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I don't understand your terminology, we don't have that kind of system in the UK. If you search the forums, you will find quite a lot about successfully using the emon system in the USA. A c.t. measures the vector sum of the currents flowing through the core. If those currents sum to zero because they are equal and opposite, it will read zero. Also bear in mind that a c.t. is a current source, not voltage, so there's a fundamental flaw in your thinking there. One American user has reported that the voltage balance is good, so simply measuring one of the 120 V lines may be adequate. However, you probably need 2 c.t's, because there are 3 currents to measure (L1, L2, N) and there's no guarantee which way they split.

This is not as simple as it seems. If you want to measure apparent power then you can simply use two CT's, one on each Hot wire coming in. If you want to measure Real Power, it gets a little trickier.

When I measure voltage, CT1, and CT2, I get a positive voltage, then one positive wattage, and one negative wattage. This is due to the way the neutral works. When you sample a 120 volt system you are measuring either the top 120, or the bottom 120 voltage and therefore one power shows negative and one positive. My Real Power is the sum of the two absolute powers.  abs(power1) + abs(power2).

Since I am not running any PV this works fine. I allow the emonTx to send the negative and positive powers, then use the emonCMS to clean up the negative number and then sum the powers.

If anyone else has experience or input on this, please chime i! The North American household power system is a little confusing..

It's not too confusing, it's just a 2-phase system and so is a bit simpler than the 3-phase system here in the UK (although most residential properties only receive one phase). If you search for 3 phase on this forum you'll find lots of useful information.

The only way to measure true power in a multi-phase system is to measure the current and voltage on each phase, calculate the individual powers and add them.

An approximation can be made by assuming the voltage is the same for each phase - which is what you are doing. In order to make this work you have to time-shift the voltage by the phase difference between the phases. For a 3-phase system this is 120 degrees but for a 2-phase system it's 180 degrees so an approximation can be made by simply inverting the voltage - which is effectively what you are doing.

The problem with this approximation is that, at least with our 3-phase system, there can be significant difference in voltage between the phases.This is often>5% here so that will limit the accuracy you can achieve without measuring the second voltage. Any difference in distortion in the 2 voltages will also add to the error.

Thanks  a lot. I appreciate the clarification. 

One question - technically the 2 "phases" we get in Canada and the US are not out of phase. They should be in phase, although they reference different "neutral points" so I guess they act as if they are out of phase? 

You mention that there is a often quite a different in voltages between the phases, so I guess an ideal North American emonTX would have two CT's and two Voltage Inputs. Does anyone here feel there would be sufficient interest to build a modified emonTX for our power system. 

Is their any real limits on the chip that would make measuring 2 VA simultaneously difficult? 

One question - technically the 2 "phases" we get in Canada and the US are not out of phase. They should be in phase, although they reference different "neutral points" so I guess they act as if they are out of phase?

It's probably just another way of saying the same thing. If they were in phase you wouldn't get much heat out of your 230V cooker!

There's no issue chip wise with measuring 2 voltages and 2 currents simultaneously it's just that the emonTx board only has provision for one voltage input so you would have to modify if, maybe using the CT3 input, and write your own sketch.

As I understand it, you are talking about a 3-wire system. Irrespective of the names you give the conductors, basic theory tells you that you need one wattmeter less than the number of wires. For a wattmeter, you will use one voltage input and one c.t. You use one wire as the reference (typically you'll use the neutral, but you don't have to) and measure the voltages of the other two conductors with respect to that, and the currents in those two conductors. Real power (which is the quantity you pay for!) is the sum of the two powers.

It shouldn't be too difficult to replicate the voltage input circuitry of the emonTx on a small external circuit board and feed an unused analogue input, in order to have a second voltage channel. The standard emonTx sketch measures voltage and current for one c.t. for a number of cycles, then repeats this for the two other c.t's, using the same voltage input. A small modification to the standard library should allow you to measure current and voltage with one pair of sensors, then the other pair, in succession, which should be reasonably accurate if the loads are relatively constant. If you require truly concurrent measurements, then you'd need to modify one of the interrupt-driven sketches to measure all 4 sensors in sequence. With that method, you should get about 40 sets of measurements per cycle, which should still give adequate accuracy.

It shouldn't be too difficult to replicate the voltage input circuitry of the emonTx on a small external circuit board and feed an unused analogue input, in order to have a second voltage channel.

Much easier to convert the CT3 input to a voltage input though. Just swap the burden resistor, R16 for a 10k one then cut the track from R16 to the jack socket tip and solder a 100k surface mount resistor across the cut, job done - well apart from putting a 3.5mm plug on the a/c adaptor.

Does the PHASECAL have units? Is it in degrees or millis? I would like to infer the phase of the reciprocal phase, from the one being measured.  Any way to do this without an oscilloscope? 

This would allow me to get both phases positive on the North American split phase system.

The PHASECAL value is multiples of the sample period so a value of 1.7 means advance the voltage by 0.7 samples.

