Chapter Six, a three phase per unit example. Let's consider a simple three phase circuit. In this case it's a single line diagram will show you the circuit we're going to study. It shows a generator connected to a Delta Wye transformer, ultimately connected to a load. The generator is rated at 300 MVA at 23 kV, the transformer is 330, MVA. Its line to line voltage ratio is 23 kV to 30 kV and it has a reactance of 11% on its own base.

This is connected to a load. It's an inductive load that is rated at 230 kV, it is pulling 240 MVA from the system at 0.9 power factor. Our task will be to find the current supplied by the generator. We are now going to analyze this circuit using a per phase equivalent circuit of course, and the first step of converting all our values to per unit values is to establish the base values. And the first base value we're going to arbitrarily pick is the VA base or MVA base, and it's going to be a three phase base value and not is going to be Essbase equal to 100 MVA. The next step is to establish our voltage basis and I'm going to again arbitrarily pick the voltage base to be 230 kV on the high side, I'm going to call that V base two.

And I'm going to say that that is equal to the line two line voltage of 230 kV. Once I've done that, of course, I've established what my low voltage base is going to be, I'm going to call it v base one. And if I've picked 230 kV for the high side base voltage, the low side base voltage is going to be 23 kV. And again, once we've picked our base values line, the line voltages, the line to neutral voltages are already established, and I have to divide the line the line voltage by root three to get Give us the line to neutral voltage. And I do the same thing on both sides. So that our V base two line to neutral is 230 kV divided by root three and our V base one line to neutral is 23 kV divided by the root of three.

And I want to change the reactance of the transformer, I'm not actually going to change the reactance I'm just going to change it from a percent to a per unit value, which means it's going to be 0.11 per unit. We are now at the point where all our arbitrary selections have been selected. And we want to come up with the impeding spaces and the current basis which have to be calculated from the M MVA base and the voltage basis that we've already established. So starting on the V base, one side or in the red zone, Zed base one is going to be given by the V base squared all over the space, and the V bases line two line base voltage that we're going to use. And that's all over the three phase s base, which is 23,000 squared, all over 100 times 10 to the sixth power, and that works out to 5.29 ohms.

In calculating the the current base, we will be using this formula where we're going to take the three phase base and divide it by root three times line two line voltage base. So working in the V base, one zone We're going to take Essbase, which is a three phase base, and that is 100 times 10 to the sixth power, all over root three times 23,000. And that works out to 2,510.2 amps. And of course, we have a impedance base and a current base per each voltage level. So now in the base two zone or the Blue Zone, I'm going to use the same formulas for calculating the impedance base, and we're going to call it Zed base two and that's going to be again, the voltage line two line squared all over the VA or MBA base, which is 20 230,000 squared all over 100 times 10 to the sixth power, which equals 529 ohms.

And calculating the current base in the Blue Zone, again, we're going to use the same formula that we use before for the other zone. And that is the S base three phase s base all over root three times the line to line voltage base, which is 100 times 10 to the sixth power, all over root three times 230 times 10 to the third power, which equals 251.02 amps. Not going to convert the transformer per unit value to per unit values that we'll be using our new s base of 100 MVA because the manufacturer calculated point one one per unit for his transformer impedance reactants based on 330 MVA. And the voltages as they work out will be the same. However, we'll still use the conversion formula that we've already developed for our new per unit value for the transformer which is given by this formula. The subscript M is for the old or the manufacturers base values, which are 330, MVA and the voltage is being 223 kV and 230 kV, the new values that we're going to use are going to be 100 MVA for the Essbase and the voltages actually are going to be the same but we're going to use them in the formula anyway just to make sure everything works out.

So in calculating the new per unit value for the transformer on the red zone or in the red zone using voltage base one I'm going to have this formula or this calculation, these figures of point one, one and 100 over 330 and 23 over 23. They come from the formula that is in the yellow box, and it's starting with the point one, one that is the, the old or the manufacturers per unit value. The 100 is our new MVA value, our new s base that we're using, and the 330 comes from the manufacturers s base, which was three 330, MVA, and of course, the new voltages and all voltages happened to be the same. So there's no no contribution to our formula there because that fraction works out to one. So if you solve point one, one times 100 over three 330 we get point 033 per unit for our new per unit value for our transformer.

