Wye or Star vs the Delta Electrical System

kwired

Electron manager
Location
NE Nebraska
Such a sleight of hand, first you mention transformer connections, then you jump to voltage differences.
What "power" do you consider is wasted with:
Delta system at 240V
Wye system at 240Y/138V
I was thinking 1.17something for the square root of 3 instead of 1.7something. Brain fart.

But the difference between any two low voltage distribution voltages is just a difference in the windings ratio of a transformer supplying the end user. It doesn't mean that anything is being wasted.
Both points I seem to unsuccessfully be trying to convince to Ron from early on.

So the voltage is 32 volts less, nothing is "wasted" though. Some changes in the advantages and disadvantages have been created but nothing is lost.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
For my take on this, please consider a model system consisting of two _single_ phase generators sharing a common shaft, with some way of adjusting the phase angle between the outputs of the generators. The details are not important, the key features are:
1) shared shaft
2) matched output voltage (for our example, call it 120V)
3) matched excitation (perhaps these are permanent magnet rotors)
4) ability to adjust phase angle
5) the two coils are placed electrically in series.
6) the common shaft is connected to a constant speed prime mover (to run the generators)
7) we have a way of measuring the mechanical output of the prime mover
8) the series output of the two generators is connected to a variable resistive load, which we adjust for our tests

It is pretty clear that by appropriate adjustment of the relative phasing of these two generators, we can get any voltage between 0 and 240V.

Let us start with the phase angles adjusted to give 240V, and let us adjust the resistive load for a current of 100A. The load is consuming 24000W. Each generator is producing 12000VA and 12000W. The losses in each generator are those associated with producing 100A of output. The mechanical input to the generator set is the sum of the 24000W output and the losses in the generator set.

Now adjust the phase angles so that the output is very low, say 1V. Each generator is still producing 120V, but the phase angles have been adjusted so that the voltage of one 'bucks' the voltage of the other, leaving a net output of 1V. Adjust the resistive load for a current of 100A. Now the load is only consuming 100W. But each generator is still producing 120V, and still carrying 100A. Each generator is still producing 12000VA, and the losses in each generator are still those associated with producing 100A. Yet the electrical output of the _system_ is vastly reduced. Where does it go?

As gar mentions above, you need to consider power factor. What is the phase angle of the current with respect to the voltage. I am not going to go into the derivation, but in the system described above, the phase angle of the two generators is about 179.5 degrees, and the phase angle of the current flowing through each generator about 89.75 degrees out of phase of the voltage. (I'll leave it as an exercise to figure out 'leading' or 'lagging' and where :) When you have a power factor in a single frequency AC system, the voltage is not in phase with the current, and for part of the cycle the power flow gets reversed. In the example above, each generator sees 12000VA, with most of this 'apparent power' circulating back and forth between the two generators.

In this example the mechanical input to the generator is the sum of 100W of output and the losses in the generator set. Since the hugely poor power factor means that each generator still has the losses associated with 12000VA (perhaps 400 - 1600W of loss), you can see that having this power factor does impact the efficiency of the generator system. But you have not 'thrown away' the full 11900W of input; you have just 'thrown away' the _losses_ associated with having 100A of current flowing through 240V of coils, when you only have an output of 1V.

Getting back to the original question: if you have a wye system consisting of 120V coils, and you connect a 100A single phase resistive load across two of the terminals, then you have 12000VA being delivered by each coil, but only 20800W being delivered to the load. Even though there is a resistive load, and the current through the resistor is exactly in phase with the line-line voltage, each coil experiences a power factor because the current is _not_ in phase with the line-neutral voltage.

A similar power factor issue is seen in delta transformers. Consider a single phase resistive load connected between terminals A and B of a delta secondary. Some of the current for this load will be supplied by the A-B coil, but some of the current will be supplied by the C-A and B-C coils operating in series. (If you prefer, consider a load connected across the open side of an open delta source....) The current through the C-A and B-C coils will not be in phase with the voltage produced by these coils. The net result is that you see a power factor on each of these other coils, even with a pure resistive load.

The answer in both of these cases is to balance the three phase loading. With a balanced three phase resistive load connected to a three phase source (either wye or delta), the current flowing through the coils will be in phase with the voltage produced by the coils, and you don't 'lose' VA.

