Hi all,
As I recall there are a few people here with the expertise to
answer these questions easily. Simple 4 u, not 4 me! g
Yes, I've done a bunch of googling and recalling the "old
school days", but it's not getting me where I need to be g.
Neglecting small interferences/insertion losses, etc.:
--------------------------------
Short description:
Here's an actual example of measurements/calcs:
120Vac measured
0.29A rms measured
24W measured
35 VA measured
PF = W / VA, or 24 / 35 = 0.686..., or about 68%. Right?

-- What numbers do I use to get kWH? Is it VA / W?

-- How many kWH do you calculate from those figures, assuming it
can be done? If it can't be done, what's missing?
-- How did you get to your result?

-- At 10 cents/kWH, how much would it cost me per hour?

---------- end short descrip -----------

You wouldn't believe the amount of work and research I've done to
get my head around this! And how confused I am at the moment!

All I started out to do was to calculate what some of the major
device costs around the house are in order to make a point to
some people about the cost of, say, leaving the lights on in an
unoccupied room over night, or never turning off say a coffee
maker, computers, radio, stereo, TV, holiday lights; things like
that. And I ended up with a brain-ache so I next decided to go
where there might be some brighter brain cells than my own! And
here I am!

Thanks for your hopefully understandable responses; it's been
over 4 decades since I was in college, so be kind please g!

Wellll, one more question while I have your attention: I've
always heard and read that residential homes never required power
factor adjustments of any kind because the power factors would
never get very low. If I'm interpreting my numbers right
however, I'm seeing PF numbers that are surprisingly low. Most
every home is full of motors and other inductive appliances.
How low IS a "low" power factor number?
Or do power companies account for power factors at the
facility? Just curious.

Regards,

Pop

wow

"Pop" wrote in message
...
Hi all,
As I recall there are a few people here with the expertise to
answer these questions easily. Simple 4 u, not 4 me! g
Yes, I've done a bunch of googling and recalling the "old
school days", but it's not getting me where I need to be g.
Neglecting small interferences/insertion losses, etc.:
--------------------------------
Short description:
Here's an actual example of measurements/calcs:
120Vac measured
0.29A rms measured
24W measured
35 VA measured
PF = W / VA, or 24 / 35 = 0.686..., or about 68%. Right?

-- What numbers do I use to get kWH? Is it VA / W?

-- How many kWH do you calculate from those figures, assuming it
can be done? If it can't be done, what's missing?
-- How did you get to your result?

-- At 10 cents/kWH, how much would it cost me per hour?

---------- end short descrip -----------

You wouldn't believe the amount of work and research I've done to
get my head around this! And how confused I am at the moment!

All I started out to do was to calculate what some of the major
device costs around the house are in order to make a point to
some people about the cost of, say, leaving the lights on in an
unoccupied room over night, or never turning off say a coffee
maker, computers, radio, stereo, TV, holiday lights; things like
that. And I ended up with a brain-ache so I next decided to go
where there might be some brighter brain cells than my own! And
here I am!

Thanks for your hopefully understandable responses; it's been
over 4 decades since I was in college, so be kind please g!

Wellll, one more question while I have your attention: I've
always heard and read that residential homes never required power
factor adjustments of any kind because the power factors would
never get very low. If I'm interpreting my numbers right
however, I'm seeing PF numbers that are surprisingly low. Most
every home is full of motors and other inductive appliances.
How low IS a "low" power factor number?
Or do power companies account for power factors at the
facility? Just curious.

Regards,

Pop


Sounds like you have a lot of time on your hands.

Here is a link to some useful charts:
http://www.mrelectrician.tv/conversi...electrical.htm

24 watts divided by 1000 equals .024 KW times 1 hour equals .024 KWH times
10 cents per hour would cost you .0024 cents to operate for one hour. I
think.

The power company puts power factor correction equipment on their lines.

I cant answer you but a meter that is fun and usefull is a Kill-A-Watt
apx 25$, it lets you plug in an apliance and measures very accuratly
time, watt, amp, power factor, Kwh used and the hours, V. Hz and more,
you will easily be able to audit usage of anything 120v you plug into it
and find the hogs to show the hogs. It is also very good on low draw
equipment on standby, such as the tv off. It measures over a 100 hr
period. You would be suprised how older things can cost 1$ a month un
used but newer "energy star" rated can cost 1 penny to keep plugged in a
month, It helped me get my electric to 14-20 a month from 50. It also
made me get a new frige that uses 1/6th the power. They have been
independantly tested very accurate, somethimes Radio shack has them.

Pop writes:

What numbers do I use to get kWH? Is it VA / W?

