I was wondering if there is a simple, correct, and safe way to draw 120V
equally from both legs of US standard two leg household service. What I
want to do is reduce my electricity bill by eliminating uneven draw from
each leg. US meters run as if the maximum draw from either leg is being
pulled from both legs. When your refrigerator compressor runs, you pay for
the same amount of current being pulled from the other leg as well.
Most electricians will say that by putting an even number of 120V breakers
on each leg will balance your draw on both legs, but I think that is a
crock. Draw on either leg will almost never be balanced with this method.
What do I do, wait for the refrigerator compressor to turn on one leg so I
can run my shop vac on the other leg?
I thought that if I could change the phase of one leg 180 degrees, I could
connect it in parallel with the other leg.
How would I go about this?
Help please
WRONG! You have been fed a pack of lies and have believed them. You can
draw all of the current from one leg, half from each, or any other
combination and the meter will measure the correct usage. Anyone with any
metering knowledge at all knows this and anyone with the simplest meter test
equipment can prove it.
The way to lower your bill is to use less electricity.
Charles Perry P.E.
who is never amazed at the outrageous utility meter stories he hears, but is
always amazed that people believe them
OK. I guess I'll do some experimenting. I know that I sould take
everything with a grain of salt, but the source that's telling me it's bunk
(usenet) is the same place that I read this (mis)information in the first
place.
The truth is out there, but where?
"Handbook for Electricity Metering", Edison Electric Institute, 1992
(although there is a newer version now). It is the bible for electric
metering.
Also, for about $1000 a day, I will be more than happy to test your theory.
I always warn customers ahead of time when it is obvious that the testing
will not get the answer they want, but if they insist, I take the money and
do the testing.
Charles Perry P.E.
google thread thl3419501793d
Could be (I didn't actually read Dan Lanciani's spiel in your link - though
if it makes you more comfortable you can consider that Dan Lanciani
acknowledges having little knowledge while Mr. Perry and others here are
very much knowledgeable). I'll second the debunking. The meter knows how
much energy you are using and you get billed accordingly. Certainly as far
as unbalanced load is concerned anyways. A simple way to verify it is to
just call the utility. I'm sure they'll happily tell you on the phone that
it works that way, and they may be able provide some document, or website,
to that effect. I am assuming you would be content that the utility would
not outright lie to you on the matter.
j
> google thread thl3419501793d
By the way, unless you have some reason to believe otherwise, your home
probably has reasonably balanced load. Utilities like for loads to be
reasonably well balanced so that their lines, transformers, generators, etc.
can carry maximal loads and provide balanced voltages. If your loads were
not all that well balanced you probably wouldn't have much to worry about.
j
to draw 120V
service.
sure you can, thats how most residential services are set up.
Commercial service will often be three phase with one line
(the stinger) higher than 120v to neutral..be careful with
those.
uneven draw from
leg is being
runs, you pay for
as well.
120V breakers
that is a
with this method.
on one leg so I
Thats about it.
degrees, I could
Thats tricky... its never done...but it might work. its
probably not legal though..and screw ups would tend to blow
the utilitity company transformer off the pole...they just
HATE when that happens.
You probably shouldnt..but yer mind is sharp...maybe you can
come up with another approach a way to isolate yer strategy so
if it gets crossed up it wont take out the pole transformer.
What you could do is fit a 240 v X 120 volt transformer
across the incoming line.. then the output, 120 volts would be
all from the common balanced source. See a wiring diagram on
the bonding and neutral issues... those are important.
You would loose about 3% in the transformer... there should be
a net savings.
Phil Scott
to draw 120V
service. What I
uneven draw from
leg is being
runs, you pay for
as well.
them. You can
other
Anyone with any
simplest meter test
There are harmonics issues with unbalanced loads that can
screw up the power factor and run the tab up on yer bill
dramatically... not something I was overly familiar with but
recent review has turned up some good books on the subject..
primarily applicable to large commcl or industrial
installations.. not dramatically to the typical home.
Phil Scott
he hears, but is
> google thread thl3419501793d
your link - though
Lanciani
others here are
meter knows how
Certainly as far
verify it is to
the phone that
document, or website,
utility would
I tend to agree...however Ive had them lie to me regarding
demand charge and power factor adjustments at industrial
facilities nation wide, repeatedly...many dont even have a
clue what those factors are... the companies in some cases
have even changed the terms on the billing statement to
occlude demand charges.
