3 phase acting like single phase?

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I don't see anything wrong with the initial measurements.

First, there is only ONE phase. (Hook a scope up and see for yourself, there will only be one sine wave) The neutral that exists between L1 and L2, will read as a center tapped single phase system because that is exactly what it is. Only when the third line is added, making a three phase system, does phase angle come into play.

A three phase system is three single phase pulses sent at different time intervals. A phase consists of a combination PAIR of hot wires. L1,2 one phase. L2,3 another phase. L3,1 third phase. Each pair is only ONE phase.
 
K8MHZ said:
The neutral that exists between L1 and L2, will read as a center tapped single phase system because that is exactly what it is. Only when the third line is added, making a three phase system, does phase angle come into play.
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In this case, however the third line _is_ present, at least at the transformer. You have a three phase system, from which you have tapped L1, L2, and N.

As you say, any pair of wires is just a single phase. L1 to L2 is a single phase, L1 to N is a single phase, and L2 to N is a single phase.

In the system being considered, those three 'single phases' are all different. In the system being considered, the equation used to calculate the neutral current is the same as that of a three phase wye system where one of the legs is always zero.

In a true single phase center tapped system, those three 'single phases' will all be the same or related by inversion.

-Jon
 
The tough concept to accept is that although the two 120 volt legs don't add up to 208 volts, the amount of current will be exactly additive on the neutral.

Current is a measure of electrons *in any direction* passing a *single* point over time.

Voltage is a measure of potential of electron flow in a determinable direction with respect to two distinct points and has nothing to do with the movement of electrons.
 
K8MHZ said:
The neutral that exists between L1 and L2, will read as a center tapped single phase system because that is exactly what it is.
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This case is not like a center-tapped transformer. Winnie explained it but I did not want this quote to slip by an unsuspecting reader and have them think it was true.
K8MHZ said:
...the amount of current will be exactly additive on the neutral.
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Please elaborate. You are not meaning a scalar sum as opposed to a vector sum are you? In this type system, I can prove by example that you can have:
Ia + Ib = |Ia| = |Ib|

oops: corrected typo
 
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mivey said:
This case is not like a center-tapped transformer. Winnie explained it but I did not want this quote to slip by an unsuspecting reader and have them think it was true.Please elaborate. You are not meaning a scalar sum as opposed to a vector sum are you? In this type system, I can prove by example that you can have:
Ia + Ib = |Ia| = |Ib|

oops: corrected typo
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I tried to go back and edit my post but was too late. That is why I added the second post.

RE: Vector vs. scalar, the sums I was referring to were those measured by the OP. I probably missed my mark, but I was attempting to explain why the OP's measurements were not in error.
 
K8MHZ said:
...I was attempting to explain why the OP's measurements were not in error.
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But there's the problem. With the wye system and the 2-pole breakers off, you would initially expect about 7.2 amps on the neutral.

The 6 amps would indicate meter inaccuracy, reactive loads, or a parallel neutral path.

I'm eliminating the standard "don't believe the data" clause because I am trusting the OP. If we disbelieve the data, we can have the simplest explanation: One 2-pole breaker was left on that was supplying a 2 amp load. This would leave 6 amps for the remaining line-neutral loads and would match the neutral current.
 
As long as I'm throwing out the data, how about another simple explanation: This is not a wye system and the voltages are odd and it really is a center-tapped neutral. Then we have a different mystery to solve: Why do you have 120/208 voltage readings?
 
mivey said:
But there's the problem. With the wye system and the 2-pole breakers off, you would initially expect about 7.2 amps on the neutral.

The 6 amps would indicate meter inaccuracy, reactive loads, or a parallel neutral path.

I'm eliminating the standard "don't believe the data" clause because I am trusting the OP. If we disbelieve the data, we can have the simplest explanation: One 2-pole breaker was left on that was supplying a 2 amp load. This would leave 6 amps for the remaining line-neutral loads and would match the neutral current.
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You lost me there....2 amps on each leg would cancel out and result in 0 neutral volts as there would be no imbalance.

24 amps on one leg and 34 amps on the other would return 10 amps on the neutral.

22 amps on one leg and 32 on the other (the suspected 2 pole breaker w/ 2A load now turned off) still returns 10 amps on the neutral.

Even with your explanation, I see no problem with the OP's measurements.
 
K8MHZ said:
You lost me there....2 amps on each leg would cancel out and result in 0 neutral volts as there would be no imbalance.

24 amps on one leg and 34 amps on the other would return 10 amps on the neutral.

22 amps on one leg and 32 on the other (the suspected 2 pole breaker w/ 2A load now turned off) still returns 10 amps on the neutral.

Even with your explanation, I see no problem with the OP's measurements.
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I'm talking about the 8/6/2 amp case with the 2-pole breakers turned off except for the "overlooked" 2 amp 208 volt load:
L1 = 6 amp L-N load + 2 amp L-L load = 8 amps
L2 = 2 amp L-L load = 2 amps
N = 6 amp L-N load = 6 amps

As for your data:
On a wye system, without reactive loads (the initial, simplified expectation), 24 + 34 = 30.3 amps, NOT 10 amps. And, 22 + 32 = 28.4 amps, NOT 10 amps.

You keep treating this as a center-tapped transformer, which it is not (according to the OP).

add: I left off the vector notation on the 24, 34, 22, 32 amps but it makes the point stand out IMO
 
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mivey said:
As long as I'm throwing out the data, how about another simple explanation: This is not a wye system and the voltages are odd and it really is a center-tapped neutral. Then we have a different mystery to solve: Why do you have 120/208 voltage readings?
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The voltages are odd?

As far as current is concerned, the measured components are a single phase system.

