Help with 1ph load on 3px TX calc 2

[RichCarnell]

Could someone please explain to me what the primary side currents in a wye-wye unearthed TX would be for a given phase to neutral load on the secondary. I understand that there would be a sharing of current to some degree across all of the primary windings, but how could they be determined?

 

[ijl]

No current sharing in a wye-wye transformer. Think it as three separate single-phase transformers.

 

[davidbeach]

ijl, you would be correct for the grounded-wye case, but the OP said "unearthed".

The voltage unbalance caused by the lack of a source side ground will cause the wye-point to move away from the "neutral point". The effect is much greater for line-line loads on the load side.

 

[ijl]

As someone said: If I am learning from my mistakes, I am getting a great education.

 

[ijl]

How is this: Source and primary side not grounded, thus no zero sequence current on the primary side. Thus no zero sequence current on secondary. Thus no current at all. (And no current sharing More education?

 

[davidbeach]

Maybe it depends on the definition of current sharing, but with no "neutral" connection, current for a single phase load has to flow in more than one winding. There was a long and involved discussion of this not long ago in this forum.

 

[ijl]

Here is how I understand the question: There is a wye wye transformer with floating neutrals. On the secondary side, an impedance is connected between (say) phase a and the neutral. Phases b and c are open. The question is: What are the currents on the primary side?

Assume that there is some current through the impedance and the secondary winding in phase a. Thus, there must be a current also in the winding of phase a on the primary side. Because the neutral is floating, this current must split and go through the b and c windings on the primary side. Thus, there must be some current also in the b and c windings on the secondary side. Contradiction, because the b and c windings were assumed to be open. Thus, no current in phase a.

 

[RichCarnell]

Thanks guys for the comments.
ijl, we seem to be on the same track with your last comment, only if we assume that the primary side is connected to a 3 phase supply voltage then current would flow in a phase primary. Because of the ungrounded wye connection current would also flow, in a reverse direction, in b and c phase primary windings. With these windings acting as inductors to some degree, but also inducing some flux into the magnetic path or steel of the TX (assume 3 limb core type) and b and c secondary windings with no load on them, how can I determine the currents in the primary windings?

 

[ijl]

I am on thin ice now, transformers are not my speciality (if anything is I can try to sketch the solution method as I see it.

Write the transformer equations for all windings that carry currents, taking the mutual inductances into account. The equation for the primary winding in phase a would be something like

VA = jw LA IA + jw LAB IB + jw LAC IC + jw LAa Ia + RA IA

where

VA = the voltage over the primary winding in phase a
w = the angular frequency (omega)
LA = the self-inductance of the primary winding in phase a
LAB = the mutual inductance between the primary windings
in phases a and b
LAC = the same for phases a and c
LAa = the mutual inductance between the primary and
secondary windings in phase a
IA, IB, IC = the currents in the primary windings
Ia = the current in the secondary winding of phase a
RA = the resistance of primary winding in phase a

Note that some of the mutual inductances may be negative, depending on the structure of the transformer.
Similar equations must be written for the other windings. Equations for current balance must be added. Something like IA + IB + IC = 0. And also an equation for the current and voltage of the impedance on the secondary side.

But where does one get the inductances, especially the mutual inductances? I do not know. Transformers are not my speciality, as I told.

 

[RichCarnell]

Thanks ijl,
any ideas where I could look for more info?

 

[ijl]

Not actually. But if you happen to use the transient simulator program ATP, or know someone who uses, you might try their discussion forum (for members only). They are more theory oriented. Just camouflage this problem as an ATP-problem.

 

[RichCarnell]

Thanks for the tip ijl

 

[jghrist]

See The Whys of the Wyes, The Behavior of Transformer Wye Connections, by GE

http://www.geindustrial.com/publibrary/checkout/38652.30055.7762.44019/generic/GET-3388B.pdf

 

[waross]

How about a word description?
To start we will discus the primary voltages. The ratios will hold for the secondary voltages.
Consider the transformers as three equal impedances connected in star. The voltage drops across the three equal impedances will be equal. The current will be the charging current of the transformers.
Now connect a load impedance in parallel with one of the transformers and construct a vector diagram. (For now, we will ignore the phase angles.)
As the load increases, relative to the charging current of the unloaded transformers, the load voltage will approach zero and the voltage on the unloaded transformers will approach line voltage.
But, these are transformers and they are voltage sensitive. The voltage on the unloaded transformers will actually be the voltage at which the unloaded transformers saturate.
Now we can draw a new vector diagram.
Draw a delta to represent the line voltage and draw an unequal star inside the delta. Two legs of the star will represent saturation voltage of the two unloaded transformers and the third leg will represent the load voltage.
A further complication;
The impedance of the unloaded transformers will be primarily reactive with a high resistance in parallel up to saturation. As the transformers approach saturation by increasing load current, the apparent resistance of the unloaded transformers will drop. In this way, the increasing load current will be reflected by an increasing in-phase component of the supply current.
I hope this gives you an understanding of the complex interactions involved.
I leave it to you to resolve the transformer impedances and the load impedance into their respective resistive and reactive components so as to be able to plot the actual real and reactive currents, the resulting actual impedances and the resulting actual voltages.
Hint. I would assume saturation at about 120% or 130% of rated voltage and work from there.
Your vector diagram of a star using the saturation voltages of the unloaded transformers will be a fairly accurate representation of the voltages, but know that a soluton for the currents and the current phase angles will be a little more complex.
In a three phase transformer, there may or may not be a further complication due to the leakage flux that develops in some configurations of three phase transformers when the fluxes are unequal.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[Marmite]

There's some information on this in the ABB Book Short Circuit Duty Of Power Transformers, Section A13.4. I don't think it's downloadable, but you can order it. Assuming a unit turns ratio, a single phase load current I on the secondary corresponds to a current at the primary side of 2/3 I in the loaded phase and 1/3 I in each of the unloaded phases.
Regards
Marmite

 

[RichCarnell]

Thanks guys for responding.

