CHAPTER FOUR be a continuation of three phase per unit analysis and transformer configurations. We're now going to look at the star Delta or Y delta connected transformer. And this transformer is a little bit more complex than the previous two that we've studied. And we're going to look at those complexities in some detail and kind of remove the mystery around them. Now if we are we're going to look at the primary connections first, and the primary connections of a wide Delta transformer are connected the same as a Y, Y connected transformer and we've already gone through that process. So I'm going to speed through it right here and end up with the phasers on are in the upper right hand corner labeled as they were in the past.

Previous diagram. So, we have the primary coils of this three phase bank or three phases of the transformer connected in this fashion and the phasers of the various voltages are shown in the upper right hand corner here. Now, the secondaries of the Y delta transformer are connected up the same way or in the same configuration as the secondaries of a delta delta transformer and the phasers are shown to the right of the transformer there in our diagram. And they take on this same shape as as the Greek letter Delta, hence the Delta terminology for the connection of the secondary of the transformer. So we've got connections in the primary connection to the secondary are pretty much the same as what we've looked at before. But what is Significant here in this transformer is the fact that the red to white voltage of the red phase and we're going to talk about only the red colored transformer For now, we will go on to talk about the other phases as well, but the red to white or the phase two phase voltage of the red secondary is in phase with the red to neutral voltage of the primary side and you can see that if I move the phaser down, it lines up perfectly with the secondary red to white voltage.

So, the phase two phase voltage of the red transformer is in phase with the red to neutral voltage of the primary. Similarly, if we look at the green coils of the transformer which has the white phase, the white phase vector or phaser of the primary is in phase with the white to blue phasor of the secondary, because the two coils are linked magnetically, and you can see that they have to have the same phaser same direction, not necessarily the same magnitude because that depends on the turns ratio. However, for simplicity purposes here, it looks like it's the same magnitude, but it could be different magnitude. However, that just makes the triangle of secondary bigger, but the relationship to the other phases are the same. And if we look at the blue colored transformer, the blue to red phaser of the secondary voltage is in phase with the blue to neutral of the primary.

These facts are what make this transformer very significant. We're now going to compare the voltages of this white Delta transformer. And when we look at the phase two phase voltage of the primary, it's indicated on my diagram there with a black arrow, and we call that the red to white voltage. And we'll look at the red to white voltage on the secondary and I've indicated there with the dotted lines as well. And we call that the lowercase r w. In in our comparison, in looking at the primary voltages, we know from our phase relationships of a balanced system, that the magnitude of the red the white voltage is going to be equal to root three times the red to neutral voltage and I've indicated the red to neutral voltage with just a capital R here for simplicity sake And the other thing we know from our phase relationship of a balanced system is that the red to white voltage is going to lead the red to neutral voltage by 30 degrees.

And that is to say in general terms at the magnitude of the phase two phase voltage is equal to root three times the magnitude of the rate of the phase two neutral voltages and all phase two phase voltages lead their equivalent phase to neutral voltages by 30 degrees. Now, looking at the secondary side, the phase two phase voltage, our w in this case is definitely in phase and we already saw that in the last slide, last slide in phase with the primary red to neutral voltage So, you can say that the red To white voltage is leading the red, the red to white primary voltage is leading the red to white secondary voltage by 30 degrees. Now, if you look at our connection drawings there, you can see that the turns ratio of the transformer and we look at just the read phase transformer for now, that the the turns ratio of the transformer is given by the capital letter are all over the lowercase red to white, which says that our red to white voltage on the secondary in our phasor diagram, the magnitude of the secondary red the white voltage is equal to the red to neutral voltage all divided by the turns right Or a, we can rewrite that equation such that the red to neutral primary voltage is on the left.

And the turns ratio times the red white voltage on the secondary is on the right hand side of the equation I just manipulated the equation. So now if we look at the equation in the on the primary side, we already saw that the magnitude of the red the white voltage was equal to red root three times the red to neutral voltage. Well I can substitute now for the red to neutral voltage or the capital R from our diagram of the secondary voltages with involving the turns ratio. And I can make that substitution right now. And I can come up with the fact that the red the magnitude of the red new red toy sorry, the magnitude of the red to white void voltage on the primary is equal to root three times the turns ratio times the red to white secondary voltage. And that can also be rewritten such that the red to white phase to phase voltage on the secondary side is equal to the red to white voltage on the primary side divided by root three times a turns ratio.

