Calculation of Unbalanced Transformer Loads 1

[7JLAman4]

I'm attempting to calculate the Line currents (with no sucess) of a 120V Delta transformer secondary with 3,4 & 5 kVA on phase windings A, B & C respectively with a phase sequence ABC. I originally started out with a 15kVA 3Ph load (5kVA per phase) and knowing how to calculate for a balanced system, tried to achieve the currents using nodal analysis which I was then going to apply to the problem above. I can't seem to calculate the line currents from the phase currents using vectors. I had started out with:
Ia=3kVA/120V=25A@0deg, Ib=4kVA/120V=33.3A@-120deg, Ic=5kVA/120V=41.7A@120deg. IA=Ia-Ic, IB=Ib-Ia, IC=Ic-Ib. IA=(25cos(0)+i25sin(0)) - (41.7cos(120)+i41.7sin(120))
I'm not sure if this is the correct method. However using a sample that I know what the outcome should be, Ican't seem to get near close enough to the answer.

 

[waross]

Hi stevenal;
I agree with your statement,
"Possibly an unbalanced source voltage on the wye side could also create circulating current, but the problem assumed the transformed voltage was 120V across each winding"
I would add. In North America, the primary neutral is left floating on a wye-delta bank. This alows the primary voltages and phase angles to rearrange themselves so as to prevent circulating currents in the delta secondary.
In other parts of the world, I have seen standard practice of grounding the primary neutral on a wye-delta bank.
The circulating curents do flow. Fuses blow. Transformers burn out. With the loss of one or two phases, the resulting low voltage backfeed burns out a lot of refrigerators and freezers.
It is common to see a transformer bank with one fuse blown, or missing altgether.
respectfully

 

[stevenal]

IA=60.2<-33.6 deg
IB=41.6<-126.8deg
IC=71.2<110.7deg

 

[jghrist]

Circulating currents or not, there is no reason to require that the winding currents sum to zero. Why can the originally proposed winding currents (magnitude and angle) not exist? It isn't like there is nowhere for the currents to go. There are three branches at each node, including the lines.

I just threw in the circulating currents to show that the winding currents are indeterminate given only the line currents. In the original post, the winding currents are given, not the line currents.

Stevenal,

IA=60.2<-33.6 deg
IB=41.6<-126.8deg
IC=71.2<110.7deg

Are IA, IB, and IC winding currents or line currents? The original posting used lower case for winding currents and upper case for line currents. I've been trying to keep with this convention even though it seems backwards.

 

[waross]

Hi fellas;
A couple of points;
The OP said a delta transformer bank.
The load current is not the same as the winding current.
This may sound like insignificant nit picking.
It may be nit picking but it is not insignificant.
I have seen several situations where there was zero load on a bank of distribution transformers, and zero current in the secondary lines, but the magnitude of the circulating winding currents was sufficient to either burn out the transformers or blow the primary fuses.
respectfully

 

[stevenal]

By definition, the sum of the three winding currents is equal to 3IO. Since these currents cannot go to the load, they are trapped to circulate in the delta if they exist. To even exist, they must have corresponding currents in the primary. The OP has given no indication that there is any source of zero sequence voltage in the primary to drive zero sequence current around the delta secondary.

The original problem did not start with winding current, it started with winding kVA and balanced voltage. His first assumption, which I challenge is that the winding currents, although unbalanced in magnitude happen to be evenly displaced in phase. This cannot occur without zero sequence current, which cannot occur with balanced voltage. It is self contradictory.

The currents I calculated above are line currents derived from winding currents. The winding currents were derived from winding kVA and balanced voltage using trig to ensure they sum to zero.

Waross,
Its no nit, I've tried to be clear when speaking of winding current versus line or load current.
The condition you speak of was most likely caused by primary voltage unbalance or transformer impedance mismatch or both. Either one of these situations would change the 120V the OP used initially to obtain winding current.

 

[jghrist]

When you're dealing with sequence components, you are generally dealing with line currents, not delta winding currents. I'm not sure if the concept applies. Do you need to have a zero-sequence voltage to get zero-sequence currents? Let's look at a delta connected load with balanced voltages.

Voltages:
V_AB = 120V < 0°
V_BC = 120V < -120°
V_CA = 120V < 120°

Connect:
A 4.8 ohm resistor across A and B phases (3kVA at 120V)
A 3.6 ohm resistor across B and C phases (4kVA at 120V)
A 2.88 ohm resistor across C and A phases (5kVA at 120V)

Current in the resistors:

I_AB = 25A < 0°
V_BC = 33.33A < -120°
V_CA = 41.67A < 120°

The sequence components would be:

I0 = 4.81A < 150°
I1 = 33.33A < 0°
I2 = 4.81A < -150°

Where does the zero-sequence current come from?

 

[stevenal]

Jghrist,

Sequence component theory is perfectly valid within the windings.

