Determining Wattage Consumed by a Hair Dryer????

[NYC Doc]

Recently, while perusing Ebay, I noticed an auction (actually several) for a
2800 watt "Volcano" hair dryer. This was 800 more watts than I had ever
seen. My initial reaction was that it couldn't be possible because this
would lead to triggered circuit breakers. So I purchased a hair dryer in
addition to a non-contact current meter.

Using a split extension cord, it was registering 9.8 amps, slightly less
than my 1800 Conair dryer. So, it is safe to say that [most of the]
voltage is in phase with current? I would assume that 99% of the energy
goes to the heating element.


If this is this case, then the power should be

9.8 * 2 *120 (approx) * 0.99 ???

= 2156 ?

 

[John Popelish]

NYC Doc wrote:

Recently, while perusing Ebay, I noticed an auction (actually several) for a
2800 watt "Volcano" hair dryer. This was 800 more watts than I had ever
seen. My initial reaction was that it couldn't be possible because this
would lead to triggered circuit breakers. So I purchased a hair dryer in
addition to a non-contact current meter.

Using a split extension cord, it was registering 9.8 amps, slightly less
than my 1800 Conair dryer. So, it is safe to say that [most of the]
voltage is in phase with current? I would assume that 99% of the energy
goes to the heating element.


If this is this case, then the power should be

9.8 * 2 *120 (approx) * 0.99 ???

= 2156 ?

For a resistive load (and this is a good example of one of those)
watts is RMS Amperes times RMS volts, so 9.8 amperes times 120 volts =
1176 watts. They lied.

 

[Ralph Mowery]

"NYC Doc" <nd@nd.com> wrote in message
news:sHoTe.29363$%w.4613@twister.nyc.rr.com...

Recently, while perusing Ebay, I noticed an auction (actually several) for
a
2800 watt "Volcano" hair dryer. This was 800 more watts than I had ever
seen. My initial reaction was that it couldn't be possible because this
would lead to triggered circuit breakers. So I purchased a hair dryer
in
addition to a non-contact current meter.

Using a split extension cord, it was registering 9.8 amps, slightly less
than my 1800 Conair dryer. So, it is safe to say that [most of the]
voltage is in phase with current? I would assume that 99% of the energy
goes to the heating element.


If this is this case, then the power should be

9.8 * 2 *120 (approx) * 0.99 ???

= 2156 ?

The total wattage is the current times the voltage. 9.8 *120 = 1176 watts.
I don't know how you managed to put the 2 in your equation. The dryer is
purely resistance (not counting the motor which should be a very small draw)
so all the current would be converted to heat.
If it was a 2800 watt unit , it would draw 23 amps. That is 3 more than
than the wall sockets are rated at the most. I doubt that any dryers would
really be rated over 1500 watts if that was their true value. That is
almost the limit of most house circuits, especially if they have anything
else on them.

I did a quick check of one at the house. It is rated 1875 watts. It
actually uses 11 amps on the highest heat. The motor uses about 1.5 amps.
This only leaves 9.5 amps for the heater. At most it could be rated 120 *
11 = 1320 watts. This is for the motor and heater.

I think they just pull numbers out of the air for the dryers. I guess they
could multiply it by 1.4 to give it a fake peak wattage instead of a RMS
wattage. That would make it about right.

 

[NYC Doc]

Isn't AC current doubled? Based on various tests, it appeared that the
current measurement only represents 1/2 of the current. For a 1000 watt
device, I was getting 5 amps, for a 2000 watts device, I was getting 10
amps.




"Ralph Mowery" <rmowery28146@earthlink.net> wrote in message
news:NtpTe.8043$_84.6921@newsread1.news.atl.earthlink.net...

"NYC Doc" <nd@nd.com> wrote in message
news:sHoTe.29363$%w.4613@twister.nyc.rr.com...
Recently, while perusing Ebay, I noticed an auction (actually several)
for
a
2800 watt "Volcano" hair dryer. This was 800 more watts than I had ever
seen. My initial reaction was that it couldn't be possible because this
would lead to triggered circuit breakers. So I purchased a hair dryer
in
addition to a non-contact current meter.

Using a split extension cord, it was registering 9.8 amps, slightly less
than my 1800 Conair dryer. So, it is safe to say that [most of the]
voltage is in phase with current? I would assume that 99% of the energy
goes to the heating element.


If this is this case, then the power should be

9.8 * 2 *120 (approx) * 0.99 ???

= 2156 ?

