The meter, is an electromeechanical integrator. It measures,integration
calculus [from 0 to 2pi]tan (omega * t) dt where omega=2*pi*f (frequency in
HZ=f)
So, it's not to be fooled by trivial methods, like putting all loads to a
high leg, etc. like some poster said, is like the elusive 100mpg carburetor,
or fuel additives that increase mileage, etc.

--
Tzortzakakis Dimitrios
major in electrical engineering

What you heard from the" another electrician" was, purely and simply, BS.
Balancing the load between legs is optimal with respect to losses but
doesn't actually increase the KW capacity (not KWH) of the system. It just
means that you can use the capacity slightly more efficiently.
If you have a load of 15KW on one leg and 0 on the other, you will have more
losses and poorer voltage regulation than if you have 7.5 KW on each leg.
The meter simply measures what you actually use (+ losses on your side of
the meter). Now the high leg metering X2 would be a cute but highly illegal
trick that, along the way, would become apparent and the cost to the
utility in class action suits would be prohibitive. Honesty in metering is
something that is actually practiced.
As to your torque without friction leading to infinite rpm- take a look at
your meter- see those magnets which provide drag proportional to the speed
of the disc? The result is that at any given torque, there will be a speed
at which this torque is balanced by the drag. Adjustments to the meter does
involve the positioning of these magnets.
--
Not a wannabe
-----
Don Kelly
cross out to reply

So if you have a 100A service I don't understand how having 100A X 120V is
the same power as having 100A drawn from each leg, 100A X 240V. I would
think at any given power factor 100A X 240V is more than 100A X 120V. I
would think balancing the load so that you use 100A from each leg would
increase your KW capacity of the service.

I certainly wasn't referring to you as a wannabe, you've always had good
reasoning and explanations. But you've seen the others, perhaps they don't
understand the question and give insulting smart-alecky replies and it's
obvious from their reply that they either didn't understand the question or
the subject they are replying to. Rereading James Sweet's posts I think
he's right on too, but I don't agree that using the same KW's from each side
of the line in is anything at all like magic bullets or 100mpg carburetors.
But I think his point may be that those that sell misinformation about
balancing the load saving money are in the same category as the 100mpg carb
rip offs.
If meters measure true power then they are measuring VA X Power factor
integrated over time, I was surprised no one took issue with krw stating
that meters just measure true power and not VA X Power factor, since they
are equal, but seems not according to krw. Also as explained with power
producing torque in the disk I didn't understand how it controlled the speed
of the disk unless their was also a drag or friction that controlled the
speed. If you just apply a constant current to a motor the speed varies a
lot according to load, doesn't seem to be useful for a meter unless, as you
explained, there is a controlled drag or friction. If you make a constant
current source and use it to supply a motor, the speed will change to keep
the torque constant. If you have a constant voltage, torque (and current
draw) will change to try to hold the speed constant. I know this varies
according to type of motor, but even induction motors with variable
frequency drives can use volts per hertz. Also stepper motors, though their
speed is controlled by the step rate, they are capable of higher speeds with
higher voltage.
RogerN

It isn't, that's not what he said. Think in terms of Watts, if you draw
100A at 240V that is twice the wattage as drawing 100A from 120V because
with a 240V 100A load it is drawing 100A from each side. If your loads
are so badly balanced that you are drawing 100A from one side and 20A
from the other and it is a 100A service, you cannot add any more 240V
loads. The capacity is there, but you are already at the max of 100A on
one side. If you can shift some of the load over to the under utilized
side then you will have freed up capacity, but you will not affect the
reading of the meter by doing so. The important part, which is beating a
dead horse at this point is that there is a coil on each leg, and both
coils create torque on the same disk so whether you draw 100A on one
side and 0A on the other, 50A on each side, 10A on one side and 90A on
the other, it doesn't matter, in any of these situations the disk will
see the same torque and turn the same speed, the only thing changing is
which of the two coils supplies more of the total amount of torque. It
doesn't matter if one box of rocks weighs 500 lbs and the other box of
rocks weighs 100 lbs, if both boxes of rocks each weigh 300 lbs, or if
one box weighs 600 lbs and other other is empty, if you put them both on
a scale, it will read 600 lbs either way. Balancing the panel is taking
rocks from one box and placing them in the other, it is not changing the
total number or weight of rocks and will not affect the reading on the
scale.

