OpenEnergyMonitor Community

US split phase monitoring requirements

And the unfortunate reality is:

  1. The emonTx/emonPi has only one voltage input,
  2. The US a.c. adapter is a 120 V beast.

Yup, as shown, having only CT1 on the circuit will miss this load, while the meter, monitoring both sides, will catch it. You’d need CT2 in order to get a complete reading. And while circuits will be staggered so that they roughly balance out, if a constant load, which is what will have the greatest effect on the bill, happens to be on the side the CT isn’t, you’ll miss that usage.

Regarding #2, you could get a European adapter and wire up a 240V circuit for it easily enough.

Regarding #1, yeah, there’s often a few volts difference between the two legs depending on load, and that can introduce inaccuracies. In my particular case, since I’m using multiple emonTx devices in even quantities, I can have them monitor separate legs of the service and hand them voltages from those specific legs. :slight_smile:

Durpity, how did I accidentally delete the post?

It looks as if you misunderstood me when I said:

So your final “channel” count will be determined by the number of 240 Volt and 120 Volt circuits you
want to monitor and whether or not you’re OK with ignoring small errors monitoring your 240 Volt loads with a single CT. If not, you’ll need two CTs for each 240 Volt circuit you want minimal measurement errors on.

as the above drawing illustrates exactly what I was trying to convey.

I don’t think anyone has ever disputed that. If you’re talking of monitoring the total consumption, everyone here agrees that you need to measure the current in both legs.

If you’re measuring one 120 V circuit, we agree you need only one c.t for that.

If you’re measuring one pure 240 V circuit with a two-wire connection, I think we agree that you need only one c.t for that.

Where I think there is some debate is when considering the inaccuracy when measuring a 240 V appliance that has a 3-wire connection. The assumption that means only one c.t. is needed is that the neutral wire carries an insignificant current compared to the two line conductors (‘hot’ wires), because that is the current which is drawn by the control electronics or similar, and it’s much smaller than the normal load.
If the assumption is valid, the error from using just one c.t. is insignificant. If it’s not, the error will be significant and to remove the error, two c.t’s are needed.

Well, the answer is simple IMO; buy one CT and stick it on the neutral wire first. If you see more than that insignificant usage, well, you’ll need to pick up a 2nd one! :slight_smile:

Oh! Would it be a good idea to use Rpi 4s for running emonHub? Or would it be overkill, or cause problems?

A post was merged into an existing topic: Can I hang 4X emonTx via wire on one emonPi/Base?

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. :slight_smile:

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.