You can save an industrial firm a lot of money in many cases
by getting rid of a bad harmonics situation (current wasted to
the neutral and between lines due to wave distortions
etc...back emf from motors and transformers...and such as
large scale transormer stage lighting). I dont do a lot of
this. when I do I spend $500 and rent a set up that clamps to
all 3 lines for amperage, reads voltage and the neutral... and
computes the power factor issues hour to hour over the
duration of the test.. I've seen some horrendous losses due to
these factors...some back fed distortions from other nearby
facilities. Desk top computer power supplies can backfeed
serious distortion into a buildings power grid. Not single
home though..but a problem in a large building.
I had been entirely unware of the problem until a client
raised the issue and I studied up... it can get complex. But
there are books on the situation and instrument rental outfits
that specialize in the test eq (25k for a real good set
up)...rental is much less.
Phil Scott
The presence of unbalanced loads does not create, or indicate, the presence
of harmonics. Also, residential customers are only billed for kWh usage, so
power factor is not an issue. Severe cases of harmonic distortion can
increase I^2 R losses, but our testing and research shows that these losses
are small.
Charles Perry P.E.
Back emf from motors causing harmonics?
You do realize that harmonics have a negligable effect on kWh, and no affect
on displacement power factor? I really don't know of an industrial metering
setup that penalizes for harmonics. Perhaps you could point one out.
Charles Perry P.E.
The effects of one house with an unbalanced load on a utility system
stop at the first distribution transformer that serves that house, it
is not passed along to the rest of the system.
The utility itself may have an unbalance problem with the load or
number of houses on each of the 3 three legs of a three phase feeder
but that is the problem of the utility (not customer caused).
BTW, 20 or 30 years ago I recall hardware stores selling "light-bulb
savers". These were nothing more than small diodes that you would
insert in the light socket (incandescent, versus compact fluourescent
being the standard at the time), and then screw the bulb in over the
saver. The claim was that the bulb would last a lifetime and energy
would be saved.
Well, it was true, the bulbs, did last longer, but the lights only
burned at half brightness because they received half-wave
rectification of the current.
This discussion of accurate meters with no neutral got me to thinking.
Would a standard meter accurately measure the power consumed by one of
these things (because of the pulsating DC component)?
There is no reason to look at the neutral. If x current is flowing in phase A
and y current is flowing in phase B they simply add x+y to get the load. Think
of it as a 120v equivilent. If x is significantly different than y the
difference is coming back on the neutral so you only get charged once. The part
that is coming back on y represents a 240v load and you get charged twice for
that.
It may be an oversimplification but the model works.
Harmonics caused by non-linear loads do *not* affect the kwh meter used
energy metering. Having a screwed up power factor does not 'run the tab up
on yer bill dramatically.' Only in very rare situations, when a very poor
pf requires the utility to take extraordinary measures do they add
*additional* metering and bill for it. PF correction is usually an issue
for large installations so they can reduce the size of the service
equipment. But a residential kwh meter will only register the true power
component, a poor pf does not affect your bill (it can affect other
appliances if it is harmonics and you have a relatively high source
impedance, but that's another story)
daestrom
google thread thl3419501793d
Having read through 'Dan Lanciani's discussion, I can say that what he's
describing is *not* a metering problem. He discusses if the voltage drop
from the meter to the load is significant, how a balanced or unbalanced
120/120 system would behave. But the difference in the two situations he
discusses (50A load on each leg vs. 100A load on one leg and neutral) is not
a metering issue. It is a *line loss* issue. And the line that has the
losses is not the utility's, it is in the homeowner's with a voltage drop
from the meter to the load that is 'significant'.
In the one case of a balanced 50A load on each leg, he assumes a voltage
drop of 0.5 volt in each leg (resistance of 0.01 ohm). So the power
delivered to a resistive load is 239V * 50A = 11.95 kW. But then he
incorrectly calculates the load as seen by the meter as 239V * 50A. This is
wrong because he assumed the meter was *upstream* of that 0.5 volt drop so
the meter registers 240V * (50A +50A)/2 = 12.0 kW. The difference is simply
the line losses (I^2*R) 50*50*.02 = 50 watts (or put another way, E^2/R
1.0*1.0/.02 = 50 watts)
Then, he calculates the condition of 100A on one leg and 0A on the other
leg. In this situation, the voltage drop on one leg is 1V and in the
neutral is 1V. So the power delivered to a resistive load is 118V * 100A =
11.8kW. But again, he incorrectly calculates the load as seen by the meter
as 239V*50A. This is wrong for the same reason as before. It should be
calculated as 240V * (0A + 100A)/2 = 12.0 kW. Same metered load as before
(just 100A on one leg instead of 50A on two legs). The line losses this
time are 100A*100A* .02ohms = 200watts.