So we know what the relationship primary the secondary as far as the voltages are concerned in regard to the magnitude. And we also know that the secondary lags the primary by 30 degrees. So far we've been looking at the voltages in respect to the various configurations of the transformers. But now I want to start looking at the current that flows through these Transformers in their various configurations. I'm going to start with the star Delta connected transformer. And I'm going to connect a load up to this transformer, but I'm going to just connect a purely resistive load because that'll make life a little bit easier for us, we won't have to worry about phase shifts due to the load.

So I'm going to make that load a Delta connected load because it will again make our life easier, I mean, doesn't really matter because I can convert a Delta load to a star load, but let's leave it as a Delta load for now. And as I said, I'm going to connect it up to the secondary of the transformer. Now, this load is going to draw current from the secondary coils of each one of those red, green and blue transformers. The red phase transformer is going to supply current to the load out of its spot terminal which is the x one terminal of the second area and that current is going to be denoted by the lowercase r. And because the that transformer is magnetically linked to the primary, there is going to be current flowing through the coil of the primary and that is going to be current draw current into the spot terminal of the primary transformer primary of the transformer.

And I'm going to denote that as capital R for the phaser. And because they are magnetically linked, those two phasers are going to be in phase with each other. In other words, they will be pointing in the same direction as I've indicated in my diagram. That magnitudes may be different depending on the the ratio of the transformer. However, we're only going to say for now that they're in phase with each other and the way Phase is going to draw current, or supply current to the load from the green transformer and it's going to be supplying current out of its spot terminal into terminal one x one. And it's going to cause current to flow into the spot terminal on the primary side coming into h one, and those two voltage or those two currents are both going to be in phase as well.

And I'm going to denote them either with a capital W and a lowercase w for primary and secondary denotation respect respectively. And finally, the blue transformer is going to supply current to the load and it's going to flow out of its spot terminal into the x one terminal. And it's going to draw current into the spot terminal on the primary, which I'm going to call a capital B and the capital B will be in phase with the lowercase b or the primary current will be in phase with the secondary current. Now the current flowing into the red transformer through the h1 connection to the bus is going to be drawing current from what I call I line r dash prime. And that current is going to flow directly into the red transformer and it's going to equal the capital R because all of the current flowing into the red bus is going to flow into the red transformer.

Similarly, the current flowing into the white phase transformer is going to be coming in from the bus and we will call that I line w prime and all of the current coming into the bus from the line if you will, is going to be equal to the current flowing in the white face. And finally, the same thing with the blue face, all of the current flowing into the bus will flow into the h1 terminal and it will be called I line B dash primed and is equal to the current in the blue transformer on the primary side. If we are going to look at current now flowing in the secondary to the load, current is going to flow out of the busbars into the line which supplies the load. And this time I'm going to call a current that's flowing into the load along the eyeline our secondary that's the current that's going to be flowing to the load.

Now, if you look backwards along the rail line there, through the bus, you can see it's connected to the red transformer and so the current flowing out of the red transformer will be flowing into the line. And it is equal to the current that is flowing through the red transformer. However, it doesn't it's not all of the current in eyeline are secondary, because some of that current is going to be bled off into the B phase. In other words, it's going there's going to be a negative amount of current flowing through the B phase, or we might call a minus b current, which is equal in magnitude to the secondary blue current, but it's opposite because it's minus B and the eyeline read, secondary current that's flowing to the load is made up of the read Minus the B. So the resultant will be, as you see here, a triangle is formed by the secondary red phase current and the minus b phase current and the eyeline are secondary.

Now I'd like to determine the angles of the triangle, in other words that contained angles. And I'm going to do this by moving minus b back over to the secondary Delta voltages of the transformer. And you can see that the minus b forms a straight line with a plus b. And because the currents are all equal, because it's a balanced system, then the sides of that delta for me of the phasers there is an equal lateral triangle and the angles that contained angles are all equal. And they have to add to 180 degrees so that I know the angle contained by the blue face and the red phase vectors is 60 degrees. And because B and minus b forms a straight line, the contained angle between the red face and a minus b vector, or phaser has to be 120 degrees.

