3 phase Delta High Leg Single Phase Load

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mivey

Senior Member
I think I see what you mean. Instead of connecting the load between x4-x2 and x3-x1 you mean connect the load between x3 and x1. In that case you have zero volts:

X2-X4:
..mmmmmmmm....mmmmmmmm
.|.........\/.........|
.|.-120+.../\..-120+..|
x4________x2.x3.......x1
.............|........|
.............|........|
.............|__-0+___|

Same as if you did this:

X1-X3:
..mmmmmmmm....mmmmmmmm
.|.........\/.........|
.|.-120+.../\..-120+..|
x4........x2.x3_______x1
.|........|
.|........|
.|__-0+___|


Add: My office AC just went out. I think I'll send out for some free repair estimates and bids over the phone. :grin:
 
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ohmhead

Senior Member
Location
ORLANDO FLA
Well i was going to ask Mivey if his ac unit was on the high leg .



Larrys point made on the opposing currents in c winding kinda is my way of thinking but how would one test this ?

I also think it would be part of the phase shift of b/ cn to a or c in there place during ther cycle on single phase loads on this common point also not just the split in c winding but where there at in time by polarity to each end of each coil winding polarity wise .

Mivey you have cleared up the the phase part for me now .

Also i think most think that voltage is instantaneous i maybe be out of line but does it not rise in voltage thur windings meaning start at zero and rise to full voltage is this why 208 is found on B ? Meaning at that instant in phase shift we see just 208 because of shift in voltage of opposite windings and there polarity at that point .

I always wondered why its just 208 volts but not 1/2 the voltage of c winding and full voltage of b winding ?

If its 120 volts on c to center tap and when you add b winding to this its 208 volts why not 460 volts is it the phase angle or is it partly the reactance of the tap a loss of voltage or because its truely a neutral point of any phase and two paralleled windings a/b out of phase make 208 volts but why do we not get then 240 volts on B just 208 volts ?
 
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jim dungar

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PE (Retired) - Power Systems
I always wondered why its just 208 volts but not 1/2 the voltage of c winding and full voltage of b winding ?

We know it is 208V because of Pythagoras.

You have a right triangle with the short leg = 120V and the hypotenuse = 240V. The Pythagorean formula says that the hypotenuse of a right triangle is equal to the square root of the sum of the squares of the other two legs so with a little math we can get:

V?bn = V?ab-V?an = (240)?-(120)? = 43,200. Now we take the square root to find Vbn = 207.85 = a nominal 208V.
 

mivey

Senior Member
Well i was going to ask Mivey if his ac unit was on the high leg .
It was working fine until I moved it over.:grin:
Larrys point made on the opposing currents in c winding kinda is my way of thinking but how would one test this ?
How about a reasonability test? You have two parallel sources that are connected at two common terminals. Current goes out the first common terminal, through the load, and into the other common terminal.

Now, in between these two terminals, we have two parallel sources with the exact same voltage from common terminal to common terminal. Both sources have identical impedances. Current division logic says the current will evenly divide between the two sources and flow from one common terminal to the other common terminal. That means 1/2 of the current goes left and the other 1/2 goes right.
I also think it would be part of the phase shift of b/ cn to a or c in there place during ther cycle on single phase loads on this common point also not just the split in c winding but where there at in time by polarity to each end of each coil winding polarity wise .
I don't get what you are saying.
Also i think most think that voltage is instantaneous i maybe be out of line but does it not rise in voltage thur windings meaning start at zero and rise to full voltage is this why 208 is found on B ? Meaning at that instant in phase shift we see just 208 because of shift in voltage of opposite windings and there polarity at that point .
Did not follow all of that but the voltage does rise across the winding and if you tap into the winding somewhere in the middle, you will get a voltage that is somewhere between the voltage at the ends of the coil. We often do that by design.
I always wondered why its just 208 volts but not 1/2 the voltage of c winding and full voltage of b winding ?
Well, that is exactly what you get at any one instant in time. Don't get confused by the rms value of the waveform which considers a whole bunch of instants in time.

