Im currently planning to add energy monitoring to my house. I live in sao paulo, brazil and here we have a 3 phase system but we only get two of the phases on our house. In the learn section of the website, it only shows loads connected to a a phase and neutral on the 3 phase example but here in brazil we have some things connected to a phase and neutral (127v) and other things connected between the two phases (220v). In the north america example two CTs are used and to calculate power its the used the following expression:
V × ICT1 + V × ICT2 = V × (I1 + I3) + V × (I2 + I3) = V × I1 + V × I2 + 2 × V × I3
But because the system here is 3 phase it should be:
V × I1 + V × I2 + 1,73 × V × I3
There is no way to achieve this expression using only ct1 and ct2. It would be possible if there was a third CT measuring current in the neutral wire. The problem is several brazilian friends pointed out that most national monitoring solutions only measure current on the phases including some oficial meters used by the electric company, so I assume there is something wrong with my logic but I cant figure out what.
If you already have an emonTx V3, the standard 3-phase sketch should work for you, but only if you want the total power/energy across your two phases. You cannot have power/energy individually for each phase, because although the sketch will give you a number - each number on its own is meaningless - only the total is correct. The “3-wire” option in that sketch will automatically give you the correct ‘whole house’ power and energy, because it treats one wire as its neutral - and that’s exactly the situation you have.
If you are thinking of buying an emonTx, I suggest that you should wait until the new emonTx V4 becomes available - this has 3 voltage inputs (although you will use only two) and 6 (expandable to 12) current inputs.
It works because classical wattmeter theory says that you need one less wattmeter than the number of wires. So with 2 phases and a neutral, you have 3 wires and you need only two wattmeters - or two channels of the emonTx.
If you need to monitor individual loads, then the situation becomes a little more complicated, but it is still possible because you already have the two voltages and you need only one c.t. per load - even when the load is connected line-line (by the same wattmeter theory).
For a 127 V load, you simply use the voltage of the line the load is connected to and the load’s c.t.
For a 220 V load, the emonTx needs to calculate two powers - using the two voltage inputs but the same c.t. - and add them.
You wrote
What I think you have forgotten is the emonTx measures real power taking the phase angle between voltage and current into account. It does not measure rms current and rms voltage and then multiply the two together, it makes pairs of measurements of current and voltage, multiplies the individual values to get the instantaneous power, and then averages those values.
You cannot compare your system to that used in North America because that is a single phase system, one phase is split in the middle to give you 120 - 0 - 120 V. We call that single phase, split phase.
Hello, thanks for the answer. Now that I think about it the 1.73 doesnt make much sense. But im still confused. I dont see how simply adding voltage between live and neutral times current measured on CT for each phase will get the correct instantenous power.
Lets say v1 is voltage between phase 1 and neutral, v2 is voltage between phase 2 and neutral and v3 is voltage between phase 1 and phase 2
I would assume the correct instantaneous power should be:
v1i1 + v2i2+ v3*i3
So with two voltage channels and 2 current channels I would have v1 and v2, and v3 is just v1-v2. But the current channels would give me i1+i3 and i2+i3, there is no way to calculate each individual current. With a third current sensor I could calculate i1, i2 and i3 and therefore calculate the instantaneous power.
Regarding firware:
it says on emontx-3-phase-userguide.pdf:
Use 3-WIRE where no neutral is available and the a.c. adapter and all loads are
connected Line – Line
In my case I have loads connected Line - Neutral, would this 3 wire option still work?
I think you need to construct a spreadsheet and verify that it does indeed give the correct result. You need one column with time (or angle), in the second column you have the first phase voltage (= 179 × sin(angle)), in the third the second phase voltage (= 179 × sin(angle + 120°)). Column 4 will be the voltage across the load (col2 - col3), and column 5 the load current, calculated from the voltage and the load resistance (col5 ÷ R). Column 6 will be the load power (col4 × col5).
Increase the angle by (say) 5° for each row, repeat for one cycle. Now calculate the average of the numbers in column 6, and check that it is the same as you calculate from the load resistance, that is 220² ÷ R.
It will work, but you will need to make a small change to the sketch. The sketch assumes, because it treats the 3rd line as a neutral, that phases 1 & 2 are 60° apart (yes, they really are when viewed from the third), you will need to change the manufactured second phase voltage to be 120° shifted from the first.