Re: Balancing the Breaker Box

[daestrom]

No, the meter measures *power*. As others have been trying to tell
you, the torque (hence RPM) of the disk is proportional to the current
times the voltage in the respective coils. No "calculating" at all
and *certainly* there is no power factor calculated or measured.

I disagree. Analog computers were around long before digital (the
speedometer in a vintage car is an obvious one).

The three coils create a torque on the disk proportional to real power.
The arrangement of the coils and inductance of the voltage sensing
coil are such that the power factor *is* implicit in the development of
the torque. A load current that is 90 degrees out of phase with the
applied voltage results in no torque developed. (which is why many
claims that 'power factor correction' will save you big money, are bogus)

If that were the only torque applied to the disk, it would spin rather
rapidly because the mechanical friction is pretty small. And it would
be hard to calibrate since the torque needed to drive all the gearing is
somewhat variable. So another permanent magnet called a "drag magnet"
is positioned next to the disk creates a counter-torque that is
proportional to the disk's speed. Obviously the disk stops
accelerating/decelerating when the torque produced by the sensing coils,
friction and the drag magnet balance. The result is that the disk speed
is proportional to the power in the sensing circuits.

(As an aside, some criminals have tried to cheat the power company by
positioning additional 'drag magnets' above/below the disk. This tends
to increase the counter-torque developed when the disk is in motion,
slowing the speed of the disk for a given power level. The law does
provide sanctions for such meter tampering.)

All the gears and wheels form an integrator (as in 'calculates the
integral of the disk speed with respect to time'). Because the disk
speed is 'revolutions / time', counting the revolutions is the calculus
operation of integration with respect to time.

It's a meter that calculates the energy usage by sensing voltage,
current and the phase relationship between them (i.e. power factor) and
integrating the results over time.

daestrom

 

[krw]

I disagree. Analog computers were around long before digital (the
speedometer in a vintage car is an obvious one).

Certainly, but this doesn't change the fact that PF is not the
measured quantity.

The three coils create a torque on the disk proportional to real power.
The arrangement of the coils and inductance of the voltage sensing
coil are such that the power factor *is* implicit in the development of
the torque. A load current that is 90 degrees out of phase with the
applied voltage results in no torque developed. (which is why many
claims that 'power factor correction' will save you big money, are bogus)
Yes.

If that were the only torque applied to the disk, it would spin rather
rapidly because the mechanical friction is pretty small. And it would
be hard to calibrate since the torque needed to drive all the gearing is
somewhat variable. So another permanent magnet called a "drag magnet"
is positioned next to the disk creates a counter-torque that is
proportional to the disk's speed. Obviously the disk stops
accelerating/decelerating when the torque produced by the sensing coils,
friction and the drag magnet balance. The result is that the disk speed
is proportional to the power in the sensing circuits.

(As an aside, some criminals have tried to cheat the power company by
positioning additional 'drag magnets' above/below the disk. This tends
to increase the counter-torque developed when the disk is in motion,
slowing the speed of the disk for a given power level. The law does
provide sanctions for such meter tampering.)

All the gears and wheels form an integrator (as in 'calculates the
integral of the disk speed with respect to time'). Because the disk
speed is 'revolutions / time', counting the revolutions is the calculus
operation of integration with respect to time.

It's a meter that calculates the energy usage by sensing voltage,
current and the phase relationship between them (i.e. power factor) and
integrating the results over time.

Yes, others have clarified this better than I. I have no clue why
he's still pursing this.

 

[daestrom]

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.

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

 

[daestrom]

No, there is a difference between the physical quantities measured and
the conversion factor between them. You can't measure power factor,
only derive it from power and VA.

Nonsense. Ever see a 'power factor meter'?? I have, on old switchgear
systems. These only measured the phase-based power factor (harmonic
content wasn't a problem in these old systems). But they simply
developed a position based on the phase difference between the applied
voltage and current (one input polarized a moving vane in a magnetic
field created by the other input). Because the position of the needle
actually showed the phase angle, the meter face was marked in a cosine
pattern (i.e. the distance between 0.9 lagging position and the 0.8
lagging position was small, but the distance between 0.1 lagging and 0.2
lagging was larger).

