Hoping someone can help with this. If so, I think it will help me get over a hurdle in my understanding of the calculations involved with CTs in parallel.
My setup: I have my emonTx v3.4 with default (loaded by the Shop) sketch. I am in the US, so I will need to modify the sketch to set "USA=TRUE", and the other changes mentioned in Building Blocks. But I am still waiting for my UART cable, so I am trying things out while I wait.
I have the US AC-AC adapter from the shop plugged into the the emonTx.
I have two circuits running from my breaker box, one to the PV array on my upper roof, one to the PV array on my lower roof. Both circuit are connected to the "hot legs" in my breaker box, so they are 240V.
I put a YHDC SCT-013-0000 from the shop on each circuit. Then I plugged both of them into a y-splitter and plugged the y-splitter into input 3 of the emonTx. (I'd like to use inputs 1 and 2 for the mains, and 4 for my air conditioning, so I need to learn about getting all the solar on input 3).
The resulting graph of daily power is the right shape. But the power is consistently off by a factor of about 4 (relative to the Envoy monitor that was installed with my array).
I think I can explain a factor of 2. The emonTx is detecting the 120 volts from the AC-AC and using that in the power calculation, but the circuits I am measuring have twice that voltage.
But I can't explain the other factor of two. I'm wondering if it is just coincidental calibration error, or if it is something that would be expected based on my setup.
Thanks!
Brandon
Re: CTs in Parallel Question
We're missing some important information.
Which way is it "off" - which is 4 times the other?
Are you only talking about input 3, and are you measuring the current in one leg of each array?
What is the actual current (roughly)?
What voltage do you get reported by your emonTx? (i.e. what sort of adapter are you using - is it 120 V : 9 V USA type?)
If that's the case, the adapter will be giving the emonTx the same voltage as a UK adapter gives it on 240 V, and in the absence of changing the voltage calibration "USA=TRUE" then you should be reading the correct power. Are your CTs the same way round - are you by any chance subtracting the two currents rather than adding them? (Does each CT on its own give the correct power?)
Can you pass both wires through one CT - what do you get then?
(The BB articles "CT and AC power adapter installation and calibration theory" and "EmonTx - Use in North America" might help.)
Re: CTs in Parallel Question
The actual power is four times what I am seeing on the emonTx input 3. So if the emonTx is reporting 1000 W, the actual reading from the Envoy monitor will be roughly 4000 W.
Yes, I'm only focusing on input 3 right now. And yes, I am measuring the current in one leg of each array.
The peak output of the two arrays together is 4.9 kW, and the peak seen on most sunny days is more like 4. 2 kW. So, at 240 V, that would be 4900/240 = 20.4 amps. There are 6 panels in one array and 16 in the other, so a guess would be that one has peak current around 5 amps and the other has peak current around 15 amps.
Yes, I'm using the 120 V: 9 V USA adapter sold in the OEM Shop. The mean voltage reporting through the emonTx is about 125.6 V.
I will try putting them on input 3 and input 4 to see if that helps. And I think I can get both legs through one CT and try that as well.
Re: CTs in Parallel Question
So you've changed "USA=TRUE", as the emonTx is reporting the correct voltage (subject to calibration of course). Therefore you are reading low by a factor of 2 due to the voltage.
I think favourite is one CT is reversed compared to the other, so you are reading 15 A - 5 A = 10 A instead of 15 A + 5 A = 20 A when you "sum" the currents, and that gives you your second factor of 2.
Flip a CT on its cable and see if the power doubles (approximately). If it does, forget putting 2 cables through 1 CT (which would add the currents provided you used the same leg), as you've found the problem.
If not, I'll need to think a bit deeper.
Re: CTs in Parallel Question
Robert was correct. The reading of the smaller array was subtracting from the one of the larger array: 15 A - 5 A = 10 A.
Thanks!
Brandon