The reason you only get 208 volts is that the sinusoidal waveforms are shifted relative to each other. The maximum values for these two voltages occur at different points in time. If we look at a bunch of points in time, we will see that the biggest sum we obtain is plus and minus 293.94 volts. Plot all of these sums out and you will have plotted out a sinusoidal waveform that has an rms value of 207.85 volts.
If its 120 volts on c to center tap and when you add b winding to this its 208 volts why not 460 volts is it the phase angle or is it partly the reactance of the tap a loss of voltage or because its truely a neutral point of any phase and two paralleled windings a/b out of phase make 208 volts but why do we not get then 240 volts on B just 208 volts ?
See above
 

mivey

Senior Member
120<0? and 240<120? are just shorthand. Consider the following
120<0? = sqrt(2)*120*cos(ωt+0?) = sqrt(2)*120*cos(ωt+0?*pi()/180?)
240<120? = sqrt(2)*240*cos(ωt+120?) = sqrt(2)*240*cos(ωt+120?*pi()/180?)
where ω = 2*Pi*freq = 377

At t=0:
120<0? = sqrt(2)*120*cos(377*0+0?*pi()/180?) = 169.71
240<120? = sqrt(2)*240*cos(377*0+120?*pi()/180?) = -169.71
The sum equals zero.

You can do this as time continues on. Here are the results for t = 0 to 0.0175 sec:

Time=0.000000 sec, 169.71 + -169.71 = 0.00
Time=0.000833 sec, 161.40 + -252.23 = -90.83
Time=0.001667 sec, 137.29 + -310.07 = -172.78
Time=0.002500 sec, 99.75 + -337.55 = -237.81
Time=0.003333 sec, 52.44 + -331.99 = -279.56
Time=0.004167 sec, -0.01 + -293.93 = -293.94
Time=0.005000 sec, -52.45 + -227.10 = -279.55
Time=0.005833 sec, -99.76 + -138.03 = -237.79
Time=0.006667 sec, -137.30 + -35.46 = -172.76
Time=0.007500 sec, -161.40 + 70.59 = -90.81
Time=0.008333 sec, -169.71 + 169.73 = 0.02
Time=0.009167 sec, -161.40 + 252.25 = 90.85
Time=0.010000 sec, -137.29 + 310.08 = 172.79
Time=0.010833 sec, -99.74 + 337.56 = 237.82
Time=0.011667 sec, -52.43 + 331.99 = 279.56
Time=0.012500 sec, 0.02 + 293.92 = 293.94
Time=0.013333 sec, 52.46 + 227.08 = 279.54
Time=0.014167 sec, 99.77 + 138.01 = 237.78
Time=0.015000 sec, 137.31 + 35.43 = 172.74
Time=0.015833 sec, 161.41 + -70.61 = 90.79
Time=0.016667 sec, 169.71 + -169.75 = -0.04
Time=0.017500 sec, 161.39 + -252.27 = -90.88

So we can see that in real time, the voltage is a summation. The shorthand version uses vector addition so we can work with the rms values so we can see what happens on average, not just at one instant in time.
 
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ohmhead

Senior Member
Location
ORLANDO FLA
Well i understand better now Mivey & Jim thanks !

Heres a question?

Art 240.85 nec in relation to this subject kinda.



Slash ratings on breakers a crt brk that has a dual 208 /120v rating the smaller of the ratings is for over current at line to ground voltages .

meaning to be cleared by one pole of the device .

The higher of the two slash# /# ratings is for a line to line fault or overcurrent at line to line voltage this means a single pole breaker on a high leg would have to be fully rated at the lower level at 208 volts not 240 volts it would have to be at 208 volts or it would not operate correctly in a overcurrent condition properly meaning the time curve to trip would be effected so thats a problem with high leg use.

Its in the NEC code iam i out of line again or is this making any sense ?
 
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lefty

Member
Location
Oklahoma
What application was it that used the high leg to serve single phase loads and how was the feeder taken from the panel?

It was used to serve high bay lights.

Not sure what you mean. Are you referring to mixing 120/208 wye and 120/240 delta? What post were you responding to?
I was responding to this one, the neutral coming back from the highbay lights would have different voltage on it than say a neutral from a 120v single phase load.
 

ohmhead

Senior Member
Location
ORLANDO FLA
Well if you read my post above you can not use a dual rated voltage breaker on a open delta or a closed delta its not NEC code and its not electrically safe thats a fact .

Subject is voltage on high leg used for single phase load or loads off of high leg this is the reason you can not do it is not safe and try and purchase a 208 volt single pole breaker with low rating of just 208 volts i have not found one yet .



Single phase single pole breakers only come with a dual rated voltage listed on breaker if you find a 208 volt show me ill buy it .
 
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ohmhead

Senior Member
Location
ORLANDO FLA
a straight rated breaker of 240 should be fine for the 208 loads.