The position didn't vary with the magnitude of the voltage or current,
only the phase relationship (of course if current was below a certain
threshold, the meter wasn't reliable)

daestrom

 

[krw]

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.

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.

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.

Interesting. How do they deal with harmonics?

 

[krw]

Nonsense. Ever see a 'power factor meter'?? I have, on old switchgear
systems. These only measured the phase-based power factor (harmonic
content wasn't a problem in these old systems). But they simply
developed a position based on the phase difference between the applied
voltage and current (one input polarized a moving vane in a magnetic
field created by the other input). Because the position of the needle
actually showed the phase angle, the meter face was marked in a cosine
pattern (i.e. the distance between 0.9 lagging position and the 0.8
lagging position was small, but the distance between 0.1 lagging and 0.2
lagging was larger).

Exactly. They don't deal with harmonics, thus are really phase angle
meters.

The position didn't vary with the magnitude of the voltage or current,
only the phase relationship (of course if current was below a certain
threshold, the meter wasn't reliable)

So what? Without measuring harmonic content, they only measure phase
angle not PF. ...unless you assume the world is a sine wave.

 

[krw]

YOU ARE TOO FULL OF YOURSELF TO UNDERSTAND WHAT IS GOING ON NOR DO YOU
HAVE A CLUE OF ANYTHING THAT I KNOW

I know that you're the stupidest hack on the Usenet, and that is
pretty damned stupid, Roy.

TAKE YOUR CHEAP SHOTS TO THE SHOP YOU HACK !

Cheap shots for dumb asses.

I AM PROTEUS

No, you are Roy the Retard.

 

[Guest]

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.

---------------------

It appears that you may be confusing capacity and utilization. Substitute
utilization for capacity and you'll be OK.
You have a 100A, 240V system which is center-tapped so that you can get 100
A at 120 V on each leg. If the load is balanced, the center-tapped or
neutral connection has 0 current. The capacity of the system then is 100A
at 120/240V which works out at 12KW per leg for a total capacity of 24KW.
If you put all the load on one leg, then you are (current) limited to 12KW .
The other leg is doing nothing. The capacity of the system is unchanged but
is only half used.

Here's a little chart: (120V per leg)
Leg1 leg2
total neutral utilization
current 100A 0
100A 100A 50%
KVA 12 0
12
losses 1 units 0
2 units 1unit

current 50A 50A
50A 0 50%
KVA 6 6
12
losses 0.25units 0.25 units
0.5units 0 units

current 100A 100A
100A 0 100%
KVA 12 12
24
losses 1 unit 1
unit 2 units

current 75A
25A ---
50A 50%
KVA 9
3 12
losses 0.56units 0.06
units 0.88units 0.25units

(unit taken as single wire I^r loss at 100A)

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.

I could quibble with you a bit with regard to the motors- it is the load
that determines the torque requirement at any given speed. Yes, speed and
voltage are related as are torque and current.-you have that right- but the
intersection of the torque speed curves of motor and load, determines the
operating point.
Note that the watthour meter is a type of induction motor and thus is
limited to less than synchronous speed. In fact, it would sepf destruct at
that speed and is operating in a region where it is nearly at standstill so
the torque is effectively constant at a given power, within the operating
range. The drag torque is linearly speed dependent.

However krw may have been referring to the fact that the inherent
measurement doesn't deal with power factor as electromechanical wattmeters
(dynamometer) and watthour meters inherently measure the instantaneous
product of voltage and current and average this to get the true power
without any consideration of power factor. So does a digital meter. They
don't use power factor because of this-effectively they go back to
fundamentals to make measurements. If we are using voltmeters and ammeters
then we don't have this multiplication and averaging of the instantaneous
values so have to account for phase. Power factor is one way to do this.
We also use rms voltage and current- but these don't actually exist - what
exists are sinusoidal (hopefully) voltages and currents. and their
instantaneous product.

 

[Guest]

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.

