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In old times calibrating a power meter was as easy as connecting 1x, 2x, 3x high power halogen bulbs to get realistic data, free of reactive power or strange waveforms with strong harmonics.

Nowadays simple heaters are not common (I have no electrical room heaters, for example) and light bulbs with LEDs produce harmonics which may affect the accuracy of a power meter, not to mention that calibrating a 2+ kW meter with 5-15 W is unrealistic.

I read online that motors are not suitable for the calibration, therefore hairdryers don't work.

What can I use, easily available in households in 2020 and which power I can accurately estimate with other means, to calibrate my power meter based on (according to Tasmota codes) "HLWBL Energy Monitoring" and "BL0937 Single Phase Energy Monitor"?

Concerning the tools, I have a multimeter which I can use to measure resistance, voltage and current values.

Concerning the devices, besides normal household ones, I also have more unusual devices like an about 80 W terrarium heating cable and a 750 W silicone heater for 3D printer, which however, contrary to the terrarium heater, is not meant to be operated without temperature control (it is meant to be overpowered).

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  • \$\begingroup\$ Most household hair dryers are a reasonable load for calibrating a power meter. The heating element current is usually much, much larger than the motor current. \$\endgroup\$
    – Dwayne Reid
    Commented Sep 23, 2020 at 23:01
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    \$\begingroup\$ Calibration (and I mean calibration and not checking if something works) requires a calibrated and repeatable load. That can be done with calibrated equipment and not some random load. \$\endgroup\$
    – Andy aka
    Commented Sep 23, 2020 at 23:37
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    \$\begingroup\$ Don't use a multi-heat hair dryer or hot air paint stripper, they often achieve the lower power setting with a series diode. \$\endgroup\$
    – Neil_UK
    Commented Sep 24, 2020 at 4:46

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Electric kettle, clothes iron, rice cooker, toaster, etc., are all resistive loads but will suffer rising resistance with temperature so you need to let them stabilise. You can then measure voltage and current with a calibrated multimeter and calculate the power for calibration. This may be enough to get you within 5% to 10%. Calibration is expensive because precision references are expensive.

Don't forget that if you are using a current transformer (CT) that you can use multiple turns and a small load to do the calibration.

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    \$\begingroup\$ just a thought to add to this answer ... you can increase the precision of your readings by using the electrical company power meter readings to determine the actual power used during your measurement \$\endgroup\$
    – jsotola
    Commented Sep 24, 2020 at 1:34
  • \$\begingroup\$ @jsotola Can you? Really? \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Sep 24, 2020 at 17:43
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    \$\begingroup\$ Sure. Switch off everything in your house, switch on the test load, take a meter reading, run until the meter has incremented by 100 counts (typically 0.1 kWh) while monitoring the meter being calibrated, record the end time, switch off the test load and start calculating. \$\endgroup\$
    – Transistor
    Commented Sep 24, 2020 at 17:48
  • \$\begingroup\$ I found that the smart meter and another power meter I found agree within 40 W for a 2 kW kettle. I can use the average of the two values and that's it. I measured the voltage with a tester and the smart meter seems to be more similar to my measurements. \$\endgroup\$
    – FarO
    Commented Oct 12, 2020 at 16:06
  • \$\begingroup\$ That's 2% and while not traceable back to an NIST standard, or similar, it's cheap and probably quite adequate for your application. Good job. \$\endgroup\$
    – Transistor
    Commented Oct 12, 2020 at 16:11
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You are correct in using resistive loads to measure ir calibrate power combined with precision shunts and voltmeters.

There is spreadsheet based calibration handling of uncertainties, factoring temperature changes in the shunt and volt meters, voltmeter error (from specification data) etc.. but lets skip that.

I use a combination of old 200 Watt light globes and the oil filled upright heaters.. which are cheap at appliance stores.. 200 Watt light globes are probably rare, but worth looking for.

You mentioned the hairdryer.. when set to max heat and low fan speed, the power factor will be so close to 1, that it shouldn't bother your measurement. If you are chasing better than 1% accuracy, then the oil heater or bar radiator solution is all you have left.

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    \$\begingroup\$ Why the down-vote? It is helpful to explain your thought process so that the original poster can learn. \$\endgroup\$
    – Dwayne Reid
    Commented Sep 23, 2020 at 23:02
  • \$\begingroup\$ @CPEng please proofread your post ... typos are rampant ... lol \$\endgroup\$
    – jsotola
    Commented Sep 24, 2020 at 1:31
  • \$\begingroup\$ But light bulbs and room heaters are exactly what I mentioned I don't have... also incandescent light bulbs are not sold anymore by law, except for more uncommon formats (which indeed 200 W halogen may be). \$\endgroup\$
    – FarO
    Commented Sep 24, 2020 at 9:08
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    \$\begingroup\$ @DwayneReid The down-vote is due to the extremely sloppy nature of the post. Difficult to read and take seriously. \$\endgroup\$
    – JYelton
    Commented Sep 24, 2020 at 15:04
  • \$\begingroup\$ @JYelton: Awesome! Feedback like that is exactly what the OP needs to improve his/her questions (and answers). \$\endgroup\$
    – Dwayne Reid
    Commented Sep 24, 2020 at 17:27

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