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  • Power factor

    Hello
    If I have a device running on an induction motor, should I divide over the PF to get the actual daily consumption of electricity in KVA not in watts?
    Example a 1500w AC running 4 hours a day with a power factor of 0.8
    Actual consumption in KVA=1500*4/0.8=7500
    ...... then use this figure of 7500 to size the solar panels...
    instead of 1500*4=6000w

  • #2
    An inverter, whether grid tied or battery powered, will consume input power almost entirely based on the wattage of the load (in kW), not the kVA.
    But the current output maximum of an off grid inverter will have to be high enough to satisfy the kVA of the load.
    Your energy calculation for the panels should be based on 6000W not 7500W.

    In the typical residential scale GTI the entire reactive current will be supplied by the grid, and you will pay nothing for it since it will not move the meter.
    SunnyBoy 3000 US, 18 BP Solar 175B panels.

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    • #3
      thanks for the answer inetdog.
      suppose for example that AC has a power factor of 0.3 at initial operation of the induction motor, thus giving a maximum output current of 30 amps when starting, but a continuous current of 8 amps.
      Does the inverter have to be sized to 25 amps output current?
      this means to power a 1500 watt AC at 220 volts alone, i have to install a 7kw inverter?

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      • #4
        Depending on the duration of time that AC draws the higher current rating will determine the size of the inverter and the peak current or watts it can handle. I do not believe you need to go to a 7kw inverter but you will need something greater than 3000 watts.

        My bigger concern is the voltage of the battery system. At 12volts that 1500watt AC will be pulling about 125 amps and more when it starts up. That will require very big wires between the battery and inverter as well as a very big battery system.

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