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Residence Power Meter And Reactive Loads Question ?

A

Andrew Gabriel

Jan 1, 1970
0
You really do not get it. The torque is LESS when the two coils that
drive it are out of phase. That means that the same POWER will yield
LESS metered reporting of it. It does not get much more simple than
that, K-Tard.

You have Power and VA mixed up. If you said the same VA will yield
LESS metered reporting, you would be right. That's because as the
current drifts further out of phase (and assuming the current amplitude
remains constant), the Power drops. This is because the meter is
measuring Power (or more correctly, the integral of Power, Energy).

Alternatively, if you want the Power consumption to remain the same
as you shift the current further out of phase, the current (and VA)
will increase to compensate. Regardless, the meter measures the
[integral of] Power, and will spin at the same rate for constant
power consumption regardless of phase angle. If you want to try this
second case for yourself, play with the first java applet example
(Phase Shift) on http://www.cucumber.demon.co.uk/PowerFactor/
which shows how the various parameters such as current and VA change
as you change the phase shift, whilst drawing constant power.

K

krw

Jan 1, 1970
0
The retarded **** has not ever once said a goddamned thing about it,
you stupid ****.

You don't read. said:
Now, IF he comes back with some brains, instead of his stupid, peanut
gallery dumbfuck crap, MAYBE your remarks would have some merit.

Now go back and read what you snipped, AlwaysWrong. It's in there.
Since all he does is parrot the same old, tired, retarded horseshit,
both he, and his retarded butt buddy (you) can shove it up each others
asses for all I care.

You don't read, AlwaysWrong. Did I say that you're always wrong?
So, his CRAP is meaningless without stating what his refutation is.

You don't read well, AlwaysWrong. Try again, if you can keep you
blood pressure below 300. If not, just take a week off and relax. A
month would be better.
And since you do not even know what fucking torque is, I doubt
seriously that you have any real depth of knowledge as it relates to ANY
mechanical device in existence. This meter is a mechanical device,
dipshit.

Nice rant, AlwaysWrong, but you're still always wrong.

K

krw

Jan 1, 1970
0
Dumbass. The field IS APPLYing the torque force.

Yes, it IS applied. Whenever a shaft is rotated by a force, torque has
been applied. And whenever a shaft is turned, torque IS the force that
was applied to the shaft, by whatever means, to turn it. Go back to
basic physics classes, dimwit.

No one said anything different, DimBulb.
So, if the shaft turns at one speed under an in phase resistive load,
and it does... and IF the shaft gets slightly less torque applied to it
by the eddy currents when "looking at" an inductive load, then ANY IDIOT
can declare that the number of turns will be less, despite the fact that
the same power was consumed, if the torque is less due to the two coils
being out of phase, and as it happens, that IS the case with inductive

Yes, but the inductive load doesn't dissipate the same power as the
resistive load, now does it AlwaysWrong? The torque is PROPORTIONAL
to the *real* power of the load. There is no error for an inductive
So, same power consumed in both cases, and the meter will read the
inductive load as having been less. Period.

Because the power in the inductive load is *LOWER*, AlwaysWrong.
The difference was not
enough for them to be concerned about, and it was made up for ON the FEED
side (utility) by putting cap banks up on the poles to shift the phase so
that an inductive load would make the metering a bit more accurate on

No, ALwaysWrong, the capacitors are there to correct their VARS so
they aren't supplying current that isn't used, with the losses that
entails. The capacitors don't affect the meter _at_all_. They
*can't*, they're before the meter.
It is not just for reducing line losses, it is also for reducing the
error in the meters on inductive loading.

AlwaysWrong, is always wrong.
You all need to go back to school... very basic school.

You never went to school, DimBulb.

A

Archimedes' Lever

Jan 1, 1970
0
No one said anything different, DimBulb.

What do you think "nor is the torque applied" refers to, you fucking
retard?

Whenever ANY shaft is made to rotate by ANY force, the force acting on
the shaft is torque.