It uses linear interpolation/extrapolation when we're actually dealing with a sine wave so it's only intended for small phase adjustments. I believe it can actually increase the error it is intended to remove even for small adjustments so definitely not suitable for a 180 degree adjustment.

Phasecal provides a time shift - not quite the same as a phase shift - but more importantly when the voltage wave is not a true sine wave, which in most cases it isn't because of the flattening of the peaks that is a result of rectifier loads, it does distort the shape of the waveform a little. If you look very hard at the algorithm, it's accurate at zero and 1, it undershoots between these points and overshoots everywhere else.

But don't confuse phase shift with inverting the wave. I believe the American system is what I would call 115 - 0 - 115 V, where the final supply transformer measures 230 V between the ends of the windings and the neutral centre tap is connected to earth. In that case, what you are calling a 180° phase shift isn't really a phase shift at all, but a mirror image of the voltage. Put another way, if you think of the classic explanation of a sine wave as being the shape drawn by a point on the rim of a rotating wheel seen edge-on, earth/neutral is the centre of the wheel and the two lines (phases) can best be represented by two points opposite each other.

So if you want to use the one voltage as a representative measurement of the voltages of both lines all you need to do for the second line is turn the voltage upside down in the maths with a minus sign.

when the voltage wave is not a true sine wave, which in most cases it isn't because of the flattening of the peaks that is a result of rectifier loads

Can you elaborate on that a little more?  Are you saying the grid V as it arrives at your house is not a true sine wave?  Which rectifier loads are you referring to?

Correct. The rectifier loads are all those on the system! Every d.c. power supply of the conventional sort (mains isolating transformer, bridge rectifier and reservoir capacitor) draws all its current around the peak of the supply (and many switched-mode ones will too). With the result that it lowers the peak voltage. See the pictures about half-way down this report.

Wow, that's a pretty impressive flat-top.  If I'm reading your graphs correctly, it looks to be about 2msec wide.  I think that works out at about a 5% drop in your Vpeak compared to nominal,  so 323V instead of 339V at 240VRMS.

I wonder how much that varies by geography.   I recall seeing a slightly deformed V here in Aus, I don't recall it being quite that flat.

I think it's likely to be reasonably representative. As you'll appreciate, the factors that affect it are the impedance of the supply and the types of loads that are connected to the system. I would imagine that, depending on where you are in the country and the proportion of loads that don't draw a distorted current wave (say resistive heaters or air conditioning loads, i.e compressors driven by induction motors) to rectifiers (computers, TVs, battery chargers) then the degree of flattening will change, but I'd expect it to be present. There's no industry near me (especially not on my local sub-station) so I might well have slightly more flattening than you would see elsewhere. You can easily get your own graph by using just the low voltage half of the test rig (the bottom half, NOT the direct connection to the mains, and of course you don't need the variable transformer), a computer sound card and the free software mentioned in the report.

Thanks Robert. I appreciate that. I see what you mean. It's confusing since so often the 3-wire North American system is compared to a polyphase system, but is really isn't since yes, you are correct both are in phase. So I will simply use the abs value of the two powers and add them. 

I'm not sure what would happen with a PV system since I could easily get lost trying to ensure the sign of the power is correct, but for me - this will work nicely. I'll simply ensure the negative 'phase' or 'side' of the 240 is made positive. 

Marquis27

When considering PV, don't ignore the little point I made above about "you need one wattmeter less than the number of wires" - and it doesn't matter which wire you use as a reference. So if your PV is connected only across the two lines at 230 V, it makes sense to regard one line (L1) as the reference and measure two voltages (L1 - N and L1 - L2) and three currents, N, L2 and PV.  Then if your L2 c.t is upstream of the PV feed-in, your nett power is total of neutral power (I_N x V_L1-N)  and L2 power (I₂ x V_L1-L2). If the voltages and currents are measured consistently, the signs are taken care of, and your generated power is (I_PV x V_L1-L2).
If you're approximating the second voltage, then the two voltages are the same sign, but one is half or twice the other - depending on which you measure!  (I wrote that I regard your system as 115 - 0 - 115 V. If you call one of the lines your reference point, it's 0 - 115 - 230 V.  Does that make it easier to visualise?)

Hello,

I'm from Brazil and we have 2 hot 110v and 1 neutral with only one voltage meter on both 110v hot(your ideal power 220v wall adapter). We have 60hz here.

I measured both hot voltages and they are almost the same(<2% difference across it).

Do you have any code to measure correctly this 2 phase system? I don't care for sleeping or energy saving because I don't use batteries on emontx.

Thanks,

Wagner Sartori Junior

If you are happy to accept that error in voltage, then you can use the standard emonTx sketches with only a small change.

You will need 2 current transformers, one on each line (not on the neutral) and your voltage a.c. adapter connected across the two lines.

In the sketch, you need to use two c.t's, and you must add the powers then divide by two to give you the total power used by the house. You divide by two because the voltage is twice the value it should be so each power on its own is twice the value it should be.

(Ensure the c.t's face the "right" way - the convention used here is imported power is positive. If one line shows as negative when importing, reverse the c.t. on its wire.)