No, kWH = kilowatts multiplied by time in hours.

Residential power meters measure watts, not reactive VA. Power factor
problems are paid for by the utility, not you.

"Richard J Kinch" wrote in message
Residential power meters measure watts, not reactive VA. Power factor
problems are paid for by the utility, not you.



In this case, I'm talking a 480V 3 phase commercial meter.
What is the power factor charges on the bill? . Aside from the normal
dials to read the kWh used, does the power factor get interpreted into it?
The meter has a needle indicator for power factor also and stays pretty much
in the same place.

Fluorescent lights and large electric motors receive part of their power by
shifting the voltage and the current slightly out of phase. A residential
meter ignores this and usually the effect is small for a household. A
commercial meter measures power factor as well as kilowatts and the customer
is charged accordingly.

try http://hyperphysics.phy-astr.gsu.edu...ic/powfac.html

On Sat, 12 Nov 2005 02:56:29 GMT, "Edwin Pawlowski"
wrote:

"Richard J Kinch" wrote in message
Residential power meters measure watts, not reactive VA. Power factor
problems are paid for by the utility, not you.



In this case, I'm talking a 480V 3 phase commercial meter.
What is the power factor charges on the bill? . Aside from the normal
dials to read the kWh used, does the power factor get interpreted into it?
The meter has a needle indicator for power factor also and stays pretty much
in the same place.


These charges are not uniform and often vary greatly from utility to
utility.

1st - Check to see if your utility has a website and has published
their rate tariffs online. If they do, there should be a section on
charges for commercial reactive power. If you are paying a lot for a
significantly low power factor (well below 0.8), you can take steps to
correct it at your own expense. Mostly these involve adding shunt
capacitors to the line possible with a timer control.

or 2nd - Contact the billing specialist at your utility company and
request a copy of the tariffs. If they have the ability to explain
it to you, in addition, consider yourself lucky.

Beachcomber

"Beachcomber" wrote in message

If you are paying a lot for a
significantly low power factor (well below 0.8), you can take steps to
correct it at your own expense. Mostly these involve adding shunt
capacitors to the line possible with a timer control.

This was done a few years ago. I'll have to dig out some bills ot see how
well it worked. There were three or four capacitors at different locations
in the plant.. This is allegedly better than one big one at the 800A main
panel.


or 2nd - Contact the billing specialist at your utility company and
request a copy of the tariffs. If they have the ability to explain
it to you, in addition, consider yourself lucky.

The tariffs are easy enough to find. Explanation is a whole other scenario.

Edwin Pawlowski writes:

In this case, I'm talking a 480V 3 phase commercial meter.
What is the power factor charges on the bill? . Aside from the
normal dials to read the kWh used, does the power factor get
interpreted into it? The meter has a needle indicator for power factor
also and stays pretty much in the same place.

OK, then you apparently do get charged a penalty for reactive power use.

The physics of reactive power requires calculus to understand.

In simple terms, a motor or other inductive load can draw a portion of
costly current and generating capacity from the utility, beyond the power
realized by the device, even though you are not realizing that power in
your facility. A simple power meter does not measure that loss, so a
factor is measured by a more complicated meter to charge you for reactive
power (what the utility sent to you) instead of real power (what you
actually used). This is reasonable, and should encourage you to fix your
installation to properly reduce the reactive power component.

On Sat, 12 Nov 2005 03:49:19 GMT, "Edwin Pawlowski"
wrote:


"Beachcomber" wrote in message

If you are paying a lot for a
significantly low power factor (well below 0.8), you can take steps to
correct it at your own expense. Mostly these involve adding shunt
capacitors to the line possible with a timer control.

This was done a few years ago. I'll have to dig out some bills ot see how
well it worked. There were three or four capacitors at different locations
in the plant.. This is allegedly better than one big one at the 800A main
panel.


You have to match the applied reactive power correction to the
different times of day when your motor loads are creating a low power
factor. Otherwise, an overcorrection (too much capacitive reactance)
is just as bad, if not worse and may make your line voltage levels
fluctuate all over the place.

It matters not if it is done at different locations or the main panel
(assuming the main panel feeds the entire plant). A good main panel
installation will have matched banks of capacitors that can be added
in stages to correct the power factor. This can be automatic,
timer-driven, or manually controlled.

It all depends on the load and mostly the motor load for that part.
Is your plant idle at night, weekends, holidays? Are all motors
running continously or do you have a lot of start-stop operations?
You may have a base load (say of pumps and air blowers) that are on 24
hours a day, hence there might be the need for a certain base value of
power factor correction.

