On completion of a 3 phase converter and after adjusting the voltages
with caps what effect does adding the power factor corrections cap have
on the commercial power meter reading.Does it add more or less power to
the power bill?
My guess is that it would add more power to the bill as the power
factor approached zero(voltage in phase with the current)
Hence according to my ramblings it would be better not to correct power
factor and let the power company worry about it.

wrote:
On completion of a 3 phase converter and after adjusting the voltages
with caps what effect does adding the power factor corrections cap have
on the commercial power meter reading.Does it add more or less power to
the power bill?
My guess is that it would add more power to the bill as the power
factor approached zero(voltage in phase with the current)
Hence according to my ramblings it would be better not to correct power
factor and let the power company worry about it.

I have a friend who bought an electric forklift for his collision shop.
He has a 3 phase converter in his shop for his milling machine. When
his electrician came out to set up the charger (which is three phase)
he told him "you're going to wear your meter ot with this charger"
I would ask the power company, may just be a matter of how much juice
your pulling.
(disclaimer: all this was hear-say and not my experience, just passing
along what i heard... and we all know how "they" can be)

walt
ps. do you need an electric forklift with the charger? i know a guy who
has one and is willing to sell it cheap : )

According to :
On completion of a 3 phase converter and after adjusting the voltages
with caps what effect does adding the power factor corrections cap have
on the commercial power meter reading.Does it add more or less power to
the power bill?

It depends on what you mean by the phrase "commercial power
meter". If you mean the average home meter, it would result probably in
a tiny reduction -- corresponding to the extra power wasted heating the
wiring from the higher out-of-phase current.

However -- at least *some* commercial *users* of power --
especially those with three phase to their shops -- are charged "demand"
metering, which can be sensitive to the out-of-phase current so they are
charged based on that -- and on the *peak* demand during a given period.
So -- run a high-current heat-treat oven (for example), for an hour or
two, and you wind up paying a *lot* more *per* *month* for several
months, until they are sure that it will not repeat. (You are being
charged for the worst-case power which you *might* use, not what you are
actually using on the average.)

Since you've built a rotary converter, I presume that you don't
have an industrial power contract, so you will get a very small
improvement in your power bill -- but you will get fewer nuisance trips
of your circuit breakers. (Depending on how close to the trip point the
"imaginary" power takes you.) The breaker works on the vector sum of
the real and imaginary power, while the home meter works only on the
real, so it is mostly a good thing to add the power factor correction,
especially if you are going to be running other (single-phase) equipment
off the same breaker.

My guess is that it would add more power to the bill as the power
factor approached zero(voltage in phase with the current)

Not so. The power which you draw doing actual work is the real
power, and you'll be charged for that in any case. ("Actual work" can
include things like the heating of wiring by the out-of-phase component
of an uncorrected power factor.)

Now -- another factor comes into the game here. The power
factor correction is only good for a single level of load. As you
increase or reduce the load level, the power factor will be somewhat
less well corrected.

So -- if you are likely to spend more time with the rotary
converter idling than actual machining time (e.g. you turn on the
converter when you come into the shop, and turn it off when you leave
for the day), but only run the machines say 20% of that time, you are
better off with the power factor correction tuned to the unloaded
converter. But, if you turn on the converter just before turning on the
machine, and then turn it back off just after you stop the machine, it
would be better tuned for the converter plus the machine running a
typical cutting load. (However, you draw more current when starting the
converter, so you will probably save more running with the converter on
most of the time you are in the shop, ready to start the machine.

Hence according to my ramblings it would be better not to correct power
factor and let the power company worry about it.

My opinion is that it would be better to correct the power
factor to save wear and tear on your breakers and wiring -- *and* heat
buildup in the idler motor as well.

Good Luck,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

wrote:

On completion of a 3 phase converter and after adjusting the voltages
with caps what effect does adding the power factor corrections cap have
on the commercial power meter reading.Does it add more or less power to
the power bill?
My guess is that it would add more power to the bill as the power
factor approached zero(voltage in phase with the current)
Hence according to my ramblings it would be better not to correct power
factor and let the power company worry about it.


No. Uncorrected power has a real (resistive) component, which you pay for, and
an imaginary (inductive or capacitive) component, which your meter ignores. If
your power bill is the only issue, there is no reason to use power factor
correction caps. However, both kinds of current really flow in your cabling, so
if your wire is anywhere near its rated current you can feel much safer if you
use PF correction caps.

Grant

DoN. Nichols wrote:


Now -- another factor comes into the game here. The power
factor correction is only good for a single level of load. As you
increase or reduce the load level, the power factor will be somewhat
less well corrected.


My opinion is that it would be better to correct the power
factor to save wear and tear on your breakers and wiring -- *and* heat
buildup in the idler motor as well.

Good Luck,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

Don,
I would disagree with you here. I think the power factor correction is
good for correcting the imaginary current. And the imaginary current
does not change with load. The real current changes with load.

I am open to any arguments you have on why the imaginary current would
change with load. I do agree the power factor changes with load, but
only because the real current changes, not because the imaginary
current changes.

Dan

According to :

DoN. Nichols wrote:


Now -- another factor comes into the game here. The power
factor correction is only good for a single level of load. As you
increase or reduce the load level, the power factor will be somewhat
less well corrected.


My opinion is that it would be better to correct the power
factor to save wear and tear on your breakers and wiring -- *and* heat
buildup in the idler motor as well.

Good Luck,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

Don,
I would disagree with you here. I think the power factor correction is
good for correcting the imaginary current. And the imaginary current
does not change with load. The real current changes with load.

