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Computer Power Usage

Discussion in 'Electronic Basics' started by paul, May 4, 2005.

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  1. paul

    paul Guest

    Hello,

    Today I plugged in a power meter I bought a while ago, and decided to
    check out how much power my old Pentium 233MMX was using, here are the
    results (UK Power):

    Normal Operation
    244 Volts, 0.21 Amps, 31 Watts, Power Factor = 0.60

    But if I disconnect the hard drive, I get:
    250 Volts, 0.17 Amps, 41 Watts, Power Factor = 1.00

    How could it be using more power when the hard drive is disconnected?
    41 Watts is how much power it is using, right? Which one (if any) is
    accurate?

    Thanks!

    paul
     
  2. Bob Masta

    Bob Masta Guest

    Just a thought, but I wonder if your power meter is able to handle
    switching power supplies (which are used by all computers).
    The power meter may be expecting simple resistive loads, or
    maybe motors where the current and voltage are a bit out of
    phase, but not be able to handle loads where the waveforms
    are chopped up by a switcher.

    Best regards,


    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
    Home of DaqGen, the FREEWARE signal generator
     
  3. paul

    paul Guest

    If that is the case - that it can't handle switching power supplies,
    would the data be completely wrong? I'm trying to work out how long I
    could power it from a 12v battery on a full charge, but need to know
    how much power it is drawing. Is there a standard/cheap way of working
    out exactly how much power the pc is using? What equipment is needed?

    Cheers,

    paul
     
  4. Bob Masta

    Bob Masta Guest

    This is a very good question, but I unfortunately have no
    answer. Note that if you were actually measuring the
    current from the battery the task would be much simpler,
    since you could assume that the voltage was pretty
    much a constant during the current draw. Then all that
    would be needed would be to integrate the total current
    in amp-hours. But if you already had the battery, the
    inverter, and the computer you could skip all that and
    just time it directly!

    I think the power meter has to integrate the product of
    instantaneous voltage times instantaneous current,
    averaging over one or more cycles of the mains frequency.
    I don't know what typical commercial power meters do.
    Anyone else have any info on that?

    Best regards,


    Bob Masta
    dqatechATdaqartaDOTcom

    D A Q A R T A
    Data AcQuisition And Real-Time Analysis
    www.daqarta.com
    Home of DaqGen, the FREEWARE signal generator
     
  5. paul

    paul Guest

    I wanted to work out the power usage, before obtaining the battery &
    inverter, so I would know exactly what size/types I'd need to get :)

    Regards,

    paul
     
  6. Wayne Farmer

    Wayne Farmer Guest

    Based on my limited knowledge (having just finished an AC circuit theory
    course at my local community college), I believe there are 3 ways to measure
    the power drawn by a load from an alternating voltage source:

    1. Apparent power. This is RMS AC Volts x RMS AC Amps. It is reported not
    in Watts, but in "VA" = Volt-Amperes.

    2. True power. This is the resistive component of Apparent Power, and is
    reported in Watts. If the load has no capacitive or inductive elements,
    then all the load is resistive. True power is the power that is dissipated
    in heat through your load.

    3. Reactive power. This is the inductive and capacitive (= reactive)
    component of Apparent Power, and is reported in "VAR" = Volt-Amperes
    Reactive. Reactive power isn't dissipated in heat, but cycles back and
    forth between you and the power company as alternating current flow.

    Apparent power is the vector sum of True Power and Reactive Power: the
    square root of (True Power**2 + Reactive Power**2).

    The ratio of True power to Apparent power is the Power Factor.

    Motors (hard drives, fans, etc.) are typically inductive loads.

    Now let's look at the figures your meter is reporting:
    244 V x .21 A = 51.2 VA apparent power. Of that, the 0.60 Power Factor x
    51.2 VA = the 31 Watts true power being consumed by the resistive loads in
    your computer and hard drive. At the same time, there is a reactive
    (inductive and capacitive) power draw of sin(arccos 0.60) x 51.2 VA = 41
    VAR. The square root of (true power**2 + reactive power**2) is indeed 51.4
    VA, which matches your figure of 51.4 VA.
    250 V x .17 A = 42.5 VA apparent power. Since the Power Factor is 1.00, all
    of that apparent power is due to the resistive loads in your computer and
    hard drive; the true power is 42.5 Watts, or approximately the 41 Watts you
    reported. There is little or no inductive and capacitive load. That makes
    sense, because the hard drive motor was a major part of the inductive load
    when it was connected.

    So, you can see that the apparent power draw does decrease, from 51.2 VA to
    42.5 VA, when you disconnect the hard drive.

    Can somebody else explain why the true power (resistive load) -increases-
    from 31 Watts to 41 Watts when the hard drive is disconnected?

    Wayne
     
  7. Back in the ancient days, and thus this information may or may not apply
    to the current day, computer power supplies had big heavy expensive iron
    transformers. When the slightly more modern, but still ancient, "pc was
    invented" transformers were replaced by lighter cheaper and often much
    poorer quality switching supplies. The widespread knowledge of those old
    days was that you NEVER turned on one of these supplies without some load,
    typically having at least a 5" full height hard drive attached to do this.

    Not having a load on a cheap low quality switching supply can lead to wild
    and unhappy oscillation within the supply, often coming to an end in a few
    seconds when you go and get a replacement for your supply.

    Perhaps the supply operating without a load is demonstrating its unhappy
    state by increasing dissapation by 35%+. Adding resistive load in small
    increments and seeing whether the true power decreased before increasing
    might help test this guess.

    I once destroyed one of these supplies just by switching it off and then
    back on too quickly.
     
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