Chapter Seven, another normalization example. Our normalization example is going to be a three phase transformer bank, which is connected y to delta y on the primary Delta on the secondary, the primary voltage is going to be 2200 volts and the secondary voltage is going to be 220 volts. These of course are line to line voltages. This is what the single line schematic would look like and that'll help us keep track of our results for our base voltages. The impedance of the transformer is all query we're going to consider it all reflected to the primary side and is given by the resistance is equal to 10.4 ohms and the reactance is J 31.3 ohms. Our task will do will be to find the short circuit current for the transformer both primary and secondary.

If we short circuit the secondary, and we're going to short circuit all three phases, which will make this a balanced situation. Our first step will be to establish a VA base, which we will call s base. And that's equal to 10,000 va an arbitrary choice on our part, but it'll make the calculations a lot easier. We're going to pick the primary voltage as our base voltage, again to help facilitate the calculations and that is good I'm going to maintain a color coordination through this this calculation just to make it a little bit clear where we are. So the green side is the primary side and I've designated the base voltage there as VP for The primary voltage base is equal to 2200 volts and if so is as indicated on the diagram. Once we've established the primary base voltage then the secondary base voltage is already established because of the ratio of the transformer and the secondary base voltage will be vs base which were which is equal to 220 volts.

Continuing on with our calculations, the next step in the per unit process for analysis is to establish our current bases and the current bases are calculated using the established vote. voltage and VA basis that we have just done. So I base is equal to s base all over root three of the voltage line, the line base. In this case, it's 10,000, all over the root of three times 2200, which works out to 2.62 amps. in calculating the base current for the secondary side, we take the S base, which is the same for both sides of the transformer 10,000. And we put it over top of Route three times the V base on the secondary side, which is an order of magnitude different than the primary side.

So it's not surprising that the answer is going to be an order of magnitude bigger than the primary side. So the I base on the secondary side is 26.24 amps. Next we want to calculate the impedance base or the Zed base. And the Zed base can be found by putting the or squaring the voltage base and putting it over top of the the S base. And that is equal to 2200 squared all over 10,000 or 484 ohms. Next we want to calculate if we want to calculate which we do the per unit quantities and a per unit quantities are calculated simply by dividing the actual quantity by the base quantity.

And in this case, it's 10.4 plus j 31.3 divided by 484 which is equal to 0.0. 215 plus j is 0.0647 per unit or in polar terms, it's 0.068 at 71.6 degrees per unit. Now that we have all our component terms in per unit values, we can now draw our equivalent circuit which would look like this. And we said at the beginning that the secondary side was short circuit adult three phases, which is not unusual if the maintenance people have left to center ground clubs on the secondary side, which would give us a short circuit and a balanced three phase fault. The equivalent circuit has to have a 30 degree operator shift in it, which is the equivalent circuit that we developed for a why to delta transformer in few chapters previous to this one. The only impedance in the circuit is that of the transformer and it's on the primary side which doesn't matter really, if we have it in per unit terms which we do.

We're assuming that the transformer is connected to an infinite bus which means that the primary voltage of 2200 volts will be maintained throughout the fault and that is represented by a generator with zero internal impedance. And we're assuming that the per unit value for the for the generator voltage is one per unit at zero degrees. So if we wanted to calculate the primary current and prepare unit terms, we just have to use ohms law which takes the voltage in per unit value over the impedance in per unit value and that is one at zero degrees. Over point 068 71.6 degrees, which results in a primary current of 14.667 at minus 71.6 degrees per unit. The secondary current is the same as the primary curd as far as the magnitude is concerned because we're dealing with per unit values. So the current on the primary is equal to current on the secondary side of the transformer which doesn't exist in our equivalent circuit of course, however, we know that the secondary current is the magnitude is the same as the primary current.

However, we have to go through a 30 reset 30 degrees phase shift because of the wide delta delta Wye connection of the transformer, and because we're going from the Y to the Delta side, we have to subtract 30 degrees. So that would give us a secondary current in per unit terms of 14.66 Seven at minus 101.6 degrees. in calculating the actual quantities for this circuit, we'll take the actual current values and multiply by their equivalent base value at their various voltage levels or in the various voltage zones. So we'll take the primary current first and multiply by the base current at that voltage, which is 14.667 times 4.55 at minus 71.6 degrees, and that's equal to six 6.7. at minus 71.6 degree amps. The secondary current will be given by the same formula but this time, we use a different Bass current because of the different voltage zone and that will work out to 666.6 at minus 101.6 degree amps.

This ends chapter seven