So I know that the contained angle of the triangle has to be 120 degrees. Now, as I said, the magnitude of the red phase and the magnitude of the blue phase as well as the magnitude of the minus blue phase are equal. So what we see here in this triangle that we just created, is an isosceles triangle means that that angle contained opposite the hundred and 20 degrees have to be adding up to 60 degrees and you split that between the two angles and that is equal to 30 degrees each. So what does that tell us that tells us that the current the eye line red dashed secondary current is lagging the red phase primary current by 30 degrees because the red phase primary current is in phase with a red phase, secondary current. In other words, the secondary lags the primary by 30 degrees. We are now going to look at the magnitudes or the links of the phasers Our resultant triangle that we just formed there.

And you can see that this triangle that's formed by the red phase minus the blue phase and the resultant current is made up of a triangle that is actually made up of two triangles. And those two triangles are 30 6090 degree triangles. And we know that the sides of those triangles 30 6090 are made up of the ratios of one, two and root three to being the hypotenuse of the triangle and the red face and the minus blue face form the high pot news of each of those two triangles. So if we wanted to calculate the length The phasor of our result in current, then we could, it's made up of two sections, one section being read phase divided by two, then multiplied by root three. And the other section would be made up of minus b times root three, all over two. Now, we're looking at only the magnitudes or the lengths of the arrows.

So we're no longer considering the, the angles, we're just going to talk about the magnitudes or the length of the line. So we're going to look at the absolute values of those phasers or vectors that are there and I've done that by placing the absolute bars on either side of the of the vectors are phasers and if I do that, then The absolute value of the read phase is equal to the absolute value of the negative beef phasor because I'm only looking at the length of each of those phasers, and they're equal in magnitude, so their absolute values have to be equal. So I can replace the minus b by the red phasor length or absolute value, and I get this equation, which is the resultant current is equal to root three times the quantity red plus red absolute values all over two, which is also equal to root three times two times the absolute value of are all over to and of course the two twos cancel out.

You're left with just the fact that the resultant current is equal to root three times the absolute value of the red face current. Now I'm going to define a term that's more generic to our problem here. And instead of looking at only the red phase are I want to look at the red phase, the white phase and blue phase, I'm going to define the current flowing in each one of those coils. As I look in lowercase letters, I line dash line, and that's going to be equal to the absolute value of R or the length of the red face. It's going to be equal to the length of the white phase. And it's going to be equal to the length of the blue face but generic in nature, it doesn't have any direction it only has length and the length of I lined a line is equal to the length of any one of those current phasers.

So I'm going to say that the eye line secondary our result and current is going to be equal to root three times this I like to lie. Now, if I want to look at the ratio of each one of those transformers, then the ratio which I define is a is given by if you look at our connection, the voltage line to neutral over the voltage line to line and the voltage line to line bridges the coil that gives us the current The red phase or the white phase or the blue phase. So it's really the turns ratio of the transformer if you would, but we're defining it as voltages. And I'm only going to be talking magnitudes here. And I'm not going to be keep putting the lines in because that gets confusing, but I'm only talking magnitudes here I'm not talking about direction. So if I'm looking at the currents flowing in each one of those coils of the transformer, then the inverse is true, because the line kaline current of the secondary over the eyeline primary is also equal to the turns ratio of the transformer.

So it's like saying I line two line is equal to the turns ratio, the transformer times the line current of the primary. So if I move and replace my eye line to line in the equation I just developed, then you can see that the eye line secondary is equal to root three times that turns ratio times I line primary, which means if I want to find the line current in the secondary, I have to take the primary line current, multiply by the turns ratio and multiply by root three. Now I'm going to say let, let's define a number or letter here. That equals a times root three, and I'm going to call that a prime. And I'm going to rewrite my equations so that I line second magnitude is equal to A prime times, eyeline primary. And if I want to change that to a phaser.

Now from the fact that we are talking about just magnitudes, then I have to subtract 30 degrees from the primary current. The takeaway from all this mathematical manipulation is that the line currents on the secondary are equal to the line currents on the primary multiplied by a factor of a primed and a prime is equal to the turns ratio of the transformer times root three and Got the secondary currents like the primary by 30 degrees. Let's take a closer look at this wide Delta connected transformer configuration and come up with a per phase equivalent circuit and a per unit equivalent circuit. We know from the phase relationship of a balanced system that the primary side of this transformer the voltages are going to be related as the second equation there, which is the line two line is equal to root three times v line to neutral at 30 degrees, meaning the line to line voltage will be leading the line to neutral voltage by 30 degrees.

