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7JLAman4 (Electrical) (OP)
7 Jun 06 14:51
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.
jghrist (Electrical)
7 Jun 06 15:42
Your method seems correct assuming that Ia is the current in winding A, Ib is the current in winding B, etc.; the line currents are IA, IB, and IC; and the transformer connection is DAC (top of winding A connected to bottom of winding C).

If the connection were DAB (top of winding A connected to bottom of winding B), then IA=Ia-Ib, IB=Ib-Ic, and IC=Ic-Ia.
cflatters (Electrical)
7 Jun 06 15:42
Sounds like a college exercise to me !
7JLAman4 (Electrical) (OP)
7 Jun 06 16:08
College didn't explain unbalanced systems.  Work is driving this goal of mine to better understand certain theories and formula shortcuts.  When using the aboe calculations on a "balanced" system, Line current IC always somehow comes out much higher than line currents IA or IB.
jghrist (Electrical)
7 Jun 06 16:33
I get IA=58.3A, IB=50.6A, IC=65.0A using your equations for a DAC winding.  IC doesn't seem inordinately high considering the unbalance in the winding kVA.

Using 5 kVA in each phase gives 72.1A in each line, the same as winding current time sqrt(3).
stevenal (Electrical)
7 Jun 06 16:35
Why are you assuming currents are evenly displaced by 120 degrees?
7JLAman4 (Electrical) (OP)
7 Jun 06 16:40
If only given kVA of the loads connected to the transformer, how would the phase angles be calculated then? and is it necessary?
stevenal (Electrical)
7 Jun 06 17:09
You can't. Necessary? Not if you have a meter.
waross (Electrical)
7 Jun 06 19:03
Further to stevenal's comments:
If the power factors of the different loads are different, you add another level of complexity to the problem.
I suggest that you draw some vectors to scale and double check your calculations.
Another suggestions to help you.
On a balanced system:
50 amps phase current will result in 50 x 1.73 = 86.5 amps line amps.
65 amps phase current will give 65 x 1.73 = 112.4 amps.
When you combine 50 amps on one phase with 65 amps on another phase, the result will be more than 86.5 amps and less than 112.4 amps.
I suggest that you use scale vector drawings and the high and low limits to self check your calculations, until you develop confidence in your solutions.
7JLAman4 (Electrical) (OP)
8 Jun 06 9:24
Thankyou for the replies.
stevenal (Electrical)
8 Jun 06 11:43
Even without differing pf loads, this is not an easy problem. A balanced delta can be converted to an equivalent wye, and solved on a per phase basis. Unbalanced systems are generally handled by breaking the problem down to its balanced sequence components, each of which can be solved on a per phase basis and recombined.

Note that on a delta system, all the line currents must add to zero since there is no neutral return path. Even if the loads are 100% resistive, unequal magnitude line currents cannot be evenly displaced by 120 degrees and still sum to zero.

If this is really necessary for work, I suggest you look for a course on sequence components.
jghrist (Electrical)
8 Jun 06 15:24
Solving for the line currents with the winding currents known is an easy problem.  It is just nodal analysis with two currents known going into each node.  The equations are as jlamann gave in the original post.

The more difficult problem is solving for winding currents with the line currents known.  In this case, you know one current going into each node and need to find the other two.
stevenal (Electrical)
8 Jun 06 19:11


Solving for the line currents with the winding currents known is an easy problem.

Agreed, but in this case all that is known is the KVA per winding. From this jlamann got a set of winding currents that don't close the delta graphically. The equations are good, but the inputs into them don't work. GIGO.
stevenal (Electrical)
8 Jun 06 21:18
Another thought, though. The winding current magnitudes calculated above can only describe one triangle. You should be able to use the laws of cosines and sines to find the relative angles of the three phasors. Then use the formulas to find the line currents. Pf and sequence components avoided.
jghrist (Electrical)
8 Jun 06 23:33
The winding currents do not have to add to zero.  The line currents do.  In the original example:

IA = Ia - Ic = 58.33@-38.21°
IB = Ib - Ia = 50.69@-145.28°
IC = Ic - Ib = 65.09@93.67°

IA + IB + IC = 0
tinfoil (Electrical)
9 Jun 06 8:56
Further to jghrist:

The line currencts must sum to zero

The nodal currents (i.e. the sum of currents at each corner of the delta) must sum to zero

The winding currents are a function only of the load. Imagine a delta service with only single phase load on one leg.  Assuming ideal transformers, only one winding will have non-zero current.
jghrist (Electrical)
9 Jun 06 9:33
Or consider a delta-wye transformer with a single phase-to-ground fault on the wye secondary and no load on the unfaulted phases.  There will be current in only one of the windings (primary and secondary).  Obviously, with current in only one winding, the sum of the winding currents will not add to zero.

