CFLs use more energy than indicated

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Would make not a wit of differemce because the motor/compressor is a sealed unit. Heat loss in the motor is directly transfered to the refrigerant.

Not induction, conduction.

Most water heaters are located in an unfinished basement or garage. How much of the heat that escapes from the water heater do you think will go towards heating the living space? With concrete all around that's at ground temp, even in the case of a basement, I'd say it's going to be not much. Which makes it very inefficient.
Even if the water heater is in the living space, typically it's in a utility closet with an outside wall and frequently also contains a furnace, which is sucking in combustion air. Bye bye to most of the heat from the water heater there too.
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On Apr 11, 10:23 am, snipped-for-privacy@optonline.net wrote:

Actualy most of the lost energy goes up the chimney in regular Ng tanks. Thats why most are no better than 60 EF.
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On Apr 9, 9:36 pm, snipped-for-privacy@privacy.net wrote:

It is not; ALL electricity coming into a house is turned into something; all of which ends up as heat. Even if it starts off as light or sound or a TV picture etc. it is all absorbed by the house, it's furnishings and the people in it! (Well Ok a little light escapes through the indows!) We have a bathroom that is almost entirely heated by the six 40 watt bulbs (240 watts) above the vanity mirror. When additional heat is needed (cold weather) the 500 watt electric baseboard with it's thermostat cuts in. Doesn't matter where the electrically made heat comes from, it's all 100% efficient. A few homes here are using (air) heat pumps; but gather that at lower temps, well below freezing for long intervals they don't pump enough heat from the cold air and the auxiliary electric heaters then cut in. So there are long periods when the heat pump is not very effective. (Or efficient!).
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stan wrote:

Stan,
Please cite a reference for your statement that heat pumps are less efficient that incandescent bulbs or resistance heat??
For one kilowatt of energy consumed, incandescent bulbs will provide very nearly one kilowatt of heating (the same goes for resistance heaters). On the other hand a heat pump will provide something between three and four kilowatts of heat for one kilowatt consumed. This means that, compared to heat pumps, resistance heat is very inefficient. If you don't understand this go to:
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/heatpump.html
Where it says:

in that they can use one unit of electric energy to transfer more than one unit of energy from a cold area to a hot area. For example, an electric resistance heater using one kilowatt-hour of electric energy can transfer only 1 kWh of energy to heat your house at 100% efficiency. But 1 kWh of energy used in an electric heat pump could "pump" 3 kWh of energy from the cooler outside environment into your house for heating. The ratio of the energy transferred to the electric energy used in the process is called its coefficient of performance (CP). A typical CP for a commercial heat pump is between 3 and 4 units transferred per unit of electric energy supplied."
Simply stated this says that for one kilowatt of electricity consumed a heat pump can provide between three and four kilowatts of heat. Put another way, from a heating standpoint heat pumps are from 300% to 400% efficient. I believe that this author is somewhat conservative and modern CP's are more like 4 to 6.
EJ in NJ
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wrote:

Except for most USA ones 22 watts or less.
- Don Klipstein ( snipped-for-privacy@misty.com)
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HeyBub wrote:

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. <G>
Jeff
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HeyBub wrote:

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!)
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wrote:

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.)
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mm wrote:

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.
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HeyBub wrote:

Power factor is real power (watts) divided by apparent power (volt-amps for linear loads). You can't have a power factor over 1.
If you connect an "ideal" capacitor (no resistance) to a watt-hour meter it will register zero, just like an ideal inductor. Like an inductor, a capacitor takes energy from the source twice per cycle and returns the same energy to the source twice per cycle.
As you said above, mechanical watt-hour meters mechanically integrate the volts times amps continuously. For an ideal cap or inductor that is negative twice per cycle. I assume a Kill-o-watt meter measures volts and amps in very short intervals and multiplies. It then integrates the readings for watts and watt-hours.
Power factor does not measure efficiency.
--
bud--

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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)
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HeyBub wrote:

Would you kindly explain what PF is? And what is reactive component?
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Briefly, in AC circuits, if you have non-resistive loads, then the current waveform and the voltage waveform may be out of sync. The upshot is that some of the current flowing in the circuit doesn't perform any work, it is just "circulating". The power factor is the ratio of the work the given current is actually doing to the work it could do if voltage and current were in sync.
The reason this matters is that our electrical distribution system (both inside and outside the home) aren't built from superconductors. So anytime current flows, there is some loss from the resistance of the wires. Thus the "circulating" current doesn't do any work, but it does cause some losses.
In the case of compact flourescents, here's the upshot: A reasonable estimate of the power factor of a CFL is 0.5. And CFLs are about 4 times as efficient lighting wise as incandescent bulbs. So if you replace an incandescent with an equivalently sized CFL, you'll be using 1/4 the power for lighting. However, the current in the circuit will only drop by 1/2, so the distribution losses won't decrease quite as much you might think. But overall it is a huge win. [Unless you live in a heating-only climate and use electric resistive heating, in which case you have much bigger efficiency problems than your lighting.]
One important implication of all this, though, is that when sizing branch circuits for CFLs, after adding up the power of the CFLs, you need to correct by the power factor to determine the current draw. For example, if you have a 15 amp branch circuit, maxed out with 18 100W incandescent bulbs on it, and you want more light, you can safely convert to CFLs with a power factor of 0.5 and use 18 50W CFLs (200W equivalent), doubling your light output. But if you were to use 18 100W CFLs (400W equivalent), then you would be drawing 30 amps on the circuit, and you'll have a big problem.
Cheers, Wayne
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On Thu, 09 Apr 2009 21:35:28 -0600, Tony Hwang wrote:

