3 phase acting like single phase?

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Barjack

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St. Petersburg, FL
The establishment in question has a 3 phase 120/208 wye service. The main disconnect is feeding a 3 phase panel. A 2 pole breaker is feeding from this panel with a neutral to a single phase sub panel in a pool pit. This panel is feeding 2 single phase 208-230V pump motors, 120V pool light, 120V pit light, 120/208-230 ice machine, and 120V area lighting and convenience receptacles.

After repairing a bad connection in the junction box containing the feeders to this sub panel, I put my amp probe on the feeders to check current.
With everything running:
L1=34A
L2=24A
N=10A
Then shut off all 2 pole 208-230V loads:
L1=8A
L2=2A
N=6A
Shouldn't the neutral current be more like 7A? My thinking is that since this sub panel is using 2 of 3 phases sharing a neutral, shouldn't the wye neutral calculation work using L1=8A, L2=2A, and L3=0A?

I realize I'm probably missing something very basic here, but I just can't wrap my head around it. I even temporarily rearranged the breakers to change the loads, and they still acted like single phase.
 
I agree that your neutral current should be about 7.2 amps, what are the decimal readings on the phase currents?

Roger
 
Last edited:
Thanks for replying.
L1=8.3A
L2=2.1A
N=5.9A
The neutral should be about 7.5A
And, to clarify, there is no phase converter that I am aware of between the main panel and sub panel.
Also, L-L Voltage=208V, and L-N Voltage=120V
 
There is one thing you have to remember about the three phase supply you have to know the phase / current shift on wye system and Yes I understand you have single phase load there but you will not get the same reading as true single phase set up will be reading on the line and netrual format and the three phase shift will be 120? sorta like wide triangle format somehow I don't have the formaila with me ATM.,

I am sure someone may have a chart or formila here and I think it was disscused before few time.

Merci,Marc
 
Frenchelectrician, I completely agree. This feeder, as long is it supplies L-N loads, should never be balanced.

310.15(B)(4)(a) even brings up this exact scenario as to why you shall count the neutral as a current carrying conductor.

There is more to this situation that I am not seeing, and I will post back if I find out more.
 
Definitely 120/208Y. The transformer is on the property on the ground clearly marked. I also read with my meter at the main disconnect, and everywhere along the line I could
L-L 208V and L-N 120V.
 
Without multiple measurements and measurements of power factor as well as phase current, I think that the most you can do is say 'hmm'.

1) With just the numbers that you've posted, I'd call those numbers close enough to what theory predicts to simply be measurement error.

2) If you really want to dig deeper you may need to adjust your theoretical understanding:

You are absolutely correct that in the case of 'single phase' service consisting of two legs from a wye supply, you calculate neutral current as if you had a three phase wye system with one leg at 0 amps.

The normal equation used for these calculations ( see http://forums.mikeholt.com/showthread.php?t=110225) is just an approximation. To correctly calculate neutral voltage, you need to take the vector sum of all of the phase currents. The equation given in the link operates on the assumption that all the phases have phase angles displaced by exactly 120 degrees. Since the supply voltage is displaced by 120 degrees, the current phase angles might be different from 120 degrees if the loads have different power factors.

3) Is it possible that there is a bootleg neutral somewhere, and that some of the expected neutral current is returning on the EGC? Did you try placing an amp clamp around the entire feeder (excluding the EGC) into the panel?

-Jon
 
Jon, thanks for the detailed reply.

Without multiple measurements and measurements of power factor as well as phase current, I think that the most you can do is say 'hmm'.

1) With just the numbers that you've posted, I'd call those numbers close enough to what theory predicts to simply be measurement error.
Click to expand...

I've been saying 'hmm' a lot and am tempted to let this go as I can't see any real danger or problem with what I am observing. As I said, though, I temporarily rearranged the loads so L1 and L2 would be closer to equal, expecting the neutral load to be between these. The neutral load got smaller the more even they were.