The total wattage is the current times the voltage. 9.8 *120 = 1176
watts.
I don't know how you managed to put the 2 in your equation. The dryer is
purely resistance (not counting the motor which should be a very small
draw)
so all the current would be converted to heat.
If it was a 2800 watt unit , it would draw 23 amps. That is 3 more than
than the wall sockets are rated at the most. I doubt that any dryers
would
really be rated over 1500 watts if that was their true value. That is
almost the limit of most house circuits, especially if they have anything
else on them.

I did a quick check of one at the house. It is rated 1875 watts. It
actually uses 11 amps on the highest heat. The motor uses about 1.5 amps.
This only leaves 9.5 amps for the heater. At most it could be rated 120 *
11 = 1320 watts. This is for the motor and heater.

I think they just pull numbers out of the air for the dryers. I guess
they
could multiply it by 1.4 to give it a fake peak wattage instead of a RMS
wattage. That would make it about right.

 

[John Popelish]

NYC Doc wrote:

Isn't AC current doubled?

AC current measured by almost any meter is the RMS value. The peaks
are the square root of 2 (1.414) times that. Same with voltage. The
meter readings are useful, because you simply multiply the meter
current by the meter voltage (for a resistive load) to get average
power. The instantaneous power is popping up to twice that, but also
going to zero, twice per cycle. But the normal use of Watts is the
average value of that swing.

Based on various tests, it appeared that the
current measurement only represents 1/2 of the current. For a 1000 watt
device, I was getting 5 amps, for a 2000 watts device, I was getting 10
amps.

Then your measurement is off (for whatever reason) and you may need
that factor of approximately 2 to get the true RMS current. But your
meter is not right. Is it a clamp on ammeter? Are you passing only
one wire through it only once, and keeping the other conductor well
away from the clamp? How do you know the other loads are nay more
honest than the hair dryer?

By the way, it is illegal in the US to sell any appliance to be
plugged into a household 15 ampere receptacle that draws more than 12
amperes (except for a brief burst at turn on). That limits the power
to no more than 12*120=1440 watts.

 

[Bob Eldred]

"NYC Doc" <nd@nd.com> wrote in message
news:qXsTe.29388$%w.9520@twister.nyc.rr.com...

Isn't AC current doubled? Based on various tests, it appeared that the
current measurement only represents 1/2 of the current. For a 1000 watt
device, I was getting 5 amps, for a 2000 watts device, I was getting 10
amps.

There is no "2" or doubling of the current. I don't know where you got that
idea? It's just volts times amps in RMS values or in DC values in a DC
circuit.
Bob

 

[Ralph Mowery]

"NYC Doc" <nd@nd.com> wrote in message
news:qXsTe.29388$%w.9520@twister.nyc.rr.com...

Isn't AC current doubled? Based on various tests, it appeared that the
current measurement only represents 1/2 of the current. For a 1000 watt
device, I was getting 5 amps, for a 2000 watts device, I was getting 10
amps.


On a standard 60 hz circuit the current and voltage should be measured with

a meter that is calibrated in RMS. You multiply them together toget the
wattage. Either your meter is wrong or the loads you are looking at are
mislabled.

I suspect the hair driers must be rated like some of the computer speakers I
have seen. They will be rated at 40 watts and driven by a small wall cube.
If you open them up they may be marked for 2 watts. Someone came up with
the advertising hype that for a milisecond or so they could have a peak of
the 40 watts. Pure BS for advertising.
Just some more of the junk advertising like the ads you see on TV for some
cars and trucks. Saw one ad where a 4 wheeler comes up to the barn and goes
up the outside wall..try doing that in the real world.

 

[Kitchen Man]

On Tue, 06 Sep 2005 23:33:31 -0400, John Popelish <jpopelish@rica.net>
wrote:

By the way, it is illegal in the US to sell any appliance to be
plugged into a household 15 ampere receptacle that draws more than 12
amperes (except for a brief burst at turn on). That limits the power
to no more than 12*120=1440 watts.

You didn't factor in the SQRT-2 advertisement allowance

1440 * 1.414 ~= 2037, so I can see why the 2800W rating would be 800W
higher than the OP had ever previously seen advertised. Reckon the
"advertisement allowance" is a might higher than usual for Volcano
brand. Do you think they might be using a peak-to-peak value?

Lessee, OP said his Volcano drew 9.8Arms. 9.8 * 120 * 1.414 ~=
1660Wpk, or 3320Wpp. Subtract 520Wpp (183Wrms, or ~1.5Arms) for the
blower motor, and we may have arrived at the marketing genius behind
the 2800W Volcano. Brilliant!