Yes they do measure true power, and no, they do not measure VA * PF over
time, they are different. I think what is happening here is the common
failure to differentiate between *equal to* and *equivalent to*. Volts *
Amps * PF is *equivalent* to Watts (true power) but it is not *equal*
to, it is not the same thing. Power can be measured directly, if you
then know any two of the three other variables, Volts, Amps, and PF, you
can calculate the third, but any one can be measured on its own without
knowing any of the others.
Look at it this way, the speedometer in a car indicates vehicle speed in
miles (or kilometers) per hour. This is equivalent to miles traveled
divided by trip time, but it is not the same. Sure you could make a
speedometer that recorded the distance traveled and time the vehicle was
in motion and use that to calculate the speed in MPH and the answer
would be right, but that isn't how real speedometers work. A mechanical
speedometer, not to be confused with the odometer, has no concept of
time or distance, rather it measures speed directly by magnetically
coupled torque acting against a known friction, in this case a spring.
In the same way, a mechanical power meter measures true power by energy
passing through coils, causing a proportional amount of work to induce
torque on the disk which rotates against a known amount of friction. In
either case you have to have a known quantity of friction in order for
the reading to have any meaning. Ignoring real world issues like
insulation breakdown, a mechanical kWH meter doesn't know or care
whether the voltage is 10V or 10,000V, rather it directly measures the
amount of work being done by the power by acting on the disk with a
proportional amount of work.
As for motors, there are many different types of motors, each with their
own characteristics, and the sort of motor in a power meter is unlike
any you're likely to find anywhere else. The reason stepper motors need
higher voltage to achieve higher speeds is that the windings are
inductive and it takes time for the current, and hence the magnetic
field to build up. The faster the motor is turning, the less time you
have per step, so the higher the voltage needs to be in order to create
a magnetic field of a given strength in the time available. Similar
reasons dictate the need to vary the voltage with the frequency of
induction motors.

But if I read it correctly, that's what Don Kelly said. I think it's right
in terms of KW but not in terms of 100A mains.
Quote Don Kelly "Balancing the load between legs is optimal with respect to
losses but doesn't actually increase the KW capacity (not KWH) of the
system. It just means that you can use the capacity slightly more
efficiently. If you have a load of 15KW on one leg and 0 on the other, you
will have more losses and poorer voltage regulation than if you have 7.5 KW
on each leg."
I think Don Kelly was replying in terms of not exceeding the limit of 1 leg.
If you had 100A main and a PF of 1 the limit should be 12KW from one leg or
24KW Total. So his example of 15KW would overload on one leg but be fine is
balanced.

That's how I understand it.

I understand this now, by original question was based on misinformation of
the meter running at a speed based on only the high leg (X2). I had never
heard that the meter ran based on the high leg (X2) before so I asked here
and was informed of the misinformation.

But if the scale took the weight of the heavy box and doubled it (that would
be like the misinformation I heard) then balancing the box would give you
the true weight without cheating.

If they measure true power and not true power over time then how to they
come up with KWH?

I realize it's not measuring everything independantly and performing
calculations, but if it's measuring true power, the results would be
equivalent VA X PF. In the formulas I've seen they use the equal sign (=)
is their a different sign for equivalent?

The speed of the disk would be proportional to KW and counting the
revolutions would be KWH. I don't see any direct way of reading KW unless
you measure the speed of the disk. The electric meter would be more like an
odometer accumulating total KWH used if I understand correctly.
RogerN

-------------
I made no reference to the current rating, I just picked a power out of the
air. It might have been clearer to you if I had said 12KW. as I wasn't
considering overload conditions.