There is no 0 sequence current in your example. All three windings are involved in supplying current to each one of the resistors you're loading. I would begin by converting your delta connected load to it's equivalent wye. Loop equations might work also. If I get time I may work out the actual line and winding currents. Just with inspection, though, with no neutral return path there can be no 0 sequence current in the lines, and with balanced voltages none can exist in the windings.

 

[tinfoil]

@waross:
An ideal xmfr is only ever encountered at school when they are trying to teach theory: It has no iron or copper losses, and can deliver infinite current with no regulation voltage drop. I referred to it as a shorthand way of saying "ignoring losses in..."

@stevenal:
In my utility experiences, we used to generate delta banks using two (open-delta) or three (full delta, preferred) discrete single phase two-primary bushing xmfrs. It was a common practice to hook up a single can line-to-line to service single phase loads. Obviously, with only one can, only one primary winding has nonzero current. If you hooked up a bank of three such transformers, and still only placed single phase load on one leg, only one WINDING would have non-zero current (again, assuming ideal transformers). I agree that two of the LINES would provide this current, but that is not what I said in my post. Once you start using real rather than ideal xmfrs, SOME currents will flow in the other two transformers' primary windings, but not enough to cause the sum of these currents to be zero.

 

[jghrist]

Why can there be zero-sequence current in the delta load, but not in the delta windings? If sequence component theory is perfectly valid within the winding, why not within a delta connected load? I used standard equations to calculate the sequence components of the current in the delta connected resistors and I found a zero-sequence component with balanced voltages applied.

Yes, you are correct that the line currents will have no zero sequence component. The line currents will be:

IA = 58.33A < -38.2°
IB = 50.68A < -145.2°
IC = 65.08A < 93.6°

Sequence components of line current:

I0 = 0
I1 = 57.73 < -30°
I2 = 8.33 < -120°

If you convert the delta connected load to a wye equivalent, you will get the same currents that I got in the line. How will that show that there is no zero-sequence current in the delta connected load?

 

[stevenal]

Sorry Jghrist, I read over your post too quickly. You were saying nothing about the winding currents, only the delta connected load currents. I concede that this unbalanced loading creates zero sequence circulating current within the load delta.

 

[jghrist]

Now, given these load currents, what is the current in the delta windings serving the load?

I don't know how to calculate this. There may be more than one answer, depending on the primary source. My assumption would be that if the source was balanced, the winding currents would equal the currents in the delta connected loads, which is where the OP started.

If you had a wye primary and one phase was open, then there would be no current in the corresponding secondary winding. The load could still be served, much as in an open-wye, open-delta connection, but the secondary winding currents would certainly be very different.

tinfoil,
I think that as long as all three primary connections were made to your three-phase bank, the other secondary windings would share some of the single phase load.

 

[waross]

Hi tinfoil;
Consider a single phase transformer compared to the open end of an open delta on the other two phases.
The terminal voltages, and voltage regulation of the single transformer will be equal to the open delta combination. The phase relationship will also be the same.
Put a 2 KW load on the single transformer and a 2KW load on the open delta.
Now if you connect the single phase transformer in parallel with the open delta, you have a closed delta with a load of 4 KW. There was no voltage difference between the two systems before you made the connection.
The current to support a single phase load on a delta bank divides equally between the single in phase transformer and the open delta equivalent.
This assumes similar transformers and equal primary voltages.
For a look at a similar division of load, look at the double delta connection that is the standard connection to convert a three phase generator to single phase use.

 

[stevenal]

Tinfoil,

Assume a secondary delta abc with a single phase load across terminals a and b and no other connection. You have the series combination of windings b to c and c to a in parallel with winding a to b feeding this load. All three windings have current. You can even remove winding a to b without disturbing the current going to the load.
Note that this situation is similar to the l to l fault I simulated in Aspen.

Jghrist,

This situation is much easier to analyze. Obviously the one load (or fault)current cannot sum with two zeros to make zero. But the delta winding currents feeding this load (fault) do sum to zero, or I would be seeing zero sequence primary current in my fault study.

 

[tinfoil]

waross:
I agree with you until "the current to support a single phase load DIVIDES EQUALLY between..."

The current division is a function of the winding impedances (worked vectorically), once you create more than one current path. The vector sum of one path in this case will not be the same as the other, if you start with three identical xmfrs

 

[jghrist]

stevenal,

In tinfoils example, the single phase load is connected phase-to-phase (it's a delta winding, so there is no neutral to connect it to). The line currents will sum to zero. Current in one line is opposite to the current returning in the line.

Your fault study, if it is like most, will not deal with the winding currents, only the line currents. There is no dispute that the line currents sum to zero. I am not convinced that the winding currents have to sum to zero.

What would you calculate as the winding currents in my delta connected resistor example if it was fed by a wye-delta transformer (or delta-delta for that matter) with balanced primary voltages?

I think tinfoil is correct that the current division in the delta will depend on the winding impedances. How about the ideal case of zero impedance?

 

[stevenal]

I meant that the line to line load currents (only one present in this case) do not sum to zero just like in your example. The line currents do. I thought it was a simpler example than yours to illustrate the point.

