I guess the big unknown is the PF, I assumed 60% (based on Googeling), the
is a power supply, would it not be higher for a transformer load?

Stefan

On Tue, Oct 18, 2011 at 12:06 PM, Ben Scott <[email protected]> wrote:

> On Tue, Oct 18, 2011 at 11:37 AM, Stefan Jafs <[email protected]>
> wrote:
> > I'm, trying to figure out how much it cost to run some of our Medical Gas
> > Alarm, current draw is 290 mA at 120 Volts, any ideas?
> > This are my guess after some searches:
> >
> > .290 * 120 = 34.8 VA * 60% (PF) = 20.9 Watts I think that's correct ...
>
>  Looks good to me, assuming your figure for the power factor is correct.
>
> >... but how do I get to kWh?
>
>  Run it for an hour.  ;-)
>
>  20.9 watts is 0.0021 kilowatts.  Run it for an hour, and it's 0.0021
> killowatt*hours.  Run it for half an hour, and it's 0.0011
> killowatt*hours.  Run it for two hours, and it's 0.0042
> killowatt*hours.
>
>  Run it for about 476 hours, and it's 1 killowatt-hour.  (1 / 0.0021 =
> 476.190)
>
>  Water makes a good analogy for electricity:
>
> voltage = pressure (how hard is it pushing?)
> amps = pipe diameter (how much can the wire/pipe carry?)
> watts = gallons/hour (how much does it use to run?)
> killowatt*hours = gallons (total amount of water/power)
>
>  High pressure (voltage) in a small pipe (wire, amps) still yields a
> high gallons/hour (watts).
>
>  How long you run the hose (equipment), times the rate of flow
> (watts) determines the amount of water (power, killowatt*hours) you
> use.
>
> -- Ben
>
> ~ Finally, powerful endpoint security that ISN'T a resource hog! ~
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-- 
Stefan Jafs

~ Finally, powerful endpoint security that ISN'T a resource hog! ~
~ <http://www.sunbeltsoftware.com/Business/VIPRE-Enterprise/>  ~

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