformer with a ratio of 220 to 11 kV and 6% internal impedance on its own base. Again purely reactive point 06 per unit t four is a 40 MVA transformer same ratio same configuration as T two and it has an internal impedance of 8% on its own base which converts 2.08. The loads on these two Transformers one is a motor load which is a three phase motor 66.5 MVA and at 10.45 kV, it has an internal impedance of 18.5% on its own base which converts 2.185 again purely reactive the three phase load which is not defined other than the fact that it absorbs 57 MVA at 10.45 kV and point six power factor like Now, having defined all of the characteristics of the various components in this circuit, we are going to draw the per unit equivalent circuit, including all the components in terms of their per unit values.

The first thing we're going to do in our analysis process is we're going to convert the three phase diagram that we had just seen to a single line schematic that you see here. The next thing that we're going to do is go through the normalization process. And the first step in normalization is to specify the base or in this case, an MVA base. And it's entirely up to our choice. It's arbitrary and we are going to pick 100 MVA as our base, and we're going to call it s subscript base. The next step or step Number two in the normalization process is to identify the voltage basis in the system.

And we have four Transformers in the system and the way they are connected will generate for individual voltage basis. But what you have to remember the voltage basis in a system are determined by the transformer line to line voltage ratios. So, we get to pick one, and that's the first voltage base and all the rest of them will be determined by the voltage ratios or the ratios of the transformers. So, the first one we're going to pick is the voltage at or of line one and the high side of Transformers T one and T two and that's going to be 220 kV and we will call that voltage base one. And I've color coded the voltage basis so we can identify various steps in our calculations as well. So the voltage base tool will be determined by the ratio of the Transformers T one and T three.

And it'll it tells us that voltage base two will have to be 22 kV and that will determine the voltage base of voltage base three which is the high side of transformer t three and T four as well as the voltage of line two which is 110 kV. The last voltage base will be the voltage of the low side of Transformers T two and T four which we will call voltage base for and that's going to be 11 kV. Step number three is to determine the impedance basis and because there are four voltage zones or voltage bases in our system and the impedance space is calculated you Using the voltage base, there's going to be four impedance bases for each voltage zone. And the equation that you use to calculate the impedance base is the voltage base squared all over the S base. Or you can just take the actual kV and square it and put that all over the MVA base.

And just to remind you of why you can do that, then you have to be careful when you're using terms like kV and MV and MVA and kV a, because there's orders of magnitude that have to be considered. In the case of the V base, and the Essbase. The V base is kV base times 1000. And if you square that you actually get one million kV. And in the case of the MVA, you end up with 1 million MVA as well. So you have to take care of the orders of magnitude and they will cancel out numerator and denominator, leaving you with the KV base squared all over the MV MVA base.

That's just a reminder of how we got to the KV v squared all over the MVA base. So let's go through the process right now at 22 kV, the impedance base will be 22, which is 22 k v squared, all over 100, which is the MVA. So it's 22 squared all over 100, giving us 4.84 ohms. At the 220 kV level, the impedance base will be 220 squared over 100, which will equal 484 ohms. After 110 kV level, we get 110 squared over 100, giving us 121 ohms. And at the 11 kV, we have 11 squirt 11 squared times all over 100, which gives us 1.21 ohms.

Now, we will just tuck that over to the side for now, and we'll go through the rest of the process but we'll keep it there so that we can refer to the voltage basis in the impedance spaces as required. We now have selected or calculated all the various base values that we need for moving on. So we will now go to Step Four of the normalization process and that is to calculate what the per unit values are for the various components in our circuit and the calculation of a per unit value is to take the actual value and divide by the base value. Or we have to take the actual per unit values that were calculated using the manufacturers own base and convert them to the base values that we want to use for this analysis. So step number one, we will look at a line one and the impedance of each phase in that line one was given as 48.4 ohms.

So we'll divide that by the impedance base that we have at that voltage level which is 484. And that generates a point one per unit. For line one line to get has a different voltage base and consequently a different impedance base. And the impedance spaces 121 ohms and the we will take the actual ohms of the line to 65 65.43 and divided by 121 which gives us point five per unit. Now the generator we already have the per unit value for the generator. So we will use this formula which we already know and calculate the new value of the per unit impedance of the generator using our own bass values that we selected.

And that would generate this equation given us the reactance of the generator to be point one eight times 100 over 90 the given us the new per unit our new base value of s base value of 100 all over the manufacturers s base of 90 and the voltage levels that the manufacturer uses the same voltage that we are using for our base value. So that really doesn't contribute anything to the equation but we plugged in the values there and that gives us point two per unit for the reactance of the generator. We can calculate now or change the impedance of the transformer it was given as point one on its own base and the base that the manufacturer used was 50 MVA and we are using 100 and the voltage levels are the same at the low side of the transformer which is 22. And that doesn't again contribute anything to the conversion.

