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RayV
 
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Default How does an electric meter work?

I did not check this out on snopes but have heard that the eletric
meters on our houses always read the higher of the two lines coming in
to determine the amount of juice used. The rumor goes on that you
should *balance* your panel so that the same number of constant users
are on either side of your panel. An example might be that your fridge
should be on one side and your furnace on the other, or electric range
on one side and electric dryer on the other.

Is there any truth to this or is it one of those tall tales based on
some obscure Tesla theory?

Would this gadget really save me money on my electric bill?

http://www.renaissance-pacific.net/E...oduct_Info.pdf

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Default How does an electric meter work?

the only thing this does is bring more money to the merchant.

Yes both sides of the power line should be balanced, but BOTH items
must be on at the same time. So the neutral doesnt carry current.

SAVE YOUR MONEY!

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kevin
 
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Default How does an electric meter work?

Wow.

I actually read that whole PDF just now. It's kind of interesting.
Other than not explaining in technical terms, looking silly, and having
a lame brochure, it actually doesn't make any wild claims like I would
have expected.

It makes a claim of 3-5% reduction. It specifically says it doesn't
work for digital meters, and it is not a magnet, and it won't cause
inductive heating problems. It mentions a handfull of other caveats too
even. Since when does a scam make such modest claims and have so many
caveats? What is the world coming to? Will I start getting "Enlarge
your ***** by 3%!", or "Lose 1 to 3 lbs and keep it off as long as you
keep exercising and maintaining a healthy diet!" emails?

So can anyone answer the op's original question? Does a mechanical
meter overestimate when the load is imbalanced? And if so, can anyone
think of a plausible scenario, in any possible situation, real or
imaginary, that would let the device in the PDF picture (I can't tell
what it is -- just some kind of metal C-shaped thing I guess) do
anything that could even have a remotely possible chance of having even
a miniscule, undetectable, insignificant but still non-zero effect on
the meter? I.e., is this just a not-very-useful and very poorly
marketed invention, or a not very useful and poorly hyped scam?

-Kevin

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Default How does an electric meter work?

I would look into power factor correction first.

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Default How does an electric meter work?

I have wondered why certain appliances (espically washers and dryers)
didn't already have power factor correction already built in.

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Jeff Wisnia
 
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Default How does an electric meter work?

wrote:
I would look into power factor correction first.


There are companies marketing power factor correcting boxes to
homeowners. They are just boxes with a bunch of capacitors in them and
instructions on how to connect up the "right" number of capacitors to
your incoming power conductors to try and bring your home's average
power factor closer to unity. I think the guys promoting them are just
"Sharp cookies selling Girl Scouts." G

AFAIK none of those boxes do an automatic correction when the overall
power factor of the loads being used changes as different appliances get
turned on and off.

Also, the big joke is that very few utilities install meters which can
measure power factor at peoples homes so nearly all the residential
meters in use now measure only the "real power" consumed anyway.

So, using one of those power factor correction boxes won't do much to
directly reduce your electric bill. It will however help reduce the
power company's losses on THEIR lines, which in a perfect world could
allow them to pass those savings back to their customers about the same
time as pigs start flying.

Installing power factor correction capacitors directly at the larger AC
motors used home appliances would reduce by a tiny amount the power
"wasted" in heating wiring within your house, but I wouldn't expect the
resultant savings to be worth the effort.

A bit more effective is the addition of capacitors arcoss the starting
winding switches on induction motors to make them work in a "capacitor
run" mode. That was shown by some US Navy researcher back in the 70s
(IIRC) to improve the efficiency of those kind of motors by about 10%.
AFAIK nothing much ever came of that, but maybe the skyrocketing cost of
fuels will bring that idea to the fore again.

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"Truth exists; only falsehood has to be invented."
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unknown
 
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Default How does an electric meter work?

On 24 Apr 2006 06:59:12 -0700, "RayV" wrote:

I did not check this out on snopes


Try doing so. That claim doesn't make sense.

but have heard that the eletric
meters on our houses always read the higher of the two lines coming in
to determine the amount of juice used. The rumor goes on that you
should *balance* your panel so that the same number of constant users
are on either side of your panel. An example might be that your fridge
should be on one side and your furnace on the other, or electric range
on one side and electric dryer on the other.