Marmite, I have seen this 2/3 and 1/3 ratio when I first came across this problem and found the reasoning behind it hard to accept. This is actually the reason I started this thread, to find a solution that made sense to me.

Waross, thanks for your short story. I am very interested in what you are saying only I am having trouble following. You say in the second paragraph that as the load increases that the voltage will drop to zero across the impedance. Is this due to the load being to great for the transformer for example - a short circuit?
... and I think that you are right ... this is starting to look quite complex.

 

[jghrist]

You say in the second paragraph that as the load increases that the voltage will drop to zero across the impedance. Is this due to the load being to great for the transformer for example - a short circuit?

The voltage drops to zero because the neutral is floating and will move toward the loaded phase. The ungrounded wye transformer connection is not suitable for serving single-phase-to-neutral loads.

 

[waross]

Hi RichCarnell;
Let's simple up for the sake of understanding.
Consider two single phase transformers in series with a floating neutral or center point on the primary.
For the sake of understanding, disregard inductance and consider the current to be in phase.
As load is connected to a transformer, the impedance drops.
So start by assuming that both transformers are represented by 100 ohms each.
The supply is 200 volts. The voltage across each transformer will be 100 volts.
Now suppose that the impedance of one transformer has a small load connected and the resulting primary impedance drops to 50 ohms.
Now there will be 133 volts across the unloaded transformer and 67 volts across the loaded transformer.
Increase the load and the impedance drops further;
20 ohms gives a 167:33 volt division.
10 ohms gives a 182:18 volt division.
BUT transformers have non-linear impedance.
At the knee of the saturation curve, the impedance drops to limit the voltage at the saturation point.
This means that a heavily loaded transformer in series with an unloaded transformer will have the voltage across the unloaded transformer limited at about the knee of the saturation curve.
The apparent impedance of the unloaded transformer will start to drop below 100 ohms and the current will increase.
If the transformer saturates at 130% voltage, the voltage
across the unloaded transformer will increase to about 130 volts and not much more.
Play with some examples on single phase to get the feel of voltage drops across resistors in series, and with non-linear impedances. When you are comfortable with this, try adding in the complication of phase angles.
The individual impedances must be resolved into inductance and resistance. The impedances representing the loaded and un-loaded transformer will have different apparent X/R ratios. You must determine the phase angle and use vector addition.
Then add the complication of three phase calculations.
Then remember that some three phase transformers have a restricted path for the return of unbalanced flux and the actual voltages will be distorted with quite a bit of harmonic component.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[RichCarnell]

Thanks waross for that detailed explanation.
I now understand what would happen in an overloaded situation.
If we were to then look at a 3 phase wye winding, using your impedance of 100 ohms per winding in the secondary, and adding a load to A phase so that the total A phase impedance were 25 ohms, then we would get a 67 volt drop across A phase and a 133 volt drop across B and C phases as they would appear as 2 parallel inductors (assuming no load on them). The current through the A phase impedance would be 2.68A lagging at some angle due to the inductance of B and C phases. This current would then be divided between B and C phases and equal 1.34A lagging.
Does this sound right to you?

 

[waross]

Three phase is a little more complicated than that.
First, substitute the word impedance where you have used the word inductance.
Second, if you draw a vector sketch, The phase angle through the loaded phase will mostly be determined by the X/R ratio of the load. With a resistive load, the transformer inductive reactances will be swamped by the load resistance.
The current in one unloaded phase will be leading and in the other unloaded phase it will be lagging.
As the voltage changes on the loaded phase, the wye is no longer symetrical and the neutral point is dragged away from the center of the delta.
That's why I used a single phase example.
I suspect that a rigorous solution of a three phase example may involve a challenging transposition of the quadratic equation. Just when you think that you have a solution, it may be further complicated by the non-linear inductance characteristics of a transformer.
It is good to understand the nature of the effects associated with an open neutral.
However, it is not a natural real world problem.
When it happens accidentally, consider that the first limit is voltage rise on the unloaded phases of about 173% (from line to neutral voltage to line to line voltage.)
This limit may be mitigated by transformer saturation. For commercial transformers, estimate a limit of about 120% to 140% of normal voltage to account for saturation.
For a last visualization, drive three nails in a workbench in an equilateral pattern, about 12 inches on a side. Now lop a rubber band around each nail and connect the other ends together. Now move the center point of the rubber bands up, down and sideways. Observe the change in ratios of the length of the rubber bands and the change in the angles between them. Now calculate the ratios and angles.
If this is a class assignment, buy a book called "1-2-3 infinity" and innocently ask your prof. for some help solving one of the unsolvable problems.


Bill
--------------------
"Why not the best?"
Jimmy Carter