Dan Lanciani notes the drop in delivered power (150 watts less power
delivered in the second case), and wonders "where did it go??" The simple
answer is that the second scenario quadruples the power losses in the
customer's (not the utility's) long line from the meter to the load. It
went from 50watts to 200 watts (difference of 150watts). This is reasonable
since the line resistance is the same, yet the current is doubled, and since
I^2*R losses are proportional to current squared, double current means
quadruple the losses.
But the losses that are increased are in the customer's system, not the
utility's. The power delivered *at the point of trade* is the same in both
cases, 12.0kW. The later case is just the customer wasting more of their
purchase in 'long lines from the meter to the load'.
True, the losses in the utility equipment are different also, but they are
upstream of the meter and we assume the voltage at the meter is constant.
If we don't assume a constant voltage at the meter the numbers change only
slightly, but the results are similar. The difference between what is
registered at the meter and what is delivered to the load is the line losses
in the customer's long line.
And if you have an average distance between the meter and the load, the
resistance becomes much smaller and so does the voltage drop and the losses.
If you have a particular service in mind, calculate the resistance in the
two legs between the meter and the load distribution point. Then you can
calculate for yourself the difference in line losses between a perfectly
balanced and unbalanced system.
daestrom
Short answer?? YES. The modern kwh meter (either electromechanical or
electronic) is just so d___ good at doing what it is designed for you'd be
amazed. High harmonics (non-linear loads), half-wave, displacement power
factor, unbalanced legs, voltage variations, you name it, it still comes out
right.
daestrom
message
regarding
cases
wasted to
Oh yes... like a barber shop quartet.
Searching googles web tab I got 45,000 hits on the following
key word string.
'Electric, harmonics, motors, line' You could trim the
list down by adding in EMF,
phase distortion, and power factor.
This one was typical..it includes a section deeper in the
body of text that addresses these issues.
formatting link
clamps to
and
due to
nearby
single
But
outfits
kWh, and no affect
No I don't. I have actually an entirely oposite view. But
no use in hassling each other, there is plenty written on
these issues..its not something new at all, with industrial
applications at least... it might be new to the residential
sector though.
I will decline to comment further in the interests of
peace and harmony and recommend that anyone interested in
these issues has available to them limitless information on
the internet by searching Googles 'web' tab with the key word
string mentioned above.
formatting link
hit the 'web' tab on the top left of the
box... then type in yer key words.
one out.
There is not such an animal to my knowledge, nor should there
be... but the power factor is calculated and the bill based on
that. Power factor deviations stem from harmonic
disturbances (and a few other things). Utility companies
have a concern with EMF impedance, and the repercussions
beyond a facility getting to the larger grid, and will address
that with owners in number of ways... power factor correction
capacitors included...and those costs are passed on to the
customer.
To address one of your earlier remarks.. I dont know if this
is suitable but it is quite common to install power factor
correction capacitors on industrial motors, boosting the net
motor efficiency and reducing utility costs, not just the cost
of running the motor but also providing a better pf to the
rest of the equipment in the facility.
These issues are not commonly known but the information is
available as mentioned above.
Phil Scott
Well, now, there are utilities that systematically meter and charge
penalties for power factor (for non-residential), such as my local utility.
In these locations, pf correction / filtering can be installed with paybacks
as low as 1 year. The utility may do this in part because they get a lot of
power via dc link. They have a certain amount of capacitance in the filters
at their inverter, but don't have the luxury of increasing a field current
to compensate for more inductive loading.
But a residential kwh meter will only register the true power
For interest: I heard harmonics may hurt (or help) to a very minor extent.
The +ve sequence components of the harmonic currents could cause a small
additive torque to the spinning disk, and the negative sequence components
could cause a small parasitic torque to slow the disk. I can't confirm
this. I believe the utility here had some students look into it for a
thesis. They were, as I recall, wondering about revenue loss due to 5th
harmonic current in the increasingly electronic load of residences.
j
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