50W fan same as 50W light?

Hi.

Probably a really simple question, but I just don't know the answer. I just
bought an oscillating floor fan that says 120V-60Hz 50W on the label on the
back. Does that mean it consumes the same electricity as a 50W lightbulb?

Thanks

Tim

Thanks for the information. Just curious, though, as to why it will use a
little more electricity, and why won't that register for billing?

Tim

<(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> It will have the same impact on your electric bill, but technically it
> will use a little more power that won't get billed.
>
> On Sun, 10 Jul 2005 11:57:29 -0400, "Tim Green"
> <(E-Mail Removed)> wrote:
>
>>Hi.
>>
>>Probably a really simple question, but I just don't know the answer. I
>>just
>>bought an oscillating floor fan that says 120V-60Hz 50W on the label on
>>the
>>back. Does that mean it consumes the same electricity as a 50W lightbulb?
>>
>>Thanks
>>
>>Tim
>>
>

On Sun, 10 Jul 2005 19:10:14 GMT, TokaMundo <(E-Mail Removed)>
wrote:

>On Sun, 10 Jul 2005 13:44:53 -0400, "Tim Green"
><(E-Mail Removed)> Gave us:
>
>>Thanks for the information. Just curious, though, as to why it will use a
>>little more electricity, and why won't that register for billing?
>>
>>Tim
> It has to do with the way your meter read inductive loads, and a
>thing called power factor.

It's called reactive power and it has to do with the fact that a 50
watt motor will actually draw more current than the 50 watt light
bulb. The extra current draw is a sine wave and will be slightly
lagging the voltage across the motor.

Because the motor is mostly composed of coils of wire. A certain
amount of power is used to set up a magnetic field to make the motor
work. Since this is alternating current... This field builds up and
collapses 60 times a second (the line frequency). While the field is
building up, it is sucking current from your power company. When the
field collapses, the power is added back to the electrical system (and
goes back in the direction of the power company). The net effect is
that it is not metered, or metered at a net of zero, either way, you
don't pay for it). Heavy commercial users of electric motors have
special electric meters that measure reactive power and they DO pay
for it, but not most homes.

Thus, separate from the actual power that is delivered from the motor
(which you do pay for), you have a magnetic field building up and
collapsing 60 times a second which is required to make the motor run.
Even though you don't pay for this out-of-phase current, it gets added
to the current flowing to the wires in your house.

There is the concept of true power (which your electric meter reads)
and apparent power for reactive (transformer, motor, or capacitor
loads). If you have any type of load other than incandescent lights
or electric radiant heating, the reactive power will always be greater
than the true power. The ratio of true power/apparent power is
called the power factor or p.f. If they can't measure, most
utilities will assume that the power factor is 0.80

Thus, a power company serving a residential subdivision with a 0.80
power factor must provide capacity of 1 / 0.80 or about 25% more
current (or reactive power) than they are billing for. Somebody has
to pay for this additional capacity in transmission and distribution
transformers, and increased wire size, even though the power company
can only charge for true power consumed.

This is why transformers are rated at reactive power levels (VA and
KVA) i.e. volts x amps instead of watts.

It's a difficult concept to understand and others can probably explain
better than me, but I took a stab at it.

Beachcomber

On Sun, 10 Jul 2005 19:10:14 GMT, TokaMundo <(E-Mail Removed)> wrote:

>On Sun, 10 Jul 2005 13:44:53 -0400, "Tim Green"
><(E-Mail Removed)> Gave us:
>
>>Thanks for the information. Just curious, though, as to why it will use a
>>little more electricity, and why won't that register for billing?
>>
>>Tim
> It has to do with the way your meter read inductive loads, and a
>thing called power factor.
The fan was rated in watts, not volt-amperes.

A 50W fan and a 50W lamp will both draw 50 watts as measured by
a watt-hr meter. However, due to power factor, the fan will draw
a higher current at the same voltage due to the power factor.

Power factor is the ratio of watts to volt-amperes.
Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
(E-Mail Removed)

If energy is actually transferred from the (load) back into the voltage
source when the field collapses does that mean if you had a generator hooked
up to your electrical system you could push power back in to the lines and
have a negative electric bill?

