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.

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.
:

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

....
:
: 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.
:

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.

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!

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).

"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

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.

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

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.

"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? 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.
:

Richard J Kinch wrote:

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

A little trig is sufficient, IMO.

Nick

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