2) If you really want to dig deeper you may need to adjust your theoretical understanding:

You are absolutely correct that in the case of 'single phase' service consisting of two legs from a wye supply, you calculate neutral current as if you had a three phase wye system with one leg at 0 amps.

The normal equation used for these calculations ( see http://forums.mikeholt.com/showthread.php?t=110225) is just an approximation. To correctly calculate neutral voltage, you need to take the vector sum of all of the phase currents. The equation given in the link operates on the assumption that all the phases have phase angles displaced by exactly 120 degrees. Since the supply voltage is displaced by 120 degrees, the current phase angles might be different from 120 degrees if the loads have different power factors.
Click to expand...

I am thinking that phase angles and power factor may have a lot to do with what I am seeing. With all the motor loads off, what power factors should I expect to see from 120V incandescent lighting loads? One of the loads is a 120V-12V transformer for a 300W 12V pool light.

3) Is it possible that there is a bootleg neutral somewhere, and that some of the expected neutral current is returning on the EGC? Did you try placing an amp clamp around the entire feeder (excluding the EGC) into the panel?
Click to expand...

As far a I know, the feeder goes directly from the panel room to the pool pit, and there are no other connections that I can see for the neutral. I did not put my amp clamp on the entire feeder, but I did check the EGC and got between 0.00A and 0.01A and assumed that was meter error.
 
Barjack said:
I am thinking that phase angles and power factor may have a lot to do with what I am seeing. With all the motor loads off, what power factors should I expect to see from 120V incandescent lighting loads? One of the loads is a 120V-12V transformer for a 300W 12V pool light.
Click to expand...

Incandescent lighting should have a unity power factor, but the transformer can introduce a lagging power factor because it is drawing magnetizing current. Remember that line-line loads cannot put current on the neutral, so if you have any line-line loads in your 'phase' measurements then they will introduce error in your calculated neutral measurement.

-Jon
 
You can also get current phase angle differences with unbalanced voltage drops. The neutral will shift and the voltages will not be 120? apart so with 100% pf loads, the currents will not be 120? apart.
 
1st off hand helds are not going to be that accurate.
2nd measure the A and B currents simultaneously-reading should match neutral readings.
3rd take peak readings and see what the form factor is.
4th is this causing you any issues?
5th measure current on the neutral ground bond.
 
3 phase acting like single phase?

The formula you reference does not work for Unbalanced loads in a 3 phase system. When the load is unbalanced (severely in this case, Ia=8, Ib=2, Ic=0) the resultant currents in the lines (Ia and Ib) are not 120 degrees out of phase, which the formula you reference is based on.

When 1 phase is opened (Ic=0) the 2 other phase currents are 180 degrees out of phase. The equation is

In=Ia + Ib + Ic where these are vectors. For this case
Ia=8 at 0 degrees = 8*cos(0) + j*8*sin(0) = 8*1 + j*8*0 = 8
Ib=2 at 180 degrees = 2*cos(180) + j*2*sin(180) = 2*(-1) + j*2*0 = -2
Ic =0

Then In= 8 - 2 = 6 which is what you measured.
 
markstg said:
The formula you reference does not work for Unbalanced loads in a 3 phase system. When the load is unbalanced (severely in this case, Ia=8, Ib=2, Ic=0) the resultant currents in the lines (Ia and Ib) are not 120 degrees out of phase, which the formula you reference is based on.

When 1 phase is opened (Ic=0) the 2 other phase currents are 180 degrees out of phase. The equation is

In=Ia + Ib + Ic where these are vectors. For this case
Ia=8 at 0 degrees = 8*cos(0) + j*8*sin(0) = 8*1 + j*8*0 = 8
Ib=2 at 180 degrees = 2*cos(180) + j*2*sin(180) = 2*(-1) + j*2*0 = -2
Ic =0

Then In= 8 - 2 = 6 which is what you measured.
Click to expand...
Your post sounds like you are saying opening one phase causes the other currents to jump to a 180 degree phase displacement, which is simply not true, and probably not what you meant. They may be 180 degrees apart because of reactive loads, but not because we only use two phases.