Now, I can rewrite that equation which I want to do because I'll be using it later on. And I'm going to bring the line to neutral voltage to the left side of The equation and everything else to the right hand side. So I'm going to have the line to neutral voltage on the primary side equal to the line to line voltage on the primary side all over root three all over 30 degrees. The phase relationship between the primary and the secondary on a per phase basis. And I'm going to look at the the ratio between the primary coil and the secondary coil and the primary coil is a line to neutral voltage on the primary side and it's aligned to line voltage on the secondary side. So the turns ratio is given by the line to neutral on the primary all over V line to line on the secondary and I'm going to let that ratio equal a I'm going to rewrite that equation just because it'll fit into our analysis a little bit further, that the line to neutral voltage on the primary is going to equal the turns ratio times the line to line voltage.

Now I'm going to want to come up with a per phase equivalent circuit for this transformer, especially for the secondary. And I'm going to follow the procedure for per phase analysis, which means the first step is to convert all Delta loads and sources to their equivalent y connections. Now there isn't a neutral on the secondary, but I can hypothesize one by using the phase relationship, have a balanced system again and like say that the line two line voltages and the line to neutral voltages are related by this equation, which is the line to line voltage on the secondary is equal to root three times the line to neutral voltage. And the line to line voltage is leading the line to neutral voltage by 30 degrees. I would like to rewrite this equation like I did for the primary voltages, and I want to bring the line to neutral voltage over to the left hand side of the equation.

So that means that the line to neutral voltage is going to equal the line to line voltage all over root three all over 30 degrees. Now, I'd like to take these two equations that I've just circled here and I'm going to divide the poppy equation by the bottom equation, and I have to do that on in terms of both the left hand side and the right hand side. So I'm going to have to take the line to neutral primary voltage and put it over the line to neutral secondary voltage. And I'm going to have to do the same thing with the terms on the right hand side of the equation. And that's gonna give me something that looks fairly complex. But I can mathematically reduce this fairly quickly by removing the division sign in the denominator.

And when I do that, I have to move the denominator of the denominator aside, all over greater and it's just a mathematical operation, and it would look like this, which is a simpler equation. Now, I can take it another step further. Because there's two like terms in the numerator and denominator, and those like terms are the voltage line to line. So I can cancel those two terms out of the equation. And I can go further than that in simplification. I'm going to take the turns ratio a and the root three, and I'm going to replace them with a term that I call a prime and a prime is going to be equal to a times the root of three, which will now give me the equation v line to neutral primary voltage over V line to neutral secondary voltage equal to A prime at 30 degrees and What that means is that the secondary lags the primary by 30 degrees.

And you can see that more readily when I rewrite the equation. And you can see that if I rewrite the primary line to neutral voltage in terms of the secondary line to neutral voltage, I have to add 30 degrees to the line to neutral voltage because it's lagging the primary voltage. So far, we've been dealing with line to neutral voltages. And the last equation that we were dealing with had the line to neutral primary voltage over the line to neutral secondary voltage. And although we have a neutral in the primary, we do not have a neutral in the secondary. So I'd like to replace these line to neutral voltages with their line to line equivalence.

And in order to do that, I have to use already established phase relationships, which I'm showing here, then numerator is, instead of having the voltage line to neutral primary, I have that line to line voltage of the primary all over root three times 30. And in the denominator, instead of having the line to neutral secondary voltage, I will have the line to line secondary voltage all over root three times 30 degrees. Now this is a fairly complex or cumbersome fractional equation, but I can simplify it just by moving terms around mathematically. And when I'm finished moving the terms around two of the fractions are disappearing. And I'm left with this fraction here that has like terms in the numerator and denominator and those like terms are root three times 30 degrees, and you can create Cancel out like terms and a fraction. And I'm left with just the line to line voltage over the line to line voltage primary to secondary.

And that is equal to the line to neutral voltage over the line to neutral voltage primary to secondary. Now, I can rewrite this equation as well, and I can end up with the primary line to line voltage is equal to the turns ratio times the line to line secondary voltage times 30 degrees. And this equation also shows that the secondary lags the primary by 30 degrees and you may have wondered why I've left off the designations in our black box as to whether I was talking about lying to neutral or lying to line or currently Because they all that is all of the secondary values lag, they're like primary values by 30 degrees, the line to line lags the line to line, the line to neutral legs aligned to neutral, and the line current leg the line currents by 30 degrees. We now have enough equations to calculate or build our per phase equivalent drawing.