240V appliances are always on both sides. They have 2 hot wires, 1
connected to each side.

Is there any truth to this or is it one of those tall tales based on
some obscure Tesla theory?

Would this gadget really save me money on my electric bill?

http://www.renaissance-pacific.net/E...oduct_Info.pdf


Sounds like BS.
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Mark and Gloria Hagwood
 
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Default How does an electric meter work?

My understanding is that large, commercial, installations
such as factories, supermarkets, refrigerated warehouses
etc. use capacitor banks to adjust the power factor of their
motors. In fact, power companies also have capacitor banks
installed alongside the street, beside switches, to adjust
the "average" power factor of loads to neighborhoods. On
the individual scale, where a commercial consumer has a
meter that measures amperes against time, that can be a
benefit to both the consumer and the power company.

The average homeowner has a meter that's called a VAR meter-
Volt Amp Reactive. The cumulative power factor of the
home's A/C, fan motor, refrigerator etc. are not measured by
the meter and even a professionally calibrated and installed
capacitor bank would make no difference in the 'Kilowatt
Hours" consumed each month. To take advantage of this, the
consumer would have to install a meter such as found in
commercial installations. On some home-type of
installations, such as an A/C compressor, pump, air
compressor or possibly a refrigerator, a run capacitor to
correct the power factor of the individual motor might
benefit from a longer life or cooler operation, but it would
be minimal and not seen on the electric bill.

I'm sure that there are some engineers out there who can
correct what I've written or expand on it. I'd sure like to
learn more about this.

Mark
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Beachcomber
 
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Default How does an electric meter work?

On Mon, 24 Apr 2006 16:51:21 -0500, unknown
wrote:

On 24 Apr 2006 06:59:12 -0700, "RayV" wrote:

I did not check this out on snopes


Try doing so. That claim doesn't make sense.

but have heard that the eletric
meters on our houses always read the higher of the two lines coming in
to determine the amount of juice used. The rumor goes on that you
should *balance* your panel so that the same number of constant users
are on either side of your panel. An example might be that your fridge
should be on one side and your furnace on the other, or electric range
on one side and electric dryer on the other.


240V appliances are always on both sides. They have 2 hot wires, 1
connected to each side.

Is there any truth to this or is it one of those tall tales based on
some obscure Tesla theory?

Would this gadget really save me money on my electric bill?

http://www.renaissance-pacific.net/E...oduct_Info.pdf


Sounds like BS.



Not only that, there is no explanation as to how this thing works that
conforms to the known laws of physics.

Any little piece of material that you wrap around your service
entrance conductors is going to have little if any effect on
"equalizing" the load between the hot wires. It is the current
flowing that creates the magnetic field.

Beachcomber

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kevin
 
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Default How does an electric meter work?

Thanks for answering (partially) my question. This thread is
interesting -- it seems to be filled with more BS and hype than the
original link even was.

I don't entirely understand your answer, though. I wasn't aware that
there was a "neutral on the utility's side of the meter". And if there
were, and I had a perfectly balanced load, then the neutral would have
no current, right? Which means... no losses, and hence no cost?
Obviously wrong... can you explain a little more?

And moreover, there was a lot of BS in this thread about balancing the
panel to begin with. To say that 240V devices are on both legs is
irrelevant to the question of whether balancing a panel makes economic
sense for the homeowner. Does a homeowner with a horribly imbalanced
panel get charged more than the same homeowner if they just rearranged
the breakers on the panel?

-Kevin

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Jeff Wisnia
 
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Default How does an electric meter work?

Dan Lanciani wrote:
In article .com, (kevin) writes:

| So can anyone answer the op's original question? Does a mechanical
| meter overestimate when the load is imbalanced?

Yes, in the sense that the typical 4-terminal (1.5 element) meter most
often used in residential split-phase service charges the customer for
150% of the losses in the neutral on the utility's side of the meter.
(That's the simple case of one service drop from transformer to customer.
For multiple connections the analysis gets more complicated.) This seems
to surprise many people. Google for Blondel's theorem before trying to
come up with an argument that the meter does not make this error.



I believe you, and If I understood you, when speaking of mechanical
meters the 1.5 element meter you describe has one voltage field across
the 240 volt line and two current fields, one in series with each "hot"
conductor.

I think I see what you're referring to.