"Beachcomber" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> On Sun, 10 Jul 2005 19:10:14 GMT, TokaMundo <(E-Mail Removed)>
> wrote:
>
>>On Sun, 10 Jul 2005 13:44:53 -0400, "Tim Green"
>><(E-Mail Removed)> Gave us:
>>
>>>Thanks for the information. Just curious, though, as to why it will use a
>>>little more electricity, and why won't that register for billing?
>>>
>>>Tim
>> It has to do with the way your meter read inductive loads, and a
>>thing called power factor.
>
> It's called reactive power and it has to do with the fact that a 50
> watt motor will actually draw more current than the 50 watt light
> bulb. The extra current draw is a sine wave and will be slightly
> lagging the voltage across the motor.
>
> Because the motor is mostly composed of coils of wire. A certain
> amount of power is used to set up a magnetic field to make the motor
> work. Since this is alternating current... This field builds up and
> collapses 60 times a second (the line frequency). While the field is
> building up, it is sucking current from your power company. When the
> field collapses, the power is added back to the electrical system (and
> goes back in the direction of the power company). The net effect is
> that it is not metered, or metered at a net of zero, either way, you
> don't pay for it). Heavy commercial users of electric motors have
> special electric meters that measure reactive power and they DO pay
> for it, but not most homes.
>
> Thus, separate from the actual power that is delivered from the motor
> (which you do pay for), you have a magnetic field building up and
> collapsing 60 times a second which is required to make the motor run.
> Even though you don't pay for this out-of-phase current, it gets added
> to the current flowing to the wires in your house.
>
> There is the concept of true power (which your electric meter reads)
> and apparent power for reactive (transformer, motor, or capacitor
> loads). If you have any type of load other than incandescent lights
> or electric radiant heating, the reactive power will always be greater
> than the true power. The ratio of true power/apparent power is
> called the power factor or p.f. If they can't measure, most
> utilities will assume that the power factor is 0.80
>
> Thus, a power company serving a residential subdivision with a 0.80
> power factor must provide capacity of 1 / 0.80 or about 25% more
> current (or reactive power) than they are billing for. Somebody has
> to pay for this additional capacity in transmission and distribution
> transformers, and increased wire size, even though the power company
> can only charge for true power consumed.
>
> This is why transformers are rated at reactive power levels (VA and
> KVA) i.e. volts x amps instead of watts.
>
> It's a difficult concept to understand and others can probably explain
> better than me, but I took a stab at it.
>
> Beachcomber
>
>
>

Bill wrote:
> If energy is actually transferred from the (load) back into the voltage
> source when the field collapses does that mean if you had a generator hooked
> up to your electrical system you could push power back in to the lines and
> have a negative electric bill?
>

Yes. Joe homeowner can't just hook up his generator
willy-nilly and do that. But if the generator & hookup
meet all the applicable rules, the utility here is required
by law to pay for the electricity fed to the grid. That may
apply across all the electric utilities - I don't know.

Ed

On Thu, 14 Jul 2005 03:09:09 GMT, ehsjr <(E-Mail Removed)>
wrote:

>Bill wrote:
>> If energy is actually transferred from the (load) back into the voltage
>> source when the field collapses does that mean if you had a generator hooked
>> up to your electrical system you could push power back in to the lines and
>> have a negative electric bill?
>>
>
>Yes. Joe homeowner can't just hook up his generator
>willy-nilly and do that. But if the generator & hookup
>meet all the applicable rules, the utility here is required
>by law to pay for the electricity fed to the grid. That may
>apply across all the electric utilities - I don't know.
>

Using a gas, propane, or Desiel generator, it is unlikely that you
would be able to manufacture electric power at a cost cheaper than the
power company could sell it to you (remember to include cost of the
maintenance on your genset in addation to the fuel cost).

In principal though, it is possible to sell power back to your utility
through a net metering agreement (your electric meter runs in reverse
when you are feeding power to the grid.)

Here is one example link:
http://www.oksolar.com/inverters/sunny_boy.html
Google Solar Power Inverters for more

Beachcomber