I do not expect you to get it, though, K-Tard.

K

krw

Jan 1, 1970
0
What do you think "nor is the torque applied" refers to, you fucking
retard?

Whenever ANY shaft is made to rotate by ANY force, the force acting on
the shaft is torque.

You really are retarded, DimBulb. Of course a torque is applied to
the disk. The point is that the torque is proportional to the *REAL*
power, not VA.
I do not expect you to get it, though, K-Tard.

You clearly don't get any physics, DimBulb. You're AlwaysWrong about
everything else, too. No surprises for any here, though.

G

Guest

Jan 1, 1970
0
Archimedes' Lever said:
Dumbass. The field IS APPLYing the torque force.

Yes, it IS applied. Whenever a shaft is rotated by a force, torque has
been applied. And whenever a shaft is turned, torque IS the force that
was applied to the shaft, by whatever means, to turn it. Go back to
basic physics classes, dimwit.
------
Torque has been applied to the shaft- true- the torque itself is not
applying a torque- which is due to the interaction of fields produced by
both the stator and the rotor part of the motor- classical electromagnetic
energy conversion concepts apply (as taught at the junior levels of EE).
So, if the shaft turns at one speed under an in phase resistive load,
and it does... and IF the shaft gets slightly less torque applied to it
by the eddy currents when "looking at" an inductive load, then ANY IDIOT
can declare that the number of turns will be less, despite the fact that
the same power was consumed, if the torque is less due to the two coils
being out of phase, and as it happens, that IS the case with inductive
------------
Again, as others, such as Gabriel, whom I respect, have pointed out, and as
you would find out for yourself if you cared to do a little bit of research
, you are wrong. The meter inherently reads real power- not volts*amperes.
So, if you have a resistive load drawing 10A at 100V rms, the power will be
1000Watts and if you have a reactive load , inductive or capacitive at 0.8
pf at the same voltage and current magnitude, the real power will be 800
watts. So, the KWH meter will run slower in the latter case because the
real power component is lower. The meter is designed to measure the real
power (actually energy) even if the load is inductive or capacitive. It does
mechanically what an electronic meter does by multiplying and averaging
sampled data. This has been beaten to death in an earlier thread.
-----
So, same power consumed in both cases, and the meter will read the
inductive load as having been less. Period. The difference was not
enough for them to be concerned about, and it was made up for ON the FEED
side (utility) by putting cap banks up on the poles to shift the phase so
that an inductive load would make the metering a bit more accurate on

It is not just for reducing line losses, it is also for reducing the
error in the meters on inductive loading.
----------
It is amazing in that in the short sentence above, there are several
incorrect concepts. Tell me, how the meter "down stream" of the capacitors
will be more accurate due to the capacitors. The meter sees the load
"downstream" from itself and even if it's accuracy is affected by the
inductance of the load (and it isn't), the capacitance will not change this
because the corrective effect of the capacitance is only seen "upstream" of
the capacitors. Kirchoff reigns!
You all need to go back to school... very basic school.

If the ideas that you present are based on what you learned in "school" ,
you have a good case for suing your teachers for misleading you.
Do you want my CV ( or some reference textual material ( I don't have much
at the "basic" level, but have some at the junior to graduate level in my
back closet)?

I'm not bluffing.

G

Guest

Jan 1, 1970
0
Archimedes' Lever said:
The retarded **** has not ever once said a goddamned thing about it,
you stupid ****.

Now, IF he comes back with some brains, instead of his stupid, peanut
gallery dumbfuck crap, MAYBE your remarks would have some merit.

Since all he does is parrot the same old, tired, retarded horseshit,
both he, and his retarded butt buddy (you) can shove it up each others
asses for all I care.

So, his CRAP is meaningless without stating what his refutation is.

And since you do not even know what fucking torque is, I doubt
seriously that you have any real depth of knowledge as it relates to ANY
mechanical device in existence. This meter is a mechanical device,
dipshit.

