Page 2 of 3
chocolatemalt writes:

No.

chocolatemalt wrote:

In a motor. the current creates a magnetic field which varies in magnitude and direction during the cycle. Some energy is stored in the magnetic field for part of the cycle and sent back toward the utility on another part of the cycle. The 300 watts of reactive power is from energy being stored and returned 120 times per second and is not power that disappears. Because of the properties an inductor the current is shifted and the current flow sine wave is later than the voltage sinewave and the reactive power is 90 degrees out of phase with the real power.

A lightly loaded motor will have a lower PF than fully loaded motor because the real power of the fully loaded motor will be higher and will be a higher percentage of the total power. My memory of dim teachings is that the reactive power doesn't change much with loading.

The higher current caused by the magnetic field interchange does cause real I squared R losses in wire as detailed in other posts.

In a capacitor energy is stored in an electric field and is also stored and released 120 times per cycle. But in a capacitor the current leads the voltage. Power factor correction capacitors cancel the effects of inductance.

Over correction - too much capacitance to balance the inductance can cause resonance (if I remember right), which can cause the voltage to rise as in the post by Beachcomber. This can be made worse by harmonics which can be generated by VFDs, power supplies, dimmers.

bud--

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.

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.

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

No. They measure real watts.

Nick

#### Site Timeline

- posted on November 14, 2005, 4:40 am

No.

- posted on November 14, 2005, 5:07 pm

In a motor. the current creates a magnetic field which varies in magnitude and direction during the cycle. Some energy is stored in the magnetic field for part of the cycle and sent back toward the utility on another part of the cycle. The 300 watts of reactive power is from energy being stored and returned 120 times per second and is not power that disappears. Because of the properties an inductor the current is shifted and the current flow sine wave is later than the voltage sinewave and the reactive power is 90 degrees out of phase with the real power.

A lightly loaded motor will have a lower PF than fully loaded motor because the real power of the fully loaded motor will be higher and will be a higher percentage of the total power. My memory of dim teachings is that the reactive power doesn't change much with loading.

The higher current caused by the magnetic field interchange does cause real I squared R losses in wire as detailed in other posts.

In a capacitor energy is stored in an electric field and is also stored and released 120 times per cycle. But in a capacitor the current leads the voltage. Power factor correction capacitors cancel the effects of inductance.

Over correction - too much capacitance to balance the inductance can cause resonance (if I remember right), which can cause the voltage to rise as in the post by Beachcomber. This can be made worse by harmonics which can be generated by VFDs, power supplies, dimmers.

bud--

- posted on November 13, 2005, 12:54 am

message
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!

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!

- posted on November 13, 2005, 2:05 pm

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

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

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).

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

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).

- posted on November 14, 2005, 2:14 am

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

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

- posted on November 12, 2005, 2:41 am

Pop writes:

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.

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.

- posted on November 12, 2005, 2:56 am

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.

- posted on November 12, 2005, 3:13 am

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.

- posted on November 12, 2005, 3:16 am

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

- posted on November 12, 2005, 3:22 am

wrote:

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

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

- posted on November 12, 2005, 3:49 am

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.

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

- posted on November 12, 2005, 7:43 am

wrote:

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

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

- posted on November 12, 2005, 4:58 am

Edwin Pawlowski writes:

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.

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.

- posted on November 12, 2005, 8:35 am

"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.

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.

- posted on November 12, 2005, 1:45 pm

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 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. :

Pop

: "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. :

- posted on November 12, 2005, 1:55 pm

"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.

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.

- posted on November 12, 2005, 2:11 pm

No. They measure real watts.

Nick

- posted on November 12, 2005, 4:14 pm

Pop writes:

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.

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.

- posted on November 12, 2005, 7:48 pm

: 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.

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

- posted on November 12, 2005, 4:27 pm

Pop wrote:

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.")

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)

Jeffry Wisnia

(W1BSV + Brass Rat \'57 EE)

Click to see the full signature.

- Kitchen faucet chatter
- - next thread in Home Repair

- Propane space heater / regulator question
- - previous thread in Home Repair

- OT: Happy Inauguration Day!
- - newest thread in Home Repair

- OT ish I just got an offer from Spectrum
- - last updated thread in Home Repair

- CH Controller/Timer
- - the site's newest thread. Posted in UK Do-It-Yourself Forum

- LED lamps, source and value
- - the site's last updated thread. Posted in UK Do-It-Yourself Forum