# US split phase monitoring requirements

Hmmm. After thinking about that, the conclusion I come to is:
The neutral current on a 240 Volt load (one with some 120 Volt devices as part of the total load) should be small, otherwise the device that’s “eating” so much current so as to cause said imbalance should be a 240 V device instead of a 120 V device.

Looking at the typical load center, the largest part of the branch circuit wiring at AWG 12 which means
a max current of 20 Amps in that circuit. Due to derating, it’s actually 16 Amps unless a breaker designed for continuous use is used. (There will no doubt be some AWG 10 used for circuits that need it, but that tends to be the exception rather than the rule. AWG 10 is rated at 30 amps max)

Wires used in 240 Volt circuits are usually AWG 8 or larger, depending on the load size.

Given the utility’s meter doesn’t use the neutral leg, how could any imbalance affect the bill?

Here’s a thought experiment to prove it:

Install your own, perfect, lossless transformer between the supply company’s transformer and your house. It has a 240 V primary winding and a 120 - 0 - 120 V secondary winding.

What is the power that’s fed into that transformer, and provided the total power on the secondary side remains the same, how does the power on the primary side change when the load on the two secondary legs is not balanced?

Because the utility does monitor both legs and add together. The current on the neutral leg comes from one of the two hot legs and the usage is caught there.

The current entering one leg of the 240 Volt secondary winding of the transformer is the same current
leaving the other leg of the transfomer. (minus any difference in the neutral leg)
If they were added together, we’d be paying double what we should.

Here’s another thought experiment for you; take the 120-0-120 service to your house, disconnect it, install a transformer with a 120V primary and a 120-0-120 secondary and connect the primary between one hot leg and neutral to the service and the secondary to your house. Now the utility has all the power going through one hot leg and back through the neutral. The other hot leg would have no load. Bam! You’re now getting free electricity from the utility! … Not.

I’m a professionally qualified electrical engineer here in the UK. I don’t need your fanciful ideas, and I rather hope anyone else reading this thread recognises them for what they are.

I appended that with “not” because it simply wouldn’t work . It would work if @Bill.Thomson’s concept of electrical metering were true… which it isn’t.

I’ve been an Electronics technician since 1974.

I showed a picture of a CL200 form factor 2S kWh meter with two pics of the interior of the meter.
Same type of meter used in a typical US residence.

How is my electrical metering concept flawed?

Okay, I may have misunderstood you. An imbalance wouldn’t affect your electric bill; the meter, as configured, monitoring current on the two legs and voltage between the two legs and neutral, will capture the correct watt-hour usage.

It will affect your own monitoring if you only monitor one of the two legs. And if you’re utilizing the service heavily, and the difference between usage of the two legs is significant, you can trip the master breaker before you get close to maximum utilization of the service.

Hmm. Actually… Re-reading what I said, and what was said earlier… If the meter has no neutral connection, it can’t measure the separate voltages between the legs and neutral. Meaning that if the voltages and loads on the separate legs aren’t identical, the measurement won’t be accurate.

Since the voltage will tend to be lower on the more heavily loaded leg, it would mean the meter would tend to measure a higher usage than actual. Probably by less than a percent, but still interesting.

Quite right. That’s why I qualified it by saying

The pole mounted fuse in the fuse door will pop long before that happens.
Even if your transformer were rated at 37.5kVA
(as it might be for a large house with 320 or 400 Amp service)
The maximum continuously available current is 156.25 Amps.
Given a 25kVA pot = 104.2 Amps continuously available current and
a 15 kVA pot = 62.5 Amps continuously available current,
it’d be mighty hard to trip a 200 Amp main breaker. Now a 100 amp breaker, that would be “doable.”
But given that most if not all US houses built after 1970 have 200 Amp service
(a term which is a source of confusion in and of itself) they’re likely to be fed with a 25kVA pot.
Easily verified by looking at the number on the transformer that feeds the service in question.

Really. Makes me wonder how main breakers can trip in the first place. (And I’ve seen it happen.)

So given your example, you’d need an imbalance of nearly 100 amps between the two legs in order to pop the main breaker without first blowing the fuse to the service transformer. Interesting.

BTW this new building has a 400 amp service, and the transformer that’s still to be put in will also be supporting a 2nd 200 amp service. I suspect they’ll use a pretty big transformer.

I just did a test on my dryer… in Airing mode (no heat) it uses 120W just tumbling away, in full blown heating mode it uses 2020W, so the drum motor is about 6% of consumption. If those numbers are representative of US dryers (big if) then with just one CT on that circuit (assuming you’re doing per-circuit monitoring) I think you’d count that 120W either as 240W or 0W depending on which leg you put the CT around. That’s a reasonably significant error in my mind, although I have been accused of being an accuracy extremist.

I wish my dryer was that economical!
A dryer circuit in a US home is 240 V @ 30 A.
My dryer eats ~6kW and given the 30 A circuit, I’d say that’s the “norm” in a residential setting.