Hope this helps.

-Jon
 

kwired

Electron manager
Location
NE Nebraska
For my take on this, please consider a model system consisting of two _single_ phase generators sharing a common shaft, with some way of adjusting the phase angle between the outputs of the generators. The details are not important, the key features are:
1) shared shaft
2) matched output voltage (for our example, call it 120V)
3) matched excitation (perhaps these are permanent magnet rotors)
4) ability to adjust phase angle
5) the two coils are placed electrically in series.
6) the common shaft is connected to a constant speed prime mover (to run the generators)
7) we have a way of measuring the mechanical output of the prime mover
8) the series output of the two generators is connected to a variable resistive load, which we adjust for our tests

It is pretty clear that by appropriate adjustment of the relative phasing of these two generators, we can get any voltage between 0 and 240V.

Let us start with the phase angles adjusted to give 240V, and let us adjust the resistive load for a current of 100A. The load is consuming 24000W. Each generator is producing 12000VA and 12000W. The losses in each generator are those associated with producing 100A of output. The mechanical input to the generator set is the sum of the 24000W output and the losses in the generator set.

Now adjust the phase angles so that the output is very low, say 1V. Each generator is still producing 120V, but the phase angles have been adjusted so that the voltage of one 'bucks' the voltage of the other, leaving a net output of 1V. Adjust the resistive load for a current of 100A. Now the load is only consuming 100W. But each generator is still producing 120V, and still carrying 100A. Each generator is still producing 12000VA, and the losses in each generator are still those associated with producing 100A. Yet the electrical output of the _system_ is vastly reduced. Where does it go?

As gar mentions above, you need to consider power factor. What is the phase angle of the current with respect to the voltage. I am not going to go into the derivation, but in the system described above, the phase angle of the two generators is about 179.5 degrees, and the phase angle of the current flowing through each generator about 89.75 degrees out of phase of the voltage. (I'll leave it as an exercise to figure out 'leading' or 'lagging' and where :) When you have a power factor in a single frequency AC system, the voltage is not in phase with the current, and for part of the cycle the power flow gets reversed. In the example above, each generator sees 12000VA, with most of this 'apparent power' circulating back and forth between the two generators.

In this example the mechanical input to the generator is the sum of 100W of output and the losses in the generator set. Since the hugely poor power factor means that each generator still has the losses associated with 12000VA (perhaps 400 - 1600W of loss), you can see that having this power factor does impact the efficiency of the generator system. But you have not 'thrown away' the full 11900W of input; you have just 'thrown away' the _losses_ associated with having 100A of current flowing through 240V of coils, when you only have an output of 1V.

Getting back to the original question: if you have a wye system consisting of 120V coils, and you connect a 100A single phase resistive load across two of the terminals, then you have 12000VA being delivered by each coil, but only 20800W being delivered to the load. Even though there is a resistive load, and the current through the resistor is exactly in phase with the line-line voltage, each coil experiences a power factor because the current is _not_ in phase with the line-neutral voltage.

A similar power factor issue is seen in delta transformers. Consider a single phase resistive load connected between terminals A and B of a delta secondary. Some of the current for this load will be supplied by the A-B coil, but some of the current will be supplied by the C-A and B-C coils operating in series. (If you prefer, consider a load connected across the open side of an open delta source....) The current through the C-A and B-C coils will not be in phase with the voltage produced by these coils. The net result is that you see a power factor on each of these other coils, even with a pure resistive load.

The answer in both of these cases is to balance the three phase loading. With a balanced three phase resistive load connected to a three phase source (either wye or delta), the current flowing through the coils will be in phase with the voltage produced by the coils, and you don't 'lose' VA.

Hope this helps.

-Jon
I think you are going too deep, OP thinks because voltage of a 208 volt system is 32 volts lower than that of a 240 volt system that something was lost. Yes voltage is lower, but all that means is current has to go up to deliver same power. If that doesn't happen then we either have a reduced output or losses for reasons other than that the voltage is lower to begin with.

This is all disregarding line losses and transformer losses, which when finding those differences between 208 and 240 volt systems are not going to be that much different anyway.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
I think you are going too deep, OP thinks because voltage of a 208 volt system is 32 volts lower than that of a 240 volt system that something was lost. Yes voltage is lower, but all that means is current has to go up to deliver same power. If that doesn't happen then we either have a reduced output or losses for reasons other than that the voltage is lower to begin with.