Okay, try this on for size: Look at any zero sequence equivalent circuit for a wye delta transformer. You will see an open circuit between the two sides. So even if you could get zero sequence current on the delta lines, it would be blocked before getting to the wye side. The fact that no zero sequence can exist in the delta lines doubly makes sure there is none on the wye side. I'm assuming only balanced sources on the wye, and loads on the delta. It's a different story if they are reversed.

Now look at the actual single phase winding transformation. Phase shifting and zero sequence blocking does not occur at the transformation, but in the delta connection. Only difference from primary to secondary winding is the turns ratio. Phase angle, percent +, -, and 0 sequence are all the same from primary to secondary. O sequence percentage in the primary lines is the same as that in the primary winding which is the same as that in the secondary winding. All are zero in this case. This is why I used a wye primary in my fault study, although the OP did not say how it was connected. In this way I could see winding current even while Aspen was showing line current. (delta delta also block zero sequence)

Nodal analysis in this case gives one less equation than there are unknowns. Knowledge of sequence components makes the difference.

How to solve? Write and solve a system of node equations and include Ia+Ib+Ic=0.

An infinite number of answers satisfy the nodal equations without that last equation. Why would you choose the one answer that happens to have currents at 120 degrees? If you draw the phasors, my solution, by unbalancing the angles, is more balanced over all. All points lie on the same circle.

 

[stevenal]

Oops, I'll take back that last sentence, seems any three points can describe a circle.

 

[stevenal]

What I should have said is: The phasors I came up with for the winding currents are more balanced than those that are displaced by 120 degrees. This can be determined graphically by measuring the distance from the phasor origin to the center of the circle described by the endpoints. Smaller the distance means better balance.

As far as transformer impedance goes, it won't have much effect on load current. With the fault case, it has a big effect. With balanced impedances the effect is in the positive and negative sequence results. Zero sequence current is zero everywhere for that line to line fault. Of course the balanced source voltage assumption goes away for this case.

 

[waross]

Say fellas, I work a lot with generators and converting from three phase to single phase is common. The preferred connection to convert a 12 lead generator to single phase is the double delta for 120/240 volts.
The same principles apply to loading a generator winding as to loading a transformer winding.
The last time I had this discussion was several months ago. A salesman used the wrong conversion to size a single phase set.
The company who supplied the generator was an honorable company. When they realised what had happened, they refunded the difference in price between the price of a 45 KW diesel generator and the price of a comparable 30 KW generator. I don't think that the technical department would have agreed with my figures unless I had it right.
What I mean is the customer got a 45 KW set and the company sent a refund check so he only paid for a 30 KW set. I never heard what happened to the salesman.
To demonstrate the division of a single phase load on a delta winding please consider the following excersize.
Use three identical transformers. The primary voltage is balanced.
Assume any percent impedance that you want as long as it is equal for all three transformers.
Calculate the voltage regulation of the single transformer with a 2 KW resistive load. With a resistive load, there will be no phase shift.
Now connect the other two transformers in open delta and calculate the voltage regulation with a 2 KW load. You will find the terminal voltage to be equal to the single phase transformer. The current will equal the current in the single phase transformer. When you do a vector addition of the voltage drops and/or the terminal voltages you will find that the resultant is equal to the corresponding figures for the single phase transformer.
Three 25 KVA transformers with a delta secondary will support a 75 KVA three phase load.
Three 25 KVA transformers with a delta secondary will support a 50 KVA load single phase load.
Do not confuse this with an open delta connection.
Respectfully

 

[jghrist]

Stevenal,

The problem with using sequential component analysis is that the basic transformations are written in terms of phase current. Once you set up your sequence diagrams for whatever the situation is, you calculate I0, I1, and I2. You get the phase currents from:

Ia = (I0 + I1 + I2)/3
Ib = (I0 + I1·a + I2·a²)/3
Ic = (I0 + I1·a² + I2·a)/3

These define the line currents, not currents in the delta winding of a transformer.

I'm going to have to ponder a while whether it makes any sense to talk about sequence components in individual windings of a 3-phase transformer. Sequence components are a mathematical representation of a 3-phase system.

The system of node equations is:

IA = Ia - Ic
IB = Ib - Ia
IC = Ic - Ib

The equation IA + IB + IC = 0 is not an independent equation because it can be derived from the first three equations. Add the three equations to get:

IA + IB + IC = Ia + Ib + Ic - Ic - Ia - Ib
IA + IB + IC = 0
regardless of what Ia, Ib, and Ic are.

The first three equations are three equations and there are three unknown winding currents Ia, Ib, Ic. What do you get when you solve for Ia, Ib, and Ic? Hint: the coefficient matrix is singular.

In any case, the load situation I set up in my earlier post with resistors and balanced voltages does result in line currents the same as you would get by solving for line currents with the OP winding currents. I see no reason why you could not physically get the line currents in my example. The question is, what are the winding currents with those line currents.