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Fan Power Consumption

V

Vey

Jan 1, 1970
0
Bill said:
Hey, I found the "electrical info plate" on the Lasko. It was stamped
into the sheet metal on the bottom of the fan. After being stamped, the
sheet metal was painted, which obscured the stamped info so completely
that the only way I can see it only with a bright light held at an
angle. As near as I can tell, it says
"Model 3723 (last two digits very unclear, could be almost anything)
Style EC437 (again very unclear)
Type 1
E20739
Listed 154C (UL symbol)
120v 2.2amp 60hz
1922.91"

Bill

120v 2.2amp = 264 watts

Compare that to the wattage you measured (>>> 99/136/196 watts) and you
can see why I said that whatever the plate says is not very accurate.
 
T

The Phantom

Jan 1, 1970
0
120v 2.2amp = 264 watts

This is not watts; it's volt-amps.
Compare that to the wattage you measured (>>> 99/136/196 watts)

You should compare it to the volt-amps he measured. Then it's not so far
off.
and you
can see why I said that whatever the plate says is not very accurate.

You are making a mistaken assumption. If you measure the current draw
(amps) and the applied voltage separately, and multiply them, you get the
(real) watts consumed *only* if the load is a pure resistance. If the load
has a reactive component (inductance or capacitance; in this case the motor
is an inductive load), then the product of volts and amps is the "apparent
power" (volt-amps), not the "real power". See:
http://en.wikipedia.org/wiki/Power_factor

An incandescent light bulb, electric stove, electric toaster or electric
blanket would be an example of a pure resistance load. Most household
equipment other than heating devices, if they have motors (refrigerator,
for example) or non-PF corrected power supplies (older computer or
television) will have a power factor of less than 1, and will require a
special type of meter (a wattmeter) to measure their real power consumption
(because the load has a so-called "wattless" component):
http://en.wikipedia.org/wiki/Wattmeter

As they say on that page, "On an ac circuit the deflection is
proportional to the average instantaneous product of voltage and current,
thus measuring true (real--my addition) power, and possibly (depending on
load characteristics) showing a different reading to that obtained by
simply multiplying the readings showing on a stand-alone voltmeter and a
stand-alone ammeter in the same circuit."

Go back and look at Bill's earlier post where he gives the result of his
actual measurements:

"99/136/196 watts
129/175/248 volt-amps
at 121 volts"

You'll see that volt-amps and watts are substantially different. This is
the beauty of the Kill-A-Watt. It measures both the apparent power and the
true power. The apparent power (the product of separately measured amps
and volts, remember) will be larger than the true power with a fan motor
load, but the true power reading is what you pay for on your electric bill
(if you're a typical residential customer). Large industrial customers pay
a penalty if the apparent power they consume is larger than the true power,
but ordinary residences don't.
 
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