Well ... when you switch on a load (typically a machine motor),
you are adding predominantly an inductive load, and it will probably
work out to close to resistive only when the motor is near full rated
power -- that is heavily loaded -- so you would need more capacitance to
tune out that additional inductance during most of the time. (Do you
always push your machines to the maximum that their motors can handle?
Even when taking finishing cuts?

I am open to any arguments you have on why the imaginary current would
change with load.

It is with the switching in of an additional motor, which (as
above) will be predominantly inductive -- thus worsening the power
factor somewhat -- even if you have already tuned the rotary converter
to have a neutral power factor with just the idler connected and
running.

If your motor in the machine were in the circuit full time, and
you just had a clutch to connect it, and a variable speed pulley (or
step belts) to change speeds, then as long as the motor was connected
and spinning, I would expect the power factor to be pretty much
constant. But most of us have machines in which the motor is switched
to start and stop the spindle. (Yes -- an old Monarch 10EE with the
motor-generator setup would probably qualify, and some versions of the
Clausing lathes (though not mine) have a variable speed belt and a
clutch.

The electronic versions of the 10EE might well improve the power
factor when switched on but not spinning the spindle, as the power
supply would probably look more capacitive. The same for a VFD hung on
a rotary converter.

I do agree the power factor changes with load, but
only because the real current changes, not because the imaginary
current changes.

I could accept this if the load motor were connected full time.
But I don't think so with the load motor being switched.

Enjoy,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

In article . com,
says...

On completion of a 3 phase converter and after adjusting the voltages
with caps what effect does adding the power factor corrections cap have
on the commercial power meter reading.

For residential meters which measure only the in-phase component,
they have no effect whatsoever.

Jim


--
==================================================
please reply to:
JRR(zero) at pkmfgvm4 (dot) vnet (dot) ibm (dot) com
==================================================

According to Ignoramus25850 :
If the meter measures actual power consumption, then hear output of
the running idler (and wires) should be equal to measured power
consumption, without a load that is. Let's keep that in mind.

Yes -- which is why I was predicting a slight improvement in
power costs for a tuned converter over an untuned one. How slight is a
function of how much of that energy is going to heat the wires.

Enjoy,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

"jim rozen" wrote: For residential meters which measure only the in-phase
component, they have no effect whatsoever.
^^^^^^^^^^^^^^^^^^^^^
The I^2R losses in the circuit are based on the vector sum of the in-phase
component and the quadrature component. If the power factor is not one,
this resultant current is larger, so the meter reading goes up accordingly.
The electric company is not going to let you heat your wires for nothing.

In article , Leo
Lichtman says...


"jim rozen" wrote: For residential meters which measure only the in-phase
component, they have no effect whatsoever.
^^^^^^^^^^^^^^^^^^^^^
The I^2R losses in the circuit are based on the vector sum of the in-phase
component and the quadrature component. If the power factor is not one,
this resultant current is larger, so the meter reading goes up accordingly.
The electric company is not going to let you heat your wires for nothing.

The ohmic losses in the wires are small compared to the (real) windage
losses in the converter, which are themselves small. To put this in
perspective, the real power consumed by my 5 hp idler is about 200 watts.
This is bearing and windage loss in the idler itself.

The magnitute of the reactive current is about 7 amps, so for a loop
resistance of, say, one ohm, that would be another 50 watts. Not much.
But it makes sense to over-size those wires anyway, true.

Jim


--
==================================================
please reply to:
JRR(zero) at pkmfgvm4 (dot) vnet (dot) ibm (dot) com
==================================================

Thanks for explaining your thoughts. I agree that switching in an
inductive load would worsen the power factor. But if you correct the
power factor for a motor while it is idling, and then load the motor,
only the real current changes and there is no need to change the amount
of capacitance used for power factor correction.

I don't think it is important to correct power factor in most cases.
The exception in my thoughts is fairly large welders which have very
low power factor when energized and no welding is going on. In that
case one can get by on a smaller breaker. Breakers over 60 amps jump
in price.

Dan



DoN. Nichols wrote:

Well ... when you switch on a load (typically a machine motor),
you are adding predominantly an inductive load, and it will probably
work out to close to resistive only when the motor is near full rated
power -- that is heavily loaded -- so you would need more capacitance to
tune out that additional inductance during most of the time. (Do you
always push your machines to the maximum that their motors can handle?
Even when taking finishing cuts?




I could accept this if the load motor were connected full time.
But I don't think so with the load motor being switched.

Enjoy,
DoN.

--
Email: | Voice (all times): (703) 938-4564
(too) near Washington D.C. | http://www.d-and-d.com/dnichols/DoN.html
--- Black Holes are where God is dividing by zero ---

In article , DoN. Nichols says...

Yes -- which is why I was predicting a slight improvement in
power costs for a tuned converter over an untuned one. How slight is a
function of how much of that energy is going to heat the wires.

All the improvements will be tough to see in a converter that is
actually loaded with machines being used.

In my case, combine that with the de-humidifier running in the
shop and it's pretty much down in the grass.

Jim


--
==================================================
please reply to:
JRR(zero) at pkmfgvm4 (dot) vnet (dot) ibm (dot) com
==================================================

"Leo Lichtman" wrote in message
...

"jim rozen" wrote: For residential meters which measure only the
in-phase component, they have no effect whatsoever.
^^^^^^^^^^^^^^^^^^^^^
The I^2R losses in the circuit are based on the vector sum of the in-phase
component and the quadrature component. If the power factor is not one,
this resultant current is larger, so the meter reading goes up
accordingly. The electric company is not going to let you heat your wires
for nothing.
Yes they are, unless you are a commercial customer paying for KVAR (KVA
reactive). They would like every customer to have a better power factor but
it's just not cost effective to meter the out of phase component in
residential and small commercial accounts.