3-phase to single phase current relatioship 1

[T100]

Sorry if this seems like a simple question to this group, but I am struggling with this.

We have an instrument that we are planning to run on single phase power. In the past the manufacturer has only used unbalanced 3-phase (Wye) configurations. They just use the three phases and the neutral to run three groups of devices. One of the service engineers tells us that when he measured current in each leg of the Wye he got something like 22A, 11A, and 7A, and that therefore we have to expect around 40A if we run on single phase. However, logic tells me this should actually be divided by (3)^0.5, and that we therefore should expect under 25A.

Whose right?

 

[jghrist]

If I understand this right, the devices are independent, i.e. they are single-phase loads each served phase-to-neutral from a different phase of the supply. If this is the case, then you should expect a current draw equal to the arithmetic sum of all the phase currents when operated on a single phase. Your service engineer is right.

The only way I can see the sqrt of 3 getting involved is if the devices are connected phase-to-phase in which case the voltage across them would be different than the phase-to-neutral voltage.

 

[twosockets]

jghrist,

If the three independent phase to neutral currents are of equal power factors, and the phase to neutral voltages are also equal, you can aritmetically sum the three currents to obtain a total when operated in a phase to neutral single-phase configuration. If they are of unequal power factors and/or voltages, the actual sum may differ from the arithmetic sum.

Incidentally, there are a couple of watt-hour meter manufacturers that do no understand this principal.

 

[T100]

Thanks for your replies, but I have read some stuff that seems to contradict this. Specifically have a look at:

http://www.du.edu/~jcalvert/tech/threeph.htm

Herein he mentions that voltage and current are not in phase in a wye-connected 3-phase system, even for a purely resistive load. Obviously they are in a single-phase system, so this would lead me to believe that one would have to take current measured in a phase-neutral leg of a 3-phase system and divide it by root 3 to get the comparable draw in a single phase system.

Or am I making this more complicated than it needs to be? Is there truly no phase to phase current when the devices are running independently in a 3-phase system?

 

[jghrist]

The currents divided by sqrt(3) in your link are "currents through a delta load," i.e. connected phase-to-phase. If your devices are connected phase-to-neutral, you don't need the sqrt(3).

 

[DanDel]

Lets assume that the existing loads are single-phase(phase to neutral), and they are 22A, 11A, and 7A, as you have stated. It doesn't really matter what the neutral current is(if any).
Let's also assume that the system voltage is 480/277V(it doesn't really matter, just change the values in the equations).
The total apparent power (VA) used would be:

22A x 277V = 6094VA
11A x 277V = 3047VA
7A x 277V = 1939VA

total VA used is 6094 + 3047 + 1939 = 11080VA = 11.08kVA

If you run single phase 277V, just divide the VA by 277 to get the current: 11080/277 = 40A, (which, by the way, is the total of the three phase currents at 277V).

If, however, you are running single phase 480V, the current would be: 11080/480 = 23.08A

 

[DanDel]

Oh, and PF has nothing to do with VA or amperage, and since your measurements(and concerns) were with amperage, it is of no use to consider it.

 

[peebee]

Yup, 40A is correct.

I'm just glad nobody added up the phase voltages, ie, 3x277 = 831 volts. You wouldn't believe how many times I've seen that done. Scary.

 

[busbar]

As long as 3ø systems remain a mystery to some, there will still be money to be made in our command of “Wizardry.”

 

[twosockets]

DanDel,

Frankly, I am stunned to see you say that you can add VA arithmetically without regard to Power Factor in an AC system. Are you sure that you want to say that?

Let's keep it simple. In a three-wire single phase transformer with a grounded neutral connected to the center tap, the current existing in the neutral is the PHASOR difference of the currents that exist in the two hot phases. In THEORY, it is possible to have a current of a higher magnitude in the neutral than exists in either of the two hot phases. Please don't refer me to The National Electric Code as it is error.

