ECN Forum Forums Technical Discussion Electrical Theory and Applications 2 phase neutral current?

This is a very interesting thread. I have been out of town doing a round of inspections in one of my communities, if I had of had acces to this forum I would have jumped in.

Let me take a stab at this...

Given... a three phase 120/208 volt 4 wire system (wye) system. What we have is...

Three separate sources whose instantaneous voltage levels are all 120 degree out of phase relative to each other. One end of each source is tied together to form the star point of the wye configuration.

Given a 1000 watt resistive load connected across phase A and phase B, what is it the resistor is actually connected across?

It is connected across two voltage sources connected in series, namely... the a phase coil and the b phase coil.

But what if any phase difference exists across the resistive load? By definition "phase difference" must describe the difference in instaneous voltage levels of "two" voltage sources. As such, the term "phase difference" can not apply here as the resistive load sees only one voltage source and not two sources.

What our resistive load sees is one voltage source whose "steady state" voltage is equal to 208 volts. The 208 volts is the sum of the instaneous voltage of the two sources connected in series.

If we look further at the two seried sources that forms the voltage source our resistive load sees, what we see is a difference in the rate of change of the voltage potential in one source relative to the other source of 60 degrees for a total of 120 degrees acros the pair.

As such... the only "phase difference" that exists in this circuit is the difference between (Phase A to Neutral) relative to (Phase B to Neutral).

I hope I have explained this in way that helps the thread.

[This message has been edited by Rick Kelly (edited 03-11-2006).]

Hey guys, hope I didn't upset anyone! Thanks for the answers

The closest thing I could relate it to in Australia is some times we have centre tapped control transformers that come out with 2 55v legs and one 0v in the centre (I would assume both 55's are out by 120 )
This gives us 110v between them , but generally we make one 55 leg the neutral and earth it as other wise we have a 'live' neutral of 55v all the time

this picture is similar to a question I am trying to answer, if it makes it easier to understand .....so will both formulas work?
I got 4.7A using the 3 phase formula
(don't have a scientific calculator on me right now to work out the other!!)



[This message has been edited by Sparkee (edited 03-11-2006).]

[This message has been edited by Sparkee (edited 03-11-2006).]

Sparkee... you first have to identify to us exactly what you have.

On first glance what you show us in the diagram is a single phase 120 volt three wire system, or what is commonly refered to as a "split phase" system

But then you mention "This gives us 110v between them , but generally we make one 55 leg the neutral and earth it as other wise we have a 'live' neutral of 55v all the time". This sounds to me like a three phase corner grounded delta system.

If it is in fact a three phase corner grounded delta system then yes... the three phase calculation is the correct one to apply to get the proper neutral current.

If it is not three phase and is in fact single phase then, no, the three phase caluclation should not be used to calculate the neutral current. If it is single phase the neutral current would be the unballanced draw from both sides of the system, or 2.08 amps.

Another thing... if it is single phase, there is no "phase difference" between the two 120 volt sources. What they are is two "in phase" voltage sources connected in series so that their voltages are additive. This is the only way possible for the combined voltage level across both sources to equal 240 vac (or the 110 volts from the text of your last post). Some people like to say that split phase legs are 180 degrees out of phase with each other, but that is not a proper or true description of what we have. Think DC for a minute. Take two 12 volt car batteries, to get 24 volts, you connect the batteries together so that the negative pole on one battery is connected to the positive pole on the other battery and you get 24 volts. There is no "phase difference" here. The exact thing applies on AC. It just looks like there is a difference.

And do not worry about upseting anyone, this is just another part of the learning process inherent in our trade, someone taught me, if I can spread around what I know, so much the better.

(Winne)
"If you look at this from the other side: consider a 120/208V wye system, and place a single phase 16 ohm resistive load between phase A and phase B. 13 amps will flow through this resistive load, and 2704W will be delivered to the load. But for 13A to flow through the load, 13A has to flow through both phase A and phase B. This means that 1560VA is being supplied by each phase. If 3120VA is being supplied, but only 2704W delivered to the load, then there has to be a power factor somewhere "

(Bob)
I do not know where you get the 1560 VA or the 3120 VA. None of the values in the problem when properly used can produce this figure.
Power = I²xR = 13²x 16 = 2704 watts.
Power = V x I = 208 x 13 = 2704 watts.

Where did you get these numbers?

(Winnie)
The phase A wye connected transformer coil is delivering 120V phase to neutral. 120V * 13A = 1560 VA. Same for phase B.
------------------------

Guys... stop thinking of a single phase load as being connected across two phases. It is not so. It is connected across two voltage sources in series. The current flow is supplied by both sources equally.

Winnie... in this discussion the face that both phase a and phase b is delevering 120 volts phase to neutral is not important as the load is not connected to the neutral.

Think of this from the perspective of DC for a second. Two 12 volt car batteries connected in series so that positive pole is connected to negative pole to give us 24 volts... both batteries supply the full current.

Ohhh... I love these threads...

Rick,

Please note that there are now two discussions going on. The original poster's discussion of two phase to neutral loads on a three phase wye system (as I understand his post; I could be wrong) and how to calculate the neutral current in this case. The second discussion is what happens when you have a single phase load connected between two phases of a three phase wye system.