One could argue that such a speedometer 'calculates' the instantaneous
speed by taking the derivative of distance. The turning of the speedo
cable is a measure of distance. It drives the odometer directly. The
magnet coupling develops a torque on the needle that is proportional to
how fast the speedo-cable is turning (i.e. the derivative of distance
traveled). The spring provides a counter-torque that converts the
torque created in the coupling into a position (i.e. for a given
coupling torque, the spring provides an exact counter-torque when the
spring is stretched to one particular position). Just another example
of an analog computer.
The way a kWh meter works is the reverse of this. It develops a torque
proportional to VA*pf. Vary the power factor and the torque developed
in the disk varies. A drag magnet develops counter-torque proportional
to disk speed. The result is disk *speed* is proportional to VA*pf.
Your argument that the disk somehow measures power directly is false.
The magnetic fields are created by a voltage sensing coil and two
current sensing coils. Period. With these two magnetic fields, you
don't have 'power', the best you have is VA. The meter shifts the phase
of the voltage sensing field by 90 degrees by using a highly inductive
coil. Then the three magnetic fields are arranged around the disk such
that a torque is developed only when the fields sequence properly. That
torque is maximized when the current is in phase with the applied
voltage (one current coil's magnetic field leads the voltage coil's
magnetic field by 90 degrees and the other current coil's magnetic field
lags the voltage coil's magnetic field by 90 degrees).
Some commercial meters can be modified to measure VAR (V*A*sin(angle)
instead of V*A*cos(angle)). They too only have current and voltage
sensing coils, but by careful arrangement their magnetic fields interact
with different timing.
<snip>
daestrom

One could argue that the Earth is at the center of the universe too,
and that the universe indeed exists inside the shell of the Earth. The
math gets annoying and Occam doesn't like the idea, but it could be
argued.

-----------------
I have to disagree here, the instantaneous torque is proportional to
instantaneous va or power. Inertia averages this to get the average power
over each cycle (or longer). VA*pf implies that it measures the product rms
voltage and current and applies a bugger (oops power) factor. The meter
"sees" none of these because they are not physically there. the equivalence
is there in that the time varying instantaneous values can be represented by
their rms frequency domain equivalents in steady state. Analysis of the
meter uses the rms approach as you have done (it avoids a nasty mess of
non-linear differential equations) but the meter doesn't.

But wouldn't you say that the torque developed is a function of the
phase angle between current and voltage? If the current lags 90 degrees
from the applied voltage, then the magnetic fields of the current coil
and voltage coil are almost exactly in-phase (owing to the high
inductance in the potential coil). With the two magnetic fields pulsing
'in-phase', there is no torque either forward or reverse developed.
The simple fact that power flow in the opposite direction develops
torque in the opposite direction shows that phase-angle between current
and applied voltage is 'built-in' to the device.
Yes, of course you're right that the inertia *averages* out the torque,
but it's not 'instantaneous va', it's 'instantaneous power'. But
'average power' *is* rms-volt * rms-current * power-factor in a
sine-wave only system (i.e. no harmonic content)
daestrom

---------------
I have no problem with this. The meter doesn't measure phase angle.
Note that the induction disc motor is essentially a form of single phase
motor. The disc wont start but if already rotating there will be a torque
bias in that direction. A single phase motor depends on this bias which is
greatest near synchronous speed. However, in the case of the induction disc
there will be very little unbalance torque -if any as the torque speed curve
is essentially (and desirably) constant torque in the operating region
(just about standstill) There will be a pulsating torque.