Turns out I misunderstood my superintendent friend. I got in touch with him again and he told me a
main breaker would indeed trip before the PMF would blow. But, the main breaker is derated same as
a branch breaker is. so a 200 A Main will trip at 160 A, unless it’s one that’s rated for continuous use.

Which should mean a 50kVA pot. Possibly larger if the 200 A subpanel is not counted as part of the
400 A main panel.

A 50 kVA pot would be capable of supplying 208.3 Amps which sounds undersized.
But considering you don’t normally run everything at once, it works.
Just as a 25kVA pot does with 200 Amp service.

The term 200 Amp service is a source of confusion. Although it sounds like 200 Amps should
be available form a 200 Amp service, the term refers to the load center ampacity.

Man that’s a lot of heat! I’m not sure my threadbare clothes would survive that. I assume that’s on a thermostat and that a lot of the time it’s only the drum motor and maybe a fan that’s running?

Our dryers just plug into a normal 10A outlet here, so 2200-2400 W is about the limit.

Spot on.

I’m gonna weigh in n this discussion because I think there is a different perspective that hasn’t been emphasized enough.

While much is made of the differences between the North American Power System (NAPS) and the system widely in use in many other parts of the world (EUPS for lack of a better term), my take is that they are more alike than they are different, and I think this thread gets to the heart of it with the discussion on metering. What follows applies only to residential and light commercial service. For industrial use, they both are three-phase and are pretty much identical.

The EUPS is commonly referred to as 230V single-phase. The NAPS can be referred to as 240V split-phase. They both have a base voltage in common, and I think the UK is actually still 240V in some areas that are not yet in harmony.

The way I see it, the advantage offered with the center tap is primarily a safety benefit. It makes 120V available which is more than adequate for most household appliances. There is a maximum 120V potential between any single conductor and earth. While still potentially lethal, it’s nowhere near the whallop of a 240V shock.

I believe the energy measured using the current on each leg, and the voltage between the two legs, is valid regardless of whether there is a slight imbalance between the two, or in the extreme case there is only a single 120V load on one of the legs. I don’t believe the voltage between either leg and the neutral is significant from the perspective of the electric utility that is selling you the power. While the meter plug was originally designed for a mechanical analog meter, the formula for an electronic meter would be power = V x (IL1 + IL2) / 2 (sorry, no math editor, real power calculation is assumed).

The voltage swing created by an imbalance between IL1 and IL2, while a theoretical problem for a home energy meter, is of little interest to the utility. The voltage imbalance is overwhelmingly a consequence of the resistance of the neutral line between the pole transformer and the meter. It reflects energy that is being consumed by that conductor in the form of heat. The Utility is simply charging you for that energy along with everything that is being consumed on your side of the meter. To de-rate a 120V load by using the voltage at the meter would require that the utility pay for the heat generated by your neutral line. The way this system works, with voltage between the two hot leads used for billing, they are already paying for the much greater heat being generated by those two conductors. It seems only fair that you pay for the lowly neutral line heat.

To aspire to be absolutely correct, a home energy meter should use a 240V transformer and apply it in equal proportion to the two legs, ignoring the voltage swing to neutral.

Whether or not you elect to go the extra distance to use a 240V reference, the measurement of a so called three-wire 240V appliance can be easily and accurately accomplished by passing both conductors through a single CT, taking care to pass one in the opposite direction of the other.

This works just like the electric meter and if you use a 240V transformer, is oblivious to any voltage variation caused by leg imbalance.

All that said, I’ve done many studies of NAPS voltage and power measurement using both 120V and 240V reference simultaneously. I am of the firm opinion that for the vast majority of users, there is no practical advantage to using the 240V transformer.

Edited power formula to add rather than multiply the current on the two legs.

Here is the analysis and the maths.

The error is half the “difference power” - the difference power being the voltage difference × the current difference.

BALANCED:

P = ½V × I1 + ½V × I2 + V × I3

which can be rearranged to

P = ½V × ((I1 + I3) + (I2 + I3))

where of course (I1 + I3) & (I2 + I3) are the currents seen by each c.t. in the meter.

UNBALANCED:

Say D is the voltage difference between legs, so that Leg 1 sees ½(V+D) and Leg 2 sees ½(V-D).

Substituting that in the first equation above,

P = ½(V+D) × I1 + ½(V-D) × I2 + V × I3

which can be rearranged to

P = “balanced power” + ½D × (I1 - I2)

So even with 100 A on one leg only, the error is probably going to be less than a couple of hundred watts in 12 kW - most likely around 1 - 2%.

But the important thing here is D will be proportional (and if not directly proportional, that’s good enough to a first approximation) to the imbalance in the currents, so the error term will follow a square law - at 50 A out of balance, the percentage error is half that at 100 A.

On that basis, I estimate the absolute error will normally be in the low tens of watts.

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Exactly, and the power dissipated in the neutral leg between the transformer, where D=0, and the meter is that ½D × (I1 - I2).