And the per phase equivalent drawing is just a single phase transformer whose voltages on the primary and secondary side are given by the line to neutral voltages. And so we would use the equation that's highlighted in yellow on the right hand side of the slide there. And you can see that our single phase trains Former has two operators one operating on the phase of the of the voltage phasor and one operating on the magnitude and how it works and what our equation describes is that if we have a voltage on this secondary and we wanted to see what the voltage on the primary would be, we would have to multiply that secondary voltage by the term a prime. Now remember a prime is the turns ratio of the transformer times root three. So we have to multiply the secondary voltage by a magnitude of A prime in order to get the primary voltage phase to neutral also, the primary phase voltage, phase two neutral voltage leads the secondary phase to new voltage by 30 degrees.

So the burgundy operator tells us that we have to shift the line to neutral secondary voltage by 30 degrees, so that it would be equivalent to the primary line to neutral voltage. Now, there is current flowing if there is a load on this transformer, and if there is a load on that on this transformer, we've already established the fact that we have the line currents of this equation. And I'm going to rewrite the equation because I want to keep the primary line current on the left side so that we can actually compare it to the voltage phasers. And that would give us this equation where the primary line current is equal to the secondary line current all over a prime but it's still times plus 30 degrees. So if we were to put our line currents on the diagram, our operators would look like this. Which tells us that if we are to go from we got the secondary line current, we would have to divide it by a prime in order to find out what the primary line current is.

But when it comes to the phase shift, the phase shift going from the secondary to the primary is the same as the phase shift for the voltage. So the current is little bit different than what the voltage is on our per phase diagram. We are now going to build our per unit equivalent circuit and in order to build a per unit equivalent circuit, we have to establish our base quantities and one of the base quantities that We're going to have to establish is the voltage base quantities and there are two of them here because there's two voltages or voltage levels when we have a transformer and in the case of the per phase equivalent circuit, the one base voltage would be established using the base quantity on the primary side and we'll call that the base one and the second base quantity would be the voltage base on the secondary side of the car the transformer and we will call that the bass tube.

Now, we can convert or if we converted all of the quantities, the voltages and the currents in the impedance and etc to per unit quantities. We would then get rid of the coils of the transformer In fact, we get rid of the transformer altogether in our equivalent circuit, and we would be left with just two straight wires, because by converting to per unit values, you get rid of the transformers. So in a single phaser, the per phase equivalent circuit, it would lead to a per unit equivalent circuit of two wires. And we would have a phase shift going from the secondary to the primary. And we have to keep that in mind. And I'll talk a little bit more about that in a few minutes.

Now, we still have the voltage base zones and the V base one is in the same location and the V base two is in the same location as its per phase equivalent. Only there are no coils of a transformer to worry about. But we Still have to keep track of the phase shift going from the secondary to the primary, because the per unit values take care of the turns ratio of the transformer, but it doesn't take care of the phase shift. So we have to keep that operator in our equivalent circuit. Now the other thing is, I'm showing the V base values as line two line voltage values. You could use the line to neutral base values, but as you convert from per unit values to actual values after your calculations, if you use line to neutral base quantities, all your voltages will be lying to neutral.

And if you want line to line you'll have to do a second step of making the phase relationship conversion but we we would probably probably want to use line to line voltage base values because then you'd have directly here line to line values when you calculated your actual values from your per unit values. So your voltages in the per unit equivalent circuit in both zones are going to be equal because they're there in per unit and they will be the same. So they will be equal other than that phase shift that we're talking about. So I'm showing them as magnitude quantities. However, the voltage quantities or the voltage scalar quantities are the same. On doesn't matter which zone you're in.

It does matter once you start to calculate the actual values because then you have to use the base voltage value that is associated with that. So but if you're in per unit values, your per unit values are the Same. And it's the same thing with the currents, your current values are the same as well. So the doesn't matter which zone you're in your per unit currents are going to be the same. Now if you are looking at the internal impedance of the transformer, you would have to be careful in your per phase equivalent circuit, which side of the transformer you're on, because there's a factor of the square the turns ratio to worry about if you're in per phase. But if you are in per unit values, it doesn't matter which zone your internal impedance is, because all of your impedances are the same if they're per unit values, regardless of which zone that you're in.

This ends chapter four B