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
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.
On Apr 9, 9:36 pm, email@example.com 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.
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:
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
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>
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
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
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
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
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.)
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.
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.
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 ( firstname.lastname@example.org)
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
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.
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.
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.
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.
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.
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