50W fan same as 50W light?

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

On Sun, 10 Jul 2005 13:44:53 -0400, "Tim Green" Gave us:

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

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 snipped-for-privacy@pvri-removethis.biz

On Mon, 11 Jul 2005 00:58:33 GMT, not snipped-for-privacy@xxx.yyy (Beachcomber) Gave us:

Yes, which is why a power meter wont see all of it.

it is in VARS instead of Watts, and a Wattmeter doesn't "catch" all of it. Just like I said.

I think you did that very well. That kind of explains why there is a penalty for low power factor.

On 11 Jul 2005 09:26:58 -0700, "Kilowatt" Gave us:

That is why it is always in the best interest (if not demanded) of companies to make sure that the electrical products they use (particularly high demand items) are fully power factor corrected.

One would not see companies such as Philips or LANL asking for a power supply that isn't power factor corrected.

It would cost them quite a bit more to operate with poor PF devices as they pay more for their service to begin with, and they are more closely monitored by the power company as well as it is in their best interests to keep their losses as low as possible.

I can't understand why they let leaky high tension lines go for so long though. That has to account for 20% of their losses. I hear sparky insulators all over the place here in SOCAL. In Ohio, I never noticed such appalling leakage.

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

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:

Beachcomber

On Thu, 14 Jul 2005 03:27:21 GMT, not snipped-for-privacy@xxx.yyy (Beachcomber) Gave us:

Build a huge water tower that is fairly high up. Fill it in your spare time by a bicycle pump or hand pump arrangement.

When it is filled, have it drain through a water mill type generator... at night, and feed the grid back for free cash!

I figure one or two hundred thousand gallons should make for a nice bit of the action every couple of weeks. Kind of like collecting cans.

Electricity, whatever it is, does not get consumed. Electrical power also does not get consumed; merely transformed. For practical purposes, electrical power gets converted almost entirely into heat. In the great scheme of things, energy is conserved. Thus the power dissipated by any device can be measured by the number of watts it consumes whether a lamp or a fan.

Bill