This is all disregarding line losses and transformer losses, which when finding those differences between 208 and 240 volt systems are not going to be that much different anyway.
Perhaps; but I think that OP was specifically asking about the difference in efficiency of wye versus delta systems, not about the resulting voltage difference. I was trying to approach this question by trying to point out exactly what _is_ lost, so that Ron could then quantify it and see that it really is not all that much even in the worst case.

Because I like talking though this stuff, I will try to rephrase my points without the exercise :)

Jim's question of
What "power" do you consider is wasted with:
Delta system at 240V
Wye system at 240Y/138V
was a much more succinct way of probing the issue than my own exercise.

If you consider a single phase load connected to a wye source, then there is (of necessity) a non-unity power factor placed on the coils of that wye source. This is true even with a resistive load, and means greater losses in the coils than would be expected if you connected the same load, at the same voltage, to a corresponding single phase source.

Ron (OP) was under the impression that the reduced voltage directly corresponded to a waste of energy. This is not true. What is true is that the reduced voltage causes a reduction in real power delivered for a given apparent power flowing in the coils, with a much smaller increase in losses associated with the larger apparent than real power flow.

Additionally, the _same_ sort of power factor and apparent power losses is in fact seen in a delta system, because a while a portion of this single phase load current flows 'in phase' with one of the source coils (assuming a resistive load), a portion of the load current flows through the other two coils, where it is not in phase with the induced voltage.

If you consider other systems (open delta, for example) you can see that for _imbalanced_ loads different source configurations will see different coil power factors, and if we design transformers with the same amount of copper and steel these different configurations (wye, delta, open delta...) will result in different losses.

Finally, if you have a balanced three phase unity power factor load placed on a wye _or_ delta source, then the current flowing through the source coils is in phase with the voltage induced in those coils, eliminating any additional coil power factors caused by wye or delta configuration.

-Jon
 

kwired

Electron manager
Location
NE Nebraska
Ron (OP) was under the impression that the reduced voltage directly corresponded to a waste of energy. This is not true. What is true is that the reduced voltage causes a reduction in real power delivered for a given apparent power flowing in the coils, with a much smaller increase in losses associated with the larger apparent than real power flow.
That is kind of what I have been trying to clear up with Ron from very early in the thread. I also think he was under the impression that because the voltage in the wye system is lower that something is wasted, which simply is not true. Now there may be some relatively minor differences in efficiency but that really has nothing to do with the fact the voltage is 32 volts lower, which to me seemed to be the main focus he was getting after.

First thing is you can not say lower voltage equates to less power. They are not the same thing. If you disregard power factor and efficiency for now to keep it simple, you still need to factor in current which was not done in the OP. If you have 100 KVA on both systems, the difference between them is the current drawn, but you still have same power on both.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Perhaps; but I think that OP was specifically asking about the difference in efficiency of wye versus delta systems, not about the resulting voltage difference. I was trying to approach this question by trying to point out exactly what _is_ lost, so that Ron could then quantify it and see that it really is not all that much even in the worst case.

Because I like talking though this stuff, I will try to rephrase my points without the exercise :)

Jim's question of

was a much more succinct way of probing the issue than my own exercise.

If you consider a single phase load connected to a wye source, then there is (of necessity) a non-unity power factor placed on the coils of that wye source. This is true even with a resistive load, and means greater losses in the coils than would be expected if you connected the same load, at the same voltage, to a corresponding single phase source.

Ron (OP) was under the impression that the reduced voltage directly corresponded to a waste of energy. This is not true. What is true is that the reduced voltage causes a reduction in real power delivered for a given apparent power flowing in the coils, with a much smaller increase in losses associated with the larger apparent than real power flow.

Additionally, the _same_ sort of power factor and apparent power losses is in fact seen in a delta system, because a while a portion of this single phase load current flows 'in phase' with one of the source coils (assuming a resistive load), a portion of the load current flows through the other two coils, where it is not in phase with the induced voltage.

If you consider other systems (open delta, for example) you can see that for _imbalanced_ loads different source configurations will see different coil power factors, and if we design transformers with the same amount of copper and steel these different configurations (wye, delta, open delta...) will result in different losses.