Here is the deal. MEASUREMENTS indicate 120 V L/N and 208 V L/L and the current measurements the OP posted. Plain and simple. I don't see an issue with it. Perhaps I am reading too much into your replies but it seems to me you are saying the measurements have to be wrong or were somehow taken incorrectly.

My lack of a universally understandable explanation does not negate the facts.

The facts are the measurements. I see no reason to try to prove they are wrong. If I wanted to try to prove they were correct, I would simply use another meter to verify the measurements.
 
K8MHZ said:
The voltages are odd?

As far as current is concerned, the measured components are a single phase system.

Here is the deal. MEASUREMENTS indicate 120 V L/N and 208 V L/L and the current measurements the OP posted. Plain and simple. I don't see an issue with it. Perhaps I am reading too much into your replies but it seems to me you are saying the measurements have to be wrong or were somehow taken incorrectly.

My lack of a universally understandable explanation does not negate the facts.

The facts are the measurements. I see no reason to try to prove they are wrong. If I wanted to try to prove they were correct, I would simply use another meter to verify the measurements.
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My original premise was to NOT say the data was wrong but to try to explain why the data is different from what one might expect from a simple load. There is insufficient data to answer that question. I thought that was the point of the thread.

If you had a 120/240 volt 3-wire service, and read L1 = 8 amps, L2 = 2 amps, and N = 3 amps, wouldn't you be curious as to why?

There is no single course of trying to disprove the data, but rather to explain the data. The OP was curious as to why the data looked curious, as are many of us here.

You do not think the data is curious, which is curious in itself.
 
This establishment, which happens to be a hotel, is most certainly and without a doubt fed by a 120/208Y transformer. There are two transformers that I could see on this property and both are 120/108Y's. At the main disconnect that feeds the panel that feeds the sub panel, I read (nominal)120V L-N on each leg, and 208V L-L. At the sub panel in the pool pit, the voltage measurements were slightly less due to a small voltage drop, but extremely close to the MD readings. I could live with the 'meter error' explanation if I simply went with the 8/6/2 readings, but, like I said, I rearranged the loads, and the trend that followed behaved like a single phase system. The neutral current should have gotten closer to the line currents. The neutral current got closer to zero the more even the line currents were. My meter might be inacurrate, but I believe it to be at least consistent.

FWIW, I have observed this exact same scenario on another job behaving exactly as I would expect.
 
K8MHZ said:
This is incorrect as your meter has indicated.

The more balanced the load, the less current on the neutral.
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You're absolutely right: I can't argue with the numbers.

But, for an example, lets change the situation a bit.

Let's say that instead of a 3 wire feeder, it is now a 4 wire feeder to this sub panel, and the sub panel is 3 phase. Let's also say that the 120V L-N loads happen to be arranged so that L1=8A, L2=6A, and L3=0A, there is unity power factor, and that there are no 2 or 3 pole loads on. Should I expect the neutral load to be closer to 7A or closer to 2A? Does the absence of current on L3 mean that L1 and L2 are now single phase?
 
Barjack said:
Should I expect the neutral load to be closer to 7A or closer to 2A?
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7a. The lines would have to be at 180 deg. for the neutral current to be 2a.

Does the absence of current on L3 mean that L1 and L2 are now single phase?
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No. They always were 1-ph loads. In a Y system, the 1-ph loads are basically independent, except for affecting the neutral current.


Simple explanation: Suppose you have three 10a 1-ph loads from your 3-ph panel. L1=10a, L2=10a, L3=10a, N=0a.

Now reduce the load on L3 to 5a. Now, L1=10a, L2=10a, L3=5a, N=5a. Each amp L3 reduces increases the neutral by 1a.

Now say you have no load on L3. Now, L1=10a, L2=10a, L3=0a, N=10a, and not 0a.
 
This would be a fun project to vary the loads on a 120 degree system and measure the resultant currents and see how they add up. From there I think we may be able to create a formula that would explain it to all of us.

The guy that come up with this was born over 150 years ago. His 100 plus year old patents are still electrical engineering wonders to this day.

One of my bosses prefers a more simple explanation. PFM.

Engineers attempt to predict the outcome of our attempt to control nature and it's forces. Technicians become the litmus test and often the messenger is shot for passing on the truth. For some reason glorified inaccuracies are easier to accept than proven truth.

Did you know, that according to the label on spray vegetable oil there is 0 percent fat? Nice job for a tube full of 100 percent fat.

Sorry about the food reference, I am getting hungry.

I ran across a straight forward simple formula for figuring three phase neutral loading that was in excel format and I can't find it now.
 
K8MHZ said:
This is incorrect as your meter has indicated.

The more balanced the load, the less current on the neutral.
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Your statements make it appear you have a fundamental misunderstanding of how a 3-wire network and a 3-phase wye work.

I will be glad to help you gain clarity if you are interested. There are probably many here who would volunteer as well, and probably already have some nice graphics and text prepared as this is not the first time this topic has come up. No slight to you at all, as this can be confusing for a lot of people.

K8MHZ said:
This would be a fun project to vary the loads on a 120 degree system and measure the resultant currents and see how they add up. From there I think we may be able to create a formula that would explain it to all of us.
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No physical project is needed as you can run the numbers for a wye system using math or a combination of math & graphics. The formulas were created long ago and don't need to be re-invented.
 
K8MHZ said:
The more balanced the load, the less current on the neutral.
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As Mivey pointed out in post # 30, this is simply not true of a wye connected neutral when only using two phases.

If A-N is carrying 10 amps and B-N is carrying 10 amps the neutral is carrying 10 amps as well

SQRT (I?A + I?B) - (IA x IB)

SQRT (10 x 10 + 10 x 10) - (10 x 10)

SQRT (100 + 100) - (100) = 100

SQRT 100 = 10

Roger
 
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