Well no actually not line to ground must be 208 volts not 240 volts ill find this in print for you if you like but its a fact the low end must be that L G voltage not above that voltage point line to line voltage must be full voltage of phase to phase .

This in nec code and also common from most breaker manufactures ill find you the web page .
 
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mivey

Senior Member
What application was it that used the high leg to serve single phase loads and how was the feeder taken from the panel?

It was used to serve high bay lights.
So you served the single-phase high bay lights from the 208 volt high-leg of a 120/240 three-phase panel? I've never seen it done. What breaker did you use?
the neutral coming back from the highbay lights would have different voltage on it than say a neutral from a 120v single phase load.
Not sure what you mean. It is tied to the same neutral point at the panel.
 

ohmhead

Senior Member
Location
ORLANDO FLA
a straight rated breaker of 240 should be fine for the 208 loads.

Well Nakulak i must say you are correct i was only looking at it from one side most single breakers are dual rated there really common and easy to get but looking at it more closely yes you can but it must be at a higher voltage and i also found its not all breakers generally used but then we cant load it with 16 circuits not a good thing just wanted to say you were right in your post i was wrong i guess i was thinking to make everyone aware of the difference and not thinking that they do make single phase 240 volt breakers .
 

a.bisnath

Senior Member
a quote of interest

a quote of interest

The book "Electrical power and measurements" by Alan Symonds,chapter 6 -transformers page 108 under heading ,Delta Connection

"The delta winding is more effective for unbalanced loads"
 

ohmhead

Senior Member
Location
ORLANDO FLA
The book "Electrical power and measurements" by Alan Symonds,chapter 6 -transformers page 108 under heading ,Delta Connection

"The delta winding is more effective for unbalanced loads"

Well to me it can be done but is it the best use of doing electrical work in my limited time with delta work no .

Your using a leg at a lower voltage but other equipment used on ABC will be 240 volts or should be so now to me its a load or resistance thing and a higher voltage to ground i feel loads should match voltage for balance on each phase .

Your taken up empty space and thats good but there can be a unbalance if you get out of control loading it up what electrician does that after 10 years so the next guy adds more to it .

Your now installing a breaker with a higher voltage rating on a lower voltage system which to me is code but i personally dont like its not at 32 volts rms .

As breakers are tested at rated voltage and current to trip oh yes it will trip but how long will it take to trip meaning thermal time to voltage time slash ratings were design to be used in common panels with nominal voltage to ground points .

And 208 volts to 240 volts talk about voltage drop lets send that in to the engineer for your fault study calculations i know we have many transformers that change voltage thur tapping but .

What i wonder about is conductors in conduit AC opposite phases cancel out so if you had this B high leg running with your other phases and its 90 degs out of phase with the other phases how does it cancel out in long runs of conduit would it be a issue ?
 
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neity

Member
You know now that we're on page ten I'd like to at least say thankyou everyone for all the thought that has gone into this. I've been closely watching this thread and I didn't want to jump in because I've been trying to hash this one out since a few weeks before I posted with my own conclusions on the issue. This is like getting a second set of eyes on the problem
One thing I want to clarify especially regarding the breaker/ratings posts is that this is probably a bad idea in the first place, simply because yes the next electrician who works on that panel may not be expecting this. I of course would say it shall be clearly marked but whens the last time you trusted a panel schedule you didn't personally create after you just did the work? I trust my fluke. So yes, currently, the situation is dangerous.
I really want to get this issue resolved and I think it belongs in the code. My thoughts on this:

1. You're going one and a half windings, but we're not trying to get a leed point for efficiency. just build a bike rack.

2. So far no clear and concise argument has been made to me that it won't "work"

3. This should be in the next edition of the code.

This forum is great because we have all sides of the industry here. Obviously some very skilled engineers and electricians. I'm dying to know who blows up their air conditioner/PowCo pole mount first though.

I still can't wrap my head quite around the flux issue though. I believe as mivey clearly shows in his plot that we will get a sustained voltage of 208rms on the high leg...not counting those collapsing fields on the windings...

We all know that if we had a delta 3w that we get nice revolving current. We center tap a phase we still have that as long as we don't load anything 1ph. We know that we can go 120 legs to neutral across half of either winding. The current will obviously flow linearly to neutral from (ok for conventions b is high today) A or C (or conversely). It takes the path of least resistance (well impedance in our case) to get back home. Here is where the "stability" factor steps in. I get worried because to me I see two seemingly parallel paths. Manufacturing tolerances are just that. Tolerances.