-------------------------------------
Vembu Gourishankar & Donald H. Kelly "Electromechanical Energy Conversion"
Second edition, Intext, 1973.

It appears that I have a somewhat deeper knowledge than you have
suggested---and I don't substitute invective for rational thought.

happy new year!

A

Archimedes' Lever

Jan 1, 1970
0
the torque itself is not
applying a torque

Pretty hard since "torque" is not a "thing".

Got any more silly horseshit?

A

Archimedes' Lever

Jan 1, 1970
0
which is due to the interaction of fields produced by
both the stator and the rotor part of the motor-

These meters are NOT a typical motor, dumbass.

There is no "rotor" other than the Al disc, and it exhibits NO field.
It responds to the field around it. Period.

A

Archimedes' Lever

Jan 1, 1970
0
classical electromagnetic
energy conversion concepts apply (as taught at the junior levels of EE).

If your previous remark was true, sure. But it was not a true remark.
There is no "rotor", and there is no coil or magnet on the part that a
dope like you would call the rotor.

The Aluminum is what responds to the field. That is why this "motor"
type will never be seen as a mechanical power source. The torque applied
by the field is enough to turn it, and the gear resistance, but there is
no way one could perform any real "work" with such a "motor".

It, in fact, can be argued as being incorrectly termed as a motor.

A

Archimedes' Lever

Jan 1, 1970
0
It does
mechanically what an electronic meter does by multiplying and averaging
sampled data. This has been beaten to death in an earlier thread.

It is mechanical. It does not perform ANY "sampling of data".

It responds ONLY to the resultant force applied to it.

That force is less for an out of phase reading, despite the power
(energy) being the same. RESULTANT FORCE.

So, even if we use your term "sample", the "sample" is what ends up
doing less turning for the same amount of use. Why? The resultant force
applied to this type of meter has an inherent error which is due to its
physical construction that you are apparently unable to grasp and were
obviously never educated about. It is physical. There is nothing you can
do about it with words. Hell, you cannot even grasp how capacitance
upstream side can play into how such a meter reads. It is all one big
circuit, dude. So it doesn't matter where they are located, they play
into the operation of the reactive device, as well as that type of
meter's accuracy when reading such a live circuit.

A

Archimedes' Lever

Jan 1, 1970
0
The meter sees the load
"downstream" from itself

The meter sees the load in the circuit loop it is in. It is a circuit.
Upstream or downstream doesn't matter.

A

Jan 1, 1970
0
A

Archimedes' Lever

Jan 1, 1970
0
I'm not bluffing.

Like I give a fat flying **** what you did. I was making the point
that whatever it was, you missed something. Doh!

A

Archimedes' Lever

Jan 1, 1970
0
and I don't substitute invective for rational thought.

No. You make stupid remarks like "no torque is applied", when the topic
is about the turning of a shaft. It doesn't get much more stupid than
that.

I would say that despite you publishing a book, you definitely missed a
few things along the way.

K

krw

Jan 1, 1970
0
No. You make stupid remarks like "no torque is applied", when the topic
is about the turning of a shaft. It doesn't get much more stupid than
that.

I would say that despite you publishing a book, you definitely missed a
few things along the way.

AlwaysWrong once again (and again, and again, and...) proves the
validity of his given name.

K

krw

Jan 1, 1970
0
Like I give a fat flying **** what you did. I was making the point
that whatever it was, you missed something. Doh!

Don, consider yourself among the elite! You scored a DimBulb *seven*
bagger! That's the best I've ever done and I've been correcting
AlwaysWrong's, always wrong, crap for years! Way to go!

A

Archimedes' Lever

Jan 1, 1970
0
That's the best I've ever done and I've been...

You're a fucking retard.

The best thing your'll ever do is feed worms.

K

krw

Jan 1, 1970
0
You're a fucking retard.

Why thank you, AlwaysWrong. Since you're *always* wrong...
The best thing your'll ever do is feed worms.

AlwaysWrong, that'll be your only contribution to the world, but
you'll somehow get that wrong too, as always.