Yes it is, in the same way it is built into a conventional electromechanical
(dynamometer) wattmeter. The torque at any instant depends on the product
of instantaneous voltage and instantaneous current. On this basis, it simply
averages the instantaneous torque. If the voltage and current are 90 out of
phase, the instantaneous voltages and currents result in a double frequency
power with 0 average
Mathematically we can say -for sinusoids:
p(t)=Vmcos(wt)*Imsin(wt+phi) =(VmIm/2){cos(phi) +a second harmonic power
term with 0 average]
The average over a period is (VmIm/2)*cos phi +0 =Vrms*Irms* cos(phi)
Physically the meter simply produces a torque which is proportional to the
instantaneous power and inertially averages it. Specifically, it doesn't
measure phase angles, rms voltages or find pf -that is my point.
If the voltage and current coils have high R/X values then the meter will
be able to handle distorted waves with reasonable accuracy (and without
doing a fourier analysis). A digital KWH meter will simply do the same
v(t)*I (t) and averaging as a mechanical meter but with a few bells and
whistles can be made to measure KVAH (pf*H is possible but meaningless) as
well but such measurements aren't required or necessary for determination
of energy.

instantaneous power = v(t)i(t) This happens to be the same as instantaneous
va but I should not have used that term. The concepts of VA and VAR's are
related to phasor analysis which is a mathematical model which gives us the
pertinent information without the labor of solving a mess of differential
equations. At your 120V outlet- there is no actual 120V source as can be
seen if you examine the voltage with an oscilloscope.
Certainly, use of rms volts(magnitude)*rms current magnitude *power factor
will give the same average power. That's part of the reason we use this
model- it works.
Also, this "model" allows us a reasonable chance of solving not-so-simple
circuits. Think of what the situation would be in solving load flows for
large systems using differential equations! Phasor analysis essentially
replaces these differential equations by algebraic equations.
Another convenient model is the use of symmetrical components for fault
studies.
Another is the use of forward/backward fields to model a single phase
machine as two opposing machines.
In these cases there are direct relationships between the model quantities
and the actual quantities present.
You know all this. The point of this long winded diatribe is that, too
often, people think that the model is the actual thing.

(you're too kind :-)
I understand that the 'true' measure of power is instantaneous V *
instantaneous I and that that can simplify to simpler terms in certain
specific situations (such as DC or sine waveforms).
I guess I just can't 'wrap my head around' the meter responding to
instantaneous V*I when the magnetic field from the potential coil is
delayed nearly 90 degrees from V.
The only way I can make sense of it is if the eddy currents in the disk
are highly inductive (much like they are in a conventional single-phase
induction motor near stalled conditions). If the eddy currents lag the
air-gap flux by nearly 90 electrical degrees, I can see a torque
developed proportional to real power.
But the lag in eddy current would mean the torque pulse also is delayed.
That would mean that a 'peak' of instantaneous power (peak V and I for
a resistive load) creates a 'peak' of torque a quarter of a cycle later
when the instantaneous power has actually dipped to zero. And that
momentary zero instantaneous power that occurs when V and I are
zero-crossing is not 'sensed' by the disk until another quarter cycle
when the eddy currents in the disk are zero-crossing and developed
torque is momentarily zero.
I know that for revenue purposes, the meter's response to harmonics
caused by non-linear loads is acceptable enough, but with both the
inductance of the potential coil and the disk, I can't help but wonder
just how accurate they can be with large loads of this type. Seems the
currents induced in the disk by a non-sine current through the current
coil would not be a perfect match and thus not perfectly accurate.
daestrom

--
The voltage coil and current coils are not co-axial. If the voltage coil is
displaced from the current coil, then when the current is maximum, the flux

Yes, yes, I'm familiar with two phase motors and the 'shift' of the
flux. But since the disk is not ferromagnetic, that alone doesn't
create a torque (i.e. it's not a reluctance motor). So we have to have
an eddy current induced in the disk (a 'rotor current') and to my mind
it's the interaction of that eddy current with the magnetic field that
produces a torque. Much like the current in the squirrel cage of a
conventional single-phase motor.
A disk that was composed of numerous 'pie slices' insulated from one
another would not work. Right?

R/X near 1 would certainly work, but an R/X near infinity might not
(i.e. there has to be some inductance). With the eddy currents exactly
in phase with one winding's flux, the currents would decay before the
other winding's flux increased very much. I see your point about R/X
near unity, after all as you said class D motors have some of the best
starting torques and that is how they achieve this.