For the latter discussion: It is entirely reasonable and correct to consider this to be a single phase load supplied by two voltage sources connected in series. However it is essential when doing the addition to consider the relative phase angle of these two voltage sources. If the two voltage sources are in phase, then you can simply add the two voltages together, and you have the 'split phase' system that you described. But if there is any phase angle between the two sources, then the resulting total voltage will be less than the sum of the individual voltages.

In a conventional 120/208V wye system, the voltage produced by the individual transformer coils are _120V_. I connect these two 120V sources in series, and low and behold I get 208V placed across my single phase load. Since both sources are in series with the load, both sources supply the full load current. This means that the total VA supplied by the two sources is greater than the total W consumed by the load. Entirely irrelevant as far as the load is concerned, but clearly an issue when sizing the sources.

To the original poster: in the example diagram that you just gave, you have 400W at 120V on one leg and 150W at 120V on a separate leg. As you've drawn things, this could either be a single phase split system (240V split to 120V each leg) or a 208V wye system (208V three phase with 120V each leg). In the latter case, use the formula that JBD posted at the top of this thread. First calculate the current in each phase, then plug those number in to the formula. The result is rather less than 4.7A.

-Jon

Hey Jon...

(Winnie)

For the latter discussion: It is entirely reasonable and correct to consider this to be a single phase load supplied by two voltage sources connected in series. However it is essential when doing the addition to consider the relative phase angle of these two voltage sources. If the two voltage sources are in phase, then you can simply add the two voltages together, and you have the 'split phase' system that you described. But if there is any phase angle between the two sources, then the resulting total voltage will be less than the sum of the individual voltages.

In a conventional 120/208V wye system, the voltage produced by the individual transformer coils are _120V_. I connect these two 120V sources in series, and low and behold I get 208V placed across my single phase load. Since both sources are in series with the load, both sources supply the full load current. This means that the total VA supplied by the two sources is greater than the total W consumed by the load. Entirely irrelevant as far as the load is concerned, but clearly an issue when sizing the sources.

-------------------------

I completely agree with the above parragraphs.

But... remember that both sources do not supply the same number of volt-amps. The level of the voltage in each source determines how many VAs that source contributes to the total supplied to the load.

[This message has been edited by Rick Kelly (edited 03-11-2006).]

Winne 1ST posted
"If you look at this from the other side: consider a 120/208V wye system, and place a single phase 16 ohm resistive load between phase A and phase B. 13 amps will flow through this resistive load, and 2704W will be delivered to the load. But for 13A to flow through the load, 13A has to flow through both phase A and phase B. This means that 1560VA is being supplied by each phase. If 3120VA is being supplied, but only 2704W delivered to the load, then there has to be a power factor somewhere"

Winnies 2nd post
(Winnie)
The phase A wye connected transformer coil is delivering 120V phase to neutral. 120V * 13A = 1560 VA. Same for phase B.

Winnie and all
You MUST understand that yes there is 13 amps on phasse A and on phase B. But the 16 ohms is connected phase to phase and not phase to neutral. You could connect it phase to neutral but then you have to deal with the currents being out of phase. You CAN NOT add the 1560 VA together. It is a VECTOR ADDITION.
I do not know of any other wy to explain it to you. I suggest you get with some one in your area who can go through the technical aspects of vector addition. If you will just look at the site I gave you you will see how the phase look and how they add together. Simple addition does not work here.

Bob,

You have me very confused here.

You keep saying that vector addition is necessary.

You keep quoting me and saying that I can't use simple addition.

But I _am_ using vector addition. I _said_ that 1560 VA is being delivered by leg A to the example load, and that 1560VA is being delivered by leg B, yet only 2704W is being delivered to the load. That answer comes from doing the vector analysis of the situation. The whole concept of volt amperes being different from watts is deeply rooted in the vector approximation of AC circuits.

The whole point of saying "2x1560VA being supplied by the transformer, yet only 2704W being delivered to the load" is an explicit example that phase angle differences are present in this system and must be accounted for using vector math. Clearly, 13A is flowing through leg A transformer coil, and 13A is flowing through the leg B transformer coil. Thus the heating experienced by the _transformer_ is what one would expect from having 1560VA per phase supplied by that transformer. But the power being delivered to the load is only 2704W, and the _difference_ between 3120VA and 2704W is explicitly caused by phase angle differences.

One must with the _voltages_ supplied by the leg A and leg B transformer coils. To determine the composite voltage from leg A to leg B you must use vectors (or some other analysis technique that accounts for the phase difference). The voltage between legs A and B is a vector quantity. Once you've calculated the voltage from leg A to leg B you can calculate the current flowing through the load.

Do the math yourself.

Define the leg A to neutral voltage as your phase zero current.
The leg B to neutral voltage is at phase angle 120 degrees.

Now calculate the phase angle of the leg A to leg B voltage. It is neither 0 nor 120 degrees.

The current through the resistive load will be in phase with the leg A to leg B voltage calculated above.

This means that the current in the leg A transformer coil is _not_ in phase with the voltage across the leg A transformer coil.

-Jon

Rick


sorry to confuse but I was just relating the 110v centre tapped control transformers we sometimes have here in Australia as the closest thing I could relate the 120/208 system (even though they are both different)

the diagram above should read 208 between both 120v legs

it didn't have anything to do with the diagram or original post

[This message has been edited by Sparkee (edited 03-11-2006).]

winnie I was told the 2 phase 120/208 system was common in most residences in canada if that helps clarify (assuming it is the same in the states)

I grabbed this from another thread


[This message has been edited by Sparkee (edited 03-11-2006).]