Finally, if you have a balanced three phase unity power factor load placed on a wye _or_ delta source, then the current flowing through the source coils is in phase with the voltage induced in those coils, eliminating any additional coil power factors caused by wye or delta configuration.

-Jon
I don't think power factor had anything to do with the OP's question.
 

ronaldrc

Senior Member
Location
Tennessee
I don't think power factor had anything to do with the OP's question.





I took a trip yesterday and could not reply.

Thanks all for the replies in regard to this question, opinions are what I am after, not just defense on you being right and me being wrong. Like a lot of things taught in the Electrical Trade the students just take it for granted that it is right because the instructors and current literature say so, and know very well most of this is theory.

How many on this forum are willing to admit that before I brought this 32 volts up, that it had never entered your minds?

I said from the start I agree with all on here according to Electrical theory and laws, according to the Power Law nothing is
gained nor loss from the power of the loss of the extra 32 volts.

I said that I think the Electrical system has to be oversized to compensate for the loss of this 32 volts. Your own post are validating this. You say although the voltage is lower the out come of the voltage to current ratio takes care of this 32 volt difference by the load pulling more current to compensate which means the power supply has to be larger to compensate, But what if the 32 volts where not lost then the power supply would not have to be oversized.

We know that the 32 volts is lost because of the wye configuration.

I don't think its just me , but I think when 120 volts + 120 volts in series doesn't equal 240 volts there is a deficiency some where no matter how small which has to be compensated for.

Thanks Ronald :)
 

GoldDigger

Moderator
Staff member
Equipment does need to be larger to deliver more current. It is larger to deliver more power.
And if you are comfortable with vectors, the missing 32 volts is a non-issue.
Do you think that there is something missing from the perimeter of a circle because the circumference is 2 pi r instead of 8 r?

Sent from my XT1080 using Tapatalk
 
Last edited:

ronaldrc

Senior Member
Location
Tennessee
Equipment does need to be larger to deliver more current. It is larger to deliver more power.
And if you are comfortable with vectors, the missing 32 volts is a non-issue.
Do you think that there is something missing from the perimeter of a circle because the circumference is 2 pi r instead of 8 r?

Sent from my XT1080 using Tapatalk
I asked a very simple question do you think the Wye system is as Efficient as the Delta system?

I think no matter how small the defficiency is, in systems as large as our electrical systems they need to be addressed no matter how small if we know what causes them.


Do you think that there is something missing from the perimeter of a circle because the circumference is 2 pi r instead of 8 r?
No , but I do think if 120+120 =208 instead of 240 something is wrong.

Ronald :)
 

Besoeker

Senior Member
Location
UK
I took a trip yesterday and could not reply.
Trust that your trip was enjoyable/successful.

Thanks all for the replies in regard to this question, opinions are what I am after
The differences between a system designed to provide 240V and a system designed to give 208V are fact, not opinion.

How many on this forum are willing to admit that before I brought this 32 volts up, that it had never entered your minds?
Maybe not many because it's an irrelevance. It's not comparing apples with apples.

I said from the start I agree with all on here according to Electrical theory and laws, according to the Power Law nothing is
gained nor loss from the power of the loss of the extra 32 volts.
The 32V isn't a loss. One system operates at 240V. A different system at 208V

I said that I think the Electrical system has to be oversized to compensate for the loss of this 32 volts.
See above.


Your own post are validating this. You say although the voltage is lower the out come of the voltage to current ratio takes care of this 32 volt difference by the load pulling more current to compensate which means the power supply has to be larger to compensate, But what if the 32 volts where not lost then the power supply would not have to be oversized.
The 32V isn't a loss..............


We know that the 32 volts is lost because of the wye configuration.
The systems produce different voltages because they are different systems.

I don't think its just me , but I think when 120 volts + 120 volts in series doesn't equal 240 volts there is a deficiency some where no matter how small which has to be compensated for.
Perhaps this is the fundamental point you are missing.
With 240V, it is usually end to end on one single 240V transformer winding. The centre tap gives 120V on each half.
For 208V, you get 120V phase to neutral on each of three different windings. You can't add them arithmetically to get 240V because they are 120deg out of phase.

I don't know how to explain......but I'll give it a shot......