Barjack,
I'm in the parallel return path camp and/or meter inaccuracy camp. I still think it would be interesting to get a line+neutral current reading to see if there is a non-zero reading. This would be like having a GFI breaker. You might could even try reading closer to the loads in addition to the feeder to see if you can find a ground fault.
 
I have not yet been back to this job, so everything I say at this point is based on memory. When I took the neutral readings, only single pole 120V loads were running, with the only inductive load being the pool light transformer. With the loads rearranged, so that L1 was as close as possible to L2, the current on the neutral was almost zero. Is the transformer load enough to pull the phases 180 degrees apart?

Barjack,
I'm in the parallel return path camp and/or meter inaccuracy camp. I still think it would be interesting to get a line+neutral current reading to see if there is a non-zero reading. This would be like having a GFI breaker. You might could even try reading closer to the loads in addition to the feeder to see if you can find a ground fault.
Click to expand...

At this point, I am also in this camp, and I agree that a L-N reading could explain what I am seeing. The work was completed that this job, so I may not be back there for a while. If I can get back there, I will get those readings and report back.

Thanks, everyone for your replies.

-Chris
 
mivey said:
Your post sounds like you are saying opening one phase causes the other currents to jump to a 180 degree phase displacement, which is simply not true, and probably not what you meant. They may be 180 degrees apart because of reactive loads, but not because we only use two phases.
Click to expand...

So if you have 2 phases of a 3 phase wye system, and you have matching power factor line to neutral loads, then the current flow in the two legs will be 120 degrees out of phase.

But if you take the same system, and instead _line to line_ (208V) loads, then the two currents will have a 180 degree phase displacement. This is true even if the line to line load is a pure resistive unity power factor load.

To keep the 120 degree phase displacement, you need another path for the current to follow, eg. the neutral. If you only have a two wire circuit (line-line without the neutral) then the current has to be exactly the same everywhere in the circuit; and because we generally look at wye systems referenced to the neutral, you get an apparent inversion.

-Jon
 
winnie said:
So if you have 2 phases of a 3 phase wye system, and you have matching power factor line to neutral loads, then the current flow in the two legs will be 120 degrees out of phase.

But if you take the same system, and instead _line to line_ (208V) loads, then the two currents will have a 180 degree phase displacement. This is true even if the line to line load is a pure resistive unity power factor load.

To keep the 120 degree phase displacement, you need another path for the current to follow, eg. the neutral. If you only have a two wire circuit (line-line without the neutral) then the current has to be exactly the same everywhere in the circuit; and because we generally look at wye systems referenced to the neutral, you get an apparent inversion.

-Jon
Click to expand...
...and in the 8/6/2 amp case he had opened all of the 2-pole breakers so we were left with line-neutral loads which makes me like your bootleg neutral theory.

There was another thread a while back that had some nice circuit diagrams showing what happens with and without the neutral on a wye. Lose the neutral and you lose the "bowstring" offset conductor that allows you to derive three phase power using the two line conductors of the wye. The neutral point then moves from the off-set "bowstring" point to a point that lies along the axis between the line voltages (i.e. no "bowstring" = 180 degree displacement).

In the line-line case, I guess you could still have a high-impedance fault to ground on one of the line conductors which would give you different phase currents. What would be even more interesting would be if both phases had a fault to ground with different fault impedances.
 
roger said:
Mivey, nice to see you back.

Roger
Click to expand...
Thanks. I went through a busy spell and hobby time was limited. This site can be too addictive at times.

I did miss the fun as I'm always impressed how the members here can keep coming up with new and interesting things to discuss. Some have a great sense of humor as well.
 
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