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Household KWH usage monitor - Page 2 — Parallax Forums

Household KWH usage monitor

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Comments

  • kwinnkwinn Posts: 8,697
    edited 2010-10-12 21:55
    Erik, you are absolutely right. If the current and voltage were measured simultaneously (or close to simultaneously) we could calculate the instantaneous power and the average power over one cycle. With enough samples this would be the most accurate way to compensate for power factor and waveform distortion. As usual the devil is in the details. Refer to the attached diagram and explanation.
    I think the simplest approach would be to use a low end micro with several channels of ADC (8+ channels), use one channel to measure voltage and the others for the current measurement using a shunt resistor.
  • Erik FriesenErik Friesen Posts: 1,071
    edited 2010-10-13 07:13
    Doing some more thinking about this;

    I can't really see the shunt thing working, for example, on the oven circuit.

    I don't think the incoming voltage sine wave would be distorted sufficiently to create much error. 5% max? I could either store the wave in an equivalent khz buffer to be applied (using some peak/zero crossing detection algorithm) or use a lookup table of some sort.
  • Erik FriesenErik Friesen Posts: 1,071
    edited 2010-10-13 07:33
    Another thing to consider -

    I talked to my power company today, and they basically told me their meters don't figure in a power factor, and since they are electronic meters, I assume they have a 100% power factor in the calculation.
  • kwinnkwinn Posts: 8,697
    edited 2010-10-13 16:01
    Erik, I agree with you regarding the shunt thing on the oven circuit and the distortion not being much of a problem. Most household loads are either resistive (heaters, ovens, incandescent lights) or inductive (AC, furnace fan motors). The switcher type loads (ccfl's, computer, electronics) are relatively low power and/or not on for much of the time, so the power factor error is not much relatively speaking.
    If I were to tackle such a project I would use one ADC input to measure the input voltage and use the rest for current measurements.
  • gunstarg10gunstarg10 Posts: 4
    edited 2010-10-20 17:01
    Hi Everyone,

    So I've done some initial testing with a setup that I've made. I'm using these CT's http://www.magnelab.com/products/Current-Sensors/Split-Core-AC-Current-Sensor-SCT-0400. I amplified the output voltage of the CT's and then perform a half wave rectification, so the voltage is all positive. I believe this output should be ready to use with the adc and the propeller. I have done some testing and not all small appliances result with a sin wave when I read the output on an oscilloscope, and thus I thought I would need to compute the rms of the signal using the propeller. This is where I am not sure how to move forward. Does anyone have any sample code or references on how to sample a signal and possibly compute the rms. Thanks.
  • Erik FriesenErik Friesen Posts: 1,071
    edited 2010-10-20 18:16
    Wiki says - The RMS value of an alternating current is also known as its heating value, as it is a voltage which is equivalent to the direct current value that would be required to get the same heating effect. For example, if we applied 120VAC RMS to a resistive heating element it would heat up by exactly the same amount as if we had applied 120V DC.

    It seems that if you track your voltage and amps with enough bandwidth, you could sum them up. Is it really necessary to compute true RMS?

    Those are nice looking split CT's. What do you have to give for them?

    I haven't given up on this project, just have other things on the burner yet.
  • FranklinFranklin Posts: 4,747
    edited 2010-10-20 18:44
    Here's what the catalog says.http://www.magnelab.com/catalogue
  • Erik FriesenErik Friesen Posts: 1,071
    edited 2010-10-20 18:48
    Yes, maybe he can get those cheaper somewhere?
  • kwinnkwinn Posts: 8,697
    edited 2010-10-20 21:13
    At $30.00 each monitoring 30 circuits is a pricey project. As for converting a sine wave to RMS, the peak voltage x 0.707 gives you the RMS voltage. In this case though that voltage must then be converted to the current as well.
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