---------------------

It appears that you may be confusing capacity and utilization. Substitute
utilization for capacity and you'll be OK.
You have a 100A, 240V system which is center-tapped so that you can get 100
A at 120 V on each leg. If the load is balanced, the center-tapped or
neutral connection has 0 current. The capacity of the system then is 100A
at 120/240V which works out at 12KW per leg for a total capacity of 24KW.
If you put all the load on one leg, then you are (current) limited to 12KW .
The other leg is doing nothing. The capacity of the system is unchanged but
is only half used.

Here's a little chart: (120V per leg)
Leg1 leg2
total neutral utilization
current 100A 0
100A 100A 50%
KVA 12 0
12
losses 1 units 0
2 units 1unit

current 50A 50A
50A 0 50%
KVA 6 6
12
losses 0.25units 0.25 units
0.5units 0 units

current 100A 100A
100A 0 100%
KVA 12 12
24
losses 1 unit 1
unit 2 units

current 75A
25A ---
50A 50%
KVA 9
3 12
losses 0.56units 0.06
units 0.88units 0.25units

(unit taken as single wire I^r loss at 100A)

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.

I could quibble with you a bit with regard to the motors- it is the load
that determines the torque requirement at any given speed. Yes, speed and
voltage are related as are torque and current.-you have that right- but the
intersection of the torque speed curves of motor and load, determines the
operating point.
Note that the watthour meter is a type of induction motor and thus is
limited to less than synchronous speed. In fact, it would sepf destruct at
that speed and is operating in a region where it is nearly at standstill so
the torque is effectively constant at a given power, within the operating
range. The drag torque is linearly speed dependent.

However krw may have been referring to the fact that the inherent
measurement doesn't deal with power factor as electromechanical wattmeters
(dynamometer) and watthour meters inherently measure the instantaneous
product of voltage and current and average this to get the true power
without any consideration of power factor. So does a digital meter. They
don't use power factor because of this-effectively they go back to
fundamentals to make measurements. If we are using voltmeters and ammeters
then we don't have this multiplication and averaging of the instantaneous
values so have to account for phase. Power factor is one way to do this.
We also use rms voltage and current- but these don't actually exist - what
exists are sinusoidal (hopefully) voltages and currents. and their
instantaneous product.

 

[Guest]

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.

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.
[/QUOTE]

 

[Guest]

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.

-----------------
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.

 

[daestrom]

Exactly. They don't deal with harmonics, thus are really phase angle
meters.


So what? Without measuring harmonic content, they only measure phase
angle not PF. ...unless you assume the world is a sine wave.

In the 30's and 40's, when these were often used, the world of
electrical loads was a sine wave

daestrom

 

[krw]

In the 30's and 40's, when these were often used, the world of
electrical loads was a sine wave

I know. I used to have such meters (a mechanically resonant Hz meter
too). My father was a power engineer and *collected* stuff (several
tons of such stuff in the attic when he passed).

 

[krw]

What percent error is typically caused by ignoring harmonic content?

30%, 40%, maybe more of the VA is in harmonic content, particularly
with electronics. Switching power supplies are often *very* bad.

Do the
harmonics fool the utility meter or how does my power meter deal with
harmonics?

They deal with it perfectly. We keep telling you that they measure
*POWER*.

It just seems logical that if a utility power meter can measure
true power, and a meter can measure apparent power (VA), then why should it
be so difficult to measure power factor?

You can calculate PF from VA and power, sure, but that's not the same
as measuring PF. Your logic is backwards.

 

[daestrom]

-----------------
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

 

[Guest]

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.

---------------
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.

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 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.

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)

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.

 

[krw]

Motors, inductive loads, voltage leads current, capacitors, current leads
voltage. Is a switching power supply more like a capacitive load or
inductive load?

Neither. It's a harmonic load. ;-) Uncorrected it has a very crappy
current waveform.