The problem of calculating winding currents from the line currents arises because the set of equations

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

cannot be solved for winding currents Ia, Ib, or Ic.  If a constant is added to each winding current, the equations still hold.  The constant adder (magnitude and phase) might be recognized as a zero-sequence current circulating in the delta.
waross (Electrical)
9 Jun 06 10:54
Hi tinfoil;
I don't undestand what you mean by an ideal transformer.


Imagine a delta service with only single phase load on one leg.  Assuming ideal transformers, only one winding will have non-zero current.
What about two transformers in parallel with a single phase load. Won't ideal transformers share the load?
When a single phase load is applied to "A" phase of a delta transformer bank, "B" phase and "C" act as an open delta transformer. The open delta may be resolved to a single phase transformer in parallel with "A" phase. The "B" phase and "C" phase transformers share 50% of the load.
jghrist (Electrical)
9 Jun 06 11:42
Let's say you have a wye-delta transformer with A phase of the primary open.  You could still serve load between the 'a' and 'b' terminals (ends of the delta A winding).  There would be zero current in the A winding and the currents in the B and C windings would be equal.  This current would flow out of the 'b' terminal (top of B winding, bottom of A winding), through the load, and back into the 'a' terminal, which is connected to the bottom of the C winding.  This would be equivalent to an open-wye, open-delta connection.

Then again, if B and C of the primary were open, the currents in the delta B and C windings would be zero and all of the load current would flow in the A winding.  This would be equivalent to a single-phase transformer.

It just shows that if the delta winding currents are known, you can calculate the line currents, but not vice versa.

stevenal (Electrical)
9 Jun 06 11:43
I agree that in the general case, winding currents do not necessarily sum to zero. However in this case, with a delta secondary feeding the load, winding currents will sum to zero. There is no source given for the circulating zero sequence current.

For jghrists example to occur; there must be a source on the delta side and a wye winding with unbalanced loading or fault. 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.


You cannot have a single phase load on a delta that involves only one leg unless you are speaking of that 4 wire delta that center taps one of the windings. Given only the three nodes of the original post, I assume this is a 3 wire delta. Single phase loads can only be connected l-l and therefore involve all three windings. Current consists of + and - sequence only. Winding current sums to zero to give no 0 sequence current.

With eng-tips MVPs challenging me on this, I've double checked my memory using Aspen. Grounded wye source with l to l fault on the delta side. No zero sequence current on either side. Memory confirmed.
waross (Electrical)
9 Jun 06 12:14
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.
stevenal (Electrical)
9 Jun 06 12:43
IA=60.2<-33.6 deg

jghrist (Electrical)
9 Jun 06 13:00
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.



IA=60.2<-33.6 deg
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 (Electrical)
9 Jun 06 13:20
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.
stevenal (Electrical)
9 Jun 06 14:05
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.

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 (Electrical)
9 Jun 06 16:14
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.

VAB = 120V < 0°
VBC = 120V < -120°
VCA = 120V < 120°

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:

IAB = 25A < 0°
VBC = 33.33A < -120°
VCA = 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 (Electrical)
9 Jun 06 16:32

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 (Electrical)
9 Jun 06 16:41
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..."

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 (Electrical)
9 Jun 06 16:51
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 (Electrical)
9 Jun 06 16:51
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 (Electrical)
9 Jun 06 17:10
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.

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 (Electrical)
9 Jun 06 17:13
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 (Electrical)
9 Jun 06 17:20

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.


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 (Electrical)
9 Jun 06 17:27
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 (Electrical)
9 Jun 06 18:13

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 (Electrical)
9 Jun 06 19:16
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 (Electrical)
9 Jun 06 19:29
Oops, I'll take back that last sentence, seems any three points can describe a circle.
stevenal (Electrical)
9 Jun 06 20:30
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 (Electrical)
9 Jun 06 21:50
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.
jghrist (Electrical)
10 Jun 06 0:18

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.  

waross (Electrical)
10 Jun 06 2:22
A single phase load current will divide equally between the two current paths in a delta transformer bank.
jghrist (Electrical)
10 Jun 06 7:18
Well, I've slept on it.  Forget what I said about solving the equations.  You meant to force the sum of the winding currents to zero, not the line currents.  Problem with the original PO terminology.  Actually, I tried this and get:

Iaw = (2·IA + IC)/3
Ibw = -(IA + 2·IC)/3
Icw = (IC - IA)/3

where IA, IB, IC are the line currents.