Beachcomber

"The physics of reactive power requires calculus to understand. "

It's really pretty simple to understand the basics. Instantaneous
power is always voltage times current. With an AC circuit and a purely
resistive load, the voltage and current are always in phase with each
other. Place a graph of voltage over a graph of current and they line
up perfectly. So simply multiplying RMS Voltage times RMS Current
gives power.

With a load that has capacitance or inductance in addition to
resistance (eg a motor), the voltage and current are no longer in
phase. Place a graph of one over the other and they appear shifted.
So when voltage is at it's peak, current is not, hence the power
consumed will be less. How much less depends on how far out of phase
voltage and current are. With a purely capacitive load or a purely
inductive load, the power will be zero.

Richard J Kinch wrote:

... The physics of reactive power requires calculus to understand.

A little trig is sufficient, IMO.

Nick

....
:
: Sounds like you have a lot of time on your hands.
Yeah, may be: It happens when one is suddenly disabled, thrown
out of work because of it, housebound and not allowed to drive or
even do the checking ;-( any longer.
:
: Here is a link to some useful charts:
: http://www.mrelectrician.tv/conversi...electrical.htm
Just what I needed, I think! Believe it or not I'm an EE but the
concussion has pretty badly beat up my memory. I'm never sure
what I remember is real or a made-up memory. It's going on 6
years now so I'm just getting out of the learning disabled stage
but a long way to go; probably never get it all back.
:
: 24 watts divided by 1000 equals .024 KW times 1 hour equals
..024 KWH times
: 10 cents per hour would cost you .0024 cents to operate for one
hour. I
: think.
:
: The power company puts power factor correction equipment on
their lines.
:

Thanks; looks like a decent deal, actually. I'll likely try
that. Still leaves me wondering how it can do that though g.

Thanks again & Regards

"m Ransley" wrote in message
...
:I cant answer you but a meter that is fun and usefull is a
Kill-A-Watt
: apx 25$, it lets you plug in an apliance and measures very
accuratly
: time, watt, amp, power factor, Kwh used and the hours, V. Hz
and more,
: you will easily be able to audit usage of anything 120v you
plug into it
: and find the hogs to show the hogs. It is also very good on low
draw
: equipment on standby, such as the tv off. It measures over a
100 hr
: period. You would be suprised how older things can cost 1$ a
month un
: used but newer "energy star" rated can cost 1 penny to keep
plugged in a
: month, It helped me get my electric to 14-20 a month from 50.
It also
: made me get a new frige that uses 1/6th the power. They have
been
: independantly tested very accurate, somethimes Radio shack has
them.
:

Below is part of what I was wondering about: Doesn't that mean,
in reality, that, unless the PF is corrected, that the customer
is the one with the advantage? As in, "free" power? IE of
shifted waveforms is going to be less than in-phase IE over time.
Therefore, the correction equipment is to "correct" the numbers
so the power company isn't delivering power it isn't charging
for? If so, why would anyone voluntarily install a capacitor
system?

Pop


wrote in message
oups.com...
: "The physics of reactive power requires calculus to understand.
"
:
: It's really pretty simple to understand the basics.
Instantaneous
: power is always voltage times current. With an AC circuit and
a purely
: resistive load, the voltage and current are always in phase
with each
: other. Place a graph of voltage over a graph of current and
they line
: up perfectly. So simply multiplying RMS Voltage times RMS
Current
: gives power.
:
: With a load that has capacitance or inductance in addition to
: resistance (eg a motor), the voltage and current are no longer
in
: phase. Place a graph of one over the other and they appear
shifted.
: So when voltage is at it's peak, current is not, hence the
power
: consumed will be less. How much less depends on how far out of
phase
: voltage and current are. With a purely capacitive load or a
purely
: inductive load, the power will be zero.
:

"Below is part of what I was wondering about: Doesn't that mean,
in reality, that, unless the PF is corrected, that the customer
is the one with the advantage?"

I guess that depends on exactly what the electric power meter installed
at your home is measuring. I don't know, but I would guess that for a
home it's not sophisticated and is likely only measuring RMS amps and
basing it on that? If that's true, then the billing advantage is to
the power company. But I would think the cost delta due to the power
factor issue is pretty small in the typical home.

wrote:

I guess that depends on exactly what the electric power meter installed
at your home is measuring. I don't know, but I would guess that for a
home it's not sophisticated and is likely only measuring RMS amps and
basing it on that?

No. They measure real watts.

Nick

Pop writes:

Doesn't that mean,
in reality, that, unless the PF is corrected, that the customer
is the one with the advantage? As in, "free" power?

No, it is just wasted in the transmission lines and/or in reduced plant
generating capacity. The point of the reactive meter is to get the culprit
customer to pay for it.