G

Guest

Jan 1, 1970
0
Archimedes' Lever said:
These meters are NOT a typical motor, dumbass.

There is no "rotor" other than the Al disc, and it exhibits NO field.
It responds to the field around it. Period.

---------
Wrong:
the rotor has eddy currents which produce a field interacting with the
"stator field" to produce torque. Essentially it is a form of induction
motor and can be analyzed as such.
If your previous remark was true, sure. But it was not a true remark.
There is no "rotor", and there is no coil or magnet on the part that a
dope like you would call the rotor.

The Aluminum is what responds to the field. That is why this "motor"
type will never be seen as a mechanical power source. The torque applied
by the field is enough to turn it, and the gear resistance, but there is
no way one could perform any real "work" with such a "motor".

It, in fact, can be argued as being incorrectly termed as a motor.

_----------------
Wrong again- according to your concept the drag cup motor won't work because
it doesn't have a coil or magnet. The fact that the "stator windings" are
set up to produce a rotating field and the aluminum disc is conductive, is
sufficient. I have made demo motors with beer cans or paper clips as
rotors- anything conductive works.
Is it a good design for a mechanical power source? No. but it does provide
motor action.

---------------
It is mechanical. It does not perform ANY "sampling of data".
------
, You have a reading problem. The electronic meter multiplies and averages
sampled data. The mechanical meter multiplies and averages instantaneous
values of power achieving the same result.
---------------
Hell, you cannot even grasp how capacitance
upstream side can play into how such a meter reads. It is all one big
circuit, dude. So it doesn't matter where they are located, they play
into the operation of the reactive device, as well as that type of
meter's accuracy when reading such a live circuit.
---------
Let's see, the circuit as a whole does affect the load voltage and current-
that is true. Changes in the load will be affect the circuit as a whole-
also true. However, the meter measures the voltage across the load and the
current through the load, whatever they may be. Accordingly the meter will
measure the power delivered to the load and at any given load impedance and
voltage, it cannot tell the difference between the situation of an ideal
source connected above it or a long line connected to a grid system with or
without capacitances. If a capacitor is connected on the load side of the
meter, it will change the current and voltage at the metering point and the
meter will measure accordingly- but as the capacitance only compensates for
inductive reactive VAR's , it will not correct the meter and the only effect
it will have on the meter will be a change in losses in the wiring on the
load side of the meter.
An industry may put capacitors downstream of the meter in order to improve
power factor- not because of any attempt to correct the meter but to reduce
the peak KVA demand which is not registered by a KWH meter but is reflected
in demand metering charges.
A utility putting a capacitor upstream of the metering point (and it is
always a parallel capacitor for very good reasons) doesn't correct the meter
downstream because it only modifies the upstream current (KCL applies). Do
you want some circuit analysis of this?
----------------
The meter sees the load in the circuit loop it is in. It is a circuit.
Upstream or downstream doesn't matter.
----
So, are you saying that, if the meter is moved to a different location in
the circuit, it reads the same thing? If not, then what are you implying?
Have you ever used a wattmeter in a circuit?

I don't know where you got your circuit or motor concepts but somewhere,
somehow, something is badly amiss in your understanding. There are many
references out there Shaum's outline on circuits is quite good for circuits
and a sophomore text such as Basic Electrical Engineering by Fitzgerald,
Higginbotham etc has been resurrected many times in different editions.
Also try:
http://fourier.eng.hmc.edu/e84/lectures/ch3/node1.html

http://nptel.iitm.ac.in/courses/Web... Technology/pdf/L-44(GDR)(ET) ((EE)NPTEL).pdf

considers the interaction between the eddy currents and the flux of the
stator- in general, this interaction is actually better described in terms
of the interaction between the stator flux and the flux produced by the eddy
currents. In any case it is simply a form of shaded pole induction motor.

If you wish to give some tangible circuit models to present your points for
discussion- please feel free to do so. If you wish me to give some examples,
let me know.

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