Well without getting into 'gawdoffal' calculations, it just seems that
if the eddy currents in the disk lag behind the flux that produces them
(i.e. the disk has some inductance) then the torque pulses would seem to
lag behind the instantaneous power 'pulses' that occur at twice line
frequency.
So I wonder how close the disk is operating near 'synchronous speed'.
Where I think the synchronous speed in this case is function of
mechanical shift between current and voltage coils (i.e. the equivalent
'number of poles' arranged around the rotor). But also the phase shift
between current in the two coils (which of course depends on connected
load and R/X of potential coil). That is, the time shift between the
flux of one pole and the other.
If the maximum power to be registered is anywhere near a fraction of
this 'sync speed', then you effectively have a significant variation in
slip of the rotor. Of course some motors have a pretty flat torque
curve at the lower end of speed when starting and for a given applied
voltage/current, that's what we want here. So that torque produced is
*not* a function of %slip but only the measured power. Or variations in
%slip are tiny because the 'sync speed' is so much higher than expected
speed of the rotor.
(ouch, my head's starting to hurt ;-) )

All of the references I've found on the matter seem to say that the
potential coil does have a low R/X so that magnetic flux is nearly 90
lagging from applied voltage. If it's something 'in between', wouldn't
that give you some calibration issues? For example, if R/X is 1 in the
voltage coil, then a connected load with a current lagging by 45 degrees
would be exactly in-phase with the voltage coil's current. No torque
produced but the customer is getting power (at 0.707 pf)
daestrom

<snip>
I have an Ohaus balance beam type of scale, and also a reloading scale that
have a sheet aluminum flat tip, the tip runs between two magnets. Somehow
the magnets produce a drag and dampen the scale. Is this effect related to
how the motor in the meter works?
RogerN

----------
I'm not even considering a ferromagnetic disk - and a good example of an
unbalanced two phase motor is an drag cup control motor. A beer can stuck on
a nail in a rotating field will run as a motor. All that is needed is a
conductor and a component of the stator field that "shifts". (A paper clip
will also rotate- done both)

--
Not necessarily but I would expect poor performance. If the slices were
connected only at the center and periphery then it may actually work better

I was thinking of completely insulated, so there was no 'complete
circuit' in the rotor.
Thanks for the discussion, it helped refresh my mind how how these
buggers work. Of course the new digital ones are a whole different story.
daestrom

If no rotor currents cannot flow (or are restricted as in the situation you
mention- which is analogous to the concept of laminations to minimize eddy
current- then there will be little or no torque.
I have since run across a site which indicates that the flux from the
voltage coil is adjusted , apparently by a shading coil, so that it is , at
unity pf, 90 degrees out of phase with the current coil flux. Beyond this
the site really didn't get down to specifics other than considering the
meter as a type of 2 phase meter where torque at standstill (or at near
standstill) will be essentially directly proportional to current at a given
voltage. One could take a 2 phase control motor, add gears and come up with
a watthour meter. It is interesting that the "Ferranti effect" which is
related to induction machines, was known before Tesla came up with the
polyphase induction motor.
You have the term "bugger" correctly applied but in polite society it is
the "Bougerre Factor".

Don
Do you have any more information on the second book by White and
Wilson? I used AddAll and found quite a few copies of the book by
Krause at very reasonable prices but did not find any thing on the
second book.
Thanks
Dave Foreman

Oops. Correction:
It is White & Woodson (of MIT), "Electromechanical Energy
Conversion"Wiley 1959, Lib. of Congress No. 59-5874
It is the original book that covers machines from a generalized basis and is
quite mathematical , by page 30 it has covered electromagnetic force
production basics and is starting on the Hamiltonian. The developments go
on to general basic models which can be used, depending on the restraints
applied, to represent any machine. Years ago, Westinghouse capitalized on
this to make a generalized lab machine set where, depending on connections
could be a DC machine, a 2 phase induction machine, a synchronous machine,
etc-- unfortunately is was atypical and somewhat inferior machine in any
configuration but did illustrate the theory.
Amazon lists one copy at $165 -it's not worth that! Krause,, while sticking
to more conventional machines is probably a better bet.

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