A purely reactive load would be consuming all kinds of current down the
line, and be loading the utility power plant, but show zero watts on a real
power meter.

Pop wrote:

Below is part of what I was wondering about: Doesn't that mean,
in reality, that, unless the PF is corrected, that the customer
is the one with the advantage? As in, "free" power? IE of
shifted waveforms is going to be less than in-phase IE over time.
Therefore, the correction equipment is to "correct" the numbers
so the power company isn't delivering power it isn't charging
for? If so, why would anyone voluntarily install a capacitor
system?

Pop

The answer to that question is.....

If it were a perfect world full of honorable people (particularly at the
utilities) then .... if every homeowner installed equipment to make
their home's power factors as close to unity as practically possible
(Probably that could just be a capacitor right across each significantly
sized motor.), the utility would have less line losses and be able to
pass the savings back to their customers in the form of lower rates.

Remember, I said "perfect world"...

Jeff (Another old phart EE who's forgotten more than he learned, but one
thing I'll never forget are the words of that Brit professor who tought
our "rotating machinery" lab who said, "You boys will never become good
electrical engineers until you learn to "take" a shock.")

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"Truth exists; only falsehood has to be invented."

In article ,
"John Grabowski" wrote:

24 watts divided by 1000 equals .024 KW times 1 hour equals .024 KWH times
10 cents per hour would cost you .0024 cents to operate for one hour. I
think.

Just a nit: You multiplied by 0.1 to get the final answer, which works
for dollars but not cents. So, .0024 dollars/hr, or .24 cents/hr.
Small change either way.

For general electric costs rule-of-thumb, I use the 100W lightbulb, at
$0.10/kWH (common rate in the U.S.), and 1 month (electric bill
frequency), to come up with:

0.1kW * 1 month * 30 days/month * 24 hrs/day * $0.10/kWh ~= $7/mo.

So, $7/mo. to run a 100W device all the time. Most appliances and duty
cycles can be scaled to this benchmark pretty easily.

"Richard J Kinch" wrote in message
. ..
: Pop writes:
:
: Doesn't that mean,
: in reality, that, unless the PF is corrected, that the
customer
: is the one with the advantage? As in, "free" power?
:
: No, it is just wasted in the transmission lines and/or in
reduced plant
: generating capacity. The point of the reactive meter is to get
the culprit
: customer to pay for it.
:
: A purely reactive load would be consuming all kinds of current
down the
: line, and be loading the utility power plant, but show zero
watts on a real
: power meter.

That's what I thought; just checking myself. Thanks.

chocolatemalt writes:

So, $7/mo. to run a 100W device all the time.

More conveniently, 1 watt-year costs 1 US dollar. But those days are
passing.

for those of you who want to "geek out" on power factor correction &
reactive power,

here are couple of links that give understandable explanations

http://home.earthlink.net/~jimlux/hv/pfc.htm
http://www.ambercaps.com/lighting/po...n_concepts.htm
http://www.nepsi.com/powerfactor.htm

the "best" power factor correction is achieved by adding "balancing"
capacitors at each inductive load (motor)

cheers
Bob

BobK207 wrote:

for those of you who want to "geek out" on power factor correction &
reactive power,

here are couple of links that give understandable explanations

http://home.earthlink.net/~jimlux/hv/pfc.htm
http://www.ambercaps.com/lighting/po...n_concepts.htm
http://www.nepsi.com/powerfactor.htm

the "best" power factor correction is achieved by adding "balancing"
capacitors at each inductive load (motor)

cheers
Bob

Hi,
Nothing new if one paid attention in his/her HS physics class.
In real world MOST electrical load is inductive which makes voltage lead
current by certain amount. Reactive power is false power(wasted power)
Tony

"chocolatemalt" wrote in
message
...
: In article ,
: "John Grabowski" wrote:
:
: 24 watts divided by 1000 equals .024 KW times 1 hour equals
..024 KWH times
: 10 cents per hour would cost you .0024 cents to operate for
one hour. I
: think.
:
: Just a nit: You multiplied by 0.1 to get the final answer,
which works
: for dollars but not cents. So, .0024 dollars/hr, or .24
cents/hr.
: Small change either way.
:
: For general electric costs rule-of-thumb, I use the 100W
lightbulb, at
: $0.10/kWH (common rate in the U.S.), and 1 month (electric bill
: frequency), to come up with:
:
: 0.1kW * 1 month * 30 days/month * 24 hrs/day * $0.10/kWh ~=
$7/mo.
:
: So, $7/mo. to run a 100W device all the time. Most appliances
and duty
: cycles can be scaled to this benchmark pretty easily.

