Twice as much.
"Every CFL light contains a small ac-dc power supply with reactive components in it that will affect the CFL's power factor (PF) - that is, the load presented to the ac line. The closer the PF is to 1, the better. A load with low power factor (
ese
The good point is I save the money running cfls.
ese
that is pretty crappy, even old style fluorescent fixtures are generally 0.8 or better.
Now I'm going to have to try that when I get home to satisfy my curiosity.
nate
The lousy power factor will show up in your bill if you are using them in a business or industrial location where that factor is measured and considered in computing your bill. You'll still come out ahead on cost for the same number of lumens though.
And I've heard tell that utilities in some parts of the country are beginning to meter power factor for residences too. I think it's where there's significant use of air conditioning.
Maybe there's a market out there for some "bulb socket extenders" with power factor correction capacitors inside them.
Jeff
These
One reservation we have about the use of CFLs is that since we heat most homes here with electricity (hydro generated) anyway and never need, in this climate AC, the so-called wasted heat from cheap (25 cent) incandescents (when on at night for example) is merely an alternative to our electric heating! One place that CFLs do make sense is outside, where they are sometimes left on at night for safety and insurance reasons. But CFLs in very cold climates do not seem to be always the best choice and ot you have to buy expensive ones to get good starting and colour! Also CFLs do not seem to be a good or necessary choice for locations where they are flipped on for a short time, such as stairs, cupboards etc. they supposed to be used (like strip fluorescents) where they will be left on continuously. We have a bunch of those, with electronic ballasts, (from a school renovation) in our workshop. Interesting finding; what about switching power supplies also?
It is true that the old incandescent bulbs do provide heat, however, like resistance heat, they are very inefficient. IIRC a good heat pump will provide about four to six times as much heat as a resistance heater or incandescent for the same power consumption.
EJ in NJ
Ernie Willson wrote: ...
... While I'm not going to say lighting is the ideal way to heat, and there's truth to the heat pump, it achieves such efficiencies only when source temperatures are relatively high whereas the resistance heater is the same irregardless.
--
These
Is the reactive portion of the load capacitive or inductive?
Jimmie
Here's a Kill-A-Watt reading from a CFL bulb at my place:
Volts: 119.0 Hz: 59.9 Amps: 0.18 VA: 21 Watts: 10 PF: 0.49 (eek!)
I think what you meant to say is that using electricty only** is a very expensive way to make heat. That doesn't meant that incandescent light bulbs are an inefficient way to provide heat. I believe that they are 100% efficient, in that all the electric power that is used is converted to heat and light, and the light is converted to heat when it lands on a surface (except for the light that that gets out through a window.)
Light is absorbed by a black surface and converted to heat at that time. Light is partially reflected from a white surface, so part of the energy is converted and part is reflected. You can tell that not all of the light is reflected because if it were, when the light source is turned off in an all white room, there would still be light inside the room, when in fact it goes dark almost instantaneusly.
**As opposed to using electricity to power a heat pump or an oil furnace.That's because the heat pump brings heat from the outside to the inside, and the electricity just powers the process. An electric powered coal stoker, that brought coal from a coal pile to a coal furnace would generate even more heat per KWHour, although I don't know that people would call a coal stoker an even more efficient means of heating. Although maybe they would.
You're quoting someone here and this seems wrong. Why would a load be measured as a numerical fraction? A load should be in amps or watts or joules or watt-hours, or whatever, I"m not sure what, but it has to include some unit of measurement. A load is not a ratio. A power factor is a ratio, a numeric value with no units (because the unit of measurement of the numerator is the same unit of measurement of the denominator.)
I'm dubious. Clearly an accruate watt meter and tell how much power is being used at a given moment, and can average those values or otherwise calculate how much power is being used during any period of time.
But where does the numerator in the fraction 0.57 come from. I presume it's the amount of intended use that is coming from the appliaces, also measure in watts (or watt-hours if the denominator is.)
How does a Kill-o-watt meter find out how much light is coming from from a CFL bulb? How does it learn such a value regarding anything that uses electricity?
How could a Kill-o-watt meter measure efficiency since it doesn't know how much work is being accomplished? It only knows how much power is being used.
Plus she doesn't seem to have considered the possiblity that a Killowatt meter that actually measures watts well might not be designed to or able to measure the efficiency of a CFL.
For all these reasons, I wouldn't trust what she says about CFL's using more power than "indicated".
(I haven't read any of the comments on the url yet.)
A load with power factor of even .5 does not draw twice as much power from the generator as one of same wattage with power factor of 1. The requirement to turn the generator is only slightly more.
The main concern of the power company is larger transformers and wires to carry extra current.
- Don Klipstein ( snipped-for-privacy@misty.com)
With the usual low power factor electronic-ballasted CFLs, most of the "non-real" portion of the current is neither inductive nor capacitive, but in the form of harmonics.
- Don Klipstein ( snipped-for-privacy@misty.com)
Except for most USA ones 22 watts or less.
- Don Klipstein ( snipped-for-privacy@misty.com)
I thought it was 2-3 times as much.
- Don Klipstein ( snipped-for-privacy@misty.com)
Which is probably 95% of the time here in Seattle.
15 or 20 years ago that was the range IIRC.
How is it inefficient if that heat is kept inside the house?
It integrates the amp load and the voltage load. If the two are completely out of phase, the PF is 0.
Theoretically, you are correct. As a practical matter, however, that's not how killowatt-hour meters work.
KWH meters assume a resistive load (PF=1). Anything less than a PF of 1 reduces the reading while a PF > 1 (capacitance load) will cause the meter to spin faster (rare). It's possible for a sufficiently large reactive load to burn up bags and bags of energy while registering almost nothing on a standard mechanical KWH meter.
You can look at an incandescent bulb as being something like 99% efficient compared to an ideal resistive heater, meaning that 99% (and I'm pulling that number out of the air) of the electricity that passes through the bulb is eventually converted to heat. Some of it is initially radiated as visible light, but most of that is eventually absorbed by some surface and converted to heat (sort of). I think the only real loss is any visible light that goes out through a window, for instance.
But he's comparing it to a heat pump which, rather than converting the electricity into heat, is using the electricity to move (pump) existing heat from outside the building to inside. Basically, an air conditioner in reverse. Supposedly, under the right conditions, this can bring in more heat than what would be produced by converting 100% of the electricity it uses directly into heat.
I personally know nothing about how much more or under what conditions.
HomeOwnersHub website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.