I can't comment on whether the "pro" library you mention in another thread is suitable. The standard library is well understood and most of us can offer support for it.

Disclaimer: rank beginner here, with a few questions italicized in the text. Thanks in advance for any help.

The US grid is like Brazil's, apparently. Here, it's common for the load on the L1-N (120V) side being quite different from that of the L2-N (120V) side, in spite of the electrical system designer's attempt to balance them. Then there's the  L1-L2 (240V) load added to those, to give the total consumption load. If I'm understanding this correctly, two CTs? will suffice to measure the total load, but there may be a small error introduced if the L1 and L2 voltages are unequal.

On my solar side, several array strings are combined and their total output (240V) is connected to L1 and L2 between the meter and the circuit-breaker box. I think this means I have a "Type 1" system, but again I will need 2 CTs? on those lines. I have a 200A service; it's possible the total load could exceed 100A. How will the 100A CTs react to that? If  the system will allow, I'd like to also monitor the household N line to monitor the degree of balance between L1 and L2.

My solar equipment does a good job of monitoring and reporting energy production every 5 minutes. I don't have real-time access to their data, but it will be interesting to see how my data compares to theirs after the fact. Integrating their 5-minute energy data gives total power for the day to within 0.93% of their reported total power (I  don't know how that's derived).

An attractive alternative for me would be to read the electric meter directly via an optical coupler, but I'm doubtful that the utility will give me access --even read only -- to their meter, which has a very comprehensive security protocol to limit access to various groups for various purposes. Does anyone do this already?

There's a new article in Building Blocks about your system, which might help. You don't need to use, and probably can't use, our standard 100 A CTs, but there are alternatives listed in there too. The YHDC 100 A won't like much more than it's official rating, you can look at the curves in the report - also in B.B. Maybe you haven't seen that yet.
Some members of the community here have decoded a couple of inverters - whence you could get quite a bit of useful data, so if yours is one of those you should be well away. If not, then unless you're up to finding out what the protocol is and decoding it, measuring the powers yourself is the easiest answer. If there are only two wires (excluding the protective earth) to your PV inverter, you only need 1 CT (because the second one would see the same current as there's nowhere else for it to go!). If your meter has a flashing LED (visible or IR) that might be of some use, or it might use a published protocol.

So welcome to the community.

I'm also a newbie here in the USA.  I have 2 CTs, one around L1 and one around L2.  As noted in the building block, one CT is opposite direction of the other, to account for the negative voltage of the other line.

My question is, if the voltage adapter is connected to L1-N, and both L1 and L2 power is reading positive (one CT is opposite of the other), why does L1 and L2 power not both read negative if connecting the voltage adapter to L2-N?

"to account for the negative voltage of the other line"
That's a funny way of looking at it - CTs have no knowledge of the voltage on the conductor they're sitting on.

Two reasons why it might not show a negative power:
1. You are looking at apparent power, not real power.
2. Your ac adapter is connected to N-L2, not L2-N.

If you'd told us which sketch you're using, it might help. My crystal ball isn't working well these days.

[And we didn't need 2 copies of your question.]

You're not the first I've heard with a faulty crystal ball.  That's why I never bought a crystal ball myself; the reliability of those things are terrible!

There must be some delay after posts because I didn't see my first post so I reposted.  Thanks for removing the duplicate.

Your reason #2 is interesting especially since the Ideal brand voltage adapter I bought from the shop does not have a polarized plug.  I performed a test with the voltage adapter as L1-N and then reversed the orientation as N-L1 and sure enough the sign of the power flipped!  I repeated my test on L2-N and got opposite results, which is expected.  You even had me wondering if my L2 receptacle was wired backwards, but fortunately that checked out.  So my conclusion is that I had the adapter as N-L2 (flipped), as you predicted.

By the way, I'm using an emonTx with the DiscreteSampling sketch version 2.1 from github.  Based on my results and getting negative power values when flipping the voltage adapter, I'd say that this sketch performs real power calculations.  Can you confirm this?

Regarding my comment on "to account for the negative voltage of the other line", yes, the CTs don't have voltage information directly, but from the perspective of L1-N the L2 current is reversed because the L2-N voltage is reversed..  My preference is to put the CTs with the dot facing the voltage source, but because we only have one voltage input to the emonTx, we have to flip something somewhere regarding the CT that is on the opposite line from the voltage adapter to have the power be positive: flip the CT, flip the polarity of the secondary of the CT, do a [(-1)*x] operation in the software...many ways to go about it all for the same result.

Thanks again for your insights and helping me learn a little bit tonight!

The delay was because it went into the moderation queue - see "Read this before posting"

Most, if not all, sketches default to calculating and reporting real power if the voltage is available, but they do calculate apparent power also, and default to that if the voltage isn't detected at power-up. So it was quite conceivable that you were seeing apparent power.

The ac adapter with the UK plug is polarised, hence the instructions say to orient the CT so that it follows our convention of imported power is positive. As long as you're consistent, it doesn't matter what your convention is (but be prepared for misunderstandings if it's not the same as ours and you don't mention it).