Using this with the line currents calculated in my resistor example, I get:

Iaw = 29.27 < -4.72°
Ibw = 33.68 < -111.79°
Icw = 37.58 < 116.33°

As far as sequential component analysis goes, I think it would apply as you say theoretically.  The tools available are inadequate for an arbitrary unbalanced phase-to-phase connected load, however.  I don't think Aspen could model this.  There are some unbalanced load sequence diagrams in the Westinghouse T&D Manual, but they do not allow all three phases to be different.
stevenal (Electrical)
10 Jun 06 9:56
No special tools are needed like the sequence connection diagrams. You just proceed this way: Do a delta wye conversion to find the wye equivalent of your delta load to find the line currents. Find the positive and negative sequence components of the line currents. To  move the sequence components into the delta windings, divide each component by sqrt(3) and shift it by 30 degrees. The positive sequence will shift in the normal direction while the negative sequence shifts in the opposite direction. Then reassemble the winding currents from the sequence components. I was required to do this so many times in college, that I can recite the procedure years later on a Saturday morning before coffee. For some reason I lack the desire to jump in and actually do the math. The same assumption applies, no zero sequence current circulating in the delta windings.
stevenal (Electrical)
12 Jun 06 19:31
I need to correct my previous postings. I wanted to use a delta wye conversion to make finding the line currents easier. This fails to work as intended, since the unbalance causes the wye point to shift away from neutral. Once I gave up on this plan and used node equations at the load end to find the line currents, shifting the sequence components as stated above yielded jghrist's results from his last post.
stevenal (Electrical)
13 Jun 06 12:31

Jghrist and I took it offline for a bit. We agree that the OP's starting point of winding kVA is suspect. A more likely starting point would be the line to line load connected kVA. If this was actually the starting point, my trig method of June 8 and 9 won't work.

Once the sequence component method to work jghrist's example is in a worksheet, it's easy enough to alter the problem to to solve for a single line to line connected load. Just raise two of the input resistances to a very high value, and the solution falls out: the in phase current splits evenly between the two paths. If currents are forced to be 120 degrees out in each winding, one transformer will carry the load as if the others were not present.

waross (Electrical)
13 Jun 06 23:27
Hi stevenal
I don't quite understand this statment.
"If currents are forced to be 120 degrees out in each winding, one transformer will carry the load as if the others were not present."
It is my understanding that the current in an open delta transformer connection is not forced 120 degrees out in each winding.
The voltages are 120 degrees out, but the current is in phase with the resultant voltage and that voltage is the vector sum of the 120 degree displaced winding voltages.
The proof is to solve the phase angle and voltage drops and resulting terminal voltage of a loaded open delta transformer bank and compare the result with a single phase transformer in place of the missing phase.
Rather than combining impedances, it is more understandable to compare voltage drops.
Just as the vector sum of the voltages in an open delta are equal to the original voltage but at a different phase angle, the voltage drops due to resistance, reactance and impedance, if added vectorily will result in equal voltage drops at different phase angles. The resultant phase angles will be found to be the same as the corresponding phase angles in the single phase transformer.
The open delta transformer will be found to be equivalent to the single transformer and if connected in parallel, (A closed delta bank) the current will divide equally.
The joker is that although the open delta is equivalent to the single transformer for the purposes of paralleling and/or current sharing the losses are double. This is one of the factors that limits this connection to 2/3 of the three phase KVA rating when a three phase generator is used on single phase.
stevenal (Electrical)
14 Jun 06 11:33

You need to read back to the beginning of the thread. A false assumption was made that the delta winding currents feeding an unbalanced load were unequal yet 120 degrees apart in phase. I was accepting that assumption briefly in order to disprove it. There is no magical force that keeps them at 120 degrees. I agree with you that current divides equally with the single line to line connected load.

stevenal (Electrical)
14 Jun 06 14:06
For you sequence component fans, here's how it's done:
The zip file contains an mcd and also an rtf for those without Mathcad. Jghrist's delta connected resistances are used, but any value of impedance may be substituted. I've deviated somewhat from the convention of the thread, and used both nodes to describe phase to phase quantities. I avoided the use matrices.
stevenal (Electrical)
14 Jun 06 14:19
Guess I should have tested the link. This one seems to work:
waross (Electrical)
14 Jun 06 19:10
Thanks stevenal
BlownUp (Electrical)
23 Jun 06 9:36
I found your post by looking for information on or a tool to do something very similar (mostly because I hate vector math).  I was unable to find anything and wound up putting together a spreadsheet to do the calculations for what I needed.  Here is that spreadsheet:

Maybe it will do what you need.  If not, you should be able to modify it.
7JLAman4 (Electrical) (OP)
23 Jun 06 10:17
I have also created a "panel transformer" worksheet that will calculate the line currents for each of the four types of transformer configurations (D-D, D-Y, Y-Y, Y-D) based on individual loads (Hp and/or KW single and three phase). I have received so much information from this post and emails directly that I had to create this.  The only limitation this worksheet has at the moment is not incorporating any centertaps from any windings.

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