Basically true for an incandescent light bulb. I went out and
bought a watt/VA meter one of the guys here suggested - and you'd
be surprised how far off that same 100W calc is if the load is
inductive. Depending, I'm seeing power factors so far as low as
58% to around 80%, which will throw off your calcs over the space
of months or a year.
That meter's a nice little gizmo for $30 and seems to be
pretty accurate to boot. No specs with it, but I did check it
against some calcs, plus what my UPS measures the line stuff at -
they lined up very nicely; less than 4% diff and I'm sure the UPS
ain't all that accurate as a rule either. How's that for a
scientific calibration check g?
Also, if you're playing with duty cycle, you don't multipy by
24 x 7 etc.; that's a 100% duty cycle on your assumption of
everything having a power factor of 1.00.
For an electric bulb though, you'd be real close. But
refrigerator, furnace, flourescent, things like that it's quite a
different story.
It's been interesting if nothing else, and might save a thou
or two over a year; making it worthwhile.

Cheers!

"Tony Hwang" wrote in message
news:j8vdf.487401$oW2.225661@pd7tw1no...
: BobK207 wrote:
:
: for those of you who want to "geek out" on power factor
correction &
: reactive power,
:
: here are couple of links that give understandable
explanations
:
: http://home.earthlink.net/~jimlux/hv/pfc.htm
:
http://www.ambercaps.com/lighting/po...n_concepts.htm
: http://www.nepsi.com/powerfactor.htm
:
: the "best" power factor correction is achieved by adding
"balancing"
: capacitors at each inductive load (motor)
:
: cheers
: Bob
:
: Hi,
: Nothing new if one paid attention in his/her HS physics class.
: In real world MOST electrical load is inductive which makes
voltage lead
: current by certain amount. Reactive power is false power(wasted
power)
: Tony

I don't think "new" has anything to do with the dialogs that have
gone on. And it's not HS physics if you want to really get into
it. You'd also need trig, calc and Field Theory at college level
to have a really good go at it! As in, there is no such thing as
a purely resistive or reactive load. Even a resistor has a
capatcitive/inductive component if you want to get picky enough.
I think the prevailing idea here was to keep the numbers going
in ways that the outcomes were perceptible, not negliglbie, and
interesting to boot. If it bores you, don't read it. Pretty
simple.

In article j8vdf.487401$oW2.225661@pd7tw1no,
Tony Hwang wrote:

BobK207 wrote:

for those of you who want to "geek out" on power factor correction &
reactive power,

here are couple of links that give understandable explanations

http://home.earthlink.net/~jimlux/hv/pfc.htm
http://www.ambercaps.com/lighting/po...n_concepts.htm
http://www.nepsi.com/powerfactor.htm

the "best" power factor correction is achieved by adding "balancing"
capacitors at each inductive load (motor)

cheers
Bob

Hi,
Nothing new if one paid attention in his/her HS physics class.
In real world MOST electrical load is inductive which makes voltage lead
current by certain amount. Reactive power is false power(wasted power)
Tony

But of course that power goes *somewhere* right?

In the interest of conservation of energy, even if that power is doing
no useful work in your electric motor, it's doing work somewhere, right?
I'm sure it's an obvious point but the answer isn't evident to me.

If you have a PF 70% motor chewing up 700 watts, then 300 watts goes...
into heat loss of the inductive windings? Perhaps the constant building
up and tearing down of the magnetic flux is causing the friction loss
via atomic realignments in the inductor itself? And similarly if you
have a capacitive reactance device, the power loss goes into... what?
Heat loss of the electrons rushing into and out of the capacitive
reservoirs?

If anyone has an understanding of this, I'd love to hear it... been
wondering about this one for a while.

chocolatemalt wrote:

If you have a PF 70% motor chewing up 700 watts...

The current flow is higher, so the real I^2R power loss in the wiring
is slightly higher, but not much, compared to the 700 watts.

Nick

On Sat, 12 Nov 2005 08:34:10 -0500, Pop wrote:

Thanks; looks like a decent deal, actually. I'll likely try
that. Still leaves me wondering how it can do that though g.

I have one of the same (I think) rebranded as a Seasonic PowerAngel. It
works quite well (my PC is now drawing ~130-140W, 200-220VA, PF=.61 .
They can do it because of the magic of microprocessors. Measure current
and voltage, multiply the instantaneous values and average for power.
Measure current and voltage, calculate RMS voltage and current and
multiply the result for VA. Divide the two and get PF. The math is quite
simple. I'm amazed there is a big enough market to get the price down to
the $30 range though.

