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I need to measure and log power, and I want to do it mostly correctly. I would greatly appreciate a listing of the errors I have made in my method. I want to measure (and log) the power output of a small 3-phase generator.

Components:

200 w, 24 v, 200 rpm - 3 phase delta generator

200 w buck/boost mppt controller

-10 v to 10 v DATAQ DI-149 data logger

Three 2-ohm, 25 w. resistors

I think logging the voltage is not a problem: measure across the 75k of a 75k/91k pair. The DI-149 can internally convert to rms and log same.

The DI-149 measures the 20 v. span in increments of 0.02 v. My plan was to put a 2 ohm resistor between each of the 3 outputs of the generator and the inputs of the MPPT and log the rms voltage drop and convert to current.

I will multiply the rms voltage and calculated current to arrive at power (and then throw in I^2 R to account for the power lost in the resistor?)?

I would be satisfied with +/- 10% accuracy. Is this going to get me there?

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    \$\begingroup\$ No. Power is not calculated by multiplying voltage with current. Also your current sensing resistor wastes way too much energy. Why not just buy a logger that logs power directly, or a power meter and a logger and connect them together? \$\endgroup\$
    – user3528438
    Commented May 15, 2017 at 20:21
  • \$\begingroup\$ Multiplying Vrms with Irms does not give you power in Watts - it gives you apparent power in VA. For each sample period you need to multiply the V and I samples to get a power value and then rms that. \$\endgroup\$
    – brhans
    Commented May 15, 2017 at 21:01
  • \$\begingroup\$ @brhans I'm afraid power value samples need to be averaged not RMS'ed. \$\endgroup\$
    – carloc
    Commented May 15, 2017 at 22:01
  • \$\begingroup\$ RMS power requires that each V*I is done in real time or filtered and sampled subcycle time then averaged between readings. \$\endgroup\$
    – D.A.S.
    Commented May 15, 2017 at 22:13
  • \$\begingroup\$ but if V is reasonably constant over say 10 cycles then you can compute Rms V and I separately but you must measure the RMS phase difference to extract real power from apparent power. So you see a true RMS meter from Fluke is much better. To my knowledge DATAQ has no builtin way to compute V*I(t) real power only Vrms. \$\endgroup\$
    – D.A.S.
    Commented May 15, 2017 at 23:20

2 Answers 2

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I will multiply the rms voltage and calculated current to arrive at power (and then throw in I^2 R to account for the power lost in the resistor?)?

AC power is more complicated. If you multiply like DC, P=U*I, then you know S, the apparent power in VA. You have to multiply S with the phase angle φ to get the Real Power P in Watts.

I suggest you read the AC power wiki page, and use those formula's to compute your values.

Since you can sample all required 6 channels (voltage and current per phase) at 10 Khz you should be able to get at least some accurate results. However, your buck/boost might cause an inaccuracy in the measurement due to distortion.

You will be in for some serious software work though.

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Since you don't know how true power is measured, and you cannot use a true RMS voltmeter to do this, I would suggest you have the wrong datalogger for this simple job.

Perhaps you can rent a Fluke 1736.

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