I'm beginning to think (and I may be totally wrong) that given a certain size transformer with a certain impedance and a combination of loading would completely accept that 208V single phase load. I think with larger transformers, the windings Hysteresis curve might lead it to decay too slow, leaving a higher voltage potential from other loads on said winding which would possibly then cause a massive voltage drop and subsequent current surge.

We can always use superposition to calculate known values, but has anybody ran into a ferroresonance problem before? I had some "high efficieny" underground fed 35kV rated transformer blow up... and then the replacement blew up. There is really no practical way to calculate a solution or solve the problem besides precautionary measures.
So I guess my question is, can it be proven that in any situation a 208V single phase load on a delta high leg transformer COULD fail. Only one example is needed. We avoid ferroresonance but it is never a known issue just more or less likely to happen. Because this is more of a myth/mystery, and no electrician would do it anyway, I think it deserves its place in the NEC, with proof of a possible cause of failure within normal operating conditions.

That being said, I thank everyone again for helping me figure this out. And sorry for the lengthy and or completely wrong post.

-Jeff
Philadelphia, PA
 

ohmhead

Senior Member
Location
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Well Neity it was fun looking at the problem check this out http://ecmweb.com/mag/electric_understanding_circuit_breaker/?smte=wl

I get going and cant stop now i wonder about this b phase i have seen it three times in the trade closed and one time i worked with a open delta but today we dont see it much .

You have a good question ?

If i can find a good point with facts ill try but i have not found solid information yet .

The only point is slash breakers our mostly single phase .

But i dont give up easy . Take care as a electrician i try to understand what they didnt teach in school!
 

ohmhead

Senior Member
Location
ORLANDO FLA
Well Jeff look at it this way
100_0333.jpg





We know you can not hook up a delta source to a wye load its a unbalance condition thats why the closed delta can only be a load on only the c winding its not a neutral its just a path of return current to phases and splits by hooking up a single ph load to A/B to C .

But when we connect b phase we see a wye connection do you agree comments ?


And a wye attached to a delta would effect current flow and create and uneven balance thur out the transformer as a delta is not in phase its 120 deg out of phase with other phases its really a 3 deal seperate transformer current must return on opposite ends of windings its a circle of flow .

If one looks at the picture sorry but there resistors kinda you can see a WYE connection from hi leg to a /c split winding a common tap but a wye thats not made to be use with a delta the hi leg.
Its off voltage and uneven return path is the key to why its not normally used . And Larry was correct its a reactance and can heat up winding c
 
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jim dungar

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Location
Wisconsin
Occupation
PE (Retired) - Power Systems
But when we connect b phase we see a wye connection do you agree comments ?
No.

The source configuration is not dependent on the load configuration.

Yes, a B-N connection might look like a wye, but it does not function as a wye.

Most of your discussion and questions seem to forget that there is a 3-phase angular relationship created by the transformer primary and secondary side connections. This 3-phase relationship of the A-B winding and A-N-C windings is not changed by the way the load is configured. All of the 3-phase circuit analysis formulas (Kirchhoff, Norton, and Thevinan) still apply and must be used.
 

ohmhead

Senior Member
Location
ORLANDO FLA
Well i dont agree with load does not effect phase to voltage and transformation from a delta source to a wye which is commonly not done ever in electrical .

Kirchoffs law i have read its kinda phasor sum of three line currents if[ equal ] key word these line currents if equal and displace by 120 deg the phasor sum of three equal phasors 120 deg apart equals zero. Neutral is zero

In a delta system current does not circulate around the delta .

The current in each leg is traveling in a different direction at 120 deg . do you agree ?

A electric circuit has to be complete at a start and a finish end point for flow of current a delta to me has no neutral point the return is the opposing phase there is no return to ground point and in some cases its a higher voltage .

By using the connection to the high leg your creating a unbalance which effects voltage current and its reactive to currents passing thur other phases .

Dont get me wrong iam here to learn but also to explain my thoughts on how i think it works and why its not proper to hook to the high leg we do it but its not correct in any load condition .


Also if you hook up a wye winding motor to a delta source it will burn up motor and even heat up the transformer to damage the winding insulation by reactance and its a drop on winding voltage output .

C phase is for single loads 1/2 ab because they are common to return on that winding but b is not its out of place .



No one has talked about Jeffs question on ferroresonance yet but in a delta its a bigg issue and thats another post i like to be at .

So comments let me buckle up my seat belt :D
 
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