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AC frequency and power of an electrical appliance

  1. Jan 1, 2016 #1
    So the frequency in the power grid is 50 - 60 Hz depending on country's electrical standard.
    If I have a ~2000 watt heater that runs on 220 volts ~ 9 amps at 50 Hz, would increasing or decreasing the frequency effect the power output of the device. For example if I make the device run at 75 Hz without changing voltage or current would the output power of the device be increased. I haven't learned any mathematical expressions involving frequency when calculating power (W = I * V). If anyone could provide such it will be helpful.
     
  2. jcsd
  3. Jan 1, 2016 #2

    CWatters

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    Small changes to the mains frequency won't significantly change the power dissipated in something like a resistive heating element. They might change the power consumed by appliances that use certain types of motor. Some types of programmable time clock rely on the mains frequency so appliances fitted with those may run fast and operate more frequently meaning a greater average power consumption.

    Are you asking just for curiosity or do you have an application?
     
  4. Jan 1, 2016 #3
    I was thinking that by increasing the frequency without changing other values like volts and current could lead to more heat production without change in power consumption. Maybe I am wrong to assume this but it seems to make sense.

    AC > Frequency Converter > Device. Please explain if I am wrong why and how does it actually work.
     
  5. Jan 1, 2016 #4

    CWatters

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    If it worked as you think it would mean that more power was coming out of the frequency converter than was going in. That would violate conservation of energy.

    For DC you are probably familiar with the equation..

    Power = Volts * Amps

    In the case of AC the instantaneous power is also given by that equation but you are normally interested in the average power. In the case of AC the voltage and current changes with time. You could sample both and multiply all the samples together to calculate the instantaneous power, then calculate the average power from that. However there is a way to "cheat" and multiply the "average" voltage by the "average" current to get the average power.

    I put average in quotes because it's actually the rms voltage and current you use. If you aren't familiar with rms see...
    http://www.nuffieldfoundation.org/practical-physics/explaining-rms-voltage-and-current

    The rms voltage or current doesn't change if the frequency changes so the average power doesn't change either.

    In the case of resistive loads that's about all there is to say. However if the loads are inductive or capacitive then things can get more complicated but perhaps leave that until you are really happy with the above.
     
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