I'm curious, I installed a 10HP variable frequency drive on
my retrofitted CNC lathe. I ran the motor at idle and the drive showed 13A,
but a meter only measured 5A on the line side. I have heard that the reason
for this is that the drive handles the power factor problem, drawing closer
to true power from the supply. If I understand correctly, a VFD is sort of
like a 3 phase SMPS with the motor being the inductors in the circuit. Just
wondering how this affects my household power factor versus other motors
running from the line.

A fully loaded motor should have a PF pretty close to unity. What any
particular VFD does to it is anyone's guess. I suggest you buy a
"Kill-A-Watt" or "PowerAngel", or some such. These things are about
$25 and will report V, A, VA, W, PF, F, kWH, and H (did I leave
anything out?). You can then play with various loads to get a feel
for what they're about.

 

[krw]

YOU ARE A TERRORIST

Roy, you are the one why has been spraying graffiti all over this
newsgroup for the past year.

ROY HAS NOT EVEN POSTED HER IN MONTHS

You're a liar, Roy.

HOPEFULLY HE FOUND A BETTER GROUP VOID OF FAGGOTS TROLLS AND
TERROSRISTS AS YOURSELF FOOL

You've been visiting the gay web sites again, Roy. Mommy doesn't like
that.

I AM PROTEUS

You are LIAR.

 

[daestrom]

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.

---------------
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.

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 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.

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)

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

 

[Guest]

[email protected] wrote:






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.
-----------------
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.

---------------
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.

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 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.

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)

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 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
of this coil is maximum but that of the voltage coil is 0. When the
voltage coil produces maximum flux , the current coil flux is 0. So , as
seen from outside, there is a shift of flux .
Ideally, if the voltage coil is 90 degrees apart in space from the current
coil, you would get a "2 phase" motor with only a rotating field . Since
the displacement is not 90 degrees, you get a weaker rotating field as well
as a pulsating field. Another example is a shaded pole motor as used in
small fans.

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.

I think you have part of it but a conventional single phase motor without a
starting winding will, at start, have 0 average torque but a healthy
pulsating torque. If you have a second winding which is ideally 90
degrees electrically (physically 90 degrees for a two pole machine) AND the
flux due to the second winding is delayed 90 degrees in time, then you have
a situation where you have a true rotating field with no pulsating torque-
as in a 2 phase motor, 3 phase motor or a single phase motor with a
dominant capacitor on the second winding.
A single phase meter lies somewhere in between -think of the voltage coil as
a starting winding which is always in the circuit. It may not be in
physical quadrature (e.g the shaded pole doesn't shift flux by 90 degrees)
The torque will have pulsations as well as a steady component but the
pulsations have a 0 average.
Should the disk R/X be low? Since the "start winding" is always in service
the physical position of this winding with respect to the other winding
provides the needed bias. It seems, on a first consideration, that a R/X
ratio near 1 in the disk may be desirable in that the torque will be nearly
constant and maximized near standstill- just as in the case of a class D
induction motor. --

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.

The disk doesn't see the voltage and current directly but does respond to
the total of currents in the v and I windings interacting with the currents
in the disk. You can then consider a model with two stator coils and two
rotor coils and wade through a bloody "gawdoffal" mess of math (task given
to grad students) to come up with an expression for torque dependent upon
both the position of the coils and the currents in them to find an
instantaneous torque expression. Then you can consider steady state
(sinusoid) at some given speed ( in any motor the torque has nothing to do
with Ldi/dt but is affected by I*dL/dx* dx/dt or the speed voltage) and come
up with a model using rms voltages and currents and the phase between them.
The fact that this model typically agrees with simpler approaches used in
typical steady state handwaving analysis of induction machines is
comforting.
Krause "Analysis of Electric Machinery" deals with some of this (at near
graduate level) and an old book by White and Wilson "electromagnetic energy
conversion" is another- this is definitely a graduate level text. Many
other texts simply jump into the steady state models- a bit more practical
but losing some concepts.

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.

--
The voltage coil may not have a low R/X ratio. However, you don't want it
producing flux in phase with the current -
High harmonic content is still a problem because you do want the torque
independent of winding impedance- variations which comes back to the
desirability of an R/X that is on the high side.