--
Keith

On Sat, 12 Nov 2005 19:54:52 -0500, Pop wrote:


"chocolatemalt" wrote in
message
...
: In article ,
: "John Grabowski" wrote:
:
: 24 watts divided by 1000 equals .024 KW times 1 hour equals
.024 KWH times
: 10 cents per hour would cost you .0024 cents to operate for
one hour. I
: think.
:
: Just a nit: You multiplied by 0.1 to get the final answer,
which works
: for dollars but not cents. So, .0024 dollars/hr, or .24
cents/hr.
: Small change either way.
:
: For general electric costs rule-of-thumb, I use the 100W
lightbulb, at
: $0.10/kWH (common rate in the U.S.), and 1 month (electric bill
: frequency), to come up with:
:
: 0.1kW * 1 month * 30 days/month * 24 hrs/day * $0.10/kWh ~=
$7/mo.
:
: So, $7/mo. to run a 100W device all the time. Most appliances
and duty
: cycles can be scaled to this benchmark pretty easily.

Basically true for an incandescent light bulb. I went out and
bought a watt/VA meter one of the guys here suggested - and you'd
be surprised how far off that same 100W calc is if the load is
inductive. Depending, I'm seeing power factors so far as low as
58% to around 80%, which will throw off your calcs over the space
of months or a year.

At least in the US, residential customers are charged for energy consumed.
They are not peanalized for crappy PF. Many corporate customers are.

That meter's a nice little gizmo for $30 and seems to be
pretty accurate to boot. No specs with it, but I did check it against
some calcs, plus what my UPS measures the line stuff at - they lined up
very nicely; less than 4% diff and I'm sure the UPS ain't all that
accurate as a rule either. How's that for a scientific calibration
check g?
Also, if you're playing with duty cycle, you don't multipy by
24 x 7 etc.; that's a 100% duty cycle on your assumption of everything
having a power factor of 1.00.
For an electric bulb though, you'd be real close. But
refrigerator, furnace, flourescent, things like that it's quite a
different story.
It's been interesting if nothing else, and might save a thou
or two over a year; making it worthwhile.

A thou or two over a year? Your bill must be mighty big! ;-) Again, you
are only charged for watts. The PF is irrelevant here (not so for your
UPS though).

On Sat, 12 Nov 2005 05:32:14 -0500, nicksanspam wrote:

Richard J Kinch wrote:

... The physics of reactive power requires calculus to understand.

A little trig is sufficient, IMO.

Simple algebra is enough, depending on what you're starting with. If you
have the instantaneous voltage and current and a four-function calculator,
you have all you need.

--
Keith

Kind of hard to explain in words and without knowing whether you
have any exp with electriclal theory; maybe someone will come up
wiht a link.

: But of course that power goes *somewhere* right?
Sort of. Your assumptions will sort of work, but they're not
what's really happening.
In a resistor ckt, current and voltage are in phase. When the
ac sine wave is at its max point, so is current. Voltage drops,
current drops accordingly.
:
: In the interest of conservation of energy, even if that power
is doing
: no useful work in your electric motor, it's doing work
somewhere, right?
: I'm sure it's an obvious point but the answer isn't evident to
me.
In an electric motor, the windings are a big coil. It sounds
like you understand that a little bit. Coils resist changing
currents. So, if the voltage jumps to its max, the current rises
slower than the voltage can rise because it has to create the
building magnetic field.
But in an ac motor, the voltage begins to fall (passes the
peak) before the current has made it all the way to the max it
would have reached if the voltage had stayed there. But the
voltage is falling toward zero now, and as the voltage falls, the
magnetic field begins to collapse. But, since it's a coil, it
cannot fall as fast as the voltage is falling. The voltage
passes zero now an continues on toward its negative peak, with
the current still trailing it, passes that peak, befoer the
current catches up, and starts toward zero again, and so on as
long as the power is applied.
P=IE but p does not= ie. (lower case means ac, upper DC). At
any point in time, where the voltage is max, the current is NOT
yet at max, and thus the power (p=ie) will be less than P=IE.
Current never gets to max, in fact for motors. So a straight
p=ie formula gives a lower wattage than if the current had
reached the max it COULD have reached, fi the voltage had stayed
there long enough.



Capacitors are just the opposite. The don't resist current
change, but they do resist voltage change. It takes time to
charge up to and discharge from a known voltage.
:
: If you have a PF 70% motor chewing up 700 watts, then 300 watts
goes...
: into heat loss of the inductive windings?
Sort of. The "lost" energy does create heating in the windings.