Reducing it to absurdity, if you only load one side of the line, all the
return current flows through the neutral, but the voltage field still
"sees" the full 240 volts at the meter terminals even though the power
you are using isn't the product of the current drawn time half of the
240 that voltage, it's the current times half of that voltage MINUS the
voltage drop in the neutral.

So, you get charged for the power equal to that current times the "full"
voltage, when you are actually using power equal to that current times a
slightly reduced voltage.

Izzat it?

Jeff



N.B. This has nothing to do with the meter's being mechanical. You can
build a 5-terminal (2 element) mechanical meter that does not have the
error.


I'm guessing that kind of meter must have TWO voltage fields?

You can build a 4-terminal (1.5 element) electronic meter that
does have the error.

| And if so, can anyone
| think of a plausible scenario, in any possible situation, real or
| imaginary, that would let the device in the PDF picture (I can't tell
| what it is -- just some kind of metal C-shaped thing I guess) do
| anything that could even have a remotely possible chance of having even
| a miniscule, undetectable, insignificant but still non-zero effect on
| the meter?


Maybe it's the AC equivalent of the "cow magnet" to be placed on the
fuel ine of your car. :-)

Jeff


--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"Truth exists; only falsehood has to be invented."
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Bud--
 
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Default How does an electric meter work?

kevin wrote:

Thanks for answering (partially) my question. This thread is
interesting -- it seems to be filled with more BS and hype than the
original link even was.


I agree with almost everything said in this thread.

I don't entirely understand your answer, though. I wasn't aware that
there was a "neutral on the utility's side of the meter". And if there
were, and I had a perfectly balanced load, then the neutral would have
no current, right? Which means... no losses, and hence no cost?
Obviously wrong... can you explain a little more?


There is a neutral from the power transformer to the service panel (then
beyond). A balanced load would result in zero service neutral current.
An unbalance would result in neutral current, a small voltge drop and
negligible power loss (small volt drop times current). If this were a
big deal, all wire sizes including to circuits inside your house would
be increased in size for lower voltage drop/lower losses.

And moreover, there was a lot of BS in this thread about balancing the
panel to begin with. To say that 240V devices are on both legs is
irrelevant to the question of whether balancing a panel makes economic
sense for the homeowner. Does a homeowner with a horribly imbalanced
panel get charged more than the same homeowner if they just rearranged
the breakers on the panel?

-Kevin


Far as I know meters accurately measure power used, balanced or not. The
gizmo is a scam. (And if a steel (EMT) service entrance pipe is used the
pipe shields the gizmo from the magnetic field.)

Another scam, as others have said, is power factor correction - unless
power factor is measured on residential meters, which I have never heard
of. Even if PF is measured on residential some of these units are scams.

bud--

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Dan Lanciani
 
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Default How does an electric meter work?

In article . com, (kevin) writes:

| I don't entirely understand your answer, though. I wasn't aware that
| there was a "neutral on the utility's side of the meter".

A typical split-phase residential service will have three wires from the
transformer to the meter: two hots and a neutral (or grounded conductor
if you prefer).

| And if there
| were, and I had a perfectly balanced load, then the neutral would have
| no current, right?

correct

| Which means... no losses, and hence no cost?

With a balanced load there would be no loss in the neutral and hence no
cost for power wasted in that neutral. There would still be cost for
the power you actually use, and that swamps the effects we are discussing
in most real cases.

| Obviously wrong... can you explain a little more?
|
| And moreover, there was a lot of BS in this thread about balancing the
| panel to begin with. To say that 240V devices are on both legs is
| irrelevant to the question of whether balancing a panel makes economic
| sense for the homeowner. Does a homeowner with a horribly imbalanced
| panel get charged more than the same homeowner if they just rearranged
| the breakers on the panel?

Yes, but the effect is very small. Again, the homeowner with a typical
split-phase service and 4-terminal (1.5 element) meter will pay for 150%
of the power lost in the neutral on the utility's side of the meter. A
perfectly balanced load would eliminate this loss. But unless you have
a long undersized neutral and a horribly unbalanced load (and possibly
even if you do) it (a) isn't worth worrying about and (b) is very hard
to fix.

Dan Lanciani
ddl@danlan.*com
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Dan Lanciani
 
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Default How does an electric meter work?