Now if we work things out on the, the base two side, we would have this formula, which again is using the same formula but different figures in the formula. And the factors come from again the old per unit value of point one, one. The new s base is 100. The old Espace is 330. And the old and the new voltages are 230. But again, it doesn't matter even if they're squared, they're both the same.

So that fraction works out to one, which means we still have point 111 hundred over 330, which works out to zero Point 033 per unit, which should not be a surprise to you because in previous calculations that we've done before, we found that the equivalent circuit for the four per unit values, it doesn't matter which side of the transformer you're on for the transformer impedance, it is the same either on the primary or the secondary side. So the value of point 033 per unit, that is what we have for the new value for our transformer impedance or reactants. Using our new base values. We are now going to calculate the per unit current that's drawn by the load. But before doing the calculations for the per unit current, we have to actually calculate the current and then convert it to a per unit value and in calculating the actual current being drawn, we have all the information right here.

We know it's an inductive load. And we know it's 240 MVA at 230, kV, and it's point nine power factor. So we are going to have to decide what that angle of the current isn't it has to be at an angle relative to something. And we're going to use the actual voltage on the load itself as our reference angle. Which means that the load voltage is going to be the, the magnitude the at at zero degrees. We don't have to calculate really what the voltage is.

All we want to know is, what it is what its angle is relative to something and we've established that we're going to take it relative to the load angle. And again, we know it's an inductive load. Now the magnitude of the load is given by this formula, which means it's equal to the MVA of the load the magnitude of the three phase, MVA all over root three times the magnitude of the line to line voltage, which is 240 times 10 to the six, all over root three times 230 times 10 to the third, which gives us 602.45 amps. And coming up with the angle of the current, we've been given the fact that the power factor is point nine. So if we take the cosine of the angle of the voltage to the current, we know that that is point nine. So what we have to do is find the inverse of the cosine.

So we find the inverse cosine of point nine And that'll give us minus 25.842 degrees. So we've just calculated our load current to be 602.45 at minus 25.842 degree amps because that's actual current that's flowing to the load. Now we'd like to make that into a per unit value. And we do that by taking the actual value and dividing by the base value. And the load in per unit then will equal like I said, the load the actual load itself divided by the base value of the current in the base to or in this voltage zone, which is 602.45 and minus 25.842 degrees. divided by 251.02.

And that gives us a per unit value of 2.4 at minus 25.842 per unit. Now that we've converted all of the components to per unit values, and the load is drawing a current load of 2.4 at 25.8 for two per unit in our equivalent circuit, that means that the generator has to be putting out that amount of current. Now, the generator does have a an internal impedance or reactance, which really doesn't come into play for what we're looking for. We were asked to find the current flowing from the generator. And if the load is pulling that amount of current in per unit value, or in any value then that's What it has to be drawing from the from the generator as far as magnitude is concerned. Now, if you had to calculate the voltage that the generator had to push to or obtain to push that amount of current with that type of that internal impedance, that's a different question.

But we are only asked to find the generator current. So the actual generator current as far as magnitude is concerned is the same as what is passing through the load. Now, because we have a Delta Wye transformer, we have to add or subtract 30 degrees from the actual load current because there's a phase shift as we go from the secondary to the primary or the primary to the secondary. There is a 30 degree phase shift and you have to subtract 30 degrees from the load current to find out what the actual generator current is. You can see our operator here and you might have recognized it from the equivalent circuit that we developed for a Delta Wye transformer. And when we are calculating the load the generator current, we would have to multiply by one at minus 30 to give us the phase shift that we are looking for.

So we would take the actual current load current being delivered to the load on the load side of the transformer and subtract 30 degrees or multiply by one at minus 30 degrees and that will give us 2.4 at minus 55.8 for two per unit. Now, if we want to find the actual value, very simple step all we got to do is multiply by the base value at that voltage level or in that zone. And the base value add that in that zone for the current is 2,510.2 multiply multiplying that with 2.4 at minus 55.842 degrees gives us 6,024.2 amps at minus 55.842. So this is the per phase equivalent circuit we almost identical to the per unit equivalent circuit, except that the current is now in actual value rather than per unit value, and all of the other components in the circuit can be found if so desired.

Although we weren't challenged with that we only need to come up with a current. All you need to do is take the individual components and and multiply by their base values at their very various voltage levels. And you would have your equivalent per phase diagram, which is a single phase diagram as well. And then you just use your phase relationships in case of a three phase, a balanced three phase load to get the actual three phase equivalent circuit, what you could do as well. This ends chapter six