http://en.wikipedia.org/wiki/Power_factor
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On a head to head competitive basis, where one could buy either, who knows which would win, or if both would survive? Some people say they save money, but others are not convinced, especially considering the up front cost of the CFL. Some say they last longer, but it has been my experience that they have a greater failure rate than incandescents.
At any rate, I have purchased enough incandescent bulbs to last me a life time, and it didn't cost much more than buying one full set of bulbs for the house. Also, if an incandescent breaks, I don't have a health issue with mercury.
But most important of all is that I bow my back against any government rule that tells me how I have to live.
Bob-tx
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A lot of this assumes the Kill-a-Watt Gizmo knows what the hell it's doing. I'm not so sure anymore. I know a lot of people on here have used them. I borrowed one from a friend and used it to measure my old vs new fridge last year. So, on a recent trip to Costco, when I saw they had the EZ model for $26, I bought one.
The EZ model lets you enter your cost of electricity per KWH and can then display the cost of electricity used per day, month, year, etc. Upon trying to enter the cost of my electricity, I discovered the "set" key just didn't work at all. So, I took it back and exchanged it. This one, the "set" key works, but the "up" key doesn't. But until I take it back, I figured I'd try it out a bit. I put it on the cat water fountain, which uses a small 2.5W plug-in transformer. It's been on there for 5 days and still reads 0.00$ used per year. It also doesn't register any current or watts either. The transformer is small, rated at onely 2.5W. Just checked it for PF and it shows 1.0
Apparently, this draws so little current it can't measure it. But after having 2 bad units in a row, and this thing being made in China, I'm beginning to wonder about how accurate or reliable it really is.
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On Apr 10, 7:49 am, snipped-for-privacy@optonline.net wrote:

You probably have defective units, I measure power supplies of as little as 4w with the old unit ok. [ I didnt try 2.5w]
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They save 75% on electricity, its a easily proven fact, they last years I have 50 , many run on photocell and after 2 years 2 failures. HD has a 7 yr warranty and top color rendition rating at www.popmechanicsmag.com If you live where AC is used you will really benifit during AC season, Incandesants are actualy electric heaters only outputting about 6% of the energy consumed as light you actualy see, the rest is heat, so run 11, 100w incandesants and you run a 1000w electric heater, and you then run the AC longer to remove that extra 1000w heat. You do and will see a lower electric bill, every place ive been in Ive cut costs 50% overall by just using Cfls. HDs soft white and good and cheap, near 1$ a bulb. Mercury, because so much electricity is generated by coal is it stated using a cfl reduces Mercury air emissions 2-3 times that over burning incandesants over a bulbs lifetime, Airborn Mercury is what poisons water, and fish. That thermostat you threw away has 10,000x more mercury in it.
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Yeah. We had a phone-in programme recently with a temporary host replacing the usual one, who was 'Ooing and Ahing' about various crackpot ideas that people were phoning in to save electricity. Some were laughable.
The host obviously had no idea about the physics involved. And was 'jumping on any band wagon' on the basis that if it was being touted it must be good idea!
Really worrying; it was like some of the debates about oil versus, gas versus electric heating being 'better'. When it all comes down to matter of cost of fuel, cost of installing and maintaining the heating equipment etc. etc.
The latest 'thing' here is wood pellet stoves! But all they can burn of course is pellets! Great some people say and in house about the size of this one, having bought and installed a $1000 stove, with vent, they burn about one bag per day at a cost of say $5 per bag. And there have been some shortages of pellets while the opening of a proposed pellet making plant is now on hold! And the pellets have to be bought and driven home in a gas using vehicle; a pallet of bags of pellets is not a lightweight matter! Try saying a "pallets of pellets" quickly three times!
Hmm; one bag a day is roughly $150 per month and there is other heat in any house from lights, cooking, hot water, TV on, maybe small heater in bathroom etc. etc. so their light bill is say $100. ($150 + $100 = $250).
That brings their total energy bill to more than my total for this all electric now 39 year old and therefore not that well insulated by today's standards, house.
Nothing magic about; you make heat by some (several) means. That costs money and/or effort, the heat escapes from the house, depending how fast is function of the structure.
Gas, except expensive propane, not vailable here.
Stay warm!
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