120V from both legs

[John Doe]

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

 

[Charles Perry]

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.

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

 

[John Doe]

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?

 

[Charles Perry]

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.

 

[John Doe]

Sounds like this may be an urban legend created as a perversion of the
following reasoning:
http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]
google thread thl3419501793d

 

[operator jay]

Sounds like this may be an urban legend created as a perversion of the
following reasoning:
http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]
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

 

[operator jay]

http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]

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

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

 

[operator jay]

http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]

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

P.P.S. "Top-posting" is generally frowned upon, FYI.

 

[Charles Perry]

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

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.

 

[Charles Perry]

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).

Back emf from motors causing harmonics?

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

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.

 

[Beachcomber]

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.

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)?

 

[daestrom]

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.

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

 

[daestrom]

Sounds like this may be an urban legend created as a perversion of the
following reasoning:
http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]
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

 

[daestrom]

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)?

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

 

[operator jay]

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.

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

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

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

 

[Charles Perry]

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.


http://www.advancedenergy.org/progressenergy/black_boxes.html

That link does not support your claims. Motors create very low levels of
harmonics. Filtering these harmonics is NOT generally worth the cost in
energy savings. Correcting power factor (displacement power factor) with
capacitors is a good idea since it reduces the magnitude of the current
flowing to the motor, reducing losses in the entire system.

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.

Harmonics have no effect on displacement power factor. Period. Quite simple
actually. Displacement power factor is defined as the phase angle between
the voltage and current at the fundamental frequency. By definition,
harmonics are not even included.

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).

Displacement power factor excludes harmonics. Poor displacement power
factor is caused by inductive, or capacitive, loads. Yes, I have seen
facilities with leading power factors. It is just as troubling as lagging,
sometimes even worse.

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.

True, utilities are concerned with the power factor of large customers, the
displacement power factor. This is corrected using capacitors. Harmonic
filtering is sometimes required in such installations, but not for the
reasons you state. It is done to avoid resonances that can damage the
capacitor bank or other equipment.

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.

They are quite commonly known in the power quality and power metering
fields.

BTW, you might be surprised to find the source of much of the information in
the link you provided ;-)

Charles Perry P.E.

 

[Dan Lanciani]

|
| | > Sounds like this may be an urban legend created as a perversion of the
| > following reasoning:
| >
| http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]
| > 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).

Umm, minor nit. I said that I do not have full knowledge of the operation
of the meter's clever magnetic circuit. I said nothing about having "little
knowledge." I suggest that you read the spiel before dismissing it.
The example requires only a very basic level of electrical knowledge to
follow. (Which is not to claim that the conclusion is correct. As I said,
maybe they do something clever to compensate to a first order. But in the
5 or so years since I gave the example I've yet to have anybody explain *how*
such compensation could work.)

| I'll second the debunking. The meter knows how
| much energy you are using and you get billed accordingly.

But how does it know? To compute power you need both the current and the
potential (voltage). The product can then be accumulated/integrated over
time to give energy. Every split phase meter I've seen is a 4-terminal
device with no connection to the neutral. It can measure the current in
each leg and it can measure the potential between the legs, but it has no way
to measure the potential between the neutral and either leg. The neutral's
potential cannot be assumed to sit half way between those of the legs. An
unbalanced load anywhere on the same transformer can drive it closer to either
leg.

For the simple case where the load being served by the meter in question
is the only load on the transformer it appears to me that you pay for the
energy you use on your side of the meter plus 150% of the energy wasted as
heat in the neutral on the utility's side of the meter. For more complicated
cases where somebody else is pulling the neutral off center you might be
charged for less energy than you actually use. Note that we are talking
about rather small numbers here. It is not, in any case, something to get
excited about.

Dan Lanciani
ddl@danlan.*com

 

[Dan Lanciani]

|
| | > Sounds like this may be an urban legend created as a perversion of the
| > following reasoning:
| > http://groups.google.com/groups?q=g...n&lr=&ie=UTF-8&[email protected]
| > 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,

No, I clearly stated that the long run is from the meter to the transformer.
The voltage drop that is significant is from the transformer to the meter,
not from the meter to the load. You can't change the example circuit and
then complain that the math is no longer correct.

[analysis and "corrections" deleted since they do not apply to the example
as posed]

| 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.

But that assumption is false. Not only are the voltages not constant but
the two line to neutral voltages likely aren't even equal.

| If we don't assume a constant voltage at the meter the numbers change only
| slightly, but the results are similar.

Except that the customer pays for some of the utility's line loss, absent
some clever correction that has yet to be described.

Dan Lanciani
ddl@danlan.*com

 

[Dan Lanciani]

|Dan the meter doesn't need to measure neutral current.

I never said it does. What it needs to know is the potential of the neutral
relative to a leg. To compute power you need both the current *and* the
voltage.

|Mr Kirchoff says any
|current going in will come out somewhere. If it comes out the opposite phase
|that is a 240v load, any unbalance is a 120v load. You have the current and the
|voltage.

No, you don't have the voltage. You are assuming that the potentials
between each leg and neutral are equal (much as the meter must). But if
there is any current in the neutral, those potentials are not equal. That's
where the error comes from. You cannot just take the leg-to-leg potential
and divide by 2 to get the leg-to-neutral potential.

Dan Lanciani
ddl@danlan.*com

 

[Charles Perry]

<snip

I think you might wish to read a little more on the issue.

I think you may need to get a copy of IEEE std 100. I will give you a
little clue, there is more than one kind of power factor.

Charles Perry P.E.