Perhaps the constant building
: up and tearing down of the magnetic flux is causing the
friction loss
: via atomic realignments in the inductor itself?
Yup. It takes time for the flux field to build and time to
collapse, so it can't change as fast as the voltage does that's
being applied to it.

And similarly if you
: have a capacitive reactance device, the power loss goes into...
what?
: Heat loss of the electrons rushing into and out of the
capacitive
: reservoirs?
Capacitors store electrons. So, they spend time collecting
electrons while the voltage is applied, and then spend time
losing the electrons when the voltage is removed.

In both cases the speed of collection/loss of electrons depends
on the DC resistance components in the ckt. A resistor basically
passes current instantaneously since there is no reactive element
involved.
Capacitor stores electrons.
Inductor creates current flow from a collapsing field, resists
them during the building of hte field. Limited by the resistance
component.

HTH
:
: If anyone has an understanding of this, I'd love to hear it...
been
: wondering about this one for a while.

In article ,
"Pop" wrote:

Kind of hard to explain in words and without knowing whether you
have any exp with electriclal theory; maybe someone will come up
wiht a link.

I've got a CE degree (hybrid of EE & CS) so I should be able to handle
the math and theory.

: But of course that power goes *somewhere* right?
Sort of. Your assumptions will sort of work, but they're not
what's really happening.
In a resistor ckt, current and voltage are in phase. When the
ac sine wave is at its max point, so is current. Voltage drops,
current drops accordingly.
[...]

I understand the theory of PF, but the question is much simpler: If the
power company is feeding you 1000W on a straight V*A basis, and your
motor is seeing just 700W of useful work on a PF*V*A basis, there are
300W of energy that have "disappeared". I guess the question is so
simple that the answer is obvious: The energy is consumed within the
motor as non-useful heat.

It seems that clever residential customers in cold climates might prefer
to find electric devices with horrible PF's just to get free heat from
the power company. And that raises the question -- would you be better
off adding some inductance to a space heater to help produce "free"
heat? The purely resistive component is what you get billed on, yet the
inductance produces heat as well and is a non-billable component for
non-commercial customers.

chocolatemalt wrote:

... If the power company is feeding you 1000W on a straight V*A basis,
and your motor is seeing just 700W of useful work on a PF*V*A basis,
there are 300W of energy that have "disappeared". I guess the question
is so simple that the answer is obvious: The energy is consumed within
the motor as non-useful heat.

No. If your motor used 700 W with a power factor of 1, 700/120 = 5.83 amps
would flow in your wiring. If the wiring resistance were 0.1 ohms, it would
dissipate 5.83^2x0.1 = 3.4 watts of heat, and you would pay for 703.4 watts.

With a 0.7 PF, 5.83/0.7 = 8.33 amps flow, so the wiring would dissipate
8.33^2x0.1 = 6.9 watts, and you pay for 706.9 watts. Your only penalty
is the difference, 6.9-3.4 = 3.5 watts. The power company is less happy
because they lose more power in their wiring, but they usually don't
complain, in the case of houses.

Nick

In article ,
wrote:

chocolatemalt wrote:

... If the power company is feeding you 1000W on a straight V*A basis,
and your motor is seeing just 700W of useful work on a PF*V*A basis,
there are 300W of energy that have "disappeared". I guess the question
is so simple that the answer is obvious: The energy is consumed within
the motor as non-useful heat.

No. If your motor used 700 W with a power factor of 1, 700/120 = 5.83 amps
would flow in your wiring. If the wiring resistance were 0.1 ohms, it would
dissipate 5.83^2x0.1 = 3.4 watts of heat, and you would pay for 703.4 watts.

With a 0.7 PF, 5.83/0.7 = 8.33 amps flow, so the wiring would dissipate
8.33^2x0.1 = 6.9 watts, and you pay for 706.9 watts. Your only penalty
is the difference, 6.9-3.4 = 3.5 watts. The power company is less happy
because they lose more power in their wiring, but they usually don't
complain, in the case of houses.

Nick

Ah, I see. The V*A apparent power that the power company sees your
house consuming can in fact be zero watts of real consumption if you
have a perfect inductor and negligible line resistance. No energy is
disappearing from the Universe.

But with the added current that you are not being billed for, when PF
1, the power company is heating the atmosphere with line losses and
burning real coal or uranium to do so, and so they hate you. Perhaps
with superconducting transmission lines someday (assuming it's
achievable), they will no longer care if your PF deviates since their
own energy expenditure will be equal to your real wattage, and reactance
will be irrelevant.

wrote:
chocolatemalt wrote:


... If the power company is feeding you 1000W on a straight V*A basis,
and your motor is seeing just 700W of useful work on a PF*V*A basis,
there are 300W of energy that have "disappeared". I guess the question
is so simple that the answer is obvious: The energy is consumed within
the motor as non-useful heat.