In article , (Jeff Wisnia) writes:
| Dan Lanciani wrote:
| In article .com,
(kevin) writes:
|
| | So can anyone answer the op's original question? Does a mechanical
| | meter overestimate when the load is imbalanced?
|
| Yes, in the sense that the typical 4-terminal (1.5 element) meter most
| often used in residential split-phase service charges the customer for
| 150% of the losses in the neutral on the utility's side of the meter.
| (That's the simple case of one service drop from transformer to customer.
| For multiple connections the analysis gets more complicated.) This seems
| to surprise many people. Google for Blondel's theorem before trying to
| come up with an argument that the meter does not make this error.
|
|
| I believe you,

That's comforting. I've had some incredibly long threads (last in the
alt.electrical.engineering group I think) trying to convince people that
the meter could be wrong.

| and If I understood you, when speaking of mechanical
| meters the 1.5 element meter you describe has one voltage field across
| the 240 volt line and two current fields, one in series with each "hot"
| conductor.

Yes, and I suspect most electronic meters are the same. After all,
they plug into the same meter base as the mechanical meter they replace
and there is usually no way for them to connect to ground/neutral
reliably.

| I think I see what you're referring to.
|
| Reducing it to absurdity, if you only load one side of the line, all the
| return current flows through the neutral, but the voltage field still
| "sees" the full 240 volts at the meter terminals even though the power
| you are using isn't the product of the current drawn time half of the
| 240 that voltage, it's the current times half of that voltage MINUS the
| voltage drop in the neutral.

Yes, though there will be some measured drop in the hot line as well so
if hot-to-hot started out as 240V at the transformer it won't be 240V
at the meter. It works out to your paying for 150% of the loss in the
neutral: the sum of all the loss in the neutral itself plus half the
loss in the loaded hot leg.

Now on a multi-drop setup you could in theory make this work to your
advantage. Take note of which hot leg appears higher with respect to
the neutral (on account of your neighbors' unbalanced loads) and shift
your own loads to this leg (but no more than will tend to pull the
neutral to the other side of center). Now you are getting free power
to the same extent of the metering error and at the expense of your
neighbors. The power company always wins (or at worst breaks even if
everything is perfectly balanced), though, which may be part of the
reason they like undersized neutrals.

| N.B. This has nothing to do with the meter's being mechanical. You can
| build a 5-terminal (2 element) mechanical meter that does not have the
| error.
|
| I'm guessing that kind of meter must have TWO voltage fields?

Yes, the terminology seems to be (1 element) = (1 voltage field plus 1
current field). I don't really like this since it seems to me that
"1.5 elements" could be ambiguous. Is the extra element a voltage or
a current winding?

Dan Lanciani
ddl@danlan.*com


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Calvin Henry-Cotnam
 
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Default How does an electric meter work?

Edwin Pawlowski ) said...

Power correction factor is different. It can be done inside, after the meter
and is perfectly legal. I don't know if it is a viable option for
residential, but for industrial use, there are savings to be had.


It is not a viable issue for residential customers as they are charged
for actual power used (kWh).

Industrial users tend to have heavy motor loads that create a lagging
power factor. With a power factor of 1, kVA is equal to kW, so the
facilities needed to deliver 1000 kWh only has to be able to carry
1000 kVAh, and the losses due to heating that equipment is at its minimum.

If the customer's power factor was 0.5, then the equiment need to deliver
that same 1000 kWh needs to support 2000 kVAh! This means twice the current
for the same power. Since heat losses are proportional to the SQUARE of the
current, you lose FOUR times the power just heating the infrastructure.

Industrial customers are charged for kVAh if they exceed a certain
threshold, so it is in their interest to keep the power factor close to
1. They maintian large capacitor banks to do so.

Residential customers do not generally run at a lagging power factor. In
fact, it is likely they have a slightly LEADING power factor resulting
from the combined capacitance of all the wiring in the home. I know this
from when I used to work as a watchman in a plastics moulding plant. During
the weekends when the plant was shut down, it was not unusual for the
power factor to be about 0.9 on the leadinig side (there was a PF meter at
the service entrance).

As the plant was started up and various motorized equipment was started,
the power factor would shift towards the lagging side and capacitor banks
would be switched in.

--
Calvin Henry-Cotnam
"I really think Canada should get over to Iraq as quickly as possible"
- Paul Martin - April 30, 2003
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