Our conversion will be point two per unit. Now if we decided to use the high side of the transformer It doesn't matter. As we've discovered before, you can have your per unit impedance of a transformer on either side. But the the equation still tells us that we can do that because the per unit value doesn't change regardless of whether you're working from the high side of the transformer or the low side, we still end up with point two per unit for transformer one. And transformer two is the same process. Point 06.

With a new SPS of 100 and the old Espace of 40, the voltages are still the same. And it doesn't matter whether you're in the high side or the low side, the impedance, the per unit impedance is going to be the same and it is point one five per unit. We'll go through the process for t three two same equation. same reasoning doesn't matter where the high side of the low side, the per unit value for the transformer will still be point one six per unit. Finally the transformer T for you can go through the same process and end up with point two per unit for transformer for going through the transformer logic of calculating its impedance, the transformer can use the voltage base on either side of words connected into the system because remember, way back when we were looking at the characteristics of a transformer even a single phase transformer, the per unit impedance on the low side was equal to the per unit impedance on the high side.

So it doesn't matter which side you're choosing to calculate. the impedance works out in per unit values works out to be the same thing. So in the case of T one, it is connected to voltage 22 k v bus and the 220 kV bus. So we can use either be base two or V base one in our calculations and transformer two, it's connected to both 11 kV and 220 kV, so we can use the base for the base one and transformer three, because of its connection to the 22 kV and 110 kV, we can use v base one or V base three and four is connected to both the 11 kV and 110 kV. So we can use v base four or the base three. Another way of reasoning yet is we have to pick a voltage base for the manufacturers voltage base we already know which base we're going to use but the manufacturer went through this same process when he was they were calculating the Manufacturers impedance base, and they too went through this process.

So it doesn't matter which voltage base they use, they still came up with the same per unit value. But in making the conversion, we have to pick the same voltage base as what the manufacturer would have picked. So in cases of the V base two, we both had to pick 22 k v in V base one we both had to pick 220 k v. So the second term in our conversion equation would always work out to one. We now have the per unit impedance values for both lines, the generator and all four of the transformers. We're now going to move on to the load side of our circuit and Starting with the motor load itself, we've been given the per unit impedance of the motor, but it's based on a trend on the manufacturers chosen base values, which are different than the ones we have chosen.

So this is going to mean we'll have to use our conversion formula again. But this time the voltage is the voltage base that the manufacturer use is different than the voltage base that we are using. So that will mean we'll have to use every element of our equation, which means that we're going to take the manufacturers impedance point 185 multiply it by the ratio of our new MVA base over the manufacturers MVA base 100 over 66 point Five and that's going to be times the manufacturers voltage base which was calculated at 10.45 kV and we'll square that and put it all over the square of our new voltage base of 11 kV, which works out 2.25 per unit. Now lastly we will go to the three phase load and the three phase load is going to be made up of a resistance plus reactance component, and we're going to calculate those right now.

The three phase load has a power factor of point six les. So that means we will take the inverse cosine of point 06 which works out to 53.13 degree lag. So that means our s load is going to be 57 at 53.13 degrees. and if we calculate the load using that the values that we have been given the rated voltage that we've been given as the manufacturer's rated voltage of 10.45. If we take the voltage squared over the load, that will give us our impedance of our load, which is 10.45 squared all over 57 at 53 point 13 degrees and in rectangular coordinates after you've done the, solve the fraction and then converted the polar notation to rectangular notation, you end up with 1.1495 plus j at 1.53267 just to be accurate. Now, if we want to find the per unit value of the the load that is going to be just taking the actual value of the impedance and dividing it by our base value for impedance which is 1.21 and that gives us 0.95 plus j at 1.2667 per unit.

So, these are the connections and per unit and impedances for our per unit equivalent circuit. However, notice it said the connections This is not quite the per unit equivalent circuit just yet. Because the Transformers have a delta y or y Delta Connection. So, we have to take into consideration the phase shift when going through the transformers and in the case of T one and T three that connection is why on the generator side Delta on the line side. So, the primary is considered on the generator side the secondary is on the line side, the secondary legs the primary by 30 degrees so, in a Dell ally to Delta Connection and a transformer in going from the delta to the Y side You have to add 30 degrees on going from your y to your delta side, you'd have to subtract 30 degrees. Now the T two and T four connections, the delta is on the high side.

Why is on the low side. The secondary leaves the primary by 30 degrees, the primary is considered on the line side, the secondary is on the load side. So we'd have to have a 30 degree phase shift in going from the Y to the Delta side, but that would be a minus 30 degree phase shift. So as we move from the delta to the Y side, we would have to add 30 degrees to our calculations. So now this is the per unit equivalent circuit for balanced three phase operation. This ends chapter eight