Oh dear. By default, the emonPi calculates apparent power also, but only transmits real power into emonHub (and thence into emonCMS).
There’s a way out of this, but it involves reprogramming the Atmel AVR chip inside the ‘emon’ part of the emonPi, which you need to do via the “Raspberry Pi” part of the emonPi. Then when you’ve done that, you must remember never to update the emonPi as an emonPi (which will cause your modified sketch to be overwritten back to our standard version), but update it instead as an emonBase.
Before I spend a lot of time with the details, would you be prepared to edit the sketch in the Raspberry Pi using a text editor there, then compile it, then use another software tool to load it into the Atmel chip? (I thought this procedure was documented in ‘Resources’, but I can’t find it.)
I gave details of the procedure here but obviously you want to do this with slightly different variables.
That does not seem like something I’d be ready to do. At the moment I am able to use the emonPi to find the run time of a machine, which is what I am most concerned with at the moment. I think it would be very useful to be able to calculate power usage as well, but I am probably not skilled enough to mess with the AVR chip. If I find I want to move forward with this at a latter date I will reach out again, but thank you for all the help thus far.
In that case, the best I can offer is to take the a.c. adapter (the 9 V a.c.) that measures the voltage off the emonPi, and let it calculate the apparent power based on your nominal voltage. As long as you understand the limitations of that, you could adjust the voltage calibration to give a “best guess” for the overall power of the machine, based on the ‘typical’ voltage and power factor.
If you leave the a.c. adapter connected, then it will be using a voltage reference that is shifted by 120° or 240° from the current, which will introduce an error of around a half, but which will vary significantly with the power factor. (That is: if the power factor is exactly unity, the power is exactly half what it should be (cos 120° = -0.5) . If the power factor is not 1, then the 120° gains an apparent shift - I’m dangerously assuming pure sine waves here - that makes the factor greater or less than 0.5)