No. If your motor used 700 W with a power factor of 1, 700/120 = 5.83 amps
would flow in your wiring. If the wiring resistance were 0.1 ohms, it would
dissipate 5.83^2x0.1 = 3.4 watts of heat, and you would pay for 703.4 watts.

With a 0.7 PF, 5.83/0.7 = 8.33 amps flow, so the wiring would dissipate
8.33^2x0.1 = 6.9 watts, and you pay for 706.9 watts. Your only penalty
is the difference, 6.9-3.4 = 3.5 watts. The power company is less happy
because they lose more power in their wiring, but they usually don't
complain, in the case of houses.

Nick



For business customers, the disparity in amps is registered
by the "demand meter". And the utility company adds a
multiplier to your bill based on that number. It's to cover
the larger wire and transformers required to deliver you
8.33 amps instead of 5.83 amps.

In Dallas, half the business bill is the demand surcharge.
Which means we pay much more for that "phantom" power than
the "real" power. Copper and real estate is expensive,
especially when NIMBY is applied.


-larry / dallas

"keith" wrote in message
news : On Sat, 12 Nov 2005 19:54:52 -0500, Pop wrote:
:
:
: "chocolatemalt" wrote in
: message
:
...
: : In article ,
: : "John Grabowski" wrote:
: :
: : 24 watts divided by 1000 equals .024 KW times 1 hour
equals
: .024 KWH times
: : 10 cents per hour would cost you .0024 cents to operate
for
: one hour. I
: : think.
: :
: : Just a nit: You multiplied by 0.1 to get the final answer,
: which works
: : for dollars but not cents. So, .0024 dollars/hr, or .24
: cents/hr.
: : Small change either way.
: :
: : For general electric costs rule-of-thumb, I use the 100W
: lightbulb, at
: : $0.10/kWH (common rate in the U.S.), and 1 month (electric
bill
: : frequency), to come up with:
: :
: : 0.1kW * 1 month * 30 days/month * 24 hrs/day * $0.10/kWh
~=
: $7/mo.
: :
: : So, $7/mo. to run a 100W device all the time. Most
appliances
: and duty
: : cycles can be scaled to this benchmark pretty easily.
:
: Basically true for an incandescent light bulb. I went out
and
: bought a watt/VA meter one of the guys here suggested - and
you'd
: be surprised how far off that same 100W calc is if the load
is
: inductive. Depending, I'm seeing power factors so far as low
as
: 58% to around 80%, which will throw off your calcs over the
space
: of months or a year.
:
: At least in the US, residential customers are charged for
energy consumed.
: They are not peanalized for crappy PF. Many corporate customers
are.
:
: That meter's a nice little gizmo for $30 and seems to be
: pretty accurate to boot. No specs with it, but I did check
it against
: some calcs, plus what my UPS measures the line stuff at -
they lined up
: very nicely; less than 4% diff and I'm sure the UPS ain't all
that
: accurate as a rule either. How's that for a scientific
calibration
: check g?
: Also, if you're playing with duty cycle, you don't multipy
by
: 24 x 7 etc.; that's a 100% duty cycle on your assumption of
everything
: having a power factor of 1.00.
: For an electric bulb though, you'd be real close. But
: refrigerator, furnace, flourescent, things like that it's
quite a
: different story.
: It's been interesting if nothing else, and might save a
thou
: or two over a year; making it worthwhile.
:
: A thou or two over a year? Your bill must be mighty big! ;-)
Again, you
: are only charged for watts. The PF is irrelevant here (not so
for your
: UPS though).

THOU?!?!?!? Who said that!! Me????? Jeez, I don't recall what
I meant to say but if I could do that, I'd go into business
selling the idea to others!!
Typo obviously g. Err, not, it wasn't either! Send me
$11.95 in a SASE and I'll tell y'all how ta do it! ;-} Now,
where'd I lay that gizmo? Hmmm ...

Regards,

Pop

"keith" wrote in message
news : On Sat, 12 Nov 2005 05:32:14 -0500, nicksanspam wrote:
:
: Richard J Kinch wrote:
:
: ... The physics of reactive power requires calculus to
understand.
:
: A little trig is sufficient, IMO.
:
: Simple algebra is enough, depending on what you're starting
with. If you
: have the instantaneous voltage and current and a four-function
calculator,
: you have all you need.
:
: --
: Keith

Umm, then how do you get the instantaneous voltage/current over
time?

You know, the phase difference? It varies with time too.