 

[DanDel]

twosockets, of course, in theory you are correct, you can't arithmetically sum VA unless the PF of each load is the same. My initial response was simply to straighten out the square-root-of-3 problem T100 had by showing him a quick way to find out the currents in reference to the voltages he is using(which he never did provide), while my remark about no PF in VA was to dismiss the Watts-VA discussion before it got started. I should have mentioned what you pointed out in your first post; for that I'll give you a star.
To carry your point to the extreme, consider two loads of the same 100A current(and voltage), one purely inductive and the other purely capacitive(representing completely lagging and leading PFs). If you put these two loads in parallel, the total current from the source is zero, because the 100A is now cycling between the two loads(remember, this is in theory). But in reality, even if you could get two loads like this, to put them in parallel, they would be connected to two C/Bs in the same panel. Each circuit would still carry 100A, as would each C/B. And the 100A would be traveling through the panel bus, so that would have to be rated for at least that load, not zero amps. You and I know(theoretically) that the transformer supplying the load can have a rating of zero VA, as long as there is voltage present.
You mentioned that the NEC is in error. The NEC requires that the total load amps be summed and the cabling, overcurrent protection, and source transformer be based on this sum(plus 25%, etc).
You may feel that this theory makes the NEC in error, but the important thing with the NEC(and with my too-simple analysis) is that it would be the conservative and safe recommendation, because, in reality, I'm sure that the loads T100 described do have very similar PFs.
And don't forget, any self-respecting electrical design engineer (and the wiring inspector) would never allow it any other way.

 

[twosockets]

DanDel,

The NEC is in error when it states that the current in the neutral of a three-wire single phase system cannot exceed the highest current in either hot phase. In THEORY, the current in the neutral could be twice as high as the highest current existing in a hot phase.

Believe it or not, this issue actually came up in a court case that I was testifying in. The plaintiff (I was an expert witness defending a large IOU from the charge that the electric meter, then a poor neutral, caused a fire) asked me to state the highest THEORETICAL current that could exist in the neutral. When I stated that the current in the neutral is the phasor difference between the currents in the two hot phases, the plaintiff provided an example. He said: “if the current in one hot phase was 200 amps and the current in the other was 200 amps, what is the maximum theoretical current in the neutral?” I responded that the answer was 400 amps. I further stated that this was impossible under real world conditions as there is no perfect inductor, nor perfect capacitor, which would be required to match the theory. The IOU was outraged as it appeared that I was agreeing with the plaintiff that the neutral could have contributed to the fire. I told them that I had no choice but to be truthful, regardless of whose case was helped or hurt.

An engineer with the IOU referenced the NEC to buttress his contention that the neutral current cannot exceed the highest current in either hot phase. He further stated that this is why the neutral is not required to be fused. The theory he sited was based on the Edison three-wire system. There is only one problem: the Edison three wire system was based on DC, and not AC. Obviously, DC power factor is always unity. If you use a three wire DC system, the NEC is correct.

By the way, we won the case in spite of the controversy.

 

[Shortstub]

T100,
If the 3 instrument loads are resistive then your service engineer is correct! If they aren't... then, it's anyones guess as to the max current when the 3 instruments are connected to the same source.

I propose a simple test. Measure the neutral current. If its about 13-14 Amp, buy the man or woman dinner! If not, and you're still interested in a figure < 40 Amps, you can solve it the hard, easier, or easist way, i.e., Vectors, Ohm's Law (if valid), or Watts, respectively.

 

[DanDel]

twosockets, I believe that 400A in the neutral connection of the circuit you describe will be the least of your worries, since what you describe appears to be a series resonant LC circuit, which, in theory, has an impedance of zero, causing fault current levels to flow in the line conductors.
I don't think you can look at it by vector analysis, whereby you simply combine two 120V, 60Hz sources, one with a 0.6 ohm capacitive load and the other with a 0.6 ohm inductive load, into a 3-leg circuit and sum the vector currents. Your circuit as described will be a resonant LC combination in series with two 120V sources in series(or 240V). At this point, 'theory' using perfect inductors, capacitors, conductors, and sources go out the window, since your line current can only be calculated by division by zero.
I believe any circuit approaching your example in the real world would first cause so much line fault current that the neutral current you speak of as the vector sum of the two separated single phase circuits would not get a chance to cause any damage.