Suppose you have a couple of rods, each 120 inches long. Put them end to end like in a straight line like 12 and 6 on a clock. The measured distance between the two ends will be 240 inches.
Now take the one pointing to 12 and move it round to 2. The measured distance (in a straight line) between the ends will now be less than 240 inches. It fact it will be 208 inches.
Nothing is lost from either of the rods - they are still each 120 inches. But they don't add as a single straight line.

Thanks Ronald :)
I'm trying......at least Mrs B says I am....
:p
 

GoldDigger

Moderator
Staff member
No , but I do think if 120+120 =208 instead of 240 something is wrong.

Ronald :)
And so if I have a triangle whose sides are 120, 120, and 208, the side which is 208 has something missing?
The voltage numbers are just the lengths of the vectors. The "real" vectors include the phase information.

Here is one for you to think about in return:
In a three phase system with a balanced load the energy per unit time ( the power) is constant throughout the cycle instead of being pulsed. That has to be more efficient.
So why does anyone ever use single phase 120/240?



Sent from my XT1080 using Tapatalk
 

ronaldrc

Senior Member
Location
Tennessee
I'm trying......at least Mrs B says I am....
:p
You know I always appreciate all the little inserts.

And in your Diagrams you don't explain how we loose the 34 volts in the 208 example
I guess you think that would be senseless though.

A simple explanation though would be is that two of the 120 volt voltage windings are in series and additive producing 240, but at the same time the other 120 volt winding is subractive which results in 208 volts.

Thanks :)
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
A simple explanation though would be is that two of the 120 volt voltage windings are in series and additive producing 240, but at the same time the other 120 volt winding is subractive which results in 208 volts.
Could you explain your math logic? 120+120-120 equals 120, not 208.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
I took a trip yesterday and could not reply.

Thanks all for the replies in regard to this question, opinions are what I am after, not just defense on you being right and me being wrong. Like a lot of things taught in the Electrical Trade the students just take it for granted that it is right because the instructors and current literature say so, and know very well most of this is theory.

How many on this forum are willing to admit that before I brought this 32 volts up, that it had never entered your minds?
I think that lots of people have wondered about the why of the square root of 3 in three phase systems. The particular issue of the 'missing' 32V came up in a discussion several years ago, about the 'oregon fudge factor'.

I said that I think the Electrical system has to be oversized to compensate for the loss of this 32 volts. Your own post are validating this. You say although the voltage is lower the out come of the voltage to current ratio takes care of this 32 volt difference by the load pulling more current to compensate which means the power supply has to be larger to compensate, But what if the 32 volts where not lost then the power supply would not have to be oversized.

We know that the 32 volts is lost because of the wye configuration.
Here I disagree with you.

I specifically brought up power factor and showed an extreme case where something is lost by having two coils, not in phase, feed a single phase load. What is lost _must_ be understood in terms of the power factor of the loads on the individual coils. And I absolutely agree: if you connect a single phase load to two legs of a wye system, because of the power factor seen by the individual coils, the equipment will need to have larger VA capability than if the coils were in phase.

But once you understand that this 'loss of capability' is created by power factor in the individual coils, you will see two other important points:

1) You see the exact same loss of capacity in a delta system. In a delta system the output voltage is the coil voltage, so no voltage is 'lost'. But add up the _currents_ in the various coils. You will see the exact same percentage loss of total current versus coil currents. In a wye system the voltage is lower than the sum of the individual coil voltages, but terminal current equals coil current. In a delta system the voltage is the coil voltage, but the terminal current is less than the sum of the coil currents. (In both cases the 'deficit' is explained using a vector sum.)

2) If, rather than considering single phase loading, you consider three phase loading, when you have a balanced three phase load then nothing is lost, and the power factors in the coils is the same as the power factor of the load. With a balanced three phase load a balanced three phase source doesn't create any additional 'coil power factors'.

-Jon
 

Besoeker

Senior Member
Location
UK
You know I always appreciate all the little inserts.

And in your Diagrams you don't explain how we loose the 34 volts in the 208 example
You don't. It was never there in the first place.

A simple explanation though would be is that two of the 120 volt voltage windings are in series and additive producing 240,
Yes, in the first diagram.

but at the same time the other 120 volt winding is subractive which results in 208 volts.
No. They add. Just not in a straight line. I thought the rods analogy might have explained that.
 
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